inequalities AMC AIME USA(J)MO USAMO function calculus

In korea, for those who prepare for olympiads and other math contests, 7th graders start studying up to precalc, and additional theorms such as Jensen and Chebyshev's inequality

so before their high school years, they've already covered all high school material in addition to some college theorms and yet some still struggle to get awarded.

Although the website states that AIME and USAMO can be solved by pre-calc methods, I know that the vast mjaority of them knows further extended materials.

My question is, how useful do you think it is to study ahead?  and which materials do you recommened in addition to the precalc stuff to solve some thought provoking Qs?  For example, convex functions, chinese remainder theorm, Minkowski inequalities?

It depends on what olympiad levels are you thinking about. If you're talking about state/aime levels what tokenadult suggested is more than ok / true.

If we're talking about usamo/national/international levels, then knowing calculus can turn out to be a serious advantage.

I think Arthur Engel's Problem-Solving Strategies might also be helpful

the Korean approach that you describe, sounds
like over-training at an early age.  I think what one
wants to do early is practice thinking, solving problems
in a serious way, develop a few things deeply rather
than a lot of what would amount to trivia at that
stage.  It makes no sense to learn Jensen's
inequality prior to a good understanding of
geometry and convex figures (and inequalities
for convex functions go hand-in-hand with
second derivatives in calculus, as well as
convexity for sequences).  Inequalities
in general are somewhat pointless if you don't
have something that you want to estimate,
such as sums of series or probabilities.

Clearly there are hundreds of facts that might
be relevant much later, but that doesn't mean
it makes sense to introduce them at age 12.
Of course this is all relative to what people
know at that age --- what is taught by that
age to strong students in Korea?

I should add that, anecdotally at least, most IMO gold/silver medalists are not very advanced by the end of high school.
They know little or nothing about basic "higher" mathematics
such as linear algebra, complex analysis, etc. (Maybe this is changing in very recent years, I don't know.)
So presumably it is also safe to not know
everything at age 12 either, even if your goal is to win
the olympiad.   Some IMOers deliberately choose to solve
more olympiad problems rather than learn advanced
material or take university courses --- in a sense most/all
choose this route as it is hard to do both without
spending your whole youth on mathematics alone.

I've splitted the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=22124]discussion about the wrapping[/url], it's new location is in the support forum.

[quote="tokenadult"]Yes, this would be more appropriate for Other Problem Solving Topics, but I can't move threads out of this forum.

I think there is much to be said for learning the standard precalculus topics THOROUGHLY. My son takes his precalculus course this coming semester in the United States (at seventh grade age), but I expect I will encourage him to review all that he "already knows" in the [url=http://gcpm.rutgers.edu/books.html]Gelfand Correspondence Program[/url] materials, in Serge Lang's book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0387967877/learninfreed]Basic Mathematics[/url], and of course in [url=http://www.artofproblemsolving.com/Books/AoPS_B_About.php]The Art of Problem Solving volumes 1 and 2[/url]. I think most of the best Olympiad participants [b]over[/b]learn all of their "basic" mathematics in lots of different contexts.[/quote]

tokenadult,

You mentioned a book called Basic Mathematics, and I was wondering, how good is it? Does it cover the topics it teaches in a thorough manner (or for that matter, how far in depth does it go i.e all the way from really basic math to say the math level of USAMO?)

Thanks!

[quote="fleeting_guest"]  Some IMOers deliberately choose to solve
more olympiad problems rather than learn advanced
material or take university courses --- in a sense most/all
choose this route as it is hard to do both without
spending your whole youth on mathematics alone.[/quote]

Have you been on IMO or involved in IMO training that you can evaluate how advanced the problems and theory is that they cover ? Are that just assumptions, is that a general statement ? You say it may have changed recently ? Which time are you referring to ?  :)

And which olympiads do you like best at the national or international level ?  :)

[quote="Valentin Vornicu"][quote="fanaticsm"]
tokenadult,

You mentioned a book called Basic Mathematics, and I was wondering, how good is it? Does it cover the topics it teaches in a thorough manner (or for that matter, how far in depth does it go i.e all the way from really basic math to say the math level of USAMO?)

Thanks![/quote]I think it's extremely hard to go from basic math to USAMO level in just one book. Just think what segment (somewhat to 1/2 to 2/3) of this the two AoPS volumes cover, and they are quite big (in size :D).[/quote]

That is also true as well :D, but I meant to say, what are the level of some of the topics it teaches :lol:...

[quote="fleeting_guest"]what is taught by that
age to strong students in Korea?[/quote]

Well Korea doesnt realli have a strong system where top students are allowed skip grades and take higher level maths.  They just have to be stuck with their normal curriculum

In 7th grade, they study

algebra (not simultaneous)
y=ax+b basic function
calculating areas of 3D objects
statistics (including histograms)
sets (including De Morrgans law)
the base numbers: like base 2,3 and 7
factors of a number and the multiples (this is basically basics of number theory learning about GCD, LCD, determining whether if a number is divisible by 2,3,4...., and so on)
and lastly, some geometry stuff

All are covered quite extensively solving up to AMC lvl Q's instead of just solving simple straigtforward put the numbers into the equation and solve questions.

[quote="fanaticsm"]tokenadult,

You mentioned a book called Basic Mathematics, and I was wondering, how good is it? Does it cover the topics it teaches in a thorough manner (or for that matter, how far in depth does it go i.e all the way from really basic math to say the math level of USAMO?)[/quote]

Serge Lang's book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0387967877/learninfreed]Basic Mathematics[/url] is not at all a book for preparing for competitions. But in my view, the one thing that more kids preparing for competitions should do is learn some math that isn't completely embedded in competition problems. Lang's purpose in writing Basic Mathematics was to write a suitable textbook for strong students in prep schools so that they could be ready for a strong calculus course as soon as possible. He also wrote the book to introduce some ideas of elementary and secondary mathematics from a higher mathematics point of view. I first learned about Lang's book from an [url=http://math.berkeley.edu/~wu/fractions2.pdf]online article about teaching fraction arithmetic[/url] by Professor Hung-hsi Wu of UC Berkeley. There is a lot of very interesting mathematical thinking in Lang's book, and some challenging proof-based exercises, but nothing in it much like competition problems. The book covers strictly precalculus mathematics, but it covers topics related to undergraduate mathematics, and definitely covers topics unknown to most United States high school graduates. I like the book a lot, but my son and I haven't had enough time to work through all the chapters yet.

[quote="orl"][quote="fleeting_guest"]  Some IMOers deliberately choose to solve
more olympiad problems rather than learn advanced
material or take university courses --- in a sense most/all
choose this route as it is hard to do both without
spending your whole youth on mathematics alone.[/quote]

Have you been on IMO or involved in IMO training that you can evaluate how advanced the problems and theory is that they cover ? [/quote]

orl,  without shedding all privacy let me mention that
the answer to your questions is yes, absolutely.

Probably like you and many others on this board,
I know quite a few people who collectively earned
several dozen IMO medals or similar prizes, and it
is not unknown for us to compare notes.
It is simply a fact that many IMO folk choose to
spend their time more (or entirely) on competition
training rather than learning more mathematics in
high school.  It's not uncommon for
such people to have some light exposure to, say,
abstract algebra or complex analysis, but this knowledge
is shallow compared to what they know
at the competition level and even compared to what
might be required at the university courses.
Many have essentially no exposure to the "advanced"
material, they wait until university.

It's just a matter of opting for a definite known
reward (such as upgrading one's IMO caliber
from silver to gold) rather than a vaguer or nonexistent
reward for learning more mathematics earlier.  It is in
the nature of olympiads that many of the participants
are competitive people who are motivated by such
considerations. Learning university material far beyond
the level relevant to competitions is an activity in zero-sum competition with solving olympiad problems.

[quote]
Are that just assumptions, is that a general statement ? You say it may have changed recently ? Which time are you referring to ?  :)
And which olympiads do you like best at the national or international level ?  :)[/quote]

My information is more from the 80's and 90's. There are
some trends that happened, or I guess have been happening
since then, that I am no longer in a position to evaluate directly
by asking others (well, maybe this board will give some idea).
I think the general preparedness level of olympiad competitors
has increased.  On the one hand this means some explicit
introduction of some theory as part of the training; many
people know it anyway and it doesn't hurt to provide the
vocabulary.  (I have in mind basics of group theory, linear algebra, algebraic number theory, various parts of combinatorics).
On the other hand the availability of problem databases
on the Internet means that the formerly scarce resource,
olympiad training problems, is now inexhaustible, and
I am guessing that this leads to a sort of arms race where
even more problems need to be solved to reach the
top IMO level; this would leave less time for studying
advanced material.

[quote="orl"]

And which olympiads do you like best at the national or international level ?  :)[/quote]

To keep this on topic, let me mention that it was
assumed that any traditional Euclidean geometry
problem at the IMO could be solved routinely by any
contestant from Bulgaria, Romania, USSR, Vietnam,
China etc.   So the national olympiads from those countries,
if one could find the problems, were a good place to
judge one's level.  Now, maybe geometry has declined
in some of those places but the problems are easier
to get.

I like the simply stated natural-sounding problems,
the contests in Russia, Hungary and the UK especially
come to mind as examples of this.  Also problems that
are related to some higher mathematics beyond the
IMO level.

My mom says Korea's schools were insane back when she was a child o_o .

She says the teachers whipped you and tortured you if you got bad grades o_o .

I'm studying Calc BC. I feel special :P .

[quote="tokenadult"][quote="Valentin Vornicu"]If we're talking about usamo/national/international levels, then knowing calculus can turn out to be a serious advantage.[/quote]

Hi, Valentin, it's interesting that you should say that. In an [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10869]earlier thread on this forum[/url], I learned that although Olympiad contests are designed so that precalculus mathematics is sufficient to solve them, postcalculus mathematics can nonetheless be fair game. In your experience, did you find that knowledge of university-level mathematics helped you see "easy" solutions to Olympiad problems?[/quote]Yes.

But to give a different example, Mihai Manea (41p / 42p at the IMO 2001 - Washington USA) solved problems 2 from both days using (a lot of) calculus ( those problems were an inequality and a geometry problem!) and solved problem 6 using Gauss' Integers representations (far away from any high-school program, even the advanced one in Romania!). So basically, half of his gold medal that year was earned using higher-maths.

I also solved the geometry problem that year using some calculus :) and many others using advanced knowledge in the team selection tests during the years. It does help but only in certain situations, and knowing advanced stuff may not be an advantage if you don't know them well enough.

that tendence to not knowing much advanced material is still there. however, we learned at IMO training last year in oberwolfach about linear algebra in order to solve some combinatorics problems easily as well as about finite fields; but that's all not real theory but more the application how to use it in olympiads.

nevertheless, i'm a bit out of that principle since i still got 2 and a half year of school in front of me(germany got 13 grades :( - but therefore i can participate still 3 times at IMO) and already study e.g. algebraic geometry in the 5th semester. i just read the book "algebraic geometry" by robin hartshorne where topics like schemes and comohology are excellently covered - e.g. i'm just doing the exercise in that book to prove the analogue of the riemann hypothesis for curves. furthermore, i got the book "introduction to cyclotomic fields" by washington as a christmas present - that book deals with (p-adic) L-functions and similar things of algebraic number theory. these all are very beautiful theories but they will hardly help me in any olympiad and even if i could the theorems about them sure are not accepted as known theorems.

however, i dare say that in the time of IMO training(the first half of each year), i rarely got much time to work on advanced mathematics since there's still some school besides - i cannot completely forget about it.

In our IMO training for Greece, we observed for in the IMO 2003 Shortlist, the comments for N6, N7, N8 all implied that the problems are rather trivialized through some higher abstract algebra results. In fact, Jacob noted that the solution for N7 is basically field theory stated in elementary terms.

Also, in your opinion, which branch of university math do you think is more useful for Olympiads?

And how does one use calculus to solve a geometry problem ?!?!  :what?:

[quote="Peter Scholze"]i just read the book "algebraic geometry" by robin hartshorne where topics like schemes and comohology are excellently covered - e.g. i'm just doing the exercise in that book to prove the analogue of the riemann hypothesis for curves.[/quote]

Wow! Congratulations. Hartshorne's book on algebraic geometry is legendarily difficult. And you are reading it (and writing to us about it) in a second language too. I think you are getting an impressive education.

I love Hartshorne's book Geometry: Euclid and Beyond. I wonder how many years it will be before the young people I know can go "beyond" that book and study algebraic geometry.

thanks! :lol:

what else than reading that book could one do during the christmas and new year holidays :D

concerning your statement about the language: it is not very hard to read something mathematical in english - there are no very formal words used and the only hard terms are those of mathematical nature - but those you learn during reading. it even led so far that i know some mathematical terms only in english, not in german :D

btw do you think that book is that hard? i recognized that it wasn't easy, but if one really keeps on trying to understand, it's ok. all the mathematical books i really like are of that level...  :P

furthermore, you considered my education very impressive. concerning my mathematical education, that may actually be correct. however, all germans learn english in school - i may be one of the better ones out of my class, but i don't think that this shows that i'm very educated. and i don't want to talk about non-scientific subjects - i used to have mark 1 or 2 everywhere(mark 1 = A, mark 2 = B, but some of my class went a year to the USA and got mark A everywhere, though they had even 3's in germany), but since i only bother about mathematics and don't learn anything for school and do nearly no homeworks at the moment, my achievements get worse.

[quote="billzhao"]In our IMO training for Greece, we observed for in the IMO 2003 Shortlist, the comments for N6, N7, N8 all implied that the problems are rather trivialized through some higher abstract algebra results. In fact, Jacob noted that the solution for N7 is basically field theory stated in elementary terms.[/quote]

yes, i hate those NT problems from the ISL 2003. they were either ugly or depended too much on theory. and yes, i think number theory and combinatorics are the fields where applying advanced mathematics (but not necessarily calculus) is most helpful.

[quote="billzhao"]And how does one use calculus to solve a geometry problem ?!?!  :what?:[/quote]

probably it was about proving that the function $f\left(x\right)=\frac{\sin\left(u-x\right)}{\sin x}$ is injective for $x\in\left[0;\pi\right[$... that's actually trivial using the addition formulas and also easy to prove using geometric considerations (a post of mine about this was lost during the hacker attack) but many people try to prove it using calculus (sorry peter but i'd say that it is one of the most idiot ways to misuse calculus, although i have seen at least two books where it is done...). anyway, i don't really know any other "applications" of calculus to olympiad geometry problems, except of some combinatorial geometry ones.

ah yes, about school marks, some time ago i did aim at getting as many 1's as possible, but now i'm so indifferent towards this all that e. g. i have a 5 in chemistry, a 5 in biology and my physics marks is also falling down the mountain...

  darij

[quote="darij grinberg"](sorry peter but i'd say that it is one of the most idiot ways to misuse calculus, although i have seen at least two books where it is done...).[/quote]

i know, i know...but i was during an IMO paper! it was just a little lemma i still had to prove and thought for it a few seconds and gave the first simple proof that came to my mind! of course, it is idiot, but i got 7 points!

yes, i actually just wanted to note it as a tip to everyone for further competitions... you probably also remember that i used half a page of epsilontics with lots of estimates to prove that

{for all x < a we have x < b} implies {$a\leq b$}

while this is completely trivial...

  darij

Regarding school marks I think it depends what counts more for top universities: top grades, first class competition prizes or a mix of it.

And I agree with Darij. The former Vietnam IMO member Lhvietbao for instance already studied the combinatorics book [url=http://www.amazon.com/exec/obidos/tg/detail/-/0387984887/qid=1104616657/sr=1-1/ref=sr_1_1/103-1541914-5567015?v=glance&s=books]Modern Graph Theory by Bela Bollobas[/url] when he still went to school. And he also actively read AMM and probably still does. He also solved lots of problems of the Hungarian math Miklos Schweitzer competition. :)

tokenadult,

most Euclidean geometry problems can be solved algebraically
by "turning the crank" using vectors, or trigonometry, or coordinates.

Similarly, most any IMO-level inequality can be proved using
calculus, particularly Lagrange multipliers. You would have to
be careful about non-smooth boundaries (e.g. the variables
in the inequality lying in a triangle or a cube or similar domain
with edges and corners), but in principle if you are careful enough
you can just turn the crank.

To show that some function is increasing in one of its variables
you can use calculus.

The advantage is that you know in advance that pure
steady calculation will deliver an answer, and can proceed to carry
out the calculation rather than thinking long about a cleverer method.
The disadvantage is that sometimes these brute force methods
take too long, or require too much care in the calculations, while
"conceptual" solutions may be more robust.

Some of the recent IMO proposals were linear algebra over
finite fields (just integers modulo 2 as I recall).

Also, it definitely helps to have a more comprehensive or advanced
outlook even if just a way of knowing which directions are NOT likely
to produce a solution.


Ramsey Theory
combinatorics
graph theory
IMO
IMO 1992
Coloring
Extremal combinatorics

Consider $9$ points in space, no four of which are coplanar. Each pair of points is joined by an edge (that is, a line segment) and each edge is either colored blue or red or left uncolored. Find the smallest value of  $\,n\,$ such that whenever exactly $\,n\,$ edges are colored, the set of colored edges necessarily contains a triangle all of whose edges have the same color.

Answer is $ 33$.
Suppose we can color $ 33$ edges and have no monochromatic triangle. Then look at any vertex, $ X$. If $ X$ has $ 5$ blue edges coming from it, then, calling them $ A_{1} ... A_{5}$ - if any edge $ A_{i}A_{j}$ is colored, then it must be colored red, since otherwise $ XA_{i}A_{j}$ is a blue triangle. But within $ 5$ points, if more than $ \frac {5^2}{4}$ are colored red, then we will have a red triangle (since if we have more than $ \frac {n^2}{4}$ edges in a graph with $ n$ edges, there must be a triangle). That means at most $ 6$ of the $ 10$ vertices $ A_{i}A_{j}$ are colored, so there are at least $ 4$ uncolored edges - contradiction, since we have $ 33$ edges.
Thus $ X$ has at most $ 4$ blue edges and at most $ 4$ red edges. Suppose it has $ 8$ edges emanating from it - then it has exactly $ 4$ blue edges, connecting to points $ B_{1} ... B_{4}$, and exactly $ 4$ red edges, connecting to points $ R_{1} ... R_{4}$. To prevent triangles, all edges $ B_{i}B_{j}$, if colored, must be red. In $ 4$ points, if more than $ \frac {4^2}{4} \equal{} 4$ of them are colored, there will be a red triangle (since if we have more than $ \frac {n^2}{4}$ edges in a graph with $ n$ edges, there must be a triangle). Thus there are least $ 2$ uncolored edges amongst $ B_{i}$, and similarly, there are at least $ 2$ uncolored edges amongst $ R_{i}$, so there are at least $ 4$ uncolored edges, a contradiction again.
Thus no vertex can have degree $ 8$, so each vertex has degree at most $ 7$. But then, there are at most $ \frac {9 * 7}{2}$ edges, so there are at most $ 31$ edges, which is a final contradiction.

To prove it is possible to have $ 32$ edges without a triangle: Consider one vertex $ V$, suppose it has $ 4$ blue edges to $ B_{1}, B_{2}, B_{3}, B_{4}$ ; and $ 4$ red edges to $ R_{1}, R_{2}, R_{3}, R_{4}$. Color edges $ B_{i}B_{i\plus{}1}$ red, and $ R_{i}R_{i\plus{}1}$ blue, for $ i\equal{}1,2,3,4$ (and $ B_{5} \equal{} B_{1}, R_{5}\equal{}R_{1}$). Color $ B_{i}R_{j}$ blue iff $ i\plus{}j$ is odd. Now we have graph with $ 32$ edges, which we easily verify contains no monochromatic triangles.

A simpler way to establish that 33 is the answer:

Same example for 32.

Suppose that we can color 33 edges with no monochromatic triangles. Now, we are done if we prove that there are 6 points such that all the possible segments connecting them are drawn (because R(3,3)=6).

Consider the complementary of the original graph, that is, the graph with only 3 lines. It's obvious that each of these lines only does the job of disallowing 2 vertices connected to be in those 6 points mentioned above together. That means that 3 lines can ''put away'' at most 3 points, but 9-3=6. It's now trivial.

[hide="Sorry to revive, but look at this"] It is clear that the minimum will occur when the graph is 6-free. By Turan's theorem, this occurs only when there are 32 vertices, else, the graph will contain $K_6$. Same example for 32. [/hide]

Precisely; that is what Bugi has done - prove Turán for this particular case ...


And I mean really, really minor:

In the AoPS skin, when I view a multiple-page thread, the buttons to move to previous pages are above the buttons to go to previous topics -- that is, they are further away from the actual posts.  When there is a poll, in fact, the option to change pages appears above the poll, far away from the text, while the option to switch topics is between the poll and the posts.  This seems very counter-intuitive to me, and I have gone to the next topic instead of the next page countless times as a result.  My suggestion is that you switch the locations of the two links.  I understand that this is not crucial, but if you ever do run out of things to change on the site ... :)

[quote="JBL"] I understand that this is not crucial, but if [u]you ever do run out of things to change on the site[/u] ... :)[/quote]This sentence only should be in the jokes section :D

As a matter of fact this current system is much better. Why do I say that? Because it is natural that when you start reading a topic you go from up to bottom right? Then you shouldn't go up again to go to next page but use the pagination below (which is conveniantely placed so that you can immediately read the next page).

However if you have already read most of the topic, and you ended up there from a search for example, you know you want to go the last page when the page loads, so you need the pagination to be as high as possible.

This is just like not using the lower nagivation bar to move to the upper forum after reading a topic which I also used to do a lot untill I got used with the much more simpler way (before I used to scroll all the way up to the upper navigation bar :D).


when they have a problem where they have a square/circle with a inscribed and circumscribed circle/square, the ratio between them is always one half.  can you prove this?  It is actually quite simple

what about a square with a circle circumscribed and another circle inscribed??

we encountered this problem in a mathcounts sprint and i was the only one who knew the formula.  Everyone else did it the normal way, because the side of the large circle was 4.


invariant
geometry
geometric transformation
rotation
complementary counting

In how many ways can 7 people sit around a round table if no two of the 3 people Pierre, Rosa, and Thomas can sit next to each other?

We can use complementary counting and PIE to get [hide]$$ \frac {7!}7 \minus{} \left(3\cdot\frac {2!6!}6 \minus{} \frac {3!5!}{5}\right) \equal{} \boxed{144}.$$[/hide]

Hm since apparently I found a different solution I will revive the thread.  Sorry :wink:
[geogebra]ddcd1242399b55d7fc20cca86467bfe268dc95cd[/geogebra]

Number the seats P, A, B, C, D, E, F in that order.

Since round tables are invariant around rotation, assume WLOG that P is the location where Pierre is seated.   By our givens, neither of A nor F can be Rosa or Thomas, so we have $4(3) = 12$ ways to choose our locations for A and F.

The remaining seats B-E can be attacked with some brute force to find $12$ ways to seat the remaining individuals.  Therefore, our answer is $12(12) = \boxed{144}$.

[quote="math154"]We ca n use complementary counting and PIE to get $ \frac {7!}7 \minus{} \left(3\cdot\frac {2!6!}6 \minus{} \frac {3!5!}{5}\right) \equal{} \boxed{144}$.[/quote]
@math154, I'm not quite sure how you arrived to counting $\left(3\cdot\frac {2!6!}6 \minus{} \frac {3!5!}{5}\right)$ for the scenarios which do not work.  May you please explain?   :wink:

I think I have a "Think About It" method. You know that the only way for Rosa, Pierre, and Thomas is to sit with one seat in between two of them, one seat between another two, and two seats between the last two. So I had 3!x4!=144.

Yay, finally got Eek! Gross! from this problem...

If anything else on this problem... you can draw a circle and perform casework. Just let Pierre pick first...(he only has 1 choice) because everybody will sit in relation to him. Next, you perform the casework to see where Thomas and Rosa can sit => 6 ways.

Then there are $4*3*2*1$ ways for the rest of the people to sit down..

So then you get $6*4*3*2$ = $12*12$ = $144$...just another method for solving.

[quote=mamis511]In this problem why do you not have to take into account rotations and reflections?[/quote]

Yes, if you're going to do it casework style then you have to account rotations and reflections.

[hide=My Solution (Smart, no casework)]
We can place a "buffer" to the right of Pierre, Rosa, and Thomas consisting of one person. There are $4\cdot 3 \cdot 2=24$ ways to do this. This guarantees that the 3 will never sit together, and each choice of buffers generates a unique seating configuration.
Now we have 4 "groups" we must seat on a round table with 4 seats: 3 of the buffers and the 1 person who remains. There are $3!=6$ ways to do this. Thus our final answer is $6\cdot 24 =\boxed{144}$

[quote=Mathpluspianoequalslife]i hav question, what's so important about 144 that it gets an achievement?
same with 42[/quote]

144 is a "gross" and 42 is the answer to life, the universe, and everything

I did complementary counting - I counted the amount of ways for any two to sit next to each other which also counts the case where all three sit next to each other. Then there are $720$ ways total (assuming reflections don't count, as the problem only specifies rotations) and $3$ ways to pick two people, so $3 \cdot 2 \cdot \frac{4!}{4} = 36$ undesirable ways. Subtracting gives a result that's way off. Why is this?


Sorry if i'm posting in the wrong section, please advise if i am.
I needed some help with the following question, if someone could help me out:

Prove (1 + x)^n >= 1 + nx for all real x > -1 and all natural numbers n.

Thanks.

The "High School --> Intermediate Section" might be a good place to post this problem. Kindly use Latex to post your problems; else they are almost unreadable.

This is the Bernoulli Inequality; a proof can be found at [url=http://mathworld.wolfram.com/BernoulliInequality.html]MathWorld[/url].

Put $y=x+1$, so we must prove $y^{n}-1\geqslant n(y-1)$ for $y>0$.

[b]Case 1.[/b] $y>1$. Start from $y^{n-1}+y^{n-2}+\dots+y+1\geqslant n$, which obviously holds since all the terms are not less than $1$ and there are $n$ of them. Multiply with $y-1$, which is positive, hence the direction of the inequality remains the same. That gives $y^{n}-1\geqslant n(y-1)$.

[b]Case 2.[/b] $y=1$. Then $0\geqslant 0$.

[b]Case 3.[/b] $y<1$. Start from $y^{n-1}+y^{n-2}+\dots+y+1\leqslant n$, which obviously holds since all the terms are not greater than $1$ and there are $n$ of them. Multiply with $y-1$, which is negative, hence the direction of the inequality reverses. That gives $y^{n}-1\geqslant n(y-1)$.

I'll re-word the question so its a little clearer:

Prove ((1 + x)^n)  >= 1 + nx for all real x > -1 and all natural numbers n.

Or alternatively, you can read it at:
http://members.optusnet.com.au/gujju_style/maths.JPG

Farenhajt, are you sure you havent made a mistake which that??

Case 1 you have "Multiply with y-1, which is positive,"
Then Case 3 you have "Multiply with y-1, which is negative"

[quote="gollum000"]
Farenhajt, are you sure you havent made a mistake which that??

Case 1 you have "Multiply with y-1, which is positive,"
Then Case 3 you have "Multiply with y-1, which is negative"[/quote]
He did those things but also took care of the change in inequality sign.

I gave a proof of this [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=inequality&t=105449]here[/url].

Yes, I believe so.  Nice proof! :)

Masoud Zargar


trigonometry
calculus
calculus computations

Let the sequence $ a_{n}$ be defined according to $ a_{2}> 0$ and $ a_{n+1}= a_{n}\sin\left(\frac{\pi}{n}\right)$ for $ n \ge 2$. Ascertain the nature of the series

$ S = \sum_{n=2}^{+\infty}\arctan a_{n}$.

Due to $ \sin x \le x$ and $ \arctan x\le x$ for all positive reell numbers the series converges.

$ a_{n+1}=a_{2}\prod_{k=2}^{n}\sin \frac{\pi}{k}\le a_{2}\frac{\pi^{n-1}}{n!}$ for $ n=1,2,3,... \qquad S=\sum_{n=2}^\infty \arctan a_{n}\le a_{2}\sum_{n=2}^\infty \frac{\pi^{n-1}}{n!}<\infty$


When I need to do my homework in math I needed to create a 3D graph,
I had come along this simple and easy web site which allows to create 3D graphs in seconds,
Just fill in the values and the numbers and click "Generate" button

The web site is:

[url]http://chartgen.blogspot.com[/url]


calculus
derivative
function
integration
real analysis
real analysis unsolved

Let $f$ be a real valued function defined for all $x \geq 1$, satisfying $f(1) = 1$ and
\[ f'(x) =\frac{1}{x^2 + f^2(x)}. \]

Prove that $f(x)$ has a limit as $x \rightarrow \infty$, and that this limit is $\leq 1 + \frac{\pi}{4}$.

Since f' is positive, f is increasing. So, $f(x) > f(1)$ for  x > 1.
We have
$f'(x) \leq \frac{1}{x^2 + 1}$ for all x >1
Then,
$f(x) - f(1) \leq \int_{1}^{x} \frac{1}{t^2+1} dt < \int_{1}^{\infty} \frac{1}{t^2+1} dt = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$
This implies $f(x) \leq \frac{\pi}{4} +1$ for all x


what is the equation of a line parallel to the line 2x-3y=7 and contains the points (-6,-5)?

I'm not sure it is advanced level.
I move it to getting started forum.

Pierre.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose we have a non-commutative group $ G$ with $ |G|\equal{}2^nm$ with $ m$ odd and there exists an element $ h \in G$ of order $ 2^n$. Is the subgroup generated by $ H$  the only subgroup of $ G$ having $ 2^n$ elements?

I tried to prove this by choosing another subgroup of $ G$, $ J$ having the same order as $ <h>$. Then pick an element $ j \in J$, which has order a power of 2. Then I looked at the subgroup generated by $ j$ and $ h$. Can I conclude that this subgroup has order a power of 2? (the group is not commutative)

[quote="benimath"]Suppose we have a non-commutative group $ G$ with $ |G| \equal{} 2^nm$ with $ m$ odd and there exists an element $ h \in G$ of order $ 2^n$. Is the subgroup generated by $ H$  the only subgroup of $ G$ having $ 2^n$ elements?

I tried to prove this by choosing another subgroup of $ G$, $ J$ having the same order as $ < h >$. Then pick an element $ j \in J$, which has order a power of 2. Then I looked at the subgroup generated by $ j$ and $ h$. Can I conclude that this subgroup has order a power of 2? (the group is not commutative)[/quote]

H is almost never the only subgroup having 2^n elements.  In fancy language "G is 2-nilpotent, so the Sylow 2-subgroup H is normal iff G is nilpotent".

A concrete example might help: take n=1, m=3, |G|=6.  In other words, G is non-abelian of order 6.  H is any subgroup of order 2.  However there are three different subgroups of order 2.  If G is the symmetric group on 3 elements, { (), (12), (23), (13), (123), (132) }, then H = { (), (12) } is one choice, but there is also J = { (), (23) } and K = { (), (13) }.


It 's not very difficult.

The inequality is equivalent to

$ \sum_{i\equal{}1}^{n} {1\minus{} \frac{x_i^2}{x_i^2\plus{}x_{i\plus{}1}x_{i\plus{}2}}} \geq 1$

or $ \sum_{i\equal{}1}^{n} {\frac{x_{i\plus{}1}x_{i\plus{}2}}{x_i^2\plus{}x_{i\plus{}1}x_{i\plus{}2}}} \geq 1$

We can show that there exists i, j such that

$ x_{i\plus{}1}x_{i\plus{}2} \geq x_i^2$ and $ x_{j\plus{}1}x_{j\plus{}2} \geq x_j^2$

by using proof by contradiction technique.

Thus, the problem is solved.

QED

[quote="mathangel"]We can show that there exists i, j such that

$ x_{i \plus{} 1}x_{i \plus{} 2} \geq x_i^2$ and $ x_{j \plus{} 1}x_{j \plus{} 2} \geq x_j^2$

by using proof by contradiction technique.

Thus, the problem is solved.

QED[/quote]
Could you write more in detail?

My solution:

Denote $ y_i\equal{}\frac{x_{i}^2}{x_{i\plus{}1}x_{i\plus{}2}}$
with $ x_{n\plus{}i}\equal{}x_i$

One may rewrite the inequality as:
$ \sum \frac{1}{1\plus{}y_{i}^2}\geq 1$
with: $ \prod y_i\equal{}1$

wich is is easy:
since: $ \prod y_i\equal{}1$ we have
There exists $ i,j$ such that $ y_{i}y_{j}\leq 1$
This give: $ \frac{1}{1\plus{}y_i}\plus{}\frac{1}{1\plus{}y_i}\geq  1$
and we're done  :D

[quote="FOURRIER"]Is my proof ok?[/quote]
It's ok, but it's mathangel's proof also. :wink:


AMC USA(J)MO USAMO induction AIME quadratics algorithm

Everyday, I'm going to post up one math problem that can be either short-answer or proof. I don't really have time to do whole lot of problems everyday so I thought this will be a good idea for people like me and obviously, anyone else who wants to help out.

I'll always post problems and my solutions. In some cases, my short-answer or proofs may be right. In another case, they may be wrong. Either way, post your solutions so I can see other ways or at least know I"m doing right or not. Also, if you can come up with slick solution, that would be better!

Problem 1
Date: 6/18/06
Level: Easy USAMO
Source: CMO 1973 #5
Problem:

For every positive integer $n,$ let

\[ h(n) = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}  \]

For example, $h(1) = 1, h(2) = 1 + \frac{1}{2}, h(3) = 1 + \frac{1}{2} + \frac{1}{3}.$ Prove that for $n = 2,3,4...$

\[ n + h(1) + h(2)+h(3)+ \cdots +h(n-1) = n h(n)  \]

[hide="My proof"]

Let's use induction here.

The base case for when $n = 2$

$2 + h(1) = 2 + 1 = 3$ which is the left side
$2 h(2) = 2\left(1 + \frac{1}{2} \right) = 3$
So this is complete $\boxed{}$.

Assume that the equation holds true for arbitary $k$ such that $k > 2$. This lets us to say that

$k + h(1) + h(2) + \cdots + h(k-1) = k h(k)$
This "assumed to be true" equation is marked $(\star)$

Then all is left to show that it works for $k+1$. When $n = k+1$ (I'll use equal signs in here but until at the end, this equal "sign" doesn't mean LHS and RHS are equal):

$k +1 + h(1) + h(2) + \cdots + h(k) = (k+1) h(k+1)$

$(k + h(1) + h(2) + \cdots + h(k-1)) + 1 + h(k) = (k+1) h(k+1)$

Using $(\star)$:

$k h(k) + 1 + h(k) = (k+1) h(k+1)$

Expanding right side and $k h(k)$ of left side by the given problem's equation, we get:

$k + h(1) + h(2) + \cdots + h(k-1) + 1 + h(k) = k+1 + h(1) + h(2) + \cdots + h(k)$

But these two are exactly same so our assumption for $n = k+1$ is also true and our proof is complete. $\blacksquare$ [/hide]

Induction's been used, so...
[hide="my solution"]
We want to show that
$n+h(1)+h(2)+...+h(n-1)=nh(n)$
Let's directly transform the LHS.
There are going to be $n-1$ 1's, $n-2$ 2's, $n-3$ 3's, and so on because 1 is going to be added once for each of h(1) to h(n-1), 2 added once for each of h(2) to h(n-1), etc.  The LHS will become
$n+(n-1)(1)+(n-2)\frac{1}{2}+...+(n-(n-1))\frac{1}{n-1}=\\ n+\frac{n}{1}+\frac{n}{2}+...+\frac{n}{n-1} -(\frac{1}{1}+\frac{2}{2}+...+\frac{n-1}{n-1})= n+nh(n-1)-(n-1)=nh(n-1)+1=nh(n-1)+\frac{n}{n}=n(h(n-1)+\frac{1}{n})=nh(n)\Box$
[/hide]

[hide]
Here's a non-induction solution:

$h(1) = 1$
$h(2) = 1+\frac{1}{2}$
$h(3) = 1+\frac{1}{2}+\frac{1}{3}$
...
$h(n-1) = 1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}$

Adding up column-wise, we have $n-k$ of each $\frac{1}{k}$ term to get the LHS equal to

$n+ \sum_{k=1}^{n-1} \frac{n-k}{k} = n+ \sum_{k=1}^{n-1} \left(\frac{n}{k}-1\right) = \sum_{k=1}^{n-1} \frac{n}{k} + 1 = \sum_{k=1}^n \frac{n}{k} = nh(n)$.[/hide]

EDIT: Oops, quantum leap beat me to it.

Problem 2
Date: 6/19/06
Level: Easy AIME
Source: BMO 1994 Round 2 #1
Problem:

Find the first integer $n > 1$ such that the average of

\[ 1^2,2^2,3^2,....n^2 \]

is itself a perfect square.

[hide="My solution"]

Solution 1:

Recall the useful formulas:

$1^2+2^2+ \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$

$1 + 2 + \cdots + n = \frac{n(n+1)}{2}$

Average is then (after simplified) $\frac{2n+1}{3}$ and the problem stated that this will equal to a perfect square $k^2$ for some integer $k$.

Clearly, since we want to find a positive integer $n$ after dividing by 2, it is clear that $k^2$ must be odd (after multiplying by 3 (odd -> odd, even -> even) and subtracting 1 (odd -> even, even -> odd). Since $n > 1,$ we try an odd perfect square greater than 1, which is 9, and this yields answer of 13.

Solution 2:

This is similar of first solution except that it doesn't require immediate try of the numbers. In this case, it was nice that we're dividing by 2 so parity simplified the answer. But if the problem was more tricky, say division by 37, then we might need more clever way to do this. Anyway, here is other way.

First, cross-multiply 3 to get the equation without fraction:

$2n + 1 = 3k^2$

Seeing how we have $2n$ on the LHS, let's separate $3k^2$ as $2k^2$ and $k^2$:

$2n + 1 = 2k^2 + k^2$

$2n - 2k^2 = k^2 - 1$

$2(n - k^2) = (k-1)(k+1)$ (I just factored in case it will come useful)

$n - k^2 = \frac{(k-1)(k+1)}{2}$

$n = \frac{(k-1)(k+1)}{2} + k^2$

From here, it's still a case work but it's a lot easier than dealing with squares like we had to do in the first solution. [/hide]

I don't know what I was thinking to use $1+2+...+n$ instead of just $n$.

Anyway, I got one Q:

[hide="Question"]

I don't get this part:

[quote]Then do modulo by 6 (the quadratic residues of 6 are 0, 1, 3, and 4) you find that $n \equiv 1,5 \mod 6$[/quote]

So, do you take perfect squares and take them in base 6? I'm not familiar with quadratic residues. [/hide]

Exactly.  Since any square of a number is congruent to the square of a number 0<n<6, there are only certain numbers that can be perfect squares modulo 6 (or any mod in general).  $1^2 \equiv 1, 2^2 \equiv 4, 3^2 \equiv 3, 4^2 \equiv 4, 5^2 \equiv 1, 6^2 \equiv 0$ so the only quadratic residues (perfect squares in a mod) are 0, 1, 3, and 4. (see aops v. 2, chapter on number theory)

Oh okay.

So I get where 0,1,3, and 4 came from. I remember reading that there are only finite numbers from AoPS book anyway.

I think there was even formula that calculated how many but I do not remember.

Anyway, how do you go from there to $n \equiv 1,5 \mod 6$?

We get $n\equiv1,5\mod6$ because $(n+1)(2n+1)$ must be divisible by 6. So we check $n=\{0,1,2,3,4,5\}\mod6$ and only $n\equiv1,5\mod6$ work.

[quote="paladin8"][quote="pkerichang"][quote="paladin8"][hide]
Hmm, I'm not sure why you divided by $\frac{n(n+1)}{2}$ to find the average, instead of just $n$... I think the average of $1^2, 2^2, \ldots, n^2$ should be

$\frac{(n+1)(2n+1)}{6}$.

From some computations, I can't make it a square... I might be missing something though.
[/hide][/quote]

I agree 100%:
[hide="my progress"]
you get the number to be $\frac{(n+1)(2n+1)}{6}$, which is a square.  Then do modulo by 6 (the quadratic residues of 6 are 0, 1, 3, and 4) you find that $n \equiv 1, 5 (\bmod\ 6)$.  Plugging $n = 6k+1, 6k+5$ back we find that the expression is equivalent to $(3k+1)(4k+1)$ or $(k+1)(12k+11)$, depending on $n$ mod 6.  doing eclidean algorithm, the factors are all relatively prime, therefore each factor is a perfect square.

For the second one, plug in $k+1 = q^2$, we have the expression to be $q^2(12q^2-1)$, so we have $12q^2 = k^2+1$, mod both sides by 4 we have $0 \equiv (1, 0) + 1$, which is an impossibility.

Therefore, we must have $(3k+1)(4k+1)$ to be a perfect square, with each factor be a perfect square.  Then someone continue...
[/hide][/quote]

Well, through rather brute force methods, I found that $k = 56$ is the smallest value for which this is true. This would give us an average of $(3 \cdot 56+1)(4 \cdot 56+1) = 169 \cdot 225 = 13^2 \cdot 15^2 = 195^2$ and $n = 337$.[/quote]

I know you brute force this, but can you just tell me how?? Through a program?? or you found some systematic way to calculate faster??

[hide="one thought of shortening the brute force"]
we know that $4k+1$ is greater than $3k+1$, and each consecutive square differ by an odd number, so we have $(4k+1) - (3k+1) \ge 2\sqrt{3k+1} + 1$, and solving this using quadratic I have $k \ge 15$.  Then I don't know. Start plugging them in one by one??
[/hide]

I was pretty much like look for a $k$ such that $3k+1$ is a square and work from there... you just skip over multiples of $3$. So I just tested like $49, 64, 100, 121, 169$ and yeah, got it then. There's probably a more effective way of narrowing down possibilities, but oh well.

Continueing on with Shyong's idea, clearly, the Pell equation has minimum solution $(m,k)=(7,1)$.  The second solution does not give positive integer solutions.  We proceed to the third solution.  By cubing $7+\sqrt{48}$, we have $1351+195\sqrt{48}$.  Using $m=1351$ in the quadratic equation, we obtain $\frac{-3\pm 1351}{4}=337$.  We need not consider the negative case.

Masoud Zargar

Yeah, that's what I got (except the first time, when I forgot to raise 2 to the nth power :( )

Matt, I agree. I had to post something but I didn't know what. Maybe you can work it out and give me a solution here.

I've been busy and I still am. So, I'm just going to post problem if I can and when I get time, I'll post solution to all the unanswered problems.

Problem 12
Date: 7/1/06
Level: AIME
Source: ARML 1989
Problem:

A convex hexagon is inscribed in a circle. If its successive sides are 2,2,7,7,11, and 11, compute the diameter of the circumscribed circle.

I don't really want to post the solution (since we gone over this one during ARML), so:
[hide="hint"]
in a circle, you can change the order of the sides of a polygon
[/hide]
[hide="more hints"]
Ptolemy and Thale's
[/hide]

Rearrange sides in order $2-7-11-11-7-2$. By symmetry, we will get a cyclic quad with sides $2-7-11-d$ where $d$ is diameter. Let the quad be $ABCD$. $BD=\sqrt{d^{2}-2^{2}}.AD=\sqrt{d^{2}-11^{2}}$. Use ptolemy's theorem to get:
$\sqrt{d^{2}-2^{2}}\times \sqrt{d^{2}-11^{2}}=7d+22$.
Solving this equation, get $d=14$.

Since no one posted I guess I will...

Problem 13
Date: 7/4/06
Level: hard AIME - USAMO (I guess because of the source below)
Source: Romanian? (or Russian) Olympiad
Problem:

find all integral solutions in to:

$x^{6}+3x^{3}+1 = y^{4}$

Replace $x^{3}$ with $z$. The equation becomes:
$z^{2}+3z+(1-y^{4})=0$
Use the quadratic formula to calculate $z$:
$z = \frac{-3(+-)\sqrt{4y^{4}+5}}{2}$
Because $x$ is an integer, $z$ is an integer, so $4y^{4}+5$ is a perfect square, let it be $q^{2}$.
Replace $4y^{4}$ with $m^{2}$, where $m=2y^{2}$. Have:
$m^{2}+5=q^{2}$. The only perfect squares whose difference is $5$ are $9$ and $4$, therefore $m^{2}=4$.
$4y^{4}=4$
$y^{4}=1$
$y=1$ or $y=-1$.
Substitute the y-value into the quadratic equation we got to get:
$z=-3$ or $z=0$.
$x^{3}=-3$ or $x^{3}=0$.
If $x^{3}=-3$ then $x$ is not an integer, so $x^{3}=0$ so $x=0$.
The solutions are $(0,1)$ and $(0,-1)$.

Because I'm really busy this summer, I'm just going to post very large amount of problems at once and post the all the solutions later in a single PDF document. You can help me out by answering problems. Just label the problem #. And please use HIDE!!!

Problem 14
Date: 7/5/06
Level: AIME
Source: TJHSST
Problem:

Solve for all real $x$: \[\sqrt[3]{38-x}+\sqrt[3]{38+x}= 4\] Problem 15
Date: 7/6/06
Level: AIME
Source: TJHSST
Problem:

Compute the exact numerical value of $\cos (\frac{\pi}{9}) \cos (\frac{3 \pi}{9}) \cos (\frac{5 \pi}{9}) \cos (\frac{7 \pi}{9})$

Problem 16
Date: 7/7/06
Level: AIME
Source: ARML 1994
Problem:

Regular decagon $P_{1},P_{2},...P_{10}$ is drawn in the coordinate plane, with $P_{1}$ at $(1,0)$ and $P_{6}$ at $(3,0)$. If $P_{n}$ is the point $(x_{n},y_{n}),$ compute the numerical value of the product

$(x_{1}+y_{1}i)(x_{2}+y_{2}i)(x_{3}+y_{3}i) \cdots (x_{10}+y_{10}i)$

Problem 17
Date: 7/8/06
Level: AIME
Source: ARML 1993
Problem:

Compute the number of real values of $x$ such that:

$x^{100}-4^{x}\cdot x^{98}-x^{2}+4^{x}= 0$

Problem 18
Date: 7/9/06
Level: AIME
Source: ARML 1991
Problem:

One angle of a triangle is twice another, and the sides opposite these angles have lengths 15 and 9. Compute the length of the third side of the triangle.

Followings are two USAMO level problems and also the last two problems.

Problem 19
Source: IMO 1998/5

Let $I$ be the incenter of triangle $ABC$. Let the incircle of $ABC$ touch the sides $BC,CA,$ and $AB$ at $K,L,$ and $M$, respectively. The line through $B$ parallel to $MK$ meets the line $LM$ and $LK$ at $R$ and $S$, respectively. Prove that angle $RIS$ is acute.

Problem 20
Source: IMO 1983/6

Let $a,b,c$ be the lengths of the sides of a triangle. Prove that

$a^{2}b( a-b)+b^{2}c ( b-c)+c^{2}a (c-a) \geq 0$

and determine when the equality occurs.

Good luck!

I think this is problem 2 - sorry if I have posted it in the wrong place!
; We have (n+1)(2n+1)/6 = $k^2$ for some k
Hence $ 2n^2 + 3n + (1- 6k^2) = 0 $
Its discriminant must be a square, hence $9 - 8 + 48k^2$ is a square, so
$1 + 48k^2 = m^2$
So we have the Pell equation $m^2 - 3t^2 = 1$ where t = 4k
solve this, we get (1351, 780) is the smallest solution that gives us an answer that works.  Hence $k = 780/4 = 195$ and substituting in the quadratic equation we get n = 337.

If you use the solution of (97,14) to the pell equation to solve for n, you get that n= 23.5 not 23, therefore it cannot be the answer.  However, the next biger solution of the pell equation gives a good integer value of n.

Yeah sorry my bad. I'll edit my solution now.


If $ x^2\plus{}2x\minus{}3$ is a factor of $ x^4\plus{}6x^3\plus{}2ax^2\plus{}6x\minus{}3a$, evaluate $ a\plus{}b$.

I need some practice problems on remainder theorem, factor theorem, and synthetic division.

Oh, that was a typo. Below is the correct question.

If $ x^2 \plus{} 2x \minus{} 3$ is a factor of $ x^4 \plus{} 6x^3 \plus{} 2ax^2 \plus{} bx \minus{} 3a$, evaluate $ a,b$.

Another question:

Solve for $ 0<\theta<180^{\circ}$: $ 8\tan^2 \theta \minus{} 2 \tan^2 \theta\minus{}5\equal{}\cot \theta$.

[quote="ThetaPi"]Oh, that was a typo. Below is the correct question.

If $ x^2 \plus{} 2x \minus{} 3$ is a factor of $ x^4 \plus{} 6x^3 \plus{} 2ax^2 \plus{} bx \minus{} 3a$, evaluate $ a,b$.　[/quote]

Let $ P(x)\equal{}x^4\plus{}6x^3\plus{}2ax^2\plus{}bx\minus{}3a$, remark that $ x^2\plus{}2x\minus{}3\equal{}(x\minus{}1)(x\plus{}3)$, now you can use factor theorem.

[quote="ThetaPi"]

Another question:

Solve for $ 0 < \theta < 180^{\circ}$: $ 8\tan^2 \theta \minus{} 2 \tan^2 \theta \minus{} 5 \equal{} \cot \theta$.[/quote]

Let $ \tan \theta \equal{}x$, you can rewite the equation as some cubic equation and you would use factor theorem.

Oh ya! I was thinking too much about synthetic division that I forgot about the factor theorem..

[quote="ThetaPi"]

Solve for $ 0 < \theta < 180^{\circ}$: $ 8\tan^2 \theta \minus{} 2 \tan^2 \theta \minus{} 5 \equal{} \cot \theta$.[/quote]

Shouldn't it be $ 0 < \theta < 180^{\circ}$: $ 8\tan^2 \theta \minus{} 2 \tan  \theta \minus{} 5 \equal{} \cot \theta$?


Let $P(x) = x^{6}+x^{3}+1$ and $Q(x)=x^{2}+x+1.$ If $x_{i}$ are the roots of $P(x)$ then compute $\sum Q(x_{i}).$

let:

$\displaystyle a = \sum_{i=1}^6 x_i$

$\displaystyle b = \sum_{1 \le i < j \le 6} x_ix_j$

then the sum in question is just $a^2-2b+a+6$ but by viet's,
$a = b = 0$

so the sum is just 6.


A man leaves his house in the morning at the same time every day. He rides his bicycle to work along the train tracks with a speed of 12 km/h, so he arrives to a railroad crossing every day at the same time as the train going in the same direction as he does. The train always goes by the same speed and is always on schedule. One morning the man did not wake up on time, and left his house 50 minutes late. Therefore, he met the train 12 km before the railroad crossing. How much longer does it take the train to get to the railroad crossing?

hm... i read the problem but don't rreally get what it meant, in other words i have no idea how to start the problem

Most of us don't know how to set the problem up. (D=RT aside) That's what deterred me. For one thing, which way is the train heading? Towards the biker or away from it?

I assume it's heading the way the man is biking.

It's good to know that one person is actually serious about something  :)

:D yes.
I think that we shoudl have a permanent annoucement on this forum just saying: NO SPAMMING! AoPS is for MATH, not for those ikkle stars!!!
And just because MC season's over doesn't mean we can :spam:
anyways...
[hide]D=RT
R[t]=rate of train
D[t]=starting distance from train to biker's house
Biker: D=RT
Train: D+D[t]=R[t]*T
D-12=...
Hold on, WHAT?!?!
"How much longer does it take the train to get to the railroad crossing?
"
From where? If it's the train's starting position, then there's 0! But if it's form the biker's house, then...hold on, is the trian travelling slower or faster than the bike? :huh: ARGHHH! I'M NOT GETTING THIS PROBLEM!!!!! :wallbash:  :surrender:  :wacko:  :help:

Don't get mad. I tried everything from A to Z, and nothing works yet.

Don't worry, even though we don't post it, II have tried many things, but this problem just does not make sense

This problem isn't impossible (I don't think). In fact, I can give you the source of this problem.
http://www.gcschool.org/pages/program/abacus/8_1_03.html

sorry, not a hyperlink ;)

The train and the man on the bike are going in the same direction along a parallel course.
[hide="hint"]If the train goes by the man 12 km before the man gets to the crossing, how long will it take the man to get to the crossing.[/hide]

I have like 20 variables but two equations, so I need some serious elimination (substution).

One thing that does help is the fact that he still has one hour when he met the train on the day he was late.

I think you guys are making this problem much harder than necessary.
[hide="hint 2"][quote="236factorial"]...he still has one hour when he met the train on the day he was late.[/quote]
Now you have all you need to solve the problem.  Just draw a timeline and mark the relevant entries on the timeline.[/hide]

I don't understand. If they're traveling in the same direction, then how can the following statement be true?

[quote="The Problem"]One morning the man did not wake up on time, and left his house 50 minutes late. Therefore, he met the train 12 km before the railroad crossing.[/quote]

He left later, but met the train earlier?  :huh:

[quote="chess64"]I don't understand. If they're traveling in the same direction, then how can the following statement be true? . . .[/quote]
Assume the train is travelling faster than the man, but both are travelling in the same direction.

The man needs a certain head start to reach the crossing at the same time the train gets there.

If the man starts late, he gets less of a head start, so the train will overtake him before he reaches the crossing.

Note:  These Abacus problems have not undergone the same quality-controls as MATHCOUNTS problems.  Most Abacus problems look pretty reasonable to me.  But, we have found some Abacus problems that are worded poorly and some that are just plain wrong.  This problem seems pretty reasonable, although the wording is a little awkward.  The word "met" may be misleading in the problem statement.  [Replace "met" with the phrase "is overtaken by".]

Let's also attempt to clarify the final question of the problem.  Replace "How much longer does it take the train to get to the railroad crossing?" with "How many minutes elapse between the moment the train overtakes the man and the moment the train passes the crossing?"

Okay, that makes sense now. "Overtake" certainly makes it easier to visualize.

It's very simple.  Forget formulae.  Just take it ad hoc, and logically.  Many formulae are time-wasting, if you know what you are doing.  Besides, what helps you more, thinking it through, or sticking one number in a "magic box" and getting another one without any explanation?
My interpretation :
[hide]In $\frac{5}{6}$hr, the train gains 12 km on a man who is travelling at 12 kph.  Therefore, the train is travelling at $\frac{12}{\frac{5}{6}}=\frac{72}{5}$ kph faster than the man; in other words, at $\frac{132}{5}$ kph.  Now, how long will it take the train to travel the remaining 12 km?  Well, $\frac{132}{5}$ kilometers per hour equals $\frac{5}{132}$ hours per kilometer.  Multiply that by 12 kilometers to get $\frac{5}{11}$ hours.  It isn't very pretty, but I think that $\frac{5}{11}$ is the answer.[/hide]

It's been several days since anybody attempted to make any progress on this problem.  So, here's my solution, assuming I understand the problem...

[hide="Solution -- spoiler"]
The only variable you need is time.  I drew a timeline.  (A number line, representing the time.)

Mark time X on the line, when the man and the train usually are at the crossroads at the same time.

Mark time Y, when the man will arrive at the crossing on the day he slept late.  [Since the man overslept by 50 minutes, time Y will be 50 minutes to the right of time X.]

Mark time Z, when the man and the train are at the same place on the day the man slept late.  [Since we know this is 12 km before the crossing and the man rides at 12 km/hr, we know this is 60 minutes to the left of time Y.]

The question wants to know how long it will take the train to get from time Z to time X.

[code]
----------------------------------------
     Z    X                        Y
          |<--------- 50 --------->|
     |<----------- 60 ------------>|
[/code][/hide]


The final scores for the math jam are as follows
1. stupitidyismygam(9th grade) 10 points
2.xxm0rg07hxx(me)(insanemorgoth)(8th grade) 9 points
3.biffanddoc 6 points
After that,
we had a final 1vs1 showdown between me and stupitidy.
The final score of that was 5-2 in my favor
so that means I'm Klebian math jam winner for tonight!!!!!
 :trampoline:  :wow:  :trampoline:

easyaspi got 4th place, btw, with 5 points.

[quote="easyas3.14159..."]and i'm a seventh grader so :trampoline: yay!!i thought id get like 0.[/quote]
then i PWNED you in the 1vs 1
it was gonna be a sweep, but i answered too fast and it was 5-3
(4-0 then easy as won 3 then i won 1)

[quote="Zmastr"]What's up with you and PWNED?  Well, good job to you and everyone else who made nats in the past week.  I have yet to make my mark at state.  MWAHAHAHA!![/quote]
It's because he get's PWNED so often.

you wish.

klebian i will PWN you at countdown
>.<

*cough* because you were too good *cough*

Come on, what really defines noob?  I think we're all noobs.  I mean, even though math is the greatest subject ever, don't you think we're kind of losers for getting up early to do math problems on a Saturday.  It seems natural to us, but others probably think we're crazy.  Oh, and did you say you'll pwn me at countdown?  Bring it.

[quote="xxm0rg"]Ever since I called biffanddoc a noob for saying owned instead of pwned he has been putting in caps so he can pose as a pro(which he still is not).[/quote]
fine then. I OWNED easyaspi

ill use OWNED PWNED instead so you don't call me a poser. and yeah lets see who does better at nats... :D

But oh no! Texas intervenes and beats the crud out of both!
Everyone knows that if I had actually PARTICIPATED in the math jam, I would've owned morgoth.
Er, I mean, I would've OWNED PWNED morgoth.

And if I even had AIM, I would have to be forced to open a can of serious kick-butt math glory in everyone's face in the name of Virginia.  Notice that the words owned and pwned appeared nowhere in the previous sentence.

[quote="Zmastr"]Fight, Fight! Today we have two very worthy states in the ring: Delaware and Arkansas.  Let's chill out and let the math do the talking in the end.[/quote]
Delawares gonna whoop arkansas' ass at nats

I'm sorry, but I just have to say what I'm thinking about some of the people in this thread right now...

biggest.losers.ever

(Now you know I don't mean that, but comon... some of this stuff is really stupid or braggety...

"Funny how I'm the only non-NOOB who's posted in this topic."

"I'm sorry, but they had to take me out cause I won too much."

"then i PWNED you in the 1vs 1
it was gonna be a sweep, but i answered too fast and it was 5-3
(4-0 then easy as won 3 then i won 1)"

Also, we're having a fight on pwned vs. owned/

Maybe I'm just cranky after state -_-.

"Fight, Fight" Delaware vs. Arkansas? Next up, Rhode Island vs. Wyoming...

hahaha.
Well said.
The thing is, after the first 3 or 4 posts, the only point of this topic is to tell people the score of the mj.
BUT, if you were at the mj then it's kinda useless. And if you weren't, then you won't understand what we're really talking about anyway.
So Biff brings up his "uber pwnage" in countdown, and everything from there becomes a really crummy flame war.
That's why mathmeticians should never be assigned role of humorist -> The only people that find mathmeticians funny are other mathmeticians.

ya i came at like the last second so i didn't do anything.
anyways michigan is gonna OWN PWN all ur @sses at nats :D
well i dunno cuz i didn't even make countdown :(
and yea we're all dorks and noobs and nerds.
That's y we're on this site
Carry on the fight now.

Hey, who says math whizzes are nerds?  I'm not a nerd.  But I still love math. Being a noob is just a state of mind.


find all polyminals P(x) with real coefficients such that

if ab+bc+ca=0 then P(a)p(b)+p(b)p(c)+P(c)p(a)=0

Obviously $ b\equal{}a\neq0$ and $ c\equal{}\minus{}\frac a2$ satisfies $ ab\plus{}bc\plus{}ca\equal{}0$.

Let $ a$ be positive and is not the root of $ P(x)$, then we have

$ P(c)\equal{}P(\minus{}\frac a2)\equal{}\minus{}\frac{P(a)}2$. Hence $ P(\frac a4)\equal{}\frac{P(a)}4$.

Assume $ P(x)$ has degree of $ n$ and $ a_n$ is the leading coefficient of $ P(x)$.

We have $ \lim_{a\rightarrow\infty}\frac{P(a)}{a^n}\equal{}a_n$

Also,

$ \lim_{a\rightarrow\infty}\frac{P(a)}{a^n}\equal{}\lim_{a\rightarrow\infty}\frac{4P(\frac a4)}{a^n}\equal{}\lim_{a\rightarrow\infty}\frac{4P(\frac a4)}{4^n(\frac a4)^n}\equal{}a_n4^{1\minus{}n}$.

Compare the two formulae above, we have $ n\equal{}1$. So $ P(x)\equal{}a_1x\plus{}a_0$. It is easy to get $ a_0\equal{}0$.


trigonometry
number theory unsolved
number theory

Show that:
$\sum_{k=1}^{n-1}(-1)^kcos^n(\frac{k\pi}{n})=\frac{n}{2^{n-1}}$
I think it's really hard and I have no solution.

i remember that i solved this problem same time ago.
i think that you can use complex numbers.  :)

I check with $n=2,3$ and they don't seem to hold  :huh:

$n=2$ , $-\cos\frac{\pi}{2}+\cos\pi=-1\neq 1$

$n=3$ , $-\cos\frac{\pi}{3}+\cos\frac{2\pi}{3}-\cos\pi=0\neq \frac{3}{4}$

 :roll:

uhm... you just can re-write your sum as \[ \frac12\left(\sum_{k=1}^{2n-1} (-1)^k\cos\left(\frac{k\pi}{n}\right) + (-1)^n\cos(\pi)\right) = \frac12\left(\sum_{j=1}^{n-1} \cos\left(\frac{2\pi}{n}j\right) - \sum_{j=0}^{n-1} \cos\left(\frac{2\pi}{n}j + \frac{\pi}{n}\right) - (-1)^n\right)  \] and notice that two sums are the sum over all non-real roots of $1$ and $-1$ of their real parts, just taking taking care of parity of $n$.
i think it's working, but i've just got up, so i can be making some mistakes :D

by the way, first sum makes always $-1$, second one gives always $0$, so it seems like $\displaystyle\sum_{k=1}^{n}(-1)^kcos(\frac{k\pi}{n})=-\frac{1+(-1)^n}2$, which is ok with shyong's computations...

uhm...
by the way, your result (as it's stated) is false :)
but, if you edit it once again, setting the sum from $k=0$ to $n-1$ or from $k=1$ to $n$, you get the right thing..

my proof:
let me call $\omega = e^{\frac{i\pi}{n}}$. since $\omega^k = 1$ iff $2n|k$, we can state these two identities:
$\displaystyle \sum_{k=0}^{2n-1} w^{2hk} = 0$ for all $n\nmid h$, and $\displaystyle \sum_{k=0}^{2n-1} w^{2hk} = 2n$ otherwise.
now, we can write \[ \displaystyle \sum_{k=0}^{n-1} (-1)^k\cos^n\left(\frac{k\pi}{n}\right) = \frac12\displaystyle \sum_{k=0}^{2n-1} (-1)^k\cos^n\left(\frac{k\pi}{n}\right)  \] morover, $\cos x = \frac{e^{ix}+e^{-ix}}{2}$, so we can write: \[ \sum_{k=0}^{n-1} (-1)^k\cos^n\left(\frac{k\pi}{n}\right) = \frac1{2^{n+1}}\sum_{k=0}^{2n-1}(-1)^k\left(\sum_{h=0}^n \binom{n}{h}\omega^{kh}\omega^{nk-h}\right)  \] but $\omega^{nk} = (-1)^k$, so we can write \[ \sum_{k=0}^{n-1} (-1)^k\cos^n\left(\frac{k\pi}{n}\right) = \frac12\sum_{k=0}^{2n-1} \left(\sum_{h=0}^n \binom{n}{h}\omega^{2kh}\right) = \frac12 \sum_{h=0}^n \binom{n}{h} \sum_{k=0}^{2n-1} \omega^{2hk}  \] but the two identities i've written allow me to conclude that this last sum reduces to the two last terms, so finally \[ \displaystyle \sum_{k=0}^{n-1} (-1)^k\cos^n\left(\frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}  \]


algebra polynomial AMC AIME LaTeX search function

[quote="t0rajir0u"]This looks like an AIME problem that's been posted before, and that has a solution somewhere.

[hide="Hint"] Cotangent addition formula.  Turns out that the expression is a quotient of two symmetric sums.  This problem is quite ugly. [/hide][/quote]
I got it unsourced. If someone could give me the year so I could kalva up a solution, that would be appriciated. Latex expressions do not work well in the search function, alas.

It's from the mandelbrot.

The AIME one is simpler, it's like 1986 #13 or something (#16 on one of the early AIMEs though).  Same concept though.

Here's a reasonably nice approach using complex numbers.  Note that the complex number $z_{k}+i$ has polar angle $\cot^{-1}z_{k}$ (modulo $\pi$).  Thus the product
\[(z_{1}+i)(z_{2}+i) \cdots (z_{20}+i) \]
has polar angle $\sum_{k=1}^{20}\cot^{-1}z_{k}$ (again, modulo $\pi$).  Thus we just need to evaluate that product.

Here's a quick way to evaluate the product.  Note that the numbers $z_{1}+i$, $z_{2}+i$, ..., $z_{20}+i$ are roots of the shifted polynomial $f(x-i)$.  Thus to evaluate the product of those numbers, we just need to compute $f(-i)$.


In how many ways can five children sit in a line if order matters?

[hide]$5$ choices for first child, $4$ for 2nd, etc, so $5\times4\times3\times2\times1=5!=120$[/hide]

In how many ways can five children sit in a circle if order matters? :D

[quote="daermon"]does it count as a unique permutation if it can be rotated to get another?[/quote]
No, otherwise we would get the same answer.

[quote="daermon"]does it count as a unique permutation if it can be rotated to get another?[/quote]
[hide](5-1)!= 24 [/hide]

Why does this work: There are $(n-1)!$ ways for $n$ people to sit in a circle?

I think you have to "fix" one of the people in a certain position before arranging the rest, or you could just rotate the circle n times and this would be counted as different permutations.

[hide="solution for 3rd problem"]We arrange the people in a straight line. There are $n!$ ways to arrange them. Now, if they were in a circle, we will have to count for permutations, aka, divide by n. so we get $(n-1)!$ ways to arrange them in a circle.[/hide]

its not still $5!$ b/c $abcde=eabcd$.  permutations of $n$ people in a circle is $(n-1)!$ b/c each permutation can be rotated $n$ times


Time allowed : 3 hours

1. The digits of a 3-digit number are interchanged so that none of the digits retain their original position. The difference of the two numbers is a 2-digit number and is a perfect square. Find the difference.

2. Consider the nonconvex quadrilateral $ABCD$ with $\angle C>180$ degrees. Let the side $DC$ extended to meet $AB$ at $F$ and the side $BC$ extended to meet $AD$ at $E$. A line intersects the interiors of the sides $AB,AD,BC,CD$ at points $K,L,J,I$ respectively. Prove that if $DI=CF$ and $BJ=CE$, then $KJ=IL$

3. Let $S$ be a subset of $\{1,2,3,...,24\}$ with $n(S)=10$. Show that $S$ has two 2-element subsets $\{x,y\}$ and $\{u,v\}$ such that $x+y=u+v$

4. Determine whether there exists a positive integer $n$ such that $n!$ begins with $2005$. Justify your answer.

#3
Let $a_1,...,a_{10}$ be the members of S.  there are 45 pairs among these numbers. If we have tep distinct pairs say $(a_1,a_2)$ and $(a_3,a_4)$ such that $a_1-a_2=a_3-a_4$ we are done since the pairs $(a_4,a_1)$ and $(a_3,a_2)$ have the desired property unless $a_3=a_2$. We say $a$ is  bad for the pairs $(a_1,a),(a,a_2)$ if $a_1-a=a-a_2$. note that if $a$ is bad for two pair of pairs then we can find the desired pairs and we are done. So assume each $a_i$ is bad fo at most one pair of pairs. For each such pair of pairs we take one pair out of consideration. Now there are no bad numbers anymore and we have at least 45-10=35 pairs left. The diffrences in these pairs range from 1 to 23 so by pigeonhole some of these differences have the same value, so we are done.

1:

I'm assuming that the digits of the three-digit number are all distinct or the proof will be more difficult (if two digits are the same).

If the three digit number was originally $abc$, the only possible permutations to make sure no digit retains its original position are $bca$ and $cab$. Then the only possible differences are $100a + 10b + c - (100b + 10c + a) = 99a - 90b - 9c = 9(11a - 10b - c)$, or $100a + 10b + c - (100c + 10a + b) = 90a + 9b - 99c = 9(10a + b - 11c)$, both of which are multiples of 9.

Since the difference is a two-digit number and a perfect square, it can only be $9^2 = 81$ since it must also be a multiple of $9$. So the answer is $81$.

in 484-448 the first 4 retains it's original position.

An example would be 213 - 132

no, the switches are like (1,2,3) -> (3,1,2)

[quote="Jan"]Here's a stupid try for #4, and probably wrong, but anyway, give me your opinion:

We are searching for a positive integer $n$ so that $2.3.4.5. ... .n$ starts with $2005$.
It's clear that $n$ is not $2$ or $3$. When we divide $n!$ by $2$, we would still have an integer (being $3.4.5. ... .n$). Divided by $2$, the number would begin with $10025$, followed by a certain ammount of numbers.
When we divide $n!$ by $3$, we would still have an integer. (being $4.5.6. ... .n$)
When we do so, we see that the number starts with $3341666666...$, going on infinitely. This shows that there exist no such $n$, because a number can not be infinitely long, unless it's not an integer. (e.g.:with numbers after the comma)

A little too easy to be right, I guess. Please tell me what you think of it....[/quote]

I think by your logic $\frac{20052}{6} = 3342$ is not an integer...

Ow yes, that's right Paladin. Makes my method complete rubbish. Well, lousy try, better luck next time...

For 1, I got 36 and 81... but no rigorous method was involved..just random playing around with numbers. here are some of them.

when doing $abc - cab$

for 36: 151, 262, 373, 484, 595, 706, 817, 928.... it seems you add 111 each time.

for 81: 201, 312, 423, 534, 645, 756, 867, 978.... same thing.. adding 111.

when doing $abc - bca$

for 36: 107, 218, 319, etc... add 111

for 81: 102, 213, 324, etc... add 111

edit:

here's just a way to get those integers, but it's not a solution to the problem

$abc-cab = 9(10a+b-11c) = k^2$

because 9 is a perfect square, we see that we should just examine $10a+b-11c$.  finding a base case to use, we see $a=2, b=0, c = 1$ gives use a perfect square.  To generate more of them, adding 1 to each of them will preserve the perfect square property because 10+1-11 = 0 (coefficients of a,b,c) - this accounts for the pattern of 111's.

i basically found one base case by trial and error, and then from there, you just take note of the characteristics of the equation.

this method also worked for $abc-bca.$

[quote]1. The digits of a [b]3-digit [/b]number are interchanged so that none of the digits retain their original position. The difference of the two numbers is a 2-digit number and is a perfect square. Find the difference.[/quote]

The answer would be same for 4 digit  (or 10 digit for that matter) as long as difference is a 2 digit number :)

[hide="because"]( The difference, after [i] any [/i] rearrangement should be a multiple of 9 .:)
 Any rearragement of the digit does not change its value modulo 9, so difference is 0 modulo 9.[/hide]
Last  problem has been discussed [url=http://artofproblemsolving.com/Forum/viewtopic.php?highlight=mantissa&t=23417]here[/url]

And By the way ..  $4968!$ does begin with [b]2005[/b]981717....:D

(Sorry but Admins might get mad if I write all 16,008 digits of 4968! :)


$ \left\{\begin{array}{l}
\angle BAC \equal{} 90^\circ \\
\angle ABC \equal{} 45^\circ \\
\text{midpoint of }\overline{AC}\text{ is }D \\
\text{midpoint of }\overline{BC}\text{ is }E \\
BA \equal{} BF \\
G \equal{} \text{arc }AF \cap \overline{ED}
\end{array}\right.$

Find $ \angle GCB$.

[asy]size(300);
pair A=(0, 10), B=(-10, 0), C=(10, 0), D=(5, 5), E=(0, 0), F=(-10+10*sqrt(2), 0);
draw(A--B--C--cycle);
draw(D--E);
draw(Arc(B, 10*sqrt(2), 0, 45));
pair G=intersectionpoint(D--E, Arc(B, 10sqrt(2), 0, 45));
draw(C--G);
label("A", A, N);
label("B", B, SW);
label("C", C, SE);
label("D", D, NE);
label("E", E, SW);
label("F", F, SE);
label("G", G, NW);
fill(C--Arc(C, 2, 150, 180)--cycle, gray);
label("?", (8,1), SW);[/asy]

[hide="Answer"]$ \angle GCB \equal{} 30^\circ$[/hide]

I think that there is a lot of different solution.
Have Fun! :jump:

[hide="First solution"]On the bisector of angle ABG, let K inside triangle KBG such that BK=AG=GC=m. Because $ BK\perp AG, \angle ABK\equal{}\angle GAC$, so $ \triangle GAC\equal{}\triangle KBA\equal{}\triangle KBG\Rightarrow KA\equal{}KG\equal{}m\equal{}AG$. Because of the equilateral triangle AKG, $ \angle KAG\equal{}60^o$.
Thus, $ \angle KAB\equal{}\angle GAC\equal{}15^o, \angle GCB\equal{}30^o$.[/hide]

[hide="Second solution"]$ BG^2\equal{}BA^2\equal{}BE.BC\Rightarrow \triangle BGE\sim\triangle BCG\Rightarrow\angle ABG\equal{}\angle BGE\equal{}\angle BCG\equal{}x$
Because of the isosceles triangles ABG and AGC, $ \frac{180\minus{}x}{2}\plus{}45\minus{}x\equal{}90\Rightarrow x\equal{}30^o$[/hide]


Let $ a,b,c >0$ such that $ a.b.c\equal{}1$
$ \frac{a}{{ka^2  \plus{} 1}} \plus{} \frac{b}{{kb^2  \plus{} 1}} \plus{} \frac{c}{{kc^2  \plus{} 1}} \le \frac{3}{{k \plus{} 1}}$

It's same to ineq 3 of you.
And they are the old problems. (it isn't true with all k)
See it here: http://www.maths.vn/forums/showthread.php?t=10982


A regular polygon with n sides is divided into n isosceles triangles by segments joining its center to the vertices. Initially,      n + 1 frogs are placed inside the triangles. At every second, there are two frogs in some common triangle jumping into the
interior of the two neighboring triangles (one frog into each neighbor). Prove that after some time, at every second, there are at least $ [ (n+1) / 2 ]$ triangles, each containing at least one frog.

What happens if three frogs occupy a triangle?
That's true for the square after 2 seconds.


There seems to be a problem when one tries to use the ' sign with the url or the hide command.

[url=http://www.mathlinks.ro]click[/url]
[url=http://www.math'links.ro]click[/url]
[url="http://www.math'links.ro"]click[/url]

[hide="Gogule, probleme?"]$\textrm{blah}^{3}$[/hide]
[hide="Gogule' probleme?"]$\textrm{blah}^{3}$[/hide]
[hide="Gogule' probleme?"]$\textrm{blah}^{3}$[/hide]

Is it solvable?

i don't think it's a bug, i think it just messes up the BBCode


A 30-power telescope uses an 18-millimeter eyepiece. How long is the tube?

20. Did you get this from Physics the Easy Way?


I  use    asymptote 1.90 + MiKtex +Texniccenter.

1

I  try  to  use  the cse5  package,  for  example  ,  the following  code ,  but  it   doesn't  work.
why?
[url]http://www.mathlinks.ro/Wiki/index.php/Asymptote:_CSE5[/url]

[code]
import cse5;
 size(200);
pair A=dir(-40), B=dir(0);
for(int k=0;k<9;++k)
{
pair C=dir(40*(k+1));
D(A--D(MA(k*40+90,"\beta_{"+string(k)+"}",12,black,C,B,A,0.1,k,orange))--C);
A=B;B=C;
}[/code]

the  code  from    the following  doesn't  work, too!
http://www.mathlinks.ro/viewtopic.php?t=149650


Hi, I think this question is flawed, but I'm not sure.

(Heat Equation. use method of separation to solve the equation)

u_t=17 u_xx for 0<x<PI and t>0; given the boundary condition

u(x,0)={0   0<=x<=PI/2,   2   PI/2<=x<=PI

u(0,t)=u(PI,t)=0

My concern is that u(PI,0)=2, but u(PI,t)=0 for t>0
so u doesn't seem to be continuous as t varies for x=PI.

can anyone advise me whats going on?

Thanks

Yes, you'll have a discontinuity at $ (\pi,0)$ but this shouldn't bother you any more than the fact that you have one at $ (\pi/2,0)$. It'll disappear immediately and the solution will be nice and smooth for $ t>0$. As $ t\to 0\plus{}$, you'll have just a couple of points $ x\in[0,\pi]$ (namely $ \pi/2$ and $ \pi$) where it is not true that $ u(x,t)\to u(x,0)$. So, just disregard this problem for now.

Thanks, I was looking at a similar problem I found on the internet where f(x,0)=20

but u(0,t)=u(L,t)=0

I did some numerical tests on calculator with the series, which vanishes at x=0, and x=L, and noticed that however close to 0 or to L you go, the series approaches the value 20 with more terms added.

But clearly you are summing a series of continuous functions, but as I suspected, an infinite sum of continuous functions is not necessarily continuous, so thats really hit the nail on the head.

But its a little annoying since the solution u(x,t) won't therefore match the boundary condition f(x) at the endpoints x=0,x=L for some particular BVP problems.

[quote="colin_java"]
But clearly you are summing a series of continuous functions, but as I suspected, an infinite sum of continuous functions is not necessarily continuous, so thats really hit the nail on the head.[/quote]

You need a stronger property, uniform convergence, to make this true.


Solve the equation: $2x^{2}-6x-1=\sqrt{4x+5}$

[hide="Solution"]Substituting $t=\sqrt{4x+5}\iff x={t^{2}-5\over 4}$, we get

$2\cdot\frac{t^{4}-10t^{2}+25}{16}-{6\over 4}(t^{2}-5)-1=t$, or

$t^{4}-22t^{2}-8t+77=0$

This factors into

$(t^{2}+2t-7)(t^{2}-2t-11)=0$

[size=59]For those who wonder: The lack of cubic term is a tell-tale sign for (t^2+Ct+a)(t^2-Ct+b) type of factoring. The free term leads us to a=7,b=11 or a=-7,b=-11, and after some easy trial and error, we find C=2, a=-7, b=-11[/size]

The solutions of the last equation are

$t_{1,2}=-1\pm 2\sqrt{2}, t_{3,4}=1\pm 2\sqrt{3}$

Because of the definition of $t$, we can take only positive values:

$x_{1}=\frac{(2\sqrt{2}-1)^{2}-5}{4}=1-\sqrt{2}$

$x_{2}=\frac{(2\sqrt{3}+1)^{2}-5}{4}=2+\sqrt{3}$[/hide]

Well done! Here I show an other[hide="trick"]Denoting $a=4x^{2}-12x+9, \ b=4x-6$ we have

$a-11=2\sqrt{b+11}$

Considering it as equation upon "variable" $11$ we have after evident calculations

$11^{2}-2(a+2)\cdot 11+(a^{2}-4b)=0$

This gives $11=a \pm 2\sqrt{a+b+1}=(4x^{2}-12x+11) \pm 4(x-1)$ and so

$11=4x^{2}-8x+7$ with roots $x=1\pm \sqrt2$

or

$11=4x^{2}-16x+15$ with roots $x=2\pm \sqrt3$

To kill extra roots we have to use $a-11=4x^{2}-12x-2\ge 0$[/hide]

Very nice :)  This called parameterization, I think, and I think a chapter of ACoPS is devoted to it.

The basic concept is that you have an equation in one variable involving one unknown constant (the "parameter"); you solve for the parameter and use that to solve for the variable.  It's a really clever insight to use it on a concrete number, though.

Once, when I was still at school my teacher gave me "very hard equation", somewhat of $\sqrt{5-x}+x^{2}=5$
It was after physic lesson where I always changed concrete parameters by letters then solved the problem and substitude. So it was easy for me to change $5$ by $a$ and to say - nothing is hard here, the equation is quadratic.
Why? It isn't - surprised the teacher. It is quadratic if we cosider 5 as variable - I answered. That was the way I knew for myself this trick.

However the teacher meant the following trick:
$t=\sqrt{5-x}$ yielding the nice system (with restriction $t\ge 0)$:

$t+x^{2}=5$
$x+t^{2}=5$

with evident $(t-x) = (t-x)(t+x)$ which devides problem into 2 parts $t=x$ and $t+x=1$
This is similar to [b]Farenhajt[/b]'s solution.

Thanks Digger for supply us a method to kill the such equations. Very nice way

Yes, the method is very elegant, but the search for the suitable $a$ and $b$ can be tricky (a lot needs to be spotted).

2hien. The method is not new. The coincidence of circumstances has helped me to reopen it.

[quote="Farenhajt"]... but the search for the suitable $a$ and $b$ can be tricky (a lot needs to be spotted).[/quote]
Yes, in general it demands to find one root of some 3d degree equation which depends on given equation of 4th degree only and leads  to something similar (but not the same) to Ferrary method - the 1st step is differrent.

But it is easy [b]to construct [/b] such "hard equations".
Try for example this one looking simple:

$x^{2}-4x+2=\sqrt{x}$

P.S. O-o-p-s, it's easy. So I need a little time to change.

P.P.S. The improved version:

$x^{2}-8x+12=\sqrt{x}$


quadratics
search
number theory theorems
number theory

Could somebody please explain what the jacobi symbols are?

I want to know their properties and usages( possibly, examples of their usage in solving a problem)

I heard that they are the generalizations of legendre symbols

thank you all :)

http://en.wikipedia.org/wiki/Jacobi_symbol  check there
Hope it is useful

Daniel

Their main properties are Euler's theorem (for Legendre smybols only), multiplicativity and quadratic reciprocity.

A non-problem usage is the fast computation of Legendre symbols (using quadratic reciprocity). Some other usage can be found in http://www.mathlinks.ro/viewtopic.php?t=64322 where it simplifies the proof (away from Dirichlet's theorem). You can find some problems using the search on the forum.


If you have a number of faces in a prism, how do you find the number of edges in that prism. For example, a cube has 6 faces and has 12 edges. THen how do you find the number of edges in a 10 faced prism. Is there a formula to find this out?

tHANKS

First, you have to know how many edges are on each face.  Then, find the total number of edges i.e. a prism with 6 squares and 4 triangles has $6\cdot4+4\cdot3=36$ total edges.  When you join 2 faces, you connect 2 edges, and come out with 1 edge, so you divide the total by 2.

A 10 faced prism would have $n$ on each face, making a total of $10n$ edges.  Then, divide by 2 to get only$5n$ actual edges.

No matter how you do it, you still have to know how many sides are on each of the faces.

If you only need to know one of the three, you can use the given formula. If you only know faces, the best you can do is attempt to visualize how many vertices it has and find the edges from there, as the vertices are generally easier to count. If you know what shape each face is, you can count the number of vertices from each shape and divide the total by $3$, because each vertex is counted $3$ times, then find the edges using the $V+F=E+2$ formula. I think that would be the easiest way.

It is Euler's fomula. Vertices + Faces - Edges = 2


Find limit of sequence $ {u_{n}}$ : Lim $ C^{n}_{2n}$

hello, do you mean $ \lim_{n\to \infty}\frac{(2n)!}{(n!)^2}$?
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, do you mean $ \lim_{n\to \infty}\frac {(2n)!}{(n!)^2}$?
Sonnhard.[/quote]
yes

[quote="SimonM"]$ \frac {4^n}{2n \plus{} 1} < \binom{2n}{n}.$[/quote]
Sorry, I dont understand your idea

[quote="MathVietNam"][quote="L_Euler"][quote="SimonM"]$ \frac {4^n}{2n \plus{} 1} < \binom{2n}{n}.$[/quote]

If that's right then $ \lim \equal{} \plus{} \infty$? But i do'nt see your solution, can you post a detail solution!!!^^[/quote]
Oh, thanhk u, L_Euler[/quote]
$ \binom {2n} {n} \equal{} \frac{(n\plus{}1) * (n\plus{}2) * ... * (2n)}{1*2*...*n} > n\plus{}1$ (Since $ n\plus{}2>2, n\plus{}3>3, ..., 2n > n$)
But $ \lim_{n \to \infty} n\plus{}1 \equal{} \plus{} \infty$ so $ \lim_{n \to \infty} \binom {2n} {n} \equal{} \plus{} \infty$


geometry
angle bisector
geometry unsolved

[quote="orl"]Prove: In an acute triangle $ ABC$ angle bisector $ w_{\alpha},$ median $ s_b$ and the altitude $ h_c$ intersect in one point if $ w_{\alpha},$ side $ BC$ and the circle around foot of the altitude $ h_c$ have vertex $ A$ as a common point.[/quote]
[hide="Solution"]
Let $ s_b$ meet $ AC$ at $ S$, $ w_{\alpha}$ meet $ BC$ at $ W$, and $ HC$ meet $ AB$ at $ H$.  Notice that since the circle with center $ H$ and radius $ HA$ passes through the point where $ w_{\alpha}$ and $ BC$ meet, which is $ W$, we have that $ WH\equal{}WA\implies \angle WAC\equal{}\angle HAW\equal{}\angle HWA\implies HW\parallel AC$.  Yet, this means that $ \frac{AH}{BH}\equal{}\frac{CW}{WB}$, so $ \frac{AH}{BH}\cdot \frac{BW}{WC}\equal{}1$.  Therefore, if we let $ P$ be the intersection between $ CH$ and $ AW$ and let $ S'$ be the intersection between $ BP$ and $ AC$, then Ceva's Theorem yields that $ \frac{AH}{BH}\cdot \frac{BW}{WC}\cdot \frac{S'C}{AS'}\equal{}1$, so $ S'C\equal{}S'A$, which implies that $ S'$ is the midpoint of $ AC$, so $ S\equal{}S'$.  Thus, $ SB$, $ AW$, and $ CH$ concur. [/hide]

[quote="orl"] [color=darkred]In an acute triangle $ ABC$ denote $ \left\|\begin{array}{ccc} D\in (BC) & ; & \widehat {DAB}\equiv\widehat {DAC} \\
\ E\in (CA) & ; & EC \equal{} EA \\
\ F\in (AB) & ; & CF\perp AB\end{array}\right\|$ . Prove that $ AD\cap BE\cap CF\ne\emptyset\ \Longleftrightarrow\ DF \equal{} DA$ .[/color] [/quote]
[hide="Proof."]
$ AD\cap BE\cap CF\ne\emptyset$ $ \Longleftrightarrow$ $ \frac {DB}{DC}\cdot\frac {EC}{EA}\cdot\frac {FA}{FB} \equal{} 1$ $ \Longleftrightarrow$ $ \frac {DB}{DC} \equal{} \frac {FB}{FA}$ $ \Longleftrightarrow$ $ DF\parallel AC$ $ \Longleftrightarrow$ $ \widehat {FDA}\equiv\widehat {CAD}$ $ \Longleftrightarrow$ $ \widehat {FDA}\equiv\widehat {FAD}$ $ \Longleftrightarrow$ $ DF \equal{} DA$ .[/hide]


vector
algebra
combinatorics
combinatorial geometry
Inequality
IMO Shortlist

Let $ u_1, u_2, \ldots, u_m$ be $ m$ vectors in the plane, each of length $ \leq 1,$ with zero sum. Show that one can arrange $ u_1, u_2, \ldots, u_m$ as a sequence $ v_1, v_2, \ldots, v_m$ such that each partial sum $ v_1, v_1 \plus{} v_2, v_1 \plus{} v_2 \plus{} v_3, \ldots, v_1, v_2, \ldots, v_m$ has length less than or equal to $ \sqrt {5}.$

[quote=P-H-David-Clarence]This bound can be improved to $\sqrt2$.[/quote]

I think it isn't

First we consider the question in one-division, we can prove the bound is 1.
By finding the maximal length of subsum of $v_i$ (say $v=v_1+\cdot\cdot\cdot+v_k$), we can dissect $v_i$ into to vectors which are orthogonal to each other. Considering the maximal condition, the projections of vectors on $v$ is at most 1, which of $v_i(i<k)$ is positive, and negative otherwise.
Applying the same process for two orthogonal directions, we can reach the bound $\sqrt{1^2+2^2}=\sqrt{5}$

I think post #5 is missing some detail.

[hide =Solution]

WLOG suppose that $u_1 + u_2 + \cdots + u_n$ is the subset of $\{u_1, u_2, \cdots, u_m\}$ with their sum of maximal magnitude. WLOG, suppose that $u_1 + u_2 + \cdots + u_n$ is a vector in the $+x$ direction, by simply rotating all the vectors. Let $u_i = \left<x_i, y_i\right>$ for each $1 \le i \le m.$ Observe that for $1 \le i \le n$ we have $x_i \ge 0$ and for $n+1 \le i \le m$ we have $x_i \le 0.$ This is clear from the maximality of $u_1 + u_2 + \cdots + u_n.$

Now, let $\sigma$ be a permutation of $1, 2, \cdots, n$ such that $|y_{\sigma(1)} + y_{\sigma(2)} + \cdots + y_{\sigma(i)}| \le 1$ for each $1 \le i \le n.$ Similarly, let $\rho$ be a permutation of $n+1, n+2, \cdots, m$ such that $|y_{\rho(n+1)} + y_{\rho(n+2)} + \cdots + y_{\rho(i)}| \le 1$ for each $n+1 \le i \le m.$

From here the finish is easy. Simply start by taking $u_{\sigma(1)}$ to be $v_1$, then take $u_{\rho(n+1)}, \cdots$ until the $x-$component becomes negative, and then take $u_{\sigma(2)}$, etc. This guarantees that, at any point, the $x-$component of a partial sum is at most $1$ in magnitude and the $y-$component is at most $2$ in magnitude. We are done.

$\square$

[/hide]

@above can u tell what $<x,y>$ means ? I tried searching but couldn't find anything
Thanks @below

[quote=Pluto1708]@above can u tell what $<x,y>$ means ? I tried searching but couldn't find anything[/quote]

For any vector $u_i$, the notation $u_i=\left<x_i,y_i\right>$ means that the $x$-component of $u$ is $x_i$, and the $y$-component of $u$ is $y_i$.


algebra
polynomial
quadratics
modular arithmetic
Galois Theory
number theory

Just to begin...

A polynomial $x^4+ax^2+b$ is reducable modulo $p$ for all $p$, $a,b\in\mathbb{Z}$. Take now $x^4+1$. It is irreducable in $\mathbb Z$.
Maybe we should extend this idea further.

Indeed, $x^4+1$ is also an example of Hilbert. What is nice about this problem is that for prime n we cannot find such polynomials. There would be so nice to have an explicit example of such polynomial for all composite n, but this cannot be easy.

I proposed the following problem for the first Kolmogorov's Cup. It has some realition to the case $n=2$. I remember vess gave some refferences to the strong results in this area.

Given a quadratic polynomial $f(x)=ax^2+bx+c$ s.t. $\forall\,m\in\mathbb N\ \exists n:\ m\mid f(n)$. Prove that $f$ has rational roots.

No one solved it, though it is quite clear.

We may even restrict to prime values of m. Do you by any chance remember where are the references given by Vess?

My fully elementary solution uses composite $m$.
Hmm, I am not an information bureau... I don't remember those topic.

I am sorry, I do not consider you an information bureau. This is why I said "by any chance".

Is this about Myth's problem? Can't we just take a large prime so that the discriminant of the equation is not a quadratic residue? :?

He said elementary, Grobber! Or you do have an elementary short proof for your assertion?  ;)

Did I do an absolutely stupid mistake, Myth?  :?

I meant that grobber's assertion is quite unclear. I think, not all large primes satisfy it  :?

I meant this (sorry, I admit it's a bit unclear): if the discriminant is not a perfect square (this is what we assume), then there is a prime (I said large because I didn't want it to divide any of the coefficients) so that the discriminant is not a quadratic residue modulo that prime.

Then the equation $ax^2+bx+c$ will have no solution modulo the respective prime.

As Harazi said, this is not elementary, although I'm pretty sure this is what he thought about as well when he said that we could restric our attention to primes.

[quote="harazi"]Indeed, if $n$ divides $ x_n^2-a$ for all $n$ then if for a certain prime $p$ the exponent of p in a is odd, let's say $2k+1$ then take $ n=p^{2k+2}$. Hope I did not make stupid mistakes.[/quote]You solved it for the case $a>0$. There is a possibility that you got $x^2+k^2$. Though it is easy too ($x^2+k^2$ is not divisible by $3k^2$).

[quote="harazi"]Prove that for any composite n there exists an irreducible polynomial (in Z[X]) f of degree n, which is reducible mod p for any prime p.[/quote]

This result is actually due to R. Brandl, ``Integer polynomials that are reducible modulo all primes'', Amer. Math. Monthly 93, (1986), p. 286--288.  More generally, Guralnick, Schacher and Sonn prove in their article ``Irreducible polynomials which are locally reducible everywhere'' ([url]http://www.math.ucla.edu/~mms/poly.pdf[/url]) that for any composite $n$, there exist irreducible $f(x) \in \mathbb{Q}[x]$ of degree $n$ which are reducible over $\mathbb{Q}_p$.

Both proofs are not constructive and use some deep results in Galois theory, including Dedekind's theorem ([url]http://www.math.niu.edu/~beachy/aaol/galois.html[/url]) and Frobenius-Chebotarev Density Theorem
([url]http://www.ias.ac.in/resonance/Dec2003/Dec2003p33-41.htm[/url], [url]http://www.mathworld.wolfram.com/ChebotarevDensityTheorem.html[/url])

Here is an explicit construction, which works only in case where $n$ is composite and not squarefree.  Suppose that $n ={p_1}^{\alpha_1} \cdots {p_s}^{\alpha_s}$ with at least one of the $\alpha_i$'s strictly greater than $1$. Define the degree $n$ polynomial \[P_n(T) = (T^{p_1}-q_{11}) \oplus\cdots\oplus (T^{p_1}-q_{1 \alpha_1}) \oplus\cdots\oplus (T^{p_s}-q_{s1}) \oplus\cdots\oplus (T^{p_s}-q_{s \alpha_s}),\] where the $q_{ij}$ are distinct prime numbers and where the operation $\oplus$ is defined on $\mathbb{Z}[T]$ as follows: if $f,g \in \mathbb{Z}[T]$, then $f \oplus g$ is the polynomial in $\mathbb{Z}[T]$ whose roots are all the $\deg(f)\deg(g)$ sums $\alpha_i + \beta_j$, where $\alpha_i \in \mathbb{C}$ (resp. $\beta_j \in \mathbb{C}$) are the roots of $f$ (resp. of $g$).

This class contains (for $p_1 =2, n=2^{\alpha}$) the so-called [b]Swinnerton-Dyer polynomials[/b] defined as $S_n(T)=\prod{(T\pm \sqrt{q_1}\pm \dots\pm \sqrt{q_\alpha})},$ where $q_1, \dots, q_\alpha$ are
distinct primes.

[b]Claim.[/b] $P_n(T)$ is irreducible in $\mathbb{Z}[T]$ and reducible in $\mathbb{Z}/p\mathbb{Z}[T]$ for each prime $p$.

[b]Examples[/b]. The polynomial $(T^2-2) \oplus (T^2-3) = T^4-10T^2+1$ (which is a Hilbert-type example) and, similarly, the polynomial \[(T^2-2) \oplus (T^2-3) \oplus (T^3-5) =
T^{12}-30T^{10}-20T^9+ 303T^8-910T^6- 2760T^5+2553T^4+8300T^3+8220T^2-4560T -23624\] are both irreducible in $\mathbb{Z}[T]$ and reducible in $\mathbb{Z}/p\mathbb{Z}[T]$ for each prime $p$.

Alin.

Thank you very much, Alin, it is always a pleasure to read your excellent proofs.  :)

This is a proof of the fact that if $n$ is not a perfect square, then there are infinitely many primes $p$ such that $\left(\frac{n}{p}\right)\neq 1$.

Since quadratic symbols are $\pm 1$, we may assume that $n=p_1p_2\cdots p_s$ is square-free. Also, assume that $n$ is odd. Then, by the Dirichlet theorem on primes in arithmetic progressions, there are infinitely many primes $p$, such that \[p\equiv \begin{cases}1\pmod 4\\
1\pmod{p_i},i>1\\
\alpha\pmod{p_1}\end{cases}\]
where $\alpha$ is not a quadratic residue $\pmod{p_1}$. (This is clear since half the residues are non quadratic, if $p_1>2$.)

To see why this can be done, by the Chinese Remainder Theorem, there exists an $N$ such that the above congruences hold, and then simply take the primes $p$ in the arithmetic progression $4n\mathbb{Z}+N$.

Now, by quadratic reciprocity,
\[\genfrac{(}{)}{}{}{p_i}{p}\genfrac{(}{)}{}{}{p}{p_i}=(-1)^{(p_i-1)(p-1)/4}=1,\]
so
\[\genfrac{(}{)}{}{}{n}{p}=\prod \genfrac{(}{)}{}{}{p_i}{p}=\prod \genfrac{(}{)}{}{}{p}{p_i}=\genfrac{(}{)}{}{}{\alpha}{p_1}\prod_{i>1}\genfrac{(}{)}{}{}{1}{p_i}=-1\]

Andrei


ratio geometry rectangle perimeter search

1. Equilateral triangles ABQ;BCR;CDS;DAP are erected outside of the (convex) quadrilateral ABCD. Let X; Y;Z;W be the midpoints of the segments PQ;QR;RS; SP, respectively.
Determine the maximum and minimum values of $ \frac {XZ \plus{} YW}{AC \plus{} BD}$.

2. Two square pieces of paper are made subject to the constraint that the sum of their areas is 2 square units. These pieces of paper are to be put inside a rectangle such that the sides of the sqauares are parallel to the sides of the rectangle. What is the least possible value for the area of a rectangle if we want to be sure that this rectangle will enclose these two pieces of paper?

3. Prove that of all triangles inscribed in a given triangle, the one with least perimeter connects the feet of the given triangle.

You might want to consider a search and or moving these problems to a higher forum because if problem 1 is any indication, the other problems may have also been discussed in more advanced forums.
See Here:[url]http://www.mathlinks.ro/viewtopic.php?p=1236864#1236864[/url]


Cool problem.  You forgot to put the ^ when you're giving the expanded of 3^2, 5^2, 7^2.  You just put 32, 52, 72.

Well Rep123max i wrote it like that just because of the formula of the geometric progression which is S=a1*(q^n-1/q-1) where a1 is the first term pf the progression, q is the factor with each each term is multiplied to get the following term and n is the number of the terms.

Wumbate you're right,it's just a typing mistake.And yes it's a cool problem,i think it was the hardest one in the contest.


If $ a$ and $ b$ are two roots of $ x^4\plus{}x^3\minus{}1\equal{}0$, prove that $ ab$ is a root of $ x^6\plus{}x^4\plus{}x^3\minus{}x^2\minus{}1$.

This has been posted many times before.

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=550693413&t=229173


I was wondering if I could make a thread for challenges at an intermediate level.  As soon as someone has given a nice answer to the last question posted, I will come up with a problem and post it the next day.  I hope this goes on for a long time.

[b]Questions 1:[/b]

Let $x>-1$ and $y=2$.  Prove: $x^2+2x\ge xy$.

EDIT: If this is easy for the majority of you, tell me to increase the level.:)

Masoud Zargar

Sorry, this problem is so easy that now I'm laughing at myself.  Anyway, I plan to change Question #1 to a new question.

[b]Question #1:[/b]

If $x>-1$ and $y>1$, prove that $6x+15x^2+20x^3+15x^4+6x^5+x^6\ge xy$.

Masoud Zargar

um, shouldnt there be some terms of y on the LHS since it seems to me that this isnt true when x=1 and y=300

Sorry, this was teh first time I made up a question, however, teh original question was meant to be somethign around:

If $x>-1$ and $y>1$, prove that $(1+x)^y\ge 1+xy$.  I manipulated the problem, hoever, it was not proposed in teh correct way.  Anyway, I'll ahev a new problem up (and correct this time) in a few hours.

Masoud Zargar

Square both sides and subtract xy to get

(x-y)^2/2 >= 0

which is true by the trivial inequality

I already know how to solve it...anyone want to prove it?

Masoud Zargar

[quote="boxedexe"]Okay, I have another problem.

[b]Problem #2:[/b]
Prove: If $x\ge 0$ and $y\ge 0$, then
\[ \sqrt{xy}\le\dfrac{x+y}{2}  \]
Masoud Zargar[/quote]
$(\sqrt{x}-\sqrt{y})^2\geq0$
$x+y\geq2\sqrt{xy}$
$\dfrac{x+y}{2}\geq\sqrt{xy}$
Proved :D

[quote="K81o7"][quote="chess64"][quote="boxedexe"][b]Question 1:[/b]

Let $x>-1$ and $y=2$.  Prove: $x^2+2x\ge xy$.[/quote]

I'll do it for fun anyway :D

[hide]
If $x=0$, then we have $0 \ge 0$, which is true. If $x\ne 0$, then we can divide both sides by $x$ to get $x+2 \ge y \implies x+2 \ge 2 \implies x \ge 0$. And because $x>-1$, $x \ge 0$ so we are done$. \ \ \ \ \ \ \ \ \blacksquare$
[/hide][/quote]

However, if you divide each side by a negative number, the sign in the middle will flip.  But there's an even simpler way.  Substitute for y.  You get $x^2+2x\ge2x \iff x^2 \ge 0$.  $\mathbb{QED}$[/quote]

Oops, sorry about that :blush: