I have no problem cell phones driving our cars. Or giving the cell phone commands on how to drive the car. :lol: ;)

Would you if I told you that almost 3,000 people die each year from this?  :huh:

Wait... let me clarify.
Not the cell phone driving the cars... driving while using cell phones!

3000 people DIE from this?!?!?!

Wow...

I am totally against it. It is totally unsafe to call while driving. I like the idea of cell phones driving cars, though...

:)

Yeah, 3000 people die and more than 10,000 are injured.  :(

I'm against the idea of people driving with cell phones, but I wouldn't try to ban it.

I've never driven while talking on a cell phone and doubt I ever will. Come to think of it, I can't even remember the last time I used a phone ... I stick with IM mostly.

but it has already been banned, hasn't it? :o

In some places yes, but not others. It's still legal where I live. But I'm against it, though hands-free talking seems to be safer, as long as the driver is competent. Many accidents, even those not your fault, can be avoided if you're aware of your surroundings.

Should it be banned? No. We already have enough boundaries set on our "freedom" in this country thanks to the few, the ignorant, the stupid. However, I do think that at least a fine should be instated, considering the extreme danger involved with careless people "talking and driving."

[quote="Drunken_Math"]Both my parents use that thingie were you can talk on the phone and not pick up the phone. It's like an ear attachment thingie. It seems safer.  :)[/quote]

Ah, yes. There is also such thing as voice activated, hands-free calling. But I still say just pull over or wait until you get there.

With the development of Bluetooth technology, and the fact that almost all new phones are bluetooth enabled, and the prices for bluetooth headsets has decreased dramatically I strongly support the cell-phone talking and driving at the same time being banned. As a matter of fact it is banned already in Romania, although nobody here really cares about the law anyway :)

It's illegal in almost every other country but the US...  :D
And speaking of the handless phones, studies show that those are just as dangerous...

I think that banning it is a violation of freedom but it is more rational for an individual to not talk on a cell phone in a car.

[quote="hello"]I think that banning it is a violation of freedom[/quote]No, you got it wrong. It's talking at the cell phone without using any handsfree equipmanet (either wired, or bluetooth-based) that it's banned in the European Union, and should be everywhere also, not complete banning of the cell phone usage in the car.

The problem with the cell phone talking while driving is that if you are speaking at the phone (which you hold with one hand) you have basically only one hand to operate the vehicle with (which is bad considering that usually some of those drivers also have a couple of beers aboard :D).

This banning is totally logical. If you are driving with cell phones, you may not be paying attention to the road ahead. Let's say you crash. Not only you get hurt (even die...), but the person you crashed might get hurt, too. If the road is very busy, multiple cars may even collide.

The probability of this happening is very low. It might happen, though, and I don't want to take any chances.

What is the punishment here in the US for illegal cell phone usage? (Not for stealing or that stuff, but for driving while talking.)

Valentin, I disagree. What if a person is a great driver and can drive with one hand? Does this also mean we should ban driving with one hand? If a person is not capable of doing it, they should not. But I think it is a violation of personal rights to force people not to talk on their cell phones even though I agree its bad to do so. Some people are capable of coping with the task though I am sure.

Btw, how much does that thing cost so that one can talk without picking up the phone?

[quote="hello"]Btw, how much does that thing cost so that one can talk without picking up the phone?[/quote]
95% of the phones today come with a wired handsfree solution, so it's basically free. The more advanced Bluetooth system varies from 15$ up to 200$ (the most performant models which also let you listen to stereo sounds etc. :) ).

[quote="hello"]Valentin, I disagree. What if a person is a great driver and can drive with one hand? Does this also mean we should ban driving with one hand? If a person is not capable of doing it, they should not. But I think it is a violation of personal rights to force people not to talk on their cell phones even though I agree its bad to do so. Some people are capable of coping with the task though I am sure.[/quote]
No person is a "great driver". Not even Michael Schumacher is a great driver if he is not paying attention to the road:!:. I wonder, do you own a driver's license? ..  :?

Why then let's unban drinking (excesively) and driving. I am sure that there are a few people that (think that they) can drive after consuming a large quantity of alchool. Do you want to be on the road when they drive by? I certainly would not!

PS I see no violation of personal rights here. You can always use a handsfree device, or stop the car to make the call ;)

I agree with the banninng of cell phones while driving but the only problem with this idea is... How will we inforce this?

If you see a cop and you're on the phone, just hang up and put it in your cup holder. Whether the cop has seen you or not, he has no proof to give you a ticket. If you deny it, it's your word vs. the cops. There is no "alcohol test" or "speed gun" for proof, making this proposed cell-phone law very very hard to inforce.


inequalities
triangle inequality
geometry proposed
geometry

Let M an interior point of the triangle ABC.
$AM\cap BC=\{A'\},BM\cap AC=\{B'\},CM\cap AB=\{C'\}$
Prove that with $[AA'],[BB'],[CC']$ can be constructed a triangle, if and only if M is the centroid of ABC.

:maybe: I don't believe this.
If we take an equilateral triangle $ABC$ and point $M \rightarrow A$, then clearly the triangle inequality is satisfied for every segment.

Am I misreading something?

If three segments are given, then can be constructed a triangle, taken these segments as it’s sidelines, if and only if, each of the given segments is not bigger of the sum of the other two segments. So, we have many cases of arbitrary cevians in a triangle, with this property.

By the way, there is well known problem, which states that can be constructed a triangle, taken as it’s sidelines the medians of every given triangle. We see a ''constructive'' proof in the figure.

Kostas Vittas.


calculus
integration
limit
inequalities
real analysis
real analysis unsolved

$f:[0,+\infty[\rightarrow ]0,+\infty[$ continous
$\lim_{x\rightarrow +\infty}\int_{0}^{x}f(t)dt=K$ where $K$ is a real number
Let $a,b>0$ such that $\frac{1}{a}+\frac{1}{b}=1$
Prove that $\lim_{x\rightarrow +\infty}\frac{1}{x^{\frac{1}{a}}}\int_{0}^{x}f^{\frac{1}{b}}(t)dt=0$

Probably the author Radu Miculescu thought about this version.

Hmm..., isn't it pretty much the same as the problem with $a=b=2$ posted before and solved by grobber?

Yes Fedja here Hölder inequality in grobber proof.


Show that the condition that dF(a) be non-singular is necessary in the inverse function theorem, by showing that if a function F from a neighborhood of a in $\mathbb{R}^{p}$ to $\mathbb{R}^{p}$ is differentiable at a and has an inverse function at a which is differentiable at F(a), then dF(a) is non-singular.

Let $G$ be the inverse of $F$. By the Chain Rule $(G\circ F)'(a)=G'(f(a))F'(a)$. On the other hand, $(G\circ F)'(a)$ is the identity matrix.


Let $a,b,c$ be positive real numbers such that $abc \leq 8$

Prove that

$\frac{1}{a^2-a+1} +\frac{1}{b^2-b+1} +\frac{1}{c^2-c+1} \geq 1$

Put $a=\frac{1}{x} , b =\frac{1}{y} , c=\frac{1}{z}$ and apply Cauchy Schwarz 's ineq .

see also in http://www.mathlinks.ro/Forum/viewtopic.php?p=251670&highlight=#251670


Find the [u]smallest[/u] constant $s$ such that the inequality holds:
\[ \sum_{k=1}^{\infty} \frac{1}{kn^k} < s\ \ \forall n,  \]
If $n$ is natural number.

When $n=1$ LHS =$1$ .When $n>1$ then LHS is decreasing , so the maximum is $1$ .

Ah, It was lack of something when I posted.

But the true LHS seems to disverge?
\[ \frac{1}{n}+\frac{1}{2n^2}+\frac{1}{3n^3}+\cdots +\frac{1}{n \cdot n^n}+ \cdots < s\ \ \forall n,  \]

You need $n > 1$.  Then making the substitution $x = \frac 1n$, $\sum_{k=1}^{\infty} \frac{1}{kn^k} = \sum_{k=1}^{\infty} \frac{x^k}{k}$.  From here, it is just generating functions.  If you really want $n$ to be integral, there will be so such $s$ because the set on the left-hand side achieves a maximum.


i think it is a direct consequence of holder's inequality.. :?


Three integers, $ e$, $ f$, and $ g$ are distinct primes such that $ e\equal{}f^2\plus{}g^2$.  What is the greatest possible value of $ e$ less than 100?

$ e$ is odd.
So, $ f$ is even and $ g$ is odd. or $ f$ is odd and $ g$ is even.
If, $ f<g$, $ f\equal{}2$.

Therefore,
$ 2^2\plus{}3^2\equal{}4\plus{}9\equal{}11 \in \text{Prime}$.
$ 2^2\plus{}5^2\equal{}4\plus{}25\equal{}19 \in \text{Prime}$.
$ 2^2\plus{}7^2\equal{}4\plus{}49\equal{}53 \in \text{Prime}$.
$ 2^2\plus{}11^2\equal{}4\plus{}121\equal{}125\equal{}5^3$.
$ 2^2\plus{}13^2\equal{}4\plus{}169\equal{}173 \in \text{Prime}$.
$ \vdots$

Thus, $ e\equal{}\boxed{53}$.


trigonometry
integration
calculus
calculus computations

(a) Show that $ \frac{d}{dx}(\cos 2x \sin^{n\minus{}1} 2x) \equal{} 2(n\minus{}1)\sin^{n\minus{}2} 2x \minus{} 2n \sin^{n} 2x$.

(b) It is given that $ I_{n}  \equal{} \int_{\frac{\minus{}\pi}{2}}^{\frac{\pi}{2}} \sin^{n} 2x \ dx$  $ (n\ge 0)$.

Use the result in (a) to show that $ I_{n} \equal{} (\frac{n\minus{}1}{n})I_{n\minus{}2}$   $ (n \ge 2)$. Hence find $ I_{8}$, leaving your answer in the form $ k\pi$, where $ k$ is a fraction in its lowest terms.

For (a), use $ (xy)'\equal{}x'y\plus{}y'x$. And for (b), apply integral by parts. (b) is often called Wallis's product formula.


Let $ABCD$ be a cyclic quadrilateral. Suppose that $O$ is the circumcenter of $ABCD$ and $P=(AC,BD), E=(AD,BC), F=(AB,CD)$. Prove that $O$ is the orthocenter of the triangle $PEF$.

See at: http://www.mathlinks.ro/Forum/viewtopic.php?t=110887

Kostas Vittas.

Brocart's theorem... for proof you can use pole and polar(prove that $PE$ is polar of $F$ and so on) or radical axises if you take the Miquel's point of $(ABP),(CDP),(FBD)$ and $(FAC).$


[quote="jhredsox"][quote="1=2"]My little brother is matching letters. There is a grid of 4*3 letters, and he picks 2 at random. What is the probability that he matches letters on his first, second, and third try?[/quote]
[hide]11/12,9/10,7/8[/hide]?

jorian[/quote]

Isn't the probability for [i]not[/i] matching letters??  :huh:

[hide="Solution"]
There is $\frac{1}{11}$ ways to pick the first time correctly. He replace it back. He has a $\frac{1}{11}$ chance of getting it right the third time. Same for the third time. So, the answer is $\frac{1}{11^{3}}=\boxed{\frac{1}{1331}}$ chance he matches it.[/hide]

Extension:
What is the probability that he will get 2 out of the 3 matches if he puts it back if he gets it wrong and takes it out of the grid if he gets it right?

Answer to Extension:

[hide]
Is it $2/231$??
[/hide]


Hello people,

This is the link to the first of three papers describing an approach to tackling most prime number conjectures. I hope you will spare the time to critique my contribution. Thanks

[url]http://addo.esmartdesign.com/3e.htm[/url]

Lemma 3.2 is trivial.
Theorem 4.1 is trivial (it follows from the infinitude of the primes).
Theorem 4.2 is trivial (when $ A \in \psi_1$ the corresponding "prime space" is finite if either some $ x \plus{} k \in A$ or $ M(A)$ covers the set of residue classes modulo some $ m$.)

Conjecture 4.3 is unsupported by the rest of the discussion.  In particular, the structure of $ N(A)$ for $ A \in \psi_{k}, k \ge 2$ is much more complicated than the structure of the spaces you actually proved results about.  This paper is just a mass of jargon.

interesting. I guess I said this is the first of three papers. Yes it is true that the space in the conjecture is much more complicated than the previous ones, but that is exactly what the next paper was going to take care of. I guess you have already failed to see through this one.

Thanks for the time though. But never the less, what do you think about the conjecture? :wink:

It would be interesting if true, but I don't think the ideas in this paper contribute much to it.


geometry
perimeter
analytic geometry
puzzles

Find all sets of solutions to the following.**

Given that you have a right triangle with a perimeter of 5 units,
two of its sides have lengths that add up to 11/4 units.  Work out the lengths for all three sides of the triangle. **

(If certain solutions for lengths are rational, then give the exact simplified fractions for them.  If certain solutions for lengths are irrational, you may report them to the nearest thousandth's place instead of using the exact radical form.)

Scale the triangle up by a factor of $4$ to simplify computations. Let $a,b$ be the legs and $c$ the hypotenuse. Then there are essentially two cases:

[hide="(1) the sum of the two sides given refers to the two legs"]Assume $a\ge b$. Also, $a+b+c=20=11+c\implies c=9$.
Furthermore, \[(a+b)^{2}=121=a^{2}+b^{2}+2ab=81+2ab\implies a^{2}+b^{2}-2ab=41\implies |a-b|=a-b=\sqrt{41}\]
Combining with $a+b=11$ yields: \[(a,b,c)=(\frac{11+\sqrt{41}}{2},\frac{11-\sqrt{41}}{2},9)\][/hide]
[hide="(2) one of those two sides is the hypotenuse"]Suppose that the leg adding up with the hypotenuse is $b$. Then, $20=a+(b+c)=a+11\implies a=9$. Thus $81=c^{2}-b^{2}=11(c-b)\implies c-b=\frac{81}{11}$. Since $c+b=11$ we find that \[(a,b,c)=(9,\frac{\frac{81}{11}-11}{2},\frac{\frac{81}{11}+11}{2}) \][/hide]
Of course, for either of the cases the legs $a,b$ can be flipped so in total there are four solutions: \[\boxed{(9,\frac{\frac{81}{11}-11}{2},\frac{\frac{81}{11}+11}{2}) \qquad (\frac{\frac{81}{11}-11}{2},9,\frac{\frac{81}{11}+11}{2}) \qquad (\frac{11+\sqrt{41}}{2},\frac{11-\sqrt{41}}{2},9) \qquad (\frac{11-\sqrt{41}}{2},\frac{11+\sqrt{41}}{2},9)}\]
Dividing each of these by four and simplifying the fractions like you asked for would yield the answers. (Of course, I don't see the point in doing this...)

Let's face my rules.

You didn't present the solutions in the form so indicated, so your candidates are wrong.  Complex fractions as coordinates don't fit, for one.  You take on the problem as a whole for credit, or it's not right.  Your strategy is ONE of the correct ways of working toward what would have been a completed solution.

Also, there are not four solutions; there are two distinct solutions.
For instance, in the case where you are flipping two leg lengths, that produces the same triangle.  It still has the same length hypotenuse as before.

I addressed all that in my solution; I don't really care if I follows your "rules" for the solution because having presented the ideas its trivial to carry out the computations. What's the reason for these rules anyway?


If a number eight times as large as $x$ is increased by two, then one fourth of the result equals

$\textbf{(A)} \ 2x + \frac{1}{2} \qquad \textbf{(B)} \ x + \frac{1}{2} \qquad \textbf{(C)} \ 2x+2 \qquad \textbf{(D)} \ 2x+4 \qquad \textbf{(E)} \ 2x+16$

[hide="solution"]
$\frac{1}{4}(8x+2)=2x+\frac{1}{2}\implies A$
[/hide]

[hide]8 times as large means 8x
 $\frac{8x+2}{4}=2x+\frac{1}{2}$ [/hide]

[size=75]comment: It's hard to imagine that this is from AMC 12[/size]

[hide][color=blue]The origional equation is $8x$. Add 2 and divide by 4 gives $2x+1/2$. So the answer is A.[/color][/hide]

[quote="Silverfalcon"]If a number eight times as large as $x$ is increased by two, then one fourth of the result equals

$\text{(A)} \ 2x + \frac{1}{2} \qquad \text{(B)} \ x + \frac{1}{2} \qquad \text{(C)} \ 2x+2 \qquad \text{(D)} \ 2x+4 \qquad \text{(E)} \ 2x+16$[/quote]

[hide] $1/4*(8x+2)=2x+1/2$, so the answer is $\boxed {A}$ [/hide]

[hide]The equation is 8x+2, if you divide by 4, you'll get $2x+1/2$ which is $A$[/hide]

[quote="Silverfalcon"] Anyway, it's early questions and it's easy. Everybody likes easy questions so you can't complain.  :D[/quote]

Yes!! I think you're right.

This is the answer:

8x+2/4 = 2x+1/2 = A

[hide="Quotes hidden too big"][quote="Silverfalcon"][quote="JesusFreak197"][quote="236factorial"][size=75]comment: It's hard to imagine that this is from AMC 12[/size][/quote]

I wouldn't say that. It's a #2. Remember that #1 from the 2006A was simple multiplication that any third grader could do.

Also, while it's generally okay for people to post their own solutions for a problem even when it's already been answered the same way, it's probably going a little overboard to have 8 people post the exact same solution, and for a #2.[/quote]

Let me add more to this.

Until 2000, AMC had 30 questions which meant that earlier questions are much easier than what they are now. But in 1980's (especially prior to 1983, the year of first AIME), the AHSME was only contest to determine USAMO people. Therefore, while beginning questions were still easy, they weren't that easy as compare to earlier ones in 1990's AHSME. Anyway, it's early questions and it's easy. Everybody likes easy questions so you can't complain.  :D[/quote][/hide]The first few questions has to be easy because they want to encourage beginners to dwell into mathematics competitions. This is AMC's mission: to promote math education through competitions. At the same time, they want to select the best math students by putting the hardest multiple choice questions at the end. Think about this: if you were given a really hard math question on your first math competition and you can't solve it right away, more than likely you give up. That's not AMC's mission.

PS Edit: One comment I want to add to JesusFreak197's post: I agree to the issue brought up by him in that there are already a couple of similar answers posted for one problem (not caring difficulty of problem). As a reader's point of view, I already got bored from reading all these similar answers over and over again and posts that only says "answer is X" and that's it. At this point, I wouldn't even waste 5 minutes of my time to continue reading. Whether you liked what I said or not, it's up to you (just a 'heads up', not meant to target at anybody or group in particular) but I wanted to point this out since the responsibilities of authors in this forum are really going down the drain (in my opinion). Good day.


videos
AwesomeMath
summer program
calculus
rich tag
Support
MATHCOUNTS

Post an embedded youtube music video that you believe to be superior to the previous one. Use [youtube]and[/youtube] tags around the URL. I shall begin with something rather hard to top:
[youtube]vc2OX-ERL3Y[/youtube]

[quote="Treething"]
aye aye[/quote]
dude that stuff sucks

lets get some real thrash metal in here

[youtube]E4udzab0qV4[/youtube]

[youtube]9USb_OQza44[/youtube]
better than the original

[quote="junggi"]
better than the original[/quote]
um no

you lose

however

the following wins

[youtube]alPFV9rR4wc[/youtube]

hmm thats pretty good
however

[youtube]7J_HyA5p33E[/youtube]

[quote="junggi"]hmm thats pretty good
however

[/quote]

hmm that will be tough to beat

let me pull my trump card

[youtube]icAzyx8EsKU[/youtube]

i can do cheesy pop

observe

[youtube]Ywxm6zLEjFY[/youtube]

[quote="Treething"]meatspin > dragonforce

the gravity of this statement is so large that it doesn't need an accompanying youtube video[/quote]
um no you lose

possibly the best video evar

[youtube]icAzyx8EsKU[/youtube]

ahhh sum 41

BAD MUSIC

[youtube]YDBIbJKjAZQ[/youtube]

the art sucks but the music is AWESOME

[youtube]https://m.youtube.com/watch?v=YoWNYzwDYvs[/youtube]

bruh moment
I mean, NGGYU might be better than 1000 doors, but just marginally

that's it, juliankuang win--

w a i t

there is another

[youtube]https://www.youtube.com/watch?v=PfYnvDL0Qcw[/youtube]

[youtube]https://www.youtube.com/watch?v=3SCBYUE_X1U[/youtube]

im 3 hours in [verbatim]:D[/verbatim]

[quote=ItzARubix][quote=g1zq]this thread has officially evolved into a meme war[/quote]

the cubic formula isn't a MEME[/quote]

Oh my god stop these songs are making me die of laughter  :rotfl:

[quote=ItzARubix][quote=g1zq]this thread has officially evolved into a meme war[/quote]

the cubic formula isn't a MEME[/quote]

y e s


Let $ n,k$ be natural numbers. On an $ n\times n$ chessboard, each entry has one of the numbers $ 0,1,2,...,k$. Each row and each column has a button which, when pressed, will increase each of the $ n$ entries by $ 1$ (but $ k$ will go to $ 0$). In the beginning, the entries were all $ 0$'s. After pressing the buttons several times, the entries are now random. Prove that you can always change it back to the original state with no more than $ kn$ pressings.

Let $ \{a_1,a_2, \ldots, a_n\},\{b_1,b_2, \ldots, b_n\}$ be the sets of number of button presses for each of the rows and columns, respectively (the button-pressing operation is commutative; and the grid elements are $ \{ij\} \equal{} a_i \plus{} b_j$). If any of $ a_i,b_i > k$, then we can reduce them to their remainder upon division by $ k$.
[hide="Any constructible board can be constructed in less than equal to nk moves."]
Suppose there exists a constructible board which cannot be constructed in $ \le nk$ moves. Then $ S(\{a_i\}) \plus{} S(\{b_i\}) > nk$, where $ S(\{\})$ denotes the sum of the set (for the board to be constructed in $ \le nk$ moves, we require $ S(\{a_i\}) \plus{} S(\{b_i\}) \le nk$).

Now note that the board generated by $ \{a_i\},\{b_i\}$ is equivalent to each of the $ k \plus{} 1$ boards $ \{a_i \plus{} m \mod {k \plus{} 1}\},\{b_i \minus{} m \mod {k \plus{} 1}\}$ for $ m \equal{} 0,1,2, \cdots, k$. The average value of $ \overline{S(\{a_i \plus{} m\}) \plus{} S(\{b_i \minus{} m\})}$ is $ 2n$ times the average of the set $ \{0,1,\cdots,k\}$, which is $ \left(\frac {k(k \plus{} 1)}2\right) \left/ (k \plus{} 1) \right. \equal{} \frac k2$; hence $ \overline{S(\{a_i \plus{} m\}) \plus{} S(\{b_i \minus{} m\})} \equal{} nk$. By assumption $ S(\{a_i \plus{} 0\}) \plus{} S(\{b_i \minus{} 0\}) > nk$, so there must exist some $ l$ such that $ S(\{a_i \plus{} l\}) \plus{} S(\{b_i \minus{} l\}) \le nk$ (contradiction). [/hide]
The [i]inverse board[/i] generated by $ \{k \plus{} 1 \minus{} a_i\},\{k \plus{} 1 \minus{} b_i\}$ is then constructible in $ \le nk$ button presses. If we now perform the sequence of button-pressing which constructs this inverse board on the original board, we then get the $ 0$-board, as desired.

Post 600!


Find all complex numbers of the form $ \frac{t\plus{}1}{t\minus{}i}$ where $ t$ is any real number.

[quote="B_Gasperov"]Find all complex numbers of the form $ \frac {t \plus{} 1}{t \minus{} i}$ where $ t$ is any real number.[/quote]
Isn't it obvious?
$ \frac{t\plus{}1}{t\minus{}i}\equal{}\frac{(t\plus{}1)(t\plus{}i)}{t\plus{}1}\equal{}t\plus{}i$.
So the answer is $ t\plus{}i$,where $ i\equal{}\sqrt{\minus{}1},t\in\mathbb{R}$.

[quote="Erken"][quote="B_Gasperov"]Find all complex numbers of the form $ \frac {t \plus{} 1}{t \minus{} i}$ where $ t$ is any real number.[/quote]
Isn't it obvious?
$ \frac {t \plus{} 1}{t \minus{} i} \equal{} \frac {(t \plus{} 1)(t \plus{} i)}{t \plus{} 1} \equal{} t \plus{} i$.
So the answer is $ t \plus{} i$,where $ i \equal{} \sqrt { \minus{} 1},t\in\mathbb{R}$.[/quote]

No Erken, $ \frac{t \plus{} 1}{t \minus{} i} \equal{} \frac {(t \plus{} 1)(t \plus{} i)}{t^2 \plus{} 1}$  :wink:


group theory
abstract algebra
superior algebra
superior algebra unsolved

suppose G is a group.H and K are the two subgroups of G.and H is normal.if H and K be both solvable then proove HK is solvable too.

$ HK/K \cong H/(H \cap K)$ and $ K$ are solvable, thus also $ HK$.


function
limit
real analysis
real analysis unsolved

Let $f(x):M\to \mathbb{R}$ be a continuous function, $M\subset[0,1]$. Prove that we can define $f$ on $[0,1]\setminus M$ s.t. $f$ will stay continuous on $M$.

[i](3 points)[/i]

I'll post what I had on this one. I hope it's correct.

For $x\in [0,1]\setminus M$, let $d_x$ be the distance between $x$ and $M$. Define $t_{\varepsilon}(x)$ to be $\sup\{f(u)|u\in M\cap ((x-d_x-\varepsilon,x-d_x+\varepsilon)\cup(x+d_x-\varepsilon,x+d_x+\varepsilon))\}$ (it can also take the value $\infty$). Now define $f(x)$ to be $\lim_{\varepsilon\to 0}t_{\varepsilon}(x)$ if this limit is finite, and $0$ otherwise (if the limit is infinite; the limit always exists).


number theory
least common multiple
greatest common divisor
Divisibility Theory

To prove equality, we will look at the exponent of any prime $ p$. If this is equal for any $ p$, both sides are equal.

Terms $ a_{i}$ that do not divide $ p$ clearly don't affect this on both sides. So we can assume either only terms divisible by $ p$, or no terms divisible by $ p$. Since $ \sum_{i\equal{}1}^{r}(\minus{}1)^{i}\binom{r}{i}\equal{}1$, cancelling a factor $ p$ from all $ a_{i}$ leaves the equality invariant. So we can assume that no terms divide $ p$.

But then there stands $ 1\equal{}1$ for the exponent of $ p$, so we're done.

[quote="Peter"]Suppose that $S=\{a_{1}, \cdots, a_{r}\}$ is a set of positive integers, and let $S_{k}$ denote the set of subsets of $S$ with $k$ elements. Show that \[\text{lcm}(a_{1}, \cdots, a_{r})=\prod_{i=1}^{r}\prod_{s\in S_{i}}\gcd(s)^{\left((-1)^{i}\right)}.\][/quote]

ler r=2,then $[a_1,a_2]=(a_1,a_2)(a_1a_2)^{-1}$. That's wrong! I think there is an error.

The formula should be $\mathrm{lcm}(a_1,..., a_r)=\prod_{i=1}^r\prod_{s\in S_i}\mathrm{gcd}(s)^{(-1)^{i+1}}$.

[quote=Peter] $ \sum_{i=1}^{r}(-1)^{i}\binom{r}{i}=1$[/quote]

I don't think this is true? For example, when $r=6$ yields that this quantity is $-1$.

Does anyone have a correct solution to this?


[quote="bigman"]Two players, A and B, play a game with a fair six-sided die. The goal is to roll a 2
or a 5, and whoever does so first wins the game. The players take turns rolling the die, with player A
going first. They keep rolling until someone rolls a 2 or a 5. What is the probability that player A
wins the game?[/quote]

P(Player A wins at his first roll)=$ \frac{1}{3}$
P(Player A wins at his second roll)=$ (\frac{2}{3})^2 \times \frac{1}{3}$
.......
P(Player A wins at his n th roll)=$ (\frac{4}{9})^{n\minus{}1} \times \frac{1}{3}$

So the probability required $ \equal{}\frac{1}{3}(1\plus{}\frac{4}{9}\plus{}(\frac{4}{9})^2\plus{}...)$
$ \equal{}\frac{1}{3}(\frac{9}{5})\equal{}\frac{3}{5}$

[hide="solution"]Let $ \ A$ be the required probability

$ A\equal{}\frac{1}{3}\plus{}(\frac{2}{3})(\frac{2}{3})A$

$ \ A\equal{}\frac{3}{5}$

or alternatively

We see that the shortest path for A to win is for him to roll a 2 and 5 at his first rolling, which has a probability of $ \frac{1}{3}$

The shortest path for B to win has a probability of $ (\frac{2}{3})(\frac{1}{3})\equal{}\frac{2}{9}$

If neither of these happens, the game is repeated and therefore their chances of winning in the ensuing rollings should be "allocated" according to the ratio of the probabilities of their shortest "paths".

Hence the probability of A winning : the probability of B winning
$ \equal{}\frac{1}{3} : \frac{2}{9} \equal{} 3: 2$
And the probability of A winning
$ \equal{} \frac{3}{5}$[/hide]

NICE SOLUTIONS.

Never would of thought of that.  I'd just brute forced it. :P


Consider an object at the base of a tank of water, and it is
 being viewed from a point not near the normal . then where
 will the object appear to be w.r.t the actula position?
 That it will appear to be raised is certain, but
 will it appear to the observer to be nearer to him than it
 actuall is in terms of horizontal co-ordinates? i.e. will the image also lie along the normal or will it be displaced?

You mean the refracted image of the object?

[quote="madatmath"]No, not the refracted image. The apparent position of the
 object ........ for example, look into a tank full of water,
 the base will appear to you to be much closer than it actually
 is. So, if an object is placed there, it will appear to be higher
 up than it actually is. This apparent position is what i'm talking about.[/quote]

Yeah, uh, isn't that because the light from the object is refracted through the water, and thus it appears to be in a different position, i.e. closer to the top. Unless I'm totally missing something...

yep, that's correct. If viewed from near the normal , then the
 apparent depth $h$ is given by $\frac{H}{\mu}$
 My question is if not viewed from near the normal, where will the image lie? will it be above or below the image in the above case? Will it be shifted to the right or to the left?

 EDIT : Here, $H$ represents the true depth of the object.


Let triangle $ABC$, prove that:
$\[\sin{nA}+\sin{nB}+\sin{nC} \le \frac{3\sqrt{3}}{2}$
for $n\in N^*$

How's this solution?

sin x is a concave down function.
Therefore, applying, Jensen's Inequality for concave functions gives us,

(sin nA + sin nB + sin nC)/3 <= sin [ (nA + nB+ nC)/3]
                                              =  sin ( n.60)

Now, sin (n.60), where n belongs to set of naturals, can take the following values,
0, sqrt. 3/2, -sqrt 3/2.

Hence, the proof.

[quote="thealchemist"]How's this solution?
sin x is a concave down function.
Therefore, applying, Jensen's Inequality for concave functions gives us...[/quote]
You're wrong here, $\sin{E}$ is a concave function, if $E$ is an angle of a triangle, [color=red]are you sure $\boxed{\sin{nA}}$ is concave function?[/color] :)

Man, try...
the problem is not simple as you think!

$nA = X$ $nB = Y$ $nC = Z$ $X + Y + Z = n\pi$

Say $S = \sin nA + \sin nB + \sin nC$
For $n \equiv 0, 1, 2, 3 \bmod{4}$ we get different identities, but we can find all easily by splitting into cases.

$n = 4k$
$\Rightarrow S = -4 \sin \frac{nA}{2} \sin \frac{nB}{2} \sin \frac{nC}{2}$

$n = 4k + 1$
$\Rightarrow S = 4 \cos \frac{nA}{2} \cos \frac{nB}{2} \cos \frac{nC}{2}$

$n = 4k + 2$
$\Rightarrow S = 4 \sin \frac{nA}{2} \sin \frac{nB}{2} \sin \frac{nC}{2}$

$n = 4k + 3$
$\Rightarrow S = - 4\cos \frac{nA}{2} \cos \frac{nB}{2} \cos \frac{nC}{2}$

I'm tired, it's 12:38 [b]A.M.[/b] but these will help significantly for anyone wishing to solve completely, it's not too bad from here.

P.S. these can be easily derived from [hide="these."]
$S = \sin X + \sin Y + \sin Z = 2\sin \frac{X+Y}{2} \cos \frac{X-Y}{2} + 2\sin \frac{Z}{2} \cos \frac{Z}{2}$
(from sum-to-product and double/half-angle trig idents)
$= 2\sin \frac{X+Y}{2} \cos \frac{X-Y}{2} + 2\sin \frac{n\pi - (X+Y)}{2} \cos \frac{n\pi - (X+Y)}{2}$
then, based on what $n$ is, we get either a plus or a minus for:
$= 2\sin \frac{X+Y}{2} \cos \frac{X-Y}{2} \pm 2\sin \frac{X+Y}{2} \cos \frac{X+Y}{2} = 2\sin \frac{X+Y}{2}\left(\cos \frac{X-Y}{2} \pm \cos \frac{X+Y}{2}\right)$
from there, use the sum-to-product identity on
$cos \frac{X-Y}{2} \pm \cos \frac{X+Y}{2}$ and use $\sin \frac{X+Y}{2} = \sin \frac{n\pi - Z}{2}$ to get it back in terms of $Z$ and everything cleans up real nice.[/hide]

Fix A.  It fixes B+C, since $A+B+C = \pi$.

Then $\sum \sin nA = \sin nA + 2\sin (\frac{n(B+C)}{2}) \cos (\frac{n(B-C)}{2}) \le \sin nA + 2\sin (\frac{n(B+C)}{2})$. (*)

Obviously cosine function has maximum 1, obtained when argument is a multiple of $2\pi$.  Thus, $\frac{n(B-C)}{2}$ is a multiple of $2\pi$.  One such multiple is zero.  Thus, when B = C, we will maximize $\sum \sin nA$.  Similarily, A = B and A = C maximizes the sum.

From (*), and using $A = \frac{\pi}{3}$,  it reduces to proving $\sin \frac{n\pi}{3} \le \frac{\sqrt{3}}{2}$.  It is periodic, so there are only 6 values to check:  checking, the LS can assume values $\frac{\sqrt{3}}{2},\frac{\sqrt{3}}{2},0,-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2},0$ and we are done.

Here is my proof.
Easily see that:
a) If at least one of three terms $\sin{nA},\sin{nB},\sin{nC}$ is less than or equal to $0$, the ineq holds.
b) Wlog, assume that $1 \ge \sin{nA} \ge \sin{nB} \ge \sin{nC} \ge 0$, it means $A\in[0,\frac{\pi}{n}]$, then applying Jensen's, we get:
$\sin{nA}+\sin{nB}+\sin{nC} \le 3\sin{\frac{n\pi}{3}} \le \frac{3\sqrt{3}}{2}$
Q.E.D.


If rx-2=5 and qx+3=12then what is r divided by q?

[hide="Solution"]You have a system:
\[ rx\minus{}2\equal{}5\\qx\plus{}3\equal{}12\]Thus: \[ rx\equal{}7\\qx\equal{}9\]Since all of $ r, q, x$ are nonzero, we can divide the first equation by the second, and obtain \[ \boxed{\frac{r}{q}\equal{}\frac{7}{9}}\][/hide]

ok solving for rx and qx we have rx = 7 and and qx = 9 when whe divid rx by qx we have $ \frac{rx}{qx} \equal{} \frac{7}{9}$ cancelling x we have $ \frac{r}{q} \equal{} \frac{7}{9}$ therefor r\q = $ \boxed{\frac{7}{9}}$


"How many different 3 number pairs can you select from 1-100 such that the largest number is greater than the product of the smaller two numbers?"

The intro to counting and probability text book doesn't explain the solution to this problem very well(it just gives you a table :( ), can any one explain the solution to this problem well?

You could use casework.
for example, 1,2,3
1,2,4 so on.

Or you could use algebraic case work... But I need an explanation, not an idea...


Hay VIETNAM from ARMENIA :wink:
Please give me those names who partisipated in IMO 2006 with their e-mails or their names in this forum. :)
And special greetings to Pham Kim Hung for his excelent book Secrets in Inequalities
(IMO Vietnam team gave this book to me in IMO 2006, Slovenia  :wink: )

Thanks and good luck for your team, Armenia. I know Ticks and really like him. I hope to meet him, and also Armenia team this summer.

Vietnam team has 6 members with names here

1, Nguyễn Xuân Chương, lớp 12, THPT chuyên Vĩnh Phúc
2, Đỗ Xuân Bách, lớp 12, Khối THPT Chuyên ĐHKHTN - ĐHQG Hà Nội
3, Phạm Duy Tùng, lớp 11, Khối THPT Chuyên ĐHKHTN - ĐHQG Hà Nội
4, Lê Ngọc Sơn, lớp 11, THPT chuyên Bắc Giang
5, Phạm Thành Thái, lớp 12, THPT chuyên Nguyễn Trãi Hải Dương
6, Đặng Ngọc Thanh, lớp 12, THPT chuyên Quảng Bình.

Actually, I dont know their names and nicks.

Could you send me those who participated last year?
Good luck with your team too. :wink:

[quote="hungkhtn"]I know Ticks and really like him. I hope to meet him, and also Armenia team this summer. [/quote]

Thank you .

Actually, his nick is "Tiks" and not "Ticks". Didn't you know it, or it's a kind of joke between you (you always write Ticks instead of Tiks).

To ARMENIA
[hide]
Narekin es linki tex@ asa. Tox Hung in shnorhakalutyun asi. Mek el hankarts chases vor motd et grqi xeroxna :rotfl: [/hide]

It is not a joke at all, but as I see it Hung just used to write my nickname that way and I asked him several times to correct himself. And, by the way, Vardanch you shouldn't write the nick incorrectly too! :mad:

[quote="N.T.TUAN"]Don't angry as that :P[/quote]

You will get angry too if they misspell your nick name intentionally(I mean Vardanch). :wink:

hungkhtn misspell my nick name , too. He is joke :P

I know that he is not doing that intentionally but someone from above does(see here [url]http://www.mathlinks.ro/viewtopic.php?t=156418[/url] ).


analytic geometry
trigonometry
vector
geometry
perpendicular bisector

Let P be a point on the circle with center (2,2) and radius 1.  Also let A and B have coordinates A(0,2) and B(0,0).  As P moves around the circle, let $\alpha$ be the largest value attained by the measure of angle $APB$.  Compute $\sin{\alpha}$

Please Help

Thanks

If this is hard, please move this to an appropriate subforum.

[quote="mathgeniuse^ln(x)"][quote="Altheman"]hint: consider a circle that is tangent to the given circle and passes through A, and B, the point of tangency is the desired point P, then just find sin APB using vectors or something[/quote]

Okay, thanks.

Is there proof for why this is true?[/quote]

if A, B are fixed points, consider the locus of points, C, such that ACB is constant...[hint it is two arcs of a circle joined together], once we have this, we just make the circles smaller, and the angle larger, until the circles are tangent

Oh yeah, then that makes sense.

[hide]So, I call the center $(x,y)$.

I know that $y=1$ since it must be on the perpendicular bisector of any two points on the circle.

So the point is $(x,1)$

Now, the distance from $(0,0)$ to $(x,1)$ is $\sqrt{x^{2}+1}$

Next, I have to find the point of tangency to the circle.

I get that the radius will be $\sqrt{(2-x)^{2}+1}-1$

I equate the two distances.[/hide]

Am I doing this right?


You and a friend are playing the number guessing game. He thinks of a number between 0 and 36 (inclusive). You must guess the number. After each guess, he tells you if your guess is too high, or less than or equal to the correct answer (if you guess the number, he will tell you if your guess was less than or equal to the number). You win the game if you can tell him the correct number WITHOUT going over more than twice. How many guess do you need before you know the answer?

do you mean the minimum number? the maximum number? or what? depending on the approach, it could vary.

It's like uhm, if you guess 18, I would say it's too high or too low. If I said it's too low, you keep guessing. If i said it's too high, you can still keep guessing. But if you guess this time and I say it's too high, then you'll lose the game (because you can't go over the number twice). If I said it's too low, then you just keep guessing. So how many guesses do you need (I think at least) before you know the answer?

Do you mean that if I guess too high TWICE or MORE THAN TWICE will I lose the game?

[hide="hint (assuming it means TWICE)"]If you have guessed [i]a[/i] and it was too low and you have guessed [i]b[/i] as and it is too high, you [b]MUST[/b] recite a+1, a+2, ... until you get the number or else there is a probability of losing the game, which we don't want!![/hide]

It's twice. But uhm, I think the question here is how many guesses.  :)


calculus
function
integration
limit
vector
geometry
3D geometry

1.  Let $ f(x)$ be a continuous decreasing function from the positive reals to the positive reals.  Put:

$ a_{N}\equal{}\sum_{n\equal{}1}^{N}f(n)\minus{}\int_{1}^{N}f(x)\hspace{1 mm}dx$

(a)  Is it true that $ \lim_{N\rightarrow\infty}a_{N}$ always exists?  Prove it or give a counter-example.

(b)  Take $ f(x)\equal{}\frac{1}{\log(x\plus{}1)}$.  Let $ L$ be the limit of the sequence $ a_{n}$ as $ n\rightarrow\infty$.  Estimate what the minimal value of $ n$ is so $ |a_{n}\minus{}L|<10^{\minus{}6}$.

2.  Let F be the vector field on $ \mathbb{R}^{3}$ defined by:

$ F(x,y,z)\equal{}(2x\minus{}y^{2}\minus{}x^{3},3y\minus{}y^{3},\minus{}x\minus{}z^{3})$.

For a closed surface $ S$ in $ \mathbb{R}^{3}$, consider $ \int_{S}F\cdot n\hspace{1 mm}dA$, the flux of $ F$ through $ S$.  Here $ n$ is chosen to be an outward normal.  For what choice of $ S$ will $ \int_{S}F\cdot n\hspace{1 mm}dA$ be maximal?  Explain your answer and compute $ \int F\cdot n\hspace{1 mm}dA$ in that case.

3.  (a)  Suppose that $ f(x)$ is defined by the rule:  $ f(x)\equal{}1\minus{}\frac{|x|}{\pi}$ for $ x\in[\minus{}\pi,\pi]$.  Compute the integrals:  $ a_{n}\equal{}\frac{1}{2\pi}\int_{\minus{}\pi}^{\pi}f(x) e^{\minus{}2\pi i n x}\hspace{1 mm}dx$.  (These give the "Fourier coefficients" of $ f(x)$.)

(Note from edoo:  actually they don't; this was a type-o in the exam)

(b)  Suppose now that we take:  $ f(x)\equal{}(1\minus{}\frac{|x|}{\pi})^{4}$.  What is:  $ \lim_{n\rightarrow\infty}n^{3}a_{n}$?  You do not need to give a detailed proof, but explain briefly the reasoning behind your answer.

4.  Find, using Lagrange multipliers, the minimum and maximum values taken by $ x^{4}\plus{}y^{4}\plus{}z^{4}$ subject to the constraint $ x^{2}\plus{}2y^{2}\plus{}3z^{2}\equal{}1$.  Briefly explain why the method of Lagrange multipliers works.

5.  (a)  Evaluate the integral $ \int_{S}(x^{2}\plus{}y^{2})\hspace{1 mm}d\sigma$ where $ S$ is the sphere of radius $ 1$ centered at $ (0,0,0)$ and $ \sigma$ is surface area.

(b)  Does the series $ \sum_{n\in S}\frac{1}{n}$ converge, where $ S$ consists of those positive integers whose decimal expansion does not contain the digit 1?

I am not actually looking for any solutions, but I thought some of the problems were kind of fun, so I wanted to share them.  Enjoy!

Btw, great thanks to AoPS for helping me hone my skills for these exams.  I am quite happy to admit that I solved problem 5(b) on this forum just a few weeks ago.   :lol:  (though it was always an old favorite of mine)

[quote]In fact, the problems look like simple   :roll: [/quote]

Really, rolling eyes? Ouch.

I just said the problems were "fun".  I still think that they are interesting to work out.  For example, in problem 2 the surface is not given -- I never saw a problem like that before -- but it is certainly doable.  And the answer to problem 1 is dependent, in general, on the assumptions on $ f$.

Again, enjoy.


parameterization
parametric equation
complex analysis
complex analysis unsolved

Let $ f(z) \equal{} \frac{az\plus{}b}{cz\plus{}d}$, c not 0 and ad-bc =1

a) find the image of the line {Re(z)=-d/c} union infinity
b) the image of the line {Im(z)=1} union infinity

I would appreciate this a lot,

Thanks

I suppose $ a,b,c,d$ are real constants.

(a) we have $ z\equal{}\minus{}d/c\plus{}it$, where $ t\in\mathbb R$ (think of this as parametric equation of the line). It's easy to find the image of something given by a parametric equation: just plug the equation into the formula for $ f$. You get $ f(\minus{}d/c\plus{}it)\equal{}\frac{a(\minus{}d/c\plus{}it)\plus{}b}{cit}\equal{}\frac{at}c\plus{}\frac{ad\minus{}bc}{c^2t}i$. This is again a line (put $ t\equal{}1/s$ to make $ s$ a new parameter), except it's horizontal and is missing one point. But that point is $ f(\infty)$, so it's covered as well.

(a) can be done more easily: the image contains $ \infty$, hence is a line. We also know that it contains $ f(a/c)$ and intersects the real line (which is preserved by $ f$) orthogonally.

(b) The image is a circle $ C$ which is tangent to the real line at $ a/c$. Consider the circle tangent to the real line at $ \minus{}d/c$ and tangent to $ \{\mathrm{Im}\,z\equal{}1\}$. Its image is a line parallel to the real line having precisely $ f(\minus{}d/c\plus{}i)$ in common with $ C$, so the diameter of $ C$ is $ \mathrm{Im}\,f(\minus{}d/c\plus{}i)\equal{}1/c^2$. owk


The pentagon ABCDE in the picture is regular side $ l$ . Knowing that the segment $ \overline{AF}$ is a equal measure of $ l$ , calculate the angle $ B\hat{FE}$.

[img]http://img2.imageshack.us/img2/7562/clsa.jpg[/img]

Triangle $ AEF$ is isosceles, with base angles of $ 72$ degress (external angle of a regular pentagon) so angle $ EAF\equal{}36$ degrees.

We know that angle $ BAE\equal{}108$ degrees, so angle $ BAF\equal{}108\plus{}36\equal{}144$ degrees. Triangle $ BAF$ is also isosceles so angle $ BFA\equal{}\frac{180\minus{}144}{2}\equal{}18$ degrees.

We already know that angle $ AFE$ is 72 degrees, so angle $ BFE\equal{}72\minus{}18\equal{}54$ degrees.

[b][size=150][color=darkred]it has another solution in trigonometry[/color][/size][/b]

Hello almonkez.

Yes ? well ... please, tell me how to do using trigonometry.

KrazyFK, I don't understand very well the angles of triangle. Internal angle of pentagon is 108 degree ? And triangle is isosceles, then 72+72+36 ?

Thanks. ;)

Ah ok Andrew, I understand. Thanks again  :lol:

Luiseduardo, I've edited the above post since you last posted to include a trigonometric solution. Hope it's OK.

[quote="AndrewTom"]As KrazyFK is not on line, I'll try to help.

AEF looks like it's meant to be a straight line ($ 180$ degrees).

The external angle is $ \frac {360}{5} \equal{} 72$.

The internal angle is $ 180 \minus{} 72 \equal{} 108$.

$ 72 \plus{} 72 \plus{} 36 \equal{} 180$

I hope this is of some help.

We could also use trigonometry as follows:

As above, triangle $ AEF$ is isosceles and so angle $ EAF \equal{} 36$.

Then angle $ BAF \equal{} 144$ and triangle $ BAF$ is isosceles.

Denoting $ BF$ by $ d$, the sine rule gives

$ \frac {d}{\sin 144} \equal{} \frac {l}{\sin 18}$

Therefore $ d \equal{} \frac {l \sin 36}{\sin 18}$

$ \equal{} 2 l \cos 18$ (since $ \sin 2\theta \equal{} 2 \sin \theta \cos \theta$)$ \equal{} 2 l \sin 72$

Now from triangle $ ABE$ where the midpoint of $ BE$ is denoted by $ M$,

$ BM \equal{} l \cos 36$

$ BE \equal{} 2 l \cos 36 \equal{} 2 l \sin 54$

therefore, using the sine rule in triangle$ BEF$, we have

$ \frac {\sin BFE}{2 l \sin 54} \equal{} \frac {\sin 72}{2 l \sin 72}$

$ \sin BFE \equal{} \sin 54$

Therefore angle $ BFE \equal{} 54$.[/quote]
 :coolspeak:
thanks AndrewTom
this is what i mean by the trigonometric solution
almonkez


algebra function domain LaTeX blogs \/closed

Signatures have grown too large and distracting.  We will soon be instituting policies regarding signature lengths, etc.  In the meantime, we ask that you shorten your signature to 3 lines of normal-sized text, maximum.  We'll start enforcing this, and other signature restrictions, soon by deleting signatures that violate the policy.

A few guidelines (not yet complete list)

- no more than 3 lines
- at most 1 image tag [img
- at most 2 smilies (an [img tag is also counted as a smilie)
- only internal links allowed ( artofproblemsolving.com or mathlinks.ro domains)
- no sizing allowed [side =small] or other means to size text
- no more than 256 characters
- you [b]can[/b] use $ \text{\LaTeX}$ in your signature, but only in-line equations (no double dollars)
- no "extreme" colors (red, yellow, cyan, violet)
- you cannot use quote, code or list tags

Generally, your signature should not be obtrusive and very distracting.

is my signature too long?   :)

I don't think it's that terribly annoying, but who knows.

[quote="anirudh"]Is my signature too long? Just asking if stuff under hide tags count as your 3 lines...[/quote]
it probably won't count as 3 lines, but...um... hide tags in sigs are...even more annoying than long sigs...so don't use hide tags?

awesome.. i like it anirudh

ACtually now that I think about it, shortened signatures would look less spamish

[quote="anirudh"]OK. Now it is three lines:
A bouncing emoticon, two buttons, and then a wink. :)[/quote]
did you NOT get what I was saying?
PLEASE DO NOT use hide tags in your signature.
If you do, it doesn't display what time you posted, how much and when you edited the post, and ,in the aops board style, the quote button.
It gets REALLY annoying if you have to click on the thing everytime to see those things.
So please don't use hide tags....

[quote="now a ranger"]wow junggi, that was really mean... have patience... poor thing.. :rotfl:

jk[/quote]
OK WTF?
I just said that it is annoying for ppl to have hide tags in their signatures, and told anirudh to not use them, and he ignored me, so I said it again. What's wrong with that?
Oh and if you're going to have "jk" at the end of your post, why don't you just not make that post at all?

[quote="now a ranger"]i just didn't like your tone of voice you use sometimes. You show different colors sometimes.

 :o

oh well.... pretend I never said anything[/quote]
It's probably the tone I use when talking to ppl I don't like.

[quote="now a ranger"]Yeah. So if you like someone you are nice, and if you don't you are mean. And the sad thing is, you don't know most of those people you are mean to. They might be a nice person inside.[/quote]
But nevertheless, they are annoying.

Don't use hide tags in signatures, it's as simple as that.  If you do and sometime tells you it's annoying and you don't do anything...and then someone else comes along and agree that it's annoying then you're very inconsiderate.

Anyway, the bug that makes that stuff be hidden by the hide tag will hopefully be fixed in which case you could have your hide tag in you signature :D (without being called a jerk...)

[quote="axcqwerty"]Oh ok
Can someone specify the exact limitations? since the ones on the signature control panel and the ones in the first post of this topic are different :?:[/quote]

no more than 255 characters
no more than 3 lines
no images
i believe only internal URLS allowed (blogs, etc.)

i think that covers most of it =)

[quote="axcqwerty"]Ok thanks, but one more question:
What about smileys?
one place it says limit 2 and on the panel it says 3 maximum[/quote]

like i said assume the worst but i think it's 3 max...but I heard if they were too annoying the admins might ban them in sigs alltogether?

[quote="kstan013"]No, I mean some of those animated ones like jackdillon and bigboy3236 used to have.[/quote]
If my old one was distracting, is this one?

Not really...it's acceptable i think  :D

You don't have one. Go [url=http://www.artofproblemsolving.com/Forum/usercp.php?mode=editprofile&sub=sig]here[/url] to create/edit it.

is my signature okay?
Also, will we be noticed/warned about our signatures if they're not okay before they're deleted?

I contacted Mr. Vornicu asking whether my signature is acceptable, and I have not yet received a reply. Maybe I'll get an answer here?

[quote="Temperal"]Well, let's see.

[list]
[*]No images - Check
[*]No external links - Check
[*]3 smilie max - Check
[*]No more than 255 characters - Check[/list]

I honestly don't know why you're worried; your signature is one of the least intrusive I've seen.[/quote]
Then I guess mine is okay...

[quote="Temperal"]Well, let's see.

[list]
[*]No images - Check
[*]No external links - Check
[*]3 smilie max - Check
[*]No more than 255 characters - Check[/list]

I honestly don't know why you're worried; your signature is one of the least intrusive I've seen.[/quote]
[quote="Signature CP"]This is a block of text that can be added to posts you make. There is a 255 character limit on 3 line(s), font size 10 to 12. In your text, no more than 75 characters without space too. No images. No more than 1 image(s). No more than 2 url(s). Only internal links allowed. Only 3 smilies are allowed. No code tags. No hide tags. No quote tags. No list tags. [color=red][b]No colors.[/b][/color] No centered LaTeX tags. No youtube tags.[/quote]

[quote="rrusczyk"]no "extreme" colors (red, yellow, cyan, violet)

Generally, your signature should not be obtrusive and very distracting.[/quote]

I wouldn't consider dark green as an "extreme" color.  I think you're okay.  It's not distracting.

Ahh didn't see that... I was just going by the stuff written in the CP.


[hide="Solution"][quote="Inequalities Master"]Consider the triangle $ ABC$, s.t. $ AB \equal{} c, BC \equal{} a$ and $ CA \equal{} b$. The point $ P$ found inside the triangle $ \triangle ABC$ has the property that
for each line that passes through $ P$ and intersects $ AB$ and $ AC$ at the distinct points $ E$ and $ F$, satisfies $ \frac {1}{AE} \plus{} \frac {1}{AF} \equal{} \frac {a \plus{} b \plus{} c}{bc}$. Prove that $ P$ is the incenter of the triangle $ ABC$.[/quote]
BP meets AC at F. We have $ \frac {a \plus{} b \plus{} c}{bc} \equal{} \frac {1}{AB} \plus{} \frac {1}{AF} \equal{} \frac{1}{c} \plus{} \frac {1}{AF}$
$ \Rightarrow \frac {1}{AF} \equal{} \frac {a \plus{} c}{bc}\Rightarrow AF \equal{} \frac {bc}{a \plus{} c}$
$ \Rightarrow FC \equal{} b \minus{} \frac {bc}{a \plus{} c} \equal{} \frac {ab}{a \plus{} c}\Rightarrow \frac {AF}{FC} \equal{} \frac {c}{a}$
So BF is the bisector of angle ABC. Similarly CF is the bisector of angle ACB.
Hence result[/hide].

yeah, very easy indeed.
I was wondering if the converse of this problem is true, that means proving that for each line passing through $ I$- the incenter of the $ \triangle ABC$ and intersecting $ AB$ and $ AC$ at $ E$ and $ F$, respectively, we have that $ \frac{1}{AE}\plus{}\frac{1}{AF}\equal{}\frac{a\plus{}b\plus{}c}{bc}$

[hide="Solution to the converse problem"][quote="Inequalities Master"]yeah, very easy indeed.
I was wondering if the converse of this problem is true, that means proving that for each line passing through $ I$- the incenter of the $ \triangle ABC$ and intersecting $ AB$ and $ AC$ at $ E$ and $ F$, respectively, we have that $ \frac {1}{AE} \plus{} \frac {1}{AF} \equal{} \frac {a \plus{} b \plus{} c}{bc}$[/quote]
Sure!
Let H on AB, K on AC such that IH || AC, IK || AB, then AHIK is a rhomb.
Easy to show that $ \frac{1}{IK}\equal{}\frac{1}{AE}\plus{}\frac{1}{AF}$ for every positions of E on AB, F on AC.
IK is a constant, so $ \frac{1}{AE}\plus{}\frac{1}{AF}$ is a constant
When E=B,  $ \frac{1}{AE}\plus{}\frac{1}{AF}\equal{} \frac{1}{AB}\plus{}\frac{1}{AF}\equal{}\frac {a \plus{} b \plus{} c}{bc}$[/hide]


If $k\in (0,4)$  and $a,b,c>0$ with  $a^2+b^2+c^2+abc=k$  prove that $a+b+c<3$

is easy to see that we have to prove only for $k = 4$.
(if $a^2+b^2+c^2+abc = t < 4$, we have one positive real $d$ such that $a^2+b^2+d^2+abd = 4$ and $d>c$).

let $k = 4$.

suppose that $a = 2cosA$, $b = 2cosB$ and $c = 2cosC$.
we have that, by our initial equality, $A,B,C$ are angles of a triangle.

by jensen, the result follows.


Some of the numbers $ a_{1}, a_{2}, ..., a_{n}$ are $ \plus{} 1$, the others are $ \minus{} 1$. Prove, that:
$ 2\cdot sin(\frac {\pi}{4}\cdot (a_{1} \plus{} \frac {a_{1}\cdot a_{2}}{2} \plus{} \frac {a_{1}\cdot a_{2}\cdot a_{3}}{4} \plus{} ... \plus{} \frac {a_{1}\cdot a_{2}\cdot ... \cdot a_{n}}{2^n}))\equal{} a_{1}\cdot \sqrt {2 \plus{} a_{2}\cdot \sqrt {2 \plus{} a_{3}\cdot\sqrt {2 \plus{} ... \plus{} a_{n}\cdot \sqrt {2}}}}$.

Hint: 2sin(a/2)=sqrt(2+cosa) , a_ (0, pi\2)

[quote="Yuriy Solovyov"]Some of the numbers $ a_{1}, a_{2}, ..., a_{n}$ are $ \plus{} 1$, the others are $ \minus{} 1$. Prove, that:
$ 2\cdot sin(\frac {\pi}{4}\cdot (a_{1} \plus{} \frac {a_{1}\cdot a_{2}}{2} \plus{} \frac {a_{1}\cdot a_{2}\cdot a_{3}}{4} \plus{} ... \plus{} \frac {a_{1}\cdot a_{2}\cdot ... \cdot a_{n}}{2^n})) \equal{} a_{1}\cdot \sqrt {2 \plus{} a_{2}\cdot \sqrt {2 \plus{} a_{3}\cdot\sqrt {2 \plus{} ... \plus{} a_{n}\cdot \sqrt {2}}}}$.[/quote]
It is quite easy . You can prove by induction.

[quote="Yuriy Solovyov"]Some of the numbers $ a_{1}, a_{2}, ..., a_{n}$ are $ \plus{} 1$, the others are $ \minus{} 1$. Prove, that:
$ 2\cdot sin(\frac {\pi}{4}\cdot (a_{1} \plus{} \frac {a_{1}\cdot a_{2}}{2} \plus{} \frac {a_{1}\cdot a_{2}\cdot a_{3}}{4} \plus{} ... \plus{} \frac {a_{1}\cdot a_{2}\cdot ... \cdot a_{n}}{2^n})) \equal{} a_{1}\cdot \sqrt {2 \plus{} a_{2}\cdot \sqrt {2 \plus{} a_{3}\cdot\sqrt {2 \plus{} ... \plus{} a_{n}\cdot \sqrt {2}}}}$.[/quote]
Obviosly, it is not true.

It is true.
Here is general problem :
$ e_i\in\{0,1,\minus{}1\}$
Then $ e_1\sqrt{2\plus{}e_2\sqrt{2\plus{}...\plus{}e_n\sqrt{2}}}\equal{}2\sin(\frac{\pi}{4}\sum_{k\equal{}1}^n\frac{e_1...e_k}{2^{k\minus{}1}}$
Prove by induction.


Hello

Let $ k$ be a positive integer. Prove that:

$ (1\plus{}x)(1\plus{}x^2)*....*(1\plus{}x^{2^{k}})\equal{} 1\plus{}x\plus{}x^2 \plus{} ....\plus{}x^m$, where $ m$ is a positive integer that needs to be determined.

[hide="Proof"]
To show:
$ (1\plus{}x)(1\plus{}x^{2})*....*(1\plus{}x^{2^{k}})\equal{} 1\plus{}x\plus{}x^{2}\plus{}....\plus{}x^{m}$

Multiply the left side by $ \frac{1\minus{}x}{1\minus{}x}$.  Then, we get:

$ \frac{1\minus{}x^{2^{k\plus{}1}}}{1\minus{}x}\equal{}1\plus{}x\plus{}x^{2}\plus{}...\plus{}x^{2^{k\plus{}1}\minus{}1}$

$ m\equal{}2^{k\plus{}1}\minus{}1$[/hide]


Will "stupid" ways to solve a question, such as counting, be awarded full points? (even if its correct?)

If you wrote a ridiculously long proof of an answer that is pure brute force, maybe, but you'd probably miss something.

Anyway, the points are awarded on mathematical correctness. If you did something brute force and missed a single case, they'd probably take off a lot.

I think that some questions actually lend themselves to counting.  All I can say is count smartly, and you'll be fine.


What is the largest subset of the integers 1, 2, ... n
such that no two elements sum to a third element?

--Dan

[hide]Let's call sum-free a subset of the integers 1, 2, ... n

such that no two elements sum to a third element.

. If n=2N, the answer it is N :

* 1,3, ..., 2N-1 is a sum-free subset (odd+odd=even).

* if 1 <= a1 < a2 ... < a(N+1) <= 2N then A(N+1)-a1, A(N+1)-a2, .. A(N+1)-aN takes N different values so one of the a(N+1)-ak must be one ai (by pigeon hole) and so the subset is not sum-free.

. If n=2N+1, the answer is N+1 :

1,3, ..., 2N+1 is a sum-free subset (odd+odd=even).

Same thing to prove that N+1 is the max.[/hide]

Magnara wrote:[hide]Wouldn't it just be 2?  Otherwise you have 1+2=3[/hide]  Or does it not have to start at 1 and not have to be consecutive?

No, they don't have to be consecutive, or start at 1.  For example,

if n=7, you could get the set 1, 2, 4, 7 or the set 3, 4, 5, 6 or ...

Try to both find the biggest set possible and also

give a proof that you can't have a bigger set.

--Dan

Belenos is [hide]right[/hide]. [hide]Nice job.[/hide]

--Dan


Let $k \in \mathbb{N} , k \ge 2$.Prove that the each natural number is showable by
sum  of the [b]constant[/b] number of power $k$th numbers.(By power $k$th, i mean , a number like $A$ which is like $A=B^k,B \in \mathbb{Z})$.for clarifying the meaning of problem , let $g(k)$ be the smallest value of this [b]constant[/b].then $g(2)=4$(we can show each natural number by sum of 4 perfect squares)

Hope this may help somehow:  http://mathworld.wolfram.com/WaringsProblem.html

and I suppose it is really hard problem - good luck  :D

:read:  :fool:  :fool: Mathematication would be able to duel whit it . Sorry for posting it (jesus, what is this :o ?)


There are two cities on the same river. There is a boat going back and forth between the two cities. Its engines always work by the same power, so the boat always travels by the same speed relative to the water in the river. It takes the boat 2 hours downstream, and 4 hours upstream to get from one city to the other. If you kick a soccer ball into the river in the city farther up the river, how long does it take the ball floating down the river to get to the other city?

This confuses me.

[hide]$2(b+w) = D$

$4(b - w)=D$

$4b - 4w = 2b + 2w \Rightarrow b = 3w$

Substituting in

$8w = D \Rightarrow \frac Dw = \fbox {8}$[/hide]

[hide][color=green]x = speed of boat (without currents)
c = the current
d = the distance between the cities
d/c = how long the ball takes
2(x+c) = d
4(x-c) = d
2x+2c = 4x-4c
6c = 2x
3c = x
using the first line and the last
2(4c) = d
8 = d/c
it takes the ball[/color][color=indigo] 8[/color][color=green] hours to get to the next city
[/color][/hide]

Another similar problem:

At the airport, The Antarctican team's coach said Cuddles could aimlessly walk the moving sidewalk, but only in the right direction.  Cuddles was dissapointed since he wanted to know how fast he could walk in the wrong direction.  Suddenly, he had an idea. Standing still, it took him 120 seconds to get from one end to the other.  By waddling in the right direction, it took just 40 seconds.  How long would it take to get from one end to the other if he waddled in the wrong direction?
After 15 min of walking, the moving sidewalk broke. How long would it take to waddle from one end to the other?

[hide][color=indigo]l = length of sidewalk
s = speed of sidewalk
c = Cuddle's speed
c-s = walking in wrong direction
l/s = 120
l = 120s
l/(s+c) = 40
l = 40s+40c
120s = 40s+40c
80s = 40c
2s = c
l/c-s = l/2s-s = l/s
so it would take him 120 seconds in the wrong direction[/color][/hide]


Equilateral triangle $ ABC$ has side length 400 cm and a perimeter equal to 100 times the perimeter of equilateral triangle $ DEF$. How long is each side of triangle $ DEF$?

If an equilateral triangle has side length $ 400$, then its perimeter is $ 3(400) \equal{} 1200$.  This divided by $ 100$ is $ 12$.  $ \frac{12}{3} \equal{} \boxed{4}$


[quote="shortstack"]A man just finished painting his house and needs something more.  At the hardware store the clerk shows him what he wants and says “One is $ \$1$.” “Fine” says the man. “I took 600, so here’s $ \$3$.”  What had he bought?[/quote]

[hide]He was buying digits for the house number, right?  :lol: [/hide]

BTW, questions like this should be posted in Puzzles & Brainteasers, IMHO.

HaHa he was buying number 4 his house :D


geometry
3D geometry
number theory unsolved
number theory

Let $ n$ be a natural number. A cube of edge $ n$ may be divided in 1996 cubes whose edges length are also natural numbers. Find the minimum possible value for $ n$.

First notice that $n^3\geq 1996$ which implies that $n\geq 13$.
Now we just find a configuration for $n=3$.
If we use $1984$ cubes of edge $1$, $11$ of edge $2$ and $1$ of edge $5$ then the we have $1996$ cubes which form a cube of edge $13$.

[hide = Solution]
[color=#f00]Claim 1[/color] \\
$ n \ge 13$ \\
Proof:
If we want to divide a cube into the maximum number of cubes, we want the edge to be $1$. So each cube with edge $n$ can be divided into at most $n^3$ cubes.
If $n = 12$ can be divided into at most $12^3 = 1728$ cubes. So $n \ge 13$ \\ \\
[color=#f00]Claim 2[/color] \\
$n = 13$. \\
Proof: \\
Let $e$ be the length of edge of the cube.\\
We divide the main cube with $n=13$ into $2197$ cubes. Then we have $201$ cubes more than we want. Hence we make $25$ cubes with $e=2$ of $200$ cubes with $e=1$. Then we have $200-25 = 175$ cubes less, hence $26$ cubes more than we want. So we make a cube with $e=3$ of $27$ cubes with $e=1$: So we have $27-  1= 26$ cubes less. So we have $1996$ cubes.
[/hide]


search
probability
ceiling function
probability and stats

Two players play a game as follows:

Player 1 chooses a number between 1 and $ n$.  Player 2 has $ q$ questions to try to determine what the number is.  In stage 1, Player 2 asks questions of the form, "Is the number less than $ k$?" and Player 1 answers truthfully.  Stage 1 ends when either Player 2 uses all $ q$ questions or when Player 1 answers "yes" to one of Player 2's questions.  After Stage 1 ends, Player 2 gets the opportunity to guess which number Player 1 had selected; Player 1 wins if Player 2 guesses incorrectly.

What strategy for Player 2 yields the best worst-case probability of winning (and what is this probability)?
If Player 1 chooses the number uniformly at random, what strategy for Player 2 yields the best average-case probability of winning?

(Feel free to move this if it fits better elsewhere.)

It's obvious that the best strategies come from asking an increasing sequence; any $ k$ that isn't greater than all preceding values is completely wasted. This effectively divides the possible values into $ q+1$ groups. Once player 1 knows which group the number is in, she guesses randomly in that group.

These strategies all result in exactly a $ \frac{q+1}{n}$ probability of winning with the uniform distribution. To maximize the worst case, make these groups approximately equal in size by asking if the number is $ \le \frac{mn}{q+1}$ at the $ m$th guess. The worst-case probability is then $ \frac1{\lceil n/(q+1)\rceil}$.


logarithms
number theory unsolved
number theory

Denote by $\gamma (k,p)$ the smallest number $s$ with the property, that if $a_{1}, a_{2}, ..., a_{s}$ are not divisible by prime $p$, then congruence:
\[a_{1}x_{1}^{k}+...+a_{s}x_{s}^{k}\equiv 0 \ (\mod p) \]
has a solution in which not all numbers $x_{i}$ are divisible by $p$.

1) Estimate $\gamma (k,p)$
2) Show that for $k=\frac{p-1}{2}$ we have $\gamma(k,p)=1+[\frac{\ln p}{\ln 2}]$.

For the first one:
We will prove that $s=k$ works.
Let us consider the sets $A_{i}=\{a_{i}.0,a_{i}.1^{k},...,a_{i}.(p-1)^{k}\}.$ Because of the fact that the equation $x^{k}=c$ has at most $k$ solutions in $Z_{p},$ it follows that $|A_{i}|\geq 1+\frac{p-1}{k}$(because $0\in A_{i}$).
[b]Theorem(Generalization of Cauchy-Davenport)[/b]
\[|A_{1}+...+A_{n}|\geq min\{p,|A_{1}|+...+|A_{n}|-(n-1)\}. \]
By this theorem it follows that $|A_{1}+...+A_{k}|\geq min\{p,|A_{1}|+...+|A_{k}|-(k-1)\}=$
\[=min\{p,k(1+\frac{p-1}{k})-k+1\}=p. \]

For the second.
$x_{i}^{k}\equiv \pm 1$
then Consider the $2^{k}$ sums:
$a_{i_{1}}+a_{i_{2}}+a_{i_{t}}$ where $t\leq k,i_{1}<\dots<i_{k}$
then if $2^{k}>p$ then there must exist two of the sums above be equvilent $(\mod p)$
so the difference of them has the form$\sum b_{i}a_{i}$where $b_{i}\in \{-1,0,1\}$ and not all of $b_{i}$ be $0$.then we get the result.
if$2^{k}<p$ then let $a_{i}=1$ ,then there is no soluton for the equation.
Sorry for the bad english.


nu stiu sa folosesc latexu dar sper ca se intelege ceva din ce vreau sa zic
1. x,y,z Nr reale pozitive
x(y+1)+y(z+1)+z(x+1)>=6 ori radical din xyz

2. a,b,c nr reale pozitive a.i a+b+c+d=1 atunci sa se arate ca
radical din8a+1 + radical din 8b+1 + radical din 8c +1 + radical din 8d+1<8

pls pana maine daca se pote sa ma ajutati

$ {Prima}$ $ {inegalitate}$:
$ x, y, z > 0\Rightarrow$ $ x.(y \plus{} 1) \plus{} y.(z \plus{} 1) \plus{} z.(x \plus{} 1)\ge\sqrt [6]{xyz}$

$ {A}$ $ {doua}$ $ {inegalitate}$:
$ a, b, c, d > 0 \land a \plus{} b \plus{} c \plus{} d \equal{} 1 \Rightarrow \sqrt {8a \plus{} 1} \plus{} \sqrt {8b \plus{} 1} \plus{} \sqrt {8c \plus{} 1} \plus{} \sqrt {8d \plus{} 1} < 8$.

$ {Iata}$ $ {solutia}$ $ {primei}$ $ {inegalitati}$, $ {folosind}$ $ {inegalitatea}$ $ {MA}\ge{MG}$:
$ x.(y \plus{} 1) \plus{} y.(z \plus{} 1) \plus{} z.(x \plus{} 1) \equal{}$
$ \equal{} (xy \plus{} yz \plus{} xz) \plus{} (x \plus{} y \plus{} z)\ge3[(\sqrt [3]{xyz})^2 \plus{} \sqrt [3]{xyz}]\ge\sqrt [6]{xyz}$.   ($ c.c.t.d.$)

$ {A}$ $ {doua}$ $ {mi}$-$ {se}$ $ {pare}$ $ {mult}$ $ {prea}$ $ {slaba}$, $ {deoarece}$ $ {potrivit}$ $ {inegalitatii}$
$ {MA}\le{MP}$:   $ \frac {x \plus{} y \plus{} z \plus{} t}{4}\le\sqrt {\frac {x^2 \plus{} y^2 \plus{} z^2 \plus{} t^2}{4}}$,
$ {avem}$:
$ a, b, c, d > 0 \land a \plus{} b \plus{} c \plus{} d \equal{} 1 \Rightarrow$
$ \Rightarrow \sqrt {8a \plus{} 1} \plus{} \sqrt {8b \plus{} 1} \plus{} \sqrt {8c \plus{} 1} \plus{} \sqrt {8d \plus{} 1}\le$
$ \le4.\sqrt {\frac {(8a \plus{} 1) \plus{} (8b \plus{} 1) \plus{} (8c \plus{} 1) \plus{} (8d \plus{} 1)}{4}} \equal{} 4.\sqrt {2.(a \plus{} b \plus{} c \plus{} d) \plus{} 1} \equal{}$
$ \equal{} 4.\sqrt {3} < 4.2 \equal{} 8$.  ($ c.c.t.d.$)

$ {Numarul}$ $ 8$ $ {din}$ $ {membrul}$ $ {drept}$ $ {al}$ $ {acestei}$ $ {inegalitati}$ $ {poate}$ $ {fi}$ $ {inlocuit}$ $ {cu}$ $ {7}$, $ {fiindca}$:
$ 4\sqrt{3}\equal{}\sqrt{48}<\sqrt{49}\equal{}7$.


If $ a,b,c$ are non-negative real numbers prove that $ a^{2}\sqrt{b^{2}\minus{}bc\plus{}c^{2}}\plus{}b^{2}\sqrt{c^{2}\minus{}ca\plus{}a^{2}}\plus{}c^{2}\sqrt{a^{2}\minus{}ab\plus{}b^{2}}\leq a^{3}\plus{}b^{3}\plus{}c^{3}$

After using amgm $ a^2\sqrt {b^2 \minus{} bc \minus{} c^2}\le \frac {a^3 \plus{} ab^2 \plus{} ac^2 \minus{} abc}{2}$, it's Schur.

A similar solution by Cauchy -

$ (LHS)^2 \leq (a(b^2 \minus{} bc \plus{} c^2 ) \plus{} b(c^2 \minus{} ca \plus{} a^2 ) \plus{} c (a^2 \minus{} ab \plus{} b^2 ))(a^3 \plus{} b^3 \plus{} c^3 ) <\equal{} (a^3 \plus{} b^3 \plus{} c^3 )^2$
(Schur inequality of degree 3 is used)


Dennis Phillips just got owned but hes still in 8th place! GO DENNIS!

[hide="cough cough"]I have a royal flush.[/hide]

i have a royal flush that pwns urs more :D

All royal flushes are the same, smarty pants.

Not if its a royal flush that goes JQKACHUCKNORRIS!


linear algebra
matrix
algebra
system of equations
IMO Shortlist

Solve the following system of equations, in which $a$ is a given number satisfying $|a| > 1$:

$\begin{matrix} x_{1}^2 = ax_2 + 1 \\ x_{2}^2 = ax_3 + 1 \\ \ldots \\ x_{999}^2 = ax_{1000} + 1 \\ x_{1000}^2 = ax_1 + 1 \\ \end{matrix}$

An shortened solution for this one (I spent too much time on it and I'm a bit fed up with it now ;) ) :

Wlog, $a > 1$, by transforming $a \rightarrow -a$ and $x_i \rightarrow -x_i$

It's not difficult to prove (once you have guessed it) that either $0 > x_i > -1 \ \forall i$ either $x_i \geq 1 \ \forall i$.

In the first case, if $x_i \geq x_j$ then $x_{i+1} \leq x_{j+1}$ from which we have then
$x_1 = x_3 = .... = x_{999}$ and $x_2 = x_4 = .... = x_{1000}$

or $x_{2i+1} = (-a \pm \sqrt{4-3a^2})/2$ and $x_{2i+1} = (-a \mp \sqrt{4-3a^2})/2$

In the other case, they are all equal $= (a + \sqrt{a^2+4})/2$


Letters A, B, C, D, E, F are arranged in order such that A is not at 1st, B is not at 2nd and F is not at 6th. How many ways are there of arrangement?

[hide="answer most likely to be wrong"]

$27000$ because

there are 5 different letters that can be in the first slot, and 5 in the second...

there are 6 different letters that can be in the 3rd, 4th, and 5th...

and 5 different letter possibilities in the 6th slot

multiply

$5*5*6*6*6*5= 125*216= 27000$

[/hide]

That reasoning is wrong. Just notice that the total number of permutations w/o any restrictions is 6!=720. So we know the answer x is less than 720.
One way to think of it is to count the number of ways that do not satisfy the condition, and then subtract this number from the total. This is useful because the number of ways that do not satisfy the condition is easier to find.
We have 6 ways that do not satisfy the condition, because A must be 1st, B must be 2nd, F must be 6th, and the last 3 letters can be placed in 3!=6 ways. The total number of permutation is 6!=720, so the answer should be 720-6=714.
The reason your reasoing is wrong is because you are ignoring the fact that your argument is conditional: different letters in the 1st and 2nd slots will affect how many choices you have later. For example, if B is the 1st letter, we have 5 choices for the 2nd letter. However, if F is the 1st letter, we have 4 choices for the 2nd letter.
That is why it is not always a good idea to use "conditional" combinatorics arguments.

[quote="drunner2007"]We have 6 ways that do not satisfy the condition, because A must be 1st, B must be 2nd, F must be 6th, and the last 3 letters can be placed in 3!=6 ways. The total number of permutation is 6!=720, so the answer should be 720-6=714.[/quote]

No the number of ways that it DOES NOT satisfy the conditions is if A is first OR b is 2nd OR F is 6th.

This is what i got (which is probably wrong)

[hide]5*4*3 * 3! = 360[/hide]

Shoe factory, your right about the "does not" condition. I got 426 though, accounting for the "or" cases with PIE

[hide]If A is the first letter, there's 120 ways to arrange the remaining. The same goes for B and F. However, in this counting, we've overcounted the number of ways to arrange the letters such that A and B are in the first and second places respectively, etc. each once, and each of these combinations (AB, BF, AF) have 24 ways to arrange the remaining letters.

So far we have 120(3)-24(3)=288.

Note also, however, that we've counted the ways to arrange the letters such that A, B, and F are in the 1st, 2nd, and 6th places respectively (which is 6 ways) twice when we subtracted the 3 24's. Thus we add 6 back again.

So we have 288+6=294.

These are the ways that we don't want to find, so since there's 720 ways to arrange the letters ABCDEF, there are 720-294=426 ways to arrange the letter with the specified conditions. [/hide]


Here is a problem that i cant really figure out how to solve:

     Three points are chosen randomly and independently on a circle.   What is the probability that all three pairwise distances between the points are less than the radius of the circle?

[hide]Firstly, if any two points to have a distance of less than the radius of the circle they must fall on an arc of less than 60 degrees.

Now fix one point on the circumphrence of the circle.  The probability that both of the other points fall within 60 degrees of this point is (1/3)^2 or 1/9.

Given that both of the points are within 60 degrees of the first one there is a 3/4 chance that they fall within 60 degrees of each other (this can be shown by probabilistic geometry -- if you are not familiar with this I will post it tomorrow).

Thus the chance that the distances between all three points are less than the radius is 1/9 * 3/4 or 1/12.[/hide]

[edit] put in spoiler


Putnam
invariant
combinatorics unsolved
combinatorics

$ \sum_{i\equal{}1}^n |a_i \minus{} b_i| \equal{} \sum_{i\equal{}1}^n [a_i \plus{} b_i] \minus{} 2 \times \text{min}\,(a_i,b_i) \equal{} n(2n\plus{}1) \minus{} n(n\plus{}1) \equal{} n^2$, since the minimal $ n$ elements are $ a_1, \ldots, a_i$ and $ b_{i\plus{}1}, \ldots, b_n$.

Very nice!  When this problem came up in MIT's Putnam seminar, there were two other solutions presented:  one showing that the given quantity is invariant under a move that generates all possible configurations, and the other by Pigeonhole.


integration
trigonometry
calculus
function
calculus computations

Could someone tell me what is wrong with this?

We attempt to evaluate $ \int (e^{ix} \sin x) \,dx$

Using $ \int u \frac{dv}{dx} \,dx \equal{} uv \minus{} \int v \frac{du}{dx} \,dx$    with $ u\equal{}\sin x$ and $ \frac{dv}{dx}\equal{}e^{ix}$:

$ \int (e^{ix} \sin x) \,dx \equal{} \minus{}ie^{ix} \sin x \minus{} \int (\minus{}ie^{ix} \cos x) \,dx$

Integrating by parts again on this second integral with $ u\equal{}\cos x$ and $ \frac{dv}{dx}\equal{}\minus{}ie^{ix}$ gives:

$ \int (e^{ix} \sin x) \,dx \equal{} \minus{}ie^{ix} \sin x \plus{}e^{ix} \cos x \plus{} \int (e^{ix} \sin x) \,dx$

Cancelling $ \int (e^{ix} \sin x) \,dx$ from both sides:

$ \minus{}ie^{ix} \sin x \plus{} e^{ix} \cos x \equal{} 0$

$ e^{ix}(\cos x \minus{}i\sin x)\equal{}0$

Using Euler's identity:

$ (\cos x \plus{} i\sin x)(\cos x \minus{} i\sin x)\equal{}0$

$ \cos^2 x \plus{} \sin^2 x \equal{} 0$

$ 1\equal{}0$

Suppose we have functions $ f, g, h : \bf R \to R$ such that $ \frac{d}{dx} f(x) \equal{} h(x) \equal{} \frac{d}{dx} g(x)$.  What does this imply about $ f(x) \minus{} g(x)$?

That $ f(x) \minus{} g(x)$ is constant. I see the fallacy now, thank you.


[url=http://usamts.org/Register/U_Register.php]Here[/url], it says to put the grade you were in during the 08-09 school year. Is that supposed to be 09-10 school year?  :maybe:

I believe the site hasn't been updated for year 21, the 09-10 school year.  I think (although I am not 100% sure) that that registration form is for the 08-09 USAMTS.  Once the round 1 problems are posted, it will probably be changed to say 09-10.

Does that mean I should wait a while before registering? If not, should I put what grade I'll be in during the 09-10 school year?

I would be safe and wait until the problems are posted (and all the years changed to 09-10, year 21, etc.) before registering.

It actually doesn't matter, because you can change your personal info even after you register on the website (just click "My USAMTS" in the sidebar, then "My Profile" in the top bar).

If you want to register now for the website, that's fine.  We do not yet have the forms set up for 2009-10 (that will hopefully happen in the next couple of weeks).  All students will have to update their information for 2009-10 once the contest starts.


[quote="CheeseIsGood"]It's the last day before school starts.

I feel like I'm going to be executed at sunrise.  :([/quote]
Hahahahhaha today wasn't that bad for me, I didn't like getting up early and going to school but otherwise its fine. I think over the winter break I was able to get rid of some of my procrastination problems by forcing myself to do a lot of math (well math is fun of course as everyone knows :)). I think I've developed a new outlook on school, school days are the same things as regular vacation days just missing 8 hours (time for homework is gained by waking up early, as I wake up at around 12 everyday during breaks :D)

[quote="CheeseIsGood"][quote="serialk11r"][quote="Ubemaya"]Uh... school only takes up 5-6 hours if you use your lots of spare time effectively (unfortunately, I prefer starting at the hot chicks).[/quote]
Hmmmm....well lets see, school ends 3 ish so that's about 3-4 hours (sometimes i'm up by 11 :D), homework is pretty much always under 1.5 hours... Okay crap my estimation skills suck. If I think about it, that's about right. I go to chess for like 2-3 hours on tuesdays and thursdays, I go to math circles for 2 hours on fridays, I play piano 2-3 hours a day, play trombone like 40 minutes (yea i know its a hella wierd instrument for asians), then I get distracted, read physics crap, chat, do math, read random interesting stuff, takes up the rest of the time until 12 am woot! (I start homework at 10 everyday actually, and yes including today, I'm doing all that crap until 10)[/quote]

Does this qualify as ultimate Asianness?  :P[/quote]
Dunno...lol


geometry
calculus
Princeton
college
linear algebra

Hey,
I was just curious to find out how many hours you guys or gals devote to homework per night?

[quote="joml88"]Probably at least 5.  I'm in college though.[/quote]

Wow, that is alot.

[quote="PTynan89"]Maybe 1/2 hour per night.

I'm only taking 5 APs and 2 college courses (1 per semester), though, along with 2 honors classes, so I guess that's why.[/quote]

Yeah, really, what ever happened to the days when kids actually took challenging course loads.

And actually, I change what I said earlier.  That should say "on average 5 hours" not "at least".

8th grade= slack time

absolutely zero ( not really)

actually more like 20 to 30 minutes mmmk  :)

I actually spend more time on geometry for fun then work from teachers since I can do it in school, and they don't assign much daily.

[quote="tokenadult"][quote="PTynan89"]Maybe 1/2 hour per night.

I'm only taking 5 APs and 2 college courses (1 per semester), though, along with 2 honors classes, so I guess that's why.[/quote]

Which AP courses are those? What is most of the assigned work, reading?[/quote]

My schedule this year is:
1) Discrete Math (honors)
2) Modern Physics (also honors, although most ppl take AP before or while taking this)
3) AP Stats (mostly done in homeroom)
4) AP Computer Science (mostly classwork)
5) Organic Chemistry (through University of Delaware; ochem I is first semester, ochem II is second)
6) AP English Lit (most of this is reading)
7) AP Economics (micro is covered first semester, macro during the second, which is why this one counts as 2)

I do about 0-1 hour amount of homework. I'm in 8th grade.

I also spend 0-1 hour on homework. The problem is that I need to spend at least an hour everyday,and I don't, so the work sometimes piles up.

[quote="PTynan89"]My schedule this year is:

1) Discrete Math (honors)

2) Modern Physics (also honors, although most ppl take AP before or while taking this)

3) AP Stats (mostly done in homeroom)

4) AP Computer Science (mostly classwork)

5) Organic Chemistry (through University of Delaware; ochem I is first semester, ochem II is second)[/quote]

So most of the reading you can keep up with during the school day? What is the textbook for each class I left listed here? What kind of homework or lab work is involved in the O Chem class?

I'm also taking 5 AP classes (AP Latin, Comp. Sci., Calculus, U.S. History, and Econ.) and i have tons of work everyday......

[quote="tokenadult"]So most of the reading you can keep up with during the school day? What is the textbook for each class I left listed here? What kind of homework or lab work is involved in the O Chem class?[/quote]
When I was in high school, I took almost every AP class my school offered (except languages, art, music), which meant 5ish both junior and senior years, and I also took a university class in grade 12. I almost never did homework at home, except when I was assigned essays.

Now, it's hard for me to tell how much time I spend on homework 'on average' because I tend to do a lot of work some days and very little on other days. On average, maybe 4, 5, 6 hours a day? But really some days it's like 0 and some days it's like 10.

University is way more fun and interesting than high school. But you have to be willing to put in a bit more effort if you want a good education. ;)

All this so far makes me wonder how much writing is assigned in some classes. I can well believe that a student can do all the reading assigned in high school or beginning college classes before going home (I did that, in my day), but my wild guess would be that writing takes longer. And my dad, the chemistry major from half a century ago, said he used to be in the lab OVERNIGHT to run some experiments. How time-consuming are lab science courses these days in high school and in college?

[quote="tokenadult"]All this so far makes me wonder how much writing is assigned in some classes. I can well believe that a student can do all the reading assigned in high school or beginning college classes before going home (I did that, in my day), but my wild guess would be that writing takes longer.  [/quote]
I think AP has poisoned our education system in that way. On the AP test, you have 90 minutes to write two essays, or 2 hours for three essays, or whatever. That's all the writing they test you on. (I mean for English and for the humanities, like history, geography, government.) So (in my AP classes at least) the writing assigments are geared towards preparing you for AP. We wrote essays in class fairly often, but we rarely had to write long essays at home.

It's nice and useful to be able to write a clear and interesting 2-4 page essay in 30 minutes, but I don't think I was taught at all (not even poorly, just not at all) good long essay writing skills. I had to struggle a bit in my freshman writing seminar at Princeton, because I didn't have much practice with crafting a long and engaging work, and with the art of polishing my words (which you don't need to worry about on a short timed essay).

[quote="tokenadult"]And my dad, the chemistry major from half a century ago, said he used to be in the lab OVERNIGHT to run some experiments. How time-consuming are lab science courses these days in high school and in college?[/quote]
I don't know where PTynan89 goes, but chemistry classes have very long labs at Princeton. I don't know if they ever have to stay overnight in lab, but I know that some of my friends spend many hours (two digits easily) every week in lab even just for the first year of organic chemistry.
In high school, we rarely have long labs. I guess in biology and chemistry I had a few labs that were on the order of 3-4 hours (of work -- I had some that were 2 days but most of it just consisted of things sitting around, and it was still just 3-4 hours of work total). In physics at Princeton our labs were usually around 3 hours, once per week. But those were intro classes.

For OChem, Charter has a special arrangement with UD where the professor comes to Charter 5th period on Monday, Wednesday, Friday and teaches.  (There is a similar program for Calc 3/ODE, one semester each, both of which I took last year.)  However, there is no lab involved.  None of the homework is collected, although I usually do most of it.  The book is just a standard college OChem book, and the only grades we have are 4 tests per semester + a final.  We have had one test so far and are having our second one this coming Wednesday.

And to Xevarion: I did already say that the course was through University of Delaware.

Also, there are sometimes longer essays, but since they are not assigned on a TYPICAL NIGHT, I am disregarding them for the sake of this discussion.

I also want to point out that I had much more hw last year, due to the courses I was taking, but not many options were left for me this year.

[quote="neelnal"]I have multivariable calc, linear algebra, and then 1 honors, and 5 ap's.

5 hours per night at least. about 40 hours a week for hw.

i hate hw.

the worst was sUMMER work, which burned me out:
AP Bio:
158 open-ended questions - i typed 63 pages single space, 10 font, times new roman

1 essay, - 2 pages isngle space 10 font

1 lab- about 10 or 11 worksheets

234 multiple choice questions- i got a 70%ish on that

one letter to hte teacher- pshh half a page max?

read 18 chapters in bio book+prepare for 2 tests on first week of school - 350+ pages

Ap English:

critically analyze and write on 25 essays - 15 pages typed 1.5 spacing

read the 25 essays lol- 102 pages

read East of Eden - 500 pages? 600 pages? yea 600

write essay on east of eden- 5 pages single spaced

AP European:

write an essay on renaissance and read the Prince exceprt - 20 pages reading, 4 pages double space

Multi variable, linear algebra:

Read 6 chapters - 100 pagesish
Do around 300 problems

So, yea. That was my summer wasted. I hardly even able to log on during the summer to AOPS.[/quote]

Wow. Look at the bright side, you are now a better rounded student.

I'm only better rounded cuz they burnt me like a hot dog  :mad:

LOL.

I hate AP Biology too much memorizing, very little skill or intelligence involved. Same goes for AP European, but that class is fun. Learning about how all the Christians wanted to kill each other in the 1500's and then how Ivan the Terrible of Russia killed his son. So fascinating, note the sarcasm.

And AP Statistics!!!!! I abhore that class ferociously.

Good thing I have really nerdy friends that like to do math rather than boring school work.

[quote="Xevarion"]University is way more fun and interesting than high school. But you have to be willing to put in a bit more effort if you want a good education. ;)[/quote]
I learned that the hard way in college.  I had no clue how to study.  I was used to getting by with good grades without having to do homework.  I started out that way at college, but eventually I figured out that if I sat down and read the books and wrote out the homework, 1) the material was really interesting, and 2) I learned a lot more.  :-)

Grad school -- it all depended on how burnt out I was feeling.  Some days I didn't do as much hw as I should have; other days I did a lot, catching up.  The burn out was a shame, because the material by that time was $really$ interesting.  I loved that stuff (my doctorate is in chemical physics, specifically quantum scattering theory, which is all applied math).

Best,
Catherine


Derive Brahmagupta's Formula, the general formula for the area of a cyclic quadrilateral.[url]http://mathworld.wolfram.com/BrahmaguptasFormula.html[/url](3)


probability
expected value
probability and stats

there are real random variables X,Y and Z.
Calculate:

Var(X) = Cov(X,X)

Cov(X,Y) = Cov(Y,X)

Cov(aX + bY,Z) = a*Cov(X,Z) + b*Cov(Y,Z)       for a,b E R

greez

There's nothing difficult or tricky about any of these. Just write out the definitions to see what you have, and maybe occasionally use the linearity of the expected value - that is, $ E(U\plus{}V)\equal{}E(U)\plus{}E(V)$ and $ E(cU)\equal{}cE(U).$


[quote="ryan22"]Rationalize the denominator: I know this is easy, i'm just brushing up on old algebra thanks:

$\displaystyle\frac{16} {\sqrt 6}$[/quote]
[hide]
$\frac{16}{\sqrt{6}}$

$\frac{16}{\sqrt{6}}*\frac{\sqrt{6}}{\sqrt{6}}$

$\frac{16\sqrt{6}}{6}$

$\boxed{\frac{8\sqrt{6}}{3}}$[/hide]

[quote="ryan22"]Rationalize the denominator: I know this is easy, i'm just brushing up on old algebra thanks:

$\displaystyle\frac{16} {\sqrt 6}$[/quote]

[hide="rationalization"]multiply numerator and denominator with ${\sqrt 6}$ and u get

$\frac{16 \sqrt 6}{6}$

simplify and ur done[/hide]

[quote="ryan22"]Rationalize the denominator: I know this is easy, i'm just brushing up on old algebra thanks:

$\displaystyle\frac{16} {\sqrt 6}$[/quote]

$\frac{16} {\sqrt 6}$

Multiply numerator and denominator by $\sqrt 6$: $\frac{16*\sqrt 6}{6}=\frac{8}{3}\sqrt 6$


LaTeX
greatest common divisor
number theory unsolved
number theory

Hi..
how can i proof that 2^n - 1 =/= k.n for all values of n? *k and n are naturals...
tkz for help : )

Take $p$ to be the smallest prime factor of $n$. You then have $p|2^n-1$. At the same time, you know that, since $p$ is odd, $p|2^{p-1}-1$. Now use the result $(2^m-1,2^n-1)=2^{(m,n)}-1$ (where $(x,y)$ is the greatest common divisor of $x,y$) for $m=p-1$. Since $p$ is the smallest prime factor of $n$, you get $p|(2^{p-1}-1,2^n-1)=2^{(n,p-1)}-1=2-1=1$, which is clearly a contradiction.

Welcom Igor! Just so you know, you can use LaTeX on this site. If you do not know what latex is, a full-tutorial is available at
http://www.mathlinks.ro/LaTeX/AoPS_L_About.php

Ok Valentin, i will try to learn about LaTeX( but i dont now english very much)...
Grubber, i understand your solution, but dont know the result of greatest common divisor that you used... i will try to prove it with my hands now, but suggesting very difficult to me... tkz


In collatz conjecture if we call A the operator and if n=4k+3 show that we can find s such that
A^(s+1)<A^s

I do not see the problem. Were you trying to reply?

Define $f(n) = n/2$ if $n$ is even, and $f(n) = 3n + 1$, if $n$ is odd. Collatz conjecture claims that regardless of the starting point the iterations settle eventually into a 3-cycle: 4, 2, 1, 4.

So I think, senouy asks us to prove that if starting point is $n=4k+3$, then the sequence cannot be non-decreasing.

Can someone prove this please. I would also like to see the proof for it. Thank you.

Take an odd naturel number n,calculate 3n+1 wich is even,divide then by the greatest power of 2 in it.You will get an other odd number.Do the same thing to this number and so on.Collatz conjectured that after a finite number of iterations ,no matter what  n is , you will always get 1.
We can see that if n has a 4k+1 form then the next step we will get a number less than n (((3k+1)))
 take n=4k+3,
 Apply the process.You get a sequence of naturel odd numbers.Show that the sequence is not increasing.


Find the smallest value of the expression
\[ A \equal{} \sqrt {1 \plus{} \sqrt x }  \plus{} \sqrt {1 \plus{} \sqrt y } \]

in which x, y is the number two : \[ {x^2} \plus{} {y^2} \equal{} 1\]

Is it assumed $ x \ge 0$ and $ y \ge 0$?

Yes, that is necessary for both $ \sqrt x,\sqrt y\in\mathbb R$.


I have a question, I remember seeing somewhere long ago that a solution to a system of linear equations can contain infinite solutions as long the equation's variables are in a ratio, for ie: "The equation satisfies the system as long as a is 2b and b is 3c."

I would I be able to recognize this were the case when I am solving a system?

[quote]The equation satisfies the system as long as a is 2b and b is 3c.[/quote]
For that system you could just write $a=6c, b=3c, c=\text{(whatever)}$ and get infinitely many solutions. But here are three variables in only two equations. For simpler systems (like linears or this ratio) substitution should be enough to figure out whether there are infinite, finite, or no solutions.

Does this only occur when number of equations < number of variables?

You won't have trouble solving a system when the number of distinct equations = number of variables.  You will have trouble when you have more equations than variables (checking is required), less equations than variables (infinite solutions), or unhelpful non distinct equations like: $x+2y = 2$ and $2x+4y = 4$.

You don't necessarily get infinite solutions when you have less equations than variables. That's only true for homogenous systems. However, if there is one solution there are infinitely many even to an inhomogenous system with less equations than variables.

Also, even if you have as many equations as variables you can have 0, 1 or infinite amount of solutions.


inequalities
logarithms
integration
calculus
calculus computations


[quote="M4RI0"]If $c \geq b \geq a \geq 0$. Prove that:

$(a+3b)(b+4c)(c+2a) \geq 60abc$[/quote]

This inequalies is:
-2ba^2 + 4ac^2 +8ca^2 +3cb^2 + 6ab^2 + 12bc^2 >= 60abc - 25abc

-Now whith AM>=GM ( For example:a^2b + b^2c + c^2a >= 3abc).

In End we have:
2ac^2 + 2ca^2 + b^2c + 6c^2b >=11abc. Here i have problems!

[quote="NkMAsTeR"]Remenber that $c \geq b \geq a \geq 0$[/quote]

Yeah!   :wink:
 I forget!

But I think your solution is already right  :wink:


See point 5 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]this post[/url].  (By the way, you could have easily found that thread by using a search.)  Neither eqnarray nor align support multiple similarly-spaced equations in the same horizontal line, but that's okay because it would probably be ugly and difficult to read.  If you really, really, really want to do it this way, you could use an array (but you'll have to tinker to make the spacing around the equals signs look right), but I really suggest you put each equation on its own line.  Also, use \cdot instead of * for multiplication.


Let $ f$ be a stricly increasing function, defined in $ [0,1]$ such that $ f(0)=0$ and $ f(1)=1$.

If $ \frac{1}{2}\leq \frac{f(a+b)-f(a)}{f(a)-f(a-b)}\leq 2$; for all $ a, b$ such that $ 0 \leq a+b \leq 1$

Show that $ f(\frac{1}{3}) \leq \frac{76}{135}$


topology
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

Prove that $ R^{n}$ and $ R^{m}$ are not homeomorphic, for $ n$ and $ m$ different positive integers.

You'll need a higher-dimensional version of the Jordan curve theorem:
[quote]If $ f\colon S^{n\minus{}1}\to\mathbb R^{n}$ be an embedding (i.e. homeomorphism onto its image), then $ \mathbb R^{n}\setminus f(S^{n\minus{}1})$ has two connected components.[/quote]
Suppose that $ m>n$ and $ f\colon\mathbb R^{m}\to\mathbb R^{n}$ is a homeomorphism. The sphere $ S^{n\minus{}1}$, considered as a subset of $ \mathbb R^{m}$, has connected complement. Hence $ \mathbb R^{n}\setminus f(S^{n\minus{}1})\equal{} f(\mathbb R^{m}\setminus S^{n\minus{}1})$ is connected, contradicting the theorem.