function LaTeX integration logarithms trigonometry calculus derivative

Fiind the primitive of ln(1+sqrt(3)*tg x) supposing this function is well defined.

In $\LaTeX$:
\[\int \ln{(1+\sqrt3\cdot\tan{ x})}dx \]

[quote="color"]In $\text{\LaTeX}$:
\[\int \ln{(1+\sqrt3\cdot\tan{ x})}dx \]
[/quote]
Use this trick:
$1+\sqrt{3}\tan{x}$=$2\frac{(\frac{1}{2}\cos{x}+\frac{\sqrt{3}}{2}\sin{x})}{\cos{x}}$=
$\frac{2\cos{(x-\frac{\pi}{3})}}{\cos{x}}$and therefore:$\ln{(1+\sqrt3\cdot\tan{ x})}=\ln{2}+\ln{\cos{(x-\frac{\pi}{3})}-\ln\cos{x}}$
and integral will be equal to
$x\ln{2}+C$

[quote="Kent Merryfield"][quote="Extremal"]and integral will be equal to
$x\ln{2}+C$[/quote]
Which would imply, by taking the derivative, that $\ln(1+\sqrt{3}\tan x)=\ln2$ for all $x.$

Since I don't believe that is true, that means I don't believe your answer, either.[/quote]
Yes you are right.
Qe must consider few case: to find primitive for
$\int\ln\frac{\cos(x-\frac{\pi}{3})}{\cos{x}}dx+x\ln{2}$

But $\int \ln(\cos x)\,dx$ isn't expressable as a finite sum of elementary functions.

Just try mathematica. Who said the primitive is to be simple?  :maybe:

Yes. There must be some interval. because there not exist primitive


AMC
USA(J)MO
USAMO
geometry
trigonometry
AoPS Books
analytic geometry

You mean the AoPS books? For the AMC 12? Late AoPS I/Early II, I think.

Aren't the chapters in the AoPS books arranged by topic and not difficulty? So how can you say late AoPS I and early AoPS II?

erm... Art of Problem Solving... you're on the website for it...

 :rotfl:

click "bookstore" on the toolbar on the left of your screen; it's the first link there. You'll be able to buy the AoPS books there.

why don't u just do all of it? u have a whole year...

yeah reading aops books 1 and 2 thoroughly = really good understanding of all amc material before usamo
good understanding = pwns at amc10/12/aime

[quote="davidyko"]Nevertheless, it's undisputable that the beginning of the book is easier than the end of the book. But I suppose you have a point.[/quote]
Not necessarily.
In volume two, I found chapter 2 much harder than chapters 24, 23, 20, 19, 18, 17, 16, 15, 14, 13, and 12.
Chapter 2 was one of the hardest ones for me.

[quote="davidyko"]I understand about all the other ones, but you found the Analytic Geo easier than chapter 2?
But yeah, never mind, chapter 2 is a pretty hard chapter.[/quote]
I didn't find chapter 2 that hard because we had pretty much done all of it in my precalc class. The hardest for me were the analytic geometry ones.

Yeah, I started chapter 2 before we covered all that trig in Precalc, so it was really confusing. I'm rereading it now, and it makes a lot more sense, so I'm finishing up the problems I couldn't do at the end of the chapter. It's rather easy now.


algorithm
continued fraction
number theory theorems
number theory

Does anyone know how to get (by hand) the periodic continued fraction form for square roots? for example $ \sqrt{2}$ = [1;2,2...] etc.
I can't find a non-calculator method anywhere.

[url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion]This algorithm[/url] is extremely easy to run by hand in tabular format, but it's difficult to describe without a diagram.


Let $ A$ = (0, 0), $ B$ = (1, 2), $ C$ = (6, 2), $ D$ = (2, -1), and $ E$ = (1, -3). Show that angle CAB is congruent to angle EAD.

Note:  I haven't learned trig yet, so I'd appreciate a solution not requiring trig.

Thanks.

Well, i don't really care much for doing the tedious algebra, but this is the process. So you take lines $ AB$ and $ AC$ and choose arbitrary points $ B'$, $ C'$ ($ B'$ on $ AB$ and $ C'$ on $ AC$) such that $ AB'$=$ AC'$ and $ \angle B'AC'$ is acute. Do the same thing with $ EA$ and $ AD$ but defining points $ E'$ and $ D'$ on $ AE$ and $ AD$. Then triangles $ C'AB'$ and $ E'AD'$ are congruent if and only if $ \angle CAB \equal{} \angle EAD$. To prove this, you just prove that $ \frac {B'C'}{AB'} \equal{} \frac {D'E'}{AD'}$.


function
calculus
integration
trigonometry
real analysis
real analysis unsolved

if we have a differential equation of the form of y'/y=f(x) the solution is simple as it is the exp of F(x) but if we have y''/y=f is there a direct expression to find y in function of f or its integrals

If $f(x)$ is a constant call it $k^{2}$, then we have $y''-k^{2}y = 0$,
so that if $y' = p$, then we have $p\frac{dp}{dy}-k^{2}y =0$.
Then you have $p dp = k^{2}ydy$. Then integration yields, $y' = \frac{dy}{dx}= p = \pm k\sqrt{a^{2}-y^{2}}$. And integrating this one more time should give $y$ as linear combination of sine and cosine.


The NY state MathCounts competition was held last week at RPI in upstate NY.  Below are the top 10 schools.  (Because this is the NYC forum, I have listed the schools from NYC in bold.)

 1. Home Educators Enrichment Group, Capital District
 2. Calkins Road MS, Monroe County
 [b]3. Salk School of Science, Manhattan[/b]
 4. Twelve Corners MS, Monroe County
 5. Bethlehem MS, Capital District
 [b]6. Hunter College HS, Manhattan[/b]
 [b]7. George J. Ryan MS 216, Queens[/b]
 8. La Grange MS, Dutchess County
 9. Ardsley MS, Westchester
[b]10. Saint Ann's, Brooklyn[/b]

The coach of HEEG, Mary O'Keeffe, will be the coach of the NY team at the National competition in Fort Worth, Texas.  Congratulations, Mary!

Below are the top 21 students.

1. Dan Mendelsohn, Progressive School of LI, Nassau County
2. Matthew Babbitt, HEEG, Capital District
3. Felix Sun, Acadia MS, Capital District
4. Schuyler Smith, HEEG, Capital District
5. Thomas M., Spry MS, Monroe
[b]6. Kevin P., Hawthorne MS 74, Queens[/b]
[b]7. Bohao Z., Hunter, Manhattan[/b]
[b]8. Taylor S., Salk, Manhattan[/b]
9. Sameer F., Union Vale MS, Dutchess
10. Kelby S., Calkins Road MS, Monroe
11. Andrew Y., Calkins Road MS, Monroe
12. Jay W., HEEG, Capital District
[b]13. Stephen C., Hunter, Manhattan[/b]
[b]14. Meena B., Hunter, Manhattan[/b]
[b]15. Yichi Z., Salk, Manhattan[/b]
[b]16. Mason Z., Edward Bleeker JHS 185, Queens[/b]
17. Jacob L., Albert Leonard MS, Westchester
18. Joseph M., Dumb Ox Club (Home School), Saratoga
19. Max L., Herricks MS, Nassau
[b]20. Dong R., Marie Curie MS 158, Queens[/b]
[b]21. Genghis C., Hunter, Manhattan[/b]

The top 4 students will form the NY team at the National competition in May.  Congratulations to Dan, Matthew, Felix, and Schuyler!

As top boy and girl, Dan M. and Meena B. each received a 1000 dollar scholarship to RPI.

GOOD JOB MEENA

good job everybody else too

For completeness, here are the schools that finished 11th to 20th:

[b]11. Riverdale, Bronx[/b]
[b]12. Lab School, Manhattan[/b]
13. Great Hollow MS, Suffolk
14. Wheatley, Nassau
[b]15. Marie Curie MS 158, Queens[/b]
16. Rye MS, Westchester
17. Farnsworth MS, Capital District
18. Herricks MS, Nassau
[b]19. Mark Twain IS 239, Brooklyn[/b]
20. Orville Todd MS, Dutchess County


[quote="silouan"]Υπάρχει συνάρτηση  $f: R \mapsto R$  έτσι ώστε  $f(f(x))=-x^3$.[/quote]

H $f$ είναι 1-1 και επι.
επισης $f(-x^3)=-f^3(x)$             $(*)$

Mετα θέτουμε διαδοχικά στην $(*)$
$x=1$ και $x=-1$ και παίζουμε με τις προκύπτουσες και την αρχική  $f(f(x))=-x^3$

είμαι μέσα??

 :)

[quote="silouan"]Μέσα είσαι αλλά τι βρήκες υπάρχει συνάρτηση ή όχι ?[/quote]

οχι.
Εσυ?

[quote="silouan"]Ακριβώς δεν υπάρχει .Σκεφτόμουν για το εξής. Αν $n>0$ είναι φυσικός αριθμός υπάρχει συνάρτηση τέτοια ώστε $f(f(x))=-x^{2n+1}$  ?

Επίσης για $n=0$  είναι γνωστό ότι υπάρχουν συναρτήσεις(μη συνεχείς) ώστε

$f(f(x))=-x$.  και νομίζω ότι όλες τους είναι κλαδικές.[/quote]

καλη σκέψη. τη συζητάμε!
 :)

Σωστά!

 Αν  $f(f(x)) = - x$ και η $f$  ήταν συνεχής , τότε , ως  1-1 , θα ήταν γνησίως μονότονη. Αλλά τότε η

     $f(f(x))$ , δηλαδή η $- x$ θα ήταν γνησίως αύξουσα  , άτοπο.

  Άρα μόνο μη συνεχείς συναρτήσεις αυτού του είδους υπάρχουν.
 Τώρα , [b]Σιλουανέ [/b], υπάρχουν και πολλές συναρτήσεις , μη συνεχείς  , που δεν είναι  πολλαπλού τύπου( ακέραιο μέρος , κλασματικό μέρος κλπ).Έχεις βρει καμία συνάρτηση που να επαληθεύει τη σχέση αυτή ;Εγώ δεν έχω προσπαθήσει.

[u] Μπάμπης[/u]

Ναι έχω βρει . Βρήκα πρώτα μία που δεν ήταν συνεχής στο μηδέν και μου δημιουργούσε προβλήματα (δείτε πρώτη σελίδα attachment).Μετά βρήκα μια αρκετά πολύπλοκη (όχι εντελώς μόνος μου(δεν είμαι και εξπέρ στην ανάλυση)) που είναι και συνεχείς στο μηδέν(δείτε δεύτερη σελίδα στο attachment .

BTW, έχω βρει κάποιες ωραίες ιδιότητες σε αυτή τη συνάρτηση κάποιες μπορεί να είναι και γνωστές. Αν θέλετε μπορώ να τις βάλλω αν σας ενδιαφέρουν(απευθύνομαι σε όλα τα μέλη του φόρουμ)


I have a pretty basic question could someone please explain to me how to figure out unknown exponents with LOGs
easy example: 2^x = Log...
This one is pretty basic but if someone could give me a general overview of how it works that would be really good
thank you very much
Magic Man

If $ 2^{x} \equal{} 64$, for example, then this has a logarithmic form $ \log_{2}{64}$, which equals $ 6$ (which is the exponent $ x$ in $ 2^{x} \equal{} 64$ that gives you $ 64$).

A similar example is $ 3^{x} \equal{} 99$. To find $ x$ we write $ \log_{3}{99} \approx 4.18...$, and $ 3^{4.18} \approx 99$.

More generally:

$ a^{x} \equal{} b \leftrightarrow \log_{a}{b} \equal{} x$.

But as far as I'm aware this is only for $ b > 0, b \neq 1$.

This obviously isn't All There Is To Know About Logarithms. There's various other identities to be aware of that arise due to the logarithm going hand in hand with exponentation. I won't list them all here, despite having kind of used them in this post, because I won't cover all the exceptions properly in all likelihood and I don't know in what context you're learning this, but I expect that this topic is covered adequately in your textbook, or, alternatively, [url=http://en.wikipedia.org/wiki/Logarithm]the Wikipedia article[/url].

Logarithm answers to question - "what is the exponent we need to use?"

there's also a few log properties
$ \log_{a}{b}\plus{}\log_{a}{c}\equal{}\log_{a}{bc}$
$ \log_{a}{b}\minus{}\log_{a}{c}\equal{}\log_{a}{b/c}$
$ \log_{a}{b^{x}}\equal{}x\log_{a}{b}$
$ \log_{a}{b}\equal{}\log_{10}{b}/\log_{10}{a}$

[quote="kh-"]
But as far as I'm aware this is only for $ b \neq 1$.
[/quote]
If $ b\equal{}1$ then the answer is $ 0$. Just a small error  :lol: .

Oops: I meant $ a > 0, a \neq 1$, and I guess $ b \geq 0$. I was confusing $ b$ with 'base', which in my example it isn't. I'll fix that.
[b]EDIT:[/b] No I won't, because I can't. But good catch regardless.


solve for N: (7!8!9!)/7!+8!+9!)=8!N!/9

no calculator allowed

correct. in specific, you factor out a [hide]7! from the top and bottom[/hide]

pretty simple, just manipulate to see that N! = 8.  key step is multiply both sidse by 7! to help reduce stuff.

I don't know why, but it took me a while. Maybe it was because of the stupid mistake I made in the middle?  :?:


Please help out a non-Canadian with the following:

What is in the water in Ontario such that the vast majority of the top-level Canadian math problem solvers come from Ontario?  In particular,  7 of the 10 winners on the CMO and 5 of the 6 Canada IMO team members are from Ontario (and all of the rest are from Alberta, another interesting factoid).  Ontario only has about one-third of Canada's population, right?  So what are they not doing in, say, BC or Quebec that they're doing in Ontario?

We see some examples of these phenomena here in the US as well -- for example, here in California, the Bay Area seems to produce more strong students than the LA area, but I've always attributed that in part to the fact that the high-tech industry is concentrated in Northern California.  I find it hard to believe that the entire explanation for Ontario is socio-economic (is Toronto really that much different than Montreal or Vancouver?).  I'm curious as to what your theories are.  Or is it just a statistical anamoly this year?

Quebec has never had a large participation in contests, and historically very few Canadian IMO team members and CMO qualifiers have come from there.

British Columbia doesn't really have an excuse for not doing well though.  I think its largely because there is a complete lack of programs for mathematically gifted youth here.  In Ontario they have their math circles, in Alberta I think they have some sort of similar thing (I could be mistaken), in BC all they have is 'Euclid Day', where everyone who scores sufficiently high on the Euclid gets invited over to UBC for a luncheon and prizes.  It's big on congratulations and short on actual mathematical content.  Actually, I tend to think that its pretty inevitable that the divide between Ontario and everyone else will just get bigger, as there really aren't any strong students in BC right now (that is, they are all graduating this year)

One of the things that I've given some serious thought to is trying to run some meetings for bright kids (that is, kids who are at the level where they can start solving olympiad problems, as there really aren't any who can solve olympiad problems currently in BC) at SFU, seeing as I'm a student there.  I dunno - in the meantime, I'm going to be a volunteer at  the problem-solving sessions at the ESSO-PIMS math camp at SFU (for grades 9/10), so I'll be able to give you a better feel for where the brighter kids are in BC in a month or two.

Actually, there's a well-known fact that states that a Canadian student's chances of making the IMO team are inversely proportional to their distance from the University of Waterloo.  So Kitchener/Waterloo students have the highest chance, but there are many more students from Toronto, with a decent chance as well... then the other 20 million canadians have a crappier chance, but maybe 1 or 2 will make it...

Also, Lake Ontario water has certain toxins that cause brain mutations, which result in decreased social activity and increased obsessive problem-solving disorder.

;)

Haha, I was actually going to write something about the Waterloo influence, but forgot.

Yeah, it definitely depends on what the profs do in their area.  Edmonton and Calgary have a couple great professors who do all they can to nuture growth, and we've consistantly seen an Albertan on the IMO team.  (Go David!)  However, Ontario has always been the strongest, they have some great programs out there.

Then again, it could be the water.  It might explain why all you Ontario people are a bit weird  ;)

[quote="Elyot"]
Also, Lake Ontario water has certain toxins that cause brain mutations, which result in decreased social activity and increased obsessive problem-solving disorder.

;)[/quote]I used to go swimming in Lake Ontario many times as a kid (I grew up in Western New York), so maybe that does explain something!

[quote="DPatrick"]and all of the rest are from Alberta, another interesting factoid[/quote]
I'm the IMO member from Alberta. I immigrated to Canada 3 years and 2 days ago (May 29th was the day I came :D  ). My parents chose to come to Edmonton for some unknown reason. Most Canadian IMO team members are immigrants. Maybe the reason is because immigrants prefer those places, Ontario and Alberta.

I'm the only white guy on the IMO team this year!

Actually, I think I'm probably the only Canadian-born IMO-team member in quite a few years.

[quote="Elyot"]
Actually, I think I'm probably the only Canadian-born IMO-team member in quite a few years.[/quote]
Robert Barrington-Leigh of 2003 IMO team was Canadian-born. He was also from Edmonton.


# Given an acute $ \triangle ABC$

1. $ M$ is midpoint of $ BC$
2. $ N$ is trisection point of AC close to C
3. $ ray NM\cap AB \equal{} P$
4. $ 2\angle C\geq\angle PBM\minus{}\angle PMB$

Prove that: $ \frac{c^{2}}{2}\plus{}\frac{ab}{3}\geq AM.BN$

[hide="Solution"]Let $ Q$ be the trisection point of $ AC$ close to $ A$.
Lemma
$ PN$ is parallel to $ BQ$.
Proof:
$ BC \equal{} 2MC$, $ CQ \equal{} 2CN$, and $ \angle C \equal{} \angle C$. By SAS, triangle $ BCQ$ is similar to triangle $ MCN$. $ MN$ is parallel to $ BQ$, so $ PN$ is parallel to $ BQ$.
--end lemma--
We substitute $ \angle PBM \equal{} \angle A \plus{} \angle C$ and $ \angle PMB \equal{} \angle CMN \equal{} \angle ANP \minus{} \angle C$ (both true by the exterior angle theorem) into the inequality
$ 2\angle C\geq\angle PBM \minus{} \angle PMB$
to get
$ 2 \angle C \ge \angle A \plus{} \angle C \minus{} (\angle ANP \minus{} \angle C)$
which is equivalent to
$ \angle ANP \ge \angle A$.
This implies that in triangle $ ANP$,
$ AP \ge PN$.
By the Lemma, triangles $ AQB$ and $ ANP$ are similar. Thus
$ AB \ge BQ$.
Since triangles $ BCQ$ and $ MCN$ have similarity ratio $ \frac {BC}{MC} \equal{} \frac {CQ}{CN} \equal{} 2$,
$ \frac {BQ}{MN} \equal{} 2$.
Thus
$ BQ \equal{} 2 MN$, and
$ AB \ge 2MN$.
$ \frac {c}{2} \equal{} \frac {AB}{2} \ge MN$, which implies
$ \frac {ab}{3} \plus{} \dfrac{c^2}{2} \ge \frac {ab}{3} \plus{} c \cdot MN$ [1].
By Ptolemy's inequality
$ BM \cdot AN \plus{} AB \cdot MN \ge AM \cdot BN$, which implies
$ \frac {a}{2} \cdot \frac {2b}{3} \plus{} c \cdot MN \ge AM \cdot BN$,
which is
$ \frac {ab}{3} \plus{} c \cdot MN \ge AM \cdot BN$ [2].
Combining inequalities [1] and [2] we get
$ \frac {c^{2}}{2} \plus{} \frac {ab}{3}\geq AM \cdot BN$
as desired.[/hide]

A/Q
1.$ 2\angle C \geq \angle PBM \minus{} \angle PMB$
2.$ 2\angle C \plus{} \angle B \geq \pi \minus{} \angle CMN$
3.$ 2\angle C \plus{} \angle B \geq \angle C \plus{} \angle CNM$
4.$ \angle C \plus{} \angle B \geq \pi \minus{} \angle MNA$
5.$ \angle MNA \geq \angle A$

Let, $ X$ be midpoint of $ AC$. In $ \triangle MNX$ we already have $ \angle MNA \geq \angle A$
$ \rightarrow$ $ MX \geq MN$
             or, $ \frac {c} {2} \geq MN$
By. Ptolemy's inequality in quad. $ ABMN$
$ AB.MN \plus{} BM.AN \geq AM.BN$
$ c. \frac {c} {2} \plus{} \frac {a} {2} . \frac {2b} {3} \geq AM.BN$
$ \frac {c^2} {2} \plus{} \frac {ab} {3} \geq AM.BN$


[quote="doicu"]hi,
determine all polynominals $ P$, with real coeffitients , and s.t. $ P^{n}(x) \equal{} P(x^{n}), \forall x \in \mathbb{R}$, where $ n$ is a natural fixed number $ n > 1$.[/quote]
we take $ S\equal{}\{a\in \mathbb{C}: \ P(a)\equal{}0\}$
if $ a\in S$,if $ a\equal{}r.exp(ib)\neq 0$ where $ b\in]0,2\pi]$, we have $ \forall k: r^{n^{\minus{}k}}.exp(ibn^{\minus{}k})\in S$ but $ S$ must be finite
then $ a\equal{}0$
gives $ S\equal{}\{0\}$ then $ P(X)\equal{}X^m$ for some $ m$


1)Let $ x,y,z > 0$ and $ xyz\equal{}1$. Prove that:
\[ \frac{\sqrt{1\plus{}x^3\plus{}y^3}}{xy} \plus{} \frac{\sqrt{1\plus{}y^3\plus{}z^3}}{yz}\plus{} \frac{\sqrt{1\plus{}z^3\plus{}x^3}}{zx} \ge 3\sqrt 3\]
2)Prove that:
\[ \sqrt{x^2\plus{}(1\minus{}y)^2} \plus{} \sqrt{y^2\plus{}(1\minus{}z)^2} \plus{} \sqrt{z^2\plus{}(1\minus{}x)^2} \ge \frac{3\sqrt2}{2}\] for all $ x,y,z$

1) apply am-gm as much as you can  :)
2) Minkowski

[hide]$ \sqrt {x^2 \plus{} (1 \minus{} y)^2} \plus{} \sqrt {y^2 \plus{} (1 \minus{} z)^2} \plus{} \sqrt {z^2 \plus{} (1 \minus{} x)^2} \ge \
frac{{\left| {x \plus{} 1 \minus{} y} \right|}}{{\sqrt 2 }} \plus{} \frac {{\left| {y \plus{} 1 \minus{} z} \right|}}{{\sqrt 2 }} \plus{} \frac {{\left| {z \plus{} 1 \minus{} x} \right|}}{{\sqrt 2 }}$ :lol: [/hide]

Can someone post the Minkowski's inequality?

http://en.wikipedia.org/wiki/Minkowski_inequality

But in this case it suffices to use
$ \sqrt{\sum_{i\equal{}1}^{n}a_{i}^{2}}\plus{}\sqrt{\sum_{i\equal{}1}^{n}b_{i}^{2}} \geqslant \sqrt{\sum_{i\equal{}1}^{n}(a_{i}\plus{}b_{i})^{2}}$


If there is one, is it too late to sign up?  :maybe:

I have no idea, but you shouldn't use your name on the internet because anyone can Google it. So you have to act responsibly now. Or decide to create a new account I guess. Maybe you should Google that competition.

uhh... i really don't think you should sign up by yourself... plus it's just one tournament... i dont' think a two hour drive is really worth briarwood

bus rides are FUNN lol

and going to an extra tournament ( not enough math tournaments for geometry people in AL ) would be fun too.

well maybe not the test itself

Have you been to their tournament before?

i went to briarwood last year.  i dunno if you think grissom is a quality tournament who knows what else you would like...  its actually probably worse than grissom but hey i personally think teen wolf too is worse than killer klownz from outerspace.

$\text{BRIARWOOD CHRISTIAN HIGH SCHOOL}$
JV MATH TOURNAMENT
$\text{Saturday, Feb. 17, 2007}$

Dear Math Team Coaches,

Briarwood will be hosting a Junior Varsity Math Tournament (two subjects: Geometry and Algebra II) on Feb. 17, 2007.  It is our intention to focus on training; therefore every entrant will cipher as well as take the written test.  Each individual score will be the sum of their correct written test answers (40 questions in one hour) and their ciphering correct answers (20 non-stop problems at 1 minute per problem).  We have no penalty for guessing.  The written test will be a multiple-choice test taken in several classrooms around the school.  The ciphering test will be one 20 question session given in the auditorium.

The format and award requirements are listed below.  We do not hold an awards ceremony.   Scores will be mailed to the coaches along with copies of the tests and keys after the tournament.  We hope you will bring a team to the tournament and that each student will gain something valuable from the experience.

Sincerely,

Jana Mays
Briarwood Christian School

FORMAT:   During ciphering and written testing, math team coaches are invited to enjoy refreshments in the lounge located outside the auditorium.  The classroom and auditorium are only open to students who are testing during that session.
	8:45-9:15	Team arrival and registration in High School Gym Lobby
	9:30 to 10:30	Algebra students take written test
			Geometry students cipher in auditorium

	10:30-10:45	Students will take a break with snacks in High School Gym Lobby
	11:00 - Noon	Geometry students take written test
			Algebra students cipher in auditorium

AWARDS:

INDIVIDUAL AWARDS:  certificates for the top two scores from each school

TEAM AWARDS:  Plaques for 1st three places in the Large School Division (5A and 6A)
		    Plaques for 1st three places in the Small School Division (1A – 4A and other schools)

For schools to be eligible to receive a team award your school must enter at least:
		     8 players (minimum) if a 5A or 6A school
		     4 players (minimum) if any other school

Remember, every score counts!  You can be in the tournament with fewer students but cannot receive a team award.  The team score will be the average of every entrant on your team.  Since this is a training tournament, we would like to encourage each school to bring more than the minimums listed above in order to put some pressure on the students to do their best.  We will not have an awards ceremony.  Results will be e-mailed to the designated sponsor.  Awards will be mailed to your school.

REGISTER BY E-MAIL

To register, please fill in the attached form.
Registration Deadline:  Jan. 31, 2007

Cost:  5.00 per student
Please bring your check (made out to Briarwood Christian School) to registration on Feb. 17, 2007.  We regret that we cannot reimburse a school for students who are registered but who do not attend the tournament.

Jana Mays
jmays@briarwood.org
School: 205-776-7900
Home: 205-620-5548

What do you (people) think? I wish there were individual trophies or something that would make people more motivated to do well. Only certificates to top two schools.... and no award ceremonies..  But then again, that means arriving home earlier...

I got the form today.  :|


geometry
geometric transformation
rotation
dilation
3D geometry
topology
real analysis

If A and B are nonempty convex subsets in $\mathbb{R}^n$. And $C = A + B = \{x+ y \ | \ x \in A, y \in B\}$. How to prove int(C) = int(A + B) = int(A) + int(B)?

"int" is the interior ?

If yes, I find this result surprising : what if A and B have empty interiors (two lines in $\mathbb{R}^2$ for instance) but if $int(A+B) \neq \emptyset$ (again two lines in $\mathbb{R}^2$ for which $A+B=\mathbb{R}^2$).

 :?

I should have included the condition - 'A and B have nonempty interiors'

We have $int(A)+int(A) \subset A+B$ and since Minkowski sum of two open sets is open we have $int(A)+int(A) \subset int(A+B)$. That's the easy part  :P

I still wonder if the inclusion may be strict in some case  :?

uhm.. take your same counterexample, just adding to the two lines the ball centered in $(0,0)$, with radius $1$...
sum of the interiors is the ball centered in $(0,0)$ with radius $2$, while interior of the sum is $\mathbb{R}^2$..
so i guess something's not working :)

Are you [i]really[/i] sure the sets you describe are convex ma_go ?

Let's assume:
http://www.stanford.edu/class/ee364/lectures/sets.pdf
and let's assume further the following two propositions are true:

[b]P1[/b].
If $A$ is a convex set with _nonempty_ interior, then $bd(int(A)) = bd(A)$.
(bd means boundary, int means interior)
(I don't have a proof of this. Is this a well-known result?) (Note that this proposition is false for convex sets with _empty_ interior.)

[b]P3[/b]
If $C, D$ are _disjoint_ _open_ convex sets and $p$ is their common boundary point, ie, $p \in bd(C) \cap bd(D)$, then there is a point $a \neq 0$ s.t. for all $x \in C, y \in D$ we have $a\cdot x \leq a\cdot p \leq a\cdot y$.
(dot means dot product)  (Again no proof)

Suppose $A, B$ are convex sets with nonempty interiors. Suppose $0 \in int(A+B)$. We want to show $0 \in int(A)+int(B)$.

Since $0 \in A+B$, there is a point $x$ s.t. $x \in A$ and $-x \in B$, i.e., $x \in -B$. WLOG assume $x = 0$ (we may translate A and B in opposite direction). Now, 0 is contained in all of $A+B, A, B$.

We have either $0 \in bd(A)\cap A$ or $0 \in int(A)$. Same is true for $B$. So we have four cases to examine.
If $0 \in int(A)$ and $0 \in int(B)$, then $0= 0+0 \in int(A)+int(B)$, done. If $0 \in bd(A)$ and $0 \in int(B)$, then apply P1, we get point $u \in int(A)$ arbitrarily close to 0 so that $-u \in (B)$, then again we have $0 = u + (-u) \in int(A)+int(B)$, done.

Now, only one case remains: the case $0 \in bd(A) \cap bd(B)$.  Suppose $0 \not\in int(A)+int(B)$ and we want to deduce a contradiction. Consider two auxiliary sets $C = int(A), D = int(-B) = -int(B)$. $C,D$ are disjoint open convex sets by hypothesises. Applying P1, we get $0 \in bd(C) \cap bd(D)$. Then now we can apply P3 to deduce that there is $a \neq 0$ s.t. for all $x \in C, y \in D$ we have $a\cdot x \leq 0 \leq a\cdot y$. Let $P = \{x : a\cdot x \leq 0\}$. we have $int(A) = C \subset P$. Since $P$ is closed, we also have $bd(A) = bd(C) \subset P$, so we have $A \subset int(A)\cup bd(A) \subset P$. Similarly, $B \subset P$. And, $A+B \subset P$. And, $0 \in int(A+B) \subset int(P)$. But $int(P)$ does not contain 0. Contradiction. done the last case.

Now, it remains to prove P1 and P3.
If one of P1 and P3 turns out to be false, my proof crashes.

key in proof: Considering auxiliary sets C and D to construct a hyperplane so that A and B are on the same side of the hyperplane.

_________________
Prayer for Peace on Korea. If another war happens, my life crashes.

Funny. Isn't that $int(A) \cap bd(A) = \emptyset$. Also why $A \subset int(A) \cap bd(A)$. Do you mean $A \subset int(A) \cup bd(A)$?

You wrote $A \subset P$ and $B \subset P$, then conclude $A+B \subset P$. I reckon this is not true.

[quote="kaos"]Funny. Isn't that $int(A) \cap bd(A) = \emptyset$. Also why $A \subset int(A) \cap bd(A)$. [/quote]
$A \subset int(A) \cap bd(A)$ is clearly false.
But there is no part in my proof that use it. Which part are you refering to?

[quote="kaos"]
You wrote $A \subset P$ and $B \subset P$, then conclude $A+B \subset P$. I reckon this is not true.
[/quote]
If you think this is not true, take a look at the definition of $P$.

[quote="redkimchi"][quote="kaos"]Funny. Isn't that $int(A) \cap bd(A) = \emptyset$. Also why $A \subset int(A) \cap bd(A)$. [/quote]
$A \subset int(A) \cap bd(A)$ is clearly false.
But there is no part in my proof that use it. Which part are you refering to?
[/quote]

See here:
[quote="redkimchi"]... Let $P = \{x : a\cdot x \leq 0\}$. we have $int(A) = C \subset P$. Since $P$ is closed, we also have $bd(A) = bd(C) \subset P$, so we have $A \subset int(A)\cap bd(A) \subset P$. Similarly, $B \subset P$. And, $A+B \subset P$. And, $0 \in int(A+B) \subset int(P)$. But $int(P)$ does not contain 0. Contradiction. done the last case.[/quote]

[quote="redkimchi"][quote="kaos"]
You wrote $A \subset P$ and $B \subset P$, then conclude $A+B \subset P$. I reckon this is not true.
[/quote]
If you think this is not true, take a look at the definition of $P$.[/quote]

Right.

Below are more of my comments.

[quote="redkimchi"]
Suppose $A, B$ are convex sets with nonempty interiors. Suppose $0 \in int(A+B)$. We want to show $0 \in int(A)+int(B)$.

Since $0 \in A+B$, there is a point $x$ s.t. $x \in A$ and $-x \in B$, i.e., $x \in -B$. WLOG assume $x = 0$ (we may translate A and B in opposite direction). Now, 0 is contained in all of $A+B, A, B$.
[/quote]

You made the supposition that $0 \in int(A+B)$. This is not true for arbitrary sets A, B. Even if $0 \in A+B$, this does not guarantee that 0 is contained in $A$ and $-B$ as what you claimed

Ok $A \subset int(A) \cap bd(A)$ is a typo. Thank you for that finding. I'll correct it to cup $\cup$.

For 0 things, it is false that $0 \in int(A + B)$ for arbitrary sets A and B, but it can be assumed without loss of generality.
$0 \in A$ and $0 \in B$ can also be assumed without loss of generality, although it is false for arbitrary sets A and B.

If P1 and P3 is true, then even if we consider the 2 WLOG-assumptions as not appropriate, what the proof has shown is at least the following:

[b]P5[/b]
If $A$ and $B$ are convex subsets of $R^n$ with nonempty interiors satisfying:
$0 \in A$, $0 \in B$, and
$0 \in int(A+B)$,
then we have $0 \in int(A)+int(B)$.

On the other hand, what we really have to accomplish is the following proposition:

[b]P7[/b]
If [b]E[/b] and [b]F[/b] are convex subsets of $R^n$ with nonempty interiors, then we have [b]int(E+F) = int(E)+int(F)[/b].

Proof of P7:

The part [b]int(E)+int(F) $\subset$ int(E+F)[/b] is already done by tµtµ.

So suppose a point [b]x $\in$ int(E+F)[/b]. We want to show that [b]x $\in$ int(E)+int(F)[/b].
Since [b]x $\in$ int(E+F) $\subset$ E+F[/b], we can find two points[b] y $\in$ E[/b] and [b]z $\in$ F[/b] s.t. [b]x = y + z[/b].
Let [b]A = E + (-y)[/b] and [b]B=F+(-z)[/b]. Then we have:
[b]0 $\in$ A[/b],
[b]0 $\in$ B[/b],
[b] 0 $\in$ int(E+F)+(-x) = int((E+F) + (-x)) = int((E+(-y)) + (F+(-z))) = int(A+B)[/b],
and that [b]A[/b] and [b]B[/b] are convex subsets with nonempty interiors.
So we can use P5 to conclude that [b]0 $\in$ int(A)+int(B)[/b]. Thefore,
[b]x $\in$ (int(A)+int(B))+x = (int(A)+y)+(int(B)+z) = int(A+y) + int(B+z) = int(E)+int(F)[/b].
QED.

In fact, this argument also shows why the 2 WLOG-assumptions are indeed WLOG-assumptions.

[quote="redkimchi"]
[b]P7[/b]
If [b]E[/b] and [b]F[/b] are convex subsets of $R^n$ with nonempty interiors, then we have [b]int(E+F) = int(E)+int(F)[/b].
[/quote]
First prove the
[b]Lemma[/b]: If $K \subset \mathbb R^n$ is convex with non-empty interior and $O \subset \mathbb R^n$ is open, then $K \cap O$ has non-empty interior.
Indeed, let $x \in K \cap O$ and let $B \subset K$ be an open ball, then $(1 - \epsilon) x + \epsilon B$ is an open ball that is contained in $K$ for $0 \le \epsilon \le 1$ and contained in $O$ if $\epsilon$ is sufficiently small.

Now to the proof of P7:
Consider the mapping $T: \mathbb R^n \times \mathbb R^n \ni (x,y) \mapsto (x+y,x-y) \in \mathbb R^n \times \mathbb R^n$. This is linear and bicontinuous (it's a rotation followed by a dilation), thus $T(E \times F)$ is convex and has non-empty interior, and $T(int(E) \times int(F)) = T(int(E \times F)) = int(T(E \times F))$. Then let $P: R^n \times \mathbb R^n \to R^n$ be the projection on the first component.  Thus $P\circ T(x,y) = x+y$.

Now consider the preimage of an open subset of $E+F$ under $P$. This is relatively open in $T(E \times F)$ and hence the intersection of an open set in $\mathbb R^{2n}$ with $T(E \times F)$. By the lemma, this has a non-empty interior, whose preimage under $T$ is a non-empty open set contained in $E \times F$. Hence any open  subset of $E + F$ is the image of an open subset of $E \times F$, which is what we wanted to prove, I guess.

[quote="redkimchi"]
[b]P1[/b].
If $A$ is a convex set with _nonempty_ interior, then $bd(int(A)) = bd(A)$.
[/quote]

Proof:
First claim: $bd(int(A)) \subset bd(A)$
Suppose $x \in bd(int(A))$ and let $N$ be an arbitrary neighborhood of $x$. Then there exists $y \in N \cap int(A)$ and $z \in N \cap int(A)^c$, where $int(A)^c$ means the complement of $int(A)$. Now $y \in N \cap A$. If $z$ is in $A^c$ we're done. Otherwise, $z$ is in $bd(A^c) = bd(A)$, then take a ball $B$ with $z \in B \subset N$, then $B \cap A^c$ is nonempty, hence $N \cap A^c$ is nonempty, done.

Second claim: $bd(A) \subset bd(int(A))$
Suppose $x\in bd(A)$ and let $N$ be an arbitrary neighborhood of $x$. Then $N\cap A$ and $N\cap A^c$ are nonempty. In particular, $N\cap int(A)^c$ is nonempty. There is a point $x \in N \cap A$ and a ball $B \subset A$. Then $(1 - \epsilon) x + \epsilon B$ is a ball that is contained in $A$ for $0 < \epsilon < 1$ and is also contained in $N$, if $\epsilon$ is sufficiently small. so, $N \cap A$ contains a ball, and hence $N\cap int(A)$ is nonempty.

QED.

[quote="redkimchi"]
[b]P3[/b]
If $C, D$ are _disjoint_ _open_ convex sets and $p$ is their common boundary point, ie, $p \in bd(C) \cap bd(D)$, then there is a point $a \neq 0$ s.t. for all $x \in C, y \in D$ we have $a\cdot x \leq a\cdot p \leq a\cdot y$.
[/quote]

Proof:
By Separating Hyperplane Theorem, there is a point $a\neq 0$ and a real number $b$ such that $a\cdot C \leq b \leq a\cdot D$.

Since $p$ is a limit point of $C$ and $a\cdot C \leq b$, we have $a\cdot p \leq b$. But $p$ is also a limit point of $D$, so $a\cdot p \geq b$. Therefore, $a\cdot p = b$. QED.

Yes, the second one is essentially the same thing as the Separating Hyperplane Theorem (with a slight twist :)), but you could make the proof of the first one simpler:

Let $a$ be a point on the border of our convex set $A$, and let $o$ be a point in its interior. Also let $\mathcal B$ be an open ball around $o$ contained in $A$. Since the closure $\overline A$ of $A$ is convex, it contains the segment $[oa]$. For every point $x\in[oa),\ \overline A$ contains the interior of the cone generated by the point $\frac{x+a}2$ and $\mathcal B$, so the interior of $A$ contains it as well. This means that every $x\in[oa)$ is an interior point of $A$, and we're done.


Prove that

$ \sqrt{c(s\minus{}c)}\plus{}\sqrt{b(s\minus{}b)}>\sqrt{a(s\minus{}a)}$

If a triangle has sides of length a ,b and c  then ,

1.$ \sqrt{a}$ ,  $ \sqrt{b}$ ,   $ \sqrt{c}$ are sides of another triangle      ($ \sqrt{a}$ +$ \sqrt{b}$ >$ \sqrt{(a\plus{}b)}$ >$ \sqrt{c}$)

2.$ a(s\minus{}a) ,   b(s\minus{}b) ,   c(s\minus{}c)$ also form the sides of a triangle because $ a(s\minus{}a) \plus{}b(s\minus{}b) >c(s\minus{}c) \leftrightarrow  a(b\plus{}c\minus{}a) \plus{}b(c\plus{}a\minus{}b) >c(a\plus{}b\minus{}c) \leftrightarrow c^2 >(a\minus{}b)^2$.

It follows that $ \sqrt{a(s\minus{}a)}$ , $ \sqrt{b(s\minus{}b)}$ , $ \sqrt{c(s\minus{}c)}$ are also the sides of a triangle.

Nice SQUING!!

[quote="sqing"]WLOG $b\ge c$,hence $b(c+a-b)+c(a+b-c)>a(b+c-a)\iff c+a>b$

$\implies (\sqrt{b(c+a-b)}+\sqrt{c(a+b-c)})^2$

$>b(c+a-b)+c(a+b-c)>a(b+c-a)$

$\implies \sqrt{b(c+a-b)}+\sqrt{c(a+b-c)}>\sqrt{a(b+c-a)}$

$\iff \sqrt{c(s\minus{}c)}\plus{}\sqrt{b(s\minus{}b)}>\sqrt{a(s\minus{}a)}.$[/quote]


inequalities
function
inequalities proposed

If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \frac{4(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}\plus{}\frac{ab}{a^2\plus{}ab\plus{}b^2}\plus{}\frac{bc}{b^2\plus{}bc\plus{}c^2}\plus{}\frac{ca}{c^2\plus{}ca\plus{}a^2} \ge \frac{7}{3}.$

[quote="peine"][quote="arqady"][quote="can_hang2007"]If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \frac {4(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {ab}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {bc}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {ca}{c^2 \plus{} ca \plus{} a^2} \ge \frac {7}{3}.$[/quote]
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2$ and $ abc \equal{} w^3.$ Hence, your inequality is equivalent to
$ f(w^3)\leq0,$ where $ f$ is a convex function. Thus, for the proof we need to check only two cases:
1) $ b \equal{} 1,$ $ c \equal{} 0$
2) $ b \equal{} c \equal{} 1.$[/quote]
please Mr.araqady, can you explain your solution,[/quote]
With pleasure! :)
Let $ a,$ $ b$ and $ c$ are non-negatives such that $ ab \plus{} ac \plus{} bc\neq0.$
$ \prod_{cyc}(x^2 \plus{} xy \plus{} y^2) \equal{} 27(3u^2v^4 \minus{} u^3w^3 \minus{} v^6).$
$ \sum_{cyc}ab(a^2 \plus{} ac \plus{} c^2)(b^2 \plus{} bc \plus{} c^2) \equal{} \sum_{cyc}(a^3b^3 \plus{} a^4bc \plus{} 3a^3b^2c \plus{} 3a^3c^2b \plus{} a^2b^2c^2) \equal{}$
$ \equal{} 27(v^6 \plus{} u^3w^3 \minus{} uv^2w^3).$
Hence, the original inequality is equivalent to
$ \frac {4(3u^2 \minus{} 2v^2)}{3u^2} \plus{} \frac {3u^2v^4 \minus{} u^3w^3 \minus{} v^6}{v^6 \plus{} u^3w^3 \minus{} uv^2w^3}\geq\frac {7}{3},$ which can be written as $ f(w^3)\geq0,$ where $ f$ is a linear function of $ w^3.$
But a linear function gets a minimal value when $ w^3$ gets an extremal value,
which happens when two numbers from $ \{a,b,c\}$ are equal or when maybe $ w^3 \equal{} 0.$
These cases very easy to check.
I exaggerated with the convex function. :D

[quote="crazyfehmy"]Mr. Arqady;

"But a linear function gets a minimal value when $ w^3$ gets an extremal value,
which happens when two numbers from $ \{a,b,c\}$ are equal or when maybe $ w^3 \equal{} 0.$"

Why $ w^3$ gets an extremal value when two numbers from $ \{a,b,c\}$ are equal? :huh:[/quote]
Because $ a,$ $ b$ and $ c$ are real roots of the following equation:
$ w^3\equal{}x^3\minus{}3ux^2\plus{}3v^2x$ and the maximal value of $ w^3$ holds when two roots are equal.  :wink:

I am interested in seeing a non-$ uvw$ solution. Can someone find it?

[quote="can_hang2007"]If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \frac {4(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {ab}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {bc}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {ca}{c^2 \plus{} ca \plus{} a^2} \ge \frac {7}{3}.$[/quote]
I have just found that this one holds for any real numbers $ a,$ $ b,$ $ c$ such that $ (a\plus{}b\plus{}c)(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2) \ne 0.$

[quote="can_hang2007"][quote="can_hang2007"]If $ a,$ $ b,$ $ c$ are positive real numbers, then
$ \frac {4(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {ab}{a^2 \plus{} ab \plus{} b^2} \plus{} \frac {bc}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {ca}{c^2 \plus{} ca \plus{} a^2} \ge \frac {7}{3}.$[/quote]
I have just found that this one holds for any real numbers $ a,$ $ b,$ $ c$ such that $ (a \plus{} b \plus{} c)(a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2) \ne 0.$[/quote]
See here: http://can-hang2007.blogspot.com/2010/02/inequality-77-v-q-b-can.html

Dear luofangxiang,

your solution is very nice, but I don't uderstand why

$\sum a(\frac{b+c}{b^2+bc+c^2} +  \frac{a+c}{a^2+ac+c^2}) \geq 4$ implies

$4(a^2+b^2) \geq (\sum a)^2 (\frac{b^2}{b^2+bc+c^2} +\frac{a^2}{a^2+ac+c^2} )$

Thank you very much.

Really BEAUTIFUL your proof by SOS of this very hard inequality.

Thank you very much.


integration
calculus
function
absolute value
calculus computations

Suppose $ z(t) \equal{} (t \minus{} 1)^2 \plus{} 2i(t \minus{} 1)^{3}, \ \ 0 \leq t \leq 2$. Then $ L \equal{} \int_{0}^{2} \left|\frac {dz}{dt} \right| \ dt \equal{} \int_{0}^{2} \left|2(t \minus{} 1) \plus{} 6i(t \minus{} 1)^2 \right| \ dt$. We then get: $ L \equal{} \int_{0}^{2} 2(t \minus{} 1) \sqrt {1 \plus{} 6i(t \minus{} 1) \minus{} 9(t \minus{} 1)^{2}} \ dt$.

How do we integrate this to get a numerical answer?

Your L is wrong;- You should not have i in your expression after you have taken moduli.

This family of curves - 2/3 powers or 3/2 powers - is notorious for having arclengths that can be computed in closed form, which is a rarity.

One note: after you fix the absolute value (as akech said), you should the split the integral into two integrals, with $ t\in[0,1]$ and $ t\in[1,2].$ Why? For one thing, when you factor $ (t \minus{} 1)^2$ out from under the square root, it comes out as $ |t \minus{} 1|,$ not $ (t \minus{} 1).$

I'd probably also as some point make the "little" substitution $ u\equal{}t\minus{}1.$ Then $ u$ goes from $ \minus{}1$ to $ 1,$ which makes it worth your while to check whether the thing you're integrating is either an even function of $ u$ or an odd function of $ u.$ (Hint: it is one of those.)


function
integration
limit
real analysis
real analysis unsolved

Let $f: [0,\infty) \to [0,\infty)$ be a differentiable function for which
$\int_0^\infty f(x) dx < \infty$ and $f'(x) \le g(f(x))$ for all $x$,
where $g$ is a continuous function. Prove that $\lim_{x \to \infty}f(x) = 0$.

If the conclusion is false then there is a sequence $x_{n} \to \infty$ such that $f(x_{n})=a>0$ for all $n$. It is clear that there is an $\alpha > 0$ such that
\[ a-\alpha \sup_{t \in \left[\frac{a}{2},a\right]}{g(t)} = C>0. \]
Let us fix $n$ and take $I=[x_{n}-\alpha_{n},x_{n}]$. I'll prove that
\[ \int_{x_{n}-\alpha}^{x_{n}} {f(x)} dx \geq \min \left\{\frac{a\alpha}{2},\frac{aC}{2S}\right\} \]
which is enough to get a contradiction. If this is false, then $\inf f(I) < a/2$, so $a/2 \in f(I)$. Since $f(x_{n})=a$, it is easy to prove using the continuity of $f$ that there is some $[u,v] \subseteq I$ such that $f(u)=a/2$, $f(v)=a$ and $a/2<f(x)<a, \forall x \in (u,v)$. Hence $f([u,v])=[a/2,a]$. Then, for any $x \in [u,v]$ we have
\[ \frac{a-f(x)}{\alpha} \leq \frac{f(v)-f(x)}{v-x} \leq \sup_{t\in [x,v]}{f'(t)} \leq \sup_{t\in [u,v]}{g(f(t))} = \sup_{t \in \left[\frac{a}{2},a\right]}{g(t)}: =S, \]
which means that $f(x) \geq C$ for all $x \in [u,v]$. It is clear that $S>0$. We have
\[ \frac{f(v)-f(u)}{v-u} \leq S \Rightarrow v-u \geq \frac{a}{2S}, \]
which means that
\[ \int_{x_{n}-\alpha}^{x_{n}} {f(x)} dx \geq \int_{u}^{v} {f(x)} dx \geq \frac{aC}{2S}, \]
which establishes our claim and the proof is complete.

Another proof:

Assume that there exist $x_n$ s.t. $f(x_n)\to \ell> 0$. (WLOG). Denote by $S=f^{(-1)}([\frac \ell 2,\ell])$.
We have that $\int_S \frac \ell 2 \le \int_S f< \infty$. Hence $\mu S<\infty\Rightarrow \forall \epsilon , \exists \delta$ s. t . $\mu (S \cap (\delta,\infty))<\epsilon$.

Assume that $[a,b]$ is an interval included in $S \cap (\delta,\infty)$  (S is a union of intervals since f is continuous). $|f(b)-f(a)|<\sup_{x\in [\frac \ell 2,\ell]}g(x)(b-a)\le \epsilon M$.

For sufficiently small $\epsilon$ we have that $b-a<\frac \ell 4$ for any interval $[a,b] \in S \cap (\delta,\infty)$.
For each $x_n$ we define ${y_n=\inf M= \{x: x>x_n,f(x)=\frac \ell 2}$. $M\neq \Phi$ since else we would have $f(x)\ge \frac \ell 2,\forall x\ge x_n$,  hence $\infty > \int_{x_n}^\infty f(x)dx \ge\int_{x_n}^\infty \frac \ell 2 dx = \infty$ false. Hence there exists $[x_n,y_n]$ such that $f([x_n,y_n])>\frac \ell 3$ for sufficienty large n. Contradiction .


Let $A=\{x\in\mathbb{R}| \ x^{2}-(1-m)x-2m-2=0 \}$ and $B=\{x\in\mathbb{R}| (m-1)x^{2}+mx+1=0 \}$. Find all $m$ such that $M=A\cup B$ has 3 elements.

We have $\Delta_{A}=(1-m)^{2}+4(2m+2)=1-2m+m^{2}+8m+8=m^{2}+6m+9=(m+3)^{2}$, and $\Delta_{B}=m^{2}-4(m-1)=m^{2}-4m+4=(m-2)^{2}$. Therefore, for $m\neq 1$ we have

$A=\{\frac{1-m+m+3}{2},\frac{1-m-m-3}{2}\}=\{2,-m-1\}$ and $B=\{\frac{-m+m-2}{2(m-1)},\frac{-m-m+2}{2(m-1)}\}=\{\frac{1}{1-m},-1\}$

Now we have the following possibilities:

$1.\qquad 2=\frac{1}{1-m}\iff m={1\over 2}$
$2.\qquad-m-1=\frac{1}{1-m}\iff m^{2}-1=1\iff m=\pm\sqrt{2}$
$3.\qquad-m-1=-1\iff m=0$
$4.\qquad 2=-m-1\iff m=-3$
$5.\qquad \frac{1}{1-m}=-1\iff m=2$

(Options 1, 2 and 3 cover for equal elements between the sets, and 4 and 5 for double roots.)

For all those values we manually check that $A\cup B$ indeed has $3$ elements.

It remains to be seen what happens with $m=1$. Then $A=\{x\in\mathbb{R}|x^{2}-4=0\}=\{2,-2\}$ and $B=\{x\in\mathbb{R}|x+1=0\}=\{-1\}$. We see that in this case $A\cup B$ clearly has $3$ elements too.

Therefore, $\mathcal{M}=\{-3,-\sqrt{2},0,{1\over 2},1,\sqrt{2},2\}$

My results are the same as yours. Since I was very tired and the book results were different ($A=\{-2,m+1\}$) I was dissapointed because usually my computations are awful. Fortunately this time they were correct! Thank you!  :)


Wow, there are proofs in the Getting Started forum.

Can you use Mathematical Induction to prove the following?

Well Ordering Principle Every nonempty set of positive integers contains a smallest member.

This is not a hint:

[hide]

If you can prove it, then try to prove the reverse. If you can do both ways, you are pretty good at proofs!

[/hide]

[quote="fuzzylogic"]Can you use Mathematical Induction to prove the following?

[b]Well Ordering Principle[/b] Every nonempty set of positive integers contains a smallest member.
[/quote]

and

[quote="fuzzylogic"]Oh, please. I didn't post it here. It was under the thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13443]Mathematical Induction/Deduction?![/url] in the [b]Getting Started[/b] forum.

This is not a challenge. I changed the subject title accordingly. I thought it's something important to know if you want to really understand the Mathematical Induction well because it's equivalent to some very obvious principle.[/quote]

The proof is too difficult for the [b]Getting Started[/b] section. The nonempty set of positive integers mentioned in the principle could be infinite. Working with infinite sets can be tricky. And there is such a temptation to use the Well-Ordering Principle in the middle of proving the Well-Ordering Principle.

Hidden under spoiler code, fuzzylogic said,
[quote="fuzzylogic"]If you can prove it, then try to prove the reverse. If you can do both ways, you are pretty good at proofs!
[/quote]

I think the reverse is easier to prove. That would be, "You have an infinite number of statements, numbered with the positive integers. Prove that if you have shown that statement #1 is true and that statement #k can be used to prove statement #(k+1) for all positivie integers k, then statement #k is true for all positive integers k. Hint: Use the Well-Ordering Principle."

Erin Schram


find all fuctions $f$:R-R such that:
$f(x+y)$=$f(x)$+$f(y)$ for all $x$ in R
$f(x)$$f(1/x)$=1 for all $x$ in R  different 0

First, you must prove that: $f^{2}(x)=f(x^{2}) \forall x$
Then you have $f$ an increasing function

There is another solution: Assume that $|x|\geq 2$. We will prove that $|f(x)|\geq 2$. It follows that the graph of $f$ is not dense in the plane; hence, as $f$ is linear in the sense that $f(x+y)=f(x)+f(y)$, it follows that $f(x)=ax+b$ for some constants $a,b$. The rest of the problem is trivial.

So, if $|x|\geq 2$, then there is some $t$ such that $x=t+\frac{1}{t}$. Therefore $|f(x)|=|f(t+\frac{1}{t})| = |f(t)+\frac 1{f(t)}|\geq 2$ by AM-GM. This proves the claim.


If $x,y,z>0$ and $xyz=1$ then prove or disprove that

$\frac{1}{2z+x}+\frac{1}{2y+z}+\frac{1}{2x+y}\leq 1$

[hide]inequality obviously doesn't hold...we can have $x,y$ be as close to 0 as we want, so that $\frac{1}{2x+y}$ is as large as we want... for instance, making $x=\frac{1}{10}=y, z = 100$, $\frac{1}{2x+y} = \frac{10}{3}>1$[/hide]


geometry
circumcircle
projective geometry
angle bisector

[quote="trejos"]Let $ ABC$ be a triangle and let $ P$ be a point on the angle bisector $ AD$, with $ D$ on $ BC$. Let $ E$, $ F$ and $ G$ be the intersections of $ AP$, $ BP$ and $ CP$ with the circumcircle of the triangle, respectively. Let $ H$ be the intersection of $ EF$ and $ AC$, and let $ I$ be the intersection of $ EG$ and $ AB$. Determine the geometric place of the intersection of $ BH$ and $ CI$ when $ P$ varies.[/quote]
Nice problem. :)
I shall prove that the locus of the intersection point (say $ X$) is the $ A$-median of $ \triangle ABC$.

We have, $ \angle PAH\equal{}\angle PFH\equal{}\frac A2$. Which implies that $ APHF$ is cyclic. Analogously, $ \angle PAI\equal{}\angle PGI\equal{}\frac A2$ and $ AGIP$ is cyclic.
Indeed, $ \angle AHP\equal{}\angle AFP\equal{}C$ and $ \angle PIA\equal{}\angle PGA\equal{}B$ implies that, $ PH\parallel BC$ and $ PI\parallel BC$.

So, $ H,I,P$ are collinear and also $ AH\parallel BC$ and $ \frac{AI}{IB}\equal{}\frac{AH}{HC}$

Let, $ AX \cap BC\equal{}M$. Then, from Ceva's Theorem, \[ \frac{AI}{IB}\cdot\frac{BM}{MC}\cdot\frac{CH}{HA}\equal{}1 \; \iff BM\equal{}MC \]

by Pappus Theorem I,P,H are collinear
by angle chasing IH//BC
by Ceva Theorem BK=KC

[quote="campos"]i'm glad you liked it, this problem is mine  :D  however you should prove that it varies along all the median, and not just a part of it...[/quote]
Congratulation! This is indeed a nice problem. :)

Thanks for the remainder. I always forget this is locus problems :blush: .
So, here goes the rest of the solution.
We take any point $ X$ on the $ A \minus{}$median.
Let, $ CX\cap AB \equal{} I,BX\cap AC \equal{} H, AD\cap IH \equal{} P,$$ CP\cap \omega \equal{} G,BP\cap \omega \equal{} F$.
Where $ \omega \equal{} (O)$, indeed, $ IPH \parallel BC$
Now we assume that, $ GE\cap AB \equal{} I',FE\cap AC \equal{} H'$ and $ I \not \equiv I', H \not \equiv H'$
But we can indeed show (like- we did before) that $ AGPI'$ is cyclic and $ IPH$ collinear and parallel to $ BC$.
Which completes the proof. :)


function
search
geometry proposed
geometry

Find out all real function $f$ of the plane such that for every $n$ regular polygon $A_1A_2...A_n$, we have
\[ f(A_1)+f(A_2)+...+f(A_n)=0 \]

If $n$ was not a fixed number:

Let A be an arbitrary point in the plane. We'll prove that $f(A)=0$. Consider a regular hexagon ABCDEF and let O be its center. Triangles AOB and COB are equilateral, so $f(A)+f(B)+f(O)=0=f(C)+f(B)+f(O)$. It follows that $f(A)=f(C)$. In the same way $f(C)=f(E)$, so
    (1) $f(A)=f(C)=f(E)$
But triangle $ACE$ is also equilateral, hence
    (2) $f(A)+f(C)+f(E)=0$
From (1) and (2) we get $f(A)=0$.

Was n a fixed number or not?

Of course n fixed is more difficult. Can you solve with this hypothesis?

Just search for it people. It's been posted at least twice before. It's from a Romanian TST.

Sorry, grobber, you're right ! :oops:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=function+regular+polygon&t=51933

If I had known it's from a Romanian TST, I wouldn't have asked whether n was fixed or not.  :)


Find polynomial f(x,y) $\in$ R[x,y] deg f(x) min $\in$ N* satisfy:
$\forall$ x,y $\in$ R we have
    f(x,y) + f(y,x)=0 and f(x,x+y) + f(y,x+y) =0

I don't exactly understand your 2nd constraint:

deg f(x) min is an element of N*


What is the least positive integer $ n$ such that $ n^2 \minus{} n$ is divisible by some but not all integer values of $ k$ when $ 1\leq k \leq n$?

hmm?  $ 3^2\minus{}3\equal{}6$, 1,2, and 3 all divide 6...

so 3 doesn't work, $ 4^2\minus{}4\equal{}12$, which is divisible by 1,2,3, and 4

however, $ 5^2\minus{}5\equal{}20$, and 3 doesn't divide that, so $ \boxed{5}$

Sorry, I thought of all values up to $ n^2\minus{}n$.

A less trial and error solution:

$ n^2\minus{}n\equal{}n(n\minus{}1)$

Thus it is clear
1 will always divide
2 will always divide
$ n\minus{}1$ will always divide
$ n$ will always divide
3 will divide if n is not 2mod3
etc.

For there to be a number n such that 1,2,n-1,n are all distinct, $ n \geq 5$

Fortunately, when we try 5, we realize 5 is 2mod3, so 3 doesnt divide $ 5^2\minus{}5$. so without plugging in, we know 5 is our number.


abstract algebra
group theory
superior algebra
superior algebra unsolved

Prove by an example, that we can find 3 groups E c F c G, where E is normal in F, F is normal in G, but E is not normal in G.

(the c denotes contained in)

Dihedral group $ D_8\equal{}<r,s|r^4\equal{}s^2\equal{}1, sr\equal{}r^3s>$.
$ <s>$ is normal in $ <r^2,s>$, $ <r^2,s>$ is normal in $ D_8$.
But $ <s>$ is not normal in $ D_8$.

We just got a hint for this problem:

Consider
G=A_4
F={Id, (12)(34), (13)(24), (14)(23)} and
E={Id, (12)(34)}

I want to use the fact that A_4 is the kernel of S_4 so automatically it is normal in S_4.  But how is F related to S_4?

F is the commutator subgroup of $ A_4$. Bradleys example is with those E and F, but $ G\equal{}D_8$

How did you see that F is the commutator subgroup of A_4?

Is E related to F in the same manner?


compute  the  limit  of  $ a_n$, where  $ a_n$  is

$ a_n \equal{} \sqrt {1 \plus{} 2 \sqrt {1 \plus{} 3 \sqrt {1 \plus{} 4 \sqrt {1 \plus{} ...   \plus{}   n  \sqrt {1 \plus{} n}}}}}$

Ramanujan's problem very popular lately!

See [url]http://www.mathlinks.ro/viewtopic.php?search_id=1112658757&t=302868[/url]

Also [url]http://www.mathlinks.ro/viewtopic.php?search_id=1080987350&t=308725[/url]


There is a cone with the angle shown $\alpha$.A small ball is jumping from the sides of cone.It stays at the same height during the motion because of the reaction force of the cone.The speed of the mass is $v$.The maximum in the vertical direction is changing between $\frac{gT}{2sin\alpha}$ and $\frac{gT}{2}$.T is the time between two collisions.Find  $\omega$ of the mass.

Hint:Try to divide the system into three planes.


geometry
incenter
trapezoid
geometry unsolved

Hi!

Given triangle ABC with incenter I, prove that CA + AI = BC iff A=2B.

Any help is appreciated.

/Kon

I think Im correct  :P .

First, assume that $A = 2B$ Then since $A > B$ we have that $BC>AC$ So there exist a point $X$ in $BC$ such that $XC = AC$. If $I$ is the incenter of $ABC$ then triangle $ACI$ and triangle $CIX$ are congruent. Then angle $IXC = IAC = A/2 = B$ and so lines $XI$ and $AB$ are parallel. And then $AI = BX$, since $ABXI$ is an isosceles trapezoid with parallel bases. And so $BC = BX+XC = AI+AC$.

Now, if $BC = CA+AI$ let $X$ be the point on $BC$ such that $BX = AI$, and $XC = AC$. Again, triangles $AIC$ and $CIX$ are congruent and so angle $IXC = IAC = A/2.$ Now Since $BX = AI = IX$ we get that triangle $IBX$ is isosceles. and so angle $XIB = XBI$ but since $BI$ is bisector we get that $ABI=IBX=BIX$ which implies that lines AB and IX are parallel. And so angle $B = IXC = IAC = A/2$ and conclusion follows.

Am I right?

The construction of the point X was the idea that somehow didn't come to (my) mind.  :oops:

Thanks, Camus, I appreciate it!

/Kon

[hide]Metrical solution
Let $AL$ be the angle bisetor of $\angle CAB$. For any triangle $BL=\frac{ac}{b+c}$ as $\Delta ALB$ is isosceles $AL=BL=\frac{ac}{b+c}$.
Let I be the incenter. For any triangle $\frac{AI}{AL}=\frac{b+c}{a+b+c}$.So $AI=\frac{b+c}{a+b+c}\frac{ac}{b+c}=\frac{ac}{a+b+c}$.
Then $AI+b= \frac{ac}{a+b+c}+b=\frac{ac+ba+b^{2}+bc}{a+b+c}$.
It is a well known fact and i think was posted here that when $A=2B \Rightarrow a^{2}=b^{2}+bc$ and this gives $AI+b=\frac{ac+ab+a^{2}}{a+b+c}=a$[/hide]


The 12 USAMO Winners and 12 Honorable Mentions are posted at the AMC website now.
Congratulations to all.

USAMO scores by student number should be posted later today (Mon, May 12) or tomorrow (Tues, May 13) at the AMC website.

The USAMO scores and invitation criteria for invitation to MOSP are:
Black Group:  USAMO Score: 28-42
Blue Group: USAMO Score 20-27 and not a graduating senior from high school.
Red Group: USAMO Score, 10-15 and 9th grade student
China Girl's Math Olympiad: Female, any grade and score greater than 11

Steve Dunbar
MAA Director, American Mathematics Competitions

Calls are supposed to be going out today.

Mailed invitations, with detailed information, and many forms to fill and return will
be prepared tomorrow and mailed to home addresses by Wednesday.

Steve Dunbar
MAA Director, American Mathematics Competitions

This seems to indicate that no freshman got a score of 16, 17, 18, or 19, but around 24-30 got between 10 and 15. Interesting... I guess no freshman got two and a half problems.

Also, they usually list perfect scorers separately, so it looks like there aren't any this year.

It's pretty hard to get half a problem worth of credit. The most likely way to get 16-19 involves partial credit on multiple problems- three solutions with at least two mistakes at the high end, or two solutions plus two partials at the low end.

[quote]
Will the CGMO team be a different mop group (other than red/blue/black)?[/quote]

Probably not.  This was discussed early on among the organizers and it was finally decided that we would mix the CGMO inivitees in at the level where they and the instructors felt they were getting the best mathematical learning.
On some days however, we will gather the CGMO team to discuss practical and logistical matters and we may have some team training classes and times.

In another posting soon, I will have more explanation about the connection between CGMO and MOSP.

This brings me to another point that I have said before:  The color groups are a convenient large scale organizational structure that allows us to schedule instructors, classrooms and topics. The color groups are not an absolute structure whose boundaries may never be crossed.  On any given day, on the advice of the instructors, a student may be moved from one group to another, based on experience, confidence, demonstrated ability or any other good reason.

I got a 21, but have not received a MOP call yet. Was I supposed to? :maybe: (I heard some other 21s received calls already...)

Perhaps I put the wrong phone number on the form?

At least one phone number was not working and several others had no answering machine/voicemail.

Send me a personal message, be sure to include your real name, address, working phone number, state, email and school.

Steve Dunbar
MAA Director, American Mathematics Competiitons

I also got a 21 but I haven't received a call either .. have they all been sent out?

[quote="CatalystOfNostalgia"]Were Red MOPpers called at all?[/quote]

Not yet.

The phones were made in the following order:

(1) TST participants were contacted. (But we have not reached everyone yet.)
(2a) GMO members. (But we have not reached everyone yet.)
(2b) Blue group members. (But we have not reached everyone yet.)
(3) Red group members. (Will call tonight.)

Zuming

Yay!
that means my lack of an answering machine wont make a difference!

[quote="k8reindeer"][quote]China Girl's Math Olympiad: Female, any grade and score greater than 11
[/quote]

I recieved a call for this, I believe.
But I'm pretty darn sure I got only a seven.
What's going on?[/quote]
at least a couple people rejected the offer (to go to rsi and such) so i guess the cutoff was lowered

Have any MOPpers received an email or been mailed anything about MOP yet?  I'm just wondering because the call mentioned emails and I haven't received anything yet.

Thanks!

The emails were sent approximately 10:00 AM CDT on Thursday, May 15, 2008.
Everything will also be sent by postal mail to the best available address we have, usually the parent's address.

Steve Dunbar
MAA Director, American Mathematics Competitions

If we haven't received an email, should we just wait for the stuff in the mail?

Thanks!

EDIT: Never mind, my parents got the email.

I didn't get an email and my parents didn't either. Do CGMOers not count?

[quote="Aryth"]I didn't get an email  :( does Red get emailed?[/quote]

Yeah, I'm in Red and I got emailed. Actually, it was my mom that got emailed.

It was my parents who were emailed and not myself, as well.

To confirm and expand on my my previous message:

The invitation emails were sent to parents last Friday, about 10:00 am.  We used the parents'
email address(es) from the USAMO S-III forms.  The emails were sent to parents because
the attachments include a number of legal, medical and travel release forms which must be filled out
and returned to our office.  The student MOSP participants must have parents fill out the forms and return them
to us.  In order to assure that parents actually receive the emails, we sent the information directly
to parents.

We have followed up with paper copies sent by postal mail to the students home address.

If you (or more precisely, your parents) did not receive the email, the instructions for contacting the AMC
office are posted elsewhere on this forum.  That is the most efficient way to address concerns about not
receiving information.

Steve Dunbar
MAA Director, American Mathematics Competitions


There are twelve coins, all look the same, but one of them has a weight slighly different than the rest and it is not known if it is heavier or lighter than the others. Using only a balance of two plates, and having only three oportunities to use such balance, find the coin that is different.

Arg! I think the answer is: there is no solution, unless you can just weigh them in your hands...

Think out of the box, there is a solution and you don't have to weight them with your hands, only the use of the balance is allowed  ;)

i know the way for 9 coins, and i think solution for 12 is same but im too lazy to think of it now   :P

I figured it out!

[hide="Hint"]
Divide it into 3 piles of 4 coins each.[/hide]

i did that but then what????  I couldnt figure out from there... The thing that made it hard was that we don't know if it is lighter or heavier.

[quote="probability1.01"]The trick is that at some point you can figure out whether it's heavier or lighter.[/quote]
Well, I would say that it is part of the solution, not a trick. Think out of the box, this is the only one that hasn't been solved yet.

Best,

[quote="djimenez"]a hint: You don't eliminate a single coin until after the last step![/quote]

doesnt taht mean we have to eliminate every regular coin after the LAST STEP?????

So you know the first weighing is 4 vs. 4.  If those are balanced then it is trivial, so assume they are unbalanced.  NOTE CAREFULLY which way the balance tips.  Now try to figure out the rest yourself.

[quote="xxreddevilzxx"][quote="djimenez"]a hint: You don't eliminate a single coin until after the last step![/quote]doesnt taht mean we have to eliminate every regular coin after the LAST STEP?????[/quote]Yes, it does!

[quote="probability1.01"]Well, you actually eliminate a lot before then...[/quote]Depending on the cases, you can eliminate a few before the last step, but it is not something consecuent! At least not in the solution I have in mind!

You split it into into three piles, each of four coins.  Take two piles and put them on the balance.  If they way the same, It wasnt on the balance, If they way different, we know which pile it was in.  After one use, we have narrowed it down to 4 coins.  Next we split the pile into two groups of 2.  Balance, find which group it is in, and then balance again to find which one it is.  Done in three tries

[quote="gopherhole112"]If they way different, we know which pile it was in. [/quote]

Actually, you don't. Because you don't know if the coin is heavier or lighter. I can't seem to get past this step either.

[quote="probability1.01"]I will assume the worst luck.

1. Weigh 4 vs. 4.  They are unbalanced.  Let's say left side is heavier.

2. Weigh 2 from the left and 1 from the right vs. 2 from the right and 1 from the left.  The left side is heavier.

3. Choose the two coins that were on the left side both times.

You should be able to take it from there.[/quote]You are not assuming the worst luck!. What if the different coin is one of the two that you are not weighting from the first step to the second one, then, you are not finding the different. And even if your steps go as you say, what happens there, you are not told that the different coin is heavier, it can be lighter and be in the right!

[quote="djimenez"]Well, as it hasn't been solved yet and it has been almost a week since the last post, then, a hint: You don't eliminate a single coin until after the last step!
Actually, I believe that it can be done even with 13 coins, but usually what I write after midnith is nonsence! (hehe, that affirmation can be understood in several different ways :P)[/quote]

Actually, I think it's easier with 13 coins.

[quote="djimenez"][quote="probability1.01"]I will assume the worst luck.

1. Weigh 4 vs. 4.  They are unbalanced.  Let's say left side is heavier.

2. Weigh 2 from the left and 1 from the right vs. 2 from the right and 1 from the left.  The left side is heavier.

3. Choose the two coins that were on the left side both times.

You should be able to take it from there.[/quote]You are not assuming the worst luck!. What if the different coin is one of the two that you are not weighting from the first step to the second one, then, you are not finding the different. [/quote]
Well then it's down to two and it's trivial... (which I thought would be better luck)
[quote]And even if your steps go as you say, what happens there, you are not told that the different coin is heavier, it can be lighter and be in the right![/quote]
I didn't say what the last weighing was.  I just said to take out the two possibilities--someone who hasn't done it should figure out what to do from there.

[quote="probability1.01"](which I thought would be better luck)[/quote]

When solving a problem such as this, you really should provide for the actions given every possible outcome, not just what you consider to be the worst.

[quote="probability1.01"]Well then it's down to two and it's trivial... (which I thought would be better luck)[/quote]
If you don't previusly know or figure out if the coin that is different is heavier or lighter, then to reduce the problem to two coins is totally useless, and it is far to trivialize the problem.

If you know that you have to coins (according to your scheme, only one more time to weight them) but you don't know if the different coin is heavier or lighter, then what do you do? you weight them? so what? I mean, You already know that one is heavier and one is lighter, because one is like all the others and one is different, to weight them is not going to tell you anything different!

[quote="polymorphic"]Actually, I think it's easier with 13 coins.[/quote]Really? How?

[quote="probability1.01"]Ehh sorry I was hoping to guide someone who hasn't done the problem to a solution  :blush: .[/quote]Are you sure you have a correct solution? I know that what you say is that you don't post one but, do you have one?

so.
you first put 4 coins on each plate. if they are equal it means the fake one is among the other 4 left. you then measure two of the 4 coins left with 2 good ones (you know the first 8 you used are good).Then you already know that eather these 2 coins or the 2 coins left unmeasured contain the fake one, so in the third try you compare one of them with a good one.
suppose the two plates with 4 coins don't remain equal.
let's denote by a,b,c,d the coins on plate that goes lower and x,y,z,t the coins on plate that goes higher, and also p,q,r,s the four coins which are good.
you then measure p,q,r,z on the left with x,y,c,d on the right and we have 3 cases:
1. they stay equal=> eather one of a or b is havier eather t is easier so in 3rd measurement you measure a and b. if they stay equal=>t is fake, if not you chose the one that goes lower.
2. if left goes upper it means eather z is easier or one of c or d is heavier and you do just as in case 1.
3.if left goes lower it means eather x or y is easier so you just compare one of them with a good one.
that's all folks. hope i didn'd do any mistakes. :)

it is weird.
I knew the answer of this since like 3-4 years ago,
but my math teacher brought this to class last week, and I saw two similar posts here and there!!

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=241507#p241507

so, now that we are speaking about it i have another one. :)
you have 10 bags of nuts; the nuts in 9 of the bags weight 10 grams each except for the nuts in one bag that measure 9 grams each.
also you have a weighting device which tells you the weight of the object you put on its plate( so the weighting device is not a balance like in the previous problem; this one tells you teh exact weight of the object you measure). doing just one weighting determine the "fake" bag.

[hide="I knew this also"]take out 1 nut from bag#1
take out 2 nuts from bag#2
take out 3 nuts from bag#3
...
take out 10 nuts from bag#10

Then, the total weight should be 550g.
However, one bag has small nuts, so it will make the total weight somewhat less than 550g.

If bag#1 contained light nuts, the total weight must have been 549g.
If bag#2 contained light nuts, the total weight must have been 548g.
....
So, if bag#n contained light nuts, the total weight must have been (550 - n)g

[/hide]

Congrats, i have to say, this is the best solution:you dont even need to do your second weighting depending on the first./
 ////////////////left////////////////////////right
1st weigh  ABCH //////////////////////  DEFK
2nd weigh ABDI //////////////////////   CEGH
3rd weigh  ACDJ /////////////////////    BFGI

a reason why this is true is because $3^3>12\cdot2$

EDIT: yes, apparently i made a mistake but at least this is "possible"

Heh, just provide a single link to [url=http://en.wikipedia.org/wiki/Balance_puzzle]Wikipedia[/url].

Also, where is coin L in your solution? Sure, if all turns out balanced then you know L is the different one, but you can do better: You can determine whether L is heavier or lighter too. Your current solution doesn't state so.

Every weighing lets you sort 3 groups, so you could find 1 coin out of 27 if you really wanted to.

Wait.. isn't this problem like http://www.artofproblemsolving.com/Forum/viewtopic.php?f=149&t=523236


[quote="alexrait1"]suppose we have:
$x_1+x_2+x_3+x_4 = 30$;
and
$x_1 \le 5$ ;
$x_2 \le 10$;
$x_3 \le 15$;
$x_4 \le 21$;

How many solutions there are?

Thanx for your help.
I don't know what properties I can define here...[/quote]

assuming you're talking about non-negative integer solutions...

we find the number of solutions where at least one of the four conditions doesn't hold, and then subtract from the total number of solutions, which is $ C(33, 3) $.

let $ S_i $ denote the set of solutions where condition $ i $ is false.  then by inclusion-exclusion,

$ | S_1 \cup S_2 \cup S_3 \cup S_4 | = |S_1| + |S_2| + |S_3| + |S_4| - |S_1 \cap S_2| - |S_1 \cap S_3| - |S_1 \cap S_4| - |S_2 \cap S_3 | - |S_2 \cap S_4| - |S_3 \cap S_4| $.

(the rest of the terms, like $ |S_1 \cap S_2 \cap S_3| $ are 0 -- i.e., no more than 2 of the conditions can be false -- since we're talking about non-negative integer solutions.)  note also that $ |S_3 \cap S_4| = 0 $ because conditions 3 and 4 can't both be false (because, again, we're talking about non-negative solutions).

now, we show how to find, for instance $ |S_1 \cap S_2| $ (computing the rest of the terms is similar...  for conditions 1 and 2 to both be false, $ x_1 $ must be of the form $ 6 + y_1 $, and $ x_2 $ must be of the form $ 11 + y_2 $, where $ y_1, y_2 $ are non-negative.  the equation then becomes $ y_1 + y_2 + x_3 + x_4 = 13 $, for which there are $ C(16, 3) $ solutions...

I have a somwhat similar problem...
Consider n students in a queue. Each knows who stands in front of him. How many options do you have to resettle them so that no one sees the same student in front of him.
So we have $p_{i}$ - student i sees the same student he saw previously.
$W(p_{i}) = (n-1)!$
$W(1) = n!$
$W(p_{i}, p_{i+1}) = (n-2)! $
$W(p_{i}, p_{j}) = (n-3)!$ where $ i+1 \ne j $
$W(2) = (n-1)(n-2)! + [ \binom{n-2}{2} - (n-1)](n-3)!$
.
.
I have some difficulty to continue.. And probably I am doing something wrong here.

[quote="pleurestique"]"Each knows who stands in front of him"

you mean just the student right in front of him, next to him, not *all* students in front of him in line, right?[/quote]
exactly...
studnet number i must not see student number i+1 after the permutation.


[quote="MCrawford"]The average of five numbers is 6.  If one of the five numbers is removed, the average of the four remaining numbers is 7.  What is the value of the number that was removed?[/quote]
[hide]
The total is 30, so..
4a + n = 30
4a = 28
n = 2
[/hide]

[quote="MCrawford"]The average of five numbers is 6.  If one of the five numbers is removed, the average of the four remaining numbers is 7.  What is the value of the number that was removed?[/quote]

[hide]
The total of the original group is 30 (5*6) and the total of the changed group is 28 (4*7). Therefore, the number removed must be 2 (30-28).
[/hide]

[quote="MCrawford"]The average of five numbers is 6.  If one of the five numbers is removed, the average of the four remaining numbers is 7.  What is the value of the number that was removed?[/quote]

[hide]the sum of all the numbers is 30

when there is 4 left, the sum is 28

so the number is 2[/hide]


inequalities
logarithms
inequalities proposed

solve the inequalitiy bellow:
$ {\log_2\left(\ {x}^{12}\plus{}3x^{10}\plus{}5x^8\plus{}3x^6\plus{}1\right) <1\plus{}\log_2\left(\ x^4\plus{}1 \right) }$
[hide]This question is from a Competition for Chinese  senior students  called [全国高中数学联赛][ just the name in Chinese  :wink: ] and the question is raletively easier than CMO [It is closer to National College Entrance Examination [some of the problem from college entering examination in China can be even harder than problems from math competition. :wink: ] ]. the students who get high mark in this competition can take part in CMO.[/hide]
What about this inequality? dear mathlinkers , can you solve it ? :?:

The inequality is equivalent with
$ \log_{2}(\ 2x^{4} + 2 = A) > {\log_{2}(\ {x}^{12} + 3x^{10} + 5x^{8} + 3x^{6} + 1 = B )}$
$ 2 > 1$ , so $ x$ must satisfy the conditions:
$ B > 0$ and $ A > B$
I can't evaluate $ x$ from these conditions.Does anyone ?

[quote]The inequality is equivalent with  $ \log_{2}(\ 2x^{4}+2 = A) >{\log_{2}(\ {x}^{12}+3x^{10}+5x^{8}+3x^{6}+1 = B )}$
[/quote]
well done Obel1x  :clap2: ! and just give you a hint :[hide] because $ {f(x)=\log_2 x}$ is increasing on (0,+∞) ,so $ {\Longrightarrow}$ $ {x^{12}+3x^{10}+5x^8+3x^6+1<2x^4+2}$
just solve this inequality :$ {x^{12}+3x^{10}+5x^8+3x^6-2x^4-1<0.}$ [/hide]

wolframalpha gives

$ x^2 < \frac{1}{2} (\sqrt 5 \minus{}1 )$

nice  :wink: ,right answer enndb0x ,but that's just wolfalpha's answer $ {x\in \left( \minus{}\sqrt{\frac{\sqrt{5}\minus{}1}{2}}, \sqrt {\frac{\sqrt{5}\minus{}1}{2}} \right)}$ ,can you give a proof , dear enndb0x ?

With the $ y \equal{} x^2$ substitution
$ y^6\plus{}3y^5\plus{}5y^4\plus{}3y^3\minus{}2y^2\minus{}1 \equal{} (y^2\plus{}y\minus{}1)(y^4\plus{}2y^3\plus{}4y^2\plus{}y\plus{}1)$, with the second factor positive, so one needs $ y \in \left ( \minus{}\frac {\sqrt{5} \plus{}1} {2}, \frac {\sqrt{5} \minus{}1} {2} \right )$, so $ x^2 < \frac {\sqrt{5} \minus{}1} {2}$.


[quote]Suppose that $ {a_n}$ is an arithmetic sequence with

\[ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_{100} \equal{} 100\] and \[ a_{101} \plus{} a_{102} \plus{} \cdots \plus{} a_{200} \equal{} 200\].

What is the value of $ a_2\minus{}a_1$?[/quote]

I need help solving this. And I have trouble with sequences in general...

[hide="too much latex"]
do partial sum equations on both equation
just in case you don't know, here it is
$ S_n\equal{}\frac{n \times (a_1\plus{}a_n)}{2}$

So you get $ \frac{100 \times (a_1\plus{}a_{100})}{2}\equal{}100$
and  $ \frac{100 \times (a_{101}\plus{}a_{200})}{2}\equal{}200$

Simplifying gives us
$ a_1\plus{}a_{100}\equal{}2$
$ a_{101}\plus{}a_{200}\equal{}4$
Set d as the common difference, the question is asking for d
We know $ a_1\plus{}100d\equal{}a_{101}$
and $ a_{100}\plus{}100d\equal{}a_{200}$
Therefore, $ a_1\plus{}a_{100}\plus{}200d\equal{}4$
              $ a_1\plus{}a_{100}\equal{}2$
Subtract to get 200d=2
$ d\equal{}\frac{1}{100}$
[/hide]


Everyone knows a calculator is a mathematician's best friend. Where would I be if I didn't have my TI-89? But a calculator is more than being a help on tests/competitions. It's really valuable to have when you are bored and have nothing else to do. So does anyone have games on their calculator? If so, what are they? Where did you download them? I have Phoenix right now and I'm addicted to it. :P Got caught playing it in Physics class... :blush:

I have something called PuzzPack and it can be downloaded from [url=http://www.detachedsolutions.com]www.detachedsolutions.com[/url].  I use a TI-83 + SE so it should work on yours.

i have a pretty cool game (i think its called bomberkids) its a one/two player link/no link game and you just collect bombs and blow the other dude up. its pretty fun. i got it from a friend so i can't tell you where to get it though.

Those are all Ti-83+ games. The TI-83's use ZiLOG Z80 processory, and the TI-89's use a motorola 6800. They also have very different OS's. If you buy a link cable, and go here: [url=http://www.ticalc.org]www.ticalc.org[/url], you should be able to find all the games you want. I warn you though, games are very addicting and will compete with math for your time. The only thing that prevented me from being completely enslaved to calculator games was losing my calculator.

[url=http://www.ticalc.org]This[/url] is probably the best calculator game resource on the Internet.  Click 'Archives.'

BomberDude is probably one of the best games there  :lol:

I used to have a Command and Conquer-type game for my TI-89.  The controls are horrible, though.  But my friend made a mouse for the TI-89!

a mouse?
That's awesome!
I got worms on my TI-83+
but they're not actually worms they're stick figures...

[quote="jli"]can you downlad games free without getting it from your friend.

can you download a game from a friend onto a TI-83?[/quote]

I got Phoenix from a friend at school but you can just download from the internet. Just google it up. However, you need to have that link cable thing. If you don't have one, you should get one because it's useful. You can download stuff other than games
ex. periodic table on to your calculator.

No. I wish, that would be awesome to have games on my calculater, but nope. I am gameless.

[quote="bleumoose"]the BEST calc game is most definitely fall down. it is totally mindless and addicting, i must admit, but you just. keep. playing it. because you keep thinking you'll beat that last score you got.[/quote]

have u played fallup (or i think thats what its called)? its pretty addicting and is really weird cause its falling up  :P .

Anybody have simple TI-82 code for any program? I can't find any code, just programs

One of my personal favorites is Block Dude.  It is most commonly found in the PuzzPack for the TI-83, but there are versions for the -86 and -89 as well.  (BTW, has anyone else beat all 11 levels?)


let ABC be a triangle .and AM be the bisector of the side BC.

prove that:  b-c/2 <AM< b+c/2

Play around with Stewart's Theorem:

We call the length of $AM=d$ and we set the lengths of $BM=CM=x$

Isolating for $d^2$ in Stewart's Theorem:

$d^2=\frac{b^2+c^2-2x^2}{2}$

Combined with the inequality:

$x \le \frac{b+c}{2}$ (triangle inequality)

x is maximized in the degenerate triangle when $x=\frac{b+c}{2}$, which gets us

$d=\frac{b+c}{2}$ as an upper bound.

Lower bound is found similarly using the lower part of the Traingle Inequality.


Most likely here most people are more concerned with the mathematical and functional aspects of $ \text{\LaTeX}$ rather than obscure typographical refinements, and sorry for not providing any links... but...

I'm looking for a nice, non-modern style serif font package (preferably Palatino or Garamond, since I'm already acquainted with those) complete with ligatures and kerning (like any other decent font package) [b]that provides old-style numerals and real small caps[/b] (and of course an accompanying set of math characters), and is fully compatible with amsmath, etc. kp-fonts has really nice features and is quite compatible, but it looks kind of weird unfortunately (Es, Fs, and Ts have their top-bars' right corners shaved, and the italic [i]w[/i]s, [i]v[/i]s, and [i]y[/i]s are a bit strange).

Secondly, is there a free calligraphic font for $ \text{\LaTeX}$ that doesn't look as modern as Calligra, but more along the lines of Zapfino (perhaps this is asking a little much)? By modern, I mean with heavy vertical strokes and a high contrast between thick and thin lines (as in Calligra or Computer Modern). Of course, there doesn't need to be millions of variations for each letter and letter combination as in Zapfino.

And also, I'm having trouble downloaded packages straight from zipped files (I've only downloaded packages via the MikTeX Package Manager) on the internet. How exactly does one do this? Perhaps it's the poor documentation or nonstandard-ness of non-'official' packages, but I've never successfully downloaded a package this way. Where do you put the files? Online tutorials usually point to a directory that I don't have. How about font packages?

And a side-question: what program to publishers use to print books? Books, of course, are well-typeset, contain ligatures, kerning, and end-of-line hyphenation, and thus could not have been done with something like MS Word or OpenOffice Writer. I have spotted incorrect usage of ligatures in some of my textbooks, so I know there must exist (mediocre) typesetting programs other than $ \text{\LaTeX}$, which would never make such a mistake.

[b]Fonts[/b]
A couple of places that give sample LaTeX fonts:
[url=http://www.tug.dk/FontCatalogue/]The LaTeX Font Catalogue[/url].
[url=ftp://tug.ctan.org/pub/tex-archive/info/Free_Math_Font_Survey/survey.pdf]A Survey of Free Math Fonts for TeX and LaTeX[/url]
Lots more information at [url=http://www.tug.org/fonts/]Fonts and TeX[/url]

[b]Packages[/b]
All packages come with installation instructions. Some will need compiling first before putting them in the relevant place called the texmf tree. In your case this is likely to be C:\Program Files\MiKTeX 2.7. A better method is to use a localtexmf tree where you put your downloaded packages which won't be affected by any upgrades to MiKTeX. Here for example are some instructions I wrote for installing the mathabx fonts which are not included with MKTeX. Other packages will be similar.

[quote]1. Decide on a name and location for your local latex files. Mine is C:\localtexmf but the name doesn't matter, though it helps to avoid spaces in file names. Create the directory then tell MiKTeX you've done so (so it knows to look there for files) in MiKTeX's Settings, Roots, Add

2. Download the mathabx files from CTAN [url=http://tug.ctan.org/get/fonts/mathabx.zip]mathabx.zip[/url]

3. Extract the files. You can put things where you like but my setup mimics the main distribution and has:
a) put all the font files *.mf in c:\localtexmf\fonts\mathabx\source
b) put all the remaining files in c:\localtexmf\tex\latex\mathabx

4. Now the important bit - tell MikTeX what you've done by going to MiKTeX, Settings, Refresh FNDB[/quote]

Are you just wanting [url=http://www.tug.dk/FontCatalogue/palatino/]\use{palatino}[/url]? It's been preinstalled on every LaTeX distro I've used.

(And, unrelatedly, you probably are going to use a bibliography sometime; thus let me introduce you to BibTeX if you don't already know each other :). The only flaw in it is the single extant bst file that provides annotated bibliographies; but for ordinary bibliographies nothing beats BibTeX.)

How does one use old style numerals in, say, Palatino?

And where can I find a Garamond with old style numerals, real small caps, and math symbols? (If I'm not mistaken, urw-garamond with mathdesign provides neither old style numerals nor real small caps.)

(Concerning BibTeX: I'm having a horrible experience with it right now; unfortunately it's not so humanities-style-referencing-friendly: none of the styles conform to MLA. And the define-your-own style dialogue in the command prompt is... awful. Perhaps someone can help?)

http://www.ctan.org/tex-archive/biblio/bibtex/contrib/mla/ - mla.bst in same directory as file - \bibliographystyle{mla} - mla funness! generally googling for bst files for your style works well. (Unless you need an annotated bibliography, as I said.)

\usepackage[osf]{mathpazo} plus \usepackage{palatino} for your desired palatino with old-style?

Thanks stevem and solafidefarms!

\usepackage[osf]{mathpazo} doesn't seem to be valid... so I won't post the log unless someone asks for it.

Does \usepackage{mathpazo} work without osf?

Yes, it works without osf (but with normal numerals, not old-style); I'm not even sure what the osf is.


Έστω ΑΒΓ τρίγωνο, Ο  το κέντρο του περιγεγραμμένου κύκλου και Η το ορθόκεντρο .Δείξτε ότι οι  κύκλοι των 9 σημείων (του Εuler) των τριγώνων ΟΗΑ , ΟΗΒ και ΟΗΓ έχουν δύο  κοινά σημεία

Λοιπόν αυτή του Ράπανου έχει πολύ  καιρό άλυτη . Μιχάλη τι λες ?Θα την σκοτώσεις και αυτή (εγώ πάντως έχω σύντομη λύση :P )

Θα δείξουμε ότι οι κύκλοι Euler των $ \triangle OHA,\triangle OHB,\triangle OHC$ τέμνονται εκτός από το σημείο $ S$ πάνω στον κύκλο Euler του $ \triangle ABC$. Είναι εύκολο να δείξουμε ότι ο κύκλος Euler του $ \triangle OHA$ τέμνει τον κύκλο Euler του $ \triangle ABC$ σε σημείο $ Q$ έτσι ώστε $ \angle DSQ \equal{} |180\minus{}2\angle AHO|$, αφού $ \angle DKS \equal{}\angle DQS \equal{}\angle QDS \equal{}\angle QKS$, οπότε $ \angle DSQ \equal{} 180\minus{}\angle DKQ \equal{} 180\minus{}2\angle DKS \equal{} 180\minus{}2\angle AHO$. Αντίστοιχα για τα σημεία $ Q', Q''$ θα ισχύουν $ \angle ESQ' \equal{} |180\minus{}2\angle BHO|,\angle FSQ'' \equal{} |180\minus{}2\angle CHO|$. Άρα για να αποδείξουμε ότι  $ Q\equiv Q'\equiv Q''$, αρκεί να δείξουμε ότι $ \angle ESQ \equal{} |180\minus{}2\angle BHO|,\angle FSQ \equal{} |180\minus{}2\angle CHO|$. Όμως \[ \angle FSQ\equal{}\angle DSF\minus{}\angle DSQ\equal{}2\angle DA'F\minus{}180\plus{}2\angle AHO\equal{}2\angle B\plus{}2\angle AHO\minus{}180\equal{}2\angle B\plus{}360\minus{}2\angle B\minus{}180\minus{}\angle CHO\equal{}180\minus{}\angle CHO\], αφού $ \angle AHO\plus{}\angle CHO \equal{}\angle C'HA' \equal{} 180\minus{}\angle B$. Για το $ Q''$ η απόδειξη είναι ανάλογη.

Προκύπτει έτσι ότι ο τρεις αυτοί κύκλοι είναι ομοαξονικοί (coaxal) και συνεπώς τα κέντρα τους είναι συνευθειακά.


inequalities
three variable inequality
algebra

Ah,let's see this.
First,let us assume $x,y,z$ are non-negative,you can easily
turn other conditions into this.
Then let's assume $x \leq y \leq z$
And we claim that $x \leq y \leq 1$
If $z \leq 1$,
then expand $(x-1)(y-1)z \geq 0$
You'll get $xyz +2 \geq xz+yz+2-z$
And $xz+yz+2-z \geq x+y+z \iff 2(1-z) \geq (x+y)(1-z)$,which is trivial.
If $z \geq 1$,
expand $(z-1)(x-1)(y-1) \geq 0$
You'll get $xyz+2 \geq xy+yz+xz+3-(x+y+z)$
And we have $\displaystyle xy+yz+xz+3=2+ \dfrac{(x+y+z)^2}{2} \geq 2(x+y+z)$
Thus our ineq is proven.
About when this ineq occurs,I haven't think about it,sorry.

Apply Cauchy's ineq :
 $[x(1-yz) +(y+z)]^2 \leq (2+2yz)(1-2yz+y^2z^2)$
 We need to prove : $(1+yz)(1-2yz+y^2z^2) \leq 2 \iff yz \leq 1$ , which is obviously true because $y^2 +z^2 \leq 2$
 The equality holds for $(x,y,z) = (1,1,0)$

[quote="nttu"][quote="Megus"]Nttu, can you please be more intructive? I cannot see how you grouping terms in your Cauchy to get left side. Please introduce exact brackets (or tell it) so that everybody knows what you multiply with what. Sorry if I am annoying.[/quote]

   $x+y+z \leq 2+xyz \iff x(1-yz)+(y+z) \leq 2$[/quote]

I meant this one

 $[x(1-yz) +(y+z)]^2 \leq (2+2yz)(1-2yz+y^2z^2)$

I have no idea how it follows from Cauchy...

Might be. But still, no rigorous proof has been given yet!!

Actually, we have
$x+y+z \leq 2+xyz \iff x(1-yz)+(y+z) \leq 2$

$[x(1-yz) +(y+z)]^2 =[\sqrt{2-y^2-z^2}(1-yz)+y+z]^2 \le [2-y^2-z^2+(y+z)^2][(1-yz)^2+1]=2(1+yz)(2-2yz+y^2z^2)\le 4.$

The last inequality is true due to:
$2(1+yz)(1-2yz+y^2z^2) \leq 4 \iff yz \leq 1$ , which is obviously true because $y^2 +z^2 \leq 2$
 The equality holds for $(x,y,z) = (1,1,0)$

I compiled posts (in Vasc's post) by Vasc and Nttu to get rigorous proof. Please try next time to give a rigorous proof as it is made for Resources project.

See also

http://www.mathlinks.ro/Forum/viewtopic.php?t=149554
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32399

  darij

This quation is not a very hard. :lol:

1.   XYZ>=0

Because this inequality is simmetric we can think that X<=Y<=Z

WE can write KAUSHI SWARTS inequality and we have
   (1+1+xyz)(x*x+y*y+z/xy)>=(x+y+z)^2
x*x+y*y+z*z=2 => xy<=(1/3) and xyz<=(sqrt(2*2*2/3*3*3))
  => that  xyz+xy<1 => z/xy>z*z+xyz => x*x+y*y+z/xy>= 2+xyz  (because x*x+y*y+z*z=2) =>
(1+1+xyz)^2>=(1+1+xyz)(x*x+y*y+z/xy)>=(x+y+z)^2  =>

    2+xyz>=x+y+z
With the same way we can prove the inequality when  XYZ<0

[quote=Megus]If $x, y, z$ are real numbers satisfying $x^2 +y^2 +z^2 = 2$, prove the inequality
\[ x + y + z \leq 2 + xyz \]When does equality occur?[/quote]
If $x, y, z$ are real numbers satisfying $x^2 +y^2 +z^2 = 2$, prove the inequality
\[ |x| + |y|+|z| \leq 2 + xyz \]

[quote=sqing]
If $x, y, z$ are real numbers satisfying $x^2 +y^2 +z^2 = 2$, prove the inequality
\[ |x| + |y|+|z| \leq 2 + xyz \][/quote]
Try $x=y=\sqrt{\frac{2}{3}}$ and $z=-\sqrt{\frac{2}{3}}$.

I don't think that will hold since xyz can go negative...

[quote=Megus]If $x, y, z$ are real numbers satisfying $x^2 +y^2 +z^2 = 2$, prove the inequality
\[ x + y + z \leq 2 + xyz \]
When does equality occur?[/quote]
1989 Chinese National Team test question.
By C-S ,
$| {x + y + z - xyz} | = | {x(1 - yz) + y + z}|$
$ \le \sqrt{\left( {{x^2} + {{(y + z)}^2}} \right)({(1 - yz)^2} + 1) }= \sqrt{2(1 + yz)(2 - 2yz + {y^2}{z^2}) }\le 2 $
$\left((1 + yz)(2 - 2yz + {y^2}{z^2}) \le2 \Leftrightarrow yz\leq 1=\frac{x^2 +y^2 +z^2}{2}\right)$.
http://www.artofproblemsolving.com/community/c6h550371p3194028

[quote=sqing][quote=Megus]If $x, y, z$ are real numbers satisfying $x^2+y^2+z^2=2$, prove the inequality
$$x+y+z\leq xyz+2$$
When does equality occur?[/quote]
[url=https://artofproblemsolving.com/community/c6h365364p2009035]1987 IMO Longlist[/url][/quote]
 Let $a,b,c\geq0,k\geq1$  such that$a^k+b^k+c^k=2.$ [url=https://artofproblemsolving.com/community/c6h1585431p9859534]Prove that[/url]
$$a+b+c\leq abc+2$$
[url=https://artofproblemsolving.com/community/c6h506554p2845215]here[/url]

If $a,b,c$ are real numbers satisfying $a^2 +b^2 +c^2 = 2.$  Then
$$a+b+c \leq abc+2$$
If $ a , b, c $ be nonnegative number satisfying $ a^2+b^2+c^2+abc = 4.$   [url=https://artofproblemsolving.com/community/c6h506554p2845215]Then[/url]$$a+b+c \geq abc+2.$$

[quote=Megus]If $x, y, z$ are real numbers satisfying $x^2 +y^2 +z^2 = 2$, prove the inequality
\[ x + y + z \leq 2 + xyz \]
When does equality occur?[/quote]
If $x, y, z$ are real numbers satisfying $x^2 +y^2 +z^2 = 2$, [url=https://artofproblemsolving.com/community/c6h2121044p15437795]prove the inequality[/url]
\[ x + y + z \leq 2 + xy \]
 The equality holds for $x=1,y=0,z= 1$ or $x=0,y=1,z= 1.$
Let $a, b, c$ be real numbers with $a^2+b^2+c^2=9$. Prove that $$2(a+b+c)\leq abc+10$$
Let $a, b, c$ be real numbers with $a^2+b^2+c^2=1$. Prove that $$a+b+c\leq 2abc+\sqrt{2}$$

[quote=sqing]Let $a, b, c$ be real numbers with $a^2+b^2+c^2=1$. Prove that $$a+b+c\leq 2abc+\sqrt{2}$$[/quote]
Let $a, b, c$ be real numbers with $a^2+b^2+c^2=1$. Prove that$$|a+b+c- 2abc|\leq\sqrt{2-5a^2b^2c^2}$$

Let $a, b, c$ be positive real numbers with $ab+bc+ca=3$. [url=https://math.stackexchange.com/questions/181121/inequality-abc-geqslant-abc-2]Prove that[/url]$$a+b+c \geq abc+2$$
[hide=Nice. Thanks.][quote=NaPrai][quote=sqing]Let $a, b, c$ be real numbers with $a^2+b^2+c^2=1$. Prove that$$|a+b+c- 2abc|\leq\sqrt{2-5a^2b^2c^2}$$[/quote]

First, we recall the identity from #22 which is
$$2-(a+b+c-2abc)^2 = 4a^2b^2c^2+\prod_{cyc}\left(1-2ab\right).$$
Notice that \begin{align*}\prod_{cyc}\left(1-2ab\right) &= \prod_{cyc}\left((a-b)^2+c^2\right) \\&\ge \prod_{cyc}c^2 \\&= a^2b^2c^2.\end{align*}
Therefore, $$2-(a+b+c-2abc)^2 \ge 5a^2b^2c^2$$ which we may reduce it as $$| a+b+c-2abc| \le \sqrt{2-5a^2b^2c^2}.$$
Equality holds for $a=b=c=\frac{1}{\sqrt{3}}$ or $a=b=\frac{1}{\sqrt{2}},c=0$ or its permutation.[/quote]

[/hide] [url=https://artofproblemsolving.com/community/c6h1184132p5744371]h[/url]

[hide=1st sol by an well known idea]
WLOG assume that $x\leq y\leq z$
By considering graph, there exists $k\geq 0$ such that $(t-x)(t-y)(t-z)+k = (t-x_1)(t-y_1)^2$ as a polynomial of $t$.
In other words, $y_1=z_1$, $x+y+z=x_1+y_1+z_1$, $xy+yz+zx=x_1 y_1 + y_1 z_1 + z_1 x_1$, $xyz \geq x_1 y_1 z_1$.
Therefore, it suffices to prove the case $x\leq y=z$. In this case,
$$xyz+2-x-y-z = xy^2 + 2-x-2y = (1-y)(2-x-xy) \geq 0$$
Because $x^2+2y^2=2\Longrightarrow y\leq 1$ and $2x(y+1)\leq x^2 + (y+1)^2 = -y^2 +2y+3=-(y-1)^2 +4\leq 4$.
Equality holds iff $(x,y,z)=(0,1,1)$ or permutaion of it.
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[hide=2nd sol by Lagrange Multiplier]
$g(x,y,z)=x^2+y^2+z^2=2$ is a sphere, hence a compact surface in $\mathbb{R}^3$.
By Lagrange multiplier method, $f(x,y,z)=xyz-x-y-z$ attains minumum only if $\nabla f = \lambda \nabla g$. More explicitly,
$$ \begin{pmatrix} yz-1 \\ zx-1 \\ xy-1 \end{pmatrix} = \lambda \begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix}$$
$\Longrightarrow xyz = (2\lambda x+1)x = (2\lambda y+1)y = (2\lambda z+1)z$
$\Longrightarrow (2\lambda x + 2\lambda y+1)(x-y)=(2\lambda y + 2\lambda z+1)(y-z)=(2\lambda z + 2\lambda x+1)(z-x)=0$
WLOG $y$ is the median of $x,y,z$.
[b] Case 1. [/b] If $x\not=y$ and $y\not=z$, then $x+y=y+z=-1/2\lambda \Rightarrow x=z$. Impossible!
[b] Case 2. [/b] If $x=y=z$, then $x=y=z=\pm\sqrt{\frac{2}{3}} \Rightarrow f(x,y,z)=\mp \frac{7}{3}\sqrt{\frac{2}{3}}$
[b] Case 3. [/b] If $x\not=y$ and $y=z$, then $x+y=x+z=-1/2\lambda$.
$\Longrightarrow (yz-1)+(zx-1) = 2\lambda (x+y)=-1 \Longrightarrow z(x+y)=1$
$\Longrightarrow x^2 + 2y^2 = 2 = 2y(x+y)=2xy+2y^2 \Longrightarrow x^2 = 2xy \Longrightarrow x=0$ or $x=2y$
$\Longrightarrow (x,y)=(0,\pm 1)$ or $(\pm\frac{2}{\sqrt{3}},\pm\frac{1}{\sqrt{3}}) \Longrightarrow f(x,y,z)=\mp 2$ or $\mp \frac{10}{3\sqrt{3}}$
Among $\mp \frac{7}{3}\sqrt{\frac{2}{3}}, \mp 2, \mp \frac{10}{3\sqrt{3}}$, the minimum is $-2$.
Equality holds iff $(x,y,z)=(0,1,1)$ or permutaion of it.
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Let $a_1,a_2,\cdots,a_n  (n\ge 2)$ be non-negative real numbers such that $a_1+a_2+\cdots+a_n=a_1a_2\cdots  a_n+n-1.$ Prove that$$a_1^2+a_2^2+\cdots+a_n^2\geq n-1.$$
[hide=n=2,3]Let  $a,b$ are non-negative real numbers such that  $a+b=ab+1.$ Prove that
$$a^2 +b^2  \geq 1$$
Let  $a,b,c$ are non-negative real numbers such that  $a+b+c=abc+2.$ Prove that
$$a^2 +b^2 +c^2 \geq 2$$
[url]https://artofproblemsolving.com/community/c4h1813778p18017844[/url]
[url]https://artofproblemsolving.com/community/c4h1999552p16049721[/url] [/hide]

[quote=sqing]Let $a, b, c$ be real numbers with $a^2+b^2+c^2=1$. Prove that $$a+b+c\leq 2abc+\sqrt{2}$$[/quote]
[url=https://artofproblemsolving.com/community/c6h2226564p16998623]here[/url]
$$a+b+c-2abc=a(1-2bc)+(b+c)\leq \sqrt{[a^{2}+(b+c)^{2}][(1-2bc)^{2}+1]}=\sqrt{2+4b^{2}c^{2}(2bc-1)}\leq \sqrt{2}$$