Prove that for any integers $ a$ and $ b$, $ a\neq 0$, there exists a unique pair $ (q,r)$ of integers such that $ b\equal{}aq\plus{}r$ and $ 0\leq r < |a|$.

How is this an extension of the division algorithm?  This [i]is[/i] the division algorithm.

[hide="Hint"] The set $ \{ b\minus{}ax : x\in\mathbb{Z}, b\minus{}ax\ge 0\}$ has a minimal element. [/hide]

As I posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=929359#929359]here[/url], the Division Algorithm, I belive, was $ r<a$, but this one has $ 0\leq r<|a|$.

The distinction is casework on the signs.  It's hardly enough to qualify as a different statement.


Isosceles right triangle $ ABC$ encloses a semicircle of area $ 2\pi$. The
circle has its center $ O$ on hypotenuse $ \overline{AB}$ and is tangent to
sides $ \overline{AC}$ and $ \overline{BC}$. What is the area of triangle
$ ABC$?
[asy]pair a=(4,4), b=(0,0), c=(0,4), d=(4,0), o=(2,2);
draw(circle(o, 2));
clip(a--b--c--cycle);
draw(a--b--c--cycle);
dot(o);
label("$C$", c, NW);
label("$A$", a, NE);
label("$B$", b, SW);[/asy]

if the area of the semicircle is 2pi then the whole circle is 4pi meaning the radius is 2, that mans that the side lengths of the square is 4 so 4*4 = 16/2 = [b]8[/b]


For a given positive integer $ n$, prove that if it is possible to find an ordered pair of positive integers $ (a, b)$ such that $ a^2 \plus{} b^2$ is divisible by $ ab \plus{} n$, it is possible to find such an ordered pair with $ a \ge b$ and $ n \ge \tfrac{b^3}{2a}$.

$ a\equal{}n^2\minus{}n\plus{}1$ and $ b\equal{}1$ seems to work...

Sorry, by attempting to simplify the problem, I seem to have made it different.  Here is the original problem I intended to post:

For a given positive integer $ n$, prove that if it is possible to find an ordered pair of positive integers $ (a, b)$, with $ a \ge b$ such that $ a^2 \plus{} b^2$ is divisible by $ ab \plus{} n$, then the ordered pair with the smallest value of $ a \plus{} b$ satisfies $ n \ge \tfrac{b^3}{2a}$.


Describe connected components of $ Q$ in topology induced from $ R$.

Is $ Q\equal{}\mathbb{Q}$ and $ R\equal{}\mathbb{R}$ ? Then $ \mathbb{Q}$ is totally disconnected, i.e. the components are the one-point-sets.

This correct answer above is listed (without proof) here [url]http://en.wikipedia.org/wiki/Connected_space[/url]

It is a little counter-intuitive. You cannot separate a rational point $ x$ from $ \mathbb{Q} \setminus \{x\}$, but you can separate any $ x \neq y$ by choosing some irrational $ z$ between them, and covering $ \mathbb{Q}$ with $ (\mathbb{Q} \cap (\minus{}\infty,z)) \cup (\mathbb{Q} \cap (z,\infty))$, therefore $ x,y$ belong to different connected components.


geometry
3D geometry
topology
real analysis
real analysis unsolved

Let $e = \{(x_{n}); |x_{n}| \leq \dfrac{1}{n}$, $\forall \;\; n \geq 1\}$ $\subset$ $(l^2)_R (N)$

Show that e is a compact subset of $(l^2)_{R}$

[quote="julien_santini"]Note $p_{k}((x_{n})_{n}) = |x_{k}|$, which is clearly continuous. Now $e = \cap p_{k}^{-1}([0,\dfrac{1}{k}])$ whence $e$ is closed. But $e$ is also bounded (trivial) which solves the question.[/quote]

I see. But is it true that in an infinite dimensional metric space, "closed and bounded" and "compact" are not equivalent?

[quote="swag"]
But is it true that in an infinite dimensional metric space, "closed and bounded" and "compact" are not equivalent?[/quote]
Yes, that's true: the unit ball in $\ell^2$ is not compact. So, julien_santini's solution is not valid.

Sorry about that I replied too fast :(. I'll post an answer later (or tomorrow). Well, ok:

$\epsilon >0$. Consider a sequence $((x_{n,i})_n)_{i}$ of elements of $e$. For each fixed $n$, since $(x_{n,i})_{i}$ is bounded, one can extract a converging subsequence $(x_{\phi_{i}(n),i})_{i}$ (with limit $l_{i}$) such that $|x_{\phi_{i}(n),i}- l_{i}|<\epsilon/2^{i}$. Iterating the argument one can assume for each $i$, $(x_{\phi_{i}(n),i})_{i}$ is converging for every $n \leq i$ in the previous way. Now using the classical diagonal argument we consider the sequence $(x_{\phi_{n}(n)})_{n}$. We must show this sequence converges to $(l_{n})_{n}$ in $l_{2}$, which is verified by: ${(\sum_{i \geq 1}}{|(x_{\phi_{n}}(n)-l_{n}|^{2}})^{\dfrac{1}{2}} < \dfrac{\sqrt{2}\epsilon}{2}$

Don't we need to show that the sequence converges in $e$ and not $(l^2)_{R}$?

The space is a topological product of compact spaces, so, according to Tychonoff's Theorem, it's compact.

Swag: the elements were taken in $e$ and the limit necessarily belongs to $e$ since $e$ is closed. If you know Tychonov theorem, then the result becomes trivial. In passing, the Tychonov theorem uses the axiom of choice in its proof, and I am aware that it is actually *equivalent* to the axiom of choice, but I'd be curious to see a proof.


I am getting my feet wet for the convergence in distribution and CLT topics.  Could anyone please provide me with a little hint on this one.

Let $ Y_{s}$ be a Poisson random variable; that is $ P[Y_{s} \equal{} k] \equal{} e^{\minus{}s}\frac {s^{k}}{k!}$.

Show that $ \frac{Y_{s} \minus{} s}{\sqrt s}\rightarrow N(0,1)$ in distribution.  N(0,1) = standard normal distribution.

Any help is greatly appreciated! Thanks.

We can decompose $ Y_n$ and write it as a sum $ Y_n\equal{}\sum_1^n X_i$ where the equality means equality of distribution functions and $ X_i$ has Poisson distribution with mean 1. Then apply CLT.


I got 197  :D  Yay
I mean its my highest ever percentile (though it might look idiotic to vihang and anit sahu and ofc sachin) i got 97.5 percentile  :D
And seemingly i got all the subjectives right and this time i shall get all wrong bcos i changed my way of writing subjectives bcos i thought i was wrong last time  :mad: .
My split up is:
Math : 92
Chem : 66
Physics : miserable if u wanna know subtract from the rest

I am terribly upset my first paper did not have any score in it. :mad:
Probably they thought that I would be a potential threat to the No.1 mark 316 so they didn't evaluvate it :rotfl:  :rotfl:

Anyway I  got 100 in paper 2


I have a task where i get the following info on a triangle:

c=15, b=7 and the triangle has the size of 35.

I found this that i consider what i needed, but didn't get a result that made any sence:

T=0,5*b*c*sin(A)

This should be a pretty easy thing to do, but i couldn't. Can anyone tell me what i do wrong:

35=0,5*7*15*sin(A)
sin(A)=35/(0,5*7*15)
A=sin(35/(0,5*7*15))^-1

(For my own usage: 114)

[quote="Sporally"]
A=sin(35/(0,5*7*15))^-1
[/quote]
This line should be $A=\arcsin \frac{35}{0.5\cdot 7\cdot 15}$, not $A=\frac{1}{\sin \frac{35}{0.5\cdot 7\cdot 15}}$.

Another approach is to try to solve the third side by Heron's formula.

Think i will go with puuhikki way to do it. The other one is too strange to me at the current time. What is 'ArcSin' and how do you write it on a calculator (i have the TI-84 plus).

Sporally:  ArcSin is just a more precise way of writing $\sin^{-1}$.  That's the button you press on your calculator.

AH, now i know what was wrong. I had it all right from the beginning (did use the arcsin at the beginning). When i checked to see if i could get the right size showing 35 i thought i just had to put in the 41,81 instead of 'sin(41,81)', but of course i have to do it this other way.


In a convex quadrilateral $ABCD$, point $M$ is the midpoint of the diagonal $AC$. Prove that if $\angle BAD=\angle BMC=\angle CMD$, then a circle can be inscribed in quadrilateral $ABCD$.

I have question we should prove that abcd is cyclic or we can put a circle in it


How many digits are there in the expansion of 13^18?

now i figured out a way to solve this...but not sure if it is the right way???

binomial expansion

(10+3)^18 = 10^18 + .......

= 1,00,000,... upto 18 zeros....

hence 19 digits is the answer...

am i right???? any flaw in the method???
interestingly i found a solution on the web which says it is 21 digits.... i dont see how??? :?:

[quote]log(13^18) = 18 log 13 = 18*1.114 = 20.05

So there are twenty-one digits in the number.
[/quote]

This is the solution i found on the web

What about the rest of the terms of the binomial expansion? Why do you assume the other terms do not have more digits?
[hide="explanation of web solution"]$20<\log 13^{18}<21\Rightarrow 10^{20}<13^{18}<10^{21}$.
$10^{20}$ has 21 digits and $10^{21}$ is the first(least) number with 22 digits. Therefore $13^{18}$ has 21 digits.[/hide]

Bictor is correct. Another way to approach this problem:

$\lfloor \log 13^{18}\rfloor+1 = \lfloor 18\log 13 \rfloor+1$

$\log 13 = 1.11394335\ldots$

$\lfloor 18 \cdot 1.11394335\rfloor+1 = 20+1 = 21$

so how do you solve it without using log, calculators and in under 60 secs? :oops:

I don't know any other way other than logs.

its a good idea to memorize base ten prime logs up to about 20.  and also log e (.434 i think)

There must be a way using Binomial Expansion.... what if the question was 87^ 21???? any more ideas anyone?

I don't know... i would love to know.... :maybe:

Perhaps this - $13^{18}=(13^{3})^{6}=2197^{6}$ (it's easy enough to multiply $169\times 13$ by hand) and that's approximately $2200^{6}$. Therefore, we have $2.2^{6}$ and $18$ more zeroes. Now $2.2^{6}=(2.2^{3})^{2}=10.648^{2}$, (again, $4.84\times 2.2$ is easy), hence we get three more digits for a total of $21$.


Dear friends,
This problem has been troubling me from a long time.
problem: There are 21 objects.They are 7 objects each repeating thrice.How can these 21 objects be arranged such that any  two identical objects are separated by two objects(obviously different) atleast?

I don't think I understand the question. What's wrong with 1,2,3...7,1...7,1...7?

the friend above:you didn't understand the last sentence.
for example: if there are 9 objects.and they are 3 objects each repeating thrice
1  2 3 1 2 3 123 because there are two numbers between "1"and "1","2"and"2","3"and"3".

let me solve the question:if a_1,a_2,?,a_21  satisfy the condition.we can get a_1=a_4=a_7,a_2=a_5=a_8,a_3=a_6=a9.also a_10=a_13=a_16,a_11=a_14=a_17,a_12=a_15=a_18.so we can get contradiction a_19=a_20=a_21.
it seems easy if I did it right.but how bother my solution is.
we can also prove if there is n kind objects,3|n,the arrange is exist.otherwise is nt

reply to Zhaobin,
Why do you think it is necessary that a_k=a_k+3 ;k=1,2,....,18.
I had put in the word "atleast " in the question.Please review.

      thanking you in anticipation,
                                                        srikanth

I'm afraid I must be wrong.
do you mean that there are one pair 1 and 1 separated by two objects  at least??

reply to Zhaobin,
Yes, separated by atleast two objects.
thanking you in anticipation,
srikanth


[color=darkblue]Here are the problems.
Reaaly nice ones. Good choice for BMO!!!!

[url=http://www.mathlinks.ro/viewtopic.php?t=274318]Problem 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=274320]Problem 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=274321]Problem 3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?t=274322]Problem 4[/url][/color]

That's my first BMO and I really liked it. We are having good time here!!!

I actually really don't like it. Third problem is really weird. I have spent most of my time on it for making an elegant coloring or something like that in order to use all the information given - which is too much and specifical. But it turned out that a really easy idea without using any of the informations which i got easily while thinking -and didn't think they deserve writing - can solve it. I know i'm not good at questions such as 4th one but i am still on the side that it wasn't a good problem either. The exam was really bad in my point of view.

Seems it was very strange olympiad.Questions 1,2 are very easy (as usual) and 4 is not hard for whom who knows a bit about functions. I have no idea about 3rd one because my combinatorics is awfull. I only got some lines and nothing else :( but managed to solve rest 3 questions less than 2 hours :)


I'm using

[code]\begin{minipage}{3in}
text goes here.
\end{minipage}
\begin{minipage}{3in}
\includegraphics[scale=0.8]{problem1.png}
\end{minipage}[/code]

I want to move the graphic to the right. Every time I try to modify {3in} in the second minipage, nothing happens. Also, the graphic appears a bit upper than the text. How to fix this?

[code]\begin{minipage}[t]{3in} % [t] align via top of minipages
\vspace{1ex} % push the text down slightly
\hspace{1in} text goes here. % can use \hfill if right-aligned needed
\end{minipage}
\begin{minipage}[t]{3in} %[t] align via top of minipages
\vspace{0pt} % required as otherwise will align with bottom of graphics
\includegraphics[scale=0.8]{image5.png}
\end{minipage}[/code]
You may also wish to use one of the packages recommended at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=textflow]Flowing text around figures in LaTeX[/url].

I've changed my preferences, I'm now using \parbox.

For example, I have
[code]
\noindent\parbox{9cm}{text goes here}
\noindent\parbox{1cm}{\includegraphics[scale=0.8]{problem2.png}}[/code]

Everytime I change the {1cm}, namely 9 or 5, the graphic doesn't move. The graphic is pretty close to the text, and that's why I want to move it to the right, but I can't.

parboxes and minipages are very similar so you can use the same code
[code]\parbox[t]{5cm}{\vspace{1ex}\hspace{1cm} text goes here}
\parbox[t]{5cm}{\vspace{0pt}\includegraphics[scale=0.8]{image.png}}[/code]

[quote="stevem"][code]\begin{minipage}[t]{3in} % [t] align via top of minipages
\vspace{1ex} % push the text down slightly
\hspace{1in} text goes here. % can use \hfill if right-aligned needed
\end{minipage}
\begin{minipage}[t]{3in} %[t] align via top of minipages
\vspace{0pt} % required as otherwise will align with bottom of graphics
\includegraphics[scale=0.8]{image5.png}
\end{minipage}[/code]
[/quote]
I was using this, but I can't yet move the graphic to the right. If I change the number (3in) in the second minipage (for example, I set it up with 10in), the graphic didn't move anyway.


Prove that any square of an odd prime number can be expressed of the sum of three squares, no two of which are 0.

Maybe its the one exeption because it is the only even (2 is the wierd prime)

i'm working on proof

the numbers that cannot be written as a sum of 3 squares are in the form $4^{k}(7+8n)$, clearly, k=0, furthermore, there are no squares 7 mod 8,

Using Legendre's is cheating :P !
(By the way, I have yet to see a proof of Legendre's theorem; if you knew it, could you post here?)

Any number not of the form $4^{n}(8n+7)$ is the sum of three squares. And that every number is the sum of four squares as a corr.

You have proven that square can be written as a sum of three squares (not necessary $\neq 0$).
So you have just shown a very trivial statement ;)

PS: proofs of the three-squares-theorem are long or use a lot of not so trivial statements.

[quote="Altheman"]what is legendre???[/quote]
Legendre's is the theorem you just invoked to trivialize my problem

[quote="rem"]What about $4$? This one does not work. :([/quote]

[quote="Farenhajt"]But it is a SQUARE of a prime, and yet it can't be split into three squares.[/quote]

Umm the question says "any square of an [b] odd prime number[/b], so it doesnt matter if 2^2 doesnt work cuz its not an odd prime...btw does anyone know the proof for "legendre"

It used to say the square of any prime, until the original poster edited it :wink:


inequalities
trigonometry
calculus
derivative
geometry
angle bisector
geometry unsolved

Let $ \alpha$ and $ \beta$ be the angles of a non-isosceles triangle $ ABC$ at points $ A$ and $ B$, respectively. Let the bisectors of these angles intersect opposing sides of the triangle in $ D$ and $ E$, respectively. Prove that the acute angle between the lines $ DE$ and $ AB$ isn't greater than $ \frac{|\alpha\minus{}\beta|}3$.

Suppose $ \alpha > \beta$. Let the acute angle we are looking for be $ \angle AME \equal{} \theta$. We have that $ CD \equal{} \tfrac{ab}{b \plus{} c}$ and we also have $ CE \equal{} \tfrac{ab}{a \plus{} c}$. Therefore,
\[ T(\theta) \equal{} \frac {\sin (\alpha \minus{} \theta)}{\sin (\gamma \plus{} \alpha \minus{} \theta)} \equal{} \frac {a \plus{} c}{b \plus{} c}
\]
Since $ T(\theta)$ is decreasing in $ \theta$, (the derivative is $ \minus{} \tfrac{\sin(\alpha \plus{} \beta)}{\sin^2(\beta \plus{} \theta)}$) it is enough to prove that
\[ L \equal{} \frac {\sin\tfrac{2\alpha \plus{} \beta}{3}}{\sin\tfrac{\alpha \plus{} 2\beta}{3}} \leq \frac {\sin\alpha \plus{} \sin\gamma}{\sin\beta \plus{} \sin\gamma} \equal{} R
\]
Notice that
\[ R \minus{} L \equal{} \frac {1}{2(\sin\gamma \plus{} \sin\beta)\cos\tfrac{\alpha \plus{} 2\beta}{6}}\left(\sin\tfrac{7\alpha \plus{} 8\beta}{6} \plus{} \sin\tfrac{5\alpha \plus{} 4\beta}{6} \minus{} \sin\tfrac{9\alpha \plus{} 6\beta}{6} \minus{} \sin\tfrac{3\alpha \plus{} 6\beta}{6}\right)
\]
and this is nonnegative when $ 9\alpha \plus{} 6\beta\leq 6\pi$ because $ (9\alpha \plus{} 6\beta,3\alpha \plus{} 6\beta)$ majorizes $ (7\alpha \plus{} 8\beta, 5\alpha \plus{} 4\beta)$, it is nonnegative when $ 9\alpha \plus{} 6\beta > 6\pi$ because then (as $ \alpha \geq \beta$) we have $ \sin\tfrac{7\alpha \plus{} 8\beta}{6} \minus{} \sin\tfrac{9\alpha \plus{} 6\beta}{6} \geq 0$ and $ \sin\tfrac{5\alpha \plus{} 4\beta}{6} \geq \sin\tfrac{3\alpha \plus{} 6\beta}{6}$.

Does the bisector of external angle at $ C$ of triangle $ ABC$ go through the point of intersection of $ DE$ and $ AB$?

Yep, this follows from Menelaus' theorem.

[quote="communist"]We have D belongs to BC and E belongs to AC and the given conditions.
Let the intersection point making the specific angle be M.
We have alpha=A and Beta= B
And BAD=DAC=A/2 and ABE=CBE=B/2
And now we suppose that angle made at M is (A-B)/3 + k for some k. and suppose that k is positive and we can prove this is impossible.
Then we get BDM= 180-(180-B + (A-B)/3 +k)
                             = (4B-A)/3 + k
Similarly we get DEC= (4A-B)/3 + k
Finally we get angle C to be 180-(BDM or EDC + DEC)
= 180-(4B-A+4A-B)/3 + 2k
= 180-A-B-2k
But angle C is 180-A-B and we specified k is positive thus this is impossible.thus proved
in sunken rock's style. synthetic and only synthetic[/quote]

There is something wrong. I think among the two angles $ \widehat{CED}$ and $ \widehat{CDE}$, there is one equal to $ \frac{|\alpha \minus{} \beta|}{3}\plus{}|k|$ and another is equal to $ \frac{|\alpha \minus{} \beta|}{3}\minus{}|k|$ :oops:

I think your calculation may be right, but the rest of the work is quite hard. Is there any fully synthetic proof for this problem?

In my opinion, it is really challenge to think of a synthetic way to express $ \frac{|\alpha \minus{} \beta|}{3}$ :(

[quote="communist"]i am a fool and it is a foolish mistake possible only by a fool. i am extremely sorry.but i have by some trigonometry able to deuduce the problem
i.e. tan(i)=(sinB-sinA)/(1+cosA+cosB)
so we have to prove tan((B-A)/3)> (sinB-sinA)/(1+cosA+cosB) is this possible.[/quote]

This is possible; I got to the expression with external trig ceva.  To solve it, we can clear the denominators and use basic sine and cosine sum formulas to simplify it.

so we i the exp. by men. and trig. and it i am sure it is the simplest form. can we not work on that exp. and try to prove it. mashimaru, is it possible? maybe we should post a new topic with this trig. ques?

[hide="Solution"]

Let $ \{M\}=AB\cap DE$. As by Menelaus' theorem, external bisector of angle $ C$ passes through $ M$.  Now we easily see that $ \angle AMC=\frac{\alpha-\beta}{2}$. If the angle we need is greater than the value given, we see that $ \angle BMD>2\angle DMC$, or ${ \frac{\sin\angle BMD}{\sin\angle DMC}>\frac{\sin2\angle DMC}{\sin\angle DMC}=2\cos\angle DMC}$

But, the same expression $ \frac{\sin\angle BMD}{\sin\angle DMC}=\frac{\frac{BD\sin\beta}{MD}}{\frac{CD\cos\frac{\gamma}{2}}{MD}}=2\sin\frac{\gamma}{2}$, because of sine law in triangles $ DMC$ and $ DMB$, and angle bisector theorem.

So, $ \sin\frac{\gamma}{2}>\sin(90^\circ-\angle DMC)$, so $ \angle DMC>\frac{\alpha+\beta}{2}$, but that's impossible. [/hide]


[quote="silouan"]if $x,y,z$ are positive reals prove that
$(x+y+z)^5-x^5-y^5-z^5\geq 60xyz(xy+yz+zx)$[/quote]

Nice inequality!!

I think the following sharper ineq holds:

$(x+y+z)^5-x^5-y^5-z^5\geq 80xyz(xy+yz+zx)$

We have
$(x+y+z)^5-x^5-y^5-z^5=5(x+y)(y+z)(z+x)(x^2+y^2+z^2+xy+yz+zx)\geq 5 \cdot 8xyz \cdot 2(xy+yz+zx)=80xyz(xy+yz+zx)$

What about your solution??

 :)

[hide="edit: just 2 minutes later than you..."]$(x+y+z)^5-x^5-y^5-z^5\geq 80xyz(xy+yz+zx)$
Since $(x+y+z)^5-x^5-y^5-z^5=5(x+y)(y+z)(z+x)(x^2+y^2+z^2+xy+yz+zx)$
So we only need to prove
$(x+y)(y+z)(z+x)(x^2+y^2+z^2+xy+yz+zx)\geq 16xyz(xy+yz+zx)$
This follows directly from
$(x+y)(y+z)(z+x)\geq 8xyz$ and
$x^2+y^2+z^2+xy+yz+zx\geq 2(xy+yz+zx)$[/hide]

You work  exactly as me. Now try to prove the two new ineqs I posted . :)


integration
calculus
derivative
real analysis
real analysis unsolved

I am trying to determine the optimal Q for the following expression by using first order conditions. The expression is :

$P = r * min(x,Q) -pQ - b(x-Q)$

The expectation of P is calculated as :

$E[P] = r\int\limits_{0}^{Q}x f(x)\; dx + rQ\int\limits_{Q}^{\infty} f(x)\; dx - pQ - b\int\limits_{Q}^{\infty}(x-Q) f(x)\; dx$

Now, the next step is to use first order conditions. After taking the first derivative of $E(P)$ with respect to Q and equating the expression on the right hand side to $0$, I get

$\int\limits_{Q}^{\infty} f(x)\; dx = \frac{p}{r+b}$

However the answer provided is :

$\int\limits_{0}^{Q} x f(x)\; dx = \frac{r-p+b}{r+b}$

Can you please help me see what i am doing wrong?

thank you

Is $\int_0^\infty f(x)\,dx=1$? If so, you agree with that answer.

Well, this is sad. I have now corrected the provided answer in my post. There is an extra $x$ in the provided solution.

Can you please help me figure out how to get to the answer provided?

Sorry for that typo there

I do not see anything wrong with your answer if the functional is as you wrote it (I'm puzzled a little bit over why you integrate from $Q$ to $\infty$ rather than from $0$ to $\infty$ in the last term, but, probably, you meant $(x-Q)_+$ instead of $(x-Q)$ in the definition of $P$). So, in my opinion, your answer is right and the one in the book is just wrong.

By the way, there is no need to post the same thing twice. It doesn't increase your chances of getting a reply but only confuses and irritates people. So, I'll delete your other identical post :)


topology
function
advanced fields
advanced fields unsolved

LEt <S, T) be a topological space and Φ ≠Fc S. We say that A c F is closed IN  F iff F - A € τIF. Show that A is closed in F iff there exists a closed subset C of S such that A = er. F. Show also that if F is closed, then A is closed in F)ff A is closed .
	. 	.

2. Letf: (S, t1) → <T, t2) be continuous: Show thatJ f/A: .‹A·, t1/A› → (T,T2) is continuous for all Ac S.
3. Let <s,t› be a topological space and Φ ≠C e c AỤ B c S. Show that C €t/AUB iff CU A €t/A and C∩B€tlB.

4. Show that a function f: (T, t) → ∏{(Sa,ta) I a €A} is continuous iff ∏. F:(T.t) → (Sa,ta) is continuous for all a€ A.

I can't see why is this thread here... IMO and topology?


Hi,

For my school project of the year in the world history class, I need to write a paper about a nation. Through random picking, I got Israel. I don't know much about Israel and wondered if there were any Israel person (I don't know what do you call person from Israel) in AoPS. I'm not sure if I can interview person through internet but if there are any, can you post here? It would be really nice to have some interviews for my paper!

Thank you.

"Israeli."  I suggest, before you start interviewing anyone, that you do a basic search for information -- for example, go read your encyclopaedia entry on it.

That is not a nation I would wanna pick out randomly.  There is so much controversy about it.

that's the same view my teacher of geography gave us in third year of high school
not having internet or any other good source, i kept believing it for many years

Look up King David Hotel Bombing!

Silverfalcon, i could help you there. I'm not full israeli (only half) but.. i know a bit about its history. What.. in specific do you need to know?

[quote="oooooh ya"]israel was started after wwII by the allies for somewhere to go for displaced jews after the holocaust. they picked the homeland of the jews, israel, and kicked out the people who lived there.

you'll probably have to pick a side to be able to write a paper, the jews say they palestinians want the land, and the israelis say they have the right of return, because it was thier land before it was the palestianians.  the problem now is there are israelis being born in israel, that didnt move there.  also palestian terrorism knocks them out of the international community.  also the u. s. has never deturred in full support of israel (israeli m-16's say property of the united states governemnt on them), the israelis are doing everything they can to get rid of the palestinians and there have been a lot of human rights violations brought up.  the u. s. have given the israeli major loans that dont have to be paid back, and the full swing of united states political support.  look up NOAM CHOMSKY!!![/quote]

I seriously hope you're joking.  It's sad to see people so misinformed.

What is this obsession with Noam Chomsky's politics? The man is a linguistics professor and researcher. His self-hating anti-Zionist politics are misinformed.


this is probably not the best place to put this question but still all those who gave KVPY,what are the expectations(like expected marks and expected cut offs)

I think I attemptef something like 55-60 questions, and I'm quite confident about the ones I have attempted. Though I don't know the cutoff, and really want to know how many marks you need to qualify (that is, to be in the top 600-800).

last time they had around 542( :maybe:)  students selected.i think a good maths and physics attempt and decent bio and chemistry should do.
how many questions were this time?

yes similar to last time i think normally the highest hovers around 70-75 and normally there is a huge gap between the 1st and the last i thing i read that in some thread here


Choose six points $ D;E;F;G;H;K$ in that order,on a circle wiht radius $ R$ and center at $ O$ such that $ DE\equal{}FG\equal{}HK\equal{}R$.$ KD$ and $ EF$ meet at $ A$;$ EF$ and $ GH$ meet at $ B$ and $ GH$ and $ KD$ meet at $ C$.Prove that:
$ OA.BC\equal{}OB.CA\equal{}OC.AB$

The topic mustn't be stopped. This problem is mentioned in "Problem in This Issue" (issue 371) of Math & Youth Magazine, which is one of math magazine in Vietnam. :mad:


Let $ a,b,c$ be positive real numbers satisfying $ a\plus{}b\plus{}c\equal{}3$.Prove that
        $ a^{2}\plus{}b^{2}\plus{}c^{2}\geq\frac{1\plus{}2a}{4\minus{}a}\plus{}\frac{1\plus{}2b}{4\minus{}b}\plus{}\frac{1\plus{}2c}{4\minus{}c}$  :oops:

[quote="toanIneq"]Let $ a,b,c$ be positive real numbers satisfying $ a\plus{}b\plus{}c \equal{} 3$.Prove that
        $ a^{2}\plus{}b^{2}\plus{}c^{2}\geq\frac{1\plus{}2a}{4\minus{}a}\plus{}\frac{1\plus{}2b}{4\minus{}b}\plus{}\frac{1\plus{}2c}{4\minus{}c}$[/quote]
$ a^{2}\plus{}b^{2}\plus{}c^{2}\geq\frac{1\plus{}2a}{4\minus{}a}\plus{}\frac{1\plus{}2b}{4\minus{}b}\plus{}\frac{1\plus{}2c}{4\minus{}c}\Leftrightarrow\sum_{cyc}\left(a^{2}\minus{}\frac{1\plus{}2a}{4\minus{}a}\right)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\left(a^{2}\minus{}\frac{1\plus{}2a}{4\minus{}a}\minus{}(a\minus{}1)\right)\geq0\Leftrightarrow\sum_{cyc}\frac{(a\minus{}1)^{2}(3\minus{}a)}{4\minus{}a}\geq0.$

[quote="toanIneq"]Let $ a,b,c$ be positive real numbers satisfying $ a \plus{} b \plus{} c \equal{} 3$.Prove that
        $ a^{2} \plus{} b^{2} \plus{} c^{2}\geq\frac {1 \plus{} 2a}{4 \minus{} a} \plus{} \frac {1 \plus{} 2b}{4 \minus{} b} \plus{} \frac {1 \plus{} 2c}{4 \minus{} c}$  :oops:[/quote]
  $ \sum({a^2 \minus{} \frac {1 \plus{} 2a}{4 \minus{} a}) \equal{} \sum{\frac {(a \minus{} 1)(3a \minus{} a^2 \plus{} 1)}{4 \minus{} a}} \equal{} \sum{\frac {(a \minus{} 1)((a \minus{} 1)(3 \minus{} a) \plus{} 4 \minus{} a)}{4 \minus{} a}} \equal{} \sum({\frac {(a \minus{} 1)^2(3 \minus{} a)}{4 \minus{} a} \plus{} a \minus{} 1}}) \equal{} \sum{\frac {(a \minus{} 1)^2(3 \minus{} a)}{4 \minus{} a}}\ge 0$

Here's my solution:

We look for $ k$ such that

$ a^2\ge \frac{1\plus{}2a}{4\minus{}a}\plus{}k(a\minus{}1)$ for every a
<=>
$ (a\minus{}1)(\frac{a^2\minus{}3a\minus{}1}{a\minus{}4}\minus{}k)\ge 0$

We set $ k$ as:
$ \frac{a^2\minus{}3a\minus{}1}{a\minus{}4}\minus{}k\equal{}(a\minus{}1)B$  :lol:

(with $ B$ some variable depending on a)

This is give: $ k\equal{}1$,  $ B\equal{}\frac{a\minus{}3}{a\minus{}4}>0$

Thus finally:
$ a^2\ge\frac{1\plus{}2a}{4\minus{}a}\plus{}(a\minus{}1)$ for every a positive
Now just sum cycly we get what we need

(Big Thanks To quangpbc :D )

[quote="super_math"]
$ \frac {9}{4 \minus{} a} \plus{} \frac {9}{4 \minus{} b} \plus{} \frac {9}{4 \minus{} c}\le\frac {(3 \plus{} 3 \plus{} 3)^2}{4 \minus{} a \plus{} 4 \minus{} b \plus{} 4 \minus{} c} \equal{} 9 \equal{} 6 \plus{} 3\le a^2 \plus{} b^2 \plus{} c^2 \plus{} 6$[/quote]
I think the sign is reversed. :wink:

[quote="stephen38"]
I think the sign is reversed. :wink:[/quote]
 Yes, incorrectly solved


The admission fee for an exhibition is $ \$25$ per adult and $ \$12$ per child.  Last Tuesday, the exhibition collected $ \$1950$ in admission fees from at least one adult and at least one child.  Of all the possible ratios of adults to children at the exhibition last Tuesday, which one is closest to $ 1$?

[hide="Seems fairly straightforward I think"]
Let's start with the best case scenario. If we had the same number of adults and children, then the ratio would be exactly. Division of 1950 and 37 yields 52 + 26/37.

Thus, with 52 children and 52 adults, the total cost is at -26 of what we want. Consider the following statements:

1. If we add an adult and subtract two children $ \implies$ net gain of 1.
2. If we add an adult, $ \implies$ net gain of 25.

Thus, if we add 2 adults and subtract two children, we will get a net gain of 26 which is what we wanted, and it is safe to assume (or check) there are no smaller cases that seem to work.

Thus, the ratio must be $ \boxed{\dfrac{54}{50} \equal{} \dfrac{27}{25}}$

[/hide]

Edited for accuracy.

It's a plausible method, but check your final answer.

[hide]I looked at it by imagining what would happen if the tickets were all adults.  There would have been 78 tickets sold.  Obviously, 78/0 is not the ratio we are seeking.  But, we would maintain an integer solution (as we can't have a number of people with an integer part) if we took away 12 adults, and added in 25 children.  This would not change the overall amount of money obtained.

78- 0
66-25
54-50 (probable)
42-75 (okay, the one above is now definite.)
...

54 adults, 50 children.  54/50=27/25.[/hide]


I don't remember where I saw this question, but it sounds rather natural: is there a $\sigma$-algebra on $P(\mathbb{R})$ which is strictly bigger than Lebesgue's and on which can be constructed a measure which coincides with the Lebesgue measure on the Lebesgue-measurable sets ?

The answer is 'yes', and in fact more is true: given [i]any[/i] one set $ A\subset\mathbb R$, Lebesgue measure can be extended to the $ \sigma$-algebra generated by the measurable real sets and $ A$.

After taking a suitable measurable subset out of $ A$, we can assume that $ A$ has inner measure zero (that is, it contains no measurable sets with positive measure). Let $ \mathcal M$ be the $ \sigma$-algebra of measurable sets, and $ \mathcal I$ the $ \sigma$-ideal of all sets with zero measure. Finally, let $ \mathcal J\subset\mathcal S$ be the $ \sigma$-ideal generated by $ \mathcal I$ and $ A$.

Because the intersection between $ \mathcal J$ and $ \mathcal M$ is precisely $ \mathcal I$ (this follows from the fact that $ A$ has zero inner measure), the inclusion $ \mathcal M\subset\mathcal S$ induces an isomorphism of boolean $ \sigma$-algebras $ \mathcal M/\mathcal I\to\mathcal S/\mathcal J$. The Lebesgue measure $ \mu$, which can be regarded as a measure on $ \mathcal M/\mathcal I$, can thus be transported onto $ \mathcal S/\mathcal J$. This measure now induces an extension of $ \mu$ to $ \mathcal S$ which vanishes on $ \mathcal J$, and we're done.


geometry
trapezoid
geometric transformation
homothety
perpendicular bisector
geometry proposed

Consider the triangle $ ABC$ with $ \angle A\equal{}60^{\circ}$. Inside the triangle take a point $ O$ so that $ \angle{AOB}\equal{}\angle{BOC}\equal{}120^{\circ}$. Pick a point $ D$ on $ [CO)$ so that $ \triangle AOD$ is equilateral. The perpendicular bisector of $ AO$ intersects $ BC$ at $ Q$. Prove that $ OQ$ bissects $ BD$.

[color=darkred][i][b]Lema.[/b][/i] Let $ ABCD$ be an trapezoid with $ AB\parallel CD$. Let $ \left\|\begin{array}{cc}O \in AC \cap BD \\
P \in AD \cap BC \end{array} \right\|$. If $ Q \in OP \cap AB$, then $ QA \equal{} QB$.

[i][b] Proof.[/b] [/i]By Thales $ \dfrac{PD}{AD} \equal{} \dfrac{PC}{BC}$. By Ceva: $ \dfrac{PD}{AD}\cdot\dfrac{AQ}{QB}\cdot\dfrac{BC}{PC} \equal{} \dfrac{QA}{QB} \equal{} 1$[/color]

[color=darkblue]Observe that $ \angle{OAC} \plus{} \angle{OCA} \equal{} 180^\circ \minus{} \angle{AOC} \equal{} 60^\circ \equal{} \angle{OAC} \plus{} \angle{OAB}$. So $ \angle{OAB} \equal{} \angle{OCA}$. By analogy $ \angle{OAC} \equal{} \angle{OBA}$. So $ \triangle{OAC}\sim \triangle{OAB}$ and $ \dfrac{AO}{CO} \equal{} \dfrac{BO}{AO}$. [/color]  [color=indigo] [i](*1)[/i][/color]

[color=darkblue]It is clear that perpedicular bisector of $ AO$ passes through $ D$. Denote $ E \in BO \cap DQ$. Then triangle $ AOE$ is equilateral and $ OE \equal{} OA \equal{} OD$. So we can rewrite[/color] [color=indigo][i](*1)[/i][/color] [color=darkblue]$ \dfrac{EO}{CO} \equal{} \dfrac{BO}{DO}$. Also $ \angle{BOD} \equal{} \angle{COE}$ , so $ \triangle{BOD} \sim \triangle{COE}$ and $ \angle{BOD} \equal{} \angle{CEO} \Rightarrow BD \parallel EC$. From [i][b]Lema[/b][/i]  $ OQ$ bisects $ BD$,q.e.d.[/color]

[quote="freemind"]Consider the triangle $ ABC$ with $ \angle A \equal{} 60^{\circ}$. Inside the triangle take a point $ O$ so that $ \angle{AOB} \equal{} \angle{BOC} \equal{} 120^{\circ}$. Pick a point $ D$ on $ [CO)$ so that $ \triangle AOD$ is equilateral. The perpendicular bisector of $ AO$ intersects $ BC$ at $ Q$. Prove that $ OQ$ bissects $ BD$.[/quote]
Solution:
Let $ \triangle ABC'$,$ \triangle ACB'$ be externally constructed equilateral triangles.It is clear that $ C',A,B'$ are collinear,name the line through them as $ l$.
Suppose $ l$ intersect $ BC$ at $ Q'$.Since $ \widehat{BAC} \equal{} 60^{\circ}$,it follows that $ Q'$ is homothety center that turns $ \triangle AB'C$ to $ \triangle ABC'$.

Also if $ O_B$,$ O_C$ are centres of these triangles,then $ O_1,O_2,Q'$ are collinear,but as we know $ AOBC'$ and $ AOCB'$ are cyclic quadrilaterals,from where we conclude that $ O_1O_2$ is perpendicular bisector of $ AO$,therefore,$ Q \equal{} Q'$.
Suppose $ QO$ intersects $ AD$ at point $ Y$.As we know $ \widehat{QOA} \equal{} \widehat{QAO}$ and $ \widehat{QOA} \equal{} 60^{\circ} \plus{} \widehat{B'OQ} \equal{} 60^{\circ} \plus{} \widehat{OAC} \equal{} \widehat{QAO}$,it follows that $ \boxed{\widehat{BOY} \equal{} \widehat{B'OQ} \equal{} \widehat{OAC}}$.

On the other hand,$ 60^{\circ} \equal{} \widehat{DAO} \equal{} \widehat{BAC}$,thus $ \widehat{YAB} \equal{} \widehat{DAB} \equal{} \widehat{OAC}$.Combining it with boxed equality,we conclude that $ \boxed{\widehat{YAB} \equal{} \widehat{YOB}}$.But as we know $ BO\|AY$,it follows that $ AYBO$ is  trapezium with $ \boxed{AO \equal{} BY}$.Also we know that $ \triangle ADO$ is equilateral,hence $ BY \equal{} AO \equal{} DO$ and $ DO\|BY$,finally $ DOBY$ is parallelogram.$ \blacksquare$


Steve has an isosceles triangle with base 8 cm and height 10 cm . He cuts it into eight equal area pieces.
What is the perimeter of the largest piece?

[hide="solution"]Since the area and height of all the triangles are equal, we know that the base of each smaller triangle is $1$.  We only need to find the triangle with the greatest sum for the lenghts of the other two sides.  Using Pythagorean Theorem, we can find all of the sides for the triangles. The outermost triangles are congruent and thus have the same perimeter of $1+\sqrt(116)+\sqrt(109)$, which is the greatest.[/hide]


let ACBD be a cyclic quad

let p=AB, q=BC, r=CD and s=AD

PROVE: $AC/BD$=$(pq+qr)/(pq+rs)$

I don't believe this equation is correct.  Substitute Ptolemy's on the denominator of the RHS, and it seems that you can vary r while the other stuff stays constant.


When I try to upload an avatar from my computer, it tells me I haven't selected any avatar. I tried this with multiple files, but to no avail... Any suggestions?

It'll say that if you have a non-supported file type. Check the FAQ for file types. (strangely, the FAQ is nothing but a list of allowed file types)

Go find a picture on google or somewhere, and save it to your desktop or anywhere on the computer. then go to   your profile and click avatar. there'll be two lines there. One has a box that says browse beside it, and the other has URL something something beside it. Click the browse button and and a new box will pop up with all the files folders and systems on your computer. Go to the place you saved your picture and say OK, and click submit on the bottom and you will have a new avatar.


Find all pairs $ (a,b,c)$ of positive integers such that

$ \frac{ab\minus{}1}{c},\frac{bc\minus{}1}{a}$ and $ \frac{ca\minus{}1}{b}$

are all integers.

I think I have a good start:

$ \begin{cases}\frac{ab\minus{}1}{c}\equal{}k\\
\frac{bc\minus{}1}{a}\equal{}l\\
\frac{ca\minus{}1}{b}\equal{}m\end{cases}$

Multiplying all equations we get $ \frac{a^2b^2c^2\minus{}a^2bc\minus{}ab^2c\minus{}abc^2\plus{}ab\plus{}bc\plus{}ca\minus{}1}{abc}\equal{}klm$
$ \frac{ab\plus{}bc\plus{}ca\minus{}1}{abc}\equal{}klm\plus{}a\plus{}b\plus{}c\minus{}abc$

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\minus{}\frac{1}{abc}$ is an integer.

can someone complete the solution?  :wink:

This problem is similar to BMO 2003 #5, although the problem asks the value of $ abc$ for primes $ a,\ b,\ c$.

[quote="kunny"]The problem is posed in Junior High school mathematics Olympiad at Novosibirsk in Russia 1963.

In Japan, two problems have been posed in the past in university entrance exam.

Here are:

[color=blue] Waseda University entrance examination 1990

Let $ a,b,c\ (a < b < c)$ are distinct positive integers such that the product of any two of $ a,b,c$ has the residue of 1 when divided by the third number.

(ⅰ)　Find the quotient when $ a \plus{} b$ is divided by $ c$?

(ⅱ)　Find the quotient when $ a \plus{} c$ is divided by $ b$?

Find the values of $ a,\ b,\ c.$[/color]

[color=green] Tokyo Institute of Technology entrance exam 1978

Let $ a,\ b,\ c\ (1 < a < b < c)$ be integers such that  $ (ab \minus{} 1)(bc \minus{} 1)(ca \minus{} 1)$ is divisible by $ abc.$

(1) Show that $ ab \plus{} bc \plus{} ca \minus{} 1$ is divisible by $ abc.$

(2) Find $ a,\ b,\ c.$ [/color][/quote]

See here. [url]http://www.mathlinks.ro/viewtopic.php?search_id=404564430&t=171830[/url]

very nice problem!
for the first part it's the same of Bachukas's solution, then I complet it,
we have $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {1}{abc} \equal{} k$ where $ k$ is an integer
suppose that one of $ a,b,c$ is equal to $ 1$ for example $ c$ then,
$ \frac {ac \minus{} 1}{b}$ is integer then $ \frac {a \minus{} 1}{b}$ is integer wich implies $ b\leq a \minus{} 1$ (*)
$ \frac {bc \minus{} 1}{a}$ is integer then $ \frac {b \minus{} 1}{a}$ is integer wich implies $ a\leq b \minus{} 1$ and from $ a\leq b \minus{} 1 \leq a \minus{} 2$ wich is false. therefore $ a,b,c \geq 2$ thus
$ k \equal{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {1}{abc}\leq \frac {3}{2}$ therefore $ k \equal{} 1$ or $ k \equal{} 0$
if $ k \equal{} 0$ then $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} \frac {1}{abc}$
$ \Leftrightarrow ab \plus{} bc \plus{} ca \equal{} 1$ wich is false, thus, $ k \equal{} 1$
$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \minus{} \frac {1}{abc} \equal{} 1$
$ \Leftrightarrow \frac {1}{bc}\left(\frac {bc \minus{} 1}{a} \plus{} b \plus{} c\right) \equal{} 1$
and we have $ a\geq 2$ therefore,
$ 1 \equal{} \frac {1}{bc}\left(\frac {bc \minus{} 1}{a} \plus{} b \plus{} c\right)\leq \frac {1}{bc}\left(\frac {bc \minus{} 1}{2} \plus{} b \plus{} c\right)$
$ \Leftrightarrow bc \leq bc \plus{} 1 \leq 2(b \plus{} c) \Leftrightarrow \frac {1}{2} \leq \frac {1}{b} \plus{} \frac {1}{c}$
suppose that $ b\geq c$ therefore, $ \frac {1}{2} \leq \frac {1}{b} \plus{} \frac {1}{c}\leq \frac {2}{c} \Leftrightarrow 2\leq c\leq 4$
if $ c \equal{} 4$ then, $ \frac {1}{a} \plus{} \frac {1}{b} \minus{} \frac {1}{4ab} \equal{} \frac {3}{4}$
$ \Leftrightarrow (3b \minus{} 4)(3a \minus{} 4) \equal{} 13$ wich don't have any integer solution.
the $ c \equal{} 2$ or $ c \equal{} 3$
if $ c \equal{} 2$
$ \frac {1}{a} \plus{} \frac {1}{b} \minus{} \frac {1}{2ab} \equal{} \frac {1}{2}$
$ \Leftrightarrow (b \minus{} 2)(a \minus{} 2) \equal{} 3$
then $ S \equal{} \{2,3,5\}$
if $ c \equal{} 3$
$ \frac {1}{a} \plus{} \frac {1}{b} \minus{} \frac {1}{3ab} \equal{} \frac {2}{3}$
$ \Leftrightarrow (2b \minus{} 3)(2a \minus{} 3) \equal{} 7$
then $ S \equal{} \{2,3,5\}$
finally the only solution for the problem is ,$ S \equal{} \{2,3,5\}$ with all permutation possible.


In a three-digit number, the hundreds digit is greater than 5, the
tens digit is greater than 4 but less than 8, and the units digit is
the smallest prime number. How many three-digit numbers
satisfy all of these conditions?

i think the definition of "smallest prime number" is unclear here...

The smallest prime number is 2. Although it doesn't matter as you only have to realize that it's 1 possible digit. $ (4)(3)(1) \equal{} 12$

well it could be the smallest prime number [i][b]in the 3-digit number[/b][/i]
and yes it does matter

There's no sol yet, so here it is: [hide=sol]There are $4$ numbers for hundreds($6, 7, 8, 9$), $3$ for tens ($5, 6, 7$), and $1$ for the ones ($2$). $4\cdot3\cdot1=\boxed{12}$[/hide]


In a country there are 80 cities and there are some conjugate flights between some cities. It is possible to travel from any city to any city. There are at least seven flights from each city. Find the least $ k$ such that in we can go from any city to any city with at most $ k$ flights.

Sorry for my bad english.

I guess $ k \equal{} 27$.

Let's take 10 complete graphs on 8 vertices each and arrange them in a row: $ G_1, G_2, \dots, G_{10}$.
Select two distinct vertices $ u_i, v_i$ in $ G_i$ for $ i \equal{} 1,2,\dots,10$.

Remove edge $ (u_i,v_i)$ from each graph $ G_i$ for $ i \equal{} 2,3,\dots,9.$

Now connect $ v_i$ with $ u_{i \plus{} 1}$ with an edge for $ i \equal{} 1,2,\dots,9.$

It is clear that any shortest path between vertices $ u_1$ and $ v_{10}$ contains $ 27$ edges.


calculus
integration
real analysis
real analysis unsolved

Oh yes :D  My fault.. a>0.

Another one:
Prove that for any $a>0$ we have:
\[ \sum_{n=0}^\infty \frac{(-a)^n}{(a+1)(a+2)...(a+n)}=\int_0^1e^{x^a}dx  \]

For a = 1 it seems it's not working, so i smell a rat :).
My computations give a different right hand member.
However tell me pls where i am wrong :).

$\frac{n!}{(a(a + 1)...(a + n))} = \sum_{k = 0}^n \frac{a_k}{a + k}$ and from here $a_k = (-1)^k * \left(\begin{array}{c} n&k \end{array}\right)$

From $(-x^{1/a} + 1)^n = \sum_{k = 0}^n a_k * x^{k/a}$, integrating one can get:

$\int_0^x (-t^{1/a} + 1)^n dt = \sum_{k = 0}^n \frac{x^{k/a} + 1}{(k/a) + 1}$,

for  x = 1, $\int_0^1 (-t^{1/a} + 1)^n dt = \frac{n!}{(a + 1)...(a + n)}$,

so $\sum_{n \geq 0} \frac{a^n}{(a + 1)...(a + n)} = \int_0^1 [\sum_{n \geq 0} (-t^{1/a} + 1)^n * a^n /n!] dt = \int_0^1 e^{a(1 - t^{1/a})}dt$

[the left series is convergent, easy to prove].


Definition of a Hypersonic Transport (HST): a vehicle going at 5-8 times the speed of sound.

Do you believe that hypersonic travel will make life easier?

A list of pros and cons:

PROS:
-faster travel
-increased trade

CONS:
-noise pollution
-air traffic (30 min. flight, 30 min. waiting to land)
-increased fuel costs
-expensive to build

What do you think?

I'm hoping it won't be landing back-ups.  I'm thinking this is well worth the disadvantages.

Yeah, that's a con. If a window breaks or something , certain death.

[quote="frankthetank"]Yeah, but if a window breaks on a regular plane death will also be pretty certain, especially for the people sitting next to it.

In the 70s, people thought supersonic transports were a good idea, but then there was that oil crisis.  Unfortunately, there will probably be another [url=http://en.wikipedia.org/wiki/Hubbert_peak]oil crisis[/url] in the near future, so I don't think hypersonic transports are going to fly (har har).[/quote]

If a window breaks, why would death be certain? I'm thinking there would be strong "wind", but that wouldn't necessarily kill people. Also, I doubt one window would cause the airplane to lose altitude/speed significantly enough to crash. Then again, I may be wrong.


Let $ x, y$ and $ z$ be positive real numbers.

(i) If $ x\plus{}y\plus{}z \ge 3$, is it necessarily true that $ \frac{1}{x} \plus{}\frac{1}{y} \plus{} \frac{1}{z} \le 3$?

(ii) If $ x\plus{}y\plus{}z \le 3$, is it necessarily true that $ \frac{1}{x} \plus{} \frac{1}{y}\plus{}\frac{1}{z} \ge 3$?

$ a\plus{}b\plus{}c \equal{} m$

$ \sqrt[3]{abc} \leq \frac{m}{3}$

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c} \geq \frac{3}{\sqrt[3]{abc} } \geq \frac{9}{m}$
 :lol:


[quote="ashwinkrishnan"]Let $\Delta$ = $abc+2fgh-a{f^{2}}-b{g^{2}}-c{f^{2}}$.Then the parametric conditions that ensure that the equation
       $a{x^{2}}+b{y^{2}}+2hxy+2fy+2gx+c = 0$
does not represent any curve in the cartesian plane are-
  (A)$\Delta$ = 0,${h^{2}}= ab$
  (B)${h^{2}}< ab$, $a{\Delta}> 0$ AND $b{\Delta}> 0$
  (C)${h^{2}}< ab$ ,$a{\Delta}> 0$ OR  $b{\Delta}> 0$
  (D)none of the above
(likely to be [b](B) [/b]or [b](C)[/b] but i am not clear as to which one)[/quote]

[b]Ans: (C)[/b]

[hide="Solution:"]

As $a$ or $b$ can be $0$ and both can't be $0$ simultaneously (then it will represent a real curve),we will take following cases:

[b]Case 1:[/b]

Consider $a\ne 0$

Rewrite the equation as quadratic in $x$

$ax^2+2(hy+g)x+(by^2+2fy+c)=0$

As per condition $\forall\ y\in\mathbb{R}\ ,$ we need Discriminant of above quadratic less than $0$

$\implies (hy+g)^2-a(by^2+2fy+c)<0\ \forall\ y\in\mathbb{R}$

$\implies (h^2-ab)y^2+2(hg-af)y+(g^2-ac)<0\ \forall\ y\in\mathbb{R}$

$\implies (h^2-ab)<0$ and $(hg-af)^2-(h^2-ab)(g^2-ac)<0$

$\implies (h^2-ab)<0$ and $a(af^2+bg^2+ch^2-2fgh-abc)<0$

$\implies (h^2-ab)<0$ and $a(abc+2fgh-af^2-bg^2-ch^2)>0$

$\implies (h^2-ab)<0$  and $a\Delta>0$

[b]Case 2:[/b]

Consider $b\ne 0$

Rewrite the equation as quadratic in $y$ and follow the process of Case 1

we will get $(h^2-ab)<0$  and $b\Delta>0$

Now combining the two cases we get ${h^{2}}< ab$ ,$a{\Delta}> 0$ OR  $b{\Delta}> 0$

[/hide]


Hi!

If a,b,c>0 and s=(a+b+c)/2, how do you show:

[tex]
$
\begin{gathered}
  \sqrt {\frac{{2b^2  + 2c^2  - a^2 }}
{{bc}}}  + \sqrt {\frac{{2a^2  + 2c^2  - b^2 }}
{{ca}}}  + \sqrt {\frac{{2a^2  + 2b^2  - c^2 }}
{{ab}}}  \geqslant  \hfill \\
   \geqslant 6 \cdot \sqrt[3]{{\frac{{s \cdot \sqrt {s(s - a)(s - b)(s - c)} }}
{{abc}}}} \hfill \\
\end{gathered}
[/tex]

Alot of the expressions can be simplified to shorter forms.

Well I'm sure the LHS can be related to cosines, and the RHS can be related to areas.  Maybe you should double check it, since there are many terms that would be almost nice.  I just wish to make sure we have the right problem.

[quote="probability1.01"]I just wish to make sure we have the right problem.[/quote]

The problem is stated correctly  :)

Sorry, it turns out I can't read :blush: .

[quote="fuzzylogic"]Hint:

[hide]
$2b^2  + 2c^2  - a^2 \ge (b+c)^2 - a^2 = 4s(s-a)$
[/hide][/quote]

Thanks, that solves it

[quote="nolachrymose"][quote="Fedor Petrov"]Are not $a,\,b,\,c$ the sides of a triangle?[/quote]

Not necesarily. We are not given the relationship between the sums of the three variables, just that each is >0.[/quote]

No, he is correct, the condition that $a,b,c$ are sides of a triangle is implied, because if one of $s-a$, $s-b$, $s-c$ is $<0$, bad things happen.


How many solutions in positive integers does the equation $ 3x\plus{}5y\equal{}1008$ have?

[hide="Solution"]
Note that if $ (x_1,y_1)$ is a solution then so is $ (x_1\plus{}5t,y_1\minus{}3t)$.

We can easily find the solution $ (1,201)$ and from there we see that for any $ t$, $ (1\plus{}5t,201\minus{}3t)$ is a solution. Since we want $ x$ and $ y$ to be positive we need to find the biggest $ t$ such that $ 201\minus{}3t$ is bigger than $ 0$, in other words:
\[ 201\minus{}3t>0\implies 201>3t\implies 67>t\]
Where it's easy to see that the biggest value is $ 66$, making $ 66$ solutions, but we also need to take in count when $ t\equal{}0$, so there are a total of $ \boxed{67}$ solutions.
[/hide]

A super simple solution

3x+5y=1008

3x and 1008 are both divisible by 3.

so 5y must be divisible by 3 too, making y a multiple of 3.

the biggest possible value for 5y is 1005, so y = 201.

we see that y can be 3,6,9,12,...,201.

Divide 201 by 3, bingo.

[quote="tonypr"]We can easily find the solution $ (1,201)$ and from there we see that for any $ t$, $ (1 \plus{} 5t,201 \minus{} 3t)$ is a solution. Since we want $ x$ and $ y$ to be positive we need to find the biggest $ t$ such that $ 201 \minus{} 3t$ is bigger than $ 0$, in other words:[/quote]
And $ t$ can't be negative because then $ 1\plus{}5t$ wouldn't be an integer or it wouldn't be positive.

[quote="AndrewTom"]How many solutions in positive integers does the equation $ 3x \plus{} 5y \equal{} 1008$ have?[/quote]
hi all
i will post asolution soon
mrmohamed

[quote="AndrewTom"]How many solutions in positive integers does the equation $ 3x \plus{} 5y \equal{} 1008$ have?[/quote]
hi all :pilot:
[url=http://www.servimg.com/image_preview.php?i=426&u=12899977][img]http://i99.servimg.com/u/f99/12/89/99/77/almonk10.jpg[/img][/url]
greeting


group theory
abstract algebra
superior algebra
superior algebra solved

I went to the City Round of the NMO today (I study in Constantza) and here's one of the problems (it's kind of easy, but I liked it):

Let G be a group s.t. |G|=2n with n odd. Let H be a subgroup of G s.t. |H|=n and axa-1=x-1 for all x \in H and all a \in G\H. Show that:

a)H is commutative;
b)ord<x>=2 for all x \in G\H.

a) a in G\H, y in H so ay in G\H, so
ayxy'a'=x'=axa', so yxy'=x for all x,y in H, so yx=xy for all x,y in H

b)Let a, not in H. then G=H \cup aH, so G=aG=aH \cup aaH, so
aaH=H, so a^2 is in H for all a not in H.

So aa^2a'=a^(-2), so a^4=e. But |<a>||2n and |<a>||4 so |<a>|=2 because n is odd. a<>e becasue e \in H. So a^2=e for all a in G\H.


limit calculus derivative algebra polynomial AMC AIME

Evaluate:
\[ \sum_{n \equal{} 1}^{\infty} \frac {4n}{3^{(n \plus{} 1)}}
\]
Meh, it's not geometric or arithmetic, so how would you solve it?

$ \sum_{n \equal{} 1}^{\infty} \frac {4n}{3^{n \plus{} 1}} \equal{} \frac {1}{3}\lim_{n\to\infty} \sum_{k \equal{} 1}^n \frac {4k}{3^k} \equal{} \frac {1}{3}\lim_{n\to\infty}\sum_{k \equal{} 1}^n\left\{\frac {2(k \minus{} 1) \plus{} 3}{3^{k \minus{} 1}} \minus{} \frac {2k \plus{} 3}{3^k}\right\}$

$ \equal{} \frac {1}{3}\lim_{n\to\infty} \left(\frac {2\cdot 0 \plus{} 3}{3^0} \minus{} \frac {2n \plus{} 3}{3^n}\right) \equal{} 1$.

[hide="Method 2"] Let the summation be $ S$ and consider $ S \minus{} \frac {S}{3}$.  [/hide]
So much for the elementary methods.  There are two very powerful methods to bring to bear on problems like this.

[hide="Method 3 (very general)"] Write as $ \sum_{n \equal{} 1}^{\infty} \sum_{m \equal{} 1}^{n} \frac {4}{3^{n \plus{} 1}}$ and exchange the order of summation.  Equivalently, use [url=http://en.wikipedia.org/wiki/Summation_by_parts]summation by parts[/url].  Both techniques are very general but a little stronger than what you need. [/hide]
[hide="Method 4 (also very general)"] Let $ F(x) \equal{} \sum_{k \equal{} 1}^{\infty} kx^{k \minus{} 1}$.  It turns out that $ F(x) \equal{} \frac {1}{(1 \minus{} x)^2}$, either by differentiation or by General Binomial Theorem. [/hide]

$ \sum_{n \equal{} 1}^{\infty} \frac {4n}{3^{(n \plus{} 1)}} \equal{} \frac{4}{3} \sum_{n \equal{} 1}^{\infty} \frac {n}{3^{n}}$,

which is now in the form of a fairly common series type. Here's how to solve the general form of that series type.

[hide="General solution"]Let $ S\equal{}\sum_{n \equal{} 1}^{\infty} \frac {n}{b^{n}}$

We can add the term $ 0 \equal{} \frac{0}{b^0}$ to it without changing the sum, so this gives:

$ S\equal{}\sum_{n \equal{} 0}^{\infty} \frac {n}{b^{n}}$

Now note that if we raise all the values of $ n$ by $ 1$ when we define the bounds, and then subtract it back out in each summand, we don't change the sum. For example, $ \sum_{n\equal{}2}^{11} (n\minus{}1) \equal{} 1\plus{}2\plus{} \ldots \plus{}10 \equal{} \sum_{n\equal{}1}^{10}$. Doing this gives:

$ S\equal{}\sum_{n \equal{} 1}^{\infty} \frac {n\minus{}1}{b^{n\minus{}1}} \equal{} \sum_{n \equal{} 1}^{\infty} \frac {n\minus{}1}{b^{n\minus{}1}} \cdot \frac{b}{b} \equal{} b \sum_{n \equal{} 1}^{\infty} \frac {n\minus{}1}{b^{n\minus{}1}}$

Now, we split that last summation to get:

$ S \equal{} b\sum_{n \equal{} 1}^{\infty} \frac {n}{b^{n}} \minus{} b \sum_{n \equal{} 1}^{\infty} \frac {1}{b^{n}}$

But, the first summation is what we started with! So now we have:

$ S \equal{} bS \minus{} b\sum_{n \equal{} 1}^{\infty} \frac {1}{b^{n}} \implies (b\minus{}1)S \equal{} b\sum_{n \equal{} 1}^{\infty} \frac {1}{b^{n}} \implies S \equal{} \frac{b}{b\minus{}1} \sum_{n \equal{} 1}^{\infty} \frac {1}{b^{n}}$

Now, we have $ S$ in terms of an infinite geometric series. We can solve this to get:

$ \boxed{S \equal{} \frac{b}{b\minus{}1} \cdot \frac{1/b}{1\minus{}1/b} \equal{} \frac{b}{(b\minus{}1)^2}}$[/hide]

[hide="Specific solution"]In this particular case, your series summation is equal to $ \frac{4}{3} \cdot \frac{3}{(3\minus{}1)^2} \equal{} \boxed{1}$[/hide]

The logic used in the general solution can be used recursively to evaluate the summation of any polynomial divided by an exponential. Each time the index shifting is done, the degree of the polynomial in the numerator is reduced by 1.

This is from AoPS Vol. 2. I'd suggest you not post such problems.

It is?
It's just a problem that was on Alan's mock MC test.

Alan tells me that this isn't the exact question from AoPS vol 2.

Incidentally, what's wrong with posting problems from AoPS?

The same thing is true of AIME problems . . .

[quote="Temperal"]Most people would know and recognize them already.[/quote]
What about those who don't have these books?

I'd be willing to bet that those are in the minority by far. Please, no more petty "what ifs".

to tell the truth, I just went like, hmm, let's make the numerator $ 4n$ and the denominator $ 3^{n\plus{}1}$... besides, I haven't read that far in Volume 2 anyways

Temperal, your claims are pretty ridiculous. This is a well-known idea, and a completely valid topic to ask about. Saying that it "comes from AoPS 2" is pretty much nonsense, because it was by no means the original idea of the authors of the book. People have been posting problems from the AoPS books since they have been in circulation.

No more, and probably quite a bit less, than your utterly useless post. Of course such series which you have to expres int tersm of an AM or GM (or alternately write a double summation) but that exact form is hardly common. It, of course, wasn't originally by the authors - obviously you haven't even bothered to look up the problem in the book, since they actually source it to another competition - but since an extremely similar problem exists in the book, the chances are that such discussion is redundant. Your last sentence is particularly ridiculous - since when is "people have been doing it for a certain length of time" at all related to whether they should continue to do it?

Redundant, eh? So what should ZhangPeijin have done? I'll shut up if you find me an easily accessible resource he could have easily found the answer to his question in, without knowing terms such as "arithmetico-geometric series" (hmmm that may not be the correct term).

An accusation that "Alan stole from AoPS 2" is absurd, I have used this type of a problem on AoPS 2, because I had thought about this on my own in 7th (?) grade, and come up with a solution myself. Why would you have the assumption that such a common idea [i]had[/i] to have come from AoPS 2, which, we have just seen, is clearly not true?

Also, I still don't have the slightest idea why you would suggest that problems from a book that maybe many of us have should not be posted here.

Guys, turn off this fighting before I have to lock this topic.

I'll state the compromise: You guys are not both wrong but you guys are not both 100% right. Generally speaking, posting problems without permission from authors are prohibited.  However, with AoPS Volume 1 and 2, I don't see any problem posting the problems because you don't understand the solutions or simply want to challenge others. The problem arises when major or most or ALL parts of the books are put up on the book.

As with similar problems, this issue occurs everywhere and it is not a big deal. Math competitions, as far as I know, do not necessarily copyright individual problems (some might but not every one). Also, it is perfectly fine to take ideas from other competitions and make a similar one (or even changing numbers). They are all perfectly fine.

Of course, only exception might be if you try to make money out of them but that's not the case here.

[b]t0rajir0u[/b], can you elaborate more on the method 3? Here's what I got:
$ \sum_{n \equal{} 1}^{\infty}\sum_{m \equal{} 1}^{n}\dfrac{4}{3^{n \plus{} 1}} \equal{} \sum_{k \equal{} 1}^{\infty}\sum_{n \equal{} k}^{\infty}\dfrac{4}{3^{n \plus{} 1}}$
It seems to work for this problem. However, I only came up with this "exchange" by some intuitive guess, not knowing the general rule or motivation behind it. Also, if the numerator is some polynomial in $ n$, how would you split it?

Besides, for method 4, I don't see how you get the simplification. To me, it's rather that we can use General Binomial Theorem to multiply out the closed form into the given summation in the question, not "factoring" it back...

The general advice when switching the order of a sum/integral over a non-rectangular region: Draw the domain.
In this case, the domain is a wedge $ m\le n$; switching the order makes that $ n\ge m$.

Higher degree expressions require either more new variables, or a reduction-style argument that goes down one degree at a time.

If you don't already know the answer, Method 4 works best either by proving using Method 1 or by integrating.  General Binomial Theorem is more a tool to help us work out the generalization, which I give [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=184510]here[/url].


ceiling function
floor function
arithmetic sequence
number theory unsolved
number theory

A quadruple $ (a,b,c,d)$ of positive integers such that
$ 1) a \leq b \leq b \leq d$
$ 2)a,b,c,d$ are consecutive terms in an arithmetic sequence
$ 3)a^3 \plus{} b^3 \plus{} c^3 \equal{} d^3$, and
$ 4)$ at lease one of $ a,b,c,d$ between $ 1867$ and $ 2008$ inclusive
is called a $ \text{quixotic quadruple}$. How many quixotic quadruples are there?

This is the last question on a small contest I just got home from. I couldn't figure out how to do it (I had run out of time while trying some stuff, but I still am not sure how). But I'll pose it as a fun problem anyways.

[hide]
Let $ a = x - 3y$ and let the common difference be $ 2y$. Then,
\begin{align*} & (x - 3y)^3 + (x - y)^3 + (x + y)^3 = (x + 3y)^3 \\
& \Longrightarrow (x - y)^3 + (x + y)^3 = (x + 3y)^3 - (x - 3y)^3 \\
& \Longrightarrow 2x(x^2 + 3y^2) = 18y(x^2 + 3y^2) \Longrightarrow x = 9y \end{align*}
Thus $ a: b: c: d = 6y : 8y : 10y: 12y = 3: 4: 5: 6$; hence $ d \ge 6 \left \lceil \frac {1867}6 \right \rceil = 1872 \Longrightarrow a = 936$ or $ a \le 3 \left \lfloor \frac {2008}3 \right \rfloor = 2007$, so the answer is the number of multiples of $ 3$ from $ 936$ to $ 2007$, or $ 358$.
[/hide]

Mhmm its a multiple choice question and $ 179$ is not any of the answers given. >_<

I realized I made a mistake as soon as I went upstairs; edited.  :oops:

Edit: Before I post another wrong answer, could you post the answer choices?


algebra
function
domain
integration
real analysis
real analysis unsolved

If $f$ is subharmonic in the domain G and attains its supremum in G, then if is constant in G

the basic property of subharmonic functions is that they satisfy $f(z)\leq \frac{1}{2\pi}\int f(z+e^{i\theta})$,so what claim is immediate.


Congratulations to Roman Kogan and Dimitriy Karabash, the two New York City Math Team veterans who were Honorable Mention on this year's Putnam exam.  (It is possible there are others I missed -- if so, please add their names.)

yes.. you missed the crazy character Joel Lewis :)

Congratulations :)

Congratulations!!
Now I kinda regret that I did not take it this year.. :P


limit
algebra
functional equation
real analysis
real analysis unsolved

Let's define $ \parallel{}x\parallel{} \equal{} \min_{m \in \mathbb{Z}} |x\minus{}m|$. Show that $ \sum_{n\equal{}0}^{\infty} \dfrac{\parallel{}2^{n}x\parallel{}}{4^{n}} \equal{} \parallel{}x\parallel{}(1\minus{}2\parallel{}x\parallel{})$.

First a small correction: the series has to start from $ n\equal{}1$, otherwise $ x\equal{}1/2$ is an easy counterexample.
Denote $ \displaystyle f(x) \equal{} \sum_{n\equal{}1}^{\infty} \dfrac{\parallel{}2^{n}x\parallel{}}{4^{n}}$ and $ g(x) \equal{} \parallel{}x\parallel{}(1\minus{}2\parallel{}x\parallel{})$.

Now $ \displaystyle f \left (\frac {x} {2} \right ) \equal{} \sum_{n\equal{}1}^{\infty} \dfrac{\parallel{}2^{n\minus{}1}x\parallel{}}{4^{n}} \equal{}$ $ \dfrac {\parallel{}x\parallel{}} {4} \plus{} \frac {1} {4} \sum_{n\equal{}2}^{\infty} \dfrac{\parallel{}2^{n\minus{}1}x\parallel{}}{4^{n\minus{}1}} \equal{}$ $ \dfrac {\parallel{}x\parallel{}} {4} \plus{} \frac {1} {4} \sum_{n\equal{}1}^{\infty} \dfrac{\parallel{}2^{n}x\parallel{}}{4^{n}} \equal{}$ $ \dfrac {\parallel{}x\parallel{}} {4} \plus{} \frac {1} {4} f(x)$, therefore $ 4f \left (\frac {x} {2} \right ) \equal{} \parallel{}x\parallel{} \plus{} f(x)$. We hope the same relation holds for $ g(x)$, i.e. $ 4\parallel{}\frac {x} {2} \parallel{}(1 \minus{} 2\parallel{}\frac {x} {2} \parallel{}) \equal{} \parallel{}x\parallel{} \plus{} \parallel{}x\parallel{}(1 \minus{} 2\parallel{}x\parallel{}) \equal{} 4\frac {\parallel{}x\parallel{}} {2}(1 \minus{} 2\frac {\parallel{}x\parallel{}} {2})$. Indeed, separately checking the cases $ 0 \leq \{x\} < \frac {1} {2}$ and $ \frac {1} {2} \leq \{x\} < 1$ does it (in fact, since $ \parallel{}x\parallel{}$ is the distance from $ x$ to the closest integer, we may assume $ 0 \leq x < 1$).

Define now $ \delta(x) \equal{} f(x) \minus{} g(x)$. Then $ \delta(x)$ satisfies the functional equation $ \delta(x) \equal{} 4\delta(\frac {x} {2})$, so (substituting $ 2x$ for $ x$) $ \delta(x) \equal{} \frac {1} {4} \delta(2x)$, and by iteration $ \delta(x) \equal{} \frac {1} {4^n} \delta(2^nx)$ for all positive integers $ n$.

Now, $ 0 \leq \parallel{}x\parallel{} \leq 1/2$ for all reals $ x$, therefore $ 0 \leq g(x)$ is bounded by $ \frac {1} {8}$ , and also bounded is $ \displaystyle 0 \leq f(x) \equal{} \sum_{n\equal{}1}^{\infty} \dfrac{\parallel{}2^{n}x\parallel{}}{4^{n}} < \sum_{n\equal{}1}^{\infty} \dfrac{1/2}{4^{n}} \equal{} \frac {1} {6}$, hence $ \delta(x)$ is also bounded. It follows that $ \displaystyle \lim_{n \to \infty} \frac {1} {4^n} \delta(2^nx)$ exists and is equal to zero, so $ \delta(x) \equal{} 0$, i.e. $ f(x) \equal{} g(x)$.

This method of proof is reminiscent of "the Herglotz trick" (see [Aigner and Ziegler, Proofs from the  Book]).


Is there an official IRC channel dedicated to AoPS? If not, does anyone think it would be a good idea?

I think there is an "unofficial" one

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=830717#830717[/url]


[quote="drunner2007"]Find the sum of all k digit palindromic numbers (k is odd).[/quote]

Let $k = 2n + 1$. Then we have $n + 1$ distinct numbers for the palindrome

Each number in the first digit place from the left will appear $n!$ times, each number in the second digit place will appear $(n-1)!$ times and so forth, on to the $n+1$ digit from the left which will appear only once. Therefore, the summation should be:

$n!(10)^{2n+1}(45) + (n-1)!10^{2n}(45) + \ldots + 10^{n+1}(45) + 10^{n}(45)2! + \ldots (45)n!$


We have -
$ \lfloor 44 \rfloor \equal{} 6$ and $ \lfloor 4444 \rfloor \equal{} 66$
Can u generalise ?  :wink:

Uh, $ \lfloor 44\rfloor \equal{} 44$.

EDIT: Ha, you meant $ \lfloor\sqrt{44}\rfloor$. Okay then.

[hide="Solution"]$ \sqrt{444\ldots 4}\equal{}2\sqrt{111\ldots 1}$.

Let $ 2n$ be the number of digits. We have $ \sqrt{111\ldots 1}\equal{}\sqrt{\frac{10^{2n}\minus{}1}{9}}\equal{}\frac{1}{3}\sqrt{10^{2n}\minus{}1}$.

It's obvious that $ \frac{1}{3}\sqrt{10^{2n}\minus{}1}<\sqrt{10^{2n}}\equal{}\frac{10^n}{3}$. Here, we have $ n$ digits.

$ \frac{10^n}{3}\equal{}333\ldots3.\overline{3}$, so if we can show that the square root is greater than or equal to $ 333\ldots 3$, we should be able to conclude our proof.

$ 333\ldots 3\equal{}3\cdot\frac{10^n\minus{}1}{9}\equal{}\frac{1}{3}(10^n\minus{}1)\leq \frac{1}{3}\sqrt{10^{2n}\minus{}1}\implies 10^n\minus{}1\leq \sqrt{10^{2n}\minus{}1}$

Squaring both sides, we get $ \minus{}2\cdot 10^{n}\plus{}2\leq 0$, which is fairly self-evident.

Therefore, the floor of the square root must be $ \underbrace{333\ldots 3}_{n\text{ digits}}$, and the conclusion follows.[/hide]

Also, if we examine $ \lim_{n\to\infty}|\underbrace{\sqrt{444\ldots 4}}_\text{2n\text{ digits}}-\underbrace{666\ldots 6}_{n\text{ digits}}|$, we see that:

\begin{eqnarray*}{
\lim_{n\to\infty}|2\sqrt{\frac{10^{2n}-1}{9}}-6\cdot\frac{10^n-1}{9}|&=&\frac 23\lim_{n\to\infty}|\frac{\sqrt{O(10^{2n})}-1}{O(10^n)}}|\\
&=&\frac 23=.66666\ldots
\end{eqnarray*}

So, as $ n$ gets sufficiently large, the difference between the square root and its floor value tends toward $ .6666\ldots$, which kind of shows that the square root wants to converge to $ 6666\ldots$ for even $ n$.

I don't know if this is a rigorous proof of the generalization (though I believe the proof I provided earlier is), but it's a cool illustration nonetheless. If anyone can tell me if this approach can be made rigorous, that'd be nice.

i am sorry about the typo.
the question ofcourse involves $ \sqrt{44}$ ,$ \sqrt{4444}$ and similarly
@grnturtle
thanks for your replies
my solutions too was similar but involved only observing the number of digits


I'm going as oralist and j/s 2 person from Oak Park River Forest.  Anyone else on this site going?

i'm going as a coach and judging orals, but probably in division 1A or 2A (OPRF is in division 4AA).

Okay, well, I'll likely see you at ARML tryout tomorrow anyway.

16.  I was the kid who got number 19 right  :)

Wow, I wish we found out that long ago.  We don't find out which events we're doing till this Friday.  naperville north

Well, our whole school didn't qualify.  Only the sophomore team, j/s 2, and orals qualified.  I'm not a sophomore, and I did j/s 2 and orals, so it's not that much of a question.

hey Magnara if u go to OPRF what Middle School did you got to (Brooks, Julian, Roosevelt etc.)
Do u no E. Hallman i think he got 5th at MC in '06


Given polynomial $f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$, with roots $r_1,r_2,\ldots,r_n$, show that the polynomial with roots $\displaystyle\frac{1}{r_1}, \frac{1}{r_2}, \ldots , \frac{1}{r_n}$, is $f(x) = a_0x^n + a_1x^{n-1} + \ldots + a_{n-1}x + a_n$.

[hide]let $g(x)=f(\frac{1}{x})$, clearly if $r$ is a root of $f(x)$, then $\frac{1}{r}$ is a root of $g(x)$
$g(\frac{1}{r})=f(\frac{1}{\frac{1}{r}})=f(r)=0$

so we have

$g(x)=f(x) = a_nx^{-n} + a_{n-1}x^{-(n-1)} + \ldots + a_1x^{-1} + a_0$

clearly, if we multiply by $x^n$, that will not change the roots, so we have

$h(x)= a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0$, where $h(x)$ and $g(x)$ have the same roots, and they have the recipricals of the roots of $f(x)$[/hide]

In problem solving it is also beneficial to know for example, that:

$f(x-3)=0$ has the the roots of $f(x)=0$ all shifted up by 3.


calculus
derivative
logarithms
calculus computations

[color=red]From the Central Examination for Greek Universities 2006, May 27[/color]

If  $f(x) = e^x$   and  $g(x) = \ln x$ , prove that the diagrams([color=red]graphs[/color])  $C_f , C_g$  of     $f,g$  have exactly two common tangents.

[u] Babis[/u]

stergiu, what does it mean by a "diagram"? Does it mean exactly the same as the "graph of a curve"?

By $f'(x)=e^x,\ g'(x)=\frac{1}{x},$ the tangent line at the point $(t,\ e^t)$ on  $y=e^x$ is $y=e^t(x-t)+e^t\Longleftrightarrow y=e^tx+(1-t)e^t \cdots [L_1]$ and the tangent line at the point $(s,\ln s)\ (s>0)$ on $y=\ln x$ is $y=\frac{1}{s}(x-s)+\ln s \cdots [L_2].$ Assume that there exist the common tangent line of $C_f$ and $C_g,$ the two tangent lines $L_1,\ L_2$ coincide, there exist the real numbers $s,t$ for which $\frac{1}{s}=e^t$ and $\ln s-1=(1-t)e^t \Longleftrightarrow s=e^{-t}$ and $\ln s -1=(1-t)e^t.$ simultaneously.
Elliminating $s$ from these equations we have $e^t=\frac{t+1}{t-1}.$ Consider the graphs $y=e^t$ and $y=\frac{t+1}{t-1}$ on $t-y$ plane we can see two intersection points of these graphs, which shows the existence of two pair of  real numbers $(s,t).$


The values in the table are generated by a rule of the form $ a\equal{}B*c\plus{}D$, where $ B$ and $ D$ are constants. Find $ a$ when $ c\equal{}19$.

The thing is, if there's no table, the problem's impossible. :)

Was there a table, or was it a glitchy problem? You should report it if it was a glitch.


How can we algebraically find the intersection of an exponential function y=a^x with the straight line y=x?
In other words, what is the solution to the equation a^x = x ?

This equation cannot be solved algebraically by elementary methods (i.e. with elementary functions - multiplication, division, addition, subtraction, powers, roots, exponential functions, trigonometric functions, or their inverses).

If a is known, however, it can be solved numerically (by Newton's Method or with a TI graphing calculator for example) to any desired level of precision.


integration
calculus
calculus computations

What is
\[ \int_0^{\infty}x^{\minus{}x}dx?\]
A numerical integrator gives approximately 2.

It is well-known that $ x^{\minus{}x}$ does not have an elementary anti-derivative.

Related: the [url=http://mathworld.wolfram.com/SophomoresDream.html]Sophomore's dream[/url].


inequalities
function
rearrangement inequality
inequalities unsolved

Hello everybody!

Here is a nice cyclic inequality out from Thomas Mildorf's pdf ( unsolved problem 42 )

Let $ a,b,c$ be positive real numbers such as $ a \plus{} b \plus{} c \equal{} 3$
 Prove that $ \frac {a}{ab \plus{} 1} \plus{} \frac {b}{bc \plus{} 1} \plus{} \frac {c}{ca \plus{} 1}\ge\frac {3}{2}$

After clearing denominators and reusing the condition you get $ a^{2}c \plus{} b^{2}a \plus{} c^{2}b \plus{} 3abc \plus{} 3 \ge a^{2}b \plus{} b^{2}c \plus{} c^{2}a \plus{} 3a^{2}b^{2}c^{2} \plus{} a^{3}bc \plus{} b^{3}ac \plus{} c^{3}ab$

I don't really know how to face cyclic inequalities , i tried Lagrange multipliers but i couldn't solve the condition...

I'm sure that there is a solution which is much easier than my solution, but better than no solution.  :roll:

[quote]Let $ a,b,c$ positive real numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
$ \sum_{cyc} \frac {a}{ab \plus{} 1} \geq \frac {3}{2}$[/quote]

Let $ \{p,q,r\}$ be a permutation of $ \{a,b,c\}$ such that $ p \ge q\ge r$. With the rearrangement inequality we obtain that
$ \sum_{cyc} \frac {a}{ab \plus{} 1} \geq \sum_{cyc} \frac {p}{qr \plus{} 1}$
From the condition $ p \plus{} q \plus{} r \equal{} 3$, it follows immediatly that $ 1\ge pqr$ and $ \frac {1}{p} \ge qr$. Same for the other two variables. Therefore,
$ \sum_{cyc} \frac {p}{qr \plus{} 1} \ge \sum_{cyc} \frac {p}{\frac {1}{p} \plus{} 1} \equal{} \sum_{cyc} \frac {p^2}{p \plus{} 1}$
The function $ f(x) \equal{} \frac {x^2}{x \plus{} 1}$ is convex on $ [0,\infty)$. With Jensen
$ \frac {1}{3} \sum_{cyc} f(p) \ge f(\frac {p \plus{} q \plus{} r}{3}) \equal{} f(1) \equal{} \frac {1}{2}$
and the inequallity is proved.

[quote="svega"]

Let $ a,b,c$ be positive real numbers such as $ a \plus{} b \plus{} c \equal{} 3$
 Prove that $ \frac {a}{ab \plus{} 1} \plus{} \frac {b}{bc \plus{} 1} \plus{} \frac {c}{ca \plus{} 1}\ge\frac {3}{2}$

[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21689

Ok, thanks a lot !

It's really nice using $ a^2\plus{}b^2\plus{}c^2 \geq \sqrt {3(a^3b\plus{}b^3c\plus{}c^3a)}$ for solving it  :D

Event though I think FelixD's proof is a bit wrong because rearrangement would more likely be $ \sum_{cyc} \frac {a}{ab \plus{} 1} \le \sum_{cyc} \frac {p}{qr \plus{} 1}$ seeing as $ p\ge q\ge r$ implies $ \frac{1}{qr\plus{}1}\ge\frac{1}{pr\plus{}1}\ge\frac{1}{pq\plus{}1}$

Well it's in the link arquady provided...

you have $ \frac{x}{xy\plus{}1}\equal{}x\minus{}\frac{x^2y}{xy\plus{}1}$

so you have to prove $ \sum \frac{x^2y}{1\plus{}xy} \leq \frac{3}{2}$
and seeing that $ 1\plus{}xy \ge 2\sqrt{xy}$ so what we want to prove is
$ \sum_{cyc}x^{3/2}y^{1/2}\le 3$

which is a corollary of the inequality mentioned in my last post.
Proof of that inequality could be given as $ (a^{2}\plus{}b^{2}\plus{}c^{2})^{2}\minus{}3(a^{3}b\plus{}b^{3}c\plus{}c^{3}a)\equal{}\frac{1}{2}\sum_{cyc}(a^{2}\minus{}2ab\plus{}bc\minus{}c^{2}\plus{}ca)^{2}$ though I dislike that proof because it comes out from nowhere :)


The brochure says: "There are six tracks. Each student takes one track and each track consists of two courses."  It then says that the lectures will focus on six subjects (number theory, etc).  My questions are:   what is the relationship between tracks and subjects?  For each track, are the two courses in the same subject?  If they are not, what are the combinations?

A track has two subjects I think. If there are 6 tracks then the subjects are probably number theory, geometry, combinatorics, and algebra.

I was wondering about this as well. My impression was that tracks were equivalent to difficulties. This conjecture is somewhat useless since I have not been to the camp and the brochure is unclear....hence:

Could somebody explain the track/course process more clearly?


function
real analysis
real analysis unsolved

Hello,

Let $f$ be a real function such that: if $\sum a_n$ converges, then $\sum f(a_n)$ converges too (the $a_n$ are taken real).

Prove that there exist $\delta>0$ and a real number $\alpha$ such that $f(x) = \alpha x$ if $x \in ]-\delta,\delta[$.

I don't know wether this is a ``classic'' or not.

Enjoy it !

This problem has been discussed at least three times on this forum. See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31422[/url], for example... It is a nice classical problem, indeed, and at some moment I thought of posting it myself but then I found out that other people had already done it.   :)


Prove or disprove that there exists an integer which is doubled when the initial digit is transfered to the end.

$ n \equal{} a 10^k \plus{} x$ with $ 1 \le a \le 9$ and $ x$ with $ k$ or fewer digits
then $ 10x \plus{} a \equal{} 2(a 10^k \plus{} x)$

$ 8x \equal{} a (2 \cdot 10^k \minus{} 1)$ notice that $ (2 \cdot 10^k \minus{} 1)$ is odd then $ 8|a$ then $ a \equal{} 8$

$ x \equal{} 2 \cdot 10^k \minus{} 1 > 10^k$ then $ x$ has more than $ k$ digits. Contradiction.


point set
projection
combinatorics
combinatorial geometry
IMO
3D geometry
IMO 1992

Let $\,S\,$ be a finite set of points in three-dimensional space. Let $\,S_{x},\,S_{y},\,S_{z}\,$ be the sets consisting of the orthogonal projections of the points of $\,S\,$ onto the $yz$-plane, $zx$-plane, $xy$-plane, respectively. Prove that \[ \vert S\vert^{2}\leq \vert S_{x} \vert \cdot \vert S_{y} \vert \cdot \vert S_{z} \vert,  \] where $\vert A \vert$ denotes the number of elements in the finite set $A$.

[hide="Note"] Note: The orthogonal projection of a point onto a plane is the foot of the perpendicular from that point to the plane. [/hide]

Let us use a combinatorial reasoning. Let us denote by $ V_{ij}$ the set of the points of $ V$ of the form $ (x, i, j)$, that is, the set of the points of $ V$ who are projected on the plane $ yz$ at the point $ (i, j)$.
We thus have : $ V \equal{} \bigcup_{(i,j) \in V_x} V_{ij}$, hence :
\[ |V| \le \sum_{(i,j) \in V_x} |V_{ij}|,
\]
and, thanks to the Cauchy-Schwarz's inequality :
\[ |V|^2 \le \sum_{(i,j) \in V_x} 1^2 \cdot \sum_{(i,j) \in V_x} |V_{ij}|^2 \equal{} |V_x| \cdot \sum_{(i,j) \in V_x} |V_{ij}|^2.
\]
We look for a set $ X$ such that $ |X| \equal{} \sum_{(i,j) \in V_{ij}} |V_{ij}|^2$. It is plain that the set :
\[ X \equal{} \bigcup_{(i,j) \in V_x} (V_{ij} \times V_{ij})
\]
does the trick.
Moreover, the application $ f : X \to V_y \times V_z$ defined by :
\[ f((x, i, j),(x', i, j)) \equal{} ((x, j), (x', i))
\]
is one-to-one, so $ |X| \le |V_y| \cdot |V_z|$, hence the result.

For reals $x,y,z$, define $a_{xy}=[(x,y,0)\in S_z]$, $b_{yz}=[(0,y,z)\in S_x]$, and $c_{zx}=[(x,0,z)\in S_y]$. Then $S\subseteq\{(x,y,z)\in\mathbb{R}^3: a_{xy}b_{yz}c_{zx}=1\}$ (and this "maximal" set can be achieved), so we need to show that
\[\left(\sum_{\mathbb{R}^3}a_{xy}b_{yz}c_{zx}\right)^2\le \left(\sum_{\mathbb{R}^2}a_{xy}\right)\left(\sum_{\mathbb{R}^2}b_{yz}\right)\left(\sum_{\mathbb{R}^2}c_{zx}\right).\]But by Cauchy-Schwarz and the fact that $t=t^2$ for $t\in\{0,1\}$,
\begin{align*}
\left(\sum_{x,y,z}a_{xy}b_{yz}c_{zx}\right)^2
&=\left(\sum_{x,y}a_{xy}\sum_{z}b_{yz}c_{zx}\right)^2 \\
&\le\left(\sum_{x,y}a_{xy}^2\right)\left(\sum_{x,y}\left(\sum_{z}b_{yz}c_{zx}\right)^2\right) \\
&\le\left(\sum_{x,y}a_{xy}^2\right)\left(\sum_{x,y}\left(\sum_{z}b_{yz}^2\right)\left(\sum_{z}c_{zx}^2\right)\right) \\
&=\left(\sum_{x,y}a_{xy}\right)\left(\sum_{y}\left(\sum_{z}b_{yz}\right)\sum_{x}\left(\sum_{z}c_{zx}\right)\right) \\
&=\left(\sum_{x,y}a_{xy}\right)\left(\sum_{y,z}b_{yz}\right)\left(\sum_{z,x}c_{zx}\right),
\end{align*}as desired.

In fact, as in [i]The Cauchy-Schwarz Master Class[/i] (by Steele), we can use the same method to prove the more general inequality
\[\left(\sum_{\mathbb{R}^3}\sqrt{a_{xy}b_{yz}c_{zx}}\right)^2\le \left(\sum_{\mathbb{R}^2}a_{xy}\right)\left(\sum_{\mathbb{R}^2}b_{yz}\right)\left(\sum_{\mathbb{R}^2}c_{zx}\right),\]where $a_{xy},b_{yz},c_{zx}$ are any nonnegative reals; this is a special case of the [url=http://en.wikipedia.org/wiki/Loomis%E2%80%93Whitney_inequality]Loomis-Whitney inequality[/url].

This can easily be extended to $n$- dimension.

Problem: [color=#4000FF]Let $\,S\,$ be a finite set of points in  $\mathbb{R}^n$. Let $\,S_{1},\,S_{2},\, \ldots, S_{n}\,$ be the sets consisting of the orthogonal projections of the points of $\,S\,$ onto the  $n$  cartesian planes respectively. Prove that  $|S|^{n-1} \le \prod_{i=1}^{n} |S_i| $
[/color]

Solution: We induct on number of points.

For  a point $a = (a_1, \ldots , a_n) \in S$, we have $(0,a_2 , \ldots , a_n) \in S_1$, $(a_1, 0,\ldots , a_n) \in S_2$ and so on.

Take some $t$, and create two nonempty  set $A$ and $B$ with elements of $S$ that have $a_i> t$, and with elements of $S$ have $a_i \le t $ respectively. (there exists at least one such $i$ )

So $|A| + |B| = |S|$, and $|A_j| + |B_j| = |S_j| $ , for all $j \ne i$, and obviously $|A_i|, |B_i| \le |S_i|$

By inductive hypothesis  $|S|= |A| + |B| \le \prod_{i=1}^{n}|A_i|^{\frac{1}{n-1}}+ \prod_{i=1}^{n}|B_i|^{\frac{1}{n-1}}$

So $|S| \le \prod_{i=1}^{n}|A_i|^{\frac{1}{n-1}}+ \prod_{i=1}^{n}|B_i|^{\frac{1}{n-1}}$   $ \le |S_i|^{\frac{1}{n-1}} \left ( \prod_{\substack{j=1 \\ i \ne j}}^{n-1}|A_i|^{\frac{1}{n-1}}+ \prod_{\substack{j=1 \\ i \ne j}}^{n-1}|B_i|^{\frac{1}{n-1}} \right )$  $ \le |S_i|^{\frac{1}{n-1}} \cdot  \prod_{\substack{j=1 \\ i \ne j}}^{n-1}\left ( |A_j| + |B_j| \right )^{\frac{1}{n-1}}$  $ = \prod_{i=1}^{n} |S_i|^{\frac{1}{n-1}}  $

The last inequality follows from Hölder's inequality.

So we get $|S|^{n-1}\le\prod_{i=1}^{n}|S_{i}|$, and the induction is complete.

Let us bound the set $S$ between two horizontal planes (i.e. planes parallel to $xy$ plane) and begin to move down the upper one plane. Every time it meets some point/poins we denote that set of points by $S^k\,,\,k=1,2,\dots,m $, where of course $m\leq |S|$. Obviously, we have: $S_x=\sum_{i=1}^m S^i_x\,;\, S_y=\sum_{i=1}^m S^i_y\,;\, S_z\geq \max (S^i,i=1,\dots,m)\,,\, S^i \leq S^i_x\cdot S^i_y$. Here and below we may omit, for simplicity, the vertical lines $|\cdot|$ when denoting the number of elements of some set.
Using it, we consecutively obtain:
\begin{align*} |S_{x}|\cdot |S_{y}| \cdot |S_{z}| &\geq \sum_{i=1}^m S^i_x \cdot \sum_{i=1}^m S^i_y\cdot \max_{1\leq i\leq m} (S^i) \\
&\geq \left(\sum_{i=1}^m \sqrt{S^i_x S^i_y}\right)^2 \max (S^i) \geq \left(\sum_{i=1}^m \sqrt{S^i}\right)^2 \max (S^i) \\
&\geq  \left(\sum_{i=1}^m S^i\right)^2 = |S|^2
\end{align*}
(jumping from first to second line we used the Cauchy-Schwarz).

There is a very nice proof of this inequality, using information-theoretic techniques, together with Han's inequality (subadditivity of entropy).
[b]Lemma:[/b] For random variables, $X,Y$, and $Z$, let, $H(X,Y,Z)$ denote their joint entropy, namely,
$$
H(X,Y,Z)=\mathbb{E}\left[\log\frac{1}{P_{X,Y,Z}}\right],
$$
where, here, $P_{X,Y,Z}$ is treated as a random variable, taking the value, $\mathbb{P}(X=x,Y=y,Z=z)$; with probability, $P_{X,Y,Z}(x,y,z)$; and define the pairwise entropies similarly. Then,
$$
H(X,Y)+H(X,Z)+H(Y,Z) \geq 2H(X,Y,Z).
$$

Now, consider $n$-points in the space; a uniform probability measure (namely, each atom in the space, is charged with, $\frac{1}{n}$) on these $n$-points; and let, $(X,Y,Z)$ be the joint random variable, describing the coordinates of a single point. Using the entropy of a uniform random variable, we have, $H(X,Y,Z)=\log n$. Next, let, $n_1$, $n_2$; and $n_3$ be the cardinalities of the sets, obtained via projecting these $n$-points onto $yz$, $xz$, and $xy$-planes, respectively.

Then, clearly, $H(X,Y)\leq \log n_3$, $H(X,Z)\leq \log n_2$, and, $H(Y,Z) \leq \log n_1$, since, a uniform distribution achieves the maximum entropy, subject to no constraints, and we do not know, whether we have a uniform or not. Now, observe what we have:
\begin{align*}
2H(X,Y,Z) &=\log n^2 \\
& \leq H(X,Y)+H(X,Z)+H(Y,Z)\\
&\leq \log n_1 + \log n_2 + \log n_3\\
&=\log n_1n_2n_3\\
&\implies n\leq \sqrt{n_1 n_2 n_3},
\end{align*}
done. $\square$

Here is a funny solution which doesn't make use of Cauchy or Holder.

First we will WLOG assume that all of $S$ is in the first octant, meaning that all points of $S$ have all coordinates strictly positive. Now, we will turn on gravity, which acts in the $-x$ direction. This means that all points which share $y-, z-$ coordinates will end up "stacked" on top of each other. The reason this is fine is that it preserves $|S|$ (number of points remains the same) and $|S_x|$ (number of stacks is the same), while $|S_y|$ and $|S_z|$ do not increase. Similarly, we can turn on gravity in the $-y$ and $-z$ directions. In this way, we can solve the rest of the problem under the assumption that if $(a, b, c) \in S$ for $a, b, c \in \mathbb{N}^3$ then $(a', b', c') \in S$ as well, whenever $1 \le a' \le a, 1 \le b' \le b, 1 \le c' \le c.$ Now, for each $(y,z) \in S_x$, let $x_{y,z}$ be the number of $x$ with $(x, y,z) \in S.$ Observe that from our assumption, $a \le c, b \le d$ implies that $x_{a, c} \ge x_{b, d}.$

Let $a, b \in \mathbb{N}$ be the least positive integers such that $S_x \subseteq [a] \times [b].$ This means that our inequality in question reduces to:

$$(\sum_{(i, j) \in [a] \times [b]} x_{i, j})^2 \le t \cdot (\sum_{i = 1}^{a} x_{i, 1}) \cdot (\sum_{j=1}^{b} x_{1, j}),$$

of course subject to the conditions obtained via gravity, where $t$ is the number of $(i, j)$ with $x_{i, j}$ nonzero. We will prove that this inequality holds via induction on $|S|.$ Consider the effect of decreasing each of the $x_{i, j}$'s which are nonzero by one. We know by the inductive hypothesis that:

$$(|S| - |S_x|)^2 \le |S_x| \cdot (|S_y| - a)(|S_z| - b).$$

Hence, if we could show that:

$$|S|^2 - (|S| - |S_x|)^2 \le |S_x| (a \cdot |S_z| + b \cdot |S_y| - ab)$$

then we would be done by summing the two equations. After dividing by $|S_x|$, we wish to show that:

$$2 |S| - |S_x| \le a \cdot |S_z| + b \cdot |S_y| - ab \Leftrightarrow (a \cdot |S_z| - |S|) + (b \cdot |S_y| - |S|) \ge ab - |S_x|.$$

It's immediate that $b \cdot |S_y| - |S| \ge 0$ from our prior assumption, so it suffices only to show that:

$$a \cdot |S_z| - |S| \ge ab - |S_x|.$$

Note that $ab - |S_x|$ counts the number of $(i, j) \in [a] \times [b]$ with $x_{i, j} = 0.$ Hence, since we can write $a \cdot |S_z| - |S|$ as:

$$\sum_{(i, j) \in [a] \times [b]} (x_{1, j} - x{i, j}),$$

we're done because it's clear that each $x_{i, j} = 0$ contributes at least one to the sum above.

The induction is complete, and hence the problem is solved.

$\square$

Note that if there are points $(x,y,0)$, $(x,0,z)$, $(0,y,z)$ between the projections, then we may consider that $(x,y,z)\in S$. Let $T_x$ be the set of all $x$-coordinates of points in $S$, and define $T_y$, $T_z$ similarly. Consider a $3$-partite graph $G$ with parts $T_x$, $T_y$, $T_z$ and the edge between two vertices $x\in T_x$ and $y\in T_y$ iff $(x,y,0)\in S_x\cup S_y\cup S_z$, and similar condition for edges between $T_x$, $T_z$ and $T_y$, $T_z$. Note that the number of edges between $T_x$, $T_y$ equals $|S_z|$, similarly for $T_x$, $T_z$ and $T_y$, $T_z$, thus the number of triples $(e_{xy},e_{yz},e_{zx})$, where $e_{ij}$ is the number of edges between $T_i$ and $T_j$, equals $\vert S_{x} \vert \cdot \vert S_{y} \vert \cdot \vert S_{z} \vert$, and every triangle in this graph corresponds to a unique vertex in $S$. Thus, we need to prove that the square of the number of triangles in $G$ is not greater than $\vert S_{x} \vert \cdot \vert S_{y} \vert \cdot \vert S_{z} \vert$. We prove this by induction on the number of vertices in $G$. The base $|G|=1,2,3$ is trivial. Now, let the statement be proved for $|G|\le k$, and we need to prove it for $|G|=k+1$. Let $T_x=T_x'\cup{\nu}$, $G_1=G\setminus\nu$, $G_2=G\setminus T_x'$. By the induction hypothetis:
\begin{align*}
S_1^2&\le \vert S_{x} \vert \cdot \vert S_{y}' \vert \cdot \vert S_{z}' \vert\\
S_2^2&\le\vert S_{x} \vert \cdot \vert S_{y}'' \vert \cdot \vert S_{z}'' \vert,
\end{align*}
where $S_i$ is the numbet of triangles in $G_i$, $S_x$ is the number of edges between $T_y$, $T_z$ in $G_1$, $S_x'$ is the number of edges between $T_y$, $T_z$ in $G_2$, similarly for other constants (S_x=S_x'=S_x''). Then
\begin{align*}
S^2=(S_1+S_2)^2=|S_1|^2+|S_2|^2+2|S_1|\cdot |S_2|
&\le\vert S_{x}\vert \cdot \vert S_{y}' \vert \cdot \vert S_{z}' \vert+\vert S_{x}\vert \cdot \vert S_{y}''\vert \cdot \vert S_{z}''| +|S_x|\cdot2\sqrt{ \vert S_{y}' \vert \cdot \vert S_{z}' \vert\cdot \vert S_{y}''\vert \cdot \vert S_{z}''|}\le\\
&\le \vert S_{x} \vert \cdot \vert S_{y}' \vert \cdot \vert S_{z}' \vert+\vert S_{x}'' \vert \cdot \vert S_{y}''\vert \cdot \vert S_{z}''| +|S_x|\cdot \vert S_{y}' \vert \cdot \vert S_{z}''|+|S_x|\cdot \vert S_{y}'' \vert \cdot \vert S_{z}'|=\\
&=|S_x|(|S_y|'+|S_y''|)(|S_z'|+|S_z''|)=|S_x|\cdot |S_y|\cdot |S_z|,
\end{align*}
what was required to be proved.

I had posted the general inequality here https://artofproblemsolving.com/community/c7h1835173

We use strong induction on $n$. The idea is to split $S$ into two smaller subsets $A$ and $B$ along a plane parallel to, say, the $xy$-plane. Without loss of generality let $|A_z| \geq |B_z|$. Then, we know that $|A_x| + |B_x| = |S_x|$, $|A_y| + |B_y| = |S_y|$, and $|S_z| \geq |A_z|$. Thus, using the strong inductive hypothesis on $A$ and $B$,
\begin{align*}
|S_x| \cdot |S_y| \cdot |S_z| &= (|A_x| + |B_x|)(|A_y| + |B_y|) |S_z| \\
&\geq |S_z|(\sqrt{|A_x||A_y|} + \sqrt{|B_x||B_y|})^2 \\
&\geq |S_z|\left(\sqrt{\frac{|A|^2}{|A_z|}} + \sqrt{\frac{|B|^2}{|B_z|}}\right)^2 \\
&\geq |A_z| \left(\frac 1{\sqrt{|A_z|}}(|A|+|B|)\right)^2 \\
&= |S_z| \cdot \frac 1{|A_z|} \cdot |S|^2 \\
&\geq |S|^2,
\end{align*}which completes the induction.


Let $S = \{A_1,A_2,\ldots ,A_k\}$ be a collection of subsets of an $n$-element set $A$. If for any two elements $x, y \in A$ there is a subset $A_i \in S$ containing exactly one of the two elements $x$, $y$, prove that $2^k\geq n$.

[i]Yugoslavia[/i]

To each element $x\in A$, associate the set $S_x=\{j\in\overline{1,k}\ |\ x\in A_j\}$. It's easy to see that the hypothesis implies the fact that $x\mapsto S_x$ is one-to-one, hence $n$ is at most as large as the number of subsets of $\overline{1,k}$, and this number is $2^k$.

Path to Combo for Undergrads

I will prove this problem the 'other way', ie. I will show that for a fixed $k$, the maximum $n$ can be is $2^k$, and indeed show a construction for $n=2^k$ which will prove the problem

Count the number of times each element occurs in $W=A_1+A_2+....A_k$ (so basically take the "multiset" formed by taking union of all these sets with multiplicity. So if $a \in A_6$ and $a \in A_9$, then a is counted twice)

Say $c_r$ elements occur r times. Notice we only need to care about elements that occur the same number of times because PHP takes care of the rest.

Now clearly, $c_r \le \binom{k}{r}$ because for each element that occurs r times, there are $\binom{k}{r}$ ways of choosing which $r$ sets it can go to, so if $c_r > \binom{k}{r}$, then two elements must have occurred in exactly the same $r$ sets.

Now note that $n=|A| = \sum_{r=0}^k c_r \le \sum_{r=0}^k \binom{k}{r} = 2^k$.

Now the construction for $n=2^k$ is just trivial, indeed for each of the $\binom{k}{r}$ elements choose some random $r$ subsets which that element should be in.

It is interesting that in the $n=2^k$ case, we can count exactly how many possible $S$ there are...

[hide=motivation]Just did what you would expect, for a fixed k, tried to work with n=2^k, because if we could construct S for 2^k, we could do for smaller $n$ also. I realised that we only cared things that occurred the same number of times, and when trying to construct for k=4, finally realised the bound for $c_r$[/hide]

All The Way Up


Let a be a positive integer. Find all positive integers n such that $ b= a^{n}$
satisfies the condition that $ a^{2}+b^{2}$ is divisible by $ ab+1$.

Hmm, i think that you can't post this problem in Pre-Olympiad box.
[hide="A hint"]From IMO 1988 problem 6 and Canada MO 1998 problem 5 we have for some $ j,m$ : $ b=a_{j+1},a=a_{j}$, here $ a_{0}=0,a_{1}=m,a_{i+1}=m^{2}a_{i}-a_{i-1}(i=1,2,...)$ [/hide]


Consider the system \[x+y=z+u,\] \[2xy=zu.\] Find the greatest value of the real constant $m$ such that $m \le \frac{x}{y}$ for any positive integer solution $(x, y, z, u)$ of the system, with $x \ge y$.

$ z,u$ is the root of the equation:
$ t^2 \minus{} (x \plus{} y)t \plus{} 2xy$
$ \Delta \equal{} (x \plus{} y)^2 \minus{} 8xy \equal{} x^2 \plus{} y^2 \minus{} 6xy \equal{} m^2$
$ \Rightarrow (\frac {x}{y})^2 \minus{} 6\frac {x}{y} \plus{} 1 \equal{} (\frac {m}{y})^2$
$ a \equal{} \frac {x}{y},b \equal{} \frac {m}{y}$
So we has :
$ a^2 \minus{} 6a \plus{} 1 \equal{} b^2$
$ \Leftrightarrow (a \minus{} 3)^2 \minus{} b^2 \equal{} 8$
$ (a \minus{} 3 \minus{} b)(a \minus{} 3 \plus{} b) \equal{} 8$
It imply that the the equation has rational solution for all $ b$ less enough.
Now Let $ b\to0$ then $ a\to 3 \plus{} 2\sqrt {2}$
So the max $ m$ is $ m \equal{} 3 \plus{} 2\sqrt {2}$

A little more analysis shows that all solutions $(x,y,z,u)$ such that $\gcd(z,u)=d$, that $\frac{z}{d}$ is odd, and that $x>y$ are in the form $d\left(m(m+n),n(m-n),m^2-n^2,2mn\right)$, where $m$ and $n$ are coprime positive integers with different parities such that $m>n$.  Thus, if $q$ is the ratio $\frac{m}{n}$, then $\frac{x}{y}=\frac{q(q+1)}{q-1}$.  As the rationals is dense in the reals, we may seek to optimize this ratio taking $q$ to be an arbitrary real greater than $1$.  Calculus or AM-GM yields the best lower bound of $3+2\sqrt2$.

Это задача Герона Александрийского.
Система:

$X+Y=Z+D$
$XY=qZD$

Решения можно записать:

$X=qs^2-qps$
$Y=qps-p^2$
$Z=ps-p^2$
$D=qs^2-ps$

Или ещё:

$X=qp^2+qps$
$Y=qps+(q-1)s^2$
$Z=qp^2+(2q-1)ps+(q-1)s^2$
$D=ps$

Где $q$ любое целое число. Видно, что система всегда имеет решения. Числа же $p,s$ задаются нами и могут быть любыми.


In a square there are four congruent right angle triangles. All five circles has the same radius $ r$. Determine $ r$ if $ AB\equal{}1$.

The problem is a classical of Sangaku, and it is included in the book "Japanese Temple Geometry problems" by Fukagawa and Pedoe with the number 3.1.4, page 38.
The answer is (3^(1/2) - 1)/4.
The problem is also discussed and solved in the bulgarian journal "Matematyka & Informatika", 6/1992

you can also have a look to :
"Sangaku : Le mystère des énigmes géométriques japonaises" by Géry Huvent, ed.Dunod
Sangaku N° 7

 :roll:

I don't have those books, so could someone post a solution.

Let $ AP\equal{}BQ\equal{}CR\equal{}DS\equal{}a$, $ AQ\equal{}BR\equal{}CS\equal{}DP\equal{}b$.
So, $ PQ\equal{}QR\equal{}RS\equal{}SP\equal{}b\minus{}a$.
Hence, $ \left\{\begin{array}{l}\frac{a\plus{}b\minus{}\sqrt{a^2\plus{}b^2}}{2}\equal{}\frac{b\minus{}a}{2} \\ \sqrt{a^2\plus{}b^2}\equal{}1\end{array}\right. \iff (a, b)\equal{}\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.
Thus, $ r\equal{}\frac{a\plus{}b\minus{}\sqrt{a^2\plus{}b^2}}{2}\equal{}\frac{b\minus{}a}{2}\equal{}\boxed{\frac{\sqrt{3}\minus{}1}{4}}$.


calculus
integration
function
calculus computations

Need help with this, Suppose f:[a,b]->R is an integrable function, and g:[0,1]->R by g(x)= f(a+b(b-a)x)

1. Show that g isintegrable
2. Show that  (b-a)$\int_{0}^{1}$g $=\int_{a}^{b}$f

The problem is not defined correctly I think.

What julien_santini meant: the second line should be

"... and define $g: [0,1]\mapsto\mathbb{R}$ by $g(x)=f(a+(b-a)x).$"


geometry
circumcircle
analytic geometry
trigonometry
conics
ellipse
parabola

[color=blue]Let two fixed circles  $(O_{1}),(O_{2})$ on the plane and $\epsilon$ be a variant tangent to $(O_{1})$, which intersects $(O_{2})$ at the points $A$ and $B$.
Find[/color] [color=darkred]the locus[/color]  [color=blue]of the circumcenter of the triangle $AO_{2}B$[/color]

[i][b]Nick Rapanos[/b][/i]

Here is my solution .
Let a point $M$ on the circle with center $O_{1}$ and the tangent at $M$ intersects the other circle at $A,\ B$.
Let $K$ be the circumcenter of $AO_{2}B$ . We have that $O_{2}K\bot AB$ .
But we also have $O_{1}M\bot AB$ so $MO_{1}//KO_{2}$ .
So while $M$ varies on the first circle $K$ lies on a line from $O_{2}$ parallel to $O_{1}M$ . So let $S$ the locus of $M$ .
Then S lies on two lines .

[quote="silouan"]Here is my solution .
Let a point $M$ on the circle with center $O_{1}$ and the tangent at $M$ intersects the other circle at $A,\ B$.
Let $K$ be the circumcenter of $AO_{2}B$ . We have that $O_{2}K\bot AB$ .
But we also have $O_{1}M\bot AB$ so $MO_{1}//KO_{2}$ .
So while $M$ varies on the first circle $K$ lies on a line from $O_{2}$ parallel to $O_{1}M$ . So let $S$ the locus of $M$ for which the tangent at $M$ intersects the other circle (because we could have no intersection) . From the above $K$ lies on a locus $S'$ similar to $S$ (it is an arc or a circle )  :wink:[/quote]

 :huh: Sorry but this is wrong  :wink:

Let $r_{1},\ r_{2}$ be radii of $(O_{1}),\ (O_{2})$ and $d = O_{1}O_{2}.$ Let M be the midpoint of AB, $O_{2}M \perp AB.$ Let $(O_{1})$ meet the center line $O_{1}O_{2}$ at C, D, such that C is closer to $O_{2}$ than D. Put the center of polar coordinates at $O_{2}$ and let $\phi = \angle MO_{2}D.$ The locus of M is a limacon with polar equation $h = O_{2}M = r_{1}+d \cos \phi$ (or a cardioid if $d = r_{1},\ O_{2}\in (O_{1})$).  From the sides $O_{2}A = O_{2}B = r_{2}$ and altitude $O_{2}M = h$ of the isosceles $\triangle AO_{2}B,$ its circumradius is $R = \frac{O_{2}B}{2 \sin \widehat{O_{2}AB}}= \frac{r_{2}^{2}}{2h}= \frac{r_{2}^{2}}{2(r_{1}+d \cos \phi)}.$ This is polar equation of a conic with focus $O_{2}$, main axis line $O_{2}D \equiv O_{2}O_{1}$ and excentricity $\epsilon = \frac{d}{r_{1}}.$ Ellipse, if $d < r_{1},\ O_{2}$ is inside of $(O_{1}),$ parabola if $d = r_{1},\ O_{2}\in (O_{1})$ and hyperbola if $d > r_{1},\ O_{2}$ is outside of $(O_{1}).$


[i]$Question 1$[/i][/u]
A "Magic" number $M$ is an integer such that $M+1$ is a perfect square, and $M+2$ is twice another perfect square. For example,
$M = 48:$
$48+1 = 49 = 7^{2}$
$48+2 = 50 = 2 \cdot{5^{2}}$

Similarly for $M = 1680$.

An Antimagic number is magic by subtraction, ie if $A$ is an antimagic number, then $A-1$ is a perfect square, and $A-2$ is twice another perfect square.

Examples are:
$A = 290$ and $A = 9802$.

Find a general method for generating all Magic and Antimagic numbers.

[i]$Question 2$[/i][/u]
Consider the integer $6$. The sum of all integers less than $6$ is $15$ and the sum of the next two integers, that is $7$ and $8$ is also $15$.

Thus, we call $6$ as a balance  number.

Find a method to generate balance numbers.

[i]$Question 3$[/i][u]
Find a recurrence relation which ( with appropriate initial terms) generates the perfect squares and with different initial terms generates all the perfect cubes.
[/u]

[hide="Is Question 2 correct?"]
The sum of all integers less than $n$ is $\frac{(n-1)n}{2}$ and the sum of the two integers after $n$ is $n+1+n+2=2n+3$.  So we have

$\frac{(n-1)n}{2}=2n+3\Rightarrow n^{2}-5n-6=0$

This factorizes as $(n-6)(n+1)=0$.  $-1$ is extraneous so the only balance number is 6.
[/hide]

[quote="E^(pi*i)=-1"][hide="Is Question 2 correct?"]
The sum of all integers less than $n$ is $\frac{(n-1)n}{2}$ and the sum of the two integers after $n$ is $n+1+n+2=2n+3$.  So we have

$frac{(n-1)n}{2}=2n+3\Rightarrow n^{2}-5n-6=0$

This factorizes as $(n-6)(n+1)=0$.  $-1$ is extraneous so the only balance number is 6.
[/hide][/quote]

I think it is correct...
[hide]In the question, for any positive integer $n$, take the inetegrers that are less than $n$. You have said that the sum is $\frac{(n-1)n}{2}$, but this includes $n$ itself.

Does this clarify?[/hide]

[quote="xxxyyyy"]
You have said that the sum is $\frac{(n-1)n}{2}$, but this includes $n$ itself.

Does this clarify?[/quote]

uhh, no it doesn't, we're summing the numbers from 1 to n-1, so we plug in n-1 for n in $\frac{n(n+1)}{2}$, do you not? Am I missing something here?