Each of the circles has a radius of 8 units. $ RP$ is a common tangent to both circles. $ QA\equal{}\dfrac{1}{6} QP$. How many units are in the perimeter of $ \triangle POQ$?

[img]http://img98.imageshack.us/img98/3246/1997statesprint9nh0.gif[/img]

[hide="Answer... with Part of Solution"]I'm assuming that Q and O are the centers...

Since $ PR$ is tangent to both circle, $ OP\equal{}PQ$
$ 48\plus{}48\plus{}16\equal{}112$[/hide]

I got 112, but the answer book says 48?

[hide][img]http://img168.imageshack.us/img168/2294/1997stateanswersfy8.gif[/img]

NO THAT WAS NOT PHOTOSHOPPED or altered[/hide]

it lies to you... States on Saturday. oh no.  T.T I should start practicing

[quote="mathcrazed"]it lies to you[/quote]

ok good since when i was checking over iw as like "how?"

[hide]

ok so since the radius is 8, two times the radius is the smaller side, so 16 is one side. 8*6=48 so 16+48+48=112

[/hide]

/me agrees with mathcrazed.


[quote="shobber"]Oley!
The Hope Math Contest is coming! In Beijing, it will take place on Sunday morning and i am going to take it.
ooo this is my first math contest! Wish me good luck!

Besides, any one here going to take that?? What grade do you get??

I shall post the problems and togather we shall have a look.[/quote]

good luck!

希望杯是一个全国性的竞赛吧?

哦？中文系统没问题了么?

哦？中文系统没问题了么?

以前的帖子還是有問題..

再观察一段时间吧,保不齐又出什么问题...

对于此次希望杯,我只想对主办方说:"你们不给我发奖也就算了,好歹给我留份试卷作纪念.可怎么连卷子都不给我呢!?无法理解!"

我曾参加过希望杯数学竞赛 (初一年级的时候..... --_-- )
我进入了复赛,但未能如愿进决赛,最终得了一枚铜制奖牌.
不久后我就移民到了加拿大.
.......

[quote="11th dimension"]我曾参加过希望杯数学竞赛 (初一年纪的时候..... --_-- )
我进入了复赛,但未能如愿进决赛,最终得了一枚铜制奖牌.
不久后我就移民到了加拿大.
.......[/quote]

你不错啦，还能进复赛！我参加过物理，化学的２次竞赛，结果颗粒无收．．．．．．最近北京有４项竞赛，我只希望能在其中的一项中拿到一个最次的奖，这样我也就知足了．．．


geometry
cyclic quadrilateral
geometry proposed

Let a cyclic quadrilateral ABCD, AC meets BD at E. Prove that the sum of two sides of the quadrilateral ABCD equals to the sum of the sum of other two if $ \angle BAD\equal{}60^o$ and $ AE\equal{}3CE$.

Hello!

$ AE\equal{}3\cdot CE\Rightarrow AB\cdot sin\angle ABE\equal{}3\cdot BC\cdot sin\angle CBE\Rightarrow$
$ \Rightarrow\frac{AB}{BC}\equal{}3\cdot\frac{sin\angle CBE}{sin\angle ABE}\Rightarrow \frac{AB}{BC}\equal{}3\cdot\frac{sin\angle CAD}{sin\angle ACD}\Rightarrow$

$ \Rightarrow\frac{AB}{BC}\equal{}3\cdot\frac{CD}{AD}\Rightarrow AB\cdot AD\equal{}3\cdot BC\cdot CD$          $ \left(1\right)$

$ BD^{2}\equal{}AB^{2}\plus{}AD^{2}\minus{}2\cdot AB\cdot AD\cdot cos60^{\circ}\equal{}AB^{2}\plus{}AD^{2}\minus{}AB\cdot AD\equal{}$
$ \equal{}\left(AB\minus{}AD\right)^{2}\plus{}AB\cdot AD$     $ \left(2\right)$
$ BD^{2}\equal{}BC^{2}\plus{}CD^{2}\minus{}2\cdot BC\cdot CD\cdot cos120^{\circ}\equal{}BC^{2}\plus{}CD^{2}\plus{}BC\cdot CD\equal{}$
$ \equal{}\left(BC\minus{}CD\right)^{2}\plus{}3\cdot BC\cdot CD$     $ \left(3\right)$
$ \left(1\right),\left(2\right),\left(3\right)\Rightarrow \left(AB\minus{}AD\right)^{2}\equal{}\left(BC\minus{}CD\right)^{2}\Rightarrow \left|AB\minus{}AD\right|\equal{}\left|BC\minus{}CD\right|$
If $ AB\minus{}AD\equal{}BC\minus{}CD$ then $ AB\plus{}CD\equal{}BC\plus{}AD$
If $ AB\minus{}AD\equal{}CD\minus{}BC$ then $ AB\plus{}BC\equal{}CD\plus{}AD$ :)


A mouse starts at point A and travels this maze according to the arrows. At point A and point B,C,or D, the mouse randomly chooses one of the outbound paths.  It exits the maze at one of the points E,F,G, or H. What is the probability that the mouse exits the maze at point F? Express your answer as a common fraction.

Billy

[hide]
There is a 1/3 chance that he will go to each of B,C, or D.
From B, there is a 1/2 chance that he will go to F. So, 1/2*1/3=1/6
From C, there is a 1 chance that he will go to F. So, 1/3*1=1/3
From D, there is a 1/3 chance that he will go to F. So, 1/3*1/3=1/9
1/3+1/6+1/9=6/18+3/18+2/18=11/18
11/18
[/hide]

Dashern's way is the exact same way I did it; since you have to consider that the probability of each outcome is DIFFERENT... the mistake commonly made on problems like this is just counting the number of pathways... see the 2004 State Sprint (I think It was number 29 or thirty...)

Yes, it was number 30 on the 2004 State sprint round, it was something about ponds and probability.

That was a simple problem that many got wrong b/c they didn't factor in the weightage of different ponds..


Let $p$ be a prime number. Show that the integer

$\binom{2p}{p}-2$ is a mutiple of $p$.

Bother ... too early in the morning.  (Squaring things.  Feh.)


I am from a IMO week country, and I already participated twice at IMO getting a HM and a bronce... in which, I should probably get a silver this time and it would be the first silver ever for the country...
However, I already finished the highschool and the university here starts at May, so going to IMO will imply loss of a year of at least a semester.... and mayority of my HS class goes to the same University and all them starts in May so I will missed out to be at the same class with all old friends....
If you are in my position, will you skip a year for IMO trying to make something historical for your country(taking the risk of failing to get silver and lose a year for nothing)??

Can you elaborate on what you mean by losing a semester? IMO is just a two week event if I remember it right :? How being absent for two weeks or even a month can disrupt the entire semester? It'll be not completely trivial to catch up, of course, but I believe that all professors will do their best to help you (after all, it is their country too :P).

well I was told that if I am formally registrated as university student then I cannot go to imo... so I need to wait until IMO ends to register to university in case that i go....

That's, indeed, a problem. I wonder if you university will consider formally admitting you starting the second semester under such circumstances and allow you to attend courses for the first semester without formal registration, etc. I mean it may be hard to convince the IMO organizers to make exceptions but within your country you should probably expect that every possible effort will be made to accomodate your case.

So, I think you should first thoroughly explore all possibilities for doing both things you wish to do. If nothing can be done, and you absolutely have to choose, I would choose going to university with your friends. After all, it is not wartime and nobody can demand your sacrificing your personal happiness for the glory of the country. Of course, it is just my personal opinion :).

thank you dude! I will try to find out if the university could make an exception....

my initial plan was go to a us college then it could solve the problem as they start at september... but I am rejected... so the plan failed hehe....

I also want to know if there exists someone who skipped a college year for an extra IMO participation or not, and do that worth it?

hey not to sound rude but we people here in india are fighting to even go to the national level olympiad and its a dream of many to go to the imo here actually and you want to say that you are actually trying to omit going to the imo!!! All i can say is Wth? It'll be a honour if anyone gets to go from here  :D

But if that univ of yours is the only univ u wanna go, i see a snag there  :D

The fact that some people are in different circumstances than you is not a good reason to omit capitalization and punctuation :P

More generally, please don't use chat-speak on this forum.

Sorry! But i was in such a shocked state that i would have strangled him to death if he had come near me.  :D   :rotfl:   :rotfl:

Well, in USA it's also very hard to go to IMO, but I wouldn't have stayed an extra year out of college in order to go. But this is because I would have had to miss a whole year; if you could enter in the spring and even audit classes during the fall, then it doesn't cost you too much. So, I agree with the previous people who said you should find out what your university will allow.

Maybe my opinion would be different but nevertheless most univs would agree bcos it is a honour to go to the IMO really  :D

[quote="probability1.01"]If you must choose between the two, I would say attend the university. To some extent, going to IMO multiple times yields diminishing returns. Going just for the shot at a good result doesn't seem worth the disruption, although I don't know what other reasons you might have.[/quote]

isn't that slightly hypocritical coming from you.  :wink:

I have other reasons, like going to MOP and not having that much else to do. Well, there are other good ways of spending the summer, but I would be seeing [i]fewer[/i] friends that way.

mm... I'm considering not going to MOP even if I qualify, unless I make black or something incredibly unlikely like that

because although it's a great math experience, I honestly don't want to spend three weeks with a lot of (male) blue MOPpers :\ . I mean sure, last year it was pretty cool, but I was in red MOP, and math skill * sanity = constant.

[quote="Lazarus"]but I was in red MOP, and math skill * sanity = constant.[/quote]All right now it's personal. When I was at MOP back in '01 are you saying that I was either dumb or crazy? :)

Anyway back on topic, why don't you consider going to a US university if you can afford it?

[quote="Lazarus"]mm... I'm considering not going to MOP even if I qualify, unless I make black or something incredibly unlikely like that
[/quote]

...C'mon, really?  I mean, because you were in red, we didn't meet much from a mathematical perspective at MOP, but...you didn't enjoy it?  Seriously, if you don't enjoy MOP, I'm sure there are people who would be grateful to have the space freed up...

As for the math skill * sanity = constant thing, I think what you mean is
amount of math done each day * sanity = constant
This is probably true (or rather that the two inversely correlate), but that constant on the right is different for each person.  For example, I knew some people there who were seriously being pushed to the edge with the math we were doing, but I, for example, was doing more math in the time between and after classes, and was really having a blast (a part of that math was preparing ELMO problems...).  If you take, say, Jacob...he was also doing math nonstop.  Although you may not understand, some people actually really love to think about math.

thanks to everyone!

after talking to the local university, they said that I may first be a unofficial student(no homework and no credit) at first quarter and then do the make-up examination to recover the credit when back of IMO. then I don't lose nothing but get harder work when back. it is a good solution after all...

valentin: I missed the december sat registration date and that basically limited me out of majority good us universities...


integration
function
calculus
algebra
system of equations

As it says, find the shortest approach:

A capacitor of capacitance C along with a series resistor R is connected in parallel, to 2 sources: one with emf E and internal resistance R/2 (operated by switch S1), and the other with emf 2E and internal resistance R/2 (operated by switch S2). Note that the positive terminal Of E and the negative terminal of 2E and a plate of the capacitor are shorted together.
S1 is closed and S2 is kept open for a long time. Then at time t=0, both switches S1 and S2 are closed.
Find
(1) Heat dissipated across the resistor R from t=0 to t=infinity
(2) Energy supplied by each battery.

It doesn't look like S1 is doing anything, could you draw a quick diagram to clarify?

[hide="Hint"]
[b]serialk11r[/b] is correct. Switch S1 is superfluous. It starts closed at $ t \equal{} \minus{}\infty$, and stays closed until $ t \equal{} \plus{}\infty$. It is for all practical purposes a connecting wire, i.e. a "short circuit."

For those who know how, I suggest performing a mesh analysis. The result will give you the current through the resistor $ R$ immediately. You may then square that, multiply by $ R$, and integrate over time to find the heat dissipated by $ R$.

For the second part of the problem, the energy supplied by each battery, find the current through the given battery, and calculate:

$ \text{Energy} \equal{}\int_{0}^{\infty} V\cdot I\,\mathrm{d}t$.[/hide]

Hmmm...my idea was to find the current through the resistor, and since $ W \equal{} RI^2$ we can get the power dissipated by the resistor. Since we assume t=infinity we can just say $ E \equal{} \frac12CV^2$ for the capacitor.
We can pretty much assume any exponential function for I, in the form of $ I(e^\minus{}x)$
Since I suck at calculus (haven't taken it) I'm not going to bother trying to figure out the integrals.
EDIT: I made a diagram, at least I think it looks correct..

[quote="serialk11r"]Since I suck at calculus (haven't taken it) I'm not going to bother trying to figure out the integrals.
EDIT: I made a diagram, at least I think it looks correct..[/quote]

Your diagram looks close enough. (I think your battery voltages are off by a factor of 2, but that is not that critical a thing. You'll get an answer that has the correct form either way.)

It would help to know the calculus but you don't have to.

I drew the circuit, and wrote down the system of equations as a system of Laplace transforms (you can do this by inspection if you know how). Solved said system for the Laplace transforms of the loop currents. I then found the inverse transforms of these (if you're not comfortable with the calculus, you can look these up in a table). The inverse Laplace transforms give you the loop currents as functions of time, and the rest is easy -- as you pointed out, the loop currents have solutions of the form: $ i(t) \equal{} e^{\minus{}\alpha t}$.

Another hint, $ \int e^{\minus{}\alpha t}\, \mathrm{d}t \equal{} \minus{}\frac{1}{\alpha}e^{\minus{}\alpha t} \plus{} K$, where $ K$ is a constant whose value is determined from the initial conditions.

[quote="Immanuel Bonfils"]When the current in the capacitor branch becomes null (after an theorically infinite time, the currents in the batteries doesn't, so the answer for (2) should be infinity (no need of calculations).[/quote]
How? Internal resistance is like an extra resistor added to the circuit, so once the capacitor is at maximum voltage no current flows period.

In the final steady state, after switches S1 and S2 being closed.for an theoretical infinite time,  battery 2 $ (2E,S_2)$ is supplying an electric power $ {{3E^2\over{2r}}}$ to battery 1 $ (E,S_1)$ , so the answer to the second question of the problem," (2) Energy supplied by each battery " , is infinity , positive for battery 2 and negative for 1.

For the net energy supplied from both, we'll have $ {{21}\over{40}} E^2 C \;$.


This is probably the wrong forum, but the following problem is being a pain in my $ 455$ for a longer time now:

I want to download and install [b]all[/b] cygwin packages (at least, their most recent versions). Is there a better way to do this than to click on every single package in setup.exe? (What most often doesn't even work because setup.exe crashes before one is done.) Or, even better, is there a way to select a lot of packages without having to mark them one by one? The normal installation has only base and dependencies, and this is absolutely not what I need.

Thanks!
  darij

I have a related question - what do people use cygwin for?

Using some advantages of linux under windows, such as bash scripting, Singular, Macaulay, etc...

Okay, the question is probably why people are using windows. And it's a good question. Does anybody know whether Scientific WorkPlace runs under WINE? Because it, along with a few CRPGs, is my main obstruction against installing linux.

  darij

[quote="solafidefarms"]Is your computer moderately new? You could virtualize a copy of Windows, install whatever on it, and know that it will work.[/quote]

Notebook from 2004. No, I fear I'm not going to VM it... But probably my next one will have linux with wine only.

[b]EDIT:[/b] Issue resolved.
http://www.cygwin.com/ml/cygwin/2003-07/msg01117.html
Sorry for being an idiot...

  darij

Ooh, that looks interesting, I might actually want to bookmark this thread... or rather that webpage :)... thanks for bringing it up.


A set contains 10 distinct positive integers, all of them smaller than 118. Prove that there exists at least two subsets A and B such that A union B is empty set, and the sum of elements in A is same as that in B.

Finally I know where I made the mistake: :blush:
I wrongly add up 1+2+...+10 and 108+109+...+117. Actually it should be 1+2+3+...+9 and 109+110+...+117.


on each of the faces of a cube a real number is written. For each of the faces, the number written on it equals the product of the numbers written on all the faces that have a common side with this face. What is the sum of all the numbers that are written on the faces of the cube? Find all possible answers.

[hide="Hint"]
So, a face (let's call it face 1) will be a multiple of an adjacent face (call it face 2). Face 2 will also be a multiple of face 1. [/hide]

How about starting with the obvious ones: all faces being 1, or all faces being 0.

[quote="AIME15"]
[hide="Hint"][hide="Hint"][hide="Hint"]

[/hide][/hide][/hide][/quote]

By the way, can you please not spam like that? If you don't have anything useful to say, please don't say it...

I got the 1 and 0 cases. However, the headache part is the real number parts. I feel that the numbers are less than 1 which look like a decimal. Another clue I found was that every opposite faces have the product of the numbers. For example, the following is a net of the cube and A-F are numbers on each face.

   E
A B C D    As you can see, A = ExBxFxD and C, the opposite face of A, = ExBxFxD.
   F

You can repeat the same for B, D, E and F. From this onwards, I become puzzled.

And you can also notice that since A and D have the same four faces adjacent to them, they must be equal. The same applies for B, E and C, F.

As a result, we reduced the number of variables from 6 to 3. Now we can express $ a, b, c$ as follows:

\begin{align*} a & = b^2c^2 \\
b & = a^2c^2 \\
c & = a^2b^2 \end{align*}

Now look at what happens when we substitute (trying to put everything in terms of $ a$):

\begin{align*} a & = b^2c^2 \\
a & = (a^2c^2)^2(a^2b^2)^2 = a^8b^4c^4 \\
a & = a^8a^2 = a^{10} \end{align*}

Now that we find that $ a = a^{10}$, the only possible values of $ a$ (limited to real numbers) are $ 0$ and $ 1$. Then it follows that the possible triples $ (a, b, c)$ are $ (0, 0, 0)$ and $ (1, 1, 1)$.

Therefore, the possible answers are $ \boxed 0$ and $ \boxed 6$.


If you select 3 cards from a pack of 52 playing cards, what is the probability you will select a king queen and jack of the same suit?

[hide]Chance you get a royal on the first go = 3/13. Then, chance you get a suitable card on second go = 2/51. and on third = 1/50.
Therefore net probability = 1/5525[/hide]


New to the forums, so not sure if number theory problems even belong here. If anyone has time, I would appreciate your help.

p is an odd prime. a is the smallest positive integer with the LeGendre symbol (a/p) = -1. Prove a is prime.

There's a "Number Theory" section, so this belongs there.

Anyway, assume $a=bc,\ b,c<a$. We have $-1=\left(\frac ap\right)=\left(\frac{bc}p\right)=\left(\frac bp\right)\left(\frac cp\right)$, which means that at least one of $b,c$ has Legendre symbol equal to $-1$. We thus have a contradiction.


function
arithmetic sequence
BritishMathematicalOlympiad

Let $f: \mathbb{N}\mapsto\mathbb{N}$ be a permutation of the set $\mathbb{N}$ of all positive integers.
[list][*] Show that there is an arithmetic progression of positive integers $a, a+d, a+2d$ where $d > 0$, such that \[f(a) < f(a+d) < f(a+2d).\] [*] Must there be an arithmetic progression $a, a+d, \cdots, a+2003d$, where $d > 0$, such that \[f(a) < f(a+d) < \cdots < f(a+2003d)?\] [/list]
(Note: A permutation of $\mathbb{N}$ is a one-to-one function whose image is the whole of $\mathbb{N}$; that is, a function from $\mathbb{N}$ to $\mathbb{N}$ such that for all $m \in\mathbb{N}$ there exists a unique $n \in\mathbb{N}$ such that $f(n) = m$.)

[hide="First Part"]

Pick $a$ so that $f(a)=1$.  Now find the smallest $k$ so that $f(j)\neq k$ for all $j<a$.  Set $d$ so that $f(a+d)=k$.  $f(a+2d)$ must be $>k$, since $k$ is the smallest number such that $f^{-1}(k)>a$.  So $f(a)<f(a+d)<f(a+2d)$.    [/hide]


algebra
difference of squares
special factorizations

[u]How[/u] do you do this problem? (not just the answer)

3^(2x) + 19 = 10^x

Thanks

$ 9^x \plus{} 19 \equal{} 10^x$

$ x$ must be even so the last digit adds to the last digit of $ 19$.

If you plug in $ 2$, you see that it works.

$ 9^2 \plus{} 19 \equal{} 10^2$
$ 81\plus{}19 \equal{} 100$

So, $ \boxed{x\equal{}2}$.

Or you could notice that $ 10^x\minus{}9^x\equal{}19$, and since $ 10\plus{}9\equal{}19$ (because of difference of squares), $ x\equal{}2$.

What do you mean by how do you do 3^(2x)?

Edit: Oh I see.  :blush:

$ 3^{2x} \equal{} (3^2)^x \equal{} 9^x$

It's basic algebra. In general, there are a couple of rules about exponents:

i) $ (a^m)^n \equal{} (a^n)^m \equal{} a^mn$

ii) $ a^m \cdot a^n \equal{} a^{m\plus{}n}$

iii) $ \frac{a^m}{a^n} \equal{} a^{m\minus{}n}$

iv) If $ a^m \equal{} a^n$, then $ m\equal{}n$. (some exceptions are when $ a\equal{}\minus{}1,0,1$)


calculus
integration
limit
derivative
real analysis
real analysis unsolved

Let $F(x)=1-x^{2}$ and $|x|<1$ and $F(x)=0$ elswhere. Moreover let $g$ be continious on $[-1,1]$.
Determine this limit
\[\lim_{n\to\infty}\frac{3}{4}\int_{-1}^{1}nF(nx)g(x)dx \]
Explain carefully please (the answer is $g(0)$)

Similar problem
\[\lim_{n\to\infty}\int_{-1}^{1}n^{2}G'(nx)f(x)dx \]

But here $G(x)=\frac{15}{16}(x^{2}-1)^{2}$ and $|x|<1$ and $G(x)=0$ elswhere. Also $f$ has continious [b]derivatives[/b].

Explain carefully please (the answer is $-f'(0)$)

[quote="mlok"]For $\epsilon>0$ there is $\delta$ such that $|g(x)-g(0)|<\epsilon$ when $|x|<\delta$. Choose $n$ so large that $F(x)=0$ for $|x|\ge\delta$. Estimate the integral from below and from above using $g(0)-\epsilon<g(x)<g(0)+\epsilon$.

The second problem reduces to first after integration by parts.[/quote]

Are there any theorems one should use?
Could you show some more details, please?

[quote="mlok"]For $\epsilon>0$ there is $\delta$ such that $|g(x)-g(0)|<\epsilon$ when $|x|<\delta$. Choose $n$ so large that $F(x)=0$ for $|x|\ge\delta$. Estimate the integral from below and from above using $g(0)-\epsilon<g(x)<g(0)+\epsilon$.

The second problem reduces to first after integration by parts.[/quote]

Maybe you made a typo but F(x) does not depend on n :?:


Just upto 3 places.I do not see any practical use.If I forget, I calculate it as required.

This belongs in the games forum, and there may be related topics already.

I just memorize the first few, like 3.1415926. I see no reason for memorizing any greater amounts of numbers.

Wow, I can't believe so many people have so many digits of pi memorized. I'm just a 3.14159 kind of guy.

Lulz, if you guys are going to copy paste (to winner2009 and professordad) you should at least get rid of the spaces from the website.

As for me...

3.14159265358979323

Not too much, but definitely enough for any practical purpose :P

enough to know when its pi day. about 3.14159

(breathes in slowly) ... 3.14159265358979323846264

I read Secrets of Mental Math (Arthur Benjamin) which is a very good book.

I used Phonetic Code:

Based on sounds:
1 - t or d   6 - j, ch, or sh
2 - n         7 - k or hard g
3 - m        8 - f or v
4 - r         9 - p or b
5 - l         0 - z or s

Translate this:
"My turtle will, my love, pick up my new mover, Ginger."

3.1415926535897

I memorized using the pi song.

My feeling is that this poll has too few options.

I have over 50 digits memorized. one day i had nothing to do so i just opened up the computer and started memorizing pi digits.

I have about 300 digits memorized, every year our teacher has a Pi day contest to see who knows the most digits of pi. However I like never win, because someone else in our class knows like 500 digits.  :!:

I used to know a lot, then I got older and don't know as much  :(

[quote="lightbluemathangel"]i have 90 memorized[/quote]

Oh my gosh... I dare you to post all of them... You don't have too!

wow winner009

u just copied from like wiki right?

I've memorized 422 digits (including 3) so far... Why did I do that? There was a pi competition at my school... :D I memorized it in Mandarin Chinese though...

uh
I only know like 3.141592653589
...enough for now

3.14159265335897932384626433832795028
that's what I have. I memorize a lot more for pi day cause we have a competition at school.

I know just under 100. Not sure why I memorized it; probably just bragging rights. And then, the bragging rights don't have much of a good base, so...

i've memorized 1729 digits(exactly) of pi. i could type it all up if you wanted me to.

[size=84][color=red]Moderator note: This thread is ridiculously spammed beyond all sanity. Pasting digits over from other websites and demonstrating pseudo-pwnage-memorization is seriously not funny, is an inconvenience to others, and clogs up space. I deleted the posts of the four users who have considered the act as a sport, as it is a revival after all. If you wish the thread to continue (of which I see no reason for it to, but I'm sure there's all sorts of reasons out there), make a revival thread. (Chances are that that one ends up like this.) [/color][/size]


geometry
circumcircle
arithmetic sequence
geometry unsolved

Ascertain all triangles (others than equilateral !) for which $ s^2\equal{}9r(2R\minus{}r)$ ,

where $ s$- semiperimeter, $ R$ - circumradius and $ r$ - inradius.

The sides of a triangle are in arithmetic progression if and only if
\[ (b + c - 2a)(a + c - 2b)(a + b - 2c) = 0.\]
This is equivalent to
\begin{align*}
(2s - 3a)(2s - 3b)(2s - 3c) &= 0 \\
\Leftrightarrow \quad 8s^3 - 12(a + b + c)s^2 + 18(ab + ac + bc)s - 27abc &= 0.
\end{align*}

But $ a + b + c = 2s$, $ ab + ac + bc = s^2 + r^2 + 4Rr$, and $ abc = 4KR = 4Rrs$, so the equation above becomes
\begin{align*}
8s^3 - 24s^3 + 18(s^2 + r^2 + 4Rr)s - 108Rrs &= 0 \\
\Leftrightarrow \quad 2s^3 + 18r^2 s - 36Rrs &= 0 \\
\Leftrightarrow \quad 2s(s^2 + 9r^2 - 18Rr) &= 0,
\end{align*}
which is equivalent to $ s^2 = 18Rr - 9r^2 = 9r(2R - r)$.

[i]Crux Mathematicorum, 213[/i]: Prove that the sides of a triangle are in arithmetic progression if and only if
\[ s^2 = 18Rr - 9r^2.\]

The answer are all triangles with one side as arithmetic mean of the others two


invariant
induction
combinatorics proposed
combinatorics

In the world the set of towns $V$ is represented by a graph $G = (V,E)$. The edge set $E$ contains the edge $\{x,y\}$ iff two capitals $x,y \in V$ are connected by a street. You drive on the street in both directions. Due to financial matters there shall be at least streets as possible such there is no ring distributor in the graph. Candidates for being a capital are those towns which are centrally located and whose biggest distance to any other town is possibly small. What is the maximum number of tows that might be considered as capital if we define the distance $d_g(x,y)$ as the distance of two towns $x$ and $y$ in the graph $G$.

I think a tree has either one or two centers. I also think that this follows from the fact that any two centers must be connected, since if they're connected by a chain longer than $1$, then the vertices situated on the same chain between the two centers are, in fact, "better qualified" to be centers :). On the other hand, if each two centers are connected, then we can have at most two, since three would already form a cycle (and a tree has no cycles).

What is a 'ring distributor'?
What does 'centrally located' mean?  :(

Pierre.

[quote="pbornsztein"]What is a 'ring distributor'?
Pierre.[/quote]

Maybe you call it internal circle. Starting at one vertex you cannot reach the same vertex anymore if you just go into one egde direction, e.g. you always go from vertex A to B and not vice verca. Is it OK ?

[quote="pbornsztein"]
What does 'centrally located' mean?  :(

Pierre.[/quote]

It is just informally supporting the statement that the distance of the capital to any other town is possibly small.

[quote]
internal circle[/quote]

Do you mean a cycle?

Pierre.

[quote="pbornsztein"]
Do you mean a cycle?

Pierre.[/quote]

Yes, that is right. I forgot the proper word.  ;)

Ok, now I understand the statement of the problem. Forgetting about towns or whatever, the question is :
'What is the center of a tree'?
(You may note that I could have read more carefully the title, ...  :blush: ).

For those who do not know :
The center of a graph $G$ is defined as the subgraph induced by the vertices of minimum eccentricity.
And the eccentricity of a vertex $v$ is $\max d(v,M)$ where $M$ varies over all the vertex set of $G$, and $d$ denotes the distance along the graph.

And a famous theorem (Jordan, Sylvester, according to Orl) states, as Grobber noticed, that :
the center of a tree is a vertex or an edge.

Pierre.

[quote="pbornsztein"]Ok, now I understand the statement of the problem. Forgetting about towns or whatever, the question is :
'What is the center of a tree'?

Pierre.[/quote]

You are right, Pierre. As the graph has not any cycles, the graph is a tree. Thus it suffices to prove:

In each tree the centre $Z$ consists from one vertex or two vertices joined by an edge. The proof should be a nice example for the surprising application of an invariant during the process, e.g. in finding the centre of a tree.

[quote="pbornsztein"]

And a famous theorem (Jordan, Sylvester, according to Orl) states, as Grobber noticed, that :
the center of a tree is a vertex or an edge.

Pierre.[/quote]

The proof of this theorem is a remarkable milestone in the development of the graph theory. During the mid 1870s C. Jordan and later A. Caley used the centre for the solution of certain counting problems regarding trees. Before 1874 J.J. Sylvester worked with A. Caley on several fields, especially the invariant theories.

[quote="pbornsztein"]Forgetting about towns or whatever, the question is :
Pierre.[/quote]

Introducing the issue of towns or something similar is very important when establishing a math olympiad problem...  ;)

Is wat I said wrong? Every two centers must be connected. Assume otherwise. There is a chain connecting the centers (because a tree is connected). Take any vertex strictly between the two centers. It's almost clear that any such vertex is, in fact, a better candidate for a center: for any path from this vertex (call it $x$) to another vertex $y$, there is a path from one of the two centers to $y$ which is longer that the path $xy$.

On the other hand, if each to centers are connected, there can be at most two because of the lack of cycles in a tree.

No, no.
Is the same thing that the theorem I stated above.
But, in 'my' version what is called a center is a graph. The theorem states that in terms of vertices, what you are calling centers is either a single vertex or two vertices joined by an edge.
The classical proof is to use induction on the number of vertices in the graph, and to delete leaves during the induction hypothesis (which is quite the same idea that you used).

Pierre.


hello, can you prove that:

$ \sin 54^\circ \plus{} \cos 108^\circ \equal{} \frac{1}{2}$

$ \sin 54^\circ \equal{} \frac{1\plus{}\sqrt{5}}{4}$
$ \cos 108^\circ \equal{} \minus{} \sin 18^\circ\equal{} \frac{1\minus{}\sqrt{5}}{4}$

Thus, $ \sin 54^\circ \plus{} \cos 108^\circ \equal{} \frac{1}{2}$

$ sin 54 \plus{} cos108 \equal{} sin54 \minus{} sin18 \equal{} 2cos36sin18 \equal{} \frac{2sin18cos18cos36}{cos18}$
$ \equal{} \frac{sin36cos36}{cos18} \equal{} \frac{sin72}{2cos18} \equal{} \frac{sin72}{2sin72} \equal{} \frac{1}{2}$

how is $ \sin 54^\circ \equal{}\frac{1\plus{}\sqrt{5}}{4}$ come ?


how do you find the square root of a square root
like
\[ \sqrt{4\sqrt{2}}\]

This is essentially the same as our exponent property that says $ (ab)^c\equal{}a^{c}b^{c}$
$ \sqrt{4\sqrt{2}}\equal{}(2^{\dfrac{1}{2}}4)^{\dfrac{1}{2}}$
$ 2^{\dfrac{1}{4}}4^{\dfrac{1}{2}}\equal{}2*2^{\dfrac{1}{4}}$

[quote="DaReaper"]how do you find the square root of a square root
like
\[ \sqrt {4\sqrt {2}}
\]
[/quote]

I don't know if this is the best way, but...

$ \sqrt{4\sqrt{2}} \equal{} \sqrt{32} \equal{} 4\sqrt{2}$

Interesting...

You squared the $ 4\sqrt2$ and then took the square root of it. :wink: The only real solutions to $ \sqrt x \equal{} x$ are $ x \equal{} \{0,1\}$ (obviously) :P .

The best we can do is $ \sqrt {4\sqrt2} \equal{} 2\sqrt {\sqrt2} \equal{} 2\sqrt [4]2 \equal{} 2^{\frac54}$.

However, to simplify something of the form $ \sqrt {a \plus{} b\sqrt c}$, just set it equal to $ \sqrt m \plus{} \sqrt n$ (or $ \sqrt m \minus{} \sqrt n$ if it's of the form $ \sqrt {a \minus{} b\sqrt c}$), square both sides, and equate.

[quote="007math"][quote="DaReaper"]how do you find the square root of a square root
like
\[ \sqrt {4\sqrt {2}}
\]
[/quote]

I don't know if this is the best way, but...

$ \sqrt {4\sqrt {2}} \equal{} \sqrt {32} \equal{} 4\sqrt {2}$

Interesting...[/quote]

There's a mistake going from $ \sqrt{4\sqrt{2}} \equal{} \sqrt{32}$ it should be $ \sqrt{4\sqrt{2}} \equal{} \sqrt{\sqrt{32}} \equal{} 2\sqrt[4]{2}$

The answer is not $ 8\sqrt{2}$, $ (8\sqrt{2})^{2} \equal{} 128 \ne 4\sqrt{2}$

where are u getting the m and the n, what isthe m and n of this prob?

There is no m and n in that problem. M and N refers for when you have $ \sqrt{4\plus{}2\sqrt{3}}$ then you want to rewrite it as $ \sqrt{(\sqrt{m}\plus{}\sqrt{n})^2}$ where m+n=4 and mn=3. Obviously m=3 and n=1 or m=1 and n=3. Then the rest of the square root is easy.

That's correct. :) And thanks for explaining, mathswimmer.

meh..Sorry guys, I forgot to square root my 32....this is why I am a phailure...


vector
calculus
geometry
geometric transformation
rotation
integration
Princeton

I am scheduled to take Regents (NY state test) Physics next year, but I was hoping I could study on my own time and take the AP Physics test. What is the difference between AP Physics AC and BC? Which course is closer to a Regents Physics course? How hard is it to learn the additional material? Are there very many labs I won't understand without doing them?

First, it's Physics B and Physics C.  Second, since I went to HS on Long Island, I will give you some info.

The regents in physics is something of a joke.  If you have any reasoning skills whatsoever and you know how to use a calculator, there is no reason not to get a 90.  If you can understand the physics, then there is no reason not to get a 95.  The material is rather basic and covers the following: Vectors, Forces/Laws of Motion, Momentum, Energy/Work, Electrostatics, Basic DC Circuits, Basic Waves, Very Basic Optics, and Basic Nuclear/Atomic Physics.  This stuff is not advanced, the equations are not calculus based, and solving problems is almost always a matter of recognizing what data you're given, what data you need, and what equation to pull from your reference tables.  Easy stuff.

Physics B is considered a "survey" course.  It covers a wide range of topics, but none of them in a particularly rigorous manner.  There is no calculus, and the math is only marginally harder than on the regents.  Some of the topics will require that you actually understand the physics behind it, and then you will be able to go to your reference table and pick out the equation you need.  B covers: Vectors, Forces/Laws of Motion, Momentum, Energy/Work, Fluids, Basic Thermophysics, Electrostatics, DC Circuits, Basic Capacitors, Basic Magnetism, Waves, Optics, Nuclear/Atomic Physics.  As you can see, there is a lot of stuff, but none of it is covered in rigorous detail.  Basically, this is the course to take if you just want to get a general knowledge.

Physics C is divided into Mechanics and E&M.  C covers substantially less material but B, but is substantially harder.  Calculus is widely used.  The math is tougher, and to do well, you will have to have a basic physics intuition.  Meaning, it isn't plug-and-chug physics.  You will have to know how and why certain laws work or else they will be completely useless to you and you'll get the wrong answer.  This is the closest to a first-year college sequence, although it is a bit easier, particularly since it uses only single-variable techniques.  Mechanics covers: Vectors, Forces/Laws of Motion, Momentum, Energy/Work, and Rotation (including torque, energy and angular momentum).  Not too much more material (sans rotation) but it is more difficult.  E&M covers: Electrostatics, DC Circuits, Very Basic RLC, RC, LC, and AC Circuits, Capacitance, Magnetostatics, Magnetic Fields, Maxwell's Equations.  You can clearly see that this is more rigorous than the previous tests.

So all and all, what should you do?  Answer: whatever you are comfortable with?  Do you know calculus already (key word is already)?  Are you motivated enough to self-study physics C on your own?  Maybe self-study B next year, and take C the year after, if you have that option.  Hope this helps.

if you are considering any sort of self study method i highly recommend the ap physics by Barrons. I received a 5 on ap physics b my junior year and then self studied with the barrons for the C exam this year and felt very comfortable with the material. another good thing about the barrons is that it assumes you have little to no previous interaction with physics, only a survey of calculus course such as ap calc ab.

The curve on the Physics B test tends to be very generous because the questions cover a wider range of material than you will actually have time to cover in any one-year course, so as long as you're familiar with the easier and more common parts of the curriculum you can safely ignore, say, nuclear physics.

Mechanics is fairly mathematical.  Your grasp of single-variable calculus will need to be fairly strong, and like JRav says you'll need physical intuition to understand why certain things are defined the way they are or else you'll never remember them.  Fortunately, if you understand the math and have good intuition there isn't much else to do.  I can't comment on E&M, though.

ap physics c e&m focuses on much of the fundamentals and concepts covered in physics b, except they must be comprehended on a more mathematical basis. the most expanded on concept on the ap physics c would have to be circuits and electro-magnetic statics in my opinion. for instance, for circuits you now need to review all the different types of circuits including rc, rl, and sometimes rcl's (yes ive seen some). as statics is concerned, the main concept you must know is how to set up integrals, and comprehend the methods for setting them up. all in all, the physics c exam covers less material, yet it covers the material in much more depth. it is one of the least taken tests i think because it combines both a 'higher' level mathematics AND science ability but in my experience i thought it was one of the lesser challenging exams as long as you are prepared for it is very easy to practice for the free response section as compared to other exams.

In response to the post on review books, I have heard bad things about Barron's.  I highly recommend Princeton review, for whichever test you choose to take.  I actually used Princeton review as a reference for my college physics course and it was very good.

Another point: In E&M, there are many times (especially in magnetism) in which there are very small things that are easy to miss, or little algebraic tricks that may make a seemingly impossible problem possible.  Not so much the case with Mechanics.  With mechanics, it usually comes down to a free-body diagram in some way.

i must agree with you that Barron's is a poor choice for about 90% of the ap exams. in my opinion though, the Princeton book for ap physics is a complete waste of time. #1 they combine the ap physics b and c curriculum into one book and while the book may do physics b justice, in no way does it cover enough physics c material. Princeton cuts out, and shortens a lot of important consequences of the calculus approach to physical phenomenon by which the c exam is suppose to cover. Princeton is a very superficial approach to the c test and i don't recommend it.

if there is one bad thing to be said about the Barron's, however, is that the problems are significantly tougher than those on the ap test. Barron's forces you do understand it on a deeper level, which I found very beneficial for the exam.

I'm pretty sure that this year, Princeton review now puts out 2 books, 1 for B and 1 for B.  I felt that the book did a very good job.  Which aspects are you referring to?  Did you want a complete derivation of all of Maxwell's equations or a proof of something because I felt it covered everything very well.


algebra
polynomial
induction
vector
linear algebra
matrix
algebra proposed

Prove that polynomial $x^{n}+4$ factorability into product of two polynomial   integer coefficient  with degree less than $n$ iff $4|n$.

I'll prove the converse.

If n is odd: the polynomial has the unique real root $-\sqrt[n]{4}$. This root isn't the root of any integer polynomial of degree less than $n$ (this is quite obvious... maybe I'll prove it). So, $x^{n}+4$ can't be factored.

If n is even, the polynomial has no roots. Then it must be factored in two even-degree polynomials (otherwise one of them would have a root), so 4|n.

[quote="edriv"]I'll prove the converse.

If n is even, the polynomial has no roots. Then it must be factored in two even-degree polynomials (otherwise one of them would have a root), [b]so 4|n[/b].[/quote]

Why?

Ooops  :blush:

Even + Even is not always a multiple of 4
2 + 4 = 6  :rotfl:

:D
[quote="edriv"]the polynomial has the unique real root $-\sqrt[n]{4}$. This root isn't the root of any integer polynomial of degree less than $n$ [/quote]

What is your prove? Post it  :P

One can prove a much more general result by using induction on chains of fields $E\subseteq F \subseteq K$. But to do so, one needs to define fields, degree of extension fileds, algebraic elements and their minimalpolynoms, and the main properties of norms between fields. To go through all these basic algebraic tools, would be far to troublesome here, and I suggest a good text book on Galois Theory. (Lang is my favorite)

But I'll do the special case (and initial step of the general proof)
$x^{p}+4$ for odd primes $p$.
Take $\alpha=-\sqrt[p]{4}$ and $V=\{r_{0}+r_{1}\alpha+r_{2}\alpha^{2}+\cdots+r_{p-1}\alpha^{p-1}\text{ : }r_{0},\cdots,r_{p-1}\in\mathbb{Q}\}$.
Then $V$ is a $\mathbb{Q}$-vector space with $\alpha V=V$.
Now $f: V\rightarrow V, v\mapsto \alpha v$ is a $\mathbb{Q}$-endomorphism of $V$ with $f^{p}(v)=\alpha^{p}\cdot v=-4v$.
So $f^{p}=-4\cdot\text{id}_{V}$ and applying the determinant we get
$(\text{det}f)^{p}=\text{det}(f^{p})=\text{det}(-4\cdot\text{id}_{V})=(-4)^{d}$ with $d: =\text{dim}_{\mathbb{Q}}V$ and $\text{det}f\in\mathbb{Q}$.
But $1,\alpha,\cdots,\alpha^{p-1}$ span $V$, so $d=\text{dim}_{\mathbb{Q}}V\leq p$.
And if we write $\text{det}f=\frac{m}{n}$ with $m,n\in\mathbb{Z}$, we get $m^{p}=(-4)^{d}n^{p}=\pm 2^{2d}n^{p}$ and $p\text{ }|\text{ }d$.
Therefore $p=d=\text{dim}_{\mathbb{Q}}V$ and $1,\alpha,\cdots,\alpha^{p-1}$ are linear independent over $\mathbb{Q}$.
Now if $x^{p}+4=gh$ with $g,h\in\mathbb{Q}[X],\text{deg}g,\text{deg}h<p$, then $0=\alpha^{p}+4=g(\alpha)h(\alpha)$.
Say $0=g(\alpha)=a_{0}+a_{1}\alpha+\cdots+a_{r}\alpha^{r}$ with $r=\text{deg}g\leq p-1$ and $a_{0},\cdots,a_{r}\in \mathbb{Q},a_{r}\not=0$.
But then $1,\alpha,\cdots,\alpha^{p-1}$ are linear dependent over $\mathbb{Q}$. Contradiction!

Remarks:
(1) the determinant of an endomorphism is defined as the determinant of the corresponding matrix for some basis of the vector space.(it doesn't depend on the basis)
(2) the determinant is multiplicative: $\text{det}(AB)=\text{det}A\cdot\text{det}B$
(3) $-4\cdot\text{id}_{V}$ has corresponding matrix (for any basis) $-4\cdot 1_{d}$, where $1_{d}$ is the $d\times d$ unit matrix and $d=\text{dim}_{\mathbb{Q}}V$
(4) with a few algebraic tools this whole proof comes down to 4 lines, and the general result is a short and very beautiful induction. So go read it!

Thank for all and olorin, I underrstand your solution  :wink:

This is a solution I know:

Suppose $x^{n}+4=P(x)Q(x)$ for $P,Q\in\mathbb{Z}[x]$ and $\deg{P},\deg{Q}<n$. We have roots of $P$ and $Q$ also are roots of $x^{n}+4$ and modul of they equal $\sqrt[n]{4}$. By Viete's theorem we have $4^{\frac{\deg{P}}{n}}\in\mathbb{Q}$ and $4^{\frac{\deg{Q}}{n}}\in\mathbb{Q}$, and therefore $n$ is even. If $\frac{n}{2}$ is odd then $\deg{P}$ and $\deg{Q}$ are odd, there they have got real root, therefore $x^{n}+4$ has real root, cotradiction because $n$ is even.


I got this problem from the trig book I'm using during my spare time to learn the trig independently. But I'm stuck on this one. :(

A surveyor wishes to find the distance across a swamp (see figure). The bearing from A to B is N 32  :^o: W. The surveyor walks 50 meters from A, and at the point C the bearing to B is N 68 :^o: W. Find (a) the bearing from A to C and (b) the distance from A to B.

I attached a diagram for this problem. I drew it from Microsoft Paint :D

Thanks for your help! :P

For (a), just draw a north-south line through A.

[hide="a"]for (a), draw a ray from A extending through BC. call it ray AD, such that angle BAD = 68. then BCD = 32. since angle BAC is a right angle, and angle DAC is the angle we are looking at (because it is the angle between the north bearing line and AC), subtract 68 from 90... and we get 22.

I wasnt sure if you had the little corner drawn on the A, notifying that BAC was a right angle. It doesnt look clear. But i assumed that you drew that there to tell us that BAC was a right angle.[/hide]

[hide="b"] this time we are going to draw the same north bearing line AD, except it is gonna end at D, such that DC is perpendicular to AC.

since earlier i have established the fact that BCD was 32 degrees, then BCA = 58.
we can find the distance AB by using tangent function.

$\tan BCA = \frac{AB}{50}$
$\tan 58 = \frac{x}{50}$
$x = 50 \tan 58 = 80.02$

so about 80 meters. [/hide]

i did it wrong. oops. i read 68 and 32 backwards....

so for (a) it is 90 - 32= 58, not 90 -68 = 22....

and for (b) i will post it soon... cuz i gotta go do the painful hmwk... sorry silverfalcon


calculus
derivative
calculus computations

Let $ z(u,v)$ such that
$ u\equal{}x^2\plus{}y^2$
$ v\equal{}xy$

Find $ z_{xx}$

Use the chain rule for partial derivatives.


Complete with numbers (do not spell the numbers, write them down, that means, don't write five, write 5)

In this sentence, the number of 0's is __, the number of 1's is __, the number of 2's is __, the number of 3's is __, the number of  4's is __, the number of 5's is __, the number of 6's is __, the number of 7's is __, the number of 8's is __ y and the number of 9's is __.

[b]REMARK[/b]: I don't know if it has solution. It just came to my mind.

[quote="JBL"]What about other bases?[/quote]
[hide="Base 2"]In this sentence, the number of 0s is 11 and the number of 1s is 100.[/hide]
For n>2:
[hide="Base n"]In this sentence, the number of 0s is 1, then number of 1s is 11, the number of 2s is 2, the number of 3s is 1, the number of 4s is 1, the number of 5s is 1, the number of 6s is 1,..., and the number of (n-1)s is 1.[/hide]

[quote="JBL"]This is a very interesting question!  I think it may not have a solution, but I am not sure yet.

Edit: Okay, I have found a solution, but I'm not quite sure why it exists yet :)
[hide]In this sentence, the number of 0's is 1, the number of 1's is 7, the number of 2's is 3, the number of 3's is 2, the number of 4's is 1, the number of 5's is 1, the number of 6's is 1, the number of 7's is 2, the number of 8's is 1 and the number of 9's is 1.
[/hide]

My challenge: solve the same problem in base 2:

In this sentence, the number of 0s is __ and the number of 1s is __, where all numerals are written in base two.

What about other bases?[/quote]Hmmm!, nice! I found a solution

[hide="My Solution"]In this sentence, the number of 0s is 11 and the number of 1s is 100, where all numerals are written in base two.[/hide]

I don't know if it is unique, but I think so! I have the idea that it should have solution on every base, but I am not so sure. I will try base 3... may be tomorrow!

There is at least one solution for every base, and it is not unique, at least not in general!.


complementary counting
number theory unsolved
number theory

What is the total number of natural numbes divisible by 9 the number of digits of which does not exceed 2008 and at least two of the digits are 9s?

This is really more of a counting problem.
[hide]
Using complementary counting, we count the desired natural numbers by first counting the total number of natural numbers divisible by 9 and with no more than 2008 digits, and then subtracting off those that have <2 digits of 9.

First, it is easy to see that the number of positive integers divisible by 9 and with no more than 2008 digits is $ \frac {10^{2008} \minus{} 1}{9}$.

Next, we count the number of these that have no digits equal to 9. Each of the first 2007 digits can be any of 9 digits 0-8, and the last digit is determined by the first 2007 digits, so there are $ 9^{2007} \minus{} 1$ natural numbers without any digits equal to 9. (Note that we must subtract one so that 0 is not counted).

Lastly, we count the number of positive integers divisible by 9 and with no more than 2008 digits that have exactly one digit of 9. To do this, we count the number of natural numbers with $ \le 2007$ digits that have no 9s and then add the 9 in one of 2008 places. Hence, the total number of these is $ 2008\cdot9^{2006}$.

Finally, the answer is $ \frac {10^{2008} \minus{} 1}{9} \minus{} 9^{2007} \plus{} 1 \minus{} 2008\cdot9^{2006} \equal{} \boxed{\frac {10^{2008} \minus{} 9^{2008} \minus{} 2008\cdot9^{2007} \plus{} 8}{9}}$.[/hide]


calculus
integration
calculus computations

Prove that

$ \int_0^{\infty}\frac{\ln(x)}{1\plus{}x^s}dx\equal{}\minus{}\left(\frac{\pi}{s}\right)^2\cdot{\cot\left(\frac{\pi}{s}\right)\cdot{\text{cosec}\left(\frac{\pi}{s}\right)}}$

$ s > 1$

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208212

use a similar argument here.


Determine the interval of convergence of the series.

$ \frac{x^2}{2}\minus{}\frac{x^6}{6}\plus{}\frac{x^{10}}{10}\minus{}\frac{x^{14}}{14}\plus{}\frac{x^{18}}{18}\plus{}\frac{x^{22}}{22}\plus{}\cdots$

Calculate the sum for the $ x\equal{}\frac{1}{\sqrt[4]{3}}$

hint :
direct calculus show that :
$ \sum ... your serie ... \equal{} \sum _{k\equal{}0}(\minus{}1)^{k}\frac{x^{4k\plus{}2}}{4k\plus{}2}\equal{}\frac{1}{2}\sum _{k\equal{}0}(\minus{}1)^{k}\frac{(x^{2})^{2k\plus{}1}}{2k\plus{}1}$
denote :
$ x^{2}\equal{}A$ then we need to find :
$ \frac{1}{2}\sum _{k\equal{}0}(\minus{}1)^{k}\frac{(A)^{2k\plus{}1}}{2k\plus{}1}$
next fact is wellknown:
$ \frac{1}{1\plus{}A^{2}}\equal{} \sum_{k\equal{}1}(\minus{}1)^{2} A^{2k}$ (when $ |A|<1$)
after integration we have :
$ arctan (A) \equal{} \int\frac{1}{1\plus{}A^{2}}dA\equal{} \sum_{k\equal{}1}(\minus{}1)^{2} \frac{A^{2k\plus{}1}}{2k\plus{}1}\equal{}2 *your serie$


Game birds such as ducks and pheasants that are raised on a farm need a brooder house temperature of $ 95^\circ$F to start.  The temperature is then decreased five degrees per week until the outside temperature is reached.  How many weeks would it take for the brooder house temperature to reach $ 60^\circ$F?

so eventually it decreased by 35 degrees.

it decreased 5 degrees per week, so 35/5=7.

answer : 7


AMC AIME geometry search USA(J)MO USAMO trigonometry

Ok, well, I've looked around a bit and everyone seems like a math geek, so I'm sure I'll get some help on here :) Last year I took the AMC 10 and didn't do that well (I think I got around an 86) but it really got me interested in math.  So I did some studying, basically just on the official amc website looking at past problems, for the AMC 12.  And surprisingly enough I did well on it; at least, well for me.  Probably sucky to all you geniuses.  But yeah I got 109.  But I'm kind of freaking out about the AIME because a) you can't use a calculator, b) it's in a week and a half, c) I need help.

I know you're thinking "Oh, just look around on the forum, blah blah blah."  Obviously I intend to do that (hrmph it would have been nice if I'd realized this was on the site before today!), but I'm also wondering if anyone is willing to take me under their wing and be a tutor until the test to prep me for it.  Otherwise just point me in the direction of helpful free online resources or the most helpful threads on these boards.

One more thing...I never took geometry, so that's probably an area of particular weakness for me.

Thank you so much!

First of all, I know you used the word "geek" in good way but it's not polite to call people "math geek." Here is a definition of the word "geek" from dictionary.com.

[quote]A person who is single-minded or accomplished in scientific or technical pursuits but is felt to be [b]socially inept[/b].[/quote]

Second, it is good that you have interest on math but I'm very positive that you will not score very high if you don't know any geometry. You might get 1~4 if you're lucky I say. Third, as for tutor, AIME material cannot be taught in short amount of time. But if you really need one, AoPS is offering a short seminar with AIME strategies and discussion of past problems. You can find more about it in [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php#class1]here[/url]. And if you need more help, use the nice button called "Search" button. Type in something like "AIME Tip" or "AIME studying" or etc.. It will be much faster to get any results that way because most things going on AoPS are repeated many times.  :D

hmm...not taking geometry yet will definitely hurt you, since I think the easiest AIME problems are the Geometry ones, but go to the contests button on the top and go check out past AIME problems and try to understand the solutions. Since you haven't been exposed to AIME problems, try sticking to the first five or so and understanding them.
Remember, to get over 3 or 4 on the AIME for your first time, and as a Junior (you are a junior right since you took the 12?), is a great accomplishment. Good luck! :)

Lol, I've been on the college confidential forums and there are actually a number of people who view "geek" with positive connotations (and like discussing the nuances between geek and nerd).  You can't take a single definition and look no further.  Honestly, to me it is a compliment.

The short seminar looks like it'd be helpful but unfortunately I have no money to fund it with :( Actually around September I was planning on purchasing the art of problem solving books but I never got around to it - I was so busy with theater and everything that I didn't want to spend the money if I wasn't sure I'd take the time to get through them.  Obviously now I'm kicking myself...

Well, I'm not expecting a high score on the AIME at all because when I look at the practice problems and solutions I tend to find them very confusing...of the ones I've looked at I've only thoroughly understood a few.  I'd just like to understand SOMETHING on the test and not feel like I'm sitting there wasting my time and that it was a total fluke that I got to it.  Well, it's not really - I didn't find the AMC 12 problems that I answered too hard.  The only one I answered which I missed was because I didn't know whether 0 was an integer...and duh, I should have known that!

Anyways I'm just happy that I did well on the AMC; in my school 7 people qualified for the AIME.  I was one of the 2 juniors (the rest were seniors) and the only girl.  So yay! :)

What I really need, if anyone's willing, is really simple explanations of some solutions.  For instance, I have no idea what "relatively prime" is supposed to mean...

Relatively prime means that the two numbers have no common factor except for $1$. For example, $15=3\cdot 5$ is relatively prime to $28=2^2\cdot 7$, but it is not relatively prime to $21=3\cdot 7$. Basically, if you put the two numbers into a fraction, it won't reduce at all.

That's good that you're asking questions, though. That's the best way to learn when you don't understand the solutions. What I would recommend is, if you really want to do your best on the AIME, spend as much free time as you can on this site reading solutions to past problems (you can find these in the Contests link near the top of the page if you're using the AoPS skin, or Resources if you're using MathLinks; click on USA, then AIME). First, try to solve them on your own; if you can't, read the solutions that others have; if you don't understand the solutions, ask questions (but try not to bump [b]really[/b] old threads ;)).

Also, don't expect to do really well on this one. You'll probably get around 2 or 3, although you might do better, depending on what kinds of problems show up and how easy the test is. However, take my case as an example. I scored a 3 on the AIME last year - one of the easiest AIME's in recent years - without ever having seen an AIME before. Now, after spending 10 months on this site, I can solve at least 11 or 12 on the same AIME. I would expect to see a similar change between this year and next year if you're serious about it. ;)

Finally, don't worry about not being able to use a calculator. The reason that calculators aren't allowed is because the problems are specifically designed so that you don't need a calculator if you're approaching it in the right way. AIME problems are generally designed to not be brute-forced, although some do work out that way; rather, you should try to find the finesse solution. (Don't let this discourage you from brute-forcing some if you really can't find a better way and know that you can do it in time, but look for the finesse solutions first.)

I would just second what JesusFreak said...you can also look up problems on kalva or actually get AIMEs/solutions from past years (they have them).

Otherwise, just do math.  Do these kinds of problems, and look at other solutions.  When you encounter a similar problem to one you've faced before, try to reconstruct the thinking in the solution.  You'll, over time, learn many, many useful techniques.  I know I'm sounding like a broken record, but really, it helps a lot.  When I first made the AIME two years ago, I could barely solve a single problem (I found out at a similar time, a couple weeks prior, and only got time to look at about one AIME).  I ended up getting a 4 or something.  The year after that, I was somewhat better, but still made careless mistakes and ended up with an 8, one below the cutoff.  Then I joined AoPS in April, and now can solve most AIME problems (with careless mistakes factored in, maybe 12-13 on an average AIME).  There are even people who have gone from barely/not qualifying for AIME to making MOP, or so I've heard on this site...Heck, my time on this site has probably improved my USAMO skills dramatically as well.

I'm a late starter too, but i got a 125.5 on the amc 12 a. I didnt take the 10 last year or start doing a little practice until a few weeks before the test.  I didnt take the 12 b but after testing was completed i took it and wouldve gotten 132.5 if i hadnt made a dumb mistake (wasnt using calculator at the time and was doing it at a relative's house).

So now here i am semi-cramming for the AIME.

Oh, and get the AoPS books vol 1 and 2; they delivered in 3-4 days although on the order sheet, it said 3-4 weeks. They've got all you need for the AIME.

[quote="lavendercloud"]Well, I'm not expecting a high score on the AIME at all because when I look at the practice problems and solutions I tend to find them very confusing...of the ones I've looked at I've only thoroughly understood a few.  I'd just like to understand SOMETHING on the test and not feel like I'm sitting there wasting my time and that it was a total fluke that I got to it.  Well, it's not really - I didn't find the AMC 12 problems that I answered too hard.  The only one I answered which I missed was because I didn't know whether 0 was an integer...and duh, I should have known that![/quote]

Maybe you should just try some of the beginning AIME problems, like 1-3, and try to understand those really well.

I will also be taking it as a junior but im not sure if i will do so well.

If i take it next year, how will colleges know, if at all, if i qualified for USAMO if i do because its past the college app time?

the aime is not really a test that you cram for...all the stuff you learn from class should be obvious, like say when to use the law of cosines, and i don't think you can really get that kind of mastery by cramming, as for taking the test, it is fine to just spend a while on the first few problems, if you spend like 45 mins on the first 3 problems, you should be able to get them, just be careful with your reasoning, and try not to miss little cases, etc

If the aops books do ship in 3-4 days then learn the geometry chapters in volume one. No knowledge of geometry is basically ommitting 6-7 questions. Other than that go through past tests and see how much you can learn and understand.

also remember, only do the work if you enjoy it, you don't want to burn yourself out, but i suppose that i enjoy doing math, so that is why i do problems all the time

Speaking of relatively prime numbers, our coach told us of a time when one of the people taking the AIME went up to him during the test and asked, "What does relatively prime mean?"  :rotfl:

It's [i]hard[/i] to answer some questions...


calculus integration geometry function derivative logarithms limit

If I evaluate the integral of (x)(ln(x)) from 0 to 1, will I get a definite area?

I tried writing it all out and I get .5x^2ln(x)-.25x^2 from 0,1. But ln(0) is undefined, so I thought that the area is undefined, but if I do it on my TI-89, I get -1/4. Which one is right? Thanks.

What do you think? My calculator tells me the absolute value function is differentiable at x=0, which I know isn't true. I'm pretty sure the TI calculates derivatives with the limit definition formula using a pre-set value for h that is significantly small (say $ 1\times10^{\minus{}4}$), and a similar approach using riemann sums is used (in that case, some arbitrarily large n). Thus, a calculator can give derivatives and integrals for discontinuous points when it shouldn't, simple as that.

End-points are completely irrelevant for the existence of the integral. In this particular case, the original function is continuous on $ [0,1]$ if you let it take the value $ 0$ for $ x\equal{}0$. Similarly, the integral can be evaluated with your primitive if you understand it as having the unique continuous extension value in $ 0$

A primitive function is $ \frac {x^2 \ln x}{2} \minus{} \frac {x^2}{4}$ as you wrote, defined as 0 for $ x \equal{} 0$ ($ \lim_{x \to 0} x^a \ln x \equal{} 0$). So the answer is this function evaluated in $ 1$ which is $ \frac { \minus{} 1}{4}$

Kalle: That only works since you have removable discontinuity. If you don`t redefine the function as having continuity over the interval it`s evaluated over, then you can`t integrate.

Any bounded function over an interval $ [a,b]$ whose set of discontinuities is of measure zero (in particular, all countable sets) is Riemann-integrable, and moreover the integral is the same no matter what the values are on those discontinuities.

In particular, it doesn't matter what the values of the function are in $ a$ and $ b$. This is also immediate from the definition of the Riemann integral. So a simple result is that all bounded functions continuous in $ (a,b)$ are riemann-integrable on $ [a,b]$.

I understand the problem now..with the stuff we're doing, we are suppose to express it as a limit so that we can evaluate it.

And ur right, considering I'm only a sophmore in high school taking AP Calc B, we're only at the surface of integrals.

Is $ \int_0^1x\ln x\,dx$ an improper integral?

Kalle has argued that is not actually improper because the function $ x\ln x$ can be extended to be continuous on $ [0,1].$

Kalle is correct, but the issue is a moot point computationally. Because we can't simply "plug in" a value into a formula at zero, we wind up treating things as limits anyway, which means that the we act as if this were an improper integral. And no harm comes of that. Paraphrasing what t0rajir0u said, the we only know that the functions in question can be extended to be continuous at zero after we take the appropriate limits as $ x$ tends to zero.

But loud and clear: if we see $ \lim_{x\to0^\plus{}}x^a\ln x$ for some $ a>0,$ we should just immediately know that that limit is zero. Yes, it's a $ 0\cdot\infty$ indeterminate case, but we also know (because we've done it before and memorized the result) that the logarithm goes to infinity "slower than any power of $ x.$"

So, $ \int_0^1x\ln x\,dx\equal{}\left.\left(\frac{x^2}2\ln x\minus{}\frac{x^2}4\right)\right|
_0^1.$

We treat that evaluation at zero as the appropriate limit. That is, we act as if this were an improper integral whether it is or isn't. But the values at zero are zero, at least in the limit, because the power of $ x$ goes to zero faster than the logarithm grows. So the result is $ \left(0\minus{}\frac14\right)\minus{}(0\minus{}0)\equal{}\minus{}\frac14.$

We isolate and take limits at singular points of an integral to define and compute improper integrals. In this case, is $ x\equal{}0$ a singular point? Kalle says it is not, but whether it is or isn't, it's a point about which suspicions have been cast. It's perfectly all right to act on those suspicions.

[quote="t0rajir0u"]I think the issue is it's not clear [i]a priori[/i] that this function is bounded unless something is said about its value at (or near) $ 0$.[/quote]
[quote="Kalle"]I think the issue is that because of high school shenanigans, Prime and Logg are not clear over what an integral is[/quote]

oh man this is such a great reply.

in any event this kind of shows that the riemann integral, in the most basic sense, is kind of a useless object - the requirement that $ f$ be defined [i]everywhere[/i] is extraneous and unnecessarily restrictive in light of 'better' methods of integration.

An approach with simple double integration.

\begin{eqnarray*}\int_0^1x\ln x\,dx&=&-\int_0^1\int_x^1\frac xu\,du\,dx\\&=&-\int_0^1\int_0^u\frac xu\,dx\,du\\&=&-\frac12\int_0^1u\,du\\&=&-\frac14.\end{eqnarray*}

Okay, maybe I was thinking of something else.

Let $ f(x) \equal{} 18$ for $ x\equal{}3$ and $ x^{2}$ otherwise. If you go by your Riemann sum, we have $ \int_{a}^{b} f(x)dx \equal{} \lim_{n\rightarrow \infty} \triangle x \sum_{i\equal{}1}^{n} f(a \plus{} \triangle xi)$ where $ \triangle x \equal{} \frac{|b\minus{}a|}{n}$. Now, wouldn't the discontinuous point give you an extraneous rectangle height?

But that one bad point is isolated within one rectangle that can be made very skinny. If the difference between the value at that one point and what it "should" be is $ H,$ then the bad effect that that point has on the Riemann sum is no more than $ H\Delta x,$ and that can be made arbitrarily small by making $ \Delta x$ arbitrarily small. Your function has the same integral as the function which is simply $ x^2$ over that interval.

A bounded function on a finite interval with finitely many inequalities is Riemann integrable. Changing a function at finitely many places does not change its integral. (Neither of these statements is as general as it could be.)

[quote="Prime factorization"]Okay, maybe I was thinking of something else.

Let $ f(x) \equal{} 18$ for $ x \equal{} 3$ and $ x^{2}$ otherwise. If you go by your Riemann sum, we have $ \int_{a}^{b} f(x)dx \equal{} \lim_{n\rightarrow \infty} \triangle x \sum_{i \equal{} 1}^{n} f(a \plus{} \triangle xi)$ where $ \triangle x \equal{} \frac {|b \minus{} a|}{n}$. Now, wouldn't the discontinuous point give you an extraneous rectangle height?[/quote]

I object to you making $ \triangle x \equal{} \frac {|b \minus{} a|}{n}$. If b < a, $ \triangle x$ should in fact be negative.


let $a,b,c >0$ and $a>6$ such that :$abc=1$.
prove that : $a^{2}/3+b^{2}+c^{2}\geq ab+bc+ca$

Let $f(a,b,c)=\frac{a^{2}}{3}+b^{2}+c^{2}-(ab+bc+ca).$ Eliminating $a$ from the given condition,and setting $b+c=s,\ bc=t,$ we have $f(a,b,c)=\frac{1}{3t^{2}}+s^{2}-\frac{s}{t}-3t\ \cdots [*].$

Now the condition for which there existed positive numbers $b,c$ such that $b>0\ ,c>0,\ a>6$ is $s>0,\ t>0,\ s^{2}\geq 4t,\ t<\frac{1}{6}\Longleftrightarrow s>0,\ 0<t<\frac{1}{6},\ t\leq \frac{1}{4}s^{2}.$

Sketch this distribution on $b-c$ plane, we obtain $0<s<\frac{\sqrt{6}}{3}.$ Thus from $[*],$ fix $t$ as the constant, we have

$f(a,b,c)\geq \frac{1}{3t^{2}}+4t-\frac{s}{t}-3t=-\frac{1}{t}\cdot s+\frac{1}{3t^{2}}+t$

$>-\frac{1}{t}\cdot \frac{\sqrt{6}}{3}+\frac{1}{3t^{2}}+t=\frac{3t^{3}-\sqrt{6}t+1}{3t^{2}}.$

Varying $t$ as variable for $0<t<\frac{1}{6},$ let $g(t)=3t^{3}-\sqrt{6}t+1,$ we have $g'(t)=9t^{2}-\sqrt{6}<9\left(\frac{1}{6}\right)^{2}-\sqrt{6}<0.$

$\therefore g(t)>g\left(\frac{1}{6}\right)=\frac{3}{216}+1-\frac{\sqrt{6}}{6}>0.$ which completes the proof.

I think you should consider the range of $x.$

Inequality, \[\left(\frac{a}{2}-b-c\right)^{2}+\frac{a^{2}}{12}-3bc > 0.\] this is obviously true.


let $ d(n)$ denote the sum of digits of a natural number $ n$
then solve for all natural numbers satisfying
$ d(n)\equal{}n^2 \minus{}10n\minus{}35$

[hide="hint"]$ n\ge\ d(n)$ use this.[/hide]

I think you made a typo.
 use $ n\geq d(n)^{*}$

since every natural number is greater or equal with sum of digits,$ \Rightarrow n^{2}\minus{}11n\minus{}35\leq 0$

enndb0x started well.
First solve $ n^2\minus{}11n\minus{}35\equal{}0$
$ n\equal{}\frac{22\pm\sqrt{261}}{2}$
For n less than that value, it's less than 0. For n greater, it's bigger than zero.
So $ n\in{1,2,\ldots,19}$

$ n\equiv d(n)\pmod{9}$, so we easily get $ n\equal{}3k\plus{}1$

So $ n\in{1,4,7,10,13,16,19}$

Now check these values.

I claim that [hide="answer"]
13 is the only value n can take.
[/hide]

[quote="Bugi"]enndb0x started well.
First solve $ n^2 \minus{} 11n \minus{} 35 \equal{} 0$
$ n \equal{} \frac {22\pm\sqrt {261}}{2}$
For n less than that value, it's less than 0. For n greater, it's bigger than zero.
So $ n\in{1,2,\ldots,19}$

$ n\equiv d(n)\pmod{9}$, so we easily get $ n \equal{} 3k \plus{} 1$

So $ n\in{1,4,7,10,13,16,19}$

Now check these values.[/quote]

Can you explain why take

$ \boxed{n \equal{} 3k \plus{} 1}$

[hide="Solution part 1"]
First using the statements stated above
$ n \ge n^{2} \minus{} 10n \minus{} 35$

and this gives us $ n \le 13$
[/hide]

[quote="harekrishna"][hide="A hint"]
First using the statements stated above
$ n \ge n^{2} \minus{} 11n \minus{} 35$

and this gives us $ n \ge 13$
[/hide][/quote]

It's 10n, not 11n. And $ n^2$ is rising faster than $ n$, so your $ n\ge 13$ is absolutely impossible. It's $ n\le k$, for some k, and I proved it's 19 (in integers).

@vinskman:

$ n\equiv n^2\minus{}10n\minus{}35\pmod{9}$
$ n^2\minus{}11n\minus{}35\equiv0\pmod9$
$ n^2\minus{}2n\plus{}1\equiv0\pmod9$
$ (n\minus{}1)^2\equiv0\pmod9$
$ n\minus{}1\equiv0\pmod3$

[hide="Solution - part 2"]

$ d(n) > 0$
so, we get

$ n \ge 13$

Thus n should be 13

[/hide]

Ignore my try for $ n\le 19$.

It's $ n\le 13$, my bad try.

harekrishna finnaly edited  :)

well i think we can solve it in this way
like we know that $ n\equiv 1\mod 3$
then we can substitute $ n \equal{} 3k \plus{} 1$
substituting that in the given condition
and also $ d(n)\equiv 1\mod3$
and having  $ d(n) \equal{} 3a \plus{} 1$
then with the condition that $ d(n)\le\ n$
we get a diophantine in $ a,k$
i think from that we can solve it
but i am not too sure with what i got .i have to verify
pls anyone post their solution in the way i have directed if possible as i am not sure with that.
i hope that the sum  would have a   good solution in that case
pls forgive if i am wrong

well i think i got it in the way i directed
from where i had left
we have
$ 3a\plus{}1\equal{}9k^2 \minus{}24k\minus{}44$
$ 3a\equal{}9k^2 \minus{}24k \minus{} 45$
$ a\equal{}3k^2 \minus{}8k\minus{}15$
since $ d(n)\le\ n$,we have $ a\le\ k \boxed{1}$
$ a\equiv 0\mod2$ or $ a\equiv 1\mod2$
case1 when $ a\equal{}2l$
we have $ n\equiv 1\mod2$ ,say $ n\equal{}2m\plus{}1$  from $ \boxed{1}$ we have that $ 2l\le\ 2m\plus{}1$ which is clearly not possible.also l,m belong to whole numbers
so $ a\equiv 1\mod2$ say $ a\equal{}2l\plus{}1$ ,accordingly k is congruent to $ 0 mod2$
from $ \boxed{1}$ we have $ 2l\plus{}1\le\ 2m$
we have that $ 2l\plus{}1\equal{}24m^2\minus{}16m \minus{} 15$
$ 2m\ge\ 24m^2\minus{}16m \minus{} 15$
$ 0\ge\ 24m^2 \minus{}18m \minus{} 15$ clearly equality does not hold good
$ 5> 8m^2\minus{}6m$
from this we have only $ m\equal{} 2$ and $ l\equal{}0$ to be the only solution
so we have only $ n\equal{}13$


algebra
polynomial
quadratics
superior algebra
superior algebra unsolved

show that the polynomial (x-1)(x-2).....(x-n)-1 is irreducible over Z for every n bigger than or equal to 1.
Hint: if the polynomial factors, consider the values of the factors at x=1, 2,....,n.

The hint seems only can show that there is no monomial factor. how about quadratic factors or high degrees' factor???

Thanks so much for the help

The key lemma: If $p$ is a polynomial with integer coefficients, $(x-y)|(p(x)-p(y))$.
Use this to show that each factor takes only one of the two possible values $\pm1$ (for large enough $n$). The first few cases should be handled separately.

[hide="solution"]We have $f(i)+g(i)=0\forall i=1,2,...,n$ and $\deg{(f+g)}<n$ therefore $f(x)\equiv-g(x)$, and we obtain $(x-1)(x-2)...(x-n)-1=-g^{2}(x)$, contradiction(note: coefficients of $x^{n}$).

P/S: By http://www.kalva.demon.co.uk/imo/isoln/isoln746.html we have $n\leq 2+\deg{f}$ and $n\leq 2+\deg{g}$ we have $n=\deg{f}+\deg{g}\geq (n-2)+(n-2)=2n-4$, therefore $n=1,2,3$ or $4$.[/hide]


what are the chances, that if you pick two integers between 1 and 100, their quotient will be terminating.

Terminating decimal, that is.

Are you counting picking the same number twice?  Also, does order matter or do we always put the smaller over the larger (or the larger over the smaller, for that matter).  By the way, this is a pretty messy problem as far as I can see.

no you cant put a smaller number over a larger number.  No you cant use the same number.

Tare wrote:Assuming you do count them twice if two fractions equal the same and that the fractions does not have to be greater or smaller than 1: [hide]Using the fact that the denominator of the fraction must be in the form (2^x)(5^y) where x and y are non-negative integers in order for the fraction to be terminating, the denominator can be 1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40, 50, 64, 80, or 100. (15 possibilities for the denominator) There are 100 possible numerator for each of these, so 100 possibilities for the numerator. 15*100 is 1500 out of a total of 100^2 or 10000, so the probability would be 1500/10000 or 3/20. (15%)[/hide]

Addendum: hmm, guess the assumption was incorrect...

even if your assumption was correct, you forget that you can do 9/9 or 11/11 or 15/30 and stuff like that.

You just told me both that you can't put a smaller number over a larger and that you can't use the same number in the numerator and denominator, but in the three examples you just gave you violated both those principles!

[quote="white_horse_king88"]Is that a real problem from a book? It should have specified perameters. However Scrambled, those examples violate the rules you gave, but 30/15 doesn't. Others are 46/23, 68/34, or even 88/11. There are many more. The ratio of the numerator vs denominator should be in form (2^x)(5^y), not the denominator itself. Plus, if the fraction can't be x/x then there aren't 100 possible numerators for a denominator y anyway, Tare. There would only be 99 in that case, so this problem is overall very confusing.[/quote]n it isnt.  The first examples i gave were using tare's assumption.


The book says that electrons as arranged into subshells like this:

$ 1s < 2s < 2p < 3s < 3p < 4s<3d<4p<5s<4d<5p<6s<4f<5d$, or $ sspspsdpsdpsfd$, but then it gives this for Rubidium:

$ 1s^22s^22p^63s^23p^63d^{10}4s^24p^65s^1$

Why is it that rubidium (and some others) are $ sspspdsps...$? It's completely mucked up the order! Thanks.

yes the first one is hund's rule that's the order the electrons arrange themselves with evry new one being added.
But while writing we write according to increasing distance from nuclie and not the order in which they are fillled.
Even if say 3d and say 4s has two electrons we write $ ...3d^{0}4s^{2}$.
also the energy of an orbital depends on the sum n+l more it is more energetic the orbital is .if n+l value is same for two different orbital then one with higher n is more energetic

Wow thanks for the fast reply, I understand the notation now, but I don't quite get what you mean in your last two sentences.

Rather than remembering the whole thing, in general:

$ ns<(n\minus{}2)f<(n\minus{}1)d<np$

[quote="Identity"]Thanks, this is only for $ n \geq 6$ right?[/quote]

Well you should know that things like $ 3f$ or $ 2d$ etc. don't exist and reason it from there.

i meant if the sum of shell + subsell number is same for two different orbitals then the one with greater principal quantum number is energetic

Actually, the reason we write the $ 3d$ orbitals first is that after they are filled with electrons, they decrease their energy relative to the $ 4s$ orbitals.  So while for empty orbitals, $ 4s < 3d$, for occupied orbitals $ 3d < 4s$.  I'm not sure exactly why this is true, to be honest, but I do know that it is.

[quote="gregx007"]Actually, the reason we write the $ 3d$ orbitals first is that after they are filled with electrons, they decrease their energy relative to the $ 4s$ orbitals.[/quote]
Doesn't the same apply if they're half-filled (chromium an example)?

Sorry for the abuse of notation.  I meant that as soon as they are occupied by one or more electrons, not as soon as they are fully filled.


Hi.. whoever you are, let's bring up Oklahoma to where it belongs..(30th to 6th or 1st...)

hi
ahhhhhhhhhh its all these jenks people!!! ahhhhhhh!

This forum exists because of those Jenks People!

Hello Mr. Beaty. Remember me? I'm Yichen Wu (Mathcounts nats 2005) and I've moved back to Oklahoma (Norman High).
I was just kidding about the "jenks people":)

Yes, I remember you, Yichen, I still have a picture of you with the Nat team on my wall.

I case you are wondering, I am the admin for this forum.

Welcome back to Oklahoma!

Jenks people rock!

All of you better tell me who you are at the next MC practice. I'm dying to know.  :D

[quote="aznotaku13"]i dont care which school is better[/quote]
Well we DO!

[quote="DonkeyKong"][quote="abacadaea"]jenks people?!?!?!?[/quote]

Yes, apparently, I think no one else in the Oklahoma uses AoPS except for Jenks...[/quote]

I'm not on much any more because I'm not taking an AoPS class this year, but I'm not Jenks. I'm a homeschooler.


Circle $A$ has radius $r_{1}$ and Circle $B$ has radius $r_{2}$.  $M$ is the midpoint of $AB$.  Circle $M$ has radius $\frac{r_{1}+r_{2}}{2}$.  Let Locus $L$ me the set of all points equidistant from circles $A$ and $B$.  Prove that Locus $L$ and Circle $M$ will always intersect at exactly $2$ points.

[quote="me@home"]How do you define distance from point to circle? Do you connect the point to the center of the circle and subtract $r$ from the length of the line?[/quote]
That is exactly what I was thinking in this problem.

[hide="Hintish thing"]
Think conics.
[/hide]


function
geometry
geometric transformation
reflection
analytic geometry
graphing lines
slope

Find all one-to-one functions f, such that f and its inverse are lines with the same slope.

If we have a function, $ f(x)\equal{}ax\plus{}b$, then $ f^{\minus{}1}(x)\equal{}\frac{x\minus{}b}{a}$.

Thus, we can write the equation:

$ a\equal{}\frac{1}a$
$ a^{2}\equal{}1$
$ a\equal{}\pm1$

From this, we get that all such functions are of the form $ x\plus{}b$ or $ \minus{}x\plus{}b$.

Isn't it easier to just think that an inverse is a reflection over the line $ y\equal{}x$, so the only slopes that remain the same after being reflected are $ 1$ and $ \minus{}1$?

[quote="Icy"]Isn't it easier to just think that an inverse is a reflection over the line $ y \equal{} x$, so the only slopes that remain the same after being reflected are $ 1$ and $ \minus{}1$?[/quote]

My method finds all solutions; how do you know you have found all solutions or even that your solutions are valid?  You need to be more rigorous than that.

Haha, alright :D I guess that's a pretty useful tip I forgot to think about.


function
integration
complex analysis
calculus
calculus computations

hy!!

i have one problem in math.this is TRANSFORMATION DE LAPLACE.
ex1:
i speak in french,
determiner la transformée de laplace de la fonction suivante:

g(x)= x si 0<x<T
g(x)=0 si x<0 ou x>>T

Calculate this directly from the definition. Integrate by parts.

The Laplace transform of $g$ is $G$.

$G(s)=\int_0^{\infty}g(x)e^{-sx}dx.$

$G(s)=\int_0^Txe^{-sx}dx= -\frac xse^{-sx}|_0^T+\frac1s\int_0^ {\infty}e^{-sx}dx$

$= -\frac xse^{-sx}-\frac1{s^2}e^{-sx}|_{x=0}^{x=T} = -\frac{Te^{-Ts}}s-\frac{e^{-Ts}-1}s = \frac{1-e^{-Ts}(1+Ts)}{s^2}.$

As one would expect from the fact that $g$ is compactly supported, $G$ is a smoth function - in fact, G is an entire function (complex analytic on the whole complex plane).


On a board are drawn the points $A,B,C,D$. Yetti constructs the points $A^\prime,B^\prime,C^\prime,D^\prime$ in the following way: $A^\prime$ is the symmetric of $A$ with respect to $B$, $B^\prime$ is the symmetric of $B$ wrt $C$, $C^\prime$ is the symmetric of $C$ wrt $D$ and $D^\prime$ is the symmetric of $D$ wrt $A$.

Suppose that Armpist erases the points $A,B,C,D$. Can Yetti rebuild them?

$\star \, \, \star \, \, \star$

[b]Note.[/b] [i]Any similarity to real persons is purely accidental.[/i]

[quote="jrshoch"]What does it mean for a point $A$ to be the symmetric of point $B$ with respect to point $P$?[/quote]

Join $A$ with $P$, and extend this line beyond $P$. the point $B$ on this extension which satisfies $AP=PB$ is the symmetric of $A$ wrt $B$.

In vectors terms, $B$ is defined by the equality $\displaystyle \overrightarrow{AP} = \overrightarrow{PB}$.

Or, more simply, $A$ and $B$ are symmetric with respect to $P$ iff $P$ is the midpoint of $\overline{AB}$.

[hide="Construction, no proof"]Let $M$ and $N$ be the points on $\overline{D'A'}$ and $\overline{B'C'}$ such that $A'M: MD' = C'N: NB' = 1: 2$.  Then $B$ and $D$ are the points on $\overline{MN}$ such that $MB: BD: DN = 1: 3: 1$.  An identical construction gives $A$ and $C$.[/hide]

[hide="One way to prove, once you know the answer"]Mass points work nicely -- take A, B, C as your triangle to define coordinates, make D arbitrary, find the coordinates of A', B', C', D', use these to find the coordinates of M, N, and show that the result of this construction spits back out B, D.[/hide]

$\overrightarrow r_{A} = \frac 12 \overrightarrow r_{D} + \frac 12 \overrightarrow r_{D^`}$
$\overrightarrow r_{B} = \frac 12 \overrightarrow r_{A} + \frac 12 \overrightarrow r_{A^`}$
$\overrightarrow r_{C} = \frac 12 \overrightarrow r_{B} + \frac 12 \overrightarrow r_{B^`}$
$\overrightarrow r_{D} = \frac 12 \overrightarrow r_{C} + \frac 12 \overrightarrow r_{C^`}$
Replacing that we`ll get $\overrightarrow r_{A}$ $\overrightarrow r_{B}$ $\overrightarrow r_{C}$ $\overrightarrow r_{D}$ in function of $\overrightarrow r_{A^`}$ $\overrightarrow r_{B^`}$ $\overrightarrow r_{C^`}$ $\overrightarrow r_{D^`}$


algebra polynomial calculus integration induction

Find all polynomials $p(x)$ such that $1+p(x) = \frac{p(x+1)+p(x-1)}{2}$

For these type of problems, notice that it is in essence a difference equation with the integral value n replaced with the continuas variable x. So look for solutions in the form $a^x$

let note $q(X)=p(X)-p(X-1)$
the relation becomes $2+q(X)=q(X+1)$
so for each integer n we have by induction $q(n)=2n+q(0)$
so $q(X)=2X+q(0)$
after that just sum the consecutif numbers q(N),q(N-1),.....,q(0)
u'll find that p(N)=(N+1)(N+q(0))+p(-1)
solution are:
X²+AX+B: A and B are random!

Anyone try mine? No? Im fairly sure it'd work, though im too lazy to solve the difference equation, ill pop mathematica on it later

I'll do it for you.  Since all polynomials $p$ are determined by their values at $\deg(p)+1$ points, it's enough to find an infinite number of values of $p$.

Write

$p(x+1)-2p(x)+p(x-1)=1$.  The characteristic equation is $x^2-2x+1=0\implies x=1\implies$ all solutions are of the form $p(x)=a_0+a_1x$

We need a third linearly independent solution to take care of the homogeneous part - we add a term $a_2x^2$

So $p(x)=a_0+a_1x+a_2x^2$

Substituting back, we get $a_2=1$

I don't understand your method. Why do we need to add $a_2$?

Because otherwise there aren't enough solutions.

Well, so you subbed $a_0+a_1x$ back into the original equation and there aren't enough solutions so you add $a_2$, right?


i seriously think this book is too hard for my brain. im only in 7th grade and it seems that every lesson i have to ask for help. its beyond mt expectations of how hard it would be. does anyone feel the same way haha?

We need to know which book you are talking about...

I'm also in 7th grade.

um yah the book would be helpfull, algebra II no i dont feel that way, analysis-- heck yah!!!

Not really, but I'm in 7th grade and doing Algebra II Honors at the high school.

If this is an advanced class, you might want to move down a level.

And if this is a normal class, it might be embarrassing but maybe you should still move down a level if you've tried everything but nothing works.


I'm taking the ACT and was wondering how many breaks there are? When (after which sections)?

Well there are only four sections. From what I remember, one long break in the middle and two stretch breaks between the other sections.

Of course the official Web site for the ACT gives official information.

http://www.actstudent.org/faq/answers/numbquestions.html

Good luck on the test.


When I view an post in a blog, I get this above where it says Home-->Blogs-->BLOG NAME:

[quote]Warning: array_merge() [function.array-merge]: Argument #1 is not an array in /var/www/Forum/includes/class_template.php on line 366

Warning: array_merge() [function.array-merge]: Argument #1 is not an array in /var/www/Forum/includes/class_template.php on line 366
[/quote]
What is happening. I use Firefox on a Windows XP.

EDIT: I just noticed something that doesn't have much to do with this: Above where I am typing this message, it says "Font Colour: Default". This is the Canadian way of spelling color, should it be color?

I don't see any problems on your blog ...  :maybe:

Here is a screen shot. Also, it says "Have something to say?
One comment has been made" repeatedly and makes it a long page. The comments are way at the bottom.

Again, I use Firefox on a Windows XP.

Always, always include a link to the page with the problem ...

Sorry, here:

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=285311


geometry
parallelogram
trigonometry
geometry proposed

Let ABCD be a parallelogram. If a circle passes through
A and cuts segments AB, AC, and AD at points P, Q, and R
respectively, then prove that
AP × AB + AR × AD = AQ × AC.

Let AS be a diameter. As ABCD is parallelogram $ \overrightarrow{AB}\plus{}\overrightarrow{AD}\equal{}\overrightarrow{AC} \Rightarrow \overrightarrow{AB}.\overrightarrow{AS}\plus{}\overrightarrow{AD}.\overrightarrow{AS}\equal{}\overrightarrow{AC}.\overrightarrow{AS}$(. is the dot product).
But $ \overrightarrow{AB}.\overrightarrow{AS}\equal{}|AB\parallel{}AS|\cos(\overrightarrow{AB},\overrightarrow{AS})\equal{}|AB\parallel{}AP|$ and simmilary for the other two.


In a plane are considered $ n$ $ (n\geq3)$ points s.t. the number $ T(n)$ of segments of lenght $ 1$ is maximum.Prove that:
                                                      $ T(n)$$ \leq\dfrac{n^2}{3}$.


analytic geometry
graphing lines
slope
arithmetic sequence

Find a formula for the $ N^{th}$ term of the arithmetic sequence: $ \minus{}2, \minus{}8, \minus{}14, \minus{}20, .....$

[hide="Starting with term $ 1$"]
$ 4\minus{}6N$.
[/hide]

my way

[hide]

-6(N-1) - 2

[/hide]

Edit: I realized my way is the same as astro's

Okay, so my way is pretty simple. [hide="Solution"]First of all, the formula for the nth term is n= a1 + (n-1) *d

so we substitute: -2 + (n-1)*-6

-6n+6-2

= -6n+4[/hide]  :lol:

The formula for the nth term of an arithmetic sequence is:
\[ A_{n}\equal{}A_{1} \plus{} d(n\minus{}1)\]
Where $ A_{n}$ is the nth term of sequence, and $ d$ is the common difference.

you can also make a linear slope intercept method that would give you the same answer by assuming that -2 is the 1st term and -8 is the second term, we can find two points (1,-2) and (2,-8), find the slope anf y-intercept and that would be your formula.


The letters A, B, C, and D represent four distinct integers from 0 through 9. When A is added to B, the result is C. When B is subtracted from A the result is D. How many possible ways are there to assign values to A, B, C and D?

[hide] All you need to do is find values of A and B such that, when added, are no greater than 9 and when subtracted are no less than 0. So let's count up possible values for a given various values of B:
1 - 7
2 - 5
3 - 3
4 - 1

Add them together and we get $\fbox{16}$ [/hide]

[quote="humblemastermind"]0 is still a possibility so you can have 2 1 - I guess I just miscounted somewhere...[/quote]

0 is not a possibility.


inequalities
geometry
perimeter
Cauchy Inequality
inequalities unsolved

Let a,b,c>0 such that $ a\plus{}b\plus{}c\equal{}1$.Prove that
$ \frac{b^2}{a\plus{}b^2}\plus{}\frac{c^2}{b\plus{}c^2}\plus{}\frac{a^2}{c\plus{}a^2}\geq \frac{3}{4}$

[quote="natural"]Let a,b,c>0 such that $ a \plus{} b \plus{} c \equal{} 1$.Prove that
$ \frac {b^2}{a \plus{} b^2} \plus{} \frac {c^2}{b \plus{} c^2} \plus{} \frac {a^2}{c \plus{} a^2}\geq \frac {3}{4}$[/quote]
We have
\[ \frac{{b^2 }}{{a \plus{} b^2 }} \plus{} \frac{{c^2 }}{{b \plus{} c^2 }} \plus{} \frac{{a^2 }}{{c \plus{} a^2 }} \ge \frac{{\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)^2 }}{{\left( {a^4  \plus{} b^4  \plus{} c^4 } \right) \plus{} \left( {ab^2  \plus{} bc^2  \plus{} ca^2 } \right)}}
\]
Hence it suffices to prove that
\[ \frac{{\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)^2 }}{{\left( {a^4  \plus{} b^4  \plus{} c^4 } \right) \plus{} \left( {ab^2  \plus{} bc^2  \plus{} ca^2 } \right)}} \ge \frac{3}{4}
\]
\[ \Leftrightarrow 4\left( {\sum {a^2 } } \right)^2  \ge 3\left( {\sum {ab^2 } } \right)\left( {\sum a } \right) \plus{} 3\sum {a^4 }
\]
\[ \Leftrightarrow 4\sum {a^4 }  \plus{} 8\sum {a^2 b^2 }  \ge 3\sum {a^4 }  \plus{} 3\sum {\left( {a^2 b^2  \plus{} abc^2  \plus{} a^3 c} \right)}
\]
\[ \Leftrightarrow \sum {a^4 }  \plus{} 5\sum {a^2 b^2 }  \ge 3abc\left( {\sum a } \right) \plus{} 3\sum {a^3 c}
\]
Since we always have
\[ 3\left( {a^3 c \plus{} b^3 a \plus{} c^3 b} \right) \le \left( {a^2  \plus{} b^2  \plus{} c^2 } \right)^2  \equal{} \left( {a^4  \plus{} b^4  \plus{} c^4 } \right) \plus{} 2\left( {a^2 b^2  \plus{} b^2 c^2  \plus{} c^2 a^2 } \right)
\]
Therefor it suffices to prove that
\[ 3\left( {a^2 b^2  \plus{} b^2 c^2  \plus{} c^2 a^2 } \right) \ge 3abc\left( {a \plus{} b \plus{} c} \right)
\]
which obviously true.

[quote="dgreenb801"][quote="Honey_S"]Since we always have
$ 3\left( {a^3 c \plus{} b^3 a \plus{} c^3 b} \right) \le \left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^2$ [/quote]
Why? (Sorry if it's easy, I've never seen it). Also, is a+b+c=1 needed?[/quote]

It's Vasc inequality. Since it's homogenous then the condition $ a \plus{} b \plus{} c \equal{} 1$ isn't needed here.
Moreover, it's true for all $ a,b,c$ which are real numbers. For proving it we can use those below identities
\[ 2\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^2 \minus{} 6\left( {a^3 b \plus{} b^3 c \plus{} c^3 a} \right) \equal{} \sum {\left( {a^2 \minus{} 2ab \plus{} bc \minus{} c^2 \plus{} ca} \right)^2 }
\]

\[ 6\left( {a^2 \plus{} b^2 \plus{} c^2 } \right)^2 \minus{} 18\left( {a^3 b \plus{} b^3 c \plus{} c^3 a} \right) \equal{} \sum {\left( {a^2 \minus{} 2b^2 \plus{} c^2 \plus{} 3bc \minus{} 3ca} \right)^2 }
\]

We also have a similar result, which was proposed by Vo Quoc Ba Can.
Let $ a,b,c$ be the sidelenghts of a triangle with perimeter $ 1$, then prove that
\[ 1 < \frac{b}{{\sqrt {a \plus{} b^2 } }} \plus{} \frac{c}{{\sqrt {b \plus{} c^2 } }} \plus{} \frac{a}{{\sqrt {c \plus{} a^2 } }} < 2
\]

$ (\sum\frac {b}{\sqrt{a \plus{} b^2}})^2\sum b(a \plus{} b^2)\geq(\sum a)^3$
$ \Rightarrow(\sum\frac {b}{\sqrt{a \plus{} b^2}})^2\geq\frac {(\sum a)^3}{\sum b(a \plus{} b^2)}$
that $ (\sum a)^3 \equal{} \sum a^3 \plus{} 3\sum ab(a \plus{} b) \plus{} 6abc \equal{} \sum a^3 \plus{} 3\sum ab(1 \minus{} c) \plus{} 6abc \equal{} \sum a^3 \plus{} 3\sum ab \minus{} 3abc$
      $ 2\sum ab \minus{} 3abc\geq 2\sum ab \minus{} (\sum ab)^2\geq 0$
$ \Rightarrow\sum\frac {b}{\sqrt{a \plus{} b^2}} > 1$

[quote="karis"]$ (\sum\frac {b}{\sqrt {a \plus{} b^2}})^2\sum b(a \plus{} b^2)\geq(\sum a)^3$
$ \Rightarrow(\sum\frac {b}{\sqrt {a \plus{} b^2}})^2\geq\frac {(\sum a)^3}{\sum b(a \plus{} b^2)}$
that $ (\sum a)^3 \equal{} \sum a^3 \plus{} 3\sum ab(a \plus{} b) \plus{} 6abc \equal{} \sum a^3 \plus{} 3\sum ab(1 \minus{} c) \plus{} 6abc \equal{} \sum a^3 \plus{} 3\sum ab \minus{} 3abc$
      $ 2\sum ab \minus{} 3abc\geq 2\sum ab \minus{} (\sum ab)^2\geq 0$
$ \Rightarrow\sum\frac {b}{\sqrt {a \plus{} b^2}} > 1$[/quote]

It's not complicated like that, karis. Let's try another way.

honey,
while using cauchy's inequality you multiplied $ a^2,b^2,c^2$ with denominators and numerators....my question is why we cant get the desired result if we apply cauchy inequality at the original inequality..in other language how the multiplication works????please help...and when we should be inspired to apply cauchy after making such kind of little changes??

[quote="matex (L)(L)(L)"]Do there exist a easiest way to prove Vasc's Inequality?

And I have a question adressed to  Honey_S

 Could You  write some general method how did You get this?[/quote]

For more you can read the below article of Vo Quoc Ba Can.  :)

CAN YOU  say the motivation of using cauchy inequality after this change???pls pls


I'm making a website with Microsoft Front Page.  What's the easiest way to implement $\LaTeX$ on it so that I can have math equations appear on the website?

I suggest using the TeXer on this website and saving the images on your web space.

Simon is right that's the easiest way possible when working with MS Frontpage.


geometry
parallelogram
geometry unsolved

Let ABCD be a parallelogram .Suppose a line passing through C and lying outside parallelogam meets AB and AD produced at E and F respectively .Show that

                                  AC^2+CE+CF=AB.AE+AD.AF

... Maybe CE+CF -> CE*CF  :?:

If I`m right, just use stewart`s theorem to evaluate AC.


inequalities
integration
calculus
function
geometry
inequalities proposed

Ok, this one is so wierd that I'm convinced it has an absolutely easy solution:
   Let $x_1,...,x_n, y_1,...,y_n$ positive real numbers with the property that for any positive real number $t$ there are at most $\frac{1}{t}$ couples of indices $(i,j)$ with $1\leq i,j\leq n$ such that $x_i+y_j\geq t$. Prove that
   $(x_1+x_2+...+x_n)(y_1+y_2+...+y_n)\leq max_{1\leq i,j\leq n}{x_i+y_j}$

Indeed, it is sort of weird. Probably, the most straightforward solution is the following.
Let $N(t)$ be the number of pairs $(i,j)$ such that $x_i y_j>t$. Since this inequality implies $x_i+y_j>2\sqrt t$, we have $N(t)\leqslant \frac 1{2\sqrt t}$ for all $t>0$. Let $X=\max_i x_i$, $Y=\max_j y_j$. Opening the parenthesis on the left, we see that the LHS equals $\int_0^{XY}N(t)\,dt\leqslant \int_0^{XY}\frac{dt}{2\sqrt t}=\sqrt{XY}$ while the RHS is just $X+Y$. But $\sqrt{XY}\leqslant X+Y$ for all $X,Y\geqslant 0$.

[quote="fedja"]Opening the parenthesis on the left, we see that the LHS equals $\int_0^{XY}N(t)\,dt\leqslant \int_0^{XY}\frac{dt}{2\sqrt t}=\sqrt{XY}$ [/quote]

Sorry that i am dumb, may i know why LHS equals that expression?

You aren't dumb at all. This is just a general formula you might not know yet. If you have an arbitrary finite (or even infinite) sequence of positive numbers $a_1,a_2,\dots$ and, for each $t>0$, $M(t)$ is the number of elements in the sequence $a_k$ that are greater than $t$, then $\sum_k a_k=\int_0^\infty M(t)\,dt$. If you know that $a_k\leqslant A$ for some $A>0$, the integral can be replaced by $\int_0^A$ (merely because $M(t)=0$ fo $t>A$). The easiest way to explain the formula is to consider the function $f(s)=a_k$, $k-1\leqslant s< k$ and to note that $\sum_k a_k$ is just the area of the set $E=\{(s,t)\, : s>0, 0<t<f(s)\}$ (i.e., the set between the graph of $f$ and the horizontal axis). On the other hand, the area of the same set can be obtained as the integral with respect to $t$ of the length of the horizontal cross-section of $E$ at the level $t$, which is exactly $M(t)$.