calculus
integration
algebra
polynomial
Ring Theory
advanced fields
advanced fields unsolved

Let $S=<R,0,1,+,X,< >$ be a model. Show that there exists an extension $S'$ of $S$ satisfying the same sentences as $S$, that contains an infinitessimally small element ($a>0$ such that for all $n$, $na<1$). Does such $S'$ contain infinitely large elements?

Well, you should at least tell us, [i]of what[/i] your S is a model of.

It seems you are using the first order language of ordered rings, i.e. "+" as addition, "x" as multiplication, "0" and "1" as neutral elements and "<" as the order.

So I assume S is an ordered (commutative) ring.

You can easily prove that S is an integral domain. Set S' := S[X], the polynomial ring, equipped with the order induced by setting $f > 0$ iff the lowest nonzero coefficient of $f$ is positive, i.e. for $f = a_{n}X^{n}+... a_{m}X^{m}$ with $n \geq m$ and $a_{m}\ne 0$ we have $f > 0$ iff $a_{m}> 0$. Then S' contains the infinitesimal element $X$. S' contains no infinite elements.
EDIT: I see, that S' does not necessarily satisfy the same sentences as S. Hm...

But if you are not talking about rings but about fields, then you have to take for S' the field of fractions of S[X]. This S' contains infinite elements.

LiMa


Let $ R_n$ be the graph with $ n$ points arranged to form a concave polygon with all vertices connected.

Suppose that an arbitrary $ R_k$ is colored with three colors; for the sake of convenience, imagine they are blue, red, and green.

What is the smallest value of $ k$ such that $ R_k$ must contain at least one monotonic $ R_4$?

Standard notation is $ K_n$ for the complete graph on $ n$ vertices; the "concave polygon" part is irrelevant.  The number you're asking for is $ n \equal{} R(4, 4, 4) \equal{} R_3(4)$, i.e. the number of vertices such that any three-colored $ K_n$ contains a monochromatic $ K_4$.  This value is unknown.  As of 2004, the best lower bound is 51; the best upper bounds are "trivial" upper bounds; for $ R(4,4,4)$, it's something over 200.  See, e.g. http://www.emis.ams.org/journals/EJC/Surveys/ds1.pdf

The only known Ramsey numbers are $ R(3, k)$ for $ k \leq 9$, $ R(4,k)$ for $ k \leq 5$ and $ R(3,3,3)$.  In addition, $ R(3,3,4)$ is known to be either $ 30$ or $ 31$.


AMC
AIME
AMC 12
trigonometry
function
AIME I
AMC 10
\/closed

What are the recommended AMC/AIME scores if you want to enroll in the AMC 10/12 or AIME Problem Series(es)?

And is the AIME Problem Series going to be offered in the Spring? And when do the Spring courses end?

I've never taken AMC or AIME (I'm in 8th grade) but I got 6 on an AIME from 1984 (probably about a 3 or 4 on a current AIME.) I'm good with all the Intro courses (I'm taking counting right now) and in the fall and spring I want to take Intermediate Algebra, Intermediate Complex Numbers/Trig, and AIME Problem Series if possible. I'm actually not sure if I'll be able to make AIME if I take AMC 12 right now, but would I be able to if I took the Intermediate courses before I took the AMC 12? Would those help much?

Thanks in advance for your help, guys. :)

If you aren't sure if you can make AIME, I would suggest taking an AMC problem series (probably AMC 12). But if you're pretty confident about making the AIME and want a little challenge, the AIME problem series might be better. I haven't taken either, so I'm getting my ideas from what other people have said. You probably want a first-hand source. I'm pretty sure any intermediate class would be good to take and would definitely help, but if you want direct contest practice the problem series are the way to go.

I would say take AIME series. First of all, u have plenty of time to prepare for AMC12, given the fact that i am  assuming you are under 11th grade. Which comes to my second point, that is, you only need cutoff score to go to USAMO. So, I would suggest doing AIME Problem Series. I have taken both, and I liked AIME better. It raised my score; AMC12 series didnt do very much help.

Hm, makes you wonder where school has actually gone to.

Oh well, you're still in middle school.

If you're borderline to make the AIME, take the AMC 10 Problem Series.

If you're likely to make the AIME, but typically solve from 0-3 AIME problems, take the AMC 12 Problem Series.

If you're completely secure about making the AIME and typically solve 3-10 problems on the AIME, take the AIME Problem Series.

That's the closest advice I can give, but it's based on the design of the courses.

Hi, I've got a question: What does intermediate algebra and trig contain?

[quote="chess64"]I've heard that Intermediate courses help with AMC 12, so if I take Intermediate Trig this fall, and Intermediate Algebra in spring (if it's offered) along with AIME problem series (if it's offered) then will I be likely to, first of all, make the AIME, and also hopefully get 5+?[/quote]

A few things:

First, most students who take Intermediate level courses and the AIME Problem Series do fairly well or very well on the AIME, but that's more a function of how hard students work than anything.  Of course, a student needs to be prepared for the material in the subject classes, so the pretests are very important indicaters.

Second, the Intermediate Trig/Complex Numbers class is a fair amount harder than the Intermediate Algebra class, but it's up to you as to the order in which you take them.  Just know that it's possible that you'll have to look back and restudy some of the Trig/Complex Numbers after you've studied it once and have spent more time focusing on the kinds of algebraic manipulations covered in the Intermediate Algebra class.


how much energy is stored in $1.00m^3$ of air due to "fair weather" electric field magnitude $150 v/m$

[color=blue]The electric potential energy is

$\large W=\omega_EV=\frac{\epsilon_0E^2V}{2}$

where $\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$[/color]


Fie  a numar real ,a>0.Sa se determine maximul expresiei:
x^3+y^3+z^3-x^2*y-y^2*z-z^2*x-x*y*z,daca x,y,z apartin intervalului [0,a].

[hide="solutie"]maximul este $a^{3}$, datorita inegalitatilor :
(fie $x$ maximul dintre $x,\,y,\,z$)
$x^{3}\leq a^{3}$
$y^{3}\leq x^{2}y$
$z^{3}\leq z^{2}x$
$0\leq y^{2}z+xyz$.
prin adunarea acestora, se obtine :
$x^{3}+y^{3}+z^{3}-x^{2}y-y^{2}z-z^{2}x-xyz \leq a^{3}$.
egalitate se obtine daca si numai daca este egalitate in toate cele patru inegalitati de mai sus :
din prima rezulta $x=a$.
din ultima rezulta $yz(y+x)=0$, dar $y+x=y+a>0$, deci $yz=0$. daca amandoua sunt $0$, atunci evident si celelalte doua inegalitati sunt adevarate, si maximul se atinge (se verifica). daca $y=0$ sau $z=0$ si cealalta variabila este nenula, atunci din inegalitatea corespunzatoare rezulta ca este egala cu $a$.
in orice caz, maximul se atinge.[/hide]


. Suppose that x, y and z are integers with  $x^{3}+y^{3}+z^{3}$ divisible by 7. Prove that at least one of $x, y and z is divisible by 7$.

in mod 7 a cube can have 0,1 or -1. If none of x,y or z is 0 in mod 7, we can not obtain 0 by using three 1 or -1(1+1+1=3,1-1+1=1,1-1-1=-1,-1-1-1=-3)


I have a problem using the blog...

I want to make this file be clicked and be downloaded for other people?

how do you do it?

Please rephrase your question. At this point it doesn't make any sense to me :)

I think he wants to know how to attach files other than images. I don't think you can.

hm....

what I'm asking for is,,

is there any way you can post things (such as words file/ texts/ adobe etc.. ) on my blog???

( not going through all the web server things )

No, and there will not be any (to prevent abuse). Try to think of it as blog spot, not as a personal briefcase ;)

You could always simply link to some other place from which you can share the files you intend to share.

[quote]No, and there will not be any (to prevent abuse). Try to think of it as blog spot, not as a personal briefcase Wink[/quote]

okay.... :(

that I think you would have to ask directors of those sites...

[quote="shinwoo"]then how do you do it in other blogs?(not the AoPS one ) :lol:[/quote]It is not a hard thing to do, but we don't want to do it because of the reason we explained above.

If you really want to share files (I presupume non-images, and non-math files - because those can easily be posted with $\LaTeX$ directly into the blog, there are numerous free sites which can serve as your personal briefcase.

For example Yahoo Briefcase (you have to register an account with yahoo first) or Rapid Share (no registering required :) ).


I have a TI 84 +/SE that I bought new in 2005. A week ago, it stopped turning on, and I changed the batteries yesterday. Nothing. I spent some time today putzing around in the garage and got the back cover off. There was light corrosion I managed to remove using compressed air, a tire pump, and a paintbrush, but the calculator still won't turn on. I've replaced the backup battery and the AAAs again, but to no avail. My mum thinks the display is hosed. Does anyone know a place that can repair calculators?

Your best bet may just be to send it to Texas Instruments and see if they can fix/replace it.

Sorry I don't know where to get your repaired...your question just made me think of just how ridiculously durable TI calculators are, since maybe that's the reason why its hard to find a repair site. I used to drop my TI-83 all the time and accidentally do other stupid damage to it and it never seemed to break. I was always amazed at that since many other electronic devices seem to break the first time you drop them.

TI calculators are mainly durable because they're built with 10-year-old technology with little regard for weight - they build them like bricks and make few structural sacrifices.


Study the convergence of the seires $\sum_{n=1}^{\infty}n^{-2-sin(n)}$

See [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=69242[/url]. Jmerry posted a brief sketch of the proof of divergence there, though it was by no means a complete argument.


induction
combinatorics unsolved
combinatorics

That argument is wrong. You need an actual proof in the induction step- [i]how[/i] do you cover the double-sized square?

[quote="ragingbullrocky"]I am trying to prove that a $2^{n}\times 2^{n}$ board with one square removed can be covered by L-shaped tiles(3 square L-shaped tile). Here is my proof:

20. Proof: We use induction on $n$. Base case: For $n = 1$, a $2 \times 2$ square grid with any one square removed can be covered by exactly one L-shaped tile. Induction step: Suppose now as induction hypothesis that for some positive integer $k$, a $2^{k}\times 2^{k}$ square grid with any one square removed can be covered using $r$ L-shaped tiles, were $r$ is some positive integer. Then a $2^{k+1}\times2^{k+1}$ board with one square removed can be covered by $2r$ L-shaped tiles. Conclusion: Hence, by induction, for a positive integer $n$, a $2^{n}\times 2^{n}$ square grid with any one square removed can be covered by $2r$ L-shaped tiles. $\square$

Is this proof correct?

Thanks[/quote]
Hint: $2^{k+1}\times 2^{k+1}=4\times [ 2^{k}\times 2^{k}]$ .

So it will be 4 $2^{k}\times 2^{k}$ squares combined to make a larger $2^{k+1}\times 2^{k+1}$ square. By induction hypothesis, we can tile three of the squares with one square removed. We remove a square from three arbitrary squares. Thus we can tile a $2^{k+1}\times 2^{k+1}$ square with one square removed?

I don't quite understand the step where you "remove a square from three arbitrary squares". Are you sure you've proven a $2^{k+1}\times 2^{k+1}$ grid can be covered when you assumed three squares away? You can refine this to show that these three removed squares can be covered by one L-shaped piece, but you need to show very clearly how this can be done.

[hide="Hint"]Your induction hypothesis allows for any one square removed from a $2^{k}\times 2^{k}$ grid. In the $2^{k+1}\times 2^{k+1}$ grid, use particular arrangements for the other three $2^{k}\times 2^{k}$ squares to create an L-shaped piece hole out of their three removed squares.[/hide]

I'm proving my profile. I think your proof (Vietnam) is incorrect. 2^n+1*2^n+1 =4(2^n*2^n) also if there were 2r pieces, then (2^n+1)^2-1 would be even which isn't.


calculus
derivative
function
quadratics
vector
linear algebra
matrix

Hi, I have a question regarding the 2nd derivative test, in the case of a two variable function, $f(x,y)$.

I understand that the point of $D=f_{xx}f_{yy}-(f_{xy})^{2}> 0$ is to make sure that all of the 'slices' of the function have the same concavity, but I'm don't know exactly why and my book doesn't prove this...can someone show me?

Also, just to clarify, it is possible use $f_{yy}$ instead of $f_{xx}$, right, since they both have the same sign?

The second derivative test is really a consequence of Taylor's theorem. In what follows, we abbreviate, for instance, $f_{x}(x,y)$ as simply $f_{x},$ and we let $r=\sqrt{h^{2}+k^{2}}$

$f(x+h,y+k)=f+f_{x}\cdot h+f_{y}\cdot k+\frac12\left(f_{xx}\cdot h^{2}+2f_{xy}\cdot hk+f_{yy}\cdot k^{2}\right)+o(r^{2}).$

We'll also assume that the function is smooth enough - that is, has enough continuous derivatives - for this to be valid.

Now we ask if $f$ could have a maximum or minimum at $(x,y).$

The first place we look is the first degree terms of this Taylor expansion: $f_{x}\cdot h+f_{y}\cdot k=\nabla f\cdot\langle h,k\rangle.$ If $\nabla f\ne 0,$ then we can find $\langle h,k\rangle$ in the direction of $\nabla f$ for which $f(x+h,y+k)-f(x,y)$ looks like a constant times $r.$ That would be positive on one side and negative on the other, and second and higher order terms can do nothing to change that.

Hence we can only have a max or min if $\nabla f=0.$ But you knew that already. First we find the critical points, then we seek to use the second derivative test to classify them.

So now we assume we have a critical point. $f_{x}=f_{y}=0$ and the linear terms in the expansion above are gone. Now our attention turns to the quadratic terms, $Q(h,k)=f_{xx}\cdot h^{2}+2f_{xy}\cdot hk+f_{yy}\cdot k^{2}.$ This is what is known as a [i]quadratic form[/i] in $(h,k)$ and our conclusions will follow from a classification of it. Note that if this is nonzero in any particular direction, then near the point in question, it is much larger than the $o(r^{2})$ terms of the expansion. The possible classifications:

[b]Positive definite:[/b] For all $(h,k)\ne(0,0),\ Q(h,k)>0.$

In that case, our function definitely has a minimum.

[b]Negative definite:[/b] For all $(h,k)\ne(0,0),\ Q(h,k)<0.$

In that case, our function definitely has a maximum.

[b]Indefinite:[/b] There is some $(h,k)$ for which $Q(h,k)>0$ and there is some $(h,k)$ for which $Q(h,k)<0.$

In that case, our function definitely has a saddle point.

[b]Positive semidefinite but not positive definite:[/b] For $(h,k)\ne(0,0),\ Q(h,k)\ge0$ but there is some $(h,k)\ne(0,0)$ for which $Q(h,k)=0.$

In that case, we have ruled out a maximum but have several other possibiltites, determined by the previously ignored $o(r^{2})$ part of the Taylor expansion. Consider the behavior of the following functions at the origin:

$f(x,y)=x^{2}$ has a non-isolated local minimum.
$f(x,y)=x^{2}+y^{4}$ has an isolated local minimum.
$f(x,y)=x^{2}-y^{4}$ has a saddle point.
$f(x,y)=x^{2}+y^{3}$ has a critical point which is neither a minimum nor a saddle point.

[b]Negative semidefinite but not negative definite:[/b] For $(h,k)\ne(0,0),\ Q(h,k)\le0$ but there is some $(h,k)\ne(0,0)$ for which $Q(h,k)=0.$

In that case, we have ruled out a minimum but have several other possibiltites, determined by the previously ignored $o(r^{2})$ part of the Taylor expansion.

[b]Zero:[/b] If $Q(h,k)=0$ for all $(h,k),$ then the test is completely inconclusive.

(Continued in the next post.)

So the question comes down to the classification of the quadratic form $Q(h,k)=f_{xx}\cdot h^{2}+2f_{xy}\cdot hk+f_{yy}\cdot k^{2}.$ This is an algebra question.

For the two-variable case, this algebra question responds to elementary methods, namely completing the square. If you find a "mainstream" calculus book that does give a proof, this is what will be there ([i]e.g.[/i] Stewart, 5e edition, the end of section 14.6).

But at heart, this algebra question is a [i]linear[/i] algebra question. Let's write $Q(h,k)$ in vector form this way:

$Q(h,k)=Q(\vec{h})=\begin{bmatrix}h&k\end{bmatrix}\begin{bmatrix}f_{xx}&f_{xy}\\ f_{xy}&f_{yy}\end{bmatrix}\begin{bmatrix}h\\ k\end{bmatrix}=(\vec{h})^{T}H\vec{h}.$

The classification of this quadratic form depends on classifying the matrix $H$ of second derivatives. As a symmetric matrix, it has real eigenvalues. (And the eigenvectors correspond exactly to the extremes of what d343seven called the concavity on slices.) In this $2\times 2$ case, there are two eigenvalues.

If both eigenvalues of $H$ are positive, then $Q$ is positive definite.

If both eigenvalues of $H$ are negative, then $Q$ is negative definite.

If the eigenvalues are of mixed sign, then $Q$ is indefinite.

If one eigenvalue is positive and one is zero, then $Q$ is positive semidefinite but not positive definite.

If one eigenvalue is negative and one is zero, then $Q$ is negative semidefinite but not negative definite.

Finally, if $H=0,$ then $Q=0,$

Note that we don't really need to know the eigenvalues themselves, just their signs. For this $2\times 2$ matrix, the product of the eigenvalues is the determinant of $H,$ namely $f_{xx}f_{yy}-(f_{xy})^{2},$ while the sum of the eigenvalues is the trace of $H,$ namely $f_{xx}+f_{yy}.$

If the determinant is negative, then the eigenvalues have mixed sign, the quadratic form is indefinite, and the function has a saddle point.

If the determinant is positive, then both eigenvalues have the same sign. We still need to determine what sign that is, so we look to the trace, $f_{xx}+f_{yy}$ But since $f_{xx}$ and $f_{yy}$ must have the same sign, we need only check one of them (it doesn't matter which one). If $f_{xx}>0,$ then the eigenvalues are positive, the quadratic form is positive definite, and the function has a minimum. If $f_{xx}<0,$ then the eigenvalues are negative, the quadratic form is negative definite, and the function has a maximum.

If the determinant is zero, then at least one of the eigenvalues is zero, and we looking at one of the semidefinite cases. The trace will still distinguish between the main possibilities.

--

That discussion was for two dimensions, but once we have introduced the matrix language, we have the tools to extend this to all dimensions. Look up such phrases as "positive definite matrix" to find tools to help with the classification.

Oh wow, thanks for the detailed response!  Very helpful.

Note that Kent Merryfield has made a post about [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=taylor&t=58360]the same topic[/url] some time ago (it took quite some time to refind it).
I remembered it because it was pretty enlightening.


In my opinion, its good and bad.  It helps you understand the material if you dont already and gives you practice for the test/exam.  Other people in my class hate homework.  Whats your opinion?
[hide=" :)"]  hehe...[/hide]

Good in limited amount. Teachers should give way less hw and more time for a student to just think.

it is not healthy the way they give hw in the usa.

the last 3 nights i got an average of 4 hours of sleep and i have the SAT's tomorow. not to mention that swim season started.

im in 8th grade, so not in high school yet.  i slept for 5 hours last night

or this morning

[quote="Ignite168"][quote="ccy"][quote="Ignite168"]There is not enough homework in the US, kids have it so easy here.[/quote]
Where do you live?[/quote]

The United States. We have it really easy here.[/quote]

The question was: WHERE in the US do you live?  A bunch of those who posted here live in the US.  You must live in a different part of the US if you find you have it really easy...

How much homework do you get in the US?

I'm in (the equivalent of) 10th grade and we are supposed to do 2 hours a night, though it would usually take most people not much longer than an hour and a half.

That already seems like enough, though sometimes we get coursework that takes up loads of extra time. Generally I prefer to work out of class, with a computer, food and music because it is easier to concentrate, so I think homework's quite good (compared with regular classwork).

The first thing you have to learn: how to avoid useless homework ;)

Honestly, in a society with kids who know what they're doing and actually strive to learn, I think that homework should be optional.

Unfortunately this society doesn't exist...

How long are all of your school days. I bet none of you have it longer than 10 hours.

[quote="neelnal"]Ok ignite, you probably don't even go to a hard school.

anyways, i get about 5 to 6 hours of homework a night. i get home at 3:00 and stay up on average to 11:30ish.[/quote]

you obviously multitask while doing your homework

i would know, so do I

I think homework is fine- in moderation.  they should just assign one or two [u]hard[/u] math problems a night, a big writing thing each week, etc... for about 30 min-1 hour and tack on a few thousand more pages of required reading  :D

I'm sorry zetaX i wasn't spamming i put a thing on how junggi always is pissed and why not command me to die again!!! I am so sorry about the pure quote!!!!!!  :oops:

[quote="ZetaX"]Funcia: stop spamming, or what is this pure quote intended to mean¿

(if some other stuff, like the "stupid" part, here is pure spam depends on the reasons, but if not given, it's spam, too)[/quote]

I think he clicked quote and then accidentally forgot to add to the message.

Now let's quit calling each other stupid and actually address the topic: What does everyone think of homework?

[quote="mustafa"]The theory of Homework is a good thing, but it is misused and overused often. History, for instance, seems to be nothing but busywork, which is not the purpose of Homework. Homework for Homework's sake must be abolished, and must be replaced by Homework for the sake of the Art of Learning. Math teachers are famous for tainting the Fine Name of Mathematics with busywork instead of Homework.

I don't know if I was being serious or not there :rotfl: . I think that I agree with what I said, but I don't know what I was thinking when I said it :huh:[/quote]

Um..All mathematics you learn in middle school, high school, etc. is "busy work" because it's APPLIED mathematics and not anything new and all homework you would get is merely practicing those concepts. Math you learn in high school is for classes like Physics, not Mathematics.

I say maximum 2 hours hw a night. The amount of hw i get, i have no time to actually sit back and think about what i am doing; i just do it.

i don't get much homework, but i listen to music and talk with friends while doing it so it stretches out to quite a bit of homework, i don't mind busywork though, it's the homework that makes me think (honors bio homework...) that pisses me off

[quote="blahblahblah"][quote="neelnal"]i bet none of you teachers have asked you to prove Scrodinger's wave equation in 3Dimensions using vector cross product.[/quote]

haha, good for you!

Too bad just about every post I've seen from you regarding math involves you using some excessively complicated method incorrectly![/quote]

BOOM.

homework is usefull and good and nice and we need homework.

come on people, have you ever been happy 'cause you had a specific homework? (note: i didn't say all the homework, but specific; you know math, physics and astronomy related)

yeah i have actually

usually when my grade's a 60 and i need points

[hide] PAN METRON ARISTON! EVERYTHING IN MODERATION! [/hide]
[hide]Of course, my teachers don't believe in that :mad: [/hide]
[hide]I honestly think that a lot of homework that makes sense to do is good, but i fry my brain over useless language arts homework Thu7 would know.[/hide]

yeah i would know soxfan34 i was up til 11 pm
I understand like homework is good for practice but in america teachers make such a big deal about it
In my opinion if u get good test grades you should get a good grade
but my teachers dont think so :mad:

[quote="neelnal"]i was kidding around man. stop being so up-tight.

but the term "private" means nothing.

i bet none of you teachers have asked you to prove Scrodinger's wave equation in 3Dimensions using vector cross product.

i was just commenting on the amount of hw u get. u said it is easy. i kno people from other countries that actually have time to play. me on the other hand? i average about 4.5 hours of sleep during weekdays and 8 hours on good weekend days. no one wants to go to my school. it has a reputation around my area of being a student smasher.[/quote]

Hm at TJ in Quantum Mechanics we have to derive polar forms of Schrodinger's equation, which are extremely complicated.

I btw agree that HW is bad.

A lot of people would say "But if you think your classes are too hard you could always take easier classes", the problem is that people seem to think that "Honors" classes mean you are supposed to get more work, so either you can be challenged mentally and completely owned by HW or you can be bored to death with very little HW, there seem to be very few other choices in most classes...at least at TJ the CS and Physics departments subscribe to the philosophy of less homework, more learning.

Yup 100% agree with JSteinhardt.

Stop giving so much homework that we don't even have enough time to think about what we are doing.

I feel that much of my hw is fruitless since I am becoming mechanical and not getting time to think. The worst is math hw. Stupid busy work that stuff is. I stopped doing math hw and my teachers started not to care anymore so I am happy.

Aye, when teachers give too much homework, it doesnt become homework anymore, it becomes a burden.  Homework in moderation is good, but it should NEVER be graded based on if you got the answers correct (pertaining to math hw).  If that happens, then you have students who arent doing the homework to learn anything, they are doing it to get the right answer.  That leads to less sleep, loss of concentration and focus, and overall poorer performance in class.  Its a nasty little snowball effect.

Homework is a dangerous tool, but inthe hands of an able teacher, can be your best friend.

I think homework is bad because it's boring and it doesn't teach me anything and it's ANNOYING!!!

I usually don't do my homework at home so if anything, homework just makes me not pay attention in class

and the fact that homework is BORING; I try it in the kitchen, with DVD, I even make my own


[quote="leewongjao"][quote="gauss1181"][quote="azn_invazn"]A peep increased by 25% is a pop. A pop decreased by 40% is a slug, and a slug increased by 100% is a slap. What percent of a peep is a slap?[/quote]
Let a be a peep, b be a pop, c be a slug, and d be a slap.

Given the information, we have:

1.25a=b
0.6b=c
2c=d

We need to find an equation that relates a and d.
d=0.12b
d=0.15a

So a slap is $ \boxed{15 \text{percent}}$ of a peep.[/quote]

Since 2c=d, and c=0.6b, wouldn't d=2c=1.2b instead of .12b?

So d=1.2b
d=1.5a

So a slap is $ \boxed{150 \text{ percent}}$ of a peep.[/quote]
oops, silly mistake :(  :blush:

I was viewing this, and then I saw 15? I was like... Hm... But then, the other guy changed it. I was like "phew"....


function
number theory proposed
number theory

By number-theoretical functions, we will understand integer-valued functions defined on the set of all integers. Are there number-theoretical functions $ f_0(x),f_1(x),f_2(x),\dots$ such that every number theoretical function $ F(x)$ can be uniquely represented in the form
$ F(x)\equal{}\sum_{k\equal{}0}^{\infty}a_kf_k(x)$,
$ a_0,a_1,a_2,\dots$ being integers?

I may be misinterpreting the problem, but as it stands what's wrong with letting $ f_k(x)$ denote the function that takes value $ 1$ at $ x \equal{} \delta(k)$ and value $ 0$ elsewhere where $ \delta(k)$ is a bijection from the non-negative integers to the integers?  Then $ F(x) \equal{} \sum_{k \equal{} 0}^{\infty} F(\delta(k)) f_k(x) \forall x$ and this representation is unique.


How do you get LaTeX (or something similar) on webpages created with HTML?

I believe you write [dollar sign]latex [LaTeX text][dollar sign]

(learned that from ak909090)

I think she's asking how to install, not how to input.

No, it most certainly does not. In fact, the way to use $ \text{\LaTeX}$ is to use something that isn't html- like pictures, as is done here.

How to put LaTeX on a page depends on exactly what exactly isabella2296 wants to do and what facilities are available (type of access etc). Could you give us some more information on this, then we can recommend a suitable method?

I'm in the process of moving my blog to a website with my own domain, so I want to have LaTeX or something similar on the pages.

I presume you don't want to use Wordpress's own blogs at http://wordpress.com as they offer LaTeX support. Does your host offer you LaTeX (plus Ghostscript and ImageMagick) or do you have root access to be able to install them on the host's server? If so, then you can use LatexRender (which is what AoPS uses) by downloading the plug-in wp-latexrender from [url=http://sixthform.info/steve/wordpress/]Using LaTeX in WordPress[/url].
If you don't have access to, or cannot install LaTeX on the server, then a good alternative is to use [url=http://www.forkosh.dreamhost.com/source_mathtex.html#webservice]mathTeX webservice[/url] with the plugin at [url=http://programmingnote.com/blog/?p=10]Using Math Formula in WordPress[/url]. This uses images generated on mathTeX's own server which may be a little slower than using your own server.

You may also like [url=http://webtopmania.blogspot.com/2008/02/mathtex-bookmarklet_17.html]Mathtex bookmarklet[/url].

Thanks, everyone, for all the great help!  :)


2004 China Western Mathematical Olympiad  Problem 8
http://www.mathlinks.ro/Forum/viewtopic.php?p=120431

http://www.mathlinks.ro/Forum/viewtopic.php?t=17498


[quote="bgbgbgbg"]show that:if a,b are rational numbers and a^5+b^5=2(ab)^2 then 1-ab is a complete square[/quote]

Let $ a\equal{}\frac pq$ and $ b\equal{}\frac rs$ with $ \gcd(p,q)\equal{}\gcd(r,s)\equal{}1$

We get $ p^5s^5\plus{}r^5q^5\equal{}2p^2r^2q^3s^3$ from which it is immediate to conclude $ s\equal{}q$

Let then $ p\equal{}cu$ and $ r\equal{}cv$ with $ \gcd(u,v)\equal{}1$. Then the equation becomes $ c(u^5\plus{}v^5)\equal{}2u^2v^2q$

So $ u^2v^2$ divides $ c$ and we have $ c\equal{}ku^2v^2$ and $ k(u^5\plus{}v^5)\equal{}2q$

Then $ 1\minus{}ab\equal{}1\minus{}\frac {pr}{q^2}$ $ \equal{}\frac{4q^2\minus{}4pr}{4q^2}$ $ \equal{}\frac{k^2(u^5\plus{}v^5)^2\minus{}4k^2u^5v^5}{4q^2}$ $ \equal{}\left(\frac{k(u^5\minus{}v^5)}{2q}\right)^2$

Q.E.D.

If $ a\equal{}0$ ,then $ b\equal{}0$. $ 1\minus{}ab\equal{}1\equal{}1^2$
If $ a\neq 0$, then $ a^5 b \plus{}b^6\equal{}2a^2b^3$;$ a^5b\equal{}2a^2b^3\minus{}b^6$
$ a^5b\minus{}a^4\equal{}\minus{}a^4\plus{}2a^2b^3\minus{}b^6$
$ a^4\cdot
 ab\minus{}a^4\equal{}\minus{}a^4\plus{}2a^2b^3\minus{}b^6$
$ a^4(ab\minus{}1)\equal{}\minus{}(a^2\minus{}b^3)^2$
$ 1\minus{}ab\equal{}(\frac{a^2\minus{}b^3}{a^2})^2$

[quote="Pirkuliyev Rovsen"]If $ a \equal{} 0$ ,then $ b \equal{} 0$. $ 1 \minus{} ab \equal{} 1 \equal{} 1^2$
If $ a\neq 0$, then $ a^5 b \plus{} b^6 \equal{} 2a^2b^3$;$ a^5b \equal{} 2a^2b^3 \minus{} b^6$
$ a^5b \minus{} a^4 \equal{} \minus{} a^4 \plus{} 2a^2b^3 \minus{} b^6$
$ a^4\cdot ab \minus{} a^4 \equal{} \minus{} a^4 \plus{} 2a^2b^3 \minus{} b^6$
$ a^4(ab \minus{} 1) \equal{} \minus{} (a^2 \minus{} b^3)^2$
$ 1 \minus{} ab \equal{} (\frac {a^2 \minus{} b^3}{a^2})^2$[/quote]

Nicer :)

second solution
$ a^2$  is a root of $ ax^2\minus{}2b^2x\plus{}b^5\equal{}0$ ;
$ \Delta \equal{}4(b^4\minus{}ab^5)\equal{}4b^4(1\minus{}ab)\equal{}k^2, k\in Q$
Hence   $ 1\minus{}ab\equal{}m^2 , m\in Q .$

_______________________________________
Azerbaijan Land of Fire   :ninja:

To  generalize problem.   $ a^{2n\plus{}1}\plus{}b^{2n\plus{}1}\equal{}2a^n b^n$,$ 1\minus{}ab$
 :D

__________________________________
$ \textbf{Azerbaijan Land of Fire }$

[quote=pco]We get $ p^5s^5\plus{}r^5q^5\equal{}2p^2r^2q^3s^3$ from which it is immediate to conclude $ s\equal{}q$ [/quote]
[quote=Αρχιμήδης 6]how you got $q=s$ ?[/quote]

Let $q=q'c$ and $s=s'c$ with $\gcd(q',s')=1$

Equality $p^5s^5+r^5q^5=2p^2r^2q^3s^3$ becomes $p^5s'^5+r^5q'^5=2p^2r^2q'^3s'^3c$
So $q'|p^5s'^5$ and so $q'=1$
And $s'|r^5q'^5$ and so $s'=1$
And so $q=s=c$


[quote="navid"]let a,b,c>0.Prove that :(2a/b+a).^(0.5)+(2b/b+c).^(0.5)+(2c/c+a).^(0.5)<=(0.5) :huh:[/quote]
If you are talking about this inequality
$\sqrt{\frac{2a}{b+a}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\leq 3$
Then try to search.It is a famous one and has been discussed in this forum long ago! :wink:

Sorry: in the right side we have:(3)
                                                                  ok :maybe:

Conviniently please link this forum there. :(

Here is one of the links where you can find this problem.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=frac+sqrt+2b+2a+sqrt&t=17247[/url]
Although it is unsolved there,but there are some other ones where it was solved(unfortunately,I can't find those links.Maybe someone else can.I am not operating very well with the searching tool :ninja: ).
If I am not mistaken this problem was proposed in a China math competition two or three years ago.Then it was written in this form
$xyz=1$(x,y,z>0)
=>
\[\sum{\frac{1}{\sqrt{x+1}}}\leq \frac{3}{\sqrt{2}}\]
You can try to search it by this expretion,maybe it will give some results.
Yeah,I just remembered that in one of the topics where this problem were discussed it was solved by $Myth$ using Jesen's inequality :lol: .
I see,I said almost everything besides providing the actual link :blush: .
Anyway,I hope,I helped,at last a little :oops: .

Edit:Eventually I have found it
[url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=sqrt+frac&t=17549[/url] :wink:
I pleased with myself ! :P  :D .

[quote="Tiks"]If you are talking about this inequality

$ \sqrt {\frac {2a}{a \plus{} b}} \plus{} \sqrt {\frac {2b}{b \plus{} c}} \plus{} \sqrt {\frac {2c}{c \plus{} a}}\leq 3$

Then it was written in this form

$ xyz \equal{} 1(x,y,z > 0)\Longrightarrow\sum{\frac {1}{\sqrt {x \plus{} 1}}}\leq \frac {3}{\sqrt {2}}.$[/quote]$ 9(a \plus{} 1)^2(a^2 \plus{} 1) \minus{} 8(a^2 \plus{} a \plus{} 1)^2 \equal{} (a \minus{} 1)^2(a^2 \plus{} 4a \plus{} 1)\geq0\Longrightarrow$

$ \sum{\sqrt {\frac {2}{a^2 \plus{} 1}}}\leq\sum{\frac {3(a \plus{} 1)}{2(a^2 \plus{} a \plus{} 1)}} \equal{} 3 \minus{} \frac {3\sum{a^2(b \minus{} c)^2}}{4\prod{(a^2 \plus{} a \plus{} 1)}}\leq3,$

where $ abc \equal{} 1.$

This problem is introduced as Exercise 2.25 on page 87 with another proof on page 211 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.golden-book.com/booksinfo/11/1181846.html]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)


In area, we have a competition with 60 multiple choice problems ranging from Algebra I to calculus, with 40 minutes to complete.  I am able to do almost 40-45 questions on average (since I am not in precalculus or calculus yet), but I am only able to complete about 35 in the 40 minute limit.  I was wondering how I could improve my speed on this exam; I only have a few practice tests and don't want to exhaust those by practicing over and over again.  I know that probably known of you have seen this test, but I was wondering if anyone could give me some tips on how to improve my speed so I can do the questions I am able to do in about 30 minutes (10 leftover for the ones I cannot do but will lean eventually).

FTW helps a lot with speed.  Could you post a few sample questions?  I could help you a lot more if I knew the range of difficulty of these questions.

FTW won't help.  As I said,  these questions range from Algebra I to calculus.  I think I am being to vague, so I am simply going to ask what is the best way to increase speed?  I know it is through doing a lot of practice tests, but again I am limited to a certain amount that is not very high, and I have a pretty good memory when it comes to recalling answers from a test taken previously.  I don't know if this helps, but we are allowed to use calculators and I use an TI-84 Plus and TI-89 Titanium (loaned by teacher).  Also, I don't like to rely on programs a lot unless it will increase the speed of doing a problem by a lot, so don't just say "make a few programs".  all replies appreciated.

I see.  If FTW won't help, you need to go through a test (if not a practice test, a test of similar difficulty) and you need to analyze your performance.

[u][b]1.  Find out where you are losing time.  Calculations?  Concepts?  etc, etc.[/b][/u]
2.  Don't spend too much time on one problem.
3.  Try not to feel nervous - if necessary, eat a banana or some chocolate.  (I did that before AMC 8 :))
4.  Know your limits.  Don't be egoistic.  It may sound silly, but it's true.  If you tell yourself that "I can and WILL solve this problem" you might overdo it and waste time in the long run.
5.  Practice.

What area are you losing time in?  If you know, that will help a lot.


show that if $ \lim max(a_{n},b_{n},c_{n})\equal{}\plus{}00$ then $ lima_{n}\equal{}limb_{n}\equal{}limc_{n}\equal{}\plus{}00$the result is obvious but i am looking for a prove

It's obviously false.

By the way, you want \infty : $ \infty$

[quote="drapt@"]show that if $ \lim max(a_{n},b_{n},c_{n}) \equal{} \plus{} 00$ then $ lima_{n} \equal{} limb_{n} \equal{} limc_{n} \equal{} \plus{} 00$the result is obvious but i am looking for a prove[/quote]
It is not true.
For example $ (a_n,b_n,c_n)\equal{}(n,1,1)$


linear algebra
matrix
trigonometry
complex numbers

These are a couple of problems which I think are interesting:

Let $A$ be a matrix over $R$ (I suppose it is true for complex numbers too)and let the eigenvectors of $A$ span $R^{n}$ then find:

1. $\cos{A}$
2. $\cos^{2}{A}+\sin^2{A}$

If you guys have similar exercies please post them in this topic. Thanx in advance!

Hi!
Does the matrix A is nxn?
If A is $||a_{ij}||$ then with cosA u mean the matrix $||cos(a_{ij})$?

[quote="RaduSorici"]the eigenvectors of  span $R^n$[/quote]

You meant that the eigenvectors are the basis in $R^n$?

Sorry dunno much your terminology. :(

Yes the eigenvectors make a basis for $R^{n}$
and $\cos{A}$ is found be expanding into a taylor series.

well actually I think that instead of expanding cos(A) in a Taylor series, is much better to rewrite it as $cos(A)=\frac{e^{i A}+e^{-i A}}{2}$, than using the eigendecomposition of A we probably obtain for the first something like $Pdiag[cos(\lambda_i)]P^{-1}$, with P being the matrix which columns are the eigenvectors of A...
Sure it's true for complex number too, since the eigendecomposition can be done anyway, i think the problem is more tricky when A has not n distinct eigenvalues....
the second exercise can be solved easily in the same fashion...
another exercise might be something like
knowing that the A has n distinct eigenvalues find

$A^A$


algebra
polynomial
real analysis
real analysis unsolved

let n be positive integer an p be the polynomial of degree at most n.show that on any nonempty finite interval$max_{x\in I}|P(x)|$ is at most $4n^2$
time of the average on $I$ of $|P(X)|$

Dear samn, you really have a passion for extremely difficult problems. I spent 3 hours on this one and didn't manage to find anything. I'm also waiting a nice idea and solution. This problem is really nice and hard.


[color=purple]The similar problem of Vu Dinh` Quy' my friend :)  :
   Let a,b,c are positive numbers satisfy $a^{2}+b^{2}+c^{2}+abc=4$. Prove that
    $\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2-c}\ge 3$
 [/color]


How many ounces of 20% acid solution must be mixed with a 30% acid solution to obtain 45 ounces of 24% acid solution?

[hide]x+y=45, .2x+.3y=10.8, solve them and you get x=27,y=18.[/hide]

[quote="easyas3.14159..."][quote="Ignite168"][hide]
The ratio of 20% to 30% must be 2:3 because the combined amount is 24%. Divide 45 by 5, to get 9. Therefore, you must have 18 ounces of 20% and 27 ounces of 30%[/hide][/quote]

I think you switched them.[/quote]

Yeah, I did, woops  :blush:

[hide]
27 ounces of 20% 18 ounces of 30%[/hide]

This is the easiest of the 3 you posted thankfully.

[hide]

So you need $45%\cdot24=10.8
$ ounces of acid

From the $20%
$ part you get $.2x$
From the $30%
$ part you get $.3y$

So $.2x+.3y=10.8$
$x+y=45$

Solve
$x+1.5y=54$
$x+y=45$
$.5y=9$
$y=18$

So $x=27$ (the 20% part)[/hide]


find all continuous function on $[0;1]$ such that \[f(x+y)=f(x)+f(y)\;\forall x,y,x+y\in [0;1].\]

The solution is $f(x) = \lambda x$, where $\lambda$ is a constant. This problem was discussed at this forum before: http://www.mathlinks.ro/Forum/viewtopic.php?t=102615 .


induction
ratio
binomial coefficients
complex numbers
geometric sequence

Is arcccos(2/3) a rational multiple of pi ?

A try (who on earth can invent such problem ????) :

[hide]

I hope it's correct this time

(*) cos (3*x) = 4 cos(x)^3  - 3 cos(x)

If arcccos(2/3) = p/q pi then 2/3 = cos(p/q pi)

Let u(n) = cos(n*p/q*pi)

It's not difficult using (*) to show by induction that u(3^n) = a(n) / 3^b(n) with b(n) strictly increasing and gcd(a(n),3) = 1.

u(n) is periodical (u(n+2q)=u(n)) so there are a finite number of u(n). But there are infinite number of u(3^n) because b(n) is increasing. Here is the contradiction.

arccccccos(2/3) is not a rational multiple of pi.

I think this result can be extended to all arccos(2/3^k) no ??[/hide]

Nice. Is bgtg 4 a rational multiple of pi ? :)

Nice. Is arctan(4) a rational multiple of pi ? :)

belenos, I have no idea what you did.  Can you explain it again, perhaps?

I hope it's ok now  :oops:  :oops:

Sorry.

I think arctan(4) isn't a rational multiple of Pi. It can be reduced to the fact that (4-i)/(4+i)  isn't a unit root. I've solved the same problem for (2-i)/(2+i) before, but I can't remember how :? I think I used some divisibility arguments: if (2-i)/(2+i) is a unit root then there exists an n s.t. (2+i)^(2n) = 5^n and then I developed the left term with binomial coefficients and all and I got a contradiction, but I can't remember which one! If I can remember that, I think the same soln could be applied to (4-i)/(4+i) (I'm not sure of that).

I think I've got it:
 Let's assume that (4-i)/(4+i) is a unit root. Then there is an n s.t. (4+i)^n = (4-i)^n => (4+i)^2n = 17^n =>                                      Sum {0<=k<=2n} 2nCk*i^k*4^(2n-k) = 17^n. Since 17^n is M4+1 and the left sum is M4+i^2n it means that i^2n=1, so 4|2n so 2|n. Now let's look at the residues of the terms on the left modulo 17. We can disregard the ones with odd k because the imaginary part of the sum on the left is 0.

Let's assume k=M4. Then 2nCk*i^k*4^(2n-k) = 2nCk (mod 17) because i^k=1 modulo 17 and 4^(2n-k) = 1 modulo 17.

If k=M4+2 then 2nCk*i^k*4^(2n-k) is also 2nCk (mod 17) because this time i^k = -1 and 4^(2n-k) = -1 modulo 17. (in these 2 cases we use the fact that n is even)

Then the sum on the left is = Sum {0<= k even <= 2n} 2nCk (mod 17) = 17^n = 0 (mod 17). But Sum {0<= k even <= 2n} 2nCk = 2^(2n-1) <> 0(mod 17) => we got a contradiction. I hope I didn't go wrong along the way. It's exactly the same for (2-i)/(2+i).

Hm, I can't really follow the part about the unit roots :oops:

Sorry it sounds so complicated. I guess what i meant by unit root is: z is an unit root iff there is an n s.t. z^n=1.

Hm. And why is

arctan(4)/:pi: is irrational <=> (4 - i)/(4 + i) is not a unit root ?

Grrrrr ... those complex numbers :evil:

Let a_1=tg(p/q*pi)=4. a_n=tg(n*p/q*Pi). Let f(x)=(4+x)/(1-4x)         a_(n+1)=f(a_n)=(4+a_n)/(1-4*a_n). It's easy to see that the sequence b_n=(a_n-i)/(a_n+i) is a geometric progression; let's call its ratio r. We must have an n s.t. a_n=0, so we must have n s.t. b_n=-1 (this is what we assume). Then we must have n s.t.        [(4-i)/(4+i)]*r^n=-1. Now try to calculate r. It can be written as k*[(4-i)/(4+i)]^2 where k is + or -1 or + or -i (i can't remember). Anyway, that's the main course of thought I followed. I don't remember every detail.


trigonometry logarithms calculus derivative integration algebra polynomial

can you find :D

$ \textbf{(i)}\;\;\;\dfrac{\textbf d^{\dfrac{1}{2}}}{\textbf dx^{\dfrac{1}{2}}}\;\left(\sin\,x\right)$

$ \textbf{(ii)}\;\;\;\dfrac{\textbf d^{\dfrac{1}{2}}}{\textbf dx^{\dfrac{1}{2}}}\;\left(\tan^{\minus{}1}\,x\right)$

$ \textbf{(iii)}\;\;\;\dfrac{\textbf d^{\dfrac{1}{2}}}{\textbf dx^{\dfrac{1}{2}}}\;\left(e^x\right)$

$ \textbf{(iv)}\;\;\;\dfrac{\textbf d^{\dfrac{1}{2}}}{\textbf dx^{\dfrac{1}{2}}}\;\left(\ln\,x\right)$

$ \textbf{(v)}\;\;\;\dfrac{\textbf d^{\dfrac{1}{2}}}{\textbf dx^{\dfrac{1}{2}}}\;\left(x^n\right)\;\;\;\quad\;\;\;(\,n\in\mathbb R,\;\;\;n>0)$

It would be nice if you had explained what you mean by $ \frac {d^{\frac 12}}{dx^{\frac 12}}$ first  :maybe:  :mad:

alright relax...  :P :D  it is [i]exactly [/i]what it says, i cannot explain it any further other than stating the fact what i had asked for was the [b]half-derivative[/b], where

$ \dfrac{\textbf d^{\dfrac{1}{2}}}{\textbf d\,x^{\dfrac{1}{2}}}\;\left(\dfrac{\textbf d^{\dfrac{1}{2}}}{\textbf d\,x^{\dfrac{1}{2}}}\;\;\textbf f(x)\;\right)\;=\;\dfrac{\textbf d}{\textbf dx}\;\;\textbf f(x)\;$

in general, fractional derivatives where $ n$ is an integer $ \geq 2$ means

$ \underbrace{\dfrac{\textbf d^{\dfrac{1}{n}}}{\textbf d\,x^{\dfrac{1}{n}}}\;\left(\;\dfrac{\textbf d^{\dfrac{1}{n}}}{\textbf d\,x^{\dfrac{1}{n}}}\;\left(\;\ldots\;\left(\dfrac{\textbf d^{\dfrac{1}{n}}}{\textbf d\,x^{\dfrac{1}{n}}}\;\left(\dfrac{\textbf d^{\dfrac{1}{n}}}{\textbf d\,x^{\dfrac{1}{n}}}\;\;\textbf f(x)\;\right)\;\right)\;\right)\;\right)\;\right)\;\right)}_{\mbox{n of them}}\;=\;\dfrac{\textbf d}{\textbf dx}\;\;\textbf f(x)\;$

In this case your question makes no sense to me. Maybe somebody else will be able to figure out what you are asking but I won't even try :P.

One definition of half-derivatives is in
[url]http://mathworld.wolfram.com/FractionalDerivative.html[/url].

Half-integrals are easier to define; then the half-derivative is the derivative of the half-integral.

[hide="Odd solutions:"]
On (iii): Since $ \frac {\text{d}^m}{\text{d}x^m} c^x \equal{} (\ln c)^m c^x$, we (still) get $ \frac {\text{d}^{\frac {1}{2} }}{\text{d}x^{\frac {1}{2}}} e^x \equal{} e^x$.
On (v): We have $ \frac {\text{d}^m}{\text{d}x^m} x^n \equal{} \frac {n!}{(n \minus{} m)!}x^{n \minus{} m}$. Now recall that $ \Gamma(z)$ is defined for $ z\in\mathbb{C}/\text{non \minus{} positive integers}$, hence $ \frac {\text{d}^{\frac {1}{2}}}{\text{d}x^{\frac {1}{2}}} x^n \equal{} \frac {\Gamma(n \plus{} 1)}{\Gamma(n \plus{} \frac {1}{2})} x^{n \minus{} \frac {1}{2}}$.
By differentiating Taylor polynomial term by term (apply $ \text{(v)}$), we have
(i): $ \frac {\text{d}^{\frac {1}{2}}}{\text{d}x^{\frac {1}{2}}} \sin x \equal{} \sum_{n \equal{} 0}^{\infty} ( \minus{} 1)^n \frac {x^{2n \plus{} \frac {1}{2}}}{\Gamma(2n \plus{} \frac {3}{2})}$.
If we differentiate again, we would get
$ \sum_{n \equal{} 0}^{\infty} ( \minus{} 1)^n \frac {x^{2n}}{(2n)!}$[/hide]

[quote="misan"]alright relax...  :P :D  it is [i]exactly [/i]what it says, i cannot explain it any further other than stating the fact what i had asked for was the [b]half-derivative[/b], where

$ \dfrac{\textbf d^{\dfrac{1}{2}}}{\textbf d\,x^{\dfrac{1}{2}}}\;\left(\dfrac{\textbf d^{\dfrac{1}{2}}}{\textbf d\,x^{\dfrac{1}{2}}}\;\;\textbf f(x)\;\right)\; \equal{} \;\dfrac{\textbf d}{\textbf dx}\;\;\textbf f(x)\;$
[/quote]This "definition" is not unique, just like the function $ x \mapsto x^2$ does not have a global inverse :( , only it's worse in this case. Try to find the fractional derivative via the Laplace transform to see why.

The situation is even worse: one can show that there is no linear operator $ L$ on $ C^\infty(\mathbb R)$ such that $ L^2\equal{}D$. Indeed, if such an operator existed, it would have to preserve the space of linear functions (because we would have $ D^2(Lf)\equal{}L^5f\equal{}L(D^2 f)\equal{}L0\equal{}0$) but there is no square root from the operator in $ \mathbb C^2$ that is given by the matrix $ \left(\begin{matrix}0&1\\0&0\end{matrix}\right)$. So, whatever linear definition of $ D^{1/2}$ you want to use, you will never have $ D^{1/2}D^{1/2}\equal{}D$ for all functions. That's why I said that misan's "axiomatic definition" makes no sense to me. The definition Ravi B was referring to makes perfect sense, of course, and all misan's questions become answerable with it but as a mod, I would strongly prefer the problem proposers to know what they are talking about themselves before posting a challenge to others  :evilgrin:


Prove that in the triangle $ ABC$ the angle $ A$ is right if and only if: $ h_a^{\minus{}2}\equal{}h_b^{\minus{}2}\plus{}h_c^{\minus{}2}$, where $ h_a,h_b,h_c$ are the lengths of the heights from $ A,B,C.$

hello, setting $ h_a\equal{}\frac{2A}{a}$, $ h_b\equal{}\frac{2A}{b}$ and $ h_c\equal{}\frac{2A}{c}$ we get
$ a^2\equal{}b^2\plus{}c^2$.
Sonnhard.


geometry
geometric transformation
reflection
Support
perpendicular bisector
angle bisector
geometry unsolved

Consider a circle $ S$ and a point $ A$ found  outside the circle $ S$. Let $ B$ and $ C$ be two points on $ S$, s.t. $ AB$ and $ AC$ are tangent to $ S$. Let $ D$ and $ E$ be two points on $ S$ s.t. $ A,D$ and $ E$ are collinear in this order.Let $ F$ be the intersection point of $ BC$ and $ AD$ and let $ M$ be the midpoint of $ BC$. Suppose that $ BF>FC$. Prove that:
$ (a)$ $ BM^2\equal{}DM\cdot ME$.
$ (b)$ $ m(\angle DME)\equal{}2\cdot m(\angle DBE)$.
$ (c)$ $ m(\angle BEM)\equal{}m(\angle DEC)$.

(a) Let $ O$ be the center of the circle in the problem. Since $ OB\perp AB$ and $ BM\perp AM$ we conclude that $ BM^2=AM\cdot OM$. Thus $ AB^2=AM^2+BM^2=AM(AM+OM)=AM\cdot AO$. But $ AB^2=AD\cdot AE$. Therefore $ AM\cdot AO=AD\cdot AE\implies OEDM$ is cyclic. Hence $ \angle AMD=\angle OED=\angle ODE=\angle OME$ which implies $ \angle AME=\angle OMD$. Combining this with $ \angle AEM=\angle DEM=\angle DOM$ we conclude that the triangles $ AME$ and $ OMD$ are similar. Thus $ \frac{AM}{ME}=\frac{DM}{MO}\implies AM\cdot MO=DM\cdot ME$ which implies $ BM^2=DM\cdot ME$.

(b) I don't think your relation is correct. The relation between $ \angle DBE$ and $ \angle DME$ is $ \angle DBE=180^\circ -\frac 12\angle DOE=180^\circ -\frac 12\angle DME$.

(c) We have $ \angle BEM=\angle DEC\iff \angle BED=\angle CEM$. Now
\begin{align*}\angle CEM&=\angle CEO+\angle OEM\\
&=\angle OCE+\angle ODM\\
&=90^\circ-\frac 12\angle COE+\angle ODM\\
&=90^\circ-\angle CDE+\angle ODM\\
&=90^\circ+\angle EAO-\angle CDE\\
&=\angle CFE-\angle CDE\\
&=\angle BCD=\angle BED\end{align*}

To nayel:
point b) of the problem is OK. If you read attentively the condition you wiil see that $ BF > FC$.But in your figure $ BF < FC$.
But however I forgot to mention that if $ BF < FC$, then the condition of $ (b)$ becomes $ m(\angle DME) \equal{} 2\cdot m(\angle DCE)$,
which becomes obvious, after seeing that $ DMOE$ is cyclic.

Btw, thank you a lot for your nice proof. :coolspeak:

I have an additional task for you, nayel.
In the same problem if we consider the point $ K$ as being the second intersection point of $ DM$ with $ S$ (of course $ \ne D$), then prove that:
$ BCKE$ is an issoscels trapezoid.

Of course that you can use the earlier proved properties.

Sorry. I had been reading the condition as $ BF < FC$ the whole time.. :blush:

Anyway for your next problem, since the line $ DM$ is the reflection of line $ EM$ on $ BM$, $ MK$ must be the reflection of $ ME$ on $ OA$. Therefore since $ OA$ is the perpendicular bisector of $ BC$, the conclusion follows. :)

Thank you very much, nayel, for the support acorded.Very nice and easy proofs!  :D

BC is the polar of A w.r.t. circle O, therefore  AD/AE=DF/FE and the circle diameter AF is the Apollonius circle w.r.t. the line segment DE, that is, MF is the angle bisector of DME. It is well known that EF, passing through the common point of the tangents at circle in B and C, is the simmedian in DBCE, i.e. angle BEF = angle MEC. It follows that triangles DMC and CME are similar, hence a)  is proved. c) this follows from simmedian property, while for b) let’s leave one the above solutions, although it was easy to be seen from the similarity of the triangles DMC and CME.

Best regards,
sunken rock


geometry
3D geometry
octahedron
tetrahedron
dodecahedron

Compute the volume and surface area of:

(1) An octahedron of side length 1.
http://mathworld.wolfram.com/Octahedron.html
(2) A tetrahedron of side length 1.
http://mathworld.wolfram.com/Tetrahedron.html
(3) A dodecahedron of side length 1.
http://mathworld.wolfram.com/Dodecahedron.html

1) The octahedron: V=2(V of tetrahedra)
V=2(1/3*BH)
B=s^2=1
(1/2*:sqrt:2)^2+ H^2=1^2
1/4*2+H^2=1
1/2+H^2=1
H^2=1/2
H=:sqrt:2/2
V=2/3*:sqrt:2/2
V=:sqrt:2/3
I think. Surface area=8* area of equilat. tri.=8(s^2*:sqrt:3)
SA=8*:sqrt:3

Tetrahedron

V=1/2(V of oct.)
V=:sqrt:2/6
SA=4 eq. tri's+ base
base=1
eq.tri's=4(s^2*:sqrt:3)
SA=4:sqrt:3+1

I'm not even going to try the dodechahdron.

I'm afraid your analysis of the octahedron = 2 tetrahedra isn't really correct.  An octahedron is two square pyramids.  You did find its volume correctly, however.  Your area formula for equilateral triangles could use some checking.


Prove that :

  n
C   (1/n)^r  <= 1/(r-1) - 1/r         for r>=2
  r

You mean
for $ r\ge2$ we have
$ \frac{1}{r\minus{}1}\minus{}\frac{1}{r}\ge\dbinom{n}{r}(\frac{1}{n})^r$?


geometry
circumcircle
incenter
geometric transformation
homothety
angle bisector

Let $ ABC$ be a triangle; $ \Gamma_A,\Gamma_B,\Gamma_C$ be three equal, disjoint circles inside $ ABC$ such that $ \Gamma_A$ touches $ AB$ and $ AC$; $ \Gamma_B$ touches $ AB$ and $ BC$; and $ \Gamma_C$ touches $ BC$ and $ CA$. Let $ \Gamma$ be a circle touching circles $ \Gamma_A, \Gamma_B, \Gamma_C$ externally. Prove that the line joining the circum-centre $ O$ and the in-centre $ I$ of triangle $ ABC$ passes through the centre of $ \Gamma$.

The center of $ \Gamma$ is the circumcenter of $ \Delta \Gamma_{a}\Gamma_{b}\Gamma_{c}$.
Also u can prove that $ A\Gamma_{a}$ is the angle bisector of $ \Gamma_{c}\Gamma_{a}\Gamma_{b}$.
and it's the same for the other angles.
It follows that $ I$ is the incenter of $ \Delta \Gamma_{a}\Gamma_{b}\Gamma_{c}$.
And also $ \Gamma_{a}\Gamma_{b}$is parrallel with $ AB$.
The same condition holds for other segments.
It means that $ \Delta ABC$ and $ \Delta \Gamma_{a}\Gamma_{b}\Gamma_{c}$ are homogeneous.
It directly follows the collinearity
(Because as mentioned $ (O)$ is the circumcenter of triangle $ ABC$ and the center of $ \Gamma$ the circumcenter of $ \Delta \Gamma_{a}\Gamma_{b}\Gamma_{c}$ and $ I$ the center of homogencity of the two triangle)

An old problem.
http://www.mathlinks.ro/viewtopic.php?t=291966

[b]Happy the Independence Day of Vietnam![/b]  :thumbup:

pretty easy
$ \Delta\Gamma_{a}\Gamma_{b}\Gamma_{c}$ and $ \Delta ABC$ are harmonious w.r.t I ,where is the incenter of $ \Delta ABC$ and $ \Delta\Gamma_{a}\Gamma_{b}\Gamma_{c}$,
notice $ \Gamma$ is the circumcenter of $ \Delta ABC$ and $ \Delta\Gamma_{a}\Gamma_{b}\Gamma_{c}$ O is the  circumcenter of $ \Delta ABC$
by using harmonious property, we get the result

Following link contains detailed discussion on INMO 2008
http://www.mathlinks.ro/viewtopic.php?search_id=1301839319&t=184278

It's an old IMO problem ;) In particular, it is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366643&sid=9087af90170afd2d61d34693c4f5ca1c#p366643]IMO 1981/5[/url].

[quote="shoki"]
It means that $ \Delta ABC$ and $ \Delta \Gamma_{a}\Gamma_{b}\Gamma_{c}$ are homogeneous.
It directly follows the collinearity
(Because as mentioned $ (O)$ is the circumcenter of triangle $ ABC$ and the center of $ \Gamma$ the circumcenter of $ \Delta \Gamma_{a}\Gamma_{b}\Gamma_{c}$ and $ I$ the center of homogencity of the two triangle)[/quote]
although its not directly related to the question but
[hide]After we have proven that a homothety exists mapping the smaller triangle to the larger one , we dont need to prove that all lines joining any point and its image passes through the homothetic center right ? [/hide]

[quote="sahilsharma94"]After we have proven that a homothety exists mapping the smaller triangle to the larger one , we dont need to prove that all lines joining any point and its image passes through the homothetic center right ? [/quote]
Of course not. It suffices just to mention that statement.

My solution involves inversion. We need to show $ O,I,O_1$ collinear where $O_1$ is the center of $\Gamma$. Conference an inversion through any of the center of three equal circles. It maps $O$ to $O^*$,$I$ to $I^*$ and $O_1$ to $O^*_1$ On drawing perpendicular from any center to  $O,I$ line you chose ,you will find a find a point on that line through which a circle can be drawn passing through $I^*, O^*, O^*_1$ and the center you chose which proves the proposition that
$the inverse of a line "l" not passing through the center is a circle passing through the center of circle.$. So those three points don't pass through the center but their inversion circle passes showing all of them lie on a line that is they are collinear. Hence proved

[hide = Definitions]
Let the centres of $ \Gamma_A, \Gamma_B, \Gamma_C$ and $\Gamma$ be $A_1, B_1, C_1, G$
[/hide]
[hide = Solution]
It is clear that $G$ is the circumcenter of $\triangle A_1B_1C_1$ . Due to equal radii condition :
$$B_1C_1 \parallel BC ; C_1A_1 \parallel CA ; A_1B_1 \parallel AB$$
Thus, by Lemma 3.10(EGMO), there is a homothety $\mathbb H$ centered at $I$ mapping $\triangle A_1B_1C_1 \rightarrow \triangle ABC \Longrightarrow \mathbb H(G) = O$
Thus, $I - G - O$ are collinear
[/hide]


topology
function
advanced fields
advanced fields theorems

Hi, I've come across the fact that a compact space is not necessarily sequentially compact, and I believe that, but I have not seen an example that I believe to both be compact and not sequentially compact. Wikipedia says that a counterexample is $ 2^{[0,1]}$, but I don't understand what this means for a couple of reasons. First, what is $ 2$ as a topological space, and how do you raise it to $ [0,1]$? Second, what would be the sequence in this space that doesn't have a convergent subsequence? Third, why is this space compact?

Would someone please explain this counterexample and/or provide another one? Thanks.

$ 2^{[0,1]}$ is more or less the set of $ f: [0,1]\to\{0,1\}$, I think.  The space is equipped with the product topology, and hence is compact by Tychonoff's theorem (since $ \{0,1\}$ is certainly compact).  For your sequence, maybe you want to take an enumeration $ (q_i)_i$ of the rationals and let $ f_i(x) \equal{} 1$ if $ x \equal{} q_i$ and $ 0$ otherwise (does this work?  I think so).

[quote="Hamster1800"]What topology is implied on $ \{0,1\}$?
I'll think about your sequence.[/quote]

the discrete one

Since convergence of a sequence in the product topology is pointwise convergence, blahblahblah's example converges to zero.

An [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=79850]old thread[/url] on the same issue.

My argument there: use $ X\equal{}\{0,1\}^C$, where $ C$ is the set of binary sequences $ \{0,1\}^{\mathbb{N}}$.
Convergence of a sequence in $ X$ is pointwise convergence. For each $ n$, define $ f_n\in X$ by letting $ f_n(x)\equal{}x_n$. No subsequence of the $ f_n$ converges, since we can always find an $ x_0$ on which the subsequence alternates values between 0 and 1.

Ah. I think I get it. Thank you very much.

However, I'd like to ask one more thing. Are you saying that convergence in the product topology is the same as convergence of each of the components? If not, what do you mean by pointwise convergence?

Yes, you've got it. I'm thinking of the product topology as a space of functions, which is way I phrased it that way.


modular arithmetic
algebra
polynomial
abstract algebra

Prove that $\binom{pa}{pb}\equiv \binom{a}{b}\pmod{p}$

[hide="solution with Identity Theorem-correct or not?"]
We fix $b$ and define a polynomial $P(a)= \frac{pa*(pa-1)*(pa-2)...(pa-pb+1)}{(pb)!}$. We note that this has $a-2b$ factors of $p$, so for $a>2b, P(a,b) \equiv 0 \pmod{p}$.

Also, we define $Q(a)=\frac{a(a-1)(a-2)...(a-b+1)}{b!}$ and we note that for all numbers where $a$ is a multiple of $p$ and $b \neq 0, a, Q(a) \equiv 0 \pmod{p}$.

We note for all numbers $a=pa'>2b \neq 0$, the two polynomials $P(x)$ and $Q(x)$ are equal. Thus, the polynomials are equal at infinitely many points, but have degree $b$, so by the identity theorem, they are equal.
[/hide]

[hide="One line"] $(x+y)^{pa}\equiv (x^{p}+y^{p})^{a}\bmod p$ [/hide]

Thanks for the solution. This problem appeared in a section about the identity theorem, so I was looking for a solution involving it.

I might not be understanding your one-liner correctly, but it would probably be easier to use $(1+x)^{pa}\equiv (1+x^{P})^{a}\pmod{p}$. Now the coefficient of $x^{pb}$ is $\binom{pa}{pb}$ on the left side, $\binom{a}{b}$ on the right.

Also- is my solution correct?

I don't think so.  You didn't show that they were equal at infinitely many points, only that they were congruent $\bmod p$ at infinitely many points.

But can't we view both equations mod p and apply the identity theorem to that field? Or does it only apply for the reals?

The thing is that we can't have "an infinite number of solutions" $\bmod p$, since the field is finite.  It is true that Lagrange's Theorem holds, but this is irrelevant if the degree of the polynomial is greater than $p$.

In short, no, the Identity Theorem does not hold.  Trivially:

$x \equiv 1 \bmod 2$

Has infinite integer solutions, but the two are obviously not equal $\bmod 2$ for all integers.


Are we to find the sum of the labels *including* the vertices $a$,$b$,$c$, or not summing $a,b,c$?

Thanks!

Billy

I don't think you're allowed to ask that.  :maybe:
[quote="DPatrick"][quote="junggi"]um.. I kind of need clarification on a question.
How do I ask?
pm one of the administrators?[/quote]You can email usamts@usamts.org, but I should warn you that it is [i]extremely[/i] unlikely that we will "clarify" any question.  However, if you believe that there is an [i]error[/i] in a problem statement, then please email and let us know.[/quote]

Read the problem carefully.  It answers it there.

One more clarification question (though I doubt I'll get an answer on this...)
Are we actually allowed to use work from a later part of a problem in an earlier part?  For example, here, could we use something from part b for our answer to part a?

These are not questions that we can answer while a round is in progress.


Of course not, it's being cool 'n spammy that counts. :P
And believe me, new people, we're all veterans of the old cool 'n spammy school of thought. :wink: About exactly a year back, random and I had a great race, each trying to outdo the other in being cool 'n spammy. We gave Dan a tough time, right Dan? :D

So guys, forget the cool club, stick up your bottoms, and be cool 'n spammy! And if anyone tries to found a club, you have bubka's endorsement. :)

NEVER

Oh wait I have... uh... CPW. Yeah. CPW.


[quote="mohd_ahsan"]which one of the following is the sum of four consecutive integers?

(a) 2000   (b)2001      (3)2002    (4)2003

Let the first number be x-1 then the numbers are

(x-1),(x),(x+1),(x+2)  if we solve this we get 2002

but if select the numbers tobe

(x),(x+1),(x+2),(x+3) and solve we get a different answer why?

why does the latter give a wrong answer?[/quote]

I fail to understand what you mean by " WRONG ANSWER".

Irrespective of whether u chose the first of  the 4 consecutive nos. to be x or x-1 , where x is a natural no., the ans. will be 2002 only.

sum((x-1),(x),(x+1),(x+2) ) = 4x + 2

Only 2002 (in the given set of options) is of the form 4x + 2.

Similarly sum((x),(x+1),(x+2),(x+3) ) = 4x + 6.

It's obvious that only 2002 (in the given set of options) fits into the 4x+6 form as well.

mod 4, the sum of 4 consecutive integers has to be 2.

Thus we get 2002.

OH yeah . Now i get it. its the same. thanx people


Let ABC be any triangle,P,Q,R points on [AB],[BC],[CA] sides respectively and these satisfy $\frac{AP}{AB}=\frac{BQ}{BC}=\frac{CR}{CA}=k<1/2$ If G point is centroid of  ABC triangle find ratio of Area(PQG)/Area(PQR)

Let $AF \cap BC =M$. So $\triangle{ABM}=\frac{1}{2}\triangle{ABC}$. Now observe triangle ABM, we see that:
\[ \triangle{ABM}=\triangle{APG}+\triangle{QMG}+\triangle{GPQ}+\triangle{BPQ} \]
Now it is just simple area ratios:
\[ \frac{\triangle{BPQ}}{\triangle{BMA}}=\frac{BP \cdot BQ}{BA \cdot BM}=2k(1-k) \]
\[ \frac{\triangle{QMG}}{\triangle{BMA}}=\frac{QM \cdot MG}{BM \cdot MA}=\frac{1-2k}{3} \]
\[ \frac{\triangle{APG}}{\triangle{ABM}}=\frac{AP \cdot AG}{AB \cdot AM}=\frac{2k}{3} \]
Thus we can find the area ratio between $\triangle{GPQ}$ and $\triangle{ABC}$. But the area ratio between $\triangle{PQR}$ and $\triangle{ABC}$ is a famous and easy conclution, so now the problem is easy to solve.


Fie ABC un triunghi isoscel cu AB=AC si D mijlocul lui (BC).Daca E este proiectia lui D pe AC si F interesectia perpendicularei din A pe BE cu dreapta DE, demonstrati ca F e mijlocul lui (DE).

Notam proiectia $L$ a punctului $B$ pe dreapta $AC$ si intersectia $M\in BE\cap AF\ .$ Prin urmare $EL=EC\ ,$ patrulaterul $ABDM$ este inscis in cercul de diametru $[AB]$ si $\triangle ADE\sim \triangle BCL\ .$ Asadar $\widehat {DAF}\equiv\widehat {CBE}$ si segmentul $[AF]$ este omolog cu mediana $[BE]$ in asemanarea mentionata. In concluzie, punctul $F$ este mijlocul segmentului $[DE]\ .$

[u]Observatie.[/u] Aceasta problema este de fapt o reciproca a problemei care apare pe ultima pagina a cartii mele de "Numere complexe" prin care ilustram aplicarea celor patru instrumente (sintetic, vectorial, analitic si numere complexe) pentru rezolvarea unei  probleme de geometrie.

Eu ma gandisem altcumva ;) .Fie M mijlocul (EC), atunci DM e linie mijlocie in triunghiul BEC, deci DM paralela cu BE si AF perpendiculara pe DM.Dar DE e perpendiculara pe AC, rezulta ca F este ortocentrul triunghiului MAD, de unde MF perpendiculara pe AD.Cum AD perpendiculara pe BC(triunghi isoscel), rezulta MF paralela cu BC, dar M e mijlocul lui (EC),atunci MF linie mijlocie in DEC.De aici avem concluzia.
   [i]Comentariu[/i]:   N-am stiut ca problema [b]nu [/b]este originala.Desi am cartea dumneavoastra, dl Nicula, nu am ajuns la sfarsitul ei, asa ca n-am avut idee de aceasta.S.S


Balance the following redox equation:

$Cr_2O_7^{2-}+HNO_2+H^+\rightarrow{Cr}^{3+}+NO_3^-+H_2O$

Did you balance using the oxidation number change method or the half reaction method (or neither?)?

I used half-reaction on this one (I find I don't mess up as much using it).


inequalities
trigonometry
inequalities unsolved

Prove that, there is allways an $ \epsilon >0$ such that, for $ x,y\in (0,1),\,|x\minus{}y|<\epsilon,\, x\neq y$ we have
\[ x^2\sin(\pi x)\minus{}xy[\sin(\alpha\plus{}\pi x)\plus{}\sin(\minus{}\alpha\plus{}\pi y)]\plus{}y^2\sin(\pi y)>0\]
where $ \alpha\equal{}\arccos\frac{x^2\plus{}y^2\minus{}\epsilon^2}{2xy}.$

[quote="cckek"]Prove that, there is allways an $ \epsilon > 0$ such that, for $ x,y\in (0,1),\,|x \minus{} y| < \epsilon,\, x\neq y$ we have
\[ x^2\sin(\pi x) \minus{} xy[\sin(\alpha \plus{} \pi x) \plus{} \sin( \minus{} \alpha \plus{} \pi y)] \plus{} y^2\sin(\pi y) > 0
\]
where $ \alpha \equal{} \arccos\frac {x^2 \plus{} y^2 \minus{} \epsilon^2}{2xy}.$[/quote]
Choose $ \epsilon \equal{} x\plus{}y$. Then $ \alpha \equal{} \arccos \minus{}1 \equal{} \pi$. Then $ \sin(\pi \plus{} \pi x) < 0$ and $ \sin( \minus{} \pi \plus{} \pi y) < 0$, so we have:
$ x^2\sin(\pi x) \minus{} xy[\sin(\alpha \plus{} \pi x) \plus{} \sin( \minus{} \alpha \plus{} \pi y)] \plus{} y^2\sin(\pi y) \equal{} x^2\sin(\pi x) \minus{} xy[\sin(\pi \plus{} \pi x) \plus{} \sin( \minus{} \pi \plus{} \pi y)] \plus{} y^2\sin(\pi y) > 0$.


geometry
perimeter
ratio
trigonometry
trig identities
Law of Cosines
power of a point

I think this is a bit easy, but enjoy!

1. In triangle $ABC$, two tangent circles are drawn, with tangent points $D$, $E$, $F$, $G$, $H$, and $J$ as shown. If $DE=3$, $CF=5$, and the perimeter of triangle $ABC$ is 28, compute $AB$.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=804[/img]

2. Find the radius of the circle that can be inscribed in a 5-12-13 right triangle.

$\hline$

3. Two externally tangent circles have radii $r$ and $R$, with $r<R$. Their common external tangent segments are extended until they meet at point $P$. If $\angle P = 60^\circ$, compute the ratio $\frac{r}{R}$.

4. Two unit circles with centers $P$ and $Q$ are externally tangent. Extend segment $PQ$ to meet the circles at $A$ and $B$. If $AC$ and $BC$ are tangents to circle $P$, as shown, and $T$ is the point of tangency, compute $AC$.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=805[/img]

$\hline$

5. Semicircles are drawn on two sides of square $ABCD$, as shown. $QP$ is a line segment with $QA=5$ and $AP=12$. Compute the area of square $ABCD$.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=806[/img]

6. Points $A$ and $B$ are on a circle of radius 10 and $AB$ intersects diameter $CD$ at $P$. If $PA=2$ and $BP=4$, how far away is $P$ from the center of the circle?

$\hline$

7. Chord $AB$ of length 12 is drawn in a circle of radius 10. Compute the distance from $B$ to the diameter containing $A$.

8. Compute the length of the tangent segment from the origin to the circle that passes through the points (3,4), (6,8), and (5,13).

i remember we had #5 on our previuos practice.

[hide]just draw BP and DQ, its quite simple[/hide]

i also remember #3(click the smilie):
[hide=";)"]
the answer is 1/3[/hide]

wow #1 is "really" easy. one can solve it in less than 10 seconds without paper and pencil

[hide]
the whole point of the problem is that the length of two tangents drawn from a point outside the circle to the cirlce is equal. so we have: AJ=AD=x and BJ=BH=y

the perimiter is: 2x+2y+6+10=28. and AB=x+y=6[/hide]

[hide="7"]
Let O be the center of the circle. We can find AOB through law of cosines to be $\frac{7}{25}$ Now to find the distance from B to the diameter from B we need to find $10\sin{180-\angle AOB}$ which we can now easily find to be $\frac{240}{25}$ after applying $\cos^2{x} + \sin^2{x} = 1$[/hide]

[hide="and also 6"]
let the distance from $p$ to the center be $x$. Then we can just apply power of a point:
$2\cdot4 = (10 + x)(10-x) \implies 8 = 100-x^2 \implies x^2 = 92 \implies x = \sqrt{92}$[/hide]

[hide="Alternate soln for 7"]
Let $AC$ be a diameter and draw segment $BC$, so that $\angle ABC = 90^\circ$. Then the distance $BD$, where $\angle ADB = 90^\circ$, is the altitude to the hypotenuse of a right triangle. Since $AC=20$, $AB=12$, (so that $BC=16$), the area of $\triangle ABC$ is $\frac{1}{2}\cdot AB \cdot BC = 96$. Also the area is $\frac{1}{2} \cdot AC \cdot BH = 96$ so that $BH = \frac{48}{5}$.
[/hide]

vvvvveeeeeerrrrrryyyyyy late reply but is #2 2? =D

I don't think so....

[hide="Problem 2"]
The area of a triangle can be expressed $ A\equal{}rs$, and in this case, the area is 30, and the semiperimeter is 30. Thus $ r\equal{}1$.[/hide]

[quote="buzzer11"]I don't think so....

[hide="Problem 2"]
The area of a triangle can be expressed $ A \equal{} rs$, and in this case, the area is 30, and the semiperimeter is 30. Thus $ r \equal{} 1$.[/hide][/quote]

[hide]$ r \equal{} 2$, because the perimeter is $ 5 \plus{} 12 \plus{} 13 \equal{} 30$, so the semiperimeter is $ 15$.[/hide]


inequalities
integration
calculus
real analysis
real analysis unsolved

Let $f: [a,b]\to \mathbb{R}$ a continuous function. Prove that:
$(\int_{a}^{b}f(x)dx)(\int_{a}^{b}\sqrt{f(x)^{4}+1}dx)\leq \frac{1}{2\sqrt{2}}(\int_{a}^{b}(f(x)^{2}+1)dx)^{2}$.

[quote="kunny"]Intuitively, can we use Cauchy-Schwarz 's inequality for Integral version?[/quote]
Yes we can...but what is your solution?

A nice solution can be obtained using Cauchy as sugested by kunny. I hope somebody will post it :)  I will put below another solution , and I would like to tell me if it is correct ( i always make mistakes on calculus problems :blush: )
[hide="solution"]
$2\sqrt{2}(\int_{a}^{b}f(x)dx)(\int_{a}^{b}\sqrt{f^{4}(x)+1}dx)=$

$2(\int_{a}^{b}\sqrt{2}f(x)dx)(\int_{a}^{b}\sqrt{f^{4}(x)+1}dx)=$

$2\int\int_{[a,b]\cdot[a,b]}\sqrt{2}f(x)\cdot\sqrt{f^{4}(y)+1}dxdy=$

$\int\int_{[a,b]\cdot[a,b]}\sqrt{2}f(x)\cdot\sqrt{f^{4}(y)+1}+\sqrt{2}f(y)\cdot\sqrt{f^{4}(x)+1}dxdy\leq $

$\int\int_{[a,b]\cdot[a,b]}(f^{2}(x)+1)(f^{2}(y)+1)dxdy=$

$(\int_{a}^{b}(f^{2}(x)+1)dx)^{2}$
[/hide]

At least pls tell me if I made stupid mistakes in my solution... :maybe:

I think it's correct.

Here's another solution: It's a very good thing to know what the equality case is. From your solution I deduced that it's $f \equiv 1$.
Write the RHS (right-hand-side) as $\frac1{2 \sqrt 2}\left[ \int_{a}^{b}(f(x)-1)^{2}\, dx+\int_{a}^{b}2 f(x) \, dx \right]^{2}$.
Thus,
\begin{eqnarray*}2 \sqrt 2 \cdot RHS &=& \left( \int_{a}^{b}(f(x)-1)^{2}\, dx \right)^{2}+2 \int_{a}^{b}(f(x)-1)^{2}\, dx \int_{a}^{b}2 f(x) \, dx+\left( \int_{a}^{b}2 f(x) \, dx \right)^{2}\\ \ &\geq& 2 \int_{a}^{b}\left( f^{2}(x)-f(x)+1 \right) \, dx \int_{a}^{b}2 f(x) \, dx .\end{eqnarray*}
This can be rearranged as follows: $RHS \geq \sqrt 2 \int_{a}^{b}\left( f^{2}(x)-f(x)+1 \right) \, dx \int_{a}^{b}f(x) \, dx$.
Note that if we replace $f$ by $|f|$ in your original inequality, it just gets stronger, so we can assume that $f$ is non-negative.
So it remains to prove that $\int_{a}^{b}\sqrt 2 \left( f^{2}(x)-f(x)+1 \right) \, dx \geq \int_{a}^{b}\sqrt{f^{4}(x)+1}\, dx$. This inequality holds [i]pointwise[/i] (I hope I didn't make any mistakes in my computations).

Nice :) And the computations are ok - $(f(x)-1)^{4}\geq 0$


induction
inequalities
combinatorics proposed
combinatorics

Suppose three direction on the plane . We draw $ 11$ lines in each direction . Find maximum number of the points on the plane which are on three lines .

This is a Russian question, most possibly a 4th round question of 9th grades - 4th question maybe. I couldn't find the year - even anything about it - on Internet about it but anyways.

Why can't it larger than 91?I find it hard to prove.

here we go, I hope it's not wrong :) :
by an affine transformation, you may assume that the three directions are these directions:

[b]1)[/b] parallel to the $x$-axis of the cartesian plane.

[b]2)[/b] parallel to the $y$-axis.

[b]3)[/b] parallel to the line $x=y$.

now fix the lines of the first two groups, so they form a $11\times 11$ table. call a point of the plane which is on three of the lines [i]special[/i], and call each line of the third group [i]good[/i]. an easy induction (omiting the first row and the first column) shows that we can have at most $1$ good line with $11$ special points on it, at most $3$ good lines with $10$ special points on it, at most $5$ good lines with $9$ special points on it, at most $7$ good lines with $8$ special points on it, at most $9$ good lines with $7$ special points on it, and at most $11$ good lines with $6$ special points on it. now using these inequalities, we obtain that there can be at most $11+2(10+9+8+7+6)=91$ special points. its example is easy to construct.


let a,b,c>0
prove that
2(ab+ac+bc)+(9/16)(1/ab+1/ac+1/bc)>=9/2+a 2 +b 2 +c 2
may be i am so stupid that i can't solve this problem
can anyone help me?

No,you aren't stupid,Master .It is easy to understand why you can't solve it.
It is wrong,for example $ a=1,b=1,c=4 $  :)


given an entire function (ie. a complex function that is differential on the entire complex plane), $ f$ and $ Re{f(z)}\le c$ for all $ z$.  Show that $ f$ is a constant function.

$ \frac{f\minus{}c\minus{}1}{f\minus{}c\plus{}1}$ is bounded; it's no greater than $ 1$ in absolute value. Liouville.

The approach I gave leads easily to "little Picard"; the range of a nonconstant entire function is dense in the plane.

Is that really Little Picard?  I thought Picard required that the range only misses at most one value, as opposed to just being dense.

We have the theorem, but I'm not completely sure on the terminology.

That the range of a nonconstant entire function is dense in the plane, or that the range of an analytic function in any punctured neighborhood of an essential singularity is dense in the plane, would be called the Casorati-Weierstrass Theorem.


This problem looks good to me!  I shall post a solution if I figure that our. ;)

From Cauchy-Schwarz inequality,
\begin{align*}\sum\frac{(c+a-b)^4}{a(a+b-c)}&\ge \frac{\left(\sum (c+a-b)^2\right)^2}{\sum a^2}\\
&=\frac{(3(a^2+b^2+c^2)-2(ab+bc+ca))^2}{a^2+b^2+c^2}\end{align*}

Let $ p=a^2+b^2+c^2$ and $ q=ab+bc+ca$. Then it remains to show that
\begin{align*}&\frac{(3p-2q)^2}{p}\ge q\\
\iff &9p^2+4q^2\ge 13pq\\
\iff &(p-q)(9p-4q)\ge 0\end{align*}
which is true since $ p\ge q$.

A stronger inequality:

For $ a,b,c$ sides of a triangle show that
\[ \frac {(c \plus{} a \minus{} b)^{4}}{a(a \plus{} b \minus{} c)} \plus{} \frac {(a \plus{} b \minus{} c)^{4}}{b(b \plus{} c \minus{} a)} \plus{} \frac {(b \plus{} c \minus{} a)^{4}}{c(c \plus{} a \minus{} b)}\geq \frac 13(a \plus{} b \plus{} c)^2
\]

Even stronger:
\[ \frac {(c \plus{} a \minus{} b)^{4}}{a(a \plus{} b \minus{} c)} \plus{} \frac {(a \plus{} b \minus{} c)^{4}}{b(b \plus{} c \minus{} a)} \plus{} \frac {(b \plus{} c \minus{} a)^{4}}{c(c \plus{} a \minus{} b)}\geq a^2\plus{}b^2\plus{}c^2
\]

Enjoy!

[quote="barasawala"]Let $ a,b,c$ be sides of a triangle, show that

$ \frac {(c \plus{} a \minus{} b)^{4}}{a(a \plus{} b \minus{} c)} \plus{} \frac {(a \plus{} b \minus{} c)^{4}}{b(b \plus{} c \minus{} a)} \plus{} \frac {(b \plus{} c \minus{} a)^{4}}{c(c \plus{} a \minus{} b)}\geq ab \plus{} bc \plus{} ca$[/quote]

Two proof from batdangthuc.net project

[img]http://i246.photobucket.com/albums/gg93/NguyenDungTN/1.jpg[/img]
[img]http://i246.photobucket.com/albums/gg93/NguyenDungTN/2.jpg[/img]

Putting $x=b+c-a$ and $y=a+c-b$ and $z=a+b-c$. The inequality become :
$\sum \frac{x^{4}}{z(x+z)} \geq \frac {\sum x^2 +3\sum xy}{8}$
But by Hölder we get $3(2\sum z)(\sum x+z)(LHS) \geq (\sum x)^{4}$ and then $LHS \geq \frac{(\sum x)^{2}}{6} \geq RHS$ which is true since it's equivalent to :$\sum x^2 \geq \sum xy$

[quote="nayel"]
Even stronger:
\[ \frac {(c \plus{} a \minus{} b)^{4}}{a(a \plus{} b \minus{} c)} \plus{} \frac {(a \plus{} b \minus{} c)^{4}}{b(b \plus{} c \minus{} a)} \plus{} \frac {(b \plus{} c \minus{} a)^{4}}{c(c \plus{} a \minus{} b)}\geq a^2\plus{}b^2\plus{}c^2
\]

Enjoy![/quote]

Let $a=x+y, b=x+z, c=y+z$. Then it's:
\[4\sum\frac{x^4}{xy+yz}\ge \sum x^2+\sum xy\]
By CS:
\[\sum\frac{x^4}{xy+yz}\ge \frac{\sum x^4+2\sum x^2y^2}{2\sum xy}\]
So, we just need to show:
\[4\sum x^4+8\sum x^2y^2\ge (2\sum xy)(\sum x^2+\sum xy)\]
\[4\sum x^4+6\sum x^2y^2\ge 2\sum_{sym} x^3y+6\sum x^2yz\]
Which is true by Muirhead's or AM-GM since:
\[2\sum x^4\ge \sum_{sym}x^3y\]
\[\sum x^2y^2\ge \sum x^2yz\]
Done.

[quote=Svejk]It was also discussed here http://www.mathlinks.ro/Forum/viewtopic.php?t=135712[/quote]
Let $a,b,c$ be sides of a triangle, show that
$$\frac{(c+a-b)^{4}}{a(a+b-c)}+\frac{(a+b-c)^{4}}{b(b+c-a)}+\frac{(b+c-a)^{4}}{c(c+a-b)}\geq a^{2}+b^{2}+c^{2}$$

Using the Ravi Substitution, the inequality is transformed into proving for positive $x,y,z$,
\[\dfrac{8y^4}{yz+z^2}+\dfrac{8z^4}{zx+x^2}+\dfrac{8x^4}{xy+y^2}\geq\dfrac{8(x^2+y^2+z^2)^2}{x^2+y^2+z^2+xy+xz+yz}=\dfrac{16(x^2+y^2+z^2)^2}{(x+y)^2+(y+z)^2+(x+z)^2}\]
Thus we need to show that
\[4(x^2+y^2+z^2)\geq (x+y)^2+(y+z)^2+(x+z)^2\iff (x-y)^2+(y-z)^2+(x-z)^2\geq 0\]

By CBS,
\begin{align*}\sum\frac{(c+a-b)^4}{a(a+b-c)}&\ge \frac{\left(\sum (c+a-b)^2\right)^2}{\sum a^2}\\
&=\frac{(3(a^2+b^2+c^2)-2(ab+bc+ca))^2}{a^2+b^2+c^2}\end{align*}

Writing in terms of standard $pqr$ notation we get:
$$\frac{(3(a^2+b^2+c^2)-2(ab+bc+ca))^2}{a^2+b^2+c^2}=\frac{(3(p^2-2q)-2q)^2)}{p^2-2q}=\frac{(3p^2-4q)^2)}{p^2-2q}$$
So remaining to prove $$(3p^2-4q)^2 \geq q(p^2-2q)$$
Or after factoring $(p^2-3q)(9p^2-22q) \geq 0)$
Which is true as $p^2 \geq 3q$

[quote=arqady][quote="barasawala"]Let $a,b,c$ be sides of a triangle, show that
\[\frac{(c+a-b)^{4}}{a(a+b-c)}+\frac{(a+b-c)^{4}}{b(b+c-a)}+\frac{(b+c-a)^{4}}{c(c+a-b)}\geq ab+bc+ca.\][/quote]
The following inequality is also true.
Let $a$, $b$ and $c$ be sides-lengths of a triangle. Show that
\[\frac{(c+a-b)^{4}}{a(a+b-c)}+\frac{(a+b-c)^{4}}{b(b+c-a)}+\frac{(b+c-a)^{4}}{c(c+a-b)}\geq \sqrt[6]{243(a^{12}+b^{12}+c^{12})}\]
 :maybe:
Dear xzlbq. Am I right? Thank you![/quote]

Dear arqady,your inequality is true!

It is this:

\[{\frac {1}{243}}\, \left( {\frac { \left( b+c-a \right) ^{4}}{c
 \left( a+c-b \right) }}+{\frac { \left( a+c-b \right) ^{4}}{a \left(
a+b-c \right) }}+{\frac { \left( a+b-c \right) ^{4}}{b \left( b+c-a
 \right) }} \right) ^{6}\geq {a}^{12}+{b}^{12}+{c}^{12}\]

[quote=barasawala]Let $a,b,c$ be sides of a triangle, show that
\[\frac{(c+a-b)^{4}}{a(a+b-c)}+\frac{(a+b-c)^{4}}{b(b+c-a)}+\frac{(b+c-a)^{4}}{c(c+a-b)}\geq ab+bc+ca.\][/quote]

In triangle,prove that

\[{\frac { \left( b+c-a \right) ^{2}}{\sqrt {c \left( a+c-b \right) }}}+
{\frac { \left( a+c-b \right) ^{2}}{\sqrt {a \left( a+b-c \right) }}}+
{\frac { \left( a+b-c \right) ^{2}}{\sqrt {b \left( b+c-a \right) }}}\geq
a+b+c\]

Since the inequality is homogenous we may assume $a^2+b^2+c^2=1$ (1).  By CS we get $\frac{(3(a^2+b^2+c^2)-2(ab+bc+ca))^2}{a^2+b^2+c^2}$. By using (1) and $a^2+b^2+c^2\ge ab+bc+ac$ we obtain the desirable result.

[quote=xzlbq]

\[{\frac { \left( b+c-a \right) ^{2}}{\sqrt {c \left( a+c-b \right) }}}+
{\frac { \left( a+c-b \right) ^{2}}{\sqrt {a \left( a+b-c \right) }}}+
{\frac { \left( a+b-c \right) ^{2}}{\sqrt {b \left( b+c-a \right) }}}\geq
a+b+c\][/quote]
By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{(b+c-a)^2}{\sqrt{c(a+c-b)}}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}\sqrt{c(a+c-b)}}\geq\frac{2(a+b+c)^2}{\sum\limits_{cyc}(c+a+c-b)}=a+b+c.$$


Find the volume of Da Vinci's Famous Tank (the one you could see rolling across on screensavers) when the height is 8 feet and the distance from the front to the back is 24 feet. Express your answer to the nearest decimal.

Could you post a picture, or something?  I do not know what Da Vinci's Famous Tank is.

here's a picture

i don's get this problem though,
there should be more given info to solve it

Hmm, Ajai, what are those measurements for?

Are you breaking it up into cylinders and cones?  If so, what are the distances?

The picture is a pretty accurate figure for the problem, but I didn't include the "chimney" portion, and the top and bottom halves are equal.

So are you saying that it is a perfect cone, with height 8 and diameter 24?

[hide="if so"]The volume would be $\frac{8\times12^{2}\pi}3=384\pi$[/hide]

If you cut out the chimney, then it's not a "complete" cone... But I think that more informations are needed. Perhaps the angle the the generatrix forms with a ray of the base...  :wink:


Im freaking out.
Does anyone know if the amcs will be held at a local building?

If your school doesn't do it, you could take it at Bergen County Academies. I'm taking the 10B on February 25th at 8:00 AM there.


A is an linear operator in real 2-dimentional space R^2. prove that there exist linear operators B and C in R^2 such that A=B^2+C^2.

Let $\lambda_{1},\lambda_{2}$ be the eigenvalues of $A$, and WLOG assume $A$ is in Jordan normal form.  Pick $c$ big enough so that $\mu_{1}^{2}=\lambda_{1}+c^{2}$, $\mu_{2}^{2}=\lambda_{2}+c^{2}$ are both positive, and let $C=\left(\begin{array}{cc}0&-c\\c&0\end{array}\right)$.

If $A$ is diagonal, then $A = \operatorname{diag}(\lambda_{1},\lambda_{2}) = C^{2}+\operatorname{diag}(\mu_{1},\mu_{2})^{2}$.  Otherwise, $A=\left(\begin{array}{cc}\lambda&1\\0&\lambda\end{array}\right)=C^{2}+\left(\begin{array}{cc}\mu&\frac1{2\mu}\\0&\mu\end{array}\right)^{2}$.


What's the point of memorizing when the Mathcounts countdown isn't going to use the same questions?

After playing a lot of games (and i mean *a lot*), you can't help but mem at least something.  I realize it doesn't help, but it's natural.  If you mem like this, it's okay.  Get over it.  On the other hand, if you deliberately try to mem something, that's bad.

I agree there are way too many people trying to mem these days.

That ties in to those five second games.

Yes, 5 second games are insane. There is a forum on these insane time limits.

[quote="aleph0"]Some people see FTW as a place to gain rating instead of a place to practice for the Mathcounts. The people who memorize questions can put the answers in faster, so they get more rating.[/quote]
I agree. There was this one guy who I cd, the guy memmed and called me "you little brat" when I got a question right. When the guy won, he/she said to me, "shut up, you're wrong cuz I beat you." The guy make threats to report me for no reason. Also, the guy kept calling me a "noob" every time I lost in a cd. Hardly anyone believes me for some reasons which is strange.

[quote="ernie"]Dude, report him/her.[/quote]
None of my reports got a person "banned". Even though I gave good reasons, the person that I reported didn't even get banned which I found strange. It's probably cause the person is a moderator's friend. My friend got muted when he was arguing with a moderator's friend which was so not fair. I saw it happen.

Reports have nothing to do with moderators.

Perhaps your report(s) was/were not detailed enough...you should provide game number and a copy-paste of the chat transcript.

Sometimes its helpful, but the chances of that happening is like 1/1000000000000000

Haha, just kidding. But mostly, it's pretty pointless. I'm sure most people dont mem on purpose. It's pretty helpful sometimes.


I completely see how you got that, but I'm a little confused on one small little detail.

What if, for example, her pen ran out of ink on the first $3$ instead of the second $3$? Then your formula wouldn't be correct....

If that happened, then the answer would still be the same.  The new formula would be $n^{2}+n-n=n^{2}$ instead of $n^{2}+n$.  Plugging in our new equation, $n^{2}\leq3000\Rightarrow n\leq54$.

Yes! I just figured it out! :lol:

[hide="solution for the case described above"] You would just have to adjust the formula a little:

$1+1+2+2+3+3...=2*\frac{(n-1)n}{2}+n=n^{2}$

Then, finding the greatest integer that satisfies the inequality
$n^{2}<3000$, we find that the answer is $54$.

I'm a bit disappointed that the answer is still the same... I guess in this case she drew the second 54, otherwise Walk Around The River's method would yeild a 53...

I think my way works for any case because if she had only drawn the first of the two equivalent segments, that is accomodated by my formula. And if she had drawn the second of the two equivalent segments (taking this case as an example), $55^{2}>3000$ while $54^{2}<3000$. So it doesn't matter if the second 54 was drawn or not.

Generalization: If Dana's pen ran out after she drew x cm, the length of the longest segment she drew is

the floor of$\sqrt{x}$

What a nice solution! :lol: [/hide]

[quote="i_like_pie"]On a large piece of paper, Dana creates a "rectangular spiral" by drawing line segments of lengths, in cm, 1, 1, 2, 2, 3, 3, 4, 4, ... as shown. Dana's pen runs out of ink when the total of all the lengths he has drawn is 3000 cm. What is the length of the longest segment Dana has drawn?

[img]http://img103.imageshack.us/img103/8650/spiral3hs.png[/img][/quote]
[hide]
$\frac{3000}{2}=1500$
$1500>(x)(x+1)/2$
$3000>(x)(x+1)$
$\sqrt{3000}$ is between $50$ and $60$
54 is the lowest value for x that doesnt go over 3000
the longest segment is length 54
[/hide]

[quote="jhredsox"]could you do this w/ triangular numbers?

-jorian[/quote]

Um, yeah. I think that's what everyone did. Formula for triangular numbers is $\frac{n(n+1)}{2}$

[quote="star99"]i think when you have triangular numbers doubled its called oblong numbers...[/quote]

Really? I've never heard of that term before. :)


MATHCOUNTS
geometry
greatest common divisor

Okay, I have 4 previous mathcounts problems here.
Please hide your answers.
The first two are sprint or target ( I can't quite remember).

1.)  A three-digit number consisting of three different digits has its digits reversed, resulting in a smaller number.  The smaller number is subtracted from the original number.  Th answer is a number composed of the same three digits in a different order.   What's the original number?

2.)  Brian has 100 feet of fencing.  He will use the fencing to enclose a play area for his puppy.  What is the maximum number of square feet he can enclose?  Express your answer to the nearest whole number.  (For this problem, find the fastest way to solve it.)

The last three are Countdown problems.  For these, find the fastest way.

3.)  If x is positive and triangle ABC with vertices A(x,x), B(2x,x), and C(0,-3x) has an area of 98 square units, what is x?

4.)  What is the least positive integer that meets the following conditions?

      a)  When divided by 7, th remainder is 5.
      b)  When divided by 5, the remainder is 3.
      c)  When divided by 3, the remainder is 1.

1
[hide]When subtracting a three-digit numbers $XYZ$ and $ZYX$, $Z$ will always be smaller than $X$ because then otherwise $ZYX$ would not be less than $XYZ$. Then that would mean $Z$ must borrow from $Y$ making the last digit of the difference of the two numbers $(Z - X + 10)$. Since $Y$ lost 10, it must borrow from $X$ so the 2nd digit would be $(Y - 1 + 10 - Y) = 9$, and thus one of the digits, $XYZ$ must be 9. Since $X$ was borrowed from, the hundred's digit would then be $(X + Z - 1)$. Since we know that one of the digits is 9, it can't be in the one's place because $ZYX$ wouldn't result in a smaller number and it can't be in the ten's place because then we can't really solve for $Z$ and $X$. So we must make $X$ equal to 9. If this is the case, then the one's place is equal to $(Z + 1)$ and the hundred's place is equal to $(8 - Z)$. We know that the final difference will be some number $Z9Y$ or $Y9Z$, but we cannot have $Z$ in the one's place because the one's place would be $(Z - 9 + 10 = Z + 1)$, and thus, cannot be $Z$. So the number must be $Z9Y$. Setting the hundred's digit, $(8 - Z)$ equal to $Z$, we get 4, and since the last digit is $(Z + 1)$, it must be 5. Because the original number was $XYZ$, the original number was $954$.[/hide]

2
[hide]Shape with largest area that he can make is a circle. Given the circumference, 100, the radius is $\frac {50}{\pi}$ and therefore the area is $\frac {2500}{\pi}$ which is approximately $796$.[/hide].

3
[hide]It is easy to see that $(x, x)$ is $x$ units away from $(2x, x)$ because it's on the same horizontal line and therefore they would be $2x - x = x$ units apart. Since the height of the triangle does not really depend on what vertical line that it is on, we can ignore the 0 in $(0, -3x)$ and just subtract $-3x - x = -4x$ which means that the height is $4x$ units. The area of the triangle is then $2x^2 = 98$ and $x = 7$.[/hide]

4
[hide]$GCF ( 7, 5, 3 ) = 105$ minus 2 is $103$.[/hide]

1.  [hide]954[/hide]
2.  [hide]You divide 100 by 4 to find the numbers closest together which in this case is 25, then square it to get 625.[/hide]

mathgeek, on #2 the question does not say that the pen has to be rectangular.


andrew is late for school, and as he opens his front door, he notice that the school bus has stopped at the front of his yard,22.0 m away. he begins to chase the bus, which begins to accelerate away from him at 1.20 m/s/s. calculate the minimum speed that Andrew  will need to catch the bus, and also calculate the distance that andrew will need to run.

Let his speed be $v$ $m/s$.
Time by which he must catch up $=$ $\frac{v}{1.20}$ $s$
Distance he travels $=$ $v\times\frac{v}{1.20}$ $m$
Distance the bus travels $=$ $\frac12\times1.20\times(\frac{v}{1.20})^{2}$ $m$
If he catches the bus,
$v\times\frac{v}{1.20}$ $m$ $=$ $\frac12\times1.20\times(\frac{v}{1.20})^{2}$ $m$ $+$ $22.0$ $m$
or, $v^{2}$ $=$ $22.0\times2.40$
or, $v=7.66$
His speed $=7.66$ $m/s$
Distance travelled $=44.0$ $m$


vector
linear algebra
linear algebra unsolved

What is the hamel dimension of $ \mathbb{R}^N$ over the field $ \mathbb{R}$

I think it should be $ 2^N$,so what I want to show is there have a independent vectors with the cardinal $ 2^N$

Can you help me?

$ X^{\alpha}\equal{}\{x_n^{\alpha}\}_{n \geq 0}$ where $ x_n^{\alpha} \equal{} e^{\alpha n}$ ?


induction
algebra
polynomial
integration
real analysis
real analysis unsolved

Hi to everybody!

I am form East, and I don`t speak english very well ( :oops: sorry :( , but stil can read very good). I hope I`ll learn someday...

I have a problem that I didn`t solve:

(a) Prove that the space $ X = C^n[a,b]$ with norm
$ \|f\| = |f(a)| + |f'(a)| + |f''(a)| + |f'''(a)| +\cdots + |f^(^n^ - ^1^)(a)| + \sup\limits_{t \in[a,b]} |f^{(n)}(t)|$
is Banach`s.

(b) Examen the separability of X.

Thanks!

About (b) - I tried by induction over n.
$ C^{n - 1}_{[a,b]}\supset C^n_{[a,b]}$
$ f\in C^n_{[a,b]} \Rightarrow f '\in C^{n - 1}_{[a,b]}$
for arbitrary $ \varepsilon > 0$ we could aproximate f ' with a polynomial P (induction hypothesis):
$ \|f' - P\|_{C^{n - 1}_{[a,b]}} < \frac {\varepsilon}{b - a + 1}$ then $ Q(t) = f(a) + \int\limits_a^b P(z)dz$ so, it is
$ \|f - Q\|_{C^n_{[a,b]}} = |f(a) - f(a)| + |f'(a) - P(a)| + ... + |f^{(n - 1)}(a) - P^{(n - 2)}(a)| + \sup\limits_{[a,b]} |f(t) - Q(t)|$
there is to be done to prove that is
$ \|f - Q\| < \varepsilon$
???

help please ...

The last term in the definition of $ \|f\|_{C^n}$ should be the supremum of $ |f^{(n)}|$, not of $ |f|$.

[color=green][I edited it for him. - K.M.][/color]

Thanks mlok... :)  I was wondering if the problem was posted correct. Thanks to moderator too. :)

The (b) can be solved in non-edited metrics/norm, with a this hint:
$ f(t)\minus{}Q(t)\equal{}\int_a^t(f'(\xi)\minus{}P(\xi))\,d\xi$

In "edited" metrics it is easier to prove that X is separable.

But I am still wondering, is there a counterexample that in "old" metrics space X isn't Banachs, or otherwise.

Can it be proved or disaproved that is
$ |f_n(a)\minus{}f_m(a)|\plus{}\sup\limits_{t\in{[a,b]}}|f_n(t)\minus{}f_m(t)|\geqslant \sup\limits_{t\in{[a,b]}}|f'_n(t)\minus{}f'_m(t)|$ where $ f_n, f_m\in X$?

In the "old" norm the space $ C^n$ is not complete. Consider $ C^1[\minus{}1,1]$ with the norm $ |f(\minus{}1)|\plus{}|f'(\minus{}1)|\plus{}\sup_{x\in[\minus{}1,1]}|f(x)|$. The sequence $ f_n(x)\equal{}\sqrt{x^2\plus{}1/n}$ is Cauchy in the above norm, but does not have a limit in $ C^1[\minus{}1,1]$.


Without using calculus, evaluate $\sum_{n=1}^{\infty} \frac{1}{n^2}$. (I read that this can be done with trig substitutions.)

http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf

here's not one, but 14 proofs.

[quote="tetrahedr0n"]Do you allow Taylor series to be used? As long as you have some standard ones memorized, they don't qualify as calculus.[/quote]

Sure.

And thanks blahblahblah, that link was really cool. :D


A round number is an integr whose base 2 representation has at least as many zeroes as ones. How many round numbers are there less than or equal to 100?

You're very welcome...
Other people in these forums have much more beautiful solutions than I do.
And we need to dig up these unsolved problems in the archive...


Has anyone done the AMC 12 P on the AMC website under the sundry section?  If anyone has can anyone tell me whether those are harder/easier/on par to the actual AMC 12?  I know the practice test should be around the actual level, but those questions seem a little harder than other AMC 12 questions I've done.

In my opinion, they are harder, probably because they are for Taiwan kids  :P

i haven't taken the amc 12p (i'm studying for amc 10) but i've taken the amc 10p and i have to say it was easier in my opinion. my score on that test was higher than all my other practice and previous years tests.

[quote="tetrahedr0n"]It is harder, but not because of who it was administered to. It was certainly easier than the 2001 AMC 12 from the year before.[/quote]

That's weird, cuz I did much better on the 2001 AMC 12 than the AMC P. I hope the AMC P is no real indication of what u'll get(I barely scraped a 104 very, very close).

I also found AMC 12P very easy...I got 21 problems correct, 3 unanswered, and 1 wrong, giving me a score of 129.5, compared to my last year's 117.5 for AMC 12B.

Good luck to everyone taking AMC.  :)

For me, the test amc 12p is harder than the amc's from practices.  Then again, it was the first amc 12 practice test that i've ever taken, so i did very poorly, i thought i got five right, then gave up, it was sad.  I would say that the mock amc 12's, for the most part, are harder than the real amc 12s.  Some that i've seen as question 15 on a real amc 12 have shown up as a question 6 on the mock amc 12.  Something insane.  Its better to have harder practice than to have it easier though.  Good Luck tomorrow guys! I'm trying to stay relaxed right now, if i dont pass, it doesnt really matter, oh well, im going to a good school anyway.  If i do pass, then it would be nice, but i shouldnt become obsessive about the whole thing.    :)


induction
combinatorics proposed
combinatorics

Let $G$ be a graph and $\chi_{\ell}(G)>t$. Prove that for each $i$ there is a list $L(v_{i})$ of colors that $|L(v_{i})|=t$ and $|\cup_{i=1}^{n}L(v_{i})|\leq |V(G)|$, and $G$ can not be colored with lists $L(v_{i})$.

[quote="Omid Hatami"]Let $G$ be a graph and $\chi_{\ell}(G)>t$. Prove that for each $i$ there is a list $L(v_{i})$ of colors that $|L(v_{i})|=t$ and $|\cup_{i=1}^{n}L(v_{i})|\leq n$, and $G$ can not be colored with lists $L(v_{i})$.[/quote]

$n$ is the number of vertices of $G$??

Can anyone solve this???
Maybe some induction works?


geometry
geometric transformation
rotation
reflection

The integers from 1 to 9 can be arranged into a $ 3\times3$ array so that the sum of the numbers in every row, column, and diagonal is a multiple of 9. Find, with proof, the number of such possible arrays. (An array that is formed from rotations or reflections from another array is considered to be distinct from that array.)

[hide="Hint"]

Well, the sum of numbers in rows and columns are 9,18,18. 9 should be achieved in 2 ways, with one same number. $ a\plus{}b\plus{}c\equal{}a\plus{}d\plus{}e\equal{}9$. 3+4+2=3+5+1,2+1+6=2+3+4,1+3+5=1+2+6

Take first: 2 should with 2 more numbers sum 18, they must be 9 and 7. Columns (or rows) are 2+7+9,3+5+1,4+6+8 and rows(columns) aren't possible

Take second: 1 must have 8 and 9 in its row(column). Rows(columns): 1+8+9,2+3+4,6+5+7 Columns(rows): 1+2+6,9+5+4,7+8+3
Now count the ways

Take third: 2+9+7 must be. Columns(rows): 2+9+7,1+3+5,6+4+8. Columns(rows): 9+5+4,2+1+6,3+7+8 - same as second

Now you got the numbers which are together in rows and columns, and now it's just counting the ways to put them, such that diagonals also be divisible by 9 [/hide]