AMC
USA(J)MO
USAMO
AIME
ARML
AMC 12
AMC 12 B

This poll is students who has an 200+ USAMO index, how many hous did you spend on math everyday during the past year(excluding the time spent over the summer)?

This might not be the best model to follow, but I do about two hours a week in the months before a math competition and then the week prior I jump to about 3+ hours a day. The day before it can get to 6 hours of cramming. My index is 213.

ehh my index is 204.5 so im not too sure about usamo yet.. it's not rly how long, but do you like what you do. i love solving aime problems, so i used to just print like 20 of them out and then do them when i was bored..

Haha, I might be a strange example. 9th, 10th and the beginning of 11th grade I didn't do too much math-wise (compared to my middle school days), but around January of junior year (so 2008) I did a double take and started prepping intensely for the AMC/AIME, burned myself out, and ended up with a 112.5/6. After that, I relaxed, but rediscovered the sheer fun in math and I've been messing around with various ideas since. I also helped found a math team in my high school, so I demonstrate problems there every week for a bit and have been going to meets once a month, so that probably kept me in good timed problem solving shape. Anyways, I walked in to the AIME with zero preparation this year, having done maybe .5-1 hours a day of math on average, and got an 11 (This being senior year). Figures.  :D

Edit: I had a 114 on the AMC, so that makes a 224 index. I don't do too well with tight time limits, I guess.

I went to a math summercamp, but the OP said to exclude the summer...
This year I really didn't practice math too regularly, but I've done several small competitions and went to ARML practices. I guess that put me in good shape. I don't really have the discipline to sit down and do problems on my own too much. I practiced more during previous years so I guess i'm getting the dividends now
But I'm a bad example.
If you want to get really far in math, you have to practice. That's my opinion. I find that a lot of the kids who get reaaly far, spent considerable time practicing.
My unofficial index: 114-AMC12B +10-AIME= 214

so doing math means you don't have a life?

I don't see how doing math and having a life are mutually exclusive. I am an example of how one can do both.

To be honest, i did hardly any preparation this year. I did a variety of mathematics competitions though, to compensate for my lack of practice. Proof competitions like USAMTS and the Pomona Talent search were really helpful.

[quote="DataBox"]Index: 250, Hours/Day: 0, Having a life: Priceless :!:[/quote]
This kid isn't talking about his true index, which doesn't exist since he didn't take the 12 :o

It's also pretty ironic that anyone trying to post this stuff would be talking about "having a life".

Basically math ability varies a lot among people, so some people may make USAMO after barely ever seeing any problems from AMC/AIME, and some people will have trouble even after working very hard on math for a long time. Don't feel bad or anything if other people don't spend time on math and are good. (this is to the OP)

I got index of 240 and I spent ~10 minutes studying per day =P

[quote="DataBox"]^Apparently, this kid doesn't know what he's talking about.  As I did take the 12 and my 12 index was still ~230.... :P[/quote]

Eh, in the above post you said your index was 250. Now it became 230?

254

Studied maybe around 30 minutes per day. Anybody can do reasonably well if they study efficiently and regularly.

I'm also calling bs on the indecisive 230/250-look-ma-no-studying guy who claims he has a life. Odds are, he's substituting math for Starcraft or something along those lines.

[quote="1=2"]AMC 10 index: 230.
AMC 12 index: 213.5.

I usually spend around 12-15 hours a week on math.[/quote]

WOW! that's like uhhh... 12/7.....13/7......14/7...uhhh.....3 hours a day!!! oh wait....

But seriously, that's alot...

About 3-4 hours a day, but only get 216.5 index because of the stupid mistakes...

Seriously, all the people who are claiming amazing indices without studying, no one believes you, and no one cares.  First off, even math class, no matter how pathetic it is, will keep your mind in shape to do difficult math, so everyone must do at least some math every day.  Also, some people who built up enough of a mathematical foundation in middle school probably won't need very much study time at all.  Again, no one cares about your USAMO index.  Come back in a few weeks, and tell us what you got on the actual USAMO.

probably 2 hours for me now. However, there was a time I did much more than this (but I think it's impossible to spend more than 4 hours a day on math for a high school student).

It's possible if you have free time in your schedule at school.
http://en.wikipedia.org/wiki/Academy_for_the_Advancement_of_Science_and_Technology

Someone here could spend 3 hours of their free time at school doing math and then do another hour or 2 at home.

2-3 hours, taking AIME II but got 218.0 index last year

227 index, 117/11

0-1 hours because i only go to math 3 days a week and i don't really practice

[quote="azndemon43"]Proof competitions like USAMTS and the Pomona Talent search were really helpful.[/quote] By this I hope you mean "Wisconsin Talent Search"  :D

But anyway I know I don't spend much time actually studying, but I do participate in the contests above and taking a couple old tests 2-3 days before their respective competitions

[quote="DataBox"][quote] Eh, in the above post you said your index was 250. Now it became 230? [/quote]

250 index when using the 10

roughly 230 index when using the 12...

[quote] Odds are, he's substituting math for Starcraft or something along those lines. [/quote]

I love how much you AoPS users like to assume things, let alone Starcraft (never played it in my eniter life), I haven't played a single video game since 5th grade....
I substitute math with weightlifting and water polo among other things...[/quote]
And look who's trying so hard to get recognition for his AMC10 150 faster  :roll: You even made a useless silly thread asking about why the pictures haven't been asked for yet.

His real index is 230, as amc10 doesn't exactly count for index. He just said 250 to make it sound more impressive.

You should stop going around and typing posts like "I totally failed" when you are trying really hard to brag about your score.

(lowest index in 3 years)

(way less time spent doing math as the two years before [2-3 -> 0-1?])

conclusion: positive correlation between time spent doing math and usamo index

very surprising

also stop the flame war please? thanks

[quote=serialk11r]
And look who's trying so hard to get recognition for his AMC10 150 faster  :roll: You even made a useless silly thread asking about why the pictures haven't been asked for yet.

His real index is 230, as amc10 doesn't exactly count for index. He just said 250 to make it sound more impressive.

You should stop going around and typing posts like "I totally failed" when you are trying really hard to brag about your score.[/quote]

Why doesn't AMC 10 count for index? and 250 is pretty impressive IMHO.


Suppose $ d$ is a digit. For how many values of $ d$ is $ 2.00d5>2.005$?

To be greater, the number has to be at least 2.0055. So, our possible choices are $ 5, 6, 7, 8, 9$, which is $ \boxed{5 \text{ choices}}$.

EDIT: Whoops, beaten to it.

Maybe you were thinking 10 was a digit. Count again. You should get 5.


function
Functional Equations
number theory
IMO Shortlist

Show that there exists a bijective function $ f: \mathbb{N}_{0}\to \mathbb{N}_{0}$ such that for all $ m,n\in \mathbb{N}_{0}$:
\[ f(3mn \plus{} m \plus{} n) \equal{} 4f(m)f(n) \plus{} f(m) \plus{} f(n).
\]

[quote=""ISL questions and solutions from 1995-2001" by A.Ajorloo"]
define the function $ G: 3\mathbb{N}_0 + 1\to 4\mathbb{N}_0 + 1$ as:

$ g(x) = 4f\left(\frac {x - 1}3\right) + 1$

now note that $ f$ is bijective and so is $ g$.

now let $ m,n\in\mathbb{N}_0$,we have:

$ \begin{eqnarray*}g\left((3m + 1)(3n + 1)\right) & = & g\left(3(3mn + m + n) + 1\right) \\
& = & 4f\left(3mn + m + n\right) + 1 \\
& = & 4\left(4f(m)f(n) + f(m) + f(n)\right) + 1 \\
& = & \left(4f(m) + 1\right)\left(4f(n) + 1\right) \\
& = & g(3m + 1)g(3n + 1)$

so for every $ x,y\in 3\mathbb{N}_0 + 1$ we have:

$ g(xy) = g(x)\cdot g(y)$

now notice that if we find a bijective function $ g: 3\mathbb{N}_0 + 1\to 4\mathbb{N}_0 + 1$ such that $ g(xy) = g(x)\cdot g(y)$,then the function $ f$ can be found as $ f(x) = \frac {g(3x + 1) - 1}4$.
now let $ P_1,P_2$ be the sets of prime numbers of the form $ 3k + 1,3k + 2$ respectively and let $ Q_1,Q_2$ be the sets of prime numbers of the form $ 4k + 1,4k + 3$,now we define the bijective function $ h$ from $ P_1\cup P_2$ to $ Q_1\cup Q_2$ such that it pairs every element of $ P_1$ with one element of $ Q_1$,and the same thing for $ P_2,Q_2$.

now we define $ g$ as:

$ g(1) = 1$ and for every number $ n\in 3\mathbb{N}_0 + 1$ greater than $ 1$,let $ n = \prod p_i$ ( $ p_i$'s are prime numbers),then let:

$ g(n) = \prod h(p_i)$

so obviously $ g(xy) = g(x)g(y)$,now that we have found $ g$,the function $ f$ can be easily obtained.

QED[/quote]

I found this problem to be quite interesting. Here's my solution: We use the following (well known) substitution- $$axy+x+y=\frac{1}{a}(a^2xy+ax+ay)=\frac{1}{a}((ax+1)(ay+1)-1)$$
Then the given condition can be re-written as $$f \left(\frac{(3m+1)(3n+1)-1}{3} \right)=\frac{(4f(m)+1)(4f(n)+1)-1}{4} \quad \forall m,n \in \mathbb{N}_0$$
This motivates us to consider the function $\psi:3 \mathbb{N}-2 \rightarrow 4 \mathbb{N}-3$ (where $\mathbb{N}$ represents the set of natural numbers) given by $$\psi(x)=4f \left(\frac{x-1}{3} \right)+1 \text{ and } \psi(mn)=\psi(m)\psi(n) \quad \forall m,n,x \in \mathbb{N}$$
We are supposed to show that there exists such a bijective function $\psi$. Note that we have $\psi(1)=1$. Talking in group theoretic terms, we have to show that the commutative monoids $X=(3 \mathbb{N}-2, \times,1)$ and $Y=(4 \mathbb{N}-3, \times,1)$ are isomorphic. As we have to find a multiplicative function, we think about the canonical representation of the positive integers which are $1$ mod $3$. Let $\zeta$ be an involutive function sending the set of primes to itself. Then we consider the function given below. $$\text{For } n=\prod_{i=1}^{k} p_i^{\alpha_i} \text{, we have } \psi(n)=\prod_{i=1}^k (\zeta(p_i))^{\alpha_i}$$
It's easy to see that this function is multiplicative. Also it is bijective (because $\zeta$ is bijective). So all we need to make sure is that the range of this function is $4 \mathbb{N}-3$. For this purpose, we include the following restriction on $\zeta$- $$\text{If } p \equiv a \pmod{3} \text{, then } \zeta(p) \equiv a \pmod{4} \text{ for } a \in \{-1,1\}$$
Combining all this information, we get that $X$ and $Y$ are isomorphic, as desired. Hence, done. $\blacksquare$

[hide=solution]
Solution. The first step is to notice that one can change the given relation into
$$
f\left(\frac{(3 m+1)(3 n+1)-1}{3}\right)=\frac{(4 f(m)+1)(4 f(n)+1)-1}{4}
$$
This has the advantage that after introducing the function $g: 3 \cdot \mathbb{N}_{0}+1 \rightarrow$ $4 \cdot \mathbb{N}_{0}+1, g(n)=4 f\left(\frac{n-1}{3}\right)+1,$ it becomes $g(m n)=g(m) g(n),$ which is much
easier than the initial relation. Because one can easily reconstruct $f$ from $g$ by $f(n)=\frac{g(3 n+1)-1}{4},$ the question becomes: is there a bijective multiplicative function $g$ between $3 \cdot \mathbb{N}_{0}+1$ and $4 \cdot \mathbb{N}_{0}+1,$ that is are the monoids $3 \cdot \mathbb{N}_{0}+1$ and $4 \cdot \mathbb{N}_{0}+1$ isomorphic? Let us introduce the analogous sets $T_{1}, T_{2}$ of positive primes of the form $3 k+1$ and $3 k+2 .$ In the same way as we proved that $P_{1}, P_{3}$ are infinite, you can prove that $T_{1}, T_{2}$ are infinite. Because they are clearly countable, there exists a bijection between $P_{1}$ and $T_{1}$ and a bijection between $P_{3}$ and $T_{2} .$ This gives us a bijection $\psi$ between $P_{1} \cup P_{3}$ and $T_{1} \cup T_{2}$ which maps $P_{1}$ onto $T_{1}$ and $P_{3}$ onto $T_{2}$ bijectively. Now, it is not difficult to construct an isomorphism $g:$ define $g(1)=1$ and if $n>1$ is in $3 \cdot \mathbb{N}_{0}+1$ write $n=p_{1} p_{2} \cdots p_{k}$ for some prime numbers $p_{i} \in T_{1} \cup T_{2},$ not necessarily distinct and define $g(n)=\psi\left(p_{1}\right) \cdot \psi\left(p_{2}\right) \cdots \psi\left(p_{k}\right) .$ We need to verify that $g$ is welldefined, multiplicative and bijective. First of all, note that there is an even.
number of elements of $T_{2}$ among $p_{1}, p_{2}, \ldots, p_{k}$. Then there is an even number of elements of $P_{3}$ among $\psi\left(p_{i}\right)$ and thus $g(n) \in 4 \cdot \mathbb{N}_{0}+1 .$ Thus $g$ is well-defined. Clearly $g$ is multiplicative (by the definition itself) and, using the properties of $\psi$ it is immediate to verify that $g$ is also bijective. This proves the existence of a function $f$ with the desired properties.

[url=https://artofproblemsolving.com/community/c146h150777p25259328]Hopefully correct PEN solution.[/url]

Let $g:3\mathbb N_0+1\to4\mathbb N_0+1$ be defined by $g(3n+1)=4f(n)+1$ for all $n\in\mathbb N_0$, then the assertion implies that $g$ is multiplicative. Since $f$ is bijective, $g$ is also bijective.

Let $h:\mathbb N\to\mathbb N$ be a multiplicative function defined on primes by:
If $p=3k+1$ for $k\in\mathbb N_0$, then $h(3k+1)=4k+1$. If $p=3k+2$ for $k\in\mathbb N_0$, then $h(3k+2)=4k+3$.

Restricting the domain of $h$ to $3\mathbb N_0+1$, the codomain is restricted to $4\mathbb N_0+1$, we have a suitable construction for $g$.


greatest common divisor
number theory
Euler
function
modular arithmetic
calculus
integration

How would one find the remainder upon division of $\frac{2^{2007}-2}{3}$ on 1000?

Thanks

find 2^2007 - 2 mod 1000, and since (3,1000) = 1, then you can divide by 3 as you would in normal arithmetic.

what if the gcf wasnt 1?-it would result in a decimal right?

Ok, if the GCD was $k$, then you'd need to compute the numerator mod $1000k$.

Like if you wanted to find $\frac{2^{2007}-3}{5}$ in mod 1000, you need to find $2^{2007}-3$ in mod 5000. Then, you can do:

$\frac{2^{2007}-3}{5}= \frac{R+5000n}{5}= \frac{R}{5}+1000n \equiv \frac{R}{5}$

So then  you would do $2^{2007}-3 \equiv 2^{7}-3 \equiv 125 \;\text{mod 5000}$ and then you get:

$\frac{2^{2007}-3}{5}= \frac{2^{7}-3+5000n}{5}= \frac{125}{5}+1000n \equiv \frac{125}{5}\equiv \boxed{25 \; \text{mod 1000}}$

wait but what if 2^2007-3 mod 5000 is greater than 1000?

What of it? You'll be dividing it by five, and you know its less than 5000, so your answer's gonna be less than 1000.

2^2007-2 mod 3000

Easier: Solve mod 3, then solve mod 1000, then combine the answers.

mod 1000 we have 2^10=24=2^3*3 so 2^7=3. We reduce 2^2007=2^(286*7+5)=3^286*2^5. That's a little simpler. Anyone want to take a crack at that?

Mod 3 we have 2^2=1  so 2^2007=2 so we know it is 0 mod 3. Therefore we take our answer mod 1000 (K) and if

K=1 (mod 3) take K+2000
K=2 (mod 3) take K+1000
K=3 (mod 3) take K

Why do they always have to put the year in the problems whenever they need a big number?

More: 9^3=729 so 3^7=2187 so 3^7=187 (mod 1000)

3^286*2^5=187^40*3^6*2^5

187=11*17.
187^2=(you could do this with pen & paper but I just used a calculator) 34969. 969=-31 so we can just write this

31^20*3^6*2^5

Still haven't gotten under scientific notation.

The problem with that approach is that a lot of random primes start popping up. It's more efficient to use the Euler-Phi aka "totient" function to find the period of repetition.

You'll find that you can disregard all multiples of 100 in the exponent, and that your $2^{2007}$ immediately collapses to $2^{7}$.

I see. Euler's generalization of Fermat's little theorem. It would help to write that down.

How can you use Euler's generalization to find the period in a modulo?

I've been fooling around with it, but I can't seem to understand how it would help you find 2^100 as a period of repetition

[quote="Bill The Maniac"]I'm not sure either.

$\phi(1000)=\phi(8)\phi(125)=(8-3)(125-3)>>100$

Basically, the equations is $a^{\phi(n)\equiv 1\mod n}$. The proof involves residue classes. $\phi(n)$ is the number of integers below $n$ that are relatively prime to $n$.

1
3
7
9
11
13
17
19
21
23
27
29...
251
253
257
259
261
I just found 101 numbers..

My method says $2^{1}00=2^{9}3^{1}1\equiv 264\mod 1000$[/quote]

Actually your phi function calculation is incorrect.
In fact its absurd since $\phi (n)=2k$ its always even...
But here is what you probably sould have done: $\phi(p^{n})=p^{n}-p^{n-1}$ so that $\phi(1000)=\phi(2^{3})\phi(5^{3})=100*4=400$.

However I don't think Euler's totient may be used because $gcd(2,1000)=2$ i.e. $2^{k}$ never equals $1$ because it must be even lol :roll:

In the first place, $2^{2007}-2 \equiv 2\pmod{3}$

Can you do non-integral mods? :huh:

Sure you can. $a \mod p/q=(aq\mod p)/q$. For reals $a \mod r=(a/r \mod 1)$.

However, this gives me an idea.

$2^{2007}-2\equiv 6\mod 8$
$2^{2007}-2\mod 125:$
$\phi(125)=100$
so $=2^{7}-2\equiv 126\equiv 1\mod 125$

$125\mod 8=5$ so $126=1\mod 125$, $126=6\mod 8$.

$126\mod 3=0$
$2^{2007}\equiv2\mod 3$ so $2^{2007}-2\equiv 0\mod 3$

By the Chinese Remainder Theorem, $126$ is the answer!

(Douglas Adams was only off by a factor of $3$)

$2^{100}\equiv 376$ in mod $1000$

Take a look at what happens to that:
Double that to get $752$
Double that to get $1504$
Double that to get $3008\equiv 8$

So while the $1,2,4$ mod $1000$ do not show up again, it does in fact cycle with period $100$ starting from $2^{3}$.

If I'm not mistaken, to find the period, we extract all multiples of the base from the mod number and then use Phi.
Like for powers of 2 in mod 1000, you'd do $\phi (125) = 1000$. For 2 in mod 100, $\phi (25) = 20$.
For mod 376, you'd do $\phi (47) = 46$.

If you wanted to do, say, 13 in mod 130, you'd have a period of $\phi (10) = 4$

My way was that, just a little more drawn out.

$(c^{a+b}-c^{a})\mod(c^{a}k)=c^{b}-1\mod k$

which is what I was doing.

Ah, now I understand

I was not removing all factors of the base
I was doing $\phi(5000)$ rather than $\phi(625)$
The latter gives a period of $400$ which properly reduces $2^{2007}$ to $2^{7}$

thank you

careful; the period can be less than what is given by the euler phi function

:maybe: How does that work? If you have $a^{b}\; (\text{mod}\; a^{k}c)$, then once $b>k$, wouldnt it repeat with the same period as the same sequence $\text{mod}\;c$? I'm not sure I follow...

The period must divide euler-phi period. Therefore "being careful" is not needed. The key thing is to keep the powers seperate ($2^{1}0\equiv 2^{3}$ but $2^{7}\not\equiv 1$). The easiest way to do this is probably CRT.

If the period is less, doesn't that make it easier to find?


It's simply something I really would like to know to see if there are some interesting shortcuts that can help get rid of the having to switch beween your keyboard and mouse situation which we all hate.  It can be from any OS/Application.

Here are some of mine:
Windows: "Winkey(Super)+M" = Minimize all windows.
Opera: "." = Inline text find.
Opera: "," = Inline link find.
Opera: "Shift+Arrow keys" = Navigate links/elements.
Launchy/Quicksilver: Everything. :rotfl:

I use alt-t for new tab a lot (firefox)
Well this isn't keyboard but I use
right-click left-click (fast) in firefox to go back pages (with extension all-in-one-gesture)
oh yeah
control-c
control-v

Yo kids, how do I keep certain tabs open in AoPS?
They change to default every time I look at them again for some reason.

My useful keybindings (fluxbox):
Mod1 = alt
Mod1-Q: workspace 1 (Workspaces allow you to have more desktop space - 4 copies instead of 1)
Mod1-W: workspace 2
Mod1-E: workspace 3
Mod1-R: workspace 4
Mod1-T: xterm -bg black -fg white (A terminal to do stuff)
Mod1-S: alsamixer (sound adjustments)
Mod1-F: firefox (wonder what that does)
Mod1-K: links -g (minimalistic browser for when ffox fails)
Mod1-F4: close window (yeah, I copied this from windows, w/e)

and the best of this group:
Mod1-X: maximize

Even better:
Mod1-#: switch to tab # (This works in a ton of apps - pidgin and firefox being the two that I use the most)

Yay for everything using Mod1!

alt+tab has got to be the best shortcut ever
it switches windows in windows

Ctrl T and Ctrl W, new tab, close window/tab :)

Alt+F4  :omighty:
when i want to escape...from all the windows i ve opened...Alt F4...is my  best friend

command -n(opens new window)
command -t(opens new tab)
command -q(quits)
command -x(cut)
command -c(copy)
command -v(paste)
command-option-escape(force quit)
command -p(print)

Ones I use most:

Alt+Tab: Changes between different applications
Alt+F4: Closes applications
Ctrl+C: Copies things
Ctrl+V: Pastes things

Ctrl+F is one that I like, especially for those research projects your English teacher assigns  :roll: .

It'd be cool if you had a Ctrl+F option in real life, then you'd never lose anything ever again  :P .

[quote="Hamster1800"]Even better than Ctrl-F is Ctrl-H.[/quote]

Lol what are you implying.  :P

Haha, nobody's mentioned Compiz yet!

Billy loves his Alt-[<-] backwards shortcut very much, which goes back in any browser that matters. Billy also enjoys Ctrl+Alt+[<-] and C+A+[->], which spin his cube, C+A+[down] which unfolds my cube, C+A+twofingersdown on my trackpad which zooms in, and command+space which brings up Run App on my computer. (Billy does not believe in using standard shortcuts.)

Favorite:
Ctrl+H: Replace

Very useful when dealing with long lists or reformatting something you pasted from somewhere else.

Hamster thinks that Compiz is silly and that real people find the eye candy a distraction.

Other useful keyboard shortcuts include:
F6 - selects address bar in firefox
Alt-L - xlock (locks screen so people can't use it while you're away - linux app)
Ctrl-C - The most important one ever. No, it doesn't copy.
Ctrl-Z - Also very useful
Ctrl-A D - Detatch a screen
Ctrl-Tab - switch tabs in firefox

Ctrl-N= New Window
ctrl-z= Fixing all my random errors
ctrl-t=New tab

A bit embarrasing, but sometimes I find myself hitting ctrl-z when ANYTHING goes wrong with my computer.

My favorite shortcut (the only one I use pretty much!!) is Ctrl-Enter; it autotypes the http://www..com in a web address.

Oh and backspace in a browser takes you back, which is pretty cool. I got a nice mouse and programmed a button to do forward and back so I don't use it anymore heh.

oh don't worry... i press ctrl-z whenever anything goes wrong and sometimes it fixes itself..

unless the computer froze...

once during English class our teacher's email message completely disappeared for some reason and she was panicking like crazy, and we said, "ctrl-z!!" and we saved her email..

then I typed up a nice chart for her and she still has it! (i gave it to her in october.. :D)

Rofl. The OS is not horrible, but I hate how they charge you like 500 bucks for a molded piece of white plastic with acrylic or polycarbonate over it.

Alt-F4 is godly

another good one that I use alot:
w/e you use to switch between languages. (varies from person to person)

My favorite, beastlier then Alt-F4

Ctrl-W

It closes windows, BUT IT DOESN'T SHUT DOWN YOUR COMPUTER!!! yay

Nobody mentioned Ctrl-Shift-Del?

<__<
>__>

Lol

I imagine that would be useful if you have parents that survey what you do on the internet/bypassing passwords.

[quote="woser456"]Apple H, anyone?[/quote]
 :rotfl:

I just tried it and it was neat.

Apple-Option-T.

Takes a picture instantly in Photo Booth (so you don't have to wait the normal 3 seconds).

[quote="Hamster1800"]Hamster thinks that Compiz is silly and that real people find the eye candy a distraction.
[/quote]

I'd agree many Compiz plugins are silly (fire? water? :|) But for me it's a faster way of interacting with multiple desktops and switching windows. (Also transparency can be fun/easier on the eyes, even if it silly.) Otherwise, besides faster multiple desktop interaction and expose-like functionality, which isn't really eyecandy, I don't find it useful. (Now, I'll admit, I [i]show off[/i] my eyecandy, but I don't use my eyecandy because it is distracting.)

Wow, did I forget about Alt-Tab.  It makes me look dumb. :)

Also, I would use Compiz Fusion if:
1. I wasn't too scared to boot with the Xubuntu CD on my shelf.
2. I wasn't too scared to install Xubuntu.
3. It was less unstable.
4. It served a purpose.

Also, I hate shortcuts that are not:
Ctrl: Tab/Window Level
Alt: Application Level
Ctrl+Alt: Desktop Level

Who agrees?

Macintosh- cmd alt app 8. Nobody knows how to turn it back.
PC- 6 key (for ghostfox, you need to download it for FF2)

apple:
apple q
apple w(terminal)

windows:
tab
alt-f4
ctrl-alt-delete

[quote="MustItAlwaysBeMe"]Xubuntu CD[/quote]
what is that??

[quote="woser456"]Apple H[/quote]

i'm scared to know what that does...

[quote="kimmystar94"][quote="woser456"]Apple H[/quote]

i'm scared to know what that does...[/quote]

Hides the window or application and returns you to what you were doing when you click on it

[quote="ak909090"]windows: window button+L

automatically has the same effect as selecting the "Switch User" button :D[/quote]...on XP or Vista, that is. :(

And one that I can't believe I forgot: Winkey-R.  If you have cmd in the textbox, it serves as an almost automatic command line hotkey, and if you add /(command line arguments) then you can already have your command inputed.  It's amazing.

Apple+ tab. Switch through programs....

Apple+H gave me a heart attack. I hope you're happy.  :dry:

Erm. F11, F10. And F7. And F12. I dunno if F11 does anything on Windows...(or F10 for that matter, either.


binomial coefficients
number theory unsolved
number theory

Stuff like this has been posted before (at least twice). Use the characteristic equation to find $x_n=a_1^n+a_2^n+a_3^n+a_4^n$, where the $a_i$'s are the roots of $x^4-x-1$. I think it works here as well.

[quote="payman_pm"]for sequence x(n) we have
x(0)=4;
x(1)=x(2)=0;
x(3)=3;
and  x(n+4)=x(n)+x(n+1);
prove that for every primes like P we have p l x(p); (x(p) is divided by P)[/quote]
Well, first, I recommend to you visit the following web
http://www.research.att.com/~njas/sequences/Seis.html
where u find all INFO about this sequence, but in the link sais that

\[
x_n=\alpha_1^n+\alpha_2^n+\alpha_3^n+\alpha_4^n, \quad \alpha_i^4-\alpha-1=0, \quad i=1,2,3,4.
\]

So, using this fact the [b]idea[/b] of the proof is the following:
[list]
1.  Use the fact $x_1=0$ to write $\alpha_1=-\alpha_2-\alpha_3-alpha_4$ and same with every index and take out this $\alpha_i$ in the expression of $x_n$ and then add the four expressions,
2. Expand every expressión obtaining binomial coefficients $\binom n k$,
3. In the case $n=p$ a prime, you know that $\binom p k=0 \ mod \ p$  for $k=1,2,\dots,p-1$,
4. Shows in the sums every factor is an integer collecting in a good way the four powers of $\alpha_i$ and use the algebraical properties of them, I mean:
\[\begin{array}{c}
\alpha_1+\alpha_2+\alpha_3+\alpha_4=0,  \\
\alpha_1\alpha_2+\alpha_1\alpha_3+\alpha_1\alpha_4+\alpha_2\alpha_3+\alpha_2\alpha_4+\alpha_3\alpha_4=0,\\
\alpha_1\alpha_2\alpha_3+\alpha_1\alpha_2\alpha_4+\alpha_1\alpha_3\alpha_4+\alpha_2\alpha_3\alpha_4=-1, \\
\alpha_1\alpha_2\alpha_3\alpha_4=1.
\end{array}
\]
[/list]
and I believe you that the result HOLDS. In fact the problem can generalize and the proof is more or less the same. But it's a nice exercise do all the steps of the proof once in your life.

Best Regards and good luck.

RX TCM


calculus
integration
complex analysis
function
topology
advanced fields
advanced fields unsolved

Let $ X\equal{}\{x \in R^2: 1\leq \|x\|\leq 2\}$.  Show that $ E\minus{}I$ is a boundary in $ X$, where $ E$  is the circle of radius $ 2$ and $ I$ is the circle of radius $ 1$.

[quote="LittleFermat"]Let $ X \equal{} \{x \in R^2: 1\leq \|x\|\leq 2\}$.  Show that $ E \minus{} I$ is a boundary in $ X$, where $ E$  is the circle of radius $ 2$ and $ I$ is the circle of radius $ 1$.[/quote]

I should clarify that I am not sure whether the statement is true since I am proving something and I need that result to be true  :blush:

It actually IS the boundary of $ X$... Ummm I know of a few ways of doing this, I think the way that relies on the least machinery of singular homology theory is just to interpret $ \mathbb{R}^2$ as $ \mathbb{C}$ and show that the integral along $ E \minus{} I$ of any holomorphic function vanishes (can be done using residue calculus).

Or you could just triangulate your space and then it becomes a simple computation. Your pick.

The idea is to find a $ 2$-chain $ D$ such that $ E\minus{}I\equal{}\partial D$.

I think you are confused  :wink:

Yeah, $ X$ is the 2-chain you're looking for =)

:huh: I guess $ X$ is the sum of $ 2$-simplices (since I do not see that $ X$ is a simplex).   Who are these simplices? .   Certainly this will clarify my point.

At least I can see that I was right and It was a boundary  :lol:

Thanks

Aha, is your problem that you are having trouble getting $ E \minus{} I$ as a boundary, rather than something pretty clearly homologous  to it? (Like $ E_1 \plus{} E_2 \minus{} I_1 \minus{} I_2$), where $ E$ subdivides into $ E_1 \plus{} E_2$ and $ I$ into $ I_1 \plus{} I_2$?)

Also, why the need to explicitly construct $ D$ such that $ \partial D \equal{} E \minus{} I$? The space $ X$ deformation retracts onto the unit circle; the induced map is an isomorphism on homology. The chain $ E \minus{} I$ (which is obviously a cycle) is mapped to $ I \minus{} I \equal{} 0$; and so is nullhomologous. Therefore $ E \minus{} I$ is a boundary.

I didn't get what the induced map is.   Thanks

If $ A$ and $ B$ are spaces, and $ f : A \to B$ is a continuous map, then there is an induced homomorphism $ f_* : H_*(A) \to H_*(B)$ on homology; in this case I am referring to the induced homomorphism from $ H_*(X) \to H_*(S^1)$.

In what context are you working on this problem? Is this for an intro alg. topology course?

My bad.   I didn't read carefully what you wrote.    I understand now.   Thank you very much all of you guys.


[quote="hotboy12"]a,b,c are 3 real numbers such that $ 1 \leq a,b,c \leq 2$.
Show that: $ (3a \plus{} 2b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\leq \frac {45}{2}$.
Have fun![/quote]

Lock it,please! This's problem T7/369 on Mathematics and Youth  :mad:


[quote="Line"][img]http://img176.imageshack.us/img176/7286/post2471168311416tr9.png[/img]

let $\triangle ABC$ be equilateral. If area of quadrilateral CDPE = area of triangle ABP, then find $\measuredangle \ APB$[/quote]

It's quite easy.
Hint:
[hide]
Area of $CDPE$ is equal to area of triangle $ABP$ iff $AP\cdot PB \sin\alpha=PD\cdot PE \sin\alpha+CE\cdot CE \cdot \sin 60^{o}$.[/hide]

yeah, with trigonometry all is easy, but, without trig.?

I have some idea:
these areas are equal iff $|\triangle ABD|=|\triangle BEC|$ so iff $BD=CE$.

[quote="nxxx"]these areas are equal iff $|\triangle ABD|=|\triangle BEC|$ so iff $BD=CE$.[/quote]

and how you prove that?

Triangle $ABC$ is equilateral....
so it has equal heights whence my equivalence :P

I've got $\triangle ABD \cong \triangle BCE$, but I dunno how prove it

another help to solve it?

These triangles are congruent due to equal 2 sides and one angle between them  :D .

But since we know that these triangles are congruent, the problem is obvious because then the angle between $AD$ and $BE$ is equal to angle between sides $AB$ and $BC$, hence $\angle APB=60^{o}$ (or $120^{o}$).


trigonometry
combinatorics unsolved
combinatorics

prove that amangue 4 reel nubers there exist two nubers wish satisfy
$\frac{a+b}{\sqrt{a^{2}+1}\cdot \sqrt{b^{2}+1}}$.
good luck. :lol:

$\frac{a+b}{\sqrt{a^{2}+1}\cdot \sqrt{b^{2}+1}}$what they satisfy??? :huh:

:blush:  ohh!!! sorry [the second line in the problem is]
$\frac{1+ab}{\sqrt{a^{2}+1}\cdot \sqrt{b^{2}+1}}\geq \frac{1}{2}$.
thinks drapt@ :wink:
pbornsztein :wink:

Still missing something, because it is obviously false if the four numbers are negative  :wink:

Pierre.

Denote these four numbers $x_{1},...,x_{4}$ and substitute $x_{i}=\tan \alpha_{i}$ ($\alpha_{i}\in (-\frac{\pi}{2},\frac{\pi}{2})$) so that $\frac{1+x_{i}x_{j}}{\sqrt{(1+x_{i}^{2})(1+x_{j}^{2})}}\geq \frac{1}{2}$ comes to $\cos (\alpha_{i}-\alpha_{j}) \geq \frac{1}{2}\iff |\alpha_{i}-\alpha_{j}| \leq \frac{\pi}{3}$ but that always must hold because, wlog, $\alpha_{1}\leq .. \leq \alpha_{4}$, then $\alpha_{4}-\alpha_{1}<\pi \iff (\alpha_{4}-\alpha_{3})+(\alpha_{3}-\alpha_{2})+(\alpha_{2}-\alpha_{1}) <\pi$


number theory
greatest common divisor
number theory proposed

Have a look here:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=337897#p337897]South African Mathematical Olympiad 1998 Problem 5[/url]


trigonometry
inequalities
inequalities unsolved

Find all $x$ for which all $n$

$\sin x+\sin 2x+\sin 3x+\sin 4x+...+\sin nx \leq \frac{\sqrt{3}}{2}$

Help me, please...

I'm still thinking.
[hide="My thought"]
prove this and then everything is finished

$\frac{cos\frac{x}{2}-cos\frac{(n+1)x}{2}}{sin\frac{x}{2}}\leq \sqrt{3}$[/hide]


Find all pairs $ (x,y)$ of positive integers with the following property:
    For all primes $ p$, we have  $ x~(mod~p)\le y ~(mod~p)$.

Only $ x\equal{}y$.
Obviosly $ x\le y$. If $ p|y$, then $ p|x$. If $ p|y\minus{}1$, then or $ p|x\minus{}1$ or $ p|x$. If $ p|x\plus{}1$, then $ p|y\plus{}1$.

[quote="Rust"]Only $ x \equal{} y$.
Obviosly $ x\le y$. If $ p|y$, then $ p|x$. If $ p|y \minus{} 1$, then or $ p|x \minus{} 1$ or $ p|x$. If $ p|x \plus{} 1$, then $ p|y \plus{} 1$.[/quote]
I have no idea what that means, but I doubt that it can be turned into a proof.
This is not an easy problem.

This is P.Erdos, P.P.Palfy, M.Szegedy theorem.


If :

A + B = A ( B (A^2) +10 ) / X ) + B ( ( B-1 ) / Y )

1)-Find all X and Y.
2)- If Y is a prime number find all X . ( only the positive intengers ) for X

The solution:

We have : A + B =  A ( B (A^2) +10 ) / X  + B ( ( B-1 ) / Y )
So: If ( B ( A^2 ) +10 ) / X  =1       and   ( B-1 ) / Y = 1
Then we Have A + B = A + B  so it is true
So we have X= B ( A^2 ) +10            Y = B-1        as a solution

Then we have      AB=R
So 1=R/AB          and  A+B+1=A+B+R/AB
so, A+B=A+B-1+R/AB       then: A+B = A ( 1 + R / B (A^2) ) + B ( A / AB - 1 )

So in final we have :
 A+B = A ( B ( A^2 ) + R ) / B ( A^2 ) + B ( B -1 ) / B
So we have another solution for X and Y
X= B (A^2)         Y=B           and        AB =10
So , the first case is for any number for A and B
the second case shows that A and B are any number that  AB=10

The second part : If Y is a prime number .

We knew that Y=B    and AB=10 so B must be 1 or 5     then A must be 10 or 2 .
So , :
Y=1   X=100
Y=5   X=20            that is not enough
We also have Y=B-1         and X=B (A^2) +10
in this case B=2 or 10          A=1 or 5
So, the final part is:
Y=1  X=60
Y=9  X=20

So the final solution that : X = 20 or 60 or 100

I want to know that if there is any mistake on my solution. And I want to know if it is hard or easy? ;)

[quote="J'adore les mathimatiques"]If :

A + B = A ( B (A^2) +10 ) / X ) + B ( ( B-1 ) / Y )

1)-Find all X and Y.
2)- If Y is a prime number find all X.

This problem is easy and I will post the solution after 1 days .

I want you to post your solution please.[/quote]

I, probably, do not understand your problem, but...

$X = \frac{A(B\cdot A^2 + 10)Y}{AY + BY - B^2\cdot Y + B}$, and

$Y = \frac{BX}{-AX + BX + A^3\cdot B + 10A + B^2\cdot X}$

Well let's wait for his solution so that we know what he was asking for in the problem.  :D  :D  :D

A + B = A ( B (A^2) +10 ) /[color=red] X[/color] ) + B ( ( B-1 ) /[color=blue] Y[/color] )

1)-Find all [color=red]X [/color]and [color=blue]Y[/color].  === All [color=red]X[/color] and [color=blue]Y [/color]are in color other than black
2)- If [color=blue]Y[/color] is a prime number find all [color=red]X[/color]. ===> All [color=red]X [/color]are colored  red.

Hope this color scheme will make all this  finding ver[color=blue]y[/color]  eas[color=blue]y[/color]

No no no Gyan. You got the wrong idea...
It's not about highlighting the variables to find them easier, it's about understanding the problem itself. First, find all X and Y? If solving for the variables, that should belong to the Getting Started forum. Find X for Y prime? We don't even know what's the nature of the variables A and B!!! We can't read his mind, you know.  :P

:D
The idea is how to know the relation between A and B

Thanks all for your written topics
 :D  :D  :D  :D  :D  :D  :D

This problem is definitely not fair to a lot of readers:

[quote="J'adore les mathimatiques"]If :

A + B = A ( B (A^2) +10 ) / X ) + B ( ( B-1 ) / Y )

1)-Find all X and Y.
2)- If Y is a prime number find all X . ( only the positive intengers ) for X[/quote]

You corrected the problem afterwards!!! You still didn't say what A and B are...

[quote="J'adore les mathimatiques"]The solution:

We have : A + B =  A ( B (A^2) +10 ) / X  + B ( ( B-1 ) / Y )
So: If ( B ( A^2 ) +10 ) / X  =1       and   ( B-1 ) / Y = 1
Then we Have A + B = A + B  so it is true
So we have X= B ( A^2 ) +10            Y = B-1        as a solution

Then we have      AB=R
So 1=R/AB          and  A+B+1=A+B+R/AB
so, A+B=A+B-1+R/AB       then: A+B = A ( 1 + R / B (A^2) ) + B ( A / AB - 1 )[/quote]

[quote="J'adore les mathimatiques"]The last line is wrong. Should be B ( 1 - 1 / B ).

So in final we have :
 A+B = A ( B ( A^2 ) + R ) / B ( A^2 ) + B ( B -1 ) / B
So we have another solution for X and Y
X= B (A^2)         Y=B           and        AB =10
So , the first case is for any number for A and B
the second case shows that A and B are any number that  AB=10[/quote]

Again this is only one solution by AB=R (by the way, this relation you introduced it out of nowhere  :?  ). In truth there are infinitely many solutions, which can be shown by any student who knows some linear algebra.

[quote="J'adore les mathimatiques"]The second part : If Y is a prime number .

We knew that Y=B    and AB=10 so B must be 1 or 5     then A must be 10 or 2 .
So , :
Y=1   X=100
Y=5   X=20            that is not enough
We also have Y=B-1         and X=B (A^2) +10
in this case B=2 or 10          A=1 or 5
So, the final part is:
Y=1  X=60
Y=9  X=20

So the final solution that : X = 20 or 60 or 100 [/quote]

What the heck? Didn't you say Y is prime? I see Y=1 and Y=9 there. 9 is not prime. Unless you consider 1 as prime, but 1 is not prime because that would contradict the unique factorization theorem.

[quote="J'adore les mathimatiques"]I want to know that if there is any mistake on my solution. And I want to know if it is hard or easy? ;)[/quote]

Hmm I think I did answer your question. Good day.

[quote]No no no Gyan. You got the wrong idea..[/quote]
Thanks. Hope  I have the right idea now..

[quote]If ( B ( A^2 ) +10 ) / X =1 and ( B-1 ) / Y = 1
Then we Have A + B = A + B so it is true
So we have X= B ( A^2 ) +10 Y = B-1 as a solution
[/quote] If you are ready to assume things like B-1/Y = 1 etc..
Why go through all that trouble, Let A=B=0 then X can be anything you want it to be, and so can  be Y!
[quote]If Y is a prime number find all X . ( only the positive intengers ) for X
[/quote]
As said about  X  can be 1, and 2 and 3.... so all you have to do is to get a large sheet of paper and lot of ink and start writing these positive integers down before anyone else claim those..

[quote]and I want to know if it is hard or easy?  [/quote]
Don't know, why don't you find it out? Once you do that  we will be ready for your next problem.. but please do not ask the next problem till you finish  writing all those positive integers down.
Good Luck.

Moderators:  Do the needful to keep this form  interesting.  You may  like to clean up and delete the above post too   along with other such posts or move it to appropiate thread such as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18316]this[/url]

Thanks to Gyan and 3x.lick

It is true that there are some mistakes in the problem and now I will fix it.

I missed telling about A and B .
I thougth I did the first topic  completly but I am not :D
For X as a prime I will change it sorry :D


Are there any properties of prime palindromes that would make them easier to find?

I need to write a script that finds all the prime palindromes between two given numbers.  The only problem is it has to execute in under 0.3 seconds.  The script I made works, but it takes 1-2 seconds when searching 500 thousand to one million numbers.

A good trick would be to generate a list with all the prime palindromes up to your required range, and include that in your program. Make the program just select those palidroms you directly from its own code :D

Let's see now, because we are talking about "small" numbers here, if the number of digits is reasonable enough, say at most n, then knowing that a prime palindrom doesn't end in 0,2,4,5,6,8 we know that it also cannot start with 2,4,5,6,8 so there are $\left \lfloor \frac {n-1}2 \right\rfloor + 1 =k $ digits that can take any value, so you only generate those numbers (of order $4 \cdot 9^k + 4 \cdot 9^{k-1} + \cdots + 4\cdot 9^{k'}$) and check their primality.

Before you do that you can also skip the palindroms with an even number of digits (they are divisible by 11 - I'll let you show that) and also when you generate the remaining palindroms, you store their digit sum and check if it's divisible by 3.

You will remain thus with a relatively small number of palindroms. Check for primality with primes under $\sqrt n$.

But I still think that including the primes directly in the file will get the smallest time :P

For small number of digits  there are very few such primes (about 113 less than a  million) For 10 million (or 100 million if you prefer  :D   no dice for even number of digits) it is still less than 800.. you might as well store all of them  :D

Faster way  for testing prime use SPRP method shown here:
http://www.utm.edu/research/primes/prove/prove2_3.html
(100 or more times faster than trial division)

For Curious minds to know about pi and such things..
pi = 3.14159265358979323846264338. .

The number happens to be prime:
31415926535897932384626433833462648323979853562951413

Enjoy.

Thanks that helped a lot, still cant make it though.
I also discovered that if you start at a prime and add 10 each time between  each million so like (1003001 <= x < 2000000) eventually you'll hit all other prime palindromes between those numbers.

Executing...
Test 1 OK [0.01 secs]
Test 2 OK [0.016 secs]
Test 3 OK [0.022 secs]
Test 4 OK [0.044 secs]
Test 5 OK [0.022 secs]
Test 6 OK [0.008 secs]
Execution error: Your program (`pprime') used more than the allotted
runtime of 0.3 seconds (it was stopped at 0.588) when presented with
test case 7, shown below.

----- Test Case 7 ------
7 10000000
----------------------------

Oh and generating them before hand is against the rules.

Hey also,
I cannot figure out how to program that test, do you think you could explain it to me?  Maybe I'm too tired to think.

Thanks

What are the maximum ranges of the numbers ?

Is this some kind of competition you are involved with?

I think it is from the USACO training pages.

Get a faster computer :P .

Seriously, though, try minimizing the number of stacked loop commands... that really slows it down...

[quote="SnowStorm"]Thanks that helped a lot, still cant make it though.
I also discovered that if you start at a prime and add 10 each time between  each million so like (1003001 <= x < 2000000) eventually you'll hit all other prime palindromes between those numbers.
[/quote]

I'm not quite sure what you are saying here, but if I am correct in assuming you are testing 1003001, 1003011, 1003021, ... etc, then yes, you will get all palindromes in that list, but its far too slow. This way you not only have to test whether it is a prime or not, but also whether it is a palindrome or not..
Instead, if you want to find palindromes between 1000000 and 2000000, then you just loop through 10 possibilities for the first digit (after the initial 1, I mean), 10 for the second digit, and 10 for the third digit. Then your number is of the form 1abccba1, and now all you have to do is test 1000 numbers for primality. If thats what you are already doing, then you can ignore this :)

I already presented him this method, plus a few extra optimizations (checking for divisibility with 3 and 11). This should make an acceptable time ;)

[quote="TripleM"]Then your number is of the form 1abccba1, and now all you have to do is test 1000 numbers for primality.[/quote]
Even better, as Valentin noted, every number of the form 1abccba1 is divisible by 11.

Yeah, I meant that in general. Noticed that while doing the program, it still works fine for in general though.

yea it was for usaco and the guy set the time wrong.. i was suppose to have a lot more time than 0.3 seconds lol


a,b,c>0
proove that
 \sum a/(a+ab+b) \geq  (a+b+c)/3

that sum is equal with \sum b3 / a2 + ab+b2 so we addem' up and we know that a2 + ab+b2 \leq 3/2 (a2+b2) and a3 + b3 \geq 1/2 (a+b)(a2+b2) (by Cebashev). So adding them up we obtain the required inequality.

We have

  a^3/(a^2+ab+b^2) \geq (2a-b)/3

which reduces to (a-b)^2 \geq 0

Summing three of these inequality we have a proof.

I don't think, as Valentin said, that

   "that sum is equal with   \sum b^3 /(a^2 + ab+b^2)"

I think Valentin's reasoning is like this:

If             \sum a^3/(a^2+ab+b^2) \geq (a+b+c)/3   (*)

 is true then, changing "a" with "b" must also be true

             \sum b^3 /(a^2 + ab+b^2) \geq (a+b+c)/3 (**)

and summing (*) and (**) we have

                 \sum (a^3+b^3)/(a^2+ab+b^2)  \geq 2/3(a+b+c)      (***)

So inequality(*) implies(***) but inequality (***) doesn't implies(*)

Instead if (*) is false then (***) is false but (***) is true so (*) must be true (reasoning by absurd).

Therefore (*)and(***) are equivalent after the reasoning by absurd.

Is it a correct reasoning,Valentin?

Have you a simple way to explain it, Valentin?

The inequality

  \sum  a^3/(a^2+ab+b^2) \geq (a+b+c)/3 (*)

is equivalent with(changing "a" with "b" )

  \sum b^3 /(b^2 + ba+a^2) \geq   (a+b+c)/3 (**)

Summing (*) with (**) we obtain

 \sum (a+b)/(a^2+ab+b^2)  \geq  2/3(a+b+c) (***)

Then we say that (*),(**),(***) are equivalent.

(  If (*) and (***) are equivalent then (*) ==> (***) and (***)==>(*))

But how can you prove directly that (***) implies(*)?

Surely my reasoning is wrong.
Can you please correct it?
Thank you very much.

\sumcyclica3/(a2 + ab + b2) - \sumcyclicb3/(a2 + ab + b2)
= (a - b) + (b - c) + (c - a)
= 0.

That's why those sums are equal, as Valentin said. :)

Hmmm... I seem to remember this. I'm sure it was solved here before. :D

arne you took the post right out of my mouth, .... i mean keyboard :D

This is the standard and quite well-known solution. Here's another one:

I used the (very fanous by now) lemma  \sum a^2/x >=
( \sum a)^2/ \sum x

We need to prove S= \sum a^4/(a^3+a^2b+ab^2) >= (a+b+c)/3. We use the lemma on S to get
S>=(a^2+b^2+c^2)^2/(a^2+b^2+c^2)(a+b+c) =
(a^2+b^2+c^2)/(a+b+c) >= (a+b+c)/3 (the last one is just Cauchy).


One thing:

[quote]For the AoPS WOOT program, a full refund will be granted will be given if a student chooses to drop the program before September 30, 2005.  No refund will be granted after that date.[/quote]

That should be updated.  Can you tell us where you saw that?

It was under the terms and conditions thing on the checkout for online classes.


algebra
polynomial
algebra unsolved
Solve equation

Let $P(x)=a_0+a_1x+a_2x^{2}+......+a_nx^{n}$ a polynomial of degree
$n; n\geq2$ such that:

$0<a_0<-\sum_{k=1}^{[n/2]}\frac{1}{2k+1}a_{2k}$

Prouve that $P(x)$ has a reel root $r$ in $]-1,1[$

I admit it is quite easy, but are you sure it is for pre-olympiad?

i don't know but i think i repost it in calcul and analysis

[quote="Myth"]Hmm... No I think you should ask the moderators to move it.[/quote]

i can't take delete it and post it calcul analysis??

even i'm how post the topic i can't delete it?? :?

Thus it is allowed to solve it now?

Assume that $P$ has no real root on $]-1,1[$. Thus $P$ has a constant sign on $[-1,1]$, and since $P(0) = a_0>0$, it follows that $P(x)>0$ for all $x \in ]-1,[.$
Therefore, $Q(x) = a_0x + \frac {a_1} 2 x^2 + \cdots + \frac {a_n} {n+1} x^{n+1}$ is increasing on $[-1,1].$
But $Q(1)-Q(-1) = 2 \sum_{k \geq 0} \frac {a_{2k}} {2k+1} < 0$ thus $Q(1) < Q(-1)$, a contradiction. Hence result.

Pierre.


I am very stuck on this problem! Someone please help:

THE BUCKLED RAILROAD TRACK

Railroad designers are always looking for ways to improve the safety and comfort of train travel.

Suppose technology made it possible to make and transport straight rails one mile long. If one winter two pairs of these rails were installed along a two mile portion of track in the desert, then each rail would expand approximately one foot in length in the summer heat. Suppose that instead of buckling, these rails keep their (straight) shape. In this case, the ends where they meet could jut upward to form a triangle. How high above the ground would the ends of the rails be?

Let H be the height (in feet) of the tracks above the ground where the two rails come together.

a) How large do you think H is? Is it big enough for you to stick your arm between the ground and the tracks? Is it big enough walk through? Could you drive a car under the buckled tracks? Make a guess.

b) Draw a picture and calculate H. (Note: a mile is 5280 feet long.)

c) How does your calculated value for H compare with your guess in part (a)?

Thanks a lot!

[hide="Solution"]Ok. Well, think of it like this.
Have you ever put two pieces of paper together, then pushed them so the edges were touching? Chances are you got a mountain shape with a wide base and a small height.

We're going to assume, for a second, that the ground under the tracks is flat, so that we can frame the problem without loss of generality.

If each piece of track gains a foot of length in the heat, then they are actually $5821$ feet long, each.

These two lengths form an isoceles triangle with the ground (again, assumed to be flat over a 2 mile stretch.)

The base of the triangle is the same amount of distance that the tracks initially covered: $2 \cdot 5280$ feet, or $10560$ feet.

The problem is asking you to find the height of this isoceles triangle with legs $5281, 5281, 10560$.

Proceeding from Geometry, we can see that you can form two congruent right triangles by dropping an altitude, call this altitude H, from the vertex of the triangle to the base.

We now see two congruent right triangles, and as such, we know we can use that great ole' theorem of Pythagoras:

$\left(\frac{10560}{2}\right)^{2}+x^{2}= 5281^{2}$

Solving this for X, we get
$X = \sqrt{10561}$ or roughly $x = \boxed{102.766}$ feet.

This corresponds to the height of the triangle formed when the tracks buckle.

You could DEFINITELY drive a car through that hole.
[/hide]

Your answer is correct, but $\sqrt{1051}$ is supposed to be $\sqrt{10561}.$

Please hide your solution. :wink:

use the following code:

[code][hide]STUFF TO HIDE[/hide][/code]

That simple  :)

@jbf13:  Wait a little while before asking again (at least more than 18 min).  Thanks.

He posted at 12:00 (a.m) and then posted again at 12:18...

(note:  these times are actually the time zone one hour less than eastern)

That's not what I meant... Never mind. :ninja:


Find a closed form for the infinite summation \[\sum_{k=0}^\infty k^{2}a^{k}, \ \ a \in (-1,1) \]

what if we make that S_2, then S_2-aS_2 gives us something+2S_1

S_1-aS_1=something +S_0

I realize this forum doesn't involve the use of calculus, but I found it helpful to take $f\left(a\right)=\sum a^{k}$ and look at $a\left(af'\left(a\right)\right)'$, knowing the closed form of $f\left(a\right)$.

... which is negative for positive $a$?

Anyway, I got $\frac{a(1+a)}{(1-a)^{3}}$ using the fact that $S=\sum_{k=0}^{\infty}k^{2}a^{k}= 0+\sum_{k=1}^{\infty}k^{2}a^{k}= \sum_{k=0}^{\infty}(k+1)^{2}a^{k+1}$

Then you can expand that last expression and write it in terms of $S$ and some other jazz.

[b]Hmmm...[/b]
Let $S_{n}=\sum_{k=0}^{\infty}k^{n}a^{k}$

$S_{n}$ can be written in the form $\frac{a \sum_{k=0}^{n-1}c_{k}a^{k}}{(1-a)^{n+1}}$ where all $c_{n}$ are integers and $c_{0}=c_{n-1}=1$

Find the remainder when $c_{1}$ of $S_{1337}$ is divided by $1000$.

[hide="Roadmap"]To do this, you will need to prove that $c_{1}$ of $S_{n}$ is $2^{n}-n-1$[/hide]

[hide="Okay to correct my mistake here is a full solution:"]
By computation, \[\left(\sum_{0}^{\infty}a^{k}\right)-a\cdot \left(\sum_{0}^{\infty}a^{k}\right)=\left(\sum_{0}^{\infty}a^{k}\right)-\left(\sum_{1}^{\infty}a^{k}\right)=1\\ \implies \qquad\left(\sum_{0}^{\infty}a^{k}\right)=\frac{1}{1-a}\qquad \text{valid only for }|a|<1\]
Thus, \begin{eqnarray*}\frac{d}{da}\left(\sum_{0}^{\infty}a^{k}\right)&=&\frac{d}{da}\left(\frac{1}{1-a}\right)\\ \implies \sum_{k=1}^{\infty}ka^{k-1}&=&\frac{1}{(1-a)^{2}}\\ \implies \sum_{k=1}^{\infty}ka^{k}&=&\frac{a}{(1-a)^{2}}\\ \implies \frac{d}{da}\left(\sum_{k=1}^{\infty}ka^{k}\right)&=&\frac{d}{da}\left(\frac{a}{(1-a)^{2}}\right)\\ \implies \sum_{k=1}^{\infty}k^{2}a^{k-1}&=&\frac{1+a}{(1-a)^{3}}\\ \implies \boxed{\sum_{k=1}^{\infty}k^{2}a^{k}}&=&\boxed{\frac{a(1+a)}{(1-a)^{3}}}\end{eqnarray*}[/hide]


calculus
integration
real analysis
function
real analysis unsolved

Prove that if $ f\in L^{1}(\mathbb{R})$, $ f$ is continuous at $ 0$, and $ \hat{f}\geq 0$, then $ \hat{f}\in L^{1}$.

In what follows, all integrals are computed with respect to the Lebesgue measure on the real line divided by $ \sqrt{2\pi}$, and the Fourier transform is defined by $ \hat f(t) \equal{}\int_{\mathbb R}f(x)e^{\minus{}ixt}\ \mbox dt$. A $ *$ between two functions denotes convolution, also defined with respect to the scaled Lebesgue measure.

Let $ \varphi(x) \equal{} e^{\minus{}\frac{1}{2}x^{2}}$, and for every positive integer $ n$ define $ \varphi_{n}$ by $ \varphi_{n}(x) \equal{} n\varphi(nx),\ \forall x\in\mathbb R$. For every $ n$ and every $ t\in\mathbb R$, the relation $ (*)\ \hat{\varphi_{n}}(t) \equal{}\hat\varphi\left(\frac{t}{n}\right)$ holds. Also, $ \int\varphi_{n}\equal{} 1,\ \forall n\ge 1\ (**)$.

Using $ (**)$, we get $ (\varphi_{n}*f)(0)\minus{}f(0) \equal{}\int\varphi_{n}(x)[f(\minus{}x)\minus{}f(0)]$. Now the fact that $ \varphi_{n}$ goes to $ 0$ as $ n\to\infty$ uniformly outside any neighborhood of $ 0$ and the continuity of $ f$ in $ 0$ imply that $ (\varphi_{n}*f)(0)\minus{}f(0)\to 0$ as $ n$ increases. What matters is that the sequence $ (\varphi_{n}*f)(0)$ is bounded. The functions $ \varphi_{n}*f$ are "well-behaved": they're continuous, integrable, and their Fourier transforms $ \hat\varphi_{n}\hat f$ are also integrable. Hence the Inversion Theorem holds, and $ \int\hat\varphi_{n}\hat f \equal{} (\varphi_{n}*f)(0)$ for all $ n$, forming a bounded sequence. On the other hand, $ (*)$ above shows that the sequence $ \hat\varphi_{n}\hat f$ increases pointwise to $ \hat f$, so the latter must indeed be an integrable function by the Monotone Convergence Theorem.


Find all functions $ f$:$ R\rightarrow R$
such that $ f(x^3 \plus{} y^3) \equal{} (x \plus{} y)(f^2(x) \minus{} f(x)f(y) \plus{} f^2(y))$

[quote="123eeee"]From CMO1996, we know f(rx)=rf(x)(r=p/q,p,q=1,2,3...)            (1)
f(x)>=0(when x>=0)                                                               (2)
f(x^3)=xf(x)^2                                                                       (3)
Then ,for x,y>=0,2f((x^3+y^3)/2)=f(x^3+y^3)=(x+y)(f(x)^2-f(x)f(y)+f(y)^2)>=(x+y)/2*(f(x)^2+f(y)^2)=(x+y)/2*(f(x^3)/x+f(y^3)/y)
or f(y)>=1/4(x^1/3+(2y-x)^1/3)(f(x)/x^1/3+f(2y-x)/(2y-x)^1/3),2y>=x        (4)
If there are two different c's such that f(x)=c[1]x and f(y)=c[2]y ,x,y>0
from (1) and (3) ,we know there are x,y>0 such that f(x)/x>2f(y)/y,and y=x+E where E>0 and it is really small
Take it into (4) ,we will see f(2y-x)<0 ,contradiction
So f(x)=cx (x>=0,c>=0) ,then we know c=0 or 1
Similarly for x<0
So f(x)=x or f(x)=0[/quote]

I look through the proof carefully,and I am strongly impressed by setting $ x,y$ so that $ f(x)/x>2f(y)/y$
I think it is very nice and intelligent :lol:
but I think there exists a flaw:we cannot get $ f(2y\minus{}x)<0$ if $ c_1\equal{}1,c_2\equal{}0$ am I right?
However,this flaw can be fixed anyway :)
I paste a more readable solution below :lol:


If   $ 0 < a < 1$, then

$ \frac{ \sqrt{1\plus{}a}}{ \sqrt{1\plus{}a} \minus{} \sqrt{1\minus{}a} } \plus{} \frac{1\minus{}a}{ \sqrt{1\minus{}a^2} \minus{}1 \plus{} a} \times ( \sqrt { \frac{1}{a^2} \minus{} 1 } \minus{} \frac{1}{a}) \equal{}$

Let $ a\equal{}\cos \theta \ (0<\theta <\frac{\pi}{2})$, the given equation is $ \tan \theta$, yielding the answer is $ \frac{\sqrt{1\minus{}a^2}}{a}\equal{}\sqrt{\frac{1}{a^2}\minus{}1}$ :lol:

hello kunny,
are you sure with this result? i have this simplified three times.
Sonnhard.

hello, now i have got $ \frac{1}{a}$ as the result.
Sonnhard.

hello, yes, i'm sure the result is $ \frac{1}{a}$.
Sonnhard.

Well, I get the answer to be $ \boxed {\minus{}1}$.

[hide="solution"]
$ \frac{\sqrt{1\plus{}a}}{\sqrt{1\plus{}a}\minus{}\sqrt{1\minus{}a}}\plus{}\frac{1\minus{}a}{\sqrt{1\minus{}a^{2}}\minus{}1\plus{}a}$

$ \equal{} \frac{ ( \sqrt{1\plus{}a}\plus{} \sqrt{1\minus{}a})^2 }{ (1\plus{}a) \minus{} (1 \minus{} a )}$

$ \equal{} \frac { 1 \plus{} \sqrt {1 \minus{} a^{2}}}{a}$

Therefore, the given expression is equal to

$ \equal{} \frac { 1 \plus{} \sqrt{1 \minus{} a^2}}{a}  \times  (\sqrt{\frac{1}{a^{2}}\minus{}1 }\minus{}\frac{1}{a})$

$ \equal{} \frac{(1\minus{}a^2)\minus{}1}{a^2} \equal{} \boxed{\minus{}1}$

[/hide]

That's wrong.

Try $ a \equal{} \frac {1}{2}$, you will obtain the value $ 2$.

You apparently don't know the usual order of operations?

[quote="harekrishna"]What's wrong with my solution i.e. where does the mistake lie?[/quote]
Multiply first, then add   :P


[quote="hm29168"]If you  have a perfect solution and explain it well, the graders should give it full credit. I don't believe repetitive casework is a problem, but if the formatting gets in the way of understanding what you are talking about, I believe that graders have a 1 point discretionary point for poor writing. However, if you explain yourself fine and just have wonky paper formatting, I think you should be okay.[/quote]

I think its decent writing, its just that it's my first time working with LaTeX, and my spacing is kind of weird.  The larger symbols cause them to be big at sometimes and not at others, if you know what I mean.

[quote="AdithyaGanesh"]Say you have a perfect solution, but your formatting is hard to read.  Can you get points taken off for that?

Also, say you have a perfect solution, but say you're doing a bunch of casework.  Say you are very repetitive in the way you present this information.  Can points be taken off?  Thanks.[/quote]

1) No, unless the formatting makes it very hard to read

2) Maybe, if it gets really long

If you have an extremely poorly written solution, the graders can take off a point.  Repetitive casework would probably not get points taken off as long as they can easily read it.  Long solutions are probably also fine (though concise and elegant is usually better).

1) The way you can avoid all of this is to make sure you write your solutions slowly and carefully, then you don't have to worry about points being docked for handwriting.  :)

2) No points should be taken off for repetitive casework. I'm not a USAMTS grader, but it seems like their grading format is quite rigorous.  Either you have the parts of the proof / justification that they are looking for or you don't.


rotation
symmetry
geometry
3D geometry
sphere
trigonometry
geometric transformation

There are some questions from oscillations that I found interesting, so here they are:

[b]1.[/b] A masless, inelastic string is wrapped around a fixed pulley. One end of the string is attached to a body of mass $m_1$. The other end is attached to a spring of elasticity $k$, with a body of mass $m_2 > m_1$ attached to the other end of the spring. The stretching of the string itself is negligible, and the length of the unstretched spring is $\ell_0$. Determine the amplitude of harmonic oscillations of this system when it is released.

[u]My comments:[/u] Determining the [i]period[/i] of small oscillations should be rather easy: I got $T = 2\pi \sqrt{\frac{m_1m_2}{k(m_2-m_1)}}$ (I may be wrong, though). As for the amplitude, I do have a complete solution, but I really don't like the method used -- it's quite strange, at least to me. I've posted this problem to see if someone might come up with a good idea.

[b]2.[/b] A hollow cyllinder of radius $R$ and mass $M$ is free to rotate about its (fixed) horizontal axis of symmetry. Another hollow cylinder (its axis of symmetry also being horizontal), of radius $r$ and mass $m$, rolls without slipping on the inside of the first cylinder. Find the period of small oscillations of this system.

[u]My comments:[/u] Well, I also have the full solution to this problem, but I simply can't find out why I'm doing it wrong  :? Expressing the non-slipping condition through equations perpetually gives me an incorrect formula and I just don't know why -- and I know its wrong, since my solution doesn't behave well in limiting cases such as $M\gg m$ :o I was hoping a fresh solution would clear things up ;)

oh man, I am so out of practice, but here goes my take on #2.

I first tried to reason out the no-slip condition:

$v_1 - r \omega_1 = R \omega_2$

where the $1$ subsripts correspond to the little cylinder and the $2$ subscripts correspond to the big cylinder.

Differentiating, we get

$a_1 - r \alpha_1 = R \alpha_2$.  (1)

Now we apply the torque equation to the big cylinder:

$F_f R =MR^2 \alpha_2$

solving for $\alpha_2$ in (1), and substituting above, we get

$F_f = (Ma_1 - Mr \alpha_1)$ (2)

Now we apply torque to the smaller sphere, since friction acts to produce an opposite torque as that found in the larger sphere, we get

$-F_f r = mr^2 \alpha_1$

solving for $\alpha_1$ in terms of $a_1$, we get

$a_1 = \frac{M-m}{M}r \alpha_1$ (3)

Finally, we apply linear forces on the smaller cylinder:

Taking the direction gravity acts to be positive, we get

$mg \sin \theta - F_f = ma_1$

substituting (3) for $a_1$, using the small angle approximation, and (2) for $F_f$, we get  (after many algebraic errors on my part and confusion about signs)

$mg \theta - Ma_1 + Mr \alpha_1 = ma_1$

this is nice, because then you get

$mg \theta = \frac{-rm^2}{M} \alpha_1$

and

$\alpha_1 = - \frac{Mg}{mr} \theta$

Now we have to find what $\theta$ is in terms of $\theta_1$.

Even though an outside force acts on the system, no outside torques act, so angular center of mass is conserved. Since both cylinders have 0 angular displacement at time 0, their displacements are related by:

$mr^2 \theta_1 = -MR^2 \theta_2$

And... I've gotta go fail ap chem, so ill just put up the answer and finish this when i get home:

EDIT 2
I made a mistake with signs, so my initial no-slip condition is off. After setting counterclockwise to be positive rotation and left to be the postive direction, I get this to be the period

$T = 2\pi \sqrt{\frac{R^2}{rg} (\frac{2M + m}{mr + MR})}$

I disagree on the period for number 2, but I think it may be because I'm reading the problem slightly differently and I'm not sure if that makes a difference. But I don't think your units check out, psychicalchemist, since your period is in $\frac{s}{m^\frac{1}{2}}$, although I might just be missing something there.

I say that you get

$-F_f r = mr^2 \alpha_1$

$mg \sin \theta_2 - F_f = ma_1$

$mg \sin \theta_2 = 2mr \alpha_1$

So using small angle approximation, $g \theta_2 = 2r \alpha_1$

But what we're really interested in is $\alpha_2$ so we need to find that. It's simple geometrically, though:

$\theta_1 r = \theta_2 (R - r)$ because the smaller mass traces out a circle of radius $(R - r)$, and differentiating we get $\alpha_1 = \frac{R-r}{r} \alpha_2$

Now we can say that

$g \theta_2 = 2r \frac{R-r}{r} \alpha_2$

$\frac{g}{2(R - r)} \theta_2 = \alpha_2$

$T = 2\pi \sqrt{\frac{2(R - r)}{g}}$

I'm not really sure though since I'm not really doing much with the outer cylinder, and I'm definitely a bit iffy on the way I used the rolling w/o slipping condition.

Let's see...

@[b]cuong[/b]: Yes, that is the correct answer for the first problem. How is it obtained in your book?

@[b]psychicalchemist[/b] & [b]blendecho[/b]: I'll post a detailed post tomorrow about the second problem ;) Btw, this problem is from David Morin's book which has been advertised even here, so if you want you may download it and see the solution for yourself...

Ah, I'll be using Morin's book next year for physics class.

Wow! I started typing my solution when I realized where I went wrong! :lol: :D

Here's the correct solution (taken more or less exactly as it is in the book):

Let $F$ be the force of friction between the cyllinders. Then the equations of their rotation would be $F = mr\ddot \phi_1$ and $F = - MR\ddot\phi_2$, where $\phi_1$ and $\phi_2$ are the angular displacements of each cyllinder with respect to their centers, and the counterclockwise direction is taken to be positive. Let $\phi$ be the angular displacement of the smaller cyllinder with respect to the center of the larger cyllinder. The equation of the smaller cylinder's translation is $F - mg\sin\phi \approx -R\ddot\phi$. The non-slipping condition is $R\ddot\phi \approx R\ddot\phi_2 - r\ddot\phi_1$. Note that in these equations we have used $R\gg r$ which isn't specified in the text -- my mistake! -- but this fact practically doesn't change anything in this problem except for an occasional simplifying. Now, expressing $F$, $\ddot\phi_1$ and $\ddot\phi_2$ through $\ddot\phi$ we get a nice equation of harmonic oscillatory motion with a period of $T = 2\pi\sqrt{R/g}\sqrt{(m+2M)/(m+M)}$.

Now, I made the error of not noting the positive direction! This is the first time my equations yielded a wrong solution due to not paying attention to the direction. Well... I've learned my lesson ;)

So, [b]psychicalchemist[/b], you nearly got it right...

Any offers for the first problem?

to Djole :i try to solve it  and i got the answer , it is wery hard one , i used central point to solve
wait for me , i will study more about LATEX :D

I solved it for stationary outer cylinder it seems :-p

Nice solution Djole.

Yeah, my solution was pretty similar to that except I accidently said

$-MR^2 \alpha_2 = mr^2 \alpha_1$

gotta watch my algebra..., that killed me on the physics olympiad.

Djole, i did it again without the assumption that $R >> r$ and got a different answer that doesnt reduce to your answer:

$T = 2 \pi \sqrt{\frac{M (R-r)}{g (M + m)}}$

this seems to work in limitting cases, since if $M >> m$ and $R >> r$, the larger cylinder doesnt move when the smaller cylinder (now a point) rolls, so the equation should simplify to the same equation as when a pendulum bob swings on a rope. My equation does this, but the one you posted doesnt. Maybe I am wrong in supposing that this occurs, since I might be wrong in assuming friction doesnt slow down the period in such a system.

[color=blue]Ok, now I post the solution to your problem (base on the idea of [b]Cuong[/b])
The center of the system will move with the acceleration $\large a=\frac{m_2-m_1}{m_1+m_2}g$
In the reference frame connect with the center of the system, the force acts on $m_1$ and $m_2$ is
$\vec{F_1}=-m_1\vec{a}$ and $\vec{F_2}=-m_2\vec{a}$
In this frame, each mass will do a SHM around the center of system with the spring constant $\large k_1=\frac{m_1+m_2}{m_2}k$ and $\large k_2=\frac{m_1+m_2}{m_1}k$
So, the period of each mass is $\large T=2\pi\sqrt{\frac{m_1m_2}{k(m_1+m_2)}}$
At the equilibrium point, we have, $F_1=k_1A$

$\large \Longrightarrow A=\frac{m_1a}{k_1}=\frac{m_2-m_1}{(m_1+m_2)^2}\frac{m_1m_2}{k}g$
Do the same to with $m_2$ and $k_2$, we get the same answer $A$
That's the answer of the problem.[/color]

Ok, thanks!!! I think that's what the solution I have was trying to do too, but it was so obfuscated that I couldn't possibly understand it...

sorry for my late:
phucvn87 is very intelligent to creat such a wonderful solution

here is mine:

take m1 as a frame:

so $m2g-kx=m2a2$
take m2 as a frame:

so $-m1g-kx=m1a1$

note that $a=a1+a2$

sum both side of the equation:

so $kx(\frac{1}{m1}+\frac{1}{m2})=a$  1

but phucvn87 prove that $T=\sqrt{\frac{m1m2}{(m1+m2)k}}$

so the effective mass would be $m=\frac{m1m2}{m1+m2}$

from 1 we get  $\frac{a}{kx}=\frac{1}{m}$

so we can image a system with spring mass and is hung horrizontally

with this imaginative system we have:

$F=ma$ and $F=KA$

but we have $a=\frac{m2-m1}{m2+m1}g$

and m is effective mass of this system

and so $A=\frac{ma}{k}=\frac{(m2-m1)m1m2g}{(m1+m2)^2k}$ :P

this way of solving problem is created by me several time ago :blush:  by the ideas of my teacher :PRO BUI QUANG HAN.

after three months i realized that our solution to proplem 1 is wrong

it 's a big trouble.

and i tried to correct it but fail.So this is another approach :(

it is easy that we have

$a12=-kx(\frac{1}{m2}+\frac{1}{m1})$

it tells us that the period is $T=2pi\sqrt{\frac{m1m2}{k(m1+m2)}}$

to find A we must have a12

but we have

m1Vector(g)+vector(F)=m1vector(a1)

m2vector(g)-vector(F)=m2vector(a2)

sum both side we get

the acceleration of central mass is g
and

$ac=\frac{m1a1-m2/a2/}{m1+m2}$

so $ac1=\frac{-m2(/a2/+a1)}{m1+m2}$

in the same way
we will get
$m1 ac1=m2 ac2$

so

$a12=ac1\frac{(m1-m2)}{m2}$

and $a12=-(/a2/+a1)(\frac{m1-m2}{m1+m2}$

but
$g= \frac{m1a1+m2a2}{m1+m2}\geq(/a2/+a1)$   (tsebychev)

when m is at the farest  distance the acceleration is max
so /a2/+a1=g

and $a12=g\frac{m1-m2}{m1+m2}$

so

A=ma/k

and
$A=\frac{m1m2(m1-m2)g}{(m1+m2)^{2}k}$ :lol:

i do it in my lecture theatre in NTu so may be it will have something wrong in the sign :oops:


algebra
polynomial
binomial theorem
calculus
calculus computations

Find the coefficient of $ x^i$ in $ e^x.\frac {d}{dx^k}\left(e^{ \minus{} x}x^k\right)$.

I found so far that when i=0, we have this coefficient equal $ k!$, and when $ i > k$, we have zero.

The general case I have no idea. How would you solve this? Or do you have another way of showing that
\[ \sum_{k \equal{} 0}^{n \plus{} 1} \binom{n \plus{} 1}{k} \frac {( \minus{} 1)^k}{k!}e^x.\frac {d}{dx^k}\left(e^{ \minus{} x}x^k\right)\ \equal{} \ x^{n \plus{} 1}\ ?
\]

The coefficient of $ x^i$ ($ i\le n$) seems to contain the binomial expansion of $ (1\minus{}1)^{n\plus{}1\minus{}i}$ as a factor, but I can't prove this. :( Any help is welcome.

There is a link to the binomial theorem: given a polynomial $ P$, we have $ e^x[P(x)e^{\minus{}x}]^{(n\plus{}1)}\equal{}e^x[(P^\prime (x)\minus{}P(x))e^{\minus{}x}]^{(n)}$, which leads to
\[ e^x[P(x)e^{\minus{}x}]^{(n)}\equal{}\sum_{k\equal{}0}^n (\minus{}1)^{n\minus{}k}\binom{n}{k} P^{(k)}(x)\]


calculus
integration
trigonometry
calculus computations

Let $ x = \tan \theta,\int_0^1 \frac {\tan ^ { - 1}x}{(1 + x^2)^{\frac 32}}\ dx = \int_0^{\frac {\pi}{4}} \theta \cos \theta \ d\theta$

$ = \left[\theta \sin \theta + \cos \theta \right]_0^{\frac {\pi}{4}} = \frac {\pi}{4\sqrt {2}} + \frac {1}{\sqrt {2}} - 1$.

hello, better is the following form of the result
$ \minus{}1\plus{}\frac{\sqrt{2}}{2}\plus{}\frac{\sqrt{2}\pi}{8}$.
Sonnhard.

hello, just rationalizing the denominators is better for further computations.
Sonnhard.


analytic geometry
Pythagorean Theorem
geometry
AMC

Two parallel chords in a circle have lengths 10 and 14, and the distance between them is 6. The chord parallel to these chords and midway between them is of what length?

I had a problem just like this in my PM contest but I changed the answers around.  All I did was use abc/4R=[ABC] to find out the radius, and then I put it on a coordinate plane.  It was easy from there.  I remember it was on an AMC if I'm not mistaken.  I'll do it anyways but with a different method.

x - distance from center to chord of length 10
y - distance from center to chord of length 14

Using the Pythagorean theorem and the property of radii lengths being equal, x :^2: + 25 = y :^2: + 49 which becomes x :^2: - y :^2: = 24.  The two chords can either be in the same half of the circle or in different halves.  If the two chords are on the same side, x - y = 6.  If they are on different sides, x + y = 6.  If x - y = 6, then x + y = 4, meaning that x = 5, but y can't be negative.  On the other hand, x + y = 6 and x - y = 4 and x = 5 while y = 1.  Solving for the radius, we get :sqrt: 50 or 5:rt2:  The chord midway between the two mentioned chords has to be 2 units away from the center.  :sqrt: (50-4) = :sqrt: 46, but that's only half of the chord.  The chord length is 2:sqrt: 46.


Let $x_1=1$ and $x_{n+1}= \frac{x_n}{n}+\frac{n}{x_n}$ prove that $[x^2]=n$ for all $n \geq 4$

It's quite a hard problem.

Try to prove that $\sqrt{n}\leq{x_n}\leq\frac{n}{\sqrt{n-1}}$.


LaTeX induction ratio factorial USAMTS AMC

The sequence $ \{a_n\}$ is defined by
\[ a_0 \equal{} 1,a_1 \equal{} 1, \text{ and } a_n \equal{} a_{n \minus{} 1} \plus{} \frac {a_{n \minus{} 1}^2}{a_{n \minus{} 2}}\text{ for }n\ge2.
\]The sequence $ \{b_n\}$ is defined by
\[ b_0 \equal{} 1,b_1 \equal{} 3, \text{ and } b_n \equal{} b_{n \minus{} 1} \plus{} \frac {b_{n \minus{} 1}^2}{b_{n \minus{} 2}}\text{ for }n\ge2.
\]Find $ \frac {b_{32}}{a_{32}}$.

[hide="how i did it (might be wrong)"]

I computed the first few terms of the sequences, and got

$ \frac {b_0}{a_0} \equal{} 1$

$ \frac {b_1}{a_1} \equal{} 3$

$ \frac {b_2}{a_2} \equal{} 6$

$ \frac {b_3}{a_3} \equal{} 10$

So I made a conjecture that

$ \frac {b_n}{a_n} \equal{} \frac {(n \plus{} 1)(n \plus{} 2)}{2}$

Plugging 32 in, we get $ \frac {b_{32}}{a_{32}} \equal{} 33\cdot17 \equal{} \boxed{561}$

[/hide]

Edit: messed latex up

[hide]It is easy to prove by induction that $ a_n\equal{}n!$ and $ b_n\equal{}\frac{(n\plus{}2)!}{2}$.[/hide]

[hide="something i kinda figured out AFTER i took the test"]
Looking at sequence $ \{a_n \}$, we can divide by $ a_{n \minus{} 1}$, getting $ \frac {a_n} {a_{n \minus{} 1}} \equal{} 1 \plus{} \frac {a_{n \minus{} 1}} {a_{n \minus{} 2}}$. Let $ r_k \equal{} \frac {a_k} {a_{k \minus{} 1}}$; we get $ r_n \equal{} r_{n \minus{} 1} \plus{} 1$, so the ratio between consecutive terms increases by one. This implies a factorial sequence, and we can see by induction that $ a_n\equal{}n!$. For sequence $ \{ b_n \}$, it's the same thing, except the ratio starts at 3 instead of 1, so $ b_n \equal{} 3*4*5.....(n \plus{} 2) \equal{} \frac {(n \plus{} 2)!} {2*1} \equal{} \frac {(n \plus{} 2)!} {2}$. so $ b_{32} \equal{} \frac {34!} {2} \equal{} \frac {33!*34} {2} \equal{} 33!*17$.
$ \frac {b_{32}} {a_{32}} \equal{} \frac {33!*17} {32!} \equal{} 33*17 \equal{} \boxed{561}$.[/hide]

I figured out the problem and concluded that the answer is (index-2) choose 2 instead of (index+2) choose 2.... Really really stupid mistake --- especially considering that I already checked... man i wish i got a time machine...

[hide="Thoughts"]I thought this was a sort of silly problem - it's basically pattern recognition. Nothing else. I can't come up with a way to prove the conjecture other than induction anyway, so I didn't really like this problem.

Edit: abacadaea's solution is really nice, didn't think of dividing by $ a_{n\minus{}1}$.[/hide]

[hide="Answer"]
I can say that this was the easiest problem on the test. Testing out the first few terms of $ a$, we can see that the term $ a_n$ is just $ n!$. For sequence $ b$, we can see that it is the same, except it starts at $ 3$. So, we have $ \dfrac{(n\plus{}2)!}{2}$. Dividing, we have $ \dfrac{17\times 33!}{32!}\implies\boxed{561}$.[/hide]

[quote="Mersenne"]I can say that this was the easiest problem on the test.[/quote]

Completely disagree.  #3 and #5 were easier - #5 took me perhaps 30 seconds, as I quickly recognized the similarity to one of last years USAMTS problems.

Let $c_n=\frac{a_{n+1}}{a_n}$. The first equation divided by $a_{n-1}$ implies that

$c_n=1+c_{n-1}$.

$c_0=1$, so $c_n=n+1$, the rest is easy.

I think this trick comes up often enough.

after i eat i'll do writeups for the hint list stuff i did today

p8: Computing a few terms for $a_n$ we get $1, 1, 2, 6, 24 \ldots$ and it's clear the $n$th term is $n!$. Similarly computing a few terms for $b_1$ gives $1, 3, 12, 60$ and it's clear the $n$th term is $\frac{(n+2)!}{2}$. The answer is hence $\frac{\frac{34!}{2}}{32!} = \frac{34 \cdot 33}{2} = 33 \cdot 17 = \boxed{561}$.


Could someone explain Power of the Point to me?

[url]http://www.cut-the-knot.org/pythagoras/PPower.shtml[/url]

This website kind of explains it.

obviously you've seen the theorems. they work because of similar triangles. remember that two angles inscribed in the same arc are congruent, and find some similar triangles. the ratios the triangles give you are equivalent to the power of a point theorem equations.

I have read through the Power of the Point section of the AoPS Volume 1 Book, and it gives a complicated(to me) explanation.

Basically draw all the lines connecting the points and look for similar triangles.


geometry
vector
AMC
AMC 12
AIME
analytic geometry
3D geometry

Aidan, please dont post the same topic twice. Its just spam. All we need is ONE, kay?

I don't remember asking this question before? I only asked where i could learn this 3D analytic geometry, not whether it was important for the AMC. Sorry if that's considered a reposting of the same topic  :)

Well, a mod might have probably deleted the other one of this same topic

[quote="aidan"]Is 3D analytical geometry (e.g. vector multiplication, equations of planes, etc) important for the AMC12?[/quote]

Vector multiplication (and vectors in general) isn't absolutely required for solving problems, but if they have #25 as a 3D geometry problem, it can most probably be solved using analytic geometry. These topics are usually reserved for the AIME, and are seen (if at all) among the harder (or hardest, ie 23-25) on the AMC.

Ok thanks.. Then i'll invest a little time in it during the December holidays. gauss1181, i may  :P  have double-clicked on "Post", sorry if i did  :P


logarithms
inequalities
inequalities unsolved

notice that $f(x) = \log_x k$, where $k > 1$ is constant, is decreasing on $(1, \infty)$, and it's increasing on $(1, \infty)$ if $k < 1$.

then since $a < b$, $\log_a b > 1$, and therefore $\log_a \log_a b > \log_b \log_a b$.  and since $\log_c a < 1$, $\log_b \log_c a < \log_c \log_c a$.

so the left side of the given inequality is $> \log_b \log_a b + \log_b \log_b c + \log_b \log_c a = \log_b [\log_a b \log_b c \log_c a] = \log_b 1 = 0$


Pythagorean Theorem
geometry
power of a point

A line is drawn through  a fixed point $ P(\alpha,\beta)$ to cut the circle $ x^{2}\plus{}y^{2}\equal{}r^{2}$ at $ A$ and $ B$,Then find $ PA.PB$  in terms of the given variable.

I assume you meant $ PA*PB$.

[hide="solution"]
By Power of a Point, the desired product is the same as square of the length of the tangent from $ P$ to the circle. Let one of the tangents be $ PT$, and let $ O$ be the origin.  From the equation, the circle is centered at $ O$.

Now, since radii are perpendicular to tangents, we have $ PTO$ is a right triangle with hypotenuse $ PO$. $ |PO|\equal{}\sqrt{\alpha^{2}\plus{}\beta^{2}}$, and $ |OT|\equal{}r$.  By Pythagorean Theorem, $ |PT|^{2}\equal{}|PO|^{2}\minus{}|OT|^{2}\equal{}$[b]$ \alpha^{2}\plus{}\beta^{2}\minus{}r^2$[/b], which is the desired answer.
[/hide]


function
limit
trigonometry
calculus
derivative
real analysis
real analysis unsolved

I have my exam for Analysis I in a few days. We've covered properties of the real numbers, sequences, continuity and the derivative. I've exhausted the questions in my textbook and looked through this forum for some good exercises, but if you all could think of any reasonable exercises for me to chew on it would be much appreciated  :D .

Prove that the convergence of $ \sum a_n$ implies the convergence of $ \sum\frac{\sqrt{a_n}}{n}$ if $ a_n\geq$.

This isn't quite what I was looking for, but similar enough.

Suppose $ f$ is a real function defined on $ \mathbb{R}$ which satisfies $ \lim_{h\to 0}[f(x\plus{}h)\minus{}f(x\minus{}h)]\equal{}0$ for every $ x\in\mathbb{R}$. Does this imply that $ f$ is continuous?

My analysis I exam was very counterexample based:

What properties can you use these functions for in counterexamples?

$ f(x)\equal{}\begin{cases} x^2\sin(1/x) ~\text{if} ~ x\neq 0 \\
0 ~\text{if} ~x\equal{}0\\ \end{cases}$

Actually, I just realized I have to run. So only one counterexample for now.

[quote="HilbertThm90"]Suppose $ f$ is a real function defined on $ \mathbb{R}$ which satisfies $ \lim_{h\to 0}[f(x \plus{} h) \minus{} f(x \minus{} h)] \equal{} 0$ for every $ x\in\mathbb{R}$. Does this imply that $ f$ is continuous?[/quote]

No, the function $ \frac{1}{x^2}$ is a counterexample.

As for the $ x^2 \sin(\frac{1}{x})$ example the derivative exists at 0 but is not continuous at 0. You can show from the definition of the derivative that f'(0) exists, but if you take $ x_n \equal{} \frac{1}{\pi n}, y_n \equal{} \frac{2}{\pi n},$ then $ \lim_{n \to \infty}x_n \equal{} \lim_{n \to \infty}y_n \equal{} 0, \lim_{n \to \infty}f'(x_n) \ne \lim_{n \to \infty}f'(y_n)$, so f'(x) is not continuous at 0.


algebra
polynomial
calculus
calculus computations

(1) Let $ l,\ m,\ n$ be positive integers such that $ _{l\plus{}n}C_l>_{m\plus{}n}C_{m}$. Examine the size of $ l$ and $ m$.

(2) Let $ R(x)$ be the remainder when polynomial with respect to $ x,\ F(x)$ is divided by $ (x\minus{}a)^3$. Express $ R(x)$ in terms of $ F(a),\ F'(a),\ F''(a).$

for 2) we can write $ F(x)\equal{}\sum_{k\equal{}0}^{n} \frac{F^{(k)}(a)}{k!}(x\minus{}a)^k$.
so $ R(x) \equal{} \sum_{k\equal{}0}^{2} \frac{F^{(k)}(a)}{k!}(x\minus{}a)^k$

[quote="anik_andrew"]for 2) we can write $ F(x) \equal{} \sum_{k \equal{} 0}^{n} \frac {F^{(k)}(a)}{k!}(x \minus{} a)^k$.[/quote]

Why $ F(x)$ can be witten as such?

[quote="kunny"](1)  Let $ \ell ,\ m,\ n$ be positive integers such that $ _{\ell \plus{} n}C_ \ell > _{m \plus{} n}C_{m}$. Examine the size of $ \ell$ and $ m$.[/quote]
\[ _{\ell \plus{} n}C_ \ell > _{m \plus{} n}C_{m}\Rightarrow \prod_{k \equal{} 0}^{n \minus{} 1}\frac {\ell \plus{} n \minus{} k}{m \plus{} n \minus{} k} > 1\Rightarrow \ell > m\]

[quote="kunny"]

Why $ F(x)$ can be witten as such?[/quote]
Taylor's formula for polynomial.... isn't it??  :?:

Regreattbly, Japanese high school students don't study Taylor expansion.

ok ;)
Let's expand $ F(x)$ by powers of $ (x\minus{}a)$. I mean $ F(x)\equal{}\sum_{k\equal{}0}^{n} c_k (x\minus{}a)^k.$
So, what $ c_k$ are ?
$ F^{(p)}(x) \equal{} \sum_{k\equal{}p}^{n} k(k\minus{}1)...(k\minus{}p\plus{}1) c_k (x\minus{}a)^{k\minus{}p}$, and it  follows that
$ F^{(p)}(a) \equal{} \sum_{k\equal{}p}^{n} k(k\minus{}1)...(k\minus{}p\plus{}1) c_k (a\minus{}a)^{k\minus{}p} \equal{} p! c_p$. and we get $ c_p\equal{}\frac{F^{(p)}}{p!}.$ Thus
$ F(x)\equal{}\sum_{k\equal{}0}^{n}\frac{F^{(k)}(a)}{k!}(x\minus{}a)^{k}$


a,n and m are integers,and HCF(n,m)=1

m^2=an

Prove that the only possible values of n are 1 and -1.

I know very simple but how do you explain it well.

By HCF do you mean the GCF, or greatest common factor?
You can say that they are "mutually prime" if they share no common divisors except for one.

$M^{2}=AN$
$N = \frac{M^{2}}{A}$
I don't know where to go from here.
See if this helps anyone at all.

[hide]From $n \mid m^{2}$ we have that all divisors of $n$ belong also to $m^{2}$, but from $\gcd(n, m) = 1$ we have that $n$ and $m$ have no factors in common except $1$. Hence $n = \pm 1$.[/hide]

[quote="DiscreetFourierTransform"]$M^{2}=AN$
$N = \frac{M^{2}}{A}$
I don't know where to go from here.[/quote]

A good time to digress on problem-solving technique.

Algebraic manipulation won't help much here because we're dealing with a number theory problem.  The way you've approached the problem, you have neglected the condition that $gcd(n, m) = 1$, and of course without that condition the problem is not solvable.

What needs to be done here is a little thinking about factors instead.  Or, if you want to be more formal, prove the following:

[b]Lemma:[/b]  Let $a, b$ be integers such that $gcd(a, b) = 1$.  Then

$a | bc \implies a | c$

But now the problem is easy.  $gcd(n, m) = 1$ and $n | m^{2}$, ergo $n | m$.  But if $n | m$ then $n | gcd(n, m) = 1$, so $n = \pm 1$.

Aha. I'm new to this whole number theory thing. My brain is set in algebra mode from school  :lol:
Thanks for that nice example. I'll try to remember that.


What are the time limits for each of the rounds on ARML? The website hasn't said anything about time limits, only the formats of the questions/how many questions there are. And have the time limits changed since 2006, or are they the same? I used the Search function and got some info regarding time limits from 2006, but I don't know if anything has changed since.

There's been no change in time limits since 2006 (or many years before that, for that matter.) The only changes were in the number of individual questions, and some point total adjustments to go with that.

Team round: 20 minutes for 10 questions for the whole 15-member team.

Power round: 60 minutes for the whole round for the whole 15-member team.

Individual round: 10 minutes per two questions; each pair of questions completed before moving to the next.

Relay round: 3 minutes for the relay for full credit. 6 minutes for "half" credit.

So "half" in relay mode=60% :ninja:  :ninja:  :ninja:

Prior to this year, the scoring for relays was 4/2. So, yes, it was half. This year the relays expanded to a total of 5 points in the 1st interval. We still just think of it as "half".


Let $ f_n$ be a sequence of complex numbers.  Is it true that $ \sum |f_n|$ converges if and only if $ \sum \left( |\Re f_n| \plus{} |\Im f_n| \right)$ converges?

The 2-norm is equivalent to the 1-norm in a finite-dimensional space.


geometry
symmetry
trigonometry
trig identities
Law of Cosines
geometry unsolved

I think I can see that we can get a horizontal cut of with the same length as $ OB$ but not how to get a shorter oblique cut.

I get $ \frac{\sqrt{2}}{2}$ for the shortest cuts horizontally and between the two equal sides.

Suppose the triangle $ ABC$ is right-angled at $ A$ and $ AB$ is horizontal with $ A$ above$ B$. Letting a cut of length $ d$ meet $ CA$ at $ D$, a distance $ x$ below $ C$ and meet $ CA$ at $ E$, a distance $ y$ from $ C$ along $ CA$,

I get that the area of triangle $ CDE$ is $ \frac{1}{4}\equal{} \frac{1}{2}xy \sin 45^{\circ}$

$ xy\equal{}\frac{1}{\sqrt{2}}$.

Then the cosine rule gives $ d^{2} \equal{} x^{2} \plus{} y^{2} \minus{}2xy \cos 45^{\circ} \equal{} x^{2} \plus{}\frac{1}{2x^{2}} \minus{} 1$

which is minimised when $ d^{2} \equal{} \sqrt{2}\minus{}1$.

I suspect I may be wrong.

I'm still bothered about my attempt at this. I think I may have gone on the wrong track and that there is an easy way of doing it that I just can't see. Can someone help? Thanks.

I got my minimum value by differentiating $ x^{2} \plus{} \frac{1}{2}x^{\minus{}2}\minus{}1$ to get $ x\equal{}2^{\frac{\minus{}1}{4}}$.

I now think I may have got it right as I know that $ xy\equal{}\frac{1}{\sqrt{2}}$ and so AM-GM gives $ \frac{x\plus{}y}{2} \ge \sqrt{xy}$, with equality for $ x\equal{}y\equal{}2^{\frac{\minus{}1}{4}}$.

[quote="AndrewTom"]A square piece of toast $ ABCD$ of side length $ 1$ and centre $ O$ is cut in half to form two equal pieces $ ABC$ and $ CDA$. If the triangle $ ABC$ has to be cut into two parts of equal area, one would usually cut along the line of symmetry $ BO$. However, there are other ways of doing this. Find, with justification, the length and location of the shortest staight cut which divides the triangle $ ABC$ into two parts of equal area.[/quote]
There is two ways one can do this:
Either the cut is from side $ AB$ to side $ AC$ or from side $ BC$ to side $ BA$. (The case where the cut is from $ BC$ to $ BA$ is the mirror image of the case where the uct is from $ BC$ to $ CA$)

[b]Case 1:[/b] (From $ AB$ to $ AC$) Let $ P$ be a point on $ AB$ and $ Q$ be a point on $ AC$ such that the cut is $ PQ$. Let $ AP \equal{} x, AQ \equal{} y$, then $ \frac{1}{2}xy \equal{} T_{\triangle APQ} \equal{} \frac{1}{2} T_{\triangle ABC} \equal{} \frac{1}{4} \iff xy \equal{} \frac{1}{2}$. $ |PQ| \equal{} \sqrt{x^2\plus{}y^2} \ge \sqrt{2xy} \equal{} 1$, so in case 1, the shortest cut is of length $ 1$.

[b]Case 2: [/b] (From $ BC$ to $ BA$) Let $ P$ be a point on $ AB$ and $ Q$ a point on $ BC$, such that the cut is $ PQ$. Let $ BP \equal{} x, BQ \equal{} y$, then $ \frac{1}{2} \sin 45^\circ xy \equal{} T_{\triangle BPQ} \equal{} \frac{1}{2} T_{\triangle ABC} \equal{} \frac{1}{4} \iff xy \equal{} \frac{1}{\sqrt{2}}$. By the law of cosines we have: $ |PQ| \equal{} \sqrt{x^2\plus{}y^2\minus{}\sqrt{2}xy} \ge \sqrt{(2\minus{}\sqrt{2})xy} \equal{} \sqrt{\sqrt{2}\minus{}1} < 1$. Equality when $ x\equal{}y\equal{}\sqrt[4]{\frac{1}{2}}$.

So the answer is $ \sqrt{\sqrt{2}\minus{}1}$, with the location given in case 2

Thanks Mathias_DK. As you can tell, my solution is almost exactly the same as yours so now I don't feel so bad. Thanks.


linear algebra
matrix
vector
linear algebra solved

Let $f$ be a linear transformation which is determined by matrix $A= \left( \begin{array}{cc} a & b \\ c & d \\ \end{array}\right)$ on $xy$ plane.

Supposed that a certain point $P\ ,\ P\neq\ (0,0)$ is mapped to $P$ itself.
Prove that there exist the line $l$ which don't pass through origin and whose any points on $l$ is mapped  the point on $l$ by $f$.

If $(p_1,p_2)$ is a point which is mapped to itself, then the matrix formed by appending the column $(p_1,p_2)^T$ to $I_2-A$ has rank $1$, just as $A$. This means that the system $(I_2-A){\bf x}={\bf p}=(p_1,p_2)^T$ is compatible. Let ${bf x}$ be a vector which is a solution for $(*)$. The line consisting of all vectors (points, actually :)) of the form ${\bf x}+t{\bf p}$ is fixed by $A$.

Thank you for your solution, grobber.

Here is my solution,especially for senior high school students in Japan.

Let $l$ be $px+qy=1\ (p^2+q^2\neq 0)\Longleftrightarrow$.

Any points on $l$ is mapped the point on $l$ by $f$.
$\Longleftrightarrow$ any points $(x,y)$ on $l$ satisfies $p(ax+by)+q(cx+dy)=1$ ,i,e, $(ap+cq)x+(bp+dq)y=1$
$\Longleftrightarrow$ this line is coincides with the line $l$
$\Longleftrightarrow ap+cq=p$ and $bp+dq=q\ (p^2+q^2\neq 0)$

$\Longleftrightarrow T:= \left(\begin{array}{cc} a-1 & c \\ b & d-1 \\ \end{array}\right)\left(\begin{array}{c}p & q \\ \end {array}\right)=\left(\begin{array}{c}0 & 0 \\ \end {array}\right),\ (p^2+q^2\neq 0)$
$\Longleftrightarrow detT=0$
$\Longleftrightarrow A\vec{p}=\vec{p}\ (\vec{p}\neq \vec{0})$ Q.E.D.

It is said that this problem is very difficult ranked  within best 10 in the history of the entrance examination in Japan.

kunny

[quote="grobber"]the matrix formed by appending the column $(p_1,p_2)^T$ to $I_2-A$ has rank $1$[/quote]

Numerical evidence shows that this is not true. Can we correct that? Or maybe it's my mistake... (It's kind of late, and I usually make big mistakes at these hours)

[quote="kunny"]
Prove that there exist the line $l$ which don't pass through origin and whose any points on $l$ is mapped  the point on $l$ by $f$.[/quote]
I'm sorry but I don't quite understand that sentence.  Do you mean :

prove that there is a line which does not pass through the origin and all points of which are mapped onto itself by f?

or do you mean :

prove that there is a uniqueline which doesn't pass through the origin and all points of which are mapped onto p?

The first, I think. I can't see how the second could make sense. (considering, for example, $A = I_2$)


Prove that the equation
$x^{2}+y^{2}= 2007+z^{2}$
has infinitely many integer solutions.

[hide] $(x, y, z) = (1004-2t^{2}, 2t, 1003-2t^{2})$

We can get this by substituting $y = 2t$ and writing the equation as

$(x+z)(x-z) = 2007-4t^{2}$.

Let $x+z = 2007-4t^{2}$ and $x-z = 1$ to get the solution set (of course, this is a very small subset of all solutions). [/hide]


Suppose N identical objects and K sets. How many different ways can the N objects be distributed along the K sets, given that all K sets must have at least 1 object in them?

Solution:

Each set having at least 1 object, the question is how to distribute the remaining N - K objects among the K sets. This is equivalent to Choose(N-1,K-1).

Is this correct?

Best,
Guillaume

NO. Try $ N\equal{}3$ and $ K\equal{}2$ by exhaustion. Also, there is some ambiguity in your question. For the case mentioned, let the $ 3$ objects be $ a,b,c$, do you expect the solution to be

$ \{a\}$ and $ \{b,c\}$, or $ \{b\}$ and $ \{a,c\}$, or $ \{c\}$ and $ \{a,b\}$, in total $ 3$ cases, the order of the  $ 2$ sets not mattering? OR

the $ 6$ cases when the order of the $ 2$ sets matters, so we need to add $ \{b,c\}$ and $ \{a\}$, or $ \{a,c\}$ and $ \{b\}$, or $ \{a,b\}$ and $ \{c\}$. Which one?

I'm terribly sorry; I forgot to mention that the objects are all identical.

So, with N = 3 and K = 2, the two possibilities are {1,2} and {2,1}.

I think you will need a hint.

How to represent a solution of $ x_{1}\plus{}\dots\plus{}x_{K}\equal{} N$ conveniently?
Write $ N$ ones:$ \underbrace{111\dots 11}_N$ and put $ K \minus{}1$ pluses somewhere in between them.

[quote="zhoraster"]These are exactly the positive integer solutions to the equation
\[ x_1 \plus{} \dots \plus{} x_K \equal{} N.\]
[/quote]

I'm very sorry, but I do not understand your argument. In any case there is only one solution to the problem.

In the equation

$ x_1 \plus{} x_2 \plus{} \ldots \plus{} x_K \equal{} N$

$ x_K$ represents the number of items in the Kth group.

For example, with N = 3, K = 2, we might have $ \{x_1 \equal{} 1, \, x_2 \equal{} 2\}$ or $ \{x_1 \equal{} 2, \, x_2 \equal{} 1\}$

To continue with zhoraster's hint, let's write this out using 1's and pluses:

$ \{x_1 \equal{} 1, \, x_2 \equal{} 2\}$ can be represented by $ 1 \plus{} 11$
$ \{x_1 \equal{} 2, \, x_2 \equal{} 1\}$ can be represented by $ 11 \plus{} 1$

We need to [i]choose[/i] the location of our plus(es) appropriately to split the items into the K (= 2) groups.

There are N items, so N-1 places we can put the pluses.
To split into K groups, we need K-1 pluses.

Hence $ \dbinom{N\minus{}1}{K\minus{}1}$ as you said.


topology
advanced fields
advanced fields unsolved

Ok, this problem is most likely very easy, but I have some kind of brain blockage so I am not figuring out what to do.

We have a metric space $(X,\rho)$. If $\mu^{*}$ is an outer measure on $X$ such that every open set is measurable, then $\mu^{*}$ is a metric outer measure (that is, if $\rho(A,B) > 0$ then $\mu^{*}(A\cup B) = \mu^{*}(A)+\mu^{*}(B)$).

Let $A$ and $B$ so that $p(A,B) > 0$. Note that $\overline{A}$ is closed, hence measurable. Because
$\overline{A}= \{x \in X: p(x, A) = 0\}$ we have $(A \cup B) \cap \overline{A}= A$ and $(A \cup B) \setminus \overline{A}= B$. Now just use $A \cup B$ as a test set for the measurability of $\overline{A}$ to get the result.

Also the converse of this theorem holds, which is more interesting.


How many of each of the ten digits $(0,1,2,...,9)$ are needed in order to write out all the integers from $1$ to $10^9$ ?

let fi(n) denote the total number of i s used in all n digit numbers(i is not equal to 0) .Observe that fi(n)=10fi(n-1)+9.10^(n-2) ,(n>=2).An easy induction shows that fi(n)=10^(n-1)+9(n-1)10^(n-2)=(9n+1)10^(n-2),since we know that fi(1)=1.This is the same for i=1,2,3,4,5,6,7,8,9,but not 0.You can find another formula for 0.The rest is easy.

i think your method is fine , but what is the answer that you got??

the answer is 9*(10^8+10^7+10^6+10^5+...+10^1+10^0)-1
i'm not sure if it is correct. can somebody confirm with me?


superior algebra
superior algebra unsolved

Find \[ \left[\mathbb{Q}(\sqrt{3}, i): \mathbb{Q}\right].  \]
 Prove that \[ \mathbb{Q}(\sqrt{3}, i)=\mathbb{Q}(\sqrt{3}-2i)  \]
 and find \[ \min(\sqrt{3}-2i, \mathbb{Q}).  \]

$\left[\mathbb{Q}(\sqrt{3}, i): \mathbb{Q}\right] = 4$ (the method was already given at the older ones).

${\frac{7}{\sqrt{3}-2i}=\sqrt{3}+2i}$, thus as sometimes before.

$x^4+2x^2+49$, also as before ;)


Prove that if p(x) is polynomial with integer coefficients and p(square root of 2) = 0 then p(-square root of 2) =0.

I don't understand your hieroglyphes.Can you use $ \text{\LaTeX}$.
$ p\sqrt {2}=0$ then  $ p -\sqrt {2}=0$
Am I OK ?

No, the question asking you to prove that if $ p(x)$ with integer coefficients has one root $ \sqrt {2}$, then another root is $ \minus{} \sqrt {2}$.

[quote="SMCount"]So, the question asking you to prove that if $ p(x)$ with integer coefficients has one root $ \sqrt {2}$, then another root is $ \minus{} \sqrt {2}$.[/quote]

Which is not true

Yeah... I was thinking that too, doesn't make sense.

Edit: Forgot the integer coefficients part :p

It certainly is true - [url=http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem]conjugate root theorem[/url].  :wink:

If $ P \equal{} \sum a_kx^k$, then $ P\left(\sqrt{2}\right) \equal{} \left[\sum_{2i} a_{2i} 2^i\right] \plus{} \sqrt{2}\left[\sum_{2i\minus{}1} a_{2i\minus{}1} 2^i\right] \equal{} b\plus{}c\sqrt{2}$, where $ b,c$ are integers, so $ b\equal{}c\equal{}0$; then $ P(\minus{}\sqrt{2}) \equal{} b\minus{}c\sqrt{2} \equal{} 0$ (because only odd-powered terms become negative).

By the way, where are these from? Someone asked me to solve these a couple of days ago.

its  always  true.   for  a polynomial  wuth real coefficients    iraational  or complex   roots occur   as conjugates