nope.
For more info, bye a algebra book.

[quote="now a ranger"]in my school, the toilets don't flush. So you can always see Vitamin C in the toilets, as well as human waste. not to mention the ferocious odor.[/quote]

i dont think ive ever been in my school bathroom over the past 3 years.

are you guys serious

most of the time i gotta take a dump or leak at least once during the school day

IRT the post above me:
do it at home in the morning

General Reply:
I always flush...not flushing creates a very bad smell...sometimes when i enter a very bad-smelling restroom, i try to flush the un-flushed toilet, and most of the time i fial, in which case i just do whatever i have to do fast and never use that restroom again for 2-3 days.

Yes, just because that other person stank up the place is no reason for you to do so as well. Be considerate! I flush, in case you can't figure it out.

No - In some restrooms here there is a motion detector or something - you cannot push the flush button, it flushes as soon as you move

same with the water taps - put your hand over this red dot, and it starts pouring water

I tend to avoid public bathrooms whenever possible

i go to sleep, or go into the world of tuning off  :blush:

[quote="Treething"]where else would you guys go in the middle of class =p

aside from study hall[/quote]
i walk around the hall, etc...jus take a walk...

Study halls are for people who are too uncool for free blocks (neither of which I have had for a while now...)

I can't flush in our P.E. bathroom ever, so it smells like poo.

I'm like... Poo!!!! Wrinkles nose

Then my friend is like, uhhh wow it does smell! POO!!! U A POO!

Anyways, I guess I don't flush, if the flush is on drugs.  :lol:

??? was that ???

i avoid public restrooms as much as i can
the only reason i know where they are at school is because they for some unknown reason are conveniently placed near the water fountains

I went once when I was dressing up to be a smurf. Wow it was SMELLY.

A friend came and said to me 5 minutes later , "You smell funny."

i was like omg

wuts a public bathroom? :rotfl:

I flush when I can. if all of the toilets are bad, I take the cleanest one.


floor function
algorithm
combinatorics unsolved
combinatorics

Let $ S$ be a subset of the set $ \{1,2,...,2002\}$.Suppose that for any (not necessarily) elements $ a,b$ in $ S$ their product $ ab$ is not in $ S$. Determine the maximal number of elements of $ S$

[hide="Solution"]We will prove that the maximal number of elements in $ S$ is $ 1958$, removing $ 44$ elements from $ \{1,2,\dots,2002\}$, which is achieved for $ S\equal{}\{45, 46, \dots, 2002\}$ (where clearly $ ab\geq 45^2>2002$, and numbers $ >2002$ are indeed not in $ S$), by assuming that an "optimal" set exists with a lower or higher minimal element.

First, start out with the set $ K\equal{}\{1,2, \dots, 2002\}$. Let us denote $ k<45$ as the minimal element that we will keep in $ S$. Thus, we must eliminate the elements lower than $ k$ ($ 1$, $ 2$, ..., $ k\minus{}1$) from $ K$ to form $ S$ (that is, $ k\minus{}1$ elements) and also any elements that are multiples $ kb$ of $ k$ for $ b\in\{k, k\plus{}1, \dots, \lfloor \frac{2002}{k}\rfloor\}$ (that is, $ \lfloor \frac{2002}{k}\rfloor \minus{}k\plus{}1$ elements). Note that as $ \lfloor \frac{2002}{k}\rfloor > k$ (because by assumption $ k<45$) and $ \left(\lfloor \frac{2002}{k}\rfloor \plus{}1\right)k > 2002 \geq \left(\lfloor \frac{2002}{k}\rfloor\right)k$, we are not overcounting or removing too many elements from $ K$ to form a set $ S$ with minimal element $ k$--the set $ S$ that is generated by this algorithm is a superset of the set with the greatest possible number of elements that has minimal element $ k$ (it is only a superset because we might need to remove, say, multiples of $ k\plus{}1$ for certain values of $ k$).

In total, we removed $ k\minus{}1\plus{}\lfloor \frac{2002}{k}\rfloor \minus{}k\plus{}1\equal{}\lfloor \frac{2002}{k}\rfloor$ elements from $ S$. But as $ k<45$, $ \lfloor \frac{2002}{k}\rfloor\geq 45$, so at least $ 45$ elements must be removed if $ k<45$ is to be the least element of $ S$, implying that this least element does not yield a maximal number of elements in $ S$. If we set the least element of $ S$ to be higher than $ 45$, say $ j$, then we will simply be removing at least (actually, exactly, in the best case) $ j\minus{}1\geq 45$ elements as well. We conclude that the set $ S$ shown in the first paragraph, with minimal element $ 45$ and all higher elements included, must be the set $ S$ with the most possible elements and thus the most elements that $ S$ can have is $ 1958$.[/hide]


geometry
circumcircle
trigonometry
function
logarithms
inequalities
angle bisector

it's maybe easy but i want a solution in maximum 2 lines!
let ABC be a scalene triangle. denote by D and M the intersection of the angle bisector <BAC with BC and circumcircle (O) (of triangle ABC), respectively. a line l passes through D and cuts MB and MC at P and Q respectively. prove that <PAQ is greater or equal to <BAC.

If my figure is correct, I think the statement is wrong. So please check my figure.

oh sorry i forgot to mention that C must be between M and Q and P between M and B!!

A bit longer than 2 lines....

Let the circumcircle of $ MPQ$ intersect $ MA$ again at $ K$.  In order to prove that $ \angle PAQ \ge\angle BAC$, we must prove that $ A$ lies between $ K$ and $ D$.  From Power of a Point we have $ BD\cdot CD \equal{} MD\cdot DA$ and $ MD\cdot DK \equal{} PD\cdot QD$.  We must show that $ BD\cdot CD\le PD\cdot QD$, or $ \frac {BD}{PD}\cdot\frac {CD}{QD}\le1$.  Law of Sines shows that this is equivalent to

$ \sin\angle BPD\sin\angle CQD \le \sin\angle PBD\sin\angle QCD$

We have $ \sin\angle PBD \equal{} \sin\angle QCD \equal{} \sin\frac {A}{2}$ and $ \sin\angle BPD \equal{} \sin\left(\frac {A}{2} \plus{} \angle PDB\right)$ and $ \sin\angle CQD \equal{} \sin\left(\frac {A}{2} \minus{} \angle QDC\right)$.  From $ \angle PDB \equal{} \angle QDC$ and from the fact that the function $ f(x) \equal{} \ln\sin x$ is concave ($ f''(x) \equal{} \minus{} \csc^2x$), the inequality follows.

ok. my solution: D lies in the circumcircle of triangle ABD(because (<ADB) > (<APB) ) so it follows (<BDP) > (<BAP) or (<CDQ) > (<PAB). similarly, D lies out of the circumcircle of ACQ. it means (<CDQ) < (<CAQ). DONE!!


Let be $ a = \underbrace {11...1}_{n + 1}\underbrace {22...2}_{n + 1} - \underbrace {33...3}_{n + 1}$ and $ b = 1\underbrace {77...7}_n9\underbrace {55...5}_n2 - 5\underbrace {33...3}_{n - 1}28$ , where $ n \in \mathbb{N^*}$. Compute $ \sqrt {\frac{{\sqrt a }}{{\sqrt b }}}$
[i]Claudiu Mindrila[/i]


Let  x,y,z  be real with $ \ x^2\plus{}y^2\plus{}z^2\equal{}1$
Find max $ \ | x^3\plus{}y^3\plus{}z^3\minus{}xyz|$

[quote="lucio"]Let  x,y,z  be real with $ \ x^2 \plus{} y^2 \plus{} z^2 \equal{} 1$
Find max $ \ | x^3 \plus{} y^3 \plus{} z^3 \minus{} xyz|$[/quote]
$ 1\minus{}(x^3\plus{}y^3\plus{}z^3\minus{}xyz)^2\equal{}(x^2\plus{}y^2\plus{}z^2)^3\minus{}(x^3\plus{}y^3\plus{}z^3\minus{}xyz)^2\equal{}$
$ \equal{}\sum_{cyc}\left(3x^4y^2\plus{}3x^4z^2\plus{}2x^2y^2z^2\minus{}2x^3z^3\plus{}2x^4yz\minus{}\frac{x^2y^2z^2}{3}\right)\equal{}$
$ \equal{}\sum_{cyc}\left(x^4y^2\plus{}x^4z^2\plus{}x^2y^2(x\minus{}y)^2\plus{}z^4(x\plus{}y)^2\plus{}\frac{5x^2y^2z^2}{3}\right)\geq0.$
Id est, $ | x^3 \plus{} y^3 \plus{} z^3 \minus{} xyz|\leq1.$ But for $ x\equal{}1$ and $ y\equal{}z\equal{}0$ the equality occurs.
Thus, $ \max_{x^2\plus{}y^2\plus{}z^2\equal{}1}| x^3 \plus{} y^3 \plus{} z^3 \minus{} xyz|\equal{}1.$

[quote="pvthuan"]It is also true that for all real numbers $ x,y,z$
\[ |x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyz|\leq (x^2 \plus{} y^2 \plus{} z^2)^{3/2}.
\]
[/quote]
It is also true that for all real numbers $ x,y,z$
\[ |x^3 \plus{} y^3 \plus{} z^3 \minus{} xyz|\leq (x^2 \plus{} y^2 \plus{} z^2)^{3/2}.
\]
[/quote]


why does polymerisation occur the way it does that  is head-tail,head-tail... :D

There are two types of polimerization: addition and condensation. For the last type, the polimerization occurs by reaction between the functional groups of the monomer(s) (with the elimination of a small molecule like water), and so there is no problem about the "selectivity".
Polimerization by addition can be of three types: radical, anionic, and cationic. However, in each of these three types, the reaction between monomers occurs so as to produce the more stable intermediate (radical, carbanion, or carbocation), and so there is a preferred selectivity for the polimerization. For example, styrene ($C_{6}H_{5}CH=CH_{2}$) reacts with a radical initiator RO (a peroxide for example) to produce $RO-CH_{2}CH(C_{6}H_{5}) \cdot$ and not the other radical, because this radical is a benzylic one and so is stabilised by resonance. Then this radical will react with another styrene molecule to produce $RO-CH_{2}CH(C_{6}H_{5})CH_{2}CH(C_{6}H_{5}) \cdot$ and so on.


For the next few months, Mathew Crawford will host a series of MATHCOUNTS Math Jams. Some of these Math Jams will be a game using problems similar to those found in MATHCOUNTS. In other MATHCOUNTS Math Jams, students will be asked to share with the class problems they either enjoy or have trouble with. Mathew will then lead the class in a discussion of the problems.

We strongly encourage MATHCOUNTS students to join us and invite their teammates and even their coaches! Mathew is a former National Test Champion, twice finished in the top 2 at Nationals, and led Alabama to the National Championship in 1991.

Great! when will the math jams be?  our first competition is in two weeks, so we need to get ready

Click the Math Jams button above for a schedule.

Are the MATHCOUNTS Math Jams appropriate for MATHCOUNTS beginners or is it intended for kids more at the State/Nationals level?  (I noticed you ranekd the difficulty at 2 rather than 1.)  I am thinking specifically about the 1pm ET timeslot one this week, if that makes a difference in terms of who you expect might show up or in what you have planned...

Thanks!

Your question is best answered by an admin here of course, but my opinion is that all mathcounts students can benefit from the math jams, because their very purpose is to demonstrate the problem solving process to those who are unable to solve (or solve quickly) the problems discussed.  Once you reach nationals, your main goal becomes to solve problems quickly, efficiently, and accurately.  Plus a few little tricks you come up with or learn. :D

The question and answer sessions are certainly good for [u][b]everyone[/b][/u].  Anyone who wants to know anything about any level of MATHCOUNTS problem solving and curriculum can participate.

The MATHCOUNTS games are good for nearly all MATHCOUNTS students, but the only students with a shot at winning the game are those who will be competing at nationals or won't be only because they're from one of the huge states.  It's a highly informal game, so students treat it mostly as a fun count down round practice (though the problems are on average bit tougher than most count down round problems).

Where is the button, I cant' see. Can anyone tell me? Please, I like Mathcount too much!!! :)

It is one of the tabs at the top of the page. However, if you still can't see it, here is the linlk:[url=http://www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php]www.artofproblemsolving.com/Community/AoPS_Y_Math_Jams.php[/url]

[quote="kchande"]:)

I would like to know what does Game Math Jam on 9th March mean. Is there going to be a game or some game related problems will be discussed. What kind of game problems.

Karan Batra.
kchande@yahoo.com[/quote]

Game Math Jams are like a big free for all Count Down Round.  If you want to see what one looks like, check out the transcripts in the Math Jams area.

For the Math Jam on wednesday, how many people will you allow, and how many are expected?

I'd like to be in wednesdays math jam...is there a place where i can sign up or simething.

There is no need to sign up for a Math Jam and there is enough room in the virtual classroom for everyone to attend.  Just log in with your username and hit the Classroom button.

Does some link open up when it's time, or is there something else you're supposed to do? Also, when can you enter the Math Jam tonight?

[quote="solafidefarms"]At 7:15 or so when you click the Classroom link, it opens the classroom. Else that link just opens a test Java applet to see if your computer is Java-enabled. If you don't see it, you probably need to get the Java plugin. If you see it and it is past 7:15, just refresh.

Billy[/quote]
k thanks... see I just joined three weeks ago so I've never done a Math Jam b4... so don't mind my cluelessness at times :blush:

[quote="ln(dx/dy)"]Do we have to sign up?[/quote]

Math Jams are informal.  You just show up.

[quote="addicted2sharpiz"]sounds like fun  :D ill be there although im kinda a noob to these competitions[/quote]

New students are always welcome.

Now I get to reveal to the world my awesomeness at quick random guesses.

Yeah!  MathCounts is the next MathJam! Maybe I can keep up this time. . . :P


algebra
polynomial
function
domain
Ring Theory
superior algebra
superior algebra unsolved

So this was my reasoning so far:

""We recognize that an ideal P is prime if and only if for two ideals A and B, AB $ \in$ P implies that either A or B is contained in P. So we must find "

So anyways, I'm thinking that it consists of all the irreducible polynomials in C[x,y] (I suppose those irreducibles can form ideals by means of multiples of those with other elements in the ring).

"Hm

First of all, standard irreducibility tests don't work
(because it's a complex domain, so (x^2 + 1) is reducible in this case).

So then we have to find irreducibles over complex numbers. BUT on the OTHER hand, we have xy, so now we can have irreducible factors of X and Y (maybe, irreducible polynomials can be elliptic curves like the one on Wikipedia)."

Y^2 − X^3 − X − 1 is prime ideal from wikipedia."

==

Another question: is there a branch known as multivariate abstract algebra? It's hard to find those topics in standard abstract algebra textbooks.

Each irreducible polynomial generates a prime ideal, and there are plenty of those. These are not all of the prime ideals; the ideal generated by $ x$ and $ y$ is another. Your reasoning ignores the fact that ideals in this ring might not be principal.

An ideal of $ \mathbb{C}[x, y]$ is the set of polynomials that vanish on a particular corresponding [url=http://mathworld.wolfram.com/AlgebraicVariety.html]algebraic variety[/url] (hence the elliptic curve example).  A prime ideal is a variety consisting of one "part," which is an intersection of the zero sets of irreducible polynomials (I think).

One way to look at it is dimension: there are varieties of dimension $ 0$, $ 1$, and $ 2$. Dimension 2 gives only the whole space, and the corresponding prime ideal is the zero ideal. Dimension 1 gives you curves; the corresponding prime ideals are singly generated, by irreducible polynomials. Dimension zero gives you points; the corresponding prime ideals are $ (x\minus{}a,y\minus{}b)$ (all polynomials vanishing at $ (a,b)$).
As you add more variables, the picture gets more complicated.


geometry
rectangle
calculus
calculus computations

Suppose you are designing an aquarium whose base is a rectangle twice as long as it is
wide. The volume is to be 2 cubic meters, and the base and sides will be made of glass.
What shape of aquarium will use the least amount of glass?

Let $x$ be the length of the base. The width is $x/2$; the height $h$ is determined from $x(x/2)h=2$. Now express the area of surface ($5$ rectangles) in terms of these sides and ultimately in terms of $x$. Optimize.


linear algebra
matrix
inequalities proposed
inequalities

For positive integer $n$ prove that:

$\sum^{n}_{k=1}\sqrt{\begin{pmatrix}n\\ k \end{pmatrix}}\leq\sqrt{n\left(2^{n}-1\right)}$

(note:$\begin{pmatrix}n\\ k \end{pmatrix}=\frac{n!}{k!(n-k)!}$ )
 :)


Hey!

Would it be possible (in USA) to read grad studies in mathematics and at the same time reading undergrad studies in physics?

When doing graduate studies in mathematics, will I have to option to decide how many courses (and which one) I would like to read fall 200X. Or has the courses already been predetermined by the institute, for all graduate students?

Here in my country a student (undergrad) could either read in a program (where all the courses have been determined except some few which the student could choose between). After reading courses worth X[color=red]*[/color] points he/she gets enough points to do a little "project" (like a simple and small thesis, you could say). After this is done the student gets a bachelor degree in, say, math.

Now, if the student doesn't choose to read in a program he/she could select all the courses himself. It's exactly as above except he gets to choose the courses himself.

The same goes for taking a Master degree. Of course, some courses are a "must read" in order to take the degrees.

So if I want I could read chemistry, biology, math, cs and physics at the same time and get, e.g., B.S. in them all.

Does it work the same way in USA?

One last thing, If I get accepted into a college when applying to the math department, does it make me accepted to the physics/whatever department at the same time? Or do I have to apply again to those separately?

Thank you.

[color=red]*[/color] Of course, out of these X point, Y of them, in this case, has to be in mathematics.

[quote="kahsi"]One last thing, If I get accepted into a college when applying to the math department, does it make me accepted to the physics/whatever department at the same time? Or do I have to apply again to those separately?[/quote]
For undergraduate, almost always yes, you are accepted to the university, not to a particular department. Many universities allow you to major in multiple subjects at once, but the top universities tend to discourage it. Usually, there are certain requirements for a major and certain requirements for graduation, and if you can fulfill the requirements for more than one major and for graduation in the normal amount of time, you will graduate with multiple majors.

I'm not the best advisor for graduate school since I haven't even started looking for schools for myself, but I think it is different there. In graduate school you are accepted to a particular department. People do get multiple graduate degrees (even PhDs) but I think it is rather rare to work on more than one simultaneously.

It is possible to do a graduate program and an undergraduate program simultaneously, but I get the impression that you have to work it out pretty carefully. I have a friend in a masters program for compsci and an undergrad program for math at the same time, but he did that by joining a BS/MS program for compsci.

However, when you are a graduate student, you can still take classes in other departments. So if you are a graduate student in mathematics, you can still take undergraduate classes (or even graduate classes) in the physics department. You won't get a degree in physics that way, but you can still learn whatever you want.

Exactly what requirements would that be?
This is basically what I can find.

[quote]In the U.S., bachelors' degrees typically require four years of full-time study,[/quote]
http://en.wikipedia.org/wiki/Bachelor%27s_degree#United_States

Now, a general question. Approximately how many courses do a regular (undergrad and grad) student read per year and how much can he affect the number of courses taken if he is in a program?


4] The function f satisfies
f(x) + f(2x + y) + 5xy = f(3x - y) + 2x^2 + 1
for all real numbers x; y. Determine the value of f(10)

Ok Help. I'm still kinda new to problem solving and don't understand the answer that they gave(seems like an assumption is made). Someone please give your thought process, as I find it to be the most helpful.

If my equation looks ugly, here's what its supposed to look like\/
http://web.mit.edu/hmmt/www/datafiles/solutions/2008.shtml

If you want to look at the answer description, it's at the same link of coarse.

$ f$ satisfies the given equation [b]for all[/b] real values $ x$ and $ y$. This means that it also satisfies the equation for any given reals $ x$ and $ y$, such as $ x \equal{} 10$ and $ y \equal{} 5$, for instance. No additional assumption had to be made.

Hopefully that cleared up your difficulty.

It seems to me like the fundamental concept behind functions is to plug in values in equation(s) to form common terms which cancel. This problem, unlike the easier one I had done deal with looking at the problems characteristics before blindly plugging in values(since the only question even vaguely similar only had one variable). Looking at the equations, it seems that only f(2x+y)=f(3x-y) or that 2x+y=3x-y will lead to anything since using y as zero leads to no simplification since you still have a number of values"that haven't been canceled. The key i guess, is to look first at finding the relationships between variables, before crossing them out.

Generally, note that $ y : \equal{} \frac{x}{2}$ immediately gives $ f(x) \equal{} \minus{} \frac{1}{2} x^2 \plus{} 1$

Basic function problems like this one are a matter of finding good values to plug in. One idea can be, as you said, to try to get two things on each side to be the same so they cancel. More complicated ones require tricky substitutions, proving injectivity/surjectivity, etc. There are plenty of difficult functional equations in the Olympiad Algebra forums of this site.


linear algebra
matrix
vector
linear algebra unsolved

Let $ A$ be a $ 3\times 3$ real matrix and $ A\equal{}(a_{ij}), B\equal{}(k\minus{}a_{ij})$, with $ k$ a natural number. Let $ S(.)$ denote the sum of the entries of its argument. Prove (or disprove!):

a) $ S(adj(A))\equal{}S(adj(B))$
b) $ \det(B)\equal{}kS(adj(A))\minus{}\det(A)$

PS. Another problem on the sum of the entries of a matrix is available on the forum.

Let $ u$ be the vector $ [1,1,1]^T$. Then $ S(V)\equal{}u^TVu\equal{}trace(uu^TV)$. Moreover $ B\equal{}kuu^T\minus{}A$. Let
$ f(k)\equal{}det(B)\equal{}det(c_1,c_2,c_3)\equal{}det(c_1,c_2\minus{}c_1,c_3\minus{}c_1)\equal{}
kf'(0)\plus{}f(0)\equal{}kf'(0)\minus{}det(A).$
$ f'(k)\equal{}trace(uu^Tadj(B))\equal{}S(adj(B))$. But $ f'(k)\equal{}f'(0)$ then:
$ S(adj(B))\equal{}S(adj(\minus{}A))\equal{}S(adj(A))$ and $ det(B)\equal{}kS(adj(A))\minus{}det(A)$.

Obviously the proof works for all dimensions $ n$ (with care) and the formulas are:
$ S(adj(B))\equal{}(\minus{}1)^{n\minus{}1}S(adj(A))$ and $ det(B)\equal{}(\minus{}1)^{n\minus{}1}(kS(adj(A))\minus{}det(A))$.


geometry
inequalities
combinatorics solved
combinatorics

We have finite squares that sum of there areas is 1.
Prove we can put them in square with area 2 without overlapping.
___________________________________
No pain,no gain

I'll post the solution asap, because I have the book :), with ASU problems :)

If I remember correctly, this problem has been proposed somewhere else on ML, thus I write a solution for both.

Let $c_1,...,c_n$ be the squares, with respective sides-length $s_1,...,s_n$.
Then, $ \sum_{i=1}^n s_i^2 = 1$, (1)
and wlog we may assume that $s_1 \geq s_2 \qeg ... \geq s_n$.

We put the squares $c_i$ on the big square following the most natural way, from the left-low corner to the right :
First, we put them on an array $c_1,c_2,...$ from left to the right, until we cannot add one without going outside the big square.
Then, if any remaining, we use the top side of $c_1$ as basis to start an other array, and so on.
Let $n_k$ be the number of squares which belong to the $k^{th}$ array.
Let $H_k = s_1+s_{n_1+1}+...+s_{n_1+n_2+...+n_{k-1}+1}$ be the height of the first $k$ arrays.

From (1), we deduce that $s_1 \leq 1$ thus the first array will fit on the big square.
If only one array suffices, we are done. Otherwise $n_1 <n$, and we have :
$s_1 + s_{n_1+1} \leq s_1 + s_{n_1} \leq s_1+s_2+...+s_{n_1} \leq \sqrt {2}$
It means that $H_1 \sqrt {2}$, thus the first two arrays will fit into the big square.
Moreover :
$s_1+s_2+...+s_{n_1} + s_{n_1+1} > \sqrt{2}$
thus
$s_{n_1+1}(s_2+...+s_{n_1} + s_{n_1+1}) \geq s_{n_1+1}(\sqrt{2} - s_1)$
and using the nondecreasing of the $s_i$'s, it follows that
$s_2^2+s_3^2+...s_{n_1+1}^2 \geq s_{n_1+1}(\sqrt{2} - s_1)$

Similarly, for each $k$, if we use a $(k+1)^{th}$ array then
$s_{n_1+...+n_{k-1}+1} + s_{n_1+...+n_{k-1}+2} + ... + s_{n_1+...+n_k} +s_{n_1+...+n_k+1} > \sqrt {2}$
thus
$s_{n_1+...+n_{k-1}+2}^2 + s_{n_1+...+n_{k-1}+3}^2 + ... + s_{n_1+...+n_k+1}^2 \geq s_{n_1+...+n_k+1}( \sqrt {2} - s_{n_1+...+n_{k-1}+1}) \geq s_{n_1+...+n_k+1}( \sqrt {2} - s_1) $

Let $p$ be the total number of arrays. In reasoning above let $s_{n_p+1}=0$.
Adding all these inequalities, it gives :
$ \sum_{i=2}^{n_1+n_2+...n_p+1} s_i^2 > ( \sqrt {2} - s_1)(s_{n_1+1}+s_{n_1+n_2+1}+...+s_{n_1+...+n_p+1})$
therefore :
$1 - s_1^2\geq ( \sqrt {2} - s_1)(H_p - s_1)$
which in turn gives :
$H_p \leq \frac {1-s_1^2} {\sqrt {2} - s_1} + s_1 = \sqrt {2} - \frac {(1-s_1 \sqrt {2})^2} { \sqrt {2} - s_1} \leq \sqrt {2}$.

It means that all the arrays fit together in the big square, and we are done.

Pierre.


I was doing some mocks one day, and I decided to create my own mock AIME. Of course, I'm not too great at AIME's and it's rare that I manage to solve any problem over #11 (let alone create one).

I was wondering if anyone more experienced (feels comfortable with making hard AIME problems) would like to team up with me?

I've got a rough draft of the first 7 problems so far, but those can be changed.

Edit: Altheman is a really good person to receive help from. Since I'm busy anyway, I rescind my assistance. Good luck!

I could contribute a few problems if you want. (Just PM me with topics/approximate difficulty) I'll look over the final test if you want. (preferred topics: geo, algebra, maybe a counting or NT)

People make them. As for which site he got all of those mock tests from, I would like to know too. Are they all around the real AIME level?

Veezbo, are you going to make your mock AIME public?

There are some Mock AIME's here: [url]http://www.artofproblemsolving.com/Wiki/index.php/Mock_AIME[/url].

it kind of seems like a long stretch for me to write really good problems (seeing as I am not that great at solving them)

but, if I somehow accumulate 15 good problems, I will make the AIME public.

And if anyone wants to help, please send me problems through PM :)


geometry
trapezoid
circumcircle
parallelogram
power of a point
radical axis
complex numbers

Given is the convex quadrilateral $ ABCD$. Assume that there exists a point $ P$ inside the quadrilateral for which the triangles $ ABP$ and $ CDP$ are both isosceles right triangles with the right angle at the common vertex $ P$. Prove that there exists a point $ Q$ for which the triangles $ BCQ$ and $ ADQ$ are also isosceles right triangles with the right angle at the common vertex $ Q$.

Assuming $ AB \parallel CD$ $ \Longrightarrow$ isosceles $ \triangle ABP \sim \triangle CDP$ are centrally similar with center $ P$ $ \Longrightarrow$ $ ABCD$ is isosceles trapezoid with diagonal intersection $ P,$ with perpendicular diagonals $ AC \perp BD,$ cyclic, and $ Q$ is obviously its circumcenter. Assuming $ BC \parallel DA$ $ \Longrightarrow$ isosceles $ \triangle ABP \cong \triangle CDP$ are congruent $ \Longrightarrow$ $ ABCD$ is isosceles trapezoid with circumcenter $ P,$ with perpendicular diagonals $ AC \perp BD$ and $ Q$ is obviously its diagonal intersection. Hence, we can assume $ AB, CD$ are not parallel and $ BC, DA$ are also not parallel. Let $ K, M$ be midpoints of $ AB, CD.$ Let $ X \equiv AB \cap CD,$ WLOG on the same side of $ KM$ as $ B, C.$ Quadrilateral $ PKXM$ is cyclic on account of the right angles at $ K, M$; let $ (S)$ be its circumcircle.

Let $ PB, PC$ cut $ (S)$ again at $ B', C'.$ By Pascal theorem for cyclic hexagon $ PKB'XC'M,$ the intersections $ B \equiv PB' \cap KX,$ $ C \equiv PC' \cap MX,$ $ L \equiv KC' \cap MB'$ are collinear $ \Longrightarrow$ $ L \in BC.$ Then

$ \angle KML \equal{} \angle KMB' \equal{} \angle KPB' \equal{} \angle KPB \equal{} 45^\circ$
$ \angle MKL \equal{} \angle MKC' \equal{} \angle MPC' \equal{} \angle MPC \equal{} 45^\circ$

Let $ PA, PD$ cut $ (S)$ again at $ A', D'.$ By Pascal theorem for cyclic hexagon $ PKD'XA'M,$ the intersections $ A \equiv PA' \cap KX,$ $ D \equiv PD' \cap MX,$ $ N \equiv KD' \cap MA'$ are also collinear $ \Longrightarrow$ $ N \in DA.$ Then

$ \angle KMN \equal{} 180^\circ \minus{} \angle KMA' \equal{} 180^\circ \minus{} \angle KPA' \equal{} \angle KPA \equal{} 45^\circ$
$ \angle MKN \equal{} 180^\circ \minus{} \angle MKD' \equal{} 180^\circ \minus{} \angle MPD' \equal{} \angle MPD \equal{} 45^\circ$

Combining, $ \triangle KLM \cong \triangle KNM$ are isosceles right and congruent $ \Longrightarrow$ $ KLMN$ is a square. Assume that $ L' \not\equiv L, N'\not\equiv N$ are midpoints of $ BC, DA.$ Since $ KLMN, KL'M'N'$ are both parallelograms $ \Longrightarrow$ $ KL \equal{} MN, KL' \equal{} MN', \angle LKL' \equal{} \angle NMN'$ $ \Longrightarrow$ $ \triangle LKL' \cong \triangle NMN'$ are centrally congruent by s.a.s. and $ (BC \equiv LL') \parallel (NN' \equiv DA),$ which is a contradiction. Therefore, $ L, N$ are midpoints of $ BC, DA$ and the square $ KLMN$ is Varignon parallelogram of the quadrilateral $ ABCD.$ Consequently, $ ABCD$ has perpendicular diagonals $ (AC \parallel KL) \perp (LM \parallel BD).$

The diagonal intersection $ E \equiv AC \cap BD$ is the 2nd intersection of the circles $ (K) \equiv \odot (ABP), (M) \equiv \odot (CDP).$ Let $ Q$ be the 2nd intersection of the circles $ (L) \equiv \odot(BCE), (N) \equiv \odot (DAE)$ $ \Longrightarrow$ the angles $ \angle BQC, \angle DQA$ are right. Radical axis $ EQ$ of $ (L), (N)$ is perpendicular to their center line, $ EQ \perp LN.$ Moreover, $ (LN \parallel BD) \perp AC$ or $ LN \perp (PC \equiv PA)$ $ \Longrightarrow$

$ \angle QBC \equal{} \angle QEC \equal{} \angle MLN \equal{} 45^\circ$ $ \Longrightarrow$ $ \triangle BCQ$ is isosceles right.
$ \angle QAD \equal{} \angle QED \equal{} \angle MNL \equal{} 45^\circ$ $ \Longrightarrow$ $ \triangle ADQ$ is isosceles right.

Here is shorter solution to the problem:
   Let's denote by $ O$ the point on $ BD$ such that $ AO$ is perpendicular to $ BD$. Than, $ \angle{AO_{1}B} \equal{} 90^\circ \equal{} \angle{APB}$. It means that $ A,P,O_{1},B$ are concyclic. So, $ \angle{AO_{1}P} \equal{} \angle{APB} \equal{} 45^\circ$ and $ \equal{} \angle{DO_{1}P} \equal{} \angle{DCP} \equal{} 45^\circ$ and points $ D,C,O_{1},P$ are also concyclic, yielding $ \angle{DO_{1}C} \equal{} \angle{DPC} \equal{} 90^\circ$ and $ \angle{AO_{1}B} \plus{} \angle{CO_{1}D} \equal{} 180^\circ$ and hence, $ O$ is the intersection of $ AC$ and $ BD$ and the diagonals are perpendicular to each other. Furthermore, $ \triangle{APC} \equal{} \triangle{BPD}$ and $ AC \equal{} BD$ (*).
      Let $ Q$ be the common point of circumcircles of triangles $ AOB$ and $ COD$, other than $ O$. $ \angle{ADQ} \equal{} \angle{COQ} \equal{} \angle{CBQ}$. Hence, triangles $ ADQ$ and $ CBQ$ are similar and $ \frac {BQ}{CQ} \equal{} \frac {QD}{QA}$, but $ \angle{BQD} \equal{} \angle{CQA}$. So, $ \triangle{BQD}$ is similar to $ \triangle{CQA}$, but (*) yields that they are equal and $ BQ \equal{} QC$, $ AQ \equal{} QD$. So there exists such a point $ Q$.
   Remark: If the circumcircles of triangles $ AOB$ and $ COD$ are tangent to each other, than we analogously deduce that $ O \equal{} Q$ satisfies the problem.

Short complex numbers solution:

Let P be the origin. A = a, B = ai, C = c, D = ci. Let Z=z the the point such that (b-z)i = c-z, or equivalently a+c-z=-zi, which can be written as (ci-z)i = (d-z)i = a-z. Thus, z = Q.


I have no idea for this one:
  For any $ n\geq 5$ we can find $ 2n+1 $ odd, distinct square-free positive integers such that the sum of their reciprocals is 1. What about you?


I am preparing an article, like this:

\documentclass{amsart}

\usepackage{amsmidx}
\makeindex{}
\makeindex{}

\begin{document}

\index{}{}
\index{}{}

\Printindex{}{}
\Printindex{}{}

But it don't work!

I also want to decide whether the indexes will be printed on a new page.

Is there a way to accomplish this ?

Thank you very much!

What does "But it don't work!" mean? What are the errors? What does the log say? Please provide a [b]complete[/b] [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] that can be compiled.

Thanks!

\documentclass[reqno]{amsart}
\usepackage{amsthm, amssymb, amsmath, stmaryrd, mathrsfs,amscd}

\usepackage{amsmidx}
\makeindex{names}
\makeindex{others}

\usepackage[pdftex,bookmarksnumbered,bookmarksopen,colorlinks,linkcolor=red,anchorcolor=black,citecolor=green,urlcolor=blue]{hyperref}

\swapnumbers
\theoremstyle{plain}
\newtheorem{theorem}[subsubsection]{Theorem}
\newtheorem*{lemma}{Lemma}
\newtheorem{cor}[subsubsection]{Corollary}
\newtheorem{prop}[subsubsection]{Proposition}
\newtheorem{conj}[subsubsection]{Conjecture}
\newtheorem{ques}[subsubsection]{Question}
\newtheorem*{notn}{Notation}
\newtheorem*{thm}{Theorem}
\newtheorem*{pp}{Proposition}

\theoremstyle{definition}
\newtheorem{defn}[subsubsection]{Definition}
\newtheorem{cons}[subsubsection]{Construction}
\newtheorem{remark}[subsubsection]{Remark}
\newtheorem{exam}[subsubsection]{Example}

\numberwithin{equation}{section}

\begin{document}
\section{}

.....arithmetic\index{names}{arithmetic}

.....Prime\index{others}{Prime}

\Printindex{names}{Contributors}

\Printindex{others}{Index}

\end{document}

I want to make an  index.
If I remove the following code , then well.

\usepackage{amsmidx}
\makeindex{names}
\makeindex{others}

\index{names}{arithmetic}

\index{others}{Prime}

\Printindex{names}{Contributors}
\Printindex{others}{Index}

If you read the [url=http://www.tug.org/texlive/Contents/live/texmf-dist/doc/latex/amscls/amsmidx.pdf]documentation[/url] for amsmidx it is [b]only[/b] for book classes so you'll need to use \documentclass[reqno]{amsbook} or similar.
Also amsmidx clashes with hyperref so you must put it after hyperref
[code]\usepackage[pdftex,bookmarksnumbered,bookmarksopen,colorlinks,linkcolor=red,anchorcolor=black,citecolor=green,urlcolor=blue]{hyperref}
\usepackage{amsmidx} [/code]This will now compile. But the problem now is that hyperref should be the last package loaded so other things may go wrong, you'll just have to try it.
If you have to have multiple indexes rather than just one you could try [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=splitindex]SplitIndex[/url] or [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=multind]multind[/url] or [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=index]index[/url] packages.

Thanks for your answer here and at http://www.cqf.info/forum/viewtopic.php?t=4420

the index comes  in a new page.

How  to tell LaTeX whether or not it will start a new page for index or References ?

amsmidx assumes that the index should start on the next odd page of a two-sided book as it issues a \cleardoublepage command. To get rid of this redefine the \Printindex command with
[code]\makeatletter
\renewcommand{\Printindex}[2]{%
  \begingroup
    % \cleardoublepage put whatever you want here
    \def\indexname{#2}%
    \raggedright
    \hyphenpenalty=10000
    \renewcommand{\seename}{see~also}
    \renewcommand{\subitem}{\par\hangindent 3em\hspace*{1em}}
    \@input{#1.ind}%
  \endgroup
  }
\makeatother [/code]

Not using \cleardoublepage still leaves one new page for the index, rather than two. You need to redefine amsbook's indextitlestyle and endtheindex commands to get rid of all new pages for the indexes:
[code]\makeatletter
\renewcommand{\Printindex}[2]{%
  \begingroup
    %\cleardoublepage
    \def\indexname{#2}%
    \raggedright
    \hyphenpenalty=10000
    \renewcommand{\seename}{see~also}
    \renewcommand{\subitem}{\par\hangindent 3em\hspace*{1em}}
    \@input{#1.ind}%
  \endgroup
  }

\renewcommand{\@indextitlestyle}{%
  \let\@makeschapterhead\indexchap
  \@xp\section\@xp*\@xp{\indexname}% use sections not chapters here (chapter would mean new page)
}

\renewcommand{\endtheindex}{} %get rid of clearpage

\makeatother [/code]

thank you very much!

If I want to make a bookmark and  an  index at the same time,  then I  have to use  hyperref  and other package about index  than amsmidx?

For instance, i could try SplitIndex or multind or index package , as you mentioned.

Bookmarks and index both work with amsmidx in:
[code]\begin{document}
\chapter{Chapter}
\section{Section}
.....arithmetic
\index{names}{arithmetic}
.....Prime
\index{others}{Prime}
\Printindex{names}{Contributors}
\Printindex{others}{Index}
\end{document}[/code]so you would have to see what happens in a real document.


What is the relation between $a_{k+1} + (k-1) \sqrt[k+1]{a_1a_2 \cdots a_{k+1}}$ and $k\sqrt[k]{a_{k+1}\cdot (\sqrt[k+1]{a_1a_2 \cdots a_{k+1}})^{k-1}}$?

Editted: Thanks for pointing it out deej21.  :)

I assume for the first one you mean $\sqrt[k+1]{a_1a_2 \cdots a_{k+1}}$ instead of what you have there.

[quote="peeta"][hide]AM-GM: $a+nb\ge (n+1)\sqrt[n+1]{ab^n}$[/hide][/quote]
Why?

It's AM GM on $\{a,b,b,\ldots,b\}$ (there are $n$ $b$'s).

\[\frac{a+b+b+\cdots+b}{n+1}\ge \sqrt[n+1]{ab^n}.\]


inequalities
trigonometry
inequalities proposed

:?:
$(cosx)^{sinx} +(sinx)^{cosx}>\cos x+\sin x >1$

$(cosx)^{sinx} +(sinx)^{cosx}>\frac{cos x}{sinx+cosx}+\frac{sin x}{sinx+cosx}=1$ :blush:

[quote="cezar lupu"]If $x, y>0$ and $x+y\leq\frac{\pi}{2}$ then $(cosx)^{sinx} +(sinx)^{cosx} >1.$[/quote]
This question similar to the following:
Prove that $x^y+y^x>1$ if  $0<x,y<1$   which you can get it from
http://www.mathlinks.ro/Forum/viewtopic.php?t=484


I Mathblitz (or eddy if u want to call me that) creator of the eleet club (join now) have decided to create the League of club makers or LCM for short this club devotes itself to suggesting clubs and creating them but not just clubs r talked about here games and  other things will also be made in this League famous ones like the public shaming game shall be invented but if u want credit JOIN NOW!

P.S.ill decide about prez and other things later

i will join. i specialize in pointless but fun clubs


wow, so many moderators... i was made one too...  which section do i moderate?  i cant find it :S

this is a beta release of the site and there are still some problems (Valentin posting here :D)  ... you are not actually all moderators, it's just a minor bug, which will be fixed as soon as we finish our complex analisys examn :D

problem fixed. hope you all happy you got your well earned stars back :D:D

Why? This is literally older than most of the AoPS community.

It is (or now at least was) the oldest non-locked topic on Fun Factory.


function
number theory
relatively prime
algebra proposed
algebra

Find the number of functions $ f: \mathbb N\rightarrow\mathbb N$ which satisfying:
(i) $ f(1) \equal{} 1$
(ii) $ f(n)f(n \plus{} 2) \equal{} f^2(n \plus{} 1) \plus{} 1997$ for every natural numbers n.

I'll write $ a_{n}$ instead of $ f(n)$.

Then the second condition gives for $ n =k, \,\, a_{k} \cdot a_{k+2} = a^{2}_{k+1} + 1997$

and for $ n = k+1, \,\, a_{k+1} \cdot a_{k+3} = a^{2}_{k+2} + 1997$.

By subtracting we get

$ a_{k} \cdot a_{k+2} - a_{k+1} \cdot a_{k+3} = a^{2}_{k+1} - a^{2}_{k+2} \Leftrightarrow \\
\\
a_{k} \cdot a_{k+2} + a^{2}_{k+2} = a_{k+1} \cdot a_{k+3} + a^{2}_{k+1} \Leftrightarrow \\
\\
a_{k+2}(a_{k}+a_{k+2}) = a_{k+1}(a_{k+1}+a_{k+3}) \Leftrightarrow \\
\\
\frac{a_{k+2}}{a_{k+1}} = \frac{a_{k+1}+a_{k+3}}{a_{k}+a_{k+2}}$ (1)

We set $ a_{k}+a_{k+2} = b_{k}$ and (1) is written $ \frac{a_{k+2}}{a_{k+1}} = \frac{b_{k+1}}{b_{k}}$ (2)

From (2) we have

 $ \prod_{k=1}^{n}\frac{a_{k+2}}{a_{k+1}} = \prod_{k=1}^{n}\frac{b_{k+1}}{b_{k}}
\Leftrightarrow \frac{a_{n+2}}{a_{2}} = \frac{b_{n+1}}{b_{1}} \Leftrightarrow \\
\\
\frac{a_{n+2}}{a_{2}} =\frac{a_{n+1}+a_{n+3}}{a_{1}+a_{3}}$ (3)

From the first condition $ a_{1} = 1$ and let $ a_{2} = m \in \mathbb{N}$.

Then $ a_{3} = m^{2} + 1997$ and (3) becomes

$ \frac{a_{n+2}}{m} =\frac{a_{n+1}+a_{n+3}}{m^{2} + 1998} \Leftrightarrow \\
\\
a_{n+2} \frac{m^{2}+1998}{m} = a_{n+1}+a_{n+3} \Leftrightarrow \\
\\
a_{n+2} \frac{1998}{m} = a_{n+1}-m\cdot a_{n+2}+a_{n+3}$

Since $ a_{n+1}-m\cdot a_{n+2}+a_{n+3} \in \mathbb{N}$ we have that $ a_{n+2} \frac{1998}{m} \in \mathbb{N}, \,\, \forall n$

Assume that $ m$ is not a divisor of $ 1998$. Then $ m| 1998 \cdot a_{n+2}, \,\, \forall n$ . (4)

We multiply the second condition by $ 1998^{2}$ and we get

$ 1998 \cdot a_{n+1} 1998 \cdot a_{n+3} = 1998^{2} a^{2}_{n+2} + 1998^{2}\cdot 1997$

and from (4)  we have that $ m^{2}| 1998^{2}\cdot 1997$ which is impossible

since $ 1997$ is a prime number.

Therefore the only accepted values of $ m$ are the divisors of $ 1998$.

But $ \tau(1998) = 16$ so there are $ 16$ functions satisfying the two conditions.

Nice problem. Excuse me for mistakes.  :)

Hello, this is my solution.

We can replace 1997 with $p$, any prime number. Now, similar to the solution above,  let $a_2 = m$, so $a_3 = m^2+p$ and $a_4 = (a_3^2+p)/m = m^3 + 2mp + p(p+1)/m$
In order for $a_4$ to be an integer, $m$ must divide $p(p+1)$.

we have:
$a_k a_{k+2} = a_{k+1}^2 + p$
$a_{k+1} a_{k+3} = a_{k+2}^2 + p$
After subtracting and rearranging, we have:
$\frac{a_{k+1}}{a_{k+2}} = \frac{a_k + a_{k+2}}{a_{k+1} + a_{k+3}}$ for all $k$.

That means, for any $n$,

\[\frac{a_2}{a_n} = \frac{a_2}{a_3} \frac{a_3}{a_4} \cdots \frac{a_{n-1}}{a_n}= \frac{a_1+a_3}{a_2+a_4} \frac{a_2+a_4}{a_3+a_5} \cdots \frac{a_{n-2}+a_n}{a_{n-1}+a_{n+1}} = \frac{a_1 + a_3}{a_{n-1} + a_{n+1}}\]

So that
$a_{n-1} + a_{n+1} = \frac{a_1 + a_3}{a_2} a_n = \frac{p+1}{m}a_n + ma_n$

Now, $m=1$ clearly works and would make $a_{n+1}$ an integer if $a_1, \cdots, a_n$ are already integers.
Let us consider the case where $m>1$. Since $p$ and $p+1$ are relatively prime, $m$ can only divide either $p$ or $p+1$ but not both.
If $m$ divides $p+1$, then it still works. $a_{n+1}$ is guaranteed to be an integer as long as $a_1, c\dots, a_n$ are already integers.
If $m$ divides $p$, since $m > 1$, then $m = p$.

But then our sequence would become:
$a_1 = 1, a_2 = p, a_3 = p^2+p = p(p+1)$
$a_4 = p^3 + 2p^2 + p + 1$
$a_5 = (a_4^2 + p)/a_3 = \frac{(p^3 + 2p^2 + p + 1)^2+p}{p(p+1)}$

But it's easy to see that the numerator of that expression is not divisible by $p$, and hence $a_5$ cannot be an integer.

So the only values of $m$ that are allowed is just divisors of $p+1$, including $1$.


geometry
analytic geometry
trigonometry
trig identities
Law of Cosines

The polar coordinates of points A, B, and C are A (2, 70 :^o: ), B(2sqrt3, 40 :^o: ), and C(3, 10 :^o: ). Compute the area of triangle ABC.

(If a point P has polar coordinates (r,  :theta: ), then P is r units from the origin, O, and ray OP makes an angle of  :theta: with the positive x-axis)

You can use Law of cosines to find the lengths of the triangle.

Angle AOB is 70 - 40 = 30, so using law of cosines in triangle AOB gives AB = 2
Angle BOC is 40 - 10 = 30, so using law of cosines in triangle BOC gives BC = :sqrt: 3
Angle AOC is 70 - 10 = 60, law of cosines gives AC = :sqrt: 7

notice the triangle is a right triangle: 4+3 = 7, so the area is (2)( :sqrt: 3) (1/2)=:sqrt: 3

You can also youse 1/2sin(C)*A*B for area of traingle

New Problem:
Find a general formula for a triangle with points at (a,A:^o:), (b,B:^o:), and (c,C:^o:).

by either the distance formula (after converting polar to rectangular) or by law of cosines you can get the side lengths... then i suppose you could apply heron's but thats really ugly, does anyone have a better way?

the sides would be:

 :sqrt: (a :^2: +b :^2: -2abcos(A-B))
 :sqrt: (b :^2: +c :^2: -2abcos(B-C))
 :sqrt: (c :^2: +a :^2: -2abcos(C-A))

kinda like how mathkid solved the first one, too

i tried doing it the way Ragingg suggested..

angle AOC = 60
area AOC = (1/2)(2)(3)(sin 60) = 3rt3 / 2

area AOB = (1/2)(2)(2rt3)(sin 30) = rt 3

area BOC = (1/2)(2rt3)(3)(sin 30) = 3rt3 / 2

area[AOC] - (area[AOB] + area[BOC]) = - rt 3

what did i do wrong?

You made the wrong calculation at the end.  If you make the picture, you'll see that you want the negative of what you calculated.


Let $x_1, \ldots, x_n$ and be positive numbers and $n$ be a natural number. Prove that

\[\sqrt{\sum^n_{k=1} x_k} + \sqrt{\sum^n_{k=2} x_k} + \sqrt{\sum^n_{k=3} x_k} + \ldots
\geq \sqrt{\sum^n_{k=1} k^2 \cdot x_k}\]

The Minkowski inequality (which actually follows from multiple application of the triangle inequality in $\mathbb{R}^n$) says that if $a_{i,j}$ are real numbers, for all natural i and j with $1\leq i\leq k$ and $1\leq j\leq n$, then

$\sqrt{a_{1,1}^{2}+a_{1,2}^{2}+...+a_{1,n}^{2}}+\sqrt{a_{2,1}^{2}+a_{2,2}^{2}+...+a_{2,n}^{2}}+...+\sqrt{a_{k,1}^{2}+a_{k,2}^{2}+...+a_{k,n}^{2}}$
$\geq \sqrt{\left( a_{1,1}+a_{2,1}+...+a_{k,1}\right) ^{2}+\left( a_{1,2}+a_{2,2}+...+a_{k,2}\right) ^{2}+...+\left( a_{1,n}+a_{2,n}+...+a_{k,n}\right) ^{2}}$.

Apply this inequality for n = k and the numbers $a_{i,j}$ defined as follows:

$a_{i,j}=\sqrt{x_{j}}$ for $i\leq j$
and $a_{i,j}=0$ for i > j.

Then you get

$\sqrt{\left( \sqrt{x_{1}}\right) ^{2}+\left( \sqrt{x_{2}}\right) ^{2}+...+\left( \sqrt{x_{n}}\right) ^{2}}+\sqrt{\left( \sqrt{x_{2}}\right) ^{2}+\left( \sqrt{x_{3}}\right) ^{2}+...+\left( \sqrt{x_{n}}\right) ^{2}}$
$+...+\sqrt{\left( \sqrt{x_{n}}\right) ^{2}}\geq \sqrt{\left( 1\sqrt{x_{1}}\right) ^{2}+\left( 2\sqrt{x_{2}}\right) ^{2}+...+\left( n\sqrt{x_{n}}\right) ^{2}}$,

so that

$\sqrt{x_{1}+x_{2}+...+x_{n}}+\sqrt{x_{2}+x_{3}+...+x_{n}}+...+\sqrt{x_{n}}\geq \sqrt{1x_{1}+2^{2}x_{2}+...+n^{2}x_{n}}$.

And this is enough. (BTW, I wouldn't call this "Pre-Olympiad", unless there is a much simpler proof which I didn't note.)

  Darij

Another approach:

Define $x_i = X^2_i,$ $\forall i \in \{1, \ldots, n\},$ and plug that all in the provided inequalities:

\[\sqrt{\sum^n_{k=1} X^2_k} + \sqrt{\sum^n_{k=2} X^2_k} + \sqrt{\sum^n_{k=3} X^2_k} + \ldots \geq \sqrt{\sum^n_{k=1} (k \cdot X_k)^2}\]

This is the triangle inequality which states that the sum of the length of the vectors $(X_1,X_2,X_3, \ldots), (0, X_2, X_3, X_4, \ldots), (0, 0, X_3, X_4, \ldots), \ldots$ is at least the lenght of the sum of these vectors: $(X_1, 2X_2, 3X_3, 4X_4, \ldots).$


[b]Problema 1713:[/b]
Pe laturile (BC), (CA) si (AB) ale triunghiului ABC se iau punctele A', B' si C' astfel ca dreptele AA', BB' si CC' sa fie concurente. Notam cu D, E, F, D', E' si F' mijloacele segmentelor (AB), (BC), (CA), (A'B'), (B'C') si respectiv (C'A'). Demonstrati ca:
[i]a[/i]). dreptele DD', EE' si FF' sunt concurente intr-un acelasi punct, care se gaseste pe o aceeasi dreapta cu punctul de intersectie al dreptelor AA', BB', CC' si cu centrul de greutate al triunghiului ABC;
[i]b[/i]). in cazul in care AA', BB' si CC' sunt inaltimile triunghiului ABC, punctul de concurenta al dreptelor DD', EE' si CC' coincide cu [i]centrul cercului lui Euler[/i] al triunghiului ABC;
[i]c[/i]). in cazul in care dreptele AA', BB' si CC' sunt bisectoarele triunghiului ABC, atunci punctul lor de intersectie([i]centrul cercului inscris[/i]), [i]punctul lui Nagel[/i](punctul de intersectie al dreptelor care trec prin virfurile triunghiului ABC si injumatatesc perimetrul sau) si punctul comun al dreptelor DD', EE', FF' se garesc pe o aceeasi dreapta;
[i]d[/i]). in cazul in care dreptele AA', BB', CC' injumatatesc perimetrul triunghiului ABC, punctul de intersectie al dreptelor DD', EE' si CC' coincide cu centrul de greutate al perimetrului triunghiului ABC.
[b]Problema 1724:[/b]
In triunghiul ABC notam cu O punctul de intersectie al inaltimilor AD si CE. Dreptele DE si AC se intersecteaza in punctul K. Demonstrati ca mediana BM a triunghiului ABC este perpendiculara pe dreapta OK.
[b]Problema 1728:[/b]
In triunghiul ABC notam cu K si cu L punctele de tangenta ale laturilor (AC) si (BC) cu cercurile exanscrise acestor laturi. Demonstrati ca dreapta determinata de mijloacele segmentelor (KL) si (AB) are urmatoarele proprietati: [i]a[/i]). injumatateste perimetrul triunghiului ABC;  [i]b[/i]. este paralela cu bisectoarea unghiului ACB.
[b]Problema 1737:[/b]
Corzile (AC) si (BD) ale unui cerc cu centrul in O se intersecteaza in punctul K. Notam cu M si N centrele cercurilor circumscrise triunghiurilor AKB si CKD. Demonstrati ca patrulaterul OMKN este un paralelogram.
[b]Problema 1747:[/b]
Notam cu A', B' si C' punctele de tangenta ale laturilor triunghiului ABC cu cercul inscris. Prin punctul P de concurenta al dreptelor AA', BB' si CC' se duc trei cercuri, fiecare dintre aceste cercuri fiind tangent la cate doua laturi ale triunghiului. Demonstrati ca: [i]a[/i]). Cele sase puncte de tangenta ale acestor cercuri sunt varfurile unui hexagon inscris intr-un cerc, iar centrul cercului circumscris acestui hexagonului coincide cu centrul  cercului inscris in triunghiul ABC;
[i]b[/i]). diagonalele principale (care unnesc varfurile opuse) ale acestui hexagon sunt concurente in punctul P;  [i]c[/i]). cel de al doilea punct de intersectie al cercurilor care trec prin punctul P se gasesc in mod respectiv pe dreptele AA', BB' si CC'.
[b]Problema 1773:[/b]
In triunghiul ABC dreptunghic in C, inaltimea CD si bisectoarea AE se intersecteaza in F. Notam cu G punctul de intersectie al dreptelor DE si BF. Demonstrati ca patrulaterul CEGF si triunghiul BDG au aceeasi arie.
[b]Problema 1788:[/b]
In triunghiul ABC notam cu I centrul cercului inscris, iar cu A', B' si C' punctele de tangenta al acestui cerc cu laturile BC, CA si respectiv AB. Dreptele AA' si BB' se intersecteaza in P, dreptele AC si A'C' in M, iar dreptele BC si B'C' in punctul N. Demonstrati ca dreptele IP si MN sunt perpendiculare.
[b]Problema 1800:[/b]
Demonstrati ca suma patratelor ariilor fetelor unui tetraedru este de patru ori mai mare ca suma patratelor ariilor celor trei paralelograme care au virfurile in mijloacele muchiilor tetraedrului.
[b]Problema 1815:[/b]
Demonstrati ca, in cazul in care perpendicularele comune ale laturilor opuse ale unui patrulater stramb sunt perpendiculare, aceste perpendiculare comune sunt concurente.
[b]Problema 1832:[/b]
Cercul inscris in triunghiul ABC este tangent la laturile BC, CA si AB in punctele A', B' si respectiv C'. Notam cu Q mijlocul segmentului (A'B'). Demonstrati ca unghiurile B'C'C si A'C'Q sunt congruente.
[b]Problema 1848:[/b]
Notam cu O centrul cercului inscris in triunghiul ABC, iar cu $ A_1, B_1, C_1$ punctele sale de tangenta cu laturile. Segmentele (OA), (OB) si (OC) intersecteaza cercul inscris in mod respectiv, in punctele $ A_2, B_2, C_2$. Demonstrati ca aria triunghiului $ A_2B_2C_2$ este egala cu jumatatea ariei hexagonului $ B_1A_2C_1B_2A_1C_2$.
[b]Problema 1860:[/b]
Fie ABCD un patrulater circumscris unei elipse si F unul dintre focarele elipsei. Demonstrati ca unghiurile AFB si CFD sunt suplementare.
[b]Problema 1878:[/b]
Demonstrati ca, punctul de intersectie al tangentelor duse la cercul avand drept diametru inaltimea [CH] a triunghiului ABC, in punctele de intersectia ale acestui cerc cu laturile (AB) si (AC) se gaseste pe dreapta suport a medianei [AM].
[b]Problema 1910:[/b]
Pe ipotenuza (AB) a triunghiului ABC, dreptunghic in A, luam un punct D. Notam cu $ O_1$ si cu $ O_2$ centrele cercurilor circumscrise triunghiurilor ACD si respectiv BCD, iar cu E punctul de intersectie al dreptelor $ BO_1$ si $ AO_2$. Demonstrati ca unghiurile BCE si ACD sunt congruente.
[b]Problema 1915:[/b]
In tetraedrul ABCD, |AB|=|BC|=|CD|=[i]a[/i] si |BD|=|DA|=|AC|=[i]b[/i]. Determinati distanta dintre dreptele AD si BC.
[b]Problema 1921:[/b]
Fie (AB) cea mai lunga latura a triunghiului ABC. Pe segmentul (AB) luam punctele M si N, astfel inncat sa avem: |BC|=|BM| si |CA|=|AN|, iar pe laturile (CA) si (BC) luam punctele P si respectiv Q, astfel incat PM||BC si QN||CA. Demonstrati ca: |QC|=|CP|.
                                                                                ([b][i]va continua[/i][/b])

[b]Problema 1927:[/b]
Fie ABC un triunghi oarecare(neisoscel!). Notam cu O, I si H centrul cercului circumscris, centrul cercului inscris si ortocentrul triunghiului. Punctele O, I si H pot sa fie varfurile unui triunghi isoscel?
[b]Problema 1935:[/b]
Toate fetele unui tetraedru sunt triunghiuri asemenea. Este oare necesar ca aceste fete sa fie congruente intre ele?
[b]Problema 1946:[/b]
In triunghiul ABC isoscel cu varful in C, AH si CL sunt inaltimi, iar I este centrul cercului inscris. Demonstrati ca lungimea proiectiei CH) a laturii (AC) pe dreapta BC este egala cu lungimea segmentului (AB), atunci si numai atunci cand: IH||AB.
[b]Problema 1952:[/b]
In triunghiul ABC, AH este inaltime, BL bisectoare si CM mediana. Demonstrati ca dreptele AH, BL si CM sunt concurente, atunci si numai atunci, cand: [i]a[/i]: $ LH\parallel{}AB$;  [i]b[/i]). $ sinA \equal{} tgB.cosC$.
[b]Problema 1955:[/b]
Punctul D se gaseste pe mediatoarea unei laturi a triunghiului ABC. Demonstrati, ca punctul C se gaseste pe mediatoarea uneia dintre laturile triunghiului ABD.
[b]Problema 1964:[/b]
Cercul exanscris laturii [AB] a triunghiului ABC este tangent la latura [AB] in C' si la prelungirile laturilor (AC) si (BC) in punctele B' si A'. Dreptele AA' si BB' se intersecteaza in punctul K. Demonstrati ca punctul K se gaseste pe cercul circumscris triunghiului ABC, atunci si numai atunci, cand razele cercurilor circumscrise triunghiurilor ABC si A'B'C' sunt egale.
[b]Problema 1968:[/b]
Construim simetricul fiecarui varf al unui patrulater convex (Q), fata de diagonala care nu trece prin acest varf. Aceste simetrice determina un nou patrulater (Q'). Demonstrati ca:
[i]a[/i]). Daca (Q) este un trapez, atunci (Q') este tot un trapez.
[i]b[/i]). Aria patrulaterului (Q') nu depaseste triplul ariei patrulaterului (Q).
[b]Problema 1973:[/b]
In triunghiul ABC, |AB|<|AC|, notam cu I centrul cercului inscris, iar cu M si cu N mijlocul segmentului [AC] si respectiv mijlocul arcului ABC. Demonstrati congruenta unghiurilor IMA si INB.
[b]Problema 1978:[/b]
Bisectoarele unghiurilor BAD si BCD a patrulaterului inscriptilil ABCD se intersecteaza intr-un punct K al diagonalei (BD) si fie M mijlocul acestei diagonale. Paralela dusa prin varful C la dreapta AD intersecteaza semidreapta AM intr-un punct P situat in afara patrulaterului. Demonstrati ca: |DP|=|DC|.
[b]Problema 1985:[/b]
Patrulaterul ABCD, care nu are laturi paralele, este circumscris unui cerc cu centrul in O. Notam cu K, L, M si N mijloacele segmentelor [AB], [BC], [CD] si respectiv [DA]. Demonstrati ca,  daca punctele K, M si O sunt coliniare, atunci punctele L, N si O sunt deasemmenea coliniare.
[b]Problema 1990:[/b]
Pe prelungirea laturii [BC] dincolo de punctul C se ia un punct X. Cercurile inscrise in triunghiurile ABX si ACX se intersecteaza in punctele P si Q. Demonstrati ca toate dreptele PQ trec printr-un acelasi punct a carui pozitie nu depinde de pozitia punctului X. ([i]L. Emelianov[/i]).
[b]Problema 1993:[/b]
Notam cu H ortocentrul triunghiului ABC si fie H un punct care nu se gaseste pe niciuna dintre dreptele AH, BH si CH. Cercul de diametru [HX] intersecteaza a doua oara dreptele AH, BH si CH in punctele $ A_1, B_1$ si $ C_1$, iar dreptele AX, BX si CX in punctele $ A_2, B_2$ si respectiv $ C_2$. Demonstrati ca dreptele $ A_1A_2, B_1B_2$ si $ C_1C_2$ sunt concurente!
[b]Problema 1997:[/b]
Pe laturile triunghiului dreptunghic ABC drept baze, se contruiesc in exteriorul triunghiului trei patrate si fie D, E si F centrele lor. Demonstrati ca aria triunghiului DEF este mai mare sau egala cu dublul ariei triunghiului ABC.
[b]Problema 2012:[/b]
Fie I centrul sferei inscrise in tetraedrul ABCD. Notam cu B', C' si cu D' proiectiile virfului A pe planele (ICD), (IBD) si respectiv (IBC). Demonstratica planul (B'C'D') este parale cu planul (BCD).
[b]Problema 2031:[/b]
Dreptele suport ale medianelor triunghiului ABC intersecteaza a doua oara cercul circumscris triunghiului in punctele $ A_1, B_1$ si $ C_1$; iar paralelele duse prin varfurile A, B si C la laturile opuse ale triunghiului, intersecteaza a doua oara cercul circumscris triunghiului ABC in punctele  $ A_2, B_2$ si $ C_2$. Demonstrati ca dreptele $ A_1A_2, B_1B_2$ si $ C_1C_2$ sunt concurente.
[b]Problema 2034:[/b]
Pe laturile AB, BC si CA ale triunghiului ABC se iau punctele X, Y si respectiv Z, astfel incat triunghiul XYZ sa fie asemenea cu triunghiul ABC($ \angle A \equal{} \angle X$, $ \angle B \equal{} \angle Y$ si $ \angle C \equal{} \angle Z$). Demonstrati, ca centrul cercului circumscris triunghiului ABC este egal departat fata de ortocentrele triunghiurilor ABC si XYZ.
[b]PS. Problemele 1968 si 1990 au fost adaugate ulterior.[/b]

can you write these problems in english

thanks :blush:


AMC
AIME
AMC 12
analytic geometry
Stanford
college
AMC 12 A

I took the AMC12 A at my high school and the AMC12 B at my local university I got a 102 on A and a 103.5 on the B.

I am taking the AIMEI at my high school, when the AIME form asks for your AMC score am I allowed to write 103.5 even though I got that score at the university not at the school I am taking the AIME at?

Also, my friend took both AMCs and got 90-something at the school and a 100.5 at the university, is he allowed to take the AIME at my high school?

You definitely are allowed to put your higher score on there, but I think you need to contact AMC or someone first. Correct me if I'm wrong, everyone.

Put your higher score on the AIME form.  We use it only as a first check, and we do not rely on that hand-entered score for anything other than a simple cross-check of identity.  That is, if we find find that there are two AIME qualifying J. Smiths in your school or city, one way to help distinguish them is by scores.  We also use several other cross-checks to verify identity.  Then we use our internally stored high score for that validated idenity.  By the way, if you enter a wrong (or invalid, or mis-represented) score, it doesn't help you, it slows down the identification process.

As for your friend who qualified on the B-date at a University administration site, have the high school contest manager get in touch with the University contest adminstrator (the name and contact information is on our web site) and have them coordinate the adminstration at the high school.

Steve Dunbar
AMC Director

Following up on this -- I administer the AMC contests at Stanford for local students whose own schools don't give the AMC -- but I also had a few students take the "B" exam when their school only gave the "A" exam (we previously verified that this was allowed). A few qualified for the AIME with me, but not on the A exam, so i need to get the AIME materials to their school so they can take it there.

getting them ANSWER forms is no problem, but the QUESTION are in a single packet and I can't open it until we give the exam --so how can I deliver a single copy to a teacher?  I'm not sure what to do about this.

If they are giving the AIME to another student at the same time, then
perhaps they can use the extra copy for the student, and I can give them another one after the fact. But what if there are two such students from the same school?

Hi Ted --

Good question.  For other folks reading this thread, here's what happened:  Ted's EPGY program at Stanford signed up as a school to administer the B-date exams, as they have done for many years now.  So as a school, we send them the sealed  envelope containing as many AIME contests as we recognize from the B-date report that they will need.

For the colleges and universities who signed up to administer the B-date contest, we realized in advance that many would have B-date AIME qualifiers who would want to sit for the AIME at their high school with A-date qualifiers form the school, just as Ted is referring to.  So for colleges and universities we send each AIME contest back to the college or university in a separate sealed envelope.  Then the high school and the college can communicate as necessary and the college can send the appropriate number of individual AIME contests in sealed envelopes to the high school.

That's what Ted wants to do, but is blocked from doing by the sealed envelope.

Here's a potential simple solution:  Each high school already has an extra "courtesy" copy of the AIME in their high school envelope.  If it is just a single student at the high school, then the high school can use the extra courtesy copy (and we will replace the courtesy teacher copy later.)  If it is more than one student at a school, contact me directly and we will rush the necessary AIME copies directly to the high school.

We are recognizing that some students are now highly mobile and will be taking the contests at various venues.  We are considering alternative procedures for 2008 to eliminate or smooth out such wrinkles.

--Steve Dunbar
Director, American Mathematics Competitions

Thanks, this sounds fine.  For that matter, if the school uses the extra copy in their envelope, I can personally replace it after the test date with the extras I will have; there are just three schools that have more than one student at them, and in at least one of those cases, one of the students already qualified on the "A" exam. So there are only two tricky cases, with more than one exam needed. I'll contact you privately with the details.

Ted Alper

In a situation like this, if an AIME exam is mailed from a university to a high school, but does not arrive on time for the exam, can the high school coordinator make a photocopy of the answer sheet for a student to use, while giving the courtesy copy of the exam itself to the student?

Yes, copying the answer form is acceptable for special "emergency" situations such as this.

We don't like to accept copies as a general solution because we cannot run them through the scanner, the data has to be hand-entered and verified.  So please, don't use copies, unless you have no other recourse.

Steve Dunbar
American Mathematics Competitions

So I am a freshman that took AMC12a and AMC10b. If I put my AMC12 score, do they use the index or do they just look at my AIME score for my qualification of AIME? Or should I just take my AMC10B score?

As mentioned elsewhere, put your highest score.

As mentioned elsewhere, we use our internal records for determining eligibility for the next level of competition.  What you record is used only as a quick cross-check on your identity.

We will use the published rules for determining your eligibility for any awards or higher levels of competition.

Steve Dunbar
Director, American Mathematics Competitions.

If we took the AMC 10A (126) and AMC 12B (121.5), should we put our 12 score or our 10 score? Will both be considered when qualifying for the USAMO by floor value?

Sorry for this last minute question, but if a high-school coordinator does plan to make a photo-copy of the answer form, is there anything that he/she needs to do in order to inform the AMC that a photocopy was used? Or will that take care of itself when the grading machine doesn't turn out any results for the photocopied answer form?

The contest manager (usually the math teacher) should note the exception on the Certification Form in the Teachers' Manual (This is the form signed by the Vice-Principal when opening the contest package.  There's a space for exceptions on the page.)   We'll notice a photocopy, and pull it separately before ever putting it in the scanner.  It's part of the quality control we do.

 Steve Dunbar
Director, American Mathematics Competitions


limit
topology
real analysis
real analysis theorems

ARE they not, limit points ,accumulation points,cluster points one and the same thing??

Probably yes. Why don't you just compare the definitions?

But whatever they are called, there is a difference between "L/A/C point of a set" and "L/A/C point of a sequence". The sequence $ (2,3,2,3,2,3,2,3\dots)$ has two such points; the set $ \{2,3,2,3,2,3,2,3\dots\}\equal{}\{2,3\}$ has none.

[quote="mlok"]Probably yes. Why don't you just compare the definitions?

But whatever they are called, there is a difference between "L/A/C point of a set" and "L/A/C point of a sequence". The sequence $ (2,3,2,3,2,3,2,3\dots)$ has two such points; the set $ \{2,3,2,3,2,3,2,3\dots\} \equal{} \{2,3\}$ has none.[/quote]

.

Why the L/A/C of a sequence differ from the L/A/C of a set ,after all when we refer to the L/A/C OF A SEQUENCE  we refer to the range of the sequence which is a set.

And another thing how can one prove that :

...............$ \lim x_{n}\equal{}x$ this implies that x is L/A/C OF the sequence.................................................

[quote="τρικλινο"]when we refer to the L/A/C OF A SEQUENCE  we refer to the range of the sequence which is a set.[/quote]
Perhaps you do. I do not.

[quote="mlok"]Probably yes. Why don't you just compare the definitions?

But whatever they are called, there is a difference between "L/A/C point of a set" and "L/A/C point of a sequence". The sequence $ (2,3,2,3,2,3,2,3\dots)$ has two such points; the set $ \{2,3,2,3,2,3,2,3\dots\} \equal{} \{2,3\}$ has none.[/quote]

I am sorry to ask but ,if you wish, can you tell me please according to what definition do you get two limit points for the sequence and none for the set??

[quote="Roosta"]An L\A\C point of a set $ A$ is a point $ x$ s.t. given any open set $ U$ containing $ x$, there are infinitely many $ y\in A$ s.t. $ y\in U$. $ \{2,3\}$, having only finitely many points, can't have an L\A\C point.

An L\A\C point of a sequence $ \{ x_n\}$ is a point $ x$ s.t. given any open set $ U$ containing $ x$, there are infinitely many of $ i\in \mathbb{N}$ s.t. $ x_i \in U$. $ (2,3,2,3,2,...)$ has $ 2$ and $ 3$ as L\A\C points because any open set containing either contains either the odd or even elements of the sequence, both of which there are infinitely many of.[/quote]

According to your definition does the limit x of a sequence {$ \ x_{n}$} is also L\A\C OF the sequence?

If yes how do we prove that?

$ x$ is a limit point of a sequence $ \{x_n\}$ if, given an open set  $ U$ containing $ x$, $ \exists M\in \mathbb{Z}^\plus{}$ s.t. $ x_n\in U$ for $ n\ge M$. This point $ x$ is then denoted the limit of the sequence. The distinction between an L/A/C point of a set and it's limit is that a sequence either has a limit or does not, whereas another sequence can have any number of L/A/C points.

[quote="Roosta"]The limit of a sequence only exists if the set of all L/A/C points of that sequence consists of a single point $ x$. This point $ x$ is then denoted the limit of the sequence.[/quote]  What about the sequence $ (0, 1, 0, 2, 0, 3, 0, 4, \ldots)$?

Haha, good point, that isn't the first time I have made that mistake. A revised definition of a limit point of a sequence $ \{x_n\}$ is a point $ x$ s.t. given any open set $ U$ s.t. $ x\in U$, all but finitely many of the $ x_n\in U$ for all but finitely many of $ n\in \mathbb{Z}^ \plus{}$. This is also often written as: $ x$ is a limit point of a sequence $ \{x_n\}$ if, given an open set  $ U$ containing $ x$, $ \exists M\in \mathbb{Z}^\plus{}$ s.t. $ x_n\in U$ for $ n\ge M$. I fixed what I wrote above as well.


[quote="236factorial"]The product of 25 less than the square of a number and 3 more than the number is 0. Find the numbers that meets these requirements.[/quote]
$a^2+3a-25=0 \rightarrow a=\dfrac{-3 \pm \sqrt{9+100}}{2} \rightarrow a=\dfrac{-3 \pm \sqrt{109}}{2}$


floor function
graph theory
combinatorics unsolved
combinatorics

Suppose $ S\equal{}\{1,2,..,n\}$ and $ 0\le k \le \frac{n}{2}$. Suppose $ A$ is a collection of all $ k$-element subsets of $ S$ and let $ B$ be the collection of all $ (k\plus{}1)$-element subsets of $ S$. Build a bipartite graph $ G\equal{}(V,E)$ with $ V\equal{}A\cup B$ by joining $ X\in A$ to $ Y\in B$ if and only if $ X\subset Y$

Suppose $ Z$ is a collection of subsets of $ S$ such that for all distinct $ X,Y\in Z, X\nsubseteq Y$ and $ Y\nsubseteq X$. Show that $ \mid Z\mid \le \binom{n}{\lfloor \frac{n}{2} \rfloor}$.

Hint 1: First show $ G$ has a matching hitting every element of $ A$
Hint 2: If $ Z$ consists of all subsets of $ S$ of size $ \lfloor \frac{n}{2} \rfloor$ then this upper bound is reached.

[quote="pbornsztein"]Pb 1 : I do not see where the question is.

Pb. 2 : This is Sperner theorem.

Pierre.[/quote]

Right. I found out about the Sperner theorem but how do I build a bipartite graph with the above mentioned conditions.


superior algebra
superior algebra solved

prove or disprove: if $[A:H]=n$ then $x^{n} \in H$ for every $x \in A$

wondering if this would work: $(xH)^{n}=H$?

Well, this surely works if $H$ is normal, but otherwise it's wrong, for example take $A=S_3, H=\{ e , (12) \}$


algebra
polynomial
calculus
integration
number theory
Rational Root Theorem

I have been trying to use the rational roots theorem and other properties of roots to solve this problem, but so far I'm still stuck.

Let a1, a2, and a3 denote integers each having the prime number c as an integral factor.

a.) Prove that any integral root of the equation x^3 + (a1)x^2 + (a2)x + a3 = 0 must have c as an integral factor.

b.) Use the result of part (a) to find the rational roots of x^4 + 24x^3 -52x^2 + 26x -52 = 0.

I'm not really sure if this question belongs in this section since it is a pre-calc question.

Um... my algebra is really, really, really pathetic... but as far as I can see, (a) doesn't require the Rational Root Theorem, just a little basic number theory $ \bmod{c}$ to show that $ x\equiv 0 \bmod{c}$. (although if this is homework your teacher will probably not accept a number theory solution...) And then for (b), we quickly realize that $ 2$ is a root, and then divide out $ x \minus{} 2$, which gives $ x^3 \plus{} 26x^2 \plus{} 26$... and then by (a) any integral root of this is divisible by $ 13$, so we only have to check $ x \in \{ \minus{} 26, \minus{} 13, 0, 13, 26 \}$ and clearly none of these work. So the only integral solution to the given quartic is $ x \equal{} 2$.

And I think that this question could be either in this forum or the HS Basics forum.


I need help on this chemistry problem.

What is the value of Kc for the reaction below?

HNO2(aq)+NH3(aq)-> NO2-(aq)+ NH4+(aq)

Ka=7.2x10^-4 for HNO2
Kb=1.8x10^-5 for NH3
Kw=1.0x10^-14

Is the Answer

A 1.3x10^6
B 7.7x10^-7
C 1.3x10^-8
D 7.7x10^7
E 1.3x10^-22

Thank you in advance.


A convex hexagon $ ABCDEF$ satisfies the following conditions:
1) $ AB\parallel DE$, $ BC\parallel EF$, and $ CD\parallel FA$.
2) The distances between these pairs of parallel lines are the same.
3) $ \angle FAB \equal{} \angle CDE \equal{} 90^\circ$
Prove that the diagonals $ BE$ and $ CF$ of the hexagon intersect with angle $ 45$ degrees.

$ \bullet$ Thank you dear [b]Babis Stergiou[/b] for your translation.  :P

Let AB, DE meet CD, FA at P, Q. APDQ is a square. Let BC, EF meet QA, PD at R, S and AP, DQ at U, V. RCSF, UBVE are rhombi, CF bisects $ \angle RCS \equiv \angle BCD,$ BE bisects $ \angle UBV \equiv \angle ABC.$ CF, BE are external angle bisectors of the $ \triangle PBC$ with the right $ \angle BPC,$ $ \angle(CF,BE) \equal{} 90^\circ \minus{} \frac {BPC}{2} \equal{} 45^\circ.$

[quote="April"]...................................................................
$ \bullet$ Thank you dear [b]Babis Stergiou[/b] for your translation.  :P[/quote]

 We must thank Professor [b]Theodor Bolis[/b] in Ioannina University, for years IMO  team leader of Greece , who traslated the problems from the russian text.I just asked from him to do this work for  April and me !!!

 [b] Babis[/b]

oldest source is Austrian-Polish 1996 Individual 2


group theory
abstract algebra
number theory
least common multiple
relatively prime
superior algebra
superior algebra unsolved

If an abelian group has subgroups of orders m and n, respectively then how do we show it has a subgroup whose order is lcm of m and n?

It is a ** problem in Herstein  :| and that itself discourages me :( . Any hints appreciated
Thank you  :lol:

How hard this problem is depends on how much you already know about [url=http://en.wikipedia.org/wiki/Abelian_group]abelian groups[/url].

The two given subgroups generate a finite group (of order $ N\le mn$). Both $ m$ and $ n$ divide $ N$, therefore their lcm divides $ N$ as well. And an abelian group of order $ N$ contains a subgroup of any order that divides $ N$ - this is an immediate consequence of the fundamental theorem of finite abelian groups.

[quote="mlok"]
The two given subgroups generate a finite group (of order $ N\le mn$). Both $ m$ and $ n$ divide $ N$, therefore their lcm divides $ N$ as well. [color=red]And an abelian group of order $ N$ contains a subgroup of any order that divides $ N$[/color] - this is an immediate consequence of the fundamental theorem of finite abelian groups.[/quote]

Is there an elementary way to prove that? I know only Lagrange's theorem, so can I avoid that theorem you are talking about?

Thanks  :)

[quote]$ c^h \equal{} e$ (the identity element) for any $ c\in C$ [color=red]and so[/color] $ k\mid h$. [/quote]
No. Consider $ C \equal{} A\oplus B$, where $ A \equal{} B \equal{} \mathbb Z_2$

[EDIT: What I wrote here was correct but irrelevant.]

Well, the order of $ C$ and the order of $ c$ are different things.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=27505]Here[/url] is a previous discussion. The lemma I like here:
Let $ m$ and $ n$ be relatively prime, and let $ G$ be an abelian group. Then $ G$ has a subgroup of order $ mn$ if and only if it has subgroups of order $ m$ and $ n$ respectively.

I have an argument involving sylow's theorem. First let $ G$ be a finite group where all subgroups are normal (for example abelian but not conversely). Let $ H_1$ and $ H_2$ be two subgroups so that $ |H_1|$ and $ |H_2|$ have orders $ h_1,h_2$ respectively. If $ \gcd(h_1,h_2) \equal{} 1$ then $ H_1H_2$ is a group of $ h_1h_2$ elements. Now we can extend this argument to any finite group so that all subgroups are normal. Let $ |G|$ be finite, let $ m\parallel{}G|$ and let $ m \equal{} p_1^{k_1}...p_s^{k_s} > 1$ be its prime factors in its decomposition. Let $ H_1,H_2,...H_s$ be the sylow $ p_i$-subgroups for $ i \equal{} 1,...,s$. Then $ \gcd(h_i,h_j) \equal{} 1$ for $ i\not \equal{} j$ that means $ H_1H_2...H_s$ (the group product) will be a subgroup of order $ m$. We have show that if $ G$ is a finite group with all normal subgroups and $ m$ is a divisor of its order then there is a subgroup of that order. Now given two subgroups of orders $ n,m$ of an abelian group (which is a special type of this group) then $ \text{lcm}(n,m)$ is also a divisor of the order, applying the argument above we get there is such a subgroup.

Thanks everyone :lol:
To [b]sylow_theory[/b], I will come back and learn your proof once I finish learning sylow theorems. They are very hard, supposedly, so might take some time. But thanks, It is nice to learn different proofs.

To [b]jmerry[/b], Awesome!

[quote="isomorphism"]Thanks everyone :lol:
To [b]sylow_theory[/b], I will come back and learn your proof once I finish learning sylow theorems. They are very hard, supposedly, so might take some time. But thanks, It is nice to learn different proofs.[/quote]
Hard? They are pretty straightforward. You can read them [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156944]here[/url]. But there is even an easier approach to them, that I learned since then.


[hide]$y(x) = \sqrt{1 - 2x} + x$

The domain is $D = (-\infty,\frac12]$.
The range is the set of values of $a$ such that the equation $a = f(x)$ has a solution for $x \in D$.

$a = \sqrt{1 - 2x} + x$
$a - x = \sqrt{1 - 2x}$
$a^2 - 2ax + x^2 = 1 - 2x$
$x^2 + 2(1 - a)x + a^2 - 1 = 0$

The condition is:
$4(1-a)^2 - 4(a^2 - 1) \geq 0$
$4 - 8a + 4a^2 - 4a^2 + 4 \geq 0$
$-8a \geq -4$
$a \leq 1$

Hence de range is $(-\infty,1]$[/hide]

[hide]

first we look at where the domain is defined, assuming on the reals.  So x's max value is $\frac{1}{2}$.  If it is any bigger than the square root is imaginary.  Thus for $x=\frac{1}{2}$ $y=\frac{1}{2}$.  There is no minimum for $x$ and since $\sqrt{1-2x}$ increases, when $x$ is negative, slower than $x$ decreases.  Thus the range is $(-\infty,\frac{1}{2}]$.

[/hide]

[quote="demji"]since $\sqrt{1-2x}$ increases, when $x$ is negative, slower than $x$ decreases[/quote]

But your entire domain doesn't have negative $x$!

opps, forgot to account for x=0, i see it so clearly now...

correction: $range=(-\infty,1]$


calculus
inequalities
induction
factorial
algebra unsolved
algebra

What is the largest natural number k satisfying
1/2 * 3/4 * 5/6 * ... * 2003/2004 < 1/k ?

Is there any good method to solve it?

[quote="Arne"]And you can prove that \[\frac{1}{2}\cdot \frac{3}{4} \cdots \cdot \frac{2n - 1}{2n} \leq \frac{1}{\sqrt{3n + 1}}\] but that doesn't help either.[/quote]

 :read:  :coolspeak:

[quote="ehsan2004"]I think $\frac 12 \times \frac 34\times ... \times \frac {2003}{2004}=\frac {(2004)!}{2^{2004}(1002)!}$ but ... :?[/quote]
That's wrong. 2004! means a product of 2004 numbers, but there should be a product of 1002 odd numbers.

Arne, you have done a realy good job. You ineq shows that n=54 satisfies this and the realy smallest number satisfying this is 56 (by Maple).

Thanks, but I don't think a method exists which allows you to find $n = 56$ - unless you are going to use some very heavy calculus tools, because that *might* give you a good approximation.

Can you just send the proof of your inequality (a hint may be enough).

Hint
[hide][u]Induction :) It's pretty straightforward[/u][/hide]

Here's a method that's not too bad.  Let $P$ denote the original product.

By telescoping, we can get the following expression for $P$:
\[
  P = \frac{1}{\sqrt{4008}}
          \prod_{\scriptstyle i=3 \atop \scriptstyle i \textrm{ odd}}^{2003}
            \frac{i}{\sqrt{i-1}\sqrt{i+1}}.
\]
Note that each factor in the new product is bigger than 1.  Thus by truncating this product, we get lower bounds on $P$.  For example, by truncating the product at $i = 15$, we can calculate that $P > \frac{1}{57}$.

For upper bounds, we can use the following expression instead:
\[
  P = \frac{\sqrt{2003}}{2004}
          \prod_{\scriptstyle i=2 \atop \scriptstyle i \textrm{ even}}^{2002}
            \frac{\sqrt{i-1}\sqrt{i+1}}{i}.
\]
This time, each factor is less than 1.  So by truncating, we get upper bounds on $P$.  For example, by truncating at $i = 12$, we see that $P < \frac{1}{55}$.  I think I had to go up to $i = 116$ to prove that $P < \frac{1}{56}$.

Thus the answer to the original problem is $k = 56$.

Actually, now that I think about it, a much better method would be to use Stirling's formula.  We can express $P$ as $\binom{2004}{1002}/2^{2004}$.  Then we can approximate each factorial with Stirling.  That shows $k = 56$ much more quickly.

Sorry for the double post, but I realized that I missed a simpler approach above.  I truncated the two products above so that I could calculate them on a calculator.  A better idea would have been to extend each product to an infinite product.  That way, we get Wallis-like products that can be evaluated exactly.  By doing so, we get the bounds
\[
  \frac{\sqrt{4006}}{2004 \sqrt{\pi}} < P < \frac{1}{\sqrt{1002 \pi}} \,.
\]
Those bounds show that $\frac{1}{P} \approx 56.1$ to one decimal place, which shows that $k = 56$ is the solution to the original problem.


[quote="GameBot"]A tiny pyramid whose faces are triangles is sliced off at each vertex of a large cube. How many vertices does the resulting polyhedron have?[/quote]

There are 8 vertices originally. I assume that the tiny pyramids don't intersect or touch. Each slcied off pyramid gives 3 new vertices, and destroys one. So $ 8 \cdot 3\equal{}24$.


algebra polynomial quadratics function domain calculus integration

Why do we not have unique factorization over the ring of cyclotomic integers? In particular, how do they not satisfy the conditions of the theorem on uniqueness of factorization, which says that: In $ F[x]$ where $ F$ is a field, if $ f\equal{}p_1 p_2 \cdots p_s\equal{}q_1 q_2 \cdots q_t$ are two factorizations of the polynomial $ f$ into a product of irreducible polynomials in $ F[x]$, then $ s\equal{}t$ and there is a 1-1 correspondence between the factos $ p_1,p_2,...,p_s$ and $ q_1,q_2,...,q_t$ ?

In particular, the fact that cyclotomic polynomials dont meet the theorem would mean that we can't factor $ z^n\plus{}y^n$ as $ (z\minus{}y)(z\minus{}\omega y)(z\minus{}\omega^2y)\cdots (z\minus{}\omega^{n\minus{}1} y)$ where $ \omega\equal{}e^{\frac{2\pi i}{n}}$ and $ z\subset \mathbb{Z}$

You shouldn't expect unique factorization.  Unique factorization doesn't even hold in integer rings of quadratic number fields; for example,

$ (1 \plus{} \sqrt { \minus{} 5})(1 \minus{} \sqrt { \minus{} 5}) \equal{} 2 \cdot 3$

is a factorization of $ 6$ into two distinct sets of prime factors in $ \mathbb{Z}\left[\sqrt { \minus{} 5}\right]$.

Is there a way to absolutely decide if factorization is unique? Maybe there is something in the above theorem but I cant see anything.

You seem to mix unique factorisation of polynomials with factorisation of algebraic integers.
Please do what I said in http://www.mathlinks.ro/Forum/viewtopic.php?t=183583 .

If you want, I can post a proof that e.g. $ \mathbb Z [\zeta_{23}]$ doesn't have unique factorisation, but that still would require some knowledge about this stuff.

[quote="ZetaX"]You seem to mix unique factorisation of polynomials with factorisation of algebraic integers.[/quote]

But can't you just consider the integers to be polynome grad. 0? Maybe it isn't entirely correct, but i know the proof for unique factorization of polynomials is similar to the one for integers!

The standard proof for both is showing they are Euclidean domains, implying unique factorisation.
But note that in polynomials of algebraic integers, we do not have unique factorisation. The uniqueness holds for polynomials with coefficients in some field only (a simple ring does not suffice!), and some other cases, but not in general.

Why polynomial unique factorisation does not imply it for algebraic integers is possible easiest seen by this example:
We have $ x^2\minus{}y^2\equal{}(x\plus{}y)(x\minus{}y)$ as a polynomial (and this factorisation is unique). But setting $ x\equal{}1, y\equal{}\sqrt {\minus{}5}$, you get the example from above which is not unique. Or by setting $ x\equal{}3,y\equal{}1$, you get $ 8\equal{}4\cdot 2$, showing that you can often factor further, too.

[quote="JSteinhardt"]If you want a criterion for unique factorization, unique factorization in integral domains is equivalent to the assertion that the prime elements and the irreducible elements are the same (note that all prime elements are irreducible in an arbitrary integral domain, but not necessarily the other way around).[/quote]

No, in addition to atoms are prime, you also need to include some finiteness condition on factorizations, e.g. ACCP (ascending chain condition for principal ideals). Or consider Kaplansky's elegant characterization: a domain is a UFD $\iff$ every prime ideal contains a prime.


c and z are complex numbers, could anyone help to show that (z^c)^n= z^cn

Well, I start with z^c= exp (c logz) then... I am mess up with it to show that result. Need some help, Thx

Use $e^a e^b = e^{a+b}$ for $(e^{c ln(z)})^n$ written as product.


trigonometry
ratio
quadratics
LaTeX
HMMT
geometry
trig identities

I know I need more practice on trig, and so do other people, I''m sure.  So this is a thread for trig problems only.  Someone wanna post the first problem?

[b]Problem 1[/b]

Find the value of  $\frac{\cos 15^\circ+\sin 15^\circ}{\cos 15^\circ-\sin 15^\circ}.$

Do you know the sum and difference formulas sean?

I suggest you learn all your identities before doing problems ;). This thread is pretty pointless if you don't even know where to begin.

I have a solution now.  Let me post it.  Don't answer this one!

$\text{Problem 2}$

Let $x$ be a real number such that $\sec x - \tan x = 2$.  Find the value of $\sec x + \tan x$.

A) $0.1$
B) $0.2$
C) $0.3$
D) $0.4$
E) $0.5$

Alright I have a trig problem for ya.  Its not that hard but if someone who could use some trig practice wants to give it a try.

$\text{Problem 3}$
What is the exact value of $\cos 7.5$?

Yes, that is the correct answer.  Does anyone want to post a solution, or should I?

[quote="mysmartmouth"]Ok.  I did it.  Please tell me if I am correct, or where I made my mistake(s).

[hide="Problem 1"]
Problem: $\frac{\cos 15^\circ+\sin 15^\circ}{\cos 15^\circ-\sin 15^\circ}$

We can determine the values of $\sin 15$ and $\cos 15$ by using the identities $\sin (x-y) = \sin x \cos y - \cos x \sin y$ and $\cos (x-y) = \cos x \cos y + \sin x \sin y$.  So we get $\sin 15=\sin (45-30)=\frac{\sqrt{6}-\sqrt{2}}{4}$ and $\cos 15=\cos (45-30)=\frac{\sqrt{6}+\sqrt{2}}{4}$.  Now we just plug these values into the original equation.  This yields an answer of $\boxed{\frac{\sqrt{3}}{2}}$.
[/hide][/quote]

Your answer is incorrect.  :(

Can you tell me where I made by mistake?  Chamoo got the same answer as me.  Was  there a mistake in one of my identities, or was there a mistake in my calculations?

You seemed to make a mistake in calculation for last step. :)

[hide="Problem 1 Reworked"]
Problem: $\frac{\cos 15^\circ+\sin 15^\circ}{\cos 15^\circ-\sin 15^\circ}$

We can determine the values of $\sin 15$ and $\cos 15$ by using the identities $\sin (x-y) = \sin x \cos y - \cos x \sin y$ and $\cos (x-y) = \cos x \cos y + \sin x \sin y$.  So we get $\sin 15=\sin (45-30)=\frac{\sqrt{6}-\sqrt{2}}{4}$ and $\cos 15=\cos (45-30)=\frac{\sqrt{6}+\sqrt{2}}{4}$.  Now we just plug these values into the original equation.  So we have $\frac{\frac{\sqrt{6}+\sqrt{2}}{4} + \frac{\sqrt{6}-\sqrt{2}}{4}}{\frac{\sqrt{6}+\sqrt{2}}{4}-\frac{\sqrt{6}-\sqrt{2}}{4}}$.  This is equal to $\frac{\frac{2\sqrt{6}}{4}}{\frac{2\sqrt{2}}{4}}$  Simplifying this, we get $\frac{\sqrt{6}}{\sqrt{2}} = \boxed{\sqrt{3}}$
[/hide]

$PROBLEM$ $9$

In a quadrilateral $ABCD$: $\measuredangle ABD=37^\circ$, $\measuredangle DBC=45^\circ$, $\measuredangle ADB =53^\circ$, $\measuredangle BDC=53^\circ$. If $\measuredangle DAC=\alpha$, find $tan \alpha$.

This may be stupid, but...

[hide]

If we know $\angle{BCA}$, we know $\angle{DAC}$.

If $AB = 1$, we have

\[ AD = \tan(53), BD = \csc(53)  \]

and by the law of sines,

\[ \frac{BC}{\sin(53)} = \frac{\csc(53)}{\sin(82)}  \]

\[ BC = \frac{1}{\sin(82)}  \]

By the law of cosines,

\[ CA^2 = (\frac{1}{\csc^2(82)})^2 + 1 - 2\cot(82)  \]

Use the law of sines again, and we get

\[ \sin(\angle{CAB}) = \sqrt{\frac{1}{sin^2(32)} + 1 - 2\cot(82)}  \]

which gives $\angle{CAB} \approx 50^\circ$, so $\angle{DAC} \approx 40^\circ$.

Actually, after further thought, $\angle{DAC} = 40.6771325$, and

$\tan(\angle{DAC}) = .8594591653$
[/hide]

Let's revive this thread.

Find $\cos 75$.

$ \cos 75= \cos(45+30)= \cos45\cos30-\sin45\sin30= (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2})-(\frac{\sqrt{2}}{2})(\frac{1}{2})= \frac{\sqrt{6}-\sqrt{2}}{4}$

yes, i used cos sum identity thing...w/e

someone make this into latex cause i dont know it

That's correct Ankit, post the next problem.

err, since he didn't i guess i'll post one. here's one i did a while back, from hmmt or rice i think...not sure.

$\frac{\sin1+\sin2+\sin3+....+\sin44}{\cos1+\cos2+\cos3+....+\cos44}=$?

And of course, it's in degrees.  :wink:

I need a hint.  Could you use the identities:
[hide]
$\sin x+\sin y = 2 \sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$
[/hide]

[hide]
$\cos x+\cos y = 2 \cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$
[/hide]

Using these identities, and pairing sums of 45, I get:

[hide]
$\frac{2\sin \frac{45}{2}\cos \frac{43}{2}+2 \sin \frac{45}{2}\cos \frac{41}{2}...+2 \sin \frac{45}{2}\cos \frac{1}{2}}{2 \cos \frac{45}{2}\cos \frac{43}{2}+2 \cos \frac{45}{2}\cos \frac{41}{2}...+2 \cos \frac{45}{2}\cos \frac{1}{2}}$

Which simplifies to:

$\frac{\sin\frac{45}{2}}{\cos\frac{45}{2}}$

Which simplifies to:

$\boxed{\tan 22.5^\circ}$

[/hide]

[quote="MysticTerminator"][quote="mysmartmouth"]I need a hint.  Could you use the identities:
[hide]
$\sin x+\sin y = 2 \sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$
[/hide]

[hide]
$\cos x+\cos y = 2 \cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$
[/hide]

Using these identities, and pairing sums of 45, I get:

[hide]
$\frac{2\sin \frac{45}{2}\cos \frac{43}{2}+2 \sin \frac{45}{2}\cos \frac{41}{2}...+2 \sin \frac{45}{2}\cos \frac{1}{2}}{2 \cos \frac{45}{2}\cos \frac{43}{2}+2 \cos \frac{45}{2}\cos \frac{41}{2}...+2 \cos \frac{45}{2}\cos \frac{1}{2}}$

Which simplifies to:

?

[/hide][/quote]

I just found it funny that you're still asking for help at this point...[/quote]

At that point, I got it.  If you look in your IM window:

[quote]
[22:45] cheesybread92: I GOT IT
[22:45] cheesybread92: OMFG I DID IT
[22:45] cheesybread92: YAY!!!
[22:45] cheesybread92: OMFG
[22:45] cheesybread92: YAY
[22:45] cheesybread92: LET ME POST IT[/quote]

Based on my solution to that problem, I'll post the next problem:

Evaluate $\tan 22.5^\circ$

:spam: But I'll leave it to Joe to delete it, so Arnav won't be mad at me.

Hmmm...I was being stupid.  I see how to evaluate that now.  So I will:

$\tan 45 = \frac{2\tan 22.5}{1-\tan^{2}22.5}$
Let $\tan 22.5 =x$
Now $1=\frac{2x}{1-x^{2}}$
Simplify all the way to get the quadratic: $x^{2}+2x-1=0$.  Solve for x with the quadratic equation, and get $\boxed{\sqrt{2}-1}$.

I will note that it was probably better that Arnav let me solve it myself instead of helping me.  So thanks for nothing Arnav!  :D

$\text{Problem 12ish}$

Evaluate $\cos 36^\circ-\cos 72^\circ$.  (AHSME 1975)

[quote="MysticTerminator"]
[color=red]MODEDIT: DIALOGUE REMOVED TO MAINTAIN CLEANLINESS[/color]
[/quote]

LMAO  :rotfl:  :rotfl:

WOW...

btw, is the answer to your problem like $\frac{1}{2}$?

It's like $\frac{1+\sqrt{5}}{4}-\frac{-1+\sqrt{5}}{4}=\boxed{\frac{1}{2}}$

something like that...

EDIT:

err, maybe I should explain stuff.

$\cos{72}$ is the fifth root of unity, therefore we try to find the real value of $x^{5}-1=0$

$x^{5}-1=(x-1)(x^{4}+x^{3}+x^{2}+x+1)=0$.

$1$ is an obvious root, however $\cos{72}$ obviously is not equal to $1$.

Therefore, we must find the root of $x^{4}+x^{3}+x^{2}+x+1=x^{2}+\frac{1}{x^{2}}+x+\frac{1}{x}=0$.

Let $y=x+\frac{1}{x}$, therefore, the equation can be simplified to:

$y^{2}-2+y+1=y^{2}+y-1=0 \Longrightarrow y=\frac{-1 \pm \sqrt{5}}{2}$

So $x+\frac{1}{x}=\frac{-1 \pm \sqrt{5}}{2}\Longrightarrow x=\frac{\frac{-1+\sqrt{5}}{2}\pm \sqrt{Q}}{2}$, where $Q<0$ and therefore imaginary. The real value therefore is $\frac{-1+\sqrt{5}}{4}$ which is equal to $\cos{72}$. The result follows...

$\text{Problem 13ish}$

If $\tan{19k}=\frac{\cos{96}+\sin{96}}{\cos{96}-\sin{96}}$, find the smallest possible integer value of $k$.

This problem is cool because it uses number theory too :D

AND WHY CAN'T U DOUBLE POST ANYMORE????? 36 HOUR LIMIT????????!!!

Hm, I think you scared everyone away.  Besides, we only have 2 SCers who actually post here anymore.

[size=134]Easier problem:[/size]

Find the degree measure of the smallest positive angle x that satisfies the equation $\sin x+\cos x = \frac{\sqrt{3}}{\sqrt{2}}$

scared everyone away, wtf? shut up and try it.

Besides, an almost EXACT problem to what you just posted has been posted before on this thread.


limit
calculus
derivative
real analysis
real analysis unsolved

This is one of the actual example L'hopital used in his book "L'analyse des Infiniment Petits Pour I'Intelligence des Lignes Courbes"

Check it out!

Find the limit of x as it approaches a for

y=[[(2*a^3*x-x^4)^.5]-[a^3(a^2x)^.5]]/(a-(ax^3)^.25)

This isn't a hard problem at all if we use L'Hospital's rule "wisely." The idea is to simplify the expression a bit before applying the rule.

$\lim_{x \to a}\frac{\sqrt{2a^{3}x-x^{4}}-\sqrt{a^{3}x}}{a-(ax^{3})^{\frac14}}$   (which of course has $0/0$ form)

$= \lim_{x \to a}\frac{(\sqrt{2a^{3}x-x^{4}}-\sqrt{a^{3}x})(\sqrt{2a^{3}x-x^{4}}+\sqrt{a^{3}x})}{(a-(ax^{3})^{\frac14})(\sqrt{2a^{3}x-x^{4}}+\sqrt{a^{3}x})}$ ... (rationalizing the numerator)

$= \lim_{x \to a}\frac{x}{(\sqrt{2a^{3}x-x^{4}}+\sqrt{a^{3}x})}\cdot \frac{a^{3}-x^{3}}{(a-(ax^{3})^{\frac14})}$

$= \frac{a}{2a^{2}}\cdot \lim_{x \to a}\frac{a^{3}-x^{3}}{a-(ax^{3})^{\frac14}}$  ... (which has $0/0$ form)

$= \frac1{2a}\cdot \lim_{x \to a}\frac{-3x^{2}}{-a^{\frac14}\cdot \frac34 x^{-\frac14}}$  ... (Using L'Hospital's rule)

$= \frac{1}{2a}\cdot 4a^{2}$

$= \boxed{2a}.$

Note that I implicitly assumed the existence of the required derivatives while using the rule in my solution.

Sorry for the trouble. This is what I was trying to solve.

[img]http://www.physicsforums.com/latex_images/11/1102142-0.png[/img]


parameterization
real analysis
function
binomial coefficients
advanced fields
advanced fields unsolved

I want to sum the series $ \sum_{k \equal{} 0}^n ( \minus{} 1)^k \binom{n}{k}\binom{m \plus{} n \minus{} k}{r \minus{} k}$. I got an answer but I tested it for specific values of m,n,r with mathematica and it is wrong. Can someone check my work? I used the so-called Snake Oil method. Read combinatorics2 first.

I see I did it the hard way.

I found one mistake on the first page. I was summing from k = 0 to infinity when I should have been summing from k = 0 to n. So, the last line on the first page is wrong. However, I don't think that changes the answer because I just need to add the factor $ (1 \minus{} \left(\frac {(yz)}{1 \minus{} y(1 \plus{} z)}\right)^{n \plus{} 1})$ and the [y^n] coefficient operator will annihiliate the second term. Thus, there must be something else wrong.

I see the problem now. I interchanged the n and k summations which does not make sense because the summation over k has n as a parameter. I'll post back if I can get it to work now. Yes-it works out.

[quote="JBL"]I haven't read what you did, but this is easy to express as a convolution with generating functions: $ ( \minus{} 1)^k\binom{n}{k}$ is the coefficient of $ x^{k}$ in $ (1 \minus{} x)^n$ and $ \binom{m \plus{} n \minus{} k}{r \minus{} k}$ is the coefficient of $ x^{r \minus{} k}$ in $ \sum_{j \equal{} 0}^\infty \binom{m \plus{} n \minus{} r \plus{} j}{j} x^j \equal{} \frac {1}{(1 \minus{} x)^{m \plus{} n \minus{} r \plus{} 1}}$, so your whole expression is the coefficient of $ x^r$ in $ \frac {(1 \minus{} x)^n}{(1 \minus{} x)^{m \plus{} n \minus{} r \plus{} 1}} \equal{} \frac {1}{(1 \minus{} x)^{m \minus{} r \plus{} 1}}$, and this is just $ \binom{m}{r}$.

Okay, now having read through your solution -- you probably made some small errors.  Check each step very, very carefully.  In principle, what you did should work fine.  Also, this topic would have gone nicely in the combinatorics forum.

Here's a combinatorial argument:  $ \binom{n}{k}\binom{m \plus{} n \minus{} k}{r \minus{} k}$ is the following: you have $ m$ girls and $ n$ boys sitting around.  You choose a team of size $ r$ by choosing first $ k$ boys, then $ r \minus{} k$ people from among everyone.  Thus, $ k \equal{} 0$ counts all teams of size $ r$, $ k \equal{} 1$ counts all teams of size $ r$ with one specially-chosen boy, and so on.  Then the given expression is just PIE to compute the number of teams consisting of only girls.[/quote]

I see. I was trying to use convolution that produced my summation as the nth term, which of course led nowhere. I didn't realize that it didn't have to be the upper limit of the summation. Now I see that you can replace the limit n with an r and get the same answer. When k > n,$ \binom{n}{k}$ is 0 and when k > r, $ \binom{m \plus{} n \minus{} k}{r \minus{} k}$ is 0. So you can always stop summing when you get to r and you get nothing extra by including the terms between n and r if n is less than r. Very cool.

Also interesting is the fact that you used r as a parameter in the second multiplicand of the convolution. I have never seen this before. All the convolution problems I have done before have had the multiplicands and the product be independent of which term your desired sum is in the product. But I suppose there is no reason why they need to be. Thanks.

[quote="FibonacciFan"]I see the problem now. I interchanged the n and k summations which does not make sense because the summation over k has n as a parameter.[/quote] Ah, I didn't notice that on my read-through.  Also, it is possible to interchange sums in such a situation, it just needs to be done carefully: $ \sum_{n \geq 0} \sum_{0 \leq k \leq n} \equal{} \sum_{k \geq 0} \sum_{n \geq k}$, for instance.  Was this the type of correction you made to your argument, or did you just do the sum over $ k$ first?

[quote="FibonacciFan"]Now I see that you can replace the limit n with an r and get the same answer. When k > ,$ \binom{n}{k}$ is 0 and when k > r, $ \binom{m \plus{} n \minus{} k}{r \minus{} k}$ is 0. So you can always stop summing when you get to r and you get nothing extra by including the terms between n and r if n is less than r. Very cool.[/quote] Right, this is very much the key of the argument.  I think that often with sums involving binomial coefficients it's worthwhile to consider what the "real" limits of summation are.  In this case, all terms are $ 0$ whenever $ k$ is larger than either $ n$ or $ r$.

[quote="FibonacciFan"]Also interesting is the fact that you used r as a parameter in the second multiplicand of the convolution. I have never seen this before. All the convolution problems I have done before have had the multiplicands and the product be independent of which term your desired sum is in the product. But I suppose there is no reason why they need to be. Thanks.[/quote] My pleasure.  I suspect, as you do more of these questions, that many little tricks will begin to feel more and more natural until you don't even realize that you're using them :)

[quote="JBL"][quote="FibonacciFan"]I see the problem now. I interchanged the n and k summations which does not make sense because the summation over k has n as a parameter.[/quote] Ah, I didn't notice that on my read-through.  Also, it is possible to interchange sums in such a situation, it just needs to be done carefully: $ \sum_{n \geq 0} \sum_{0 \leq k \leq n} \equal{} \sum_{k \geq 0} \sum_{n \geq k}$, for instance.  Was this the type of correction you made to your argument, or did you just do the sum over $ k$ first?
[/quote]

I just summed over k first.