inequalities
logarithms
real analysis
real analysis unsolved

[quote="jackie-chan01"]For $ x,y > 0$
prove $ x^y \plus{} y^x > 1$[/quote]

it suffices to prove the case when $ 0 < x,y < 1$.

this shows up in Arthur Engel's [i]Problem Solving Strategies[/i].

[color=red]his solution[/color]
[hide]Let $ x \equal{} \frac {1}{1 \plus{} p},y \equal{} \frac {1}{1 \plus{} q}.$

by generalization of Bernoulli's,

$ x^{y} \equal{} \frac {1}{(1 \plus{} p)^y} > \frac {1}{1 \plus{} py} \equal{} \frac {1 \plus{} q}{1 \plus{} p \plus{} q}$
$ y^{x} > \frac {1 \plus{} q}{1 \plus{} p \plus{} q}$

adding this yields $ LHS > \frac {2 \plus{} p \plus{} q}{1 \plus{} p \plus{} q} > 1$ :D[/hide]

Let $ y > 0$ be fixed and consider $ f(t) \equal{} t^y \plus{} e^{t \log y} \minus{} 1$ on $ [0, x]$.
Then by the MVT theorem there exists $ \theta \in (0, x)$ such that

$ f(x) \minus{} f(0) \equal{} x \cdot f'(\theta) \implies f(x) \equal{} x \cdot \left(y \cdot \theta^{y \minus{}1} \plus{} y^{\theta} \cdot \log y \right) > 0$.


1. Which of the following is the same as $ \frac{2 \minus{} 4 \plus{} 6 \minus{} 8 \plus{} 10 \minus{} 12 \plus{} 14}{3 \minus{} 6 \plus{} 9 \minus{} 12 \plus{} 15 \minus{} 18 \plus{} 21}$?

$ (A) \minus{}1 \qquad (B) \minus{}\frac{2}{3} \qquad (C) \frac{2}{3} \qquad (D) 1 \qquad (E) \frac{14}{3}$


I'm pretty sure I have right answer to this so I'll post it at the bottom of this post.

40. If $a(n,r)$ and $s(n,r)$ stand, respectively for $\frac{1 - (1+r)^{-n}}{r}$ and $\frac{(1+r)^n - 1}{r}$, then $\frac{1}{a(n,r)} - \frac{1}{s(n,r)} = $?

Enjoy!

[hide="I promised so"]

$1 - (1 + r)^{-n} = 1 - \frac{1}{(1+r)^n} = \frac{(1+r)^n - 1}{(1+r)^n}$ Then:

$\frac{\frac{(1+r)^n - 1}{(1+r)^n}}{r} = \frac{(1+r)^n - 1}{r(1+r)^n}$ and the reciprocal of this is $\frac{r(1+r)^n}{(1+r)^n - 1}$ and the reciprocal of other one is $\frac{r}{(1+r)^n - 1}$.

So, the problem is:

$\frac{r(1+r)^n}{(1+r)^n - 1} - \frac{r}{(1+r)^n - 1}\\
=\frac{r[(1+r)^n - 1]}{(1+r)^n - 1} \Rightarrow r$[/hide]

My solution is at the bottom...

[hide]
I used the same logic and ended with the answer r if and only if r is not equal to 0.  If r=0, then it's undefined.[/hide]

[i]Edited by a moderator to add 'hide' tags and delete superfluous space.[/i]

Could we just use the [ Hide ] and [ /Hide ] tags to conceal our solution?

Oh....that's how you do it...meant to ask..thx though

[quote]I'm pretty sure I have right answer to this so I'll post it at the bottom of this post.
[/quote]
your answer looks right to me, cause i got the same think, (note that r cant be zero)


Prove that 1/1+1/2+...+1/n can never be an integer for n>=2.

(Hint: Bertrand's Postulate/Theorem says that for n>1, a prime must exist between n and 2n.)

You don't need Bertrand's Postulate, and this has been posted many (!) times recently, such as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=149081]here[/url].


How many distinct strings of four letters can be formed from letters of "probability"?

No, that's not correct, but I get an answer different from the key anyways.

Here's what I did:
[hide]First count the number of strings with two B's and two I's: $\frac{4!}{2!2!}=6$
Then I counted the number of strings with two B's, one I, and another letter: $\frac{4!}{2!}*7=84$ where the seven comes from the numbers of "another letter" mentioned earlier.
Using the same reasoning, there are $84$ strings with two I's, one B and a different letter.
Finally, the number of strings with zero or one B's and/or I's: $9*8*7*6=3024$
So my final sum is $3198$.
[/hide]

But the answer is
[hide]$3702$
[/hide]

towersfreak2006, your answer is not good, cause you forgot to count those with 2 B's and no I's and with 2 I's and no B :)

[hide="Oops..."]
Well, for 2 B's and no I's, there are $\frac{4!}{2!}*\frac{7!}{5!2!}=252$ strings, and another $252$ for 2 I's and no B's. Adding those to the $3198$ earlier, $3702$ is obtained. :)
[/hide]

probability is 11 letters, so...
pretend there there's only 1 b and only one i.  Then the number of possibilities is 9*8*7*6=3024.
Now let there be two b's and two i's, but you MUST pick at least one of them twice.  Let's say that you picked the b twice.  Then there are 6 ways to place those in 4 slots, and then there are 8*7 ways to pick the other two letters (assuming that you didn't pick two i's).  That makes 6*7*8.  We get another one from the possiblity of picking two i's, so we get another 6*7*8.  Add those to the original 6*7*8 and you get 11*6*7*8.  Then, if you picked b twice and i twice, there are 6 ways.  So the answer is 6(7*8*11+1)=3702


function
integration
real analysis
real analysis unsolved

$C$ the set of positive convex functions $f: [0;1] \to  \mathbb{R}^+$ that satisfy $f(0)=0$ and $f(1)=1$. Find the extrem points of $C$. (i.e. functions of $C$ that are not a non trivial convex combination of $2$ functions of $C$)

I believe they are the functions $f_c,\ c\in[0,1]$, which are $0$ on $[0,c)$ and linear on $[c,1]$ (note that $f_1$ is the only such function which is not continuous, and any function in $C$ can have at most one discontinuity point, namely $1$).

It's clear enough, I think, that $f_c$ are extreme points, so let's prove there are no others. Let $f$ be a function in $C$ such that $f^{-1}(0)=[0,t],\ t\in[0,1]$, and such that for some $x_0\in(t,1),\ f(x_0)$ is strictly below the chord connecting the points $(t,0),(1,1)$. We try to write $f$ as the mean of $g,h$, where $g|_{[0,x_0]}=\lambda f|_{[0,x_0]},\ h|_{[0,x_0]}=2f|_{[0,x_0]}-g|_{[0,t]}$, and $h|_{[x_0,1]}=1-\frac{1-(2-\lambda)f(x_0)}{1-f(x_0)}f|_{[x_0,1]},\ g|_{[x_0,1]}=2f|_{[x_0,1]}-h|_{[x_0,1]}$. It shouldn't be too hard to see that if $\lambda\in(0,1)$ is close enough to $1$, this works.

Sorry it looks so ugly, but a picture should make things clear.


How many ways are there to distribute 5 people into 3 groups, namely red, green, and blue?

I'm not so sure about my solution but am I going on the right track? If all 5 people were in red team, then there is 1 way to do this (5C5). We can also do the same for the green and blue team so there are 3 ways in total (all of them in either red, blue, or green).

If 4 were in red, 1 in green and none in blue, there are 5C4X1C1= 5 ways. We can also have 4 in red, 1 in blue, 1 in green, or 4 in green, 1 in red, and 1 in blue, etc. There are 6 ways to do this (3X1X1) so there's 30 in total with 4-1-0 arrangement.

We can also have 2,2,1 and this can be achieved in 5C2x3C2X(3!/2!) ways.

We do this for all possibilities such that the sum of of the people assigned in the three groups is 5.

Is this correct? Is there a shorter solution? Thanks!

[hide="Solution"]Let $ *$ represent a person. We're looking for configurations like these:
\[ \underbrace{*}_{\text{red}}|\underbrace{*}_{\text{green}}|\underbrace{***}_{\text{blue}}\]

So, as you can see, there are $ 7$ slots of which $ 2$ must be dividers, which happens in $ \binom 72\equal{}\boxed{21}$ ways. (This is called the balls and urns argument).[/hide]

[quote="grn_trtle"][hide="Solution"]Let $ *$ represent a person. We're looking for configurations like these:
\[ \underbrace{*}_{\text{red}}|\underbrace{*}_{\text{green}}|\underbrace{***}_{\text{blue}}\]
So, as you can see, there are $ 7$ slots of which $ 2$ must be dividers, which happens in $ \binom 72 \equal{} \boxed{21}$ ways. (This is called the balls and urns argument).[/hide][/quote]

Hmm, so let's replace the 5 people with 5 identical objects. Will the answer still be the same? Also, what's wrong with my solution? Thanks

[quote="Jakey"]Hmm, so let's replace the 5 people with 5 identical objects. Will the answer still be the same? Also, what's wrong with my solution? Thanks[/quote]

Ahh, I made the assumption, for some reason, that the people are non-distinguishable, which in retrospect doesn't make sense. What I gave above still holds, for identical objects.

For non-identical objects (people), isn't it just $ 3$ choices of a team for each person, so $ 3^5\equal{}243$? Or maybe it's just really late and I'm not thinking clearly.

[quote="grn_trtle"][quote="Jakey"]Hmm, so let's replace the 5 people with 5 identical objects. Will the answer still be the same? Also, what's wrong with my solution? Thanks[/quote]

Ahh, I made the assumption, for some reason, that the people are non-distinguishable, which in retrospect doesn't make sense. What I gave above still holds, for identical objects.

For non-identical objects (people), isn't it just $ 3$ choices of a team for each person, so $ 3^5 \equal{} 243$? Or maybe it's just really late and I'm not thinking clearly.[/quote]

hey, thanks for the help! yep, your answer's right :) i verified it with my solution above haha...anyway, why can't you use the ball and urn method for people?

[quote="grn_trtle"]You might be able to get balls and urns to work with people, but the major difference is that people are unique and objects are not.

That is, instead of working with configurations like $ **|*|**$, you're working with $ AB|C|DE$, which complicates things because you'd end up overcounting if you're not careful (for example, configurations like $ AB|C|DE$ and $ AB|C|ED$ are equivalent but would get counted twice, and so on).[/quote]

Ok, I got your point. Thanks so much! You made me understand it! :)


Let $ a$ be a natural number which is greater than $ 3$ and consider the sequence $ (u_n)(n\equal{}1,2,...)$ defined inductively by
$ u_1\equal{}a$ and $ u_{n\plus{}1}\equal{}u_n\minus{}[\frac{u_n}{2}]\plus{}1$ for all $ n\equal{}1,2,...$
Prove that there exists $ k \in N*$ such that $ u_n\equal{}u_k$ for all $ n \geq k$

@mods: Move this to temp, please (MY active problem).

let $ f(x) \equal{} x \plus{} 1 \minus{} [x/2]$ : $ u_{n \plus{} 1} \equal{} f(u_n)$
for $ u_n\ge 4$ then $ u_{n \plus{} 1} < u_n$
so $ \exists h < 4$ such that $ u_{k} \equal{} h$ for some $ k\in\mathbb{N}$
$ i)$ if $ h \equal{} 3$ then $ \forall n\ge k: \ u_{n} \equal{} fofof..n \minus{} k\ times..of(3) \equal{} 3 \equal{} u_k$ because $ f(3) \equal{} 3$
$ ii)$ if $ h \equal{} 2$ then $ \forall n\ge k: \ u_{n} \equal{} fofof..n \minus{} k\ times..of(2) \equal{} 2 \equal{} u_k$ because $ f(2) \equal{} 2$
$ iii)$ if $ h \equal{} 1$ then $ u_{a \plus{} 1} \equal{} 2\Longrightarrow ii)$
$ iv)$ if $ h \equal{} 0$ then $ u_{a \plus{} 2} \equal{} 2\Longrightarrow ii)$

Should be locked. This is the problem from M&Y of Vietnam.

[quote="April"]There are problems from the newest Mathematics and Youth Magazine (7/2008)

[b]T10/373.[/b] Let $ a$ be a natural number wich is greater than $ 3$ and consider the sequence $ (u_n)$ ($ n \equal{} 1, 2, \ldots$) defined inductively by $ u_1 \equal{} a$ and $ u_{n \plus{} 1} \equal{} u_n \minus{} \left[\frac {u_n}{2}\right] \plus{} 1$ for all $ n \equal{} 1, 2, \ldots$. Prove that there exists $ k\in\mathbb{N}^{*}$ such that $ u_n \equal{} u_k$ for all $ n\ge k$.

[And the dealine for submitting solutions is 15 September, 2008.[/quote]

Japanese Communities Moderator

kunny


calculus
integration
trigonometry
logarithms
calculus computations

Calculate the following indefinite integrals.

[1] $\int \frac{dx}{e^x-e^{-x}}$

[2] $\int e^{-ax}\cos 2x dx\ (a\neq 0)$

[3] $\int (3^x-2)^2 dx$

[4] $\int \frac{x^4+2x^3+3x^2+1}{(x+2)^5}dx$

[5] $\int \frac{dx}{1-\cos x}dx$

[quote="amirhtlusa"]5
use: $\sin{(2x)}=\frac{1-\cos{x}}{2}$ :?
answer:
$\frac{-\cot{(\frac{x}{2})}}{4}$[/quote]

[quote="amirhtlusa"]2, was good.
$\frac{e^{-ax}\sin{(2x)}}{3a}-\frac{e^{-ax}\cos{(2x)}}{6}$[/quote]

 :(

yes, i made a mistake in 1
the way to do it is to let $u=e^{2x}$ then integrate:$\frac{\sqrt{u}}{2(u-1)}$

[quote="amirhtlusa"]5
use: $\sin{(2x)}=\frac{1-\cos{x}}{2}$
answer:
$\frac{-\cot{(\frac{x}{2})}}{4}$

the answer would be:
$-\cot{(\frac{x}{2})}$[/quote]

O.K., amirhtlusa :)

[quote="amirhtlusa"]ok, i enjoyed #2 more,
ans:
$\frac{4}{4+a^2}(\frac{e^{-ax}\sin{(2x)}}{2}-\frac{a.e^{-ax}\cos{(2x)}}{4})$[/quote]

That's right! :)

[quote="adidasty"]dont murder me for this guys.

#2.

$\frac{e^{-ax}(2\sin(2x)-a\cos(2x))}{a^{2}+4}$[/quote]

 :)


parameterization
algorithm
algebra proposed
algebra

You are the boss. You have ten men and there are ten tasks. Your men have two parameters to each task, one is [b]enthusiasm[/b], the other is [b]ability[/b].Now you are to assign the tasks to your men one by one.When man $A$ has more enthusiasm about task $v$ than about task $u,$ and man $A$ has better ability in task $v$ than man $B$ does, though if you assign task $u$ to man $A$ and task $v$ to man $B,$ man $A$ feel unsatisfied.Or, if there is a more efficient way than yours that you can assign each task to men with better ability, you will be brought to account by your employer.Prove that there is a way to assign tasks without any dissatisfaction or disadvantage.

Algebra Proposed & Own Problems?
Do you have a solution kunny? If so please post it

Hmm.

First assign the ten tasks based solely on ability - that is, for each task $t_k$ we will assign it to the man with the highest ability and ignore enthusiasm.  Now, I'm not sure what the restrictions are as far as enthusiasm - is there a disadvantage if you assign task $t_k$ to man $M$ that he is unenthusiastic about but that he is the most skilled?

This is called the Stable Marriage Problem; it is well known in combinatorics. The way it is usually stated is that there are $n$ men $A,B,C,\dots$ married to $n$ women $\alpha, \beta, \gamma, \dots$, respectively ($A$ is married to $\alpha$, $B$ to $\beta$, etc.) Each man has a ranking of the desirability of all of the women, and each women has a ranking of the desirability of all of the men. The marriage setup is stable if there does not an exist a pair $A,\beta$ where $A$ finds $\beta$ more desirable than $\alpha$ and $\beta$ finds $A$ more desirable than $B$.

The proof that a set of stable marriages exists works by constructing an algorithm to match the men and the women. The men line up. Each man proposes to the top woman on his ranking who hasn't rejected him, until a woman accepts. A woman accepts (temporarily) if she hasn't been chosen yet or if her current match is less desirable to her than the new proposer, in which case she dumps her current match in favor of the new guy. The recently dumped man goes to the end of the line.

This algorithm always finishes because when a woman is first proposed to, she is never unmatched after that. It is guaranteed to give a stable set of marriages because suppose that the opposite is true; there exists a pair $A,\beta$ where $A$ finds $\beta$ more desirable than $\alpha$ and $\beta$ finds $A$ more desirable than $B$. If $A$ preferred $\beta$ to $\alpha$, he would have proposed to her before $\alpha$. So $\beta$ must have had a more desirable match when $A$ proposed. But she ended up with $B$, who is less desirable than $A$, which is a contradiction because a woman's match increases in desirablility each time.


Find intersection point of-
$3x+5y=1$
$(2+c)x+5c^2y=1$
where c tending towards 1


Given a,b,c>0,prove that:
$ 2(\frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a})\plus{}1$≥$ \frac{21(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}$


MATHCOUNTS
geometry
number theory
AoPS Books

This is specifically intended for those who have more tutoring/teaching experience than I do; if you don't fit those requirements, your opinions are certainly welcome but I would especially like to hear from teachers, MATHCOUNTS coaches, etc.

This year, I'll be coaching my school MATHCOUNTS team. There are four kids apart from myself, three of which know Algebra and are currently in Geometry classes. The last is in a pre-algebra course. All four students know nothing outside of school, though they are all bright and fairly motivated.

I recall JBoyd giving AIME15 advice, that was along the lines of giving the kids school sprints and then focusing on the weak points, such as number theory or combinatorics. This is what I will be doing tomorrow, at our first meet. Does anyone have any specific tips for teaching such broad topics that will be totally new for them, as well as coaching in general? I've never tutored such a large group outside of the Internet, so it's very new to me. I want the atmosphere to be fun, collaborative, and welcome to everyone, since the students barely know each other. Because of this, I don't want to just give the kids AoPS books and tell them to work on their own.

I'm personally tutoring the student who doesn't know Algebra one-on-one, to acquaint him with Algebra, of course with a problem solving approach. He's smart and willing to put in an hour or so a week, but probably not much more. Any tips here?

Thanks in advance!

It is extremely likely that I will be in the same situation, except all of the kids will be in 6th grade, have no out-of-school knowlege, and have not done pre-alg.

Yeah same here I have to be the leader of the next year team and I am a TA in the algebra class

although i am in 7th grade so i've been pretty much in the same situation.

The only difference is that the kids in my school aren't motivated and are forced by their parents to do mathcounts.

I haven't done any coaching/tutoring... but here are my thoughts:

1. [b]Start out easy.[/b] If they get a 20 on a school sprint, tell them the best school in some states had a highest score of 16 on the school. This will encourage them and build confidence, as well as motivate them to do better.

2. [b]Let them do the talking/learning.[/b] Give an example, like Math Jams on AoPS, and ask "How would you attack this problem?". Let the students solve it.

3. [b]Lessons[/b] Focus on C&P and NT, but give "out of school" bits of Algebra and Geometry as well. The kids in Geometry will/have probably learn the basics of Algebra and Geometry, and the Pre-Alg. student is doing it independently. I don't know about other books, but Intro to C&P has a summary at the end of each chapter. Give a short example of each main point and include the stuff in the blue boxes.

A lot of times my MathCounts coach has us do a Chapter level Sprint or Target round, checks it while we have a (really) quick break, then goes over solutions to the problems we don't understand.

In my school, we go over Chapter and State Sprint and Target rounds, and then hear the answers(not solutions), and have candy.  No wonder our chapter isn't very good because all teachers at Calumet Chapter do this.  :| (When I mean not good:  state qualifying score was 16/46 I think)

westiepaw: I was thinking about doing this, but imo, a math club should be something where people do a lot of problems together. At least, that's how I learn best.

PowerOfPi:

1. Yeah, we started doing this. I made copies of the 2007 school sprint, and we're slowly going over them. At our first meet, this past Tuesday, we got up to #13, I believe. On #9 we digressed into combinations, Pascal's, Sierpinski's, etc. :D

2. This resulted in long, awkward silences.

3. I tried this as well, but it didn't seem to work out very well. Instead, I mostly end digressing and rambling on each problem, leading into number theory and combinatorics and geometry and algebra and stuff, which oddly seems to be doing all right.

Thanks to those who gave advice! I'm still hoping for more. :(

The whole problem would be solved if schools actually bothered to teach combinatorics and number theory, but since they don't, you definitely should stress that. Unless they take some sort of outside class, they won't be able to solve those kind of problems.

Since it looks like there are a few people who want teaching tips, I will give a few.
[i]The most important rule of teaching is that you have to understand the problems you teach.  Also you should teach more than 1 method if there is multiple ways you could solve that problem, sometimes it might be easier to solve a problem one way and the next similar problem another way.[/i]


I have 2 problem which I don't understand:

If n distinct planes intersect in a line, and another line L intersects one of these planes in a single point, what's the LEAST number of these n planes that L could intersect?

A. n
B. n-1
C. n-2
D. n/2
E. (n-1)/2

If the magnitudes of vectors a and b are 5 and 12, respectively, then the magnitude of vector (b-a) could NOT be

A. 5
B. 7
C. 10
D. 12
E. 17

Thanks

[quote="juicybooty911"]I have 2 problem which I don't understand:

If n distinct planes intersect in aline, and another line L intersects one of these planes in a single point, what's the LEAST number of these n planes that L could intersect?

A. n
B. n-1
C. n-2
D. n/2
E. (n-1)/2

If the magnitudes of vectors a and b are 5 and 12, respectively, then the magnitude of vector (b-a) could NOT be

A. 5
B. 7
C. 10
D. 12
E. 17

Thanks[/quote]
[hide]
for the second one , isn't it just A because $12-5 \le x \le 12+5$?[/hide]

Number 1 seems to be [hide]B[/hide]
The line can be parallel to at most one plane, because the planes are not parallel to each other.


Daca am un corp cu 62 de fete dintre care: 20 sunt triunghiuri echilaterale, 30 sunt patrate, iar 12 sunt pentagoane regulate, atunci cate muchii are corpul nostru? ;)

Pentru ca fiecare muchie apartine la exact doua fete, numarul muchiilor este
\[ m=\frac{20\cdot 3+30\cdot 4+12\cdot 5}{2}=120\,. \]
Iar daca poliedrul este convex, putem folosi relatia lui Euler pentru a determina numarul varfurilor:
\[ v=m-f+2=60\,. \]
Daca poliedrul nu este convex(are "gauri"), numarul varfurilor este mai mic.

Aceasta problema a fost data la Lumina Math 2005 la clasa a 8a daca nu ma insel.


CEMC
number theory
Diophantine equation
Pythagorean Theorem
geometry

Good luck to everyone writing the Hypatia, Galois, and Fryer contests tomorrow. :)

Good luck everyone!!!  :thumbup:

- Nawahd Werdna

They don't take off marks for spelling on Hypatia, do they? :lol:

[quote="cynthia1991"]I just hope that it's not gonna be too hard...[/quote]
LOL I am sure you'll destroy it.

ya fgh's are real easy..like if u pwn things like AMC u can probably get 38+/40 no doubt

Do these contests always provide all the formulas used?

@lena_siz: No formulas are provided.
@SM4RT: FGH are really quite easy. I think they were meant as preparation for the Euclid.
I got 40/40/40 on the FGH series, so it's not so hard to get a perfect score (the marking is fairly relaxed, too, unlike some years for the COMC!)

@Cynthia: you went to Winter Camp. You will most certainly find the Hypatia too easy.

@Inspired By Nature: I don't believe they would, but please don't spell something like "circle" wrong :P

Good luck to everyone, and have fun doing it!

I will take that advice to heart nneonneo, thank you. :P
I have a late-start day today, so I'm going to be going to school at 10, and writing the contest at 11.
Good luck everyone. :)

I definitely need solutions for #4 hypatia

hope it will be posted tomorrow

I think I did quite well. I'm guessing I'll either make it to Gold or silver. :P
@ Anyh: Which part of 4? I found the diagram for c but ran out of time before i could write the solution. :(

i did completely horrible, I didnt get to sleep till 1:30 because my sister was yelling and screaming and freaking out about packing for her trip the next day (which she had to leave at 6am for btw..). So i wake up at 5am to more yelling and bickering.. ugh.

i did the best on 3 and 4.. I was just so clueless at 8:30 this morning  :(

Yes, I too did quite well, but I didn't do the Hypatia, I did the Galois... (such a joke!) it was really easy... I finished about 20 minutes early, and just checked my answers... anyone else do the galois?? I am anxious to compare answers, so when can we start??? :D  :D

Does anyone know when the results will be posted? Or how long it usually takes for them?

Here are the questions and answers for the Fryer contest:http://www.cemc.uwaterloo.ca/english/contests/fryer.shtml
You can see other contests by going on contests overview, and clicking on which contest you want.

Well, our school atleast got our results. I think I got a 38.

Yup, my school got them too. I think all schools should have received them by now. Except I still don't know my result because I didn't have a chance to ask.

how come i still don't know my result? so frustrated :mad:

ask your principal or the person who registered for the contest in your school.

The contest averages and cutoffs are posted on the CEMC website now. :D  :D

Yay! Thanks for telling me. Gold in my first year as a 6th grader. How did you people do?

the awards are posted on the site:
[url]https://cemc.math.uwaterloo.ca/english/contests/hypatia.shtml[/url]
                      Fryer Galois Hypatia
Average (Mean) 22.7   20.9   20.9
Gold Standard     36      34     32
Silver Standard   32      30     27
Bronze Standard  28      26    24

Sample solutions have been posted for Fryer Contest, see the link below:
http://www.cemc.uwaterloo.ca/english/contests/fryer.shtml

The results are out for the Fryer, Galois, and Hypatia contest on the CEMC website!

I got gold for Galois!  :winner_first:  :w00tb:

Gold here in Hypatia  :winner_first:   :P

Gold in Fryer! (Not very hard to get of course)

39... a silly adding error in the last question prevented me getting perfect on the Galois...


function
induction
combinatorics unsolved
combinatorics

Let S={1,2,..,2^n + 1} and T={0,1,2,..,2^(n-1)-1}.
Define the function f : S2---> T such that for any x,y,z
(S2 is the set of subsets of size 2 of S) one of f({x,y})  , f({x,z}) and
f({y,z}) is the sum of the other two. Prove that there are a,b,c such that
f({a,b}) = f({a,c}) = f({b,c}) = 0.

I’ll use induction. Case n=1 is trivial. Now I want to prove it for n. For any w in S define E(w)={x in S such that f({x,w}) is even} and O(w)={x in S such that f({x,w}) is odd}. Take a graph G with vertices coresponding to the elements of  S. The edge between x and y is drawn iff f({x,y}) is odd. By the given condition, G can’t contain any K3. Then, by Turan’s theorem it has at most 2^(n-1)*[2^(n-1)+1] edges (i.e the number of edges of  K 2^(n-1),2^(n-1)+1). Then there is a vertice v of degree d(v)<= 2*{2^(n-1)*[2^(n-1)+1] }/(2^n+1)<2^(n-1)+1 (since the sum of degrees is twice the number of edges), so d(v)<=2^(n-1). Let w be the element of  S coresponding to v. Take M={w} U E(w). Obviously, M has at least 2^(n-1)+1 elements (since E(w)= 2^n-O(w) and O(w)=d(v)). Take S’ a subset of M with exactly 2^(n-1)+1 elements. Note that if x, y belong to S’, then f({x,y}) even (to prove this, take f({x,y}), f({x,w}) and f({y,w}) if x,y!=w) . Now divide all values of f by 2 and everything follows by induction.

I saw this problem on forum earlier. In my opinion, dzeta's arguments are quite complicated.

My solution (prepared for Project 2002).
By induction. Case $n=1$ is obvious.

Let $a\in S$. Consider sets
\[A=\{x\in S\;|\;f(\{a,x\})\mbox{ is even}\}\]
and
\[B=\{x\in S\;|\;f(\{a,x\})\mbox{ is odd}\}.\]
It is clear if $x,\,y\in A$ or $x,\,y\in B$ then $f(\{x,y\})$ is even. There are two cases:

[b]1)[/b] $|A|\geq 2^{n-1}$. Consider $S'=\{a\}\cup A$, $f'(\{x,y\})=\dfrac{1}{2}f(\{x,y\})$ for all $x,\,y\in S'$. Now apply induction for $S'$ and $f'$.

[b]2)[/b] $|A|<2^{n-1}$. In this case $|B|\geq 2^{n-1}+1$. Consider $S'=B$, $f'(\{x,y\})=\dfrac{1}{2}f(\{x,y\})$ for all $x,\,y\in S'$, apply induction.


Hi all,
I want to go to Italy to learn Maths Ph.D. in English. :) Can you post some websites of Universities here?
Thank you very much! :)

Although I just ended up high school I'll answer. We are not so active here  :wink: .

I think you mean to join a school like the [url=http://www.sns.it/en/]Scuola Normale Superiore[/url], or [url=http://www.scuolagalileiana.unipd.it/en/index.html]Scuola Galieliana[/url].
I don't have enough informations about the existance of other good places where to study.

Of course, in case you don't know, it works like this: you have to study both at the "standard" university in the same city of the school, and follow some courses (don't know how to take a Ph.D exactly  :D ) at the school you choose.

I know that in the Scuola Normale Superiore are some foreign students, so you probably get the chance to take exams in English, but you have to check better at the website.  :)

Sorry if I don't have enough infos.

Thank you very much. I am finding Graduate School in Italy. But they must use English in their courses! :)


limit
algebra
polynomial
calculus
calculus computations

[i]Another hard problem :  Let $ n$ be a positive integer

   Evaluate : $ \lim_{x \ \to \ \plus{} \infty} \left( \sqrt [n]{(x \plus{} 2 )(x \plus{} 4 ).....(x \plus{} 2n )} \minus{} \sqrt [2n]{(x \plus{} 1 )(x \plus{} 2 ).....(x \plus{} 2n ) } \right)$[/i]

[color=green]New problem, new topic. - Moderator[/color]

The answer is $ \dfrac{1}{2}$.
Consider the polynomial under the first radical:
$ (x\plus{}2)(x\plus{}4)\cdots(x\plus{}2n)\equal{}x^n \plus{} n(n\plus{}1)x^{n\minus{}1}\plus{}\mathrm{terms\  containing}$
$ x^{n\minus{}2}\mathrm{\ and \ lesser \ powers}$
As such for $ x>0$,
$ \sqrt[n]{(x\plus{}2)(x\plus{}4)\cdots(x\plus{}2n)}\equal{}x\sqrt[n]{1\plus{}\dfrac{n(n\plus{}1)}{x}\plus{}\mathrm{terms\  containing}  \dfrac{1}{x^2}\mathrm{\ and \ higher\  powers\ of}\ \dfrac{1}{x}}$
Therefore, as $ x\to\plus{}\infty$, $ \dfrac{1}{x}\to0^\plus{}$ and as such, neglecting the second and higher order terms of $ \dfrac{1}{x}$, we get that as $ x\to\plus{}\infty$,
$ \sqrt[n]{(x\plus{}2)(x\plus{}4)\cdots(x\plus{}2n)}\equal{}x\sqrt[n]{1\plus{}\dfrac{n(n\plus{}1)}{x}}\equal{}x\left(1\plus{}\dfrac{1}{n}\cdot\dfrac{n(n\plus{}1)}{x}\right)$ (applying binomial approximation.)
Therefore, as $ x\to\plus{}\infty$
$ \sqrt[n]{(x\plus{}2)(x\plus{}4)\cdots(x\plus{}2n)}\equal{}x\plus{}(n\plus{}1)$
Similar, approximation gives
$ \sqrt[2n]{(x\plus{}1)(x\plus{}2)\cdots(x\plus{}2n)}\equal{}x\plus{}\dfrac{2n\plus{}1}{2}$ as $ x\to\plus{}\infty$
As such the required limit, as $ x\to\plus{}\infty$ becomes
$ (x\plus{}(n\plus{}1)\minus{}x\minus{}\dfrac{2n\plus{}1}{2}\equal{}\boxed{\dfrac{1}{2}}$


determine which of the following holds and prove the ones that are true and give counterexample for the false ones.
sup (A cap B) <= inf{sup (A), sup (B)}
sup (A cap B) = inf{sup (A), sup (B)}
sup (A cup B) >= sup{sup (A), sup (B)}
sup (A cap B) = sup{sup (A), sup (B)}

what if $ A$ and $ B$ are disjoint but nonempty?

This looks like a homework problem  :roll:


algebra
polynomial
inequalities
algebra unsolved

Let $ P(x)\equal{}x^2\plus{}ax\plus{}b$ be a real polynomial with $ a<2$. Suppose  $ P(P(x))\equal{}0$ has 4 distinct real roots and the sum of some two of them is $ \leq \minus{}1$.Prove that $ P(x\plus{}y) \geq P(x)\plus{}P(y)$.

So $ P(0)\equal{}P(x\minus{}x)\geq P(x)\plus{}P(\minus{}x)$ for all $ x$? Obviously, it is wrong.

That is not a contradiction... like $ f(x)\equal{}\minus{}x^2$

[quote="rofler"]That is not a contradiction... like $ f(x) \equal{} \minus{} x^2$[/quote]

But the leading coeficient of $ P$ is $ 1$.

[quote="aakansh92"]Let $ P(x) \equal{} x^2 \plus{} ax \plus{} b$ be a real polynomial with $ a < 2$. Suppose  $ P(P(x)) \equal{} 0$ has 4 distinct real roots and the sum of some two of them is $ \leq \minus{} 1$.Prove that $ P(x \plus{} y) \geq P(x) \plus{} P(y)$.[/quote]
$ P(x \plus{} y) \ge P(x) \plus{} P(y)$
$ (x \plus{} y)^2 \plus{} a(x \plus{} y) \plus{} b \ge x^2 \plus{} ax \plus{} b \plus{} y^2 \plus{} ay \plus{} b$
$ 2xy \ge b$

Are there any restrictions on $ x$ and $ y$ other than that they are real numbers? Otherwise, there are some reals $ x,y$ such that $ 2xy < b$
Or did I do something wrong?

[quote="praneeth"][quote="aakansh92"]oops! $ x,y$ are non negative reals.[/quote]

P(x+y)-P(x)-P(y)=2xy. If x,y are non-negative reals, the given expression is obviously greater
than or equal to zero. It doesn't need those extra conditions.[/quote]

Actually, $ P(x\plus{}y)\minus{}P(x)\minus{}P(y)\equal{}2xy\minus{}b$. This is greater than or equal to $ 0$ for all $ x,y \ge 0$ if $ b \le 0$.

1)I was able to show it for $ 0<a<2$.
Will any one continue further....

2)Can some moderator edit my first post please and add at the end :" for all nonnegative reals x,y".


trigonometry
algebra
polynomial
Vieta
geometry
power of a point

[geogebra]534a8cb6a7bcecfbe134984726910daa6bc73d71[/geogebra]

$ \triangle ABC ,AB \equal{} AC,BC \equal{} 1$

A point $ D$ is on $ AB$ such that $ AD \equal{} 1$ and  $ DC\equal{} \sqrt 2$

Find $ \angle A$.

hello, we have
$ 1\equal{}x^2\minus{}2x\cos(\alpha)$ in triangle $ ADC$ and
$ 2\equal{}(x\minus{}1)^2\plus{}1\minus{}2(x\minus{}1)\cos(\frac{\pi}{2}\minus{}\frac{\alpha}{2})$ in triangle $ DBC$
solving this equation system for $ x$ and $ \alpha$ we get searched the solution.
Sonnhard.

Dr Sonnhard Graubner is ur solution short? is it noble of solving the equation s^3-2s^2-s+1=0. is the equation solvable?

I have an ugly and lengthy trig solution, would love to see a synthetic one.

[hide]See the attached diagram for the notation.

Menelaus: $ {1\over x}\cdot 1^2 + {x - 1\over x}\cdot x^2 - 1\cdot(x - 1) = 2\iff x^3 - 2x^2 - x + 1 = 0\quad(*)$

Cosine Law for $ \triangle ACD$: $ \cos\alpha = {x^2 + 1^2 - 2\over 2x} = {x^2 - 1\over 2x}$

Denote $ c: = \cos\alpha$. Then $ x^2 - 2cx - 1 = 0\implies x = c + \sqrt {1 + c^2}$ (only the positive solution is taken)

Plugging that into $ (*)$ - after some easy but lengthy simplification, which I will skip here - we get

$ \sqrt {1 + c^2}(4c^2 - 4c) + 4c^3 - 4c^2 + 2c - 1 = 0$

$ \sqrt {1 + c^2} = - {4c^3 - 4c^2 + 2c - 1\over 4c^2 - 4c} = - c - {2c - 1\over 4c(c - 1)} = {2c - 1\over 4c(1 - c)} - c$

$ 1 + c^2 = \left({2c - 1\over 4c(1 - c)}\right)^2 - {2c - 1\over 2 - 2c} + c^2$

$ \left({2c - 1\over 4c(1 - c)}\right)^2 = 1 + {2c - 1\over 2 - 2c} = {1\over 2(1 - c)}$

$ (2c - 1)^2 = 8c^2(1 - c)$

$ 4c^2 - 4c + 1 = 8c^2 - 8c^3$

$ \boxed{8c^3 - 4c^2 - 4c + 1 = 0}$

It may be known to the reader that $ c \in\{ \cos{\pi\over 7},\cos{3\pi\over 7},\cos{5\pi\over 7}\}$ are the solutions to this equation (it's a bit "classic"), but we'll show that. Keep in mind that we'll constantly use $ 8c^3 = 4c^2 + 4c - 1$ in the following simplifications:

\begin{eqnarray*}\cos 6\alpha & = & 2\cos^23\alpha - 1 \\
& = & 2(4c^3 - 3c)^2 - 1 \\
& = & 2(16c^6 - 24c^4 + 9c^2) - 1 \\
& = & 32c^6 - 48c^4 + 18c^2 - 1 \\
& = & {(4c^2 + 4c - 1)^2\over 2} - 6c(4c^2 + 4c - 1) + 18c^2 - 1 \\
& = & {16c^4 + 16c^2 + 1 + 32c^3 - 8c^2 - 8c\over 2} - 24c^3 - 24c^2 + 6c + 18c^2 - 1 \\
& = & 8c^4 - 8c^3 - 2c^2 + 2c - {1\over 2} \\
& = & c(4c^2 + 4c + 1) - 4c^2 - 4c + 1 - 2c^2 + 2c - {1\over 2} \\
& = & 4c^3 - 2c^2 - 3c + {1\over 2} \\
& = & \left(4c^3 - 2c^2 - 2c + {1\over 2}\right) - c \\
& = & - c\end{eqnarray*}

Therefore $ \cos 6\alpha = - \cos\alpha\implies\alpha={\pi\over 7}$[/hide]

a non trigonometric approach would be solving the equation by stewart's theorem s^3-2s^2-s+1=0. and thus we get s orthe equal side of iso. triangle. and then angle A.  roots of the equation?

Actually, the synthetic solution is very simple and elegant once you find it. See the attached diagram.

Draw the circle $ K$ with center in $ D$ and radius $ 1$. By power of a point, $ CM\cdot CA\equal{}CU\cdot CV\equal{}(\sqrt{2}\minus{}1)(\sqrt{2}\plus{}1)\equal{}1$, hence $ CM\equal{}{1\over CA}$. But that implies $ MC: CB\equal{}BC: CA$, hence $ \triangle BCM\sim\triangle ABC$, which yields $ BM\equal{}1$ and $ \angle MBC\equal{}\alpha$.

Since $ \triangle ADM$ is isosceles, we get $ \angle AMD\equal{}\alpha\implies\angle MDB\equal{}2\alpha$ (external angle), but then $ \triangle DMB$ is also isosceles, which yields $ \angle DBM\equal{}2\alpha$. So we finally get $ \beta\equal{}\angle ABC\equal{}2\alpha\plus{}\alpha\equal{}3\alpha\iff 3\alpha\plus{}3\alpha\plus{}\alpha\equal{}\pi\iff\alpha\equal{}{\pi\over 7}$

[quote="PKMathew"][quote="Farenhajt"]Actually, the synthetic solution is very simple and elegant once you find it. See the attached diagram.

Draw the circle $ K$ with center in $ D$ and radius $ 1$. By power of a point, $ CM\cdot CA \equal{} CU\cdot CV \equal{} (\sqrt {2} \minus{} 1)(\sqrt {2} \plus{} 1) \equal{} 1$, hence $ CM \equal{} {1\over CA}$. But that implies $ MC: CB \equal{} BC: CA$, hence $ \triangle BCM\sim\triangle ABC$, which yields $ BM \equal{} 1$ and $ \angle MBC \equal{} \alpha$.

Since $ \triangle ADM$ is isosceles, we get $ \angle AMD \equal{} \alpha\implies\angle MDB \equal{} 2\alpha$ (external angle), but then $ \triangle DMB$ is also isosceles, which yields $ \angle DBM \equal{} 2\alpha$. So we finally get $ \beta \equal{} \angle ABC \equal{} 2\alpha \plus{} \alpha \equal{} 3\alpha\iff 3\alpha \plus{} 3\alpha \plus{} \alpha \equal{} \pi\iff\alpha \equal{} {\pi\over 7}$[/quote]

Hello ,Farenhajt.
From your diagram shown , I wonder why you are sure that the length of BD is longer than the radius of the circle (radius=1) , i.e. B is outside the circle.
Would you please explain it?[/quote]

You can employ Stewart. If $ x: \equal{}BD$, then $ {1\over 1\plus{}x}\cdot 1^2\plus{}{x\over 1\plus{}x}(1\plus{}x)^2\minus{}1\cdot x\equal{}2\iff x^3\plus{}x^2\minus{}2x\minus{}1\equal{}0$. Hence $ x^3\plus{}x^2\equal{}2x\plus{}1\iff x^2\plus{}x\equal{}2\plus{}{1\over x}$. Therefore $ x^2\plus{}x>2$, which implies $ x>1$.

[quote="PKMathew"][quote="Farenhajt"]I have an ugly and lengthy trig solution, would love to see a synthetic one.

[hide]See the attached diagram for the notation.

Menelaus: $ {1\over x}\cdot 1^2 + {x - 1\over x}\cdot x^2 - 1\cdot(x - 1) = 2\iff x^3 - 2x^2 - x + 1 = 0\quad(*)$

Cosine Law for $ \triangle ACD$: $ \cos\alpha = {x^2 + 1^2 - 2\over 2x} = {x^2 - 1\over 2x}$

Denote $ c: = \cos\alpha$. Then $ x^2 - 2cx - 1 = 0\implies x = c + \sqrt {1 + c^2}$ (only the positive solution is taken)

Plugging that into $ (*)$ - after some easy but lengthy simplification, which I will skip here - we get

$ \sqrt {1 + c^2}(4c^2 - 4c) + 4c^3 - 4c^2 + 2c - 1 = 0$

$ \sqrt {1 + c^2} = - {4c^3 - 4c^2 + 2c - 1\over 4c^2 - 4c} = - c - {2c - 1\over 4c(c - 1)} = {2c - 1\over 4c(1 - c)} - c$

$ 1 + c^2 = \left({2c - 1\over 4c(1 - c)}\right)^2 - {2c - 1\over 2 - 2c} + c^2$

$ \left({2c - 1\over 4c(1 - c)}\right)^2 = 1 + {2c - 1\over 2 - 2c} = {1\over 2(1 - c)}$

$ (2c - 1)^2 = 8c^2(1 - c)$

$ 4c^2 - 4c + 1 = 8c^2 - 8c^3$

$ \boxed{8c^3 - 4c^2 - 4c + 1 = 0}$

It may be known to the reader that $ c \in\{\cos{\pi\over 7},\cos{3\pi\over 7},\cos{5\pi\over 7}\}$ are the solutions to this equation (it's a bit "classic"), but we'll show that. Keep in mind that we'll constantly use $ 8c^3 = 4c^2 + 4c - 1$ in the following simplifications:

\begin{eqnarray*}\cos 6\alpha & = & 2\cos^23\alpha - 1 \\
& = & 2(4c^3 - 3c)^2 - 1 \\
& = & 2(16c^6 - 24c^4 + 9c^2) - 1 \\
& = & 32c^6 - 48c^4 + 18c^2 - 1 \\
& = & {(4c^2 + 4c - 1)^2\over 2} - 6c(4c^2 + 4c - 1) + 18c^2 - 1 \\
& = & {16c^4 + 16c^2 + 1 + 32c^3 - 8c^2 - 8c\over 2} - 24c^3 - 24c^2 + 6c + 18c^2 - 1 \\
& = & 8c^4 - 8c^3 - 2c^2 + 2c - {1\over 2} \\
& = & c(4c^2 + 4c + 1) - 4c^2 - 4c + 1 - 2c^2 + 2c - {1\over 2} \\
& = & 4c^3 - 2c^2 - 3c + {1\over 2} \\
& = & \left(4c^3 - 2c^2 - 2c + {1\over 2}\right) - c \\
& = & - c\end{eqnarray*}

Therefore $ \cos 6\alpha = - \cos\alpha\implies\alpha = {\pi\over 7}$[/hide][/quote]

Hello,Farenhajt.

Thanks for your friendly replies.

In fact ,  the synthetic solution shown is very nice and helpful.

Another problem  in your trigo. solution ,

$ \boxed{8c^3 - 4c^2 - 4c + 1 = 0}$

$ c \in\{\cos{\pi\over 7},\cos{3\pi\over 7},\cos{5\pi\over 7}\}$

How could you show these three roots for the above cubic eqt?

 :P[/quote]

Any idea? :)

Since the equation is irrelevant to this problem, we may skip it altogether. Besides, I think you can google all about it.

Or just practice your trig by applying Vieta and calculating $ \cos{\pi\over 7}\plus{}\cos{3\pi\over 7}\plus{}\cos{5\pi\over 7}$, $ \cos{\pi\over 7}\cos{3\pi\over 7}\plus{}\cos{\pi\over 7}\cos{5\pi\over 7}\plus{}\cos{3\pi\over 7}\cos{5\pi\over 7}$ and $ \cos{\pi\over 7}\cos{3\pi\over 7}\cos{5\pi\over 7}$


I [b]SUCK[/b] at probablilty/counting and I am preparing for nationals. Where can I get help (im poor). I have AOPS Vol 1, but that wasn't [i]that[/i] helpful. Suggestions?

Just do a lot of questions that involve probability.

I know, I was in the same kind of situation as you were, biffanddoc. I also agree with tarquin. I like to think of probability more as fractions, and I think that you should also think from the POV (point of view) of the person who wrote the question. Think of what the question is asking and ways to get there.

[quote="tarquin"]The 2004 State Team test #10 was a pretty good probability question, if you want to try it. I don't quite remember it though.[/quote]Where can I get it?

[quote="tarquin"]The 2004 State Team test #10 was a pretty good probability question, if you want to try it. I don't quite remember it though.[/quote]

Was that the one with the different types of CD's

That was the only question on the team round we missed.

[quote="mathmom"]Do you have a good "mental model" of how to think about these questions, or are you just trying to apply formulas?  (I'd recommend forgetting the formulas at first and getting used to thinking about the problems more "intuitively".)

There was a "Counting and Combinatorics" Stretch in the MATHCOUNTS handbook for 2002-2003.  If your coach doesn't have it, you can probably get a copy from http://goodlymaths.0catch.com/contests/mathcounts/mathcounts_archive.htm

I'm attaching a handout I made for my students that discusses some of those problems and begins to develop the logic behind the formula for the number of combinations that you may have memorized.[/quote]Both. But I know what the  formulas mean an why they work and use them.

[quote="biffanddoc"]Both. But I know what the  formulas mean an why they work and use them.[/quote]

Well, then the handout I posted is probably too elementary for you.

You might want to check out some of the following resources, which go a bit deeper:

[url=http://www.cs.utk.edu/~booth/311-04/notes/combinatorics.html]www.cs.utk.edu/~booth/311-04/notes/combinatorics.html[/url]
[url=http://www.themathpage.com/aPreCalc/permutations-combinations.htm]The Math Page - Permutations and Combinations[/url]
[url=http://mathforum.org/library/drmath/sets/high_perms_combs.html]Ask Dr. Math - Permutations and Combinations[/url]
[url=http://mathforum.org/dr.math/faq/faq.comb.perm.html]Ask Dr. Math FAQ on Permutations and Combinations[/url]

Hope that helps!


function
trigonometry
algebra proposed
algebra

Find all $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that

$ f(x \plus{} yf(x)) \equal{} f(x) \plus{} xf(y)$

$ f(x \plus{} yf(x)) \equal{} f(x) \plus{} xf(y)$

First if $ f(1)\not \equal{} 1$, then $ y \equal{} \frac {1}{1 \minus{} f(1)}$ $ \Rightarrow$ we get  $ f(1) \equal{} 0$ and therefore when we set $ x \equal{} 1$ we will have $ f(1) \equal{} f(y) \equal{} 0$ for every $ y\in \mathbb{R}$.
So now, we can suppose that $ f(1) \equal{} 1$ and setting first $ x \equal{} 1$ and then $ y \equal{} 1$, we'll get $ f(1 \plus{} y) \equal{} 1 \plus{} f(y)$ and $ f(x \plus{} f(x)) \equal{} x \plus{} f(x)$.

If we set $ y \equal{} \frac {1}{f(x)}$, we will get $ f\left(\frac {1}{f(x)}\right) \equal{} \frac {1}{x}$ and therefore $ f(x)$ is a bijection. Of course, before doing this, we should notice that if there exists $ x_0\not \equal{} 0$ such that $ f(x_0) \equal{} 0$, then $ f\equiv 0$, so we can assume $ f(x)\not \equal{} 0$ for $ x\not \equal{} 0$.

The rest is easy ... :roll:

grn_trtle you ignored that if $ f(0)\equal{}0$; $ f(yf(0))\equal{}f(0)$ does not imply that $ f$ ia a constant function

what about $ f(x)\equal{}x$ ????

without calcule !!!

i signaled that because you're forgot that than i will post my answer
________________________________________________________
$ lahoucine$

[quote="ThiagoSPinheiro"]grn_trtle you ignored that if $ f(0) \equal{} 0$; $ f(yf(0)) \equal{} f(0)$ does not imply that $ f$ ia a constant function[/quote]

Thanks, I fixed it.

@mathema*: See my edited post.

[quote="grn_trtle"]
Let $ y \equal{} \minus{} x/f(x)$ to get $ f(0) \equal{} f(x) \minus{} x^2/f(x) \equal{} 0\implies f(x)^2 \equal{} x^2\implies f(x) \equal{} \pm x$.

Thanks, ThiagoSPinheiro.[/quote]

I plug in $ y\equal{}\minus{}\frac x{f(x)}$ and I only get
$ f(0)\equal{}f(x)\plus{}xf(\minus{}\frac x{f(x)})$.

[quote="BG Yoda"]

The rest is easy ... :roll:[/quote]

I am wondering if you could write down "the rest" even you think it is easy. BTW, I wish you did not base your proof on a claim similar to what you made at http://www.mathlinks.ro/viewtopic.php?t=262598. There are many $ f$ such that $ f$ is bijective and $ f(x)\plus{}x$ is not surjective, e.g., $ f(x)\equal{}\minus{}x\plus{}c*\tan^{\minus{}1}(x)\plus{}a$ where $ c<0$ and $ a$ are constants.

So how do you finish the problem, xxp2000?

If $ f(a)\equal{}0$ for some $ a\ne0$,
$ P(a,y): af(y)\equal{}0$ implies $ f(y)\equal{}0, \forall y$ is one solution.
Now we assume $ f(y)\ne0,\forall y\ne0$.
It is easy to see $ f(1)\equal{}1$. So we have $ f(y\plus{}1)\equal{}f(y)\plus{}1,\forall y$.

1) If $ f(a)\equal{}a$, then $ f(y\plus{}a)\equal{}f(y)\plus{}a,\forall y$.
Suppose $ f(a)\equal{}a$ and $ a\ne0$.
$ P(a,y): f(ay\plus{}a)\equal{}af(y)\plus{}a\equal{}a(f(y)\plus{}1)\equal{}af(y\plus{}1)\Rightarrow f(ay)\equal{}af(y),\forall y$.
$ f(y\plus{}a)\equal{}af(\frac ya\plus{}1)\equal{}a(f(\frac ya)\plus{}1)\equal{}f(y)\plus{}a$.

2) $ f(yf(x))\equal{}xf(y)$.
$ P(x,1): f(x\plus{}f(x))\equal{}x\plus{}f(x)\Rightarrow f(x\plus{}f(x)\plus{}yf(x))\equal{}x\plus{}f(x)\plus{}f(yf(x))$, where we use 1).
$ P(x,y\plus{}1): f(x\plus{}yf(x)\plus{}f(x))\equal{}f(x)\plus{}xf(y)\plus{}x$
Compare the above two, we have
$ f(yf(x))\equal{}xf(y)$  (*)

3) $ f(xu)\equal{}f(x)f(u)$ and $ f(x\plus{}u)\equal{}f(x)\plus{}f(u)$.
Let $ y\equal{}1$ in (*), we get $ f(f(x))\equal{}x$,
Let $ x\equal{}f(u)$ in (*), we get $ f(yu)\equal{}f(u)f(y)$ (**)
Plugging (*) back into the f.e., we have
$ f(x\plus{}yf(x))\equal{}f(x)\plus{}f(yf(x))$
For any $ x\ne0$, $ yf(x)$ can be arbitrary value, so we have
$ f(x\plus{}u)\equal{}f(x)\plus{}f(u)$.

4) $ f(x)\equal{}x$.
(**) implies $ f(x^2)\equal{}f(x)^2>0$. So $ f(x)>0,\forall x>0$.
Combining with Cauchy equation, we know $ f$ is increasing.
Then Cauchy equation with monotonicity implies $ f(x)\equal{}x$.

Either $ f(x)\equal{}0$ or $ f(x)\equal{}x$.

[quote=grn_trtle][hide="Solution"]$ f(x \plus{} yf(x)) \equal{} f(x) \plus{} xf(y)$.

Let $ x \equal{} 0$ to get $ f(yf(0)) \equal{} f(0)$. Let $ f(0) \equal{} k$, where $ k$ is constant. We have $ f(yk) \equal{} k$ implying that $ f$ is a constant function.

Plug in $ f(n) \equal{} k$ for all $ n$ to get:

$ k \equal{} k \plus{} kx\implies k \equal{} k(x \plus{} 1)$. Assume $ k\ne 0$ to safely divide by $ k$, giving $ x \plus{} 1 \equal{} 1\implies x \equal{} 0$, a contradiction because $ x$ can be any real. Therefore, $ k \equal{} 0$.

We find the solution $ \boxed{f(n) \equal{} 0}$.[/hide]

Edit:
[hide="Case 2"]
If $ f(0)\equal{}0$:
Let $ y\equal{}\minus{}x/f(x)$ to get $ f(0)\equal{}f(x)\minus{}x^2/f(x)\equal{}0\implies f(x)^2\equal{}x^2\implies f(x)\equal{}\pm x$. However, checking, $ f(x)\equal{}\minus{}x$ does not work, so we get the other solution $ \boxed{f(x)\equal{}x}$.
[/hide]

Thanks, ThiagoSPinheiro.[/quote]

This solution is very poor... \bump.

[quote=elcinmusazade]
$(f(x),-1) \implies f(f(x))=-xf(-1)$
[/quote]

Is this possible? I think this is completely wrong.

[hide=Solution]Take $x+f(x)=a$, we have that $f(a)=a$. So $a$ is a fixed point of $f(x)$, hence $f(x)=a-x$. $$ \therefore f(f(x))=a-f(x)=x$$[/hide]


trigonometry
number theory unsolved
number theory

Find the following :

$ S \equal{} \sin 1^0 \plus{} \sin 2^0  \plus{} \sin 3^0  \plus{}...... \plus{} \sin 89^0  \plus{} \sin 90^0$

Good luck

When doing this, directly find a formula for $ \sum_{k\equal{}1}^n \sin(kx)$ (similar for $ \cos$).

[b]Hello ZetaX :

Did you mean you need from to write the formula  for sin(kx) and cos(kx).when someone solves my question .

That's will be nice task for me.

I'll try to find, But I need help to give me the 1st step.

Thanks  :)[/b]

[b]Thanks Yuriy Solovyov :

I said before you have a nice mind . :wink:

The math coffee is your cup every morning .

[/b]


factorial
Ring Theory
superior algebra
superior algebra unsolved

In a previous post, an author said $ R[x,y]/(x^2 + y^2)$ is not previous.
But, it seems: $ R[x,y]/(x^2 + y^2)$ is isomorphic to $ C'[x,y] = {f|f\in C[x,y]}$ constant term of $ f$ belongs to R}. The irreducible element is those, whose roots can't "form" a real number (except all of its roots).

Though this factorial is not unique but it seem to be a factorial ring.

$ x^2 \minus{} y^2 \equal{} (x \plus{} y)(x \minus{} y) \equal{} 2x^2 \equal{} 2x(x) \equal{} \minus{}2y^2 \equal{} 2y(y)$.

? thanks, but how do this contradict to its factorial.
I didn't say it's UFD

"Factorial ring" means UFD, at least as far as I know.  What do you think it means?

I think he means that each element can be reduced into a product of primes or something like that in which case it looks more like you're interested in primary decomposition or something like that...


function
real analysis
real analysis unsolved

Is there a real continuous function $f$, $2\cdot\pi$ periodical and such that $f*f(x)=\sum_{n\geq 1}{\frac{e^{inx}}{n^2}}$? Here $f*f$ is the convolution product of the functions.

Equivalent to the following: is it possible to find a sequence of signs $e_n=\pm1$ such that

$\sum_{n=1}^{\infty}\frac{e_n}ne^{inx}$ is the Fourier series of a continuous function?

(Harazi: are you sure you meant "real" as in "real-valued" in "real continuous function?" Alternatively, are you sure you want the Fourier series to be one-sided?)

Oh, no, the real should be forgotten. Sorry.


A certain state issues license plates consisting of six digits (from 0 to 9). The state requires that any two license plates differ in at least two places. (For instance, the numbers 027592 and 020592 cannot both be used.) Determine, with proof, the maximum number of distinct license plates that the state can use.

All of the practice problems are from old USAMO's. :)

[quote="Phelpedo"]An old USAMO.

[hide] The answer is $10^{5}$. It follows from basic pigeonhole that more than $10^{5}$ is impossible, and then a clever construction that allows you to have $10^{5}$. [/hide][/quote]

This question is just like that one I posted about in the Intermediate Topics.

So, now I get why 10^4, is the right answer.

No, the question I posted in the forum was 5 digits, so it is 10^5.

Anirudh, next time could you post the subject of the questions as well, so we don't have to clikc to see where to go.


conics
hyperbola
rotation
analytic geometry
trigonometry
ellipse
absolute value

Can someone explain why $xy = k$ forms a hyperbola?

Isn't a hyperbola modeled by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $b^2 = c^2 - a^2$

How do you get $xy = k$ from $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$?

have you ever heard about oblique conics? basically those are when the conic is rotated so the major and minor axies are not parallel to the coordinate axies, also when you have those 2 forms for hyperbolas, those are only the forms for hyperbolas with the conjugate and transverse axies parallel to the x and y axies

A conic of the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is a hyperbola only if $B^2-4AC>0$. The form you stated is true when $B=0$, but there are many conics that are 'rotated' (have an $xy$ term). You can rotate the coordinate axes by an angle of $\theta$ when $\tan{2\theta}=\frac{B}{A-C}$, and you'll cancel out the $xy$ term and be able to put the graph in the form you stated, only in a new set of coordinate axes $x',y'$.

Think about it this way:

xy=k

k is a constant.  x and y are variables.  Therefore, x and y are inversely proportional, and therefore graphs the two swooping lines of a hyperbola, only rotated 45 degrees.  Make an x and y chart and see for yourself.

So polar coordinates will be needed for the study of oblique conic sections?

One more question:

When deriving $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ from

$\sqrt{(x-c)^2 + y^2} - \sqrt{{x+c}^2 + y^2} = \pm2a$

Mathworld's derivation simply sets the difference to equal $2a$ and not $\pm2a$. Wouldn't this just give you the equation for half a hyperbola?

No, because the absolute value of both sides is the same, so you can square that and that's the same as squaring the original equation without the $\pm$.

[quote="breez"]So polar coordinates will be needed for the study of oblique conic sections?[/quote]

Technically, you could just write it in the form $Ax^2 + Bxy + Cy^2 + Dx + Ey = F$, but it's a lot easier to rotate, etc. when it's written in polar.

And remember that the most basic definition of a hyperbola is based on the set of points such that the difference of the distance between the two foci is a constant, and it can be shown that $xy = k$ fits this definition, it's just that the line between the foci is not parallel to an axis (although you need to find the foci, and I never was very good at that  :)  )

I don't like the method of proof that is on Mathworld's site, because there are several steps where you have to square things and you don't have equivalence, and surprisingly all proofs I've seen on the internet forget to make sure you can do the calculations "backwards".

It is much better to do the proof using that $(r_2 - r_1)(r_2 + r_1) = r_2^2 - r_ 1^2$, gaining an expression for $r_2 + r_1$ so that you can solve a linear system of equations. This way you have equivalence in all steps with no worries.

[quote="breez"]Isn't a hyperbola modeled by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$[/quote]

No, it's not. What you wrote is an ellipse. A hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ :)

Oh yes my bad.

btw, substituting $b^2$ for $a^2 - c^2$ gives the formula for hyperbola, except $b^2$ becomes a complex number. So in a way, the equation for real ellipse is also the equation for an imaginary hyperbola and the equation for a real hyperbola is the equation for an imaginary ellipse.

[hide="Proof that xy=k is a hyperbola"]What we need is a formula for the new coordinates when we rotate the axes $45^\circ$ counterclockwise. Take an arbitrary point $M(x,y)$ and draw the lines $y=x$ and $y=-x$. Obviously, the new coordinates will be projections of $M$ onto those lines. Let $M'$ be its projection onto $y=x$. (Careful: We're talking about [i]perpendicular[/i] projection.) Then we have:

$x'=OM'=OM\cos\angle MOM'=OM\cos(\angle MOx-45^\circ)=$
$=OM\cos\angle MOx\cdot{1 \over \sqrt{2}}+OM\sin\angle MOx\cdot{1 \over \sqrt{2}}=x{1 \over \sqrt{2}}+y{1 \over \sqrt{2}}={{x+y} \over \over \sqrt{2}}$

Similarly, $y'={{x-y} \over \sqrt{2}}$. Solving these for $x$ and $y$ we get

$x={{x'+y'} \over \sqrt{2}}, y={{x'-y'} \over \sqrt{2}}$.

Plugging this into $xy=k$ we get ${x'^2 \over {2k}}-{y'^2 \over {2k}}=1$. For $k>0$ that's obviously a hyperbola, and for $k<0$ we can write it as ${y'^2 \over {-2k}}-{x'^2 \over {-2k}}=1$, which is also a hyperbola, but mirrored over the line $y'=x'$, because the coordinates are swapped - i.e. its foci lie on the $y'$-axis.[/hide]


inequalities
function
inequalities proposed

[quote="arqady"]Let  $ \{a,b,c\}\subset[ k, \plus{} \infty)$ such that $ abc\geq1$ and
$ k$ is a real root of the equation $ x \plus{} \sqrt [3]x \plus{} 1 \equal{} 0.$ Prove that
\[ (a^2 \plus{} a \plus{} 1)(b^2 \plus{} b \plus{} 1)(c^2 \plus{} c \plus{} 1)\geq(a \plus{} 2)(b \plus{} 2)(c \plus{} 2)
\]
I posted this inequality  here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179021
but it's still unsolved.[/quote]
Here is proof when $ a,b,c > 0$
WLOG $ a \geq b \geq c$ then the inequality is equivalent to

$ (a(a \plus{} 1) \plus{} 1)(b(b \plus{} 1) \plus{} 1)(c(c \plus{} 1) \plus{} 1) \geq (a \plus{} 1 \plus{} 1)(b \plus{} 1 \plus{} 1)(c \plus{} 1 \plus{} 1)$

This is true becouse

$ a(a \plus{} 1) \geq b(b \plus{} 1) \geq c(c \plus{} 1)$

$ a \plus{} 1 \geq b \plus{} 1 \geq c \plus{} 1$ and

$ a(a \plus{} 1) \geq a \plus{} 1$

$ a(a \plus{} 1)b(b \plus{} 1) \geq (a \plus{} 1)(b \plus{} 1)$

$ a(a \plus{} 1)b(b \plus{} 1)c(c \plus{} 1) \geq (a \plus{} 1)(b \plus{} 1)(c \plus{} 1)$

Also see here
[url]http://www.mathlinks.ro/viewtopic.php?t=255469[/url]

[quote="arqady"]Let  $ \{a,b,c\}\subset[ k, \plus{} \infty)$ such that $ abc\geq1$ and
$ k$ is a real root of the equation $ x \plus{} \sqrt [3]x \plus{} 1 \equal{} 0.$ Prove that
\[ (a^2 \plus{} a \plus{} 1)(b^2 \plus{} b \plus{} 1)(c^2 \plus{} c \plus{} 1)\geq(a \plus{} 2)(b \plus{} 2)(c \plus{} 2)
\]
I posted this inequality  here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179021
but it's still unsolved.[/quote]
From the given hypothesis, we have
\[ \frac{a^2\plus{}a\plus{}1}{a\plus{}2} \ge \sqrt[3]{a^2}\]
Applied this for the cyclic terms and multiplying them all, we can get the result. :)

[quote="Nak"][quote="arqady"][quote="Nak"]and if $ b,c < 0 < a$ then $ b,c \geq k$ implies that $ |b|,|c| < 0.32 \Longrightarrow a > 9.76 \Longrightarrow \frac {a^2 \plus{} a \plus{} 1}{a \plus{} 2} > 9 \Longrightarrow$
$ \Longrightarrow \frac {a^2 \plus{} a \plus{} 1}{a \plus{} 2}\cdot \frac {b^2 \plus{} b \plus{} 1}{b \plus{} 2}\cdot\frac {c^2 \plus{} c \plus{} 1}{c \plus{} 2} \geq 9 \cdot \frac {3}{8} \cdot \frac {3}{8} > 1$[/quote]
But  $ f(x) \equal{} \frac {x^2 \plus{} x \plus{} 1}{x \plus{} 2}$ is a decreasing function on $ [k, \minus{} 2 \plus{} \sqrt3].$  :wink:[/quote]
and what?[/quote]
Why $ \frac {a^2 \plus{} a \plus{} 1}{a \plus{} 2}\cdot \frac {b^2 \plus{} b \plus{} 1}{b \plus{} 2}\cdot\frac {c^2 \plus{} c \plus{} 1}{c \plus{} 2} \geq 9 \cdot \frac {3}{8} \cdot \frac {3}{8}$ is true?

OK, Nak. I don't see a mistake in your proof.


Prove that if $ x$, $ y$, $ z$ are nonnegative and $ xy\plus{}yz\plus{}zx\equal{}1$ than
$ \frac{1}{x\plus{}y}\plus{}\frac{1}{y\plus{}z}\plus{}\frac{1}{z\plus{}x}\ge\frac{5}{2}$

Use Cauthy:
$ LHS \ge \frac{1}{x+y}+\frac{4}{x+y+2z}\ge \frac{5}{x+y}$
with the condition,we get $ x,y\le 1$
so$ +y\le 2$
so $ LHS\ge \frac{5}{2}$

But we have
$ \frac{4}{x\plus{}y\plus{}2z}\le\frac{4}{x\plus{}y}$

By Jensen  ( $ \frac{\partial ^2\frac{1}{t}}{\partial t^2}>0$ )

$ \frac{1}{x\plus{}z}\plus{}\frac{1}{y\plus{}z}\plus{}\frac{1}{x\plus{}y}\geq \frac{9}{2}\frac{1}{(x\plus{}y\plus{}z)}$

I'm not too good with inequalities. but the rigth hand side of the last have a maximum $ \frac{3 \sqrt{3}}{2}$ that is clearly greater than $ \frac{5}{2}$

can this be used?

[quote="EulerIntegral"]By Jensen  ( $ \frac {\partial ^2\frac {1}{t}}{\partial t^2} > 0$ )

$ \frac {1}{x \plus{} z} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{x \plus{} y}\geq \frac {9}{2}\frac {1}{(x \plus{} y \plus{} z)}$

I'm not too good with inequalities. but the rigth hand side of the last have a maximum $ \frac {3 \sqrt {3}}{2}$ that is clearly greater than $ \frac {5}{2}$

can this be used?[/quote]

you can try $ x \equal{} y \equal{} 1, z \equal{} 0$

well,I made a mistake.So another solution which use Muirhead:
$ \rightarrow \sum xy(\sum\frac{1}{x\plus{}y})^2\ge \frac{25}{4}$
which is equal to:
$ [(5,1,0)\minus{}(4,2,0)]\plus{}3[(5,1,0)\minus{}(3,3,0)]\plus{}(4,1,1)\plus{}14(3,2,1)\plus{}38(2,2,2)\ge 0$

[quote="j555"]Prove that if $ x$, $ y$, $ z$ are nonnegative and $ xy \plus{} yz \plus{} zx \equal{} 1$ than
$ \frac {1}{x \plus{} y} \plus{} \frac {1}{y \plus{} z} \plus{} \frac {1}{z \plus{} x}\ge\frac {5}{2}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=31377
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=101476


inequalities
trigonometry
algebra unsolved
algebra

$ \cos^{2}(\pi x) \minus{} sin^{2}(\pi y) \plus{} 2\sin(\pi x)sin(\pi y)cos(\pi(x\minus{}y))  \leqslant  \frac{1}{2}$

[quote="Iver"]sorry, I didn't add the order:

"Solve the inequality:"[/quote]
that doesn't matter,put $ x\equal{}y\equal{}0$ then your inequality is equivalent to:

$ 1\leq\frac 12$

which is incorrect.


superior algebra
superior algebra solved

well I remember having solved the following problem :

Let A Mn(R), n>=2, invertible. with d=det(A)
a) prove that det(A*)=d^(n-1)
b) proove that (A*)* = (detA)^(n-2).A
c) find out A so that (A*)*=A

this is rather an 11th grade problem! cheers!

yeah.. i remember to that this problem was solved before.. actually i solved it in one of my hout class.

heh, this is exactly the exercise in the algebra textbook which i use.

[quote="Moubinool"]3.Let A  Mn(R), n>=2, invertible. Proove that (A*)* = (detA)^(n-2).A
[/quote]
is the condition "invertible" required here?

A*=d*A^(-1)

so i guess it should be invetible!!!

ok...
If we define A* directly, (not via A<sup>-1</sup>), we shall not need the condition "invertible".

Suppose
      [a11 a12 ... a1n]
      [a21 a22 ... a2n]
A = [.....................]
      [an1 an2 ... ann]

then we define
        [A11 A21 ... An1]
A* = [A12 A22 ... An2]
        [......................]
        [A1n A2n ... Ann]
where Aij is algebraic complement of the aij.

in fact, we can prove that
[code]
           n,   if rank A = n
rank A* =  1,   if rank A = n-1
           0,   if rank A \leq n-2
[/code]


Aside from the obvious (using a Virus or Spyware scanner), I would try going to the "Add or Remove Programs" portion of your Control Panel (this is in terms of Windows XP, I'm not sure what OS you're running) and looking for anything that could be related.  There are some other, more labor-intensive methods of ransacking your computer for a potential source, but I'll leave those to the experts around here.

[quote="Rep123max"]Ya, I have virus stuff. But this is just like some folder that pops up after every time I restart my computer. I have XP and theres nothing in the Add/Remove Programs related (I wish there were, it would make it a lot easier)![/quote]
Remove the shortcut from your startup folder, silly.

It's not necessarily that obvious. A spyware may try to use one of the more esoteric methods for launching a program at system boot-up: registry (there are 4 or 5 places in the registry that you can use to run a program), system.ini (passed down from Windows 3.1), et cetera. I use [url=http://www.mlin.net/StartupMonitor.shtml]StartupMonitor[/url] and [url=http://www.mlin.net/StartupCPL.shtml]Startup Control Panel[/url] to control these programs.

Max. Install Adware. Its free I believe from download.com

Try CWShredder, Spybot - Search and Destroy, Spy Sweeper, or Hijack This!

Reformat the hard drive and install [url=http://linuxiso.org]Linux[/url]

Get a mac.  Or if you don't want to spend $800, then install linux.  Or find a way to have only a unix shell on your computer :D.  If you do any of those 3 things, you probably will never have any adware/spyware problems again since everyone writes viruses for windows.  I recommend the first suggestion most.


calculus
integration
function
inequalities
probability
real analysis
real analysis unsolved

$f(x)$ and $g(x)$ are non-negative measurable increasing functions on $[0,\infty)$. $\varphi(x)$ and $\psi(x)$ are non-negative measurable functions on $[a,b]$. Show that
$
\left(\int_a^b f(\varphi(t))g(\varphi(t))\psi(t) dt\right)\left(\int_a^b \psi(t)dt\right) \geq
\left(\int_a^b f(\varphi(t))\psi(t) dt \right)\left(\int_a^b g(\varphi(t))\psi(t)dt \right).
$

liyi: I wouldn't have used both $\phi$ and $\varphi$ and tried to distinguish between them - they are, after all, the same Greek letter.

But in any case the result appears to be false. Let $[a,b]=[0,1].$ Let $\varphi(t)=1$ and let $\phi(t)=t.$ Then the inequality is that for positive $f$ and $g$,

$\int_0^1f(t)g(t)\,dt\ge\left(\int_0^1f(t)\,dt\right)\left(\int_0^1g(t)\,dt\right).$

If we think of $f$ and $g$ as being random variables on a probabilty space, this says that the two random variables are always positively correlated. We know that can't be true.

For an example, let $f(t)=t$ and $g(t)=1-t.$ The left hand side of this expression is $\frac12\cdot\frac12=\frac14$ while the right hand side is $\frac16.$ The inequality goes in the other direction (the correlation is negative.)

i'm sorry that i forget $f$ and $g$ are increasing.
Now i have re-stated the problem.

[quote="liyi"]i'm sorry that i forget f and g are increasing. [/quote]
[i]Now[/i] we should be able to show they are positively correlated.


18. What is the maximum number of points of intersection when two circles and three straight lines intersect each other?

I kept drawing, but I only get 15..
[hide="Answer"]17 points[/hide]

Let's approach this theoretically.

The two circles intersect each other at 2 points.

One line can intersect one circle at 2 points.

So... $ 2\plus{}4*3$, but the second line can intersect the 1st, and the third can intersect both.

So... $ 2\plus{}5*3\equal{}17$

yeah drawing it out isn't the way to do these types of problems.

[quote="mathcrazed"]Let's approach this theoretically.

The two circles intersect each other at 2 points.

One line can intersect one circle at 2 points.

So... $ 2 \plus{} 4*3$, but the second line can intersect the 1st, and the third can intersect both.

So... $ 2 \plus{} 5*3 \equal{} 17$[/quote]
Sorry, I am kind of slow. Can you specify what 2, 4 and 3 come from?

So what you do is just do it step by step.

So first, with one circle, there will be no intersection points. Then, add another circle.

So one circle and another circle have 2 intersection points. This is easy to see if you draw it out. Then add a line.

So the line can intersect both circles at 2 points each. So that is 2*2 or 4 points. That added to where the circles intersect is 4 + 2 or 6.

So then, add another line. It too can intersect both circles at two points each, but it can also make an intersection point with the first line. So 2*2 + 1 is 5, then added to what we already have is 5+6 or 11.

Then, the last line also intersects the two circles at two points each, but it can also intersect both of the previous lines at one point. So 2*2 + 1+1 is 6.

and 6 + 11 is 17.

[quote="t0rajir0u"]Let's categorize each of the intersection points.

- $ 3$ of the intersections are lines with lines.  Drawing only the lines, it's easy to see that this is the maximum.
- $ 12$ of the intersections are lines with circles.  Each line intersects each circle twice, and there are three lines and two circles.  Drawing only one line and one circle at a time, it's easy to see that this is the maximum.
- $ 2$ of the intersections are circles with circles.  Drawing only the circles, it's easy to see that this is the maximum.

The problem with drawing the circles and the lines together is that it's very hard to see how everything breaks down into categories.  It's also hard to resist the temptation to symmetry - in general, maximizing the number of intersections requires an asymmetric diagram.  (The lines do not, for example, all intersect one point.)[/quote]
Oh, Okay. I see it now. Thank you.


To be frank, I'm very bad in physics...  :( but i would still like to see what's going on in PhO .. So I would like to see some materials or wtever to see how difficult it is.... anyone can give me a link?

You should also ask USAPhO participant... or the semi-finalist test participant....

since i brought it up, WHO have participated in USAPhO or been to International test???

I made the training camp this year; its pretty helpful doing the past problems on the compadre website and reading the feynman lectures.

feynman lectures??? what is that and where do we get one?

i looked it up in amazon.com... but there are so many of them i dunno which one to get?? timhsieh, any ideas??

xxreddevilzxx, are you sure there are many Fenyman lectures? I believe there are is only Volume I , II or III.

I think that you would want [url=http://www.amazon.com/exec/obidos/tg/detail/-/0201021153/qid=1113741669/sr=8-1/ref=pd_csp_1/102-0609151-1972963?v=glance&s=books&n=507846]this one[/url] or one comparable to it.  It has a description of what it is there.

[quote="timhsieh"]Volume 1 (a great deal of mechanics) and the beginning of volume 2 (electromagnetism) are the most helpful.  Volume 3 is mostly quantum mechanics and other discrete topics that aren't necessary for the olympiad.[/quote]

Are these the kinds of books that people who have done Physics C - Quantum Mechanics or  E&M courses should read??? or is it written also for beginning Physics students??

yay, Yang-mills for me. :)

The Feynman Lectures are not for 1st timers. but the adept AP Physics C student should be able to read it and not have much problems. A basic knolwedge of multivar calculus certainly helps(div, curl, and grad) and line integrals and things like it. There really is no "order" to the topics, they have pseudo-ordering. but each book as stated previously on earlier posts generally seem to concentrate over some specific things. Seriously, Get it. get it now.

thans for the info. i will certainly look into it.


The diagram consists of three nested squares. Find the ratio of the area of the smallest square to the area of the largest square. Express your answer as a common fraction.

[img]http://i96.photobucket.com/albums/l167/adasarathy/nested_rectangles.gif[/img]

[hide="Solution"]Using pythagoras once, we get a side of the middle square as 5 and using it again, we get the side of the smallest square as $\sqrt{3^{2}+2^{2}}=\sqrt{13}$. The side of the largest square is 7 so the ratio of the smallest to the largest square is $\boxed{\frac{13}{49}}$[/hide]

[hide="solution"]The length of a side of the medium square, by the pythagorean theorem, is 5. We subtract 2 from that: 5-2=3. We plug 2 and 3 into the pythagorean theorem:

$\sqrt{2^{2}+3^{3}}=\sqrt{13}$

We want the ratio of the area of the smallest square to the area of the largest square. One side of the larger square is 3+4=7. We calculate the areas: 49,13. The ratio: $\frac{13}{49}$[/hide]


Does anyone know why I'm unable to access either of the NYSML or the NYC ARML websites?

The nycmathteam site seems to have been down for the last few days.  It's possible that the nysml site shares the same host, and so is down for the same reason.

Good luck to everyone at ARML!  I won't be going, but my daughter will be there.

There might be some maintenance going on on the NYC math team site due to the restructuring of NYCIML (just a guess); both websites were constructed/are maintained by Jack Po, so I wouldn't be surprised if they were running off the same server or some-such.

Good luck indeed!  Although, given the time and the time that we're supposed to be at Stuyvesant tomorrow morning (8 AM, for those desperately trying to find out), I would hope all the actual competitors are asleep already :).  (I'll keep my eyes open for your daughter, Ravi.)

You guys are both correct!  Indeed the server was down due to a hard drive crash (and they were both on the same server).

I've recovered the data and now the sites should be really stable (and on two different servers)!


In the figure below, Triangles RTS and UTV are congruent, the measure of angle R is 36 degrees and the measure  of Angle T is 42 degrees. FInd the number of degrees in the measure of angle RQV.

[hide]Since $\angle R$ is an inscribed angle, arc $VS= \frac 12 \angle R = 18$. Since $\angle T$ is an angle formed by two secants, $m \angle T = \frac 12 (RU-VS)$. Substituting, $84=RU-18$ So $RU = 102$. Since $\angle V$ cuts off arc $RU$, it's measure is half of $RU$. We already determined that $RU$ is 102, so $m \angle V = 51$. So $m \angle RQV = 180 - m \angle V - m \angle R$. Substituting, $m \angle RQV = 180-51-36=\boxed{93}$[/hide]

That's not what I got, but of course I could be wrong. I solved it a much simpler way. [hide="hint"] Angle UVT is the same as  RST.[/hide]

Not sure how he did it, but here's my method:

[hide]Since Triangles RTS and UTV are congruent, angle R is congruent to angle U (by the CPCTC conjecture).
So angle U is also 36 degrees. We know that angle T is 42 degrees, so angle UVT must equal 102 degrees, for triangle UTV to have angles summing up to 180 degrees.
Similarly, angle RST must also equal 102 degrees for the same reason.

There, you have a quadrilateral SQVT. Angle T is 42, Angle V is 102, Angle S is 102, so angle Q must be 114 degrees for the quadrilateral have its angles summing up to 360 degrees.

Note that this is not the right angle Q we're looking for though. This is angle VQS, and we are looking for angle RQV.
However, angle RQV and angle VQS are a linear pair. Since a linear pair of angles is supplementarly, and angle VQS is 114 degrees, we can see that angle RQV is $\boxed {66}$ degrees.[/hide]

[hide]$m\angle RST=180-36-42=102$ because it and the given angles make a triangle.
Because triangle UVT is congruent to triangle RST, $m\angle UVT=102$.
$m\angle RVQ=180-102=78$ because they are a linear pair.
$m\angle RQV=180-36-78=66$ because they are all part of the same triangle.[/hide]

hmmm can someone see where I went wrong in my solution

I don't really understand arc stuff in general, but I do know you went wrong somewhere with that. Maybe you assumed something that you can't assume.

The arc should be twice the size of the angle, not half.  For example, an an arc of 90 degrees yields an inscribed angle of 45 degrees.

[hide]
36+42=78, to get angle S (or V, their congruent) you do 180-78=102
Since the sum of degrees in a quadrilateral is 360, you do 102+102+42=114
Next, to get angle Q, (since Q and the angle that's 114 degrees are on the same line) you do 180-114=66[/hide]


I need to choose one of the following to study from:

Organic chemistry by Brown/Foote
Organic chemistry by Seyhan Ege

Do any of you know anything about the quality of either?

I've never used the Ege book, but I can speak for Brown and Foote.  I used that for my first orgo course, and I absolutely loved the book.  Read it cover to cover, in fact.  Material is well-organized and logically presented.


analytic geometry
geometry
3D geometry
sphere
AMC

Consider two solid spherical balls, one centered at $(0,0,\frac{21}{2})$ with radius $6$, and the other centered at $(0,0,1)$ with radius $\frac{9}{2}$. How many points $(x,y,z)$ with only integer coordinates (lattice points) are there in the intersection of the balls?

$\text{(A)} \ 7 \qquad \text{(B)} \ 9 \qquad \text{(C)} \ 11 \qquad \text{(D)} \ 13 \qquad \text{(E)} \ 15$

[hide="incomplete"]
All I know is that it is either B or D. If you draw this from the side you can see that these spheres surround (0,0,5). Since those are the extreme points of the spheres, all lattice points in question must lie on the same horizontal plane. If viewed from the top, the center point (0,0,5) is certainly part of this amount, and I can eliminate A C and E because the spheres are radially symmetrical and if point (0,1,5) is included, then so is (1,0,5), (0,-1,5) and (-1,0,5). That means that the answer must be of the form 4n+1.

Given the choices I can tell that (1,1,5), (-1,1,5), (-1,-1,5) and (1,-1,5) are also included which brings the total to 9. The only question is whether the point (2,0,5) and its counterparts are included. If so then the answer is D, if not, then B.

Can someone explain how to find out?[/hide]

[hide]From the first sphere $z\geq \frac{9}{2}$ and from the second, $z\leq \frac{11}{2}$. The integer integer $z$ in this interval is 5.

The lattice points will all lie inside a circle centered at (0,0,5) and on the plane $z=5$. For the first sphere, the intersection with $z=5$ gives a circle of radius $\sqrt{36-\frac{121}{4}}=\frac{\sqrt{23}}{2}$. For the second, the circle has radius $\sqrt{\frac{81}{4}-16}=\frac{\sqrt{17}}{2}$.

The intersection of the circles has radius $\frac{\sqrt{17}}{2}$, so the lattice points satisfy $z=5$ and $x^2+y^2\leq \frac{17}{4}$. But $x$ and $y$ are integers so $x^2+y^2\leq 4$.

The solutions for $x$ and $y$ are given by:

(i) $x=y=0$ (1)
(ii) one of $x$ and $y$ is 0, the other is $\pm 1$ (4)
(iii) $|x|=|y|=1$ (4)
(iv) one of $x$ and $y$ is 0, the other is $\pm 2$ (4)

So there are 13 points $\Rightarrow \boxed{D}$.[/hide]


Obviously, the counting and number theory chapters will help.  Buy the intro books if you struggle with specific subjects.

All the sections, ideally. You have time, and it's better to look at the whole thing to improve your problem solving, and then work on speed for MC in particular.

[quote="mathemagician1729"]What chapters or sections are the most helpful for MATHCOUNTS.

I get about 35~38 on chapter tests and 24~28 on state tests.

I'm strongest in Algebra and Geometry but am terrible at Counting and Number Theory.

Thanks in Advance[/quote]

Eh, sorry.

[quote="ernie"]
If you don't want to buy C&P, use Alcumus.[/quote]

Posts like this are the best way to get Alcumus taken away or make AoPS decide to make you pay to use it.


If $ \frac{x}{y}\equal{}\frac{3}{5}$ then compute $ \frac{5x\plus{}7y}{9x\minus{}5y}.$

$ \frac{5x\plus{}7y}{9x\minus{}5y}\equal{}a, \qquad a\equal{}\frac{5}{9}\plus{}\frac{8.8y}{9x\minus{}5y}\qquad \frac{1}{a\minus{}\frac{5}{9}}\equal{}\minus{}\frac{5}{8.8}\plus{}\frac{9}{8.8}\cdot\frac{x}{y}\equal{}\frac{52}{8.8}$. Now you only have to calculate the valoue of a.

[hide="Cheating"]
$ x\equal{}3$, $ y\equal{}5$.

$ \frac{5x\plus{}7y}{9x\minus{}5y}\equal{}\frac{15\plus{}35}{27\minus{}25}\equal{}\boxed{25}$.
[/hide]


Prove that every positive integer is uniquely expressible in the form: $2^{j_{0}}+2^{j_{1}}+2^{j_{2}}+...+2^{j_{m}}$ where $m \geq 0$ and $0 \leq j_{0}< j_{1}< j_{2}< ... < j_{m}$

[mod.: bigger edit]

I strongly recommend using the Preview button!

And before I edit the post just what it really says: show that binary representation of positive integers exists and is unique.
[and moved to Intermediate]

[hide="hint"] Assume for the sake of contradiction that

$2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{n}}= 2^{b_{1}}+2^{b_{2}}+\cdots+2^{b_{k}}$,

where all $a_{i}$ and $b_{i}$ are distinct. Choose $\max{(a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{k})}$... [/hide]

[hide="One line"] $(1+x)(1+x^{2})(1+x^{4})...(1+x^{2^{n}})... = 1+x+x^{2}+x^{3}+...+x^{m}...$

[hide="To see why this is true..."] Multiply both sides by $1-x$.

[hide="And to see why this implies unique binary representation..."] Every integer $a$ has a single coefficient $x^{a}$ on the RHS derived from multiplying out powers of $2$ on the LHS. [/hide] [/hide] [/hide]

rawr at one line solutions...you should explain stuff more

obviously this is asking if binary represents integers uniquely...

we can show this by using an algorithm to solve it

ex. take N, mod 2, then whatever that is is teh first digit, subtract that, and divide by 2, to get N_1

rinse and repeat

this process stops because we are dealing with finite numbers
[anybody have a better proof that this always works...i don't know loop invariant stuff, etc]


logarithms
induction
inequalities
number theory unsolved
number theory

Let $ S_n$ denote the $ n^{th}$ partial prime sum. That is $ S_5 \equal{} 2\plus{}3\plus{}5\plus{}7\plus{}11$

Prove there is always a square between two consecutive prime partial sums.

That is prove for some integer $ k$, $ S_n < k^2 < S_{n\plus{}1}$.

This is true for small values of $ n$ because $ p_n \approx 2n\plus{}1$ there and it's true asymptotically because $ p_n \approx n \ln n \implies \frac{p_n}{ \sqrt{S_n} } \approx \frac{n \ln n}{n \sqrt{\ln n}} \equal{} \sqrt{\ln n}$, but I don't see a theoretical reason why this might not fail for some moderate $ n$.

What range would you consider "small"? Hundreds, billions? It seems from the $ 2n\plus{}1$ its just up to around $ n \equal{} 20$ ish. Or double digit $ n$s.

[quote="Fermat's Little Turtle"]Can anyone prove this? :([/quote]
Just let $ Q_n$ be the number such that $ S_n < (Q_n)^2 < S_{n\plus{}1}$, and prove that $ S_{n\plus{}2}\minus{}S_{n\plus{}1} > (Q_{n\plus{}1})^2 \minus{} Q_n$ by induction.. That is how i proved some time ago. (I can post full solution some other day, but im in a hurry)

$ S_1 < 2^2 < S_2 < 3^2 < S_3 < 4^2 < S_4$
assume that the statement is true up to some $ n$(grater than 4). let $ x$ be the largest such natural number that $ S_{n \minus{} 1} < x^2\leq S_n$ than by definition $ (x \plus{} 1)^2 > S_n$. we shall prove that $ (x \plus{} 1)^2 < S_{n \plus{} 1}$. and with this the induction is complete.

$ (x \plus{} 1)^2 < S_{n \plus{} 1}$ is equivalent to $ x^2 \plus{} 2x \plus{} 1 < S_n \plus{} p_{n \plus{} 1}$ where $ p_{n \plus{} 1}$ is (n+1)th prime number. thus it sufices to prove that $ 2x \plus{} 1 < p_{n \plus{} 1}$
this is further equivalent to $ 4x^2 < p_{n \plus{} 1}^2 \minus{} 2p_{n \plus{} 1} \plus{} 1$

the upper inequality holds if thisone does (because $ x^2 < S_n$)
$ 4S_n < p_{n \plus{} 1}^2 \minus{} 2p_{n \plus{} 1} \plus{} 1$
it's true for $ n \equal{} 4$. assume that it's true for integers from $ 4$ up to some $ m$. than if we go to $ m \plus{} 1$ the LHS increases for $ 4p_{m \plus{} 1}$ and RHS increases for $ 2(p_{m \plus{} 2} \minus{} p_{m \plus{} 1})p_{m \plus{} 1} \plus{} (p_{m \plus{} 2} \minus{} p_{m \plus{} 1})^2 \minus{} 2(p_{m \plus{} 2} \minus{} p_{m \plus{} 1})$ and because $ p_{m \plus{} 2} \minus{} p_{m \plus{} 1}\geq2$ we have
$ 2(p_{m \plus{} 2} \minus{} p_{m \plus{} 1})p_{m \plus{} 1} \plus{} (p_{m \plus{} 2} \minus{} p_{m \plus{} 1})^2 \minus{} 2(p_{m \plus{} 2} \minus{} p_{m \plus{} 1})\geq 4p_{m \plus{} 1}$
whis means that $ 4S_n < p_{n \plus{} 1}^2 \minus{} 2p_{n \plus{} 1} \plus{} 1$ is true for every $ n > 3$ consequently that $ 2x \plus{} 1 < p_{n \plus{} 1}$ and $ (x \plus{} 1)^2 < S_{n \plus{} 1}$ are true. this completes the starting induction. Q.E.D


function
quadratics
algebra
algebra proposed

Find all function $f: \mathbb R-{0}\to \mathbb R-{0}$ such that: \[f\left(x^{4}+{1 \over x^{4}}\right) = x^{3}+{1 \over x^{3}}.\] The problem seems to be trivial, but I have no nice solution for it.
I wonder if the general problem is possible? \[f\left(x^{p}+{1 \over x^{p}}\right) = x^{q}+{1 \over x^{q}}\ , \quad p,q \in \mathbb R.\] Sorry for my stupidness!

Put $y=x^{p}+{1\over x^{p}}$ and solve as a quadratic in $x^{p}$ to express $x$. Then you get $f(y)=\mathrm{whatever}$

hallo,
but what if the solutions of the quadratic equation are negative (or at least one of them)? we get a complex valued function . :)


Okay, since nobody answered the Round 3 i'll make a new one.  This time anybody can PM me the answers.
Here it is:

1.   There are 6 black socks, 14 green socks and 8 blue socks in a drawer.  What is the smallest number of scoks that must ve taken ouit, all at once, in order to be sure to get a matching pair (same color).

2.  If $x$ and $y$ are natural numbers and $19x+97y=1997$ then the smallest value of $x+y$ is what?

        A) 21
        B)23
        C)38
        D) 41

3.  If $a+b=5$ and $ab=2$, then $a^4+b^4$ is equal to what integer?

4.  $2^n - 2^{n-1}=$

        A)$2^{n-1}$
        B)$2^{2n-1}$
        C) $2^{2n}$
        D)$2$

5. What number does:

    $\frac{1996(1997^2-9)1998}{2000(1997^2-1)}$

equal?

            A)1994
            B)1995
            C) 1996
            D)1997
            E)1998
------------------------------------------------------------------
Anyone can answer, PM me the answers two weeks from today (sunday, July 10, is today)

Wait, so should i PM the answers anytime from today to two weeks from today or just two weeks from today?

any time.  THe deadline is two weeks from today.

These people have sent me a pm:
 JesusFreak197
 nat mc
 danne89
And they all got 5/5

no, calculators are not allowed.  I should have stated that earlier.

New results:
JesusFreak197
nat mc
danne89
G-UNIT
Xantos C. Guin
colts18
all got 5/5

Anyone else wanna try?

yay, mysmartmouth tried it...he/she got 5/5...i have decided that there will be a round 4 :lol:

[quote="math92"][quote="Xantos C. Guin"]What happened to the old Round 3?[/quote]
you won, remember?
nobody else answered the problems exept you.[/quote]

I ve solved round 3 also  :mad:
Cant wait for round 4  :P

i will try to make round 4 tommorow. :lol:

[quote="math92"][quote="Xantos C. Guin"]What happened to the old Round 3?[/quote]
you won, remember?
nobody else answered the problems exept you.[/quote]

Oh Yeah.

lol :rotfl:
Round 4 will be posted in the next hour. :)

sorry about that, you can participate in Round 4. :)

I'm about to do Round 3 and Round 4 as fast as I can. :)

[quote="math92"]well, it has been half an hour since you posted that. And you got Round 3 done and its all correct.[/quote]

Actually, I PMed you my answers at 3:23PM, which included typing them up. So I finished in less than 9 minutes. :)

oh, i wasn't sure how you did it. But whoa you are fast


geometry
congruent triangles
geometry solved

We have a quadrilateral ABCD with AD=BC. Two points K on BC and L on AD such that
LD=BK. Prove that the midpoints of the diagonals AC and BD and the midpoint of LK
are in the same line. A solution soon please without vectors.

Since the points L and K lie on the equal segments DA and BC and have the same distance from their corresponding endpoints D and B (in fact, LD = BK), you can consider the triangles DLA and BKC as two congruent degenerate triangles. And thus your problem is a particular case of the problem proposed at http://www.mathlinks.ro/Forum/viewtopic.php?t=20260 , applied to the (self-intersecting, but this doesn't matter) quadrilateral DABC with DA = BC and the two congruent triangles DLA and BKC constructed on its sides DA and BC.

  Darij

I do not understand your solution. Wright a solution only at Clasical Geometry.Please

Your problem is just a trivial consequence of the 5th problem of the IMO2005.
Consider The point of indersection of the line EF and the diagonals, and let P the intesecton of the diagonals. They define a triangle, let it be T.
Then the midpoints of the segments you are talking about define the simpson line of the triangle T, with respect to the intersection point of the perpedicular bisectors of the diagonals.

For a better explanation you may read the solutions given for the 5th problem in the forum. (Read grobber's proof.)

Dear silouan, Darij's proof is a very nice one, so it's better for you to understand it.
You can learn a lot by reading carefully others' solutions. ;)


limit
geometry
geometric transformation
calculus
calculus computations

Find the limit: x/sqrt(x^2-9) when x approaches to -3 from LHS
I tried using Squeze theorem but it doesn't work for me(just give me an upper bound). Is it solvable with "the precise definition of limit"?I didn't know how to do it in that method since I can't guess the limit. :(

$ \lim_{x\to\minus{}3^ \minus{} }\frac {x}{\sqrt {x^2 \minus{} 9}}$ - or at least that's how I read it.

There's not a whole lot to squeeze here, since the limit is $ \minus{} \infty.$

Unless it's a different problem. $ \lim_{x\to\minus{}3^ \minus{} }\frac {x \plus{} 3}{\sqrt {x^2 \minus{} 9}}$ would be different.

[quote="Kent Merryfield"]$ \lim_{x\to \minus{} 3^ \minus{} }\frac {x}{\sqrt {x^2 \minus{} 9}}$ - or at least that's how I read it.

There's not a whole lot to squeeze here, since the limit is $ \minus{} \infty.$
[/quote]
Your translation into formula is correct, Prof.Merryfield. But can you show me how to prove that it is negative infinity? Thank you in advance. :)

Well, to put it sloppily, $ \frac { \minus{} 3}{0^{ \plus{} }} \equal{} \minus{} \infty$

The denominator gives division by 0, but the square root guarantees the denominator is also $ > 0$, thus we know in this case $ \lim_{h \rightarrow 0} \frac {a}{h} \equal{} \infty$ for $ a > 0$

So essentially we're computing $ \minus{}\lim_{h \rightarrow 0} \frac {a}{h}$ with $ a \equal{} 3$


conics
ellipse
analytic geometry
geometry proposed
geometry

is it possible to solve this pb without a lot of calculation ?.

let an ellipse E with center O. Consider all lines (PQ) so that P,Q are 2 points on E and angle(POQ)=Pi/2.
What is the curve wich is tangent to all lines (PQ) ??

the ellipse x^2/a^2+y^2/b^2=1 can be transformed to
cos 2 \theta/a2 + sin 2 \theta/a2 = 1/\rho2

suppose the coordinates of P is (p1, \alpha) and Q(p2, \alpha+\pi/2). thus,

1/p12 = cos 2 \alpha /a2 + sin 2 \alpha/a2

and

1/p22 = cos 2 (\alpha+\pi/2) /a2 + sin 2 (\alpha+\pi/2)/a2

hence

1/p12+1/p22
= 1/a2+1/b2

i.e., p12p22/(p12+p22)=a2b2 / (a2+b2)

so the length of the altitude on the side of PQ in the right triangle QOP is
d = |OQ||OP| / \sqrt(|OP|2+|OQ|2) = p1p2/ \sqrt(p12+p22) = \sqrt(a2b2 / (a2+b2)) = r is a constant. this implies there is a circle with radius r tangents to the chord PQ.


Given that sq. rt of A + 1/(sq rt of A)=3 assuming that a is not equal to 0

what is the value of A-1/A=?

please help me, i don't know how and where to start. .
Thank you :)

Factor $ A\minus{}\frac{1}{A}$ to get $ \left(\sqrt{A}\plus{}\frac{1}{\sqrt{A}}\right)\left(\sqrt{A}\minus{}\frac{1}{\sqrt{A}}\right)$.

Now you are to find the value of $ \sqrt{A}\minus{}\frac{1}{\sqrt{A}}$.

Good luck! :)

thank you but there is no given value for sq rt of A minus 1/(sq rt of A)  :(

well i really never have thought of that hehe Like Kunny, i am thinking of difference of two squares or something. .  well anyway thank you so much. .  :)

Easier solution:

$ \left(\sqrt {A} \plus{} \frac {1}{\sqrt {A}}\right)^2 \equal{} 3^2 \\
A \plus{} \frac {1}{A} \equal{} 7.$

EDIT: Sorry, wrong quantity.

[quote="picture74"]$ \sqrt {A} \plus{} \frac {1}{\sqrt {A}} \equal{} 3$

$ \Longrightarrow A^2 \minus{} 7A \plus{} 1 \equal{} 0$

$ A \equal{} \frac {7\pm3\sqrt {5}}{2}$

Hence, $ |A \minus{} \frac {1}{A}| \equal{} 3\sqrt {5}$[/quote]

That's correct answer. :)

@marc1217

$ \left(\sqrt{A}\minus{}\frac{1}{\sqrt{A}}\right)^2\equal{}\left(\sqrt{A}\plus{}\frac{1}{\sqrt{A}}\right)^2\minus{}4$.

Errr mr. Kunny sir. . I can't seem to think how did you come up with that equation???    :huh:

\[ \left(x\plus{}\frac{1}{x}\right)^2\equal{}x^2\plus{}\frac{1}{x^2}\plus{}2x\cdot\frac{1}{x}\equal{}x^2\plus{}\frac{1}{x^2}\plus{}2\] and \[ \left(x\minus{}\frac{1}{x}\right)^2\equal{}x^2\plus{}\frac{1}{x^2}\minus{}2x\cdot\frac{1}{x}\equal{}x^2\plus{}\frac{1}{x^2}\minus{}2.\] You should remember these along with \[ \left(x\plus{}\frac{1}{x}\right)^3\equal{}x^3\plus{}\frac{1}{x^3}\plus{}3\left(x\plus{}\frac{1}{x}\right)\] and \[ \left(x\minus{}\frac{1}{x}\right)^3\equal{}x^3\minus{}\frac{1}{x^3}\minus{}3\left(x\minus{}\frac{1}{x}\right).\]

Here, $ x$ can be anything. Right now $ x\equal{}\sqrt{a}.$

[hide="Using kunnys first post here..."]
\[ a\minus{}\frac{1}{a}\equal{}\left(\sqrt{a}\plus{}\frac{1}{\sqrt{a}}\right)\left(\sqrt{a}\minus{}\frac{1}{\sqrt{a}}\right).\] We already know  \[ \sqrt{a}\plus{}\frac{1}{\sqrt{a}}\equal{}3,\] so \[ \left(\sqrt{a}\plus{}\frac{1}{\sqrt{a}}\right)\equal{}9\] and \[ \left(\sqrt{a}\minus{}\frac{1}{\sqrt{a}}\right)^2\equal{}9\minus{}4\equal{}5.\] That means \[ \sqrt{a}\minus{}\frac{1}{\sqrt{a}}\equal{}\sqrt{5}\] and finally \[ a\minus{}\frac{1}{a}\equal{}\boxed{3\sqrt{5}}.\][/hide]


In 2009, they will stop broadcasting to antennae. How many of you still use antennae?

I myself still use antenna. I wouldn't care THAT much if they stopped broadcasting to antennae, since I can still play video games (99 percent of the time the TV is on, it's for a video game) and watch DVDs.

i actually will like that, because we have a random chinese channel on antennae (which only broadcasts news and which i cant understand) but the cool channels are on cable. so w00t!

who cares about tv news? internet news doesn't have commercials that say "Buy this product because I do" or "This hot girl is doing something! Buy this product!"

[quote="mewto55555"]TV is the opiate of the masses...My family's TV does not have an antenna (although I wish it would stop working, how many times can one little girl watch Hannah Montana reruns?)[/quote]

1) That would make me insane after about 2 episodes
2) Sabotage the antenna and make it look like product failure ie. cause a power surge to the atenna or if your desperate throw water on it.
3) Does anyone have the "box" thing that your suppose to get if you have antenna TV so it still works?

Hmm...I heard the government was only doing this because they wanted to free up some of the airways to make more money. Is that true?


function
number theory unsolved
number theory

Let $ (a_n)_{n \geq 1}$ be a sequence of positive integers satisfying $ a_{n\plus{}1}\equal{}a_n\plus{}p[^p\sqrt{a_n}]$ for all $ n \geq 1$.Prove that $ ^{p}\sqrt{a_k}$ is an integer for some $ k \in N^*$.

Oh,I am sorry for not mentioning about $ p$.$ p$ is a prime number.The problem is clear now.Sorry for my blunder.


logarithms
calculus
calculus computations

I didnt solve :(
m=3^n
n=lnp so;
 dm/dp = ?

a) (3^lnp)/p

b) ln3/p

c) (3^lnp).ln3/p

d) (3^n+p).ln3

e) p.log3
Can you show me your solution way?
Thanks and Best regards!

[hide="Solution."]
$ m \equal{} 3^n \implies \frac {dm}{dn} \equal{} 3^n\cdot \ln 3$

$ n \equal{} \ln p \implies \frac {dn}{dp} \equal{} \frac {1}{p}$

Thus:
$ \frac {dm}{dp} \equal{} \frac {dm}{dn}\cdot \frac {dn}{dp} \equal{} (3^n\cdot \ln 3)\left(\frac {1}{p}\right) \equal{} \frac {(3^{\ln p})\cdot \ln 3}{p} \implies \boxed{C}$.
[/hide]
This does not belong in this forum, by the way. Calculus questions should be posted [url=http://www.artofproblemsolving.com/Forum/index.php?f=66]here[/url].