[quote="madskillz1"]I got 1/49  :maybe:[/quote]

nah, cheese is right. why dont you post your solution so we can see where you went wrong.

[quote="Art of Owna"]There are 50 people in a class, including John and Bob.  When the teachers randomly pair people for a project what are the chances that John and Bob are partners?[/quote]

[hide="hmmmmmm"]We know that there can be $\dbinom{50}{2}$ possibilities for people to be paired together. For John and Bob to be together, there is only $1$ possibility. Therefore, the answer is $\frac{1}{\dbinom{50}{2}}=\boxed{\frac{1}{1225}}$.[/hide]

[quote="madskillz1"]ok, if you have aops vol. 1 you can look at number 486 as an example,  it is a similar type of problem and the answer is 1/19, as it is just 1*1/19 so its 1*1/49 in this case[/quote]

someone stole my book.  but that doesnt sound like this situation...
can you post the problem?

A teacher with a math class of 20 students randomly pairs the students to take a test.  what is the probabilit that Camilla and Cameron, two students in the class, are paired with each other (MATHCOUNTS 1991)- the answer is 1/19

I believe what you are doing is this, lets say theres another student A if Bob and John are together Bob and A cannot be together because Bob and John are already together so your overcounting the denominator-other tahn that I cant see what the problem is

yes. I guess I'll have to agree with madskillz once he posts a sol.

How can it be 1/19? That doesnt even make sense, for that its 1/190 and for the first one its not 1/49.. you can list out the posibilities

Think of it this way:

Camilla has 19 parters to choose from, and Cameron is one of them.  Each partner is equally likely.  Hence the probability, $\frac{1}{19}$

[hide]
Total number of ways two kids can be chosen is
${50\choose 2}$.
Therefore the answer is
$\frac{1}{1225}$[/hide]

IF the problem said that only [b]one[/b] pair was chosen out of 20 and what is the prob that both are in the pair, then your sol. would be correct.  However, that is a different problem.

Read madskillz1's last post.

[quote="CheeseIsGood"]I am pretty bad at probability... is it
[hide]1/1225?
The denominator is C(50,2) (is that how you write it? :blush: ), and since there is only 1 way they can be together, the numerator is 1.
I can't learn LaTeX, no matter how hard I try. :huh: Actually, I don't try very hard.
[/hide]
I am a noob at math. :([/quote]

I'm a noob too, probably worse than you, so don't worry. :D

No, its 1/49 because everyone is chosen for a partner.  Let's examine a easier case with 4 people A B C D

There are 3 choices
AB CD
AC BD
AD BC

I want A and B together so thats 1/3 chance, not 1/(4C2) it is 1*(1/3)

[quote="egghead91"]isn't it
[hide]1/2450? 1/50*1/49?[/hide][/quote]

No, since it doesn't matter what the order of the people in the pair is, you have to multiply by two to get 1/1225, which other people are getting.

However, that answer is wrong!

Like madskills said, once Bob is paired with somebody he can't be paired with someone else.  There are 1225 possible combinations for [i]one[/i] pair, but only 49 possible pairings for the [i]whole class[/i].  [/i]

[quote="jasonwingofcarmelin"]Wait.. I dont get it why isnt it like this

There are 20C2 different pairs
ONly 1 pair contains both Camille and Cameron so its 1/190?[/quote]
there are 20C2 different pairs, but they cannot be all the pairs in this case. If person A and B are a pair, person A and C can't be a pair.

I hope this is team or target so I can use my calculater! Dpatricks counting class really covered this stuff well and I had a blast with it. I highly recomend it for anyone taking mathcounts.
[hide]Probability is ways to "win" over the total ways. There is 1 way to "win" (where bob and john are together) yet there are 50 choose 2 ways total. The answer is 1/(50!/2!*48!)[/hide]
Edit: Forgot to hide my work agian! *hide*

[quote="ProtestanT"][quote="nicholaskramer"]I hope this is team or target so I can use my calculater! Dpatricks counting class really covered this stuff well and I had a blast with it. I highly recomend it for anyone taking mathcounts.
Probability is ways to "win" over the total ways. There is 1 way to "win" (where bob and john are together) yet there are 50 choose 2 ways total. The answer is 1/(50!/2!*48!)[/quote]
but there is one hundred percent chance that both of them will be in at least one pair, because there are more than one pairs being picked[/quote] The problem isnt asking what the probability they both will be in a pair is. The question asks what is the chance they are paired TOGETHER is.

so much for ur counting class, everyone in the class is paired so its 1/19 if it said they chose 1 pair from all the possible pairs, then ur right, but its not asking that

[quote="nicholaskramer"]RichardNi: At the time of speaking you have one post and you joined "earlier today". I am under the impresion you make this account merely to flam me out. If you would read the past posts the majority of the AoPSers agree with my answer. the answer for the AoPS vol.1 problem is, I agree, 1/19 yet the answer to the problem this post is dedicated to is 1/(50!/2!*48!)

Welcome to the AoPS community and I wish you made a better first impression than saying my counting class was useless.

The goal of AoPS is to help students learn in a friendly and helpful community. The way to make your point wouldn't be to say that you don't think I learned anything in C&P but to say "nicholas, I disagree with your answer. This is how I got my answer *detailed solution here* . Why is there a disagreement in our results?"[/quote]
Nicholas, you understand that the teacher puts the [i]whole class[/i] into pairs, right?  As I said before, Bob has 49 partners to choose from, so our probablilty is 1/49.

[quote="nicholaskramer"]Stupid me!Everyone who used the combinitorics way (including me) is thinking too hard (or not enough) about the problem. The combinatorists are saying if the teacher makes 1 pair only (we chose 1 pair from 50 people that makes 1 pair only. He can get paired with john and the rest of the class be situated differetly. I still think everyone has been wrong. There is more than 1 way to win. As someone once said if you think to hard about a problem you can nolonger solve it. I am currently rendered useless. Hope this helps![/quote]
Yes!  Yes!  Yes!
If people still aren't convinced that it's 1/49, here is [i]another[/i] way to look at it:
Say there are 4 dudes in the class.  List out the pairs.
AB CD
AC BD
AD BC
1 pair has AB together, so the probablilty is 1 out of 3.
Say 6 kids:
AB CD EF
AB CE DF
AB CF DE
AC BD EF
AC BE DF
AC BF DE
AD ...
AD...
AD...
AE...
AE...
AE...
AF...
AF...
AF...
3 ways out of 15 = 1/5.

n kids :  1/(n-1).
Which is what I said earlier.
The End.

Ya archimedes has been right all along.. its supposed to be really easy if you read it correctly(and have good common sense)

[quote="vishalarul"]There have been many posts.

Could someone clarify what the question would read is if the answer is $\frac{1}{1225}$?[/quote]

There are 50 students in the class, including John and Bob.  The teacher chooses two students for a special project.  What is the probability that the two students he chooses are John and Bob?

[quote]
What about $\frac{1}{49}$? (Should I assume that this is the answer to the current problem?)[/quote]

There are 50 people in a class, including John and Bob. When the teachers randomly pair up everyone in the class for a project, what are the chances that John and Bob are partners?  (i.e., the current problem)

[quote]Please explain.[/quote]

[hide="my best shot at explaining"]

In the first problem, there are $\dbinom{50}{2}$ possible pairs, and only one is going to be made.  The probability that it is the required one is thus $\frac{1}{\dbinom{50}{2}}$

In the second problem 25 pairs are going to be made.  John and Bob don't have to be the "first" pair.  Once Suzy and Mary are paired, for example, neither John nor Bob is at "risk" of being paired with either Suzy or Mary, so the chances of a "bad" outcome go down.

The easy way to think about this is as Archimedes1 suggested: out of the 49 people who can be John's partner, only one of them is Bob.

If really you want to do it combinatorially, I think the following works:

There are $\dbinom{50}{2}$ ways of choosing the first pair.  Then there are $\dbinom{48}{2}$ ways of choosing the second pair.  And so on....  Since the order doesn't matter, once we've done that, we need to divide by $25!$ to account for all the different orders the 25 pairs could have been chosen in.

So... Let N represent the total number of ways the class can be paired up:

$N = \frac{\dbinom{50}{2}\dbinom{48}{2}\dbinom{46}{2}... \dbinom{2}{2}}{25!}$

Similarly, let M represent the total number of ways the class can be paired up in which Bob and John are partners.  So, take them out, and do the same thing with the 48 remaining students:

$M = \frac{\dbinom{48}{2}\dbinom{46}{2}... \dbinom{2}{2}}{24!}$

The required probability is $\frac{M}{N}$.  Most everything cancels out leaving:

$\frac{M}{N}= \frac{25}{\dbinom{50}{2}}= \frac{1}{49}$

Now... don't you like Archimedes1's way better!?!

Frost13 suggested a simpler route by saying that since 25 of the 1225 pairs were going to be selected, you could immediately say that the probability is $\frac{25}{1225}$ but I don't really follow that line of reasoning, since not all of the 1225 pairs are possible at the same time.   Which is why I used my more convoluted reasoning above which clearly accounts for the number of choices left at each stage, but which in the end comes out with the same simple answer.  :maybe:
[/hide]

[quote="vishalarul"]There have been many posts.

Could someone clarify what the question would read is if the answer is $\frac{1}{1225}$?

What about $\frac{1}{49}$? (Should I assume that this is the answer to the current problem?)

Please explain.

(I really need help on probability for MathCounts)[/quote]

[hide]Mr. Frost's explanation is simple and elegant.

Question 1: When selecting 2 students from a group of 50, what is the probability that they will be John and Bob?

Question 2:  A class of 50 students are being split into groups of 2 students. What is the probability that John and Bob will be in the same group?

To answer the first question, you have to basically ask yourself, "how many ways to select 2 people from a group of 50 people?" The answer is $\binom{50}{2}$. So the desired probability is $\frac{1}{\binom{50}{2}}= \frac{1}{1225}$.

To answer the second question, you need to look at it from, say, John's perspective and ask the question, "how many different people I can be paired with?" The answer is 49. So the desired probability is $\frac{1}{49}$.

I think the key thing is to clearly understand what is being asked. Otherwise we may likely misinterpret the question due to time pressure at the contest and end up solving the wrong problem![/hide]

[quote="frost13"]I love these problems.
[hide="my thoughts"]
While there are 1225 groups, remember that 25 of those groups are being formed. Or 25 out of the 1225 groups or 1/49 of all possible groups.

The easier logic is that John needs a partner and there are 49 people to pair him with.  The probability is 1/49 he gets Bob.  The question is slightly vague though.  If only 2 people are being chosen for a project then it is 1/1225.  If the whole class is being broken into pairs then it is 1/49[/hide][/quote]

I never thought of that.

Yes, I think the answer is $1/49$

You can choose 49 partners for John, and only one is Bob.

1225 is the number of total combos, including different ones in which Bob and Joe are together.

So the answer [i]is[/i] 1 over 49?  :wow: I get it now. I should keep that formula mentioned by archimedes in mind if i get a similar problem next year-- i seem to have issues with probability. :(

[quote][quote]Fanatic Fanatic here:

richardni is NOT me, he's an imposter

/clarification

thx for the pm bpms.

Could you please IP ban him? I know that's illogical for a first offense, but I think he's either johnz, who's been stalking me, or imaretard reicarnated (he spammed earlier this year and seemed to know who I was)[quote][/quote][/quote][/quote]

Go ahread ban the ip for richard ni.. because im not him.  But what is this about stalking you.  You wrong Im in inforat, and I asked YOU if you liked stalking people as a joke, and you went on and on about this chapter crap.  Then i said i would beat you, and you were like no ur a **** ******...

[color=blue]kyyuanmathcount:  swearing won't get you too far from an IP ban.  [/color]

Liar Liar

I bet 5000 dollars that you are richardni, johnz. Deal or no deal?

[quote]Then i said i would beat you, and you were like no ur a **** ******...

kyyuanmathcount: swearing won't get you too far from an IP ban. [/quote]

Omg i never said that!!!! fanatic said that!!! check my pms if you have to! i never swore in my pm, fanatic said i was a ***** ******* not me

[color=blue]kyyuanmathcount: yes, fanatic is an admin and can edit your posts.  I think you should look at my avatar and think about it before accusing others again.[/color]

Locked for off-topic flaming and arguing.


algebra
polynomial
trigonometry
geometric series

Find all the zeroes of $ P(x) \equal{} x^{4}\plus{}x^{3}\plus{}x^{2}\plus{}x\plus{}1.$ Hence factorise P(x) into irreducible factors over $ R$. Deduce that $ cos\frac{2\pi}{5}\equal{}\frac{\sqrt5\minus{}1}{4}$.

[hide="book's soln."]
$ (x\minus{}1)(x^{4}\plus{}x^{3}\plus{}x^{2}\plus{}x\plus{}1) \equal{} x^{5}\minus{}1$. How would you recogonise this quickly?

$ z_{1}$=1 is one fifth root of unity but it is not considered.

The non-real fifth roots of unity are $ z_{2}$ and $ \bar{z_{2}}$ where $ z_{2}$=$ cos\frac{2\pi}{5}\equal{} isin\frac{2\pi}{5}$[/hide]

$ z_{3}$ and $ \bar{z_{3}}$, where $ z_{3}$=$ cos\frac{4\pi}{5}\plus{}isin\frac{4\pi}{5}$

$ x^{4}\plus{}x^{3}\plus{}x^{2}\plus{}x\plus{}1$=$ (x^{2}\minus{}2cos\frac{2\pi}{5}x\plus{}1)(x^{2}\minus{}2cos\frac{4\pi}{5}x\plus{}1)$

These factors of P(x) are irreducible over $ R$. Further equating coeffecients of x,
$ 2cos\frac{4\pi}{5}\plus{}2cos\frac{2\pi}{5}\plus{}1 \equal{} 0$
how did they obtain this?[hide]

[hide="Brief explanation"]Note that $ x^{4}\plus{}x^{3}\plus{}x^{2}\plus{}x\plus{}1$ is a [[geometric series]], so it follows that: $ x^{4}\plus{}x^{3}\plus{}x^{2}\plus{}x\plus{}1 \equal{}\frac{x^{5}\minus{}1}{x\minus{}1}$. From here we use [[roots of unity]] to find all solutions to $ x^{5}\minus{}1, x\ne1$, which will be in terms of cosine, sine, and $ i$, of course. The book then multiplies the factors that are complex conjugates to get real factors, which will have terms in cosine, again. Then he expands the real factors, which are in terms of cosine. He equates the coefficients of this expansion with the original polynomial, since the polynomials are equal, to deduce that $ \cos{\frac{2\pi}{5}}\equal{}\frac{\minus{}1\plus{}\sqrt{5}}{4}$[/hide]

ah right I see! I like how you spotted it's a gm series. I would in a million years never have got that though its obvious

thanks for your help, appreciate it.


[quote="GameBot"]The lines $ 2x \plus{} y \equal{} 6$ and $ x \plus{} y \equal{} 10$ together with the
$ x$-axis and $ y$-axis are drawn on a coordinate grid to
form quadrilateral $ ABCD$. What is the area of
quadrilateral $ ABCD$?

[asy]draw((4,-2)--(-1,8),Arrows);
draw((12,-2)--(-2,12),Arrows);
draw((-2,0)--(15,0),Arrows);
draw((0,-2)--(0,15),Arrows);
label("$x$",(15,0),E);
label("$y$",(0,15),N);
label("$A$",(0,6),W);
label("$B$",(0,10),W);
label("$C$",(10,0),S);
label("$D$",(3,0),S);
fill((3,0)--(10,0)--(0,10)--(0,6)--cycle,gray(0.6));[/asy][/quote]

$ 50\minus{}9\equal{}41$.


I have the equation (3^(2x+1)) - (10((3)^x)) + (3) = 0 , I know I have to get all the same bases, but im not sure how.

[quote="linzk3"]I have the equation (3^(2x+1)) - (10((3)^x)) + (3) = 0 , I know I have to get all the same bases, but im not sw.[/quote]
If I am not wrong you me mean solve the equation $ 3^{2x\plus{}1}\minus{}10\cdot3^{x}\plus{}3 \equal{} 0$
Here is my solution
Solution(by galois01)
$ 3^{2x\plus{}1}\minus{}10\cdot3^{x}\plus{}3 \equal{} 0\Longleftrightarrow (3^{x})^{2}\cdot3\minus{}10\cdot3^{x}\plus{}3 \equal{} 0$
Let $ 3^{x}\equal{} z$
So we get that:
$ 3z^{2}\minus{}10z\plus{}3 \equal{} 0\Longleftrightarrow$ $ z \equal{} 3$ or $ z \equal{}\frac{1}{3}$
Finally we get:
 $ 3^{x}\equal{} 3\Longleftrightarrow x \equal{} 1$ and
 $ 3^{x}\equal{}\frac{1}{3}\Longleftrightarrow x \equal{}\minus{}1$
So the solutions of the equation are $ x \equal{} 1$ and $ x \equal{}\minus{}1$  :wink:  :)

[quote="b-flat"][quote="galois01"] $ 3^{x}\equal{} 3\Longleftrightarrow x \equal{}\minus{}1$
[/quote]

I'm pretty sure you mean $ x^{x}\equal{}\frac{1}{3}\Longleftrightarrow x \equal{}\minus{}1$  :wink:

Other than that, your solution is correct.[/quote]

I'm pretty sure you mean $ 3^{x}\equal{}\frac{1}{3}\Longleftrightarrow x \equal{}\minus{}1$  :wink:

Other than that, your solution is correct.
 :D  :D :D
EDIT: 101st Post!!!

[quote="b-flat"][quote="galois01"] $ 3^{x}\equal{} 3\Longleftrightarrow x \equal{}\minus{}1$
[/quote]

I'm pretty sure you mean $ 3^{x}\equal{}\frac{1}{3}\Longleftrightarrow x \equal{}\minus{}1$  :wink:

Other than that, your solution is correct.[/quote]
Τhank's for correcting me.You are right I mean $ 3^{x}\equal{}\frac{1}{3}\Longleftrightarrow x \equal{}\minus{}1$ I edit  :wink:  :)


Upstate New York. :)

About an hour north of Albany.

COME ON, PEOPLE! POST!


Let f(n) be a function defined on the set of all positive integers and with all its values in the same set. Prove that if

\[ f(n+1) > f[f(n)] \]
for each positive integer n, then f(n) = n for each n.

I need some hints for this problem.

one thing that might work it to define a function g(x) that is the inverse of f(x), do that to both sides, and you get

n+1>f(n), do n=1, we get
2>f(1), and since domain is positive integers, we have f(1)=1

then i don't know what to do

the only thing wrong about this is that we dont know that an inverse exists, but we do know that the > symbol stays the same because we are dealing with positives...(i think), just an idea, see if that helps at all

Altheman is absolutely correct. I don't know why he is not confident. The fact that f(f(x)) exists makes it clear that f(x) is also an integer. This is because if f(f(x)) has to exist then the range of f(x) should be in the domain of the f.
     Now read the above mail written by Altheman and understand how can you proceed? Anyway don't use induction or such stuff. You will readily get an answer for all n. Now you might ask why not f(n) = n - x where x???I. Reason why can't it be so yourself.


algebra
polynomial
real analysis
real analysis unsolved

Suppose $ f(x)$ is a polynomial of degree $ n$ such that $ f(x)\geq  0$ for each $ x \geq 0$. Prove that
$ f(x)+f'(x)+f''(x)+...+f^{(n)}(x) \geq  0$;
for all $ x \geq  0$.

I'm 100% sure it has been already posted but, since I cannot find the link, here is the solution. We have $ f(x+t)=\sum_{k}f^{(k)}(x)\frac{t^{k}}{k!}$. Now multiply this by $ e^{-t}$ and integrate with respect to $ t$ from $ 0$ to $ +\infty$.

I think this also works: Let $ g(x) = f(x)+f^\prime (x)+\ldots+f^{(n)}(x)$ and note that $ g^\prime (x) = f^\prime (x)+\ldots+f^{(n)}(x)$.
Whenever we have that $ g(x) < 0$, it also holds that $ g^\prime (x) = g(x)-f(x) < 0$.
So suppose that $ \min_{x \geq 0}g(x) = m < 0$ and let $ u$ such that $ g(u) = m$.
In that point we must have that $ g^\prime (x) \geq 0$, so we have a contradiction.

Is it good? (It is like the fourth version of this argument that I'm posting.)

Yes, it works :) The only thing that is currently missing in this argument is an explanation of why the minimum is attained (I agree that it is easy, but the entire problem is not very hard either) ;)


number theory
greatest common divisor
algorithm
Diophantine equation
Euclidean algorithm

Find all integers solutions of $ 3x\plus{}5y\equal{}12$

[hide]$ \forall n \in \mathbb{Z}, \ (x, y)\equal{}(4\plus{}5n, \minus{}3n)$[/hide]

[quote="Kouichi Nakagawa"][hide]$ \forall n \in \mathbb{Z}, \ (x, y) \equal{} (4 \plus{} 5n, \minus{} 3n)$[/hide][/quote]

It's correct ;)

what method you used for find the  solution?

Consider the equation $ Ax + By = C$. Suppose $ (x_1, y_1)$ and $ (x_2, y_2)$ are two solutions. Then:

$ \begin{array}{c} A x_1 + B y_1 = C & A x_1 + B y_1 = C & \therefore A(x_1 - x_2) + B(y_1 - y_2) = 0 \end{array}$

Thus, the difference between any two solution pairs must satisfy $ A \Delta x + B \Delta y = 0$. Note that all integer solutions to the above equation can be written as $ \left( \frac {B}{\text{GCD}[A,B]}, \; - \frac {A}{\text{GCD}[A,B]} \right) n$.

Now, note that we require $ \text{GCD} [A,B] | C$. If this is the case, divide the equation by $ \GCD[A,B]$ to get $ A'x + B'y = C'$. [i](Otherwise, there is no solution)[/i]. We can rewrite the solutions to $ A \Delta x + B \Delta y = 0$ as $ (\Delta x, \Delta y) = (B', - A')n = (B'n, - A'n)$

So now, if we have any initial solution $ (x_0, y_0)$ to the original equation, then the solution set is $ (x_0, y_0) + (B'n, - A'n) = \boxed{\left( x_0 + B'n , y_0 - A'n \right) }$

All that's left is to find an ordered pair $ (x_0, y_0)$ that satisfies the equation. In this case, $ 3 | 12$, so we can just use $ (x_0,y_0) = (4,0)$ and get $ (4 + 5n , - 3n)$. We could have used any other ordered pair that works, too, such as $ (9, - 3)$.

3x + 5y = 12 ...sooooo  x = (12 - 5y)/3 = 4 - y + (-2y)/3

 so now for some integer "A" We can say that A = -2y/3........so that

y = -3A/2 = -A + (-A/2)......now  B = -A/2  so  A = -2B

By substituting we get y = 3B and ... x = (12 - 15B)/3 = 4 - 5B

now we have (x,y) = (4 - 5B, 3B) = (4 +5B, -3B)

BTW wats a "diofantic equation" ?

A [[diophantine equation]] is of format $ ax\plus{}by\equal{}c$ where you have solutions for $ (x,y)$. It could also be in other forms, as described in the article.

We call that a [i]linear Diophantine equation in two variables.[/i]  A Diophantine equation is just any equation which asks for integer solutions.

[quote="jorgeston"]Find all integers solutions of $ 3x \plus{} 5y \equal{} 12$[/quote]

[hide="shorter(is this rigorous?) solution"]We notice that $ 12 \equal{} 3\cdot 4$, thus $ (4,0)$ is a solution. We now solve for $ y$ in terms of $ x$:

$ 5y \equal{} 12 \minus{} 3x$

$ y \equal{} \dfrac{12 \minus{} 3x}{5}$.

$ y$ must be an integer, so looking through all modular residues $ \bmod 5$, $ x\equiv 4\bmod 5$. All integers $ x$ such that $ x\equiv 4\bmod 5$ have a corresponding integer $ y$ such that $ 3x \plus{} 5y \equal{} 12$. (little white square here)[/hide]

You didn't prove that $ (x,y)$ has only those solution sets. Unless I'm not seeing something. :huh:

I proved that x must be 4mod5, and y is restricted by x.

If we have any integer solution, then $ (3)(x\minus{}4)\equal{}\minus{}5y\equal{}t$ where $ t$ is some integer. Note that $ 3|t$ and $ 5|t$, so $ 15|t$. So $ t\equal{}15k$ for some integer $ k$. Then $ x\equal{}5k\plus{}4$ and $ y\equal{}\minus{}3t$. It is easy to see that these work.

You could also use the Euclidean Algorithm (for $ 3x\plus{}5y\equal{}1$) and then multiply by $ 12$ to get solutions.

[quote="happyme"]You could also use the Euclidean Algorithm (for $ 3x \plus{} 5y \equal{} 1$) and then multiply by $ 12$ to get solutions.[/quote]

That doesn't generate all solutions, especially since $ \gcd(3, 12) > 1$.  You use the Euclidean Algorithm to generate a particular solution; it's easy to find all of them from there.

[quote="jorgeston"][quote="pelao_malo"]i know you jorgeston
that equation is easy
you can check the pdf that GlagosSa uploaded
salu2 xD[/quote]

hola, yo = te conozco jaj, y te hablare en español jerga chilena...esta no era una consulta,era un " propuesto", ..claramente es muy facil

saludos[/quote]English please. If you guys are trying to chat, use PM's or talk on your personal weblogs or talk in Community forums.

Follow-ups:

1) Show that $ 3x\plus{}5y\equal{}12$ has no positive integer solutions, ie no solutions where $ xy>0$.

2) Find the maximum value of $ xy$, where $ x,y$ are the integer solutions to the above equation.

$ xy>0$ does not imply $ x>0 \wedge y>0$

In any case:

[b]#1:[/b]

$ 3|3x \wedge 3|12 \implies 3|5y \implies 3|y$
$ \therefore y>0 \implies y \geq 3 \implies 5y \geq 15 \implies 3x \leq \minus{}3 \implies x \leq \minus{}1$

[b]#2:[/b]By AM-GM:

$ xy \equal{} \frac{(\sqrt{3x \cdot 5y})^2}{15} \leq \frac{36}{15} \equal{} \frac{12}{5}$, with equality at $ (x,y)\equal{}\left(2, \frac{6}{5} \right)$

We pick the two nearest integer solutions: $ (4,0)$ and $ (\minus{}1,3)$, and find that $ (4,0)$ produces the minimum value of $ xy$, namely $ xy\equal{}0$

[b]Alternately:[/b] note that there are no solutions with $ x,y \in \mathbb{Z}^{> 0}$, as shown in part 1, and obviously there are no solutions with $ x,y \in \mathbb{Z}^{<0}$, so $ xy \leq 0$ for all integer solutions. Checking, we see that $ (4,0)$ gives $ xy\equal{}0$, thus it is the maximum.


geometry
circumcircle
inequalities
trigonometry
logarithms
probability
number theory

http://math.furman.edu/tournament/tournament.html

I decided to do Furman Math Contest this time since they're much harder and much more fun than USC (especially the older ones!)
Need #5,9,19, 25, 27, 33, 38

EDIT: don't need #35, forgot I already did it

[hide="1"]Using the distance formula, the sides of the triangle are $\sqrt{20}, \sqrt{20}, \sqrt{40}$ and is a right triangle. Therefore, the center of the circumcircle is the midpoint of the hypotenuse which is $(3,3)$.[/hide]

[hide="2"]We have a right triangle with legs $100$ and $1100$, the hypotenuse (which is the distance) is therefore $100\sqrt{122}$.[/hide]

[hide="3"]Simplifying to $\frac{x^{1998}-1}{x-1}\leq 0$. Therefore, $x$ must be negative and $\leq-1$ since if $-1<x<0$, then the numerator and denominator are both negative, thus the inequality is positive. Therefore we must have $x: x\leq-1$.[/hide]

[hide="4"]The distance between the two points is $\sqrt{13}$. The circle will have a radius of $\frac{2\sqrt{13}}{3}=\frac{\sqrt{52}}{3}$[/hide]
.
[hide="6"]The most efficient way to do this problem is WLOG, let the lines be $y=\frac{1}{2}x+1$ and $y=\frac{1}{2}x+2$. We have a right triangle with legs $1$ and $2$, and hypotenuse $\sqrt{5}$. The altitude from the base $\sqrt{5}$ is the perpendicular distance and therefore in this case is $\frac{2}{\sqrt{5}}$. Plugging into the answer choices, $(c)$ works.[/hide]

[hide="7"]WLOG, let $t=60^\circ$ and $AB=2$. $\triangle{ABO}$ is equilateral and has area $\sqrt{3}$. The semicircle has area $\frac{\pi}{2}$ so that $E/D=\pi \cdot \frac{1}{2\sqrt{3}}$. Plugging in, $(a)$ is the correct answer choice.[/hide]

[hide="8"]Using rational root theorem, we see that $x=2$ is a root. Thus, the equation factors to $(x-2)(x^{4}+10x^{3}+26x^{2}+10x+1)$. Let $(x^{2}+ax+1)(x^{2}+bx+1)= x^{4}+10x^{3}+26x^{2}+10x+1$. Expanding this, we see that $a+b=10$ and $ab+2=26$. Thus, $(a,b)=(4,6)$. Thus, we completely factor the equation to $(x-2)(x^{2}+4x+1)(x^{2}+6x+1)=0$. From there, the answer can be found.[/hide]

[hide="10"]Let the centers of the quarter circle, the bigger semicircle, and the smaller semicircle be $O, A, B$ respectively. Let the radius of the smaller circle be $x$. Now notice that $\triangle{AOB}$ is a right triangle with legs $AO=1, BO=2-x$, and hypotenuse $AB=x+1$. We have $1+(2-x)^{2}=(1+x)^{2}\Longrightarrow x=2/3$.[/hide]

[hide="11"]We have a $45,45,90$ triangle with legs $6$ and hypotenuse $6\sqrt{2}$. The length of the poster is a cevian from the top vertex of this triangle and creates another right triangle with height $6$, base $x$, and hypotenuse $\sqrt{6^{2}+x^{2}}$. Using the law of cosines with the angle of $30^\circ$. We have that $(6-x)^{2}=(6\sqrt{2})^{2}+6^{2}+x^{2}-2(6\sqrt{2}\sqrt{6^{2}+x^{2}})\sqrt{3}/2$. After some algebra, we have that $x=3\sqrt{2}-\sqrt{6}$.[/hide]

[hide="12"]WLOG, let the square on the very left have $1$ assigned to it. The square to the right of it has $2$ choices for its number. Similarily, all other squares have $2$ possible number arrangements. So we have $2^{4}$ but we multiply by $3$ (for the three possible choices for assigning the first square a number). The answer is $48$.[/hide]

[hide="13"]Notice that the possible numbers we can use that look the same normal and upside down are $0,1,6,9,8$. Thus there are a total of $5^{5}=3125$ zip codes we can make from these numbers. However, as shown by the example, numbers which are palindromes that look the same upside down can’t be used. Since $6$ and $9$ don’t look the same upside down, the possibilities for this kind of zip codes are $5 \cdot 5 \cdot 3 \cdot 1 \cdot 1=75$. Our answer thus is $3125-75=3050$.[/hide]

[hide="14"]The center of the circle must be collinear to the perpendicular bisectors of both the chords. Thus, the center of the circle must be $2$ units from the left of the intersection of the two chords, and $1/2$ units down from it. The diameter of the circle therefore is $2\sqrt{(7/2)^{2}+2^{2}}=\sqrt{65}$.[/hide]

[hide="15"]Notice that $4^{2x}+4^{x}=y \Longrightarrow 2x\ln{4}+x\ln{4}=\ln{y}\Longrightarrow x=\frac{\ln{y}}{3\ln{4}}$. The answer is $(e)$.[/hide]

[hide="16"]Checking a few of the first powers and seeing a pattern, we find out that if we use the binomial theorem, we can do: $(100+1)^{29}=1000...000+00290...0+0006=10296...$ and the thousands digit is $2$ so the answer is $(e)$.[/hide]

[hide="17"]Simple trial and error show that $(b)$ can be true.[/hide]
[hide="18"]
By assigning a value to $m$, the answer follows.[/hide]

[hide="20"]The bag originally contains $8P$ and $1Q$ and a total of $9$ coins. The probability that we take out the quarter in four tries is $1/9+1/9+1/9+1/9=4/9$. The urn then contains $1Q$ and $15P$. The probability that we take the quarter out again in three tries is $1/16+1/16+1/16=3/16$. So $\frac{4}{9}\cdot \frac{3}{16}=\frac{1}{12}$. But we also need to add in the probability that the quarter was never taken out of the bag in the first place which is $1-4/9=5/9$. Thus $1/12+5/9=23/36$.[/hide]

[hide="21"]$\frac{6}{10}=\frac{x}{90+10}\Longrightarrow x=60$.[/hide]

[hide="22"]By creating an equilateral triangle with the radii of all the circles and WLOG, letting $r=1$, we have that the area we want is $4\sqrt{3}-\frac{3}{2}\pi-\frac{3}{6}\pi$ and the height we want is $2\sqrt{3}+2$. $(d)$ is the answer.[/hide]

[hide="23"]We have that:
$k \equiv 1\mod{2}$
$k \equiv 2\mod{3}$
$\cdots$
$k \equiv 9 \mod{10}$
Adding $1$ to each side gives us that $k+1$ is the LCM of all the numbers from $2$ to $10$. Thus $k+1=2520$. The sum of the digits of $k$ is $17$.[/hide]

[hide="26"]Using the change of base formula, we simplify the equation to $\frac{\log{y}}{\log{3}}=2 \Longrightarrow y=3^{2}=9$.[/hide]

[hide="28"]Let $a=x-1, b=x$, and $c=x(x-1)$. We have that $D=\sqrt{(x-1)^{2}+x^{2}+(x+1)^{2}}=\sqrt{x^{4}-2x^{3}+3x^{2}-2x+1}=\sqrt{(x^{2}-x+1)^{2}}=x^{2}-x+1$. This is always an odd integer for integer values of $x$.[/hide]

[hide="29"]For this problem, we work with each radical at a time and convert them into powers of $2$. $\sqrt{32 \cdot 16^{-3/4}}=\sqrt{2^{5}\cdot 2^{-3}}=2$. Now we have $\sqrt[4]{128 \cdot 2}=\sqrt[4]{2^{8}}=2^{4}$. Finally we have $\sqrt[5]{64 \cdot 2^{4}}=\sqrt[5]{2^{6}\cdot 2^{4}}=2^{2}$. $\log_{4}{4}=1$.[/hide]

[hide="30"]$PB^{2}=5^{2}+3^{2}-4^{2}=18$. $PB=3\sqrt{2}$.[/hide]

[hide="31"]Six days and $2.5 \cdot 6=15$ minutes are off on the watch at 1:00 PM on March 31. If the watch loses $2.5$ minutes per $24$ hours, it loses $5/48$ minutes per hour. So at 9:00 AM on March 31, the watch is off by $15-4(5/48)=14+\frac{7}{12}$
[/hide]
[hide="32"]Using $\sin{(a-b)}=\sin{a}\cos{b}-\sin{b}\cos{a}$, and keeping in mind the quadrants, we have $(\frac{2}{3})(\frac{3}{4})-(-\sqrt{1-(3/4)^{2}})(-\sqrt{1-(2/3)^{2}})=\frac{6-\sqrt{35}}{12}$.[/hide]

[hide="34"]Each suit contains $13$ cards and we must choose $5$ cards out of these $13$. Also, there are $4$ different suits. Thus the number of flushes are $4 \cdot \binom{13}{5}=5148$.[/hide]

[hide="35"]The probability of $(a)$ is $1/6+3/6=2/3$. The probability of $(b)$ is $(\frac{3}{6})^{3}=\frac{1}{8}$. The probability of $(c)$ is $\frac{3}{6}\cdot \frac{3}{6}=\frac{1}{4}$. $(c)$ is the answer.
[/hide]
[hide="36"]Solving the two linear equations, we have the intersection point as $(3,-1)$. Notice that $x^{2}+y^{2}+4y+1=20 \Longrightarrow x^{2}+(y+2)^{2}=23$. The center of this circle is $(0,-2)$. Using the distance formula, we see that the distance is $\sqrt{10}$.[/hide]

[hide="37"]From $2pqr$ and $rqp2$, we see that since either $r=8$ or $r=3$ can satisfy the last digit of $2pqr$ being $2$ after being multiplied by $4$, and since $r=8$ gives us a more likely correct result (since $r=3$ will be too small), we have $r=8$. Now we have $2pq8 \cdot 4=8qp2$. Since $4 \cdot 8$ makes us carry the $3$ over to $p$, we have that $4q+3 \mod{10}=p$. Also, we can’t have $p>2$ or else that will make the first digit be $>8$. Using this info and trial and error gives us $q=7$ and $p=1$, and $1+7=8$.[/hide]

[hide="39"]Notice that:
$2^{48}-1=(2^{24}+1)(2^{24}-1)=(2^{24}+1)(2^{12}+1)(2^{12}-1)= (2^{24}+1)(2^{12}+1)(2^{6}+1)(2^{6}-1)= (2^{24}+1)(2^{12}+1)(65)(63)$. $63+65=128$.[/hide]

[hide="40"]The one dressed as Ross Perot says that Harry is dressed as Bill Clinton. If this is true, then Harry is indeed dressed up as Bill Clinton. To confirm this, Harry says that Dave is dressed up as Bill Clinton, but Harry always lies. Therefore, Harry is the one dressed up as Bill Clinton.[/hide]

EDIT: Spoilers

Your gaps are so trivial Shobhit.

[hide="5"]
The taut belt consists of a bunch of line segments and arcs.  The line segments run parallel to the sides and tangent the circles directly above the vertices of the polygon.  Their sum is thus $\sum_{i=1}^{n}d_{i}$.  The arcs will add up to make up the circumference of one full circle of radius $r$.  Thus, our answer is C.[/hide]

[hide="9"]
After taking $k$ milk out of $A$ you are left with $1-k$ milk in $A$ and $k$ milk in $B$.  Then you add $1-k$ water to $B$.  The ratio of milk in $B$ to total liquid is just $k: 1$ or $k$.

To fill $A$ back up we need to take $k$ pints of this $k$ percent milk solution.  So we will end up putting $k\times k =k^{2}$ milk back in $A$.

Thus we have $k^{2}+1-k$ milk in $A$.  This is a parabola so its minimum occurs at $k=1/2$ making the amount of milk in $A$ equal to $(1/2)^{2}-(1/2)+1=3/4$. C[/hide]

[hide="19"]
Well, \begin{eqnarray*}a &=& 1+2+\cdots 100 = 5050\\ b &=& 1+2+\cdots+2^{19}=1,048,575\\ c &=& 200\cdot 365\cdot 24=1,752,000 \end{eqnarray*} Thus D is correct.[/hide]

[quote="joml88"]Your gaps are so trivial Shobhit.

[hide="5"]
The taut belt consists of a bunch of line segments and arcs.  The line segments run parallel to the sides and tangent the circles directly above the vertices of the polygon.  Their sum is thus $\sum_{i=1}^{n}d_{i}$.  The arcs will add up to make up the circumference of one full circle of radius $r$.  Thus, our answer is C.[/hide][/quote]

Oh, I didn't know what the picture would look like. At first, I thought the taut belt would've made a circle around the entire figure. Then that didn't work out, and I thought it'd simply be in the shape of the figure itself. :blush:. I should've known what it woulda looked like.

[hide="27"]
First $M-N$ is subtraction of sets.  Basically you take all the elements that $M$ and $N$ have in common out of $M$.  For example, if $M=\{1,2\}$ and $N=\{2,3\}$ then $M-N=\{1\}$.  So basically $(M-N)\cup(N-M)$ is going to be the union of $M$ and $N$ with their intersection taken out.  From this definition it is easy to see that the operation is commutative and assosciative.

Then, we need to see if there is an identity element.  But $T$ contains the null set which would act as an identity element.

Finally, each set is its own inverse since subtracting a set from itself leaves you with the null set and then taking the union of two null sets is still the null set.

So the answer is D.[/hide]

[hide="33"]
We have the inequality $|2x-1|-|x+3|<4$.  The critical points that we need to look at are when $|2x-1|=0$ and $|x+3|=0$ which are at $x=1/2$ and $x=-3$.

On $[1/2,\infty)$ our inequality simplifies to $2x-1-x-3<4 \Leftrightarrow x<8$.  Thus $1/2\le x<8$ (since we assumed $x$ was on $[1/2,\infty)$ in the first place).

On $(-3,1/2)$ we have $1-2x-x-3<4 \Leftrightarrow x>-2$.  Thus we get $-2<x<1/2$.

On $(-\infty,-3]$ we havee $1-2x+x+3<4\Leftrightarrow x>0$.  But there are no $x\in (-\infty,-3]$ such that $x>0$.

Putting together our answers for the first two parts we see that $x\in (-2,8)$ so its length is 10. D[/hide]

[hide="38"]
We have $47_{a}=74_{b}$ which is the same as saying $4a+7=7b+4\Leftrightarrow 4a = 7b-3$.  Taking both sides modulo 4 gives $0\equiv 7b-3 \mod 4$.  After simplifying this we see that $b\equiv 1 \mod 4$.  So the first couple possible values of $b$ are $1, 5, 9, 13,\ldots$.  But $b$ can't be 1 or 5 since 7 is a digit in base $b$.  So we take $b=9$.  This makes $a=15$.  We check and find that $47_{15}=67_{10}$ and $74_{9}=67_{10}$ so we're good.

Finally $9+15=24$. D[/hide]


Um, not the best way to do this?

[hide]Basically, I multiplied out the 2nd equation: xy-kx-hy+hk = c
Then, since in the 1st equation, y is on one side, I isolated all the terms with 'y' in them: xy-hy = kx-hk + c.
Some factoring gave: y(x-h) = k(x-h) + c.
Divide each side by (x-h): y = (k(x-h)+c)/(x-h) and this equation looks a lot like the first
Now we see that k = 4, h = 3, and c = 5[/hide]

Your avatar is a game of goe (go)...?

[quote="Ignite168"]The equation $y=\frac{4x-7}{x-3}$ can be reexpressed in the form $(x-h)(y-k)=c$, where h, k, and c are constants. Identify h, k, and c.[/quote]
[hide="solution"]
$y=\frac{4x-7}{x-3}$
$(x-h)(y-k)=c$
$y-k=\frac{c}{x-h}$ restriction: $h<>x$ (how do u do not equal to?)
$y=\frac{c}{x-h}+\frac{k(x-h)}{x-h}$
$y=\frac{(k)x-(kh-c)}{x-h}$
$h=3$ since $x-3=x-h$ (denominator), so new restriction: $x<>3$
$kx-(3k-c)=4x-7$ (numerator)
$k=4$ since $kx$ = $4x$
$12-c=7$
$c=5$
so $h=3$, $k=4$, $c=5$
[/hide]

[quote="ProtestanT"]how do u do not equal to?[/quote]
\ne
$\ne$

:D


That's not more and not less than an algebraic equation. There is no general solution, and for $n \geq 5$ you can't even express the solutions with roots and rational numbers.

If you want integer or rational solutions, you can use the rational-root-theorem.


Harvard
college
combinatorics unsolved
combinatorics

Let n be a positive integer, and let x1,... , xn, y1 ,....., yn be positive reals with thefollowing property: for every nonempty subset S of {1,...,n}, there exists anothernonempty subset T of {1,...,n} such that
 \sumxi(i \in T)/ \sum yi(i \in T)= \sum yi(i \in S)/ \sum xi(i \in S)
Must there necessarily exist, for each i = 1,..., n, some j such that xj = yi andyj = xi?

Let $n$ be a positive integer, and let $x_1$, ..., $x_n$, $y_1$, ..., $y_n$ be positive reals with the following property:
for every nonempty subset $S$ of $\{1,...,n\}$, there exists another nonempty subset $T$ of $\{1,...,n\}$ such that
\[\frac{\sum_{i\in T}x_i}{\sum_{i\in T}y_i}=\frac{\sum_{i\in S}y_i}{\sum_{i\in S}x_i}.\]
Must there necessarily exist, for each $i = 1,..., n$, some $j$ such that $x_j = y_i$ and $y_j = x_i$?

P.S. I hope I correctly rewrote problem.

This is one of Gabriel's Unsolved problems.
http://www.people.fas.harvard.edu/~gcarroll/math/unsolved.pdf

Gabriel Carroll, ex IMO contestant (he's too old now :)). You can find his website by running his name through Google. He has some pretty nice problem lists. Some have been discussed here.

[quote="Myth"]Who is Gabriel?[/quote]

Gabriel D. Carroll, one of the most successful olympians in the history of the U. S. IMO team, now a Harvard mathematics major.
http://www.people.fas.harvard.edu/~gcarroll/


limit trigonometry calculus derivative integration function geometry

Is there a magical change of variables to prove that

$ \lim_{x\to0}\frac{\sin x}{x}\equal{}1$?

Change of variables?  This is known as the Sandwich Theorem.  Establishing it is the standard way to establish the derivatives of $ \sin,\cos$, and as such it needs a separate proof.  It is usually proved geometrically.

[hide="Hint"] $ L \equal{} r\theta$ [/hide]

[quote="Hetidemek"]Is there a magical change of variables to prove that

$ \lim_{x\to0}\frac{\sin x}{x}\equal{} 1$?[/quote]

Formally, the definition for sine is $ \sin x \equal{} x\minus{}\frac{x^{3}}{3!}\plus{}...$
Upon division we have,
$ \frac{\sin x}{x}\equal{} 1\minus{}\frac{x^{2}}{3!}\plus{}...$
Now because of uniform convergence you can pass the limit through getting $ 1$ as the answer. (You cant really use geometric proofs, because they arent up to rigor).

In trying to calculate the derivatives of $ \sin$ and $ \cos$ by the definition (and the angle addition identities), you will inevitably run into $ \lim_{x\to 0}\frac{1\minus{}\cos x}{x}$ and $ \lim_{x\to 0}\frac{\sin x}{x}$. The first is pretty easy to handle; the second is tricky.

The way you approach this limit depends on what definition you use for $ \sin$. Most calculus textbooks use a geometric definition of $ \sin$; if we do this, the proof that $ \lim_{x\to 0}\frac{\sin x}{x}\equal{}1$ is based on showing that $ \sin x<x<\tan x$ for $ 0<x<\frac{\pi}{2}$ and using the sandwich theorem as t0rajir0u described.

We could also use a purely calculus definition of $ \sin$. We could define $ \arcsin x$ to be $ \int_{0}^{x}\frac{1}{\sqrt{1\minus{}t^{2}}}\,dt$ for $ \minus{}1\le x\le 1$ and define $ \sin$ to be the inverse of $ \arcsin$, continued to be periodic. Then this limit and the derivative of $ \sin$ are simple calculations involving the FTC.
Alternately, we could define $ \sin$ using its power series or the differential equation $ y''\plus{}y\equal{}0$. These make the derivative and the related limits very easy.

Why don't textbooks use any of those calculus-based definitions? If they did, we wouldn't be able to use trig functions when talking about derivatives; the integral tools are developed later, and power series even further on.

A previous [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=120492]discussion[/url] of some of these issues.

I just wanna mention a little on Jmerry's post.
If you read Jmerry post you will find that he gives any defintions on the $ \sin$ function. Alternatively, we can use many other approaches which he didn't mention.

Calculus books do some things a little informally. Why? Isn't it better for a math book to things in rigor? Yes, but not when it comes to Calculus. The point of Calculus is to introduce new concepts to students [b]and[/b] provide their applications to other areas of science, such as finding area, volumes, and physics .... If we did it in a formal way (like a Real Analysis book would) the reader wouldn't see the motivation for making these definitions. However, when a student studies Calculus, though informal, sees that these functions are based on properties of angles and circles and he will understand why the are important. So for studing Calculus its better to be a little informal at first.


Leap Day, February 29, 2004, occurred on a Sunday. On what day of the week will Leap Day, February 29, 2020, occur?

$ \textbf{(A) } \text{Tuesday} \qquad \textbf{(B) } \text{Wednesday} \qquad \textbf{(C) } \text{Thursday} \qquad \textbf{(D) } \text{Friday} \qquad \textbf{(E) } \text{Saturday}$

[hide]A leap day occurs once every 4 years, so the next Leap Day occurs (365 + 365 + 365 + 366) = 1461 days after the current Leap Day.

$1461 = 5 \mod 7 = -2 \mod 7$, so the Leap Day in 2008 is 2 days before the Leap Day in 2004. We see that evey next Leap Day occurs 2 days before the last, so:

2004: Sunday
2008: Friday
2012: Wednesday
2016: Monday
2020: Saturday

hence $(E)$.[/hide]

[hide] The very quick and non-mathematical way to do this is to just use the calender on the organizer app for TI graphing calculators :lol: .[/hide]

[hide="A generalization"]In general, since $365\equiv1\pmod7$ and $366\equiv2\pmod7$, we count forwards 1 weekday every calendar year and 1 extra for every leap year. Since we are going forward 16 years, 4 of which are (always going to be - until 2100) leap years, we go forwards 16+4=20 weekdays, which is the same as going back 1 weekday. Saturday comes before Sunday, so $\text{(E)}$.[/hide]

Mods also known as modulus': http://mathworld.wolfram.com/Modulus.html


geometry
3D geometry
Vieta
quadratics
algebra
quadratic formula
system of equations

Let a and b be the two real values of x for which

         (x^1/3)+((20-x)^1/3)=2

The smaller of the two values can be expressed as p - (q^1/2), where p and q are integers. Compute
p + q.

[hide]
First, note that a+b = 20 since if a is a solution, then so is 20-a. We then have the following system of equations:
a^1/3 + b^1/3 = 2
a + b = 20
If we cube the first equation and factor, we have that
(a+b) + 3 * (ab)^1/3 * (a^1/3 + b^1/3) = 8, or
(ab)^1/3 = -2, so ab = -8.
By Vieta's formulas, a and b must be the roots of the quadratic equation y = x^2 - 20x - 8.
Solving with the quadratic formula yields a = 10 - (108)^1/2, with final answer [b]118[/b].
[/hide]

[hide="faster"]Cubing the equation results in $ \left(\sqrt[3]{x}\plus{}\sqrt[3]{20\minus{}x}\right)^3\equal{}20\plus{}3\left(\sqrt[3]{x}\plus{}\sqrt[3]{20\minus{}x}\right)\sqrt[3]{x(20\minus{}x)}\equal{}8$. Simplifying results in $ x(20\minus{}x)\equal{}8$. We have a quadratic. Use the Quadratic formula to get $ x\equal{}10\pm{108}\implies\boxed{118}$.[/hide]


Do you think North Toronto Collegiate is a good high school (in terms of academics) compared to all the schools within the Toronto District School Board (TDSB)?

Would you prefer Northern Secondary School?

This is a math forum, so I'll only speak in terms of math.

From what I gather, it's clearly above average but by no means amongst the elite.

Between it and Northern, I would have no preference either way.

In terms of math, it's above average. Academically, it's among the top (eng+other stuff) I was going to go there if I didn't trabsfer from Lord Lansdowne to Zion. Musically though, it's amongst haig and *insert another musical school*.  They have some of the most talented players, I know one from Zion, the name is Henry Lau. The music program is superb I'm sure you are going to love it there. The vice principal Mr. Ferguson is an excellent conductor, he directs at NMC as you've prolly heard. Ms Goldhar's nice aswell.

Well that's good to know but still, I think North Toronto needs more caretakers. The school is a filthy mess! AHHH!... and I thought my school was dirty.  :roll:

[quote="Connie"]Sorry I know this is off topic but--d.lam.86, your picture looks funny. Is that you?[/quote]

lol, why wouldn't it be

Yeah that's me.  :blush: I don't smile that much so it's actually really not me. I barely ever smile, in real life. I think I'll put something not smiling...  :lol: I had to crop the image so much... It was a difficult task.  :mad:


Once you stop laughing, can someone please answer my question:

I have a brand new TI-89 Titanium, and I read the manual, but I can't figure out how to solve a set of equations.
For example:
$y=x-12$
$y=6-x$

I tried graphing, and hitting F5: INTERSECTION (5) but that doen't give me the right answer.  I cannot use a program, because we have to clear the memory for the exit exam on Monday.  I know there is a simple method, so can someone please answer before Monday?!!?

Edit: I just got this calculator, and I am trying to connect it to my CPU.  I intstalled all the TI software, connected the calculator, and restarted the computer.  I have gone through the troubleshooting guide, and my calculator still isn't recognized by TI's software.  Microsoft keeps "installing" drivers for a "TI educational device" every time I restart, so I know the CPU and calculator are connected properly.  Does the calculator have to be off, or on a certain screen?  HELP!!!!

[quote="bubala"]graphing is easiest.  Just make sure to set the lower and upper bounds correctly.  That could be what you're doing wrong.[/quote]

What do you mean? I am new to all of this.  Step-by-step please :blush:

You could set the y values equal and do this:

[code]
solve(x-12=6-x,x)
[/code]

Then plug that x value back in.  Or using a graph, first graph the two equations.  You have to know some knowledge of where they intersect though.  Like if you can figure that they intersect between x=0 and x=10 you are good.  I think you might even have to have the screen so that the point of intersection is showing.  Do that by changing the window properly.  Now, go to Math, Intersection.  Plug in an x value that you know is less than the point of intersection for the lower bound (once again this x value must be viewable from the window however) and then put in an x value greater than the point of intersection for the upper.

Actually, the easiest way to solve a system of equations is solve(y=x-12 and y=6-x, {x,y})

Unless you're just interested in the x-value, in which case it could simply be solve(y=x-12 and y=6-x, x)

" and " may be found as the second item in the catalog, or you could just type it in, if you really wanted to.  But remember the spaces.

The TI-89 can solve systems of equations that way? Wow! I never knew that; maybe I should read the manual... that is if I can find it.

Yea, I remember I had to read the manual to figure out how to do sum of sequences for my TI 85

one of the worst experiences of my life. Now, I am against reading manuals of all sorts...


1.Show that if x y z are positive integer ,then (xy+1)(xz+1)(yz+1) is a perfect square if and only if

xy+1,xz+1,yz+1 are all perfect squares

2.if a b c are positive integer such that  0<a^2+b^2-abc<=c

show that  a^2+b^2-abc is perfect square

Both are PEN problems: http://www.mathlinks.ro/resources.php?c=200&cid=115&year=2007


limit
superior algebra
superior algebra solved

Let (xn) be a sequence of real numbers so that:
a) |x_m-x_n|<=|m-n|/(m+n) for m,n>=1
b)lim x_(n^2)=1.
Prove that lim (x_n)=1

cheers! :D

[quote="belenos"]A Try :
. n = N  \lim  \infty : |x(m)-1| <= 1 so x is bounded
. x(f(n)) is an converging extracted sub-sequence of x(n)
. let's define g(n) such as g(n) is the closest square of f(n). then |f(n)-g(n)| <= g(n)+1
. | x(f(n)) - x(g(n)) | <= (g(n)+1) / (g(n) + (g(n)-1))  \lim 0.
. So x(f(n))  \lim 1.
. x is bounded with only one converging extracted sub-sequence, x  \lim 1.

Is it correct ?[/quote]

First from (a) it easily follows that x_n is a Cauchy sequence and thus convergent.
From (b) you get a convergent subsequnece of x_n tends to 1, so x_n, who was convergent, tends to 1.


A function f defined on the positive integers satisfies f(1) = 1996 and
f(1) + f(2)+... + f(n) = n^2f(n) (n > 1): Calculate f(1996).


SUMaC admissions/rejections were emailed yesterday. I got wait-listed. Did anybody who reads this get accepted? or knows someone who did?

I'm curious, how many people do you think will turn down a SUMaC acceptance (probably from being accepted to other camps (i.e PRMOYS, Mathcamp, etc))?
What are my chances?

or rather... how many wait-listed kiddies have gone in previous years?

I now have to wait a whole month (again) to find out my final status. |sigh|

jimmy.

I know (of) two people who were accepted. I also know that at least one waitlisted person went in 2001. That's a rather small sample size though.


Gauss
number theory unsolved
number theory

Prove that the equation,
$ x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}3(x\plus{}y\plus{}z)\plus{}5 \equal{} 0$
has no solutions in rational numbers.

It is Gauss Lemma
An positive integer can represent in the form$ a^{2}\plus{}b^{2}\plus{}c^{2}$ if and only if it is not in the form $ 4^{l}(8k\plus{}7)$
An other problem
$ Let a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\equal{}7.2^{n}$
Find $ (a,b,c,d)$

[quote="TTsphn"]
An other problem
$ Let a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\equal{} 7.2^{n}$
Find $ (a,b,c,d)$[/quote]

$ a,b,c$ are integer positives ?

$ a,b,c,d$ is integer.Forum Number theory .

Positive integer is not similar to integer  :mad:


limit
real analysis
real analysis unsolved

n is a positive integer. Show that the equation 1/(x - 1) + 1/(22x - 1) + ... + 1/(n2x - 1) = 1/2 has a unique solution xn > 1. Show that as n tends to infinity, xn tends to 4.

Could you clarify your expression? What do you mean by "n2x"? Should that be $n^2x$?

$f_n(x)= \frac{1}{x-1} + \frac{1}{2^2x-1} + ... + \frac{1}{n^2x-1 }$

is defined on $]1,+\infty[$

$f'_n(x)<0$ so $f_n$ is strictly decreasing
$\lim_{1^{+}}f_n=+\infty ,\; \lim_{+\infty}f_n=0$
$f_n$ is continous there exist one $x_n>1$ such that $f_n(x_n)=1/2$

Let $g_n(x)=f_n(x)-1/2<g_{n+1}(x)$

$g_{n}(x_{n+1})<g_{n+1}(x_{n+1})=0=g_n(x_n)$

but $g_n$ is decreasing so $x_{n+1}>x_n$

not finish...


Find all functions $f: \mathbb  R\to \mathbb R$ such that
\[ f(f(x) \plus{} y) \equal{} f(f(x) \minus{} y) \plus{} 4f(x)y
\]
for all $ x,y \in \mathbb R$.

[quote="Allnames"]Find all functions such that
\[ f: \mathbb R\to \mathbb R\]

\[ f(f(x) \plus{} y) \equal{} f(f(x) \minus{} y) \plus{} 4f(x)y\]
for all $ x,y \in \mathbb R$[/quote]
No ideas. Where are you, pco ???? :maybe:

[quote="pco"][quote="Allnames"]Find all functions such that
\[ f: \mathbb R\to \mathbb R\]

\[ f(f(x) \plus{} y) \equal{} f(f(x) \minus{} y) \plus{} 4f(x)y\]
for all $ x,y \in \mathbb R$[/quote]

Let $ P(x,y)$ be the assertion $ P(x,y)$ : $ f(f(x) \plus{} y) \equal{} f(f(x) \minus{} y) \plus{} 4f(x)y$

$ f(x) \equal{} 0$ $ \forall x$ is a trivial solution. Let us now consider that $ \exists c$ such that $ f(c)\neq 0$. Then :

Let $ a\in\mathbb R$
Let $ u \equal{} f(c) \minus{} \frac {a}{8f(c)}$ and  $ v \equal{} f(c) \plus{} \frac {a}{8f(c)}$

$ P(c,\frac {a}{8f(c)})$ $ \implies$ $ f(v) \equal{} f(u) \plus{} \frac a2$

$ P(u,x \minus{} f(u))$ $ \implies$ $ f(x) \equal{} f(2f(u) \minus{} x) \plus{} 4f(u)(x \minus{} f(u))$

$ P(v,2f(u) \minus{} f(v) \minus{} x)$ $ \implies$ $ f(2f(u) \minus{} x) \equal{} f(2(f(v) \minus{} f(u)) \plus{} x) \plus{} 4f(v)(2f(u) \minus{} f(v) \minus{} x)$ $ \equal{} f(x \plus{} a) \plus{} (2f(u) \plus{} a)(2f(u) \minus{} a \minus{} 2x)$

So $ f(x) \equal{} f(x \plus{} a) \plus{} (2f(u) \plus{} a)(2f(u) \minus{} a \minus{} 2x) \plus{} 4f(u)(x \minus{} f(u))$ $ \equal{} f(x \plus{} a) \minus{} a^2 \minus{} 2ax$

So $ f(x) \minus{} x^2 \equal{} f(x \plus{} a) \minus{} (x \plus{} a)^2$ $ \forall x,a\in\mathbb R$

So $ f(x) \minus{} x^2$ is constant and $ f(x) \equal{} x^2 \plus{} b$
Plugging back this necessary condition in the original equation, we find that this indeed is a solution.

Hence the full set of solutions :
$ f(x) \equal{} 0$ $ \forall x$
$ f(x) \equal{} x^2 \plus{} b$ $ \forall x$, for any real $ b$[/quote]
Amazing proof , dear pco. I shouldn't post mine now.

you should've really looked through the problems base, besides [url=http://www.mathlinks.ro/viewtopic.php?p=827178#827178]BMO problems[/url]are quite popular  :wink: .


geometry
3D geometry
linear algebra
matrix
symmetry
probability
expected value

Question:
A particle moves on the eight vertices of a cube in the following way: at each step the particle is equally likely to move to each of three adjacent vertices, independently of its past motion. Let i be the initial vertex occupied by the particle, and let o be the vertex opposite i.
a). Calculate the expected number of steps until the particle returns to i for the first time.
b). Calculate the expected number of visits to o until the first return to i.
c). Calculate the expected number of steps until the first visit to o.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I know answers to (a) and (c). The given Markov chain is irreducible and recurrent. The part (a) asks for the mean recurrence time for i (which is 8, by the way). The part (c) asks for the mean first passage time from i to o. But I do not know the answer to the part (b). I know that in an absorbing Markov chain the (i,j)th entry of the fundamental matrix (I-Q)^-1 is the expected number of visits to state j starting from state i, but the Markov chain in our case does not have an absorbing state... Can you please help me?

Don't bother with general formulae. Just denote by some letter the expected number of times to visit o before you visit i starting from every given vertex (due to symmetry, you'll need just 4 different letters) and set up the linear system. It'll have a unique solution from which you'll know everything.


Bryan and Nate are dirt bike racers and are brothers.
When practicing with his younger brother, Bryan gives Nate a 60 meter head start.

After t seconds, Nate is a distance 6t + 60 from the starting line and Bryan is a distance of 7t from the starting line.

a. How far ahead of Bryan is Nate after t seconds?

b. Evaluate the answer for part a when:

1.) t = 15

 2.) t = 30

 3.) t = 45

c. Does Bryan ever catch up with Nate? If so, after how many seconds?

I like your replies because you break the question into pieces.

Thanks!


Find every function $ f: (0;\plus{}\infty)\mapsto (0,\plus{}\infty)$ satisfies the condition:

$ f(f(x)\plus{}x\plus{}y)\equal{}xf\left(1\plus{}xf\left(\frac{1}{x\plus{}y}\right)\right),\forall x,y\in (0,\plus{}\infty)$

hello
you want to said $ f: ]0; \plus{} \infty[ \longrightarrow ]0; \plus{} \infty[$ ????

No, I meant $ (0,\plus{}\infty)\equal{}\{x\in\mathbb{R}|x>0\}$.

[quote="Mashimaru"]Find all function $ f: (0; \plus{} \infty)\mapsto (0, \plus{} \infty)$ satisfies the condition:

$ f(f(x) \plus{} x \plus{} y) \equal{} xf\left(1 \plus{} xf\left(\frac {1}{x \plus{} y}\right)\right),\forall x,y\in (0, \plus{} \infty)$[/quote]

  [i] Mashimaru tell me to correct the demand of the problem

   Can anybody solve it ? It's from Vietnam IMO training test

  A problem of sir Nam Dung . :lol: [/i]


counting
distinguishability
function
geometry
geometric transformation
reflection
rotation

I was looking at the ceiling fan on the kitchen (skip to the second paragraph if you dnt want to read this story), and mind you this fan is missing one blade so we never use it. As I looked at it I wondered if there was another the blades could be re-arranged, there wasn't, 3 blades could only be arranged in a circular pattern with 4 possible positions in one way. All other combinations are identical, because all the blades are [b]indistinguishable[/b]. So I thought what about if it was missing 2? 2 ways, ok,
    What if the fan had "n" blades and and "n-k" missing blades? In how many ways can they be arranged?
In the following diagram each row stands for the number of blades on the fan (first row=1blade), the position (1st, 2nd,...) is the number of blades present ("k"), and the number in the actual position - or the number in the nth row and kth place - is the number of ways that can be arranged =D
1
11
111
1211
12211
133311
1343211
1456[b]4[/b]411
I think the 4 bolded is wrong, it would look nice if it was 5, haha...but then again, there can be other errors..numberous errors...but the whole point of this post is to show the idea and [b]ask[/b]:
Has anyone seen this pattern before? or: Where can I learn more of this pattern? or: idk, help!? ={
i like it, if not i wouldn't have bothered writing this long *** post
Thanx to all that respond!!!! XD I really appreciated =D

You're counting necklaces (or maybe bracelets?) with $ k$ beads of one color and $ n \minus{} k$ beads of a second color -- it's sequence [url=http://www.research.att.com/~njas/sequences/A052307]A052307[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url], I believe, though that disagrees with your computation starting in your 7th row.  On the other hand, your 7th row is pretty clearly wrong, since it should be symmetric ($ k$ blades and $ n \minus{} k$ gaps can be arranged in exactly the same number of ways as $ n \minus{} k$ blades and $ k$ gaps).

No formula given there, and I haven't given any thought to the problem of actually trying to compute these numbers.

I give the generating function [url=http://qchu.wordpress.com/2009/06/21/gila-v-the-polya-enumeration-theorem-and-applications/]here[/url].  If you want to learn more about Polya theory, you should start [url=http://qchu.wordpress.com/2009/06/13/gila-i-group-actions-and-equivalence-relations/]here[/url].  The OEIS entry appears to be using the [url=http://en.wikipedia.org/wiki/Dihedral_group]dihedral group[/url] - that is, counting necklaces as the same up to reflection as well as rotation - whereas I think your question is about cyclic groups.

It sounded from the description like cyclic symmetry, but the errors appear a row earlier if it is: with 6 spots and 3 of each color, there are four arrangements: the blades can go in positions {1, 2, 3}, {1, 2, 4}, {1, 2, 5} or {1, 3, 5}.  (The second and third sets are similar under a reflection, but not a rotation.)  Probably I guess the O.P. was thinking of cyclic symmetry while using dihedral symmetry.

In this case, the desired sequence is [url=http://www.research.att.com/~njas/sequences/A047996]A07996[/url].

I dn't know which one is best  :(  :o ...
But thanx for introducing me to cyclic groups,..this name Poyla ...reminds me of some author that wrote some problem solving book, if i remember right...It's weird
I didnt know that deriving a formula for this seemingly simple problem could be so complicated...=D...I just asked, cuz in didn't know where to start - do you think it would have been best if, (1) I had put time and effort at finding it by myself and learning what works and what doesn't and had wasted a lot of time (but then again, i dnt think it's a waste of time because I think i learn it to a more personal level, when i do do that), (2) or asking and taking a regular scholastic approach??

It's always more efficient to ask.  You can't be expected to rediscover Polya theory on your own.


Let $a,b,c\in\mathbb{R}$ such that $a+b+c=1$ and $a^2+b^2+c^2=3$.Prove that $abc\in [-1,\frac{5}{27}]$ ;)

This is easy! Here is my solution.

$ab+bc+ca=\frac{1}{2}\{(a+b+c)^2-(a^2+b^2+c^2)\}=-1$, thus $a,b,c$ are the roots of $x^3-x^2-x-abc=0$.
Let $f(x)=x^3-x^2-x$, we are to find the condition for which $y=f(x)$ have intersection points with $y=abc$.

From $f'(x)=(x-1)(3x+1)$,the condition is $f(1)\leq abc\leq f\left(-\frac{1}{3}\right)\Longleftrightarrow -1\leq abc\leq \frac{5}{27}$. Q.E.D.


How fast is the site for you?  Same as usual?  Much slower than usual?  Much faster than usual?

Same as usual. Might have some speed changes but I don't really notice.

I guess mine's going a tiny bit faster, it still doesn't matter so much.

We're interested in slowdowns. We didn't do anything in particular to speed it up, but we are running some new projects that might slow down the site at times.

At times site a while to load but, otherwise same as it has been.

I had some temporary slowdowns (partly pretty annoying), but my internet connection is playing mad these days...

  Darij

[quote="minicon"]I think its at the same speed.  Maybe your *new program* on your site made it a teensy bit slower.
For those of you who are wondering what the *new program* is, I'll let you ponder! :P[/quote]Yes, that might be the main reason for the (sometimes) slow speed of the site if they exist at all.

My browser runs much slower on AoPS than it used to, actually... I was about to make a topic about it.

And what's the new program? A new phpbb3 mod?

[quote="Temperal"]A new phpbb3 mod?[/quote][quote="#H34N1"]I think its probably an apache update.[/quote]It's none of these.

Hmm, the blogs are acting a little slower than usual.

I have occasional slowdowns on the site, and when I go on other websites, they work fine. But sometimes its just due to some internet connection problems I'm having, and none of the sites work.

Mine has been always fast enough for me.. :lol:

[quote="Thor"]Both types of websites. I have tried with Opera and IE.[/quote]Try Firefox :D But surely when I asked the question we were doing something on the server that was putting an unusual load on it. Right now, it seems you're the only one affected by this, and it has started as you claim right when you changed your provider, so it must be something on your side only.

No, its none of those (That I know of :D)
You're asking [b][i][u]too[/u][/i][/b] many questions.

[quote="#H34N1"]Oh I think I got it. Did AoPS switch servers/upgrade processors/motherboards? Or did it change its internet connection? Or did it change platforms?[/quote]None of these.

[quote="Valentin Vornicu"][quote="#H34N1"]It upgraded its classroom software/transcripting software?
DNS change?[/quote]It's not an upgrade   :ninja:[/quote]

well, if something didn't go up, something must have come down.  :P

My patience has risen this past week from the slowness of loading the forum page.

Is it the polls forum?

It has nothing to do with the forums: it's a totally new feature. When the time is right we will make sure everyone knows about it :D

This could warrent a new topic but...

Around 20 minutes ago the server for AoPS/Mathlinks was down. Accidental or on purpose?

When I log in, I always check the "log me on automatically" box. It never works. Why is that?

If you're deleting your history (an option in Firefox, not sure in IE7 since I don't use it), it could delete the password.

[quote="Temperal"]Check cookies, 1=2.

Also, do you frequently alter use between the mathlinks and the aops domains?[/quote]

I've used AoPS since I came here. I wonder what MathLinks looks like.


david is fishing in his canoe. it springs a leak and david immediately begins bailing water at a rate of 3 gallons per minute. david knows from past fishing trips that it takes 72 gallons of water to sink his canoe. if david's canoe sinks 1.2 hours after springing a leak, and what rate, on average, did water enter david's canoe. give your answer in gallons per minute.

how would u do this problem?

[hide] Well each minute he takes on x gallons, minus the three he scoops out, and after 72 minutes he has taken on 72 gallons so we get the equation:

72(x-3) = 72

and its clear to see that x = 4  so its 4 gallons per minute[/hide]

[quote="vishalarul"]It takes $1.2$ $\bold{hours}$= $72$ $\bold{minutes}$ to sink his canoe. So the rate is $1$ $\bold{gallon}$ $\textrm{per}$ $\bold{minute}$.

Am I right? I am not so sure, since nateharman1234 disagrees with me...[/quote]
He's bailing out 3 gallons per minute.  :wink:

[hide]Yes, that is correct.

The equation is just

$72(x-3)=72$

So $x=4$[/hide]

first 1.2 hours is 72 minutes.
since he bails 3 gallons out per minute, how many gallons will he have bailed out by the time it sinks? (216)
using that, how many gallons total went in to the boat, including gallons he bailed out and gallons that sunk it? (216 plus 72 to sink it or 288)
use that to find the rate of water going in. (288 gallons in 72 minutes or 4 gallons per minute)

does it seem to anyone else that he might have given up long before 72 minutes? that's a long time to bail water...


The first of a possible series of statistics. Now with an improved computer program so it's less work for me!

Try guessing the min and max of the following statistics.
[hide="USAMO qualifiers by first letter of first name"]
A: 55
B: 17
C: 28
D: 31
E: 15
F: 3
G: 13
H: 9
I: 5
J: 77
K: 27
L: 10
M: 32
N: 15
O: 2
P: 19
Q: 7
R: 22
S: 46
T: 23
U: 0
V: 6
W: 9
X: 3
Y: 15
Z: 2
[/hide]

[hide="By first letter of last name"]A: 16
B: 19
C: 48
D: 14
E: 6
F: 13
G: 23
H: 43
I: 3
J: 17
K: 27
L: 38
M: 21
N: 6
O: 5
P: 13
Q: 2
R: 9
S: 46
T: 18
U: 0
V: 2
W: 34
X: 8
Y: 26
Z: 34
[/hide]

[hide="By the number of vowels in full name as listed on the AMC website"]
[includes Y as a vowel]
1: 0
2: 16
3: 132
4: 161
5: 102
6: 55
7: 17
8: 2
9: 5
10: 1
11+: 0
[/hide]

Enjoy and good luck to all who qualified for the USAMO!

heh. i fit the majority on only one of these...
well im gonna fail USAMO so i might as well keep wandering around the forum..

[quote="randomdragoon"]By number of vowels in full name...[/quote]
Clearly, Thai names have an unfair advantage.
The champion name (Sakulbuth Ekvittayaniphon) uses y as a consonant, so it should really only be 9. It's still 9 syllables, which is very impressive.

[quote="miller4math"]heh. i fit the majority on only one of these...
[/quote]
plurality*... sorry couldn't help myself.

I'm in the same situation though.

[quote="Phizzle"]There's only one Z first name other than me? I feel more special now.[/quote]

There would be one more if I got one more AMC 10 question right.  :P

Higher numbers from the last names list: C, H, L, S, W, Z

Examples: Chen, Hwang, Li, Song, Wang, Zhang

 :rotfl: sorry, I couldn't help it

Rawr, H last names pwn.

Not so much on the P first names though.

I've good luck based on my first name(A), but not my last name(F)...

Both my last and first name spawned mediocre results.....  :(

my name has 2 vovels, 5 letters long(first and last). im quite pround of it.

My initials are JC, the most common initials.

you seem to know this "Shawn Qian", guy really, well. as a matter of fact, i know him too, but not nearly as well as you do. I also have a friend who i am not really happy with. he is on AoPS 24/7, i think i should tell him to stop getting on so much and start working on physics or Calculus or studying for ARML or ICTM, unfortunately for my friend, he didn't make USAMO, so he should probly start studying for next year.

Lol.  Who's Shawn Qian? I also think Shawn should stop telling Youkow things he doesn't tell me -_-. xD

[quote="serialk11r"][quote="ProtestanT"]Cool there are two Q's (for last name). Who's the other Q?[/quote]
You're is Qu right? I saw a "Shawn Qian" on the list. I remember only because that's my last name too :) But I failed AMC10 by misbubbling so I didn't make it :([/quote]
Interesting...

So anyways, how did you know which Q I was?

Protestant (sorry I left out the random "accidental letters"), you posted it in my thread in G&FF ;)

I think its interesting how "Qs" are usually Chinese people, even though when you think about it that's not a surprise since there's even less white people names that start with Q eh...I gotta be on that list next year, so you Qs don't feel lonely :)


function
algebra
polynomial
integration
quadratics
real analysis
real analysis theorems

A set of real-valued functions $\{\varphi_{n}(x);n = 1,2,3,...\}$ is defined on the interval $[a,b]$.

If $\{\varphi_{n}(x);n = 1,2,...,N\}$ forms a finite orthonormal set on $[a,b]$ and

$P_{N}(x) = \sum^{N}_{n=1}c_{n}\varphi_{n}(x)$

denotes a polynomial of degree $N$, where the coefficient set $\{c_{n},n=1,2,...,N\}$ contains only constants, show that the polynomial with the best mean-square approximation to $f(x)$ on $[a,b]$ is the Fourier polynomial defined by $f_{N}(x) = \sum^{N}_{n=1}\langle f,\varphi_{n}\rangle\varphi_{n}(x)$ where $\langle f,\varphi_{n}\rangle = \int^{b}_{a}f(x)\varphi_{n}(x)dx$.

It's clear from  your notation that you intend for these functions to be real valued and you're using the real inner product. The details below would have to be adjusted slightly for a complex inner product.

Start of the proof: we wish to find $c_{n}$ that minimize:

$\left\|f-\sum_{n=1}^{N}c_{n}\varphi_{n}\right\|_{2}^{2}= \left\langle f-\sum_{n=1}^{N}c_{n}\varphi_{n}, f-\sum_{n=1}^{N}c_{n}\varphi_{n}\right\rangle.$

Distribute out  the inner product. Take advantage of the orthonormality. We have a sum of quadratic expressions in the $c_{n}.$ Complete the squares in each of those. Now what do we have?


function
linear algebra
matrix
calculus
calculus computations

At which points of $\mathbb{R}^{2}$ does the function F : $\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ defined by F(x, y) = $(x^{2}+y^{2},{x}-{y})$ have a smooth local inverse? At those points where it has one, what is the differential of the local inverse?

A crude paraphrase of the Inverse Function Theorem (and Implicit Function Theorem) is that you can do smoothly what you can do linearly - the restrictions on these theorems are the restrictions imposed by linear algebra.

The Jacobian matrix of this transformation is

$\begin{pmatrix}2x&2y\\1&-1\end{pmatrix}$

If we want a local inverse, then that needs to be an invertible matrix. The determinant is $-2x-2y,$ so we have a local inverse a every point for which $x+y\ne0$ but local invertibility fails everywhere on the line $x+y=0.$


geometry incenter circumcircle Euler trigonometry

Let $ A$, $ B$ be two fixed points and $ C$ is a variable point on the plane such that $ \angle ACB\equal{}\alpha$ (constant) ($ 0^{\circ}\le \alpha\le 180^{\circ}$). Let $ D$, $ E$, $ F$ be the projections of the incenter $ I$ of triangle $ ABC$ to its sides $ BC$, $ CA$, $ AB$, respectively. Denoted by $ M$, $ N$ the intersections of $ AI$, $ BI$ with $ EF$, respectively. Prove that the length of the segment $ MN$ is constant and the circumcircle of triangle $ DMN$ always passes through a fixed point.

[quote="bilot4"]I will briefly write my solution to this beautiful problem!

First of all it is a well known lemma that: $ BN\perp AN$, $ AM\perp BM$. Moreover we have that $ \angle{AIB} \equal{} 90 \plus{} C/2 \equal{} 90 \plus{} \frac {a}{2} \equal{} constant$.
So we have that $ M,N$ are located in the circle with diameter $ AB$ and that the arc MN is constant since $ \angle{MAN} \equal{} AIB \minus{} 90 \equal{} \frac {a}{2}$. So the chord of the circle MN must be constant!

Also if $ P$ is the intersection of $ AN, BM$ we can easily see that $ I$ is the orthocenter of $ ABP$. Thus MND is the orthic triangle and that means that M,N,D are located in the Euler circle of $ ABP$. But we know that this circle passes through the midpoint of $ AB$ (constant point)! QED[/quote]
Sorry,but you are wrong    M,N,A,B are not concyclic

And I don't think the problem is true , in my figure MN is variable

[quote="plane geometry"][quote="Allnames"][quote="plane geometry"]
Sorry,but you are wrong    M,N,A,B are not concyclic

And I don't think the problem is true , in my figure MN is variable[/quote]
No,In the truth , $ M,N,A,B$ are still  concyclic because $ BN\perp AN$, $ AM\perp BM$.
And we also get
\[ MN = AB.sin{\frac {\alpha}{2}}
\]
that is a constant
For b) We have $ \angle MTB = 2\angle MAB = \angle CAB$ (where $ T$ is the midpoint of $ AB$)
But $ E,A,D,I,N$ are located in the circle with diameter $ AI$ thus
\[ \angle FND = \angle CAB = \angle MTB
\]
It means $ T$ belongs to $ (MDN)$
It is not hard enough to be a Vietnam NMO[/quote]
I still think it is a false proposal, or the statement of April is wrong  :([/quote]
no, M,N,A,B are concyclic. solve it very easy :wink:  :)

This is wrong:
[quote="April"]Let $ A$, $ B$ be two fixed points and $ C$ is a variable point on the plane such that $ \angle ACB \equal{} \alpha$ (constant) ($ 0^{\circ}\le \alpha\le 180^{\circ}$). Let $ D$, $ E$, $ F$ be the projections of the incenter $ I$ of triangle $ ABC$ to its sides $ BC$, $ CA$, $ AB$, respectively. Denoted by $ M$, $ N$ the intersections of $ AI$, $ BI$ with $ EF$, respectively. Prove that the length of the segment $ MN$ is constant and the circumcircle of triangle $ DMN$ always passes through a fixed point.[/quote]
and this is correct version, O.K. ?

Let $ A$, $ B$ be two fixed points and $ C$ is a variable point on the plane such that $ \angle ACB \equal{} \alpha$ (constant) ($ 0^{\circ}\le \alpha\le 180^{\circ}$). Let $ D$, $ E$, $ F$ be the projections of the incenter $ I$ of triangle $ ABC$ to its sides $ \color{red}AB$, $ \color{red}BC$, $ \color{red}CA$, respectively. Denoted by $ M$, $ N$ the intersections of $ AI$, $ BI$ with $ EF$, respectively. Prove that the length of the segment $ MN$ is constant and the circumcircle of triangle $ DMN$ always passes through a fixed point.

Oh, I am sorry, my mistake!  :blush: Could you please edit my first post and delete this one for me? Yetti's version is correct. Thanks in advance!

It is not hard enough to be a Vietnam NMO, I agree with you, A student in grade 9 can solve this problem by easy way!

[quote="yetti"]Sorry, [b]April[/b], but this would make a fool of [b]plane geometry[/b], so I will not do it.[/quote]
DEAR yetti，
I am sure there is sth wrong with the problem  :)  Thank you for pointing it out

G,A,D,B are harmonic conjugate  and  notice DF⊥AM => MA is bisector of ∠GMD
=> MB⊥AM
Analogously,AN⊥BN
=> M,B,A,N are concyclic   =>  MN=AB*sin∠MBN= AB*sin(90-∠MIB)= AB*sin(∠ACB/2)
MN  is constant
∠MON=2∠MBN=∠ACB  ∠MDN=180-∠DMN-∠DNM=180-∠B-∠A=∠C
=>M,N,D,O are concyclic
Done!

[quote="April"][quote="yetti"]Sorry, [b]April[/b], but this would make a fool of [b]plane geometry[/b], so I will not do it.[/quote] I am sorry, but I didn't mean to make a fool of plane geometry. I only want to correct the problem statements. Anyway, thanks for your attempt.[/quote]
that's OK,April :wink:

Sorry plane geometry,I posted my solution without checking the April 's version,I believed him :lol:
By the way,my proof is still true,isn't it?

b, Very easy.
(MND) is the Euler's circle of triangle IAB so (MNP) passes through O :D

I spent half an hour tryin to prove that $\sin^2\frac{A}{2}\sin\frac{B}{2}$ is constant. :mad: ALWAYS check well before posting.
@Mod please correct the problem

Lemma: AM is perpendicular to MB and BN is perpendicular to AN.
Proof:Angle chasing and concyclicity

From the lemma we see that ABMN is a cyclic quadrilateralwith diameter AB.By easy angle chasing $\angle{MAN}=\frac{C}{2}$ so by applying sine rule $MN=AB sin\frac{C}{2}$ which is constant as C varies.
Also ANIF and MBFI are cyclic quadrilaterals so we have $\angle{MFN}=\angle{MFI}+\angle{NFI}=\angle{IAN}+\angle{IBM}=\frac{C}{2}+\frac{C}{2}=C$
If X is the midpoint of AB we have $\angle{MXB}=C$ as X is the center of circle MBAN.Therefore the circumcircle of MFN passes through X which is fixed as C varies.

Let $O$ be the midpoint of $AB$. We claim $(DMN)$ passes through $O$. By a well-known lemma the circumcenter of $ANMB$ is $O$, and quadrilaterals $ANID$ and $BMID$ are cyclic. Note $$\measuredangle MDN=\measuredangle MDI+\measuredangle IDN=\measuredangle MBI+\measuredangle IAN=\frac{1}{2}\widehat{MN}+\frac{1}{2}\widehat{MN}=\measuredangle MON$$
By law of sines, $$\frac{MN}{\sin\alpha}=\frac{OM}{\sin{(90^\circ-\alpha/2)}}=\frac{AB}{2\cos{\alpha/2}}\implies MN=AB\sin\frac{\alpha}{2},$$
which is clearly fixed.

I spent so much time solving the problem with wrong the problem statement :mad: mods please do correct the first post

[b]Solution:[/b] From [url=https://artofproblemsolving.com/community/c6h1719239p11113686]RMO 2018 P6[/url], $ANFID$, $BMEID$ and $ANMB$ are cyclic $\implies$ $I$ is the orthocenter WRT $\Delta ATB$, where $T$ $=$ $AN$ $\cap$ $ID$ $\cap$ $BM$ $\implies$ $\odot (DNM)$ is the nine point circle WRT $\Delta ATB$ $\implies$ it passes through the midpoint of the fixed chord $AB$.
$$MN=\frac{AB \cdot MT}{AT}=AB \cos \angle ATB$$
But $\angle ATB=180^{\circ}-(\angle TAB+\angle TBA)=90^{\circ}-\frac{C}{2} \implies MN$ is fixed

I will assume $\triangle ABC$ is not degenerate since otherwise it is trivial.

Let $K$ be the midpoint of $AB$. By Iran Lemma, we have $MNAB$ is cyclic. Let $J = EF \cap AB$.

Since $(A,B;D,J) = -1$ we have $JD \cdot JK = JA \cdot JB = JN \cdot JM \implies MNDK$ is cyclic. Thus $(DMN)$ passes through a fixed point.
We also have $\alpha = 2\angle NAM = \angle NKM$ since $K$ is the center of $(ANMB)$.

So, $MN = 2\cdot KN\cdot \sin(\alpha/2)= AB\sin(\alpha/2)$ which is fixed. We are done.


function
real analysis
real analysis solved

$f:[0;1] \rightarrow R$ continuous
$f(0)=f(1)=0$
$\forall x\in [0;\frac{7}{10}]$, $f(x+\frac{3}{10})\neq f(x)$

1/ prove that $f(x)=0$ has at least seven solutions
2/ give an example of such function

[quote="Moubinool"]
1/ prove that $f(x)=0$ has at least seven solutions
[/quote]

I know that I am like Cindirella, but after midnight instead of become ugly dressed, I become stupid!  :D . Any way, may be I am just not seeing something that is infront of my nose. Any way, if we consider 0 and 1 as part of the solutions, I can prove that the function you describe has four more roots, but somehow I cannot prove that it has to have a fifth one to have seven. To have a fifth root, I need $f\left(x+\frac14\right)\neq f(x)$ for $x\in\left[0,\frac34\right]$. Are you sure about the statement?

[quote="djimenez"] Are you sure about the statement?[/quote]
yes it was like this

Here's an example which, BTW, has exactly $7$ roots :).

Consider the piecewisely continuous function connecting the points $(0,0),(1,-1),(\frac 83,\frac 23),(4,-\frac 23),(\frac{17}3,1),(7,-\frac 13),(\frac{26}3,\frac 43),(10,0)$. This yields a function defined on $[0,10]$ s.t. $f(x+3)>f(x),\ \forall x\in[0,7]$. If we divide all the values by $10$, we get what we want.

Well, a little late, but here it is my solution.

[hide="My Solution"]
Note first that the conditions over $f$ are symmetric with respect to the sign. Now, as $f\left(x+\frac{3}{10}\right)\neq f(x)$ for all the $x$ that make sence, then the functions $g_k(x)=f\left(x+\frac{3k}{10}\right)$ defined on $\left[0,1-\frac{3k}{10}\right]$ are continous and non-zero, then, we can assume that they are always possitive. Therefore, it is not hard to see that $f\left(\frac{1}{10}\right)<0$, $f\left(\frac{3}{10}\right)>0$,  $f\left(\frac{4}{10}\right)<0$, $f\left(\frac{6}{10}\right)>0$, $f\left(\frac{7}{10}\right)<0$ and $f\left(\frac{9}{10}\right)>0$, therefore, you apply the intermediate value theorem and you have exactly 7 roots (taking in consideration 1 and 0).
[/hide]

But I still don't have an example  :(

Best,


Chinese Northern Region Mathematical Olympiad 2005 was held on July 28-29, in Shi jiazhuang, Hebei. Shobber had posted problem 2, here are the whole set of problems.

1. http://www.mathlinks.ro/Forum/viewtopic.php?t=50691

2. http://www.mathlinks.ro/Forum/viewtopic.php?t=50578

3. http://www.mathlinks.ro/Forum/viewtopic.php?t=50687

4. http://www.mathlinks.ro/Forum/viewtopic.php?t=50690

5. http://www.mathlinks.ro/Forum/viewtopic.php?t=50688

6. http://www.mathlinks.ro/Forum/viewtopic.php?t=50689

Then the pdf files.

I want  to say that the problem is really veryvery easy since I get all the solutions in 15 minutes.The style of the problem is old as well.I HAVE THE SOLUTION[not official],BUT I THINK IT IS OF NO USE. :P


I give you here a link, where you will find the photos I took in Japan :

http://membres.lycos.fr/shadowmare/photos.imo

( yes, I know, the ones which are not blurred are all the same... :? , but it's better than nothing...)

ool

Singapore team's photos: [url=http://f1.pg.photos.yahoo.com/ph/charmainesia/lst?.dir=/IMO+2003&.src=ph&.order=&.view=t&.done=http%3a//f1.pg.photos.yahoo.com/]Here[/url]

The other teddy bear in the photos is the South African teddy bear, Chebyshev...and a bit of imagination will tell you why the Polish flag (white on top, red at the bottom) is half as long as it should be... :D


calculus
combinatorics proposed
combinatorics

How many numbers which have n digits : a_1-a_2-.....-a_n satisfy for i:=1 to n-1 : a_i and a_(i+1) can be chosen in the set {1,2,3,4,5} and ( a_i - a_(i+1) )^2 = 1 ?

This one is Baltic Way 1994.

Hint: Try to use induction. Put s(n,i) number of such numbers with n digits and last digit is i then s(n,1) = s(n-1,2) ...

Namdung


In the convex  quadrilateral $ ABCD$,the points $ G_1,G_2,G_3,G_4$ are center of gravity for next triangles $ BCD,ACD,ABD$ and $ ABC$.Prove that lines  $ AG_1,BG_2,CG_3,DG_4$ have a common point.

Let $ M,N,P$ be the midpoints of $ AC,BD,DC.$ Since $ \frac {_{NG_3}}{^{NA}} \equal{} \frac {_{NG_1}}{^{NC}} \equal{} \frac {_1}{^3},$  it follows that $ G_1G_3 \parallel AC$ and $ G_1G_3 \equal{} \frac {_1}{^3}AC.$ Similarly, $ G_2G_4 \parallel BD$ and $G_2G_4 \equal{} \frac {_1}{^3}BD.$

Since $ \frac {_{PG_2}}{^{PA}} \equal{} \frac {_{PG_1}}{^{PB}} \equal{} \frac {_1}{^3}$ $ \Longrightarrow$ $G_1G_2 \parallel AB,$  $G_1G_2 \equal{} \frac {_1}{^3}AB.$ By similar reasoning, we conclude that quadrilaterals $ G_1G_2G_3G_4$ and $ ABCD$ are centrally similar with similarity coefficient $ \minus{} \frac {_1}{^3}$ $\Longrightarrow$ $ AG_1,$ $BG_2,$ $CG_3,$ $DG_4$ concur at their similarity center.


If $a,b,c\geq0$, prove or disprove the inequality
\[ \frac32(a^2+b^2+c^2)\geq\sqrt{a^4+b^3c}+\sqrt{b^4+c^3a}+\sqrt{c^4+a^3b}.  \]

The inequality is clearly false. We only need to choose $c=0$.

Hmm, what a pity. Let $c=0$, the desired one reads
\[ a^2+b^2\geq2\sqrt{a^3b}. \]
and this oneis not true.

Problem even of yours which you can't solve should be given in Unsolved Topic. It's very tired to solve your inequalities which can be peasuded true or false.

[quote="pvthuan"]While others are interested in solving problems, and I agree that it's a good and worthy thing to do. At the moment I hurry to create new problems (either original or just some inspiration from others). Sometimes, I just want to hear from different views. That is also the main reason we come to a forum like this one.

I know some forums in Vietnam and some students there are not modest enough. Just too arragant and boastful.

One thing I like about this forum is that I can meet excellent people but they are very modest and polite.[/quote]

But if you post many incorrect problems which never think about they carefully, it's really not good. Why? Because the reader is very tired to solved problems which at first he think it's true and cost a lot of thought but finally, he realize that it was false. It' very terrible (because problems posted in "Ineq Proposed and Own" need to be think cautious, or you can post in Unsolved topic. It's better for the readers, or you will make them ...). I think it's also your responsibilties for your post.

Sorry If it make you unplesant, it's my opinion.

Actually, I have tried to do so, you can find many of them in the unsolved problem section.


Obviosly $ x\equal{}\sqrt[3] 2$ is not solution and $ x\equal{}k\pi , k\in Z$ solution.
For other solutions we get $ f(x)\equal{}g(x)$, were $ f(x)\equal{}\frac{3^{sinx}\minus{}1}{sinx}, \ g(x)\equal{}\frac{2^{x^3\minus{}4}}{8\minus{}4x^3}$.
We get $ \frac 23\le f(x)\le 2$. Therefore must be $ x^3<2$. Let $ y\equal{}2\minus{}x^3>0$, then $ g(y)\equal{}\frac{2^{\minus{}y}}{16y}.$
Therefore we get one solution $ x\equal{}\sqrt[3] 2\minus{}\epsilon, \epsilon<0.1$.


How many ordered triples (x, y, z) of integers satisfy the system below?

x^2 - 3xy + 2y^2 - z^2 = 31
-x^2 + 6yz + 2z^2 = 44
x^2 + xy + 8z^2 = 100

(AHSME 1980, #29)

Add all the equations up we get
[tex]x^2-2xy+2y^2+6yz+9z^2=175[/tex]

[tex](x-y)^2+(y-3z)^2=175[/tex]

All perfect squares are 0 or 1(mod 4),  [tex]175 \equiv 3(\mod 4)[/tex] so 175 can't be expressed as sum of squares. There is no integer solution


The Northern lighthouse flashes every $ 18$ seconds. The Southern lighthouse flashes every $ 15$ seconds. The two lighthouses flash together. How many seconds will it be before the next time they both flash together?

[hide]The LCM of 15 and 18 is[b] 90[/b], which is the answer.[/hide]

:) LCM of 15 and 18.
            Which is 90


geometry
3D geometry
sphere
linear algebra
matrix
vector
linear algebra unsolved

In space $ \mathbb{R}^n$ , ($ n \in \mathbb{N}$,$ n \geq 1$) Let $ (n \plus{} 1)$ dinstinct points $ A_1,A_2,...,A_{n \plus{} 1}$ such that :
 $ \forall k_1\neq k_2,k_3 \neq k_4 , (k_1,k_2,k_3,k_4) \in \{1,2,..,n \plus{} 1\}^4,A_{k_1}A_{k_2} \equal{} A_{k_3}A_{k_4}$ (or $ A_1A_2 \equal{} A_1A_3 \equal{} ... \equal{} A_1A_{n \plus{} 1} \equal{} A_2A_3 \equal{} A_2A_4 \equal{} ... \equal{} A_2A_{n \plus{} 1} \equal{} ... \equal{} A_nA_{n \plus{} 1}$)
and lying in a sphere $ \mathbb{R}^n$ $ (O_1)$ have center $ O$ and radius $ r_1$ . A point $ M$ in space $ \mathbb{R}^n$ moves on the sphere $ \mathbb{R}^n$ $ (O_2)$ have center $ O$ and radius $ r_2$
1) Prove that the unique exist points $ A_1,A_2,..,A_{n \plus{} 1}$
 I call $ A_1A_2...A_{n \plus{} 1}$ a regular $ (n \plus{} 1)$  plane
 example : $ n \equal{} 2$ : $ A_1A_2A_3$ be an equilateral triangle
                 $ n \equal{} 3$  : $ A_1A_2A_3A_4$ be an regulare tetrahedron....
2) Let $ MA_1 \equal{} a_1,MA_2 \equal{} a_2,...,MA_{n \plus{} 1} \equal{} a_{n \plus{} 1}$
 Calculate :
 $ \det\begin{bmatrix}1 & \minus{} a^2_1 & \minus{} a^2_2 & ... & \minus{} a^{2}_{n \plus{} 1} \\
a_1^2 & 1 & 0 & ... & 0 \\
a_2^2 & 0 & 1 & ... & 0 \\
... & ... & ... & ... & ... \\
a_{n \plus{} 1}^2 & 0 & 0 & ... & 1\end{bmatrix} \equal{} ?$
[hide="Answer of me"]$ 1 \plus{} (n \plus{} 1)(r_1^2 \plus{} r_2^2)^2 \plus{} 4\frac {n \plus{} 1}{n}r_1^2r_2^2$[color=red] (Answer of me True of false ?)
In $ C^n$ Does It correct ? [/color]
I hope that you are interested it (thanks you very much  :) ) [/hide]

[quote="t0rajir0u"]Be careful; the word "super" refers to something [url=http://en.wikipedia.org/wiki/Super_vector_space]different[/url].[/quote]
 Oke I am stupid in english  . I  edit  :blush:
 I hope that you are interested it ?


1.what is the probability that a randomly chosen pair of diagonals in a n-gon intersect?

2. what is the maximum number of regions the diagonals of an n-gon make?

EDIT: CONVEX polygons is what I meant

[hide="1st hint for #1"]Number the vertices 1-N[/hide]

[hide="2nd hint for #1"]Assuming the polygon is convex, the diagonals intersect iff the segments joining the vertex-numbers on a number line overlap.[/hide]

a).
To form a diagonal , we need 2 non- adjacent vertices.The total no. of ways of selecting 2 vertices out of n is C(n,2).This number also contains the selections where two vertices r adjacent.Those selections are simply n in number bcoz the polygon has n sides.Thus,the total number of diagonals = C(n,2)-n which after solving give
u {n*(n-3)/2}.

[hide="hint for #1"]Consider any quadrilateral formed with four vertices of the polygon. The two diagonals will intersect.[/hide]

xpmath: I modeling number 1 like that, but I am having a hard time coming up with a model counts the number of quadrilaterals formed by vertices of the polygon that does not include the edges of the polygon (or should I somehow correct for this overcounting later?), so I tried counting the number of quadrilaterals with non consecutive vertices (on the polygon), but I cannot find a model for this either, since the vertices go around in a circle (it would be A LOT easier if the vertices were just in a row)

basically: xpmath and Darmani: I have beforehand noticed those hints you gave me (but thanks anyways for your help), but I have not been able to solve this with those hints.

I would greatly appreciate more hints or help.

Thanks.

EDIT: kabi, I know the formula for the number of diagonals, but there seems to be a lot more to both of these problems.

EDIT: Yea they should be convex polygons

I think number 2 is $ \dbinom n0 \plus{} \dbinom n2 \plus{} \dbinom n4 \minus{} n$

[hide="1"]
$ \dfrac{\dbinom n4}{\dbinom{\dfrac{n(n \minus{} 3)}2}{2}} \equal{} \dfrac{\dfrac{n(n \minus{} 1)(n \minus{} 2)(n \minus{} 3)}{24}}{\dfrac{\dfrac14n^4 \minus{} \dfrac32n^3 \plus{} 2n^2 \plus{} \dfrac34n}2} \equal{} \frac {(n \minus{} 1)(n \minus{} 2)(n \minus{} 3)}{3(n^3 \minus{} 6n^2 \plus{} 8n \plus{} 3)} \equal{} \boxed{\frac {(n \minus{} 1)(n \minus{} 2)}{3(n^2 \minus{} 3n \minus{} 1)}}$.
[/hide]
[hide="2"]
First, we have $ 1$ region, the polygon itself.

Then, each time we draw a diagonal, we add a region, giving us $ 1 \plus{} \dbinom n2 \minus{} n$ so far.

However, if we drew a diagonal that intersected another diagonal, we would have to add another region.

In every quadrilateral, there are two unique intersecting diagonals, so we now have

$ \dbinom n0 \plus{} \dbinom n2 \plus{} \dbinom n4 \minus{} n$ (as hurdler said), or

$ \frac{n^4\minus{}6n^3\plus{}23n^2\minus{}42n\plus{}24}{24}$

$ \equal{}\boxed{\frac {(n \minus{} 1)(n \minus{} 2)(n^2 \minus{} 3n \plus{} 12)}{24}}$.
[/hide]

oh ok thanks. I was getting confused for some odd reason. That makes sense now


abstract algebra
algebra
polynomial
number theory solved
number theory

Let $A$ be a natural number.

Prove that there exist a natural number like $M$ such that

$A \mid 1^n+2^n+...+M^n$ for every $n \in \mathbb{N}$

When A is odd, take M=3A.
When A is even, take M=12A.

Moreover, actually there exists infinitely many M satisfying the conditions.

Am i missing something? :?
hmm,let $A=P>3$ and $n=p-1$ then

we should have

$A \mid 1^{p-1}+2^{p-1}+...+(3p)^{p-1}$

but we have that

$1^{p-1}+2^{p-1}+...+(3p)^{p-1} \equiv 1+1+...+0+1+1...+0+1+1...+1+0=3p-3 (\text{module P})$
(Every thing will be one beside of $p,2p$ and $3p$ which are zeros)

but
$P \not| 3p-3$

So $3A$ doesnt work , :? ?

ooops :blush:
sorry for my stupid mistake... :(

Well i have solved it for the primes:
[hide="solution"]

Lemma:
Let $p$ be a prime and $k \in N$ and $p-1\not|k$ the
$p|1^{k}+2^{k}+...+p^{k}$ ,

Put $M=p^{2}$ ,So for the $n$s that $p-1\not|n$ then by using the lemma $P|\sum_{i=1}^{p^{2}}i^{n}$ and for the

$n=k(p-1)$ then

${1^{p-1}}^{k}+{2^{p-1}}^{k}+...+{{p^{2}}^{p-1}}^{k}\equiv 1+1+...+0+1+...+0+1+....+0=p^{2}-p (\text{module P})$
every thing will be 1 beside of the $p,2p,...p^{2}$ so

$p|\sum_{i=1}^{p^{2}}i^{k(p-1)}\equiv p^{2}-p$[/hide]

Take $M= A!\cdot A - 1$. In fact, that the condition $A | P(1) + ... + P(M)$ for all $P(x)=x^n$ is equivalent to the statement that this is true for all $P\in \mathbb{Z}[x]$ which in turn is equivalent to the statement that it is true for all $P(x) = n! {x\choose n }$. Now, this condition is easy to evaluate (for $n\geq A$ it is trivial; for $n<A$ we can show this by noting that $\sum_{i=1}^{M} n!{i\choose n} = n! {{M+1}\choose {n+1} }$ is divisible by $\frac{M+1}{n+1}$ - but $n+1$ divides $A!$ so we are done).

[quote="me@home"]I don't see how that could be wrong... that's amazing they overlooked such an answer![/quote]

No one overlooked anything , I said $M=P^{2}$ is a solution for primes , Darij rewrote it for all naturals. Peter said problem condition has been simplified.

I don't completely understand what Peter said, how can he prove that it works for all polynomials (which is what he meant by $\mathbb{Z[x]}$, right?) from the fact it works for $x^{n}$ (oh wait is he just using combinations of divisibility congruences?) also the second equivalence ($P(x) = n!{x\choose n }$)
And did Darij use any of these facts?


inequalities
Euler
inequalities proposed

Prove that

$ \frac{ab}{a^2\plus{}b^2}\plus{}\frac{bc}{b^2\plus{}c^2}\plus{}\frac{ca}{c^2\plus{}a^2}\geq\frac{3r}{R}$

.

[quote="Ligouras"]Prove that

$ \frac {ab}{a^2 \plus{} b^2} \plus{} \frac {bc}{b^2 \plus{} c^2} \plus{} \frac {ca}{c^2 \plus{} a^2}\geq\frac {3r}{R}$

.[/quote]

using CS, we have

$ \frac {ab}{a^2 \plus{} b^2} \plus{} \frac {bc}{b^2 \plus{} c^2} \plus{} \frac {ca}{c^2 \plus{} a^2}\geq \frac{(ab\plus{}bc\plus{}ca)^2}{\sum{ab(a^2\plus{}b^2)}},$

so we just need to prove that

$ \frac{(ab\plus{}bc\plus{}ca)^2}{\sum{ab(a^2\plus{}b^2)}} \ge \frac {3r}{R}$

$ \Leftrightarrow {\frac{({s}^{2}\plus{}4Rr\plus{}{r}^{2})^{2}}{2{s}^{4}\minus{}8{s}^{2}Rr\minus{}32{R}^{2}{r}^{2}\minus{}16R{r}^{3}\minus{}2{r}^{4}}} \ge {\frac {3r}{R}}$

$ \Leftrightarrow f(s^2)\equal{}(R\minus{}6r)({s}^{2})^2\plus{}(26R{r}^{2}\plus{}8{R}^{2}r){s}^{2}\plus{}104{R}^{2}{r}^{3}\plus{}49R{r}^{4}\plus{}16{R}^{3}{r}^{2}\plus{}6{r}^{5} \ge 0$,

if $ R \ge 6r$, ineq holds;

if $ R<6r$, by Gerretsen Inequality $ 4R^2\plus{}4Rr\plus{}3r^2 \ge s^2 \ge 16Rr\minus{}5r^2$, we just need to prove that $ f(4R^2\plus{}4Rr\plus{}3r^2) \ge 0$ and $ f(16Rr\minus{}5r^2) \ge 0$, actually, by Euler Inequality, we have

$ f(4R^2\plus{}4Rr\plus{}3r^2) \equal{} 8(R\minus{}2r)(3{r}^{4}\plus{}2R{r}^{3}\plus{}2{R}^{4}) \ge 0$

$ f(16Rr\minus{}5r^2) \equal{} 8{r}^{2}(R\minus{}2r)(9{r}^{2}\minus{}52Rr\plus{}50{R}^{2}) \ge 0$

proven.


Let $A;B$ two distinct 2004-digit positive integers ,both of which have one thousand 1s,eight hundred 2s;two hundred 3s,and four 4s.Prove that they are not divisible by each other

[hide="My solution"] Suppose that $A$ is divisibled by $B$
let $k=\frac{A}{B}$
It's easy to prove that $2\le k\le 4$
But both $A$ and $B$ are divisibled by 3 but not 9 so k couldnt be 3.
*)If $ k=2$ $A$ must contain four 8s.
*)If $ k=4$ $A$ must contain at least one 6.
[/hide]


group theory
abstract algebra
complex analysis
function
logarithms
topology
advanced fields

$A$ a Banach algebra with a unit element, $A^{-1}$ the group of invertible elements of $A$.
1/Show that $x \in A^{-1}$ is of the form $exp(a)$ if, and only if, there exists a connected commutative subgroup $H$ of $A^{-1}$ that contains $x$.
2/ show that if $0$ belongs to the unbounded connected component of $\mathbb{C}-Sp_A(x)$ then $x$ is of the form $exp(a)$

(2) In order to prove this properly, on would have to develop a few tools, but I'll write the short version. Given an open subset $D$ of $\mathbb C$, and the open subset $A_D$ of $A$ consisting of all those elements of $A$ whose spectra are contained in $D$, one can define "holomorphic" functions defined on $A_D$ with values in $A$ via countour integration. We get an application sending a function holomorphic in $D$ to such a function holomirphic in $A_D$. This application preserves products, sums, etc. (i.e. it's an algebra isomorphism) and also compositions.

We can do the above for an open simply connected set $D$ containing the spectrum $\sigma(x)$ of $x$ but which doesn't contain $0$. Since there is a holomorpic logarithm $\log$ defined in $D$, and $\exp\circ\log=\mbox{id}$, the functions defined on $A_D$ and on $A$ respectively corresponding to $\log$ and $\exp$ satisfy the same identity, and we're done.

There's a small section on this subject (extensions of holomorphic functions to Banach algebras) in Rudin's "Functional Analysis".

(1)

If $x=\exp a$, then we can take $H=\{\exp(ta)\ |\ t\in\mathbb R\}$

Conversely, it suffices to prove that the connected component of $e$ (the identity in $A$) of the group $A^{-1}$ is contained in the image of $\exp$ when $A$ is a commutative Banach algebra (instead of our initial $A$, we work with the Banach algebra generated by $H$, and this algebra is indeed commutative).

The application mapping the points of $A$ to their spectra is upper semicontinuous, in the sense that given $x$ and an open set $D$ containing $\sigma(x)$, there is a $\delta>0$ such that $\|y-x\|<\delta\Rightarrow\sigma(y)\subset D$. Since the spectrum of $e$ is $\{1\}$, this means that elements of $A^{-1}$ sufficiently close to $e$ (in a neighborhood $V$ of $e$, say) satisfy the hypothesis of (2), and hence lie in the range of $\exp$. The group generated by $V$ is, on the one hand, open (if it contains $x$, it contains $xV$), and on the other hand closed (if $x_n$ belong to this group and $x_n\to x$, then for large enough $n,\ x_n^{-1}x\in V$, and $x$ belongs to the group as well). Since the group contains $e$, this means that it contains the entire component of $e$. Because $A$ is commutative, $\exp a\cdot\exp b=\exp(a+b)$ holds, so the entire group generated by $V$ is contained in the range of $\exp$ (exactly what we wnated to prove).