some of you may have already observed that I have forgotten to specify n \geq 4 in the 1st problem ... otherwise the problem has no sense :)
thanks to Octavian Ganea (first to e-mail me about this)

[b]bump [/b]wow I wasn't even born when this thread was created  :|

[quote=Leo890][quote=barry.yyc]me too lol
its awkward knowing some post is older than  your entire life[/quote]
yes, imagine the person who borned now and is seen this message 14 years ago[/quote]

That's internet for you.

All of the things will remain here for thousands of years.

I was born on the day this thread was made

Today also 385,000 babies have born.

I am assuming at least 100 of them would be great in maths and join AOPS and read this thread 15 years later.

[quote=HIA2020]I was born on November 20th, 2020[/quote]

:thinking: and your account was made on May 11th 2020


trigonometry quadratics algebra polynomial AMC

acrtually i think ive seen this q. before.  Where did you get it?

Anyone solve this yet? I tried Viete's and then dividing/subtracting, but it didn't yield anything useful.

[quote="darkquantum"]If x^2-3xcosA-2=0, and x^2+6xsinA+4=0 have a root in common, then find the value of tanA[/quote]
okay, after working on this for a while, i've decided i'll use brute force if i have to...that means making cos:^2:A+sin:^2:A=1.

cos A = (x:^2:-2)/3x. sin A = -(x:^2:+4)/6x. Squaring and adding them together, I get (letting x:^2:=y): (y:^2:-4y+4)/9y+(y:^2:+8y+16)/36y=1. Multiplying all by 36y and bringing the RHS over, I get:
4y:^2:-16y+16+y:^2:+8y+16-36y=5y:^2:-44y+32=0. This is a quadratic in y, which can be solved as (5y-4)(y-8)=0, or x:^2:=4/5 or 8.

Okay, now, going back to the formulas for cos A and sin A at the beginning of the previous paragraph, I divide them to get tan A = -(x:^2:+4)/2(x:^2:-2). This gives -1 or 2 for the value of tan A. I suspect one of them might be an extraneous root, but I don't feel like checking.

hmm
i believe u did not use the fact taht the two expressions have a root in common...

wow cool. so did I do something wrong or was that unneccesary info?

They have a root in common.. thus substituting that root into both and equating them will probably get a few things cancelling..

[quote="darkquantum"]If x^2-3xcosA-2=0, and x^2+6xsinA+4=0 have a root in common, then find the value of tanA[/quote]

yeah...I just thought of that. Anyways, letting our root in common be r, we have r:^2:-3rcosA-2=0 and r:^2:+6rsinA+4=0. Adding twice the first equation to the second, we get 3r:^2:-6rcosA+6rsinA=0. Since r is not 0, we can cancel out 3r to get r-2cosA+2sinA=0. Therefore r=2cosA-2sinA. Also, we can subtract the our original equations to get 6rsinA+3rcosA+6=0, or 2rsinA+rcosA+2=0. Plugging in our value for r and dividing by 2, we have 2cosAsinA-2sin:^2:A+cos:^2:A-sinAcosA+1=0.
This simplifies to cosAsinA-sin:^2:A+2cos:^2:A=0. Dividing everything through by cos:^2:A, we have tan:^2:A-tanA+2=0, which is a quadratic in tan A and gives the same answer that I got last time.

yea
so i guess it isn't really necessary

but it might have made the problem easier

[quote]what is "Viete's" ?[/quote]

For a polynomial f(x) = a_n x^n + a_(n-1) x^(n-1) + ... + a_0, the kth symmetric sum of the roots is (-1)^k * a_(n-k) / a_n .

Not quite. For example, if we have variables a, b, and c, the symmetric sum of ab is ab+ac+ba+bc+ca+cb, which is double what you expect. Vite's formula doesn't do duplications, so the kth symmetric sum of the roots is (-1)^k*(n-k)!*a_(n-k)/a_n.

Well, Im not sure how you define a "symmetric sum", but I practically copied my definition from page 82 of AoPS v2.

hmm, i think simon might have misinterpreted ur answer...

thx. i knew about symmetric sums, i just didnt know that they were discovered by this Viete.


number theory
combinatorics proposed
combinatorics

Every prime of the form 4n + 1 is a sum of two squares.

        If p = 4n + 1, then by Fermat's Little Theorem each of the numbers 1, 2^{4n}, 3^{4n},\dots, (4n)^{4n} is congruent to one modulo p. The differences 2^{4n}-1, 3^{4n}-2^{4n},\dots,(4n)^{4n}-(4n-1)^{4n} are therefore all divisible by p. Each of these differences can be factored as

            a^{4n}-b^{4n} = \left(a^{2n}+b^{2n}\right)\left(a^{2n}-b^{2n}\right).

Since p is prime, it must divide one of the two factors. If in any of the 4n − 1 cases it divides the first factor, then by the previous step we conclude that p is itself a sum of two squares (since a and b differ by 1, they are relatively prime). So it is enough to show that p cannot always divide the second factor. If it divides all 4n − 1 differences $2^{2n}-1, 3^{2n}-2^{2n},\dots,(4n)^{2n}-(4n-1)^{2n}$, then it would divide all 4n − 2 differences of successive terms, all 4n − 3 differences of the differences, and so forth. Since the kth differences of the sequence $1^{k}, 2^{k}, 3^{k},\dots$ are all equal to $k!$, the $2nth$ differences would all be constant and equal to $(2n)!$, which is certainly not divisible by p. Therefore, p cannot divide all the second factors which proves that p is indeed the sum of two squares.

This excerpt was taken from the wikipedia article http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares.

Could someone help me understand the part where it gets into differences of differences and k!? Thanks.

you should move this to the number theory section...
this problem has been discussed plenty of times...

Oops I thought I did post it in number theory.


Well what do you think? Which one is the best and why?

(4 more options added by mod)

[quote="ZetaX"]Do not start a poll for any question that arises to your spammy head!!! (not just meant to you, but all the others that post these annoying topics)

This Forum is for serious discussion, and already the idea of this topic is not.[/quote]

no seriously, im not spamming. I just want to see what other people on AoPS think of the companies.

[quote="ZetaX"]I believe you that you want to know this. But nonetheless this will be a thread like all the other polls here...:
- it will get spammed by stupid comments (*looks at nobody, whistels*).
- from it's idea, there is not really anything to discuss (simply collecting oppinions is just spam, this forum is "for serious discussions").
- already the given four possibilities are a bit... well... random.[/quote]

hmm that is true i never thought about the comments. perhaps moving it to Games and Fun Factory? if you still think this would be spam then please delete this thread. anyway, the four possibilities were the ones that made the most popular items today.

[quote="b-flat"]Microsoft makes computers?[/quote]

yeah, they are also good with programming, like MS Paint, Word, Excel, etc...

[quote="b-flat"]Oh, I would have thought you would have put Dell above Microsoft for computers.

What about TI & their calculators  :rotfl:

Edit:  what about Google?[/quote]

oh yeah I guess I forogt google....hmmm that could get interesting...can i put it in?

[Edit: ugh i need a mod to do it never mind then]

[quote="asdf2345"]what is a compony? I tried looking it up in the dictionary but it turned out no results. Could someone please clarify? My english is not too good...[/quote]

actually it's company. look that up...if i spelled it wrong earlier i'm deeply sorry.

the texas instrunment tv's look interesting

-jorian


is this true that Classroom Math is for sixth grade and Basic is like 7th grade and math count is 8th grade math? Kristy told me and I was just wondering if it's true. You can delete this after you answer my question

What I think is the Classroom math is what it is....problems someone could find in a normal middle school class room (6, 7, and 8th grades). Then the Basics is problems meant for middle schoolers that are like word problems. They are easier thatn Mathcounts problems. Then the Mathcounts is for basically only mathcounts problems or prepretory material  for Mathcounts. So in general all three of the forum are meant for middle schoolers (6, 7, and 8th) Also all the material in these three forums should be at a lower level than the Getting Started forum.

It's some what like that Classroom math is meant to be the easiest while Mathcounts is the hardest in Middleschool.

These responses are basically correct.  Classroom math is any ordinary 6th, 7th, and 8th grade curriculum.  The Basics is meant for easier middle school math team problems -- a step above classroom math.  MATHCOUNTS is where you can discuss any level of MATHCOUNTS problems.  Easier MATHCOUNTS problems can just as easily be problems in the Basics, so the categorization isn't necessarily easiest to hardest, but that's not a bad way to think about it.


integration
trigonometry
calculus
function
derivative
calculus computations

Show that the indefinite integrand $ \displaystyle \int e^{ax}\sin{bx}\,dx$ may be found without the use of integration by parts.
[hide="Hint"]rewrite it as $ \int e^{n\theta}\sin{k\theta}\,d\theta$.[/hide]

Quite so.

$ \boxed{\cos{bx}\equal{}\frac{e^{ibx}\plus{}e^{\minus{}ibx}}{2}}$ $ \boxed{\sin{bx}\equal{}\frac{e^{ibx}\minus{}e^{\minus{}ibx}}{2}}$

$ \int e^{ax}\cos{bx}\,dx\equal{}\frac{1}{2}\cdot \frac{\bar{z}e^{zx}\plus{}ze^{\bar{z}x}}{z\bar{z}}\plus{}C$

$ \int e^{ax}\sin{bx}\,dx\equal{}\frac{1}{2i}\cdot \frac{\bar{z}e^{zx}\minus{}ze^{\bar{z}x}}{z\bar{z}}\plus{}C$

where $ z,\bar{z}$ are the complex conjugates $ a\plus{}bi,a\minus{}bi$ respectively.

[quote="Carcul"]And how about $ \int (x^3 \plus{} x^2 \plus{} x \plus{} 1) \sin x\,dx$?[/quote]

[hide="Ans."]$ \minus{} (x^3 \plus{} x^2 \plus{} x \plus{} 1)\cos x \plus{} (3x^2 \plus{} 2x \plus{} 1)\sin x \plus{} (6x \plus{} 2)\cos x \minus{} 6\sin x \plus{} C.$[/hide]

Yes Kunny, I know you are aware of how to solve it, but the purpose was for Igiul to solve it. And how about hide the answer?

Getting back to the original problem: we can use the differential equations technique called "undetermined coefficients."

In fact this is precisely the right kind of problem for that: a linear constant coefficient nonhomogeneous ODE with the function on the right hand side belonging the appropriate class of functions:
\[ y'\equal{}e^{ax}\sin bx\]
We're looking for a particular solution, to which we'll add the solution of the corresponding homogeneous equation. In this case, that's $ y'\equal{}0,$ which has solution $ y\equal{}C.$

For the particular solution, we assume $ y$ has the following form:
\[ y\equal{}Ae^{ax}\cos bx\plus{}Be^{ax}\sin bx.\]
Take the derivative of that, match it to the right hand side, and equate corresponding terms. We should have two linear equations for the two unknown coefficients $ A$ and $ B.$

We can also solve kunny's problem by the same method, but it will take more terms and more coefficients.


probability
number theory
greatest common divisor
analytic geometry

How do I prove this? I am not sure if this is true, but I think its true as I read it somewhere...

The probability that two randomly chosen integers $a$ and $b$ have a given greatest common divisor $d$ is [hide]$6/{{\pi^{2}}{d^{2}}}$[/hide]

this might help
http://en.wikipedia.org/wiki/Greatest_common_divisor#Probabilities_and_expected_value

This really occured very often before...
See for example: http://www.mathlinks.ro/Forum/viewtopic.php?t=31028

[quote="Gyan"]Some what intuitive argument for the result:
Consider $n \times n$  lattice ($n$ VERY large), (Positive quadrant only) $n^{2}$  total number of points.
Now  let  $p$  be the probability that two randomly chosen integers are  co prime to each other.
Then “number of points you can see from the origin” = $pn^{2}$   (Others would be blocked by some other lattice point – (a,b) would block (ka,kb) because they are on the same line)

Number of points whose gcd = 2 = $p(n^{2})/4$
(Think of only those points where both coordinates are even – wider grid, (n/2xn/2  lattice points)
Similarly number of points whose gcd = 3 = $p (n^{2})/9$
(and number of points whose gcd = d = $p(n^{2})/d^{2}$

But adding all these you must get total number of points ($n^{2}$)  (remember $n$ is VERY large  :)  .. we go as high counting 'd' as we want our results to be accurate..and neglect the boundry effects  )

But  $p n^{2}(1+1/2+1/4+1/9 \dots) = n^{2}$
 Or $p(\pi^{2}/6) = 1$
which gives $p$  and what you need is $p/d^{2}$[/quote]

Thanks...

[quote="Gyan"]Some what intuitive argument for the result:
Consider $n \times n$  lattice ($n$ VERY large), (Positive quadrant only) $n^{2}$  total number of points.
Now  let  $p$  be the probability that two randomly chosen integers are  co prime to each other.
Then “number of points you can see from the origin” = $pn^{2}$   (Others would be blocked by some other lattice point – (a,b) would block (ka,kb) because they are on the same line)

Number of points whose gcd = 2 = $p(n^{2})/4$
(Think of only those points where both coordinates are even – wider grid, (n/2xn/2  lattice points)
Similarly number of points whose gcd = 3 = $p (n^{2})/9$
(and number of points whose gcd = d = $p(n^{2})/d^{2}$

But adding all these you must get total number of points ($n^{2}$)  (remember $n$ is VERY large  :)  .. we go as high counting 'd' as we want our results to be accurate..and neglect the boundry effects  )

But  $p n^{2}(1+1/2+1/4+1/9 \dots) = n^{2}$
 Or $p(\pi^{2}/6) = 1$
which gives $p$  and what you need is $p/d^{2}$[/quote]

Very nice proof, I would have never thought of it this way. But is this "the right way" to do it? I mean do you have to hardcore proove it or is this fine?


I was browsing my CRC the other day (with no power, I'm sometimes at a loss of things to do), and saw a strange structure:

\[ \text{N}\equiv\text{N}=\text{N}-\text{CH}_3\quad \text{(azidomethane)} \]Um.  Nitrogen forming 5 bonds?  Obviously we're not hybridizing into the $2d$ sublevel here :lol:.  I'm assuming this is a resonance structure... maybe something like this?

\[ \left\{\underbrace{\overline{\text{N}}\equiv\stackrel{+}{\text{N}}-\stackrel{-}{\overline{\underline{\text{N}}}}-\text{CH}_3}_{\text{minor}} \quad\longleftrightarrow\quad\underbrace{\stackrel{-}{\overline{\underline{\text{N}}}}=\stackrel{+}{\text{N}}=\overline{\text{N}}-\text{CH}_3}_{\text{major}\right\} }\]I figure the structure I labeled as the major contribution would have a more stable negative charge; the minor structure would be destabilized by the electron-donating alkyl group.

Can anyone confirm or deny this for me?  It's driving me up the wall.  Thanks.

yes, your two structures are correct, (at least they make sense).
and for yes your stability of the two is correct also.  :)

Hm... wikipedia says that an azido group is an electrophile on the terminal $\text{N}$ and is electron-donating to the carbon atom.  It has a diagram showing what I labeled the minor structure.  Can anyone explain this?


Hello,

I will like to train my self in soloving differential equation but I haven't an good exemple can someone help with this?

$y''+y=8cos(x)cos(2x)$

Greets Thanks a lot.

The method I gave in the other thread works exactly the same here...


function algebra polynomial quadratics Vieta

hello,

  Let the equation $ (E): x^3\minus{}x\minus{}1\equal{}0$ , prove that $ \,(E)\,$ have three distinct solution $ \,\alpha,\,\beta\:$ and $ \,\lambda$ .

Evaluate $ S\equal{}\frac{1\plus{}\alpha}{1\minus{}\alpha} \plus{} \frac{1\plus{}\beta}{1\minus{}\beta} \plus{} \frac{1\plus{}\lambda}{1\minus{}\lambda}$

[hide="Solution"] $ (x^3 \minus{} x \minus{} 1)' \equal{} 3x^2 \minus{} 1$, which has no common roots with $ x^3 \minus{} x \minus{} 1$, so $ \alpha, \beta, \gamma$ are indeed all distinct.

Let $ g(x) \equal{} \frac{1 \plus{} x}{1 \minus{} x}$.  This function has inverse $ g^{\minus{}1}(x) \equal{} \frac{1 \minus{} x}{1 \plus{} x}$ (incidentally, what other functions have this property?), so it follows that the numerator of

$ \left( \frac{1 \minus{} x}{1 \plus{} x} \right)^3 \minus{} \left( \frac{1 \minus{} x}{1 \plus{} x} \right) \minus{} 1$

has roots $ \frac{1 \plus{} \alpha}{1 \minus{} \alpha}$, etc.  Expanded, this is (after multiplying by $ \minus{}1$)

$ x^3 \minus{} x^2 \plus{} 7x \plus{} 1$

so it follows that $ S \equal{} 1$. [/hide]

thanks t0rajir0u , but I don't understand your proof  :(  you can explain more than ? thanks

Which line don't you understand?  A specific question will be answered more promptly than a general one.

why $ \frac{1 \plus{} \alpha}{1 \minus{} \alpha}$ ... is roots of  $ \left( \frac{1 \minus{} x}{1 \plus{} x} \right)^3 \minus{} \left( \frac{1 \minus{} x}{1 \plus{} x} \right) \minus{} 1$  ?  and how you see that $ S\equal{}1$ ?? sorry :(

Because $ \frac{1 \minus{} \frac{1 \plus{} \alpha}{1 \minus{} \alpha}}{1 \plus{} \frac{1 \plus{} \alpha}{1 \minus{} \alpha}} \equal{} \alpha$.  As for the second question, the polynomial I wrote down is $ (x \minus{} \frac{1 \plus{} \alpha}{1 \minus{} \alpha})(x \minus{} \frac{1 \plus{} \beta}{1 \minus{} \beta})(x \minus{} \frac{1 \plus{} \gamma}{1 \minus{} \gamma})$, so $ S$ is the negative of its quadratic term.

Why do you only consider the numerator of $ \left( \frac{1 \minus{} x}{1 \plus{} x} \right)^3 \minus{} \left( \frac{1 \minus{} x}{1 \plus{} x} \right) \minus{} 1$?

Because its roots are when it equals 0, which is the same as when the numerator is 0. Nice solution!

[quote="t0rajir0u"]Because $ \frac {1 \minus{} \frac {1 \plus{} \alpha}{1 \minus{} \alpha}}{1 \plus{} \frac {1 \plus{} \alpha}{1 \minus{} \alpha}} \equal{} \alpha$[/quote]
I think this is a mistakes, isn't it $ \minus{} \alpha$ ? The inverse should be the negative of this, $ \frac {\alpha \minus{} 1}{1 \plus{} \alpha}$

I dont understand why S=1 ? Explain please :(

Ah - you're right, of course.

[hide="The correct answer..."] is given by the numerator of

$ \left( \frac{x \minus{} 1}{x \plus{} 1} \right)^3 \minus{} \frac{x \minus{} 1}{x \plus{} 1} \minus{} 1$

which is $ x^3 \plus{} 7x^2 \minus{} x \plus{} 1$, so actually $ S \equal{} \minus{}7$. [/hide]

As for the reason why this works, that's [url=http://www.artofproblemsolving.com/Wiki/index.php/Vieta's_formulas]Vieta's sums[/url].  Do you know how to evaluate sums like $ \alpha \plus{} \beta \plus{} \gamma$ or $ \alpha^3 \plus{} \beta^3 \plus{} \gamma^3$?


[quote="Fitim"]If $a,b,c\geq 0$ prove:
$\frac{(a+b+c)^{2}}{3}\geq a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}$[/quote]

$\frac{(a+b+c)^{2}}{3}\geq ab+bc+ca = \sum (\sqrt{ab})^{2}\geq \sum \sqrt{ab}\sqrt{ac}=\sum a\sqrt{bc}$

Another solutions:
After transfer we have:
$a^{2}+b^{2}+c^{2}+2ab+2ac+2bc\geq 3a\sqrt{bc}+3b\sqrt{ac}+3c\sqrt{ab}$

Now with AM\geq GM:
$\frac{a^{2}}{2}+\frac{a^{2}}{2}+\frac{bc}{2}+\sum\frac{ab}{2}\geq 6\sqrt[6]{\frac{a^{6}b^{3}c^{3}}{2^{6}}}= 3a\sqrt{bc}$

 :wink:


linear algebra
matrix
modular arithmetic
symmetry
induction
function
algebra

On a given $ ({2^n}\minus{}1) x ({2^n}\minus{}1)$ board, each unit square it's written a +1 or a -1.
We say that the distribution it's "heavy" if every number it's equal to the product of the numbers written on the neighbour unit squares (two squares are neighbour if they share a side).
What is the number of heavy distributions?
And if the board is of mxn?

(Excuse me for my poor english)

The number of "heavy" distributions on a $ 2^n\minus{}1 \times 2^n\minus{}1$ board is 1.
[hide="proof"]
Given integers $ m$ and $ n$, define a system of linear equations over $ GF(2)$, the field of two elements, as follows. Take $ mn$ variables $ x_{i,j}$ with $ i \in \{1,\ldots,m\}$ and $ j \in \{1,\ldots,n\}$ and consider the $ mn$ equations

$ x_{i,j} \plus{} x_{i\minus{}1,j} \plus{} x_{i\plus{}1,j} \plus{} x_{i,j\minus{}1} \plus{} x_{i,j\plus{}1} \equal{} 0$

where $ x_{k,l}$ is interpreted as 0 whenever $ k,l$ are out of range (namely $ k\equal{}0$ or $ k\equal{}m\plus{}1$ or $ l\equal{}0$ or $ l\equal{}n\plus{}1$). Let $ d(m,n)$ be the dimension of the solution space for that system. Then the number of "heavy" distributions on a $ m \times n$ board is $ 2^{d(m,n)}$. The reason is that we can replace $ \plus{}1$ and $ \minus{}1$ with $ 0$ and $ 1$, respectively, and multiplication with addition modulo 2 (namely, addition in $ GF(2)$), and the "heavy" condition is just that the system of equations is satisfied.

Let us say that $ (m,n)$ is a [b]regular [/b]board if $ d(m,n)\equal{}0$. The reason for the terminology is that this is just the definition of a regular matrix, or independent system of equations: the $ mn$ equations in the $ mn$ variables $ x_{i,j}$ only have the trivial solution $ x_{i,j}\equal{}0$ for all $ i,j$. (A board is regular iff the only "heavy" distribution is the trivial one consisting of all $ \plus{}1$s.)

The crucial fact about regular boards is this: say we change the system of equations so that the RHS, instead of always being 0, is 1 for some of the equations. Then it is still true that the system has a unique solution. This is a standard fact in linear algebra. In the homogeneous case (all equations are $ \equal{}0$), the unique solution is obviously the 0 solution. In other cases it's something else, but it always exists and is unique.

We will need a simple fact about $ m \times 1$ boards: they are regular iff $ m \not \equiv 2 \pmod{3}$. This is very easy to see (in fact it's easy to also show that $ d(m,1)\equal{}1$ when $ m \equiv 2 \pmod{3}$) and we leave this as an exercise.

We can now prove:

[b]Lemma[/b]: If $ (m,n)$ is regular and $ m \not \equiv 2 \pmod{3}$ then $ (m,2n\plus{}1)$ is regular as well.

[b]Proof[/b]: Consider the $ m \times 2n\plus{}1$ board as consisting of a $ m \times n$ board on the left (call it A), another $ m \times n$ board on the right (call it B), and a $ m \times 1$ column in the middle (call it I). Let's fill in the cells in that middle column in some arbitrary fashion. Since $ (m,n)$ is regular, this forces a unique way filling in the sub-boards A and B so that the equations are satisfied on those sub-boards.

Also note that, by symmetry, the cells in board B are just the mirror image of the cells in board A: just swap the variables left-to-right and you'll find that the variables in B satisfy the exact same set of linear equations as the variables in A.

In particular, the cells to the left and to the right of I have the same value. We now consider what the constraints tell us about the cells of I. Each such cell has to be the sum of the one above it (or 0 if it's at the top), the one below it (or 0 if it's at the bottom), and the ones to the left and right - but those last ones always add up to 0. In other words I satisfies the "heavy" condition in and of itself. Since $ m \not \equiv 2 \pmod{3}$, the board I is a regular $ (m,1)$ board, so the only way to satisfy the requirement on I is to fill it with 0's. This forces both A and B to be all 0, so the large $ (m,2n\plus{}1)$ is regular. QED

It's now easy to show by induction that $ d(2^k\minus{}1,2^k\minus{}1) \equal{} 0$. In fact, $ d(1,1)\equal{}0$, or if you don't like this degenerate case just prove directly that $ d(3,3)\equal{}0$. In general, let $ m\equal{}2^k\minus{}1$, and assume by induction that $ (m,m)$ is regular. Then $ m \not \equiv 2 \pmod{3}$ (since powers of 2 are never divisible by 3), so by the lemma $ (m,2m\plus{}1)$ is regular, so of course $ (2m\plus{}1,m)$ is regular as well, and so once again by the lemma $ (2m\plus{}1,2m\plus{}1)$ is regular (note that $ 2m\plus{}1 \equal{} 2^{k\plus{}1}\minus{}1$ is also never 2 mod 3). This finishes the proof.
[/hide]
The general question for a $ m \times n$ board seems very difficult, and apart from some periodicity properties which I mention below I don't think one can say a whole lot about it.
[hide="more information"]
We saw that $ d(1,n)$ only depends on the residue of $ n$ mod 3. This is a general phenomenon: for any fixed $ m$, the function $ d(m,n)$ is periodic. This isn't hard to prove and is left as another exercise.

Specifically, it's not hard to see that $ d(2,n)$ depends only on the residue of $ n$ modulo 4. Similarly, $ d(3,n)$ depends on the residue of $ n$ mod 6, while $ d(4,n)$ has period 5 and $ d(5,n)$ has period 24. As can be seen the periods are very erratic. Interestingly, It seems empirically that $ d(12,n)\equal{}0$, and in fact that $ d(m,n)\equal{}0$ whenever $ m \equiv \pm 2 \pmod{10}$. I've no idea why this is happening and how to prove this.

[/hide]


On a blackboard ten numbers are written: 1,0,0,0,0,0,0,0,0,0.
In each step one may erase two of the numbers and replace them by two copies of the average of the two numbers erased.
What is the smallest positive number that can ever appear on the blackboard?

Please explain to me with steps.. THZ A LOT!!

[quote="Baronov"]This will need a proof.[/quote]

how to prove ?

[quote="Baronov"]This will need a proof.[/quote]

I think it is quite obvious that the smallest positive number can only be obtained by taking the average of the two smallest numbers.

Yeah, it can be easily proved, but is very tedious. This is a finite problem after all.

[quote="JBL"]Sure it's obvious.  Can you prove it?[/quote]

I do agree with IrinaSharova. It is obvious. Here is a proof :

Let $ f(n)$, $ n\in[0,9]$ the smallest positive number which can be obtained (starting from the given situation ) in a situation where we have exactly $ n$ "0" numbers on the board.

Clearly, a step :
 - can't increase the number of zeroes
 - can't decrease the smallest positive number if the number of zeroes does not decrease
 - can't decrease of more than $ 50\%$ the smallest positive number if the the number of zeroes decreases.

In other words, $ f(9)=1$ and $ f(n-1)\geq\frac{f(n)}{2}$ and so $ f(0)\geq\frac{1}{2^{9}}=\frac{1}{512}$

And since IrinaSharova gave one example of value $ \frac{1}{512}$, we have $ f(0)=\frac{1}{512}$
Q.E.D.


Which one of the following statements is false?  All equilateral triangles are

$\textbf{(A)} \  \text{ equiangular} \qquad
\textbf{(B)} \  \text{isosceles} \qquad
\textbf{(C)} \  \text{regular polygons } \qquad
\textbf{(D)} \   \text{congruent to each other} \qquad
\textbf{(E)} \   \text{similar to each other} $

[hide]
D becuase the lengths of the sides can be different.[/hide]

[hide]
D, they can be different side lenghts.[/hide]


Let $ a_n \equal{} 2 \minus{} \frac {1}{n^2 \plus{} \sqrt {n^4 \plus{} \frac {1}{4}}}$, $ n \equal{} 1,2,...$ Prove that  $ \sqrt {a_1} \plus{} \sqrt {a_2} \plus{} ... \plus{} \sqrt {a_{119}}$ is an integer.

Rationalize the denominator and  denest the radical to get a simple and easy solution.


Let $A$ be an operator from $l_{p}\to l_{p}$, where $p=\frac{7}{3}$, $Ax = (\sqrt{2}x_{1},...,n^{\frac{1}{n}}x_{n-1},... )$. Find the norm $\parallel A\parallel $.

Let $Ax=(a_{1}x_{1},a_{2}x_{2},\ldots)$, where $\sup |a_{i}| = a <\infty.$
Then
\[\parallel A\parallel =\sup_{x\in l_{p}}\frac{\parallel Ax\parallel }{\parallel x\parallel }=\sup_{x\in l_{p}}\sqrt{\frac{\sum_{k=1}{\infty}|a_{i}x_{i}|^{p}}{\sum_{k=1}{\infty}|x_{i}|^{p}}}\le\]
\[\le \sup_{x\in l_{p}}\sqrt{\frac{\sum_{k=1}{\infty}|a|^{p}|x_{i}|^{p}}{\sum_{k=1}{\infty}|x_{i}|^{p}}}=|a|\].
Now, let $i_{j}$- is sequence such that $a_{i_{j}}\to a,j\to\infty$.
Define $\xi_{j}=(0,0,...,0,1,0...,0,....)$, where $1$ is on the $i_{j}$-th place.
Then $\parallel x_{i_{j}}\parallel =1 \quad \forall j$, and $\parallel A\xi_{j}\parallel =|a_{i_{j}}|\to a$ when $j\to \infty$. Thus \[\parallel A\parallel =\sup_{x\in l_{p}}\frac{\parallel Ax\parallel }{\parallel x\parallel }\ge \parallel A\parallel =\sup_{j \in \mathbb{N}}\frac{\parallel A\xi_{j}\parallel }{\parallel \xi_{j}\parallel }\ge a\].
So \[\parallel A\parallel =a=\sup_{i}|a_{i}|\]
In our case, i thnk $\parallel A\parallel =\sqrt[3]{3}$
($\sqrt[3]{3}>\sqrt[2]{2}$)  :)
If $x=(x_{0},x_{1},...)$ then $\parallel A\parallel =\sup \{0,\sqrt[2]{2},\sqrt[3]{3},\sqrt[4]{4},...\}=\sqrt[3]{3}$


Αν είναι $ a,b,c>0$ και $ a\plus{}b\plus{}c\equal{}1$ να αποδείξετε ότι

$ \frac{a}{1\plus{}b\plus{}c}\plus{}\frac{b}{1\plus{}c\plus{}a}\plus{}\frac{c}{1\plus{}a\plus{}b}\geq\frac{3}{5}$

Πολλαπλασιάζουμε κάθε όρο με τον αριθμητή, κάνουμε Andreescu και χρησιμοποιούμε την
$ ab\plus{}bc\plus{}ca\leq \frac{(a\plus{}b\plus{}c)^2}{3}\equal{}\frac{1}{3}$ και τέλος :)

Ωραίος bilot4 γρήγορη και ωραία λύση   :)

Πώς δεν το σκέφτηκα να χρησιμοποιήσω την Andreescu (που είναι και η αγαπημένη μου :wink: )   Η δική μου είναι μεγαλύτερη και πιο πολύπλοκη  (αφού ομογενοποιήσω χρησιμοποιώ Jensen).

Παιδιά Lagrange multipliers....... :wink:

¨Οταν λύνεται στοιχειωδω΄ς νομίζω οι πολλαπλασιαστές Lagrange ναι μεν δίνουν τη λύση, αλλά ενδείκνυνται περισσότερο για πιο δύστροπα προβλήματα...

Στέλιος

Κοίτα Στέλιο.....
 Μου αρέσουν οι πρωτότυπες λύσεις αλλά σου δίνει άλλη σιγουριά να ξέρεις ότι έχεις μια σίγουρη λύση στο τσεπάκι σου :wink:

[quote="galois01"]Αν είναι $ a,b,c > 0$ και $ a \plus{} b \plus{} c \equal{} 1$ να αποδείξετε ότι

$ \frac {a}{1 \plus{} b \plus{} c} \plus{} \frac {b}{1 \plus{} c \plus{} a} \plus{} \frac {c}{1 \plus{} a \plus{} b}\geq\frac {3}{5}$[/quote]

Εχουμε:

$ \frac {a}{1 \plus{} b \plus{} c} \plus{} \frac {b}{1 \plus{} c \plus{} a} \plus{} \frac {c}{1 \plus{} a \plus{} b}\geq\frac {3}{5}$

ή

$ \frac {a}{1 \plus{} b \plus{} c \plus{} a \minus{} a} \plus{} \frac {b}{1 \plus{} c \plus{} a \plus{} b \minus{} b} \plus{} \frac {c}{1 \plus{} a \plus{} b \plus{} c \minus{} c}\geq\frac {3}{5}$

ή

$ \frac {a}{2 \minus{} a} \plus{} \frac {b}{2 \minus{} b} \plus{} \frac {c}{2 \minus{} c}\geq\frac {3}{5}$

θετουμε x = 2 - α, y = 2 - b, z = 2 - c
και εχουμε

$ \frac {2 \minus{} x}{x} \plus{} \frac {2 \minus{} y}{y} \plus{} \frac {2 \minus{} z}{z}\geq\frac {3}{5} \leftrightarrow \frac {2}{x} \plus{} \frac {2}{y} \plus{} \frac {2}{z} \minus{} 3\geq\frac {3}{5} \leftrightarrow \frac {2}{x} \plus{} \frac {2}{y} \plus{} \frac {2}{z} \geq\frac {3}{5} \plus{} 3 \leftrightarrow \frac {2}{x} \plus{} \frac {2}{y} \plus{} \frac {2}{z} \geq\frac {18}{5} \leftrightarrow \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \geq\frac {9}{5} \leftrightarrow \frac {1}{\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}} \leq\frac {5}{9} \leftrightarrow \frac {3}{\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}} \leq\frac {15}{9} \leftrightarrow \frac {3}{\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}} \leq\frac {5}{3}$

Το οποιο ισχυει αφου απο την ανισοτητα αρμονικου - αριθμιτικου μεσου εχουμε:

$ \frac {3}{\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z}} \leq\frac {x \plus{} y \plus{} z}{3} \equal{} \frac { \minus{} a \minus{} b \minus{} c \plus{} 2 \plus{} 2 \plus{} 2}{3} \equal{} \frac {6 \minus{} 1}{3} \equal{} \frac {5}{3}$

Edit: προσθετω ακομα οτι η ισοτητα ισχυει οταν $ x\equal{}y\equal{}z$, δηλαδη οταν $ a\equal{}b\equal{}c$

Ισος η πρωτη ασκηση που λυνω σε αυτο το φορουμ που να ξεφευγει απο τα σχολικα δεδομενα :P

Ωραιος!!! Κατι παρομοιο σκεφτικα κι εγω αλλα η λυση του bilot4 ειναι πιο συντομη.

Επηρεασμένος από άλγεβρα Α λυκείου (προφανώς η Andreescu είναι η σωστότερη λύση από πλευράς συντομίας) δίνω μία όμορφη λύση:

Θέτουμε $ a\plus{}b\equal{}cos2x$ κτλ. Παρατηρήστε ότι μπορούμε να το κάνουμε αυτό αφού δεν ξεπερνούν το 1. Οπότε τελικά παίρνουμε με τύπο διπλασίου τόξου του σχολικού βιβλίου:

$ \sum(cosx)^2\equal{}\frac{5}{2}$

Με τύπο αποτετραγωνισμού του σχολικού βιβλίου:

$ \sum\frac{1\minus{}cos2x}{1\plus{}cos2x}\equal{}\sum(tanx)^2\equal{}\sum\frac{1\minus{}(cosx)^2}{(cosx)^2}\equal{}\sum\frac{1}{(cosx)^2}\minus{}3$

οπότε αρκεί

$ \sum\frac{1}{(cosx)^2}\geq\frac{18}{5}$

που ισχύει από Andreescu. :)

Μιας και παρουσιάστηκαν διάφορες λύσεις θα παραθέσω και την λύση που βρήκα με χρήση της Jensen.

H αρχική είναι ισοδύναμη με την

$ \frac {1 - (b + c)}{1 + b + c} + \frac {1 - (a + c)}{1 + a + c} + \frac {1 - (a + b)}{1 + a + b}\geq\frac {3}{5}$

Θέτουμε

$ a + b = x$

$ a + c = y$

$ a + b = z$

με $ x + y + z = 2$ οπότε αρκεί να αποδείξουμε ότι

$ \frac {1 - x}{1 + x} + \frac {1 - y}{1 + y} + \frac {1 - z}{1 + z}\geq\frac {3}{5}$

Η συνάρτηση $ f(a) = \frac {1 - a}{1 + a}$ με $ a>0$  είναι κυρτή  αφού $ f^{''}(a) = \frac {4(a + 1)}{(a + 1)^{4}}>0$ για κάθε $ a>0$ και επομένως από την ανισότητα Jensen έχουμε

$ \frac {f(x) + f(y) + f(z)}{3}\geq\(f(\frac {x + y + z}{3}) = f(\frac {2}{3}) = \frac {1}{5}$

οπότε προκύπτει το ζητούμενο. :)

Διάβασα σε προηγούμενο ποστ σου ότι έκανες jensen αφού ομογενοποίησες. Όμως βλέπω ότι δεν ομογενοποίησες. Μια αντικατάσταση έκανες. Ομογενής ανισότητα είναι μια ανισότητα που δε μεταβάλλεται αν όπου $ a$ ας πούμε θέσεις $ ka$. :wink:

Ίσως το είπες εκ παραδρομής ίσως όμως δεν το είχες κατανοήσει. :wink:

Protonie έχω καταλάβει την είναι η ομογενής ανισότητα. :wink: Απλά στο προηγούμενο post  έγραψα οτι ομογονοποιώ από αφηρημάδα.  Ευχαριστώ πάντως  για την παρατήρηση :wink:
P.S Αν θέλει καποιός από τους moderators ας το διορθώσει για να μην δημιουργείται σύγχηση. :)

μιας που υπαρχει ποικιλια ειπα και εγω να βαλω μια λυση με Holder ή με b-c-s.:P
$ \frac{5}{3} \sum \frac{a}{1\plus{}b\plus{}c}\geq [\sum a(1\plus{}b\plus{}c)][\sum \frac{a}{1\plus{}b\plus{}c}]\geq(a\plus{}b\plus{}c)^2\equal{}1$
QED


assume a complete thermal transfer
how many grams of magnesium hydroxide can be decomposed from the heat lost by 125 grams of water vapor at 115.0 degrees C cooling down to -45.0 degrees C

For this problem, we need some additional data:

1) Heat capacity of water vapor
2) Enthalpy of vaporization of water
3) Heat capacity of liquid water
4) Enthalpy of solidification of water
5) Heat capacity of ice
6) Heat absorved per mole of Mg(HO)2


function
AMC
AIME
algebra
polynomial
logarithms
algebra unsolved

Let  $ f: R\rightarrow R$ such that  $ f(x)f(2x^2)\equal{}f(2x^3\plus{}x)$ and $ f(0)\equal{}1$ , $ f(2)\plus{}f(3)\equal{}125$.
Find $ f(5)$

[quote="aktyw19"]Let  $ f: R\rightarrow R$ such that  $ f(x)f(2x^2) \equal{} f(2x^3 \plus{} x)$ and $ f(0) \equal{} 1$ , $ f(2) \plus{} f(3) \equal{} 125$.
Find $ f(5)$[/quote]

Does anybody have a solution for this problem ?

I'm quite sure it's impossible : if it exists one non constant function such that $ f(x)f(2x^2) \equal{} f(2x^3 \plus{} x)$, then it exists infinitely many. And I think we can change $ f(5)$ without changing $ f(2)\plus{}f(3)$

Moreover, the constraint $ f(0)\equal{}1$ is useless since any solution of the problem with $ f(0)\equal{}1$ implies another solution with $ f(0)\equal{}0$

Looks like an old AIME problem, but without the polynomial condition:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=141148[/url]

Oh thanks !

The polynomial conditon is very very strong and I think that aktyw19 forgot it.

Btw, using the post you indicated, I can then show that the original problem is impossible to solve :

Let $ f(x)\equal{}(x^2\plus{}1)^2$, which is a polynomial solution.

Let $ h(x)$ a non continuous solution of Cauchy's equation such that :
$ h(\ln 2)\equal{}\ln 2$
$ h(\ln 5)\equal{}\ln 5$
$ h(\ln 13)\ne \ln 13$

Such a function exists since $ \frac{\ln 2}{\ln 5},\frac{\ln 2}{\ln 13},\frac{\ln 5}{\ln 13}\notin\mathbb Q$

Then $ g(x)\equal{}e^{h(\ln f(x))}$  is also a solution of the original problem and $ g(5)\ne f(5)$

I really wonder why posters dont carefully check the problems when they copy them :( !!


the nth term is(x+n)/n.if the sum of the first 50 terms is equal to 50,find x.i got 0.
iwana see your solution.
2.this is a similar question where the nth term is 1/(k(k+1)).where you are expected to find an expression for the sum.[/hide]

Ignore what I posted.  I assumed it was for an arithmetic series, which is wrong.

For 2, I'll assume you ment $ \frac{1}{n(n\plus{}1)}$.

[hide="2"]This is a pretty common series.  You can expand it and get
$ \frac{1}{n(n\plus{}1)}\equal{}\frac{1}{n}\minus{}\frac{1}{n\plus{}1}$.
This is a telescoping function.  Now the sum from n=1 to some number k is
$ \frac{1}{1}\minus{}\frac{1}{2}\plus{}\frac{1}{2}\minus{}\frac{1}{3}\ldots\frac{1}{k}\plus{}\frac{1}{k}\minus{}\frac{1}{k\plus{}1}$.
All the terms in the middle cancel (do you see why?) and you end up with
$ \frac{1}{1}\minus{}\frac{1}{k\plus{}1}\equal{}\frac{k}{k\plus{}1}$.[/hide]

Yes 0 is right.

[hide="solution"]$ a_{n} \equal{} \frac {x \plus{} n}{n}$
We know that the sum of the first 50 terms is 50 so:
$ x \plus{} 1 \plus{} \frac {x \plus{} 2}{2} \plus{} \frac {x \plus{} 3}{3} \plus{} ... \plus{} \frac {x \plus{} 50}{50} \equal{} 50$
Its easy to see from here that $ x \equal{} 0$ is a solution.
[/hide]

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

Thus, $ \sum_{k \equal{} 1}^{50} \frac {x \plus{} k}{k} \equal{} \sum_{k \equal{} 1}^{50} \left( 1 \plus{} \frac {1}{k} x \right) \equal{} 50 \plus{} \left( 1 \plus{} \frac {1}{2} \plus{} \ldots \plus{} \frac {1}{50} \right) x$

Now, let's use the shorthand $ C \equal{} 1 \plus{} \frac {1}{2} \plus{} \ldots \plus{} \frac {1}{50} > 0$.

Our equation is $ 50 \plus{} Cx \equal{} 50 \implies Cx \equal{} 0 \implies x \equal{} 0$

There is a simple way to think about this problem. We have division of a number by itself (plus some constant) and that would be [i]at least[/i] 1 for every term. If the sum of the first 50 terms is 50, it means that every term is equal to one, so we can conclude that $ x\equal{}0$.

if the sum of the first 50 terms in that particular series is 50,it is easy to see that the sum of the first term is one
i.e x+1=1
    x=1-1=0


function
trigonometry
calculus
calculus computations

If,
$ y\equal{} a\sin \theta \plus{} b\cos \theta$
Then the range of values of $ y$ lies in is?

$ y\equal{}\sqrt{a^{2}\plus{}b^{2}}\left[\frac{a}{\sqrt{a^{2}\plus{}b^{2}}}\sin\theta\plus{}\frac{b}{\sqrt{a^{2}\plus{}b^{2}}}\cos\theta\right]$

$ y\equal{}\sqrt{a^{2}\plus{}b^{2}}\sin\left(\theta\plus{}\arctan\frac{a}{b}\right)$

After this everthing is easy. :D

hello
[hide="solution"]
Just divide and multiply by $ \sqrt {a^2 \plus{} b^2}$
range would be -$ \sqrt {a^2 \plus{} b^2}$ to +$ \sqrt {a^2 \plus{} b^2}$

[/hide]Thank u.


Στο παρακάτω σχήμα $ K$ είναι το περίκεντρο του τριγώνου $ ABC$ , $ O$ είναι το περίκεντρο τυχαίου κύκλου που περνάει από τα σημεία $ B,C$ και  $ L$ είναι το περίκεντρο του τριγώνου $ ADE$.Να αποδείξετε ότι
  [color=red]α)[/color] $ AK \parallel OL$  και στη συνέχεια ότι
  [color=red]β) [/color] $ ON \perp AN$, όπου $ N$  είναι το δεύτερο σημείο τομής των κύκλων $ K$ και  $ L$

[b] ΣΧΟΛΙΟ[/b]

[i] Στους παλαιότερους το δεύτερο ερώτημα κάτι θα θυμίσει.[/i]

Για το πρώτο ερώτημα έχουμε:
$ DE\perp{LO}$ (αφού $ LO$ η μεσοκάθετος του $ DE$). Aρα αρκεί $ DE\perp{AK}$.

Πράγματι ξέρουμε ότι $ \angle{BAK}\equal{}90\minus{}\angle{C}\equal{}90\minus{}\angle{EDA}$ αφού το $ EDBC$ έιναι εγγράψιμο.

Αρα $ \angle{BAK}\plus{}\angle{EDA}\equal{}90$ και τελειώσαμε :)

Το δεύτερο ερώτημα δεν το έχω σκεφτεί ακόμα :blush:

[quote="silouan"]Το δεύτερο ερώτημα είναι IMO 1985 pr 5 . Υπάρχουν δύο λύσεις μία με ριζικούς άξονες και μία με αντιστροφή κέντρου Α  :)[/quote]
Εμένα το δεύτερο ερώτημα μου θύμισε τις αντιπαράλληλες ευθείες και έτσι από $ AL\perp BC,$ βγαίνει νομίζω εύκολα. :wink:

Κώστας Βήττας.

Όλες οι λύσεις που ήξερα μέχρι τώρα γγια την άσκηση είχαν κάτι μετρικό μέσα τους , και έίπα να βρω μια καθαρα γεωμετρική απόδειξη .

Έχουμε ότι $ \angle{DAL} \equal{} 90 \minus{} \angle{AED} \equal{} 90 \minus{} \angle{ABC}$ άρα $ AL\bot BC$
Επομένως $ AL//KO$ και $ AK//LO$ (από το α) ερώτημα) άρα $ AKOL$ παραλληλόγραμμο και έστω $ P$ το σημείο τομής των διαγωνίων του .Έστω ότι η $ KP$ τέμνει την $ AN$ στο $ S$ . Όμως η KL είναι μεσοκάθετη της ΑΝ άρα το S είναι το μέσον του ΑΝ.
Άρα η ΡS που συνδέει τα μέσα των πλευρών στο τρίγωνο $ ANO$ είναι κάθετη στην ΑΝ άρα το ίδιο θα συμβαίνει και με την ΟΝ .

Η λύση είναι αφιερωμένη στον κ .Κώστα Βήττα  :)

[quote="silouan"]Όλες οι λύσεις που ήξερα μέχρι τώρα γγια την άσκηση είχαν κάτι μετρικό μέσα τους , και έίπα να βρω μια καθαρα γεωμετρική απόδειξη .

Έχουμε ότι $ \angle{DAL} \equal{} 90 \minus{} \angle{AED} \equal{} 90 \minus{} \angle{ABC}$ άρα $ AL\bot BC$
Επομένως $ AL//KO$ και $ AK//LO$ (από το α) ερώτημα) άρα $ AKOL$ παραλληλόγραμμο και έστω $ P$ το σημείο τομής των διαγωνίων του .Έστω ότι η $ KP$ τέμνει την $ AN$ στο $ S$ . Όμως η KL είναι μεσοκάθετη της ΑΝ άρα το S είναι το μέσον του ΑΝ.
Άρα η KS που συνδέει τα μέσα των πλευρών στο τρίγωνο $ ANO$ είναι κάθετη στην ΑΝ άρα το ίδιο θα συμβαίνει και με την ΟΝ .

Η λύση είναι αφιερωμένη στον κ .Κώστα Βήττα  :)[/quote]

Σιλουανέ , ακριβώς !

 Επειδή κυκλοφορούν μετρικές λύσεις, αυτός είναι ο λόγος που την έβαλα - λίγο σπασμένη - σε αυτή τη μορφή. Η λύση σου είναι ωραία.καθαρά , ΄μη μετρική λύση είχα δει σε ένα αυστραλέζικο βιβλίο , διότι την ίδια άσκηση την έγβαλαν και στην  ΕΟ Αυστραλίας την επόμενη χρονιά !!!
 Νομίζω ότι άξιζε να γνωρίσουμε και αυτή τη λύση.
   Μπάμπης
(Α! Τη λύση με αντιστροφή δεν την γνωρίζω. Στείλτε με κάποιο σκάνερ , αν βρεις χρόνο).Ευχαριστώ.

Γεια σας. Για το πρώτο ερωτημα, έφερα την εφαπτομενη του κύκλου Κ στο  Α , χρησιμοποίησα την γωνία χορδής εφαπτομένης=με την εγγεγραμμένη Β= με την C(εξωτερική) από το εγγράψιμο και τελικα DE// εφαπτ. και καθώς ΚΑ κάθετη στην εφαπτομ θα είναι κάθετη και στην DE

[quote="silouan"]Όλες οι λύσεις που ήξερα μέχρι τώρα γγια την άσκηση είχαν κάτι μετρικό μέσα τους , και είπα να βρω μια καθαρά γεωμετρική απόδειξη .

Έχουμε ότι $ \angle{DAL} \equal{} 90 \minus{} \angle{AED} \equal{} 90 \minus{} \angle{ABC}$ άρα $ AL\bot BC$
Επομένως $ AL//KO$ και $ AK//LO$ (από το α) ερώτημα) άρα $ AKOL$ παραλληλόγραμμο και έστω $ P$ το σημείο τομής των διαγωνίων του .Έστω ότι η $ KP$ τέμνει την $ AN$ στο $ S$ . Όμως η $ KL$ είναι μεσοκάθετη της $ AN$ και άρα το $ S$ είναι το μέσον του $ AN.$
Άρα η $ KS$ που συνδέει τα μέσα των πλευρών στο τρίγωνο $ ANO$ είναι κάθετη στην $ AN$ άρα το ίδιο θα συμβαίνει και με την $ ON.$

Η λύση είναι αφιερωμένη στον Κώστα Βήττα  :)[/quote]
Σιλουανέ σε ευχαριστώ πολύ για την αφιέρωση. Η λύση αυτή είναι ακόμα πιο απλή από αυτή που είχα υπόψη μου, όταν αναφέρθηκα στη σχέση $ AL\perp BC$ $ ,(1)$

Ας τη δούμε σαν μία εναλλακτική προσέγγιση.

Δια του $ N,$ φέρνουμε την κάθετη ευθεία επί την $ AN,$ η οποία ως γνωστό, περνάει από τα αντιδιαμετρικά σημεία $ A'$ και $ A''$ του $ A,$ στους κύκλους $ (L)$ και $ (K)$ αντιστοίχως.

Αν είναι $ O'\equiv NA'A''\cap LO,$ αποδεικνύεται εύκολα ότι $ LO' \equal{} \frac {1}{2}\cdot (AA'')$ και έτσι μας μένει να αποδείξουμε ότι $ O'\equiv O$ $ \Longrightarrow$ $ LO \equal{} \frac {1}{2}\cdot (AA'')$ $ ,(2)$

Η ισότητα όμως $ (2)$ προκύπτει από το παραλληλόγραμμο $ ALOK$ $ ($ γιατί $ AK\perp DE$ και $ AL\perp BC$ $ )$ που ανέφερες, αφού $ LO \equal{} AK \equal{} \frac {1}{2}\cdot (AA'')$ και η πρόταση έχει αποδειχθεί.

Κώστας Βήττας.


Larry, Curly and Moe estimated the distance of a trip as 320 miles, 330 miles and 335 miles, respectively. If two estimates were off by 5 miles and the third was off by 10 miles, what was the number of miles in the actual distance of the trip?

two estimates were off by 5 miles. Thus, the estimates are either the same or 10 apart. Since 320,330 are 10 apart, those are two guesses, which are both off by 5. This makes the distance 325.

I forgot to put that this works for the "off by 10" as well.


geometry
incenter
trigonometry
circumcircle
symmetry
angle bisector
perpendicular bisector

Let $I, H$ be the the incenter and the orthocenter of $\triangle ABC$. Show that $AI \bot IH$ if and only if  $\cos B + \cos C = 1$.

here is my solution. I think it is not very nice ;) .
extend $AI$ and note this intersect the circumcircle of $\triangle ABC$ at $D$
it is well know that $DI=DB=DC$,
so we can get $AI=AD-DI=AD-DB=2R(\sin (\frac{A}{2}+B)-\sin \frac{A}{2}$)
and $AH=2R \cos A $
it is obvious that
$AH \bot HI\Leftrightarrow AI=AH \cos(\frac{A}{2}+C-\frac{\pi}{2})$
$\Leftrightarrow \sin(\frac{A}{2}+B)-\sin \frac{A}{2}=\cos A \sin(\frac{A}{2}+C)$
$\Leftrightarrow (1-\cos A)\sin(\frac{A}{2}+B)=\sin \frac{A}{2}\Leftrightarrow\cos B + \cos C = 1 $
we are done

Thanks for your solution. Two remarks

$\bullet$ small typo:  $AH \bot HI$ should be replaced by [color=blue] $AI \bot HI$[/color]

$\bullet AH=2R \cos A $ is true if $A$ is an acute angle, [color=blue]while if $A$ is obtuse[/color] then $AH=-2R \cos A $

Leon

[quote="treegoner"]It is easy to notice that if IH intersects AB and AC at U and V respectively then
    $IH$ is perpendicular to $AI$ iff $AUV$ is issocelles iff $HB' + HC' = 2r$, here $B', C'$ are the feet of the altitudes.
    And this yields the result.[/quote]

Can you, Treegoner, explain your solution in a more detailed way ?

Leon

Let ${F}=AH\cap{BC}$ and ${D}=AD\cap{BC}$.  Then writing the power of point A with respect to the circle in which the quadrilateral $HFDI$ is inscribed we get:
$AH\cdot{AF}=AI\cdot{AD}$. But $AH=2R\cosh$ , $AF=\frac{2S}{a}$ where S is the area;
$abc=4RS$ and $AI=\frac{l_{a}\cdot(b+c)}{2p}$ where p is the semiperimeter;
$l_{a}=\frac{2bc\cdot\cos\left(\frac{\angle{A}}{2}\right)}{b+c}$. Hence we get:
$\cos(\angle{A})=\frac{b+c-a}{b+c}$ which is equivalent with $\cos(\angle{B})+ \cos(\angle{C})=1$ (from cosine theorem).

I also don't completely understand Treegoner's solution in post #4, but I have my own solution which seems to be pretty similar.

[color=blue][b]Problem.[/b] Let I be the incenter and H the orthocenter of a non-equilateral triangle ABC. Prove that $AI\perp IH$ if and only if cos B + cos C = 1.[/color]

[i]Solution.[/i] We will consider directed segments on the lines BC, CA, AB; here, these lines will be directed in such a way that the segments BC, CA, AB will be positive (and thus, the segments CB, AC, BA will be negative). In the following, the directed length of a segment XY will be simply denoted by XY, while the non-directed length will be denoted by |XY|.

First we note that the case when |CA| = |AB| can be left out of consideration. In fact, in this case, the triangle ABC is an isosceles triangle with base BC, so that its apex A, its incenter I, and its orthocenter H all lie on the symmetry axis, and thus the relation $AI\perp IH$ cannot hold; on the other hand, since the base angles of an isosceles triangle are equal, we have B = C (an equality of angles, not of points), so that the equation cos B + cos C = 1 becomes 2 cos B = 1, which is equivalent to $\cos B=\frac12$ and B = 60°, but an isosceles triangle with an angle of 60° must be equilateral, and we have explicitly required our triangle ABC to be non-equilateral. Hence, in the case when |CA| = |AB|, none of the conditions $AI\perp IH$ and cos B + cos C = 1 can be fulfilled, and thus, this case is negligible to the solution of the problem.

Hence, in the following, we can assume that $\left|CA\right|\neq\left|AB\right|$. Now, let the altitudes of triangle ABC issuing from the vertices B and C have feet E and F, respectively, and intersect each other at the orthocenter H of triangle ABC. Then, < AEH = 90° and < AFH = 90°, so that the points E and F lie on the circle with diameter AH. On the other hand, since < BEC = 90° and < BFC = 90°, the points E and F lie on the circle with diameter BC; thus, the quadrilateral BFEC is cyclic, so that < CBF = 180° - < CEF. In other words, < ABC = < AEF. Similarly, < ACB = < AFE. Thus, we have < ABC = < ACB if and only if < AEF = < AFE. Now, if we would have |AE| = |AF|, then the triangle EAF would be isosceles, so that we would have < AEF = < AFE, what would therefore imply < ABC = < ACB, so that the triangle ABC would be isosceles with |CA| = |AB|; but we have assumed that $\left|CA\right|\neq\left|AB\right|$, and thus we get a contradiction. Hence, we cannot have |AE| = |AF|.

Let Q be the orthogonal projection of the point H on the line AI. Then, of course, $AI\perp IH$ if and only if I = Q.

Since the point I is the incenter of triangle ABC, the line AI is the angle bisector of the angle CAB; thus, < QAE = < QAF. Since < AQH = 90°, the point Q lies on the circle with diameter AH. We also know that the points E and F lie on the circle with diameter AH. Now, if two chords in a circle have equal chordal angles, then they must be equal; hence, the chords QE and QF in the circle with diameter AH, having equal chordal angles (< QAE = < QAF), must be equal: |QE| = |QF|. So the point Q is equidistant from the points E and F. Note that Q is the only point on the line AI which is equidistant from the points E and F (in fact, if there would be two different points on the line AI equidistant from the points E and F, then both of these points would lie on the perpendicular bisector of the segment EF, so that the line AI would be the perpendicular bisector of the segment EF, but that would yield |AE| = |AF|, what is impossible as noted above). Hence, we have I = Q if and only |IE| = |IF|.

Since we know that $AI\perp IH$ if and only if I = Q, we thus have $AI\perp IH$ if and only if |IE| = |IF|.

Now, let Y and Z be the points of tangency of the incircle of triangle ABC with the sides CA and AB, respectively. Then, since I is the center of the incircle, we have $IY\perp CA$ and $IZ\perp AB$. Hence, the triangles IYE and IZF are right-angled. Also, they have equal catets |IY| = |IZ| (since the points Y and Z both lie on the incircle of triangle ABC, which has center I). Thus, if at least one of the two conditions |IE| = |IF| and |YE| = |ZF| is fulfilled, then these triangles IYE and IZF must be congruent, and thus the other condition automatically follows. Hence, we have |IE| = |IF| if and only if |YE| = |ZF|. Since we know that $AI\perp IH$ if and only if |IE| = |IF|, we thus can conclude that $AI\perp IH$ if and only if |YE| = |ZF|. Here, the segments are not directed. Using directed segments, we could restate this as follows: We have $AI\perp IH$ if and only if either YE = ZF or YE = - ZF.

On the other hand, we have YA = AZ, since the two tangents from a point to a circle are equal in length.

Now, if YE = - ZF, then AE = YE - YA = (- ZF) - AZ = - AF. With non-directed segments, this would mean that |AE| = |AF|, what is impossible as we saw above.

Thus, among the cases YE = ZF and YE = - ZF, the case YE = ZF is the only possible one. Hence, we see that $AI\perp IH$ if and only if YE = ZF.

Now, if X is the point of tangency of the incircle of triangle ABC with the side BC, then, since the two tangents from a point to a circle are equal in length, ZB = BX and CY = XC, so that ZB + CY = BX + XC = BC. Also, the right-angled triangles BEC and CFB yield $CE=BC\cdot\cos\measuredangle BCE=BC\cdot\cos C$ and $FB=BC\cdot\cos\measuredangle CBF=BC\cdot\cos B$.

Since YE = CE - CY and ZF = ZB - FB, the equation YE = ZF is equivalent to CE - CY = ZB - FB. This rewrites as FB + CE = ZB + CY. Since $FB=BC\cdot\cos B$, $CE=BC\cdot\cos C$ and ZB + CY = BC, this equation becomes $BC\cdot\cos B+BC\cdot\cos C=BC$, what is clearly equivalent to cos B + cos C = 1. Since we have $AI\perp IH$ if and only if YE = ZF, we thus have $AI\perp IH$ if and only if cos B + cos C = 1. This solves the problem.

Altogether, the solution was quite simple, but the observations necessary to exclude degenerate cases like |AE| = |AF| really took a lot of space... I don't see, however, how to simplify them.

  Darij


FIND a, b and c so that x + 5/(x - 1)(x - 2)(x - 3) = a/x - 1 + b/x - 2 + c/x - 3

Dona Corazon Salvatore has a closed rectangular jewelry box of uniform thickness x. The box has outer dimensions of 6 inches, 4 inches and 3 inches. If the box has an inside volume of 30 cubic inches, find x.

I think we're just doing your homework for you, and I'd encourage others not to answer.


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Given triangle $ABC$ with an arbitrary point $P$ inside triangle prove that
$\frac{(PB+PC)(PC+PA)(PA+PB)}{PA+PB+PC}\ge\min\{ab,bc,ca\}$
Equality hold iff $M\equiv A$ if $a\ge b\ge c$ and other cases are similar
I saw it is true in computer, but I have tried it for two month but I can't give any solution, now I post it here hope it will have nice solutions from somebody.

[quote="gemath"]Given triangle $ABC$ with an arbitrary point $P$ inside triangle prove that
$\frac{(PB+PC)(PC+PA)(PA+PB)}{PA+PB+PC}\ge\min\{ab,bc,ca\}$[/quote]

Unfortunately, this is wrong. See the attached graphic for a counterexample.

What, however, seems to be true: If P is a point inside of a triangle ABC, then

$\frac{\left(PB+PC\right)\left(PC+PA\right)\left(PA+PB\right)}{PA+PB+PC}\leq\max\left\{bc,ca,ab\right\}$.

But I don't have a proof...

  Darij

useless post [edit]
@gemath: yes it is, he pmed me back

Right I saw,too. But the following is really true
$\frac{a^{2}PA+b^{2}PB+c^{2}PC}{PA+PB+PC}\ge\min\{ab,bc,ca\}\quad (1)$ and
$\frac{a^{2}PA+b^{2}PB+c^{2}PC}{PA+PB+PC}\ge\frac{(PB+PC)(PC+PA)(PA+PB)}{PA+PB+PC}\quad (2)$
and I hoped $\frac{(PB+PC)(PC+PA)(PA+PB)}{PA+PB+PC}\ge\min\{ab,bc,ca\}$ is true to prove $(1)$ but it is false.However $(1)$ and $(2)$ is true I only prove $(2)$ and I can't do with $(1)$ any idea from other ?
@me@home maybe he used Euklides, try to search by google.


probability
real analysis
real analysis unsolved

Let $\{x_n\}_{n\geq 0}$ be a sequence of real numbers which satisfy \[ (x_{n+1} - x_n)(x_{n+1}+x_n+1) \leq 0, \quad n\geq 0. \] a) Prove that the sequence is bounded;

b) Is it possible that the sequence is not convergent?

A friend of mine nailed a nice solution by putting $y_n=x_n+\frac{1}{2}$ you get that $y_{n+1}^2\leq y_n^2$ from where the conclusion is visible.

I'd like to see some opinions on my solution... Is it correct, or fundamentally wrong?

I feel a bit frustrated right now... I got 6/28p on the olympiad. And they didn't even bother to list the number of points everybody got on each problem, like they did last year. It really makes me wonder whether my solutions were really correct. I'm starting with this one.

PS: The second time this happens. Last time I got nothing after the "contestations", and I don't think I'll get anything now. Should I give it up?

[quote="perfect_radio"](a) We have $x_{n+1} \in \left[ x_n, -1-x_n \right] = I_n$.

By some computations, it will become apparent that $I_{n+1} \subseteq I_n$.

(b) Yes. We take the sequence $3,-4,3,-4,\ldots$.[/quote]

The second part is ok but at the first one who says that $x_n\leq -\frac{1}{2}$

[quote="gnosis"]The second part is ok but at the first one who says that $x_n\leq -\frac{1}{2}$[/quote]

I took two cases:

1. The interval is like I wrote it.

2. The interval is the other way around. In the olympiad, I said you treat this case analogously. Is that wrong?

i found that the official solution, which i nailed, is the easiest. just take $y_n=(x_n)^2+x_n$

There is a big probability that nothing that I said in my previous post is correct. I got 1 point, and the example was worth 2 points.


geometry conics ellipse circumcircle vector parabola

Let X be a circle, Y a point outside the circle and Z a point on the circle (Y, Z fixed). Consider the pencil of lines through Y cutting the circle in variable points M and N. Find the locus (as M and N vary) of (i) the orthocentre (ii) symmedian point of MNZ.

(error fixed now)

The locus of the orthocenter is a disk centered at $Z$, of radius twice that of circle $(X)$: Assume $M$ is also fixed. In this case the locus of the orthocenter is a circle the same radius as $(X)$, passing throuh $Z$ and $M$. The locus we're looking for is obtained by rotating this circle around $Z$, thus covering a disk of radius twice that of the circle $(X)$.

In the second case, the locus seems to be the disk $(X)$. This is because according to my sketch, the locus of the symmedian pt when $Z,M$ are fixed is an ellipse situated inside the circle $(X)$, tangent to this circle in $M,Z$. I didn't try to prove this yet, but I will. Anyway, as we move $M$ around the circle, this ellipse will cover the entire disk and only the disk, I think we should eliminate the point $Z$ from the locus, but this is merely a detail. It's not a symmedian point of any non-degenerate triangle $ZMN$, but it's a limit point. Actually, the same goes for all the points of the circle $(X)$.

Of course, the proof of the fact that the locus of the symmedian pt is an ellipse will have to use some projective stuff, I just don't know what yet :).

Ok, I've got it:

Let $z,m,n$ be the lines tangent to $(X)$ in $Z,M,N$ respectively. It's well-known that the lines formed by $(Z,m\cap n),\ (M,z\cap n),(N,m\cap z)$ are the symmedians. When $M,Z$ are fixed, so are $m,z$. $n$ varies being tangent to $(X)$. This means that it determines a homographic map from $m$ to $z$, the map being $m\cap n\to z\cap n$. This means that the transformation from the pencil through $M$ to the pencil through $Z$ mapping the symmedian through $M$ to the symmedian through $Z$ is projective, so the locus of the intersection of two corresponding lines through this transformation must be a conic (these are all well-known properties of projective transformations) passing through $M,Z$.

Now all we need to show in order for the entire thing to be over is that the symmedian pt can't lie outside the circumcircle, and this is really easy because the symmedian point is always inside the triangle, and the triangles $ZMN$ always lie inside $(X)$.

[quote="darij grinberg"][quote="grobber"]The locus of the orthocenter is a disk centered at $Z$, of radius twice that of circle $(X)$: Assume $M$ is also fixed. In this case the locus of the orthocenter is a circle the same radius as $(X)$, passing throuh $Z$ and $M$. The locus we're looking for is obtained by rotating this circle around $Z$, thus covering a disk of radius twice that of the circle $(X)$.[/quote]

I believe this is a misunderstanding. The line MN must pass through Y, hence the sought locus is a curve (it is, indeed, a circular arc, if my dynamic geometry drawing is correct, but I have no proof).

  Darij[/quote]

Yes, it is a misunderstanding :(. For some reason, I was looking for the locus of those pts when $Z$ is fixed and $M,N$ just go around the circle. I don't know why I did that, since what Vinoth wanted is pretty clear :).

Ok, let me make up for that mistake :). I only have the answer to the first problem, though.

Let $X$ be the center of the circle $(X)$ and let $X'$ be the center of the circumcircle of the triangle formed by $M,N$ and the orthocenter $H$ of $MNZ$. It's easy to show that $X'$ is on a circle centered at $Y$. Now we prove that the vector $\vec {X'H}$ is constant, thus obtaining the result that $H$ moves on a circle obtained from the locus of $X'$ by a translation.

In order to show that $\vec{X'H}$ is constant it's enough to observe that $\vec{X'H}=\vec{XZ}$, which is constant because $X,Z$ are fixed.

We've shown that $H$ is on that circle, but not the entire circle works. However, determining the exact arc is easy. We simply consider its limit points reached by $H$ when $MN$ becomes tangent to $(X)$.

[Tex question: How do you make wider arrows over vectors?]

As for the second problem, the locus seems to be a conic, so it definitely won't be as easy to find. I don't even know if it's always the same sort of conic (sometimes it looks like an arc of an ellipse tangent to $(X)$ in $Z$ and sometimes it looks like an arc of a parabola with the same property.

Grobber, thanks for the solution of the first problem! I had a similar idea: The locus for (i) is (a part of) the circle with center T and radius TZ, where X is the center of the circle (X), and T is the point where the parallel to XZ through Y meets the parallel to XY through Z. You will easily see that this is equivalent to what you said.

[Concerning the TeX question: I have always been using $\overrightarrow{XZ}$.]

  Darij

Here's an idea for the second problem:

Assume the variable line through $Y$ cuts $(X)$ in $M,N$. The tangents in $M,N$ to $(X)$ intersect at point $T$ the locus of which is a line (it's the pole of $MN$, which always passes through $Y$, so $T$ must be on the polar of $Y$; maybe the locus of $T$ isn't the entire line, but that's not really important :)). Let's try to find the locus of $S=MN\cap ZT$. This point is the fourth harmonious conjugate of $Z,T',T$, where $T'=ZT\cap (X)$. The transformation which maps a point $P$ of the plane to the fourth harmonious conjugate of $Z,P,ZP\cap \ell$ ($\ell$ is a line) is projective, so it turns conics into conics (this can be shown by linear algebra, using the matrices of the conic and the projective transformation). We apply this for $\ell=$the polar of $Y$ and the conic we want to transform being $(X)$. This turns the points of $(X)$ into the points of the locus of $S$, which is thus a conic (call it $(S)$).

We now apply the same result to the same $\ell$, but this time trying to transform the conic $(S)$. We obtain the locus of $K$ (the symmedian of $ZMN$), which must then be a conic too.

I think it's not that hard to show that because in these cases $Z$ is on the conics we are transforming, the resulting conics will all be tangent to each other in $Z$.

The locus of $K$ is an arc of a conic tangent to $(X)$ in $Z$. Moreover, it's the arc which is situated inside $(X)$.

I said that by a projective transformation a conic turns into another conic, I don't know if this is true for transformations which don't have a non-singular matrix, but I think our transformation has a non-singular matrix.

You might have noticed that I'm not really sure about all of this, so I apologuize just in case it's wrong :D.

[Edit: [url=http://www.css.tayloru.edu/~btoll/s99/424/res/mtu/Notes/geometry/geo-tran.htm]here[/url]'s is something that's very useful]

My solution for [b](ii)[/b]:

Let $ K $ be the symmedian point of $ \triangle MNZ $ .
Let the polar of $ Y $ WRT $ \mathcal{X} $ cut $ \mathcal{X} $ at $ Y_1 $ and $ Y_2 $ .
Let $ T $ be the pole of $ MN $ WRT $ \mathcal{X} $ and $ S=ZT \cap MN $ .

It's well-known that $ K \in ZT $ .

Since $ Y $ lie on the polar $ MN $ of $ T $ WRT $ \mathcal{X} $ ,
so $ T $ lie on the polar $ Y_1Y_2 $ of $ Y $ WRT $ \mathcal{X} $ . ... $ (\star) $
From $ Y(Y_1, Y_2; T, S)=-1 \Longrightarrow $ pencil $  YT \mapsto $ pencil $ YS $ is homography ,
so combine with $ (\star) $ we get pencil $ ZT \mapsto $ pencil $ YS $ is homography $ \Longrightarrow S $ lie on a conic $ \mathcal{C}_1 $ .
Consider the case $ M \equiv N \equiv Y_1 $ and the case $ M \equiv N \equiv Y_2 \Longrightarrow \mathcal{C}_1 $ pass through $ Y, Y_1, Y_2, Z $ .

Since $ (T,K;S,Z)=-1 $ ,
so pencil $ Y_1S \mapsto $ pencil $ Y_1K $ is homography $ \Longrightarrow $  pencil $ Y_1K \mapsto $ pencil $ ZT $ is homography ,
hence $ K \equiv Y_1K \cap ZT  $ lie on a conic $ \mathcal{C}_2 $ (pass through $ Y_1, Y_2, Z $ ) $ \Longrightarrow $ the locus of $ K $ is the part of $ \mathcal{C}_2 $ inside $ \mathcal{X} $ .

Q.E.D

Another solution for (ii):

Label $(O)$ the fixed circle. Tangents of $(O)$ at $M,N$ meet at $E$ and $ZE$ cuts $(O)$ again at $F.$ $K$ denotes the symmedian point of $\triangle ZMN.$

Cross ratio $(Z,K,F,E)$ equals that of equilateral triangle $\triangle ZMN$ with center $K$ and circumcircle $(O),$ as there is a homology transforming $\triangle ZMN$ into an equilateral triangle with circumcircle the image of the circumcircle of $\triangle ZMN.$ Hence $(Z,K,F,E)=\tfrac{1}{4}$ $\Longrightarrow$ the application $K \mapsto F$ is therefore homologic, fixing the polar of $Y$ WRT $(O)$ and the line pencil through $Z \in (O)$ $\Longrightarrow$ locus of $K$ is a conic $\mathcal{C}$ tangent to $(O)$ at $Z.$


geometry
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real analysis

You have a fresh ordinary pencil and start to draw a line. Would it be
possible to draw a line as long as the earth’s equator before the pencil is finished?

Suppose the pencil have length $L$.

Assuming proportionality, After you draw $S$ meters the pencil loses $N$ meters of it's length.

So $K = \frac{S}{N}$

You can determine the constant $K$ empirically.

After you solve for $S$ and make a substituition $N=L$

Note: I am not assuming loses for another kind. Just assuming a ideal case. The real case must be a little bit less than the ideal.

This is sort of a [url=http://en.wikipedia.org/wiki/Fermi_question]fermi problem[/url] so you should
continue and develop your thoughts, and eventually be able to answer the question without
any more information.

suppose the pencil is faber castell. so it have a constant K

but if I use a poor qualit pencil, it will have another constant.

so. it's possible that i have the worst possible pencil.

and. there is a lot of constant relevants in this problem.

i think is impossible to solve this problem just with that.

we can try a simple thing.

suppose the graffit of the pencil have length L and radius r (it is a cilinder)

so the volume of it is V.

if we draw a line with thicknes approximatelly 0, and width 2 r and length S we can calculate the total area of this line ( 2 r S).

we can use infinitesimals and a good argument to proof that numerically V = K 2 r S

but, with a analogous argument we can proof that is possible to draw a infinite line. just make more thickless and thin.

the limit tending to 0

so i need more input to solve this. because it seens that the problem is a joke, sorry

No the problem is not a joke!  :mad:
It is however possible and easy to solve.

Your physical reasoning is not sufficient. Do not consider constants
and faber castell or anything. And think! Read the article I posted. What do you mean by:

[i]with a analogous argument we can proof that is possible to draw a infinite line. just make more thickless and thin.

the limit tending to 0 [/i]

This is not real analysis. Of course you cant draw an infinite line! Otherwise the question would
have been trivial. Now you do not need anything more to solve this, it is easy.

the problem is:
you didn't explain what I could do:

what is possible and what is impossible.

in theory I can draw a very very thin line with thickness very small.

if i make the limite of thickness tending to zero, i can make a infinite line.

since is possible for a function have infinity length and finite area.

because you didnt explained whats in your mind. so it's a little bit of trouble for me understand, i can make a lot of assumptions for this problem.

my solutions is Yes, we can. but without more input i cannot express is mathematically

This is not a mathematical problem! (Of course one could see it as one, but your reasoning is still
insufficient)  Also even if mathematical reasoning
is fruitful in modeling physics problems, one has to, as a good theorist, make
reasonable and physical assumptions.
Especially since this is sort of a Fermi question!
Thus no information is needed, since it could be solved with a few assumptions
and some common sense.

Just because one is familiar with real analysis, doesn't mean that one can solve
everything in that manner, and think that one has done something good since it felt
mathematical. Everything is not infinitesimal. Real analysis is itself a model frame.

Yes you are right color. And yagaron we know what kind of pencil it is, its an ordinary pencil :d.
And color, can you sharpen it on the way?

the width of the line is approximately $10^{-3}m$

the length of the line is approximatelly $10^{7}m$

the area is $10^{4}m$

with the same pencil we can draw n pages of 10x10 cm. totally black.

so, since we are using the same pencil, there areas need to be equal.

$10^{4}=n (10^{-2})$
$n = 10^{6}$

but. n is too big. so it's impossible.

we can make this argument...

Well sure you can sharpen it, but that does not contribute to the theoretical longest line.

But of course its hard to write with a pencil which is not sharp  :P .

[b]@yagaron[/b]
Well ok, but even $10^{6}$ is a large overestimate.

Also the question was about the equator, not any $n$ and sheets of paper.

-----
Check it! assume the graphite layer is 100 atoms thick.
The limit is 1 atom (not very reasonable but anyway it gives a very certain limit)
Now pencils are usualy $10^{-1}$m long.$\ldots$

er?

i made for equator!!!

the radius of earth is approximatelly 6000km

so  L = 2 pi r  :)

and i made a comparison.

if the out put were = 100. so we can think in another proof.

but, since it was too large, it make things impossible.

a pencil finish after 300 sheets of paper os less.

and using atomos is a anoying.

if the problem is to solve like that. suppose i take the graphite and break all ligations. so i will have on C.

i can make a straigth line with all atoms.

with that i can make a line that goes from here to the sun.

aff.

Lets assume that the pencil has a diameter of $1mm$. When you draw the line the graphite in the pencil sticks to the paper, and of course graphite is in layers, and the thickness of the line must at least one atomic layer.now we have to approximate the distance between the layers of the graphite. The density of graphite is about $2000\frac{kg}{m^{3}}$, the molar mass of carbon is $0,012\frac{kg}{mol}$. so $1mol$ of graphite has a volume of $6\cdot 10^{-6}m^{3}$. Now I think we can say that the graphite atoms is made of cubes, so then the sides of these cubes are: $1,8\cdot 10^{-2}m$. We also know that in $1mol$ of graphite there are $6\cdot 10^{23}$ atoms, so on one side of this cube the number of atoms: $8.5\cdot 10^{7}$ so the distance between the layers is approximately : $2.1\cdot 10^{-10}m$. the length of a standard pencil is about $15cm$ so there are $7\cdot 10^{8}$ layers in the pencil. Now since the diameter is $1mm$ each layer makes a $1mm$ line so thats totally: $7\cdot 10^{5}m$ now thats much more less than the diameter of the earth.
So do you think this approximation is correct? it really is an approximation. :D

Yes this is correct and was more or less the expected answer, very good.

As I said $10^{6}$m is an overestimate. Your answer too, since the layers are more likely to be about 100-1000 atoms thick, but the aim here was to find and upper limit, which you did.


Let $ ABC$ be right-angled triangle with $ \angle C =90 ^{\circ}$. Points $ M$ and $ N$ are chosen on  the sides $ AC$ and $ AB$ respectively such that $ CM=MN$, $ \angle MNB =\angle CBM$, $ K$ is a foot of a perpendicular from $ C$ to $ MB$. Prove that $ NK$ passes though midpoint of $ BC$.

First let $ {BC\cap NK\equal{}P}$.Notice that :$ {MC^2\equal{}MK*MB\equal{}MN^2}$  $ {\longrightarrow}$ $ {\triangle MNK}$ is similar to $ {\triangle MBN}$ $ {\longrightarrow}$ $ {\angle MNB\equal{}\angle MKN\equal{}\angle PKB\equal{}\angle CBM}$ $ {\longrightarrow}$ $ {PK\equal{}PB}$ ;easy to get that :$ {PB\equal{}PC}$;...$ {QED}$


What is the first composite number that satisfies:
${a^p}\equiv \textbf{a (mod p)}$
for a number p
were a<p

Well, in other words, you're looking for the first [url=http://mathworld.wolfram.com/CarmichaelNumber.html]Carmichael number[/url], which is : 561. :)

[b]nateharman1234[/b], I don't quite get you.. : $2^8 \equiv 0 \pmod 8$.  :?

yep you are right and i didnt really understand nateharman1234's answer either.


The day is almost over. But there is still time for starting this discussion.

So what do you think of it? Another act of benevolence of the US and another instance of the effctiveness of the USO - I mean - UNO, or what?

Give your views.

Well that is the eternal question.  Even if Hiroshima was necessary, what about Nagasaki?

Some arguments that could be added :

pro : USA often claims more Japanese lives (military as well as civilian) would have been lost if the war went on and was fought out the "normal way"

Supporters also claim Japan had threatened to execute POW (Prisoners of War) in case of an invasion of their mainland.

contra : Japan was already on the brink of surrendering, but even after Hiroshima, refused to accept the abdication of their emperor.  After Nagasaki they surrendered unconditionally... and ironically general MacArthur insisted on not putting the emperor on trial, and he was allowed to stay in power.

Another contra argument is that it was the USSR that forced them to surrender.
The USSR had signed a neutrality pact with Japan, and only broke that between both atomic bombings and launched Operation August Storm.  The operation was very succesful and may have resulted in occupation of the mainland.

Yet another theory is that the real reason to go through with it, was that the Manhattan Project, which resulted in the development of the nuclear fission bomb, would have been useless, despite its huge costs.  This way they could justify the costs and at the same time impress/scare the USSR.

I sometimes wonder myself whether or not it was a desperate attempt by the USA to beat the USSR, fearing they would lose Japan (or part of it) to communist rule.


Consider a function of the sort invented by nmtruong1986 in:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63937

That is, $f(x)$ is continuous, and for any $x$ there exists some integer $n$ such that $f^n(x)=1$.

Prove that there exists an interval $[a,b]$ containing 1, such that $f(x)=1$ for all $x\in[a,b]$.

Are you sure it holds? Can't use your ideas from http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63937 ,
since $[a,b]\backslash \{1\}$ is not an interval.
Will you please post your solution?

"An interval $[a,b]$ containing 1" means an interval $[a,b]$ such that $1\in[a,b]$, not the result when 1 is subtracted from such an interval.

To be honest, my proof is kind of unwieldy (which is why I posed the question here in the first place!)  But here's a sketch.  Transform the problem to a simpler version, where 0 rather than 1 is the "special" value and the function is restricted to nonnegative values, as follows:  Given $x\geq 1$, there exists $n$ such that $f^n(x)=1$, so there must be a least whole number $m=m(x)$ such that $f^m(x)\leq 1$.  The function $f^{m(x)}(x)$ is continuous on $[1,\infty)$, so we can define a continuous function $f_1(x)=1-f^{m(x+1)}(x+1)$ from the nonnegative reals to themselves, with $f_1(0)=0$.  Similarly taking $m(x)$ on $(-\infty,1]$ as the least whole number such that $f^m(x)\geq 1$, we can define another function $f_2(x)=f^{m(1-x)}(1-x)-1$ satisfying the same conditions.  So $g(x)=f_1(f_2(x))$ satisfies these conditions as well, and given $x$ there exists $n$ such that $g^n(x)=0$.

For the simpler function $g(x)$, by considering nested intervals we see that if there were values of $x$ arbitrarily close to 0 for which $g(x)>0$, then there would exist values of $x$ for which $g^n(x)>0$ for all $n$, a contradiction.  So there exists $\delta_0$ such that $g(x)=0$ for all $x\leq \delta_0$; that is, there is a continuous interval containing 0 such that $g(x)=0$ for all $x$ in the interval.  Similar arguments then show that
(1) for at least one of $f_1(x)$, $f_2(x)$ there is a continuous interval containing 0 such that $f_i(x)=0$ for all $x$ in the interval
(2) there exists an interval $[1,1+\delta]$ (if $f_1(x)$ satisfies the condition in (1) ) or $[1-\delta,1]$ (if $f_2(x)$ satisfies the condition in (1) ) such that $f(x)=1$ for all $x$ in the interval.


Again, a cute problem from an unknown source via Michael Albert.

Find all functions $f(x)$ from the positive rationals to themselves satisfying \[ f(x+1)=f(x)+1 \] and \[ f(x^3)=f(x)^3 \] for all $x$.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=54212

Pierre.


Francesca uses 100 grams of lemon juice, 100 grams of sugar, and 400 grams of water to make lemonade.  There are 25 calories in 100 grams of lemon juice and 386 calories in 100 grams of sugar.  Water contains no calories.  How many calories are in 200 grams of her lemonade?

$ \textbf{(A) } 129 \qquad \textbf{(B) } 137 \qquad \textbf{(C) } 174 \qquad \textbf{(D) } 223 \qquad \textbf{(E) } 411$

600 grams of lemonade contains (25 + 386) = 411 calories, so 200 grams contains 411/3 = 137 calories,

hence $(B)$.


I am here, posting in Fifth period math. Supposed to be doing online math :p. Anyone reccommend a level for my charachters to beat the Fusion Dragon, I want to beat it before buying Golden Sun 2.

well fusion dragon is fairly easy to beat. If you want to continue using your custom characters in GS 2 then i'd say level 25 or above is ok. But i'm a lazy person so i only got my characters to level 19 :? . However it's not much different with weak characters from the original GS game. Unless you want to beat the big bosses such as dullahan(or however you spell it) to receive better summons.

i have both golden sun games but have never played either. odd isnt it?

golden sun...great game??...well, I dont have it so im probly not qualified to make judgments...but from what i've heard, its pretty boring...

you're right this game is pretty boring in many aspects(a lot like pokemon) all you do is go around and fight people. However the plot (at least to me) isn't half bad (although a little weird especially in the second game). And they have a lot of lame jokes in there. I just had nothing [b]much[/b] to do over the summer.

Blah.  My brother has Golden Sun and I tried it.  I didn't like it.  Boring.  Basically a pseudo-RPG where you collect these pokemon like things that improve your stats, then you "release" them during battle.  Yeah, a lot like pokemon.

Personally, I think the makers of GameBoy have gone way overboard with their series...making lights, adapters, see-thru stuff, attachments, and then new systems like pro and advance and what not...

Can't you see Nintendo's out of ideas? From what I heard through my dad (who almost got head-hunted by Nintendo so...:wink: :wink:) Nintendo's going to file bankrupcy within the next decade...
I suggest you stick with the PS series.

What do you mean "out of ideas"...  You can always create new systems and such.

That's just the thing.  Their games are all, like, totally, all the same.  They haven't come out with anything revolutionary in terms of systems (like N64) or games (like Zelda).  They're losing to playstation and xbox and whatnot b/c only little kids want to play their games.  In short, they're doomed.  But then again, we're all doomed anyway (yep, we're all gonna die, right?).  :wink:

Nintendo will die before any of us dies...

Well...one good idea could save the company...

I mean...Chrysler was "doomed"...until they came out with the Voyager and saved the company b/c everyone loved the car...


function
algebra
polynomial
induction
algebra unsolved

(i)If $ f(x)$ is a polynomial function and $ f: (0,\infty)\rightarrow(0,\infty)$ and $ f(f(x))\equal{}6x\minus{}f(x)$, then find $ f(17)$.

(ii)If $ f(x)$ is a polynomial function such that $ f(x^2\plus{}1)\equal{}(f(x))^2\plus{}1$ and $ f(0)\equal{}0$,then find $ f(34)$.

(i)If $ f(x)$ is a polynomial function and $ f: (0,\infty)\rightarrow(0,\infty)$ and $ f(f(x)) \equal{} 6x \minus{} f(x)$, then find $ f(17)$.

Suppose that the degree of the polynomial is greater than $ 1$
Let $ deg(f)\equal{}n>1$ then $ deg(f(f(x))\equal{}n^2$ and $ deg(6x\minus{}f(x))\equal{}n$ then $ n^2\equal{}n$ contradiction

Therefore $ f(x)\equal{}ax\plus{}b$
$ a (ax\plus{}b)\plus{}b\equal{}6x\minus{}(ax\plus{}b)$
$ x (a^2\plus{}a\minus{}6)\plus{}(a\plus{}2)b\equal{}0$
$ a^2\plus{}a\minus{}6\equal{}0$ , $ a\equal{}2$ or $ a\equal{}\minus{}3$
and $ b\equal{}0$
then $ f(x)\equal{}2x$ or $ f(x)\equal{}\minus{}3x$
therefore the only possible solution is $ f(x)\equal{}2x$ and thus $ f(17)\equal{}34$

(ii)If $ f(x)$ is a polynomial function such that $ f(x^2 \plus{} 1) \equal{} (f(x))^2 \plus{} 1$ and $ f(0) \equal{} 0$,then find $ f(34)$

Let $ a_0 \equal{} 0$ and $ a_{n \plus{} 1} \equal{} a_n^2 \plus{} 1$ for all $ n \in \{0,1,2,3,\cdots\}$
By induction $ f(a_n) \equal{} a_n$ then because $ f(x)$ is a polynomial so is $ f(x) \minus{} x$ with only finite roots then $ f(x) \equal{} x$ for all $ x$ so $ f(34) \equal{} 34$.


topology
advanced fields
advanced fields unsolved

A separable, Hausdorff space $ X$ has cardinality less than or equal to $ 2^c$ (where c is the cardinality of the real numbers). If $ X$ is also first-countable then its cardinality is less than or equal to $ c$.

I don't quiet see what happens if we drop Hausdorff condition? :)

Take an indiscrete space X and consider $ X\times\mathbb{R}$. This is separable, because $ \{x\}\times\mathbb{Q}$ is dense, but it is not Hausdorff in general. So if you drop the Hausdorff condition there is no upper bound for the cardinality.

Oh yes... that was so obvious that I didn't see it ...

Happy new year :)


logarithms
real analysis
real analysis unsolved

Prove that the equation

$ e^{\minus{}x^2}\minus{}e^{\minus{}x}\equal{}lnx$ has only one solution ($ x\equal{}1$) for $ x>0$.

Alexandros

\ln ($ \ln$) looks nicer.

For $ 0<x<1$, the LHS is positive and the RHS is negative. For $ 1<x$, the LHS is negative and the RHS is positive.


integration
Support
function
limit
advanced fields
advanced fields unsolved

Let $f\in L^{1}(\mathbb{R})$. The Fourier Transform of $f$ is defined by \[\hat{f}(\xi): =\int_{\mathbb{R}}e^{ix\xi}f(x)\,dx\] If moreover $f$ has compact support, show that $\hat{f}$ is a $C^\infty$ function

I prefer to define the Fourier transform as $\widehat{f}(\xi)=\int_{\mathbb{R}}f(x)e^{-2\pi i\xi x}\,dx$

The plus/minus sign and the location of the $2\pi$ are all conventions, and can be interconverted.

As for the analyticity:

If the support of $f$ is contained within $[-M,M],$ then

$\left|\int_{\mathbb{R}}f(x)x^{k}\,dx\right|\le \|f\|_{1}M^{k}.$

Now write out the exponential as its power series and interchange sum and integral. The resulting power series in $\xi$ can easily be shown to have an infinite radius of convergence. When your professor said [i]analytic[/i] he might as well have gone ahead and said [i]entire[/i].

[quote="ma_go"]isn't usually fourier transform defined as $\hat{f}(\xi) : = \int_\mathbb{R}f(x)e^{-i\xi x}dx$?

anyway..
suppose $\text{spt}f \subset [-M, M]$, and define $g(x) =-ixf(x)$.
notice that since $f$ has compact support, also $g$ has.
we have to show that the following exist: $\lim_{h\to 0}\frac{\hat{f}(\xi+h)-\hat{f}(\xi)}{h}= \lim_{h\to 0}\int_{-M}^{M}f(x)e^{-i\xi x}\frac{e^{-ihx}-1}{h}dx$.
now, thanks to dominated convergence theorem, we have $\hat{f}'(\xi) = \hat{g}(\xi)$.
finally, since $\hat{g}$ is differentiable itself, $\hat{f}$ is $C^\infty$.

one last thing.. my analysis teacher told us that $\hat{f}$ under stated conditions, is actually [i]analytic[/i], which is pretty stronger :)
but he also told us it's much more difficult... i guess some of you have a proof: if it's acceptably easy, could you give us some hint, if you think we need them? :) thank you very much[/quote]
Just trying to understand your argument.............

So you say that the fact support of $f$ contained in $[-M,M]$ is used to bound $|e^{-ihx}-1|$ so that we can apply Lebesque Dominated Convergence Theorem?

hm.. more or less.
i use the fact that $\frac{e^{-ixh}-1}{h}$ converges (in $h$) to say that it's bounded for every (fixed) $x$. then i use the fact that support of $f$ is bounded to say that for every sequence $h_{n}\to 0$ the sequence $f(x)e^{-ix}\frac{e^{-ixh_{n}}-1}{h_{n}}$ is dominated (by, say, $|f(x)(x+2)|$), and hence apply the theorem...
have i been clear, this time? :)


function
integration
inequalities
real analysis
real analysis unsolved

Let $ 1 < p < q < \infty$ . Find a sequence of functions $ \{f_n\}$ such that $ f_n\rightarrow0$ weakly in $ L^p(R)$ and strongly in $ L^q(R)$, but $ \{f_n\}$ does not converge strongly in $ L^p(R)$.

It is easy to see that forall $ n\in \mathbb{N}^\star$ there's $ u_n \in \mathbb{N}$ such that $ u_n \geq n$ and $ 1 \leq \sum_{k\equal{}n}^{u_n} \frac{1}{k} \leq 2$

We can define $ f_n$ by $ f_n(x) \equal{}\frac{1}{n^{1/p}}$ if $ x\in [n,u_n\plus{}1[$ and $ f(x)\equal{}0$ either.

We have $ \int |f_n|^q \leq \sum_{k\geq n} \frac{1}{k^{q/p} }$, so $ f_n$ converges strongly to $ 0$ in $ L^q$ because $ p<q$.

and $ \int |f_n|^p \equal{} \sum_{k \equal{} n}^{u(n)} \frac{1}{k} \in [1,2]$, so $ f_n$ does'nt converge strongly to $ 0$ in $ L^p$.

We show that $ f_n$ converges weakly to $ 0$ in $ L^p$.  Let $ g\in L^{s}$ where $ 1/p\plus{}1/s\equal{}1$. By Hölder's inequality, it comes

$ | \int f_{n} g  | \leq \left( \int_n^{\plus{}\infty} |f_{n}|^p \right)^{1/p}  \left( \int_{n}^{\plus{}\infty} |g|^{s} \right)^{1/{s}} \leq 2^{1/p}  \left( \int_{n}^{\plus{}\infty} |g|^{s} \right)^{1/{s}} \rightarrow 0$

In fact, [b]holomorf[/b]'s example gives even more, a sequence which converges not only in one $ L_q(\mathbb R)$ with $ q > p$, but in all of them.

I wonder, if it is possible to make it converge in $ L_{q_1}(\mathbb R)$ and $ L_{q_2}(\mathbb R)$ with $ q_1 < p < q_2$, but not in $ L_p(\mathbb R)$.

No, it's not possible, because $ |f(x)|^p\le \alpha |f(x)|^{q_1} \plus{} (1\minus{}\alpha)|f(x)|^{q_2}$, where $ \alpha$ is s.t. $ \alpha q_1 \plus{} (1\minus{}\alpha) q_2 \equal{} p$.

[quote="zhoraster"]I wonder, if it is possible to make it converge in $ L_{q_1}(\mathbb R)$ and $ L_{q_2}(\mathbb R)$ with $ q_1 < p < q_2$, but not in $ L_p(\mathbb R)$.
[/quote]

Using Holder's inequality you can even estimate the $ L^p$ norm of such a function in terms of the $ L^{q_1}$ and $ L^{q_2}$ norms.

[quote="blahblahblah"]
Using Holder's inequality you can even estimate the $ L^p$ norm of such a function in terms of the $ L^{q_1}$ and $ L^{q_2}$ norms.[/quote]
[quote]
No, it's not possible, because $ |f(x)|^p\le \alpha |f(x)|^{q_1} \plus{} (1 \minus{} \alpha)|f(x)|^{q_2}$, where $ \alpha$ is s.t. $ \alpha q_1 \plus{} (1 \minus{} \alpha) q_2 \equal{} p$.[/quote]


calculus
integration
function
limit
real analysis
real analysis unsolved

Let $f$ be a continuous fonction $\mathbb{R}\rightarrow \mathbb{C}$ such that $\lim_{x \rightarrow+\infty}|x^{2}f(x)|=0$ and such that $\int_{-\infty}^{+\infty}|f(x)|dx$ exists.

Prove that $\lim_{h \rightarrow 0^{+}}\sum_{n=0}^{+\infty}hf(nh)=\int_{0}^{+\infty}f(x)dx$.

The second condition is entirely redundant; we don't care about negative $x$, and the integral from zero to $\infty$ always converges when $f(x)=o(x^{-2})$. Also, $O(x^{-2})$ is plenty. ($|f(x)|\le ax^{-2}$ for some $a$ and sufficiently large $x$)

For any large enough $M$ and $h\le 1$, $\sum_{nh\ge M}hf(nh)\le \sum_{nh\ge M}\frac{a}{n^{2}h}\le \int_{M/h-1}^{\infty}\frac{a}{hx^{2}}\,dx\le \int_{M-1}^{\infty}\frac{a}{y^{2}}\,dy=\frac{a}{M-1}$.

This goes to zero as $M\to\infty$, uniformly in $h$.

Now, $\sum_{n=0}^{\infty}hf(nh)=\sum_{0\le nh<M}hf(nh)+\sum_{nh\ge M}hf(nh)$. We look at the error in three pieces: $\left|\int_{0}^{\infty}f(x)\,dx-\sum_{n=0}^{\infty}hf(nh)\right|\le \left|\int_{0}^{\infty}f(x)\,dx-\int_{0}^{M}f(x)\,dx\right|+\left|\int_{0}^{M}f(x)\,dx-\sum_{0\le nh<M}hf(nh)\right|$
$+\left|\sum_{nh\ge M}hf(nh)\right|$
The first term goes to zero as $M\to\infty$, since the improper Riemann integral of $f$ exists by comparison. The third also goes to zero as $M\to\infty$, by our integral test estimate above. The second goes to zero as $h\to 0$ for any fixed $M$, by Riemann integrability of $f$ on bounded intervals. We can choose $M$ large enough that the first and third terms are each less than $\frac{\epsilon}3$, then find $\delta$ so that the middle term is less than $\frac{\epsilon}3$ when $0<h<\delta$. This makes the total error less than $\epsilon$ whenever $0<h<\delta$, and we have convergence.

There was nothing special about $\frac1{x^{2}}$. This argument proves that $\lim_{h\to 0^{+}}\sum_{n=0}^{\infty}hf(nh)=\int_{0}^{\infty}f(x)\,dx$ whenever $f(x)=O(g(x))$ as $x\to\infty$ for some positive and decreasing function $g$ on $[0,\infty)$ with finite integral.


What is the definition of a "proper factor" of an integer n? Mathworld says that it is all of the factors except n itself, but I have seen some old AIME problems define it as all of the factors except n AND except 1.

I'm pretty sure Mathworld is correct.  Thats what I thought too, and when I also think about perfect numbers, which are the sum of their proper factors.. 1 is included there. What exactly did the questions in AIME say? I mean, if they state a proper factor not equal to 1, then we're fine.

[quote]What exactly did the questions in AIME say?[/quote]

"...proper divisors of 1000000. (By a proper divisor of a natural number we mean a positive integral divisor other than 1 and the number itself.)"

Ah. I always get suspicious when competitions try to define words that most people should know.

[quote="MCrawford"]Would you happen to know the year and number of that AIME problem?[/quote]

That was 1986/8. There was a similar occurence on 1987/3. I am using the wording from the Contest Problem Book V, so I'm assuming that it is the original, official wording.

I just took a look at both of those problems and I have to say that those definitions baffle me.  I have never noticed the term "proper divisor" used that way.  I guess that's why they gave the definitions both times.  I don't think I've ever seen a definition on the AMC, AIME, or major Olympiad that included a definition other than what I knew/thought to be standard.

I'm not sure what to tell you, though I think on other contests "proper divisors" will include 1.

I also had a little look at those problems. Funny thing is, in both problems, it doesn't make a single bit of difference whether you use 1 or not. The first one involves a sum of the logs of the factors, and log 1 is 0, and the second one involves the product of the factors, which again multiplying by 1 does nothing.

For what it's worth, the Mathworld Web Site acknowledges that there are various definitions of this term.  [url]http://mathworld.wolfram.com/ProperDivisor.html[/url]  "To make matters even more confusing, the proper divisor is often defined so that -1 and 1 are also excluded."


A figure is constructed from unit cubes.  Each cube shares
at least one face with another cube.   What is the minimum number
of cubes needed to build a figure with the front and side views
shown?
[asy]/* AMC8 2003 #15 Problem */
draw((0,0)--(2,0)--(2,1)--(1,1)--(1,2)--(0,2)--cycle);
draw((0,1)--(1,1)--(1,0));
draw((4,0)--(6,0)--(6,2)--(5,2)--(5,1)--(4,1)--cycle);
draw((5,0)--(5,1)--(6,1));
label(scale(0.8)*"FRONT", (1, 0), S);
label(scale(0.8)*"SIDE", (5,0), S);[/asy]

[quote="r15s11z55y89w21"]I think $ \fbox{4}$. Does anyone have an explanation?[/quote]
I do :)
[hide="Solution"]
This is one example of the solution of $\boxed{\text{4}}$ cubes. The corner cube can't be removed because the cube needs to look the same from the front [i]and[/i] the side view
[asy]
import three;
defaultpen(linewidth(0.8));
real r=0.5;
currentprojection=orthographic(3/4,8/15,7/15);
draw(unitcube, white, thick(), nolight);
draw(shift(1,-1,0)*unitcube, white, thick(), nolight);
draw(shift(1,0,0)*unitcube, white, thick(), nolight);
draw(shift(1,0,1)*unitcube, white, thick(), nolight);[/asy]
[/hide]


Hi, I just thought about an interesting problem, and posed it as a challenge to this forum...  Hope you guys enjoy solving it.

Suppose there is a rocket with mass M when empty (fuel and tube only, no liquid being transported), having the shape of a hollow cylinder (that is capped). It has a radius r, height h. It is propelled by having fuel being ejected from it at a velocity v, at a constant rate R in the opposite direction of motion.  The rocket is carrying a fluid inside the cylinder of density P (rho), and initial volume Vi (the liquid contributes a mass not counted in M).

Suppose the rocket starts its engine in outer space at time t=0, with negligible gravitational fields. Unfortunately, there is a hole of of area A at the bottom of the rocket, causing the fluid to leak when the rocket accelerates.  The hole has negligible size compared to the radius of the rocket.

1. What is the rocket's velocity as a function of time?

2. What is the volume of the liquid as a function of time?

Thanks for contributing!  :D

PS: things that are not negligible:

- change in liquid volume (and mass)
- change in fuel mass carried
- acceleration variation causes variation in leaking rate
- other factors that seem not negligible and that I forgot about.

Nice problem. You have a really inquisitive brain. I will give this a thought tonight...

For now all I can say is that the mass of rocket is not constant nor the acceleration is constant and the leakage of fuel too is not constant.

Also, rockets should not leak.:rotfl:

Note: I'm going to change one of the letters of the original problem. I'm going to call the velocity of the rocket $ v$ and the velocity of the ejected fuel (with respect to the rocket) $ u$. I'm also going to call the initial mass of the rocket with fuel $ M_0$, and the mass of the liquid cargo $ M_L$.

At time $ t$ the rocket (with its fuel and liquid cargo) has a mass of $ M + M_L$ and a velocity of $ v$. An infinitessimal time later, at $ t + \Delta t$, the rocket will have ejected a fuel mass of $ R \Delta t$ and will have leaked a liquid mass of $ \Delta M_L$ (this amount to be discussed later). So its new mass will be $ M + M_L - R \Delta t - \Delta M_L$. The rocket will have a velocity of $ v + \Delta v$, the dischared fuel will have a velocity of $ v - u$, and the leaked liquid will have a velocity of $ v - v_L$ (to be discussed later).

So, by conservation of momentum, we will have $ (M + M_L)v = (M + M_L - R \Delta t - \Delta M_L)(v + \Delta v) + (R \Delta t)(v - u) + (\Delta M_L)(v - v_L)$ or $ p_i = p_f + {(R \Delta t)(v - u) + (\Delta M_L)(v - v_L)}$, where $ p_i$ and $ p_f$ are the before and after momenta of the rocket respectively.

So $ \frac {dp}{dt} = R(u - v) + \left(\frac {dM_L}{dt}\right)(v_L - v)$
By Newton's First Law, $ F = \frac{dp}{dt} = ma$
So $ (M + M_L)\frac {dv}{dt} = Ru + \left(\frac {dM_L}{dt}\right)v_L - \left(R + \frac {dM_L}{dt}\right)v$

Because the fuel is ejected at a constant rate, R, $ M = M_0 - Rt$. So this equation becomes
1.) $ (M_0 - Rt + M_L)\frac {dv}{dt} = Ru + \left(\frac {dM_L}{dt}\right)v_L - \left(R + \frac {dM_L}{dt}\right)v$

Now for the leaking cargo. The acceleration, $ a$, of the rocket will force the liquid to press against the bottom of the rocket and come out of the hole in the bottom, just as gravitational acceleration $ a$ would do. So we can model the leaking of the liquid as if it were from a fixed tank due to a gravitational acceleration $ a$.

In a fixed tank with liquid leaking due to gravity, the potential energy of the liquid is translated into kinetic energy of the liquid.
So $ (\Delta M_L)ah = \frac {1}{2}(\Delta M_L)(v_L^2)$, which means $ v_L = \sqrt {2ah}$, where $ a = \frac {dv}{dt}$ and $ h = \frac {M_L}{\pi r^2 \rho}$
So 2.) $ v_L = \sqrt {\frac {2M_L}{\pi r^2 \rho} \cdot \frac {dv}{dt}}$

Furthermore, 3.) $ \frac {dM_L}{dt} = \rho A v_L$

So you have a system of three differential equations (numbered above) in the three unknown functions $ v$, $ v_L$, and $ M_L$

[quote="gauss202"]$ F \equal{}\frac{dp}{dt}\equal{} ma$[/quote]
It's in opposition to the previous work that takes in account the variable mass.

I think you're right. Let me see if I can get around it.

The momentum of the whole system at time $ t$ is $ (M \plus{} M_L)v$, and the momentum of the whole system at time $ t \plus{} \Delta t$ is $ (M \plus{} M_L \minus{} R\Delta t \minus{} \Delta M_L)(v \plus{} \Delta v) \plus{} (R\Delta t)(v \minus{} u) \plus{} (\Delta M_L)(v \minus{} v_L)$

Since there are no external forces acting on the system, the change in momentum is zero. So
$ (M \plus{} M_L \minus{} R\Delta t \minus{} \Delta M_L)(v \plus{} \Delta v) \plus{} (R\Delta t)(v \minus{} u) \plus{} (\Delta M_L)(v \minus{} v_L) \minus{} (M \plus{} M_L)v \equal{} 0$
Expanding terms and simplifying gives
$ (M \plus{} M_L)\Delta v \plus{} ( \minus{} R\Delta t \minus{} \Delta M_L)\Delta v \plus{} (R\Delta t)( \minus{} u) \plus{} (\Delta M_L)( \minus{} v_L) \equal{} 0$

Then dividing by $ \Delta t$ and letting the differentials tend to zero gives
$ (M \plus{} M_L)\frac {dv}{dt} \plus{} 0 \minus{} Ru \minus{} \frac {dM_L}{dt}v_L \equal{} 0$

Or, $ (M_0 \minus{} Rt \plus{} M_L)\frac {dv}{dt} \equal{} Ru \plus{} \frac {dM_L}{dt}v_L$

This equation should replace equation 1.)

Really nice solution gauss!  I actually knew my problem was complicated, but 3 diff. equations and 3 functions... that's pretty nasty to solve (if even possible).


In the two concentric circles shown, the radius of the outer circle is twice the radius of the inner circle. What is the area of the gray region, in square feet, if the width of the gray region is 2 feet? Express your answer in terms of $ \pi$.

[asy]filldraw(circle((0,0),4),gray);
filldraw(circle((0,0),2),white);

draw((2,0)--(4,0),linewidth(1));

label("$2^{\prime}$",(3,0),N);[/asy]

We can easily find that the radius of the inner circle is $ 2$. The desired area is $ 4^2\pi\minus{}2^2\pi\equal{}16\pi\minus{}4\pi\equal{}\boxed{12\pi}$.


A mother wants to spend at least 40 dollars but no more than 60 dollars on school clothes for her child. Shirts cost 10 dollars and pants cost 20 dollars. How many of each could she buy?

What inequalities would describe this situation?

Let $ s$ be the number of shirts the mother buys and $ p$ be the number of pants that she buys. Then we know that she will spend $ 10s \plus{} 20p$ in total. Since that value is at least 40 dollars, we know that $ 10s \plus{} 20p \geq 40$ and since it is no more that 60 dollars, we know that $ 10s \plus{} 20p \leq 60$. So the two inequalities describing the situation are

$ \boxed{10s \plus{} 20p \geq 40 \ and \ 10s \plus{} 20p \leq 60}$

*Note: You can also just write it as just one inequality $ \boxed{40 \leq 10s\plus{} 20p \leq 60}$ but they are really just the same thing.

[quote="sheeprule"]How many of each could she buy?[/quote]

Again, let $ s$ be the number of shirts bought and $ p$ be the number of pants bought.
Then, all of the integer solutions $ (s,p)$ are:
$ (4,0), (5,0), (6,0), (0,2), (0,3), (2,1), (3,1), (1,2), (4,1), (2,2)$


Find all $ f: R \minus{} > R$ such that for each $ x$ from real $ f(x)$ is nonegative and

$ f(x \plus{} y) \plus{} f(x \minus{} y) \minus{} 2f(x) \minus{} y^2 \equal{} 0$

Let $ g(x) \equal{} f(x) \minus{} \frac{x^2}{2}$. This leads to $ g(x) \equal{} \frac{g(x\plus{}y)\plus{}g(x\minus{}y)}{2}$. We could solve for $ g(x)$ and get $ g(x) \equal{} ax \plus{} b$.

[quote="mathangel"]Let $ g(x) \equal{} f(x) \minus{} \frac {x^2}{2}$. This leads to $ g(x) \equal{} \frac {g(x \plus{} y) \plus{} g(x \minus{} y)}{2}$. We could solve for $ g(x)$ and get $ g(x) \equal{} ax \plus{} b$.[/quote]
It will true if $ f(x)$ is continuous function ,mathangel

Let $ a\equal{}f(0)$ and $ g(x)\equal{}f(x)\minus{}\frac{x^2}2\minus{}a$.
We see Cauchy equation holds for $ g(x)$ and
$ g(x)\ge\minus{}\frac{x^2}2\minus{}a$.
Now we have $ f(rx)\equal{}rf(x)$ for any rational $ r$.

Consider the case $ a>0$.
Fix $ x\ne0$. We can find positive rational $ r_n\rightarrow\sqrt{\frac{x^2}{2a}}$.
Then $ f(x)\equal{}r_nf(\frac x{r_n})\ge\minus{}\frac{x^2}{2r_n}\minus{}ar_n\rightarrow \minus{}\sqrt{2a}|x|$.
Hence, we have
$ g(x)\ge\minus{}\frac{x^2}2$ when $ a\equal{}0$;
$ g(x)\ge \minus{}\sqrt{2a}|x|$ when $ a>0$.
Since $ g$ is odd function, we also have
$ g(x)\le\frac{x^2}2$ when $ a\equal{}0$;
$ g(x)\le\sqrt{2a}|x|$ when $ a>0$.

Now we fix any number $ q$ and let $ r_n$ converge to $ q$.
We see when $ a\equal{}0$,
$ g(qx)\equal{}r_ng(x)\plus{}g((q\minus{}r_n)x)\ge r_ng(x)\minus{}(q\minus{}r_n)^2\frac{x^2}2\Rightarrow f(qx)\ge qf(x)$ ,
or when $ a>0$,
$ g(qx)\equal{}r_ng(x)\plus{}g((q\minus{}r_n)x)\ge r_ng(x)\minus{}\sqrt{2a}|(q\minus{}r_n)x|\Rightarrow f(qx)\ge qf(x)$ ,
Similarly, we can use the upper bound to show $ f(qx)\le qf(x)$.

Hence, we proved $ g$ has to be linear.
$ f(x)\equal{}\frac{x^2}2\plus{}ax\plus{}b$ with $ 2b\ge a^2$.