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LaTeX
trigonometry
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system of equations

[hide="Solution"]It is clear that $a, b, c \geq 0$.  Moreover, if $a = 0$ we have the solution $(0, 0, 0)$, while if $a > 0$ then $b > 0$ and $c > 0$.  Note that for $x > 0, x \neq \frac{1}{2}$ we have $4x^{2}-4x+1 > 0$ and so $1 \geq \frac{4x}{1+4x^{2}}$ and so $x > \frac{4x^{2}}{1+4x^{2}}$, so if $a > 0$ and $a \neq \frac{1}{2}$ then $a > b > c > a$, a contradiction.  Otherwise, if $a = \frac{1}{2}$ we get the solution $a = b = c = \frac{1}{2}$.[/hide]

[quote="bthings_2000"]Cool JBL, but:

[quote]$1 \geq \frac{4x}{1+4x^{2}}$ and so $x > \frac{4x^{2}}{1+4x^{2}}$[/quote]

How did you get rid of the equivalence possibility between those two steps?[/quote]
$1 \geq \frac{4x}{1+4x^{2}}$, so $x \geq \frac{4x^{2}}{1+4x^{2}}$, but $x = \frac{4x^{2}}{1+4x^{2}}$ only if $x=\frac{1}{2}$, so $x > \frac{4x^{2}}{1+4x^{2}}$.

Yah, actually, if you look at the equation before that, the [i]introduction[/i] of the possibility of equality was in error, not its removal :)

Also, I was a little sloppy -- I should have also shown that if $a \neq \frac{1}{2}$, then the others don't, either.  But this is pretty clear (if one of them is $\frac{1}{2}$, then the other to will be by plugging in cyclicly).

Cool that makes a lot of sense... and it's a lot cooler than the massive substitution scheme I'd have tried to pull. But I guess that's the point to these kinds of problems.

Can anyone find the thread where the generalization was considered?  Specifically, what conditions must be placed on a function $f(x)$ to guarantee that the only solution to the system

$f(a_{0}) = a_{1}$
$f(a_{1}) = a_{2}$
...
$f(a_{n}) = a_{0}$

Occurs when $a_{0}= a_{1}= ... = a_{n}$?

$f(x)$ increasing is a sufficient condition, but I don't think it's necessary...?

Suppose without loss of generality that $a\geq b\geq c\geq0$.  Using the equations, you obtain that $a=b=c$.

But the system is cyclic so you cannot assume $a\geq b\geq c$

[quote="chala"][color=blue][size=150]Solve of the system of equations
solve in (IR, IR, IR)
a=4c²/(1+4c²)
b=4a²/(1+4a²)
c=4b²/(1+4b²)[/size][/color][/quote]

I think that the IR's mean irrational....(I hope)

[hide="the brute force solution"]$\frac{4x^{2}}{1+4x^{2}}=f(x)$

$f(f(f(a)))=a$

$\frac{4x^{2}}{1+4x^{2}}$

$\frac{\frac{16x^{4}}{16x^{4}+8x^{2}+1}}{\frac{32x^{4}+8x^{2}+1}{16x^{4}+8x^{2}+1}}=\frac{16x^{4}}{32x^{4}+8x^{2}+1}$

$\frac{\frac{256x^{8}}{1024x^{8}+512x^{6}+192x^{4}+16x^{2}+1}}{\frac{1580x^{8}+512x^{6}+192x^{4}+16x^{2}+1}{1024x^{8}+512x^{6}+192x^{4}+16x^{2}+1}}=\frac{256x^{8}}{1580x^{8}+512x^{6}+192x^{4}+16x^{2}+1}$

$x=\frac{256x^{8}}{1580x^{8}+512x^{6}+192x^{4}+16x^{2}+1}$

$1580x^{8}+512x^{6}+192x^{4}+16x^{2}+1=256x^{7}$

The first thing to do is to see if there are any RATIONAL roots. Since there are no roots ≥1, we try less than 1.

The factors of 1580 are 1, 2, 4, 5, 10, 20, 79, 158, 316, 395, 790, 1580.(I got them all, don't worry)

1/2. Nope. 1/4. Nope. 1/5. Nope. But then we realize that we can see if 256x^7 is greater than 512x^6.

256x^7   512x^6
x^7   2x^6

So it's not for positive numbers less than 1. So no positive rationals work. Let's try negative numbers.

Now since that equation will be positive for all negatives, we now know that there are no rational solutions.

I'm stuck here
[/hide]

I'm pretty sure that IR stands for irrational.

[quote="1=2"]I'm pretty sure that IR stands for irrational.[/quote]

First off, since I've already solved it for real solutions, this includes the possibility of irrational solutions -- there aren't any.  (There are also some complex solutions.)  Second, "IR" is ugly pre-Latex notation for $\mathbb{R}$.  (The "I" is supposed to represent the extra bar.  Why anyone ever thought this was legible is beyond me.)

[hide]
It's obvious that a=b=c=0 is a solution.
Rewrite the system as $\frac{1}{a}=1+\frac{1}{4c^{2}}$...
and let $\frac{1}{2a}=u, \frac{1}{2b}=v, \frac{1}{2c}=t$
Add those 3 equations to get $(u-1)^{2}+(v-1)^{2}+(t-1)^{2}=0$ [/hide]

[quote="chala"][color=darkred][size=117]Solve in $\mathcal R$ the system of equations $\{\begin{array}{ccc}a=\frac{4c^{2}}{1+4c^{2}}\\\\ b=\frac{4a^{2}}{1+4a^{2}}\\\\ c=\frac{4b^{2}}{1+4b^{2}}\end{array}$[/size][/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Observe that $\{a,b,c\}\subset [0,\infty )$. If $a=0$ then $\boxed{a=b=c=0}$. If $a>0$ then $\{a,b,c\}\subset (0,\infty )$.

In this case denote the interval $I=(0,\infty )$ and the functions $\{\begin{array}{c}f\ : \ I\rightarrow I\ ,\ f(x)=x\\\\ g\ : \ I\rightarrow I\ ,\ g(x)=\frac{4x^{2}}{1+4x^{2}}\end{array}$. These functions are increasing because

$x\nearrow\Longleftrightarrow$ $1+4x^{2}\nearrow\Longleftrightarrow$ $\frac{1}{1+4x^{2}}\searrow$ $\Longleftrightarrow-\frac{1}{1+4x^{2}}\nearrow$ $\Longleftrightarrow$ $1-\frac{1}{1+4x^{2}}\nearrow$ $\Longleftrightarrow$ $g(x)\nearrow$. Thus the our system becomes $\{\begin{array}{c}f(a)=g(c)\\\ f(b)=g(a)\\\ f(c)=g(b)\end{array}$.

Suppose w.l.o.g. $a\le b$. Therefore, $\boxed{a\le b}\Longleftrightarrow f(a)\le f(b)\Longleftrightarrow$ $g(c)\le g(a)\Longleftrightarrow$ $\boxed{c\le a}\Longleftrightarrow$ $f(c)\le f(a)\Longleftrightarrow$ $g(b)\le g(c)\Longleftrightarrow$

$\boxed{b\le c}$ $\implies$ $\{\begin{array}{c}a\le b\\\ c\le a\\\ b\le c\end{array}\|$ $\implies$ $a\le b\le c\le a$ $\implies$ $a=b=c$. In conclusion, the solutions of this system are only $\{\begin{array}{c}a=b=c=0\\\\ a=b=c=\frac{1}{2}\end{array}$.

[b]Remark.[/b] Using the above idea, you can compose now many similar exercises ! For example,

[/color] [quote="Virgil Nicula"][color=darkred][size=117]Solve in $\mathcal R$ the following systems of equations : $1\blacktriangleright\{\begin{array}{ccc}2x=y+\frac{1}{y}\\\\ 2y=z+\frac{1}{z}\\\\ 2z=x+\frac{1}{x}\end{array}$ and $2\blacktriangleright\{\begin{array}{c}y^{3}-9x^{2}+27x=27\\\\ z^{3}-9y^{2}+27y=27\\\\ x^{3}-9z^{2}+27z=27\end{array}$[/size][/color][/quote]

[quote="Virgil Nicula"][color=darkred][size=117]Solve in $\mathcal R$ the system of equations $\{\begin{array}{ccc}2x=y+\frac{1}{y}\\\\ 2y=z+\frac{1}{z}\\\\ 2z=x+\frac{1}{x}\end{array}$[/size][/color][/quote]
[hide]
For nonzero real $t$, $t$ and $t+\frac{1}{t}$ has same sign, so $x,y,z$ all positive or all negative. If all negative, we can negate each one to get a solution to the system where $x,y,z$ all positive.

Now by AM-GM, $t+\frac{1}{t}\geq 2$ for positive $t$, so we find $x,y,z\geq 1$. Now adding all the equations gives
\[x+y+z=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\]
If one of $x,y,z$ is greater than 1, the left hand side is greater than the right hand side, contradiction! Therefore $x=y=z=1$.

Accounting for negative solutions as well, the solutions are $\boxed{(1,1,1),\ (-1,-1,-1)}$.
[/hide]

[color=darkred][u][b]Another idea :[/b][/u] Let $\{\begin{array}{c}f\ : \ I\rightarrow (0,\infty )\\\ g\ : \ I\rightarrow (0,\infty )\end{array}$ be two functions, the first is increasing and the second is decreasing. Then

$f(x)\cdot g(y)=f(y)\cdot g(z)=f(z)\cdot g(x)\Longleftrightarrow x=y=z$. Here is some examples ! Solve the following systems of equations :

$\begin{array}{ccccc}1\blacktriangleright\{\begin{array}{c}\{x,y,z\}\subset [0,\infty )\\\\ x+y+z=\frac{\pi}{2}\\\\ \sin x\cos y=\sin y\cos z=\sin z\cos x\end{array}& & & & 2\blacktriangleright\{\begin{array}{c}n\in N^{*}\ ,\ x_{k}>0\ ,\ (\forall)k\in\overline{1,n}\\\\ x_{1}+x_{2}+\ldots+x_{n}=1\\\\ x_{1}\sqrt{1-x_{2}}=x_{2}\sqrt{1-x_{3}}=\ldots =x_{n-1}\sqrt{1-x_{n}}=x_{n}\sqrt{1-x_{1}}\end{array}\end{array}$[/color]


calculus
Stanford
college
CMU
Carnegie Mellon
number theory

It's the end of junior year now, and I'm still not sure what I should major in. I'm not enthusiastic about applying as an undecided major, so I want to know what you guys think.

Basically, as a typical Asian-American male, I enjoy doing math and science. I have taken Calculus AB this year and thought it was pretty easy, but I'm not sure about the job outlook for math majors. I also took Physics B and SAT II Physics this year, but again, I don't know the job outlook. If anyone could comment on this, please do.

My advice is: don't be in a rush to pick a major, especially if you're undecided now.  Many college don't require you to choose a major until the end of freshman year (or perhaps even later than that).  Wait until you have a better idea what you like.

Where I went to school (Carnegie Mellon), unless you were going in Fine Arts (which is a whole different ball game), not only were you not required to choose a major on your application, you were in fact [i]prohibited[/i] from choosing a major on your application -- you only choose a college (Science, Engineering, Humanities, or Business), and even that's not a big deal because it's easy to switch colleges.  You then choose a major at the end of freshman year, or sophomore year for Humanities.  The system was also essentially the same at the University of Washington (where I was a postdoc).

My other advice is to do what you enjoy and not worry too much about the job market.  Predicting what the job market will be like 5 years from now is pretty dicey anyway.  And certainly you shouldn't do something you don't enjoy just for the sake of a "good job".

Besides, there's a lot you can do with a math degree -- see our Careers in Math forum for a lot more exploration of this topic.

Okay, I checked out the Careers forum, and the main career focused on was actuary. I looked up actuaries on the Board of Labor Statistics website, and it didn't really appeal to me.

Even though many colleges don't require a specified major during admission, personally I would like to have one. I know the first 2 years are basically GE, but I'd still like something to guide me.

I'm somewhat leaning towards Electrical Engineering, but with only a Physics B background, would you guys think I could get into a decent school, such as Stanford or UCBerkeley?

My friend had only a Physics B background and got into UC Berkeley as an EECS major (Electrical Engineering/cOMPUTER science)....He ended up going to MIT...

It depends on your grades. As far as I know, Berkeley isn't as much into ECs as many other universities.

many have tried to sway me to go into engineering when it comes time for college... "you'll find a job" they say; "what are you gonna do with a physics/math degree," they say. Well to that, I say, the power to find a job [i]easily[/i] is insignificant compared to the power of pure, unadultered math and physics!

All joking aside, though, I intend to do what I enjoy most, and that's mathematics and physics sans engineering (my few projects involving design and fabrication didn't go so well  :roll: )

I don't exactly have the experience of going through college yet, but those that do tend to advise keeping an open mind and not letting 4 year old high school biases against the humanities or the sciences keep you from exploring whatever it is you dislike.

That's exactly the way I feel, curl. I believe in doing whatever I like, but in the future, I would also appreciate financial stability.

Ankur, would EECS be a joint major in Electrical Engineering and Computer Science (as in 2 degrees) or EE with emphasis on computers?

i'm in the same position and i thought i could mend it by taking EECS as major (emphasis on computers) and either major or minor in math, depending on how many math classes i feel like taking, i dont' want to get myself tied into requirements and then not be able to fulfill them!
my choice, however, is continually evolving and is very likely to change by the time i have to declare majors.

Hey guys. I noticed a lot of you are interested in engineering courses. When I first started uni, I entered an elec. eng. and computer science degree and realised it really wasn't my thing. So I switched to a Science degree, and now I'm a final year double major in Discrete Math and Computer Science.  I'd like to share some experiences about engineering, math (and life  :) ) I hope some of you find useful  when choosing uni's :?

Engineering's not all about math and physics.  In fact, the emphasis is more on design and innovation - while using math and physics to... well, design and innovate. While in this degree, the more math I learnt, the more interested I became in math as its own discipline and the more I felt that the design bits were tedious and preventing me from exploring what I was truly interested in. Engineering is a  fascinating field - some examples of what you could learn are designing circuits, building machines, understanding digital systems (these are just some electrical applications).  But I always felt that math concepts deemed relevant to engineering applications were being drilled into me without actually fully understanding why they were used or how they were linked.  It was always "what you need to know to build...." or "this will be handy when designing.... ". Even my subject names revealed this: it was always "Mathemactics B for Electrical Engineers" and "Digital Systems 1".  Instead of "Differential Calculus 1" or "Boolean Algebra" and Number Theory" etc.

If you're interested in planning, have a strong creative side, and like using math to enhance it, engineering is a great subject. But if you're also truly interested in the why? and how? of it all... a math major shouldn't be sneezed at. You could even do applied mathematics - which incorporates a lot of theories used in engineering, but still focuses on the mathematics behind it all. As for careers, if you're interested enough in what you're studying, you'll probably make a career of it.  Money shouldn't be the only driving force behind learning.  A Math degree can be used in so many different circumstances, and you don't have to be a genius lab nerd who never sees sunlight to take it up. I go out, I surf, go to the beach, and go to parties just like other young people, and last year I landed a holiday internship with Ford Motors - not because I'm a super brain (I'm not!), just because I'm an ordinary guy who's pursuing a useful, fascinating, and rewarding field of study.

Hope this helps... give math a go. Cheers.

Well put - I think you nicely summed up what I missed about math when I studied engineering.  I most enjoyed the classes that drove to fundamentals, but many of them simply did not - only the applications.  But if you want to build things or apply your knowledge, that's pretty much what engineering is for.

Thanks everyone for replying.

Considering that I am not a creative person, I'm starting to think that engineering doesn't suit me. I was looking for a math-related major that was applicable to the real world, so Applied Mathematics somewhat appeals to me (thanks awylde!) Regarding the money issue, I wasn't really looking for a high-paying career; I only wanted to major in something that would guarantee a steady job and involved math. I will definitely look into Applied Mathematics as a possible major.

Thanks everyone!  :lol:


search
geometry
ARML
AMC 10
Pythagorean Theorem
AMC

#21: http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1193996005&t=126738

#23: http://mathworld.wolfram.com/GausssCircleProblem.html

I couldn't solve for it exactly either, but using basic geometry, it is easy to see that answer is slightly more than half of the hypotenuse, which was $ \sqrt{73}$ . $ \frac{29}{41}$$ \sqrt{73}$ , or answer C was the obvious answer based on that conclusion.

Why is #7 E???

BTW, for #22, see the solution of this years ARML 2008 power round #5

[quote="mathemagician1729"]What is the solution to that triangle one, I think it was number 6 or so.[/quote]
I solved this one, which is weird, since it was geometry... :P
[quote="Problem 6"]In the diagram below, we have $ AB\equal{}3$ and right triangles $ ABD$ and $ ADE$. The point $ D$ bisects $ BC$. If $ AC\equal{}\sqrt{73}$, what is the value of $ AE$ in simplest radical form? Note: the diagram is not necessarily drawn to scale.
[i][diagram inside hide tags][/i]
$ \text{(A)}\, \sqrt{31/2}\qquad \text{(B)}\, \frac{\sqrt{39}}{73}\qquad \text{(C)}\, \frac{25\sqrt{73}}{41}\qquad \text{(D)}\, \sqrt{55}\qquad \text{(E)}\, \sqrt{70}\qquad$[/quote]
[hide="my bad solution"]Refer to the poorly made diagram below, where I have calculated some obvious side lengths with the Pythagorean Theorem.
[asy]pair A=(0,3), B=(0,0), C=(8,0), D=(4,0), E=(200/41,48/41);
draw(A--B--C--cycle); draw(A--D); draw(D--E);
label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,S); label("E",E,NE);
label("3",(0,1.5),W); label("5",(0,1.5),E); label("4",(2,0),S); label("4",(6,0),S);[/asy]
Extend $ AD$ down to point $ F$ so that $ AF\perp CF$. (Now we have this new diagram)
[asy]pair A=(0,3), B=(0,0), C=(8,0), D=(4,0), E=(200/41,48/41), F=(164/25,-48/25);
draw(A--B--C--cycle); draw(A--D); draw(D--E); draw(D--F); draw(C--F);
label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,S); label("E",E,NE); label("F",F,S);
label("3",(0,1.5),W); label("5",(0,1.5),E); label("4",(2,0),S); label("4",(6,0),S);[/asy]
Triangles $ ABD$ and $ ACD$ have the same area, so $ [ABD]\equal{}6\equal{}[ADE]\plus{}[CDE]$. We also have $ [ADE]\equal{}\frac{5}{2}DE$ and $ [CDE]\equal{}\frac{1}{2}DE\cdot DF$, so $ 6\equal{}\frac{5}{2}DE \plus{} \frac{1}{2}DE\cdot DF \\ \iff 12\equal{}5\cdot DE\plus{}DE\cdot DF \qquad(*)$.
We also know that triangles $ ADE$ and $ AFC$ are similar, so we have $ \frac{5}{5\plus{}DF}\equal{}\frac{DE}{FC} \Rightarrow 5\cdot FC\equal{}5\cdot DE\plus{}DE\cdot DF$. Using $ (*)$, we have $ 5\cdot FC\equal{}12 \iff FC\equal{}\frac{12}{5}$. By the pythagorean theorem, we know $ FC\equal{}\sqrt{16\minus{}DF^2}$. Plugging this in and solving for $ DF$ gives $ DF\equal{}\frac{16}{5}$. Then plugging this into $ (*)$ lets you solve for $ DE$, and gives $ DE\equal{}\frac{60}{41}$. Then finding $ AE$ is just a simple use of the pythagorean theorem, and it turns out to be $ AE\equal{}\frac{25\sqrt{73}}{41}$, so the answer is C.[/hide]
[quote="pythag011"]Why is #7 E???[/quote]
It took me a while to figure out, but... both the L's must be the same letter, so it would be $ \frac{3}{4}\cdot\frac{1}{4}\cdot\frac{1}{4}\equal{}\frac{3}{64}$.

[quote="Altheman"]You should post the problems too...[/quote]
Hehe, here... (I stole them :P )

Thats a really nice solution actually (well the idea of drawing point $ F$ anyway). I used the tangent subtraction formula to find tan CAD and then it was easy, but kinda ugly.

[quote="MellowMelon"][quote="Brut3Forc3"]both the L's must be the same letter[/quote]
That's not a very well worded question then.[/quote]

i'm afraid i agree
the problem just says that "LLN, where L symbolizes a letter and N a number"
it basically means that L could symbolize any letter

bah I thought they were different letters.. and #H34N1 was explaining to me about how the order of LLN didn't matter or something? Like LLN would be the same as LNL and NLL.. wth. :roll:

can someone tell me what i did wrong on #16?

I used the last number of 15 nCr 6

The two powers don't add up to 15 so that term doesn't exist and the coefficient is 0. Last number is 0.

thank u

never thought that it was a trick

BTW

could anyone post the solution to 20+ problems?

I sort of got 22 and 23 and got them right by guessitimate but would still like to know the officical answer

Also, 20 sounded like a physics problem????

So much amc hype and there's this one dude that revives a 9 year old thread

Two things:
 - First, I don't know what "Q25" is, please post it
 - Second, it would have been easier and more convenient to make a new thread of the name "Mock AMC 10A (#H34N1) #25" rather than bumping a thread that's almost a decade old.

[quote=Torontohrb]Can anyone post the solution to Q25? Is there an easy way to solve it without using integration?[/quote]

Just use Geometric probability.


Does anyone know how to put a little question mark over an inequality sign to indicate "desired" or "possible" inequality?

Thanks!

Cheers,

variable

\stackrel{?}{<} gives $ \stackrel{?}{<}$


The following is a puzzle. Can anyone solve it.

There are two urns, one contains $ b$ dollars and the other contains $ 2b$ dollars. Suppose that we are allowed to do anyone of the following.
1. Select an urn and take the money in it.
2. Select an urn and see the money which contains, and then take the money in the other urn.

 Which one  would be a better choice? Why?

This is usually presented as the "two-envelope paradox."


[hide="hint"]Find $a,b$ such that $a_{n+2}=aa_{n+1}+ba_{n}$, and then...[/hide]

Solution (wish I came up with this during the actual test, it only took me 5 mins now!! :mad: )

First we ignore this sequence but instead consider the sequence $a_{n}/3$ which is a little nicer.
Notice the equivalent \[\frac{a_{n+2}+a_{n}}{a_{n+1}+a_{n-1}}=\frac{a_{n+1}}{a_{n}}\]
Thus, \[\frac{a_{n}}{a_{1}}=\frac{a_{n+1}+a_{n-1}}{a_{n}+a_{n-2}}\cdot\frac{a_{n}+a_{n-2}}{a_{n-1}+a_{n-3}}\cdots \frac{a_{4}+a_{2}}{a_{3}+a_{1}}=\frac{a_{n+1}+a_{n-1}}{a_{3}+a_{1}}\]
so that \[a_{n}=225a_{n-1}-a_{n-2}\]
The rest is characteristic polynomials then scaling back the series

[quote="me@home"]The rest is characteristic polynomials then scaling back the series[/quote]

I think you are done already. Just divide through by $a_{n-1}$ and then divide through by $a_{n}$ and do some adding, yeah?

EDIT: Well, it's slightly more complicated than that, but there's no need to actually find the general term with the characteristic polynomial.

[quote="Altheman"]*cough* math olympiad treasures p. 83 *end cough*[/quote]

No wonder it looked familiar...

[hide="Finishing it up"] Consider $b_{n}= \frac{a_{n+1}}{a_{n}}$.  From the recursion,

$b_{n}= 225-\frac{1}{b_{n-1}}$

It's not hard to see that $b_{n}$ is strictly increasing.  We want the value of

$b_{2006}+\frac{1}{b_{2006}}< b_{2006}+\frac{1}{b_{2005}}= 225$

Hence the largest integer less than or equal is $\boxed{224}$. [/hide]

[quote="me@home"]Oooh :oops: that's easier and nicer... actually I'm not sure but I posted my solution and after that I realized that valentin had already explained his solution in the mathjam, so I looked at the transcript and he used that exactly, n.t.tuan...[/quote]
I don't know Valentin wrote what. Above solution is easy for me.  :wink:


real analysis
function
limit
integration
calculus
derivative
real analysis unsolved

Hello:

Let $A$ be a Lebesgue measurable set of $R^{n}$ such that
$m(A)>0$ . I know that for $n=1$ , $diff(A)=\left\{x-y: x\in A\;and\; y\in A\right\}$ includes an open interval of the form (-$\epsilon$,$\epsilon$). Is this true for arbitrary $n$ ?

We need to tweak the statement slightly, but the answer is essentially yes.

Theorem: If $A$ is a measurable subset of $\mathbb{R}^{n}$ with positive measure, $A-A$ contains an open ball centered at zero.
Proof:
Let $f$ be the characteristic function of $A$; this function is bounded and measurable, hence locally integrable. If $B_{r}(x)$ is the ball of radius $r$ centered at $x$, we have $\lim_{r\to 0}\frac{\int_{B_{r}(x)}f}{\mu(B_{r}(x))}=f(x)$ for almost all $x$ by the Lebesgue theorem on differentiation of the integral. The integral in the numerator is the measure of $A\cap B_{r}(x)$, so we have that $\lim_{r\to 0}\frac{\mu(A\cap B_{r}(x))}{\mu(B_{r}(x))}=f(x)=1$ for almost all $x\in A$. Since $A$ has positive measure, this "almost all" is nonempty. Let $x_{0}$ have the property above.
Choose $r$ such that $\frac{\mu(A\cap B_{r}(x_{0}))}{\mu(B_{r}(x_{0}))}>\frac23$. We can find $s$ between zero and 1 (depending only on $n$) such that $\frac{\mu(B_{1}(0)\cap B_{1}(x))}{\mu(B_{1}(0))}\ge\frac23$ whenever $|x|<s$. Now whenever $|x-x_{0}|<rs$, we have $\mu(A\cap B_{r}(x_{0}))>\frac23\mu(B_{r}(x_{0}))$, $\mu((A+x-x_{0})\cap B_{r}(x))>\frac23\mu(B_{r}(x_{0}))$, and $\mu(B_{r}(x_{0})\cap B_{r}(x)>\frac23\mu(B_{r}(x_{0}))$. Combining the pieces, $\mu(A\cap B_{r}(x_{0})\cap B_{r}(x))>\frac23\mu(B_{r}(x_{0}))+\frac13\mu(B_{r}(x_{0}))-\mu(B_{r}(x_{0}))= \frac13\mu(B_{r}(x_{0}))$ and $\mu(A\cap B_{r}(x_{0})\cap B_{r}(x)\cap (A+x-x_{0}))>(\frac13+\frac23-1)\mu(B_{r}(x_{0}))=0$. A set with positive measure is nonempty, so in particular $A\cap (A+x-x_{0})$ is nonempty and $x-x_{0}$ is in $A-A$. Therefore $A-A$ contains the ball of radius $rs$ centered at zero, and we are done.

jmerry:

From this fact does it follow that $A$ contains an open ball ?
I'm not sure about this because I can think of the following example: $\mathbb{R}$-$\mathbb{Q}$ (irrationals) are measurable
(complement of $\mathbb{Q}$ which has 0 lebesgue measure hence measurable) and have positive measure but clearly contain no interval.

No, $A$ does not have to contain an open ball. The example you gave generalizes to $n$ dimensions. The theorem I proved does not make that claim.

jmerry thanks for your help

What I wanted to prove is that given $f: \mathbb{R}^{n}\to\mathbb{R}$ a differentiable function the
set $D_{k}$={$x: f(x)=k$} has 0 measure assuming that
$\nabla$ $f(a)$ $\neq0$ (derivative of f in a)
$\forall$ $a$ $\in$ $D_{k}$

I can show that $D_{k}$ contains no open ball so I thought that
by supposing that $D_{k}$ had positive measure I could arrive to
a contradiction (but your last post confirms no such contradiction is plausible). Do you have any ideas for this problem, which does not use the Lebesgue density theorem? Thanks in advance.


[quote="dule_00"]prove or disprove following:

If $x^{n}+x^{m}+1$, ($m,n \in \mathbb{N}$ and $n>m$) is reducible over $\mathbb{Z}[x]$ then also $x^{n}+x^{(n-m)}+1$ is reducible over $\mathbb{Z}[x]$.

edit: if $3 \mid m+n$ it is simple to see that both of them are divisible by $x^{2}+x+1$ so asume that $3 \not \mid m+n$. :)[/quote]
1. is obviosly $x^{n}+x^{n-m}+1=x^{n}(y^{n}+y^{m}+1),y=\frac{1}{x}$.
2. \[x^{n}=x^{n}-x^{n-3}+x^{n-3}=x^{n_{3}}(x-1)(x^{2}+x+1)+x^{n-3}=x^{n-3)(mod \ x^{2}+x+1)}\] .
Therefore $x^{n}=x^{n(mod \ 3)}(mod \ x^{2}+x+1)$. It give $(x^{2}+x+1)|(x^{n}+x^{m}+1)$ if and only if $m\not =n(mod \ 3)$ and $3|(m+n)$.

[quote="Rust"][quote="dule_00"]prove or disprove following:

If $x^{n}+x^{m}+1$, ($m,n \in \mathbb{N}$ and $n>m$) is reducible over $\mathbb{Z}[x]$ then also $x^{n}+x^{(n-m)}+1$ is reducible over $\mathbb{Z}[x]$.

edit: if $3 \mid m+n$ it is simple to see that both of them are divisible by $x^{2}+x+1$ so asume that $3 \not \mid m+n$. :)[/quote]
1. is obviosly $x^{n}+x^{n-m}+1=x^{n}(y^{n}+y^{m}+1),y=\frac{1}{x}$.
.[/quote]

I don't get it. $y^{n}+y^{m}+1$ for $y=\frac{1}{x}$. is not polynomial of variable $x$.


Let $ S$ be the set of ordered triples $ (x,y,z)$ of real numbers for which
\[ \log_{10} (x \plus{} y) \equal{} z\text{ and }\log_{10} (x^2 \plus{} y^2) \equal{} z \plus{} 1.
\]There are real numbers $ a$ and $ b$ such that for all ordered triples $ (x,y,z)$ in $ S$ we have $ x^3 \plus{} y^3 \equal{} a \cdot 10^{3z} \plus{} b \cdot 10^{2z}$. What is the value of $ a \plus{} b$?

$ \textbf{(A)}\ \frac {15}{2}\qquad \textbf{(B)}\ \frac {29}{2}\qquad \textbf{(C)}\ 15\qquad \textbf{(D)}\ \frac {39}{2}\qquad \textbf{(E)}\ 24$

Let $ z\equal{}0$, then $ x\plus{}y\equal{}1$ and $ x^2\plus{}y^2\equal{}10$. And the last equation becomes $ x^3\plus{}y^3\equal{}a\cdot 10^{0}\plus{}b\cdot 10^0\equal{}a\plus{}b$. So it suffices to find $ x^3\plus{}y^3$.

But $ (x\plus{}y)^2\minus{}(x^2\plus{}y^2)\equal{}2xy$, so $ xy\equal{}\minus{}\frac{9}{2}$. Then $ x^3\plus{}y^3\equal{}(x\plus{}3)^3\minus{}3(xy)(x\plus{}y)\equal{}1\plus{}\frac{27}{2}\equal{}\frac{29}{2}$, and we are done.

Alternate solution:
[hide]
Since $(x+y)(x^{2}+y^{2})=x^{3}+y^{3}-xy(x+y)$. After plugging in all the equalities and $xy=\frac{1}{2}(10^{2z}-10(10^{z}))$ and simplify, we have $2a(10^{z})+2b=30-10^{z}$. So $a=-\frac{1}{2}, b=\frac{30}{2}$ and $a+b=\frac{29}{2}$.

[/hide]

[quote="w squared parabola"]Alternate solution:
[hide]
Since $(x+y)(x^{2}+y^{2})=x^{3}+y^{3}-xy(x+y)$. After plugging in all the equalities and $xy=\frac{1}{2}(10^{2z}-10(10^{z})$ and simplify, we have $2a(10^{z})+2b=30-10^{z}$. So $a=-\frac{1}{2}, b=\frac{30}{2}$ and $a+b=\frac{29}{2}$.

[/hide][/quote]

Could you explain more about you substitution?

Um...[b]w square parabola[/b] has the equation $(x + y)(x^2 + y^2) = x^3 + y^3 - xy(x + y)$, and we know the values of $x + y,x^2 + y^2,xy,$ and $x^3 + y^3$ in terms of $z$, so we can just plug those values in.

Notice: Letting $z=0$ finishes the problem quickly :P

[hide=Fake Solve]Since it is given that it’s true for all triplets, we can plug in a particularly nice one with $z=0$.

So note that we must have $x=k+1,~y=-k$ with $k^2+(k+1)^2=10$.  Rearranging the second equation gives us gives us $k^2+k=\frac{9}{2}$.

Thus $a+b=(k+1)^3-k^3=3k^2+3k+1=3(k^2+k)+1=3\left(\frac{9}{2}\right)+1=\boxed{B~\frac{29}{2}}$


Prove that if $x,y,z\in{R}^{+}$ and $x+y+z=xyz+2$,then
\[xyz\geq 1\]

[quote="Tiks"]Prove that if $x,y,z\in{R}^{+}$ and $x+y+z=xyz+2$,then
\[xyz\geq 1 \]
[/quote]
What if $x=y=1$ and $z=\frac{1}{2}$ $?$ :wink:

[quote="arqady"]
What if $x=y=1$ and $z=\frac{1}{2}$ $?$ [/quote]

I don't know  :D :blush:.

It seems to me that there was something wrong with me yesterday :oops: ; this is the second wrong problem I posted then :( .

Please, forget about this!!


vector
induction
combinatorial geometry
linear algebra
linear algebra unsolved

We call nice a familly of vectors $v_1,..,v_k$ if every vector in $R^n$ can be written as a linear combination of $v_i$ with strictly positive coefficients.Find the best $k_n$ with the property that any nice familly with $k_n$ vectors contains a proper sub-familly which is also nice.

I think that from any nice family of $\ge 2n$ vectors we can extract a nice subfamily of size $2n$. We can perform induction on the size of the set and the dimension of the space we're working in.

Since $0$ is in the convex hull of our set, it's a convex combination of strictly fewer than $2n+1$ vectors (at most $n+1$ by Caratheodory's Theorem, for example), so it's a positive convex combination of $<2n+1$ vectors $v_1,v_2,\ldots,v_t$. It's easy to see that in these conditions, $v_1,\ldots,v_t$ form a nice family in the space $\mathcal F$ of dimension $m\le n$ generated by them, so we can use the induction hypothesis to extract $\le 2m$ out of these $t$ which form a nice family in $\mathcal F$.

Now let $\mathcal F^{\perp}$ be the orthogonal complement of $\mathcal F$, and consider the projection of our vectors on $\mathcal F^{\perp}$. It's clear that these projections form a nice family in $\mathcal F^{\perp}$, so we can find at most $2(n-m)$ vectors whose projections form a nice subfamily in $\mathcal F^{\perp}$.

The two families of vectors, of sizes $\le 2m$ and $\le 2(n-m)$ can be united to give a family which is easily shown to be nice (in the entire space).

I hope it's Ok.


calculus
integration
trigonometry
function
calculus computations

So I'll just jump into the meat of it:

For each $ f: [0, \pi] \mapsto R$, satisfying $ f(0)\equal{}f( \pi )\equal{}0$, define the integral

$ I(f)\equal{} \int^\pi_0 (f'(t))^2 \plus{} 2\sin (t) f''(t) dt$

Find a function f which minimizes the integral. (hint: integrate by parts)

---and just to clarify, that R is the set of real numbers...I wasn't sure how to get the proper symbol into Latex. Any help you can give me, I would greatly, greatly appreciate it.

so i guess you're in the same class as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227987]this guy[/url], right?

Look F(t)=sint(by intution) dy/dt= cos t & d/dt(dy/dt)=-sint

substitute the values & get the ans.

blahblahblah...I guess so. Hahaha, the happy time before problem sets are due...

Thank you both for the help.


function
integration
real analysis
calculus
limit
trigonometry
real analysis unsolved

Does there exist a measurable function....?

no

That is what I thought, but at the moment I don't see any reason for that.

There exists such function. Consider the set $S$ of all real numbers whose decimal representation contains only a finite number of "1". It is easy to see that
1) The set $S$ has measure 0.
2) The cardinality $|S \cap (a,b)|$ is continuum (for every $a<b$).

We claim that there exists a partial map $f: S\to [0,1]$ s.t. $f$ maps every $(a,b)$ onto $[0,1]$. Indeed, we construct such a map by transfinite induction. Consider the set $P=\{(r, (a,b)) : r\in[0,1],\, a,b\in{\mathbb Q}\}$. Enumerate all elements of $P$ (use the Axiom of Choice) by ordinals up to $\mathbf c$ (continuum). Now pick elements of $P$ one by one. For every pair $(r, (a,b))$, consider a point $x\in S \cap (a,b)$ for which $f$ is not defined yet. Define $f(x) =r$, and proceed to the next pair. We obtain the desired function $f$.

Finally, define $h$: $h(x) = f(x)$ where $f(x)$ is defined; $h(x)=1/2$ elsewhere. Since $h(x) =1/2$ a.e., the average value of $h(x)$ equals $1/2$ on every interval. Since $f(x)$ assumes all values from $[0,1]$ on every interval, $h(x)$ also assumes all values from $[0,1]$ on every interval.

Note that Yury's example function is equal to $\frac12$ almost everywhere; he's only tweaking it on a set of measure zero to get its local range to be all of $[0,1].$

There was no choice about that; we must have $f(x)=\frac12\ a.e.$ Why? By the Lebesgue theorem on the differention of the integral, $\lim_{\eta,\delta\to0}\frac1{\eta+\delta}\int_{x-\eta}^{x+\delta}f(t)\,dt=f(x)$ for almost every $x,$ assuming that $f$ is a locally integrable function. The hypothesis of this problem is that every such normalized integral equals $\frac12,$ so certainly the limit equals $\frac12$ everywhere.

Another (not much different) construction: let $I_{n}$, $n\ge 1$, be an enumeratation of all nonempty intervals $(a,b)\subset [0,1]$ with rational endpoints. For each $n$ choose a compact set $E_{n}\subset I_{n}\setminus\bigcup_{m<n}E_{m}$ of measure zero and cardinality continuum. (A tiny copy of the Cantor set will work; note that $I_{n}\setminus\bigcup_{m<n}E_{m}$ is open and nonempty.) There is a bijection $f_{n}\colon E_{n}\to [0,1]$. Define $f=f_{n}$ on each $E_{n}$ and $f(x)=1/2$ elsewhere.

[quote="puuhikki"]Does there exist a measurable function $h: \mathbb{R}\to [0, 1]$ such that for all $a,b$ such that $a<b,\,h|_{[a,b]}$ gets all values from $[0,1]$ and $\frac{1}{b-a}\int_{a}^{b}h(x)\,dx=1/2?$[/quote]

Here is an explicit construction. I'll let the reader verify that it works, because I'm lazy.

Let $h(x) = \lim_{n\rightarrow\infty}\sin^{2}\left(2^{2^{n}}x\right)$ if the limit exists; otherwise let $h(x) = 1/2$.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Find the differentiable $f(x)$ which satisfies the following equation.

\[f(x)=-\int^x_0 f(t)\cdot\tan tdt+\int^x_0 \tan(t-x)dt\ \left(-\frac{\pi}{2}<x<\frac{\pi}{2}\right)\ \]

We should rewrite $\int_0^x\tan(t-x)dt$ as $-\int_0^x\tan(x-t)dt=-\int_0^x\tan(t)dt.$

Differentiate the given integral equation to obtain this differential equation:

$f'(x)=-f(x)\tan(x)-\tan(x); f(0)=0.$

I'll leave this differential equation for someone else to finish solving.


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

if $ G$ is a group , $ |G|=p^n, p \ is \ prime$, and $ N$ is normal subgroup of $ G$ of order $ p$, prove that $ N\subseteq Z(G)$

My thinking about it :

$ N$ is a G-set via conjugation on the elements of $ N$, using the class equation with the center of $ G$ form, we can get $ N$ is in the center .

from :

$ |G|=|Z(G)+\sun\limits_{i=1}^p {[G: C(n_i)]}$

since if $ |G|=p^{n}$ then $ Z(G)\ne\{e\}$ , and $ N$ is abelian cyclic subgroup since it's of prime order.

my problem  ,is necessary $ C(n)=N$ if $ n\not in Z(G)$ and $ n\in N$.

since if $ C(n)=N$ , we can get the center is trivial which is contradiction >

 :)

Let $ N \equal{} < n >$ of order $ p$ and $ g\in G$. Since $ N$ is normal, we must have $ g^{ \minus{} 1}ng \equal{} n^t$ with $ p\not|t$. This implies $ g^{ \minus{} (p \minus{} 1)}ng^{p \minus{} 1} \equal{} n^{t^{p \minus{} 1}} \equal{} g$, i.e. $ g^{p \minus{} 1}\in C_G(n)$. Since $ p \minus{} 1$ is coprime to the order of $ g$ we have $ g\in C_G(n)$. Since $ g$ was arbitrary, we have $ n\in Z(G)$ and $ N\subseteq Z(G)$

how do u get $ n^{t^{p \minus{} 1}} \equal{} g$ ?

this question from Group action , it must be solved using group action

Thank you  :)

sorry, this is a typo and should read $ n^{t^{p \minus{} 1}} \equal{} n$. This equation holds, since $ p\not|t\Longrightarrow t^{p \minus{} 1}\equiv 1\bmod{p}$ and $ n$ has order $ p$.

P.S. my proof is via group action: $ G$ operates on $ N$ via conjugation or more formally $ G\rightarrow Aut(N)$, which gives $ G/C_G(N)$ is isomorphic to a subgroup of $ Aut(N)$. This implies $ C_G(N) \equal{} G$, since $ p$ and $ |Aut(N)|$ are coprime. This would have been an alternative proof, although modulo formalism they are identical   :)

P.P.S a true alternative proof would be via induction on the group order resp. the assumption of a minimal counterexample: If $ N\not\subseteq Z(G)$, we must have $ G$ non-abelian and $ NZ(G)/Z(G)\subseteq Z(G/Z(G))$. Therefore, for any $ g\in G,n\in N$ we can determine $ z\in Z(G)$ with $ g^{\minus{}1}ng\equal{}nz$. Since $ N$ is normal, we have $ nz\equal{}g^{\minus{}1}ng\in N$ and hence, $ z\equal{}1$, a contradiction to the assumption of a minimal counterexample.


induction
number theory proposed
number theory

Find all $f:\mathbb{N^*}\to\mathbb{N^*}$ satisfying the condition:
   $f(f(n)+f(n)=2n-1$ if $n=2k$ and $f(f(n)+f(n)=2n+1$ otherwise
I remember it was posted before but I don't find it.Please give me the links,thanks!

[quote="zscool"]could you fix the parantheses, im not sure if its supposed to be

f(f(n)+f(n)) or f(f(n)) + f(n)...[/quote]
$f(f(n))+f(n)...$.Sorry for my hand bad writing!

For some reason, this problem is not appealing to me, so I'll pass :).

one solution is $f(n)=n-1$ for $n$ even, $f(n)=n+1$ for $n$ odd.

in fact, it is the only one:

consider the set $M_n=\{|f(x)-x|: x\leq n\}$.
let $x\leq n$ be such that $|f(x)-x|$ is maximal in $M_n$.

we have $f(f(x))-f(x)=2(f(x)-x)-1$ or $f(f(x))-f(x)=2(f(x)-x)+1$; in any case, $|f(f(x))-f(x)|\geq 2|f(x)-x|-1$.

if $f(x)\leq x$, then we would have $f(x)\leq n$, therefore $|f(f(x))-f(x)|\leq |f(x)-x|$. but $|f(f(x))-f(x)|\geq 2|f(x)-x|-1$, giving $|f(x)-x|\geq 2|f(x)-x|-1$ => $|f(x)-x|\leq 1$.

now let $f(x)>x$.

then $f(f(x))\leq 2x+1-f(x)\leq f(x)$. consider $N_x=\{|f(f(y))-f(y)|: y\leq x\}$. let $y\leq n$ be such that $|f(f(y))-f(y)|$ is maximal in $N_x$. we have $f(f(y))-f(y)=2(f(y)-y)\pm 1$, thus $|f(f(y))-f(y)|\leq 2|f(y)-y|+1\leq 2|f(x)-x|+1$; but $|f(f(x))-f(x)|=2|f(x)-x|\pm 1$.

at least we know that $|f(f(y))-f(y)|\leq |f(f(x))-f(x)|+2$. but $f(f(f(x)))-f(f(x))=2(f(f(x))-f(x)\pm 1$ gives $|f(f(x))-f(x)|+2\geq 2|f(f(x))-f(x)|-1$ => $|f(f(x))-f(x)|\leq 3$. this at least gives $|f(x)-x|\leq 2$.

summing up, we know that $|f(x)-x|\leq 2$.

but we now $f(f(x))+f(x)=2x+1$ for $x$ odd. we have five cases:
$f(x)=x$, which immediately gives a contradiction.
$f(x)=x-1$ gives $f(x-1)=x+2$, contradiction.
$f(x)=x-2$ gives $f(x-2)=x+3$, contradiction.
$f(x)=x+2$ gives $f(x+2)=x-1$, contradiction.
thus $f(x)=x+1$ giving $f(x+1)=x$. but thats what we wanted to prove.

Or you can just induct.

Clearly, $f(1) = 2$ and $f(2) = 1$.

Suppose $f(n) = n-1$ for $n$ even, $f(n) = n+1$ for $n$ odd for $n \leq 2m, m\in \mathbb{Z}$.

Since $f(f(2m+1)) + f(2m+1) = 4m+3$, it's clear that $f(2m+1) \geq 2m+2$.  Suppose $f(2m+1) > 2m+2$.  Let $f(2m+1) = k > 2m+2$, then $f(k) < 2m+1$.  But $f(f(k))+f(k) > 4m+7$, a contradiction.  Hence $f(2m+1) = 2m+2$ and $f(2m+2) = 2m+1$.

Thank you very much, Valiowk!!!  :D  :D  :D

[quote="Peter Scholze"]but induction is so boring :D
well, your solution is indeed "a bit" nicer, valiowk :lol:[/quote]
Peter are not objective ;)


AMC AIME USA(J)MO USAMO HMMT USAMTS ARML

If your goal is just to pass AIME and make USAMO and not necesarily do well on USAMO, is it still a good idea to do proofs?

I know when you are studying for math, even if its purely computation, you should still do proofs to understand why something is.

But is it that beneficial to work on like old USAMO/IMO proofs or even basic general beginner well known proofs (like AM-GM) when your main objective is only to pass AIME which is all computation?  Will it help you on AIME (i personally dont think it will)

I would work on it obvisouly if I had time, but i dont have that much time so do you think I should spend like even 1/4 of the time i do have studying for the AIME working on proofs?  Or should I spedn the whole entire timejust doing old AIMEs and good computation problems (like HMMT)

I was wondering because i know alot of people here have made USAMO and have experience.

oh yea and its not like im very good at proofs, i probably will get a 0 on any USAMO, I dont think i can even prove most (like 1/2) of the ones in the Art of problem solving vol. 2 book

Thanks

[quote="the future"]But is it that beneficial to work on like old USAMO/IMO proofs or even basic general beginner well known proofs (like AM-GM) when your main objective is only to pass AIME which is all computation?  [/quote]

I wouldn't call it "computation"; it's all problem-solving, and working through easy USAMOs is definitely a way to sharpen your problem-solving skills.

It certainly doesn’t hurt to do proofs. Some of the tricks you could learn to do with proofs could just happen to save some of your effort on questions from the AMC/AIME. This year especially I think they’re banning calculators on the AMCs, so studying higher-level problem solving could be useful.

There’s about a one month period of time between the AIME and the USAMO, so there’s always a slight bit of cramming time left for just focusing on proofs (as if there weren’t enough to do in April!). Last year I definitely didn't expect to make the USAMO but I did, and I didn't score all too badly on it.

The [url=http://www.usamts.org]USAMTS[/url] is a proof-based contest which has a difficulty usually ranging between the AIME and USAMO level, so it’s not so bad of a transition between the two contests. Typically the first couple of questions might involve more calculation-based and wouldn’t be all that different from some questions on the AIME, while the last one can sometimes near the difficulty of some of the easier USAMO questions. The proofs are graded on a scale (though much more relaxed than Olympiad contests) and there is feedback given, albeit most of the time it being just one word.

Working out a few proofs can help make your thinking clearer, which will help ward off logical errors during the AIME (especially with tricky combinatorics problems where it's easy to miss or double-count a case).

Never mind the USAMO. How interested are you in mathematics for its own sake? Will you be a mathematician? A sophisticated user of mathematics? An interested amateur? Mathematics must be communicated, and mathematics is communicated and learned through proof.

As an aside, I [b]strongly[/b] recommend that members of the ARML team I coach participate in at least one mail-in proof contest, either USAMTS or the Pomona College MTS (which is the California version of a contest that has an identical Wisconsin version.) There are probably others. Of course, ARML itself does have one proof-based round.

And I agree with the essential thrust of the three posts above this. Those are all good points, too.


Would anyone be willing to proofread one of my short essays for Princeton? The topic is "tell us about a person who has influenced you in a significant way". If you are willing, then I will PM it to you.

I would really appreciate any assistance.

Thanks.

collegeconfidential.com  :D

(somewhat tongue in cheek)

You can send it to me also if you like but I am really busy so I can't promise I'll get back to you quickly.

I will do it to (even if the only reason is  to read a college paper


i think you are wrong!! the inequality you suggested first is wrong!


floor function
number theory unsolved
number theory

Hi ,Please help me to solve the probelm.

[hide]We can take

$ a \equal{} t(t \plus{} 2)$,
$ b \equal{} (t \plus{} 1)(t \plus{} 3)$,
$ c \equal{} t(t \plus{} 3)$
and
$ d \equal{} (t \plus{} 1)(t \plus{} 2)$,

where $ t \equal{} \lfloor \sqrt {p \plus{} 1} \rfloor$. This gives $ t > \sqrt {p \plus{} 1} \minus{} 1$, therefore $ min\{a, b, c, d\} \equal{} a \equal{} t(t \plus{} 2) > p$. Also, to get $ max\{a, b, c, d\} \equal{} d \equal{} (t \plus{} 1)(t \plus{} 3) < (2 \plus{} \sqrt {p \plus{} \frac {1}{4}})^2$, we need

$ t^2 \plus{} 4t \plus{} 3 < 4.25 \plus{} p \plus{} 4\sqrt {p \plus{} 0.25}$
(now we use $ t \le \sqrt {p \plus{} 1}$)
$ \Longleftarrow p \plus{} 1 \plus{} 4t \plus{} 3 < 4.25 \plus{} p \plus{} 4\sqrt {p \plus{} 0.25}$]
$ \Longleftrightarrow 4t < 0.25 \plus{} 4\sqrt {p \plus{} 0.25}$
$ \Longleftrightarrow 4t < 0.25 \plus{} 4\sqrt {p \plus{} 0.25}$
$ \Longleftrightarrow t < 0.0625 \plus{} \sqrt {p \plus{} 0.25}$
$ \Longleftarrow \sqrt {p \plus{} 1} \minus{} \sqrt {p \plus{} 0.25} < 0.0625$
(now we are going to "rationalize")
$ \Longleftrightarrow 0.75 < 0.0625(\sqrt {p \plus{} 1} \plus{} \sqrt {p \plus{} 0.25})$
$ \Longleftrightarrow 12 < (\sqrt {p \plus{} 1} \plus{} \sqrt {p \plus{} 0.25})$
$ \Longleftrightarrow 12 \le (\sqrt {p \plus{} 1} \plus{} \sqrt {p \plus{} 0.25})$
$ \Longleftrightarrow 12 \le 2\sqrt {p}$

So this is OK for $ p \ge 36$. The other cases can be made by hand, but I don't want to do it. For example, for $ 36 > p \ge 30$, we can choose $ a \equal{} 56$, $ b \equal{} 42, c \equal{} 48$ and $ d \equal{} 49$.

Any miscalculations?
[/hide]


function
algebra
functional equation
algebra proposed

Find all the functions $ f: R \rightarrow R$ so that $ f(f(y)\plus{}2\plus{}x)\plus{}f(f(y)\minus{}x)\equal{}yf(y)(x\plus{}1)$ is true for any real numbers $ x$ and $ y$.

Substitute $ g(x) \equal{} f(x \plus{} 1)$. The functional equation becomes
\[ g[g(y) \plus{} x] \plus{} g[g(y) \minus{} x] \equal{} (y \plus{} 1)xg(y)\hspace{0,5cm}(*)
\]
Taking $ x \equal{} \minus{} g(y)$ and $ x \equal{} g(y)$, we find that
\[ \minus{} (y \plus{} 1)[g(y)]^2 \equal{} g[2g(y)] \equal{} (y \plus{} 1)[g(y)]^2
\]
because $ g(0) \equal{} 0$. In fact, putting $ x \equal{} 0$ in $ (*)$ gives $ g[g(y)] \equal{} 0$ and putting $ y \equal{} g(x)$ shows that $ g$ is an odd function.
Thus
\[ (y \plus{} 1)[g(y)]^2 \equal{} 0
\]
and $ g(y) \equal{} 0$ for all $ y\neq \minus{} 1$. But $ g(\minus{}1)\equal{}\minus{}g(1)\neq 0$ which implies that $ f\equiv 0.$


I got 21 points with two commended...somehow...but I'm not on the leaders list. Why not?

Did you complete the optional form allowing your name to be listed on the USAMTS site?

EDIT: Oh, and also, score updating may not be complete yet.

I'm sure I sent in the permission form with the entry form, and obviously they did get my entry form. And the leaderboard looks complete to me. Hmm. Maybe my memories are playing tricks on me in regards to the permission form...

I think you should PM or email Mr. Rusczyk.

[quote="bookaholic"]Um...bump. Can an administrator please confirm that I do not, in fact, have a permission form on file? I'll send one later if that's the case.[/quote]Your profile page on the USAMTS website should indicate what forms we have on file for you.


Which one of the following four figures has the greatest area?
(a) triangle $ABC$ with $\angle A = 60^\circ$, $\angle B = 45^\circ$ and $AC = \sqrt{2}$.
(b) trapezoid with diagonals of lengths $\sqrt{2}$ and $\sqrt{3}$, and a $75^\circ$ angle between them.
(c) a circle with radius 1
(d) a square with diagonals of length 2.5

[hide]The answer is:

a) Using 30-60-90 triangles, and then using 45-45-90 triangles after extending an altitude from C to AB. You have area as $(\frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2})(\frac{\sqrt{6}}{2}) = \frac{3}{4} + \frac{\sqrt{3}}{4}$. That is a little greater than 1.

b) Let AC be the diagonal with length $\sqrt{3}$, and BD be the diagonal with length $\sqrt{2}$. Extending altitudes from D and B to AC, we can form similar triangles, and then solve. You have the area being the sum of the areas of the two large triangles ABC and ABD, which is $\sqrt{2} + \sqrt{6}$. That is about 3.85...

c) Well, [O] = :pi:$r^2$. That equates to about 3.14159

d) Notice that the area of the square is $D^2 / 4$, which is 25/8 = 3.125.

B is the answer I believe.[/hide]


geometry
3D geometry
algorithm
search
group theory
abstract algebra
graph theory

considering the diameter of the cube's cayley graph, a straightforward shortest path search is too resource intensive.

a lot of cube solvers generate a heuristics table that encompasses all states for a certain number of moves (around 10 or 12, i think?), and then use a graph search method (or a few) to finish it off.

one solver uses group theory instead of massive tables to simplify the problem. not a god's algorithm, but more interesting than one. it uses subgroup G = {U,D,R2,L2,F2,B2} to split up the solve into two parts, first solving the cube into G and then solving from there using the moves in G. the code is available, if you want to take a look (it's written in C) http://kociemba.org/cube.htm

i actually started writing one of these before, with the intention of generating human algs (nice, easy to execute ones like 3-generators) instead of god algorithms, ultimately with the ability to impose an arbitrary constraint on the generated alg so i could find some better ones to solve certain cases, but got kind of lazy. actually, i still don't know how to approach this, since you usually have to move far away from the solved state and then come back quickly.

[quote="t0rajir0u"]There are several well-known algorithms for the [url=http://en.wikipedia.org/wiki/Shortest_path_problem]shortest path problem[/url], but the issue here is that the Cayley graph of the [url=http://en.wikipedia.org/wiki/Rubik%27s_Cube_group]Rubik's cube group[/url] is pretty large.  I'm not sure if this is a tenable approach.[/quote]

In the meantime the shortest solution from every possible state has been proved to have length 20([url=http://en.wikipedia.org/wiki/God's_algorithm]also called God's number[/url]), that says something about the diameter. [url=http://cube20.org]It was proved through an intensive 35 CPU-years number crunching.[/url]

[quote="EulerIntegral"]I wanted to make a program to inspect possible rubik's cube solving algorithms. It is not intended to solve the cube, but to inspect, as said.

For example. Suposse you have a cube with some faces in it's position, and you want to insert a face somewhere in the cube without altering that faces, but you can alter others faces. The program would try to make all sort of movements to match the initial and final position.

The representation is not importante here, so let's go on. We have a set of possible moves, don't matter what. And we modify the initial state until we match the final state. Them we choose the sequence with shortest moves.

The problem is: If we have only 12 moves (left up, top rigth, etc) and we test it at a maximum deep of 9 moves we end with something near 5 billions of possible moves.

So, let we have a set of possible moves, and we search at a deep $ n$, how can we improve the efficience of the algorith to find the solution?

I thought of discarting moves or partial moves that are cyclic ( you are in a state, go to another and them came back to the original).

It is a good improvement, but not to good. How could've I made more improvements?

I thought also something like that: If the piece wanted to be inserted is not moved for 3,4 moves, the sequence is discarted. But it is discarted only on runtime, the other (below) is before runtime.

Sorry if i didn't were clear, but again: The program is to inspect (or create) solving techniques (algorithms). And as the deep goes deep (joke) we have more sequences to search, and the problem is to make this set of sequences smaller discarting redundant results.

Thanks[/quote]

If you want you can use Perl and CM::Permutation on CPAN which includes [url=https://metacpan.org/module/CM::Rubik]CM::Rubik[/url]
It's an MVC implementation of the Rubik's cube where the Model contains the actual state as a permutation of the facelets. Also you have at your disposal the API with the classical moves
F,U,R,B,L,D and their inverses so you can do any moves you want.
In my oppinion a simple solving algorithm would involve a lot of pattern-matching and cases/subcases.


Putnam
inequalities unsolved
inequalities

[quote="Butterfly"]$ a,b,c,A,B,C \in R$,and $ a,A\neq 0$.Suppose that for any real number $ x$,$ |ax^{2}+bx+c|\leq |Ax^{2}+Bx+C|$,prove that $ |b^{2}-4ac|\leq |B^{2}-4AC|$.[/quote]
We separately the three cases:
[u][i]Case 1[/i][/u]: $ Ax^{2}+Bx+C=0$ has two distinct roots.
Because $ |ax^{2}+bx+c| \le | Ax^{2}+Bx+C |, \forall x$ :we must have two roots of $ Ax^{2}+Bx+C=0$ also roots of $ ax^{2}+bx+c=0$
It follows that:
$ |\frac{a}{A}|=|\frac{b}{B}|=|\frac{c}{C}|$
Let $ x=1$ we have $ |c| \le |C|$. Then: $ |\frac{a}{A}|=|\frac{b}{B}|=|\frac{c}{C}| \le 1$
Or :
$ \frac{|b^{2}-4ac|}{|B^{2}-4AC|}\le 1$
Or: $ |b^{2}-4ac| \le |B^{2}-4AC|$
[u][i]Case 2[/i][/u]: $ Ax^{2}+Bx+C=0$ has root $ x_{0}$
Then:
$ Ax^{2}+Bx+C=A(x-x_{0})^{2}$. We have $ x_{0}$ also root of $ ax^{2}+bx+c=0$
It follows that $ ax^{2}+bx+c=a(x-x_{0})(x-x_{1})$
+If $ x_{0}%Error. "different" is a bad command.
x_{1}$ . Let $ x=\frac{x_{0}+x_{1}}{1+|\frac{a}{A}|}$
We have $ |A\parallel x-x_{0}|^{2}\ge |a\parallel x-x_{0}\parallel x-x_{1}|$ is wrong.
So $ x_{1}=x_{0}$. It follows that $ |b^{2}-4ac| = |B^{2}-4AC|=0$
[u][i]Case 3[/i][/u]: $ Ax^{2}+Bx+C >0, \forall x$. We have $ 0 \le a <A$ and:
$ Ax^{2}+Bx+C =A(x+\frac{B}{2A})^{2}-\frac{4AC-B^{2}}{4A}$
$ ax^{2}+bx+c =a(x+\frac{b}{2a})^{2}-\frac{4ac-b^{2}}{4a}$
Let $ x=\frac{-2B}{A}$:
$ |Ax^{2}+Bx+C|=\frac{|B^{2}-4AC|}{4A}$, $ |ax^{2}+bx+c|=\frac{|b^{2}-4ac|}{4a}$
We have : $ \frac{|B^{2}-4AC|}{4A}\ge \frac{|b^{2}-4ac|}{4a}$. It follows that: $ |b^{2}-4ac| \le |B^{2}-4AC|$
We are done!

[quote="NguyenDungTN"] Let $ x=\frac{x_{0}+x_{1}}{1+|\frac{a}{A}|}$
We have $ |A\parallel x-x_{0}|^{2}\ge |a\parallel x-x_{0}\parallel x-x_{1}|$ is wrong.
[/quote]
Why? Explain, please, this point.
Maybe $ |A\parallel x-x_{0}|^{2}\ge |a\parallel x-x_{0}\parallel x-x_{1}|$ is wrong for $x\rightarrow x_{0}$?
[quote="NguyenDungTN"]
Let $ x=\frac{-2B}{A}$:[/quote]
You are mean $ x=\frac{-B}{2A}$.
[quote="NguyenDungTN"]
 $ |ax^{2}+bx+c|=\frac{|b^{2}-4ac|}{4a}$
[/quote]
Why?

See this link : [url=http://amc.maa.org/a-activities/a7-problems/putnam/-pdf/2003s.pdf]2003 Putnam exam[/url]


geometry
circumcircle
incenter
geometry unsolved

The Problem is obvious,look for my solution  , [url=http://www.mathlinks.ro/viewtopic.php?t=168866&start=80]here[/url]  at the 5th page

Let $ ABCD$ be a bicentric quadrilateral(i.e there exist incircle and circumcircle of $ ABCD$),and $ O$,$ I$,$ H$ be the circumcenter,incenter,diogonals intersection point.We must prove that $ O,I,H$ are collinear.
In my proof,i've denoted $ F\in CB\cap AD,G\in AB\cap DC$.
At frist i'll prove that $ H$ lies on the polar of $ F$ and $ G$ with respect to the incircle.
Let $ M,N$  be the points of tangency incircle with $ AD$ and $ CB$ respectively.Then
From [color=green]Brianchon theorem[/color]([color=red]main diogonals in Circumscribed hexagon are concurrent[/color]),it follows that $ H\in MN$,but $ MN$ is the polar of $ F$ w.r.t incircle.
Similarly $ H$ lies on the polar of $ G$ w.r.t to the incircle.
Now using the well-known fact:
Polar of the intersection point of polars of two points is the line through this points.
You can also read about it in anybook about pole and polars.
The second fact is aslo well-known(about $ H$ lies on the polars of $ F$ and $ G$ w.r.t to the circumcircle).In the proof we use harmonic fours,but it is too long solution.I think(or sure) that we can use this fact without proof,though the proof is easy.
Any questions?

[quote="dimitrov89"]ABCD is quadrilateral. I is center of the inscribed circle, O is center of the circumcircle. If AC intersects BD in M, prove that I,O,M are collinear.[/quote]
The well-known shortest proof of this old problem I have in mind, is based on the polar theory as mentioned by [b][size=100]Erken[/size][/b].

There is also another well-known proof, I think the nicer, by inversion.

For the simplest elementary proof of the collinearity of $ I,$ $ O,$ $ M,$ you can see at http://www.mathlinks.ro/Forum/viewtopic.php?t=31693

[quote="dimitrov89"]Why H lies on the polars of F,G w.r.t (I) and (O),hence FG is a polar of
H w.r.t (I),(O) ?[/quote]
An elementary approach of this result you can see at http://www.mathlinks.ro/Forum/viewtopic.php?t=139564

 Kostas Vittas.

[quote="Erken"][quote="vittasko"]
For the simplest elementary proof of the collinearity of $ I,$ $ O,$ $ M,$ you can see at http://www.mathlinks.ro/Forum/viewtopic.php?t=31693
[/quote]
Elemantary?,what do you mean,i'm sorry,i don't know english well :blush: .
Do you talk about your proof?[/quote]
I also don’t know English well, dear [b][size=100]Erken[/size][/b] (I think that I am the worse of all [b][size=100]mathlinkers[/size][/b], in English).

When I say ''elementary'', I mean that we use basic geometric tools for the solution of a problem, comprehensible to the beginners in geometry, who probably don’t learned the polar theory and other extended chapters of geometry theory, yet.

As for the proof you asked me, really, I don’t know is it is a ''my'' one. There are so many famous geometers who preceding us and it is very probable to has been discovered something (a problem or a solution of well-known problem), before our discovery.

Best regards, Kostas Vittas.


geometry
incenter
trigonometry
geometry proposed

Let $ABC$ be a triangle with incenter $I$ and let the incircle touch sides $BC,CA,AB$
at $D,E,F$. Let lines $AI$ and $EF$ intersect at $P$; let $K$ be the intersection of $BP$ and
$DF$, and $L$ that of $CP$ and $DE$. Show that $KL$ is parallel to $EF$.

$\frac{AK}{KF}=\frac{\sin\angle CBP}{\sin\angle ABP}$. Now it should be easy to conclude.

See http://www.mathlinks.ro/Forum/viewtopic.php?t=15830 .

  darij


geometry MATHCOUNTS trigonometry function

[hide]The height of the equilateral triangle is 5*3=15
(refer to your other post for the reason)
By some calculation the sidelength of the triangle is $10\sqrt{3}$
The area required=$\frac{1}{2}\times 10\sqrt{3} \times 15 -\pi(5)^2\approx51.36$[/hide]

[i]USE HIDE! - A+MATH[/i]

[quote="pianoforte"]A circle with radius 5 is inscribed in a equilateral triangle. What is the area of the space whithin the triangle that is [b]not[/b] taken up by the circle to the nearest hundredth?[/quote]
[hide]
Since in an equilateral, the alititudes, medians, and angle bisectors are the exact same three lines... the height of the triangle would be 15.

Using special right triangle properties, you would find that each side of the triangle is $10\sqrt{3}$

So the area of the triangle is $\frac{10\sqrt{3}*15}{2}=75\sqrt{3}$

The area of the circle is $25\pi$

$75\sqrt{3}-25\pi$ or $25(3\sqrt{3}-\pi)$ is our desired area.[/hide]

This problem looks like a problem from warm up 18 in a mathcounts hand book... I'll be lenient on this one, but let's not post too diffcult problems like this one.

sorry. I did not realize it was in the Mathcounts workbook.

Hello, boxedexe,

Trigonometry isn't allowed in this forum. I understand that its' tempting to use that in problems like this, but it wasn't necessary.

[quote="boxedexe"][hide]I do not know if I am supposed o post in this section, but I'll give is a try.  Well, we said that we have a circle with a radius of 5 units embedded within an equilateral triangle. First of all, we have to draw a line from the centroid of the circle perpendicular to the base of the triangle and label it "opp."  We now have a line (opp) that is 5 units long, and that could be used to figure out the length of the base and the height of the triangle.  Then, we draw a line from the centroid of the circle, centre of the the triangle, to one of the vertices that are on either end of the base, and call it "hpp."  The adjacent side of this small triangle shall be known as "adj."  As we all know, since we have drawn a line from the centre of the circle, also the centre of the triangle, side "hpp" has divided an angle of the equilateral triangle (180/3=60) into 2, thus the angle that subtends side "opp" is 60/2=30 degrees.  As you may be currently observing, we now have constructed a small 30-60-90 triangle.  Since sin30=opp/hpp=5/hpp, hpp=5/sin30=5/0.5=10.  This represents us the rule that in any 30-60-90 triangle, side "opp" is always half side "hpp."  As a result, "hpp" equals 10 units.  This 10 unit long is the same length from the centroid of the circle to any of the corners of the equilateral triangle.  This has made me aware of the axiomatic fact that in any equilateral triangle that has a circle in it,   the height of the triangle is 3*r (radius of the circle); in this case, h=3*5=15.  Now it is time to figure out the length of the base.  Since we had a 30-60-90 triangle with opp=5 and hpp=10 constructed, it is time to figure out b/2, or as we had previously labeled, side "adj."  We know that from tan30=5/adj, adj=5/tan30=8.66 units.  Since side adj=b/2=8.66, b=8.66*2=17.32 units long.

We now know that the base of the equilateral triangle is 17.32 units long, and that the height is 15 units long.  The area of the triangle (A=bh/2) is therefore approximately 129.9 squared units.  The area of the circle (pir^2) is approximately 78.54 squared units.

In conclusion, the area of the places not occupied by the circle is 129.9-78.54= 51.36 squared units.  [/hide]

Sorry for the long explanation, but I can't help solving simple, yet beautiful problems.

Masoud Zargar[/quote]

I edited it so it was hidden, but in the future use hide. When you finish typing your solution select the text and then press the button that says "Hide".

Thank you for adding the hide function ( I did not know I had to do that).  By the way, I used trigonmetry to explain the process in a more clear way, even though it was not necessary.

Masoud Zargar


trigonometry
inequalities
inequalities open

if x,y>0 & x+y=1
proof that: x/sqrt(y) + y/sqrt(x) >= sqrt(2)

[quote="mathmanman"]
Bad title + wrong section.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=144405[/url]
[/quote]
 :D

[quote="hph"]$ \{\begin{array}{c}x>0\ ,\ y>0\\\ x+y=1\end{array}\|$ $ \implies$ $ \frac{x}{\sqrt y}+\frac{y}{\sqrt x}\ge \sqrt 2$.[/quote]
[color=darkblue][b][u]Proof I [/u] (similarly to BaBaK Ghaleby's).[/b] Exists only one $ \phi\in (0,\frac{\pi}{2})$ so that $ \tan\phi =\sqrt{\frac{x}{y}}$, i.e. $ \{\begin{array}{c}\sin^{2}\phi=x\\\ \cos^{2}\phi =y\end{array}$.
The our inequality becomes $ \frac{\sin^{2}\phi}{\cos\phi}+\frac{\cos^{2}\phi}{\sin\phi}\ge\sqrt 2$ $ \Longleftrightarrow$ $ \sin^{3}\phi+\cos^{3}\phi\ge \sqrt 2\sin\phi\cos\phi$ $ \Longleftrightarrow$ $ (\sin\phi+\cos\phi)(1-\sin\phi\cos\phi )\ge\sqrt 2\sin\phi\cos\phi\ \ (1)$.
Denote $ \boxed{\sin\phi+\cos\phi =t}$. But ${ t\in (1, \sqrt 2]}$ because $ (\sin\phi+\cos\phi )^{2}=1+\sin 2\phi \in (1,2]$ for any $ \phi \in (0,\frac{\pi}{2})$. Thus, $ \sin\phi\cos\phi =\frac{t^{2}-1}{2}$.
The relation $ (1)$  becomes : $ t\cdot (1-\frac{t^{2}-1}{2})\ge \sqrt 2\cdot\frac{t^{2}-1}{2}$ $ \Longleftrightarrow$ $ t\cdot (3-t^{2})\ge \sqrt 2\cdot (t^{2}-1)$ $ \Longleftrightarrow$ $ t^{3}+\sqrt 2t^{2}-3t-\sqrt 2\le 0$ $ \Longleftrightarrow$
$ (t-\sqrt 2)(t+\sqrt 2)^{2}\le 0$, what is truly. We have equality if and only if $ t=\sqrt 2$ $ \Longleftrightarrow$ $ \phi =\frac{\pi}{4}$, i.e. $ x=y=\frac{1}{2}$.[/color]

[quote="hph"][quote="BaBaK Ghalebi"]$ x,y>0,x+y=1$

so there exists angle $ z$ s.t $ \sin^{2}z=x,\cos^{2}z=y$

now we have to prove that:

$ \frac{\sin^{2}z}{\cos z}+\frac{\cos^{2}z}{\sin z}\geq \sqrt{2}$

$ \Rightarrow \sin^{3}z+\cos^{3}z\geq \sqrt{2}\sin z\cos z$

$ (\sin^{3}z+\cos^{3}z)(\sin z+\cos z)\geq \sqrt{2}\sin z\cos z(\sin z+\cos z)$

now by couchy we know that:

$ LHS\geq (\sin^{2}z+\cos^{2}z)^{2}=1$

so it remains to show that:

$ RHS\leq 1$

but in the other hand by AM-GM and couchy we have:

$ \sin z\cos z\leq \frac{\sin^{2}z+\cos^{2}z}{2}=\frac{1}{2}$

$ \sin z+\cos z\leq \sqrt{2}$

now from the above two inequalities we get that:

$ RHS\leq \frac{1}{2}.\sqrt{2}.\sqrt{2}=1$

as wanted...[/quote]

you cant do this
$ \Rightarrow \sin^{3}z+\cos^{3}z\geq \sqrt{2}\sin z\cos z$

$ (\sin^{3}z+\cos^{3}z)(\sin z+\cos z)\geq \sqrt{2}\sin z\cos z(\sin z+\cos z)$

maybe $ (\sin z+\cos z)$ not possetive[/quote]
read the problem carefully,I did nothing wrong...  :D

$ \sin^{2}z=x \Rightarrow \sin z=\pm \sqrt{x}$
note that $ z \in (0,\pi)$ so $ \sin z>0$ now there are two angles $ z \in (0,\pi)$ for which $ \sin z=\sqrt{x}$ one of them is $ \alpha$ and the other one is $ \pi-\alpha$ where $ \alpha \in (0,\frac{\pi}{2})$,now let $ z=\alpha$ so $ \sin z,\cos z>0$ so $ \sin z+\cos z>0$...


[b][u]Problem 1[/u][/b]:Find all $n\in N^*$ such that $\frac{2^n+1}{n^2}\in Z$
[b][u]Problem [/u][/b]:Let a paralell $ABCD: AD=DB$.Let a point $M$ such that:$MA\perp AC;MD\perp DC;N=DB\cap MC;I=AC\cap MD$.Find $\angle MNI$
[b][u]Problem 3[/u][/b]:Let 2006 positive real number $a_1,a_2,...,a_{2006}$.Prove that there exists 2006 real number$b_1,b_2,...,b_{2006}$:
(i)$a_i<b_i\forall i=1,2,..,2006$
(ii)$\frac{1}{2}<\frac{b_i}{b_{i+1}}<2\forall i=1,2,..,2005$
(iii)$\sum_{i=1}^{2006}b_i<3\sum_{i=1}^{2006}a_i$
  Prove that 3 is the best number when we replace 2006 by n.
 :)  ;)

[quote="TomciO"]What is the sense of giving so very well-known problem in the TST :O?[/quote]
Because it's very old. :)


When I was doing my math homework, i found an an interesting relation that
$\frac{d^{n}f(x)}{dx^{n}}$ is closely related to $(f(x))^{1-n}(\frac{df(x)}{dx})^{n}$ but I can't tell what exactly is the relation.

I would like to ask if there is any theorems or approximations of $\frac{d^{n}f(x)}{dx^{n}}$?

well, maybe you have your answer here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=130998


http://www.youtube.com/watch?v=cFH6y60zorE

10 out of 9!

What $ 10/9!$? that's way too low. I would give it at least $ (9!\minus{}10)/9!$. That's so  :rotfl: Hilarious!

Is there any operations with $ ?$
How bout 10 out of 9?

$ x? \equal{} \begin{cases} \frac{\sqrt{\frac{\int^{x!}_{x\minus{}2}\sum\limits_{r\equal{}x}^{x\plus{}263}(x!)!^{\minus{}34}}{5x\plus{}x!^{2x\plus{}e}}}}{\frac{dx^{e\plus{}(|x\minus{}1|)?}}{dx}\minus{}\sqrt[152]{34}} \text{ if } x \ne 0  \\
0 \text{ if } x \equal{} 0
\end{cases}$

A recursive formula!

[quote="lingomaniac88"][quote="Temperal"]$ x? \equal{} \begin{cases} \frac {\sqrt {\frac {\int^{x!}_{x \minus{} 2}\sum\limits_{r \equal{} x}^{x \plus{} 263}(x!)!^{ \minus{} 34}}{5x \plus{} x!^{2x \plus{} e}}}}{\frac {dx^{e \plus{} (|x \minus{} 1|)?}}{dx} \minus{} \sqrt [152]{34}} \text{ if } x \ne 0 \\
0 \text{ if } x \equal{} 0 \end{cases}$

A recursive formula![/quote]

*stares blankly with mouth wide open*[/quote]

*stares blankly with mouth side open at the person staring blankly with mouth wide open at a horribly ugly looking recursion*

please keep this on topic.

$ \dfrac{10!}{9}$


Let $ V$ be a $ d$-dimensional inner product space, and let $ a$ and $ b$ be two nonzero vectors.  Let $ S=\{ u \in V: (a,u)=0 \}$.  Show that the distance $ h$ of $ b$ from $ S$ is given by:

$ h= \frac{|(a,b)|}{\sqrt{(a,a)}}$.

What would be the corresponding formula for $ h$ when $ S$ is an arbitrary $ m<d$ dimensional linear manifold and $ b$ is not on $ S$?

This formula would include the original case. Let $ \{v_{1},v_{2},\dots,v_{d-m}\}$ be an orthogonal basis for $ S^{\perp},$ which is the orthogonal complement of $ S.$ (If we have a basis that isn't orthogonal, we'll have to do some more work.) Then:

$ h=\sqrt{\sum_{j=1}^{d-m}\frac{\langle v_{j},b\rangle^{2}}{\langle v_{j},v_{j}\rangle}}$

In the original case stated above, the sum has only one term.

This all belongs to the family of arguments that includes orthogonal projections and the Gram-Schmidt process.


group theory
superior algebra
superior algebra unsolved

Can u give me more description about the AUTOMORPHISM GROUP OF A FINITE CYCLIC GROUP?  I know $ Aut(\mathbb{Z}/ n\mathbb{Z} )\cong \mathbb{Z}/n \mathbb{Z}^\times$.

[quote="kan"]$ Aut(\mathbb{Z}/ n\mathbb{Z} )\cong \mathbb{Z}/n \mathbb{Z}^\times$.[/quote]
That's all.

Yea I know it. But how do we show that? Can u give more explanation?

Thanks

An automorphism of a cyclic group is determined by its value at a generator.


[quote="13375P34K43V312"]A cylindrical column is 8 feet tall and has a diameter of 6 inches. A ribbon is wrapped around the column so that it spirals from bottom on top and passes around the column exactly seven times. How many feet are in the length of the ribbon? Express your answer to the nearest tenth of a foot.[/quote]
I've seen something like this before
[hide="hint"]When you "unravel" the ribbon you get a hypotenuse of a right triangle[/hide]

[hide="Solution"]
So you know that the ribbon will go horizontally around the cylinder seven times the circumference of the cylinder, and that it will go up the height of the cylinder. The circumference is $6\pi$, and it goes around seven times for a total of $42\pi$ inches. It goes up 8 feet which is 96 inches. You now have a right triangle with legs $42\pi$ and 96, and you need to find the hypotenuse (the length of the ribbon). So, you have the length of the ribbon = $\sqrt{(42\pi)^{2}+96^{2}}\approx 98.14$. Dividing by 12, we get $\fbox{8.18 ft}$[/hide]


The Mole 3 have ended, so I'm gonna start signups for mole 4 now.

Applications for the fourth installment of The Mole on AoPS begin now! If you are applying, please do not make it public. Applications must be sent to me (pacman2812) via PM. Below, you can see the past seasons of The Mole on AoPS.

Season 1: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186949[/url] , [color=red]Mole (hm29168) wins[/color]
Season 2: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=202287[/url] , Abandoned
Season 3: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=221586[/url] , [color=green]cf249 wins[/color]

To apply, simply copy/paste the application below into a PM to me, [b]with subject heading “Mole 4 Application” (or something similar)[/b] and fill in the answers (so that the questions and answers are in the PM).

-Username:
-Gender (so I can correctly refer to you as ‘he’ or ‘she’):
-Date of Birth (MM/DD/YY):
-If it is 7 PM in New York City (EST), what time is it where you live?
-Did you participate in either of the first three mole games?
-How often did you go inactive in the previous games, and for how long?
-How likely is it that you will need to be replaced?
-How often are you willing to check the game thread?
-Do you want to be The Mole? Why?
-Which is your favorite INTEGER number? Why? (edited because I don't want to receive "pi" in an application :mad:).
-Which skin do you use (AoPS? Mathinks?)? Why?

Hmmm, I didn't win season 1, hm29168 did. He was The Mole.

[quote="zapi2007"]does skin=avatar=picture under username?[/quote]Do you use artofproblemsolving.com, or do you use mathlinks.ro? That's the skin thing.

[quote="ProtestanT"]"-Did you participate in either of the first two mole games? "

Do you mean the first [b]three[/b] mole games?[/quote]

Yeah lol. I copy-pasted it from Mole 3, and I forgot it.

Edited.

(edit: btw, I've already received as many as 15 apps in less of one day)

How many people will you be accepting to play in the game?

[quote="pacman2812"]You will not know until when the game starts... *drumroll*[/quote]

When does the game start.

[quote="pacman2812 in TM3 thread"]in early february[/quote]

OK, applications are officially closed. Thank you for applying!

How early in Febuary? I wan the game to start. The Mole is so awesome.

I believe it's soon. and lol hongsquare.

[quote="pacman2812"]Applications for the fourth installment of The Mole on AoPS begin now! [size=200][b]If you are applying, please do not make it public.[/b] [/size]
[/quote]

Bleh, so much for that rule.

I was told to "confirm in-thread" that I have been accepted.

I assume this is the thread...

/confirm

Confirm!

Are we playing in this thread?

[quote="alekim"]confirmed, kicking mz94 and BOGTRO in nuts this saturday.[/quote]

 :dry:

any reason why?

[color=blue]*mz94 will use super ninja skillz to hurt alekim after chapter mc[/color]

NOOO I CALL FIRST REPLACEMENT IF NESSECARY.  :(  :(  why didn't you choose me??

/confirm.

Oh wait, that's only 9 characters...

/confirm

The message is too small. Please make the message longer before submitting.

Confirm.
[hide=" "]
The message is too small. Please make the message longer before submitting.[/hide]

[quote="Brut3Forc3"]No new thread? Ok...[/quote]
I'm guessing that pacman will make a new thread when the game starts, and he didn't want that thread cluttered with confirmations.

grr... i got a ****** power shutdown, and I lost ALL my done job...

thread set up.

@any games mod: lock this.


Find the smallest n such that if {1, 2, ... , n} is divided into two disjoint subsets then we can always find three distinct numbers a, b, c in the same subset with ab = c.

now we shall prove that this is not always true for $ n\equal{}1,2,\ldots,47$. does anyone have any method other than checking one by one? :P


... I hope you told
AIME15
Wickedestjr
Mathelete
Ernie
cf249
mewto5555[b]5[/b]
isabella2296
and chen[b]h[/b]si

about this before randomly signing them up...

You just wasted a post. But let's count this as the first one.

minus 1 from isabella2296

AIME15: 0
Wickedestjr: 0
Mathelete: 0
Ernie: 0
cf249: 0
mewto55555: 0
isabella2296: -1
chenhsi: 0

PS. how do you send a PM to multiple people at once?

subtract 2 from Wickedestjr

AIME15: 0
Wickedestjr: -2
Mathelete: 0
Ernie: 0
cf249: 0
mewto55555: 0
isabella2296: -1
chenhsi: 0

To answer the P.S.:  You can't.  Several people in the past have requested this to be part of the forum and each time, their request has been denied.

Forget it, if you want to sign up, it is in another topic. This topic is messed up. I am waiting for 7 sign ups. You are the first. It was my mistake to randomly choose people.


function
real analysis
real analysis unsolved

$f: ]0; +\infty[ \to \mathbb{R}^+$ a function such that $lim_{x \to 0}{f(x)}= +\infty$.
Does there exists a function $g: \mathbb{R} \to \mathbb{R}$ satisfying:
$|g(x)-g(y)| > |x-y|f(|x-y|)$ for all $x \neq y$

No, there are no such functions $g$. This is because there is some $k>0$ s.t. $|g(x)-g(y)|>1,\ \forall x\ne y\in I=[0,k]$, so $g$ is injective on $I$, but $g(I)$ has countably many points.

Am I missing something obvious? :?

of course I made a mistake. I've corrected  :)

I remembered this problem while doing some random browsing. It still looks fishy to me. Are we supposed to find out if given any $f$ satisfying the required properties, there is such a $g$? Because this is almost obviously false. One could take $f(x)=\frac 1x$, and we're in the situation from post #2 :?.

I remember that I discussed with a friend the case: $|f(x)-f(y)| > \sqrt{|x-y|}$
Even a simple case like this is tricky i think ...

For this particular case and related ones see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=dilatation&t=45121]this[/url]. Even if that's not correct, I think it's the right idea :).

I've just read your proof : great idea  :)


[quote="4everwise"]Generally, if a solution has an answer, the method of finding the answer, but doesn't prove that it is the only answer, how many points are awarded? Three?

Cause I have an answer to one of this round's problems that fits this category...  :|[/quote]
Depends on the problem.

For some problems, the only way to find the answer is to construct the proof. Leaving off the proof counts simply as a bad write-up of the problem for four points. I see that Round 3 has at least one problem of this type.

For other problems, finding the answer is easy and the work is in showing that the answer is valid in all cases. A correct answer without a proof would be worth one or two points. I see that Round 3 has at least one problem of this type.

And some problems are in between. I see that Round 3 has at least one problem of this type.

And to make this response even more useless, we do not decide on the grading scale until after all the solutions are in. My statements, "I see that Round 3 has at least one problem of this type," are pure guesswork.

Erin Schram
USAMTS grader


How much would it cost to cover the entire United States (including Alaska and Hawaii) with dollar bills?

What would be the cost to each person in the United States?

Area of the United States=about $ 3.79*10^6$ square miles
Area of one dollar bill=$ 16.0254$ square inches.

One square mile=$ 4.014*10^9$ square inches, so the total area of the U.S. is $ 1.52*10^{16}$ square inches.  Dividing this number by 16 gives us approximately $ 9.5*10^{14}$, or $ 950,000,000,000,000$.  That's how many dollar bills it would take to cover the entire U.S.

There are about $ 300,000,000$ $ (3*10^8)$ people in the U.S. so it would cost each person about $ \$3,170,000$ $ (3.17*10^6)$ to supply enough money to cover the U.S. with dollar bills.[/url]

Note: I took the above population and area figures from Wikipedia.


Find the number of solutions in positive integers $ (k; a_1,a_2,\ldots,a_k; b_1,b_2,\ldots,b_k)$ to the equation
\[ a_1(b_1) \plus{} a_2(b_1 \plus{} b_2) \plus{} \cdots \plus{} a_k(b_1 \plus{} b_2 \plus{} \ldots \plus{} b_k) \equal{} 7
\]

[quote="implex"]$ a_2(b_1 \plus{} b_2) \equal{} 2,4,6$[/quote]Why? I can have $ a_1\equal{}1,b_1\equal{}2$ so that $ a_2(2\plus{}b_2)\equal{}5$ which yields $ a_2\equal{}1,b_2\equal{}3$ as a possible solution not found in your answer.

[quote="10000th User"][quote="implex"]$ a_2(b_1 \plus{} b_2) \equal{} 2,4,6$[/quote]Why? I can have $ a_1 \equal{} 1,b_1 \equal{} 2$ so that $ a_2(2 \plus{} b_2) \equal{} 5$ which yields $ a_2 \equal{} 1,b_2 \equal{} 3$ as a possible solution not found in your answer.[/quote]
yupp , right I missed a few cases

Very different approach, very different answer.

[hide="Solution"]Let $ c_n \equal{} b_1 \plus{} b_2 \plus{} \ldots \plus{} c_n$

The problem is equivalent to asking for the number of solutions to $ a_1 c_1 \plus{} a_2 c_2 \plus{} \ldots a_k c_k \equal{} 7$, where $ a$ is a sequence of positive numbers and $ c$ is a [i]strictly increasing[/i] sequence of positive numbers.

So now, we just want to find the number of partitions of $ 7$, since each partition corresponds to a solution to the above. For example, $ 7 \equal{} 3 \plus{} 2 \plus{} 1 \plus{} 1$ corresponds to $ (c_3, c_2, c_1) \equal{} (3,2,1); \;(a_3, a_2, a_1) \equal{} (1,1,2); \; k \equal{} 3$, where the values of $ c$ are the [i]different[/i] numbers we used, and the values of $ a$ are [i]how many times[/i] we used them.

We then list out the partitions of $ 7$ and find that there are $ 15$ of them:
[hide="The partitions of 7"]7
6+1
5+2
5+1+1
4+3
4+2+1
4+1+1+1
3+3+1
3+2+2
3+2+1+1
3+1+1+1+1
2+2+2+1
2+2+1+1+1
2+1+1+1+1+1
1+1+1+1+1+1+1[/hide][/hide]

Dude I get 15 too

[hide="Answer"]If we assume that those three solutions in positive integers $ (k; a_1,a_2,\ldots,a_k; b_1,b_2,\ldots,b_k)$ are solutions, then that means that $ b_1, b_1 \plus{} b_2,b_1 \plus{} b_2 \plus{} b_3,\dots , b_1 \plus{} b_2 \plus{} b_3 \plus{} \dots \plus{} b_k$ is a strictly increasing sequence.

$ a_n$ ($ n\in R$) so $ a_n$ and $ b_n$ determine a unique partition of 7. It is now quite evident that we just need to find how many partitions of 7 there are, because you can get an $ a_n$ and $ b_n$ from any partition of 7. There are $ \boxed{15}$ partitions of 7.
[/hide]

[b]@ implex:[/b] A flaw in your counting is that you don't check to see if your $ b$ values work with the previous $ a$ values.

For example, you might get that if $ a_2 (b_1 \plus{} b_2) \equal{} 6$, then $ (a_2, b_1, b_2) \equal{} (1,3,3)$ works. Notice, however, that that this forces $ a_1 b_1 \equal{} 1 \implies a_1 \equal{} 1/3 \not \in \mathbb{N}$

While a case-wise constructive counting approach may be very appropriate, I don't think there's an effective quick-fix to your particular method. It's not that it's impossible, but the checking process would probably make it too long for an effective Guts round solution  :wink:

[b]@ #H34N1:[/b] Nice to see you agree with my approach  :roll:

yupp I was confused, I got 17 first then !! anyways even the first attempt was wrong

I see.. I couldn't see the bijection to partitions of 7 at first after reading the HMMT solutions. So "Guts" is all about finding quick solutions? Haha..

In the Guts round, all the teams sit in a large auditorium, and run up to the front to get questions in packets of three. They solve them together, run them back, and get the next packet of 3 questions.

This particular problem was #5, which means that it was in the second triplet of questions. While it is important to make sure to get the first questions right (they get harder as you go on), it is also a good idea to save time for the harder questions.

It's lots of fun =)


Find all functions $ f$ defined on non-negative real numbers having the following properties:
(i) For all non-negative $ x$ it is $ f(x) \geq 0$.
(ii) It is $ f\left(1\right)\equal{}\frac 12$.
(iii) For all non-negative numbers $ x,y$ it is $ f\left( y  \cdot f(x) \right) \cdot f(x) \equal{} f(x\plus{}y)$.

Given f(yf(x)).f(x)=f(x+y) -- (1)
keep y=0 => f(0).f(x)=f(x) => (f(0)-1)f(x)=0 => for x=1 f(x) is not zero. So, f(0)=1
keep x=1 in original eq, => f(y/2)*1/2 = f(y+1) => f(2y+1)=f(y)/2 -- (2)
=> f:[0,inf) -> [1,0) is a monotonically decreasing function and so, it is a bijection
Keep y=1/f(x) in equation (1)
f(1).f(x)=f(x+1/f(x))
=> f(x)/2 = f(x+1/f(x))
from equation (2),
f(2x+1)=f(x+1/f(x))
As f is a bijection,
2x+1=x+1/f(x)
=> f(x)=1/(1+x)


Given a the sequence defined by $a_{1}=2$ and, $a_{n+1}=\left(\prod_{k=1}^{n}a_{k}\right)+1$ prove or disprove that every term in the sequence is prime.

Note that every two terms in this sequence are relatively prime.  As a consequence, we get a nice proof that there are an infinite number of primes.  (Just choose one prime factor from each term.)


Solve for x :stretcher:
$ x \plus{} \sqrt {17 \minus{} x^2 }  \plus{} x\sqrt {17 \minus{} x^2 }  \equal{} 9$ :)

Let $ t\equal{}x\plus{}\sqrt{17\minus{}x^2}\Longrightarrow x\sqrt{17\minus{}x^2}\equal{}\frac{t^2\minus{}17}{2}$ with $ \minus{}\sqrt{17}\leq t\leq \sqrt{34}\ \cdots [*]$.
Thus we can rewrite the given equation as $ t^2\plus{}2t\minus{}35\equal{}0\Longleftrightarrow t\equal{}\minus{}7,\ 5$. By $ [*]$, $ t\equal{}5\Longleftrightarrow \sqrt{17\minus{}x^2}\equal{}5\minus{}x$
$ \therefore x\equal{}1,\ 4$.

Maybe an explanation would be helpful...
Let $ A = x^2$

\begin{eqnarray*}
\sqrt{A} + \sqrt{17-A} + \sqrt{A} \cdot \sqrt{17-A} &=& 9\\
(\sqrt{A} + 1)(\sqrt{17-A} +1) &=& 10
\end{eqnarray*}

At this point, it would be easy to solve if you know that x is an integer.  Just plug in pairs of factors of $ 10$ and you'd get $ \boxed{1,4}$

[hide]it would look easier if you put $ a \equal{}x ; b \equal{} \sqrt{17\minus{}x^2}$
you should see a system of equations .
Put S=x+y , p =xy and you get the answer[/hide]

...But you are only given one equation in the problem?
Am I missing something?

when you put $ a = x$ and $ b = \sqrt{17-x^2}$ you' have $ a^2 + b^2 = 17$ and $ a+b+ab=9$
There are many ways to solve this kind of system . The official way is to put S = a+b , P = ab .
you' have $ S^2 - 2P =17$(1) and $ S+ P = 9$
replace P as 9-S into (1) and it is now easy to find S

As I see , your solution , as quoted below , is incorrect .
[quote="falconwing64"]Maybe an explanation would be helpful...
Let $ A = x^2$

\begin{eqnarray*} \sqrt {A} + \sqrt {17 - A} + \sqrt {A} \cdot \sqrt {17 - A} & = & 9 \\
(\sqrt {A} + 1)(\sqrt {17 - A} + 1) & = & 10 \end{eqnarray*}

At this point, it would be easy to solve if you know that x is an integer.  Just plug in pairs of factors of $ 10$ and you'd get $ \boxed{1,4}$[/quote]

hmm...oh i see it now,
yeah, your solution would be better for finding exact solutions.  i just assumed integers


[b]NIVEL BASICO[/b]

[b]Problema 1[/b] [Teoria de Numeros]
Sea $A=\{a^2+2b^2 / a, b \in \mathbb{N} \}$  . Si $p$,es un
n\'umero primo tal que $p^2$ pertenece a $A$ , demuestre que $p$ pertenece a $A$.
[Sugerido por:[i]Jorge Tipe[/i]]

[b]Problema 2[/b] [Combinatorio]
De $3n+1$ objetos $n$ son indistinguibles ("iguales"), y los
restantes son distintos. Mostrar que uno puede elegir $n$ objetos
apartir de estos, en $2^{2n}$ maneras.
[Sugerido por:[i]Los Organizadores[/i]]

[b]Problema 3[/b] [Algebra/Secuencias]
Sean $x_1, x_2$ las raices de la ecuaci\'on: $x^2 +px -1 = 0$, $p$ es impar,
y sea: $y_n = x_1^n + x_2^n, n\geq 0$. Mostrar que: $y_n$ y $y_{n+1}$
son enteros coprimos.
[Sugerido por: [i]Carlos Bravo[/i]]

Disfruten de los problemas  ;)
Los Organizadores

aki la version en pdf   y latex..
para que los descarguen ...

Saludos,

Cralos bravo ;)
Lima -PERU

En la definición de $A$ falta colocar un " $/$ "debería ser: $A=\left\{a^2+2b^2 / a, b \in \mathbb{N} \right\}$

$Tipe$

[b]NIVEL AVANZADO[/b]

[b]Problema 1.[/b]
$A_1,A_2,A_3...A_n$ y $B_1,B_2,B_3...B_n$ son conjuntos cada uno con $k$ elementos,y ademas $n<2^{k-1}$. Pruebe que existen elementos $a_1,a_2...a_n$ y elementos $b_1,b_2,...b_n$ tales que $a_i\in A_i$, $b_i\in B_i$ y $a_i\neq b_j$.

[b]Problema 2. [/b]
Hallar todas las funciones de los reales en los reales (sin incluir al cero) tal que $f(x)+f(y)=f(xy)\left(f(\frac{1}{x})+f(\frac{1}{y})\right)$ para toda pareja $x$ y $y$ reales distintos de $0$.

[b]Problema 3.[/b]
Pruebe que para todo entero $n$, existe un natural $a$, tal que $a^2+a+1$ es divisible por exsactamente $n$ primos.

Los Organizadores ;)

Hola,...

fabycv ....me comenta q ya le salieron los prob 02 y 03 del nivel basico ...
me los estara enviando estos .dias ..

haber si ... otros se animan a enviar sus soluciones ...
para compararlas ...

Saludos,

Carlos Bravo ;)
Lima - Peru

Para el problema 2 de avanzado: ¿ $f: (\mathbb{R}- \left\{0\right\}) \longrightarrow \mathbb{R}$?

$Tipe$

Aja .........
parece ser que es asi .... Tipe. ..

Gracias .

Carlos Bravo ;)
Lima -Peru

enviar a:
pascual2k1@hotmail.com (pascual) , carlosb.bravo@gmail.com (carlos)

Mayor informacion sobre este evento:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=72933

Carlos Bravo ;)
Lima -Peru

[quote="carlosbr"][b]NIVEL BASICO[/b]

[b]Problema 1[/b] [Teoria de Numeros]
Sea $A=\{a^2+2b^2 / a, b \in \mathbb{N} \}$  . Si $p$,es un
número primo tal que $p^2$ pertenece a $A$ , demuestre que $p$ pertenece a $A$.
[Sugerido por:[i]Jorge Tipe[/i]]

[/quote]

[b]Solución[/b]

Sea $p^2=a^2+2b^2$ , vemos que $b<p$, luego $b$ y $p$ son coprimos $\ldots (*)$
De la ecuación obtenemos que $(p+a)(p-a)=2b^2$ , como $(p+a)$ y $(p-a)$ tienen la misma paridad y su producto es par, entonces necesariamente ambos son pares, con esto tenemos que  $b$ es par.
Sea $b=2^{\beta}p_1^{\alpha _1}...p_i^{\alpha _i}...p_n^{\alpha _n}$ la descomposición en factores primos de b $(\beta \geq 1)$. Entonces \[ (p+a)(p-a)=2^{2\beta +1}p_1^{2 \alpha _1}...p_i^{2 \alpha _i}...p_n^{2 \alpha _n}  \]
Supongamos que para cierto $i$ $(1 \leq i \leq n)$ tenemos que $p_i \mid (p+a)$ y $p_i \mid (p-a)$, luego tendríamos que $p_i \mid 2p$ y, como $p_i \neq 2$, entonces $p_i \mid p$ , luego $p_i$ sería un factor común de $b$ y $p$, esto contradice $(*)$
Con esto hemos probado que $(p+a)$ y $(p-a)$ tienen distintos factores primos impares, luego podemos reordenar y  reindexar los primos $p_i$ de tal forma que se tenga que \[ (p+a)=2^c p_1^{2 \alpha _1}...p_j^{2 \alpha _j} \quad { \text y } \quad (p-a)=2^d p_{j+1}^{2\alpha _{j+1}}...p_n^{2 \alpha _n}  \] donde $c+d=2\beta +1$ , sean $m=p_1^{\alpha _1}...p_j^{ \alpha _j}$ y $n=p_{j+1}^{\alpha _{j+1}}...p_n^{ \alpha _n}$ , tenemos entonces que $(p+a)=2^cm^2$ y $(p-a)=2^dn^2$ $\ldots (**)$.
Como $(p+a)$ y $(p-a)$ son pares vemos que $c$ y $d$ son positivos, por otro lado, como $c+d$ es impar tenemos dos casos:

1) $c$ es par y $d$ impar .
  Es claro ver que $c \geq 2$ y $d \geq 1$, sumando las ecuaciones en $(**)$ y dividiendo entre 2 tenemos que $p=2\left(2^{\frac{c-2}{2}} m\right) ^2+ \left( 2^{\frac{d-1}{2}} n \right)^2$ , de donde $p \in A$.

2) $c$ es impar y $d$ par.
  Analogamente al caso anterior, como $d \geq 2$ y $c \geq 1$, tenemos que: $p=\left(2^{\frac{c-1}{2}} m\right) ^2+ 2\left( 2^{\frac{d-2}{2}} n \right)^2$, de donde $p \in A$

En ambos casos: $p \in A$  $\blacksquare$.

$Tipe$


calculus
integration
linear algebra
matrix
linear algebra unsolved

Two positive definite matrices $A,B$ with integer coefficients are equivalent iff there exists a matrix $U \in GL_n{\mathbb{Z}}$ such that : $A=U^t B U$.
Prove that there is only a finite number of classes of equivalence of positive definite matrices with integer coefficients with determinant $D$.

Isn't this just equivalent to some highly-general version of the Class Number Theorem?


[quote="jltd"]variable draught of floating object in water
=================================================================
situaton / given :
water, no waves/ no currents, a floating object (smal drum) length
1 meter, diameter 0.5 meter, weight 50 kilograms.
On specific moment water begins to move/ making waves, the situation
at the end is a harmonic wave with amplitudo of 1 meter and frequency
0.25 seconds (1 complete sin. curve lasts 4 seconds)
How do I calculate the variable draught of the "drum" in the water
density water = 1000 kilograms / cubic meter
accelration gravitation = 9.81 meters/sec*sec
damping factor = 100 N*sec / meter
=================================================================
the harmonic wave is not the problem.
but the beginning of the wave how to put this in a mathematical formula.
because at t=0 it is as if there are no currents
the buoyant force is variable because can be positive as negative
this is caused by damping factor
=================================================================
Height (s) on time t becomes  y=sin(pi/2*t)
Velocity (v )(ds/dt)) on time t becomes  -2cos(pi/2*t)/pi
Damping force = - (200*cos(pi*t/2))/pi
depth of barrel in water with no currents:
= (50*9.81) / (pi*0.5*0.5*1000*9.81)
=0.2546479 meters so far so good.
but with currents I found for downwards force:
downwards force = ((50*9.81)+(- (200*cos(pi*t/2))/pi))
and so  I found for depth:
(downwards force * 4)/ (pi*0.5*0.5*1000*9.81)
but if I use t=0  it should be 0.2546479 meters
but it isn't??
who can help me on this one :?:
Thanks John[/quote]
Mods,

Please move this to the Advanced Physics forum.

The general approach you should use for this problem is:

1. Starting with the definition of draught (or draft if you prefer): The distance between the waterline and the keel of a ship, give it a name: $ y$.

2. Model the equation of motion as a differential equation.
\[ m\ddot{y} \plus{} b\dot{y} \minus{} (F_{\text{buoyant}} \minus{} \text{weight}) \equal{} F_{\text{driver}}(y)
\]

\[ F_{\text{driver}}(y) \equal{} \minus{} ky\qquad \text{Hooke's law}
\]
\[ k \equal{}m\omega^2\]
\[ \text{weight} \equal{} mg\]

\[ F_{\text{buoyant}} \equal{} \rho_{\ce{H2O}}\,g\,(\pi r^2y)
\]

3. Solve the differential equation for $ y(t)$.


geometry
rectangle
inequalities
invariant
combinatorics unsolved
combinatorics

In a $8*8$ board we color each unit square of black and white. A $step$ is choose a rectangle of $3*4$ and change the colors of each square. A coloration of the board is $good$  if after a sequence of $steps$ we can finish  in a board with all its unit squares: white, if this is not posible the coloration is $bad$. Prove that the number of $bad$ $colorations$ of the board is at least  15 times the number of $good$ $colorations$ of the board.

It’s kind of late, but I think that I have a working solution.

Let $A$ be the set of all “good” colorings and let $B$ be the set of all “bad” colorings.  We need to show that $15\cdot |A|\le |B|$.  To prove this statement, we will show that for every “good” coloring, we can derive $15$ “bad” colorings that cannot be derived from any other “good” colorings.  This will establish the above inequality, and the result will follow.

So let $\mathbb{A}\in A$ be a “good” coloring.  Derive the new $15$ colorings from $\mathbb{A}$ by switching the color of at least of one the squares in the $4$ corners of the $8\times 8$ board, and let these colorings be called $\mathbb{A}$’s ${\emph sons}$.  There are obviously $2^{4}-1=15$ ${\emph sons}$, and if we can show these are all “bad,” then we cannot have two distinct “good” colorings that share a ${\emph son}$. (This requires a little thought, but it is not too hard to see.)

To show these colorings are in fact "bad," label the rows of the original chessboard from top to bottom $1$ through $8$, and the columns from left to right $1$ through $8$.  Also, let the value $(i,j)$ be the cell in the $i$th row and the $j$th column.  Then let the subset of the $8\times 8$ board $S$ be the set

$\{(i,j) | i,j\in \{1,3,5,7\}\}$.

Now here is the trick: notice that the number of white squares in $S$ remains invariant modulo $2$ no matter what “step” is taken.  Hence, if we have a coloring where the number of white (or black) squares in $S$ is odd, then that coloring must be “bad,” because the completely white board has an even number of white squares (namely $16$) in $S$.  This is also true for $S+(0,1)$, $S+(1,0)$, and $S+(1,1)$.

So if we change the color of at least one of the corners of $\mathbb{A}$, by the same sequence of moves that makes $\mathbb{A}$ all white, we will achieve an all-white board except for the corners whose color was switched.  But then one of the subsets $S$, $S+(1,0)$, $S+(0,1)$, or $S+(1,1)$ will contain an odd number of black squares (or in other words, the switched corners), and since the number of black squares in these subsets will remain odd whatever the steps, these colorings must be bad.  Thus all the ${\emph sons}$ of $\mathbb{A}$ are bad, so $15\cdot |A|\le |B|$ as desired.

[quote="Severius"]Another solution:

Note that there are $60$ different ways of placing a 3x4 rectangle on the board. Suppose that a sequence of steps leads from some good coloration to the all-white board. The order of the steps clearly doesn't matter, it only matters which "moves" we make (there are $60$ possible moves). Besides, making the same move more than once is unnecessary (because making a move twice is the same as not making it). So there are $60$ possible moves which we can use or not; this gives, at most, $2^{60}$ possibilities. Since any good board can be brought to the all-white board by some sequence of steps, any good board can be obtained from the all-white board by means of the same sequence; since there are $2^{60}$ distinct sequences in total, there are no more than $2^{60}$ good boards.

Since there are $2^{64}$ possible boards in total and every board is either good or bad, the result follows.[/quote]

Oops.. how did we miss this solution?


Look, that all non-positive x satisfy this inequality, so let's assume, that x >0.
Then we have :

[b]ax-2 >= bx[/b]
      or
[b]ax-2 <= -bx[/b]

The answer is:

x >= 2/(a-b)  or   x <= 2/(a+b) or x <= 0


calculus
geometry
3D geometry
sphere
function
real analysis
real analysis unsolved

The unit sphere $ U$={$ x: |x|\equal{}1$} in $ \mathbb{R}^n$ is connected.
Let $ h: U \to \mathbb{R}$ be a continuous function.
Show that $ h(a)\equal{}h(\minus{}a)$ for some $ a \in U.$

Define the continuous function $ g$ on $ U$ with $ g(a) \equal{} h(a)\minus{}h(\minus{}a)$ for all $ a \in U$.
Then the question becomes, does $ g$ have a zero.
Suppose the answer is no.
Then $ g(a)>0$ (or $ g(a)<0$) for all $ a\in U$.
But if $ g(a)>0$, then $ g(\minus{}a)\equal{}\minus{}g(a)<0$.
A contradiction, which proves your statement.


Έστω $ A$ η παράσταση τότε από AM-GM έχουμε:

$ A \le \sqrt[3]{\frac{a^3b^3c}{8\sqrt{a^3 8c b^3 (17)^3a^3 c (59)^3b^3}}} \equal{}$

$ \sqrt[6]{\frac{a^6b^6c^2}{a^6b^6c^2(2\cdot17\cdot59)^3}} \equal{} \frac{1}{\sqrt{2006}}$

Ευκολούτσικη.  :P

[quote="maniopas"]όταν $ a^3 \equal{} 8c, b \equal{} 17a, c \equal{} (59b)^3$[/quote]

Πραγματι σε αυτά που κάνεις η ισότητα ισχύει τότε αλλά αυτή η ισότητα δεν μπορεί να ισχύει  :P

Σου εξηγώ γιατί : $ c\equal{}(59b)^3\equal{}(59\cdot 17a)^3\equal{}59^3\cdot 17^3\cdot a^3\equal{}59^3\cdot 17^3\cdot 8c$ άρα $ c\equal{}0$

άρα $ a\equal{}b\equal{}c\equal{}0$ άτοπο άρα είσαι λάθος και δεν είναι και πολύ ευκολάκι τελικά  :P ...


Ross Mathematics Program
summer program
PROMYS
search
function
Mathcamp
number theory

I am aware of the fact that these two programs are extremely similar and that PROMYS is apparently run by former Ross people. For this reason, I am having trouble deciding to which program I should apply. They both involve a lot of number theory, and the curriculum looks very similar if not identical. PROMYS is 6 weeks long while Ross is 8 weeks, and this seems to be the biggest difference. The other big difference is obviously location, Ross being at Ohio State and PROMYS at Boston University. They also both offer additional advanced courses along with the number theory curriculum, so this is yet another similarity. Could those that know more about these programs tell me about other differences than could help me decide which one is the better for me?
Thanks

One other difference relevant to kids younger than fourteen is that they can only go to Ross. PROMYS can't let in kids younger than fourteen, I'm told, apparently because of B.U. dorm rules.

Im 16 so thats not really a problem for me, but thanks for the response tokenadult. Surely there must be other differences between the programs! If anyone who has attended either one could tell about what they think the differences are or which one is better suited for what needs that would be great.

I suggest you use the search function for this forum to look up some old threads -- there are probably a number of discussions on topics related to this one.  (I did not attend either program and I know people who have enjoyed both, so that is the best advice I can give.  )

also consider USA-Canada Mathcamp http://www.mathcamp.org. It's different from either Ross or PROMYS because there is a lot of choice and variety. There is a whole array of offerings, not just number theory, and participants choose whatever they are interested in.

I assure you texas137 that i am considering mathcamp as well. The purpose of this thread is just to distinguish Ross and PROMYS from each other. The programs are very similar and I can not decide which one would better suit me.

Well I attended the Ross program for the last two summers, and will most likely be returning as a Counselor this summer.  I highly recommend it, of course both programs are great