geometry AMC USA(J)MO USAMO college circumcircle inequalities

im taking it
but im not going anywhere
[sarcasm]
hopefully ill get higher than you this year, what with your apparent degernation in you math skills
YESH!!
[/sarcasm]
good luck all

My goal: 1-3 if there's no geo...(of course I'm never going to score that high but whatever)...
Good luck!

Dudes I'm so freaking pissed right now. BAMO is starting is less than half an hour, and I'm so sleepy I can barely think.

I tried this reaction-time game, which is supposed to test your sleepiness. Normally I get < 0.2 seconds, but today I was averaging over 0.4, wtf.
http://www.bbc.co.uk/science/humanbody/sleep/sheep/

I have a feeling I'm gonna do reallyyyy bad on the BAMO.

dude. that game is awesome. that's the coolest game iv played in a while.

So... I'm guessing discussion will be okay after about 7:00-8:00?  For some reason I had to take it during the school day.  my lips are sealed though  :)

[quote="DiscreetFourierTransform"]Let's wait until tomorrow to talk about it, on the off chance that people are taking it after 7:00.[/quote]

This is unlikely because then the test would go to 11:00.

[quote="Ubemaya"]So, who's going?

I'm aiming for 2 full problems and hopefully a whole bunch of pity points. That's about what I got last year, and I think with the studying I've done in the past 2 months, I'm at about at my last year's skill, so this seems reasonable.

I did the 2004 and 2005 tests for practice, and my scores vary really crazily, though. On 2005, I solved only ONE problem (#2). That one only took me 15-20 minutes. I spent another 50 minutes on #1, but I couldn't solve it for stupid reason. Granted, I only spent a little more than an hour total, so I'll probably do better than that on the actual test. On the other hand, I did 2 1/2 whole problems on the 2004 test, which is even past my goal. Hopefully that's gonna happen again tomorrow![/quote]
dudes I am so pissed
okay so this is what happened
I solved the first one like immediately
took another 10 minutes writing a solution
okay
then I skimmed the other 10 problems and varied my thoughts for 10 minutes
so 20 minutes spent
I spent another 5 minutes on the second problem and got nowhere
I spent another 2 minutes on the third and got nowhere
then I spent two hours on the fourth and got a few similar triangles that didn't help
I went back to the third one and found out that the answer was 255 in about 5 minutes (but didn't prove it)
I went back to the fourth one for another hour and got nowhere else
so like half an hour left
I spend like another 15 minutes on the second and fifth one (mostly the second) but get nowhere
then I prove that 255 is the answer to the third one within another 10 minutes
but I had 5 minutes left and couldn't write a solution in that amount of time
so I got a 7....  :(
PS: no-one in my school solved the fourth one
dang I am so pissed

arch: they grade it significantly light than the USAMO
sometimes, a non-rigorous answer missing a few cases will get a 7

For me it's pretty much the same story as QuattoMaster 3000, including finding the wobbly similar triangles that apparently did nothing, except I messed up on the Alcatraz problem (254, which I couldn't prove and then couldn't remember how I came up with).

And I had zero insight for problem 2, which seemed boring (was the answer 17?)

Does anyone want to post a full solution?

[quote="The QuattoMaster 6000"]
...
Ubey beasted it today, he got a 28
...
so I got a 7....  :(
...
[/quote]

My understanding is that the solution papers will be send to Berkeley to grade. It will take at least a few weeks for the graders to finish grading.

The QuattoMaster 6000, how can you get the scores so early right after the contest? (after all, it's not short-answer type).

james4l, I think that's assuming his solution to #1 is well-written and unflawed.  How exactly is the BAMO graded?  Is it like USAMO?

For me, I'm sure I got #1 100% right.
#2, I just bsed it and gave a mini solution on maximizing points starting from the 4x4 square. I think I got it right. I probably won't get full credit though. However, I didn't get 17 so maybe I was wrong.
#3, I failed and put down 2 answers. I said 2 teams because i was like OMG TIES but I also saw that the more reasonable answer was 255!
#4 I missed totally. I drew a pic and might recieve .5? or 1 point lol.
#5 I got it and probably got it right.

[quote="Ubemaya"]Can we start using the actual problem numbers instead of 1, 2, 3, 4, 5? That way CircleSquared and the middleschoolers won't get confused when they log on.

#3, I solved fully. Only took 5 minutes or so :) .
#4, I got the right answer (16), but my proof is kind of handwaved.
#5, I got the right answer (255), I'm not sure I rigorously proved it though. I didn't really show why 255 worked cause I thought my proof for why 254 didn't work showed why 255 worked (it's kind of a constructive proof).
#6, I saw Quatto spending 3 hours on it and swearing at his paper, so I didn't try :D .
#7, I think I gave a good proof. The answer is yes, right?

So my max is 28 as Quatto says, but more likely I got around 18, same as last year. Still pretty good :) .[/quote]

Yes I got 16 as well. My proof totally failed becuase I just gave a mini proof showing the best optimization. I got that for #7 as well but I don't think my proof was too good. I didn't even draw a good pic for #6 lol

[quote="happyme"][quote="The QuattoMaster 6000"]by the way, the number 6 BAMO problem can be nicely applied to the following IMO 1998 Shortlist problem:
[quote="orl"]Let $ M$ and $ N$ be points inside triangle $ ABC$ such that
\[ \angle MAB \equal{} \angle NAC\quad \mbox{and}\quad \angle MBA \equal{} \angle NBC.
\]
Prove that
\[ \frac {AM \cdot AN}{AB \cdot AC} \plus{} \frac {BM \cdot BN}{BA \cdot BC} \plus{} \frac {CM \cdot CN}{CA \cdot CB} \equal{} 1.
\]
[/quote]
this was number 4 on the shortlist
apparently, the BAMO liked the 1998 IMO Shortlist
because number 2 on the 1998 Shortlist was BAMO 1999 number 5

and dude, junggi, the BAMO was more messed up this year than it normally is
like normally people get 7s for proper solutions
bowei got like a 1 on number 3 (I didn't see his solution though, so I dunno)
and Ubey got a 1 on number 7, which I am pretty sure is solution is valid
also the ordering was totally messed up this year (most of the reason I didn't try number 7 was because I thought it was harder than 6)
whateves

didn't go to the seminar
how was it guys?
and yeah tall middleschoolers are cool[/quote]

There are two other BAMO questions (I think) that I know were on the shortlist. Maybe they don't think anyone looks at those on Kalva?

The seminar was okay...

The speech was on Lexicodes which have little practical use aside from Error Correction in computers. However, he had some pretty interesting things about Vector Addition. On the other hand, lets just say the Professor didn't look entirely respectable. To put things in sense, my mom was asking me who the hobo sleeping in the front row was before he gave the speech.

Yes, tall middle schoolers do rock. I think I was the only tall one there. However, I do not rock (at math).[/quote]

Rofl, that hobo happens to be one of the most respected mathematicians of the 20th/21th century.

[quote="happyme"]The speech was on Lexicodes which have little practical use aside from Error Correction in computers. [/quote]
i suggest you abandon this attitude pretty soon
math is not about applying stuff to real life; its about doing math for the sake of doing math

and yeah mr conway was a very strange man indeed
awesome nevertheless

I'm sorry that the connotation of my words implied that I believed so.

In reality, I was just following up a question someone asked at the ceremony that I don't believe got properly answered (It was something along the lines of the practical use of Lexicodes). Most of the time, I do math for the fun I get from Problem Solving.

If I only enjoyed the useful aspects of math, I don't think I would've found Vector Addition much too useful either.

Just clearing some things up.

Its also interesting to see the different ways Mathematicians have presented themselves over the last (9?) years of the BAMO. Some have been eloquent and well dressed and some have had...questionable...attire.

:lol:

He is a genius though.

The most thing I got from John Conway's lecture is:
As long as you can create a consistent system in math (even if you make unconventional changes of some conventional definitions), that's a good math.

His lecture style was pretty engaging.

This summer I heard Conway speak 4 times. It was very interesting - BTW what did Conway talk about? I couldn't go this year.

[quote="junggi"][quote="daermon"][size=200]CONWAY IS AWESOMEEEEEEE[/size][/quote]
LOL I SAW ERIC AND I WAS LIKE
HMMM I SAW THIS GUY BEFORE
THEN I WAS LIKE LOL WHATEVER
THEN I WAS LIKE LOL ITS DAERMON[/quote]
LOLOLOLOLOLOL
i recognized you right awayyyyyy
tweedle dee
tweedle dum

[quote="bookaholic"]Wait...the speaker was *John Conway*?! ...Argh. Sigh.[/quote]
It said on the yellow sheet. Oh wait maybe not everyone got one.

[quote="Fermat1601"]So Daermon, were you the one who spraypainted "CoNWAY" on the Olin Hall wall?

WHat id dCOnway talk about anyway?

(I won a prize at BAMO but couldn't go to the awards :'([/quote]
lol...no that wasnt me
it was this weird lexicode thingy...pretty cool
and im positive youre kolya now so stop denying it  :D

Yes....

But what [u]did[/u] Conway talk about????

[quote="james4l"]for the lexicode theorem this is kind of like a summary of what Conway said: [url]http://www.dpmms.cam.ac.uk/Seminars/Kuwait/abstracts/L25.pdf[/url][/quote]
at least you could check what was already posted in the topic...

I heard Conway talk about pinecones and triangles last summer :)


geometry
3D geometry
number theory solved
number theory

I've got three similar problems, but am not sure how to solve it. Got something like a solution, but don't know if it is correct.
So maybe you can come up with some solutions ?

1)Let x and y be positive integers such that xy divedes x*x + y*y + 1. Show that x*x + y*y + 1 / xy = 3

2)Determine all pairs (a,b) of integers for which a*a + b*b + 3 is divisible by ab

3)Let x, y and x*x + y*y + 6 / xy be positive integers. Show that x*x + y*y + 6 / xy is a cube of an integer.

The first one was solved on this forum before. The main idea is that if (x^2+y^2+1)/xy=k where x<=y then (a^2+x^2+1)/ax=k also, where a=(x^2+1)/y <= x. By this way we get a decreasing sequence of solns, until we reach the pair (x, y)=(1, 1), then we calculate k for this one and it's 3. The other 2 might work in a similar way. I'm thinking..

You do exactly the same for the second one, the only difference being that there are 2 possible pairs where you can stop (I mean minimal pairs): (1, 1) and (1, 2). The solns are generated by these pairs by the transformation (a, b)->(b, (b^2+3)/a) where b>=a. In the first case (a^2+b^2+3)/ab=5 and in the second case (a^2+b^2+3)/ab=4.

The problem could be stated like this: if ab|a^2+b^2+1 then prove that (a^2+b^2+3)/ab=4 or 5.

OK but that doesn't give an explicit formula
Do you think this is sufficient ?
because they ask for "all pairs of integers"

I don't know if it's enough, but it's really close to being a full soln.

 :D
For the 3'rd one:
It's the same trick as in the other 2: Let property P(k) mean that (a^2+b^2+6)/ab=k. Then if (a, b) has property P(k) then (b, (b^2+6)/a) and ((a^2+6)/b, a) are also pairs with property P(k). If a=b then it means that a^2|2a^2+6, so a^2|6 so a=1=b. Then we can consider the case b>= a+1. If a<b by the transformation (a, b)->((a^2+6)/b, a) we get another pair s.t. (a^2+6)/b <=a so we can continue this until we get (a^2+6)/b > a, which means that a<=5. We just check all these cases and we can obtain all the initial solns. From these initial pairs a<=b we get all our solns by the transf. (a, b)->(b, (b^2+6)/a). I didn't check all the cases, but I think that (a^2+b^2+6)/ab is always 8. I'm not sure, we should check all the cases a<=5. For a=1 we get b=1 (after some calculations)  Even if not all pairs have P(8) we can still check a few cases and we'll probably get perfect cubes for every pair that works (pair that works means pair for which ab|a^2+b^2+6).

You can see that this trick comes in really handy.

the official solution for the 3-rd one was looking for the min. x and for this x the min. y for what x*x + y*y + 6 = n and n is a positive integer. Then it showed that this min. solution is (1,1) and then n = 8. But which I can't now clearly see, is the fact that if n = 8 for (1,1) then why should n = 8 for all the pairs (x,y) ?

Am I right, that with these decreasing sequense of solns (in positive integers) we can show that for some integers a < b (when we took a solution (a,b) at first):  (a^2 + 1)/b = a ?


Let $ a,b,c$ be positive numbers such that $ abc \ge 1$
Prove that $ a^3\plus{}b^3\plus{}c^3\ge ab\plus{}bc\plus{}ca$

assume that $ a\geq b\geq c$ then with Chebychev:
$ a^3\plus{}b^3\plus{}c^3 \geq \frac{1}{3}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2) \geq ab\plus{}bc\plus{}ca$
because $ a^2\plus{}b^2\plus{}c^2\geq ab\plus{}bc\plus{}ca$ and by AM-GM: $ a\plus{}b\plus{}c \geq 3\sqrt[3]{abc}\geq 3$

[quote="skywalkerJ.L."]Let $ a,b,c$ be positive numbers such that $ abc \ge 1$
Prove that $ a^3 \plus{} b^3 \plus{} c^3\ge ab \plus{} bc \plus{} ca$[/quote]
We can use AM-GM :)
$ a^3\plus{}b^3\plus{}1 \ge 3ab  ,b^3\plus{}c^3\plus{}1 \ge 3bc, c^3\plus{}a^3\plus{}1 \ge 3ca$
and$ ab\plus{}bc\plus{}ca \ge 3$ :)


[quote="t0rajir0u"]
By Jensen's

$\sum_{cyc}\frac{a^{2}}{2ab+b^{2}+c^{2}}\ge \frac{3}{2+2ab+2bc+2ca}$[/quote]

I'm probably missing the obvious but what function did you use?


topology
advanced fields
advanced fields unsolved

Let $ G$ be a group acting on a manifold $ M$.   Show that if the action is free and properly discontinuous, then the quotient space $ G/M$ is Hausdorff.

[b]Remark: [/b]  The action is [b]free[/b] if there is no fixed point in $ M$ by the action.    The action is [b]properly discontinuous [/b] if for every compact set $ K$ in $ M$
\[ \# \{g\in G : g.K \cap K \neq \varnothing \} < \infty\]

Let $ O\equal{}\{(x_1, x_2)\in M\times M : \exists g\in G, \ x_1\equal{}g\cdot x_2\}$. Can you prove that $ M/G$ is Hausdorff iff $ O$ is closed in $ M\times M$?

I believe $ M$ need only be a locally compact Hausdorff space.

Note that $ O$ is the image of $ f: G\times M\to M\times M$ by $ (g,x)\mapsto (g\cdot x, x)$. Since $ M$ is locally compact Hausdorff, so is $ M\times M$, thus $ f$ is a closed map. Thus $ O\equal{}f(G\times M)$ is closed, so by the first question above $ M/G$ is Hausdorff.

[quote="HilbertThm90"]Let $ O \equal{} \{(x_1, x_2)\in M\times M : \exists g\in G, \ x_1 \equal{} g\cdot x_2\}$. Can you prove that $ M/G$ is Hausdorff iff $ O$ is closed in $ M\times M$?

[/quote]

This follows from the fact that the projection onto equivalence classes from $ M$ to $ M/G$ is a closed map, isn't it?

[quote="HilbertThm90"]

I believe $ M$ need only be a locally compact Hausdorff space.

Note that $ O$ is the image of $ f: G\times M\to M\times M$ by $ (g,x)\mapsto (g\cdot x, x)$. Since $ M$ is locally compact Hausdorff, so is $ M\times M$, thus $ f$ is a closed map. Thus $ O \equal{} f(G\times M)$ is closed, so by the first question above $ M/G$ is Hausdorff.[/quote]

I don't get why if $ M$ is locally compact Hausdorff, then $ f$ is a closed map.   I know that if  $ G \times M$ is also locally compact Hausdorff (is it?) and $ f$ is a proper map (is it?) then $ f$ is a closed map.  In fact, $ f$ would be a proper map if the projection is but I don't think that the projection is necessarily a proper map.   Should I use the fact that the action is properly discontinuous?   Also, I haven't used the fact that the action is free.

I was trying to prove the Hausdorffness of $ M/G$ by the usual definition and I think the key point (as you also stated) is that $ M$ is locally compact and Hausdorff.   What I was able to do was to separate a point $ x$ from the rest of the elements of its orbit, but I can't do it for any two of them.

I hope you can help me since this problem is making me crazy

I think two claims will clear up the questions.

Claim 1: A continuous proper map $ f: X\to Y$ where $ Y$ is a (reasonable) Hausdorff space (locally compact will suffice), then it is closed.

Proof: Let $ A\subset X$ be closed. The claim is that $ f(A)$ is closed. So we'll show that its intersection with every compact subset is closed. Let $ K\subset Y$ be compact, then $ f^{\minus{}1}(K)$ is compact. Thus $ A\cap f^{\minus{}1}(K)$ is compact, since it is a closed subset of a compact set.

Thus $ f(A\cap f^{\minus{}1}(K))$ is compact since $ f$ continuous.

Now $ f(A\cap f^{\minus{}1}(K))\equal{} f(A)\cap K$ (this is one of those weird cases where equality actually works, it doesn't depend on compact or closed, $ f(B\cap f^{\minus{}1}(C))\equal{}f(B)\cap C$ in general). Thus $ f(A)$ intersect any compact set is closed and hence is itself closed.

So now the claim that will make the original post work is if we can show the $ f$ given there is proper.

Proof: Let $ L\subset M\times M$ be compact. Then let $ K\equal{}\pi_1(L)\cup \pi_2(L)\subset M$ where $ \pi_j$ is the standard projection onto the $ j$-th factor.

Now $ f^{\minus{}1}(L)\subset f^{\minus{}1}(K\times K)\equal{}\{(g, p) : g\cdot p\in K \ and \ p\in K\}\subset G_K\times K$. Where $ G_K\equal{}\{g\in G : g\cdot K\cap K\neq\emptyset\}$. Note that properly discontinuous gives us $ G_K$ is discrete and hence compact, and $ K$ is compact by assumption, so $ G_K\times K$ is compact.

By continuity, $ f^{\minus{}1}(L)$ is closed, so it is a closed subset of a compact set and hence compact. Thus the preimage of compact sets under $ f$ are compact and hence $ f$ is proper.

Throughout writing this, I've become convinced that this is probably not the most efficient method of proving this...


linear algebra
matrix
algebra unsolved
algebra

$C_{n}=A^{n}-B^{n}$ satisfies $C_{n+1}=12C_{n}-35C_{n-1}$ for all $n\in\mathbb{N}$,
and $x^{2}-12x+35=0$ has roots $5$ and $7$.
So there exist $E,F\in\mathbb{R}^{2\times 2}$ with $C_{n}=E\cdot 5^{n}+F\cdot 7^{n}$ for all $n\in\mathbb{N}_{0}$.
$n=0,1$ gives $E,F$.

Indeed $A^{n}={7\cdot\text{Id}_{2}-A\over 2}5^{n}+{A-5\cdot\text{Id}_{2}\over 2}7^{n}$ and $B^{n}={7\cdot\text{Id}_{2}-B\over 2}5^{n}+{B-5\cdot\text{Id}_{2}\over 2}7^{n}$ for all $n\in\mathbb{N}_{0}$.


geometry
3D geometry
sphere
calculus
rectangle
prism
analytic geometry

I was helping some of my fellow students out with some Calculus problems dealing with optimization and volumes of solids of revolution the other day.

One of the problems was trying to find the maximum area of a rectangle inscribed within a circle of radius $ R$.  (I successfully explained how to do this, thankfully.)

So, it got me wondering is it feasible to find the maximum volume of a rectangular prism inscribed within a sphere?  If so, how would one do this?  Feel free to state and make any necessary assumptions, although one might just need to know that the radius of the sphere is $ R$.

In addition to that, what would one do to find the maximum volume of a cylinder inscribed within a sphere of radius $ R$?  Again, feel free to state and make any necessary assumptions; it would probably help to make it a right cylinder.  One possibility might be to deal with revolving a rectange inscribed within the coordinate axes and a quarter-circle arc in the $ xy$-plane and then revolving this around an axis and doubling the result....?

THANKS!

A cube inscribed in a sphere has a unique volume; its diagonal is the diameter of the sphere.  (Are you talking about a rectangular prism?)

The techniques in three dimensions are the same as in two dimensions.  For the cylinder problem, for example, set the radius of the cylinder to $ r$ and compute its height in terms of $ r$ and $ R$.

Yes, I meant a rectangular prism.  Sorry, brainfart. >.<  But would it end up being a cube anyways?

Ah, yes, thanks for the tip with the cylinder.  D'oh... I completely forgot about that, lol.

Thanks!

How about the maximum volume of a rectangular prism inscribed in a sphere of radius $ R?$ We can express this as follows:

Maximize $ V\equal{}8xyz$ subject to the constraint $ x^2\plus{}y^2\plus{}z^2\equal{}R^2.$

Now the problem is that we have left the realm of Calculus I (single-variable calculus) because there are too many variables. We need a multidimensional technique.

There is a beautiful and powerful technique, called Lagrange multipliers.

Briefly, if we want to maximize $ f(x,y,z)$ (or however many variables there are) subject to the condition $ g(x,y,z)\equal{}C,$ then we introduce an extraneous variable $ \lambda,$ called the multiplier, and solve the equations $ \nabla f\equal{}\lambda\nabla g,$ together with the constraint equation. This often gives is a clear understanding of the "shape" of a solution, and is often well worth using even for those problems usually stated in a single-variable calculus course. The solutions will be critical points; we still must decide whether the critical points are maxima, minima, or something else.

In this case, we get these three equations:

$ \frac{\partial}{\partial x}: \quad yz\equal{}2\lambda x.$

$ \frac{\partial}{\partial y}: \quad xz\equal{}2\lambda y.$

$ \frac{\partial}{\partial z}: \quad xy\equal{}2\lambda z.$

If any one of $ x,y,$ or $ z$ is zero, then two of those variables are zero. That gives us one collection of critical points. Of course, for these, the volume is zero, so they're not the desired maximum.

So assume that none of $ x,y,$ or $ z$ are zero. Go ahead and solve for $ \lambda:$

$ 2\lambda\equal{}\frac{yz}{x}.$

Substitute that in the second equation:

$ xz\equal{}\frac{yz}{x}\cdot y,$ which becomes $ x^2z\equal{}y^2z.$

Since we are also assuming that $ z\ne0,$ then we have $ x^2\equal{}y^2.$ We similarly get that $ x^2\equal{}z^2,$ so $ x^2\plus{}y^2\plus{}z^2\equal{}0.$ Since we need $ x,y$ and $ z$ to be positive to make it clear what the box is, we have that our box of maximal volume has $ x\equal{}y\equal{}z$ and is thus a cube - the very cube that t0rajir0u mentioned.

[quote="Kent Merryfield"]Maximize $ V \equal{} 8xyz$ subject to the constraint $ x^2 \plus{} y^2 \plus{} z^2 \equal{} R^2.$

Now the problem is that we have left the realm of Calculus I (single-variable calculus) because there are too many variables. We need a multidimensional technique. [/quote]

Or AM-GM...  :P


Companies/Websites advertise on google.  Some also pay to get their websites.  Notice when you search on google. there's a "sponsered links" section.

Yes, 90% of profits come from advertising, and advertisers will vie for the chance to get a sponsored link. Google does special 'auctions' for those spots. Google.. google is very powerful... you will see  :lol: .

The greatest company is Wal-Mart, they have the greatest revenue!!! I cant believe that Microsoft is only 40 somethingish. Why, some in the top 10 I never even heard of, I mean to p companies in revenue.

A cool side note: Wal-Mart has the most advanced computer system in the world, next to the army's. It studies automatically what people buy under all sorts of conditions, and then self programs itself to buy more of that in the future, to certain areas, all automatically - for instance it found that in areas that were going to be hit by a hurricane, people bought pop tarts cuz it doesnt go bad and doesnt need to be heated.

I wonder if advertising alone can earn enough for their survival.  :oops:  Look, when google put the Havard library stock on the web, the publishers may lose a great deal, much more than it could possibly earn by advertising!

Haha. Advertising on Google isn't like putting an Ad in the local newspaper. We're talking thousands...millions of dollars for an Ad. I mean think about it. 200 million searches a day. 200 million people could potentially see your Ad. What kind of money would you pay for that if you were a multi-billion dollar company with a hundred-million-dollar marketing budget? And outside of Google, their powerful AdSense program brings advertisements to places that Google itself doesn't even go.

According to last Friday's Wall Street Journal, Microsoft now actually looks at Google as a potential software competitor. I find that quite interesting. And who holds how many AOL shares is getting into the mess as well.

Google Earth is really cool, you can use it to spy on government  buildings and stuff. :ninja:  :?  It's also really useful, some guy was playing with at, and found some thousands-year old villa from the b.c.'s...does anyone know how often google earth images are updated?   :D  :D

According to their own [url=http://investor.google.com/fin_data.html]financial statements[/url], Google has had about 2.6 billion dollars in advertising revenue alone in the first half of 2005, and a net profit of about 700 million dollars.

Oh my, that is expensive advertising.  :huh:

So, through advertising on search engines alone, they could become an internet empire?

Wow! :D

Google Earth is amazing.

[quote="hwenterprise"]How does Google earn profits?

It's a search engine...

That won't earn enough money to buy a house!

How could it be that big of a corporation if all it is is a search engine/program company? Google Maps seems to be the only program that Google earns money off of. Does Google earn money off of Froogle?

I am confused.[/quote]

Maybe people give them small amounts of money to include them in the search engine...

Also, they make money from the advertisements. (right side of search screen)

People dont have to pay to get on the search engine, I think a lot of aopsers can be found on google, including this forum.

[quote="Valentin Vornicu"][quote="hwenterprise"]So, through advertising on search engines alone, they could become an internet empire?
Wow! :D[/quote]*cough**cough* they [b]are[/b] an internet empire in case you haven't noticed :)[/quote]

:blush: That's what I meant :blush:

Is Yahoo anywhere near overpowering Google?

[quote="hwenterprise"]:blush: That's what I meant :blush:

Is Yahoo anywhere near overpowering Google?[/quote]I don't think so. Basically the only company that is overpowering them and probably will at least in this decade is Microsoft.

[quote="samath"]Search "failure" on Google!![/quote]I've posted that already on some topic :D It's because of Google bombing ;)

"Overpowering" has a lot of different interpretations...

If you're talking about the number of searches conducted, I'd say Google has a fair lead. But if you look at the size of their "communities" (the number of user accounts they have), their international reach, their services, and even things like revenue Yahoo is definitely way way up there. Yahoo also happens to own millions of shares of Google (I forgot the exact percentage of Google's stock that they have...).

As far as revenue and stuff like that goes, Amazon and EBay are also pretty high up. And although right now it's just a community sponsored by donations, it would probably be wise not to overlook Wiki's growth.

Actually, Yahoo holds an immense share in the search engine industry. Its a misconception that microsoft is close behind google - both google and yahoo hold around the same power in the serach engine industry really, google however with a little bit more, microsoft only holds about 20-25%, and askjeeves and others hold ther est.

My parents enjoy "Googling." Lots of people use Google instead of Yahoo. That's a good thing for Google.

Ever tried Snap!? snap.com. Otherwise, Mama uses Blingo which uses Google since she *might* win a free iPod, and the rest of us use Google.

It's Google's 7th birthday... :icecream: (that's the closest to a happy-birthday emoticon they have).

Unrelated: Why don't they put the most-used emoticons at the top? They could do it dynamically so it would be kind of a pop-poll.

Billy

:!:  :?:  :idea:  :arrow: Yeah, this row is hardly ever used. They should put the star wars there instead.

lol, google has a very interesting way of symbolizing birthdays/anniversaries. (small chunksof cake around the letters and a 7 replacing the "L" to show their number of years)

I love it when Google does that :D

I missed it, though. :(


conics
parabola
analytic geometry
graphing lines
slope
calculus
calculus computations

Hey all, i've got a few tangent to curve equations which, through trial and error, i found the answers to... but i do not understand the process of showing the correct work or using the correct method to finding the answers. could anyone help me?

1.  the line y=3x+k is tangent to the curve y=x^3 when k is equal to?

[b]answer = 2 or -2[/b]

2.   the two tangents that can be drawn from the point (3,5) to the parabola y=x^2 have slopes?

[b]answer = 2 and 10[/b]

sh0ebox

[hide="Number 1"]
$y=x^3\Rightarrow y'=3x^2$
$3x^2=3\Leftrightarrow x=\pm 1\Rightarrow y=x^3=\pm 1$

So the tangent line goes through either $(1,1)$ or $(-1,-1)$.
First case: $y-1=3(x-1)\Leftrightarrow y=3x-2$
Second case: $y+1=3(x+1)\Leftrightarrow y=3x+2$

$\therefore k=\pm 2$
[/hide]

[b]EDIT:[/b] nevermind i figured it out! but i've got a series of questions now that are related and i don't know how to approach them. anyone willing to help? it has to do with the [i]motion of a particle in a plane[/i]

sh0ebox


At a chess tourney, each player played with all the other players two matches, one time with the white pieces, and one time with the black pieces. One point was given for a victory, and 0,5 points were given for a tied game. In the end of the tourney all the players had the same number of points.

a) Prove that there exist two players with the same number of tied games;

b) Prove that there exist two players which have the same number of lost games when playing with the white pieces.

I think I'm misunderstanding, because the problem seems wrong to me. Point (a) is ok:

Let's assume each player has a different number of ties. All players must have a number of ties with the same parity, because the parity of the number of ties tells us if the total score, which is common to all players, has form $n$ or $\frac {2n+1}2$. The number of ties of a player ranges from $0$ to $2n-2$. We must have $n$ different numbers from this set, all with the same parity, so they must be $0,\ 2,\ ..,\ 2n-2$. This means that there is a player $A$ who has tied in all his matches, and there is a player $B$ with no ties. What happens when $A$ plays with $B$? This obviously yields a contradiction, so there must be two players with the same number of ties.

As for point (b), I think it's very easy to give a counterexample: Let's assume we only have two players in the tournament, $A,\ B$. When $A$ plays with the white pieces he wins, and when he plays with the black pieces he loses. $A$ has $0$ lost games when playing with the white pieces, while $B$ has $1$ lost game when playing with the black pieces. However, the total score is $1$ for both $A$ and $B$, so the conditions of the problem are satisfied. This sort of counterexample can easily be extended to more players, so a condition like "at least three players" doesn't work.

What am I doing wrong?

you made the same mistake as most contestants did. probably you are not very fond of chess :P :)

anyway, the thing is that the problem ask to prove that there exist two players which have the same number of lost games when playing with the [b]white [/b]pieces.

in fact in your example, both A and B have 0 losses when playing with the white pieces. :)

Oh my! That's a very stupid mistake :blush:

Sorry..

where´s the slution of item b?
please,give me a solution  :D

Chavez, be patient!  Or, better, go solve the question yourself :)

[hide="Brief solution outline"]We know each player plays n - 1 games as white.  Thus, for each player to win (or lose) a distinct number of games as white, some player must win (n - 1), another (n - 2), and so on, and some player must win 0 as white.  Now, each player has the same number of points, so this number is (n - 1).  The player who wins 0 as white must therefor win all (n - 1) games he plays as black.  However, he plays a match as black against a player who (playing white) won all of his games.  Since they both cannot win, our assumption that each player could win a distinct number of games as white is absurd and so two players must win the same number of games playing white.[/hide]

Suppose that there are $n$ players. Note that there are $n(n-1)$ games in total, hence every player got $n-1$ points.

a) Since this is an integer, we see that every player had an even number of tied games. Since everyone played $2(n-1)$ games, there would be exactly one player with $2k$ draws for $k=0,1\ldots, n-1$ if the statement was wrong. We obtain a contradiction by considering the games between the player with $0$ draws and the one with $2(n-1)$ draws.

b) Pretty similar. If the statement was wrong, there would be exactly one player with $k$ lost white games for $k=0,\ldots, n-1.$ Consider the player with $0$ losses as white, who must have lost all the remaining games (as he otherwise would end up with more than $n-1$ points). But then he also lost his game with black against the player with $n-1$ losses at white, i.e with $0$ wins, impossible.


For any 10 digit natural number $ N$ in which

the first digit corresponds to the total no of 1's.
the 2nd digit corresponds to the total no of 2's.
.
.
.
the 10th digit corresponds to the total no of 0's.

determine, with proof, if the number of such natural number $ N$ is finite, and if proved true, find them all.

A generalization of
http://answers.yahoo.com/question/index;_ylt=ArxJYR_e8QoLHAfblB5DIp8Cxgt.;_ylv=3?qid=20080628051813AA0p296

Also, extend this to any numerical base $ M$ such that the $ M^{th}$ digit corresponds to the total number of 0's and $ (M\minus{}1)^{th}$ digit corresponds to the total number of $ (M\minus{}1)$'s for any natural number $ M$.

Well seeing as how all such numbers must be 10 digits, there are fewer than $ 10^{10}$ of them, so there are certainly finitely many solutions.  This holds just as trivially in any other base.  In base 10, it isn't too hard to show that there is exactly one solution (prove first that only one of the numbers from 3 to 10 can be in the number, and it can only appear in the 10th digit).

Though it's not true that there is a solution in every base.  For example the problem in bases 2, 3, and 6 has no solution.  The unique solution for base $ M \equal{} 4$ is 2101, and the unique solution for $ M \equal{} 5$ is 12002.  For base $ M\geq 7$ there is always at least one solution: $ 21000...0001000...000(M \minus{} 4)$, where the 1 in the middle appears in the $ (M \minus{} 4)$th position.  I'm fairly certain this is also the only solution in base $ M$, using an argument similar as for the base 10 case.


What is the sum of all integers less than 100 which have 12 divisors?

A non-brute force approach:

Ok, let's say $ x$ has $ 12$ factors.  You know that $ a$ has $ b$ factors if you prime factorize, take the exponents, add $ 1$ to each of them, and multiply.

We can do this backwards.

$ x: (0\plus{}1)(11\plus{}1)$

$ x: (1\plus{}1)(5\plus{}1)$

$ x: (2\plus{}1)(3\plus{}1)$

$ x: (1\plus{}1)(1\plus{}1)(2\plus{}1)$

Thus, the number can be in the following forms:

$ a^{11}, ab^5, a^2b^3, abc^2$

Remember that $ a, b, c \ge 2$.

For the first one, $ 2^{11}$ is obviously too big.

For the second, however $ 2^5$ is not too big, so we can experiment with this.

$ 3 \times 2^5 \equal{} 96$

However, you will not be able to make anymore in this form.

$ 2^3$ is obviously less than $ 100$.

$ 3^2 \times 2^3 \equal{} 72$

Now, let's try the third form.

$ 3 \times 5 \times 2^2 \equal{} 60$

$ 3 \times 7 \times 2^2 \equal{} 84$

$ 2 \times 5 \times 3^2 \equal{} 90$

This is all that can be done, so the answer is $ \boxed{402}$.


Say you have every number in the form 10017, 100117, 1001117, etc.. How are you supposed to prove that the $ n^{th}$ such number can be written as:

$ k_{n}\equal{}k_{n\minus{}1}\plus{}(90100 \cdot 10^{n\minus{}1})$

Sorry if my question isn't as clear as possible.

[hide="Major hint"]Find a closed form expression for $ k_n$.
[hide="Closed form"]$ k_n$ always starts with $ 1001$, followed by $ (n-1)$ $ 1$'s, and ends with a $ 7$.

So, $ k_n$ can be written as

\begin{align*}
k_n &= 1001\cdot 10^n + \underbrace{111..11}_{n-1}\cdot 10 + 7 \\
&= 1001\cdot 10^n + \frac{10^{n-1}-1}{9}\cdot 10 + 7 \\
&= 1001\cdot 10^n + \frac{10^n - 10}{9} + 7 \\
&= \frac{9009\cdot 10^n + 10^n  - 10 + 63}{9} \\
k_n &=\frac{9010\cdot 10^n + 53}{9}
\end{align*}[/hide]What do you get when you subtract $ k_{n-1}$ from $ k_n$?
[/hide]

Edit: I'm not wrong after all!

[hide="My super awesome fun way :D"]
$ k_n \equal{} 10^{n \plus{} 3} \plus{} 7 \plus{} \sum_{i \equal{} 1}^n{10^i}$
$ k_n \minus{} k_{n \minus{} 1} \equal{} 10^{n \plus{} 3} \plus{} 7 \plus{} \left(\sum_{i \equal{} 1}^n{10^i}\right) \minus{} \left[10^{n \minus{} 1 \plus{} 3} \plus{} 7 \plus{} \left(\sum_{i \equal{} 1}^{n \minus{} 1}{10^i}\right)\right]$
$ \equal{} 10^{n \plus{} 3} \plus{} 7 \plus{} \left(\sum_{i \equal{} 1}^n{10^i}\right) \minus{} 10^{n \plus{} 2} \minus{} 7 \minus{} \left(\sum_{i \equal{} 1}^{n \minus{} 1}{10^i}\right)$
$ \equal{} 10^{n \plus{} 3} \minus{} 10^{n \plus{} 2} \plus{} \left(\sum_{i \equal{} 1}^n{10^i}\right) \minus{} \left(\sum_{i \equal{} 1}^{n \minus{} 1}{10^i}\right)$
$ \equal{} 10^{n \plus{} 2}(10 \minus{} 1) \plus{} \left(10^1 \plus{} 10^2 \plus{} \cdots \plus{} 10^{n \minus{} 1} \plus{} 10^n \minus{} 10^1 \minus{} 10^2 \minus{} \cdots \minus{} 10^{n \minus{} 2} \minus{} 10^{n \minus{} 1}\right)$
$ \equal{} 10^{n \plus{} 2}\cdot{9} \plus{} 10^n$
$ \equal{} 10^n(10^2\cdot{9} \plus{} 1) \equal{} 901\cdot{10^n} \equal{} 9010\cdot{10^{n \minus{} 1}}$

Now add $ k_{n \minus{} 1}$ to both sides:

$ k_n \equal{} k_{n \minus{} 1} \plus{} 9010\cdot{10^{n \minus{} 1}}$[/hide]

As b-flat said (below), It would seem the problem statement is incorrect.

[quote="grn_trtle"]Can anyone tell me what is wrong with this (terribly painstaking) approach?
[hide="Overly complicated (and apparently incorrect) way"]
$ k_n \equal{} 10^{n \plus{} 3} \plus{} 7 \plus{} \sum_{i \equal{} 1}^n{10^i}$
$ k_n \minus{} k_{n \minus{} 1} \equal{} 10^{n \plus{} 3} \plus{} 7 \plus{} \left(\sum_{i \equal{} 1}^n{10^i}\right) \minus{} \left[10^{n \minus{} 1 \plus{} 3} \plus{} 7 \plus{} \left(\sum_{i \equal{} 1}^{n \minus{} 1}{10^i}\right)\right]$
$ \equal{} 10^{n \plus{} 3} \plus{} 7 \plus{} \left(\sum_{i \equal{} 1}^n{10^i}\right) \minus{} 10^{n \plus{} 2} \minus{} 7 \minus{} \left(\sum_{i \equal{} 1}^{n \minus{} 1}{10^i}\right)$
$ \equal{} 10^{n \plus{} 3} \minus{} 10^{n \plus{} 2} \plus{} \left(\sum_{i \equal{} 1}^n{10^i}\right) \minus{} \left(\sum_{i \equal{} 1}^{n \minus{} 1}{10^i}\right)$
$ \equal{} 10^{n \plus{} 2}(10 \minus{} 1) \plus{} \left(10^1 \plus{} 10^2 \plus{} \cdots \plus{} 10^{n \minus{} 1} \plus{} 10^n \minus{} 10^1 \minus{} 10^2 \minus{} \cdots \minus{} 10^{n \minus{} 2} \minus{} 10^{n \minus{} 1}\right)$
$ \equal{} 10^{n \plus{} 2}\cdot{9} \plus{} 10^n$
$ \equal{} 10^n(10^2\cdot{9} \plus{} 1) \equal{} 901\cdot{10^n} \equal{} 9010\cdot{10^{n \minus{} 1}}$

Now add $ k_{n \minus{} 1}$ to both sides:

$ k_n \equal{} k_{n \minus{} 1} \plus{} 9010\cdot{10^{n \minus{} 1}}$[/hide]

I can't find any mistake that would account for the 9010 instead of 90100  :huh:[/quote]

Your solution is correct; the problem statement is incorrect, I believe.

Note that for $ n\equal{}2$, $ k_2\equal{}100117\equal{}10017\plus{}9010\cdot10^1$.


modular arithmetic
number theory unsolved
number theory

Prove that for any positive integer d

that 2d-1, 5d-1, 13d-1 are not all squares.

Thanks.

$ 2d\minus{}1\equal{}x^{2},5d\minus{}1\equal{}y^{2},13d\minus{}1\equal{}z^{2}$  $ \implies$  $ 2\nmid d$ .
Because if $ 2|d$ $ \implies$  $ x^{2}\equiv \minus{}1\pmod{4}$  Contradiction!
$ 2\nmid d$ $ \implies$  $ d\equal{}2n\plus{}1$  .
$ 4n\plus{}1\equal{}x^{2},10n\plus{}4\equal{}y^{2},26n\plus{}12\equal{}z^{2}$  .
$ 10n\plus{}4\equal{}y^{2}$  $ \implies$  $ 2|n$  $ \implies$  $ n\equal{}2p$
$ 8p\plus{}1\equal{}x^{2},5p\plus{}1\equal{}l^{2},13p\plus{}3\equal{}s^{2}$  $ \implies$
$ s^{2}\minus{}l^{2}\equiv 2\pmod{4}$ $ \implies$  Contradiction!   :wink:

In other words, there are no solution (mod 16)  :)


linear algebra
matrix
superior algebra
superior algebra unsolved

Let $ K/F$ be a field extension.
Prove that any system of linear equations over $ F$ with a solution in $ K$ has a solution in $ F$.

Recast the question in matrix terms; it should be much more clear.


no-- its not that you should unseat somemoderators of their statuses, its just that moderators should not be assigned so many forums
the reason:
i am trying to report someone on the HSB, but there are 6 moderators!
i suggest you either
a) give moderators a reduced number of forums to moderate
b) set up a 'pm all moderators function' (or rather, pm all moderators of this forum function)

Why don't you just choose one or two of the moderators to private message?

i am worried that i may pm either the most pmed moderators or the inactive moerators

If after a few days whoever you pmed hasn't opened it, you could just pm another moderator.

...which brings up a good point--that's pretty time consuming and inefficient. I agree with stevenmeow.

One may also note that some forums (most notably the National and Local Communities) have no moderators.

...which sometimes allows for spamming??? :wink:  :P

1. If there are multiple mods of a certain forum - choose the one that you think is more/the most active (you can do this by checking at the bottom of the forum index to see if they've been on in the last 24 hours). If neither of the mods are active - contact and admin.

2. I am also pushing for a "Report a Post" function.


function probability algebra polynomial Vieta quadratics

a) If f(x^3)=(f(x))^3 - 10, find f(0)+f(1)+f(-1)
b) Find the inverse function of f(x)=cosh x (show workings pls)
c) In 100 lottery tickets there are 2 winning tickets, what is the probability of winning if you buy 10 tickets?
d) Choosing 2 numbers between 1 - 20, what is the probability that the larger number is a multiple of the smaller number?

Thanks.

[hide="a)"]$ f(0^3)\equal{}[f(0)]^3\minus{}10 \implies [f(0)]^3\minus{}f(0)\minus{}10\equal{}0$

$ f(1^3)\equal{}[f(1)]^3\minus{}10 \implies [f(1)]^3\minus{}f(1)\minus{}10\equal{}0$

$ f((\minus{}1)^3)\equal{}[f(\minus{}1)]^3\minus{}10 \implies [f(\minus{}1)]^3\minus{}f(\minus{}1)\minus{}10\equal{}0$

We can see that $ f(1)$, $ f(\minus{}1)$ and $ f(0)$ are the roots of the same polynomial $ p(x)\equal{}x^3\minus{}x\minus{}10$. Applying Vieta's formula we get $ f(0)\plus{}f(1)\plus{}f(\minus{}1)\equal{}\boxed{0}$.[/hide]

[hide="c)"]
There are $ 100$ tickets in total and there are $ \binom{100}{10}$ different ways to pick $ 10$ of them. There are $ 98$ not-winning tickets, so we can pick those on $ \binom {98}{10}$ different ways. Using complementary probability we get desired probability:
$ 1\minus{}\frac{\binom {98}{10}}{\binom {100}{10}}\equal{}\boxed{\frac{21}{110}}$.[/hide]

[hide="d)"]
Smaller number needs to be less or equal than 10 in order for larger number to be its multiple. So let's count how many pairs of multiples are there:

1 - 19
2 - 9
3 - 5
4 - 4
5 - 3
6 - 2
7 - 1
8 - 1
9 - 1
10 - 1

for total of 46 "good" pairs. Therefore the probability is $ \frac{46}{\binom{20}{2}}\equal{}\boxed{\frac{23}{95}}$.[/hide]

hello, for b) we can write $ y\equal{}\frac{1}{2}\left(e^{x}\plus{}e^{\minus{}x}\right)$ multiplying by $ e^{x}$ gives
$ 0\equal{}e^{2x}\minus{}2ye^{x}\plus{}1$ solving this quadratic equation we get $ e^{x}\equal{}y\plus{}\sqrt{y^2\minus{}1}$ (the other solution is senseless) and it follows $ x\equal{}\ln(y\plus{}\sqrt{y^2\minus{}1})$
Sonnhard.


LaTeX
linear algebra
linear algebra unsolved

It may sound a bit ridiculous as question but I can see it how to solve this problem...

Let U be the subspace of R^(infinite) defined by

U={x=(x1,x2, ... ,xn) belongs to R^(inf) | xn+4=5xn+2-4xn(n>=1)}

Prove that the infinite sequences x,y,z,w defined by
x[size=59]n[/size]=1, y[size=59]n[/size] = (-1)^(n-1), z[size=59]n[/size] =2^(n-1), w[size=59]n[/size] =(-2)^(n-1)

form a basis for U. You may assume as known that dim(U) =4. Use this to find a formula for the n-th term of the infinite sequence u belongs to U where u[size=59]1[/size]=1, u[size=59]2[/size]=2, u[size=59]3[/size]=3,u[size=59]4[/size]=4.

Translated into $\LaTeX$:
[quote="archmage"]Let $U\subset \mathbb{R}^{\mathbb{N}}$ be defined by

$U=\{x=(x_1,x_2, \dots ,x_n,\dots) : x_{n+4}=5x_{n+2}-4x_n\,\forall n\}$

Prove that the infinite sequences $x,y,z,w$ defined by
$x_n=1, y_n = (-1)^{n-1}, z_n =2^{n-1}, w_n =(-2)^{n-1}$
form a basis for $U$. You may assume as known that $\dim(U) =4$. Use this to find a formula for the $n$-th term of the infinite sequence $u\in U$ where $u_1=1, u_2=2, u_3=3,u_4=4$.[/quote]

If you assume the dimension as known, you merely have to show that you have four linearly independent elements of $U$.
If not, just note that $x_1,x_2,x_3,x_4$ determine everything by induction.

You can find the coefficients relative to your basis using only the first four terms of the sequence.


calculus
integration
limit
logarithms
function
derivative
real analysis

Let  $0<a<b<\infty$  and  $D: ={\mathbb R}\setminus \{1\} .$ For  $r\in D$  define  $g_r: [a,b]\to {\mathbb R}$   as  $\displaystyle g_r(x) =\displaystyle x^{r-1} .$ [b]
Suppose that   $\theta : D\to [a,b]$ satisfies  $\; \int\limits_{a}^{b}g_r(t)\; dt =\displaystyle (b-a)g_r(\theta) . \; \; \; \;$ [/b] [color=red] [b] Find the set   $\theta(D)$    [/b][/color] .
(Note: RMT=Revista Matematica din Timisoara)

We have that $\int_a^bg_r(t)dt=\frac 1 r (b^r-a^r)$. So, $\theta(r)=\frac{b^r-a^r}{r (b-a)}=\frac{a^{r-1}}{r}\frac{\left (\frac b a\right)^r-1}{\frac b a-1},\forall r\in D$. Denote $\frac{b}{a}=t>1$. We have that $\theta(r)=\frac{a^{r-1}}{r}\frac{t^r-1}{t-1}$ is continuous on $\mathbb{R}\setminus \{0\}$. Setting $\theta(0)=\lim_{r\to 0}\theta(r)=\frac{\ln b-\ln a}{b-a}$ we get a continuous function.
If at least one between $a,b$ is greater than $1$, $\theta(D)=(0,\infty)$. If $a<b<1$ limits of the function at infinity is 0 and $\theta(r)> 0 \forall r\in D$. So, $\theta(D)=[\theta(m),\theta(M)]$ where m and M are zeroes of the derivative of $\theta$ or 0. I will not caculate them now.


group theory
abstract algebra
superior algebra
superior algebra unsolved

I know the $ {S_n}'\equal{}A_n$ , I need to know how to prove that for any $ \sigma \in A_n$ is a commutator of two elements of $ S_n$.

[quote="Hidden Scofield"]I know the $ {S_n}' \equal{} A_n$ , I need to know how to prove that for any $ \sigma \in A_n$ is a commutator of two elements of $ S_n$.[/quote]

If $ \sigma$ is a permutation than $ \sigma$ and $ \sigma^{\minus{}1}$ have the same sign. So $ {S_n}' \subseteq A_n$. Another way to prove it is to consider the quotient $ S_n/A_n$

$ {S_n}'$ is a normal subgroup of $ S_n$ and for $ n\ge5$ $ A_n$ is simple and so $ {S_n}' \equal{} A_n$.

We need to consider the cases $ n<5$. If $ n\equal{}3$, $ A_3$ is the cyclic group of order $ 3$ and $ S_3$ is not abelian and so $ {S_3}' \equal{} A_3$. If $ n\equal{}4$, $ A_4$ has two normal subgroups $ {1}$ and $ V$ (the klein four-group). $ S_4$ is not abelian and the quotient $ S_4/V \cong S_3$ and so $ {S_n}' \equal{} A_n$.

thank you vict85

But , I need to prove element in $ A_n$ is just a commutator $ [\sigma,\delta]$, for some $ \sigma ,\delta \in A_n$.

 :)


To be honest, my brother and I created this problem before when we solved the inequality
\[\sum \sqrt{(a^{2}-ab+b^{2})(a^{2}-ac+c^{2})}\ge a^{2}+b^{2}+c^{2}. \]
The solution is easy by Cauchy
\[\sqrt{(a^{2}+ab+b^{2})(a^{2}+ac+c^{2})}=\sqrt{((a+\frac{b}{2})^{2}+\frac{3b^{2}}{4})((a+\frac{c}{2})^{2}+\frac{3c^{2}}{4})}\]
\[\ge(a+\frac{c}{2})(a+\frac{b}{2})+\frac{3bc}{4}=a^{2}+\frac{ac}{2}+\frac{ab}{2}+bc. \]
Adding up similar results we have done.

The inequality
\[\sum_{\rm cyclic}\sqrt{(a^{2}+ab+b^{2})(b^{2}+bc+c^{2})}\geq(a+b+c)^{2}\]
has geometric interpretation and was solved in my book.

Note that the one posted by hitek is different.


My BSNL Broadband is going dead slow...  thanks to undersea cables getting cut somewhere near egypt. What about you people?

And yeah I'm back..... After a battle with a darn virus which I finally won sans the help of any anti-virus software (all of which were useless). Cmd Prompt and Msconfig helped me.

My ortel(it uses Relaince) is crying :rotfl:  :rotfl:  :rotfl:  :rotfl:
it's dying.... :rotfl:  :rotfl:
it's all due to egypt thing..

MINE (AIRTEL) IS REASONABLY FAST  :D  :D  :D

Dont think so. Its better than yesterday but still pretty slow. :(

Was a little slow yesterday ... 75% of normal speed.

Works fine today! (BSNL best hai!)

Mine seems to work perfectly fine, except when it comes to uploading photos... Zoomr is killing me, I have been trying to upload and have succeeded in uploading 15 photos in the past 1 hour.... UGHHH!!!! Infuriating.... Picasa is much faster.. .Shre, killer sense man!

@Rohit - Gime your mail id, will mail those three or four pics to you directly... uploading wont work.....
@Shre - Wait a little while more k?

[quote="pardesi"]hey where are u uploading them....[/quote]

Dont you dare tell that fool where u uploaded it  :D  :D . Dei paradesi why did u leave  :mad:

dei the servermade my Linux crash :o  :o  :o  :rotfl:

Yah, It was dead slow two days before but now it is almost perfect.

Lol.. Dont even worry about it....yet... I still have to upload the photos someplace where other people can see it.. As of now, its just me and my computer :D


geometry
limit
combinatorics unsolved
combinatorics

A convex $ 2n$-gon has its vertices at lattice points. Prove that its area is not less than $ \frac{n^3}{100}.$

Read the following article: http://www.math.nthu.edu.tw/~tjm/abstract/9712/tjm9712_1.pdf.  It may be helpful.

It is known that if $ a(n)$ is the minimal area of a convex $ n$-gon whose vertices are integer points then $ \lim_{n \rightarrow \plus{}\infty} \frac {a(n)}{n^3}$ exists and is equal to $ L\equal{}0.01850567\cdots$.
As far as I know, the limit $ L$ hasn't been related to already known constants.

The above result has been proved by I. Barany and N.Tokushige.
I have the paper but I don't know where it has been published (I got it as a preprint :wink: ), I attach it.

Pierre.

I don't think the above statement is quite accurate.  From the paper, it seems like what is [b]known[/b] is that the limit exists, and what is [b]suspected[/b] is that the limit is 0.01850567.

On the other hand, the cited result of Rabinowitz there gives a lower bound which is larger than $ \frac{n^3}{100}$.


function
calculus
derivative
logarithms
induction
complex analysis
real analysis

No idea?
What about thinking at the function f(x)=a^x ant its derivatives,Taylor formula,etc?

I don't know, but the problem seems obvious. I mean, just expand everything as entire function and make equal the coefficients. This will give you the first statement. From here it is trivial to deduce the second statement. Just use some limits.

Exactly. Let $f(z)=\sum_i e^{z\ln a_i}$ and $g(z)=\sum_j e^{z\ln b_j}$. Both $f$ and $g$ are entire functions, and $f\left(\frac1n\right)=g\left(\frac1n\right)$. A standard theorem of complex analysis says that if $f,g$ are analytic in (simply connected) $X$ and $f(z_k)=g(z_k)$ on some sequence $z_k$ converging to an interior point of $X$, $f$ and $g$ are equal on $X$.

The desired equality is simply the equality of derivatives at zero. Since the sequences $t^n$ are linearly independent for any collection of distinct $t$, the coefficients must match for the difference to be identically zero.

The first statement can also be proved by induction using some limits.


quadratics
induction
number theory unsolved
number theory

Find all the positive integers $x,y,z$ satisfiing : $x^{2}+y^{2}+z^{2}=2xyz$

Looking modulo $4$ gives that $x,y,z$ are even and by substitution $x=2x_1,y=2y_1,z=2z_1$ you get $x_1^2 + y_1^2 + z_1^2 = 4 x_1 y_1 z_1$. But now all three numbers must again be even and so on leading to something like $x_i^2 + y_i^2 + z_i^2 = 2^{i+1} x_i y_i z_i$ with $x=2^ix_i,y=2^iy_i,z=2^iz_i$ . So there is no solution because every solution must infinitly often be divisible by $2$.

Sorry Zetax, but not all of $x,y,z$ is necessarily even
we can have $x$ even , $y$ odd and $z$ odd
and the sum of their squares is even !!!!!  :?

But then ($x$ even) the RHS and $x^2$ would be divisible by $4$ and by this also $y^2 + z^2$ and this only works when also $y,z$ are even (since for odd $y$ you have $y^2 \equiv 1 \mod 4$)

what about:
$x^2+y^2+z^2=3xyz$ :)
and finally:
$x^2+y^2+z^2=kxyz$ :)

[quote="_tequilla"]
$x^2+y^2+z^2=kxyz$ :)[/quote]

Nice ! I can prove for now that there is no solution for $x,y,z = (0,0,0); (0,0,1)$ but I gave up while trying to prove something for $(1,1,1)$ and $(0,1,1)$ because I don't like to consider too many cases and that was ahead of me when I was trying to solve it. So I'll gladly see the solution  :)

The general formula of this problem is Markov's equation  :)

Markov Equation is the one with $k=3$ [look here http://mathworld.wolfram.com/MarkovEquation.html ].

This one with $k$ is great and probably super-hard or am I wrong ?

[quote="_tequilla"]what about:
$x^2+y^2+z^2=3xyz$ :)
[/quote]
You can easy see that $x=1$, $y=1$ and $z=1$ is a solution. Suppose that $(x,y,z)$ is a solution. If you keep $y$ and $z$ fixed, you will get next quadratic equation.
\[
x^2+y^2+z^2=3xyz\Rightarrow x^2-3yzx+y^2+z^2=0
\]
Beside $x$ there exist another root $x_1$. Notice that
\[
x+x_1=\frac{3yz+\sqrt D}{2}+\frac{3yz-\sqrt D}{2}=3yz,
\]
which implies that $x_1=3yz-x$ is also an integer. This shows that if $(x,y,z)$ is a solution, $(3yz-x,y,z)$ is also a solution. By induction there are infinitie solutions.

As I remember $k$ can be $1,2,3,4,5,6,8,9$
So what is the solution for the other cases? (for $k=4,5,6,8,9$)

Let $x,y,z$ must be all even to satisfy the condition
Bcoz if taken even and odd it violates
$x=2a_1,y=2k_1,z=2f_1$ .,$a_1,k_1,f_1\in\mathbb{N} a_1,a_2,a_3 \ge 1$
$(a_1)^2+(k_1)^2+(f_1)^2=4a_1k_1f_1$
Further
$a_1=2a_2,n=2p_2,p=2c_2, v(a)\ge 1$
$(a_2)^2+(p_2)^2+(c_2)^2=8a_2p_2c_2$
Looking at this we conclude
$(x_i)^2+(y_i)^2+(z_i)^2=2^{i+1}x_iy_iz_i$
From there it follows and it is true for four variables $x, y, z, x_n$
$a^4+b^4+c^4+d^4=2abcd$
Same way solutions are taken.
$parity$ violates if we take $(o, e, o), (o, o, o), (e, o, e),... $

This is just pell's equation infinite descent!


Let $a,b,c,d,m,n$ be integers such that $ad-bc=1$  and  $mn\neq0$. Prove: $gcd(am+bn,cm+dn)=gcd(m,n)$.

$gcd(am+bn,cm+dn)\le gcd(am+bn,d(am+bn)-b(cm+dn),cm+dn)\le gcd(am+bn,m,cm+dn)=gcd(m,gcd(b,d)n)=gcd(m,n)$ and obviosly, that $gcd(am+bn,cm+dn)\ge gcd(m,n)$.


Hey!
i am recipient of AMS's Karl Menger research award (at ISEF)  :)
is any other of recepients a member of the forum?
here goes the site: http://www.ams.org/prizes/menger-award.html

nice.. did u have a nice time at isef?
hey and where are u going to study for college?
i am a senior this year so i haven't applied...hope that all the isef awards and the karl menger thing helps me in the applications... :?

i am going in to my senior year so i am in the same boat u r...don't worry, u'll be fine...

great.. hope to see you at next year's isef.. if u plan to go...
i'm also applying to siemens and intel talent search so if i qualify hope to see you there too, if you plan to participate also..


My three-digit code is 023. Reckha can't choose a code that is the same as mine in two or more of the three digit-positions, nor that is the same as mine except for switching the positions of two digits (so 320 and 203, for example, are forbidden, but 302 is fine). Reckha can otherwise choose any three-digit code where each digit is in the set $ \{0, 1, 2, ..., 9\}$. How many codes are available for Reckha?

So first of all there are 1000 codes that are possible for Rekha.

The first condition (Reckha can't choose a code that is the same as mine in two or more of the three digit-positions) can be filled in 30 ways; 10 ways to choose an x for each of the following sets:
02x
0x3
x23
But for 3 sets; 02(3), 0(2)3, and (0)23, these numbers are the same, so we must combine these 3 into 1; we are left with 28 sets.

Then for the second condition ([Reckha can't choose a code that]is the same as mine except for switching the positions of two digits (so 320 and 203, for example, are forbidden, but 302 is fine)) we have that there are 3 pairs of numbers that can be switched, so there are 3 forbidden codes.

None of these codes overlap, so there are a total of 28+3=31 forbidden codes, leaving 1000-31=[b]969[/b] codes that Reckha can use.


For $a>c>0$ and $b>c>0$, prove that: \[\sqrt{(a+c)\cdot(b+c)}+\sqrt{(a-c) \cdot (b-c)}\leq 2\sqrt{ab}\]

Is there a typo? Should the first radicand be $(a+c)(b+c)$? Because such as it is, there are counterexamples, for instance $a=3,b=2,c=1$.

[hide="Assuming so"]We can just square and expand:

$ab+ac+bc+c^{2}+ab-ac-bc+c^{2}+2\sqrt{(a^{2}-c^{2})(b^{2}-c^{2})}\leqslant 4ab$

$2ab+2c^{2}+2\sqrt{(a^{2}-c^{2})(b^{2}-c^{2})}\leqslant 4ab$

$\sqrt{(a^{2}-c^{2})(b^{2}-c^{2})}\leqslant ab-c^{2}$

$a^{2}b^{2}-a^{2}c^{2}-b^{2}c^{2}+c^{4}\leqslant a^{2}b^{2}-2abc^{2}+c^{4}$

$-(a^{2}+b^{2})c^{2}\leqslant-2abc^{2}$

$a^{2}+b^{2}\geqslant 2ab$

$(a-b)^{2}\geqslant 0$

which is the trivial inequality.[/hide]

Indeed there was a typo. I have corrected it.

why do you write $\leqslant$ and not $\le$?

I just prefer that form. And generally find it more legible - less prone to be confused with $<$ in fast reading.


Find the greatest integer that will divide 13511, 13903, and 14589 and leave the same remainder.

[hide]We can say that
$13511 = ay + n$
$13903 = by + n$
$14589 = cy + n$
where $n$ is the remainder and $y$ is the integer we're looking for. Subtracting some equations we get $392 = y(a - b)$ and $686 = y(c-b)$ so looking at the common factors of both of those numbers we get that $98$ is the greatest value of $y$.[/hide]

Wow, I'm confused, I set up the equation like this:

13511/x=y+r

13903/x=z+r

14589/x=w+r

so:

13511=x(y+r)

13903=x(z+r)

14589=x(w+r)

Did I set it up wrong? And if I didn't, how would you solve this?


Let $ p$ be a prime number and $ m$ be a pozitive integer. Show that there exist a positive integer $ n$ such that there exist $ m$ consecutive zeroes in the decimal representation of $ p^n$.

I don't know for $ p\equal{}2,5$. For other primes, exist $ l$, suth that $ p^l\equal{}1\mod 10$, then take $ n\equal{}10^{m\plus{}1}l$.


On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports cars next month. How many sedans does it expect to sell?

$ \textbf{(A)}\  7  \qquad
\textbf{(B)}\   32  \qquad
\textbf{(C)}\   35  \qquad
\textbf{(D)}\  49   \qquad
\textbf{(E)}\  112$

[hide="Solution"]
Let the expected amount of sedans sold be $ x$.  Then,

\begin{align*}
\dfrac{4}{28}&=\dfrac{7}{x}
\\ \dfrac{1}{7} & = \dfrac{7}{x}
\\ x &=49
\end{align*}

corresponding to $ \boxed{\textbf{(D)}}$.
[/hide]

Set up a ratio
X=number of sedans
4:7=28:x

4x=196
x=49

49 sedans


Find the maximum value of

sinx + cosx (for all real x)

It is Sqrt(2) i think.
 Use trigonometric identities
     :) [/code]

Yeah, i think so too.

where x = 45

but how is this proven?

Ok. Here it is.       sin(x) + cos(x) =
                           sin(x) + sin(π/2 - x) =
                           2 sin(π/4) cos(x - π/4) =
                           Sqrt(2) cos(x - π/4)

 Max occurs when x -π/4 = 2kπ , k an integer.

Now don't hate me, but you can also prove using calculus... just don't look if you don't want to :)
[hide]
$F(x)=sin(x)+cos(x)$ now find the critical values by taking the derivitive
$F'(X)=cos(x)-sin(x)$ then  set it equal to zero
$cos(x)-sin(x)=0$ so $x=\frac{\pi}{4}, \frac{5\pi}{4}...$
now just by common sense, the maximum is found at $\frac{\pi}{4}$ and minimum is $\frac{5\pi}{4}$
[/hide]

[hide="Another proof"]
(This is essentially the same as Ramanujan's proof.)

$\sin x+\cos x$
$=\sqrt{2}\left(\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}}\right)$
$=\sqrt{2}\left(\sin x\cos \frac{\pi}{4}+\cos x\sin \frac{\pi}{4}\right)$
$=\sqrt{2}\sin \left(x+\frac{\pi}{4}\right)$

So the maximum value is $\sqrt{2}$ when $x+\frac{\pi}{4}=\frac{\pi}{2}+2k \Longleftrightarrow x=\frac{\pi}{4}+2k$

[/hide]

Yet another

$(a+b)^2 \leq 2 \cdot (a^2 + b^2)$ so $\left(\cos x + \sin x\right)^2 \leq 2$


probability
geometry
symmetry
integration
probability and stats

I recently asked this of a class:

Choose a point $ A$ at random in a certain square. ("At random" denotes a uniform-in-area distribution on that square.) Choose a point $ B$ at random in the same square, independent of $ A.$ What is the probability that the circle centered at $ A$ that passes through $ B$ is contained within the original square?

Let $ U,V$ be independent uniform variables on the interval of length $ a$, then using a symmetry argument in a square of length $ a$ with regard to its 4 natural sub-squares we calculate for the likelihood of the minimum distance $ d(U,V)=\min\{U,V,a-V,a-U\}$ of the point $ (U,V)$ from the squares hull (for $ 0\leq d\leq \frac{a}{2}$)

 $ \begin{array}{l}\mathbb{P}(d(U,V)\leq d)=4\cdot \mathbb{P}(d(U,V)\leq d\cap \{U,V\leq\frac{a}{2}\})=\\[0.2cm]
4\cdot \mathbb{P}(\min\{U,V\}\leq d\cap \{U,V\leq\frac{a}{2}\})=\\[0.2cm]
\int_{0}^d\frac{a-2t}{a^2}dt=\frac{4d}{a}-\frac{4d^2}{a^2}\end{array}$

and the required likelihood for the described event is given by:

$ \int_{0}^{0.5a}\underbrace{\frac{a^2-\pi t^2}{a^2}}_{\mbox{cond'l prob}}\underbrace{\left(\frac{4}{a}-\frac{8t}{a^2}\right)}_{\mbox{density of }d(U,V)}dt=1-\frac{\pi}{24}$

That's the probability of the complementary event - that the circle does cross the boundary.


Now you see the reason why you should post it on the board and not put it online as picture....


geometry
function
calculus
calculus computations

A pebble is dropped in still water, forming a circular ripple whose radius is expanding at a constant rate of $ 10$ cm/s. Find a formula giving the area enclosed by the ripple as a function of time. When the radius is $ 20$ cm, how fast is the area enclosed by the ripple increasing?

We are trying to find the rate at which the area is increasing so we are looking for the answer the have the units $ cm^2/s$.


analytic geometry
graphing lines
slope
trigonometry

3.13) A ski tow consists of a long belt of rope around two loops, one at the bottom of a slope and the other at the top.The pulleys are driven by a husky electric motor so that the rope moves at a steady speed of 1.5m/s. The pulleys are separated by a distance of 100m, and the angle of the slope is 20degree.

Skiers take hold of the rope and are pulled up to the top, where they release the rope and the glide off. If a skier of mass 70kg takes the tow every 5s on the average, what is the average force required to pull the rope? Neglect friction.

3.17) An inverted garbage can of weight W is suspended in air by water from a geyser. The water shoots up from the ground with a speed Vo , at a constant rate dm/dt. The problem is to find the maximum height at which the garbage can rides. What assumption must be fulfilled for the maximum height to be reached?

3.13) Not sure I got all this stuff of loops, pulleys and glide(?), but
       [hide="seems to be"] 311 kgf  ?[/hide]
3.17)Assuming no other force on the watter than weight and that of the can,
      [hide="and elastic collision"] $ h= {{v_0^2 - {{W} \over2{{dm}\over{dt}}}\over{2g}}}$ [/hide]

2nd seems to be right. Dunno about first, no answer given. But I am more interested in the logic towards 1.

Imagine you have an object being carried up the slope at constant velocity. What's the force required to do this? Well, it's the same as the component of the object's weight down the slope. So if it has mass $ m$ the force needed is $ m g \sin{\theta}$, where $ \theta$ is the angle of the slope. To find the average force required to carry the skiers, we need to find the average mass being carried at any time, which we can get by finding the average number of skiers on the tow at any time.

Let $ \lambda$ be the average density of skiers per unit distance. Then $ \lambda \equal{} \frac{1}{v \Delta t}$ where $ v$ is the speed of the tow and $ \Delta t$ is the time between consecutive skiers getting on the tow.
From this it quickly follows that the average force is $ F \equal{} \frac{M g L \sin{\theta}}{v \Delta t}$ where $ M$ is the mass of each skier and $ L$ is the length of the tow.

What about the change in mass of the system when a skier hops on or leaves the rope?

For every skier that gets on, another gets off - the total mass doesn't change.

No... sometimes there are 13 skiers on the tow, and sometimes there are 14. You can find the average number of skiers by multiplying the average density of skiers by the length of the tow.

That's basically what I meant - the average total mass doesn't change, disregarding little fluctuations.  The exact total mass wouldn't change if $ L$ were a multiple of $ v\Delta t$.

BTW, it appears that Immanuel Bonfils made a slight typo for no. 1 because his answer isn't dimensionally correct - did you mean $ h \equal{} \frac {v_0^2 \minus{} \left(\frac {W}{2dm/dt}\right)^2}{2g}$?

Well here's my take.

The length of the rope is 100, so there are on average 100/7.5 skiers on the rope.  The force is mgsin20. 100/7.5*70kg*9.8m/s2*sin 20degrees= average force needed.

[quote="integration"]2nd problem:

The water has to provide a force equal to W.

Assumption: water acts like balls colliding with the can elastically, (speed of recession=speed of approach).
So when water bumps into the can, it goes backward at an equal and opposite velocity.  That means the change in velocity is 2 vf.

W=dp/dt=2 (vf*dm/dt) =>vf=W/(dm/dt)* (1/2)
vf2=vi2+2gh

h=(vf2-vi2)/(2g)

Plug in vf from the initial equation, and I guess it should be right (unless my assumption was wrong?)[/quote]
Your answer matches mine, so I think it is the correct answer (unless we both are wrong!).


function
analytic geometry
geometry
geometric transformation
reflection
algebra proposed
algebra

Find all function $f$ continuous from $[0,1]$ to itself satisfying:
\[f(f(x))=x \quad \forall x \in [0,1]\].

Since $f$ is clearly a bijection, there are two possibilities (using also the continuity of $f$): either it's increasing, or it's decreasing.

If $f$ is increasing, we can show that it's the identity: $f$ is its own inverse, so its graph is symmetric wrt the main diagonal of the square. Assuming there is some $x$ s.t. $f(x)>x$, it's easy to see that the first coordinate (call it $x'$) of the reflection of the point $(x,f(x))$ in the main diagonal is larger than $x$, but the second coordinate of the same point (which is $f(x')$) is smaller than $f(x)$, so $x<x',\ f(x)>f(x')$, contradicting the fact that $f$ is increasing. This means that $f(x)=x,\ \forall x$ (we argue just as above for the case $f(x)<x$).

If we look for decreasing functions, there are lots of them, but they all have a certain form. First of all, they must obviously have a unique fixed point (all continuous functions from $[0,1]$ to itself have a fixed point, and if our function had $\ge 2$ fixed points, this would contradict the fact that it's decreasing). Conversely, take a point $x_0\in(0,1)$, and let $g:[x_0,1]\to [0,x_0]$ be any decreasing bijection (which is automatically continuous). Now simply reflect the graph of $g$ in the first diagonal of the square $[0,1]^2$ to get the whole graph of $f$ (it's easy to see this works). We could write out the actual definition of $f$ based on $g$, but that wouldn't be very interesting; this is the main idea, I think :).

I don't know if this gonna works. Correct me if i'm wrong
f(f(x)) = x

Obviously, we have one solution f(x) = x

In case of f(x) != x:

Recall f(x) = t, we have:
f( t ) = x and f ( x) = t

So we get, x * f(x) = t * f(t)

Recall g(x) = x f(x)

We have g(x) = g(t) with all x, t belong to [1,0]

This only happens when g(x) = constant. The rest is easy.

We will get f(x) = k/x (with k is any number )

The problem is this:

f: [0,1] -> [0,1]
     x     ->  f(x)

and f (f (x)) = x. Find all f(x)

I did not check it carefully, you can add some conditions to make it work such as k=0
Cheers :lol:

Grobber
If its decreasing then $f(x)=1-x$
Are there some others :?:

did you actually read what I wrote? There are many others :). All you need is a continuous graph symmetric wrt the main diagonal of teh square $[0,1]^2$. I'm sure you can easily find such examples different from $f(x)=1-x$, whose graph is a line, so, in a certain sense, is the simplest one.

Sincerely i dont!
If $f$ is line,and continious then $f(x)=ax+b$
As its decreasing and bijection from[0,1] to itself then$f(0)=1$ and $f(1)=0$
Can you please give me others :?

I said that $f(x)=1-x$ was a line, not that the other examples should also be lines. You can take a piecewise linear function. For instance: the graph of $f$ is the segment connecting the points $(0,1),(\frac 13,\frac 13)$, followed by the segment connecting $(\frac 13,\frac 13),(1,0)$.

Is the main idea still not clear?:?

Can sombody tell me what is wrong with my solution? I will try to do it again but i don't have much time since there is a big test coming up

$f(x)x=g(x)=const=f(0)*0=0$ for all $x$ in $[0,1]$
so $f(x)=0$and thus $f([0,1])=[0]$
But we must have $f[0,1]=[0,1]$ (Surjection)!
I thing now its clear


geometry
calculus
function
area of a triangle
calculus computations

Hi,
I have a problem and I am stuck to one question:

[b]The area of a triangle is given by the formula:  A=1/2*b*c*sin(θ )

At time to, we have bo=10 inches, co=20inches, θo=pi/3[/b]

a) Find the area of the triangle at to
    [b] I found 50*sqrt(3)[/b]

b)  find the rate of change of the area with respect to b at to.
[b]I found 5*sqrt(3).[/b]

c)  Find the rate of change of the area with respect with θ  at to.
 [b]I found 50.[/b]

d) Using the rate found in part c) , calculate (by differentials) the approximate change in area if angle theta is increased by one degree.
[b]I need help for this question. I do not know what to do here.[/b]

Thank you for your help

[quote="Kent Merryfield"]Note: the calculus of trigonometric functions is always in radians. Convert one degree to $\frac{\pi}{180}$ radians.[/quote]

Please I know that this osunds silly, but please what is the definiton of rate of change Dy/Dx: Is that the change of [b]y[/b] per the amount of [b]x[/b] ([b]or unit of x[/b])?

I ask that because if I just multiply my answer I got in c) by $\frac{\pi}{180}$  , I get $5\frac{\pi}{18}$ radians which is the answer in the textbook. So this means to me that the definiton of the rate of change is the change of [b]y[/b] per the amount of ([b]unit of x[/b])?

Please make it clear for me.

Thank you

B


How many edges must a graph with $n$ vertices have in order to guarantee that it is connected.

[hide] Well obviously the complete graph with $n-1$ vertices is not connected, and that has $\frac{(n-1)(n-2)}{2}$ edges. So if we add one to that, we know it's connected. We just need to show that $\frac{(n-1)(n-2)}{2}$ is the maximum for a not connected graph.

If the graph is not connected, we can split it into two groups with $m \ge 1$ and $n-m \ge 1$ vertices which have no edges between them. Each can have at most

$\frac{m(m-1)}{2}$ and $\frac{(n-m)(n-m-1)}{2}$

edges, respectively. Some algebraic manipulation tells us that

$\frac{m(m-1)}{2}+\frac{(n-m)(n-m-1)}{2}\le \frac{(n-1)(n-2)}{2}$,

so we're done. [/hide]

[quote="Phelpedo"]This only deals with the case of two subgraphs, but it's possible that there are several unconnected subgraphs, and you need to show that the total sum is less than $\binom{n}{2}+1$ to be completely rigorous.[/quote]

My proof suffices because even if you can split it into more than two, you can still split it into two subgraphs (suppose you have three, just put two into the same one, it still satisfies all conditions I specified).

So if you have more than two subgraphs, just add more edges until you get two.

[quote="Phelpedo"]So if you have more than two subgraphs, just add more vertices until you get two.[/quote]

Hmm? You don't need to add anything. The subgraphs need not be connected. Consider the following graph:
[code] A--B--C D--E F--G [/code]
Split it into two groups, say
[code] A--B--C D--E [/code] and [code] F--G [/code]
The first has at most $10$ edges (obviously less in this case) and the second has at most $1$. Furthermore $10+1 < \frac{5 \cdot 6}{2}= 15$.

I mean to say add edges to form two connected graphs. It's basically the same idea.


During an ill-advised orientation week activity, a first-year engineering student attaches the end of a roll of toilet paper with radius 8 cm to the roof of the university library, and lets the roll fall to the ground. After t seconds, the distance in metres fallen by the roll is given by d = 4.9t^2. As the roll unwinds during its fall, its cross-sectional area decreases by 5 cm^2 for each metre it has fallen. At what rate is the radius decreasing when the roll has fallen 5 m?

i got 1.052 cm/s, can any1 verify that?

Let the distance the roll has fallen be $l$, then we have $\frac{dl}{dt} = 9.8 t$, so
\[ \frac{dA}{dt} = \frac{dA}{dl}\cdot\frac{dl}{dt} = -5(9.8)t = -49t  \]
Also, after the roll has fallen 5 meters, approximately $1.101015$ seconds have passed, so the area is approximately $64\pi - 5(1.01015) \approx 195.5569$, and thus the radius is approximately $7.8897$.
\[ \frac{dA}{dt} = -5(9.8)(1.01015) = -49.49735 = 2\pi (7.8897) \frac{dr}{dt} \Rightarrow \frac{dr}{dt} \approx -.9984  \]

(It's late, so I might have made a careless mistake somewhere :) )

i think u did, cuz i got -1.5 something, and so did a lot of my friends...but ur thing looks right, maybe some calculation erros, i made a lot of those recently...on tests too!


Two blocks of masses $ m$ and $ M$ with $ M>m$ and connected by an inextensible massless string passing over a pulley. The pulley has coefficient of friction $ \mu$  with the string (the axle is smooth).

Find the values of tension in the string.
Find the acceleration of the masses.
[hide="Note:"]Note that tension in the string on the side of $ M$ is greater than that on the side of $ m$. This problem interested me a lot but I couldn't get the appropriate approach.[/hide]

[quote="Rajiv"]Two blocks of masses $ m$ and $ M$ with $ M > m$ and connected by an inextensible massless string passing over a pulley. The pulley has coefficient of friction $ \mu$  with the string (the axle is smooth).

Find the values of tension in the string.
Find the acceleration of the masses.
[hide="Note:"]Note that tension in the string on the side of $ M$ is greater than that on the side of $ m$. This problem interested me a lot but I couldn't get the appropriate approach.[/hide][/quote]

[hide]
My approach was as follows:
See figure, consider small angle $ d\theta$ and integrating $ 0 \to \frac {\pi}{2}$ on each side.
[/hide]

[quote="aadil"]rajiv isnt frictional force mu times the normal force of contact between the bodies in contact. as the mass of the pulley and the string is 0 friction is also 0[/quote]
Agreed it is $ \mu N$. But I'm about to disagree that the Friction force itself is $ 0$.
The string does exert a force $ T_1 \plus{} T_2$ on the pulley. The pulley being fixed does not budge and exerts an [b]equal and opposite force[/b] on the string, the Normal Reaction Force $ N$. And it is this $ N$ that is responsible for friction between the pulley and the string.
I believe you are confused, drawing a Free Body Diagram (here, of the pulley) for any question can work wonders. Sometimes the FBD itself reveals the answer.


superior algebra
superior algebra unsolved

Let G be finite group and n| |G|. Suppose that L_n(G)={g\in G| g^n=1}. If for very divisor n of |G| we have |L_n(G)| divides |G|, what is the structure of G?

Hint:

Frobenius has proved that if $ n$ divides the order of an arbitrary finite group $ G$, then the number of elements in $ G$ with $ x^n\equal{}1$ is a multiple of $ n$, that is, in your notation $ |L_n(G)|\equal{}kn$.

That hint isn't needed.
Take a look to the proofs that finite multiplicatively closed subsets of fields are cyclic. Try to use the same methods here.

how to prove? i don't know, please give me more hints.

Count elements having a given order. In fact, find good bounds on their number.

The $ p$-groups satisfying that condition can't be classified! (they contain all regular $ p$-groups by [url=http://plms.oxfordjournals.org/cgi/reprint/s2-36/1/29?maxtoshow=&HITS=10&hits=10&RESULTFORMAT=&fulltext=Hall+A+contribution+to+the+theory+of+groups+of+&searchid=1&FIRSTINDEX=0&resourcetype=HWCIT]Hall[/url])
So the best we can hope for is to prove, that such $ G$ must be nilpotent.
But I have no idea if this is true, or how to prove it.

$ S_3$ satisfies the condtion but is not nilpotent.  Solvable maybe?  Where did this problem come from?

And what does ehsan2004's hint mean?  Isn't that just false?

Sorry, I thought $ |L_n(G)|$ shall divide $ n$.

But ehsan2004's comment is correct, but that's a pretty hard (but "elementary") result.

And $ S_3$ doesn't fulfill the condition, $ |L_2(S_3)|\equal{}4$ doesn't divide $ 6$.

Was thinking of order somehow.  Sorry about that.


So I got bored and made this instead of studying for the USAMO:

[hide="picture"]
[img]http://i42.tinypic.com/2vrzlzs.png[/img]
^Notice the generous amount of copy-pasting >.>
[/hide]

Unfortunately, it's largely illegible if you don't know what it says ("love" right side up and "me" upside-down). I usually don't have problems making typographical designs readable but this one really stumped me. After much fixing it still looks like "moe" instead of "me". If anyone here can give me some advice on how to fix it so that it reads slightly better, please do! I would greatly appreciate it :blush:!

This is really cool!  How exactly do you go about making these things?  I've only seen the ones in Angels + Demons (which oddly enough I was just rereading the other day)

Well, I first figure out what words/characters I'm using for that, and then I just sketch a bunch of possibilities for what it might look like. Experiment with a mix of uppercase/lowercase if needed; and sometimes the word just naturally works.

(Instant ambigram: "NOON")

And other common stuff like "u" corresponds with "n", "m" with "w" sometimes, etc. A good amount of words don't actually work, so it's rare to have a perfect ambigram with correct capitalizing.

I haven't gotten to the point where all of my glyphs have the same "style" so it actually looks like a font of some sort, like the ones in A&D.

(As for how I made it in digital format, I just used the pen tool in Photoshop to approximate what I drew on paper.)


invariant
linear algebra
matrix
search
group theory
linear algebra unsolved

Let $ f(x) \equal{} \max\left\{|x_{1}|,|x_{2}|,...,|x_{n}|\right\}$ where $ x \equal{} (x_{1},x_{2},...,x_{n})\in \mathbb{R}^{n}$
Let $ A$ be a square nxn matrix such that $ f(Ax) \equal{} f(x)$ for all $ x\in \mathbb{R}^{n}$
show that $ A^{m} \equal{} I_{n}$ for some $ m\in N$.

Any progress on this one? I've had some ideas, but I bet I'm nowhere near the solution.

First, $ A$ has rank $ n$; we can't have $ Ax\equal{}0$ and $ x\neq 0$.
Since $ A$ is surjective on $ \mathbb{R}^n$, it must be bijective on each "sphere".
Now, the hypercube $ |x_i|\le 1$ is mapped bijectively to itself. This permutes the $ 2^n$ vertices, and so some power of $ A$ fixes all vertices. That power must fix everything, and we're done.

We can reduce further; the possible matrices are the signed permutation group, which has $ 2^nn!$ elements.

indeed, the point is to show that $ A$ is a signed permutation matrix! :)

Ok, this problem is also here and there is a suggestion for a generalisation
http://www.mathlinks.ro/viewtopic.php?search_id=222853700&t=163477


linear algebra
matrix
inequalities
symmetry

Hi, if we know that $A \in R^{n \times n}$ is a positive definite matrix, and define the $A$-norm by $\Vert x \Vert_A = (x^TAx)^{1/2}$,

how would we use the fact that the $2$-norm is a norm on $R^n$ then prove that the $A$-norm is a norm on $R^n$

Most of the properties of a norm are easy here - the scaling property, that it's positive definite. What you need to work on is the triangle inequality.

Define $\langle x,y\rangle_A=y^TAx.$

Show that $\langle \cdot,\cdot\rangle_A$ satisfies all of the properties of an inner product - namely that it is a positive definite symmetric bilinear form.

From here, we can follow the abstract pathway for showing that any inner product can be used to define a norm. The first step is the Cauchy-Schwarz inequality - in this case, $|\langle x,y\rangle_A|\le\|x\|_A\|y\|_A.$

Hi, thx for helping.
I think I know what you mean by the [b]Cauchy-Schwarz inequality[/b], but I dont know how to use it to approach to provide the proof.
Could you say more about it? Thx  :blush:

$\|x+y\|_A^2=\langle x+y,x+y\rangle_A$

$=\langle x,x\rangle_A+2\langle x,y\rangle_A+\langle y,y\rangle_A$

(that used symmetry: $\langle x,y\rangle_A=\langle y,x\rangle_A$)

$=\|x\|_A^2+2\langle x,y\rangle_A+\|y\|_A^2$

(now use Cauchy-Schwarz)

$\le \|x\|_A^2+2\|x\|_A\|y\|_A+\|y\|_A^2 =\left(\|x\|_A+\|y\|_A\right)^2.$

Take square roots and that's the triangle inequality.


geometry
geometric transformation
reflection
analytic geometry

Let P be a convex polygon with n vertices. Assume that the vertices of P have integer coordinates. Prove that the area of P is at least (n - 1)/2. Can you give a better bound in case n is very large?

For the first part of the question,
[hide]We can use Pick's Theorem, which states that the area of a lattice simple polygon is:

$A=I+\frac{B}{2}-1$

where $I$ is the amount of points on the inside of the polygon, and $B$ is the amount of points on the boundary of the polygon.

We want to minimize the area of the polygons to determine that the minimum must be greater than $\frac{n-1}{2}$. To do that, we say that the vertices of the polygon are the only boundary points (to make polygons with the smallest areas). Therefore, $B=n$.

This gives us $A=I+\frac{n}{2}-1=I+\frac{n-1}{2}-\frac{1}{2}$. This means that $I>\frac{1}{2}$ in order for the condition to be true.

Since $I$ is the number of internal vertices, it can't be a fraction, therefore, $I$ must be at least 1.

Unfortunately, consider the case of the unit square and the 45-45-90 unit right triangle--they do not fulfill this requirement, therefore, you [i]can't[/i] prove that the area of $P$ is at least $\frac{n-1}{2}$ because it's not true.

On reflection, you only really need one counterexample to prove this is wrong. I can already find 2, so the conjecture is false.
[/hide]

where $I$ is the amount of points on the inside of the polygon, and $B$ is the amount of points on the boundary of the polygon

What do you mean by inside of the polygon? Do you mean there's a concave corner?

No. It's simpler than that--you just draw it out on the coordinate grid, count how many points are in the inside lol.

For instance, if you draw a 2 by 2 square the area is clearly 4. Using Pick's Theorem, there are 8 points on the boundary, and one point on the inside. Therefore, the area is:

$A=I+\frac{B}{2}-1=1+\frac{8}{2}-1=4$

which is the same answer as above.


Find all strictly monotonous functions $f: N\rightarrow N$ such that $f(f(n)) = 3n$


calculus
integration
function
logarithms
calculus computations

Evaluate the double integral:

$ \int \int_R \frac{y}{x}dA$

where $ R$ is the region bounded by $ y \equal{} \minus{}2x \plus{} 4$, $ x \equal{} 3 \minus{} y$, and the $ x$-axis.

Compute the mass in $ kg$ of a solid given that the solid is given by the region bounded by $ x \equal{} 0$, $ y \equal{} 0$, $ z \equal{} 0$, and $ x \plus{} y \plus{} z \equal{} 1$ and that its density is given by $ D(x,y,z) \equal{} x^2 \plus{} 2y \plus{} e^z$, where $ x$, $ y$, and $ z$ are in $ m$ and the function $ D$ is in $ \frac{kg}{m^3}$

\[ \iint\limits_R {\frac{y}
{x}dA} \equal{} \iint\limits_{1 \leqslant x \leqslant 2} {\frac{y}
{x}dA} \plus{} \iint\limits_{2 \leqslant x \leqslant 3} {\frac{y}
{x}dA} \equal{} \int_1^2 {\frac{1}
{x}\left( {\int_{ \minus{} 2x \plus{} 4}^{3 \minus{} x} y } \right)dx}  \plus{} \int_2^3 {\frac{1}
{x}\left( {\int_0^{3 \minus{} x} y } \right)dx} .
\]

LOL, actually, I don't know if it makes sense if $ x$, $ y$, and $ z$ are in $ m$.... the rest should make sense, though.

[hide="My Answer to First Integral"]$ 1 \plus{} \frac{9}{2}\ln{3} \minus{} 8 \ln{2}$[/hide]


search
algebra
polynomial
superior algebra
superior algebra unsolved

i don't know wheteher this is the right subforum..!
but is root(Pi) a transcendent number?

where can i find the proof of it!

thanx a lot!

Simplifying your search: $\sqrt{x}$ is algebraic if and only if $x$ is algebraic, meaning that $\sqrt{x}$ is transcendental if and only if $x$ is transcendental.

After all, the algebraic numbers are closed under both (integer) powers and (integer) roots - under rational powers, if you want to lump those together.

So if you're willing to accept that $\pi$ is known to be transcendental, you're done. If you want to see the proof of that, at least you now know what you're looking for.

Couldn't you even say more generally that $a$ is algebraic if and only if for every noncosntant polynomial over the rationals $p(a)$ is algebraic?

and isn't transcendentality of $\pi$ exceedingly difficult

The most general would be: the algebraics are closed under addition, subtraction, multiplication, division and taking roots of polynomials.

The prove for the transcendence of $\pi$ isn't that difficult (e.g. a corollary to the theorem of Lindemann-Weierstrass, which can be proved by near to elementary methods).


For what value of $ n$ is the five-digit number $ 7n933$ divisible by 33?


trigonometry
calculus
integration
algebra unsolved
algebra

Find all $k\in N$ such that:
$sink\in Q$.

I think there is only $k=0$ that satified codition of problem

[quote]I think there is only k=0 that satified codition of problem[/quote]

What about $k=90\ degrees$

No, this $k$ is $k radians$ .
 ;) .
Can you post your soluition. :(

Consider the Taylor Expansion of $\sin x$ up to order $2n+1$.
Then plug in the value of k and find a contradiction there.

Given that $k \in \mathbb{N}$.
Suppose on the contrary that $\sin k = \frac p q$ for some integers $p$ and $q$ with $\gcd(p,q)=1$ and $q>0$.
Then there exists $n \in \mathbb{N}$ such that $q\leq 4n+3$.---(*)
As $k$ is never an integral  multiple of $\pi$, $\sin k \ne 0$.
Then either one of the intevals $[k,k+1]$ or $[k-1,k]$ will lie entirely in the I,II quadrants or entirely in the III,IV quadrants.

Suppose it is the case of $[k,k+1]$ lying entirely in the I,II quadrants.( Other cases are similar as will be seen in the proof below.)

By Taylor's expansion about $k+1$ up to order $4n+3$ ,

$\sin x = \sum_{i=1}^{2n+2} \frac{(x-k-1)^{2i-1}} {(2i-1)!} + \frac{\sin (\lambda+(4n+4) {\frac{\pi}2})}{(4n+4)!}(x-k-1)^{4n+4}$ for some $\lambda$ between $x$ and $k+1$.

Sub. $x=k$ and simplify a bit on RHS:

$\frac p q = \sum_{i=1}^{2n+2} \frac{(-1)^{2i-1}} {(2i-1)!} + \frac{\sin \lambda}{(4n+4)!}$

Now, multiply both sides by $(4n+3)!$, then
LHS is an integer by (*).
RHS is the sum of an integer and the term $\frac{\sin \lambda}{(4n+4)}$
But $0< \vert \frac{\sin \lambda}{(4n+4)}\vert <1$ because $k<\lambda<k+1$.
This leads to a contradiction.
Therefore, $\sin k$ is not a rational number.


A chicken can lay three eggs in four days.  At this rate, how many days would it take three chickens to lay three dozen eggs?

If a chicken can lay three eggs in four days, then three chickens can lay nine eggs in four days.

The three chickens need to lay $ 36$ eggs, which will take them $ 4\cdot 4\equal{}\boxed{16}$

[hide=Sol]3 times the chickens to lay 12 times the eggs.
Ans = 12/3*4=16[/hide]


Prove the equation: $ \sqrt{a\plus{}\sqrt{a\plus{}\sqrt{a\plus{}\cdots}}}\equal{}\frac{1\plus{}\sqrt{1\plus{}4a}}{2}$

[quote="Obel1x"]Prove the equation: $ \sqrt {a + \sqrt {a + \sqrt {a + \cdots}}} = \frac {1 + \sqrt {1 + 4a}}{2}$[/quote]

Let $ x = \sqrt {a + \sqrt {a + \sqrt {a + \cdots}}}$

$ x^2 = a + \sqrt {a + \sqrt {a + \cdots}}$

$ x^2 - a = \sqrt {a + \sqrt {a + \cdots}}$

$ x^2 - a = x \implies x^2 - x - a = 0$

$ \therefore x = \frac {1 \pm \sqrt {1 + 4a}}{2}$

As $ \sqrt {1 + n} \ge \sqrt {1}\,\, \forall n\ge 0$

Then $ - \sqrt{1 + n}\leq - 1$

$ 1 - \sqrt {1 + n} \leq 0$

But as $ LHS > 0$,

Then $ \sqrt {a + \sqrt {a + \sqrt {a + \cdots}}} = \frac {1 + \sqrt {1 + 4a}}{2}$


function
real analysis
real analysis unsolved

Let $f: [0,1]\longrightarrow\mathbb R$ be a smooth function that $f(x_{1})=f(x_{2})=\dots=f(x_{n})=0$. Prove that for each $a\in[0,1]$ there is a $\zeta\in[0,1]$ that
\[f(a)=f^{(n)}(\zeta)\prod_{i=1}^{n}\frac{a-x_{i}}{i}\]

Apparently, $x=a$. If so, it is a classical formula. The proof is also very short: consider $g(x)=f(x)-\frac{f(a)}{\prod_{j}(a-x_{j})}\prod_{j}(x-x_{j})$ and note that it vanishes at $n+1$ points: all $x_{j}$ and $a$ (I assume that $a$ is different from all $x_{j}$; otherwise any point $\zeta$ fits). Then, by Rolle's theorem applied $n$ times, there exists $\zeta$ such that $g^{(n)}(\zeta)=0$. This $\zeta$ is the one we need.