calculus
AoPSwiki
number theory
real analysis
real analysis theorems

Does anybody have any book about distribution of primes and applications of calculus in number theory (about primes) ?
Or some another source for this ...

 :roll:  :roll:

Distribution of primes has to do with the Riemann-Zeta function.  Most introductory number theory books have something on this.

You can also take a look at [[Prime Number Theorem]] on AoPSWiki and see if you can understand the proof to begin with (It seems like the very end is missing there but you should be able to recover it if you have no trouble with the rest).

If your analytical skills are good enough to go through that proof, you can try reading Davenport's "Multiplicative number theory". If you master that book, try Vaughan's "Hardy-Littlewood method". Once you are through that, you should be able to read almost anything in classical analytic number theory.

If you have trouble with this AoPSWiki proof, then you may step back a bit and try Apostol's "Introduction to analytic number theory" or something equivalent.

There are also plenty of introductory books below that level but let's see where you stand first :).


Due to time spent improving our classes over the past year and the cost of running the online school we will be increasing the prices of our classes by about 25% for any class starting after December 31, 2004.  All classes currently listed are listed at our old prices.

We still feel strongly that we provide an excellent value for students and we always endeavor to create the best learning environment for highly motivated students of mathematics.

Is this going to apply to the independent study program as well?

The prices as listed now are the prices for 2005.


analytic geometry
trigonometry
geometry
geometry proposed

Take an origin of a complex plane at $P$. By smaller letters I'll denote corresponding complex coordinate of a point. Let ${{\epsilon=\cos (\frac{2\pi}{3}})+i \sin (\frac{2\pi}{3}})$. Then:

$a_2+a_1 \cdot \epsilon=0$
$a_6+a_5 \cdot \epsilon=0$
$a_4+a_3 \cdot \epsilon=0$.

$LMN$ being equilateral is equivalent to:

$\frac{1}{2}((a_2+a_3)+(a_1+a_6)\cdot \epsilon+(a_4+a_5)\cdot \epsilon ^2=0$ which is true and easy to check with help of above relations.

[quote="k2c901_1"]Let $P$ be a point on the plane. Three nonoverlapping equilateral triangles $PA_1A_2$, $PA_3A_4$, $PA_5A_6$ are constructed in a clockwise manner. The midpoints of $A_2A_3$, $A_4A_5$, $A_6A_1$ are $L$, $M$, $N$, respectively. Prove that triangle $LMN$ is equilateral.[/quote]

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=116839#p116839]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16908 post #3[/url].

  Darij

[hide="Here is a similar to Megus proof."]
[color=darkblue][b]A equivalent enunciation.[/b] Let $P$ be a point on the plane of the triangle $ABC$. Three nonoverlapping equilateral triangles $PAA'$, $PBB'$, $PCC'$ are constructed in a clockwise manner. Given are the middlepoints $L$, $M$, $N$ of the segments $A'B$, $B'C$, $C'A$ respectively. Prove that the triangle $LMN$ is equilateral.

[b]Proof.[/b] I shall use also the complex numbers. Denote $X(x)\ -$ the point $X$ with the affix $x$ and I shall choose the origin in the point $P$. Thus, $P(0)$, $A(a),\ B(b),\ C(c)$ a.s.o. Define the complex number $\omega =\cos \frac{\pi}{3}+i\sin \frac{\pi}{3}$. Therefore, $\omega ^3=-1$, $\omega ^2=\omega -1$, $\overline {\omega}=\frac{1}{\omega}=-\omega ^2$ and $a'=a\omega$, $b'=b\omega$, $c'=c\omega$ $\Longrightarrow$ $2l=b+a\omega$, $2m=c+b\omega$, $2n=a+c\omega$ $\Longrightarrow$ $2(l-m)=(b-c)+(a-b)\omega$, $2(n-m)=(a-c)+(c-b)\omega$ $\Longrightarrow$ $2(n-m)\omega=$ $(a-c)\omega +(c-b)\omega ^2=$ $(a-c)\omega +(c-b)(\omega -1)=$ $(b-c)+(a-b)\omega =$ $2(l-m)$ $\Longrightarrow$ $\boxed {\ (l-m)=(n-m)\omega\ }$ $\Longleftrightarrow$ $LM=NM$ and $m(\widehat {LMN})=60^{\circ}$, i.e. the triangle $ABC$ is equilateral.[/color][/hide]

See also here http://www.mathlinks.ro/Forum/viewtopic.php?t=45874

Here in the last post you can find a solution posted by me    using pure geometry!!!!  :lol: What do you think about this ?


[size=75]A bored student walks down a hall that contains a row of closed lockers, numbered 1 to 1024. He opens the locker numbered 1, and then alternates between skipping and opening each closed locker thereafter. When he reaches the end of the ahll, the student turns around and starts back. He opens the first closed locker he encounters and then alternates between skipping and opening each closed locker thereafter. The studen continues wandering back and forth in this manner until every locker is open. What is the number of the last locker he opens?[/size]

[hide]in the first time he walks, he opened 1,3,5,7,....1023

           second time           he opened 2,6,10,14.....

           third time               he opened 4,12,20...

           .........

           .........

             nth time                he opened 1*2^n, 3*(2^n), 5*(2^n)....

so the last one he opens will be 1024 because the only way to express it is 1*2^10. [/hide]

[color=white]hm... i seem to get 342[/color]

Beta, I think your solution is wrong, because your answer is for when he starts from the first door every time, but instead, he starts from the end where he finished. 1024 can't be the answer because he closes it the 2nd time he goes across...i think.

saturnlife, how did you get that answer?

beta, the student turns around after reaching the end and starts back, he doesn't start from locker #1 each time.

I solved it by writing everything out like this:

1st time(-->) doors left:512
{2,4,6...1024} equation: x/2 is an integer, where x is the door number.

2nd time(<--) doors left: 256
{2,6,10,14....1018,1022} equation: (x+2)/4 is integer

3rd time(-->) doors left: 128
{6, 14, 22....1014,1022} equation: (x+2)/8 is integer

4th time (<--) doors left: 64
{6, 22, 38... 1014} equation: (x+10)/16 is integer, note that you add 10 now to x because 1014, the last door remaining is 10 less than 1024.

5th time(-->) doors left: 32
{22,54,86...1014} equation: (x+10)/32 is integer

6th time(<--) doors left: 16
{22,86,150....982} equation: (x+42)/64 is integer

7th time(-->) doors left: 8
{86, 214...982} equation: (x+42)/128 is integer

8th time(<--)doors left: 4
{86,342,598, 854} equations: (x+170)/256 is integer

9th time(-->) doors left: 2
{342, 854} equation: (x+170)/512 is integer

10th and last time (<--) doors left: 1
{342}

Sorry, I misunderstood the problem( the problem seemed too easy when I was solving it)


MATHCOUNTS
summer program
PROMYS
Galois Theory

[quote="ajai"]Rationalize $\frac{1}{\sqrt{6}+\sqrt{5}+\sqrt{11}}$[/quote]

[hide]This is also $\frac{\sqrt{6}+\sqrt{5}-\sqrt{11}}{2\sqrt{30}}$
this is

$\frac{6\sqrt{5}+5\sqrt{6}-\sqrt{330}}{60}$[/hide]

Multiply the numerator and denominator by $\sqrt{6}+\sqrt{5}-\sqrt{11}$:

${\frac{\sqrt{6}+\sqrt{5}-\sqrt{11}}{2\sqrt{30}}}$

Multiply the numerator and denominator by $\sqrt{30}$:

${\frac{\sqrt{180}+\sqrt{150}-\sqrt{330}}{60}}$

Finally, simplify:

${\frac{6\sqrt{5}+5\sqrt{6}-\sqrt{330}}{60}}$

I've only seen problems with 2 irrationals in the denominator. However, the problem worked out nicely since $6+5=11$. :P  What would you do if the $\sqrt{11}$ was a $\sqrt{13}$?

[quote="vishalarul"]I've only seen problems with 2 irrationals in the denominator. However, the problem worked out nicely since $6+5=11$. :P  What would you do if the $\sqrt{11}$ was a $\sqrt{13}$?[/quote]

When you rationalize the denominators of the form $\sqrt{a}+\sqrt{b}+\sqrt{c}$ you just group them like $(\sqrt{a}+\sqrt{b})+\sqrt{c}$ and treat it like 2 irrationals. It doesn't matter what the a,b,c are.

True, but I just noted that the problems becomes extremely easy if $a+b=c$

I think you just need to keep simplifying it until you get a positive integer on the bottom, but yeah, it is more tricky with $\sqrt{13}$, then $\sqrt{11}$.

[quote="mathgeniuse^ln(x)"][quote="kunny"]How about this one?

Rationalize \[\frac{1}{\sqrt{11}+\sqrt{12}+\sqrt{13}+\sqrt{14}}\] [/quote]

[hide]Rearranging we get

$\frac{1}{(\sqrt{11}+\sqrt{14})+(\sqrt{12}+\sqrt{13})}$

Now we multiply the numerator and the denominator by the denominator's "conjugate" and simplify to get

$\frac{-\sqrt{11}-\sqrt{14}+\sqrt{12}+\sqrt{13}}{-2\sqrt{154}+2\sqrt{156}}$.

Now we multiply the numerator and the denominator by the denominator's conjugate to get
$\frac{-2\sqrt{1716}-2\sqrt{2184}+2\sqrt{1872}+2\sqrt{2028}-2\sqrt{1694}-2\sqrt{2156}+2\sqrt{1848}+2\sqrt{2002}}{8}$

This reduces to
$\frac{24\sqrt{13}+52\sqrt{3}+4\sqrt{462}+2\sqrt{2002}-4\sqrt{429}-4\sqrt{546}-22\sqrt{14}-28\sqrt{11}}{8}$

Simplifying once more we get

$\frac{12\sqrt{13}+26\sqrt{3}+2\sqrt{462}+\sqrt{2002}-2\sqrt{429}-2\sqrt{546}-11\sqrt{14}-14\sqrt{11}}{4}$

So this is our answer, i think

$\frac{12\sqrt{13}+26\sqrt{3}+2\sqrt{462}+\sqrt{2002}-2\sqrt{429}-2\sqrt{546}-11\sqrt{14}-14\sqrt{11}}{4}$[/hide][/quote]
Wow. I probably would've messed up somewhere in a problem like this.

yay, I got it right.

And yes, thank goodness that I got it right.

Who knows what i would have done.

That is a lot of radicals.

[quote="Ignite168"]I think both of these problems are way above mathcounts level.[/quote]
The first one might be like number 27 on state and the second one might be like number 29 or 30 on a nats.

No they are not way above it.

They involve thinking.

They involve the stuff you learned about rationalizing, in Algebra.

I learned nothing about rationalizing in Algebra, not even simplifying square roots. This might be covered in Algebra 2 though..

Try this on for size  :wink:

Rationalize

$\frac{1}{3+4 \sqrt[3]{2}+7 \sqrt[3]{4}}$

(Edit:  If you know the answer to this, you probably shouldn't be posting in the Mathcounts forum  :rotfl: )

I found an answer, I will post my solution if it is correct.

[hide]$\frac{86\sqrt[3]{2}-5\sqrt[3]{4}-47}{1023}$[/hide]

If I am correct, I still will be posting in the MathCounts Forum. :)

I checked on my calculator and it says I am correct.

Here is my work.

[hide]We can rewrite this fraction to be $\frac{1}{3+7\sqrt[3]{4}+2\sqrt[3]{16}}$.  Then we can factor the denominator to get

$\frac{1}{(3+\sqrt[3]{4})(1+2\sqrt[3]{4})}$.

This is also the same as

$\frac{1}{3+\sqrt[3]{4}}*\frac{1}{1+2\sqrt[3]{4}}$

We will rationalize the two fractions one by one.

Notice that $(a+b)(a^{2}-ab+b^{2})=a^{3}+b^{3}$

So this means that $\frac{1}{a+b}=\frac{a^{2}-ab+b^{2}}{a^{3}+b^{3}}$.

So from the first fraction, letting $a=3$, and $b=\sqrt[3]{4}$, we get that $\frac{1}{3+\sqrt[3]{4}}=\frac{9-3\sqrt[3]{4}+2\sqrt[3]{2}}{31}$.

From the second fraction, letting $a=1$, and $b=2\sqrt[3]{4}$, we get that $\frac{1}{1+2\sqrt[3]{4}}=\frac{8\sqrt[3]{2}-2\sqrt[3]{4}+1}{33}$.

Multiplying these two fractions and simplifying we get the answer to be

$\frac{86\sqrt[3]{2}-5\sqrt[3]{4}-47}{1023}$

Hope it makes sense.[/hide]

what if the denominator wasn't factorable.

well, don't ask me.

And I got it correct so I don't care.

Well, I think that is why torajirou gave that question, for a reason.

Are you serious?

Whatever, I still got the first one, so I don't care.

But yeah, you should have gotten it correct the first time. :lol:

But yeah, the answer seems weird, though it still is correct.

[quote="t0rajir0u"]:huh:

The denominator wasn't supposed to be factorable.  Change the $4$ to a $5$  :rotfl:[/quote]
Which 4?

Probably the one in the radical but I don't know.

But yeah, please be more specific next time.

Torajirou, if you were being serious, could you please tell us how to do this problem if you are changing the 4 to a 5?

You should start a topic in Intermediate Forums or whereever it belongs, this is MathCounts. And it also is the beginning of the year when most people are concentrating on School/Chapter.

You are correct, Ignite.

The problem that torajirou posted definitely is above MathCounts.

The intended solution is probably above Mathcounts level, I apologize, but here was the intended question (of course, I haven't checked if this is factorable):

$\frac{1}{3+5 \sqrt[3]{2}+7 \sqrt[3]{4}}$

In any case, does every problem proposed in a forum need to be immediately solvable?  The intention I had of proposing this problem is to give some of you some food for thought, not necessarily to provide extra Mathcounts practice.  Think about how you might do this without a factorization  :D

Yes, that is true.

But I still solved the question that you posed earlier, so I feel proud of my self.

You know the answer to your question posted later right, torajirou?

Seriously, I don't think this can be solved without any techniques that we can think of.

What is the solution to this question?
I don't even think torajirou has a solution to this, no offense or anything.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106616]here[/url] is the thread.

[quote="mathgeniuse^ln(x)"]I don't even think torajirou has a solution to this, no offense or anything.[/quote]

I rarely post problems I haven't already solved  :wink:  (And when I do, I add comments that make it pretty clear.)  This problem came out of a question on my Galois Theory homework at PROMYS which asked me to decide whether $\mathbb{Q}[ \sqrt[3]{2}]$ was a field or not.  In order to do so, I needed to be able to find the inverse of $a+b \sqrt[3]{2}+c \sqrt[3]{4}$, which is the motivation for this problem  :P

And yes, if you're wondering, Farenhajt's algebraic trick was the way I solved this problem the first time through.

I am sorry, for being that way.

Thank You very much for all of your help.


The expression $ 2 \plus{} \sqrt{2} \plus{} \frac{1}{2 \plus{} \sqrt{2}} \plus{} \frac{1}{\sqrt{2} \minus{} 2}$  equals:

$ \textbf{(A)}\ 2\qquad
\textbf{(B)}\ 2 \minus{} \sqrt{2}\qquad
\textbf{(C)}\ 2 \plus{} \sqrt{2}\qquad
\textbf{(D)}\ 2\sqrt{2}\qquad
\textbf{(E)}\ \frac{\sqrt{2}}{2}$


inequalities search AMC USA(J)MO USAMO function calculus

Does anyone know of any good books for beginning to olympiad-level inequalities?

The free ones I can think of are the Kedlaya packet, the Mildorf packet, and the various ones on advanced techniques like SOS and mixing varaibles. The inequality thread ( I can find it if search is being stupid) is very good for basics. Po-Ru Loh(???) has a packet of problems, also.

The way it has been presented to me:
Once again, start with AoPS 2 and the inequality thread. After that, Engel, as usual, plus the Kedlaya packet for general techniques. Then spend some time learning all the advanced stuff and get it down to intuition. Then go through the Mildorf Packet.

Also, the first USAMOs contain some very instructive inequality solving techniques. I found those very helpful, especially with convex functions and variable exponents.

Honestly though, in my opinion , just learn cauchy and AM-GM (QM,HM,etc) and do well on the AIME. Even if you make USAMO, it uses less high powered techniques and more intuition and creativity, which is why I reccomend it for some hard practice problems to chew on.

Good Luck and Best Regards,
PenguinIntegral and his superpowered liver

[quote="Ravi B"]I recommend [i]The Cauchy-Schwarz Master Class[/i] by Steele.[/quote]

but doesn't that have too much calculus though?

[quote="Ravi B"]I recommend [i]The Cauchy-Schwarz Master Class[/i] by Steele.[/quote]
I second this. While I have not personally read it, many people I have talked to (eg Profs) like it.

Almost every discrete inequality has a continuous version, and that's where calculus comes into play.  But you can skip the continuous versions.

[quote="13375P34K43V312"]I have the Kedlaya and Mildorf notes, thanks. I'll look up the Cauchy book (it is a book, right?). Also, I'll read the MAA book list more carefully to see what I can find.[/quote]
For the reference of those reading this thread, the book is available to be bought here:
[url=http://www.amazon.com/Cauchy-Schwarz-Master-Class-Introduction-Mathematical/dp/052154677X/sr=8-2/qid=1164161287/ref=pd_bbs_sr_2/104-6032667-1613563?ie=UTF8&s=books]Link[/url]

It's published by the MAA, so I'm now sure it's high quality.

[quote="Max300"]I heard that from now on the usamo will no longer have classical inequalities. So if thats what you are preparing for...it might be better to just focus on am-gm and cauchy.[/quote]
Woah?

What do you have as a source on this?

[quote="Max300"]I heard that from now on the usamo will no longer have classical inequalities. So if thats what you are preparing for...it might be better to just focus on am-gm and cauchy.[/quote]

I remember this being brought up before, but I believe that you're wrong. (I hope)

According to Arnav, people at MOP predicted in 2005 that there would be no more inequalities on the IMO, but there was one both years. I think it is true that, since problem-solving teaching has evolved so much (partly due to this site), more people are familiar with more powerful techniques, and this limits the effectiveness of many inequality problems because people will just bunch or "dumbass" with high-powered stuff like Jensen.

Yeah, everything mentioned above is good. I'd also like to add Hojoo Lee's Topics in Inequalities:

http://myhome.naver.com/hojoolee/orange.html

There's a rather large collection of Number Theory problems (no solutions though) there too.

About the inequalities disappearing thing; inequalities are a fundamental part of 'higher math'; as a LOT of work later on involves estimating

thazn1; you have 666 posts  :D


A semicircle with diameter $ d$ is contained in a square whose sides have length 8. Given the maximum value of $ d$ is $ m \minus{} \sqrt{n},$ find $ m\plus{}n.$

In the largest semicircle possible that fits in the given square, why must the center of lie on one diagonal of the square? I don't really understand aops wiki explanation of wiggle room to find the solution.

http://www.artofproblemsolving.com/Wiki/index.php/2005_AIME_I_Problems/Problem_11

The intuitive notion here is symmetry.  If the semicircle is not placed in a symmetric spot, presumably either one "side" or the other contains more "wiggle room" to expand the semicircle.

This is the intuitive notion.  That's not actually a rigorous explanation.  Once you've arrived at that diagram, however, you can show that no semicircle of greater size can fit into a square by considering the set of possible locations of the center of the semicircle.

Actually,... what is the center of the semicircle exactly? It doesn't look like the center of the circle that would be made of two of the semicircles in the picture.

Also, in the first solution given in the above link, why must drawing the perpendicular from O to AB, form a 45-45-90 triangle?

Sorry; by "center of the semicircle" I actually meant the center of the corresponding circle (which is also what the writer of the wiki solution meant).

The answer to your second question is also symmetry.  $ FE$ is perpendicular to $ AC$ in a symmetrical arrangement.


Αν $ a_1,a_2,a_3$ πραγματικοι αριθμοι τετοιοι ωστε $ a_1^2\plus{}a_2^2\plus{}a_3^2\equal{}1$ νδο

$ |a_1\cdot sin\theta\plus{}a_2\cdot cos\theta\plus{}a_3|\leq \sqrt{2}$

γνωριζω απο Β-C-S:
$ (a_1\cdot sin\theta  \plus{} a_2\cdot cos\theta \plus{} a_3)^2 \leq(a_1^2 \plus{}a_2^2 \plus{} a_3^2)(sin^2\theta \plus{} cos^2\theta \plus{} 1^2)\equal{}2$
<=>
$ |a_1\cdot sin\theta  \plus{} a_2\cdot cos\theta \plus{} a_3|\leq \sqrt{2}$

καθε απολυτο χ οταν ισχυει  $ |x|<\equal{}k$ mporei na graftei kai ws
$ \minus{}k<\equal{}x<\equal{}k$ an den kanw lathos.. ara sigoura tha eixe kai katw fragma(to $ \minus{}\sqrt{2}$) :)

Σωστός  :oops: , γράψε λάθος.

[quote="maniopas"]Λαμβάνοντας υπ' όψιν την άσκηση:
[url]http://www.mathlinks.ro/viewtopic.php?t=197898[/url]
(από εδώ νομίζω και η έμπνευση)

είναι:

$ (a_1 \cdot sin \theta \plus{} a_2 \cdot cos \theta \plus{} a_3)^2 \le (\sqrt {a_1^2 \plus{} a_2^2} \plus{} a_3)^2 \le 2(\sqrt {a_1^2 \plus{} a_2^2}^2 \plus{} a_3^2) \equal{}$

$ 2(a_1^2 \plus{} a_2^2 \plus{} a_3^2) \equal{} 2 \Leftrightarrow \minus{} \sqrt {2} \le a_1 \cdot sin \theta \plus{} a_2 \cdot cos \theta \plus{} a^3 \le \sqrt {2}$

έτσι έχουμε και κάτω φράγμα  :)[/quote]

Εχω μια μικρή αμφιβολία για αυτή τη λύση...(και για αυτήν του Athinaios).
Βασικά η αμφιβολία μου έγκειται στο ότι νομίζω ότι η λύση σας είναι μόνο για $ a_1,a_2,a_3$ θετικούς, ενώ η άσκηση αναφέρεται σε πραγματικούς (π.χ. η B-C-S που χρησιμοποιεί ο Athinaios είναι μόνο για θετικούς).

Οσον αφορά τη λύση σου Μανιέ το πρόβλημα μου είναι στην πρώτη ανισότητα (αυτή που χρησιμοποιείς και το link). Το πρόβλημα σε αυτήν είναι ότι αν και ισχύει ότι
$ |a_1 \cdot sin \theta \plus{} a_2 \cdot cos \theta|\plus{}a_3\leq \sqrt {a_1^2 \plus{} a_2^2}\plus{}a_3$ δεν σημαίνει ο΄τι θα ισχύει αυτή η σχέση αν τα 2 μέλη είναι υψωμένα στο τετράγωνο!!! γιατί για να ισχύει η ίδια σχέση όταν τα υψώνουμε στο τετράγωνο πρέπει το 1ο μέλος να έιναι θετικό πράγμα που δεν ξέρουμε σίγουρα :maybe:

Πείτε μου αν λέω κάτι λάθος

Η BCS ισχύει για όλους τους πραγματικούς   :D   δηλ αν $ a,b,c,x,y,z$ πραγματικοί αριθμοί τότε

$ (a^2\plus{}b^2\plus{}c^2)(x^2\plus{}y^2\plus{}z^2)\geq (ax\plus{}by\plus{}cz)^2$

:blush: Ωχ σωστά...μπερδεύτηκα τελείως  :oops:


To keep this post in line with the secular policies that may or may not exist on this forum, I would like to keep the focus off of the "religious aspect" of today, and focus primarily on the important events that happened on this day in the past millennia...

For example,... hrmmm....

George Washington crossed the Delaware on this day in 1776 I believe...

Apollo 8 successfully exited lunar orbit to come home to earth in 1968...

Mikhail Gorbachev resigned as president of the USSR (or was it the Soviet Union at the time) in 1991...

And who could forget when the Huygens probe landed on Titan 5 or so years ago.

Also, we must not forget that Computer Science majors always confuse Today's date with Oct. 31. :P

So merry Dec. 25th
And to all a good night

Cheers,

Hunter

P.S. Numeral system jokes anyone?

The occupation of Hong Kong began today, ummm, (2009 - 1941) years ago.

In 1990, there was the first successful trial run of the system which would become the World Wide Web. Thank god.

And ZOMG Newton was born on this day in 1642. No wonder scientists celebrate Christmas.


A function $ f: \mathbb{R}\rightarrow\mathbb{R}$ is such that $ f(0)\equal{}0,f(1)\equal{}1$, and
$ f(x^2\plus{}\frac{1}{x})\equal{}f(x)^2\plus{}f(\frac{1}{x})$ for all $ x\neq 0$.

Prove that $ f$ is not bounded from above

$ f(1)\equal{}1\Rightarrow f(2)\equal{}2$.
Assume $ f$ has an upper bound and take
$ c: \equal{}\sup\{f(x): x\in\mathbb{R}_\plus{}\}\ge 2$ and $ x_1,x_2,\ldots \in\mathbb{R}_\plus{}$ with $ f(x_n)\mathop{\longrightarrow}_{n\to\infty}c$.

Then $ c\ge f(({1\over x_n})^2\plus{}x_n)\equal{}f({1\over x_n})^2\plus{}f(x_n)$
and $ 0\le f({1\over x_n})^2\le c\minus{}f(x_n)\mathop{\longrightarrow}_{n\to\infty}0$.
So $ f({1\over x_n})\mathop{\longrightarrow}_{n\to\infty}0$
and then
$ c\ge f(x_n^2\plus{}{1\over x_n})\equal{}f(x_n)^2\plus{}f({1\over x_n})\mathop{\longrightarrow}_{n\to\infty}c^2$.

But $ c\ge c^2$ gives $ {1\over 4}\ge (c\minus{}{1\over 2})^2$ and $ {1\over 2}\ge |c\minus{}{1\over 2}|\ge c\minus{}{1\over 2}$,
which contradicts $ c\ge 2$.


calculus
integration
real analysis
function
geometry
trigonometry
absolute value

Show that the improper Riemann integral ....(integrate from 0 to 1)  f(x) dx of

f(x) = (pi/x) sin (pi/x)  if x not equals 0
      = 0                       if x = 0

exists and finite while the Lebesgue integral is infinite.

Hint: Integration by parts.

I have trouble showing both the Riemann integral exists and the lebesgue integral is infinite....can someone help me with it?

For the integration by parts: When you want to prove (conditional) convergence of the integral of an oscillating function, this method is standard. Integrate the oscillating part, differentiate whatever's left over. Eventually, you get something you can show is convergent by comparison.

For the other half: You want to show that the integral of the absolute value diverges. Set up a crude estimate of the area of each bump; you should get a divergent series. In this case, that series will be comparable to the harmonic series.

Expanding a little on jmerry's hint about the integration by parts:

$\int_t^1\frac{\pi}x\sin\left(\frac{\pi}x\right)\,dx =?$

Let $dv=\frac{\pi}{x^2}\sin\left(\frac{\pi}x\right)\,dx,\,v =\cos\left(\frac{\pi}x\right)$

$u=x,\,du=dx$

So the integral equals

$\left.x\cos\left(\frac{\pi}x\right)\right|_t^1-\int_t^x\cos\left(\frac{\pi}x\right)\,dx$

Note what happens in the first term as you let $t\to0^+$ and note that the second term is the integral of a bounded function over a bounded interval.

Does this mean that the Lebesgue integral of that does not exist? I recently began studying this type of integration, and I need some clarifications. All help is welcome.
 :D

There are only two ways for a Lebesgue integral to fail to exist. One is some problem with measurability - and worrying about that in this case would be like looking for whales in the Sahara - ain't gonna happen. The other way is what the real issue is: we call a function Lebesgue integrable when it is [b]absolutely[/b] integrable.

The analogue - and it's such a perfect analogue that it's really the same thing - is to the convergence of a series. If a series is absolutely convergent, then you may rearrange the terms at will and it will still converge, and to the same number. But a series that converges conditionally - that is, without being absolutely convergent - can be rearranged to add up to something different or even to diverge.

The very defnition of the Lebesgue integral essentially involves rearrangements of the function in question, or at the very least, not adding up the parts in any particular fixed order. It stands to reason that we would insist on insensitivity to rearrangement, and the condition that guarantees that is absolute convergence.

Look at jmerry's post; he indicates how we will be able to compare the integral of the absolute value to the harmonic series. We don't have to be particularly clever or particularly tight with our estimates to make that happen.

Note also that the thing that converges here is not the Riemann integral of this function on $[0,1].$ A prerequisite for even starting to talk about Riemann integrability is that we have a bounded function on a bounded interval. This is an unbounded function. What converges is the [b]improper[/b] integral, which is an limit of integrals of the function over subdomains, arranged in a particular order.


geometry
incenter
circumcircle
vector
geometry proposed

[quote="sam-n"]Given a non-equilateral triangle $ABC$. For any point $M$ inside of triangle $ABC$, we denote by $A^{\prime}$, $B^{\prime}$, $C^{\prime}$ the orthogonal projections of this point $M$ on the lines $BC$, $CA$, $AB$, respectively, and we define $f\left(M\right)=AC^{\prime}+BA^{\prime}+CB^{\prime}$.

Let $M$ and $N$ be two points inside of triangle $ABC$. Prove that $f\left(M\right)=f\left(N\right)$ holds if and only if the line $MN$ is parallel to the line through the incenter and the circumcenter of triangle $ABC$.[/quote]

Well, let O be the circumcenter and I the incenter of triangle ABC. We have to prove that f(M) = f(N) holds if and only if the line MN is parallel to OI.

We will use directed segments on the lines BC, CA, AB; hereby, the segments BC, CA, AB will be positive and the segments CB, AC, BA will be negative.

Let D, E and F be the midpoints of the sides BC, CA, AB, i. e. the orthogonal projections of the circumcenter O on the sides BC, CA, AB. Then, the vector $\overrightarrow{A^{\prime }D}$ is the projection of the vector $\overrightarrow{MO}$ on the line BC. Therefore, $\overrightarrow{MO}\cdot \overrightarrow{BC}=A^{\prime }D\cdot BC$, and

$A^{\prime }D=\frac{\overrightarrow{MO}\cdot \overrightarrow{BC}}{BC}=\frac{\overrightarrow{MO}\cdot \overrightarrow{BC}}{a}$.

Similarly,

$B^{\prime }E=\frac{\overrightarrow{MO}\cdot \overrightarrow{CA}}{b}\;\;\;\;\;\;\;\;\;\;\;\;\text{and}\;\;\;\;\;\;\;\;\;\;\;\;C^{\prime }F=\frac{ \overrightarrow{MO}\cdot \overrightarrow{AB}}{c}$.

Hence, we have

$A^{\prime }D+B^{\prime }E+C^{\prime }F=\overrightarrow{MO}\cdot \left( \frac{\overrightarrow{BC}}{a}+\frac{\overrightarrow{CA}}{b}+\frac{ \overrightarrow{AB}}{c}\right)$,

or simply $A^{\prime }D+B^{\prime }E+C^{\prime }F=\overrightarrow{MO}\cdot \overrightarrow{v}$ with the notation

$\overrightarrow{v}=\frac{\overrightarrow{BC}}{a}+\frac{\overrightarrow{CA}}{ b}+\frac{\overrightarrow{AB}}{c}$.

Now,

$f\left( M\right) =AC^{\prime }+BA^{\prime }+CB^{\prime }=\left( AF-C^{\prime }F\right)+\left( BD-A^{\prime }D\right)+\left( CE-B^{\prime }E\right) =\left( BD+CE+AF\right)-\left( A^{\prime }D+B^{\prime }E+C^{\prime }F\right) =\left( \frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)-\overrightarrow{MO}\cdot \overrightarrow{v}$.

Hence, it follows that f(M) = f(N) holds if and only if $\overrightarrow{MO}\cdot \overrightarrow{v}=\overrightarrow{NO}\cdot \overrightarrow{v}$, what is equivalent to $\overrightarrow{MN}\cdot \overrightarrow{v}=0$. In other words, f(M) = f(N) holds if and only if the line MN is perpendicular to the vector $\overrightarrow{v}$. Now, it remains to show that this vector $\overrightarrow{v}$ is perpendicular to OI; then, it follows that f(M) = f(N) holds if and only if the line MN is parallel to OI, and we are done.

How to prove that our vector $\overrightarrow{v}$ is perpendicular to OI ? Well, let M = I. Then, the points D, E, F are the points of tangency of the incircle with the sides BC, CA, AB. Hence, B'A = AC', C'B = BA', A'C = CB'. Thus,

$AC^{\prime }+BA^{\prime }+CB^{\prime }=\frac{1}{2}\cdot \left( B^{\prime }A+AC^{\prime }\right)+\frac{1}{2}\cdot \left( C^{\prime }B+BA^{\prime }\right)+\frac{1}{2}\cdot \left( A^{\prime }C+CB^{\prime }\right) =\frac{1}{2}\cdot \left( BA^{\prime }+A^{\prime }C\right)+\frac{1}{2}\cdot \left( CB^{\prime }+B^{\prime }A\right)+\frac{1}{2}\cdot \left( AC^{\prime }+C^{\prime }B\right) =\frac{a}{2}+\frac{b}{2}+\frac{c}{2}$.

But from what we have seen before,

$AC^{\prime }+BA^{\prime }+CB^{\prime }=\left( \frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)-\overrightarrow{IO}\cdot \overrightarrow{v}$.

Hence,

$\frac{a}{2}+\frac{b}{2}+\frac{c}{2}=\left( \frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)-\overrightarrow{IO}\cdot \overrightarrow{v}$,

so that $\overrightarrow{IO}\cdot \overrightarrow{v}=0$, and the vector $\overrightarrow{v}$ is indeed perpendicular to the line OI. Proof complete.

  Darij


geometry
3D geometry
tetrahedron
inequalities

Let $a,b,c,d$ be the areas of the triangular faces of a tetrahedron, and let $h_a, h_b, h_c, h_d$ be the corresponding altitudes of the tetrahedron. If $V$ denotes the volume of tetrahedron, prove that

\[ (a+b+c+d)(h_a+h_b+h_c+h_d) \geq 48V \]

Since the volume of a tetrahedron equals $\frac13\cdot\text{face area}\cdot\text{corresponding altitude}$, we have $V=\frac13\cdot a\cdot h_a$, so that $h_a=\frac{3V}{a}$. Similarly, $h_b=\frac{3V}{b}$, $h_c=\frac{3V}{c}$ and $h_d=\frac{3V}{d}$. Thus,

$\left(a+b+c+d\right)\left(h_a+h_b+h_c+h_d\right)=\left(a+b+c+d\right)\left(\frac{3V}{a}+\frac{3V}{b}+\frac{3V}{c}+\frac{3V}{d}\right)$
$=3V\cdot\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$.

But by the Cauchy-Schwarz inequality,

$\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\geq\left(\sqrt{a}\cdot\sqrt{\frac{1}{a}}+\sqrt{b}\cdot\sqrt{\frac{1}{b}}+\sqrt{c}\cdot\sqrt{\frac{1}{c}}+\sqrt{d}\cdot\sqrt{\frac{1}{d}}\right)^2$
$=4^2=16$.

Thus,

$\left(a+b+c+d\right)\left(h_a+h_b+h_c+h_d\right)=3V\cdot\left(a+b+c+d\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$
$\geq 3V\cdot 16=48V$,

and we are done.

The problem actually should be in Pre-Olympiad.

  darij


function
geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope

I'm interested in different (beautiful :?: ) examples of continuous functions $f~: ~[0,1] \to [0,1]$ s.t. the set of fixed points of $f \circ f$, which are not fixed by $f$ is not closed.

Let $G$ be the graph of $f$, let $D$ be the diagonal of $[0,1]\times [0,1]$, and let $G'$ be the reflection of $G$ about $D$. You are looking for $f$ such that the set $(G\cap G')\setminus D$ is not closed. Simply put, $G$ should contain a set that is symmetric about $D$ and touches it. Such as a segment with slope $-1$. Or two symmetric segments with a common endpoint on $D$. Or a circular arc with a center on $D$...


[quote="SnowStorm"]When $n$ standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of $S$.  What is the smallest possible value of $S$?[/quote]
[hide]
The largest number divisible by 6 above 1994 is 1998. The range of probabilities is 1988 to 1998. So then the corresponding number in the bottom half of the sums is $1988+(1998-1994)=\boxed{1992}$?[/hide]

I got
[hide]
333
Basically, almost any sum we use to get to 1994 will only be possible in one way, so if we use the least number of dice possible, we can get a minimum.  The minimun occurs with 332 sixes and 1 two.  Hence, the minimum number of dice is 333 and the minimal sum is when each die shows one.
[/hide]
But that's probably wrong, I dunno.

This was on the 1994 AHSME so if someone has it, could they verify the answer?

[hide]Let $X$ be the random variable equal to sum of all the scores on the $n$ dice.

Now it is easy to show that $\text{P}(X = k) = \text{P}(X = 7n - k)$ for any positive integer $k$.

So if $\text{P}(X = 1994) = \text{P}(X = S)$ then $S = 7n - 1994$ and we know that $n \geq 333$.

So $S \geq 7 \times 333 - 1994 = 337$.[/hide]


I took some intro/intermediate classes on aops,  , should I write on college apps that I took classes here?

I wouldn't.  Adcoms might not know what it is, and in my opinion, it doens't really demonstrate anything.

Depends on the college.  MIT certainly knows who we are.  Their Assistant Director does a Math Jam on our site each year.  Check the Math Jams schedule for this year's Math Jam!

I don't think including outside classes is helpful unless it shows education in a different or more mature field, or at an advanced level.

So if you have other things to demonstrate your interest and skill in mathematics, it will probably only add clutter. You want your most amazing points to shine and sparkle XD


topology
advanced fields
advanced fields solved

Show a seperable space with an uncountable subset without an accumulation point.

Define a topology on $\mathbb R$ as follows: the open sets are the empty set and all the non-empty sets which contain $0$. The set $\{0\}$ is dense :), so the space is separable, and the set $\mathbb R\setminus\{0\}$ is uncountable and has no acumulation points.

Now find such a space which is also Hausdorff.

I think $\mathbb{N}^{\mathbb{R}}$ with the box topology should work. :)


graph theory
combinatorics unsolved
combinatorics

Consider a chessboard $8 \times 8$. A [i]lazy rook[/i] can only move from a square to a vertical or a horizontal neighbour. Some two cells of the board are clipped (so that the lazy rook can't visit these cells). We consider the existance of Hamiltonian cycle of the lazy rook (that is a cyclic path of lazy rook which visits every cell of the board exacly once) on the board $8 \times 8$ with two clipping cells. Find all pairs of clipping cells so that such a cycle exists.

P.S. I don't know the full solution. I just know the obvious fact: if the clipping cells have the same color (in ordinary chess coloring) then the cycle doesn't exist. It's really obvious because the colors of consecutive cells in the cycle are different. If the clipping cells have different color then sometimes the cycle exists and sometimes doesn't. For example if the cells $a2$ and $a3$ are clipped then the cycle doesn't exist. And if the cells $a1$ and $a2$ are clipped then the cycle exists. What is the full necessary and sufficient conditions?

still I didn't find any answer but I can say a way wich probably approaches to the desired conditions:
it's just like tiling a chess board with Dominoes.As you say you can tile such a board when the clipping cells have different color.but when u want to complete your cycle u need to join the two ends of the path and it will do when ur tiling has the same conditions.


Find all triangles with integer sides and having at least one angle equal to $ 60$ degrees

That is, find all rational solutions of $ x^2 \minus{} xy \plus{} y^2 \equal{} 1$. I recommend secants from here.

Yes you are right... I was thinking of other thing, but I was completely mistaken :blush:


If $ a,b,c$ are positive reals and $ a\plus{}b\plus{}c \equal{} 1$
Prove that

$ \frac{1\plus{}a}{1\minus{}a} \plus{} \frac{1\plus{}b}{1\minus{}b} \plus{} \frac{1\plus{}c}{1\minus{}c} \le 2 ( \frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} )$

That's Japan Mathematical Olympiad Finals 2004!

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=31803[/url]

What's AMTI?


conics ellipse inequalities quadratics trigonometry calculus derivative

Two points $P,Q$  moves on an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\ (a>b>0)$ such that $\angle POQ=\frac{\pi}{2}$.

 Find the maximum and minimum value of $\frac{1}{OP}+\frac{1}{OQ}$.

Up to now I have not still found in elementary way the maximum value of  $\frac{1}{{OP}} + \frac{1}{{OQ}}$.

In order to find the minimal value of  $\frac{1}{{OP}} + \frac{1}{{OQ}}$  I will use the well know inequality between the harmonic and the quadratic mean:

[color=blue][i]If $u,v$ are two positive numbers we have[/i][/color]

$\frac{2}{{\frac{1}{u} + \frac{1}{v}}} \le \sqrt {\frac{{u^2  + v^2 }}{2}} $

[color=blue][i]with the equality if and only if  $u = v$.[/i][/color]

If $P,Q$ are two points of the ellipse $\frac{{x^2 }}{{a^2 }} + \frac{{y^2 }}{{b^2 }} = 1$ with $\measuredangle POQ = \frac{\pi }{2}$ then there is $t \in \left[ {0,2\pi } \right]$ such that

$ P\left( {a\cos t,b\sin t} \right) $

$ Q\left( {a\cos \left( {t + \frac{\pi }{2}} \right),b\sin \left( {t + \frac{\pi }{2}} \right)} \right) = \left( { - a\sin t,b\cos t} \right) $

and, consequently, we have

$ OP = \sqrt {a^2 \cos ^2 t + b^2 \sin ^2 t}  $

$ OQ = \sqrt {a^2 \sin ^2 t + b^2 \cos ^2 t}  $

The harmonic-quadratic mean inequality yields

$\frac{1}{{OP}} + \frac{1}{{OQ}} \ge \frac{2}{{\sqrt {\frac{{OP^2  + OQ^2 }}{2}} }} = 2\sqrt {\frac{2}{{a^2  + b^2 }}} $

with the equality iff

$ OP = OQ \ \ \   \Leftrightarrow \ \ \  $

$ a^2 \cos ^2 t + b^2 \sin ^2 t = a^2 \sin ^2 t + b^2 \cos ^2 t \ \ \  \Leftrightarrow \ \ \  $

$ a^2 \cos 2t - b^2 \cos 2t = 0 \ \ \   \Leftrightarrow \ \ \  $

$ \left( {a^2  - b^2 } \right)\cos 2t = 0  $

Since  $a > b$  we have that

$OP = OQ \ \ \  \Leftrightarrow \ \ \  \cos 2t = 0 \ \ \  \Leftrightarrow \ \ \  t \in \left\{ {\frac{\pi }{4},\frac{3}{4}\pi } \right\}$

Hence the minimum value of $\frac{1}{{OP}} + \frac{1}{{OQ}}\ \ $ is $\ \ 2\sqrt {\frac{2}{{a^2  + b^2 }}} $  and is obtained when

$P\left( {\frac{a}{{\sqrt 2 }},\frac{b}{{\sqrt 2 }}} \right),Q\left( { - \frac{a}{{\sqrt 2 }},\frac{b}{{\sqrt 2 }}} \right) \ \ $  or  $\ \ P\left( { - \frac{a}{{\sqrt 2 }},\frac{b}{{\sqrt 2 }}} \right),Q\left( { - \frac{a}{{\sqrt 2 }}, - \frac{b}{{\sqrt 2 }}} \right)  \blacksquare$

[b]P.S.[/b] Using the first derivative of the function

$f(t) = \frac{1}{{\sqrt {a^2 \cos ^2 t + b^2 \sin ^2 t} }} + \frac{1}{{\sqrt {a^2 \sin ^2 t + b^2 \cos ^2 t} }}$

it seems that the maximum value of  $\frac{1}{{OP}} + \frac{1}{{OQ}}$  is $f(0) = \frac{1}{a} + \frac{1}{b}$, but I'm searching for a synthetic solution !

Leon

[quote="kunny"]Regrettably you are wrong, Leon. :?

When you  set the coordinate of $P(a\cos \theta ,b\sin \theta)$, then that of $Q$  can't be set as yours.

Please try again. I'm looking forward to your solution.

kunny[/quote]

Sorry for the mistake Kunny :( . You are right, but until now I'm not successful to find the correct solution !

Leon

We are to analyze $f(x)=\sqrt{\frac{\cos^2x}{a^2}+\frac{\sin^2x}{b^2}}+\sqrt{\frac{\sin^2x}{a^2}+\frac{\cos^2x}{b^2}}$.
We have $f(x)=\frac{1}{ab}(\sqrt{ta^2+(1-t)b^2}+\sqrt{(1-t)a^2+tb^2})$, where $t=\cos^2 x$. We see that sum of radicands is constant.
Using Jensen's inequality for $\sqrt{u}$ we obtain $f(x)\leq \frac{2}{ab}\sqrt{\frac{a^2+b^2}{2}}$. And using Karamata's inequality we obtain $f(x)\geq \frac{a+b}{ab}$.

Therefore, answer: $\min=\frac{a+b}{ab}$ and $\max=\frac{\sqrt{2(a^2+b^2)}}{ab}$.

[quote="kunny"]Prove that $\frac{1}{OP^2}+\frac{1}{OQ^2}$ is constant.[/quote]
It is already done in my previous reply (it was an essential idea of the proof).


geometry
3D geometry
HMMT
number theory unsolved
number theory

[quote="BadrGhamdi"]Find all natural numbers $ n$ such that $ n$ equals the cube of the sum of its digits.

good luck[/quote]

Consider $ 10^p>n\ge 10^{p\minus{}1}$ and so $ 9p\ge S(n)$ and so $ 729p^3>S(n)^3\equal{}n$ and so $ 729p^3>10^{p\minus{}1}$ and so $ p\le 6$

So $ S(n)\le 54$ and we just have $ 54$ tests to do : $ 1^3,2^3,...,54^3$ and the result : $ \boxed{n\in\{1^3,8^3,17^3,18^3,26^3,27^3\}\equal{}\{1,512,4913,5832,17576,19683\}}$

[quote="r3d30m3j"]Note that the original question at HMMT asked for all 3-digit solutions, which results in a much shorter calculation.[/quote]

Ahhhh sure. With this constraint, we just have $ 5$ tests to do : $ 5^3,6^3,7^3,8^3,9^3$ and the unique result $ 512$


I paid for the 4-6 weeks shipping for my aops books, is there anyway I can get the books faster. ( i'm willing to pay for expediated shipping cost if I can get it in about 5 days. )  I want to start on the books during spring break.

Thanks

Assuming that your order hasn't already shipped, we can upgrade your shipping.

You'll need to call us at (619) 659-1612 so that we can reauthorize your credit card for the priority shippng charge.

I ordered with the long shipping and it came within 5 business days.


function
real analysis
real analysis unsolved

//G-delta set is the intersection of countable family of open sets.

a) If f is a real-valued founction on X,
show the set of points which f continuous is a G-delta set.

b) Show no function on [0,1] can be continuous just on rational points...


geometry
circumcircle
parallelogram
geometry unsolved

In an acute triangle $ ABC$, $ \omega$ is the circumcircle and $ O$ the center, $ \omega_{1}$ the circumcircle of triangle $ AOC$, and $ OQ$ the diameter of $ \omega_{1}$. Let $ M,N$ be points on $ AQ,AC$ respectively such that the quadrilateral $ AMBN$ is a parallelogram. Prove that $ MN,BQ$ intersect on $ \omega_{1}$.

[quote="radio"]OK. This problem is from IMO compendium, so it might be wrong.

So anyway, can you show your result?[/quote]
I have not in mind any proof of the result I mentioned and I don't know if I can prove it.

Kostas Vittas.

Thank you dear [b][size=100]Huyen Vu[/size][/b] and sorry dear all my friends, because of my blindness, I didn't saw the parallelogram $ AMBN$ and I thought that the problem was wrong. :blush:

Kostas Vittas.

[quote="vittasko"] .....
By the way, I can only see that if $ M,$ is the midpoint of the segment $ AQ,$ and $ N$ a point on $ AC$ such that $ BN\parallel AQ,$
then $ ($ by the drawing $ )$ it is true the result that the point $ P\equiv MN\cap BQ,$ lies on the circumcircle $ (\omega)$ of $ \bigtriangleup ABC.$[/quote]

We can prove this result based on the follows Lemma.

[b][size=100][color=DarkBlue]LEMMA. – A circle $ (O)$ is given with diameter $ AB$ and let $ (A)$ be, a circle centered at point $ A,$ which intersects $ (O),$ at points $ C,$ $ D.$ Through point $ B,$ we draw an arbitrary line which intersects $ (A)$ at points $ M,$ $ N$ and $ CD$ at point $ P$ $ ($ $ M,$ between $ P,$ $ B$ $ ).$ Prove that $ \frac{MB}{MN}= \frac{1}{2}\cdot \frac{PB}{PN}$[/color][/size][/b]

$ \bullet$ So in our configuration ( see the above figure [b][size=100]t=156167.gif[/size][/b]  ), we denote the points $ S\equiv AC\cap BQ,$ $ M\equiv AQ\cap NP,$ where $ P\equiv (O)\cap BQ$ and we will prove that $ M$ is the midpoint of the segment $ AQ.$

Based on the above [b][size=100]Lemma,[/size][/b] we have that $ \frac{PQ}{PB}= \frac{1}{2}\cdot \frac{SQ}{SB}$ $ ,(1)$

From $ BN\parallel AQ$ $ \Longrightarrow$ $ \frac{PQ}{PB}= \frac{MQ}{BN}$ $ ,(2)$ and $ \frac{SQ}{SB}= \frac{AQ}{BN}$ $ ,(3)$

From $ (1),$ $ (2),$ $ (3)$ $ \Longrightarrow$ $ \frac{MQ}{BN}= \frac{1}{2}\cdot \frac{AQ}{BN}$ $ \Longrightarrow$ $ AQ = 2(MQ)$ and the proof is completed.

Kostas Vittas.

PS. I will post here later, the proof of the Lemma I have in mind.

Sorry !
 I was mistaken. I uploaded the figure to Vitasko's note.

 Babis

[quote="Huyền Vũ"]$ D$ is the midoint of $ AB$
$ w1$ meets $ BQ$ at $ K$
$ KD$ meets $ AQ$ at $ M'$
we have $ O,K,B,D$ are cyclic ($ \angle OKB=\angle ODB=90$) so $ \angle DKB=\angle DOB=\angle M'AB$
then $ M',A,K,B$ are cyclic
so $ \angle M'BA=\angle M'KA=\angle DKO+\angle AKO=$$ \angle DBO+\angle OCA=\angle DAO+\angle OAC=\angle BAC$
so $ BM'//AC$
thus $ M'\equiv\ M$
We see $ MD$ meets $ AC$ at $ N$ then problem is done $ q.e.d$[/quote]

[b]FIFURE[/b]

[quote="vittasko"]I think that something is not correct in the notations, of this problem.

So, if I am not mistaken, please clarify your problem $ ($ The quadrilateral $ AMBN$ is non-convex and for the parallelogram $ AMNB,$ the problem has not a meaning $ ).$

By the way, I can only see that if $ M,$ is the midpoint of the segment $ AQ,$ and $ N$ a point on $ AC$ such that $ BN\parallel AQ,$
then $ ($ by the drawing $ )$ it is true the result that the point $ P\equiv MN\cap BQ,$ lies on the circumcircle $ (\omega)$ of $ \bigtriangleup ABC.$

Kostas Vittas.[/quote]

 Dear Kostas, I put your figure on the screen. Do you like this fifure as you sse it on the site?

 Babis


calculus
integration
trigonometry
real analysis
real analysis solved

Calculate:

Integral (from 0 to pi/2)  (( 1/(1+(tan x)^(pi/2)) , dx)

Gyan, the parentheses don't balance in your expression. Could you proofread it and balance the parentheses? In particular, what is being raised to the $\pi/2$ power?

I guess just the tangens x, elseway there won't be parenthesis around tan x.

So it becomes:

$\displaystyle \int^{\frac{\pi}2}_0 \frac1{1+(\tan x)^{\frac{\pi}2}}\cdot dx$

Yes the above  is the correct (+ much more clear) form. Thanks.

(Please put the answer in spoiler...)

The answer is [hide]pi/4. It would be that for any positive power of tan x; nothing special about the pi/2 power. Multiply numerator and denominator of the integrand by (cos x)^{whatever power you're using}. (In this case, the power is pi/2.) Then flip the interval around (i.e., let u = pi/2 - x) to see what happens.[/hide]

this type of integral has been posted before.
see [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=1360[/url]

[quote]this type of integral has been posted before. [/quote]

Yes .. I did suspect that . Its an old classic (You  have to know how to run fast   because there is a chance that once you tell your solution the person may not be amused to find that the problem was easy)...:lol:

Thanks for the link ... the other  problem shown there was fun too.

Kent - Yes the answer is, of course,  pi/4 :lol:


trigonometry
quadratics
algebra
quadratic formula

[quote="Hikaru79"][quote="felipesa"][1] find four factors for $x^8-2x^4.cos(2\alpha)+1$,$\alpha \in R$[/quote]

What is that $.$ there? Is that supposed to be a multiplication sign?[/quote]

Yes!:)A multiplication sign.

[quote="felipesa"][1] find four factors for $x^8-2x^4.cos(2\alpha)+1$,$\alpha \in R$[/quote]

[hide]
first we substitute:
\[ y^2 - 2\cos{(2\alpha)}y + 1 \]

using the quadratic formula we find \[ y = \frac{2\cos{(2\alpha)} \pm \sqrt{4\cos^2{(2\alpha)}-4}}{2}=\cos{(2\alpha)} \pm i\sin{(2\alpha)} \]

so we find the four factors by solving $x^4 = \cos{(2\alpha)} + i\sin{(2\alpha)}$

we find 8 roots though by demovires. so, 4 factors?
[/hide]


Find all functions $ f: \mathbb R\rightarrow \mathbb R$ satisfying $ f(x^2 \plus{} yf(z)) \equal{} xf(x) \plus{} zf(y)$ for every $ x,y,z\in \mathbb R$

$ f(x^2\plus{}yf(z))\equal{}xf(x)\plus{}zf(y)$

$ \bullet y\equal{}z\equal{}0 \Rightarrow f(x^2)\equal{}xf(x)$

$ \bullet x\equal{}0$ and $ y\equal{}1 \Rightarrow f(f(z))\equal{}zf(1) \Rightarrow f(f(x))\equal{}xf(1)$

$ \bullet y\equal{}0 \Rightarrow f(x^2)\equal{}xf(x)\plus{}zf(0) \Rightarrow zf(0)\equal{}0 \Rightarrow f(0)\equal{}0$

$ \bullet x\equal{}0 \Rightarrow f(yf(z))\equal{}zf(y)$

But $ f(f(x^2))\equal{}x^2f(1) \Rightarrow f(xf(x))\equal{}x^2f(1)$

$ f(yf(z))\equal{}zf(y) \Rightarrow f(xf(x))\equal{}xf(x)$

Finally $ \Rightarrow x^2f(1)\equal{}xf(x)$ so for $ x\not\equal{} 0 \Rightarrow f(x)\equal{}xf(1)$

Let $ f(1)\equal{}a \Rightarrow ax^2\plus{}a^2yz\equal{}ax^2\plus{}ayz \Rightarrow a^2\equal{}a \Rightarrow a\equal{}0$ or $ a\equal{}1$

In conclusion, $ f(x)\equal{}x$ or $ f(x)\equal{}0$

$ f(f(x))\equal{}xf(1) \Rightarrow x\equal{}1, f(f(1))\equal{}f(1)  \minus{}\minus{}\minus{}(1)$
$ f(yf(z))\equal{}zf(y) \Rightarrow y\equal{}1, z\equal{}f(1), f(f(1))\equal{}f^2(1)  \minus{}\minus{}\minus{}\minus{}\minus{}(2)$

From 1 and 2 $ f(1)\equal{}1$ or $ f(1)\equal{}0$
$ f(1)\equal{}0$ gives $ f(x)\equal{}0,\forall x\in R$

Consider $ f(1)\equal{}1$
In 2, $ y\equal{}1, f(f(z))\equal{}z \minus{}\minus{}\minus{}\minus{}(3)$

Say $ f(z_1)\equal{}f(z_2)\equal{}z_0\Rightarrow f(f(z_1))\equal{}f(z_0)\equal{}z_1, f(f(z_2))\equal{}f(z_0)\equal{}z_2 \Rightarrow z_1\equal{}z_2$

$ f(x^2)\equal{}xf(x) \Rightarrow x\rightarrow f(x) , f(f^2(x))\equal{}xf(x)$ from 3

So $ f(x^2)\equal{}f(f^2(x)) \Rightarrow f^2(x)\equal{}x^2 \Rightarrow f(x)\equal{}x$


inequalities
calculus
derivative
function
inequalities unsolved

If $ n$ is a positive real number greater than 1 and $ x$ positive real number prove or disprove the following inequality:

$ \frac{\sqrt[n]{2}x}{\sqrt[n]{x+1}}\ge\frac{(2n-1){x}^\frac{n-1}{n}-1}{2n-2}$

just a question

Generalization of what?

[url]http://www.mathlinks.ro/viewtopic.php?t=156238[/url]

See this topic and also solution on vietnamese forum (link was given by [b]zaizai-hoang[/b]) :wink:

Can anybody prove (or disprove this) :?:

Inequality stands for $ n{\geq}3$, which can be shown using calculus: first of all, find the derivative of $ f(x)=LHS-RHS$ by the variable $ x$, and put up the equation $ f'(x)=0$. It is now necessary to prove that the only nonnegative solution of this equation is $ x=0$, i.e. that $ f(x)$ is decrasing for $ x<1$ and increasing for $ x>1$. After a little bit of hard work, you can put in a substitution $ t=2x/x+1$ which can take values from $ 1$ to $ 2$ (not really a two, but as a limit). Now iti si possible to make a new function $ g(t)$ from which a derivative can be taken and shown that there is only one value of $ t$ for which the function takes the zero value, i.e. functions $ f(x)$ and $ f(t)$ don't "return" after touching the x-axis. I know that this solution is pretty confusing, but I hope you get the point... :lol:

The inequality doesn't hold for numbers $ n$ between one and two (example: take $ n=2, x=15$). It doesn't hold for some other values between 2 and 2.5, so another question arises: what is the least constant $ n$ for which inequality stands?


If $a=\frac{27}{13}$, then $3^a(\sqrt[9]{3^a})(\sqrt[3]{3^a})=?$

[hide="Hint"]
$\sqrt[z]{x^y}$=$x^\frac{y}{z}$

How can you apply this to the problem? :)
[/hide]


I have a question about the case environment on Mathlinks. If I try to make it more complex as below my formulae cannot be parsed (see quoted):

\[ f(x)\equal{}
\begin{cases}
1 & \text{2},\\
3 & \text{4}.
\end{cases}\]

[quote="ME"]
f(x)=
\begin{cases}
\frac{x}{2} + \frac{x}{g(x)} & \text{if $ x$ is even},\\
2^{\frac{x+1}{2}} & \text{if $ x$ is odd}.
\end{cases}
[/quote]

You have dollar signs which have been misinterpreted by the forum so as a workaround use \(...\) instead plus a latex space to compensate for the one that is swallowed:
\[ f(x) \equal{} \begin{cases} \frac {x}{2} \plus{} \frac {x}{g(x)} & \text{if \ \(x\) is even}, \\[1ex]
2^{\frac {x \plus{} 1}{2}} & \text{if \ \(x\) is odd}. \end{cases}\]
OR
Make more use of \text
\[ f(x) \equal{} \begin{cases} \frac {x}{2} \plus{} \frac {x}{g(x)} & \text{if } x \text{ is even}, \\[1ex]
2^{\frac {x \plus{} 1}{2}} & \text{if } x \text{ is odd}. \end{cases}\]
I have also separated the cases slightly.

By the way if you want to quote code which involves a dollar sign use the code tag
[code]f(x)=
\begin{cases}
\frac{x}{2} + \frac{x}{g(x)} & \text{if $x$ is even},\\
2^{\frac{x+1}{2}} & \text{if $x$ is odd}.
\end{cases}[/code]

Thanks, actually I tried to split it into more text components like your second option but it did not work!  :)

It should work but you will have to post what went wrong.


I am _not_ looking for a proof to this question.  Rather, I would like to know the origin of the problem (have you seen it before?  where?)

ABC equilateral with sidelength s; cevians from B and C seperating ABC into 4 polygons; two of which are a quad and a triangle with base BC.  Suppose these two have equal incircle radii,  find these radii in terms of s.

Thank you.

This is problem GC8, proposed in Paul Yiu's "Geometric Constructions III" and solved in his "Geometric Constructions V" on the website http://www.math.fau.edu/yiu/Geometry.html . A reference is given to

H. Fukagawa, D. Pedoe, [i]Japanese Temple Geometry Problems[/i], Wennipeg 1989.

  Darij


quadratics
geometry
geometric transformation
reflection
trigonometry
conics
parabola

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=301536]Refer here for the full set of problems[/url]. Note, there are two forms (some problems appear in only one).

- A quadratic $ f$ with real coefficients has root $ 2\plus{}3i$. Reflect $ f$ over $ y\equal{}6$, shift it two units left, and three units up to obtain $ f^\prime$. Compute the sum of the roots of $ f^\prime(2x)$.

- Consider a thirteen-sided polygon with vertices that are all lattice points when graphed in the Cartesian plane. Determine its minimal possible area.

- Find the sum of all positive two-digit integers $ x$ such that $ (x\plus{}2)(x^2\plus{}7x\plus{}13)$ is divisible by 9.

- Right triangle $ ABC$ with right angle $ C$ has the property that one of the trisection points $ D$ of segment $ AB$ is the foot of the altitude from $ C$. If $ BD<AD$, then find $ \sin {\angle CAB}$.

- An incredibly large population of $ 2009^{2009^{2009}}$ amoebas splits into groups. Every group has 42 amoebas, except for one with only

$ a. 7 b. 14 c. 21 d. 28 e. 35$

- Call a number modular if it is not divisible by any prime except for 5, 7, and/or 11 (e.g. 175 and 385 but not 363). What is the sum of all reciprocals of modular numbers?

- A spaceship is at $ (0,0,0)$. Every time it teleports, it moves to a lattice point one unit away. Let $ m$ be the number of paths the spaceship can take so that it takes 12 teleportations to reach $ (4,4,4)$, while avoiding the black holes at $ (1,1,1)$ and $ (3,3,3)$. Compute the sum of the digits of $ m$.

Problem:

"Consider a thirteen-sided polygon with vertices that are all
lattice points when graphed in the Cartesian plane.
Determine its minimal possible area."

My work and an answer are hidden here:
[hide]Let me use Pick's Theorem for this.

Area = $ \frac12b \plus{} i \minus{} 1$ where b is the number of boundary points,
and i is the number of interior points.

A 13-sided polygon has 13 vertices.  If the 13 vertices can be placed in
the xy-plane in at least one orientation so that i) there are no interior
points and ii) the vertices of the polygon form the boundary points
exclusively, then we will have the best, or minimal area, which is:

$ \frac12(13) \plus{} 0 \minus{} 1 \equal{} 6.5 \minus{} 1 \equal{} 5.5$

$ 5.5$ square units minimal area . . . my answer

An actual construction of a specific instance could be to plot the following
13 points as vertices:

A = (1,0), B = (0,1), C = (1,1), D = (0,2), E = (0,3), F = (1,2), G = (1,3),
H = (2,3), I = (2,2), J = (3,3), K = (3,2), L = (2,1), M = (3,1)

Then draw in the 13 line segments as follows:
AB, BC, CD, DE, EF, FG, GH, HI, IJ, JK, KL, LM, MA

If you can see that this example (when plotted) meets the conditions
of (i) and (ii) from above, then it should be convincing enough.
[/hide]

Hints:

- A quadratic $ f$ with real coefficients has root $ 2 \plus{} 3i$. Reflect $ f$ over $ y \equal{} 6$, shift it two units left, and three units up to obtain $ f^\prime$. Compute the sum of the roots of $ f^\prime(2x)$. [b]Consider a graphical representation of $ f$. [/b]

- Find the sum of all positive two-digit integers $ x$ such that $ (x \plus{} 2)(x^2 \plus{} 7x \plus{} 13)$ is divisible by 9. [b]Expand the polynomial.[/b]

Quadratic one:

This is almost certainly not the best solution, but it gets it done.

[hide="1."]Because of the complex conjugates theorem, $ 2 \minus{} 3i$ must also be a root. Multiplying out, we get $ f$ to be $ a(x^{2} \minus{} 4x \plus{} 13) \equal{} a(x \minus{} 2)^{2} \plus{} 9$. Because only $ \frac {b}{a}$ matters where the end equation is $ ax^{2} \plus{} bx \plus{} c$, we can eliminate the $ 9$ as it serves no purpose. Now, because we don't care about that extra added term, we can just reflect the parabola about $ y \equal{} 0$ and get the same result. So we have $ \minus{} a(x \minus{} 2)^{2}$ now. We don't need to care about moving up, because that doesn't matter, so moving two units left, we get $ \minus{} a(x)^{2}$. Substituting 2x into the equation gives the sum of the roots as 0.[/hide]

[hide="3"]
There are three ways that their product can be divisible by 9:

Case 1 $ (x \plus{} 2)$ is divisible by 9:
Obviously, all numbers 7 mod 9 work here.

Case 2 $ (x^{2} \plus{} 7x \plus{} 13)$ is divisible by 9
Trying out all 9 numbers, this doesn't work.

Case 3 they both are divisible by 3:
We find that x= 1 and 4 mod 9 work.

Going back to case one, the lowest possible number is $ 16$, while the highest is $ 97$. Summing this using the addition equation gives $ 565$. Now the 1 and 4 mod 9 parts are simply 3 and 6 less per number than the 7 mod 9 part. Because there are 10 numbers in that part, the final answer is $ 565*3 \minus{} 30 \minus{} 60 \equal{} 1605$.[/hide]

@the last one: wasn't there also a black hole at $ (2,2,2)$?

[quote="Ihatepie"]Quadratic one:

This is almost certainly not the best solution, but it gets it done.

[hide="1."]Because of the complex conjugates theorem, $ 2 \minus{} 3i$ must also be a root. Multiplying out, we get $ f$ to be $ a(x^{2} \minus{} 4x \plus{} 13) \equal{} a(x \minus{} 2)^{2} \plus{} 9$. Because only $ \frac {b}{a}$ matters where the end equation is $ ax^{2} \plus{} bx \plus{} c$, we can eliminate the $ 9$ as it serves no purpose. Now, because we don't care about that extra added term, we can just reflect the parabola about $ y \equal{} 0$ and get the same result. So we have $ \minus{} a(x \minus{} 2)^{2}$ now. We don't need to care about moving up, because that doesn't matter, so moving two units left, we get $ \minus{} a(x)^{2}$. Substituting 2x into the equation gives the sum of the roots as 0.[/hide]

[b]My [/b] [hide="Solution"]
Consider a graphical representation of $ f$. It's a parabola with vertex $ x\equal{}2$. After all these shifts (most of which don't do anything), the vertex ends up at $ x\equal{}0$, making $ f^\prime$ have two opposite roots $ \implies \boxed{0}$.
[/hide]

[hide="3"]
There are three ways that their product can be divisible by 9:

Case 1 $ (x \plus{} 2)$ is divisible by 9:
Obviously, all numbers 7 mod 9 work here.

Case 2 $ (x^{2} \plus{} 7x \plus{} 13)$ is divisible by 9
Trying out all 9 numbers, this doesn't work.

Case 3 they both are divisible by 3:
We find that x= 1 and 4 mod 9 work.

Going back to case one, the lowest possible number is $ 16$, while the highest is $ 97$. Summing this using the addition equation gives $ 565$. Now the 1 and 4 mod 9 parts are simply 3 and 6 less per number than the 7 mod 9 part. Because there are 10 numbers in that part, the final answer is $ 565*3 \minus{} 30 \minus{} 60 \equal{} 1605$.[/hide]

[hide="My Solution"]
The expression is just $ (x\plus{}3)^3\minus{}1$, so we just need to find cubes that have 1 mod 9. The above solution solves for that.
[/hide]

@the last one: wasn't there also a black hole at $ (2,2,2)$?
[b]Look at the harder problems. This one has less computation involved. [/b][/quote]

[hide="easier way to do #3"]
If $ x\equiv 0 \pmod{3}$, then both $ x\plus{}2$ and $ x^2\plus{}7x\plus{}13$ are not divisible by 3, doesn't work.
If $ x\equiv 1 \pmod{3}$, then both $ x\plus{}2$ and $ x^2\plus{}7x\plus{}13$ are divisible by 3, works.
If $ x\equiv 2 \pmod{3}$, both are not divisible by 3.
So it is just the 2 digit integers that are 1 mod 3, or $ 10\plus{}13\plus{}\cdots\plus{}97\equal{}\frac{10\plus{}97}{2}\times 30\equal{}1605$.
[/hide]

[hide="4"]
Let AD=2x and BD=x
Because ADC and CDB are similar, we have AD/DC=CD/DB, or CD^2=AD*DB=2x^2
Then AC^2=AD^2+CD^2=4x^2+2x^2=6x^2, and sinCAB=CD/AC=(sqrt(2)*x)/(sqrt(6)*x)=1/sqrt(3)[/hide]

[hide="5"]
This problem is finding $ 2009^{2009^{2009}} \pmod{42}$. Because $ 2009^{2009^{2009}}$ is divisible by 7, it is equal to 0 mod 7. Also, $ 2009^{2009^{2009}}\equiv (\minus{}1)^{2009^{2009}}\equal{}\minus{}1 \pmod{6}$. Using CRT, it is guaranteed that the only possible remainder is 35.[/hide]

Is there a nicer way to do 6 other than writing out all of the infinite series?
Also, for 7 I did PIE bash, is there a nicer solution than that?

[hide="Another way for 3"]$ (x \plus{} 2)(x^2 \plus{} 7x \plus{} 13)\equiv(x \plus{} 2)(x^2 \minus{} 2x \plus{} 4) \equal{} x^3 \plus{} 8\equiv0\pmod9$

$ \Longleftrightarrow x^3\equiv1\pmod9\Longleftrightarrow x\equiv1\pmod3.$[/hide]

For number 7 I have an attempt. I think my idea is right but I think I might have messed up calculations.

we have to move in the x, y, and z direction 4 times each. If the problem did not involve the black holes, the solution would be the number of distinct combinations of 4 x's, 4 y's and 4 z's, for example $ xxxyzzxyyyzz$ would be a combo, so this answer would be $ \frac{12!}{4!4!4!} \equal{} 34650$. Since we have a black hole at $ (1,1,1)$ and $ (3,3,3)$, we must subtract the number of routes that pass through $ (1,1,1)$ and/or $ (3,3,3)$. To get to the point $ (1,1,1)$ there are $ 3!$ ways. From $ (1,1,1)$ to $ (4,4,4)$ there are $ \frac{9!}{3!3!3!}$ ways, making $ (3!)(\frac{9!}{3!3!3!})\equal{}10080$ ways to from $ (0,0,0)$ to $ (4,4,4)$ passing through $ (1,1,1)$. From symmetry, there are $ 10080$ ways to go from $ (0,0,0)$ to $ (4,4,4)$ passing through $ (3,3,3)$ as well. If we subtract $ 34650\minus{}10080\minus{}10080$, we are overcounting the routes that pass through both $ (1,1,1)$ and $ (3,3,3)$, so we must add those back in. There are $ 3!$ ways to go from $ (0,0,0)$ to $ (1,1,1)$, $ \frac{6!}{2!2!2!}$ ways to go from $ (1,1,1)$ to $ (3,3,3)$, and $ 3!$ ways to go from $ (3,3,3)$ to $ (4,4,4)$, a total of $ (3!)(3!)(\frac{6!}{2!2!2!})\equal{}3240$ ways to go from $ (0,0,0)$ to $ (4,4,4)$ passing through both black holes.

Adding back in we get Total Routes = $ 34650\minus{}10080\minus{}10080\plus{}3240\equal{}17730$ ways

Elefont, that is the correct solution. Well done!

You can extend his work to the even more annoying PIE question in the Harder Problems section.

Does Elefont's solution account for the black hole at (2,2,2)? If not, can we just use the same method?

Pretty much, just gotta calculate the number of ways to go from $ (0,0,0)$ to $ (4,4,4)$ through $ (2,2,2)$.
And calling these three points A, B, and C, and #(N1, N2...) the number of ways to get from the origin to the destination through the points N1, N2, etc...
We use PIE: Total - #(A) - #(B) -#(C) + #(A,B) + #(B,C) + #(A,C) - #(A,B,C)

[quote="Caelestor"]- Call a number modular if it is not divisible by any prime except for 5, 7, and/or 11 (e.g. 175 and 385 but not 363). What is the sum of all reciprocals of modular numbers?[/quote]
Nobody seems to have answered this one, so here:

We note that the reciprocals of every such integer can be represented in the form
\[ \left(1 \plus{} \frac {1}{5} \plus{} \frac {1}{5^2} \plus{} \frac {1}{5^3} \plus{} \dots \right)\left(1 \plus{} \frac {1}{7} \plus{} \frac {1}{7^2} \plus{} \frac {1}{7^3} \plus{} \dots \right)\left(1 \plus{} \frac {1}{11} \plus{} \frac {1}{11^2} \plus{} \frac {1}{11^3} \plus{} \dots \right).\]
Computing the sum of each of them individually, we get that that thing is equal to
\[ (5/4)(7/6)(11/10) \equal{} 77/48.\]
ycz9000: it is.


[quote="Xaenir"]
\[ (a \plus{} 1)^7 (b \plus{} 1)^7 (c \plus{} 1)^7 > 7^4 a^4 b^4 c^4
\]
[/quote]

Do you mean $ a,b,c$ positive? If not , then we can choose negative $ a,b,c$ to make your inequality wrong.

If you mean $ a,b,c$ positive, then

$ (a\plus{}1)^7\equal{}a^7\plus{}7a^6\plus{}21a^5\plus{}35a^4\plus{}35a^3\plus{}21a^2\plus{}7a\plus{}1>(a^7\plus{}a)\plus{}7(a^6\plus{}a^2)\plus{}21(a^5\plus{}a^3)\plus{}35a^4 \ge 93a^4$

So $ (a \plus{} 1)^7 (b \plus{} 1)^7 (c \plus{} 1)^7 >93^3 a^4 b^4 c^4>7^4 a^4 b^4 c^4$

Sorry, my bad ,of course $ a,b,c>0$ :lol:
Any other approach except brute force? :wink:


number theory
prime numbers
prime factorization
AMC 8

[quote="jhredsox"][quote="tridecalogism_lover"]How many factors of 3003 are prime numbers?

***************************
Is there some sort of shortcut where you can find it out?[/quote]

well, just do prime factorization and you get[hide]4[/hide] :)

-jorian[/quote]

~ to jorian~
You made an error....it said [b]prime[/b] numbers..thanks for trying though!..thank you davidli, that seems to be correct.

All you have to do is do a prime factor tree. The numbers you get is 3, 7, 11 13. So there and 4 prime factors os 3003.

[quote="tridecalogism_lover"]oh wait,  never mind...i didn't see that you had listed the [b]number[/b] of factors :blush:[/quote]

it's ok

-jorian

[quote="Ubemaya"]Memorize that 1001 = 11 * 13 * 7[/quote]
Also you should know that 111 = 3 * 37 (for other problems...)

[quote="perfect628"]Yeah that came up on a nat. sprint. i almost forgot that 1001 was composite... It's a little late, but for questons with 1003, remember it's 17*59-that came up for me a lot last year.(2006)[/quote]

Yeah, countdown this year, abcdabcd, means 10001 is a factor.

It's nice to memorize [hide="that"]1001=7*11*13. So we have [b]4[/b] prime factors: 3, 7, 11, and 13.[/hide]

@bpms: yes, 223 is prime.

[quote="jhredsox"][quote="13375P34K43V312"][quote="Ubemaya"]Memorize that 1001 = 11 * 13 * 7[/quote]
Also know that abc * 1001=abcabc, where abc is a 3 digit number. a can be zero. Make sure you know why this works. Also,

ab*101=abab
abc*1001=abcabc
abcd*10001=abcdabcd
abcde*100001=abcdeabcde

I think you see where I'm going with this.[/quote]

the 101 one was on amc 8 i believe

-jorian[/quote]
Yes, it was. That was a pretty tricky question, I spent like five minutes on it, and then I'm like Doh, ABA=101

[quote="bpms"][quote="perfect628"]Yeah that came up on a nat. sprint. i almost forgot that 1001 was composite... It's a little late, but for questons with 1003, remember it's 17*59-that came up for me a lot last year.(2006)[/quote] this year it's going to be important that $2007=3^{2}\cdot 223$ 223 is prime I think.[/quote]
Yes, 223 is prime

[quote="bpms"][/quote][quote="jhredsox"][quote="13375P34K43V312"][quote="Ubemaya"]Memorize that 1001 = 11 * 13 * 7[/quote]
Also know that abc * 1001=abcabc, where abc is a 3 digit number. a can be zero. Make sure you know why this works. Also,

ab*101=abab
abc*1001=abcabc
abcd*10001=abcdabcd
abcde*100001=abcdeabcde

I think you see where I'm going with this.[/quote]

the 101 one was on amc 8 i believe

-jorian[/quote][quote="bpms"]
Yes, it was. That was a pretty tricky question, I spent like five minutes on it, and then I'm like Doh, ABA=101[/quote]

OH! I remember that one!! It was pretty simple when you figured that out. :)

[quote="Ubemaya"]What was the problem?

I was too noobish to take the AMC 8.[/quote]

Huh? Noobish? :huh:

I don't remember it exactly, but it was something like:

ABA*CD=CDCD

And I think you had to find the sum of A and B. Like I said, I don't remember exactly so it may have been different, but the concept was the same.

[quote="Ubemaya"]What was the problem?

I was too noobish to take the AMC 8.[/quote]

1st of all: you are not "noobish"

2nd: the problem was:

Find the value of A if ABA*CD=CDCD

i don't remember the choices....
this was the only problem that stumped me for a while
i felt that i could kill myself after i found how obvious the answer is.


Show that if $k$ is a positive integer and $x_1,x_2,...,x_n$ are positive real numbers wich sum to $1$,then

$\sum \dfrac {1-x_i^k}{x_i^k} \leq (n^k-1)^n$

Hello,

Are you sure you got the right wording ? What happens if x is arbitrarily close to 0 ?

The initial inequality is manifestly false (think $x_1$ very close to $0$). May be the inequality is $\ \sum \dfrac {1-x_i^k}{x_i^k} \geq (n^k-1)n$, and that's trivial because $\dfrac 1 {x_i^k}$ is convexe .
:cool:


$m=n=1.$
For example, if $m>n>1$, and $p$  a prime divisor of $n$ we get that $n!$ and $m!$ contain a $p-1$ in their factorizations, and if $p\neq 2,5$, $2^{m!}+5^{n!}\equiv 2(\mod p)$, from Fermat. So we must have $p$ either $2$ or $5$, which obviously is not. The same goes further for the remaining $2^{m!}+1$.


geometry
trigonometry
area of a triangle
trig identities
Law of Cosines

1. The area of a triangle is 10 cm squared. Two of the sides are of lengths 5 cm and 7 cm. What are the possible lengths of the third side?

2. A ship if 6.7 km from a lighthouse at a bearing of 308 degrees. Another ship is 8.3 km from the lighthouse at a bearing of 74 degrees. Calculate the distance from between the two ships and the degree measure in bearings from the first ship to the second ship.

EDIT:Sorry! Posted in wrong place.. :blush:

[hide="Problem 1"]Let $C$ be the common angle of the two given sides.  By the area formula $A=\frac{ab\sin C}{2}$, we have $10=\frac{5\cdot 7\sin C}{2}$.  Therefore, $\sin C=\frac{4}{7}$.  By the Pythagorean Identity, $\cos C=\pm\sqrt{1-\sin^{2}C}=\pm\sqrt{1-\frac{16}{49}}=\pm\sqrt{\frac{33}{49}}=\pm\frac{\sqrt{33}}{7}$.  By the Law of Cosines, $c^{2}=a^{2}+b^{2}-2ab\cos C$.  Substituting the values we know, we have $c^{2}=25+49\pm 2\cdot 5\cdot 7\cdot\frac{\sqrt{33}}{7}=\boxed{74\pm 10\sqrt{33}}$.[/hide]

Edited.

[quote="matt276eagles"]By the Pythagorean Identity, $\cos C=\sqrt{1-\sin^{2}C}=\sqrt{1-\frac{16}{49}}=\sqrt{\frac{33}{49}}=\frac{\sqrt{33}}{7}$.[/quote]

this line is incorrect; you are missing one of the solutions  ;)

Should it be $\cos x=\pm\sqrt{1-\sin^{2}x}$?


A school has $n$ students and $k$ classes. Every two students in the same class are friends. For each two different classes, there are two people from these classes that are not friends. Prove that we can divide students into $n-k+1$ parts taht students in each part are not friends.

For a clique with $m$ vertices, we define its value as $m-1$.

Reformulation: If all vertices of a simple graph $G$ can be partitioned into $k$ sets so that for any two sets there are two vertices joined by an edge each in one of them, then they can also be partitioned into cliques and the total value (sum of values) of all cliques is at least $k-1$.

Proof: Use induction on $k$. Pick a set $S$ arbitrarily. By induction hypothesis, the subgraph composed of $V(G)\backslash S$ can be partitioned into cliques $\{K_i\}_{i \ge 1}$ with their total value at least $k-2$. Firstly, consider each vertex in $S$ as a clique for use. We deal with two cases:

Case 1: No $V(K_i)$ is completely covered by neighbors of $S$ in $V(G)\backslash S$ (denoted by $N(S)$). Then, each $K_i$'s value is at least $|V(K_i) \cap N(S)|$. Therefore, the total value of $\{K_i\}$ again is at least $|N(S)| \ge k-1$.

Case 2: There is a $V(K_i)$ completely covered by $N(S)$. Then we can decompose this clique into small cliques and attach each small clique with its unique neighbor in $S$ respectively, forming new cliques. As a result, the total value of cliques in the whole graph $G$ increases by at least 1.

Cool problem!

Let $G$ be the obvious graph corresponding to the students in the school. Let's label the classes $C_1, C_2, C_3, \cdots, C_k$ in an arbitrary manner.

[b] Claim. [/b] For each $2 \le i \le k$, there is a way to select empty (no friendships) vertex-induced subgraphs of $G$ satisfying the following properties.

1. Each of the subgraphs contains at least two vertices.
2. Each of the subgraphs contains no vertices not in $C_1 \cup C_2 \cup \cdots \cup C_i$
3. If we consider the graph $G'$ on $i$ vertices where we connect $a$ and $b$ iff $C_a, C_b$ are in the same subgraph at least once, then $G'$ is connected.
4. The vertex sets of the subgraphs are disjoint.
5. The sums of the sizes of the subgraphs minus one is equal to $i-1$.

[b] Proof. [/b] The proof goes by induction on $i.$ When $i = 2$, this is obvious.

Suppose it holds for $i = t.$ That is, there exist complete subgraphs $G_1, G_2, G_3, \cdots, G_x$ of $G$ which satisfy the four conditions above. Consider the graph $H$ with $x$ vertices labelled $1, 2, \cdots, x$ where $a, b$ are connected iff there is some $1 \le r \le i$ so that $G_a, G_b$ both contain a vertex of $C_r.$ This graph must be a tree (left as exercise to the reader). We consider a few cases.

[b] Case 1. [/b] There is some vertex $v \in C_{t+1}$ and $w \in C_j$ for some $1 \le j \le t$ so that $w$ is not in any of the $G_i$'s and $v, w$ are not acquainted.

Then, simply add the empty subgraph induced by the vertices $v, w.$

[b] Case 2. [/b] There is not such a vertex.

Consider vertices from each of the sets $C_1, C_2, C_3, \cdots, C_t$ which don't know some vertex in $C_{t+1}.$ By assumption, these vertices must all lie in $G_i$'s. For a $1 \le j \le t$ which has vertices from multiple $G_i$'s, we will say that its [b] leader [/b] is the $G_a$ for which the common vertex of $G_a, C_j$ isn't connected to some vertex in $C_{t+1}.$ The other $G_i$'s in $C_j$ will then be said to [b] bow down to [/b] $G_a.$ Consider the "bow down" digraph. It is clearly acyclic, and so there must exist a $G_a$ which bows down to no other $G_j.$ Now, let $V$ be the set of vertices in $C_{t+1}$ which don't know somebody in $G_a.$ By assumption, the set of non-neighbors of $V$ covers $G_a$ completely. Since $G_a$ is empty, we can now easily partition $G_a \cup V$ into empty subgraphs, each of which has exactly one vertex of $V.$ Then, the new size minus one is $|G_a| - 1$ rather than $|G_a|$, so condition 5 is now satisfied. Clearly, conditions 1, 2, 4 are satisfied. It remains only to check that condition $3$ is satisfied. However this is obvious. We essentially just broke up $G_a$ and then glued them together with $V$.

As we've considered all cases, the induction is complete.

$\blacksquare$

Apply the claim when $i = k$. Now, consider splitting the vertices of the graph into the subgraphs given by the claim and singletons. This partition clearly works.

$\square$

We consider a graph $G$ where two vertices are connected iff they are not friends. We are given that vertices of $G$ can be partitioned into $k$ groups $H_1,H_2,\ldots,H_k$ such that there is an edge between any two groups. Our problem is equivalent to showing that $G$ can be partitioned into some cliques $\{C_i\}_{i \ge 1}$ such that
$$ \sum_{i \ge 1} (|C_i| - 1) \ge k-1 $$
We solve the problem using induction on $k$. Base case $k=1,2$ are direct. We perform the inductive step now. WLOG assume that between any two groups precisely one edge is there and there are no edges between the same group (as we can delete other edges otherwise).

[b]Case 1: [/b] There is a group $H$ such that at it has $\ge 2$ non-isolated vertices.

Let $v_1,v_2,\ldots,v_l$ ($l \ge 2$) be the non-isolated vertices of $H$. For each $1 \le i \le l$, let $S_i$ be the set of groups $H_j$ such that $v_i$ is connected to some vertex in $H_j$. Consider the graphs
$$ G_i = v_i \cup S_i \qquad \forall ~ 1 \le i \le l$$
These graphs form a partition of $G$. For each $i$, by induction hypothesis we can divide $G_i$ into some cliques $\{D_j\}_{j \ge 1}$ such that
$$ \sum_{j \ge 1} (|D_i| - 1) \ge |S_i| $$
Combing all such cliques $D_i$ in the main graph $G$ we get
$$\sum_{i \ge 1} (|C_i| - 1| \ge \sum_{i=1}^l |S_i| = k -1$$
So we are done.

[b]Case 2:[/b] Each group contains only $1$ non-isolated vertex.

Then there is $k$-clique in $G$, which of course suffices.

Here is a smoothing approach. Let $\overline{G}$ be the complement of $G$

WLOG there exists exactly one edge from $V_i$ to $V_j$ in $\overline{G}$, for if there are more than one we can delete the additional edges.

Let $a_{i,j}$ denote the vertex in $V_i$ adjacent to $a_{j,i}$. We have if $i,j,k$ are pairwise distinct and $a_{i,j}\ne a_{i,k}$, then $N(a_{i,j}) \cap N(a_{i,k})= \emptyset$ by the above assumption.

Suppose $a_{i,j}\ne a_{i,k}$. Consider $\overline{G'}$ be the graph where $a_{i,k}$ has no neighbours, while $a_{i,j}$ has neighbours $N(a_{i,j}) \cap N(a_{i,k})= \emptyset$ (where $N$ is the neighbourhood in $\overline{G}$). Consider a coloring of $G'$, then let $C$ be the set of vertices that are colored the same color as $a_{i,j}$ and call the color 1. This means in $\overline{G'}$ they are a clique. Also, call the color of $a_{i,k}$ 2.

We can color $G$ as follows: color the vertices in $(C\cap N(a_{i,j}))\cup a_{i,j}$ with color 1 and vertices in $(C\cap N(a_{i,k})) \cup a_{i,k}$ with color 2, and all other vertices in $G$ with the same color as that of $G'$, which must work.

It follows $\chi(G)\le \chi(G')$. Repeatedly doing this step gives $\overline{G'}=K_k$, so $\chi(G)\ge \chi(G')=n-k+1$, as needed.


let $ x_n\equal{}sin(n) ,  n\ge\ 0$

is $ (x_n)_n$ a convergent or divergent row?

[hide="it is not hard"]Let $ x_n\equal{}\sin n \to x$ then $ \sin (n\pm 1) \to x$
From $ \sin (n\pm 1)\equal{}\sin n \cos 1 \pm \cos n \sin 1$ it follows that $ y_n\equal{}\cos n$ convergents too, say $ y_n \to y$
Now from the system

$ x\cos 1 \plus{} y \sin 1\equal{}0$
$ x\cos 1 \minus{} y \sin 1\equal{}0$

we have $ x\equal{}y\equal{}0$, but on the other hand from $ x_n^2\plus{}y_n^2\equal{}1$ we have $ x^2\plus{}y^2\equal{}1$[/hide]

[quote="GeorgeCraciun"]is $ (x_n)_n$ a convergent or divergent row?[/quote]
What does this mean (specifically "row")?

wouldn't it be divergent since sin(n) is periodic?

i guess i'm not understanding the problem

In this problem you have to prove that the sequence $ x_n = \sin n$ is convergent....that means that it has a finite limit.

[hide="see here"]Supose that $ x_n$ is convergent $ \implies$ $ \sin n _ > l$, $ l\in R$
In particular $ \sin(n + 1)$, $ \sin(n - 1)$,$ \sin 2n$ have the same limit $ l$.
$ \|\begin{array}{cccc}\sin(n + 1) - \sin(n - 1) = 2\sin 1 \cos n \implies \cos 1 - > 0 \\
 \\
\sin 2n = 2\sin n \cos n \implies sin 2n - > 0\implies \sin n - > 0\end{array}$
But of course, it is known that $ \sin^2 n + \cos^2 n = 1$ and then [color=brown][b]![/b][/color][/hide]

[hide="An evident typo in my post"]
The system would be read as follows

$ x\cos 1 \plus{} y \sin 1 \equal{} x$
$ x\cos 1 \minus{}  y \sin 1 \equal{} x$

[/hide]

In fact the more strong statement is true.

$ (\forall a\in [\minus{}1;1])(\forall\varepsilon >0)(\exists n\in \mathbb N)(|\sin n \minus{} a|<\varepsilon)$

In other words the sequence $ \sin n$ contains subsequences convergenting to any $ a\in [\minus{}1;1]$


Let A be the set of all sequences whose terms are 0 or 1. Thus, a typical element of A would be {a_1, a_2, .. , a_n, ..} where each an is a 0 or 1. Show that A is not countable.

Okay suppose that the set is countable

List the sets as decimals:

$ 0.a_{1}a_{2}a_{3}...$
$ 0.b_{1}b_{2}b_{3}...$
$ 0.c_{1}c_{2}c_{3}...$
$ ...$

construct another element;

$ 0.z_{1}z_{2}z_{3}...$

where $ z_{1}\ne a_{1}$
$ z_{2}\ne b_{2}$
$ z_{3}\ne c_{3}$
...

Hence $ z$ is not counted since it has one digit different than every number in the list. I.e. there is always one set that is not counted. Hence the set is not countable.

Or we could biject from $ A$ to binary representations of $ \mathbb{R}$, letting the first bit denote sign.

Well, I just used cantor's argument...so that is really the same thing. In a sense, you are citing a proof that I provided.


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Show that a connected metric space having more than one point is uncountable.

Search: metric uncountable
Fifth one down.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1171461129&t=162201]here[/url]


Sorry. I forgot one coefficient. I edited it.

Is any one aware of when the equality holds? It's very interesting.

Yes, I can. It follows from a lemma in my coming book.

So don't say something marketing your book (before it is published) like that. If you feel free to post your lemma in your coming book, please post here, and thank you very much.  If not, stop spam!  :cleaning: Let me say that here is polite Mathlinks.


[quote="stvs_f"]let $ p,q,c \in R$ .find that Maximum of b that :
\[ p^2 \plus{} q^2 \plus{} 1 \geq bp(q \plus{} 1)
\]
[/quote]
If $ p\equal{}\sqrt {2}, q\equal{}1$, we have $ 4\ge 2\sqrt{2}b \Leftrightarrow b\le \sqrt{2}$.
and by AM-GM,
$ p^2\plus{}q^2\plus{}1\equal{}(\frac{p^2}{2}\plus{}q^2)\plus{}(\frac{p^2}{2}\plus{}1)\ge\sqrt{2}p(q\plus{}1)(\ge bp(q\plus{}1))$
So we have $ b\le \sqrt{2}$, Maximum of $ b\equal{}\sqrt{2}$.

[quote="NextPeace"][quote="stvs_f"]let $ p,q,c \in R$ .find that Maximum of b that :
\[ p^2 \plus{} q^2 \plus{} 1 \geq bp(q \plus{} 1)
\]
[/quote]
If $ p \equal{} \sqrt {2}, q \equal{} 1$, we have $ 4\ge 2\sqrt {2}b \Leftrightarrow b\le \sqrt {2}$.
and by AM-GM,
$ p^2 \plus{} q^2 \plus{} 1 \equal{} (\frac {p^2}{2} \plus{} q^2) \plus{} (\frac {p^2}{2} \plus{} 1)\ge\sqrt {2}p(q \plus{} 1)(\ge bp(q \plus{} 1))$
So we have $ b\le \sqrt {2}$, Maximum of $ b \equal{} \sqrt {2}$.[/quote]
How do you find $ p \equal{} \sqrt {2}$ and $ q \equal{} 1$?


Suppose that  $ S\equal{} \{1,2,3,...,n \}$and that $ A_1,A_2,...,A_k$ are subsets of $ S$ such that for every $ 1 \le i_1, i_2,i_3,i_4 \le k$, we have
$ \mid A_{i_1} \cup A_{i_2} \cup A_{i_3} \cup A_{i_4}\mid \le n\minus{}2$ Prove that
$ k \le 2^{n\minus{}2}$

Isn't this obvious ? $ 2^{n\minus{}2}$ gives all the subsets of $ \{1,2,\dots,n\minus{}2\}$. From now on any subset chosen invariably contains $ n$ or $ n\minus{}1$. Apart from these $ 2^{n\minus{}2}$ subsets, if we choose subsets of $ \{1,2,\dots,n\}$ including $ n\minus{}1$ or $ n$, we can always find at least one $ 4$-subset tuple whose union contains more than $ n\minus{}2$ elements.

[quote="I hate rating"]Isn't this obvious ? $ 2^{n \minus{} 2}$ gives all the subsets of $ \{1,2,\dots,n \minus{} 2\}$. From now on any subset chosen invariably contains $ n$ or $ n \minus{} 1$. Apart from these $ 2^{n \minus{} 2}$ subsets, if we choose subsets of $ \{1,2,\dots,n\}$ including $ n \minus{} 1$ or $ n$, we can always find at least one $ 4$-subset tuple whose union contains more than $ n \minus{} 2$ elements.[/quote]

But $ A_1,A_2,...,A_n$are arbitrary subsets. So,it is possible that just some subsets of $ \{1,2,\dots,n \minus{} 2\}$ can be $ A_i$ (sorry my bad english)

Let us partition the power set of $ \{1,\dots,n\}$ into $ P_1$ and $ P_2$. $ P_1$ does not contain a subset that includes $ n$ or $ n\minus{}1$. In the trivial case when we select all subsets of $ \{1,\dots,n\minus{}2\}$.It is clear that we cannot have more than $ 2^{n\minus{}2}$ elements. Let condition $ C$ be: for every $ 1 \le i_1, i_2,i_3,i_4 \le k$ ,$ \mid A_{i_1} \cup A_{i_2} \cup A_{i_3} \cup A_{i_4}\mid \le n\minus{}2$.

Let us select a few subsets $ S_1$ from $ P_1$ say $ k_1$ of them and a maximum number of subsets $ S_2$ from $ P_2$ say $ k_2$ so that $ C$ is satisfied( maximum means selecting any other subset from $ P_2$ shall override $ C$) . We shall prove that $ k_1\plus{}k_2\leq 2^{n\minus{}2}$.

If $ k_1\geq 3$ we select a ordered triple $ (a_i,a_j,a_m)$ so that $ |a_i\cup a_j\cup a_m|\equal{}max|a_i\cup a_j\cup a_m| ,\forall 1\leq i<j<m\leq k,\quad a_i,a_j,a_m \in S_1$ (Analogous cases for $ k_1\equal{}1$ and $ k_1\equal{}2$ can be similarly handled). We shall select a maximum number of subsets from $ P_2$ so that $ | a_i\cup a_j\cup a_m\cup b_q|\leq n\minus{}2$ where $ b_q\in S_2$. Let $ a_i\cup a_j\cup a_m\equal{} \{1,\dots,k\}$ wlog. $ b_q$ does not contain atleast two elements from $ \{k\plus{}1,k\plus{}2,\dots,n\}$ but does contain $ n\minus{}1$ or $ n$. Hence $ P_2$ can contain at maximum of $ 2^{n\minus{}k\minus{}2}$. $ S_1$ can contain a maximum of $ 2^k$ elements. We have $ 2^k \plus{}2^{n\minus{}k\minus{}2}\leq 2^{n\minus{}2}$ and equality occurring only at $ k\equal{}4$.

[quote="I hate rating"]$ S_1$ can contain a maximum of $ 2^k$ elements.[/quote]

Why so?

@ Petrov:
On message #4, fourth line from below I wrote "Let $ a_i\cup a_j\cup a_m\equal{}\{1,\dots,k\}$ wlog". Our objective is this - try creating most large $ S_1$ and $ S_2$ and prove that we still have $ |S_1|\plus{}|S_2|\leq 2^{n\minus{}2}$. We have ensured largest $ S_2$.The largest $ S_1$ is obtained when $ \exists$ exactly one  $ s'\in S_1$ so that $ s'\equal{}\cup s\quad \forall s\in S_1$ and $ S_1$ is the powerset of $ s'$.

[quote="I hate rating"]@ Petrov:
The largest $ S_1$ is obtained when $ \exists$ exactly one  $ s'\in S_1$ so that $ s' \equal{} \cup s\quad \forall s\in S_1$ and $ S_1$ is the powerset of $ s'$.[/quote]

But the union of all sets of $ S_1$ may contain more then $ k$ elements, yes? The union of any three sets in $ S_1$ contains at most $ k$, but the union of all of them contains more then $ k$ --- it is quite possible. What am I missing?

@ Petrov: Well, it is the other way.
We construct $ S_1$ for every $ k\leq n \minus{} 1$ so that $ |\cup s'| \equal{} k,\forall s'\in S_1$. In general,we may or may not have a three sub sets whose union gives $ k$ elements. But if we are looking at maximizing $ |S_1|$ we have to have a subset with all $ k$ elements and hence a union of some three subsets give $ k$ elements.

It is clear now ?

No, it is not clear at all. Do you use that the union of no [i]four[/i] subsets contains at least $ n\minus{}1$ element? Or you claim to prove the stronger statement, in which four is replaced by two?

[quote="Fedor Petrov"] Or you claim to prove the stronger statement, in which four is replaced by two?[/quote]

Yes, 4 can be replaced by 2.

Then it fails for $ n\equal{}4$: take all subsets with at most 1 element. There are 5 such subsets, and $ 5>2^{4\minus{}2}$.

Let us suppose you take all the subsets of $ \{1,2,3,4\}$ of size $ \leq 1$. They are $ \{1\},\{2\},\{3\},\{4\},\emptyset$. we cannot use this as a example as we have $ |\{1\}\cup\{2\}\cup\{3\}\cup\{4\}|\equal{}4$ which is not $ \leq n\minus{}2\equal{}2$

But you claim that it holds with 2 subsets instead 4

May be this failure for $ n\equal{}4$ is just because we have equality as I have pointed in the last line of #4 ?


Solve the simultaneous equations
ab + c + d = 3; bc + d + a = 5; cd + a + b = 2; da + b + c = 6;
where a; b; c; d are real numbers.

The unique solution is $a=\frac43$, $b=-2$, $c=\frac23$, $d=7$.

Let $a=1+p$, $b=1+q$, $c=1+r$, $d=1+s$. The first equation becomes $1+pq+p+q+r+s=3$; we have similarly $1+qr+p+q+r+s=5$, $1+rs+p+q+r+s=2$, and $1+sp+p+q+r+s=6$.
Subtracting the first and second, we have $q(r-p)=2$; the third and fourth give $s(p-r)=4$, so $s=-2q$. Subtracting the first and fourth equations gives $p(s-q)=3$; the second and third give $r(q-s)=3$, so $r=-p$.
Now $p+q+r+s=-q$, so $pq-q=2$ and $-pq-q=4$; adding and subtracting gives $q=-3$ and $pq=-1$, so $p=\frac13$. Substituting back gives the solution above.

I don't think the above solution is correct. I get the unique solution of $(2,0,0,3)$.

Subtracting the first equation from the second, we get $(b-1)(c-a)=2$.
Subtracting the third from the fourth, we get $(1-d)(c-a)=4$.
From this, we find that $d=3-2b$.

Similarly, by subtracting the first equation from the fourth, and the third from the second, we get $c=2-a$.

Substituting this into the first equation and factorising, we get $(a-2)(b-1)$=0.

If $a=2$, by solving, we get the solution $(2,0,0,3)$.
If $b=1$, $d=1$, but from equations $1$ and $2$, we find that $a+c=3$ and $a+c=5$ - a contradiction. Hence, we have just the one solution.


1) Outline a synthesis of 2-acetylnaphthalene from naphthalene.

2) Starting from 2-acetylnaphthalene, propose a synthesis of [i]Dehydroabietic acid[/i] (structure below).

1. Friedel crafts acylation with $ \ce{CH_3 COCl}$ in presence of a polar solvent like nitrobenzene so that substitution occurs at the 2nd position.

for the second, u cud probably start by stobbe condensation. followed by cyclization to get a tetralone like compound with one extra benzene ring(dunno how to name it). :oops:

Show your reasoning with structures. And "probably" is not good enough.

sorry :blush: . i just gave the start,not the total synthesis, and drawing structures is really boring. but ill try to explain. 2 acetyl naphthalene + stobbe condensation gives a lactone, which on hydrolysis gives an acid. which on treating with H2SO4 gives a compound which resembles tetralone, just that it has an extra benzene ring. and a methyl group. i dunno how to carry the rest of the synthesis. :blush:
p.s: how many steps?

u can start off with wittigs reaction with to get a methylene group and reduction to get isopropyl group
2). treat with conc.h2so4/120c and fuesed naoh to get beta naphthol, and birch reduction reduction so that only one benzene ring is reducd.
thinking abt the rest of the synthesis. :huh:

Very good VARUNARASU. But you must explain the selectivity of the sulphonation step.

under thermodynamic control, naphthalene shud be sulphonated at beta position not alpha, coz of steric hindrance(1,8 peri interaction) with the hydrogen on the other benzene ring. dats also the reason why alpha naphthol is more acidic than beta naphthol.

Yes, sulphonation should occur on the other ring. But in what position and why?

in position 2 for the same reason as posted above. bcoz of the 1,8 peri interaction between the hydrogen of the adj ring the so3h of the ring. so3h is a bulky group and sulphonation is reversible, and the product got under thermodynamic control, is the one whcich eq. favors. so................ :huh:

Yes that's why suggested acylation in polar medium so that 2-acyl derivative is more favoured. :)

continuing the rest of the synthesis:
Treat with lda and ch3br.
Then treat with nabh4 +conc.h2so4 to get alkene
Then tret with 1.eq. seo2, allylic hydroxylation. To get oh at sec.carbon. (reactivity towards se02- sec>pri)
Now heat with butadiene (diels alder)
Treat with PCC to oxidize oh to ketone.
Now treat with lda and PhSeBr foll by r2culi
Now again treat lda and PhSeBr foll by ch3cn/naoh/h2o.
Finalyy mozingo reduction, and barbier weiland degradation

Will draw the structures if u want but wasn’t too sure of my synthesis, dats why I dint draw any.

Structures please: otherwise I will not even read that.


The sum of three consecutive odd integers is 102.  What are the three integers?

hint:

an odd + an odd + an odd = 2n-1 + 2k-1 + 2l-1 = 2(n+k+l) - 3

you could make this an equation

(x) + (x+1) + (x+2)= 102

combine like terms:

3x + 3 = 102
   -3           -3

3x = 99
factor out a 3:

(x=33) 3 (cancel the 3)

x = 33

now plug them in:

33 + (33+1) + (33+2)

33+34+35=102
102=102

33,34,35 are the consecutive numbers

gkomudi, the problem says consecutive odd numbers :)

[hide="click me if you are really desperate to find out"]it's impossible, see Barnacle's post #2[/hide]

i can't find an answer, so i guess it is impossible. :?:

[hide]For one thing, odd + odd + odd is odd, so there must be no solution since 102 is even.

Anyway,
n + n + 2 + n + 4 = 102
3n = 96
n = 32
32 isn't odd![/hide]

okay, i'm really bad at making problems. :P


Is this question weird or am I just doing it wrong?

The revenue generated by a new product can be modeled as $ R(x) \equal{} \minus{} 5x^2 \plus{} 21x$.  The cost function is $ C(x) \equal{} 4x \plus{} 14$.  How many items are required to maximize the profit?  What is the maximum profit?

Ok, so...
Let P(x) represent the profit function.

P(x) = R(x) - C(x)
$ P(x) \equal{} \minus{} 5x^2 \plus{} 21x \minus{} 4x \minus{} 14$
Complete the square...
$ \minus{} 5 (x \minus{} 17/10)^2 \plus{} 9/20$

But since you can't have 17/10 of an item the maximum would be at 2 items... generating a profit of $ \$$0...

Thanks.

this question seems a bit weird to me, as i did it your way too...

$ P(x) \equal{} R(x)\minus{}C(x)$ as stated above, and $ P(x) \equal{} \minus{}5x^{2}\plus{}21x\minus{}4x\minus{}14 \equal{} \minus{}5x^{2} \plus{} 17x \minus{}14$

Then $ P'(x) \equal{} \minus{}10x \plus{}17,   P'(x) \equal{} 0 \equal{} \minus{}10x \plus{}17$ at $ x\equal{}\frac{17}{10}$.

$ P(1) \equal{} \minus{}5 \plus{} 21 \minus{} 4 \minus{} 14 \equal{} \minus{}2$

$ P(2) \equal{} \minus{}20 \plus{} 42 \minus{} 8 \minus{} 14 \equal{} 0$


Very easy challenge! Everyone should be able to solve it!

1) What is the probability that a two digit number will contain exactly one?

2) On a regular dice(only one dice) what is the odds of rolling a multiple of 2?

 :D  :D  :D  :D  :D  :lol:  ;)  :)

Here's what I think number one means lucy:
What's the probability that a randomly selected two digit number will contain exactly one 1? Express your answer as a common fraction (I had to add in that Mathcounts-style touch :lol: )

[hide]THere are exactly 19 numbers from 1 to 99 that contain a 1, but you have to subtract 2 b/c 11 and 1 don't count, as one contains two ones and the other isn't two digit. Since there are 90 two digit numbers, the answer is $\frac{17}{90}$.[/hide]

If the problem asked: What's the probability that a randomly selected two digit number will contain a 1? , then it be...

[hide]There are, again 19 numbers that contain a 1, but you have to subtract out 1 b/c 1 doesn't work. That makes the fraction 18/90 which equals $1/5$.[/hide]

Notice the difference between the different wordings. For anyone taking ANY competition, make sure you watch the wording, b/c test writers will always try to trick you... Esp. in mathcounts...

[quote="LuCy4EvA"]I don't get number 1, but here's number 2

[hide]The multiples of $2$ is: $2,4,6.... $ so the odd of rolling a multiple of $2$ is  ${\frac33}={1}$[/hide][/quote]

How did you get one??? 3 multiples of 2, out of 6 possibilities. 1/2 right?? Maybe im being an idiot and missing something easy.

Um.... on number one, I don't think she finished her question, it just said how many two digit number contain exactly one.... but it doesn't exactly one what.... or am I just not getting it?

for number 2, it's asking for the odd, which is like yes/no.... or the possibility of a thing happening against it not happening.... happening - 3... not happening -3... so it's 3/3.... or did I get the thing fliped over... is it yes/no or no/yes?

Lucy is correct on this one. Odds(a) is Favorable Outcomes/Unfavorable Outcomes. Therefore, 1/1 is correct. If the question asked for probability, then the answer is 1/2, which I think is what the writer meant. Remember that Odds and Probability are DIFFERENT!!!

yup that was the little trick! thats why every time i get a probability or odds question i read it twice to make sure

An even easier way of seeing the answer to the dice question is that each die has a 1/2 chance of being even or odd.  The last roll could switch the overall sum between even and odd with equal probability.

My bad....So i was being an idiot and making a dumb mistake :lol:  :blush:

IT'S OKAY, THE ONLY REASON THAT I NOTICED THAT WAS BECAUSE WE LEARNED ABOUT THE DIFFERENCE BETWEEN ODDS AND PROBABILITY IN SCHOOL THAT DAY...  :D

MCrawford's approach is absolutely correct... I just wanted to explain out my logic... anyways for clarity purposes, here are your formulas:

$Odds(E)=\frac{P{E}}{P(\overline{E})}$
$Probability(E)=\frac{P(E)}{P{(E)}+P(\overline{E})}=P(E)/Total$ $Outcomes$
Where $P(\overline{E})$ = $1-P(E)$

sry! forgot to put the one at the bak
there i corrected it!
Very easy challenge! Everyone should be able to solve it!

1) What is the probability that a two digit number will contain exactly one 1?

2) On a regular dice(only one dice) what is the odds of rolling a multiple of 2?


Ring Theory
algebra
function
domain
absolute value
superior algebra
superior algebra solved

let us consider the rings $\mathbb{Z}[\frac{1+\sqrt{d}}{2}]$ where d is bigger in absolute value than 1
and $d = 1$ mod $4$ and d squarefree
suppose the ring is a unique factorization domain

prove that $2$ is irreducible if and only if $d=5$ mod $8$

I know that this is a weird question, with allowing d to be positive or negative, and all.. still any help would be really helpful, especially for the real case

If $d \equiv 1 \mod 8$:
Let $\mathfrak{o}$ be the ring we work in.
$(2\mathfrak{o}+\frac{1+\sqrt d}{2}\mathfrak{o})(2\mathfrak{o}+\frac{1-\sqrt d}{2}\mathfrak{o})=2\mathfrak{o}$, so $2$ is a nontrivial product.

If $d \equiv 5\mod 8$:
When $2=xy$ is a nontrivial product, you get that $||x|| = \pm 2$, but both give simple contradictions $\mod 8$.

$o$ is the ring we work in , or an element? :?
that factorisation of 2 might be the key to my problem

could you please elaborate?

As I said: $\mathfrak{o}$ is the whole ring (see the second line ;) ).
We know that the ideal generated by $2$ is not prime (it's a nontrivial product of ideals), so $2$ cannot be irreducible.

oooo, products of ideals, I am afraid I am one of those guys who can only work in principal ideals domain
Is there a factorisation of the element 2 now?

Ooh, and does this argument hold too, we have UFD,
if 2 is irreducible, as it divides $\frac{1-d}{4}$ it should divide $\frac{1+\sqrt{d}}{2}$ or its conjugate which it does not? :huh:

Ok, that works too and is simpler (you even don't need UFD) :)

But the reason why I came up with the ideals is the unique decomposition into prime ideals, where this can easily be generalised.