probability
topology
abstract algebra
geometry
real analysis
Putnam
algorithm

I'm entering my senior undegrad year and have taken: Calc I,II,III, Discrete Math, Linear Algebra, Advanced Linear Algebra, Differential Equations I and II, Probability
and Statistics, Real Analysis I, Abstract Algebra I and II, Complex
Analysis, Topology I, and Differential Geometry. I am planning on
taking Real Analysis II, Special Topics in Algebra (algebras, modules,
etc.), and Topology II next semester (fall).

I was wondering if there are any books or subjects not in there that people had wished they were familiar with looking back on entering graduate school. I'm looking for something to read and work through this summer and I'd like it to be on something that other people think is important to know upon entering grad school.

Thanks for any suggestions.

Functional analysis, unless it was included in Real Analysis.
Representation theory.
Combinatorics.

I don't have any particular books to recommend: look up the subjects on books.google until you see something that you like.

[b]interesting topics[/b]
galois theory (choose a yellow book from your library)
measure theory (halmos)

[b]hardcore topics[/b]
commutative ring theory (matsumura/atiyah&mcdonalld)
homological algebra (weibel)
algebraic topology (hatcher/munkres)
functional analysis (dunford)

:D

(perhaps interesting site: http://www.ocf.berkeley.edu/~abhishek/chicmath.htm )

Just one suggestion to consider: instead of looking for a theory textbook in some area to read, you may want to look for some "problem solving book" to work through (like Polya and Szego "Problems and theorems in analysis" if you like analysis; I have no doubt that something similar has been wtitten in almost any area though I am not ready to come up with titles). The reason I'm bringing this suggestion up is that almost all the graduate students I dealt with had fairly good skills as long as it concerned reading theory but fairly poor problem solving skills. Of course, this may or may not apply to you, you'll have to decide that for yourself. The main question here is whether you've been sufficiently exposed to problems requiring creative thinking during your undergraduate classes, by which I mean the problems that could not be easily solved by just following the pattern discussed in the class a day (or a week) ago. If not, it may really give you trouble during your graduate school years. Anyway, as I said, it is just a suggestion to consider :).

[quote="fedja"] The reason I'm bringing this suggestion up is that almost all the graduate students I dealt with had fairly good skills as long as it concerned reading theory but fairly poor problem solving skills. [/quote]

There are a lot "general courses" at disposal in a bachelor education. Which ones in particular are responsible for it (do you think)? To be honest: at my college you don't need much creativity. Examples:
1) the exam of a "measure/hilbert/mulitvar"-course consisted 75% of routine exercices (33% stokes/green; and somehow some deep result of measure theory was finding out when $f: (0,+\infty) \rightarrow \mathbb{R}: f(x) = \frac{1-e^{-t}}{t^\alpha}$ was integrable  :o )
2) the beauty of abstract algebra was totally lost when our algebra professor asked us to find a nasty Jordan-decomposition and some ugly minimal polynomial).
(Some students who were/are convinced that a group has ideals had a score which was 30% higher than mine...?)

I'm interested at what level someone who hasn't participated problem-solving competitions (like the IMO) or the non-college AoPS section [i]can/must [/i]adjust his attitude at college (assuming he first started "doing" mathematics when he entered college). I'm not yet a graduate student, so I haven't experienced the troubles your students are dealing with. Neither do I know what they are. Ofcourse, mathematics is fun and I strongly believe the problem-solving analysis books mentioned above are useful for any person interested in math. But at college (unlike the 'kids' -- read as: <18 years) students have very little time... Therefore I would like to know your opinion which things to give priority. Let's consider (Big) Rudin or Halmos(his measure theory book) to keep in the analysis area. Do these book not prepare you properly to tackle problems in advanced mathematics? As you probably will know; Rudin's exercies are definitely not routine exercices (neither are Halmos'). Is it a good thing to read such books or are the problem books more fruitful, you think...? (I'm perhaps ignorant) [u]But[/u] I think these books learn you to solve a problem which its context is known (in contrary to a problem with barely any information/hints given) Furthermore I was interested at which level "abstraction" improves your problem solving skills for advanced mathematics. I'm not a neuro-scientist... but 'what if' you don't experience problems concerning abstraction and can deal with it (relative) easily. Training on abstract obliges you to follow strict/logical thinking-patterns to solve a problem (and isn't this a little bit of problem-solving too?) And does this flexibility with abstraction reduce problems in graduate courses? As I implied in my reply above: I worked myself from zero-knowledge to some exotic (although not yet very deep) areas such as functional analysis and algebraic geometry (without any help of professors) in less than 2 years... :) Nonetheless I'm sure I can find a 14 year old kid on the forum who can solve a problem that I will not be able to solve (the problem requires no preliminary knowledge). Most of the time (actually 60% I think) when I study mathematics, it is not related to school... Ofcourse that is because I'm overwhelmed by all those interesting things our professors hide for us. But is this a good baggage for long term results? Or is quantity of secondary importance?

I want to avoid the problems your students experience... that's why I'm asking.

thanks for your time :)

Wow, yeah, I'd like to know some more thoughts on this too. I didn't even think to maybe just sit with good problems for a summer. Maybe work through some sets of old Putnams. Look for other good problem books. I feel like I pick up on theory pretty quickly, yet my problem skills are lacking (there are 14 year olds that could no doubt beat me in USAMO/IMO type problems). But are these skills utilized a sufficient amount in grad school? I've heard that lots of top Putnam scorers tend to not pursue research careers in math due to the different nature of the beast.

Thoughts would be greatly appreciated.

I think that working through a textbook that has many well-chosen hard exercises (e.g. Rudin's or Folland's books on analysis) may be a better preparation for graduate school than Putnam/Olympiad kind of problems.

I should, probably, make two things clear: first, theory books like Rudin that have plenty of exercises are definitely a great reading, no questions about that. Secondly, the book by Polya and Szego I mentioned is by no means just a random collection of unrelated problems to solve; it is quite well structured and you have there many sequences of exercises leading to several quite deep theorems.

My point was just that a lot of undergraduate courses are designed in the following way: a professor does a few theorems/examples in the class and then the students are asked to just mimic these in their homework assignments. In the worst cases the homework consists of purely computational problems that should be solved following some standard algorithms, which would be better left to a computer.

This is not how mathematics is really done. I agree that the ability to absorb other people's ideas and to use them in a similar but slightly different setting is a necessary skill, but it is far from sufficient. It is true that solving exercises in, say, Rudin's book teaches you a lot. Actually, if properly done,  it can teach you everything you need to know about problem solving except one thing: solving problems for which the set of tools to use is not given in advance. And while it is true that most of the problems in mathematical research arise in some context of well-developed theory with some standard set of tools, it is also true that the most interesting questions are those for which one has to go beyond this standard set and to find a completely different approach. That is one thing that solving "non-curricular out of context" problems can teach you quite well. The aim is to develop an ability to search your vast arsenal of knowledge for the right tool without any initial hints as to where to look and to use the standard tools in the non-routine way.

The problem about continuous Hahn-Banach theorem mlok posted recently is a typical example of the kind of problem I'm talking about: despite I fancy that I know both the functional analysis and the convex geometry fairly well, it is completely unclear to me even where to start. And that was a typical situation with many things I have done during my mathematical career: the most "natural" approach coming from the context in which the problem arose was a dead end and I had to look somewhere else to find a solution. This is not how it is with the textbook problems: if a textbook problem goes after chapter 21, you may bet 10 against 1 that it is solvable if you use some tools from chapter 21 and that alone simplifies your life enormously.

Of course, the ability to search in an arsenal is useful only if you have that arsenal and good theory textbooks/courses are irreplaceable here, but I have often seen people who knew an awful lot and still were not able to put $2$ and $2$ together because the first $2$ was oranges and the second $2$ was apples for them. This ability to see apples in oranges and vice versa is something that never comes if you restrict yourself to solving only context-dependent problems.

What problems to choose? They do not need to be olympiad problems at all. The olympiad type problems are not bad to start with (and I still love thinking of them in my free time myself), but you may want to choose more "research oriented" problems. In principle, you may even choose to spend the entire Summer time thinking of some related circle of problems revolving around some open question (I think almost any mathematics professor will be able to supply you with a set like that).  You may be willing to read some (parts of various) textbooks to get some ideas and technical tools that you may find useful but instead of trying to comprehend some complete big picture laid before your eyes as it happens when you go through a typical textbook, you'll try to put some pieces of a jigsaw puzzle together (the extra difficulty will come from the fact that those  pieces will not be given but will have to be found first) and see what picture emerges from such a process.

As to what would prepare you better for the graduate school: solving Putnam problems or reading Rudin, I honestly don't know. I myself was a problem-solving type person who often preferred thinking of various questions my professors gave me (those questions were often completely unrelated to the courses I took from them) to reading extra textbooks and I had no trouble in the graduate school: my thesis was just a set of solutions to one of such circles of questions that my advisor once gave me during my undergraduate years and the only difference with what I did all the time before that was that I had to carefully write my solutions down and call them "theorems".
On the other hand, what was good for me may be bad for you and there is no way to tell now what would become of my mathematical career if I chose to spend more time on books and less on problem solving as an undergraduate. Besides, you should take into account that the regular undergraduate courses in Leningrad were of much higher level than many graduate courses in the USA nowdays (there was no nonsense about "gentle introduction" and "various levels of understanding"; we were just exposed to rigorous mathematics from the very first day of classes), so there was much less need for self-education through reading: the regular curriculum gave you a fairly decent "mathematical baggage" to start with.

Anyway, this is just a personal point of view worth the traditional 2 cents :). Mlok's or Kent's professional experience might lead them to quite opposite conclusions and they both are at least as fine mathematicians as I am, as far as I can tell.


Given three positive numbers a, b, and c, such that a/b = b/c = c/a, find the numerical value of (a + b + c)/(a - b + c).

[hide](a+b+c)/(a+b+c)=a/b, b=c=a, a-b+c=c, and a+b+c=3c, so 3.[/hide]


[quote="math92"]Find 5 consecutive even integers that total 100.[/quote]

[hide]$x+x+2+x+4+x+6+x+8=100$

$5x+20=100$

$5x=80$

$x=16$

$\text{Therefore the numbers are 16, 18, 20, 22, and 24.}$[/hide]

[hide]

5 consecitive integer's sum is 100; so you can start with 5 * 20 = 100.

To make them consecitive numbers, make it 18,19,20,21,22. Their sum is 100
But since they have to be even numbers, you have to subtract 2 more so it is 16,18,20,22,24 [/hide]


A $2^{n}\cdot 2^{n}$ square has a single $1 \cdot 1$ square missing from the top left corner.  Prove that this shape can always be tiled with "curved" triominoes.

[hide="Hint"]
Induction
[/hide]

[hide="Induction"]
[b]Base Case:[/b]
The case of $n=1$ is a single "trinomino"

[b]Inductive step:[/b]
The case of $n=k$ can be tiled as shown below (may take a few moments to load).

[img]http://img.photobucket.com/albums/v194/mukmasterxps/inductivestep.gif[/img]

If a single "trinomino" is used to fill up the space in the middle as shown, then the case of $n=k+1$ results.

[/hide]


algebra
polynomial
number theory proposed
number theory

Let $ n$ be a given positive integer. Show that we can choose numbers $ c_k\in\{\minus{}1,1\}$ ($ i\le k\le n$) such that \[ 0\le\sum_{k\equal{}1}^nc_k\cdot k^2\le4.\]

For a polynomial $ f$ define $ \Delta_c f$ by $ (\Delta_c f)(x) : = f(x + c) - f(x)$. This is a polynomial of degree one less than $ f$ (or the $ 0$ polynomial if $ f$ was constant).
Then consider $ \Delta_4 \Delta_2 \Delta_1 f$, it is $ 0$. Expanding it gives that it is $ (x + 7)^2 - (x + 6)^2 - (x + 5)^2 + (x + 4)^2 - (x + 3)^2 + (x + 2)^2 + (x + 1)^2 - x^2 = 0$. This reduces the original problem to checking those cases with $ n \leq 7$, which can easily be done.
One couls also use $ \Delta_2 \Delta_1 f$ to be $ 4$, easily reducing the problem to $ n \leq 3$, making the checking easier.
This also shows that one can replace the square by arbitrary powers $ k^n$ if one just changes $ 4$ into a big enough number (the biggest value occuring in the first $ 2^{n + 1}$ numbers is the best possible bound, a more general bound is $ n! \cdot 2^{\frac {n(n - 1)}{2}}$ or $ 2^{n^2}$).

See also http://www.mathlinks.ro/viewtopic.php?t=30998 for a related problem.


let $ a,b,c,d>0$ such that $ abcd\equal{}1$. prove that
$ \sum \frac {1}{(1\plus{}a)^2} \geq 1$.

[quote="reza1370"]let $ a,b,c,d > 0$ such that $ abcd \equal{} 1$. prove that
$ \sum \frac {1}{(1 \plus{} a)^2} \geq 1$.[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=16027
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98967


Square $ AIME$ has sides of length $ 10$ units.  Isosceles triangle $ GEM$ has base $ EM$, and the area common to triangle $ GEM$ and square $ AIME$ is $ 80$ square units.  Find the length of the altitude to $ EM$ in $ \triangle GEM$.

[hide] Let GE meet AI at X and let GM meet AI at Y. Clearly, XY=6 since the area of trapezoid XYME is 80. Also, GXY ~ GEM. Let the height of GXY be h.

By the similarity, $ \dfrac{h}{6}\equal{}\dfrac{h\plus{}10}{10}$, we get $ h\equal{}15$. Thus, the height of $ GEM$ is $ h\plus{}10\equal{}\boxed{025}$.[/hide]

I found a pretty interesting way to do this problem, if you draw it out, then you have a trapezoid and two triangles.  We are told the trapezoid has an area of 80 and the whole square has an area of 100.  So each triangle has an area of 10.  This means that the triangles have a base of 2 each, meaning the upper line where the triangle intersects the square has length 6.  6 is 2/5 of 10 (the square's base length), and makes an altitude of 10 (for the trapezoid).  Therefore the altitude of the triangle is 5/2(10), or 25.
Idk how to make diagrams :(

[quote=fidgetboss_4000]Lol this could've been a #1-#5 on the AMC 10[/quote]

what did you contribute to this thread with your bump

[hide=Solution]
We know that if $GEM$ is entirely inside the square, its area must be less than or equal to $\frac{10 \cdot 10}{2}=50$ which is less than $80$.

We know that if $G$ is outside the square such that the intersection between $GEM$ and $AIME$ is a trapezoid, then one base on $AI$ will be $6$ and the other on $EM$ will be $10$.

We know that the altitude from $G$ in the triangle with base $6$ is $\frac35$ of the altitude from $G$ in the triangle with base $10$.

We know that this means that the difference $1-\frac{3}{5}=\frac{2}{5}$ of the altitudes must be $10$.

We calculate the answer to be $10 \cdot \frac{5}{2}=\boxed{025}$
[/hide]

[hide=Solution]Let $\triangle EMG$ intersect $\overline{AI}$ at points $B$ and $C$, with point $B$ being closer to $A$ than $C$, Since we know that the area of trapezoid $EBCM$ is $80$, we can calculate $\overline{BC}$ to be $2(8)-10=6$. Next, we notice that triangles $\triangle GEM$ and $\triangle GBC$ are similar. Therefore, we have the following equation for $x$, the height of $\triangle GBC$.
$$\frac{x+10}{10}=\frac{x}{6}$$
Therefore, $x=15$. Finally the altitude form $G$ to $\overline{EM}$ is $15+10=\boxed{025}$.[/hide]


Do the graphs of $ x^{4} \plus{} y^{4} \equal{} 1$ and $ x^{16} \plus{} y^{16} \equal{} 1$ and so on lead to an equation for a square?

$ x^n \plus{}y^n \equal{} r^n$
$ x\equal{}\lim_{n\to \infty } \, \left(r^n\minus{}y^n\right)^{\frac{1}{n}}\equal{}\begin{cases}
 r & \minus{}r<y<r \\
 0 & \text{Otherwise}
\end{cases}$
Same for $ y$
The equation is clearly a square

you can also make a hipercube

Thanks EulerIntegral, but how did you evaluate that limit?

It's not the square as corner points are missing: $ 1^n \plus{} 1^n\to 2\ne 1$. To show the limit, note that $ 1^n\to 1$, $ x^n\to 0$ if $ |x| < 1$, and $ x^n\to \infty$ if $ |x| > 1$.

The limit as $ x \to 1^+$ and $ n \to \infty$ exist?


[quote="cckek"]$f^{(n)}(x)=Q(x)e^{x^{2}}$ where $Q(x)$ is a polinomal of degree $2n$ an has dominant coefficient $2^{n}a_{n}$. So if $Q(x)$ has real roots (otherwise is positive onentire R) then he has  a maximal one be this $x_{0}$. So if $x>x_{0}$ then  $Q(x)$ has a constant sign wich is $+$ thue its degree even and dominant coefficient positive. So $f^{(n)}(x)>0$ if $x>x_{0}$.[/quote]
But in my problem, $x_{0}$ is independent for $n$.


geometry
perpendicular bisector
Pythagorean Theorem

ABCD is a square.   A circle with radius AB and center A is drawn.  This circle meets the perpendicular bisector of BC at two points, of which X is the closest to C.  Find angle AXC.

This is a partial solution. I haven't proved some things yet, but I think the numbers are right.

Define 2r as the radius.
Define E as the midpoint of AD

<AEX and <DEX are 90

AXE is a 30-60-90 triangles since using the pythagorean theorem we can establish that the sides are r,2r,r*sqrt3

DXE is also a 30-60-90 triangle because of SAS congruence with AXE

<AXE and <DXE are 30

DC=AB, definition of square
XDC is isoceles
<XDC = 90-60=30
<DXC = (180-30)/2=75

<AXC = <AXE+<EXD+<DXC = 30+30+75 = 135

*image created using KSEG: download for free at http://www.mit.edu/~ibaran/kseg.html


modular arithmetic
combinatorics proposed
combinatorics

Very nice result:
Let $X$ is a set with cardinality $2p-1$.
a) Prove that there exist some $Y$ subset of $X$ such that $|Y|=p$ and $\sum_{y\in Y}y=0$ in $Z_{p}.$
b)If $m$ is the number of such sets then $m\equiv 1\pmod{p}.$
Enjoy  :P

a) is the theorem from the title and was posted before.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=57546 for example (scroll down).

b) I assume that $p$ is supposed to be prime here, otherwise take $p=4$ and $X=\{1,1,1,1,1,1,1\}$ (multiset).
For primes:
Then the proof linked above shows that the total number of solutions of the system is a multiple of $p$. Thus there are $a \equiv-1 \mod p$ solutions beside the trivial and always $(p-1)^{p}\equiv-1 \mod p$ of them give the same set, thus in total $\equiv 1 \mod p$ sets.


In latex, $\sqrt{x}$ is implemented as \sqrt{x}.

Do you mean $\sqrt[4]{3}$?  This has a simple answer - I don't think your statement does.

[quote="YooSam"][quote="shadysaysurspammed"]Solve-
$x^4-2(sqroot3)x^2-x+3-(sqroot3)=0$[/quote]

Do you mean: ${x^4-2\sqrt{3}x^2-x+3-\sqrt{3}=0}$ ?[/quote]

Yes

No solution yet from anyone?This aint so tough,its a dumb 10th grade question :lol:

:huh:  :ninja: No one is able to solve this yet...cmon,dont me take the names of the people i expect to solve this.. :rotfl:  :rotfl: just kidding,i expect $everyone$ to solve it.

[hide]Do you mean to solve:  $(x^2-x-\sqrt 3) (x^2+x+1-\sqrt 3) = 0 $ ? [/hide]

This problem's eating my brain. :wallbash_red: Any hints???

if what he gave is correct then you should be able to solve with the quadradic equation.

i would like to know how he came to that equation though.

How did you see how to factor that Gyan? Was there a method?

[quote]   Was there a method?[/quote]
Standard (Believe - is it called Descarte's method)  ,  When there is no cubic term, just assume the factors are $(x^2+kx+a) (x^2-kx+b) $ Compare the terms, if you are lucky you may guess the value of $k$  right away  (Problem setters always make sure that this  value is someting nice :)  - if not you may have to solve   it :). Once you know $k$, rest is easy.

[hide]
The mutliplying the two factors would give:
$x^4 + (a+b-k^2) x^2 -k(a-b) x + ab $ so  comparing each coefficent we get:
$(a+b)= k^2-2\sqrt 3$
$(a-b) = 1/k$
$ab = 3 - \sqrt 3 $
You are lucky enough to see that substituting these in:  $(a+b)^2-(a-b)^2=4ab$ gives $k^2=1$  etc...[/hide]

Hope that helps.

ok i understand how if factors to what you got i just dont get what you showed in the hidden content.

Check the hidden content again (above) and see if it helps-
Also FWIW you can also write it as $ (x^2-\sqrt 3)^2 - x - \sqrt 3 $
Now if you  can see to add $ x^2 $  and subtract itl   [hide] $ (x^2-\sqrt 3)^2 - x^2)+(x^2 - x - \sqrt 3) $
$ =(x^2+x-\sqrt 3)(x^2-x-\sqrt 3)+ (x^2-x-\sqrt 3) = (x^2-x-\sqrt 3)(x^2+x+\sqrt3+1) $ etc[/hide]

well since there is no x^3 term, we know that the sum of the roots is 0 (using newton's sums), so that is how we can break it down like gyan did (generally we know that the coefficient before the last term of polynomial is the opposite of the sum of the roots of the equation)


AMC
AIME
summer program
MathZoom
MATHCOUNTS

Either A or B or both...
Let me know. :)

138 on 10A and same on 10B, though I definitely should have got a perfect score on 10B.

Hoping for an 8+ on AIME :)

Argh I don't even want to know my score.  I only took 10B and I know I got less than 120 so yeah.   :(

They still encourage you to take the test so they can place you into groups more easily.

Also, I think you get grader feedback, which is always good :)

[quote="Smartguy"]If we made the AIME, then we are automatically qualified for mathzoom, correct?

I made it in.[/quote]
Right; qualifying for AIME is a good indicator of your skills, so we don't need to see your assessment test.  I'll wait for your application form then.  :D

97.5 on the AMC 10A (fail) :(
Made it into state Mathcounts at least
I'm aiming to get into Nationals next year. (probably not going to)

[quote="AIME15"]Jerry, California is beaster than you think :(

As in, our region definitely does not show the talent in CA.[/quote]

I know, thats why i said I'm probably not going to :(


[quote="probability1.01"][hide="Hint"] use that Sophie something or other identity that darktreb posted in some other thread somewhere[/hide] (probably)[/quote]
There may be an alternate solution, but mine was relatively brief using no special identities.

[quote="Magnara"][quote="diophantus"]Find all integer solutions and prove you have found all such solutions to the equation:
$x-y^4=4$ where x is prime.

[b]Please hide your solutions so that others may answer.[/b][/quote]

[hide]Note that we can assume that y is positive with no loss in generality.  $x=y^4+4=y^4+4y^2+4-4y^2=(y^2+2)^2-4y^2=(y^2-2y+2)(y^2+2y+2)$
Because y is positive, we have $y^2-2y+2<y^2+2y+2$.  Therefore, $y^2-2y+2$ must be equal to 1, which implies that $y^2-2y+1=0$, so that the only solution is $y=1$.  Testing this out, we get that the only possible answer is (5,1).[/hide][/quote]
Really close, but are you sure that is all of the solutions?
[hide="Hint"]Who said y was positive?[/hide]

[hide]
The other must be (5, -1)

Good solution Magnara
[/hide]

Oh, whatever.  I found the right value of y^4.  :)


Wait and you'll see the impresions of most people on the site :D I hope it will be better, but you never know ..  :?  :)

[quote="Dave Vather"][quote="wisemaniac"]Do you expect Mexico to be a good host (something like usa 2001)?[/quote]
Of Course!!!
I´m sure that many Mexicans might think that the question is an ofense[/quote]I don't think wisemaniac ment any offence. I think he is a contestant of IMO 2001 and he was impressed with the Organization of olympiad then. He was curious as if all IMOs are that nice. The answer is yes :) At least all 4 I have seen up to now :D

Thank you Valentin, that's very nice from you to try to avoid misunderstandings :) (you are so nice that's sometimes i can't believe you are the admin!!).To be more clear , No one doubts that mexico has a lot to show off, i asked this question just to know how people imagine imo 2005 and then compare to how it's going to be . So it's far from being an offense , and for those who think it is , i hope i find out i shouldn't have asked it after the event  :) .About me , imo2005 is going to be my first  ,but i heard a lot about imo 2001 from some contestants that's why i gave it as an example .

[quote="wisemaniac"]You are so nice that's sometimes i can't believe you are the admin!![/quote]Why? :P Is the Admin supposed to be a bad guy?  :evilgrin:  :lol:

[quote="wisemaniac"]No one doubts that mexico has a lot to show off, i asked this question just to know how people imagine imo 2005 and then compare to how it's going to be . So it's far from being an offense .[/quote]
No problem, i´m o.k. whit the question. I´m also sure that it´ll be a great IMO since the organizing comitee has been preparing everithing for a long while. Also i have to say that i´ll be very dissapointed if this is not the best IMO in some time, because if i´m not going out of my country, at least i hope that it will be a great IMO

well if imo 05 can beat imo 04, that's great!
I heard from my friends that in imo 04 it's so good to live in a very nice hotel, everything's very grand...

imo 2004 was a bit disorganized but it was great and a lot pof partying and stuff like that... maybe the disorganization was good..

[quote="siuhochung"]... imo 04 it's so good to live in a very nice hotel, everything's very grand...[/quote]
I agree. Only flaw I noticed was the long line for meals(Some people couldn't get breakfast because of long line up).
Except for that, it was perfect. :)

but meals were horrible hah no offence to the greeks its just different.. :blush: i always ate out of the hotel

but we had a great time in greece

The hotel in Greece was great! Does anyone know if we are going to live in a hotel  :clap: or some university residence  :mad: ?

There was one other problem in Greece - the lack of communication between the leaders and the students (after the exam of course).

The meals were somewhat repetitive, but were still quite good, as long as you know when they start and line up at the right time.

Yufei, I belive all of us will be staying in Hotels in Merida (at least from 14th on :D), but in more hotels (there are like 4-5 hotels for the students, leaders, observers and coordinators).

in Greece we had a great time but if we were in another city maybe we could have been better cuz Athens is like a jungle! I was thinking of a place near the beach!! :)  :oops:


calculus
integration
trigonometry
calculus computations

the integral of (2x+4)/(x^2+4x+5) dx. Maybe using partial fraction but I cant factor out the denominator here. Some suggestions??

The denominator is an irreducible (over $\mathbb{R}$) quadratic. Complete the square. What do you know about $\int \frac1{t^{2}+1}\,dt$ and $\frac{t}{t^{2}+1}\,dt$?

I think that $(x^{2}+4x+5)' = 2x+4$ must bee the idea to find the integral !
Think at $ln(x^{2}+4x+5)$, derive it... :wink:

thanks guys, I really forget about that. I got one more problem which I can solve but it's really long, the problem is:

the integral of (cos x)^2 * (sin x)^5 dx.

I solve this by changing (cos x)^2 = 1- (sin x)^2 and then I got (sin x)^5 - (sin x)^7 but counting the integral of this is a pain in the ass. I can do this but it takes almost a page to calculate this. Is there any easier way??

How about going the other way? An odd power of $\sin$ makes a cosine substitution nice.

what do you mean going the other way?? use (sin x)^2 = 1- (cos x)^2?? If I do that then I get (cos x ^2 - cos x ^4) sin x ^3


email
parameterization
AMC 10
AMC 10 A
AMC 12 A
AMC

Did that already get implemented?

I know that I got 20 correct and 1 wrong 4 blank on the AMC 12A but I got my score reported back as 126. That only makes sense if leaving a question blank = 1.5 points.

But the front of the test booklet says blank questions are worth 2.5...what's going on?

I haven't received my own score yet, since my school is on winter break right now, but I've heard of other people getting scores reported back to them that also seem to indicate that the blanks were worth 1.5 points, so your case doesn't seem to be an isolated one.

wow, your school has winter break now?? that's cool

On the web and on the front of the packet it says 2.5, maybe you mischecked your answers and really got 19 correct?

No, because that would give me 124. 19 * 6 + 4 * 2.5 = 124. Also, the other people at my school faced a similar decrease in score. 115->105 for example.

Hmm, interesting cuz I [i]think[/i] the only way to get 126 with 2.5's is 21 correct and 4 wrong.  And I doubt you accidentally filled the 4 in that you thought you left blank  :P   So. . . well was this an official score?  I didn't get anything, and stats aren't up on the website  :huh:

Officially, I remember that they said it would be implemented in 2007, maybe they were testing it on you guys  :P .

I think they made a mistake, because I've heard of other schools getting their scores twice, the second one is corrected.

Yup. My math teacher got it through email from the AMC people.

Thank god, I had a horrible day on the amc and I might not have qualified...that would have been over 5 hrs a day wasted.

Last week, the program that does the scoring and reporting of scores had a parameter of 1.5 entered where it should have been 2.5.  That parameter change occurred as we were testing the scoring system for 2007.  On Friday, Feb 17 some early email reports went out with the incorrect parameter scoring, but the mistake was discovered quickly and corrected.  Those schools receiving an incorrect report then received a second corrected reportlater that day.  The hard-copy reports currently being postal mailed have the correct scoring.  Please note that in the Teachers' Manual, on page 6, in Section V it says "Results are not official until you recieve the hard copy of your report..."

Again, I emphasize, reports emailed since approximately late Friday afternoon contain the correct scoring.  The 2006 scoring is 6 points for a correct answer, 2.5 points for a blank answer 0 points for an incorrect answer, just exactly what  item 4 on the cover of the 2006 AMC 10 A, 2006 AMC 12 A contest booklets said.

Steve Dunbar
MAA Director, American Mathematics Competitions


geometry
3D geometry
prism
conics
ellipse
geometry unsolved

Assume that $T$ is a fixed circle, and $A$ and $B$ are two fixed points such that the line $AB$ does not intersect the circle $T$.
For any point $P$ on the circle $T$, denote by $P^{\prime}$ the point of intersection of the line $AP$ with the circle $T$ different from $P$, and denote by $f\left(P\right)$ the point of intersection of the line $BP^{\prime}$ with the circle $T$ different from $P^{\prime}$. Now define a sequence $P_{0},P_{1},P_{2},....$ of points such that $P_{0}=P$ and $P_{n+1}=f\left(P_{n}\right)$ for every $n\geq 0$.

Let $k$ be some positive integer.

Show that if there exists one point $P$ on the circle $k$ such that $P_{0}=P_{k}$, then we have $P_{0}=P_{k}$ for [b]any[/b] point $P$ on the circle $k$.

I don't have a solution, but I did observe a few things while playing with the drawing in Euklides:

Assume we're trying to find stuff about the variable line $MN$, where $M=AP\cap T,\ N=BP\cap T$ when $P$ goes around the circle. I was trying to show that the line $MN$ envelops a conic. Then we could use Poncelet's porism to solve the problem.

I don't know what posessed me to draw $S=AN\cap BM, R=AS\cap MN$. Then I traced the locus of $R$. It seems to be an ellipse (sometimes a circle) which is always tangent to $MN$, so $MN$ does envelop a conic, but unfortunately I have no proof for any of this :(.

I'm trying.. I'm sure you have a nice solution sam, (unlike this desperate attempt of mine :)). Do you happen to know where it was proposed (TST, preparation for IMO, NMO)?

in my material is just mentioned "fourth round ,Bucharest,17 may,1997"

You know, I'm pretty sure that this works even if $T$ is replaced by a conic.

It is easy to show that $f(z)$ is linear-fractional mapping on $T\subset \mathbb{C}$, and, in general, on $\mathbb{C}$, moreover some power of $f$ has fixed point $z_0\in T$.
We know that if non-degenerate linear-fractional mapping has three fixed points then it is identical mapping. Where do we find remaining two points in addition to $z_0$? Actually linear-fractional mapping $f(z)$ has two distinct (I believe this) fixed points and it is geometricaly obvious they don't lie on T.
Thus some power of $f$ has three fixed points ==> $f^k(z)\equiv z$.

I didn't check, but it is possible that f(z) is linear-fractional mapping for arbitrary conic.

Here's what I did after struggling for a while:

Apply a projective transformation which takes line $AB$ to the infinity line. Then the circle is turned into another circle or an ellipse because it's a conic with no improper points (because the initial circle didn't intersect $AB$). Now project onto a plane s.t. the ellipse turns into a circle. This transformation is affine, so it preserves parallelism, so $Pf(P)$ will go around the circle and it will have constant length because $\angle PP'f(P)$ is constant. This means that $Pf(P)$ envelops a circle concentric to the circle on which $P$ moves. We prove the assertion in this simple configuration, so it must also hold in the initial configuration.

Sam-n! What can you say? Do you have another solution?

Myth, I have to admit I didn't quite understand your proof: if $f^k(z)$ has three fixed points does it also follow that it's identical? I know it's true for $f$, but is it also true for  power of $f$?

Sorry if I'm asking stupid questions :).. it wouldn't be the first time.

a power of a linear fractional mapping is again a linear fractional mapping(where power doesn't mean we take the values of f to the k-th power, but we use f k times). so all that you know about f, you know about all powers of f :)

Power $f^k$ is a superposition $f\circ f...\circ f$, and superposition of  linear-fractional mappings is again linear-fractional mapping.

In my turn, I have to admit that I don't understand your solution. Generally, I am opponent of such mediate methods which so far from school education (for example, in Russia).

I know what he meant by "power", the only problem being that I wasn't really thinking when I wrote my last post :D. It seems to happen to me a lot lately..

If $k\ge 3$, then $P_{0},P_{1},P_{2}$ are three different fixed points of the projective transformation from the circle to itself mapping $X$ to $f^{(k)}(X)$, so this projective transformation is the identity, which is exactly what we want to show.

Clearly, $k\ge 2$ (we cannot have $P=f(P)$ for any point $P$, because $AB$ does not meet the circle), so the only case left is $k=2$. It's easy to see that in this case $A$ lies on the polar of $B$ (and vice-versa), and it's easy to draw the conclusion from here.


logarithms
AMC
USA(J)MO
USAMO
geometry
3D geometry
sphere

the number of primes of the form $3n+1$ is almost equal $3n+2$ under n where n is large..is it true? and the generalization?

[hide="Well..."] The number of primes of the form $ak+d < M$ where $gcd(a, d) = 1$ is asymptotically given by

$\frac{M}{\varphi(a) \ln M}$ [/hide]

If you're saying $\frac{3n+2}{n}= 3+\frac{2}{n}\to 3$ as n goes bigger, not!
[EDIT]:t0 is faster!  :lol:

ok. because I dont want to open a new thread so concerning this ques (10th USAMO 1981): A convex polygon has n sides. Each vertex is joined to a point P not in the same plane. If A, B, C are adjacent vertices of the polygon take the angle between the planes PBA and PBC. The sum of the n such angles equals the sum of the n angles subtended at P by the sides of the polygon (such as the angle APB). Show that n = 3. in the solution at kalva I don't understand this [hide]he sum of the n angles between the planes is at least (n - 2) 180o. If we take a sphere center P. Then the lines PA, PB intersect it at A", B", ... which form a spherical polygon. The angles of this polygon are the angles between the planes.[/hide] thanks.


[quote="Ulanbek_Kyzylorda KTL"]can you post solutions please[/quote]

Why are you so on hurry ?

Since you posted in this forum, it means you already have a solution, is not it ?

no just i tried tried but i couldn't get solution of last questions and i am on hurry to know solutions of these questions
if you have solution to these questions can you post them, please?

[quote="Ulanbek_Kyzylorda KTL"]no just i tried tried but i couldn't get solution of last questions and i am on hurry to know solutions of these questions [/quote]

Just a reminder (just read the front page of Olympiad forums) :

[u]NT unsolved problems [/u]: "Problems you couldn't solve and to which you know that there is a solution (i.e. a problem from a contest, etc.) but you don't know it."

[u]NT Proposed & own Problems [/u]: "Problems [u][b]that you have already solved [/b][/u]and you are interested in second opinions or solutions."

[u]NT Open Questions [/u]: "An open question is a question that has no known solution up to this moment, and it is not known wheter the problem has one or not."

Thanks for following the rules of these forum.

[quote="Ulanbek_Kyzylorda KTL"]Prove that there exists natural number (n>1) so that product of some n consecutive natural numbers are equal to product of some
n+100 consecutive natural numbers.[/quote]

Choose for example $ n\equal{}5050$ :

$ \sum_{k\equal{}102}^{5151}k\equal{}\sum_{k\equal{}1}^{5150}k$

i didn't understand and in question asked product not sum. has anybody solution? i need it and it's necessary for me

please post your solution  i need it quickly

i've never seen anyone so desperate for a solution. are you cheating on a homework or take-home exam?

use a similar idea to the one that pco gave you for sums, for products instead. it's easy, do it yourself  :wink:


Sum of the following terms to n terms
$ sin^2\theta\plus{}sin^2\left(\theta\plus{}\frac{2\pi}{n}\right)\plus{}sin^2\left(\theta\plus{}\frac{4\pi}{n}\right)\plus{}........\plus{}sin^2\left(\theta\plus{}\frac{2(n\minus{}1)\pi}{n}\right)$

No one trying this???? I know it is easy for you but not for me. :(  :(

Please give your solution.

[quote="Euclid_great"]Sum of the following terms to n terms
$ sin^2\theta + sin^2\left(\theta + \frac {2\pi}{n}\right) + sin^2\left(\theta + \frac {4\pi}{n}\right) + ........ + sin^2\left(\theta + \frac {2(n - 1)\pi}{n}\right)$[/quote]

[hide]
[there should be an easier way] Let $ S$ be the sum in question, and by the double-angle formulas, we have $ S =$ \[ \sum_{k = 0}^{n - 1} \sin ^2 \left(\theta + \frac {2k\pi}{n}\right) = \sum_{k = 0}^{n - 1} \frac {1 - \cos \left(2\theta + \frac {4k\pi}{n}\right)}{2} = \frac {1}{2}\left[n - \sum_{k = 0}^{n - 1} \cos \left(2\theta + \frac {4k\pi}{n}\right)\right]
\]Let $ T = \sum_{k = 0}^{n - 1} \cos \left(2\theta + \frac {4k\pi}{n}\right)$; then\begin{align*}T\sin\left(\frac {2\pi}{n}\right) & = \sum_{k = 0}^{n - 1} \cos \left(2\theta + \frac {4k\pi}{n}\right)\sin \left(\frac {2\pi}{n}\right) \\
& = \sum_{k = 0}^{n - 1} \sin\left(2\theta + \frac {4k\pi + 2\pi}{n}\right) - \sin\left(2\theta + \frac {4k\pi - 2\pi}{n}\right).\end{align*}This summation telescopes to just $ - \sin\left(2\theta - \frac {2\pi}{n}\right) + \sin\left(2\theta + \frac {4n\pi - 2\pi}{n}\right) = 0$, and so $ T\sin\left(\frac {2\pi}{n}\right) = 0$. For $ n \neq 1,2$, then $ T = 0$, and we get $ S = \boxed{n/2}$. (for $ n = 1,2$ we get $ S = n\sin^2 \theta$)
[/hide]

Please say whether I am correct or not.

$ \frac{1}{2}(1\minus{}\cos2\theta)\plus{}\frac{1}{2}\left(1\minus{}\cos\left(2\theta\plus{}\frac{4\pi}{n}\right)\right)\plus{}\frac{1}{2}\left(1\minus{}\cos\left(2\theta\plus{}\frac{8\pi}{n}\right)\right)\plus{}\dots\plus{}\frac{1}{2}\left(1\minus{}\cos\left(2\theta\plus{}\frac{4(n\minus{}1)\pi}{n}\right)\right)$

$ \equal{}\frac{n}{2}\plus{}\frac{\sin\frac{4n\pi}{2n}}{\sin\frac{2\pi}{n}}.\cos\left(4\theta\plus{}(n\minus{}1)\frac{4\pi}{n}\right)$

$ \equal{}\frac{n}{2}$


geometry
geometric transformation
reflection
geometry unsolved

$A,B,C,D,E,F$ are on the circle $O$ counterclockwise. $l$ is tangent to $O$ at $E$ is parallel with $AC$, and let $DE>EF$. Let the intersections of $BC,BD,CF,DF$ and $l$ are $P,Q,R,S$, prove that $PQ=RS$ iff $QE=ER$

I think there is somethink wrong in the problem.Because, we dont use $A$ in the problem. :oops:

Since $AC \parallel l$and $OE \perp l,$ A is the reflection of C in the line EO. Let B', D' be reflections of B, D in EO. Assuming EQ = ER, B'D' meets the tangent l at R. Let DF, B'A meet at a point X.

$\angle XFR = \angle CFD = \angle CBD = \angle AB'D' \equiv \angle XB'R$

so that quadrilateral $XB'FR$ is cyclic. As $AC \parallel l \equiv QR,$

$\angle QRF \equiv \angle QRC = \angle RCA \equiv \angle FCA = \angle FB'A \equiv \angle FB'X$

It follows that

$\angle QRF+\angle FRX = \angle FB'X+\angle FRX = 180^\circ$

and the points Q, R, X are collinear, $X \in l \equiv QR.$ Since DF meets the tangent $l \equiv QR$ at S, the points $X \equiv S$ are identical, DF, BA' meet at $S \in l.$ Since B'A is a reflection of BC in OE, EP = ES. Combined with EQ = ER, PQ = RS. And the other way around.

[quote="yetti"] And the other way around.[/quote]

I can't see how to do this. It's not similar to the first case...


Do your teachers care if you use bad language in school?
Well, mine don't actually notice, except for my science teacher's assistant, who just goes "I heard that" or "watch your language" and doesn't mention it again.

I have/had some teachers who use "bad language" lol

my health teacher used a few... that class laughed when he did
the teachers usually don't hear

Oh yeah, my health teacher screams bad words at us in order to "test" our maturity.
So basically, if we don't laugh while he's saying the bad words, then we're mature.

My english teacher says "crap" and "damn" and etc. lighter words very frequently, and she said "s***" before.

PE teacher doesn't care at all; he even laughed when some guy made up a funny combination of two words. Most others notice but don't do anything, sometimes a "I heard that."

One of my classmates in Spanish is always either late or forgets to bring his books, binder, etc., and the teacher gets SO mad at him, he just waits for the kid to say something bad and then yells at him for like 5 minutes. :rotfl: And if the kid makes some smart remark, it gets really nasty. :stretcher:

Either the teachers don't hear, or they don't really care. But we try not to make a habit of cussing in front of the teacher  :P .

i once said, "oh shoot" and my woodshop teacher looks in my way and says "what did you say?" and i said, "i said oh shoot".

-_-;;

i've heard shut up a quite a few times.
i've heard crap a lot...
i've heard h-e-double-hockey-sticks a few times... sometimes referring to the place.
i've heard the d*** but it was just quoting.

and whenever we read about infrastructure out loud in history, and we read dam, our teacher tells us to watch our language...

once i said pissed and my teacher says to watch my language...

I can say anything I want to my teacher. :rotfl:

I am homeschooled and self teaching. I love posting in these topics. :rotfl:

bad words is school?  where did you learn english? :P

Most of my teachers use "crap" when they're angry. My English teacher likes to use cuss words, but isn't allowed to (she says crud instead of crap). This is why her favorite day of the year is when she reads the first chapter of [i]Day No Pigs Would Die[/i] to us (contains several cuss words, but since it's in a literature book, she's allowed to say them).

If you use words and phrases like crap then the teachers don't care much, and the nicer ones don't even care if you say things like, "OH EFF." But if you openly cuss randomly using as many cuss words as possible... well, just don't do it.

lol my teachers really dont care and neither does the principal. For instance on thursday I said D**n it and he heard me and all he did was smile. Another time I said O Sh** because I dropped all my stuff and the teacher just said she would help me pick up my stuff.

teachers generally don't care if you really don't have malicious intent when you use profanity

exceptions suck

[quote="1=2"][quote="alanchou"]Where did you learn english? :P[/quote]

In America.

I am self-teaching, so I can say whatever I want to myself.[/quote]

I wasn't talking to you...

my teachers pretend they didn't hear me correctly :|

I think high school teachers don't really care, as long as they aren't directed at them. I knew this kid that got suspended for cussing out his teacher.

I go to a very small school, so all of the teachers know me very well... I don't make a habit of cursing, but my friends all do it when the teachers aren't around. Cursing=Bad. :ninja:

I don't swear very often; I only do it when I get angry.  One of my pet peeves is hearing kids walking across the hallway talking, and every other word is a swear.

In my school, students don't swear at teachers. Occasionally, a person might secretly flick them off...
The only teachers I've heard swearing are the music teachers when they are pissed off.


Does anyone know the official site of BMO 2009?

I don't know if there is such a page at the moment. Website of Mathematical Society of Serbia (dms.org.yu) certainly doesn't mention it.

This is the site:
 http://www.dms.org.yu/bmo/

I think the Macedonian team (judging from results of Macedonian Mathematical Olympiad) is:
Bodan Arsovski
Filip Talimdzioski
Predrag Gruevski
Bojan Joveski
Stefan Stojcevski
Stefan Lozanovski (after additional qualification round).

I also think the Serbian team (judging from results of Serbian Mathematical Olympiad) is:
Teodor Von Burg
Luka Milicevic
Dusan Milijancevic
Mihajlo Cekic
Vukasin Stojisavljevic
Stefan Stojanovic.

Seems a lot of people called Stefan qualified  :lol:.

The members of BiH team:
Jelena Radović
Vlado Uljarević
(first reserve(iliti prva rezerva :) ): Stefan Tanović)

The rest of team will be determined tomorow

Bice na sajtu, ali i mene zanimaju  :D

Komentari na zadatke? Meni se cine previse klasicni (mislim Diofantove jednacine, funkcije se provlace stalno kroz BMO)

Prvo, generičko grubo upozorenje: [color=red]Ako si član hrvatske MEMO ili IMO ekipe, jel ti bar pet puta rečeno da ne čitaš ništa o BMO do sljedećeg vikenda? Jel je? Je. Dakle, izvoli ne čitati dalje i idi rješavati nešto drugo.[/color]

Drugo, kolegica Floyd (ili Pink, ako ju hoćete zvati imenom) zabranila mi je da do sutra spominjem potencijalne pragove, rezultate i slično. Stoga, žalite se njoj što vam ništa nisam napisao... ona je zla, nisam ja, ja bih vam sve rekao. :D

Treće, komentari na zadatke - prvi zadatak je relativno lagan, eventualno onaj zadnji korak i priča od djeljivosti sa 7, ali i to je ideja u granicama normale, bar po mojem mišljenju. Ostale zadatke nisam pokušavao rješavati, drugi se čini relativno razuman (po dva rješenja na Mathlinksu koja su u mojim granicama shvaćanja), treći izgleda ružno (onaj drugi uvjet pogotovo), a četvrti sam vidio i shvatio rješenje i ne izgleda čak ni toliko težak... ako izuzmemo da se ja toga nikada ne bih sjetio. :D Ali lako nama tu pričati, na natjecanju je sve puno teže nego tu gdje vidiš rješenja, imaš pristup literaturi i neograničeno vremena - mislim da je 10 bodova (iliti jedan zadatak) sasvim dobar rezultat... ja sam pesimist, pa od sebe ne bih očekivao baš puno više (niti malo više :D), da sam na natjecanju.

Objavljeni su nezvanicni rezultati 3 zadatka (bez geometrije)
[url=http://www.dms.org.yu/bmo/index.php?action=results]Rezultati[/url]
[hide="Srbija"]
Teodor von Burg 10+9+6
Luka Milićević 10+1+8
Dušan Milijančević 10+2+2
Vukašin Stojisavljević 10+4+0
Mihajlo Cekić 10+0+0
Stefan Stojanović 10+1+0
[/hide]
[hide="BiH"]
Beširević Admir 10+6+0
Bevrnja Alija 1+0+0
Radović Jelena 10+0+1
Saković Zlatan 1+0+0
Uljarević Vlado 1+0+0
Šarčević Franjo 1+0+0
[/hide]
[hide="Makedonija"]
Bodan Arsovski 7+10+0
Filip Talimdžioski 10+0+0
Predrag Gruevski 6+1+1
Bojan Joveski 9+1+1
Stefan Stojčevski 2+0+0
Stefan Lozanovski 5+0+1
[/hide]
[hide="Crna Gora"]
Nikola Milinković 0+0+1
Aleksandra Gogić 0+0+0
Rastko Pajković  0+1+0
Tanja Ivošević 1+0+0
Radovan Krtolica 0+1+0
Nevena Crnogorac 0+0+0
[/hide]

Nije baš točno da je u svakom slučaju fali geometrija - kod Srbije i Makedonije da, ali kod BiH fali 3. zadatak, a kod Crne Gore, najbitnije, 1.

Evo i rezultati [hide="Srbije B"]
Aleksandar Vasiljković 3+3
Igor Spasojević	0+0
Filip Živanović 1+0
Radomir Đoković 1+0
Stevan Gajović 3+1
Vukan Levajac 0+0
[/hide] (fale 1. i 2. zadatak).

Ali, koliko znam, još od jučer se znaju i potpuni(ji) rezultati, pa je sve ovo malo zastarjelo.

[quote="mornik"]Nije baš točno da je u svakom slučaju fali geometrija - kod Srbije i Makedonije da, ali kod BiH fali 3. zadatak, a kod Crne Gore, najbitnije, 1.
[/quote]

Hvala na ispravci!

Sad će ispasti da tersam i kritikujem preko svake mjere, ali zadatak 1 je bio poklon... (možda sam previše subjektivan jer je NT moja omiljena oblast matematike). Rezultati mi se čine prilčno očekivanim, mada je moglo bolje...

Čestitke onima koji su zadovoljni urađenim  :lol:

Ne mogu reći da se u potpunosti slažem sa uvaženim kolegom Šiljkom.  :P  Da je prvi zadatak bio lagan, jeste, slažem se. Da su manje-više svi koraci prirodni, jesu. Ali da je bio poklon (i ovo govorim također sa stajališta čovjeka kojem je teorija brojeva najdraža) - nije. Pogledaj rezultate BiH - osim Admira i Pink nitko nije dobio više od 1 bod. Ne znam po kojem kriteriju inače jaka Rumunjska bira ekipu (valjda im ovo nije prva ekipa, što li...), ali više od 50% njihovih učenika nije do kraja riješilo 1.

Ovaj zadatak jednostavno nije toliko lagan da ga [b]svi[/b] riješe bez problema za pola sata i onda sljedeća četiri sata rješavaju ostale (što bi bila definicija poklona). A i kažem ponovno, kad sjediš u učionici i imaš četiri i pol sata da nešto napišeš i moliš se da bar riješiš nešto sitno jer ćeš biti deprimiran ako jedini u ekipi budeš imao 0 bodova, a svi ostali će sve riješiti (ili ti se tako čini u tom trenutku)... možda ti u tom trenutku prvi zadatak, koliko god objektivno bio blesav, i nije baš lagan. U svakom slučaju, "makni" 1. zadatak pa pogledaj (možda bolje kad dođu finalni rezultati) prosječan broj bodova... ne bi baš bio izvanredan, pogotovo za "slabije" ekipe.

Također, čestitke svima koji su zadovoljni, ali pričekajmo još konačne rezultate.

I ja bih ipak rekao da je 1. zadatak poklon (najvise zbog toga sto je i meni TB jaca oblast), dok se cak i 4. zadatak moze smatrati delimicnim poklonom. Eto, mnogo ideja koje se ponavljaju su prosle kroz BMO u vidu razlicitih funkcionalnih jednacina i nejednacina, ali ipak ostaje tezi zadatak...

Ali, naravno, mornik je potpuno u pravu sa cinjenicom da nama nije isto kao ucenicima u ucionici...

Evo novosti, Srbija je prva na BMO (sudeci po pricama sa srpskih foruma)!!!

Wow, čestitam ako je točno! :first: U konkurenciji Bugarske, Rumunjske, Turske...

Rezultati Srbije:
Teodor Von Burg 35-zlato
Luka Milicevic 29-zlato
Dusan Milijancevic 25-zlato
Mihajlo Cekic 20-srebro
Vukasin Stojisavljevic 14-bronza
Stefan Stojanovic 11-bronza

:first: :first: :first:

Za ostale drzave jos ne znam

Čestitke  :D
Kada će se znati za ostale države ? :D

Ne znam za ostale, probacu da stavim ovde sto pre saznam (ali ne znam koliko svi prate rezultate sad kad znaju svoje medalje :D )

Čestitam!  :D (A iz BiH ima valjda sigurno dvije medalje, iz Makedonije valjda 2-3, sudeći po parcijalnim rezultatima...)

Bugi, znaš li možda koji su bili pragovi za broncu (10,11?), srebro i zlato?

Izašli zvanični  :D
Pragovi su :
Bronza - 6
Srebro - 15
Zlato - 23

:winner_first: Teodor von Burg, Luka Milićević, Dušan Milijančević
:winner_second: Mihajlo Cekić, Bodan Arsovski, Aleksandar Vasiljković, Stevan Gajović, Admir Beširević, Filip Živanović,
:winner_third: Vukašin Stojisavljavić, Filip Talimdžioski, Jelena Radović, Bojan Joveski, Stefan Stojanović, Igor Spasojević, Predrag Gruevski, Radomir Đoković

Čestitke ! :D

:coolspeak:, :clap2:!  (:winner_first:,  :winner_second:,  :winner_third:!)

[size=75]U prijevodu: Cool, čestitke svima! (Posebno zlatnim, srebrnim i brončanima!).[/size]

[size=59]By the way, što ne znam napisati u smileyima, dosta nizak prag za broncu - čak 88 od 115 ljudi (76.5%) je dobilo barem broncu.[/size]

[size=59]Ne da mi da bude više od 5 "posebnih" smileya, morah smanjiti poruku. :( [/size]

Moderator se vratio na normalnu internet konekciju pa može napisati duži osvrt na BMO...

Prvo, složit ću se sa uvaženim kolegom Ornikom u njegovom neslaganju s mojom malenkošću: naime, ni ja ne mislim 100% ono što sam napisao  :lol: Igrom slučaja, čim sam vidio zadatak i počeo svoj congruence-bashing metod došao sam relativno brzo do kraja, što me navelo na zaključak da je zadatak poprilično lagan. Međutim, na drugi pogled vide se finese za koje je vrlo moguće da ih takmičar na samom takmičenju ne uvidi... and the rest is history. Pridružujem se čestitkama svima.

Što se praga za bronzu tiče, poprilično je korektan (računajmo samo zvanični dio takmičenja). Naime, 2/3 svih takmičara dobijaju medalje, sa pragom na 7 bi to bilo 63.6%, a ovako je 68.1%.

Sad sam došao u iskušenje da ispišem nekih desetak smileya, da bih pokazao nadmoć moderatora nad običnim korisnicima, ali neću, I'll use my powers wisely  :yoda:

Istina, u brojenju nisam činio razliku između službenih i neslužbenih natjecatelja... ovako izgleda realnije. My mistake.


Can somebody post the 2004 National Target and Team answers?

I lost the answer key and I need to check my target and team.

I don't need the sprint because it's on tz's public site.

can someone post the 2004 national questions too?

Look at tz's website.

http://www.freewebs.com/tzpublicsite/helpfulwebsites.htm

[quote="SplashD"]I mean target and team[/quote]
I have them. \:/

[quote="SplashD"]can you post them please :?:[/quote]I think I already scanned them but If I haven't then It might be a while.  :|

[quote=Kevinisawesome]hey can someone post the sprint answers as well the website was taken down[/quote]

You revived a 10 year old thread lol

yea lol cuz i was doing the mock test for this at home and i couldnt find the answers. Can someone post the answers?

Mathgeniuses you next time should hide the answers...
But thanks for the answers anyway.


trigonometry
calculus
calculus computations

Define $\sin x = \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n+1}}{(2n+1)!}$.

1)Find all zero's of $\sin x$.

2)Show that the set of zero's has a least positive solution called $\pi$.

3)Show that $\sin (x+2\pi) = \sin x$

4)Show that $2\pi$ is the fundamental period.

Differentiate the series term by term to obtain $\sin''=-\sin$. Then continue as in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=148311]this thread[/url] or in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142926]that one[/url].


inequalities
function
inequalities proposed

I dont see how that leads to a proof ...

My solution is to deal with the maxmization problem  described above  -
a function of 2n real variables under three constraints

use the method of Lagrange  Multipliers.

It will work well   -   though the way to draw conclusions from the equations might be
a bit  unusual ...


geometry
circumcircle
geometric transformation
reflection
rotation
homothety
symmetry

Here's what I did, although I bet there are easier ways (I won't post all the details) :).

Let $O_1,O_2,O_3$ be the circumcenters of the three triangles, in the order you have listed them. We need to show that the reflections of $A$ in the sides of $O_1O_2O_3$ form an equilateral triangle, which is equivalent to the fact that $A$ is one of the [url=http://mathworld.wolfram.com/IsodynamicPoints.html]isodynamic points[/url] of $O_1O_2O_3$, which is equivalent (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=name+AND+points&t=6076]this thread[/url]) to the fact that the images of $O_i$ through an inversion of pole $A$ are the vertices of an equilateral triangle. This is not hard to prove (I don't think I'll do it here): these three points form a triangle similar to the one formed by the projections of $A$ on $B_1C_2,B_2C_3,B_3C_1$. Just show that this latter triangle is equilateral.

The fact that the equilateral triangle obtained from the intersections of the three circles is congruent to the three initial ones should be further justified, but it's not hard either.

Sorry for this, I don't know if it helps. We need a nice clean solution, but I couldn't stop thinking about the isodynamic points, and this yielded something nasty, as you can see :).

[Moderator edit: Yetti renames the points A, $B_{1}$, $C_{1}$, $B_{2}$, $C_{2}$, $B_{3}$, $C_{3}$ into O, $C_{1}$, $B_{1}$, $C_{2}$, $B_{2}$, $C_{3}$, $B_{3}$, respectively.]

Let $\triangle T_{1}= \triangle OB_{1}C_{1}, \triangle T_{2}= \triangle OB_{2}C_{2}, \triangle T_{3}= \triangle OB_{3}C_{3}$ be the 3 congruent equilateral triangles, $(P), (Q), (R)$ the circumcircles of the triangles $\triangle OB_{2}C_{1}, \triangle OB_{3}C_{2}, \triangle OB_{1}C_{3}$ and $E, F, G$ their pairwise intersections. Invert the figure in the circle $(O)$. The points $E, F, G$ go to the points $E', F', G'$, the circles $(P), (Q), (R)$ into the lines $B_{2}C_{1}\equiv E'F', B_{3}C_{2}\equiv F'G', B_{1}C_{3}\equiv G'E'$, the lines $EF, FG, GE$ into the circumcircles $(P'), (Q'), (R')$ of the triangles $\triangle OE'F', \triangle OF'G', \triangle OG'E'$.

The triangles $\triangle B_{1}B_{2}B_{3}, \triangle C_{1}C_{2}C_{3}$ are both similar to the triangle $\triangle E'F'G'$: From the circle $(O)$, the angle $\measuredangle B_{1}B_{2}C_{1}= \measuredangle B_{1}OC_{1}/2 = 60^{o}/2 = 30^{o}$ and similarly, the angles $\measuredangle B_{2}B_{3}C_{2}= \measuredangle B_{3}B_{2}C_{3}= 30^{o}$. Hence, the line pairs $B_{1}B_{2}$ and $B_{2}C_{1}\equiv EF$; $B_{2}B_{3}$ and $B_{3}C_{2}\equiv FG$; $B_{3}B_{1}$ and $B_{1}C_{3}\equiv FE$ form angles $30^{o}$. Similarly for the triangle $\triangle C_{1}C_{2}C_{3}$. Consequently, $\measuredangle G'E'F' = \measuredangle B_{3}B_{1}B_{2}= \measuredangle B_{3}OB_{2}/2$ and similarly, $\measuredangle E'F'G' = \measuredangle B_{1}B_{2}B_{3}= \measuredangle B_{1}OB_{3}/2$, $\measuredangle F'G'E' = \measuredangle B_{2}B_{3}B_{1}= \measuredangle B_{2}OB_{1}/2$.

From the circles $(R), (P)$ and from the isosceles triangles $\triangle C_{3}B_{1}O, \triangle C_{1}B_{2}O$, the angles $\measuredangle OEB_{1}$ and $\measuredangle OEC_{1}$ are

$\measuredangle OEB_{1}= 180^{o}-\measuredangle OC_{3}B_{1}= 180^{o}-\frac{180^{o}-\measuredangle C_{3}OB_{1}}{2}= 90^{o}+\frac{\measuredangle B_{3}OB_{1}-60^{o}}{2}$

$\measuredangle OEC_{1}= 180^{o}-\measuredangle OB_{2}C_{1}= 180^{o}-\frac{180^{o}-\measuredangle C_{1}OB_{2}}{2}= 90^{o}+\frac{\measuredangle B_{1}OB_{2}-60^{o}}{2}$

$\measuredangle OEB_{1}= \measuredangle E'F'G+60^{o}$
$\measuredangle OEC_{1}= \measuredangle F'G'E+60^{o}$

Since $\measuredangle E'F'G+\measuredangle F'G'E+\measuredangle G'E'F = 180^{o}$ are the internal angles of the triangle $\triangle E'F'G'$, $\measuredangle B_{1}EC_{1}= 360^{o}-(\measuredangle OEB_{1}+\measuredangle OEC_{1}) =$$360^{o}-(\measuredangle E'F'G+\measuredangle F'G'E+120^{o}) = \measuredangle G'E'F+60^{o}$. Hence, the points $E, F, G$ are the corresponding points of the equilateral triangles $\triangle T_{1}, \triangle T_{2}, \triangle T_{3}$ and the inversion center $O$ is the isodynamic point of the triangle $\triangle E'F'G'$:

$\measuredangle EOF = \measuredangle E'F'G+60^{o}$
$\measuredangle FOG = \measuredangle F'G'E+60^{o}$
$\measuredangle GOE = \measuredangle G'E'F+60^{o}$

The tangents to the circle $(P')$ at the points $E', F'$ form the angles $180^{o}-\measuredangle E'OF' = 120^{o}-\measuredangle E'G'F'$ with the side $E'F'$, the tangents to the circle $(Q')$ form the angles $180^{o}-\measuredangle F'OG' = 120^{o}-\measuredangle F'E'G'$ with the side $F'G'$, and the tangents to the circle $(R')$ form the angles $180^{o}-\measuredangle G'OE' = 120^{o}-\measuredangle G'F'E'$ with the side $G'E'$. Hence, the tangents to $(P'), (Q')$ at the point $F'$ form the angle $240^{o}-\measuredangle E'G'F'-\measuredangle F'E'G'-\measuredangle G'F'E' = 240^{o}-180^{o}= 60^{o}$ between each other and similarly, the tangents to $(Q'), (R')$ at the point $G'$ and the tangents to $(R'), (P')$ at the point $E'$ also form the angles $60^{o}$ with each other, respectively. Since inversion preserves angles between curves, the lines $EF, FG, GE$ also form the angles $60^{o}$ with each other, i.e., the triangle $\triangle EFG$ is equilateral.

The equilateral triangle $\triangle T_{1}$ can be transformed to the equilateral triangle $\triangle T_{2}$ and the point $E$ of $\triangle T_{1}$ into the corresponding point $F$ of $\triangle T_{2}$ by a rotation around the inversion center $O$ by the angle $\measuredangle B_{1}OB_{2}$ followed by a rotation around the center of the $\triangle T_{2}$ by the angle of $120^{o}$. In a similar manner, the triangle $\triangle T_{2}$ and the point $F$ of $\triangle T_{2}$ can be transformed into the triangle $\triangle T_{3}$ and the point $G$ of $\triangle T_{3}$, and the triangle $\triangle T_{3}$ and the point $G$ of $\triangle T_{3}$ into the triangle $\triangle T_{1}$ and the point $E$ of $\triangle T_{1}$. It follows that the equilateral triangle $\triangle EFG$ must be congruent to the triangles $\triangle T_{1}, \triangle T_{2}, \triangle T_{3}$.

[quote="paul_mathematics"]Let $AB_{1}C_{1}$, $AB_{2}C_{2}$, $AB_{3}C_{3}$ be directly congruent equilateral triangles. Prove that the pairwise intersections of the circumcircles of triangles $AB_{1}C_{2}$, $AB_{2}C_{3}$, $AB_{3}C_{1}$ form an equilateral triangle congruent to the first three.[/quote]

I have corrected a typo in the above problem - in fact, the original formulation in Kedlaya's book claimed that the pairwise intersections of the circumcircles of triangles $AB_{1}C_{1}$, $AB_{2}C_{2}$, $AB_{3}C_{3}$ form an equilateral triangle, what is wrong.

Actually, there is much more to say!

[color=blue][b]Extended theorem.[/b] In the following, the abbreviation "circle $P_{1}P_{2}P_{3}$" will mean the circle through three given points $P_{1}$, $P_{2}$, $P_{3}$.

Let A, $B_{1}$, $B_{2}$, $B_{3}$, $C_{1}$, $C_{2}$, $C_{3}$ be eight points in the plane such that the triangles $AB_{1}C_{1}$, $AB_{2}C_{2}$, $AB_{3}C_{3}$ are directly congruent equilateral triangles.

[b](a)[/b] Let the circles $AB_{1}C_{2}$ and $AB_{2}C_{3}$ intersect at a point X (apart from A). Let the circles $AB_{2}C_{3}$ and $AB_{3}C_{1}$ intersect at a point Y (apart from A). Let the circles $AB_{3}C_{1}$ and $AB_{1}C_{2}$ intersect at a point Z (apart from A). Then, the triangle XYZ is equilateral and inversely congruent to the triangles $AB_{1}C_{1}$, $AB_{2}C_{2}$, $AB_{3}C_{3}$.

[b](b)[/b] Let D, E, F be the centers of the circles $AB_{3}C_{1}$, $AB_{1}C_{2}$, $AB_{2}C_{3}$. Then, the points X, Y, Z are the reflections of the points $B_{1}$, $C_{1}$, A in the line DE. Similar results hold for the lines EF and FD.

[b](c)[/b] The triangles $B_{1}B_{2}B_{3}$ and $C_{1}C_{2}C_{3}$ are directly congruent.

[b](d)[/b] Let the circles $AB_{2}C_{2}$ and $AB_{3}C_{3}$ intersect at a point X' (apart from A). Let the circles $AB_{3}C_{3}$ and $AB_{1}C_{1}$ intersect at a point Y' (apart from A). Let the circles $AB_{1}C_{1}$ and $AB_{2}C_{2}$ intersect at a point Z' (apart from A). Also, let D', E', F' be the centers of the circles $AB_{1}C_{1}$, $AB_{2}C_{2}$, $AB_{3}C_{3}$. Then, the triangles D'E'F' and X'Y'Z' are centrally symmetric to each other and directly similar to the triangles $B_{1}B_{2}B_{3}$ and $C_{1}C_{2}C_{3}$.[/color]

[color=green][b]See graphics at the end of the post and in the next post (click on them to view them in proper size)![/b][/color]

[i]Proof.[/i] We will use directed angles modulo 180°, and we will WLOG assume that the equilateral triangles $AB_{1}C_{1}$, $AB_{2}C_{2}$, $AB_{3}C_{3}$ are directed clockwisely; this means that the angles $\measuredangle C_{1}AB_{1}$, $\measuredangle C_{2}AB_{2}$ and $\measuredangle C_{3}AB_{3}$ are 60° rather than -60°. So we have

$\measuredangle C_{1}AB_{1}=\measuredangle AB_{1}C_{1}=\measuredangle B_{1}C_{1}A=\measuredangle C_{2}AB_{2}=\measuredangle AB_{2}C_{2}=\measuredangle B_{2}C_{2}A$
$=\measuredangle C_{3}AB_{3}=\measuredangle AB_{3}C_{3}=\measuredangle B_{3}C_{3}A=60^{\circ}$.

Also, since the triangles $AB_{1}C_{1}$, $AB_{2}C_{2}$, $AB_{3}C_{3}$ are congruent and equilateral, we have

$AB_{1}=AC_{1}=B_{1}C_{1}=AB_{2}=AC_{2}=B_{2}C_{2}=AB_{3}=AC_{3}=B_{3}C_{3}$.

Now, $AB_{1}=AC_{1}=AB_{2}=AC_{2}=AB_{3}=AC_{3}$ yields that the points $B_{1}$, $C_{1}$, $B_{2}$, $C_{2}$, $B_{3}$, $C_{3}$ all lie on a circle with center A. Denote this circle by k.

The radical axis of two intersecting circles is the line joining their two points of intersection. Thus, the radical axis of the circles k and $AB_{1}C_{2}$ is the line $B_{1}C_{2}$, the radical axis of the circles k and $AB_{2}C_{3}$ is the line $B_{2}C_{3}$, and the radical axis of the circles $AB_{1}C_{2}$ and $AB_{2}C_{3}$ is the line AX. Since the pairwise radical axes of three circles concur at one point, it follows that the lines $B_{1}C_{2}$, $B_{2}C_{3}$ and AX concur at one point. Denote this point by U. Similarly, the lines $B_{2}C_{3}$, $B_{3}C_{1}$ and AY concur at one point V, and the lines $B_{3}C_{1}$, $B_{1}C_{2}$ and AZ concur at one point W.

[Of course, these points U, V, W are the inversive images of the points X, Y, Z in the circle k, but this doesn't bother us here, since we won't use inversion.]

Let U', V', W' be the midpoints of the segments $B_{3}C_{1}$, $B_{1}C_{2}$, $B_{2}C_{3}$, respectively. Since the triangles $AC_{3}B_{3}$, $AC_{1}B_{1}$, $AC_{2}B_{2}$ are equilateral and equally oriented, by the Asymmetric Propeller Theorem ( http://www.mathlinks.ro/Forum/viewtopic.php?t=16908 post #3), it follows that the triangle U'V'W' is equilateral and oriented equally to the triangles $AC_{3}B_{3}$, $AC_{1}B_{1}$, $AC_{2}B_{2}$. Hence, $\measuredangle U^{\prime}V^{\prime}W^{\prime}=\measuredangle AC_{3}B_{3}$. In other words, $\measuredangle U^{\prime}V^{\prime}W^{\prime}=-\measuredangle B_{3}C_{3}A$. Since $\measuredangle B_{3}C_{3}A=60^{\circ}$, this yields $\measuredangle U^{\prime}V^{\prime}W^{\prime}=-60^{\circ}$.

Now, since $AB_{1}=AC_{2}$, the point A lies on the perpendicular bisector of the segment $B_{1}C_{2}$; thus, as the point V' is the midpoint of this segment $B_{1}C_{2}$, we have $AV^{\prime}\perp B_{1}C_{2}$, so that $\measuredangle AV^{\prime}W=90^{\circ}$. Similarly, $\measuredangle AU^{\prime}W=90^{\circ}$. Hence, the points V' and U' lie on the circle with diameter AW. Thus, $\measuredangle AV^{\prime}U^{\prime}=\measuredangle AWU^{\prime}$. Similarly, $\measuredangle AV^{\prime}W^{\prime}=\measuredangle AUW^{\prime}$. Thus,

$-60^{\circ}=\measuredangle U^{\prime}V^{\prime}W^{\prime}=\measuredangle AV^{\prime}W^{\prime}-\measuredangle AV^{\prime}U^{\prime}=\measuredangle AUW^{\prime}-\measuredangle AWU^{\prime}$
$=\measuredangle\left(AU;\;UV\right)-\measuredangle\left(AW;\;VW\right)=\left(\measuredangle\left(AU;\;VW\right)+\measuredangle\left(VW;\;UV\right)\right)-\measuredangle\left(AW;\;VW\right)$
$=\left(\measuredangle\left(AU;\;VW\right)-\measuredangle\left(AW;\;VW\right)\right)+\measuredangle\left(VW;\;UV\right)=\measuredangle\left(AU;\;AW\right)+\measuredangle\left(VW;\;UV\right)$
$=\measuredangle UAW+\measuredangle WVU$.

On the other hand, the points $B_{1}$, $C_{1}$ and $C_{3}$ lie on the circle k, whose center is A; thus, by the central angle theorem for directed angles modulo 180°, we have $\measuredangle AB_{1}C_{1}=90^{\circ}-\measuredangle C_{1}C_{3}B_{1}$. Similarly, for the points $B_{3}$, $C_{3}$ and $C_{1}$ on the circle k, we get $\measuredangle AB_{3}C_{3}=90^{\circ}-\measuredangle C_{3}C_{1}B_{3}$. Now, since $\measuredangle AB_{1}C_{1}=\measuredangle AB_{3}C_{3}$ (in fact, $\measuredangle AB_{1}C_{1}=\measuredangle AB_{3}C_{3}=60^{\circ}$), we conclude that $\measuredangle C_{1}C_{3}B_{1}=\measuredangle C_{3}C_{1}B_{3}$, and thus $C_{1}B_{3}\parallel B_{1}C_{3}$. In other words, $VW\parallel B_{1}C_{3}$. Hence, $\measuredangle WVU=\measuredangle B_{1}C_{3}U$.

Since the points $B_{1}$, $C_{2}$, $B_{2}$, $C_{3}$ lie on one circle (namely, on the circle k), we have $\measuredangle B_{1}C_{3}B_{2}=\measuredangle B_{1}C_{2}B_{2}$. In other words, $\measuredangle B_{1}C_{3}U=\measuredangle UC_{2}B_{2}$. Hence,

$\measuredangle WVU=\measuredangle B_{1}C_{3}U=\measuredangle UC_{2}B_{2}=\measuredangle UC_{2}A-\measuredangle B_{2}C_{2}A=\measuredangle UC_{2}A-60^{\circ}$.

Substituting this into the equation $-60^{\circ}=\measuredangle UAW+\measuredangle WVU$, we get

$-60^{\circ}=\measuredangle UAW+\left(\measuredangle UC_{2}A-60^{\circ}\right)$.

This simplifies to $\measuredangle UAW+\measuredangle UC_{2}A=0^{\circ}$, so that $\measuredangle UAW=-\measuredangle UC_{2}A$. But since the point X lie on the circle $AB_{1}C_{2}$, we have $\measuredangle AC_{2}B_{1}=\measuredangle AXB_{1}$. Thus, $\measuredangle UAW=-\measuredangle UC_{2}A=\measuredangle AC_{2}U=\measuredangle AC_{2}B_{1}=\measuredangle AXB_{1}$. Hence, $B_{1}X\parallel AW$. In other words, $B_{1}X\parallel AZ$.

Now, consider the circles $AB_{3}C_{1}$ and $AB_{1}C_{2}$. These two circles have the centers D and E, and their two common points are A and Z. Clearly, the two common points of two intersecting circles must be reflections of each other in the line joining the centers of the circles. Hence, the point Z is the reflection of the point A in the line DE. Hence, $AZ\perp DE$. Thus, $B_{1}X\parallel AZ$ becomes $B_{1}X\perp DE$. Hence, the line DE is the perpendicular to the line $B_{1}X$ through the point E. But the perpendicular bisector of the segment $B_{1}X$ must also be perpendicular to the line $B_{1}X$ and it must also pass through the point E (since the segment $B_{1}X$ is a chord in the circle $AB_{1}C_{2}$, whose center is E, and the perpendicular bisector of a chord in a circle always passes through the center of the circle). Hence, the line DE is the perpendicular bisector of the segment $B_{1}X$. In other words, the point X is the reflection of the point $B_{1}$ in the line DE. Similarly, the point Y is the reflection of the point $C_{1}$ in the line DE. Finally, as we already know, the point Z is the reflection of the point A in the line DE. Of course, we could proceed in the same way to obtain analogous results for the lines EF and FD. Hence, problem [b](b)[/b] is solved.

Now, problem [b](a)[/b] becomes a piece of cake: Since the points X, Y, Z are the reflections of the points $B_{1}$, $C_{1}$, A in the line DE, the triangle XYZ is inversely congruent to the triangle $B_{1}C_{1}A$. But since the triangle $AB_{1}C_{1}$ is equilateral, being inversely congruent to the triangle $B_{1}C_{1}A$ is equivalent to being inversely congruent to the triangle $AB_{1}C_{1}$. Hence, we can state that the triangle XYZ is inversely congruent to the triangle $AB_{1}C_{1}$. Similarly, the triangle XYZ is inversely congruent to the triangles $AB_{2}C_{2}$ and $AB_{3}C_{3}$. Since these triangles $AB_{1}C_{1}$, $AB_{2}C_{2}$, $AB_{3}C_{3}$ are equilateral, the triangle XYZ must therefore also be equilateral. And problem [b](a)[/b] is solved.

Now we come to problems [b](c)[/b] and [b](d)[/b]. We stop using directed angles modulo 180° and pass over to directed angles modulo 360° (since we will use rotations). Since the triangle $AB_{1}C_{1}$ is equilateral and directed clockwisely, we have $AB_{1}=AC_{1}$ and $\measuredangle C_{1}AB_{1}=60^{\circ}$; thus, the point $B_{1}$ is the image of the point $C_{1}$ under the rotation with center A and rotation angle 60°. Similarly, the points $B_{2}$ and $B_{3}$ are the images of the points $C_{2}$ and $C_{3}$ under this rotation. Hence, the triangle $B_{1}B_{2}B_{3}$ is the image of the triangle $C_{1}C_{2}C_{3}$ under this rotation. Thus, the two triangles are directly congruent. This solves problem [b](c)[/b].

The point D' is the circumcenter of the equilateral triangle $AB_{1}C_{1}$. Thus, by the well-known metric relations in the equilateral triangle, $\measuredangle C_{1}AD^{\prime}=30^{\circ}$ and $\frac{AD^{\prime}}{AC_{1}}=\frac{1}{\sqrt3}$. Hence, the point D' is the image of the point $C_{1}$ under the spiral homothety with center A, rotation angle 30° and stretch factor $\frac{1}{\sqrt3}$. Similarly, the points E' and F' are the images of the points $C_{2}$ and $C_{3}$ under this spiral homothety. Hence, the triangle D'E'F' is the image of the triangle $C_{1}C_{2}C_{3}$ under this spiral homothety. Thus, the triangle D'E'F' is directly similar to the triangle $C_{1}C_{2}C_{3}$. Since the triangle $C_{1}C_{2}C_{3}$ is directly congruent to the triangle $B_{1}B_{2}B_{3}$, the triangle D'E'F' must also be directly similar to the triangle $B_{1}B_{2}B_{3}$.

Now, we will prove that the triangles D'E'F' and X'Y'Z' are centrally symmetric to each other. Once this will be proven, we will be able to deduce from the fact that the triangle D'E'F' is directly similar to the triangles $B_{1}B_{2}B_{3}$ and $C_{1}C_{2}C_{3}$ the fact that the triangle X'Y'Z' is also directly similar to the triangles $B_{1}B_{2}B_{3}$ and $C_{1}C_{2}C_{3}$, and problem [b](d)[/b] will be solved. So, it remains to show that the triangles D'E'F' and X'Y'Z' are centrally symmetric to each other.

We have found $\frac{AD^{\prime}}{AC_{1}}=\frac{1}{\sqrt3}$ and we can similarly find $\frac{AE^{\prime}}{AC_{2}}=\frac{1}{\sqrt3}$. Thus, $\frac{AD^{\prime}}{AC_{1}}=\frac{AE^{\prime}}{AC_{2}}$. Since $AC_{1}=AC_{2}$, this yields AD' = AE'. Similarly, AE' = AF', so that AD' = AE' = AF'. Hence, the point A is the circumcenter of triangle D'E'F'.

The two common points of two intersecting circles are reflections of each other in the line joining the centers of the two circles. Thus, the points A and X', being the two common points of the circles $AB_{2}C_{2}$ and $AB_{3}C_{3}$, must be reflections of each other in the line joining the centers E' and F' of these circles. In other words, the point X' is the reflection of the point A in the line E'F'. Similarly, the points Y' and Z' are the reflections of the point A in the lines F'D' and D'E'. So the points X', Y', Z' are the reflections of the point A in the sides E'F', F'D', D'E' of triangle D'E'F'. Since the point A is the circumcenter of triangle D'E'F', we thus see that the points X', Y', Z' are the reflections of the circumcenter of triangle D'E'F' in its sides E'F', F'D', D'E'. Now, it is well-known (and easy to prove) that if $\Delta$ is an arbitrary triangle, then the reflections of the circumcenter of $\Delta$ in the sides of $\Delta$ form a new triangle which is centrally symmetric to the triangle $\Delta$ (by the way, the center of symmetry is the nine-point center of $\Delta$). Applying this to the triangle $\Delta=D^{\prime}E^{\prime}F^{\prime}$, we see that the triangle X'Y'Z', formed by the reflections X', Y', Z' of the circumcenter of triangle D'E'F' in its sides E'F', F'D', D'E', is centrally symmetric to the triangle D'E'F'. This completes our proof.

Thus, as said above, problem [b](d)[/b] is solved. So the solution is complete.

  Darij

This post is here to contain the first graphic to the solution.

  Darij

Remark:

[b][i]The problem is true for the general case: the three triangles $\triangle OA_1B_1, \triangle OA_2B_2, \triangle OA_3B_3$ be similar, same way oriented and $\angle A_1OB_1+\angle A_2OB_2+\angle A_3OB_3=180^\circ$. Then the triangle formed by the midpoints of the segments $B_1A_2, B_2A_3, B_3A_1$ is similar to the three initial triangles.[/i][/b]

Complex numbers or vectors easily solve this problem (maybe posted already somewhere on ML?).

Best regards,
sunken rock


[quote="math92"]Show that the solutions of
$3x^{2}-2x+3=0$ are reciprocals.[/quote]
uh, I think the question is incomplete. Do you mean show that they are reciprocals of themselves?

i get..
[hide]$\frac{2+4i\sqrt{2}}{6}$ and $\frac{2-4i\sqrt{2}}{6}$ as my solutions..

meaning that
$\frac{6}{2+4i\sqrt{2}}= \frac{2-4i\sqrt{2}}{6}$

$\frac{6}{2+4i\sqrt{2}}\cdot \frac{2-4i\sqrt{2}}{2-4i\sqrt{2}}= \frac{12-24i\sqrt{2}}{36}$

$= \boxed{\frac{2-4i\sqrt{2}}{6}}$[/hide]
...so they are infact reciprocals

[quote="Glen Mann"]i get

$\frac{4i\sqrt{2}}{3}$ and $\frac{-4i\sqrt{2}}{3}$ as my solutions..

i suck at complex numbers so im not sure how to show that they are reciprocals?[/quote]Solve the equation again.

i made a stuipid mistake when I tried to simplify them >_<
edited my post above.

[quote="10000th User"]Let's play some generalization:

$a,b$ are real numbers such that $ax^{2}+bx+a=0$. Can you show that the roots are also reciprocals?[/quote]
[hide]
the roots of this equation are
$\frac{-b \pm \sqrt{b^{2}-4a^{2}}}{2a}$
meaning that
$\frac{-b+\sqrt{b^{2}-4a^{2}}}{2a}= \frac{2a}{-b-\sqrt{b^{2}-4a^{2}}}$

$\frac{2a}{-b-\sqrt{b^{2}-4a^{2}}}\cdot \frac{-b+\sqrt{b^{2}-4a^{2}}}{-b+\sqrt{b^{2}-4a^{2}}}= \frac{-2ab+2a\sqrt{b^{2}-4a^{2}}}{4a^{2}}$

simplifying we get
$\boxed{\frac{-b+\sqrt{b^{2}-4a^{2}}}{2a}}$
showing that the roots are reciprocals..

very very interesting idea, if a and c are equal then the roots will be reciprocals  :o
[/hide]

That's correct  :wink:  Just to be complete, I forgot to mention that we have to assume $a\ne0$ :P

Another interesting idea is the following:

The equation $ax^{2}+bx+c=0$ has roots that are reciprocals of the roots of $cx^{2}+bx+a=0$, for $a,c\ne0$  :wink:

For any quadratic of the second degree of the form $ax^{2}+bx+c$, the product of the roots is $\frac{c}{a}$. Here, since $a=c$, the product of the roots is 1. By definition, two numbers whose product is 1 are reciprocals.

[hide="generalization"]
Given a polynomial $\sum^{n}_{k=0}a_{k}x^{k}$, the polynomial $a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+...+a_{n-1}x+a_{n}$ has roots that are reciprocals of the first.[/hide]

is it bad that I have no idea what the last two posters said?

[hide="For the special quadratic case"] Given a quadratic equation

$P(x) = ax^{2}+bx+c$

With $a, c \neq 0$ and with some roots $r_{1}, r_{2}$such that $P(r_{1}) = P(r_{2}) = 0$, we wish to find a quadratic equation $Q(x)$ such that

$Q\left( \frac{1}{r_{1}}\right) = Q \left( \frac{1}{r_{2}}\right) = 0$

Well, a [b]function[/b] with this property is given by

$F(x) = P \left( \frac{1}{x}\right) = \frac{a}{x^{2}}+\frac{b}{x}+c$

Because

$F \left( \frac{1}{r_{1}}\right) = P(r_{1}) = 0$
$F \left( \frac{1}{r_{2}}\right) = P(r_{2}) = 0$

But this is not a polynomial.  However,

$Q(x) = x^{2}F(x) = a+bx+cx^{2}$

Is a polynomial, which therefore satisfies our requirements.

This transformation can be generalized in the way I described in my previous post. [/hide]

[quote="Glen Mann"][quote="kunny"][quote="math92"]Show that the solutions of
$3x^{2}-2x+3=0$ are reciprocals.[/quote]

Since $x\neq 0,\ 3\left(\frac{1}{x}\right)^{2}-2\left(\frac{1}{x}\right)+3=0.$
we are done.[/quote]
you didnt do anything but replace x with the reciprocal of x.. that doesnt show anything.[/quote]
he showed that if you replace it with the reciprocal, you get the exact same equation(after multiplying through by x^2)

[quote="Glen Mann"][quote="kunny"][quote="math92"]Show that the solutions of
$3x^{2}-2x+3=0$ are reciprocals.[/quote]

Since $x\neq 0,\ 3\left(\frac{1}{x}\right)^{2}-2\left(\frac{1}{x}\right)+3=0.$
we are done.[/quote]
you didnt do anything but replace x with the reciprocal of x.. that doesnt show anything.[/quote]

Actually, he divided by $x^{2}$.  ;)

Yup. I just learned all that from AoPS Vol. II, the chapter on polynomials.

[quote="t0rajir0u"][/quote][quote="Glen Mann"][quote="kunny"][quote="math92"]Show that the solutions of
$3x^{2}-2x+3=0$ are reciprocals.[/quote]

Since $x\neq 0,\ 3\left(\frac{1}{x}\right)^{2}-2\left(\frac{1}{x}\right)+3=0.$
we are done.[/quote]
you didnt do anything but replace x with the reciprocal of x.. that doesnt show anything.[/quote][quote="t0rajir0u"]

Actually, he divided by $x^{2}$.  ;)[/quote]

ermm.. that makes no sense >_>

Are you reading my posts?  Let's go over this [hide="one more time"] Let the solutions to $3x^{2}-2x+3 = 0$ be $r_{1}, r_{2}$.  Then we know that

$3r_{1}^{2}-2r_{1}+3 = 0$
$3r_{2}^{2}-2r_{2}+3 = 0$

Clearly $r_{1}, r_{2}\neq 0$, so we can divide by them.  In particular, dividing the first equation by $r_{1}^{2}$ produces

$\frac{3r_{1}^{2}-2r_{1}+3}{r_{1}^{2}}= 3-2 \left( \frac{1}{r_{1}}\right)+3 \left( \frac{1}{r_{2}}\right)^{2}= 0$

And dividing the second equation by $r_{2}^{2}$ produces

$\frac{3r_{2}^{2}-2r_{2}+3}{r_{2}^{2}}= 3-2 \left( \frac{1}{r_{2}}\right)+3 \left( \frac{1}{r_{2}}\right)^{2}= 0$

But look at what just happened here!  It turns out that $\frac{1}{r_{1}}, \frac{1}{r_{2}}$ are [b]also[/b] roots of our original quadratic!

Since we know it can only have two roots, and since neither of the roots can be $1,-1$ either, we conclude that

$\frac{1}{r_{1}}= r_{2}$
$\frac{1}{r_{2}}= r_{1}$ [/hide]


When drawing all diagonals in a regular pentagon, one gets an smaller pentagon in the middle. What's the ratio of the areas of those pentagons?

It's $\frac{7-3\sqrt{5}}{2}$.

[hide]The pentagons are similar so their area ratio is the ratio of their sides squared.
Let the side of the big pentagon be $a$ and the side of the small pentagon be $b$. Let the distance between center points of the respective (parallel) sides be $h$, the distance between side $a$ and the opposite vertex of the big pentagon be $H$.

Finding the appropriate right triangles is quite easy. We have:

$h=a\cdot\cos\frac{\pi}{10}$

$H-h=a\cdot\cos\frac{3\pi}{10}$

Thus $H=H-h+h=2a\cdot\cos\frac{2\pi}{10}\cos\frac{\pi}{10}$

Now it is: $k=\frac{b}{a}=\frac{H-h}{H}$

Using: $\cos\frac{\pi}{10}=\sqrt{\frac{5+\sqrt{5}}{8}}$ and taking $k^2$ we get the result.[/hide]

Hmm yes, that's the answer for small/large ratio.

Now a solution? :)

[quote="Peter VDD"] Now a solution? [/quote]
C'mon Peter, it's all there. :)

:blush: overlooked the hide :blush:

I guess I need some sleep. :P


function
real analysis
real analysis unsolved

For each a,b in (0,1] construct a function f:[0,1]->[0,1] s.t.

i)f(0)=0 and f(1)=1
ii)f continuous strictly increasing
iii)f(Ca)=f(Cb) where Ca, Cb Cantor sets of meausure 1-a, 1-b

Why do you post your problems here? They belong in Unsolved/Proposed.


geometry
geometric transformation
homothety
ratio
geometry proposed

Let c and c' be two circles intersecting at A and B. Let P be a point on c. Find by rurel and compass the point(s) Q('s) on c' so that  AB bisects PQ.

[quote="darij grinberg"]Am I completely idiot or is this trivial? The points Q we have to construct are the points of intersection of the circle c' with the image of the line AB in the homothety with center P and factor 2. Aren't they?

  Darij[/quote]

If there is someone stupid this is me: the problem is really trivial. I didn't reflected to much  before posting this problem because I already had the solution from [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=132711#132711]this thread[/url]

After lemmaRL151104, is enough to intersect the tangent to c at P with the Apollonious circle related to the two centers of circles and such the ratio is equal to r/r'. The tangents from this points of intersection to c' give us the points requested.


How mny scalene triangles have all sides of integral lengths and perimeter less than 13?

You h ave to remember the triangle inequality. (any two sides of a triangle must be greater than the third side).


geometry
parallelogram
geometric transformation
homothety
reflection
geometry solved

We're given two congruent, equilateral triangles $ABC$ and $PQR$ with parallel sides, but one has one vertex pointing up and the other one has the vertex pointing down. One is placed above the other so that the area of intersection is a hexagon $A_1A_2A_3A_4A_5A_6$ (labelled counterclockwise). Prove that $A_1A_4$, $A_2A_5$ and $A_3A_6$ are concurrent.

Since the corresponding sides $AB \parallel PQ,\ BC \parallel QR,\ CA \parallel RP$ of the equilateral triangles $\triangle ABC \cong \triangle PQR$ are parallel, these 2 triangles are centrally similar, which means that the lines $AP, BQ, CR$ connecting their corresponding vertices concur at their similarity center $H$. Since the 2 equilateral triangles are oppositely congruent, their similarity coefficient is -1. Therefore, for the oriented segments, we have $HA = -HP,\ HB = -HQ,\ HC = -HR$ and consequently, $H$ is the common midpoint of the segments $AP, BQ, CR$

The quadrilateral $AA_2PA_5$ is a parallelogram, because the lines $AA_2 \equiv AB \parallel PQ \equiv QA_5$ and $A_5A \equiv CA \parallel RP \equiv A_2P$ are parallel. Hence, its diagonals $AP, A_2A_5$ cut each other in half at the midpoint $H$ of $AP$.

The quadrilateral $BA_4QA_1$ is a parallelogram, because the lines $BA_4 \equiv BC \parallel QR \equiv QA_1$ and $A_1B \equiv AB \parallel PQ \equiv A_4Q$ are parallel. Hence, its diagonals $BQ, A_1A_4$ cut each other in half at the midpoint $H$ of $BQ$.

The quadrilateral $CA_6RA_3$ is a parallelogram, because the lines $CA_6 \equiv CA \parallel RP \equiv RA_3$ and $A_3C \equiv BC \parallel QR \equiv A_6R$ are parallel. Hence, its diagonals $CR, A_3A_6$ cut each other in half at the midpoint $H$ of $CR$.

As a result, $A_1A_4,\ A_2A_5,\ A_3A_6$ all pass through and concur at the similarity center $H$.

There is an even simpler way of formulating the solution:

Since the triangles ABC and PQR have parallel sides, they are homothetic, and since they are congruent, their homothetic factor is either 1 or -1 (where 1 would actually mean that there is a translation mapping one to the other). The condition that one of the triangles has one vertex pointing up and the other one has the corresponding vertex pointing down specifies that the factor is -1 rather than 1. Thus, there is a homothety with factor -1, i. e. a reflection in a point, which maps the triangle ABC to the triangle PQR. In other words: There exists a point Z such that the triangle PQR is the reflection of the triangle ABC in the point Z. More precisely, the reflection in the point Z maps the points A, B, C to the points P, Q, R.

The points $A_1$, $A_2$, $A_3$, $A_4$, $A_5$, $A_6$ were defined as points of intersections of the sides of triangles ABC and PQR, without specifying which point is the intersection of which sides; let's WLOG assume that $A_1=CA\cap QR$, $A_2=AB\cap QR$, $A_3=AB\cap RP$, $A_4=BC\cap RP$, $A_5=BC\cap PQ$ and $A_6=CA\cap PQ$. The reflection in the point Z maps the points A, B, C to the points P, Q, R. Of course, it must then also map the points P, Q, R back to the points A, B, C. Hence, it maps the point $CA\cap QR$ to the point $RP\cap BC$. In other words, our reflection maps the point $A_1$ to the point $A_4$. Hence, the line $A_1A_4$ passes through the center Z of our reflection. Similarly, the lines $A_2A_5$ and $A_3A_6$ also pass through this center Z. Hence, the lines $A_1A_4$, $A_2A_5$ and $A_3A_6$ are concurrent (namely, at the point Z).

  Darij


Propose a synthesis of 3-cyclopentylpropyne starting from:

1) Cyclohexanone
2) Acetylene

[hide="click to reveal the hidden content"]From cyclohexanone:
1.Do alpha chlorination
2.Then treat it with NaOH(favorskii rearrangement)-->cyclopentanecarboxylic acid
3.Treat it with SOCl2
4.Treat the acyl chloride with sodium acetylide(acetylene + NaNH2)-->req. product
5.Wolff kishner reduction[/hide]

Alkyne anions react sluggishly with acyl chlorides. Also, your synthesis is incomplete.

sorry finally wolf-kishner reduction :oops:
Ok after acyl chloride:
[hide="click to reveal the hidden content"]
1.Rosendmund's reduction
2.Sodium acetylide
3.Wolff-kishner reduction[/hide]

From acetylene:
[hide="click to reveal the hidden content"]
1.Probably 1 mole of NaNH2(to get sodium acetylide) and reaction with C5H9-CH2-I
[/hide]
Is it right? :maybe:  :maybe:

Yes, that's right, but from where come from the C5H9-CH2-I?

Here's a suitable method:
[hide="click to reveal the hidden content"]
1.adipic acid+$ \ce{Ca(OH)2}$-->calcium adipate
2.Heat it-->cyclopentanone
3.Treat it with $ \ce{PPh3\equal{}CH2}$(ylide)-->methene cyclopentane
4.Hydroboration followed by reaction with $ \ce{Br2}$-->Bromocyclopentyl methane(with-I reaction is slow)
5.Finkelstein reaction-->$ \ce{C5H9\minus{}CH2\minus{}I}$
[/hide]

[quote="Chemrock"]1.adipic acid+Ca(OH)2-->calcium adipate
2.Heat it-->cyclopentanone[/quote]

How exactly does the second step occur?

This occurs coz decarboxylation and dehydration take place at the same time.........

Is that so? Then can you explain the mechanism?

i don remeber the mech properrly.........

i'll just refer my book and tell ya

Yes the reaction is slow u said that a long time ago :blush:  :blush:  :blush:

I am asking that because I am not seeing at the moment how heating calcium adipate can give cyclopentanone.

Why $ \ce{CaCO3}$ elimination takes place to form cyclopentanone :huh:  :huh:

And what is exactly the mechanism of that?

Ok if u want another method of preparation of cyclopetnanone use Ruzicka reaction to adipic acid:
$ \ce{HOOC - (CH{}_2){}_4 - COOH}$ + ${ \Delta - > [ThO_2] C{}_5H{}_1{}_0}$

not sure abt the mech, but probably some intramolecular  nucleophilic attack takes place by COO- on the other COO-

Carboxylate groups ($ \ce{CO2\minus{}}$) are unreactive towards nucleophilic attack.

Or one alternative:
adipic acid + $ \ce{ThO2}$ --> cyclopentanone (Ruzcika reaction)


inequalities
function
calculus
derivative
inequalities unsolved

$1<\left(\frac{2^x-1}{x}\right)^{\frac{1}{x-1}}<2$ for $x>0$

$0<\ln(2^x-1) - \ln(x)<(x-1)\ln(2)$ for $x>1$

The left part is trivial, and for the right part we put $f=(x-1)\ln(2) + \ln(x) - \ln(2^x-1)$.

$f(1)=0$, and for $x>1$ the function is increasing, since it's first derivative has positive denominator, and numerator equal to $2^x -x\ln(2)-1 > 0$

I wonder if someone has a solution without calculus... :?


trigonometry
integration
algebra
polynomial
function
linear equation
rational function

Hello hello!
I am doing homework and have been stuck on one problem for awhile and its bugging me.
I need help on figuring out the solution doing the method of undetermined coefficients.
Thanks a lot for any help :)

$ \frac{dy}{dx}\plus{}y\equal{}\tan(x)$

First, try to solve the homogenious equation $ \frac{df(x)}{dx}\plus{}f(x)\equal{}0$.
The solution is $ f(x) \equal{} C\exp(\minus{}x)$
Then substitute $ y\equal{}C(x)\exp(\minus{}x)$ in $ \frac{dy}{dx}\plus{}y\equal{}\tan(x)$
Solve $ C(x)$ and you have a solution for $ y$.
The solution contains an indefinite integral.

Let me continue a bit:
$ \frac{dy}{dx}\equal{}C'(x)\exp(\minus{}x)\minus{}C(x)\exp(\minus{}x)$.
So the equation is
$ C'(x)\exp(\minus{}x)\minus{}C(x)\exp(\minus{}x)\plus{}C(x)\exp(\minus{}x)\equal{}\tan{x}\equal{}\frac{dC}{dx}e^{\minus{}x}$.
Rearranging a bit, we have
$ dC\equal{}e^{x}\tan{x}\ dx\Rightarrow C(x)\equal{}\int e^{x}\tan{x}\ dx$

Undetermined coefficients really only works in certain narrowly defined cases. The most important such narrowly defined case is that of a constant coefficient linear equation in which the right hand side belongs to the "exponential-polynomial class" of functions. By that, I mean the span (set of finite linear combinations) of functions of the form $ x^ke^{rx}$ where $ k$ is a nonnegative integer and $ r$ is any complex number. (If we're doing everthing in real terms, we expand out the combinations of terms like $ e^{rx}$ to get $ e^{ax},e^{ax}\cos bx,$ and $ e^{ax}\sin bx,$ including cases in which $ a\equal{}0.$)

A function $ y$ belongs to the exponential-polynomial class if and only if there exists a constant coefficient linear operator $ L$ such that $ L(y)\equal{}0.$

A function $ y$ belongs to the exponential-polynomial class if and only if the Laplace transform of $ y$ is a proper rational function of $ s.$

Note that no member of the exponential-polynomial class has any singularity anywhere. They are all entire functions.

------------------------------------

That brings us to problem with your example: $ \tan x$ does not belong to the exponential-polynomial class. Hence the method of undetermined coefficients doesn't really have anything to work with - you're not in position to guess the form of the solution.

Now, your problem is a first-order linear equation, and there is a well-established method for those - the method of multiplying factors. What JRav has done is essentially a rearrangement of this method. And in the end he arrived at an antiderivative which is not an elementary function.

I can't understand why the differential equation should be solved by Undetermined Coefficients.

Anyway, you should note : $ \boxed{e^{ \minus{} kx}\frac {d}{dx}e^{kx}f(x) \equal{} f'(x) \plus{} kf(x)}$ for some constant $ k$.


Counting with Restrictions Part 2

We count the number of ways to stack 7 different Art of Problem Solving books, such that the books are grouped by color.  We thereby learn more about dealing with restrictions in counting problems, and we see 7 outstanding books!

Wow, a shameful advertising plug even in the description of the video.


induction
number theory proposed
number theory

Find all positive integers $ n$ such that $ (n\plus{}1)!$ is divisible by $ 1! \plus{} 2! \plus{} ... \plus{} n!$

OK, here is my solution:

Lets prove that $ n\equal{}1$ and $ n\equal{}2$ are the only solutions. Lets prove that there are no solutions for $ n \geq 3$

Let

$ \frac{(n\plus{}1)!}{1! \plus{} 2! \plus{} ... \plus{} n!} \equal{} x$ where $ x$ is an integer then $ x \leq n$

But thats impossible since $ (n\plus{}1)!$ is divisible by some $ 2^p$ and therefore x must be divisible by that number and thats an obvious contradiction!

How many points did you get for that? I think that isn't correct because you have a fraction and a portion of those $ 2$-s will disappear.
My complicated solution:
We prove by induction that $ n\minus{}1<\frac{(n\plus{}1)!}{1!\plus{}2!\plus{}3!\plus{}...\plus{}n!}<n$. I don't want to write whole solution but it's correct. From this we see that the number must be between 2 integers so it's not integer himself.

[quote="B_Gasperov"]Sorry, the solution I gave is not complete, some parts are omitted but I think its correct. I got all the points on that one.

How can a portion of 2's disappear when the denominator is odd?  :mad:

Please elaborate on that[/quote]
you have a beautiful solution  :wink: .
if we suppose that $ 2^{a}\le n\plus{}1<2^{a\plus{}1}$ and $ (n\plus{}1)!\equal{}k(1\plus{}2!\plus{}3!\plus{}...\plus{}n!)$
then $ 2^{a\plus{}1}|(n\plus{}1)!\Rightarrow 2^{a\plus{}1}|k \Rightarrow 2^{a\plus{}1}\le \frac{(n\plus{}1)!}{1\plus{}2!\plus{}..\plus{}n!}\le \frac{(n\plus{}1)!}{n!}\equal{}n\plus{}1$ (impossible)

Thank you.  :D

But the problem is by no means difficult!!!


Does anyone have this years AMC10B? I'll trade for it.

Well, you can just go to the "Contests" bar at the top of the page, and under USA, and AMC 10, there's an archive of all the old contests, plus their solutions. You don't really need to ask someone to trade for AMC 10 tests, because they're basically all there in that section. Check it out, and you have it all. :)

Also, this is the MATHCOUNTS forum, so it's meant to be for discussion for Mathcounts, not AMC. Any topic regarding AMC contests should go in the American Mathematics Competitions forum. :wink:

Yes, there's an AMC forum for just that type of reason. This is the MC forum, which is for discussion on MathCounts. If you just want to look at a problem real fast or something, instead of getting it to practice with, you can find some of the tougher problems with solutions in the aops math jam transcripts.

Um...

Firstly, I agree with the others. This should have been posted in the AMC forum. Secondly, just go to AoPSWiki and type in AMC 10.

-rts2007


Evaluate the following integrals.

(1) $\int_{0}^{\frac{\pi}{4}}\frac{dx}{1+\sin x}.$

(2) $\int_{\frac{4}{3}}^{2}\frac{dx}{x^{2}\sqrt{x-1}}.$

[quote="kunny"](2) $\int_{\frac{4}{3}}^{2}\frac{dx}{x^{2}\sqrt{x-1}}.$[/quote]
Let $x = \sec^{2}u \implies x = 1+\tan^{2}u \implies dx = 2\tan u\sec^{2}udu$, which yields:

\begin{eqnarray*}\int{\frac{1}{{x^{2}\sqrt{x-1}}}dx}&=& 2\int{\cos^{2}udu}\hfill \\ &=& \sin u\cos u+u+c \hfill \\ &=& \frac{{\sqrt{x-1}}}{x}+\sec^{-1}\sqrt x+c \hfill \\ \end{eqnarray*}

Now..., evaluate...

[quote="kunny"]$\int_{0}^{\frac{\pi}{4}}\frac{dx}{1+\sin x}$[/quote]
First:

$\int{\frac{1}{{1+\sin x}}dx}= \int{\frac{{1-\sin x}}{{\cos x\cos x}}dx}= \int{\sec^{2}x-\sec x\tan xdx}$

Finally:

$\int{\frac{1}{{1+\sin x}}dx}= \tan x-\sec x+c$

Now.., just evaluate...

[quote="Extremal"]Evaluate the following integrals.

(1) $\int_{0}^{\frac{\pi}{4}}\frac{dx}{1+\sin x}.$
$\int\frac{dx}{1+\sin x}$=$\int\frac{dx}{cos^{2}x}$+$\int\frac{dcos x}{cos^{2}x}$=$\tan x-\frac{1}{cosx}$
simpl[/quote]

I'd rethink it....