trigonometry
geometry
circumcircle
inequalities
triangle inequality
geometry solved

[quote="nttu"]
  Let $ABC$ be a triangle , a point$ I$ is in the triangle $ABC$ $ <BIC = x , <AIC = y < AIB = z $ , $M$ is an arbitrary point in the plane , we have :
 $ MA.sinx + MB .siny + MC .sinz \geq IA.sinx + IB .siny + IC .sinz $

 [/quote]

[color=blue][b]Theorem 1.[/b] Let P be a point inside a triangle ABC, and denote x = < BPC, y = < CPA and z = < APB. Then, for any other point Q in the plane, we have

$QA\cdot\sin x+QB\cdot\sin y+QC\cdot\sin z\geq PA\cdot\sin x+PB\cdot\sin y+PC\cdot\sin z$;

equality holds here if and only if Q = P.[/color]

[i]Proof.[/i] Let the circumcircle of triangle CPA intersect the line BP at a point Y (apart from the point P). Then, the quadrilateral APCY is cyclic, so that < AYC = 180° - < CPA = 180° - y, and by the chordal angle theorem, < YCA = < YPA = 180° - < APB = 180° - z and < CAY = < CPY = 180° - < BPC = 180° - x. Thus, by the Sine Law in triangle CYA, we have

$YC:CA:AY=\sin\measuredangle CAY:\sin\measuredangle AYC:\sin\measuredangle YCA$
$=\sin\left(180^{\circ}-x\right):\sin\left(180^{\circ}-y\right):\sin\left(180^{\circ}-z\right)=\sin x:\sin y:\sin z$.

Now, by the Ptolemy inequality, applied to the quadrilateral AQCY, we have

$QA\cdot YC+QC\cdot AY\geq QY\cdot CA$.

In view of $YC:CA:AY=\sin x:\sin y:\sin z$, this becoms

$QA\cdot\sin x+QC\cdot\sin z\geq QY\cdot\sin y$.

Thus,

$QA\cdot\sin x+QB\cdot\sin y+QC\cdot\sin z=QB\cdot\sin y+\left(QA\cdot\sin x+QC\cdot\sin z\right)$
$\geq QB\cdot\sin y+QY\cdot\sin y=\left(QB+QY\right)\cdot\sin y$.

By the triangle inequality for the (possibly degenerate) triangle BQY, we have $QB+QY\geq BY$. Thus,

$QA\cdot\sin x+QB\cdot\sin y+QC\cdot\sin z\geq BY\cdot\sin y$.

Now, when does equality hold in this inequality? In deducing this inequality, we had applied the Ptolemy inequality to the quadrilateral AQCY; this inequality becomes an equality if and only if the quadrilateral AQCY is convex and cyclic, i. e. if the point Q lies on the arc CA of the circumcircle of triangle CYA. Also, we had applied the triangle inequality for the triangle BQY (namely in the form $QB+QY\geq BY$); this inequality becomes an equality if and only if the point Q lies on the segment BY. Hence, in the resulting inequality $QA\cdot\sin x+QB\cdot\sin y+QC\cdot\sin z\geq BY\cdot\sin y$, equality holds if and only if the point Q lies both on the arc CA of the circumcircle of triangle CYA and on the segment BY. But the arc CA of the circumcircle of triangle CYA and the segment BY have only one common point, namely the point P. Hence, in the inequality $QA\cdot\sin x+QB\cdot\sin y+QC\cdot\sin z\geq BY\cdot\sin y$, equality holds if and only if Q = P. Thus, we have

$PA\cdot\sin x+PB\cdot\sin y+PC\cdot\sin z=BY\cdot\sin y$.

And using this equation, we can rewrite our inequality $QA\cdot\sin x+QB\cdot\sin y+QC\cdot\sin z\geq BY\cdot\sin y$ in the equivalent form

$QA\cdot\sin x+QB\cdot\sin y+QC\cdot\sin z\geq PA\cdot\sin x+PB\cdot\sin y+PC\cdot\sin z$.

As we saw above, equality holds if and only if Q = P. Thus, Theorem 1 is proven.

By the way, thanks for the compliments in the other thread, Manlio...

  Darij

Thank you very much for you so timely and accurate proof :)

I have another solution :
[color=red][[b]Moderator edit:[/b] In the solution below, the angles $x$, $y$, $z$ are denoted by $\alpha$, $\beta$, $\gamma$.][/color]
 $ MA.IA \geq \overline{MA}.\overline{IA} = (\overline{IA} - \overline{IM}).\overline{IA} = IA^2- \overline{IA}.\overline{IM} $
 So $ MA.sin\alpha \geq IA.sin\alpha - \frac{\overline{IM}.\overline{IA}}{IA}.sin \alpha = IA.sin \alpha- 2.\frac{\overline{IM}.\overline{IA}.S_{IBC}}{IA.IB.IC} $
 So $ MA.sin\alpha + MB.sin\beta + MC.sin\gamma \geq IA.sin\alpha + IB.sin\beta + IC.sin \gamma + \frac{-2.\overline{IM}}{IA.IB.IC}.(S_{IBC}.\overline{IA}+ S_{IAC}.\overline{IB} + S_{IBA}.\overline{IC}) = IA.sin\alpha + IB.sin\beta + IC.sin \gamma $


function
algebra unsolved
algebra
functional equation

find all function f:Q---> Q such
       $1) f(x) f(\frac{1}{x})=1$
       $2) f(2x)=2f(f(x)) \forall x \in Q$

I think the only continuous functions are $f(x) = x$, $f(x) = |x|$, and $f(x) =-|x|$.

[quote="RDeepMath91"]1) $x = 1$ yields $f(1) = 1$ or $f(1) =-1$
$x =-1$ yields $f(-1) = 1$ or $f(-1) =-1$
For $|p| > |q|$, let $f(\frac{p}{q}) = k$ for any $k \in \mathbb{Q}$
Then it follows that $f(\frac{q}{p}) = \frac{1}{k}$ for $|p| < |q|$[/quote]
i don't think it right
 if $2p>q$then $k=f(\frac{2p}{q})=2f(f(\frac{p}{q}))=2f(\frac{1}{k})$ is it true???

[quote="hunter_cvp_ic"]nobody solve it???  :([/quote]
There are infinetely many solutions. For example $f(2^{k}p_{1}^{k_{1}}p_{2}^{k_{2}}...p_{m}^{k_{m}})=2^{k}p\prod_{i}p_{i}^{g(p_{i},k_{i})}.$
It is solution  for any functions g(p,n), satisfyed:
(i) $g(p,-n)=-g(p,n), p\in P, n\in Z$,
(ii) $g(p,n)=g(p,g(p,n)).$
For example let $\theta : P\to N$ and $g(p,n)=\theta(p)$ if $n>0$, $g(p,n)=-\theta (-n)$ if $n<0$, g(p,0)=0.


probability
expected value
probability and stats

Find $ E(X)$ if:

$ X \equal{} 2^Y$ where Y is a poisson random variable.

Can anyone help me solve this? :) Im clueless.

Thanks.

$ E(X)\equal{}E(2^Y)\equal{}e^{\minus{}\lambda}\sum_{y\equal{}0}^{\infty} \frac{2^y\lambda^y} {y!}$

$ \equal{}e^{\minus{}\lambda}e^{2\lambda}\equal{}e^{\lambda}$

where $ \lambda$ is the expected value of $ Y.$


let $ A_1,\dots,A_n$ be a finite family of $ F_{\sigma}$ sets, $ \bigcup A_i \equal{} I$ - interval.

Construct a family of $ F_{\sigma}$ sets $ B_1,\dots,B_n$ such that $ \bigcup B_i \equal{} I$ and $ B_i$ are pairwise disjoint

Any other conditions? How are $ B$ and $ A$ related?

sets $ A_i$ are given (they are subsets of interval $ I\subset\mathbb{R}$), and one should use them to construct sets $ B_i$

for example, taking their union (so that $ m\equal{}1$)? :huh:

$ m\equal{}n$, sorry for the mistake  :oops:

What do you mean "use them"? What stops me from just using [0,1/n],(1/n,2/n],(2/n,3/n], ... , ((n-1)/n,1]?

maybe he needs $ B_i\subset A_i$ for some reasons we're not given to know ;) but stating it would have been useful, maybe :p

It's certainly a more interesting question if the $ B$'s are to be a refinement of the $ A$'s, as you suggest.

[quote="primenumber"][quote="ma_go"]maybe he needs $ B_i\subset A_i$ for some reasons we're not given to know ;) [/quote]
yes - this condition is needed  :wink:

sorry for mistakes, it's not a single task, but a part of bigger theorem which I have to know with complete proof (exam), I don't understand few steps in it[/quote]

No problem. I would like to take a look at this, but would also like to sleep first :). When is your exam?

nice dreams  :wink:
exam - at the beginning of september

Seems like Nukular is still fast asleep :P, so let me try.

1) Any closed set is a $ G_\delta$ set.
2) The difference of two closed sets is an $ F_\sigma$ set.
3) A finite union of closed sets is closed.
4) A countable union of $ F_\sigma$-sets is an $ F_\sigma$-set.
5) If $ A_j\equal{}\bigcup_{k\ge 1}F_{jk}$, then
\[ \bigcup_j A_j\equal{}\bigcup_j \left[\bigcup_k (F_{jk}\setminus\bigcup_{\substack{\ell<k\text{ or }\\ \ell\equal{}k\text{ and }i<j}}F_{i\ell})\right]\,.\]

which sets do you have in this sum:
\[ \bigcup_{\substack{\ell < k\text{ or } \\
\ell \equal{} k\text{ and }i < j}}F_{i\ell}\]
?

$ F_{i\ell}$ for $ \ell < k$ ?
$ F_{ik}$ for $ i < j$ ?

so $ B_j\equal{}\bigcup_{i<j,\ l\le k}F_{il}$ ?
I don't see why these sets are pairwise disjoint ... :|
(all others conditions obvious)

Of course, not! :roll: $ B_j\equal{}\bigcup_k (F_{jk}\setminus\bigcup_{\substack{\ell < k\text{ or } \\
\ell \equal{} k\text{ and }i < j}}F_{i\ell})$.

ok, ok, I had a latex error and deleted a part of code there

assume that there exists $ b\in B_p\cap B_q$, where $ q > p$.
then
$ b\in F_{pk}\setminus \bigcup_{i < p,\ l\le k}F_{il}$ fore some $ k$
and
$ b\in F_{qm}\setminus \bigcup_{i < q,\ l\le m}F_{il}$ fore some $ m$
where $ k > m$. How to get contradiction?

Well, we have an ordering on pairs which is $ (i,\ell)<(j,k)$ if [($ \ell<k$) or ($ \ell\equal{}k$ and $ i<j$)]. The sets that are written are just $ F_{jk}\setminus \bigcup_{(i,\ell)<(j,k)} F_{i\ell}$ (you subtract from each set all previous ones), so they are all pairwise disjoint.

In your notation, $ (q,m)<(p,k)$, so $ b$ does not belong to the first set because $ F_{qm}$ is one of the sets explicitly subtracted there.

I think the possible source of misunderstanding is that awkward "... or ... and ..." construction, so I put it above again with parentheses to avoid any ambiguity.

[quote="fedja"]
I think the possible source of misunderstanding is that awkward "... or ... and ..." construction,[/quote]
Yes, it was. Now everything is clear, thanks very much :)


Put something like calculating combinations of salad dressings.

What made you think to post a poll on salad dressing?  It seems entirely random.

Great idea! Salad dressing is one of the most diverse things at the dinner table.

I like honey mustard, caesar, and ranch, but I like ranch more as a dip for vegetables and stuff. I've tried thousand island and french and italian but didn't like them as much.

i like a whole bunch of salad dressing but ranch is my favorite. i like the kraft Ranch dressing but the hidden valley stuff makes me REALLY sick... :P

I love bleu cheese, it is so tasty... and I like the cheese bleu cheese too.

i like ranch on salads and honey mustard on calculators....yum.

I actually dislike all salad dressing.  I don't know why. :P

Haha...I've liked Caesar ever since I first got it at Hard Rock Cafe...uh...4 , 5 years ago... :-D

Let's see...Ranch is good SOMETIMES. Although, that is not applicable to the cheap bottled type you get in stores.

Um...Italian is good with iced tea in my opinion :-)

Any thoughts as to why it is called 'dressing'?

I like many kinds, but mostly they should have some vinegary taste. I do not like bleu cheese.

[quote="mcalderbank"]Any thoughts as to why it is called 'dressing'?[/quote]

Well dressing is usually something to compliment or embellish something else. To that effect, I think it's a great name. Salad sauce would sound weird. Salad topping doesn't make sense. Salad juice...well it's too thick to be juice and you don't drink it (or at least sane people don't). In that manner, I think it's also like process of elimination. :D

Now...I personally think it should be called salad filling. ^-^ That makes it just that much more delightful to think about.

[quote="MithsApprentice"]Salad juice...well it's too thick to be juice and you don't drink it (or at least sane people don't).[/quote]

 :?

Of course not...

[quote="MithsApprentice"]Salad juice...well it's too thick to be juice and you don't drink it (or at least sane people don't).[/quote]
:blush:

Thousand Island with celery! Yum :)
But I'm anti-sauce, anti-dressing, etc, so I just put salt and olive oil on my salads..

Tough choice between Ranch and Italian...But Ranch wins in the end. :D

Hmmm... Looks like ranch wins it in the end.

I just had lunch *barf*  :D  Anyways raw eggs probably will make you sick.  I like ranch

elvenchamp777

this would make an interesting ad for ranch sauce

I don't really have a preference.  I'll eat them all.

I was reading down this list of posts and getting really hungry. Untill the raw eggs. I think I'll skip lunch.

[quote="MysticTerminator"]raw eggs[/quote]
Where are you from MysticTerminator?  In the United States, most people I know wouldn't eat raw eggs. I wouldn't eat eggs no matter how you cook them, but that's just me.  :)  :)  :)

I say that blue cheese dressing from this restaurant called Valentino's is the best.

I also like raspberry vinaigrette.

And that dressing they put on Big Macs at McDonalds.
 :)  :)  :)

School salad dressing is Italian i think. it's really sour.

[quote="MithsApprentice"]I think McD's uses some sort of ranch...

Anyways, Arnav is situated in North Carolina at the moment...they eat things raw over there.[/quote]

hey, not ALL North Carolinians eat raw eggs. at least i don't, i guess i can't say the same about my fellow North Carolinian AoPSers. :D


Harvard college topology HMMT geometry calculus integration

I found that the first three problem sets were deleted from the website. Has anyone downloaded them?

BTW, this semester's math 55a seems to cover more advanced topics(e.g., general topology and  categories) than previous semesters

harvard's 55's reputation is no doubt quite big, sometimes being dubbed as the 'hardest freshmen math course in the country'

however, the class i'm currently in at stanford is also quite tough imo, it is 100% proof based, and moves quite quickly,
the URL for almost all the materials is at:

http://www.stanford.edu/class/math51h

1) Typically, Math 55a and Math 25a are the same class for the first three weeks, after which the class splits.  Therefore, to find the first three problem sets, check the Math 25a website.

2) I took a quick look at this year's Math 55a syllabus.  It looks like the standard Math 55 syllabus and not especially advanced (btw, this is a good thing).

3) Math 51h at Stanford is actually more comparable to Math 25 at Harvard.  I suspect that Harvard's 25 is still tougher than Stanford's 51h, but I have a limited basis for this claim.  However...

4) Leon Simon is the man.

The syllabus of the math 55 is quite brief. The first three sets of 55 and 25 are quite different(I just got them from a friend.). Ths year's 25 prob sets are almost the same as last year's(and those of the year before last year).

After a preliminary glance at Math 25's problem sets in comparison with the Stanford Math 51H's, which I have been working a bit from, I think they are pretty close, though the Harvard class seems to have more emphasis on sets/analysis (hence the use of Rudin).

cstalg, my apologies.  I guess I should know that syllabi often lie.  The problem sets do cover some slightly fancier than usual topics.

Btw, one good reason not to use Rudin is that it is obscenely expensive.  Another is that it's far too terse for most students.

could someone explain exactly how much experience a student would need to enter that class?
for example, I have taken up to cal III, linear algebra, and some abstract algebra, but I'm sure others have had a lot more preparation.

A previous course in multivariable calculus and linear algebra and experience in writing proofs is the baseline level of preparation that gets you beyond "Don't even think about it," but AoPS moderator JBL tells a story of an eastern European IMO medalist who learned integration in Math 55.

I'm taking a course based off of the text Multivariable Mathematics, by Theodore Shifrin (Honors Professor at UGA). Has anyone heard anything about the book?

I wonder if working through that book could prepare me for that level (or at least close to that level).

I think Shifrin's book is used for a course at Stanford also. I have never seen it. The book I really like as a cram book for Math 55 is Hubbard and Hubbard, [i]Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach[/i].

I'm currently using the textbook for 51H at Stanford.
I think the exercises in the book are just that, exercises, and mostly not up to the level of theory that a course like 55 would have. The problem sets I posted in the link above do provide proof based problems, some of which are useful results. One thing tho, our course does NOT go along with the book, however I believe this first quarter we have finished chapters 1-6. Look at the schedule and you'll see what I mean :D

Princeton's math major track (for freshman I mean) uses Rudin, Hoffman & Kunze's Linear Algebra, and then Spivak's Calculus on Manifolds (that's three courses). All of the classes are proof-oriented. You can find past exams on the Princeton Math Department web page:
http://www.math.princeton.edu/undergraduate/
You've got a respectable start on undergraduate math if you get through that material. I don't know anything about Math 55, so I don't know how much farther than this it goes.

I don't know why people hype up Math 55 so much though. It sounds like they try to teach a huge volume of material really quickly so that high school math types will be impressed and go to Harvard. Learning too much math too quickly means you probably won't have a very good foundation when you get to later classes that depend on that material...

[quote="Xevarion"]
I don't know why people hype up Math 55 so much though. It sounds like they try to teach a huge volume of material really quickly so that high school math types will be impressed and go to Harvard. Learning too much math too quickly means you probably won't have a very good foundation when you get to later classes that depend on that material...[/quote]

That's not the impression I get.  The class seems brutally intense.  It's a full year course devoted to calculus and linear algebra.  I don't see why this is too huge a volume of material (breadth wise).  I don't believe it gives you credit for DE, analysis or anything else.

It seems like the class is really deep in the material it does cover.  Hearing descriptions such as "USAMO-type problems for homework" and "20 hours a week" make it seem like a MOP type of thing.  If you want to survive in the class it seems you'll be forced to work much, much harder than in other standard, freshmen math classes.  I bet a lot of people become motivated for the sheer purpose of trying to survive the class which would be quite an accomplishment.

That being said, the course might be exaggerated a little bit.  I would find this natural as the class is the most hardcore freshman-level undergraduate math class in the world (i.e. people tend to exaggerate extremes).

Of course, I'm no expert in math 55 myself.

[quote="joml88"]That's not the impression I get.  The class seems brutally intense.  It's a full year course devoted to calculus and linear algebra.  I don't see why this is too huge a volume of material (breadth wise).  I don't believe it gives you credit for DE, analysis or anything else. [/quote]
I heard that they learn about $L^{p}$ spaces and that they learn topology. I didn't learn about $L^{p}$ spaces until my third analysis class (and some people take that one as their fifth analysis class). Maybe they only get an introduction to these things, but it still seems to me like a lot.

People shouldn't be doing USAMO type problems once they've graduated from high school unless they are writing questions for contests or coaching high schoolers, imho. Olympiad math is a great way to draw talented students into the world of math, but it's like the classical mechanics is to physics. You start with it, it's good to know it, but there's a point where you say "that's history, let's move on." Unless you just meant USAMO difficulty, not topics, which is fine.

Also, I should mention that it's not necessary for a class to be on a wide range of topics in order for it to have beastly problem sets (ask pdiao about his analytic number theory class, or me about my measure theory class).

But anyway my point wasn't that they don't do enough work to properly cover those topics. I mean that you are rushed through a large amount of material without having enough time to build good intuition and understanding of the relationships between things. Personally, I think I've learned analysis too quickly, so when I find some free time I'm going to go back and read some other texts about material I already "learned," and do more problems.

I'm not trying to say I'm sure Math 55 is overall bad, or even that learning a lot of math quickly is overall bad. I'm just saying that people (especially people who are looking ahead, not back) tend to be extremely optimistic about the good qualities of the course and never mention the potential shortcomings.

"[url=http://www.thecrimson.com/article.aspx?ref=516216]Burden of Proof[/url]" (6 December 2006 Harvard Crimson article about the Math 55 class)

[quote="hypo3400"]
I am wondering about (d), since I have heard that Math 55 is only within the reach of IMO or strong USAMO participants. I feel that I satisfy the first three attributes, but was worried since I have not done particularly well at AIME/USMAO (I missed the cutoff for USAMO by one AIME question for the past three years, due to careless mistakes). Do you think Math 55 is within my reach, and if so, what can I do to better prepare? Thanks for any advice.[/quote]

I know someone who is currently a sophomore at Harvard who never managed to qualify for USAMO until senior year, and didn't do very well. However, he took Math 55 in freshman year at Harvard and did well in it. He had independently studied mv calc in 11th grade, and done a mixture of abstract algebra and number theory in 12th grade.

[quote]I also take stance against too much rigour. At my school, in the second semester multivariable calculus course they introduce lebesgue integration and prove the general statement of fubini's theorem as well as change of variable theorem for upper semicontinuous functions and the the change of variable being a diffeomorphism. All of this is done with extremely elementary tools, so it's a huge pain, and in retrospect I think it's completely pointless. In a more advanced integration theory course you will do such things in a more abstract framework where things are much easier to understand. Why should one dabble about details in a course where you don't have any tools, when there is no way you will ever refer to those proofs again.
[/quote]

I can see what you're saying. I've been doing independent study in multivariable calculus with Apostol this year, which contained many elementary proofs of some of the main calculus theorems. I ended up writing up some topology I had learned and eventually used that to prove all of the theorems. (In particular, integrability of continuous functions and the extreme value theorem are much easier to prove if you use compactness).

I also second Gauss202's post.

This is a bit of an aside, so please don't get too off-topic (I could put this in another post if need be), but how is Math 122 at Harvard? Would it be strange to take it before having done something like Math 25? And how much linear algebra does one have to know? I'm mainly planning to do it because I would take it as a high school student and it's the only one that's at a time that works for me, in addition to the fact that I already have the prerequisites for Math 122.

It looks like they have MAT 23, 25, or 121 as a prerequisite for MAT 122, and I believe those to be good suggestions.

The stuff covered in that 122 class is majorly important stuff, and a little bit hard (in my opinion). I'd only take that class if you feel you are totally ready to master it, or at least prepared to take it again if you don't. I can't recommend anything directly to you because I don't know anything about you, but you will need to have a good foundation in linear algebra and abstract algebra for later math classes. Any shallownesss of understanding you have of it will come back to haunt you later (believe me I know, because it happened to me!).

"I think AoPS needs a special Math 55 FAQ. Actually, there should be a FAQ for every top university's advanced freshman math offerings, but Math 55 desperately needs one."
DEFINITELY!!!

"Asking for information about Math 55 is one of my (the forum moderator's) Frequently Asked Questions, so I think the idea of developing a Math 55 FAQ makes sense. I'll try to do that before the national reply date, when all of this year's admitted students have to decide which college to go to."
Em, what seems to be the problem/issue here then tokenadult? Since I can't seems to find the FAQ. What are all other moderators/admins saying (I hope it is not confidential!  :rotfl: )?

This thread seems to be FAQ enough. The syllabus of Harvard's Math 55 course is not static. (It had noticeably more complex analysis this year than it had a few years ago, and Math 25 was correspondingly beefed up.) People seem to find the information they need by searching this thread up and asking follow-up questions.

http://www.ajorza.org/courses/math/m55/

is a place to look for old problem sets, etc.

^ :fool:

Link doesn't work. Are they the same psets that are posted on Elkies links?

it worked when i tried it yesterday, and had several years worth of problemsets.

Did you happen to download them all to your computer, by any chance?  :|

no but i expect that it will be up again in the near future.

Try again and see if it works; sometimes the site is down.

The problem sets posted are indeed different from the ones on Elkies' page; Andrei Jorza was Elkies' TA for 55 at least once, and perhaps didn't bother saving the sets because he knew Elkies' page had them.

(If you go to the parent directory, http://www.ajorza.org/courses, you can find a great deal of stuff from other Harvard math courses.  I don't know how he managed to take so many in just four years, unless he was getting them from other people.)

I just went through and made the complete set of all 55 stuff listed on that website in .zip form. PM me and I'll send it to you through email if you need it.  :)

Wow, just read the posts! Amazing! Anyway, can you guys list down the most famous math courses from the best university for math? Like:

Harvard : Math 55
Princeton : the four analysis sequence
...

Chicago's honors analysis course, 207/8/9, would be on the shortlist.  This class is for freshmen who test into it or sophomores who have performed very well in the (Spivak-based) honors calculus course.  Usually it has a really wild variety of topics.  (This year, the first two quarters were taught as more of a straight-up graduate real analysis class, out of Royden's book.)

I believe Stanford has a well-known honors sequence as well, but I'm not informed enough to comment on it.

http://www.math.harvard.edu/~ctm/courses.html

During the last three semesters, Prof. McMullen had posted all the course materials, including notes, readings, and solutions, on his website. It's interesting too see how the new math 55 curriculum works.

But the hw and sol links of 55a seem to be no longer active.  Has anyone downloaded the materials?


Hello

I was searching for proof of the area of this, but I couldn't find any understandable proof, so here goes.

Given a triangle ABC, and a point P inside the triangle, let the pedal triangle of P be the triangle formed by the feet of the perpendiculars of P to the sides of the triangles.

Prove that the ratio of the area of the pedal triangle of P divided by the area of the triangle ABC is $ \frac{R^2\minus{}OP^2}{4R^{2}}$

Thanks


Netherlands has a flag like that, except for it is horizontal.

[quote="mtms5467"]What country has a flag made of 3 vertical  stripes, from left to right in orange white blue?[/quote]
Netherlands flag is orange, white, blue horizontal.
Cote d'lvoire flag is orange, white, green vertical.
Ireland flag is green, white, orange vertical.

I'm pretty sure there's no orange, white, blue vertical flag.

Flags with an orange and a blue stripe would kick ass.

A flag like that would be a tricolor flag (tri=three, get it? :ninja: )
[url]
http://en.wikipedia.org/wiki/Flag_of_France#National_flags
[/url] has a pretty decent list of those flags.


3 Points $A,B,C$ lie on the spherical surface with radius [b]1[/b],center[b] O [/b].

Let $P,Q,R$ be the mid points of $BC,CA,AB$ respectively.
Prove that at least one segment of the segments $OP,OQ,OR$ in length is greater than or equal to $\frac{1}{2}$.

Well, since $\angle BOC+\angle COA+\angle AOB\le 2\pi$, we get that at least one of them is $\le 120^{\circ}$, and that's pretty much it, isn't it?


[quote="Phunky"]Okay I really don't want to waste anyones time but google failed me and I figure this would be good trivia for all you math wiz's.  In terms of blackjack, what would the average card in a deck be.  My best guess was 7 but I suck at math and would just like to know for my own information.

Thanks,
Phunky.[/quote]
If they go up to 13, then the total sum of "points" would be 7X4X13=364. 52 cards in a deck, divide to get 7. Don't post this kind of stuff here, it doesn't belong here. And this seemed a little like a homework problem...

Thanks, its actually for playing blackjack tournaments.  I'm too proud to use a guide that tells you what to do, so I needed to know what the exact average was for the next card.  Sorry if it was out of your genre here, I honestly didn't know where to go.  GOOGLE failed me, and google never fails me.  :o

Lol but really thanks, I didn't know how to calculate the aces right.  But my guess was right so thanks for confirming it.

In the interest of accuracy, aces are actually worth 11 or 1, and K, Q, J, 10 are all worth 10.  So the average is more like $ (10*4\plus{}9\plus{}...\plus{}1)/13 \equal{} 85/13 \equal{} 6.54$ (assuming aces are 1, since that's the more useful average).

[quote="Phunky"]sweet thank you :D

I really didn't know how to figure aces, so that makes a lot more sense.  I'll still average them as seven.  Now another question if ya'll don't mind. Maybe even harder than the last.

Lets say the average card is "Seven".  Dealers showing a Face card.  Do I figure the down card is a 7 or do I figure the down card would be a four given that both the two cards should add to 14 to average 7 each?

Or is there another approach I'm not even considering?

Just too add, I really appreciate your brains.  I didn't know how else I could figure this out.  Honestly I never made it past algebra.[/quote]

Each of the events are independent so the face down card should be a 6-7. You should assume the 7 because the facedown card might be an ace.
If you think about the averages, the facedown card would be a 3, because the average is 6.5

Awesome.  Thank you very much, I really appreciate it.  Hope I can get ya back some day.


function trigonometry calculus integration

For what functions $f$ is $y=\sin{(f(x))}$ periodic?

It would be periodic when either

$f(x)$ is a periodic function

or

$f(x)$ is linear

proof: (might have flaws) (requires calculus)
let $g(x) = sin(f(x))$

if $g(x)$ is periodic, then $g'(x)$ must necessarily be periodic

$g'(x) = cos(f(x)) * f'(x)$

so $f'(x)$ must be periodic
f(x) must be the antiderivative of a periodic function (which is either periodic or linear)

[quote="krustyteklown"][quote="yagaron"]for all $f(x) = f(x+T)+2. \pi$[/quote]
This is like the definition of periodicity (2pi for sin), you didn't explicitly solve for f[/quote]

Hm...

Suppose $f$ is diferentiable.

so we get $f'(x) = f'(x+T)$

So $f$ is the integral of the fourier Series.

Finished.... humpf.

[quote="junggi"][quote="rnwang2"]But consider $y=\sin{(x^{2})}$.[/quote]
$\sin(x^{2})$ isn't periodic...[/quote]
Oh right, I got that confused with $y=\sin^{2}{x}$, which is obviously periodic.


Hi I live in Canada and I am going into grade 12 after my summer break

I was wondering if anybody can add me on msn that can help me with my calculus studies it would be GREATLY appreciated.. my msn is certifiedlancer@hotmail.com

thx

If you have any calculus related questions, just post them in this forum, and I'm sure they'll be answered (As long as it's not a list of homework problems).


[quote="isomorphism"]But all numbers with ones digits as 4 or 6 have squares ending with 6. Am I missing something?[/quote]

1-100 like the perfect square is between there

[quote="the future"][quote="ajain"]Question:
             How many perfect squares have 6 as the ones digit?

The answer said one, but it should be two.

Why?

$ 4^2$ and $ 6^2$[/quote]

uhh...are you sure it was as ones digit?

becuase I remeber doing t hat question and the answer given was 2.

However, I also rememberd a question that said tens digit was 6 which was 1 (because I remembered the other questions answer and quickly put it down, getting it wrong)[/quote]

maybe it was the tens, im not sure.  but i still think it was the ones. (problems go by really  fast)


This is a variant of a problem in the MATHCOUNTS forum:
There are 100 children numbered 1 to 100 standing in a circle. Starting from the 1st one, the count off 1,2,3,1,2,3... Any child who counts the number 1 sits down. What is the number of the last child standing?

Could you tell me how to do these type of questions?

[quote="AstroPhys"]Since you know that they start counting at 1, you know that the last one standing should be in the 90s and that the pattern repeats itself so you can just look at the first 10 to see what happens in the 90s.
$1, 2, 3, 4, 5, 6, 7, 8, 9, 10$
$2, 3, 5, 6, 8, 9$
$3, 5, 8, 9$
$5, 8$
$8$
Therefore 98 should be last.[/quote]

wait why does it have to be in the 90s?

Made an assumption that is does, since the numbers with the same unit digit sit down for each ten.
That assumption can be completely invalid though  :wink: .

since the order is 1, 2, 3, we can concern our numbers in $\mod{3}$.

Since the first in the order has to sit down, then all of the numbers $n$, that are $n \equiv 1 \mod{3}$sit down. This leaves 2 and 3$\mod{3}$.

egh, this is where i get stuck, then I notice a pattern and i cant think of a way to express it  :wink:

Your assumption seems correct to me since they start counting at number one. They sit down in the order 1, 4, 7,... thus one in the 90's would be the last.  :?:

The pattern I saw was
$n: \left\{n\eq0 , n\in Z \right\}$
$3n+1\leq100$
$4n+1\leq100$
$\vdots$
$9n+1\leq100$
Those are the digits that sit down from the 100 in order, with each increasing integer coefficent of n.

Not quite. When you get to the end, the first person doesn't say 1 again, so it's not 4n+1.

Yeah, I noticed that some of the numbers are "recounted", but the pattern seems to work otherwise.
Again, I could be mistaken  :wink: .


Credits to iLord for making this thread about a year ago.

There are 9 slots. Once all 9 sign up, the game will start.

You get 3 points. You will choose to give 1, 2, or all 3 of your points to a player. You may split your points between 2 or 3 players. Once a majority is reached or all players have given their points, the player with the most points becomes the Executioner. That player chooses one player to be executed, and that player is eliminated from the game. Once there are 2 players left, the dead souls choose the winner.

SIGNUPS
1. CircleSquared
2. yaofan
3. cf249
4. ernie
5. Bugi
6. auroraak
7. eggylv999
8.
9.

i join!!!

jeez how many words do u need to make it longer...

I will join.

@yaofan:  It's a 10-character minimum.

I join. Can you give points to yourself?

@jjx: No, you may not give points to yourself.

Let the game begin!

CircleSquared (3)
yaofan (3)
cf249 (3)
ernie (3)
Bugi (3)
auroraak (3)
eggylv999 (3)
jjx1 (3)
budi713 (3)

Number in parentheses indicate the number of points you have left to give.

Points Given:
CircleSquared: 0
yaofan: 0
cf249: 0
ernie: 0
Bugi: 0
auroraak: 0
eggylv999: 0
jjx1: 0
budi713: 0

14 to reach majority

hmm...
i guess i give all of my points to... ernie

[hide]CircleSquared (3)
yaofan (3)
cf249 (3)
ernie (6)
Bugi (3)
auroraak (3)
eggylv999 (3)
jjx1 (3)
budi713 (3)

Number in parentheses indicate the number of points you have left to give.

Points Given:
CircleSquared: 0
yaofan: 3
cf249: 0
ernie: 0
Bugi: 0
auroraak: 0
eggylv999: 0
jjx1: 0
budi713: 0

14 to reach majority[/hide]

[quote="yaofan"]hmm...
i guess i give all of my points to... ernie

[hide]CircleSquared (3)
yaofan (0)
cf249 (3)
ernie (6)
Bugi (3)
auroraak (3)
eggylv999 (3)
jjx1 (3)
budi713 (3)

Number in parentheses indicate the number of points you have left to give.

Points Given:
CircleSquared: 0
yaofan: 0
cf249: 0
ernie: 3
Bugi: 0
auroraak: 0
eggylv999: 0
jjx1: 0
budi713: 0

14 to reach majority[/hide][/quote]

Lists seemed wrong, so I fixed.

List is still wrong. People don't use the points you give them, just the 3 they start out with. Nobody is necessarily going to reach 14, the Executioner is just whoever has the most points at the end.
CircleSquared (3)
yaofan (0)
cf249 (3)
ernie (3)
Bugi (3)
auroraak (3)
eggylv999 (3)
jjx1 (3)
budi713 (3)

Points Given:
CircleSquared: 0
yaofan: 0
cf249: 0
ernie: 3
Bugi: 0
auroraak: 0
eggylv999: 0
jjx1: 0
budi713: 0

14 to reach majority

3 to yaofan

CircleSquared (3)
yaofan (0)
cf249 (3)
ernie (3)
Bugi (3)
auroraak (0)
eggylv999 (3)
jjx1 (3)
budi713 (3)

Points Given:
CircleSquared: 0
yaofan: 3
cf249: 0
ernie: 3
Bugi: 0
auroraak: 0
eggylv999: 0
jjx1: 0
budi713: 0

This is somewhat of a bump, but

[b]1 point to yaofan[/b]

CircleSquared (2)
yaofan (0)
cf249 (3)
ernie (3)
Bugi (3)
auroraak (0)
eggylv999 (3)
jjx1 (3)
budi713 (3)

Points Given:
CircleSquared: 0
yaofan: 4
cf249: 0
ernie: 3
Bugi: 0
auroraak: 0
eggylv999: 0
jjx1: 0
budi713: 0

I give 2 points to CircleSquared.

CircleSquared (2)
yaofan (0)
cf249 (3)
ernie (1)
Bugi (3)
auroraak (0)
eggylv999 (3)
jjx1 (3)
budi713 (3)

Points Given:
CircleSquared: 2
yaofan: 4
cf249: 0
ernie: 3
Bugi: 0
auroraak: 0
eggylv999: 0
jjx1: 0
budi713: 0


inequalities
AMC
USA(J)MO
USAMO
inequalities solved

Let $ n\ge 2$ be an integer. Let $ a_1,a_2,\cdots,a_n$ be positive real numbers satisfying $ \sum_{i \equal{} 1}^{n}a_i\cdot\sum_{i \equal{} 1} ^{n}\frac {1}{a_i}\le (n \plus{} \frac{1}{2})^2.$ Prove that in set $ \{a_1,a_2,\cdots,a_n\},$ the greatest one is not more than $ 4$ times of the least.

[quote="Fang-jh"]Let $ n\ge 2$ be an integer. Let $ a_1,a_2,\cdots,a_n$ be positive real numbers satisfying $ \sum_{i \equal{} 1}^{n}a_i\cdot\sum_{i \equal{} 1} ^{n}\frac {1}{a_i}\le (\frac {n \plus{} 1}{2})^2.$ Prove that in set $ \{a_1,a_2,\cdots,a_n\},$ the greatest one is not more than $ 4$ times of the least.[/quote]

Is this $ \sum_{i \equal{} 1}^{n}a_i\cdot\sum_{i \equal{} 1} ^{n}\frac {1}{a_i}\le (\frac {n \plus{} 1}{2})^2$ equal to $ (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)(\frac {1}{a_1} \plus{} \frac {1}{a \plus{} 2} \plus{} \cdots \plus{} \frac {1}{a_n})$

If YES is the answer then :

By Cauchy Shwarz inequality we have that :$ (a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n)(\frac {1}{a_1} \plus{} \frac {1}{a \plus{} 2} \plus{} \cdots \plus{} \frac {1}{a_n})\geq n^{2}$

In other hand  $ \sum_{i \equal{} 1}^{n}a_i\cdot\sum_{i \equal{} 1} ^{n}\frac {1}{a_i}\le (\frac {n \plus{} 1}{2})^2$

So we have that : $ n^{2} \leq \sum_{i \equal{} 1}^{n}a_i\cdot\sum_{i \equal{} 1} ^{n}\frac {1}{a_i}\le (\frac {n \plus{} 1}{2})^2$

Then :$ (n \minus{} 1)(n \plus{} \frac {1}{3})\leq 0$ and only natural solution is $ n \equal{} 1$
 So  $ n < 2$ ...Where I'm wrong  :(

[quote="Fang-jh"]Let $ n\ge 2$ be an integer. Let $ a_1,a_2,\cdots,a_n$ be positive real numbers satisfying $ \sum_{i \equal{} 1}^{n}a_i\cdot\sum_{i \equal{} 1} ^{n}\frac {1}{a_i}\le (\frac {n \plus{} 1}{2})^2.$ Prove that in set $ \{a_1,a_2,\cdots,a_n\},$ the greatest one is not more than $ 4$ times of the least.[/quote]

There is a typo somewhere in the problem statement.
By Cauchy-Schwarz $ \sum_{i \equal{} 1}^{n}a_i\cdot\sum_{i \equal{} 1} ^{n}\frac {1}{a_i}\ge n^2$.
Hence for $ n\ge2$ there are no positive real numbers that satisfy the stated conditions.

We were thinking about the same  [b]test20[/b] ,I thought I'm misunderstanding something,but it seems there's a typo in problem

The bound in the actual USAMO problem is $ {\left( n \plus{} \frac{1}{2} \right)}^2$.

[quote="Ravi B"]The bound in the actual USAMO problem is $ {\left( n \plus{} \frac {1}{2} \right)}^2$.[/quote]

And the problem is a special case of a result stated on page 169 of Crux Mathematicorum,  Volume 32, issue 3, April 2006:
---------------------------------------------------------------------------------------
Let $ \alpha>1$ and let $ x_1,\ldots,x_n$ be positive real numbers such that
\[ \sum_{i\equal{}1}^n x_i \, \sum_{i\equal{}1}^n \frac{1}{x_i}
< \left( n\minus{}2\plus{}\sqrt{\alpha} \plus{}\frac{1}{\sqrt{\alpha}}  \, \right)^2.\]
Then $ x_i<\alpha\, x_j$ holds for any two indices $ i$ and $ j$.
---------------------------------------------------------------------------------------
The statement is best possible for every $ \alpha>1$, and the proof is a simple application of the Cauchy inequality.
The problem is closely related to IMO 2004/4.


I think AoPS played a very small role in qualifying me.  Just looking over past tests was the most effective.

i see there's no option for the aops community NOT helping you significantly, so the poll is a bit one sided.  for me personally, i found that it helped to study old tests.

AoPS Community definitely helped to qualify. Okay last year I couldn't even qualify for AIME, this year I made USAMO(barely, but considering that I screwed up bad on AIME)

Although I joined AoPs last year, I totally forgot about it and didn't become active until this year's AIME was already over, so I basically qualified on my own this year.  But next year I'm sure it'll be a big help :)

considering that last yr b4 the aime began, I forgot this site after i took the
AIME class, and considering that this yr I visited here for 4 hr everyday b4 AIME, it certainly helped me.

But i can thank more for books (AoPS v2, Art and Craft, Engel, etc.) and past problems because they helped me see things through problem-solving.

Definitely helped me a lot. I was on AoPS around AIME time last year too, but not as actively. This year it was like AoPS 10 hours a day before the AIME ;) Guess it worked. I also took the classes, but mostly Olympiad ones.


Putnam
Ross Mathematics Program
summer program
PROMYS
HCSSiM

What particular programs are recommended?

How many programs should I apply to in order to have a reasonable chance at getting into one of them? I'm a sophomore with decent grades/math background, a mentionable Putnam result, etc. I.e. somewhere between slouch and superstar.

I hear a lot of good comments about Joseph Gallian's REU in Duluth, Minnesota. I saw once on the website for that program that it appeared to help one get into the REU if one had done some secondary-level summer program, such as Ross (or, I suppose, MOP).

I doubt that high school summer programs matter much (unless it's MOP).  The deciders will care most about recommendations, grades, and coursework.

Duluth is regarded as the top REU, but the mathematical focus is narrow, so it doesn't appeal to everyone.  None of the other programs are really "elite" but Williams is pretty strong.

[quote="yenlee"]I doubt that high school summer programs matter much (unless it's MOP). [/quote]  Well, I've never been to an REU nor participated in an REU selection process, but I strongly suspect that this suggestion is wrong.  The skill set for high-level olympiad performance is quite different than the skill set for research and the people who run REUs probably know this.  Moreover, the skill set taught at (say) RSI, PROMYS, Ross, Hampshire or the like is actually much closer to the skill set for research.  If I were in charge of choosing students for an REU, I would take someone with time at one of the other programs over someone with time at MOP (all else being equal).  On the other hand, college credentials tend to trump high school credentials.  One effect of this is that it can be very difficult to get into many REUs before junior year of college.

(I also don't agre with yenlee's second paragraph, but I don't have any content to add other than that the notion of a "top" REU is rather silly.)

When applying to these REUs, how does one know if they even have the necessary math background to be able to participate in the research? With many of the REU websites, I couldn't gauge or find a description of the kind of background knowledge required. Is there a bare minimum core of coursework that a student should have before he/she applies to REUs?

In response to JBL, REU admissions criteria will be as varied as there are opinions of mathematicians, but the gist of it is that REU admissions criteria is probably like grad school admissions criteria, except with one  less year (or two) of information.  Hence that's why I emphasized recommendations, grades, and coursework.

*Personally*, if I ran an REU, I wouldn't care one lick whether an applicant went to Ross or whatever.  These are great programs, but they don't say anything useful about the applicant.  MOP at least tells you *something*.  (And this is coming from one of the least pro-competition persons on this website.)  I should have also mentioned RSI as an exception, since that program actually has a research component.  I guess if JBL ran an REU, the standards would be slightly different.

As for "top" REUs, it's nice to think that there isn't a top REU (just as it might be nice to think that there is no "top" math journal or "top" whatever), but the truth is that Duluth is most likely the top REU as measured by several standards:  highest selectivity, highest research output, success of mathematical careers of past participants.  FYI, I didn't go to Duluth.  Full disclosure: I went to Williams, which is why I felt comfortable recommending it as a "strong" program.

Well, I think your estimation of Ross, PROMYS, Hampshire, etc., is clearly wrong ;)  Probably some of the professors running these things are familiar with these programs, and for second-year undergrads who've not had the opportunity to do any research yet, they are certainly a good thing to mention.  But we definitely both agree that application data from college are going to be far more relevant, and you're not going to make up for a mediocre recommendation or a B- in analysis with a high school summer program.


Let $ P$ be a given point inside a convex quadrilateral $ O_1O_2O_3O_4$. For each $ i \equal{} 1, 2, 3, 4$, consider the lines $ l$ that pass through $ P$ and meet the rays $ O_iO_{i\minus{}1}$ and $ O_iO_{i \plus{} 1}$ (where $ O_0 \equal{} O_4$ and $ O_5 \equal{} O_1$) at distinct points $ A_i(l)$ and $ B_i(l)$, respectively. Denote $ f_i(l) \equal{} PA_i(l) \cdot PB_i(l)$. Among all such lines $ l$, let $ l_i$ be the one that minimizes fi. Show that if $ l_1 \equal{} l_3$ and $ l_2 \equal{} l_4$, then the quadrilateral $ O_1O_2O_3O_4$ is a parallelogram.


of course you can participate on the next MathLinks Contest, as long as you meet the requirements. not to discourage you or something but i was able to solve [u][b]none[/b][/u] of the problems. so you can imagine the difficulty :D


limit
real analysis
real analysis unsolved

Let $f \in C^{3}((\alpha, \beta)), x_{0}\in (\alpha, \beta)$. Assume that $f'(x_{0}) = f''(x_{0}) = 0$ and that $f'''(x_{0}) \neq 0.$
Prove that $f$ cannot have extremum at $x_{0}$.

By L'Hopital's rule, $\lim_{x\to x_{0}}\frac{f(x)-f(x_{0})}{(x-x_{0})^{3}}=f'''(x_{0})/6$. It follows that in some neighborhood of $x_{0}$ the difference $f(x)-f(x_{0})$ has the same sign as $f'''(x_{0})(x-x_{0})$.


MATHCOUNTS
geometry
HMMT
AMC
AMC 12
AMC 10
ARML

I have been doing a bunch of practice AMC's but in nearly all of them, I can't seem to break the 130 barrier. Any tips (aside from do more problems)?

First, try to determine which topic/subject is troubling you the most. Then, study that subject. If it's just difficult problems with no specific subject that you're having difficulty with, then just do more problems and make sure you understand the solutions.

Well for me, I went from 90 on last year's amc 10 (never thought there was anything past mathcounts...) to 150's on my last two practice amc 10's (115-125 amc 12's). Before the 150's, I always got somewhere in the 120's. Then, one day, I broke the barrier with a 138.

I did that by a combination of two things:

1) Most important: problems. I did all the amc 12's from 1975-2005. I also did around 5 amc 10's. To deal with the upper-end 20-25 amc 10 problems, I would recommend amc 12's, because they offer you a chance to experience these types of problems, but 10-12 of them per contest (IMO, 20-25 amc 10=10-20/2 amc 12.) .
With my experience from mathcounts for speed, I can pretty much do the first 20 in 30 min, leaving me 45 min. to tackle (they're not hard at all after doing amc 12's) the last 5 and CHECK MY ANSWERS.

2) The next most important thing that helped me was going through aops 1 (im halfway through 2, although for amc 10 u dont need it..).

Hope that helped!

My #1 question is: are you getting ALL geometry problems right? If not, do lots of HMMT geometry problems and come back to it. If you can do all geometry problems on 10, you should have no problem breaking 130.

This is assuming that you do decent in other areas of course. But geometry is the area that people often have trouble.

I sometimes have trouble with Geometry but the test I did today, I did [i]nearly[/i] all the geometry problems correctly :)
I will also start doing AMC 12's.

Doing past AMC 12's definitely helps improve your amc 10 score.
Aops 1 is useful too if you are willing to get it.
Doing more geometry also helps because there are more geometry on the AMC 10 than any other kinds of problems.

i've done all the intros and I'm starting on Volume 2.

Thank you all for the good suggestions above, especially about doing amc12 mid-section problems to nail down the last 5 problems on amc10. One question I have is if I do this now to prepare for last 5 problems on amc10, wouldn't that mess up my true score on amc12 when I do past amc12 exams next year (because I already know these problems) ? So far, I have been trying not to look at past amc12 problems at all so that when I do take the past amc12 tests later for practice, I will know where I stand exactly on the amc12 tests.

Any suggestion on this ? Thanks.

No, it is not.
In my school's math team, every year for every big competition (ARML, HMMT, PUMAC...) we do the same problems over and over again.  Believe me, you don't remember the exact solutions, but you can vaguely remember how to approach it.  But this actually helps you, because it allows you to go over the same concepts year in and year out, keeping you a well-rounded mathematician.

Also, the main thing in order to improve your score is to do the problems you couldn't do/got wrong again until you know exactly how to do them (2-3 days later, just do them again, in other words).


How did you all do? I got 13 (1-12,14), though I "cheated" on #10. I'm happy with a 271 index.  :D

You know there's already a growing thread about this....


graph theory
combinatorics unsolved
combinatorics

Given a $K_{n}$ (a complete graph with $n$ vertices) give with proof a method or formula to count the maximum number of $K_{3}$ subgraphs.

[hide="maybe this will help"]I think it's just $n$ choose $3$... but i went on math world and the whole "clique" thing confused me  :? [/hide]

I don't understand. If you start from a $K_n$ then the number of triangles is obviously $\frac{n(n-1)(n-2)} 6.$  ;)

Pierre.


admirable should be admirably in this quote:

"How can it be that mathematics, being after all a product of human thought independent of experience, is so admirable adapted to the objects of reality? - Albert Einstein"


Given integers $1 \leq a_{1}, ..., a_{m}\leq n, 1 \leq b_{1}, ..., b_{n}\leq m$, show there are integers p, q, r,s for which $a_{p}+...+a_{q}=b_{r}+...+b_{s}$

In a circular arrangelment of zeros and ones, with n terms altogether, prove that if the number of ones exceeds (k-1)n/k, there must be a string of k consecutive ones.

Suppose that the squares of  an n by n chessboard are labeled arbitrarily with numbers 1 to n^{2}. Prove that there are two adjacent squares whose labels differ in abosloute value by at least n.

[quote="D.P.L"]Given integers $1 \leq a_{1}, ..., a_{m}\leq n, 1 \leq b_{1}, ..., b_{m}\leq m$, show there are integers p, q, r,s for which $a_{p}+...+a_{q}=b_{r}+...+b_{s}$

In a circular arrangelment of zeros and ones, with n terms altogether, prove that if the number of ones exceeds (k-1)n/k, there must be a string of k consecutive ones.
[/quote]

where did the n come from in number 1?

[hide="2"]number of 0s < n/k place these on the circle.

place (k-1) 1s between each pair of zeros and therefore you have <(k-1)n/k 1s

but you'er told you have >(k-1)n/k 1s, so by the PHP, there must be k consecutive 1s somewhere on the circle[/hide]

2
[hide]
if there are more then (k-1)n/k  1's then there are less than n/k 0's

so let this be our arrangement:

0___0___0.... 0___0___  (less than n/k zeros)

where the blanks represent places to put ones, and since there are less than n/k blanks and more then (k-1)n/k 1's to place, by the pidgeon hole principle one blank must have at least k 1's[/hide]

3- I wasnt sure if diagnol squares count as adjacent, if they do then this works, if not ignore it.
[hide]
condisider where 1 and n^2 are on the board, they are at most n squares away from eachother (counting diagonals) so in those n squares the numbers must go up n^2 - 1, but if no adjacent squares differ by at least n,  this can be at most n (n-1) = n^2 - n which is less than n^2-1, so thus two adjacent squares must differ by at least n[/hide]

These are pretty good porblems. There are m partial sums for a's and n partial sums for b's. Consider all numbers which can be expressed as the sum of a partial sum of a's and a partial sum of b's. These numbers range from 2 to mn, and since there are mn ways of combining partial sums of a's and b's, there is at least one number which is expressible in two ways, which is equiv to the result.

Solution to #3 will have to wait until tomorrow.

#3: 1 is somewhere and $n^{2}$ is somewhere. There is a path from $n^{2}$ to $1$ of length at most $n-1$. Therefore, at least one of the differences is greater than or equal to $\frac{n^{2}-1}{n-1}=n+1$.

The result holds even if we require adjacent squares to share an edge. However, the solution is kind of long to write up (at least the one I found), so I'll only give a sketch. First, you can prove that if a certain arrangement does not have any adjacent squares differing by n or more, you can sort the numbers in each row from smallest to largest without changing that property. Since you can also symmetrically sort the columns, you basically keep sorting until all rows and columns are ordered (pretty easy to show this eventually happens).

Now consider what happens as you place counters on the numbers in order starting from 1. It can be shown that eventually, there will be at least n boxes without counters which are adjacent to some box with a counter. After that, it's clear that some number we place will eventually be at least n larger than an adjacent box. This also proves that for a m by n board, the minimum maximum difference is min(m, n).


A car travels downhill at $ 72$ mph (miles per hour), on the level at $ 63$ mph, and uphill at
only $ 56$ mph. The car takes $ 4$ hours to travel from town $ A$ to town $ B$. The return trip
takes $ 4$ hours and $ 40$ minutes. Find the distance (in miles) between the two towns.

Let $ x$ be the number of level miles needed to be covered. Let $ y$ be the number of hilled miles to be covered. Then we get the system of equations

$ \frac{x}{63}\plus{}\frac{y}{72}\equal{}4$
$ \frac{x}{63}\plus{}\frac{y}{56}\equal{}\frac{14}{3}$

Subtracting the first equation from the second equation gives

$ \frac{1}{252}y\equal{}\frac{2}{3}$
$ y\equal{}168$
$ \frac{x}{63}\plus{}3\equal{}\frac{14}{3}$
$ \frac{x}{63}\equal{}\frac{5}{3}$
$ x\equal{}105$

Total distance between A and B is $ 105\plus{}168\equal{}\boxed{273}$ miles.

ok thanks a lot for clearing that up... I was trying to solve the problem with 3 variables (considering uphill and downhill different)

[quote="gauss1181"]Let $ x$ be the number of level miles needed to be covered. Let $ y$ be the number of hilled miles to be covered. Then we get the system of equations

$ \frac {x}{63} \plus{} \frac {y}{72} \equal{} 4$
$ \frac {x}{63} \plus{} \frac {y}{56} \equal{} \frac {14}{3}$
[/quote]

Hi, could you explain this to me =)

Arent you assuming that all hills on the way are all down and all hills on the way home up? I dont really understand.

I think that ultimately it doesn't matter because you're counting the uphill and downhill together, but the speeds are different in each..


linear algebra
matrix
linear algebra unsolved

tou can suppose that $ A$ is triangular and everything becomes much easier ..


Find x + y where x and y are non-zero solutions of the system of equations:
$ y^{2}x$ = $ 15x^{2}\plus{}17xy\plus{}15y^{2}$
$ x^{2}y \equal{} 20x^{2}\plus{}3y^{2}$

I'm fairly sure the solution is x=19 and y=95. Just don't know how to get there.

[quote="bos1234"]Find x + y where x and y are non-zero solutions of the system of equations:
$ y^{2}x$ = $ 15x^{2}\plus{}17xy\plus{}15y^{2}$
$ x^{2}y \equal{} 20x^{2}\plus{}3y^{2}$
[/quote]
Let $ \frac{x}{y}\equal{} t.$ Hence, $ \frac{x^{2}y}{y^{2}x}\equal{}\frac{20x^{2}\plus{}3y^{2}}{15x^{2}\plus{}17xy\plus{}15y^{2}}\Leftrightarrow t \equal{}\frac{20t^{2}\plus{}3}{15t^{2}\plus{}17t\plus{}15}\Leftrightarrow$
$ \Leftrightarrow5t^{3}\minus{}t^{2}\plus{}5t\minus{}1 \equal{} 0\Leftrightarrow t \equal{}\frac{1}{5}.$
Thus, $ y \equal{} 5x$ and $ x \equal{} 19,$ $ y \equal{} 95.$

This problem has been posted before. But the general idea is to multiply the first equation by x and the second by y, then subtract them. Then you have a homogeneous system where you can solve for $ x: y$. This idea was implicitly used by arqady in his solution.


geometry
geometric transformation
rotation
MATHCOUNTS

In how many patterns can six identical L-shaped pieces, each consisting of three unit squares, be placed to perfectly cover a 3 by 6 board? One such pattern is shown.

[asy]draw((0,0)--(6,0),linewidth(2));
draw((0,3)--(6,3),linewidth(2));

draw((0,0)--(0,3),linewidth(2));
draw((2,0)--(2,3),linewidth(2));
draw((4,0)--(4,3),linewidth(2));
draw((6,0)--(6,3),linewidth(2));

draw((0,1)--(1,1)--(1,2)--(3,2)--(3,1)--(5,1)--(5,2)--(6,2),linewidth(2));[/asy]

Yes, of course you can interpret it that way. How many answers do you get that way? Infinity. You can rotate it in any way possible. The usual idea is to interpret the question with common sense. :wink:

[quote="NicksDad"]This problem is ill posed!  The problem statement doesn't preclude rotations of the 3X6 grid.  So, if we assume that rotations are not allowed, then the solution is correct:  there are 8 possible patterns.  If we assume that rotations are allowed, then there are four pairs of patterns, where each member of the four pairs is identical to the other member of its pair after rotating 180 degrees.

Is this really how the MathCounts problem was stated?  If so, shame on them![/quote]

Exactly!! :mad:  :furious:  :censored:  :starwars:

[hide=Solution since no one has posted one]For the first two cases, there is only one way to place another piece to cover the lower left corner. In the last case, there is no way to place another piece to cover the lower left corner without overlapping the first piece. In both of the first two cases, the two leftmost columns will be covered. So, we can use this logic again, on the top left square which has not yet been covered. We have two choices of how to cover the first two columns, two choices of how to cover the next two columns, and two choices of how to cover the last two columns, so there are $2\cdot2\cdot2=\boxed{8}$ total ways to cover the entire board.[/hide]

who else just listed them bc they weren't sure


algebra
polynomial
function
binomial theorem

Let $P(x)$ be a polynomial with real coefficients. Show that there exists a nonzero polynomial $Q(x)$ with real coefficients such that $P(x)Q(x)$ has terms which are all of a degree divisible by $10^{9}$.

let k be a fifth root of unity. Then P(x)*(P(-x)) has only degrees of 2 because the polynomial is even. Also, P(x)*(P(kx))(P(xk^2))P(xk^3)P(xk^4) gives us something with coefficients only x^5 because plugging in any 5th root of unity *x gives us the same value as if we plugged in x. Its easily demonstrated that it will be  a rational polynomial by some basic binomial theorem and the fact that k^(1+2+3+4)=1. So now we know P(x)P(-x)P(xk)P(xk^2)P(xk^3)P(xk^4) give us degrees all divis by 10. So substitute x^10 for y in our new polynomial and repeat to get as many powers of 10 as we want.

[quote="Phelpedo"]Hm, you think that a polynomial with all powers of $x$ divisible by $2$ times a polynomial with all powers of $x$ divisible by $5$ will multiply out to a polynomial with all powers of $x$ divisible by $10$? Unfortunately, exponent laws don't facilitate this, but it is close.

By the way, is it a nontrivial proof that $P(x)=P(\omega x) = P(\omega^{2}x) = P (\omega^{3}x) = P(\omega^{4}x)$ is enough justification that the degrees of $x$ in $P$ must all be a multiple of $5$?[/quote]
oops
well you can just apply that fifthpower function on f(x)f(-x)


[hide="Solution"]
Start from the inside out.
\begin{align*}\sqrt{3+2\sqrt{5+12\sqrt{3+2\sqrt{2}}}}&=\sqrt{3+2\sqrt{5+12(1+\sqrt{2})}}\\&=\sqrt{3+2\sqrt{17+12\sqrt{2}}}\\&=\sqrt{3+2(3+2\sqrt{2})}\\&=\sqrt{9+4\sqrt{2}}\\&=\boxed{1+2\sqrt{2}}.\end{align*}

Somewhat convenient.

Sorry, not much to explain here (either that or I'm horrible at explaining things.)[/hide]

See post 2 of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21294]this thread[/url] for a good way of solving these problems.

I didnt get it, Sorry

How did $ =\sqrt{9+4\sqrt{2}}$ turn into ${ =1+2\sqrt{2}}$ ? and the other transformations, I seem not to get it  :(

[quote="marc1217"]I didnt get it, Sorry

How did $ \sqrt {9 + 4\sqrt {2}}$ turn into ${ = 1 + 2\sqrt {2}}$ ? and the other transformations, I seem not to get it  :([/quote]

so you want $ \sqrt {9 + 4\sqrt {2}}$ ${ = a + b\sqrt {c}}$ So square both sides to get

${ 9 + 4\sqrt {2}}$ = ${ a^2+b^2*c+2ab\sqrt {c}}$ so $ a^2+b^2=9$ and 2ab=4. You know that c=2 since that is the only number in a radical.

oh thank you hehe i got it now, thank you too Mr. Ricollo and mewtwo, especially to the link you have posted hehe


salam dostan!
man 1 taze vared dar olampiad hastam va mikhastam begam shoma che ketab haii ra baraye man ke hanooz hichi halim nist va taze varedam pishnahad mikonid?
baad mikhastam beporsam be nazare shoma man ke taze mikham beram 3vom dabirestan age az alan shoro konam va sakht kar konam be marhale 2vom miresam?aye mitoonam onja ghabol besham? :?:
rasti man baram tarkibiat yekam sakhte chekar konam?
mochakeram!! :)

Salam

 ketab hayi ke be nazare man baraye to mitoonan mofid bashan :

1. Hendeseye mosatahe ( siamak ahmadpor )
2. Jabr dar olampiad haye riazy ( mahdi safa )
3. Nazariye adad ( maryam mirzakhani )
4. Tarkibiat ( Alipoor )
5. Soalate olampiad haye shoravy ( motarjem: ostad Parvize shahriary)

In ketabha alefbaye olampiadan ,agar ham soaly barat pish omad mitooni postesh kony
hameye ketabha be rahaty peida mishan
haddeagalle olampiad ineke riazit khob mishe va azash lezat mibary  :wink:

http://sampadia.ir/forum/index.php/topic,686.0.html

be nazare man adade mirzakhani vase shuru sakhte!!(baraye man ke sakht bude va hast!!)
mito0ni nazarie adade entesharate khoshkhano bekhooni ke yekam ba formesh Bshtar ashna shi

dar kol ketab haye entesharate khosh khan KHEYLI  rahat tar az ketabaye fatemian(ke albatte inam khub nis vali baraye shuru khube!)

movafagh bashi :wink:

RasT!! :D

man asio mishnasam ke az sevvom shuru kard va talaye jahani am gereft(!)

tarkibiato az roo che ketabi mikhuni ke sakhte?!! :huh:  :maybe:

salam!
dasteton dard nakone ketabaie kheili khobi moarefi kardid.
darm roshon kar mikonam!
man tarkibiat ro az ro ketab zard fatemi khondam vali chon hanoz mabahes e sale 3vom ro balad nistam baram sakhte albate 1kam!!! :lol:
rasti ketab nezari e adad fatemi kheili sakhte! 1nafar az dostam goft bara marhale 3vom e!!!!fekr konam ketab e entesharat e khoshkhan baram kheili behtar bashe!

behar hal az rahnamii shoma kheili mamnon! :)

fek nemikonam ketabe tarkibiate fatemi niazi be darsaye sevvom dashte bashe!(man taze miram dovvom!)

vali age to intor fek mikoni man ketabe tarkibiate khosh khano pishnehad mikonam ! ke 2 jelde(abbas servaT)

nazarie adadm ke hichi...!!
man 6 fasle avalesho to0 ye sal khundam!!

salam. be nazare man kheily be in gir nade ke folan ketabo bekhuni.
say kon az har matlab ye zare etelaate nesbio dashte bashi
bad say kon bishtar masale hal koni.unam bishtar masalehaye keshvar haye mokhtalef ke tu ghesmate resources hast.badesh kheily ham say nakon ke hameye masalehaye ketabaro hal koni.


You do know the mathcounts site right? [url]http://www.mathcounts.org[/url]
there's some problems there.

Um.. I can't think of anything else at the moment I'll update if I do..

Well, A can't be too high if it's going to be in the tens place.  But it won't be a zero.

To hide your answers to a problem, type it in, then highlight it by right clicking your mouse and moving over that text, then hit the hide button above the area where you type in the text.  I'll do that now.

[hide]Start with A=2 (guessing).  B and C can't be greater than 8 & 9 as those are the highest single digit numbers.  However, that gives A+B+C = 19, so A can't be 2.

So A must be 1.  If 0 can be one of B or C, then B could be 0 and C=9, which gives 1+0+9=10. We know that B+C must be at least 9, so that 1+b+c is greater than 9. However, if 0 cannot be a digit, then A+B+C must be at least 11. If A,B, and C are different digits, A+B+C must be at least 12, so that A and B are different. So 1+B+C>=12.

This means that B+C>=11. B must be one greater than the last digit of the sum(so that when 1 is added the last digit is B).Now we know that B+C are >=11 and <=18. B+C can't be the same, so we can't have B+C=18. Therefore, we could have B+C equal to 11,12,13,14,15,16, or 17. Now letting the last digit +1 be B, plug each of these values in to B+C=1(b-1). You can find the values from here.[/hide]

SDG

[quote="bryantacademy"]What do you guys use for pre-mathcounts practice? I use the olympiad book by george lencher and I am reading the Art of Problem Solving (I highly recommend it). My friend is a moderator and he helps me sometimes, but I am in pretty deep this time.
I have not learned complex algebra yet.

   A
   B
[u]   C[/u]
A B

Please help, :?:  :?:  :?:  :?:  :?:  :?:  :?:  :?:  :?:  :?:[/quote]

*cough*Art of Problem Solving Books Volume 1*cough*

 :P


trigonometry
Pythagorean Theorem
geometry

Let $ p$ and $ q$ be the radius of two circles through $ A$, touching $ BC$ at $ B$ and $ C$, respectively. Prove that $ pq\equal{}R^{2}$.

[quote="#H34N1"]Let $ p$ and $ q$ be the radii of two circles through $ A$, tangent to $ BC$ at $ B$ and $ C$, respectively. Prove that $ pq \equal{} R^{2}$, where $ R$ is the circumradius of $ \triangle ABC$.[/quote]
[hide="Hint"]Let the foot of the perpendicular through $ A$ to $ BC$ be $ D$. Let $ AD\equal{}a,\ BD\equal{}b,\ CD\equal{}c$. Then solve for $ p,q,R$ in terms of $ a,b,c$. You can use Pythagorean Theorem and $ R\equal{}\frac{abc}{4[ABC]}$.[/hide]
[hide="Answer"] $ p\equal{}\frac{a^{2}\plus{}b^{2}}{2a},\ q\equal{}\frac{a^{2}\plus{}c^{2}}{2a},\ R\equal{}\frac{(b\plus{}c)\sqrt{a^{2}\plus{}b^{2}}\sqrt{a^{2}\plus{}c^{2}}}{4\frac{a(b\plus{}c)}{2}}\equal{}\frac{\sqrt{(a^{2}\plus{}b^{2})(a^{2}\plus{}c^{2})}}{2a}$[/hide]

[quote="#H34N1"]Can someone please provide some kind of a diagram? I can't seem to draw it right...[/quote]
I think this works

[img]10498[/img]

It was easier for me to [hide="solve for"]$ b,c$ in terms of $ p,q,R$ instead of the other way around.[/hide]

Prove that: $ 2p\equal{}\frac{c}{\sin B}$ and $ 2q\equal{}\frac{b}{\sin C}$. [Think extended law of sines...]


Έστω $f(x) = ax^{2}+bx+c$ τριώνυμο με $b>a$ και $f(x) \geq 0$ για κάθε $x \in \mathbb{R}$ .

Βρείτε την ελάχιστη τιμή του $\frac{a+b+c}{b-a}$ .  :)

Επειδή $f(x)\geq 0$ για κάθε $x \in \mathbb R$  θα ισχύει $f(-2)=4a-2b+c\geq 0 \Leftrightarrow a+b+c\geq 3(b-a)$ και επειδή $b>a$ προκύπτει ότι $\frac{a+b+c}{b-a}\geq 3$.
Η ισότητα ισχύει αν και μόνο αν $b=c=4a$.

Stelios...

Σωστό φαίνεται   :)

Edit: Η τροποποιημένη απάντηση είναι πολύ καλύτερη  :D


quadratics
number theory proposed
number theory

If p is prime and an arbitrary factorization of p - 1 = D1D2...Dn so that Di | p - 1, prove that if p = 4k + 1 there are an even number of D's which are nonquadratic residues mod p.

If p = 4k + 3 there are an odd number of D's which are nonquadratic residues mod p.

In short, in an arbitrary factorization of p - 1, if n is the number of divisors which are nonquadratic residues of p, (-1|p) = -1^n.

For example p = 43

42 = 1 x 42

42 = 2 x 3 x 7

42 = -1 x 3 x 3 x -7

etc.

Because 43 = 4k + 3 there are an odd number of nonquadratic residue divisors in each of these factorizations of 43 - 1.

These factorizations must be complete but not necessarily a prime decomposition.

Bear in mind 0 is even so a particular factorization of p - 1 may have 0 nonquadratic divisors.

Example 2. p = 13

12 = -1 x 2 x -2 x 3 and there are an even number of nonquadratic residue divisors, (2 and -2), because 13 = 4k + 1.

Rust, I don't think you understand the problem.

40 = 2 x 20 so an [b]even[/b] number of these divisors are nonquadratic residues mod 41 because 41 = 4k + 1

Yes there are an even number (0) of nonquadratic residues in this particular factorization and (-1|41) = -1^0 = 1 and since 41 = 4k + 1 (-1|41) = 1 as predicted.

Let

Q = a quadratic residue
R = a nonquadratic residue

Consider the properties of the Legendre Symbol-

1.  (-1|p) = (p - 1|p) = 1 if p = 4k + 1
2.  (-1|p) = (p - 1|p) = -1 if p = 4k + 3
3. (Q1Q2|p) = (Q|p) = 1 so the product of any number of Q's is a Q
4. (QR|p) = (R|p) = -1 so the product of any number Q's and an R is an R and the Q's can be deleted in the evaluation of (n|p)
5. (R1R2|p) = (Q|p) = 1 so the product of two R's is a Q. More  to the point, the product of an even number of R's is a Q and the product of an odd number of R's is an R.

Let p - 1 = Q1Q2...QmR1R2...Rn

(-1|p) = (p - 1|p) = (Q1Q2...QmR1R2...Rn | p) = (R1R2...Rn | p) = (Q|p) = 1 if n is even which means that p = 4k + 1

If n is odd

(-1|p) = (p - 1|p) = (Q1Q2...QmR1R2...Rn | p) = ( (R1R2...Rn - 1)Rn | p) = (Rn | p) = -1 which means that p = 4k + 3

So (-1|p) = -1^n for n as defined above.

_____________________________________________

This can all be generalized as follows-

(a | p) = -1^n, where n is the number of nonquadratic residue divisors in any factorization of a.

All this is closely related to the fact that the exponent of a prime which divides a square must be even.
_____________________________________________


Let $ ABCD$ is a quadrilateral. Let $ (O)$ is circum circle of $ ABCD$. Let $ AC$ meet $ BD$ at I.

Prove that: $ \frac{AB}{CD}\plus{}\frac{CD}{AB}\plus{}\frac{BC}{AD}\plus{}\frac{AD}{BC}\le \frac{IA}{IC}\plus{}\frac{IC}{IA}\plus{}\frac{IB}{ID}\plus{}\frac{ID}{IB}$

[u][b][color=darkred]This problem is from Sanct-Petersburg 1988 Olympiad Grade 10.[/color][/b][/u]

[color=darkblue]Here is my solution:

$ \| \begin{array}{cc} \triangle{AIB}\sim \triangle{DIC} \Rightarrow \frac {AB}{CD} = \frac {IB}{IC} = \frac {IA}{ID} \Rightarrow \| \begin{array}{cc} \frac {AB}{CD} = \frac {IB}{IC} \\
 \\
\frac {CD}{AB} = \frac {ID}{IA} \end{array} \| \\
 \\
\triangle{BIC}\sim \triangle{AID} \Rightarrow \frac {BC}{AD} = \frac {IB}{IA} = \frac {IC}{ID} \Rightarrow \| \begin{array}{cc} \frac {BC}{AD} = \frac {IB}{IA} \\
 \\
\frac {CD}{AB} = \frac {ID}{IC} \end{array} \| \end{array} \Rightarrow \begin{array}{cc} \frac {AB}{CD} + \frac {CD}{AB} + \frac {BC}{AD} + \frac {AD}{BC} \\
= \\
\frac {IB}{IC} + \frac {ID}{IA} + \frac {IB}{IA} + \frac {ID}{IC} \end{array}$

Note that $ IA\cdot IC = IB\cdot ID = p(I)$.
Then we should prove:

$ \boxed{\begin{array}{cc} \frac {IB}{IC} + \frac {ID}{IC} + \frac {ID}{IA} + \frac {IB}{IA} \le \frac {IA}{IC} + \frac {IC}{IA} + \frac {IB}{ID} + \frac {ID}{IB} \\
 \\
\frac {IB + ID}{IC} + \frac {ID + IB}{IA} \le \frac {IA^2 + IC^2}{p(I)} + \frac {IB^2 + ID^2}{p(I)} \\
 \\
\frac {BD\cdot AC}{p(I)}\le \frac {IA^2 + IC^2 + IB^2 + ID^2}{p(I)} \\
 \\
BD\cdot AC \le IA^2 + IC^2 + IB^2 + ID^2 \\
 \\
x^2 + y^2 + z^2 + t^2\ge (x + y)(z + t) \end{array} }$

But $ \| \begin{array}{cc} x^2 + y^2\ge \frac {(x + y)^2}{2} \\
z^2 + t^2\ge \frac {(z + t)^2}{2} \end{array} \| \Rightarrow x^2 + y^2 + z^2 + t^2 \ge \frac {(x + y)^2}{2} + \frac {(z + t)^2}{2} \ge (x + y)(z + t)$

$ \mathrm{{\LARGE Q.E.D.}}$[/color]

[quote="Ahiles"][u][b][color=darkred]This problem is from Sanct-Petersburg 1988 Olympiad Grade 10.[/color][/b][/u]

[color=darkblue]Here is my solution:

$ \| \begin{array}{c}{cc} \triangle{AIB}\sim \triangle{DIC} \Rightarrow \frac {AB}{CD} = \frac {IB}{IC} = \frac {IA}{ID} \Rightarrow \| \begin{array}{cc} \frac {AB}{CD} = \frac {IB}{IC} \
 \
\frac {CD}{AB} = \frac {ID}{IA} \end{array} \| \
 \
\triangle{BIC}\sim \triangle{AID} \Rightarrow \frac {BC}{AD} = \frac {IB}{IA} = \frac {IC}{ID} \Rightarrow \| \begin{array}{cc} \frac {BC}{AD} = \frac {IB}{IA} \
 \
\frac {CD}{AB} = \frac {ID}{IC} \end{array} \| \end{array} \Rightarrow \begin{array}{cc} \frac {AB}{CD} + \frac {CD}{AB} + \frac {BC}{AD} + \frac {AD}{BC} \
= \
\frac {IB}{IC} + \frac {ID}{IA} + \frac {IB}{IA} + \frac {ID}{IC} \end{array}$

Note that $ IA\cdot IC = IB\cdot ID = p(I)$.
Then we should prove:

$ \boxed{\begin{array}{c}{cc} \frac {IB}{IC} + \frac {ID}{IC} + \frac {ID}{IA} + \frac {IB}{IA} \le \frac {IA}{IC} + \frac {IC}{IA} + \frac {IB}{ID} + \frac {ID}{IB} \
 \
\frac {IB + ID}{IC} + \frac {ID + IB}{IA} \le \frac {IA^2 + IC^2}{p(I)} + \frac {IB^2 + ID^2}{p(I)} \
 \
\frac {BD\cdot AC}{p(I)}\le \frac {IA^2 + IC^2 + IB^2 + ID^2}{p(I)} \
 \
BD\cdot AC \le IA^2 + IC^2 + IB^2 + ID^2 \
 \
x^2 + y^2 + z^2 + t^2\ge (x + y)(z + t) \end{array} }$

But $ \| \begin{array}{c}x^2 + y^2\ge \frac {(x + y)^2}{2} \
z^2 + t^2\ge \frac {(z + t)^2}{2} \end{array} \| \Rightarrow x^2 + y^2 + z^2 + t^2 \ge \frac {(x + y)^2}{2} + \frac {(z + t)^2}{2} \ge (x + y)(z + t)$

$ \mathrm{{\LARGE Q.E.D.}}$[/color][/quote]
Thank you very much.


[url=http://answers.yahoo.com/question/index;_ylt=AqXfXbGaduRj9Gc8659U9kkjzKIX;_ylv=3?qid=20080110113023AABYw1F]Reference.[/url]

There's my question.

ps: What approach should I take to solve the question? How long does it take to answer the question?

Thank you vary much!

That reminds of the classical [url=http://mathworld.wolfram.com/SultansDowryProblem.html]dowry problem[/url] The crucial difference is that the dowry problem requires just maximizing the chances of winning the largest prize while the problem in question requires maximizing the expected value of the prize. I'm not sure this version is well-posed at all, i.e., I doubt that there is an optimal strategy that works simultaneously for all values of prizes. The offered solution in the thread you mentioned is obviously not optimal if there is one huge prize and a few small ones. So the first question here is how to state the problem correctly, not how to solve it.


inequalities
combinatorics unsolved
combinatorics

Let $X=\{A_{1},A_{2}, \dots,A_{n}\}$ be a set of distinct 3-element subsets of $\{1,2, \dots ,36\}$ such that

i) $A_{i}$ and $A_{j}$ have non-empty intersection for every $i,j$
ii) The intersection of all the elements of X is the empty set.

Show that $n \leq 100$. How many sets $X$ are that when $n=100$?

I believe that there are ${36\choose 3}$ distinct $X$ when $n=100$, but a definite proof is too elusive for me right now.  Anyone have a soln?

[Moderator edit: See also http://www.mathlinks.ro/Forum/viewtopic.php?t=25362 .]

I will solve only the first part because the following is quite enough boring, and I don't have courage to look more precisely the equality cases... :wink:

First case : Assume, that $A_{1}=\{a,b,c \}$ meets each of the others $A_{i}$ in two elements. Then, these $A_{i}$ are formed by a choice of two elements from $A_{1}$ (3 possibilities) and for each of them, we have to complete with any element except $a,b,c$, which leaves $33$ possibilities.
Therefore, there are at most $3 \times 33 = 99$ other $A_{i}$. Adding $A_{1}$ gives the desired result.

Second case : Wlog, we may assume that $A_{2}= \{a,x,y \}$ where $\{b,c \}$ and $\{x,y \}$ are disjoints.
Now, since the intersection of all the sets is empty, we may assume that $a \notin A_{3}$ so that, wlog, $A_{3}= \{b,x,t \}$ (it has to intersects $A_{1}$ and $A_{2}$), and $t$ can be equal to $c$ or $y$. Since $c \neq y$ we may assume that $t \neq y$ (so that $t = c$ is the unique possibility).

Now, let $n_{a}$ be the number of $A_{i}$ which meets $\{a,b,x \}$ only at $a$, and with $i >3.$
The numbers $n_{b},n_{x}$ are defined in the same way.
Let $n_{ab}$ be the number of $A_{i}$ which meets $\{a,b,x \}$ only at $a$ and $b$, and with $i >3.$
The numbers $n_{bx},n_{xa}$ are defined in the same way.

Each set counted in $n_{a}$ must contain $t$ because it has to intersect $A_{3}$. Thus it has the form $\{a,t, \alpha \}$ where $\alpha \notin \{a,b,x,t\}$ (this leaves only $32$ possibilities)
In another hand, each set counted in $n_{bx}$ has the form $\{b,x, \beta \}$ where $\beta \notin \{a,b,x,t\}.$ But, $b,x \notin \{a,t, \alpha \}$ and $\beta \notin \{a,t\}$ thus to preserve the intersection propert, we must have $\alpha = \beta.$
This means that if there are at least two sets of type $n_{a}$ (resp. $n_{bx}$) then there is none of type $n_{bx}$ (resp. $n_{a}$). It easily follows that $n_{a}+n_{bx}\leq 32.$

In a similar way, we have $n_{b}+n_{ax}\leq 32$ and $n_{x}+n_{ab}\leq 32$.
Adding these three inequalities, we obtain :
$n_{a}+n_{bx}+n_{b}+n_{ax}+n_{x}+n_{ab}\leq 96$.
We have to add $A_{1},A_{2},A_{3}$ for a partial total not greater than $99$.

Now, if some of our $A_{i}$ is $E = \{a,b,x \}$ then there is none of the $A_{i}'s$ which cannot intersect $E$ which leads to a total not greater than $100$, as desired.
Otherwise, $E$ is not one of the $A_{i}'s$ and :
1) If $t \neq c$ then there is at most one of the $A_{i}'s$ which does not intersect $E$, namely $\{c,y,t\}$ because such a set have to intersect each of $A_{1},A_{2},A_{3}.$ This leads again to a total not greater than $100$.
2) If $t =c$ :
Any of the $A_{i}'s$ which does not intersect $E$ has the form $\{c,y,n\}$ where $n \notin E$. But, if there exists $n \notin E$ such that $A_{i}= \{c,y,n \}$ then we are in the previous case (that is : Second case ending with 1) ), interchanging $b$ with $c$, and $x$ with $y$.Thus, the conclusion holds again.

Pierre.

I solved it a little differently.  Assume one of the sets is $S ={a,b,c}.$  Define sets $A, B, C, D, E, F$ to be the subsets of $X$ that intersect $S$ at $\{a\}, \{b\}, \{c\}, \{a,b\}, \{a,c\},$ and $\{b,c\}$ respectively.  We will use the following lemma.

Lemma: If $Y$ is a set of 2 element sets, such that every two members share an element, and $|Y| > 3$, then all the members of $Y$ share a common element.

Proof:  Let two members of $Y$ be $\{a,b\}$ and $\{a,c\}$.  If not all members of $Y$ share a common element, then there is a member that does not contain $a$.  It must be $\{b,c\}$, and then there can be no other sets, a contradiction.

In order for $A %Error. "union" is a bad command.
F$ to have more than 4 elements, either $|F| > 2$ or $|A| > 2$.  If $|F| > 2$, every element of $F$ contains a single element other than $a$ and $b$, and so every element of $A$ must contain all those single elements for every member of $F$, which is impossible.  So $A$ must be empty.  There are 33 choices for the third element of $F$, so $|A %Error. "union" is a bad command.
F| \leq 33$.

Otherwise, suppose $|A| > 2$.  Then for any element $T$ of $B, C,$ or $F$, by the lemma, all the members of $A %Error. "union" is a bad command.
{T}$ must share a common element.  Note that the members of $A$ cannot share more than one common element other than $a$, so in fact $A$ must share this one common element with all of $B, C,$ and $F$.  Call this element $d$.  Then the only possible member of $F$ is $\{b,c,d\}$, and all members of $A$ are of the form $\{a,d,e\}$, with $e$ chosen from the remaining 32 elements.  So $|F| \leq 1$ and $|A| \leq 32$, so again we have $|A %Error. "union" is a bad command.
F| \leq 33$.

The same argument applies to the pairs $(B,E)$ and $(C,D)$, so each pair has no more than 33 combined elements, and $X$ can have no more than $1+33+33+33 = 100$ elements.

How many ways can this be achieved?  For each of the three pairs mentioned above, we have two choices:

Case I:  $(A,B,C)$ is empty and $(D,E,F)$ has 33 elements.

Case II: $(A,B,C)$has 32 elements sharing a common element, and $(D,E,F)$ has one element with that same common element.

Note that of two or more of the three pairs are in Case II, the common element must be the same for all cases.

We claim that having two of the three pairs be in Case II, and one in Case I, is equivalent to all three being in case I.  We also claim that having two of the three pairs in Case I, and the other pair in case II, is equivalent to having all three in case II.

Claim I:  We can suppose WLOG that $(A,F)$ and $(B,E)$ are in case II, and $(C,D)$ is in Case I.  Then let $T = \{a,b,d\}$, and define $A', B', C', D', E', F'$ in the same fashion that we defined $A, B, C, D, E, F$ above.  Then we claim that all three pairs are now in Case I;  indeed, all the members of $A$ and $F$ contain both $a$ and $d$, all the members of $B$ and $E$ contain $b$ and $d$, all members of $C$ and $D$ contain $a$ and $b$.

claim II: We can suppose WLOG that $(A,F)$ and $(B,E)$ are in case I, and $(C,D)$ is in Case II.  Then let $T ={a,b,d}$, and define $A', B', C', D', E', F'$ in the same fashion that we defined $A, B, C, D, E, F$ above.  Then we claim that all three pairs are now in Case II;  indeed, only one member of $A$ and $F$ contains more than one element of $T$ ($\{b,c,d\}$),  only one member of $B$ and $E$ contains more than one element of $T$ ($\{a,c,d\}$), and only one other element of $X$ contains more than one element of $T$ ($\{a,b,c\}$).

We claim that these two remaining cases are everywhere disjoint.  A simple counting argument suffices.  In the first case, $a,b,c$ are contained in 67 sets each;  the remaining elements are contained in 3 elements each.  In the second case, $a,b,c$ are contained in 35 sets each; element $d$ is contained in 99 sets;  and the remaining elements are again contained in 3 sets each.

The number of possible sets that fit in the first case is clearly ${36 \choose 3}$;  $a,b,c$ are symmetric, as are the remaining elements.  In the second case, element $d$ is distinguished, $a,b,c$ are again symmetric, and the remaining 32 are symmetric as well, so there are $\frac{36!}{3!32!}$ cases.  Thus there are

${36 \choose 3}+\frac{36!}{3!32!}$ = 242,760 possibilities in total.

Very long :)
I didn't read your proofs :)


Define the sequence $(a_n)_{n\geq 1}$ by $a_1=1$, $a_2=p$ and \[a_{n+1}=pa_n-a_{n-1} \textrm { for all }  n>1.\]
Prove that for $n>1$ the polynomial $x^n-a_nx+a_{n-1}$ is divisible by $x^2-px+1$.
Using this result, solve the equation \[x^4-56x+15=0.\]

I don't have time, but solving the characteristic equation will solve the first part. Then, the second part: p=4.


logarithms
trigonometry
real analysis
real analysis unsolved

If $ a_n$ is a real sequence such that $ |a_n\minus{}a_{n\minus{}1}|\to 0$ as $ n\to \infty$, then it should be true that $ a_n$ is a cauchy sequence. Can someone give a hint to prove this?

Absolutely not. Take $ a_n\equal{}\sqrt n.$ It is true that $ a_n\minus{}a_{n\minus{}1}$ tends to zero but it is not a Cauchy sequence. You may see by using the definition or observing that if $ a_n$ were Cauchy, it would converge to a limit but $ a_n\to\plus{}\infty.$ Same for $ a_n \equal{} \ln n$....

[quote="limsup"]If $ a_n$ is a real.....[/quote]

Yes, you are absolutely right. I didn't pose the question properly. I think the following is true.
If $ a_n$ is a real(or complex) sequence such that $ |a_n \minus{} a_{n \minus{} 1}|\to 0$ as $ n\to \infty$ and if there is a subsequence $ a_{n_i}\to a$(say), then every subsequence of $ a_n$ converges to the same limit ($ a$) and, in fact, $ a_n \to a$
Am I right now?

New example: $ \sin(\sqrt {n})$. Everything in $ [\minus{}1,1]$ is a limit point.

yes I am wrong again. Actually if the hypotheses are true then the  shifted subsequence $ a_{n_{i} \plus{} k}$ also converges to $ a$ for any fixed $ k$. (This can be easily proved and this time I must be correct :) ) Thanks jmerry, Arrigo-Sacchi.


$ e^x\equal{}kx$ for what k is there only one solution?

I think that for negative values of $ k$. :maybe:

If we graph $ y \equal{} kx$ and $ y \equal{} e^x$ we see that for some positive $ k$ there is no solution so it is with $ k \equal{} 0$, but for any other positive $ k$ there are two solutions (I think $ kx$ always catches e^x becauses this one keeps getting "verticaler" and "verticaler")(intersections of the graphs). Also we see that for any negative value of $ k$ there will be only one intersection point.

 :)

i think there is a solution for a positive value of k... for example if you draw the graph of $ y\equal{}e^x$ and that of $ y\equal{}ex$ (k=e) there appears to be one intersection as required, but how do we prove this?

EDIT: See below. Didn't think about using calculus and Mean Value...

If $ e^x\equal{}kx$ has only one soultion $ kx$ is a tangent to $ e^x.$ But all tangents to a function are given by $ f(x_0)\plus{}(x\minus{}x_0)f'(x_0)\equal{}f'(x_0)x\plus{}f(x_0)\minus{}x_0f'(x_0)$ and we get $ f(x_0)\equal{}x_0f'(x_0)\Rightarrow e^{x_0}\equal{}x_0e^{x_0}\Rightarrow x_0 \equal{} 1$ and $ k\equal{}e.$

[hide="Is this correct? :maybe: "]Differentiate both sides with respect to $ x$ to get $ e^x\equal{}k$, which combined with the original equation gives $ x\equal{}1$ and $ k\equal{}e$.[/hide]

[quote="bertram"]If $ e^x \equal{} kx$ has only one soultion $ kx$ is a tangent to $ e^x.$[/quote]
Or $ k<0.$ :wink:


integration
logarithms
calculus
calculus computations

find

$ \int \frac {1}{x^7 \minus{} x} dx$

nice solution  :clap2:

here is another one

$ \frac {1}{x^7 \minus{} x}$

= $ \frac {1}{x^4(x^3 \minus{} \frac {1}{x^3})}$

put $ \frac {1}{x^3} \equal{} u$ $ \implies$ $ du \equal{} \minus{} \frac {3}{x^4} dx$

                                     $ \frac {1}{x^4} dx \equal{} \minus{}\frac {du}{3}$.

the integral becomes

   $ \minus{}\frac {1}{3} \int \frac {1}{(\frac {1}{u} \minus{} u)} du$

= $ \minus{}\frac {1}{3} \int \frac {u}{1\minus{}u^2} du$

= $ \frac {1}{6} \int \frac {\minus{}2u}{1\minus{}u^2}du$

= $ \frac {1}{6} \ln |1\minus{}u^2| \plus{} C$

= $ \frac {1}{6} (\ln|x^6 \minus{} 1| \minus{} 6ln|x|) \plus{} C$

=  $ \frac {1}{6} \ln|x^6 \minus{} 1| \minus{} \ln |x| \plus{} C$


The access code is an integer $x \geq 0$. Knowing the sums of some of the digits of the code, find the smallest number which fulfills the given condition.

We consider the digits of $x$ numbered from right to left. Thus, given a number $\overline{\ldots a_4 a_3 a_2 a_1}$, $134$ represents $a_1 + a_3+a_4$. (the sums are encoded in increasing order, i.e. $413$ is not valid)

Input: $N$ (the number of known sums); $N$ pairs of values: the first one represents the digits we sum up, whilst the second is the value of that sum

Output: If such $x$ exists, print the smallest one; else print $-1$.

restrictions: $0<N<10$ and $x$ has at most 9 digits

e.g. $N=2$; $13 ~ 3$; $12 ~ 8$; $\Rightarrow x=\overline{053}=53$.

i got this problem on an olympiad in the 9th grade, and i still don't know how to solve it :(

Hm, would a simple backtracking solution not work? Just proceed from the highest-order digits, trying to make them as small as possible (=trying 0, then 1, 2, ... if all the possibilities of the lower digits are found to be impossible)...

I tried the backtracking algorithm, but it worked pretty slow if $x$ had $9$ digits  :(

If you reverse the backtracking order you should end up with a linear algorithm. First you try to set the last digit ($x_1$) to the greatest possible value (9) and then check if it's all right with all the rules. If you find out that it isn't you just keep decreasing it's value. Then you do $x_2$, $x_3$ and so on. For your example:

From the first rule, $x_1$ can be at most 3 so we set $x_1=3$. Then from the second rule, $x_2$ can be at most 5 so it [i]will[/i] be five. Finally, $x_3$ must be zero.

[quote="bus"]then check if it's all right with all the rules[/quote]

this seems blurry :blush: ... can you tell me how do we proceed on:

$a_5+a_3+a_1=7$

$a_4+a_2+a_1=11$

$a_5+a_2=6$

Because of rule #1, $a_1$ can be at most 7 so we set it to 7. Now, $a_1=7$ so $a_2$ can be at most 4 because of rule #2, we set $a_2=4$. Finally, $a_3$, $a_4$ and $a_5$ should be all zero. Doesn't seem to work, huh? Then I was probably wrong, sorry :).

So, how can this problem be solved? (excepting backtracking , which is slow in many cases)

You know, this is really bugging me. There [i]has[/i] to be some way to do it faster than backtracking...

I also believe(d) this because the problem was given at the lowest phase of the olympiad, and in 9th grade (we do backtracking in 10th grade)

but if no one can find a better solution, maybe the author's solution was indeed backtracking


In the figure, $ m\angle ADE \equal{} 30^\circ$. What is the number of degrees in $ m\angle CGF$?

[asy]size(150); fontsize(10pt); real angle = 40; pair A = (0,0), C = dir(angle), B = C + tan(angle*pi/180)*(dir(-(90-angle))), D = .4*dir(angle), E = (D.x,0), G = B + .70*dir(90+angle), F = (G.x,0);

draw(A--C--B--cycle); draw(D--E); draw(G--F); draw(rightanglemark(A,E,D,2)); draw(rightanglemark(A,C,B,2)); draw(rightanglemark(B,F,G,2));
label("$E$",E,S); label("$D$",D,NW); label("$C$",C,N); label("$G$",G,NE); label("$F$",F,S); label("$B$",B,S); label("$A$",A,S);[/asy]

If ADE is 30 then CDE is 150 and every other angle except the one that we are trying to find is a right angle. Since it is a pentagon its angles will add to 540, we have (90*3) + 150 + x =540
270+150+x =540
420+x=540
x=120

[quote="mathemagician1729"]Since it is a pentagon its angles will add to 540, we have (90*3) + 150 + x =540
[/quote]

In case you don't know how to find this, there's a formula. The sum of the degrees of a polygon with $ n$ sides is $ 180(n \minus{} 2)$. So when we plug $ 5$ in, we find the sum of the angles is $ 540^\circ$.

[quote="isabella2296"]
In case you don't know how to find this, there's a formula. The sum of the degrees of a polygon with $ n$ sides is $ 180(n \minus{} 2)$. [/quote]
I thought you had to divide

You only divide by n when you are trying to find the idividual angle measure of a regular polygon.

so $ 180(n\minus{}2)$ is for the total angle sum?

Yes; for an individual angle, you do $ 180 \minus{} \frac{360}{n}$ for an $ n$-gon.


1) Prove that if $ a,b,c\geq0$ then
\[ \sum a\sqrt{a^2\minus{}ab\plus{}b^2}\sqrt{a^2\minus{}ac\plus{}c^2} \geq a^3\plus{}b^3\plus{}c^3.\]
2) Let $ a,b,c$ be non-negative real numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}1$. Prove that
\[ \frac{a^2}{1\minus{}ab}\plus{}\frac{b^2}{1\minus{}bc}\plus{}\frac{c^2}{1\minus{}ca}\leq\frac32.\]
3) Let $ a,b,c,d$ be non-negative real numbers such that
\[ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}1.\]
Prove that
\[ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}\max(ab\plus{}cd, bc\plus{}ad, ac\plus{}bd)\geq1.\]
4) Prove that if $ a,b,c\geq0$, then
\[ (a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)(a^2\plus{}b^2\plus{}c^2)\leq\frac1{64}(a\plus{}b\plus{}c)^8.\]

[quote="hitek"]1) Prove that if $ a,b,c\geq0$ then
\[ \sum a\sqrt {a^2 \minus{} ab \plus{} b^2}\sqrt {a^2 \minus{} ac \plus{} c^2} \geq a^3 \plus{} b^3 \plus{} c^3.
\]

[/quote]
Try $ a\equal{}0,$ $ b\equal{}2$ and $ c\equal{}1.$ :wink:

In problem 4, set a=0 b=0 c=2 and you get that the inequalitie doesn't hold  .

:oops: I'm sorry....  :P I've been so stupid

[quote="hitek"]
4) Prove that if $ a,b,c\geq0$, then
\[ (a^2 \plus{} b^2)(b^2 \plus{} c^2)(c^2 \plus{} a^2)(a^2 \plus{} b^2 \plus{} c^2)\leq\frac1{64}(a \plus{} b \plus{} c)^8.
\]
[/quote]

Assume $ a \plus{} b \plus{} c \equal{} 1$ and setting $ q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$ then rewrite the inequality
\[ \left( {a^2 \plus{} b^2 } \right)\left( {b^2 \plus{} c^2 } \right)\left( {c^2 \plus{} a^2 } \right)\left( {a^2 \plus{} b^2 \plus{} c^2 } \right) \le \frac {1}{{64}}\left( {a \plus{} b \plus{} c} \right)^8
\]

\[ \Leftrightarrow \left[ {\left( {4 \minus{} 2q} \right)\left( {q^2 \minus{} 4r} \right) \minus{} r^2 } \right]\left( {2 \minus{} q} \right) \le 2
\]
By Schur for third degree we have
\[ r \ge \max \left\{ {0,\frac {{8q \minus{} 8}}{9}} \right\}
\]
Hence if $ q \le 1$ then $ r \ge 0$ so we have
\[ \left[ {\left( {4 \minus{} 2q} \right)\left( {q^2 \minus{} 4r} \right) \minus{} r^2 } \right]\left( {2 \minus{} q} \right) \minus{} 2 \le 2q^2 \left( {2 \minus{} q} \right)^2 \minus{} 2
\]
We just need to show that
\[ 2q^2 \left( {2 \minus{} q} \right)^2 \minus{} 2 \le 0
\]

\[ \Leftrightarrow q\left( {2 \minus{} q} \right) \le 1
\]

\[ \Leftrightarrow \left( {q \minus{} 1} \right)^2 \ge 0
\]
If $ q \ge 1$ then $ r \ge \frac {{8q \minus{} 8}}{9}$ so
\[ \left[ {\left( {4 \minus{} 2q} \right)\left( {q^2 \minus{} 4r} \right) \minus{} r^2 } \right]\left( {2 \minus{} q} \right) \minus{} 2 \le \left[ {\left( {4 \minus{} 2q} \right)\left( {q^2 \minus{} \frac {{32q \minus{} 32}}{9}} \right) \minus{} \frac {{\left( {8q \minus{} 8} \right)^2 }}{{81}}} \right]\left( {2 \minus{} q} \right) \minus{} 2
\]
So it suffices to prove that
\[ \left[ {\left( {4 \minus{} 2q} \right)\left( {q^2 \minus{} \frac {{32q \minus{} 32}}{9}} \right) \minus{} \frac {{\left( {8q \minus{} 8} \right)^2 }}{{81}}} \right]\left( {2 \minus{} q} \right) \minus{} 2 \le 0
\]

\[ \Leftrightarrow 162q^4 \minus{} 1288q^3 \plus{} 3784q^2 \minus{} 4928q \plus{} 2270 \le 0
\]

\[ \Leftrightarrow \left( {q \minus{} 1} \right)\left( {162q^3 \minus{} 1126q^2 \plus{} 2658q \minus{} 2270} \right) \le 0
\]
which is true. It's done.

By the way we also have an similar inequality is
\[ \left( {a^3 \plus{} b^3 \plus{} c^3 } \right)^2 \left( {a^3 b^3 \plus{} b^3 c^3 \plus{} c^3 a^3 } \right) \le 4
\]
with $ a,b,c \ge 0$ satisfy $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$

For solution you can see here (in the attachment file).

http://www.mathlinks.ro/viewtopic.php?t=208794

Moreover we have a stronger and nicer than yours is
\[ \left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^2 \left( {a^2 b^2 \plus{} b^2 c^2 \plus{} c^2 a^2 } \right) \le 4
\]
with $ a,b,c \ge 0$ satisfy $ a \plus{} b \plus{} c \equal{} 2$. Equality hold if and only if $ a \equal{} b \equal{} 1,c \equal{} 0$. :)

And now I am thinking about
\[ \left( {a^4 \plus{} b^4 \plus{} c^4 } \right)^2 \left( {a^4 b^4 \plus{} b^4 c^4 \plus{} c^4 a^4 } \right) \le 4
\]
with $ a,b,c \ge 0$ satisfy $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 2$.  :wink:

The solution for 4) is much simpler. Let $ a \plus{} b \plus{} c \equal{} 1$, $ ab \plus{} bc \plus{} ca \equal{} q$, and $ abc \equal{} r$. The left hand side is rewritten in terms of $ p,q,r$
\[ (1 \minus{} 2q)^2(q^2 \minus{} 2r).
\]
Since $ r\geq0$, we just need to show
\[ (1 \minus{} 2q)^2q^2\leq\frac1{64}
\]
which is equivalent to the obvious $ \minus{} (4q \minus{} 1)^2\leq0$.

[quote="hitek"]The solution for 4) presented in my coming English version book is much simpler. Let $ a \plus{} b \plus{} c \equal{} 1$, $ ab \plus{} bc \plus{} ca \equal{} q$, and $ abc \equal{} r$. The left hand side is rewritten in terms of $ p,q,r$
\[ (1 \minus{} 2q)^2(q^2 \minus{} 2r).
\]
Since $ r\geq0$, we just need to show
\[ (1 \minus{} 2q)^2q^2\leq\frac1{64}
\]
which is equivalent to the obvious $ \minus{} (4q \minus{} 1)^2\leq0$.[/quote]

Yes sorry for my stupid solution  :blush: . But mine and yours is the same, I didn't notice that with $ r \ge 0$ is enough for your inequality  :wink: .

[quote="arqady"][quote="hitek"]1) Prove that if $ a,b,c\geq0$ then
\[ \sum a\sqrt {a^2 \minus{} ab \plus{} b^2}\sqrt {a^2 \minus{} ac \plus{} c^2} \geq a^3 \plus{} b^3 \plus{} c^3.
\]
[/quote]
I have proved that
$ \sum_{cyc} a\sqrt {(a^2 \minus{} ab \plus{} b^2)(a^2 \minus{} ac \plus{} c^2)} \leq a^3 \plus{} b^3 \plus{} c^3.$  :maybe:[/quote]

My solution is
\[ \left(\sum_{\rm cyclic} a\sqrt{a^2\minus{}ab\plus{}b^2}\sqrt{a^2\minus{}ac\plus{}c^2}\right)^2\equal{}\left(\sum_{\rm cyclic} \sqrt{a^3\minus{}a^2b\plus{}b^2a}\sqrt{a^3\minus{}a^2c\plus{}c^2a}\right)^2\leq (a^3\plus{}b^3\plus{}c^3)^2.\]

[quote="hitek"][quote="arqady"][quote="hitek"]1) Prove that if $ a,b,c\geq0$ then
\[ \sum a\sqrt {a^2 \minus{} ab \plus{} b^2}\sqrt {a^2 \minus{} ac \plus{} c^2} \geq a^3 \plus{} b^3 \plus{} c^3.
\]
[/quote]
I have proved that
$ \sum_{cyc} a\sqrt {(a^2 \minus{} ab \plus{} b^2)(a^2 \minus{} ac \plus{} c^2)} \leq a^3 \plus{} b^3 \plus{} c^3.$  :maybe:[/quote]

My solution is
\[ \left(\sum_{\rm cyclic} a\sqrt {a^2 \minus{} ab \plus{} b^2}\sqrt {a^2 \minus{} ac \plus{} c^2}\right)^2 \equal{} \left(\sum_{\rm cyclic} \sqrt {a^3 \minus{} a^2b \plus{} b^2a}\sqrt {a^3 \minus{} a^2c \plus{} c^2a}\right)^2\leq (a^3 \plus{} b^3 \plus{} c^3)^2.
\]
[/quote]
If are you mean that $ \sum_{cyc}\sqrt{(a^3 \minus{} a^2b \plus{} b^2a)(a^3 \minus{} a^2c \plus{} c^2a)}\leq\frac{1}{2}\sum_{cyc}(a^3 \minus{} a^2b \plus{} b^2a\plus{}a^3 \minus{} a^2c \plus{} c^2a)\equal{}$
$ \equal{}a^3\plus{}b^3\plus{}c^3$ then it's also my proof.

5) Let $ a,b,c$ be non-negative real numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}1$. Prove that
\[ \sum_{\rm cyclic} a\sqrt{a^2\minus{}ab\plus{}b^2}\sqrt{a^2\minus{}ac\plus{}c^2}\geq\frac1{\sqrt3}.\]

6) Let $ x,y,z,w$ be non-negative real numbers such that
\[ x^2\plus{}y^2\plus{}z^2\plus{}w^2\equal{}1.\]
Prove that
\[ \frac{x^2}{1\minus{}xy}\plus{}\frac{y^2}{1\minus{}yz}\plus{}\frac{z^2}{1\minus{}zw}\plus{}\frac{w^2}{1\minus{}wx}\leq\frac43.\]

7) Let $ x,y,z,w$ be non-negative real numbers such that
\[ x^2\plus{}y^2\plus{}z^2\plus{}w^2\equal{}1.\]
Prove that
\[ \frac{x^3}{1\minus{}xy}\plus{}\frac{y^3}{1\minus{}yz}\plus{}\frac{z^3}{1\minus{}zw}\plus{}\frac{w^3}{1\minus{}wx}\geq\frac23.\]