I just have six problems that I don't understand in PSAT Writing Section. I wish I was good at Writing/Vocabulary like Math (I got perfect score without really trying!  :D ) but that's not the case..  They were supposed to be label as A,B,C,D, and E but since I don't know how to do without using Latex, I'll put them in different color.

1. Because Roberto wants [color=blue]to help preserve [/color] marine life, he intends [color=darkblue]on declaring[/color] marine biology [color=indigo]as[/color] his major when he [color=green]begins college[/color] next fall. [color=violet]No error[/color]

2. This year's benefit game [color=blue]drew[/color] [color=darkblue]nearly twice[/color] [color=indigo]as many[/color] spectators [color=green]as last year[/color]. [color=violet]No error[/color]

3. [color=blue]Opposite to[/color] the opinion of several respected literary critics, Jane Austen [color=darkblue]does not make[/color] good taste or manners in themselves sure [color=indigo]signs of[/color] virtue [color=green]in her characters[/color]. [color=violet]No error[/color]

4. [color=blue]If not[/color] in [color=darkblue]good[/color] physical condition, a bicycle excursion through the countryside [color=indigo]can be[/color] [color=green]painfully exhausting[/color]. [color=violet]No error[/color]

5. [color=blue]Setting herself[/color] up [color=darkblue]as[/color] a preserver of the Classical tradition, Clara Schumann was an [color=indigo]outspoken[/color] [color=green]critic toward[/color] Romantic composer Franz Liszt. [color=violet]No error[/color]

6. [color=blue]When looking[/color] at modern photograph of that area of Indonesia, the effects of the 1883 [color=darkblue]eruption of[/color] the volcano Krakatau [color=indigo]are[/color] [color=green]still evident[/color]. [color=violet]No error[/color]

I hope this is OK to post.. I'm taking PSAT next week and I want to make sure I understand everything in the packet..  :lol:

[hide="Answers"]
1. on declaring
2. as last year
3. Opposite to
4. If not
5. critic toward
6. When looking [/hide]

1. [hide][i]to help[/i] is an infinitive, so [i]on declaring[/i] should also be an infinitive - [i]to declare[/i].[/hide]
2. [hide][i]This game[/i] is being compared to something - it should be [i]last year's game[/i], or, more casually, [i]last year's[/i].[/hide]3. [hide][i]Opposite to[/i] is not a gramatically correct phrase, in general.  The context here suggests [i]Contrary to[/i] or [i]In contrast to[/i].[/hide]
4. [hide][i]If not[/i] makes the good condition refer to the bicycle excursion.  Perhaps, [i]If one is not[/i].[/hide]5. [hide]One is not a [i]critic to[/i] - most appropriate might be [i]critic of[/i].[/hide]
6. [hide]I'm guessing you missed a letter or a word or something around "modern photograph"?  The lack of a subject at the beginning of the sentence makes "the effects" the subject.  It might be replaced by [i]When one looks[/i].[/hide]

Hope that helps.

[quote="Silverfalcon"]Thanks. That helps a lot.

I just few questions though.

1. Why is phrase "Oppsite to" grammatically incorrect? I know phrase like "Contrary to.." or "In contrast" is used more often but still,  :? .

2. For #5, what's the problem with using "critic toward" instead of "critic of"? I'm not quite sure when do you use word "toward."

And last question (irrelevant to the above questions):

I know I have no way to prepare for PSAT this year but I want to be prepared for next year because I'll be junior (taking both PSAT and SAT!! And I heard PSAT picks national scholarships too!) and wondering how I can prove my critical reading/writing skill. What texts do you recommend?  :)

I know some people in this site got perfect score or close to on the SAT so I hope I get really good responses..  :)[/quote]

1. "Opposite to" is just not used, and thus gramatically incorrect. The correct phrase would be "opposite of."

2. I believe "toward" indicates some sort of direction, so it wouldn't work here. I'm not completely sure about that, though.

[quote="inteluser"]"Opposite of" doesn't fit in the sentence, though.  It's used in the context of, "One is the opposite of the other."  But it is correct that "opposite to" is never correct.

"Toward" isn't used only for physical direction though - "I feel derision towards his plan."  However, it is used with a verb in all cases.  Therefore, "a critic towards" is not correct.  Note that it is not always correct to use "towards" with a verb, either.

Hope that helps.[/quote]

Thanks! :)

[quote="elston"]Anybody here taking the PSAT at their school on Wednesday?[/quote]

Yup. BTW, there's another thread on this.


function
calculus
derivative
limit
logarithms
real analysis
real analysis unsolved

Let M be a large real number. Explain briefly why there must be exactly one root w of the equation Mx=e^x with w > 1. Why is log M a reasonable approximation to w? Write w = log M  + y. Can you give an approximation to y, and hence improve on log M as an approximaton to w?

$ f(x)\equal{}\frac{e^x}{x}$ is a strictly increasing function on $ [1,\infty)$ (easily shown after differentiation) and $ \lim_{x \to \infty} f(x) \equal{} \infty$.
So $ Mx\equal{}e^x$ has exactly one solution for large $ M$.
Now $ x \equal{} \log(M) \plus{} \log(x)$.
If $ x>1$ then $ x > \log(M)>2$ for large $ M$.
Thus from now on we suppose that $ x>2$.
Then $ \log(x) < \frac12 x$.
and  $ x \equal{} \log(M) \plus{} \log(x) < \log(M) \plus{} \frac12 x$.
Then $ \frac12 x < \log(M)$ and $ \log(x) < \log(2) \plus{} \log(\log(M))$
Now we have  $ x \equal{} \log(M) \plus{} \log(x) \equal{} \log(M) \plus{} O(\log(\log(M))$.
Taking on both sites logarithms:
$ \log(x) \equal{} \log( \log(M) \plus{} O(\log(\log(M))) \equal{} \log(\log(M)) \plus{} \log(1\plus{}O(\frac{\log(\log(M))}{\log(M)}))$.
$ \equal{}\log(\log(M)) \plus{} O(\frac{\log(\log(M))}{\log(M)})$.
Substitute this result in $ x \equal{} \log(M) \plus{} \log(x)$.
$ x \equal{} \log(M) \plus{} \log(x) \equal{} \log(M) \plus{} \log(\log(M)) \plus{} O(\frac{\log(\log(M))}{\log(M)})$..

Thanks.

There must be an easier way of doing this as the question is aimed at students in their final year of school who probably would not have met O-notation.

Can anyone supply an easier (to understand) solution?

Can anyone show me a simpler way of doing this?

Consider the function $ g(x)\equal{}\frac{e^x}{x}$. Now, $ g(1)\equal{}e$ and $ g'(x)\equal{}x^2e^x(x\minus{}1)$ for $ x>1$, so $ g$ is increasing. This means that for every large $ M$, there is an $ x$ such that $ g(x)\equal{}M$, hence $ e^x\equal{}Mx$.

To see why  $ \log M$ is a good approximation to $ x$, take log on both sides of the equation $ e^x\equal{}Mx$, and get $ x\equal{}\log M \minus{} \log x$. The idea is now that $ x>>\log x$ for large $ x$, and $ \log M$ is a good approximation.

The precise argument for the 3rd question is in Peter's post.

An please: One post is enough, 6 posts nagging for another solution is annpoying..


inequalities
trigonometry
inequalities unsolved

In two Triangle $ABC$ and $A'B'C'$ ,the following inequality is true(or not)?
$ \sin\frac{A}{2} \sin\frac{A'}{2} +\sin\frac{B}{2}\sin\frac{B'}{2}+\sin\frac{C}{2}\sin\frac{C'}{2} \leq \frac{3}{4}(\tan\frac{A}{2}\tan\frac{A'}{2}+\tan\frac{B}{2}\tan\frac{B'}{2}+\tan\frac{C}{2}\tan\frac{C'}{2})$

Wrong. Take $A=B'=0, C=C'=\epsilon\to 0$.

Edit: I was just told that it is not at all clear what I mean, so if you want a real counterexample, take $A=B' = 0.02^{o}$, $C=C'=50^{o}$ and$A'=B=129.98^{o}$.


Find all positive integers (x,y) such that
$7^{x}-3^{y}= 4$

Grrr, this is annoying...

the only significant thing I've come up with so far is that $x \equiv y \pmod 2$...

You can add something:  $x\equiv y\equiv1\pmod2$ :)

Re write it as $7(7^{x-1}-1)=3(3^{y-1}-1)$

[quote="manuel"]You can add something:  $x\equiv y\equiv1\pmod2$ :)

Re write it as $7(7^{y-1}-1)=3(3^{y-1}-1)$[/quote]Do you mean $7(7^{x-1}-1)$?


I have a problem. Usually whenever I try to log in, the only way I can log in is when I click the "log in automaticly button". But Iv found out that the whenevr I do click that button and log in, when i try to reply to posts, i get the message "invalid session please resubmit the form" which prevents me from posting. Whenever I wouldn't click that button and acually am able to log in(very rarly), I would not get the "invalid session" notice when posting. Iv tried this on 3 different computers each on opera,firefox,and IE , and they all work the same way. Recently, Iv tried deleting all Private data and I have been able to log in without clicking the button, which has allowed me to post on the forum. Im not very good with computers at all, but im pretty sure this is a cookie issue. But the problem is that even when my browser is on "accept all cookies," it still wont log me in unless i click the "log in automaticly button", which therefore wont allow me to post on the forum since its a "invalid session". Its a pain deleting all my private data every time i go to AoPS (which is about everyday) so does somone know about this bug( if it is a bug)? If this problem has already been solved, you can delete this topic.

This is very strange. Where were the 3 computers you tried this on, and when did you test this, first of all.

where? well these computers are all at home and use the same WLAN connection, and i tested it like a week ago on all computers, and i still test it every day on my laptop. any ideas whats wrong?

Seems like something is wrong with either your connection setup or your account settings. To make it completely sure which one is the case, one can try to login from some other place using your username and password (you can try it yourself from school or local library; any admin can also do it from his PC) and to try to login from one of your computers to a different account. If you have somebody like your sibling or neighbor or schoolmate who has an account on AoPS, just ask him to try it; the duplicate accounts are generally a strict No-No, but if no admin or somebody else who knows better answers you soon, I guess you'll not be really penalized for setting one more account and trying it for login (just do not use it for posting except, maybe, in the test forum!). If you do it, choose a shorter user name as different from your current one as possible (though I doubt this is the problem). If that other account has no problem with it on your computer, it will indicate that the problem is most likely on the AoPS side and you'll have to wait for Valentin to figure out what it is and fix it. If you have the same problem with another account (and/or if you have no problem with your account on the computers outside your home), it will almost certainly mean that the problem is on your end and you'll need to take a close look at your internet connection settings.

Definitely the problem is not on our side, otherwise a lot more people would be complaining. This is a local issue with your cookies. Try completely erasing your cookies.


By Pythagorean's Theorem, $ AC\equal{}\sqrt{AB^2\plus{}AC^2}\equal{}\sqrt{7^2\plus{}(\sqrt2)^2}\equal{}\sqrt{51}$


Here are the problems and their proof (Vietnamese version).

We haven't translated it to English version. I think it will be not easy to read, especially for the Geometrical Problems. :(

Anyway, here are they. :)

Thank you very much, Honey_S, for this very nice document :)


invariant
induction
combinatorics proposed
combinatorics

Prove that each finite set of integers can be arranged without intersection.

orl, maybe you could restate some of the problems you've recently posted? For instance, I have no idea what this one means :?. First of all, what does "arranging a finite set of integers" mean? Second of all, what does "intersection" mean in this case?

Sorry if this is common language, it's not common to me.. :)

Thanks!

[quote="grobber"]

Sorry if this is common language, it's not common to me.. :)

Thanks![/quote]

Maybe I should define it first what I want to know.  :lol:

In the sequence $\{a_1,a_2,a_3,a_4\} = \{1,3,6,5\}$ and $a_2 = \frac{a_1 + a_4}{2}$ is the arithmetic mean of $a_1$ and $a_4$. A sequence $\{a_1,a_2,a_3,a_4\}$ of integers is called free of intersections if there are no indices $i,j,k$ with $1 \leq i < j < k \leq n$ and $a_j = \frac{a_i + a_k}{2}$, e.g. that there is no element which is the arithmetic mean of its left and right neighboring element.

Hint: Find some operations which remains a the sequence invariant.

Ah, ok. Thus, I still do not see where there must be an invariant (unless you are talking about induction) but here it goes :
Let $n$ be a positive integer. And $U_n = a_1,..., a_n $ be a sequence of positive integers.
We will say that $U_n$ is good if no term  $a$ is the arithmetical mean of two other terms of the sequence one on each side from $a$.
Thus both $U_3 = -1,1,0$ and $U_6 = -2,2,0,-3,1,-1$ are good.

Now assume that $U_n$ is good.
Let $U_{2n} = 2a_1,2a_2, \cdots , 2a_n, 2a_1-1,2a_2-1, \cdots, 2a_n - 1 = b_1, \cdots, b_{2n}.$

We will verify that $U_{2n}$ is good too :
Let $1 \leq i<j<k \leq 2n$.
If $i \leq n$ and $k>2n$ then $b_i$ and $b_k$ have opposite parities so that their sum is not equal to $2b_j$.
If $k \leq 2n$, then $b_i = 2a_i,b_j = 2a_j,b_k = 2a_k$, and the induction hypothesis ensures us that $b_i+b_k \neq 2b_j.$
If $i > 2n$ then $b_i = 2a_{i-2n} - 1,b_j = 2a_{j-2n}-1,b_k = 2a_{k-2n}-1$, and again $b_i+b_k \neq 2b_j.$

This proves that $U_{2n}$ is good, as claimed.

Now, it is easy from $U_3$ and $U_6$ given above to verify that $U_{3 \cdot 2^n}$ is a permutation of $E_n = \{-2^{n+1} , -2^{n+1} + 1, \cdots, 0, 1 ,2, \cdots, 2^n \}$.
Then, since any finite set of integers is contained in such set $E_n$ for sufficientely large $n$ and since to obtain a good permutation of our given set it suffices to delete the members of $E_n$ which do not belong to our given set (so that the sequence remains good), we are done.

Pierre.

More or less the same problem was used by the Bundeswettbewerb Mathematik 2005 as the problem 4 in the 1st round:

For which positive integers n is it possible to arrange the n numbers 1, 2, 3, ..., n in a sequence such, that for any two numbers the arithmetic mean of these two numbers doesn't stand somewhere between them?

Of course, the answer is: This is possible for all positive integers n.

Another thread with a discussion of the same topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=14758 .


symmetry
superior algebra
superior algebra solved

Show that $\mathfrak{S}_{4}$ can be embedded in $\mbox{SL}_{3}(\mathbb{Z})$.

Can we generalize?

So, how could you embed $S_{4}$ ? I've searched for ten minutes or so ("randomly") and couldn't grasp the thing. Is there a general approach or do you have to "guess" the form of the image by this embedding ?

Note that $S_{4}$ is the symmetry group of the cube.
Now it should be easy.

Actually, that symmetry group is $S_{4}\times \mathbb{Z}_{2}$. You want just the rotations.

Thank you. I wouldn't have thought about that  :lol:


Battery lifetime is normally distributed for large samples. The mean lifetime is 500 days and the standard deviation is 61 days. What percent, to the nearest whole number, of batteries have lifetimes longer than 561 days?

Usually calculator commands don't suffice for answers... in any case, it's simply the probability that you're at least one standard deviation greater than the mean... around 16% I believe.

I'm just wondering if we can use Chebyshev's inequality...

[EDIT]Oh, it [u]is[/u] Chebyshev's theorem  :blush:


Devise a pair of dice with exactly the same outcomes as ordinary dice (the sum 2 comes out once, etc.), but which are not ordinary dice.

[hide="Hint"] Factor the polynomial $ (x \plus{} ... \plus{} x^6)^2$ into two factors that take the value $ 6$ at $ x \equal{} 1$ with non-negative integer coefficients.  There is exactly one other way to do this. [/hide]

[hide="Solution"][b]t0rajir0u[/b] is basically saying that the generating functions corresponding to the two individual dice must multiply to the same generating function as multiplying those for two normal dice would. Additionally, if each dice must have six sides (we're just putting a number on each face), then the coefficients of each polynomial must add to 6.

It turns out that $ (x \plus{} \ldots x^6)^2 \equal{} x^{2} (x^2 \minus{} x \plus{} 1)^2 (x \plus{} 1) (x^2 \plus{} x \plus{} 1)^2$

If we don't limit ourselves to non-negative face values, then we should write it as follows:
$ x^{1 \minus{} n} x^{1\plus{}n} (x^2 \minus{} x \plus{} 1)^2 (x \plus{} 1) (x^2 \plus{} x \plus{} 1)^2$

For now, I'll pretend we want positive face values (we can ditch this assumption by adding any constant to one die's faces and subtracting it from the other die). Thus we need the leading coefficients to be 0 in each case. We know the sum of the coefficients of a polynomial is equal to that polynomial evaluated at $ 1$. We find the following contributions:
$ x \to 1$
$ x^2 \minus{} x \plus{} 1 \to 1$
$ x \plus{} 1 \to 2$
$ x^2 \plus{} x \plus{} 1 \to 3$

We see that each dice must get one factor of $ (x \plus{} 1)(x^2 \minus{} x \plus{} 1)$, which contribute a factor of 6. We additionally need a factor of $ x$ in each so that the leading coefficients are $ 0$. Now we need to figure out how we want to distribute the two  $ (x^2 \minus{} x \plus{} 1)$'s.

We can do:
$ x(x^2 \minus{} x \plus{} 1)(x \plus{} 1)(x^2 \plus{} x \plus{} 1) \; ; \qquad x (x^2 \minus{} x \plus{} 1) (x \plus{} 1)(x^2 \plus{} x \plus{} 1)$
$ x(x \plus{} 1)(x^2 \plus{} x \plus{} 1) \; ; \qquad x (x^2 \minus{} x \plus{} 1)^2 (x \plus{} 1)(x^2 \plus{} x \plus{} 1)$

Expanding the above, we get the following pairs:
$ x \plus{} \ldots x^6 \; ; \qquad x \plus{} \ldots x^6$
$ x \plus{} 2x^2 \plus{} 2x^3 \plus{} x^4 \; ; \qquad x \plus{} x^3 \plus{} x^4 \plus{} x^5 \plus{} x^6 \plus{} x^8$

These correspond to dice with face values:
$ (1,2,3,4,5,6) \; ; \qquad (1,2,3,4,5,6)$
$ (1,2,2,3,3,4) \; ; \qquad (1,3,4,5,6,8)$

The first is just a standard pair. The only other possibility is $ (1,2,2,3,3,4) \qquad (1,3,4,5,6,8)$[/hide]


1-) Find all polynomials P(x) with (for all real x):

P(1) = 1

P(x² + 1) = [P(x)]² + 1

2-) Find all polynomials P(x) with (for all real x):

P(x²) + P(x).P(x + 1) = 0

Thanks a lot!!!  :)

I think the first one is easy. $ P(1)\equal{}1 , P(2)\equal{}2 , P(5)\equal{}5 , P(26)\equal{}26 , ...$ So $ P(n)\equal{}n$ for infinitely many $ n \Longrightarrow$

$ P(x)\equal{}x$ for all $ x$ real numbers. :wink:

The second one is a little harder. If $ P(x)$ is constant so $ P(x) \equal{} 0, \minus{} 1$

$ u$ is a complex root of $ P \Longrightarrow$

$ P(u^{2^k}) \equal{} 0$ for all positive integer $ k$. There exists integers $ m$ and $ n$ so $ u^{2^m} \equal{} u^{2^n} \Longrightarrow$

$ |u| \equal{} 0,1$

$ P((u \minus{} 1)^{2^k}) \equal{} 0$ for all positive integer $ k$. So by same reason $ |u \minus{} 1| \equal{} 0,1$

$ u \equal{} 0, 1 , \frac {1 \plus{} \sqrt {3}i}{2}, \frac {1 \minus{} \sqrt {3}i}{2}$

$ P(x) \equal{} x^{\alpha}(x \minus{} 1)^{\beta}(x \minus{} \frac {1 \plus{} \sqrt {3}i}{2})^{\theta}(x \minus{} \frac {1 \minus{} \sqrt {3}i}{2})^{\gamma}$

$ x^{2\alpha}(x^2 \minus{} 1)^{\beta}(x^2 \minus{} \frac {1 \plus{} \sqrt {3}i}{2})^{\theta}(x^2 \minus{} \frac {1 \minus{} \sqrt {3}i}{2})^{\gamma} \equal{}$

$ \minus{}x^{\alpha}(x \minus{} 1)^{\beta}(x \minus{} \frac {1 \plus{} \sqrt {3}i}{2})^{\theta}(x \minus{} \frac {1 \minus{} \sqrt {3}i}{2})^{\gamma} (x \plus{} 1)^{\alpha}x^{\beta}(x \plus{} \frac {1 \minus{} \sqrt {3}i}{2})^{\theta}(x \plus{} \frac {1 \plus{} \sqrt {3}i}{2})^{\gamma}$

But there isn't any non constant polynomial.

Remark: If $ P(x^2) \equal{} P(x).P(x \plus{} 1)$ and $ P$ is non constant so $ P(x) \equal{} (x^2 \minus{} x)^{\theta}$.

First of all, thanks for the answers!  :)

But still about the first problem, i also found that answer (which is the right one, by the way), although i couldn't prove it's unic.

I tried the following:

Q(x) = P(x² + 1)
T(x) = [P(x)]² + 1

x = P(x):

Q(P(x)) = P([P(x)]² + 1)
Q(P(x)) = P(T(x))

Since T(x) = Q(x):

Q(P(x)) = P(Q(x))

Ideed, P(x) = x is a valid answer.
But does it prove anything about the uniticity?

About the second question, i just got to thank you once again.

[quote="Valle"] Ideed, P(x) = x is a valid answer.
But does it prove anything about the uniticity?[/quote]

As crazyfehmy said : $ P(n)\equal{}n$ for infinitely many $ n$, so $ P(x)\minus{}x$ is a polynomial with infinitely many roots, so is zero. So $ P(x)\equal{}x$ $ \forall x$


geometry
3D geometry
pyramid
induction
ratio
geometry unsolved

Consider a pyramid $VA_1A_2...A_n$ with $n\geq 5$.  On the side $VA_i$ consider the point $B_i$.
Let $M=\left\{P_{ij}\right\}$, where $P_{ij}=A_iB_j \cap A_jB_i$. Prove that, if at least $\binom{n-1}{2}+1$ points of $M$ are in the same plane, then the points $B_1$, $B_2$, ..., $B_n$ are in the same plane.

[quote="S_Alex"]Consider a pyramid VA1A2...An. (n>=5)  On the side VAi consider the point Bi.
Let M={Pij / Pij=AiBj \cap AjBi. Prove that if at least C(n-1,2) +1 points of M are in the same plane then the points B1, B2, ..., Bn are in the same plane.[/quote]
Anyone any ideeas ?.......... ?.............?............. No ? ........... :D  :D  :D

[quote="S_Alex"]Consider a pyramid VA1A2...An. (n>=5)  On the side VAi consider the point Bi.
Let M={Pij / Pij=AiBj \cap AjBi. Prove that if at least C(n-1,2) +1 points of M are in the same plane then the points B1, B2, ..., Bn are in the same plane.[/quote]
Is it really that hard ????..... :D  :D

[quote="S_Alex"]Consider a pyramid VA1A2...An. (n>=5)  On the side VAi consider the point Bi.
Let M={Pij / Pij=AiBj \cap AjBi. Prove that if at least C(n-1,2) +1 points of M are in the same plane then the points B1, B2, ..., Bn are in the same plane.[/quote]
Guess so.  :D  :D

[quote="galois"]i have seen this problem earlier and dint solve it back then and cant solve it now :D and if i remember correctly its arad 1999 county olympiad[/quote]
You're right galois but that doesn't help me very much  :D  :D  :D  because the book which contains that county olympiad does not provide any proofs whatsoever.   :D

[quote="grobber"]I was asking myself this: if we know that all the Bi's are coplanar, then does it follow that all the Pij's are coplanar?[/quote]
Good question. But I don't believe that the reverse statement it's true. Why is that I can't tell (I could be wrong).

[quote="grobber"]If we can prove it for n=5, induction can continue the argument.[/quote]
Yes I know that but how do you prove it for n=5. It's hard even to draw it.
 :D (joking).

First of all we can assume that $P_{12}, P_{13}$ and $P_{23}$ lie in that plane (associated graph necesserily contains triangle). :!:
Suppose $C_{n-2}^2+1$ points lie in the plane $\pi$ and points $A_1,...,A_n$ lie in the plane $\alpha$, $(B_1B_2B_3)=\beta$. Suppose $\alpha$ and $\beta$ intersect. Then we can apply the projective transformation, which maps $\alpha\cap \beta$ to the line of infinity.
So we have now that $\alpha ||\beta$. It immediately gives that $\alpha||(P_{12}P_{23}P_{31})$, i.e. $\alpha||\pi$. Since our graph is connected it immediately follows that all ratios $VB_:VA_i$ are equal, so all $B_i$ lies in $\beta$. Hence the result.


We may say concerning the solution of
\[ |x|^2 \plus{} |x| \minus{} 6 \equal{} 0
\]
that:

$ \textbf{(A)}\ \text{there is only one root}\qquad
\textbf{(B)}\ \text{the sum of the roots is }{\plus{}1}\qquad
\textbf{(C)}\ \text{the sum of the roots is }{0}\qquad \\
\textbf{(D)}\ \text{the product of the roots is }{\plus{}4}\qquad
\textbf{(E)}\ \text{the product of the roots is }{\minus{}6}$

EDIT: got it now.

[hide="POTENTIALLY DANGEROUS CONTENT"]$ |x|^2 \equal{} x^2$.  Always.

So the equation can be rewritten as $ x^2 \pm x \minus{} 6$ if we get rid of the abs. value.

The sum of the roots in $ x^2 \plus{} x \minus{} 6$ is 1 by Veita's.

The sum of the roots in $ x^2 \minus{} x \minus{} 6$ is -1 by Veita's.

The total sum is $ 1 \plus{} ( \minus{} 1) \equal{} \boxed 0$

Our answer is C[/hide][/hide]

[hide="Alternate Solution"]$ |x|^2 \plus{} |x| \minus{} 6 \equal{} 0$

Let $ u \equal{} |x|$

Then,

$ u^2 \plus{} u \minus{} 6 \equal{} 0$

$ \left(u \minus{} 2\right)\left(u \plus{} 3\right) \equal{} 0$

$ u \equal{} 2, \minus{} 3$

But $ u\geq0$

$ \therefore u \equal{} 2$

ie $ |x| \equal{} 2$

$ \therefore x \equal{} \pm 2$

Hence the sum of the roots is 0

ie $ \boxed{C}$[/hide]

[hide="Another Solution"]
We can note that the function $ f(x)\equal{}|x|^2\plus{}|x|\minus{}6$ is even. Thus, if $ r$ is a root then $ \minus{}r$ is also a root, so the sum of the roots must be $ \boxed{0}$.
[/hide]

This problem has been solved, but I want to comment on [b]abcak[/b]'s solution:

Your solution implies that there are four roots, two from each of the two quadratics you ended up with. However, the actual function has only two roots. The problem is that absolute value asks you to pick + or - in a very particular way -- it does not give you the freedom to choose. By changing $ |x|$ to $ \pm x$, you are creating a multivalued function.

In actuality, only the positive root of $ x^2\plus{}x\minus{}6$ and the negative root of $ x^2\minus{}x\minus{}6$ are valid.


Putnam
inequalities
function
logarithms
inequalities unsolved

a1, a2, ... , an, b1, ... , bn are non-negative reals. Show that (∏ ai)1/n + (∏bi)1/n ≤ (∏(ai+bi))1/n.[/tex][/list]

I guess that (∏ ai)1/n is a root of n -th power of (∏ ai). So
  ∑ ai/(ai+bi) >= n* [ ∏ai/(ai+bi)]1/n
  ∑ bi/(ai+bi) >= n* [ ∏bi/(ai+bi)]1/n
Summing:
  n =n* [ ∏ai/(ai+bi)]1/n + n* [ ∏bi/(ai+bi)]1/n or
  (∏ ai)1/n + (∏bi)1/n ≤ (∏(ai+bi))1/n

This is Putnam 2003 A2, and prowler's solution is the nicest and cleanest such one.  Another nice solution is to write k_i = a_i/b_i and divide through by (∏bi)^(1/n) so that the inequality is equivalent to:

1 + ∏(k_i)^(1/n) <= ∏(1+k_i)^(1/n)

Now, letting f(x) = ln(1+e^x), this is interpreted as:

∑ f(k_i) / n >= f( ∑ f(k_i) / n ), but this is exactly Jensen's inequality for the convex function f(x).

You are right [b]blahblahblah[/b],it is easy to solve the problem using that the function $f(x)=\ln{(1+e^{x})}$ is convex everywhere on $R$ and
moreover we can prove more general inequality(Hyugens inequality(sorry if I do mistake)).
${a_{1}}^{\gamma_{1}}\*{a_{2}}^{\gamma_{2}}\cdots{a_{n}}^{\gamma_{n}}+{b_{1}}^{\gamma_{1}}\*{b_{2}}^{\gamma_{2}}\cdots{b_{n}}^{\gamma_{n}}\leq{(a_{1}+b_{1})}^{\gamma_{1}}\*{(a_{2}+b_{2})}^{\gamma_{2}}\cdots{(a_{n}+b_{n})}^{\gamma_{n}}$
where $a_{i},b_{i},\gamma_{i}>0(i=1,2,...,n)$ and $\sum_{i=1}^{n}{\gamma_{i}}=1 (*)$.
To prove it we have to write the Jensen inequality $f(\sum_{i=1}^{n}{\gamma_{i}\*x_{i}})\leq\sum_{i=1}^{n}{\gamma_{i}\*f(x_{i})}$,for
the function $f(x)=\ln{(1+e^{x})}$ and for $x_{i}=\ln{\frac{a_{i}}{b_{i}}}(i=1,2,...,n)$.
But there is another approach which has already used [b]prowler[/b]:
there is another nice inequality:
${x_{1}}^{\gamma_{1}}\*{x_{2}}^{\gamma_{2}}\cdots{x_{n}}^{\gamma_{n}}\leq{\gamma_{1}}\*x_{1}+{\gamma_{2}}\*x_{2}+\cdots+{\gamma_{n}}\*x_{n}$,where
$x_{i},\gamma_{i}>0(i=1,2,...,n) and \sum_{i=1}^{n}{\gamma_{i}}=1 (**)$
 $(**)$ is a simple corolarry of the following simple inequality
$a^{\alpha}\*b^{\beta}\leq\alpha\*a+\beta\*b$,for $a,b,\alpha,\beta>0$ and $\alpha+\beta=1$.
Using $(**)$ we can write
$\prod_{i=1}^{n}{(\frac{a_{i}}{a_{i}+b_{i}})^{\gamma_{i}}}\leq\sum_{i=1}^{n}{\gamma_{i}\*\frac{a_{i}}{a_{i}+b_{i}}}$
and
$\prod_{i=1}^{n}{(\frac{b_{i}}{a_{i}+b_{i}})^{\gamma_{i}}}\leq\sum_{i=1}^{n}{\gamma_{i}\*\frac{b_{i}}{a_{i}+b_{i}}}$.
Now if we add last to inequalities we get $(*)$.


Hi,

I was just wondering if AIME (alternate AIME in particular)  must be written at the same time on March 28 or it just has to be written some time that day (like 1 pm or something).
It's just that AIME I is during March Break and I am pretty sure my school won't write it at that time. AIME II is on March 28, which is the same day for Canadian Math Olympiad, and CMO has to be written at 9 am local time.

Thank you for your help.

Last year, it did not have to be administered at a specific time.  I do not believe this has changed, but I don't know for sure.


Let $ \mathrm{T}: \mathbb{R}^{3}\rightarrow\mathbb{R}$ be linear.  Show that there exists scalars $ a$, $ b$, and $ c$ such that $ \mathrm{T}(x,y,z)\equal{}ax\plus{}by\plus{}cz$ for all $ (x,y,z)\in\mathbb{R}^{3}$.  Can you generalize this result for $ \mathrm{T}: F^{3}\rightarrow F$?  State and prove an analogous result for $ \mathrm{T}: F^{n}\rightarrow F^{m}$.

Uh, by linear you mean also respects multiplication, right? Otherwise this is not true.

$ a \equal{} T(1,0,0), b \equal{} T(0,1,0), c \equal{} T(0,0,1)$, uniquely determined

Linear as in $ T(cx\plus{}y)\equal{}cT(x)\plus{}T(y)$.


Find the amplitude of the graph of $ 3\sin x\plus{}4\cos x$.

$ 3\sin x\plus{}4\cos x\equal{}5\left(\sin x \cdot \frac{3}{5}\plus{}\cos x \cdot\frac{4}{5}\right)$

in genl .
$ \minus{}\ce{sqrt(a^2\plus{}b^2)} <\equal{} a.sinx\plus{}b.cosx <\equal{} \ce{sqrt(a^2\plus{}b^2)}$

[quote="AnonAmoz"][quote="kunny"]$ 3\sin x \plus{} 4\cos x \equal{} 5\left(\sin x \cdot \frac {3}{5} \plus{} \cos x \cdot\frac {4}{5}\right)$[/quote]
err... I still don't get it  :([/quote]

He just used standard trig identities to rearrange the expression into something where the amplitude was obvious.

[quote="AnonAmoz"][quote="kunny"]$ 3\sin x \plus{} 4\cos x \equal{} 5\left(\sin x \cdot \frac {3}{5} \plus{} \cos x \cdot\frac {4}{5}\right)$[/quote]
err... I still don't get it  :([/quote]

$ 3\sin x \plus{} 4\cos x \equal{} 5\left(\sin x \cdot \frac {3}{5} \plus{} \cos x \cdot\frac {4}{5}\right)$

$ \equal{}5(\sin x\cos \alpha \plus{}\cos x\sin \alpha)$

$ \equal{}5\sin (x\plus{}\alpha)$, where $ \cos \alpha \equal{}\frac{3}{5},\ \sin \alpha \equal{}\frac{4}{5}$

I don't understand how to get from here:
[quote="kunny"]
$ 3\sin x \plus{} 4\cos x \equal{} 5\left(\sin x \cdot \frac {3}{5} \plus{} \cos x \cdot\frac {4}{5}\right)$
[/quote]
to here:
[quote="kunny"]
$ \equal{} 5(\sin x\cos \alpha \plus{} \cos x\sin \alpha)$
[/quote]
 :maybe:

$ \alpha$ is defined as an angle such that $ \cos \alpha \equal{} \frac {3}{5}, \sin \alpha \equal{} \frac {4}{5}$.  This is possible because $ \left( \frac {3}{5} \right)^2 \plus{} \left( \frac {4}{5} \right)^2 \equal{} 1$.  (In particular, $ \alpha$ is one of the angles in a $ 3 \minus{} 4 \minus{} 5$ triangle.)


calculus
integration
rotation
calculus computations

I need help setting this up as an integral:

Find the volume that are within these bounds:
y=e^(-x), y=0, x=-1, x=0

Rotate about x=1

Thanks!

I think that might indicate we're actually talking about the volume of a solid of revolution.

xboxlive2020, could you clarify your question? Use words, not just shorthand.

Fixed the question. Sorry.

Can someone please move this to the unsolved and proposed problems? Thanks!


Represent the number $ 2008$ as a sum of natural number such that the addition of the reciprocals of the summands yield 1.

$ 8\plus{}10*80\plus{}30*40\equal{}2008$

$ 1/8 \plus{} (1/80)10 \plus{} (1/40)30\equal{}1$

Train of Thought:

To figure this out, I figured that we know perfect squares work instead of 2008 (very obviously), so the idea was to take a perfect square close to 2008 and manipulate the fractions to give the sum 2008 instead of the perfect square. Then, trying many squares, I came to the realization that $ 40^2\equal{}1600$ would work out nicely, the fractions would combine nicely since 40 has many factors. So I had $ (1/40)40$ to start with, which gave a sum of 1600. But I needed 2008. So I needed to increase the 1's digit by 8. This came by taking $ 1/40\plus{}1/40\plus{}1/40\plus{}1/40\plus{}1/40\equal{}1/8$ which decreased 1600 by 200, then increased by 8. So now I had 1408 and needed to get to 2008. But that's a difference of 600, and every time you split up $ 1/40\equal{}1/80\plus{}1/80$ it increases by 120. Since 120 |600, I needed to do this 5 times. And the result then follows.

Wishful Thinking FTW

$ 1*(1*4) \plus{} 4*(4*4)\plus{} 14*(14*4)\plus{}17*(17*4)\equal{}2008$
$ 1/4\plus{}4/16\plus{}14/56\plus{}17/68\equal{}1$

Rational: $ 2008\equal{}a*b$ and $ b\equal{}$ sum of $ a$ squares.

More interesting, Which positive integers have such partition (clearly not for 2 or 3)? for a given one (Example: 2008) which partition has minimum number of terms?

Every integer greater than 77 has such a partition. It's one of those facts that is often quoted as a property of sufficiently large integers where "sufficiently large" has a peculiar precise bound. I remember reading about this some years ago but Googling quickly provides some references, like this one:

http://www.math.uiuc.edu/Bulletin/Abstracts/October/oct31-97analnotheory.html

The 77 result is given in Engel, and as such, this is not a very good competition problem.

Can anyone show if $ 23$ is or is not a lucky number? Thanks.

Finally found the paper that proves that all integers greater than 77 can be partition into distinct positive integers whose reciprocals sum to 1.
[url]http://www.math.ucsd.edu/~fan/ron/papers/63_02_partitions.pdf[/url]


Find exact solutions to $ y^3\plus{}y^2\equal{}2y\plus{}1$

Just because I was not lazy and actually looked for an algebraic solution for the 7th roots of unity doesn't mean you have to abuse your privileges! Give credit to the correct people at the times of positive rectitude!

The equation $ y^3\plus{}2y^2\minus{}y\minus{}1\equal{}0$ is totally mine!!!! I claim it!!!!

Huh wait what? This is just random, i used random number genereator on my calc for coefficients. Cubic formula mayhaps?

I am not in the mood to spam about random topics.

Stupid Problem:

Given that $ a,b,c$ are positive real numbers and $ a\plus{}b\plus{}c\equal{}3$, find the minimum value of $ \sum_{\text{cyc}}\frac{a^2}{b^2\plus{}c^2\minus{}bc}$.

Easier version:

Given that $ a,b,c$ are positive real numbers and $ a\plus{}b\plus{}c\equal{}3$, find the minimum value of $ \sum_{\text{cyc}}\frac{a^2}{b^2\plus{}c^2\minus{}1}$.

Basically, it means you cycle the set $ (a,b,c)\equal{}(a,b,c)\equiv(b,c,a)\equiv(c,a,b)$ and substitute the variables in.


limit
induction
blogs
function
inequalities
triangle inequality

I'm not sure whether this problem belongs in this section, a mod should move it if necessary.

$ a_1\equal{}a_2\equal{}1, a_n\equal{} \frac{1}{a_{n\minus{}1}} \plus{} \frac{1}{a_{n\minus{}2}}, n \ge 2$. Find the $ \lim_{n\to\infty}a_n$ and the convergence rate.

Substituting the limit $ l$ for $ a_{n}, a_{n\minus{}1}$ and $ a_{n\minus{}2}$, we get $ l \equal{} \frac{1}{l} \plus{} \frac{1}{l}$ so that $ l^{2} \equal{} 2$ and therefore $ \lim_{n \to \infty} a_{n} \equal{} \sqrt{2}$.

[quote="AndrewTom"]Substituting the limit $ l$ for $ a_{n}, a_{n \minus{} 1}$ and $ a_{n \minus{} 2}$, we get $ l \equal{} \frac {1}{l} \plus{} \frac {1}{l}$ so that $ l^{2} \equal{} 2$ and therefore $ \lim_{n \to \infty} a_{n} \equal{} \sqrt {2}$.[/quote]

that doesn't prove anything, unless you can show that the sequence converges..

[hide=" a start"]
We will show that the sequence is bounded by induction
Base: $ 1 \le a_1, a_2 \le 2$
Inductive step:
$ 1 \le a_k, a_{k \minus{} 1} \le 2 \Longrightarrow 1 \ge \frac {1}{a_k}, \frac {1}{a_{k \minus{} 1}} \ge \frac {1}{2} \Longrightarrow 2 \ge \frac {1}{a_k} \plus{} \frac {1}{a_{k \minus{} 1}} \ge 1 \Longrightarrow 2 \ge a_{k \plus{} 1} \ge 1$

So by induction $ 1 \le a_n \le 2$ for all $ n \in \mathbb{N}$

Note: this may not help because the sequence doesn't seem to be monotonic decreasing or incresing (i think it is an oscilating convergance)
[/hide]

I am a bit doubtful but is the convergence rate [hide]$ \frac {1}{\sqrt {2}}$?[/hide] :blush:

$ a_1\equal{}a_2\equal{}1 ,a_3\equal{}\frac{1}{1} \plus{}\frac{1}{1}\equal{}2$ and $ a_4\equal{}\frac{1}{1} \plus{}\frac{1}{2}\equal{}\frac{3}{2}$

We see that $ a_2<a_3$ ,and $ a_3 >a_4$

Ocha was right , this is oscilating sequence,so it is not convergent (no limit)

No, this does converge. Oscillation doesn't imply it has no limits; imagine a sequence that's powers $ \frac{1}{2^n}$ for even $ n$ and $ \frac{\minus{}1}{2^n}$ for odd $ n$.

I have a [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=297241]blog post[/url] on this. It's not that easy. I got this problem from Problem-Solving Strategies, and Engel happens to omit the hardest part too. :P

I posted a variant of this problem in the calc forum, too.

I wouldn't mind if anyone could improve/correct my solution in my blog.

Unfortunately, clicking on blue text (within someone's post) does not work on my computer so I don't know what ocha, Potla or Zhero have posted. However, I can offer the following idea:

$ a_{i}, a_{i \plus{} 1}$ in $ [a, \frac {2}{a}]$ (for $ a \leq \sqrt {2}) \implies a_{j}$ in $ [a, \frac {2}{a}]$ for all $ j \geq i$ (by substituting in the recurrence).

I would now like to be able to prove that

$ a_{i}$ in $ [a, \frac {2}{a}]$ and $ a_{i \plus{} 1}$ in $ [a \plus{} c, \frac {2}{a} \minus{} c] \implies a_{j}$ in $ [a \plus{} d, \frac {2}{a} \minus{} d]$ for some $ d$ a function of $ c$.

However, I get $ [a \plus{} d', \frac {2}{a} \minus{} d'']$ where $ d' \equal{} \frac {a^{2}c}{4 \minus{} 2ac}$ and $ d'' \equal{} \frac {c}{a^{2} \plus{} ac}$

I hoped that $ d'$ and $ d''$ would be greater than $ c$ to ensure convergence but (unless I've made some errors) they don't seem to be.

I guess this doesn't work because the sequence oscillates about $ \sqrt {2}$ and perhaps I should look at $ a_{i \plus{} 2}$ as well as $ a_{i}$ and $ a_{i \plus{} 1}$ before considering $ a_{j}$ above as the interval $ [a\plus{}d', \frac{2}{a} \minus{} d'']$ will only get narrower from one end at a time.

Yes; the fact that this sequence oscillates so much makes it harder to prove that it converges.

I'll just copy and paste what I wrote in my blog:

[hide]
With the help of leshik from the calc forums (and from fwolth's random methods) I believe I have found a solution to this following limit:

Suppose $ a_0, a_1 > 0$, and $ a_{n \plus{} 1} \equal{} \frac {1}{a_n} \plus{} \frac {1}{a_{n \minus{} 1}}$.

[hide="Step 1"]If the limit exists, the limit is clearly $ \sqrt {2}$. So we let $ b_i \equal{} \frac {a_i}{\sqrt {2}}$. Then our recurrence relation becomes $ b_{n \plus{} 1} \equal{} \frac {\frac {1}{b_n} \plus{} \frac {1}{b_{n \minus{} 1}}}{2}$. Let $ c \equal{} \frac {1}{b_i}$ now; we thus have $ c_0, c_1 > 1$, $ c_{i \plus{} 1} \equal{} \frac {2}{c_i \plus{} c_{i \minus{} 1}}$. We now wish to show that $ \lim_{n \to \infty} c_n \equal{} 1$. [/hide]

[hide="The hard part..."]Let $ x$ be the inferior limit of $ c_i$, and $ y$ be the superior limit of $ a_i$. We wish to show that $ x \geq 1$ and $ y \leq 1$; then $ 1 \geq y \geq x \geq 1$, so $ x \equal{} y \equal{} 1$, as desired.

Because of fail notation, I will here denote $ a_n$ by $ c_n$.

We will only show that $ x \geq 1$; the proof that $ y \leq 1$ will be similar. By definition, we can find a subsequence $ a_{n_1}$, $ a_{n_2}$, ... that converges to $ x$. By our recurrence definition, we have that $ a_{n_k \minus{} 1} \plus{} a_{n_k \minus{} 2}$ approaches $ \frac {2}{x}$.

Assume now for the sake of contradiction that $ a_{n_k \minus{} 1}$ does not converge to $ \frac {1}{x}$. We claim now that there exists a subseqnece of $ a_{n}$ that converges to a value greater than $ \frac {1}{x}$. We now consider an arbitrary convergent subsequence of $ a_{n_k \minus{} 1}$, $ a_{n_{m_k} \minus{} 1}$ that converges to some real number $ c$. If $ c \geq \frac {1}{x}$, we are done (since then $ c > \frac {1}{x}$, since $ c \neq \frac {1}{x}$.) Otherwise, we have $ c < \frac {1}{x}$. Consider now $ a_{n_{m_k} \minus{} 2}$; by our recurrence relation, we know this converges to $ \frac {2}{x} \minus{} c$ (because $ a_{n_{m_k} \minus{} 1} \plus{} a_{n_{m_k} \minus{} 2}$ converges to $ \frac {2}{x}$, which is greater than $ \frac {1}{x}$ since $ c < \frac {1}{x}$.)

So now, let $ a_{p_n}$ be the subsequence that converges to $ \frac {1}{t} > \frac {1}{x}$. Then we get that $ a_{p_n \minus{} 1} \plus{} a_{p_n \minus{} 2}$ converges to $ 2t$, where $ t < x$. We claim now that there exists a subsequence of $ a_n$ that converges to a value less than or equal to $ t$.

We have two cases: there are infintely many terms in $ a_{p_n \minus{} 1}$ and $ a_{p_n \minus{} 2}$ such that both terms are greater than or equal to $ t$, or that there are only finitely many such terms. In the first case, we simply pick the terms greater than $ t$ in $ a_{p_n \minus{} 1}$ as a subsequence; we easily see that it is necessary that both sequences must converge to $ t$ (they are all positive, they are always greater than $ t$, and their sum is bounded above by something that converges to $ 2t$.) Otherwise, we consider a subsequence that takes terms from both $ a_{p_n \minus{} 1}$ and $ a_{p_n \minus{} 2}$, such that every term in that subsequence is less than $ t$ (there are infinitely many such terms in the union of these sequences, by assumption.) Thus, all convergent subsequences of this sequence converge to a value less than $ t$; we can find such a subsequence, so we are done.

But now, we have that some subsequence converges to $ t < x$, which contradicts our assumption that $ x$ is the infimum. It thus follows that $ a_{n_k \minus{} 1}$ must converge to $ \frac {1}{x}$, and thus $ a_{n_k \minus{} 2}$ must also converge to $ \frac {1}{x}$. But then, by our recurrence relation, we see that $ a_{n_k \minus{} 3}$ converges to $ 2x \minus{} \frac {1}{x}$. Since $ x$ is the infimum, we have $ 2x \minus{} \frac {1}{x} \geq x$, so $ x \geq 1$, as desired.

Meh this is probably a bit more complicated than it needs to be...
[/hide]
[/hide]

Thanks Zhero. Unfortunately the only blue text on this page that does anything on my computer is blog post. Once there, clicking on (post your comment) works but clicking on Step 1 or The hard part ... has no effect. I've thought of modifying what I posted but the fact that the sequence doesn't alternate but oscillates in a way I can't predict makes it difficult for me.

P.S. I have a copy of Arthur Engel's Problem-Solving Strategies and have looked at pages 225 and 231. I had earlier considered the sequence $ \frac{1}{a_{n}}$ but got nowhere with it.

EDIT: I've cleaned up the solution

[hide="solution"]
[b]Lemma 1[/b] if $ a_n,a_{n - 1} \in \left[\frac {2}{c_n},c_n\right]$ then $ a_m \in \left[\frac {2}{c_n},c_n\right]$ for all $ m \ge n$
[hide="proof"] From the condition
$ \frac {2}{c_n} \le a_{k}, a_{k - 1}\le c_n\Longrightarrow \frac {c_n}{2}\ge\frac {1}{a_{k}},\frac {1}{a_{k - 1}}\ge\frac {1}{c_n}\Longrightarrow c_n\ge\frac {1}{a_{k}} + \frac {1}{a_{k - 1}}\ge \frac {2}{c_n}\Longrightarrow c_n\ge a_{k + 1}\ge \frac {2}{c_n}$
So by the inductive hypothesis, the statement is true for all $ n\in \mathbb{N}$ [/hide]

We wish to find a decreasing family $ \left[\frac {2}{c_n},c_n\right]$ that traps the higher order elements of $ a_n$.
So let $ c_n = \max\left(a_n,a_{n - 1},\frac {2}{a_n},\frac {2}{a_{n - 1}}\right)$

The choice of $ c_n$ assures that $ a_{n - 1},a_{n} \in \left[\frac {2}{c_n},c_n\right]$, and hence by our lemma $ a_m \in \left[\frac {2}{c_n},c_n\right]$ for all $ m\ge n$. It follows that $ c_n$ is a monotonicly decreasing sequence.

In addition $ c_n$ is bounded below by $ \sqrt {2}$ because clearly $ \max\left(a_n,\frac {2}{a_n}\right) \ge \sqrt {2}$ so by the [b]monotonic convergence theorem[/b] the sequence $ \{c_n\}$ converges $ (*)$.

Now we use inequalities to show how $ |c_{n + 2} - c_{n}|$ varies with $ a_n$

[hide="one"]$ \max(a_n,a_{n - 1}) \ge \frac {a_n + a_{n - 1}}{2} \ge \frac {2}{\frac {1}{a_n} + \frac {1}{a_{n - 1}}} = \frac {2}{a_{n + 1}}$

$ \therefore \max(a_n,a_{n - 1}) - \frac {2}{a_{n + 1}} \ge \max(a_n,a_{n - 1}) - \frac {a_n + a_{n - 1}}{2} = \left|\frac {a_n - a_{n - 1}}{2}\right| \qquad(1)$ [/hide]

[hide="two"]$ \max\left(\frac {2}{a_n},\frac {2}{a_{n - 1}}\right) \ge \frac {1}{a_n} + \frac {1}{a_{n - 1}} = a_{n + 1}$

$ \therefore \max\left(\frac {2}{a_n},\frac {2}{a_{n - 1}}\right) - a_{n + 1} \ge \left|\frac {1}{a_n} - \frac {1}{a_{n - 1}}\right|\qquad(2)$ [/hide]

It follows from $ (1)$ and $ (2)$ that $ c_n - \max(a_{n + 1},\frac {2}{a_{n + 1}}) \ge \min\left(\frac {|a_n - a_{n - 1}|}{2}, \left|\frac {1}{a_n} - \frac {1}{a_{n - 1}}\right|\right)$

Replacing $ a_n$ with $ a_{n + 1}$ in the above and putting everything in terms of $ c_n$ gives

$ c_n - c_{n + 2} \ge \min\left(\frac {|a_n - a_{n - 1}|}{2}, \left|\frac {1}{a_n} - \frac {1}{a_{n - 1}}\right|,\frac {|a_{n + 1} - a_{n}|}{2}, \left|\frac {1}{a_{n + 1}} - \frac {1}{a_{n}}\right|\right) \qquad(**)$

We know from $ (*)$ that $ c_n$ converges so by Cauchy's Lemma, $ |c_{n + 2} - c_n| \to 0$ as $ n\to \infty$. So from $ (**)$ we have ${ \min (|a_n - a_{n - 1}|,|a_{n + 1} - a_n}|)\to 0$ has $ n\to \infty$

It is now easy to show using the triangle inequality that $ |a_n - a_{n - 1}|\to 0$ as $ n\to \infty$ and it's been proven that $ a_n$ converges!

Now let $ \lim_{n\to \infty} a_n \to a$ and $ a = \frac {1}{a} + \frac {1}{a} \Rightarrow a = \sqrt {2}$

$ \lim_{n\to \infty} a_n \to \sqrt {2}$
[/hide]


superior algebra
superior algebra unsolved

G finite with p| |G| (p=prime) Prove that

p|  |{g in G: g^p =1}|

and

|{H<G : |H|=p}|  = 1  (mod p).

The first can be generalised to $n| \ \# \{ x \in G | x^{n}=e \}$ for all $n| \ \#G$.

For the second, see http://www.mathlinks.ro/Forum/viewtopic.php?t=123631 for a generalisation.

Any hint about how can we prove these statements?

Thank you!

Notation: $|A|$ is the number of elements in $A$.

The first can be done by a simple counting argument: consider the set $S$ of $p$-tuples of elements $(a_{1},a_{2},\dots,a_{n})$ in $G$ with product $a_{1}a_{2}\dots a_{n}=e$. $S$ has $|G|^{n-1}$ elements, since we can choose $a_{1},a_{2},\dots,a_{n-1}$ freely and then choose the last element to fit.
The shift operator $L(a_{1},a_{2},\dots,a_{n})=(a_{2},a_{3},\dots,a_{n},a_{1})$ sends $S$ to itself. The orbits have size $p$ or $1$; we can only have an orbit of size 1 if all $a_{i}=a$, and $a^{p}=e$. Thus $|\{a: a^{p}=e\}|\equiv |S|\equiv 0\mod p$.

This argument can be generalized to prove $n| |\{x\in G: x^{n}=e\}|$ for all squarefree $n$ dividing $|G|$. I've claimed what ZetaX stated before, but my proof had an error.


Find all integer solution of the equation: $4x+x^2-x^3=4y+y^2-y^3$


$ x \plus{} {1\over y} \equal{} 4 ,\quad y \plus{} {1\over z} \equal{} 1 ,\quad z \plus{} {1\over x} \equal{} {3 \over 7}$

Find $ xyz \equal{}$?

$ (x\plus{}\frac{1}{y})(y\plus{}\frac{1}{z})(z\plus{}\frac{1}{x})\equal{}(x\plus{}\frac{1}{y})\plus{}(y\plus{}\frac{1}{z})\plus{}(z\plus{}\frac{1}{x})\plus{}xyz\plus{}\frac{1}{xyz}$.
So, $ 4\cdot1\cdot\frac{3}{7}\equal{}4\plus{}1\plus{}\frac{3}{7}\plus{}xyz\plus{}\frac{1}{xyz} \iff \frac{12}{7}\equal{}\frac{38}{7}\plus{}xyz\plus{}\frac{1}{xyz} \iff xyz\plus{}\frac{1}{xyz}\equal{}\minus{}\frac{26}{7}$.
Thus, $ xyz\plus{}\frac{1}{xyz}\equal{}\minus{}\frac{26}{7} \Rightarrow (xyz)^2\plus{}\frac{26}{7}xyz\plus{}1\equal{}0 \iff xyz\equal{}\boxed{\frac{\minus{}13\pm2\sqrt{30}}{7}}$.

[quote="Kouichi Nakagawa"]$ (x \plus{} \frac {1}{y})(y \plus{} \frac {1}{z})(z \plus{} \frac {1}{x}) \equal{} (x \plus{} \frac {1}{y}) \plus{} (y \plus{} \frac {1}{z}) \plus{} (z \plus{} \frac {1}{x}) \plus{} xyz \plus{} \frac {1}{xyz}$.
So, $ 4\cdot1\cdot\frac {3}{7} \equal{} 4 \plus{} 1 \plus{} \frac {3}{7} \plus{} xyz \plus{} \frac {1}{xyz} \iff \frac {12}{7} \equal{} \frac {38}{7} \plus{} xyz \plus{} \frac {1}{xyz} \iff xyz \plus{} \frac {1}{xyz} \equal{} \minus{} \frac {26}{7}$.
Thus, $ xyz \plus{} \frac {1}{xyz} \equal{} \minus{} \frac {26}{7} \Rightarrow (xyz)^2 \plus{} \frac {26}{7}xyz \plus{} 1 \equal{} 0 \iff xyz \equal{} \boxed{\frac { \minus{} 13\pm2\sqrt {30}}{7}}$.[/quote]

You have to show that both answers work, i.e. there are two distinct answers .

From 1st, 2nd equation, we have $ x\equal{}\frac{4y\minus{}1}{y},\ z\equal{}\frac{1}{1\minus{}y}$.

$ \therefore z\plus{}\frac{1}{x}\equal{}\frac{3}{7}\Longleftrightarrow \frac{1}{1\minus{}y}\plus{}\frac{y}{4y\minus{}1}\equal{}\frac{3}{7}\ \cdots (*)$

$ \Longleftrightarrow 5y^2\plus{}20y\minus{}4\equal{}0$

$ \Longleftrightarrow (5y)^2\plus{}20\cdot (5y)\minus{}4\equal{}0$

$ \Longleftrightarrow 5y\equal{}\minus{}10\pm 2\sqrt{30}$, thus from $ (*)$, we have $ xyz\equal{}\frac{4y\minus{}1}{y}\cdot y\cdot \frac{1}{1\minus{}y}$

$ \equal{}\frac{4y\minus{}1}{1\minus{}y}\equal{}\frac{3}{7}(4y\minus{}1)\minus{}y\equal{}\frac{5y\minus{}3}{7}\equal{}\boxed{\frac{\minus{}13\pm 2\sqrt{30}}{7}}.$


In a right triangle with sides $ a$ and $ b$, and hypotenuse $ c$, the altitude drawn on the hypotenuse is $ x$. Then:

$ \textbf{(A)}\ ab \equal{} x^2 \qquad\textbf{(B)}\ \frac {1}{a} \plus{} \frac {1}{b} \equal{} \frac {1}{x} \qquad\textbf{(C)}\ a^2 \plus{} b^2 \equal{} 2x^2$
$ \textbf{(D)}\ \frac {1}{x^2} \equal{} \frac {1}{a^2} \plus{} \frac {1}{b^2} \qquad\textbf{(E)}\ \frac {1}{x} \equal{} \frac {b}{a}$

hello, after a well-known fact we have $ c \cdot x \equal{} a\cdot b$, from here we get  $ x^2 \equal{} \frac {a^2b^2}{c^2} \equal{} \frac {a^2b^2}{a^2 \plus{} b^2} \equal{} \frac {1}{\frac {1}{a^2} \plus{} \frac {1}{b^2}}$ and $ \frac {1}{x^2} \equal{} \frac {1}{a^2} \plus{} \frac {1}{b^2}$. So the answer is $ \text{D}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]after a well-known fact we have $ c\cdot x \equal{} a\cdot b$[/quote]What is that well known fact?
Oh yeah areas.
But please don't just say that it's a well known fact without stating what fact it is.
[hide="outline"]similar triangles destroys this problem[/hide]

hello, yes, we have $ \frac{c \cdot x}{2}\equal{}\frac{a\cdot b}{2}$, sorry for my well-known.
Sonnhard.


inequalities
trigonometry
inequalities proposed

[quote="hitek"]Prove that if $ A, B,C$ are angles of a triangle then
\[ \cos^3A\plus{}\cos^3B\plus{}\cos^3C\plus{}5\cos A\cos B\cos C\leq1.\][/quote]

Prove or disprove the following inequality:
If $ a>0,c>0,b>0, a ^ 2 + b ^ 2 + c ^ 2 + abc = 4 $,
 then $ a ^ 3 + b ^ 3 + c ^ 3 +5 abc\le\ 8 $
_______________________________________
Sandu Marin, Bucuresti, Romania

[quote="hitek"]Prove that if $ A, B,C$ are angles of a triangle then
\[ \cos^3A\plus{}\cos^3B\plus{}\cos^3C\plus{}5\cos A\cos B\cos C\leq1.\][/quote]

\[1-\cos^3{A}-\cos^3{B}-\cos^3{C}-5\cos{A}\cos{B}\cos{C} = \frac{1}{4}\frac{Q}{R^3}.\]

$Q=(3r-5R)s^2-Rr^2+20R^3+8R^2r-r^3.$

$y=4R^2+4Rr+3r^2-s^2-r^2\frac{R-2r}{R-r}\geq 0.$ is yang xue zhi INEQ.

Note this:

\[r^2(R+r)\frac{R-2r}{R-r}+y(-3r+5R)=Q.\]

and done.

[quote=sandumarin][quote="hitek"]Prove that if $ A, B,C$ are angles of a triangle then
\[ \cos^3A\plus{}\cos^3B\plus{}\cos^3C\plus{}5\cos A\cos B\cos C\leq1.\][/quote]

Prove or disprove the following inequality:
If $ a>0,c>0,b>0, a ^ 2 + b ^ 2 + c ^ 2 + abc = 4 $,
 then $ a ^ 3 + b ^ 3 + c ^ 3 +5 abc\le\ 8 $
[/quote]
Given inequality is true for all triangle. ;)
By the way, the inequality $\sum_{cyc}\frac{(a^2+b^2-c^2)^3}{8a^3b^3}+5\prod_{cyc}\frac{a^2+b^2-c^2}{2ab}\leq1$ is true for all positives $a$, $b$ and $c$.

[quote=hitek]Prove that if $ A, B,C$ are angles of a triangle then
\[ \cos^3A\plus{}\cos^3B\plus{}\cos^3C\plus{}5\cos A\cos B\cos C\leq1.\][/quote]

Let $\cos\alpha=\frac{x-2}{2}$, $\cos\beta=\frac{y-2}{2}$ and $\cos\gamma=\frac{z-2}{2}$, $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, $\{x,y,z\}\subset(0,4)$.
Since $\cos\alpha+\cos\beta+\cos\gamma\geq1$, we obtain $u\geq\frac{8}{3}$.
In another hand, $\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1$ gives $2(xy+xz+yz)-x^2-y^2-z^2=xyz$
or $w^3=3(4v^2-3u^2)$ and we need to prove that $\sum_{cyc}\frac{(x-2)^3}{8}+\frac{5(x-2)(y-2)(z-2)}{8}\leq1$ or
$\sum_{cyc}(x^3-6x^2-10xy+32x)+5xyz\leq72$ or $\sum_{cyc}(x^3-11x^2+32x)\leq72$ or $f(w^3)\geq0$, where
$f(w^3)=72\left(\frac{w^3}{3(4v^2-3u^2)}\right)^3-96u\left(\frac{w^3}{3(4v^2-3u^2)}\right)^2+11(9u^2-6v^2)\cdot\frac{w^3}{3(4v^2-3u^2)}-27u^3+27uv^2-3w^3$.
But $f''(w^3)=\frac{16w^3}{(4v^2-3u^2)^3}-\frac{64u}{3(4v^2-3u^2)^2}=\frac{48}{(4v^2-3u^2)^2}-\frac{64u}{3(4v^2-3u^2)^2}\leq\frac{48}{(4v^2-3u^2)^2}-\frac{512}{9(4v^2-3u^2)^2}<0$.
Id est, $f$ is a concave function.
Hence, $f$ gets a minimal value for an extremal value of $w^3$, which happens in the following cases:
1. $w^3\rightarrow0^+$.
Let $x\rightarrow0$. Hence, $\alpha\rightarrow180^{\circ}$ and $\beta=\gamma\rightarrow0^{\circ}$. We see that in this case our inequality is true;
2. $y=x$, which gives $\beta=\alpha$ and $\gamma=180^{\circ}-2\alpha$, which gives $(2\cos\alpha-1)^2(2\cos^2\alpha+2\cos\alpha+1)\cos^2\alpha\geq0$.
 Done!  :)


if i have $\alpha_{n}\to \alpha$ as $n\to \infty$
then why is
$ [-\infty,\alpha]=\cup_{n=1}^{\infty}[-\infty,\alpha_{n}]$

shouldn't it be intersection ?

Please post such problems in the calculus forum next time, it's not complex analysis.

It's union if the sequence $\alpha_{n}$ is always $\leq \alpha$, it's intersection if always $\geq \alpha$. So it depends, and none is correct if the sequence is arbitrary.

forgot to mention
$\alpha_{n}< \alpha$

zetax you're right, sorry,it was obvious


[hide]
Taking the toll road means that you spend 6+16=22 dollars
so 22 dollars is what you spend on gas meaning you use 22/4 = 5.5 gallons of gas
5.5*30=165[/hide]

[quote="abacadaea"]a car travels 30 miles per gallon of gas. ever gallon of gas is $ \$$4.(yes thats where its headed). a certain toll road is 120 miles long and costs 6 dollars to drive on it. another free road is x miles long.if both ways require the same amount of money(including gas), what is x?

i made this up on the spot so it may not make sense but im kinda out of it so i cant really tell...[/quote]
Here's my solution:
[hide]The 120 mile road consumes 4 gallons of gas, so it would take $ \$$16 for gas mileage. It also costs $ \$$6 more to drive on it, so the cost of driving on that road is $ \$$16 + $ \$$6=$ \$$22. So the $ x$-mile road, which is free to drive on, requires $ 22/4$ = $ 5.5$ gallons of gas, so $ x$ = $ 30 * 5.5$ = $ 165$ miles.[/hide]

I got [hide]165[/hide], just to confirm your answer.


Let $G$ be a finite group and $\phi: G\to C^\times$ be a non trivial homomorphism. Show that
\[\sum_{g\in G}\phi(g)=0\]

Also note the dual result:
\[\sum_{\phi \in G^{*}}\phi(g) = 0 \]
with $G^{*}$ being the dual group of $G$ (homomorphisms from $G$ to $\mathbb C^{*}$) and $g \in G$, $g$ nontrivial.

Both are proven by multiplying with $\phi(g) \neq 1$ (here each is one fixed).

[quote="ZetaX"]A
Both are proven by multiplying with $\phi(g) \neq 0$ (here each is one fixed).[/quote]

Can you elaborate more, please?

Sorry, it's $\neq 1$, not $\neq 0$:

For the first, take a $h \in G$ with $\phi(h) \neq 1$.
Then we get $\sum_{g\in G}\phi(g)=\sum_{g\in G}\phi(hg)=\sum_{g\in G}\phi(h)\phi(g)=\phi(h) \cdot \sum_{g\in G}\phi(g)$, thus $(1-\phi(h)) \cdot \sum_{g\in G}\phi(g) =0$, giving the result.
For the second, take a homomorphism $\psi$ with $\psi(g) \neq 1$ and use the same idea.


calculus
derivative
calculus computations

we just started calculus this year, and we done chain, product, quotient, power rules, implicit diff, second derivative...and i cant get some of the questions assigned to cover all these

Find the second derivative d^2y / dx^2

x^2-3x+y^2=0

x^3+y^3=3

I'll do the second one only, because it's quicker.

Let $y'=\frac{dy}{dx}$.

We have $x^3+y^3=3$, so $3x^2+3y^2y'=0$, or $y'=-\left(\frac xy\right)^2$

Differentiating again, we have

$6x+3y^2y''+6y(y'^2)=0$.  We already know what $y'$ is - substitute in.  Then solve for $y''$ to finish.

hm, i seemed to forgot to substitute y', lol...damn, wtf was i thinking...anyway, thx a lot! i just got the answer for the second derivative, one of the signs is opposite than the answer in the back of the book, but im sure i'll figure something out...THX a lot


LaTeX
quadratics
algebra
polynomial
Rational Root Theorem

Once I found this on a math site: A man goes into a shop and buys 4 things when the shopkeeper calculates the bill he stars to multiply the prices and obtains 19.7$. The customer tells him that he calculated wrong so he added the prices this time and he obtained 19.7$. So the question is which were the prices?
 The problem resumes to calculating the sistem:
a+b+c+d=19.7
a*b*c*d=19.7
  Is there a special method to solve it or one must ''guess" the answer?
  Can a computer programe be written to solve it, and other similar problems?

[quote="Alicea"]Once I found this on a math site: A man goes into a shop and buys 4 things when the shopkeeper calculates the bill he stars to multiply the prices and obtains $19.7$. The customer tells him that he calculated wrong so he added the prices this time and he obtained $19.7$. So the question is which were the prices?
 The problem resumes to calculating the system:
$a+b+c+d=19.7$
$a \cdot b \cdot c \cdot d=19.7$
  Is there a special method to solve it or one must ''guess" the answer?
  Can a computer programe be written to solve it, and other similar problems?[/quote]

careful w/ your dollar signs :wink:

Well, they're prices, so they all have to take the form of an integer divided by $100$.

Forgive me I'm new around so I didn't know what signs from the keyboard cause problems :blush: .

If you know how to use them, they don't cause problems.

Generally, to make a dollar sign, type:
[code]$ \$ $[/code]

This will make:
$\$$

Dollar signs represent the start of an equation, expression, etc. written in $\LaTeX$.

[hide="not quite a solution"]Let a, b, c, and d be roots of a quartic.

$x^{4}-19.7x^{3}+dx^{2}+ex+19.7=0$

$d=ab+ac+ad+bc+bd+cd$

$e=abc+abd+acd+bcd$

From the rational root theorem, the roots must be "factors" of +-19.7. But 197 is prime, so I think I did something wrong somewhere....[/hide]

You can't apply the rational root theorem with non-integer coefficients.  In fact, even after you've multiplied by $10$ you don't know whether your other two coefficients (don't call one of them $d$ when you've used that letter already!) are integer.

Really, the rational root theorem doesn't tell you anything you didn't already know.

oooooooooooooooops! :oops:

Well, 19.7 is the sum and product of the roots of the quadratic, so I got that part right.


Compute the sums: $ \displaystyle\sum_{i\equal{}0}^{n}\frac{\binom{k\plus{}i}{k}}{k\plus{}i}$ and $ \displaystyle\sum_{i\equal{}0}^{n} \frac{\binom{k\plus{}i}{k}}{k\plus{}i\minus{}1}, n,k \in \mathbb{N}, k \ge 2.$

For the first, write $ \frac {\binom{k \plus{} i}{k}}{k \plus{} i} \equal{} \frac {1}{k} \cdot \binom{k \plus{} i \minus{} 1}{k \minus{} 1}$; for the second, write $ \frac {\binom{k \plus{} i}{k}}{k \plus{} i \minus{} 1} \equal{} \frac {1}{k} \binom{k \plus{} i \minus{} 1}{k \minus{} 1} \plus{} \frac {1}{k} \cdot \frac {\binom{k \plus{} i \minus{} 1}{k \minus{} 1}}{k \plus{} i \minus{} 1}$ and then repeat the same idea again with the second summand.

In both cases, finish with $ \sum_{i \equal{} 0}^n \binom{m \plus{} i}{i} \equal{} \binom{m \plus{} n \plus{} 1}{n}$.


I just came up with this a little bit ago as I was playing around with some infinite sequences.

Given: $y=x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x...$

Solve for x.

$y=x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x...$
Given
$y=x+y$
Subsitution
$0=x$
Subtraction.

Thus any number added infinately is equal to zero, or even more simply, any number is equal to zero.   Could this be an extraneous solution?

It doesn't seem right, and it kind of reminds me of Zeno's Paradox... What are your thoughts on this?

Hmm
Wont y+x always be x greater than y so they cannot be equal even if it's an infinite sum?
So the only time it will work is when x = 0.

True, but thats the thing.  If you are to look at it that way, other problems that use this message of substitution are similarly flawed.

When $x=0$, then what you have is correct. When $x\ne0$, you have basically (using symbolism liberally) $\infty=x+\infty,$ where $x$ is a real number, but this does not imply $x=0$. You have to be careful when you're dealing with infinite sums.

[quote="Farazy"]But I want the reasoning for why this is false.

Because you can put almost any number as x and get the same exact thing.

$y=5+5+5+5+5+...$
$y=5+y$
$0=5$[/quote]$y$ is not number that actually has a value. You're attempting to do arithmetic with infinitely large numbers.

The algebraic principles that you're using to derive $x=0$ (namely subtracting $y$ from both sides) do not apply to infinite numbers. For example, if $\aleph_0$ is the cardinality of the set of all integers, then $\aleph_0=\aleph_0+C$, where $C$ is any real number, but certainly this does not imply $C=0$. The algebraic rule that you may use to derive $C=0$ simply does not apply.

I also came up with something else that proves what I said wrong...

$y=x+x+x...$

We can subsitute y in like before.

$y=x+y$ but we notice that y is x more then that...So:

$y=x+(y-x)$

Which goes back to

$y=y$

what does "countably infinite" mean?  I thought you can't count infinity?

[quote="DPatrick"]This "paradox" is also closely related to the Infinite Hotel paradox.

A hotel has a (countably) infinite number of rooms, numbered 1, 2, 3, etc.  The hotel is currently full -- every room has an occupant.

A man shows up at the front desk asking for a room.  The manager regretfully informs the man that the hotel is full.  The man then says, "Well, you can free up a room for me by just telling everybody to move to the room that's one higher!"  So the person in Room #1 moves to Room #2, the person in Room #2 moves to Room #3, etc., and then Room #1 is empty for the man to move into.

Is this a paradox?[/quote]

Wait, but where does that leave the person in the highest numbered room...where does he move into?

What highest number room?

There are an infinite number of rooms.

So, how is that countable? The hotel can't claim to be full ever - actually, practically the hotel would never know if it was full or not.

"Countably infinite" alludes to the fact that there are different sizes of infinite sets.

Basically, an infinite set is "countably infinite" (or just "countable") if we can count it 1, 2, 3, 4, ..., or more precisely, it can be put into 1-to-1 correspondence with the positive integers.

So in the problem I posted above, the hotel has a Room #n for every positive integer n.  So indeed it has an infinite number of rooms.

Other infinite sets, like the set of all points on a line segment, are "uncountably infinite" (or just "uncountable").

All of this is explained quite nicely in Section 17.7 of AoPS Volume 2 (which is inside the chapter titled "Counting in the Twilight Zone").

[quote="Farazy"]But I want the reasoning for why this is false.

Because you can put almost any number as x and get the same exact thing.

$y=5+5+5+5+5+...$
$y=5+y$
$0=5$[/quote]

It dosnt tend to a value, it just get bigger, so you cant use that technique with these kind of problems.


integration
calculus
algebra
polynomial
real analysis
real analysis unsolved

I liked the problem.  Easy to check but still a neat construction.

New problem:
Construct an integral equation in terms of n general polynomials divisible by (x + a)^n.

Any solution of the question of googleplex ?


Three girls and two boys are on the Math Team. [Note: Names do not indicate gender.]

Two students wear white shirts; three wear black shirts.
Miken and Carry wear shirts of different colors.
Bari and Jamie wear shirts of the same color.
Pita and Carry are the same gender.
Jamie and Miken are of different genders.

The boy in the white shirt scored the most points. What is his name?

a. Bari    b. Carry    c. Jamie    d. Miken   e. Pita

[hide=" click to reveal hidden content"]

I think that Bari and Miken are boys and that Carry, Jamie and Pita are girls; Miken and Pita wear white, Bari, Carry and Jamies wear black. So Miken scored the most points.

[/hide]


induction
modular arithmetic
algebra
binomial theorem
algebra unsolved

Let $ \left(x_{n}\right)$ be a real sequence satisfying $ x_{0}=0$, $ x_{2}=\sqrt[3]{2}x_{1}$, and $ x_{n+1}=\frac{1}{\sqrt[3]{4}}x_{n}+\sqrt[3]{4}x_{n-1}+\frac{1}{2}x_{n-2}$ for every integer $ n\geq 2$, and such that $ x_{3}$ is a positive integer. Find the minimal number of integers belonging to this sequence.

First, let's simplify everything a little by writing $a = \sqrt[3]{2}$ so we keep in mind that $a^3 = 2$.

$x_{n+1} = \frac{ ax_n + 2a^2 x_{n-1} + x_{n-2} }{2}$

So we can find that

$x_3 = \frac{ a^2 x_1 + 2a^2 x_1 + 0}{2} = \frac{3a^2}{2} x_1$
$x_1 = \frac{2}{3a^2} x_3 = \frac{a}{3} x_3$ for our integer $x_3$.

Write the sequence in terms of $x_3$.

$x_0 = 0$
$x_1 = \frac{a}{3} x_3$
$x_2 = \frac{a^2}{3} x_3$
$x_3 = x_3$

At this point, let's rewrite the recursion to make calculation easier.

$x_{n+1} = \frac{ 3ax_n + 6a^2 x_{n-1} + 3x_{n-2} }{6}$

It should be clear now that the exponent of $a$ in $x_n$ is $n \bmod 3$, which means for $3 | n$ we have a rational coefficient of $x_3$.

Now, let's calculate a few more terms.

$x_4 = \frac{3a + 2a^4 + a}{6} x_3 = \frac{4a}{3} x_3$
$x_5 = \frac{4a^2 + 6a^2 + a^2}{6} x_3 = \frac{11a^2}{6} x_3$
$x_6 = \frac{5.5a^3 + 8a^3 + 3}{6} x_3 = 5x_3$, an integer.
$x_7 = \frac{15a + 11a^4 + 4a}{6} x_3 = \frac{41a}{6} x_3$
$x_8 = \frac{20.5a^2 + 30a^2 + 5.5a^2}{6} x_3 = \frac{56a^2}{6} x_3$
$x_9 = \frac{28a^3 + 41a^3 + 15}{6} = \frac{153}{6} x_3$, not an integer.

Conjecture:  All $x_{3k}$ are non-integer for $k \ge 3$.  This would mean the only integers in the squence are $x_0, x_3, x_6$.  However, I am unsure if I have made a calculation mistake, and even if I haven't I lack the brainpower to complete the proof  :(

The offical solution is very very long and I am crazy to see it.The answer is 5,and $x_0,x_3,x_6,x_{12},x_{24}$ must be integers. :wacko:
The problem is created by a university professor and he's famous to create such crazy problem for MO and luckily he has retired. :bye:

:censored: Calculation up to $x_{24}$ would be very tedious, and proving non-integer for all $x_{3k}, k > 8$ more tedious still - why write a problem that does not have elegant solution?!

I've a friend who took part in this contest and failed in this problem,so he can't take part in IMO.And he hated the problem very much.

It is easy to show $x_{n}=\frac{2^{-2n/3}((1+\sqrt{3})^{n}-(1-\sqrt{3})^{n})a}{3\sqrt{3}}$ when $x_{3}=a$.
Of course, if xn is integer, then 3|n, and when number of natural integer in this sequence is minimum, then a=1
Let $y_{n}=x_{3n}, a=1$.
then $y_{n+2}=5y_{n+1}+\frac{1}{2}y_{n}$
binomial theorem and power of 2 will give you a answer.
only $y_{0},y_{1},y_{2},y_{4},y_{8}$ are integer.

Is it hard?


Let $ a,b,c>0$ . Prove that :

\[ \frac{\sqrt{a\plus{}b\plus{}c}\plus{}\sqrt{a}}{b\plus{}c}\plus{}\frac{\sqrt{a\plus{}b\plus{}c}\plus{}\sqrt{b}}{c\plus{}a}\plus{}\frac{\sqrt{a\plus{}b\plus{}c}\plus{}\sqrt{c}}{a\plus{}b}\ge \frac{9\plus{}3\sqrt{3}}{2\sqrt{a\plus{}b\plus{}c}}\]

[quote="Pain rinnegan"]Let $ a,b,c > 0$ . Prove that :
\[ \frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {a}}{b \plus{} c} \plus{} \frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {b}}{c \plus{} a} \plus{} \frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {c}}{a \plus{} b}\ge \frac {9 \plus{} 3\sqrt {3}}{2\sqrt {a \plus{} b \plus{} c}}\]
[/quote]

without loss of generality, assume that $ a \plus{} b \plus{} c \equal{} 1$, so

\[ \frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {a}}{b \plus{} c} \plus{} \frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {b}}{c \plus{} a} \plus{} \frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {c}}{a \plus{} b}\ge \frac {9 \plus{} 3\sqrt {3}}{2\sqrt {a \plus{} b \plus{} c}}\]\[ \Leftrightarrow \sum \frac {1 \plus{} \sqrt {a}}{1 \minus{} a} \ge \frac {9 \plus{} 3\sqrt {3}}{2}\]
we can prove that $ \frac {1 \plus{} \sqrt {a}}{1 \minus{} a}\ge\frac {3}{4} \plus{} \frac {9 \plus{} 6\sqrt {3}}{4}a$, thus
\[ \sum \frac {1 \plus{} \sqrt {a}}{1 \minus{} a} \ge \sum\left(\frac {3}{4} \plus{} \frac {9 \plus{} 6\sqrt {3}}{4}a\right) \equal{} \frac {9 \plus{} 3\sqrt {3}}{2}\]

my solution
it is equivalent to $\sum_{cyc}\frac{a}{b+c}+\sum_{cyc}\frac{\sqrt{a(a+b+c)}}{b+c}\ge\frac{3+3\sqrt{3}}{2}$
we only need to prove that
$\sum_{cyc}\frac{\sqrt{a(a+b+c)}}{b+c}\ge\frac{3\sqrt{3}}{2}$
by Holder inequality,we have $(\sum_{cyc}\frac{\sqrt{a}}{b+c})^{2}(\sum_{cyc}(b+c)^{2}a^{2})\ge\ (a+b+c)^{3}$
so we only need to prove that
$4(a+b+c)^{4}\ge\ 27(2\sum_{cyc}a^{2}b^{2}+2\sum_{cyc}a^{2}bc)$
hences to $4\sum_{cyc}a^{4}+16\sum_{sym}a^{3}b-30\sum_{cyc}a^{2}b^{2}-6\sum_{cyc}a^{2}bc\ge\ 0$ and that's trivial by Muirhead


Ορθογώνιο συμπλήρωμα του υπόχωρου U ως προς έναν πεπερασμένης διάστασης διανυσματικό χώρο V, ονομάζεται o υπόχωρος W, με  $ W\subset V$ ,
έτσι ώστε o V να είναι το ευθύ άθροισμα  των W και U δηλαδή $ V\equal{} U\plus{}W$ και $ U\cap W\equal{}\emptyset$.
Αποδείξτε ότι για οποιονδήποτε υπόχωρο U του V,υπάρχει ένα ορθογώνιο συμπλήρωμα του ,και ότι οι διαστάσεις δύο οποιονδήποτε ορθογώνιων συμπληρωμάτων του U ως προς τον V, είναι ίδιες.

[quote="spinos"]Ορθογώνιο συμπλήρωμα του υπόχωρου U ως προς έναν πεπερασμένης διάστασης διανυσματικό χώρο V, ονομάζεται o υπόχωρος W, με  $ W\subset V$ ,
έτσι ώστε o V να είναι το ευθύ άθροισμα  των W και U δηλαδή $ V \equal{} U \plus{} W$ και $ U\cap W \equal{} \emptyset$. Αποδείξτε ότι για οποιονδήποτε υπόχωρο U του V,υπάρχει ένα ορθογώνιο συμπλήρωμα του ,και ότι οι διαστάσεις δύο οποιονδήποτε ορθογώνιων συμπληρωμάτων του U ως προς τον V, είναι ίδιες.[/quote]

Αυτό δεν είναι απολύτως τετριμμένο για απειροδιάστατους χώρους, αλλά για χώρους πεπερασμένης διάστασης τα πράγματα είναι αρκετά απλά. Αρκεί να επεκτείνεις μία βάση του $ U$ μέχρι να εξαντλήσεις όλον το χώρο.

Η λογική είναι η εξής: αν $ U \equal{} V$, τελείωσες. Αλλιώς, θα υπάρχει τουλάχιστον ένα διάνυσμα στον $ V\setminus U$, το οποίο μπορείς να προσθέσεις στη βάση του $ U$ και να πάρεις έναν καινούριο χώρο $ U_1$. Ξανακάνεις την ίδια δουλειά με τον $ U_1$ και φτιάχνεις έναν $ U_2$ κ.ο.κ. μέχρι να φτάσεις στον $ V$. Το συμπλήρωμα του $ U$ θα παράγεται ακριβώς από αυτά τα διανύσματα που προσέθεσες.

Αν είναι απειροδιάστατοι οι χώροι, χρειάζεσαι το αξίωμα επιλογής για να έχεις βάση για κάθε δ/κό χώρο... βασικά, μπορείς να το κάνεις πάνω κάτω όπως παραπάνω, αλλά ο Dem σίγουρα θα ξέρει κάποιον καλύτερο τρόπο ;)

Cheerio,

Durandal 1707


analytic geometry
graphing lines
slope
trigonometry

[quote="THECHAMP"]Find the acute angle between two lines with slopes -2 and -1/2 to the nearest degree.[/quote]
[hide]Call the first line $ l_1$ and the second $ l_2$. The slope of $ l_1$ is $ \minus{}2$ or $ \tan\theta\equal{}\minus{}2$, thus it makes a $ \tan^{\minus{}1}(\minus{}2)$ angle with the x-axis. Similarly with the second line, it has the angle $ \tan^{\minus{}1}(\minus{}\frac{1}{2})$. Now we want the angle between these two lines so we have that $ \alpha\equal{}\tan^{\minus{}1}(\minus{}2)\minus{}\tan^{\minus{}1}(\minus{}\frac{1}{2})$. Taking the tangent of both sides and using the tangent subtraction formula we have $ \tan\alpha\equal{}\frac{\minus{}2\minus{}\frac{1}{2}}{1\plus{}(\minus{}2)(\minus{}\frac{1}{2})}\equal{}\frac{5}{4}$. Thus the angle between these two angles is $ \tan^{\minus{}1}(\frac{5}{4})$ which approximated to the nearest degree is $ 51^{\circ}$[/hide]


Given two lines $(k: l : m), (p : q: r)$ in trilinears, what is the formula for the (smaller) angle between them?


8 girls and 16 boys are standing in a line. How many of the permutations of their ordering are there such that exactly 5 of the girls are not in the spot they were before, while 3 are. (The positions of the boys don't matter, and everyone is distinguishable)

The 16 boys permute in $ 16!$ ways. There are $ 8!/5!3!$ ways to categorize the girls. Then use the derangement formula to find the number of derangements of the 5 girls. (which I believe to be $ \sum_{k\equal{}0}^5 (\minus{}1)^k/k!$)

Even if we derange the set of 5 girls, what if some of those girls take spots which were previously occupied by boys?

I see... this problem seems much more difficult than I thought it would be (I didn't think carefully.... last period of school day and I was tooo tired... needa take a nap when got home... See? I even got the derangement formula wrong  :oops:  :P )

First, let's choose the three girls that stay in their place. There 56 ways to do this.

Without regard to restrictions, the remaining 21 people may be arranged in 21! ways. There are 16! ways to place all 5 girls in the spot they were in before. There are 17! ways to place any 4 of those girls in the spot they were in before.

Using the Principle of Inclusion-Exclusion, there are $ 5(20!) \minus{} 10(19!) \plus{} 10(18!) \minus{} 5(17!) \plus{} 16!$ ways to place them so that atleast one of them are in their previous position. Simplifying, we get $ 526236(16!)$ Because we want the number of ways to place them in a position where none of them are in their previous position, we need to subtract $ 526236(16!)$ from $ 21!$
$ 21! \minus{} 526236(16!) \equal{} 2441880(16!) \minus{} 526236(16!) \equal{} 1915644(16!)$

Multiplying this by our previous answer of 56, and we get $ 107276064(16!)$
or, if you prefer $ 2,244,514,547,083,640,832,000$ ways.


geometry
circumcircle
trigonometry
inequalities
function
geometry proposed

[quote="nickolas"]In a  triangle $ABC$  we have  $\angle C>10^0$ and  $\angle B=\angle C+10^0.$We  consider  point  $E$  on side  $AB$  such that  $\angle ACE=10^0,$ and point  $D$ on side $AC$  such that  $\angle DBA=15^0.$ Let  $Z\neq A$ be a point  of interection of  the  circumcircles of  the triangles  $ABD$ and $AEC.$Prove that  $\angle ZBA>\angle ZCA.$[/quote]

Here what I did...
First put $\angle BCA=10^0+x$ then $\angle ECB=x,\angle ABE=x+20^0,\angle DBC=x+5^0$
However the cord $AZ$ is the same for both circles we will just prove that circumradius of ABD is smaller than the circumradius of AEC (this follows by the sine law) or the inequality $\frac{AB}{\sin{\angle ADB}} < \frac{AC}{\sin{\angle AEC}}(*)$
By the sine law it also follows that
$\frac{AB}{\sin{(10^0+x)}}=\frac{AC}{\sin{(20^0+x)}}(**)$ now dividing (*) and (**) side by side gives:
$\sin({\angle AEC})\cdot\sin{(x+10^0)}<\sin({\angle ADB})\cdot\sin{(x+20^0)}$ because of $\angle AEC=2x+20^0$ and $\sin{\angle ADB}=2x+15^0$ it is enough to prove that $\sin(2x+20^0)\cdot\sin{(x+10^0)}<\sin(2x+15^0)\cdot\sin{(x+20^0)}(***)$
Now I proceeded as follows:
Take $f(y)=\frac{\sin(y+x)}{\sin(y+2x)},x<90^0,f'(y)=\frac{\sin(x)}{(\sin(y+2x))^2}\ge 0$ so $f$ is nondecreasing...Hence $\frac{\sin(20^0+x)}{\sin(20^0+2x)}\ge \frac{\sin(15^0+x)}{\sin(15^0+2x)} >\frac{\sin(10^0+x)}{\sin(15^0+2x)}$ and (***) follow...
In this proof I didn't bothered about the sign of the sine functions because from the condition of the problem we have $2x+20^0<90^0$ or $x<35^0$


analytic geometry
geometry
geometric transformation
perimeter

The circle is tangent to both coordinate axes and contains the point(6,3). Find the center of the circle and the radius.

[quote="delta"]Too easy i think [hide]
Let the center of the circle is $O(x,y)$ as it touches both coordinate axis then $x=y$. Then if $A(6,3)$ $OA^{2}=(x-6)^{2}+(x-3)^{2}=2x^{2}-18x+45$
Now if $A$ is in the circle $OA \le x^{2}$ giving $x \in [3,15]$.[/hide][/quote]
Technically a circle is only the points that are a constant distance from a point, and not the points in the interior. A [i]disk[/i] contains the interior points.

[quote="randomdragoon"]Technically a circle is only the points that are a constant distance from a point, and not the points in the interior. A [i]disk[/i] contains the interior points.[/quote]
It could be a language problem.

Take a look at the following translations English/Spanish:

A disk is bounded by a circle, and its perimeter is called circumference.
Disk $\to$ Círculo
Circle $\to$ Circunferencia
Circumference $\to$  Longitud de la circunferencia

I've been told the Spanish meanings are more faithful to the original latin words and English messed it up. But that's maybe because I'm Spanish. Judge yourself:

'Deus est circulus cuius centrum ubique, circunferentia vero nusquam'


Is the code for page margin
[code]\setlength\leftmargin{1in}[/code]
correct?
Basically, my document is otherwise perfect, but it seems to be shifted to the right, so much so that if I put textwidth as more than 7, the text goes off the page on the right side. For 6, it is noticeably right aligned and for 5, it seems correct.
My preamble is this
[code]\documentclass[11pt]{article}
\usepackage{amsmath,amssymb}
\setlength\topmargin{-1in}
\setlength\textwidth{5in}

\pdfpagewidth 8.26in
\pdfpageheight 11.5in[/code]
This gives correct view, but there is too much (but symmetric) left and right margin which I can't reduce.

Page layout is very complicated with lots of variables so use the geometry package which makes it very easy indeed. For example [code]\usepackage[left=1in, right=1in]{geometry}[/code] gives you left and right margins of 1in (you can use cm if you wish)

If you use the textwidth command and want a centered page, you should set the side margins to zero. Anything else moves the page off center. For example, suppose you use a page that's 8.5 in wide and 6.5 in of text. If you set the left margin to 1 in, the page is shifted 1 in to the right of center, so its right edge is justified to the right edge of the page.
I use 6.5 in textwidth, with left and right margins set to zero; this gives me a centered page, with actual margins of 1 inch on each side.

If you're using a negative top margin, you're probably putting a lot of space into some other commands; there are at least three commands I know that add to the top margin on a standard page. From my experience, you probably want the sum slightly positive; the page looks better if the text is a little below center.

Thanks for the help. When I put all margins 0 (in the geometry package) the text is horrribly aligned towards the top-left corner. Problem is that I'm using a previous document as a 'template' for my document, i.e. I've kept the preamble same and have only added my text between \begin{document} and \end{document}. So I don't really know what settings were used by the previous author.

Zero margins is my advice without the geometry package; that package follows different rules.

Actually, now that I look at it: that leftmargin command seems to do nothing at all, and the default without it is about two inches of margin on the left.

What does work for me is the \oddsidemargin command; that gives about one inch of margin on the left if you set it to zero, plus whatever you add. It doesn't default to center.

For the margins at the top, look for \headheight and \headsep. They add to the top margin, and (at least in my system) they default to slightly positive if you don't define them. The total margin at the top comes out to about 1 inch plus the sum of the \topmargin, \headheight, and \headsep values.

I would advise deleting all the page layout settings in the preamble and only using the geometry package. The [url=http://tug.ctan.org/tex-archive/macros/latex/contrib/geometry/manual.pdf]Users' manual[/url] will give you all the control you want.


Find the only integer n such that $ \frac{(n\plus{}1)^2}{n\plus{}12}$ is an integer.

'm not sure if there is only 1, but if the denominator evaluates to 1, and the numerator evaluates to integer *which it would be if n=-11*, then the fraction is an integer

$ n \equal{} \minus{} 11$

EDIT:
wait what about denominator evaulates to -1 and n=-13?

$ n\equal{}\minus{}13$ also

sorry I meant to say find the only positive integer n. :blush:

UNLESS you mean positive in which it is 109 as follows:

$ \frac{(n\plus{}1)^2}{(n\plus{}12)}$
$ \frac{n^2\plus{}2n\plus{}1}{n\plus{}12}$
synthetic division...

$ n\minus{}10\plus{}\frac{121}{n\plus{}12}$
$ n\minus{}10$ is always an integer given n is an integer so the only question is
is $ \frac{121}{n\plus{}12}$ an integer?
or
what integer when added to 12 is a factor of 121?
the factors are -121, -11, -1, 1, 11, and 121
of these, only 121 is positive when 12 is subtracted from it
$ 121\minus{}12\equal{}\boxed{109}$


Hi,
I am stucked on a question on my homework. Here is the problem:
The diagonals of a rectangle are 8 units long and intersect at a 60 degree angle. Find the dimensions of the rectangle.

What I dont get is , what is "dimension"? This is supposed to be about the "Special Right Triangle" (30-60-90) (45-45-90) Theorems.

Any help will be appreciated. Thank you!

I am pretty sure that "dimensions" means sidelengths here.

Anyway, this is neither Olympiad Section level, nor an open question.

  darij

Thank you for the help.
Sorry about posting in the wrong forum :S It is my first time here, I don't know where else I should post. Sorry about that. Thanks again.


$\boxed{\ \left\{\begin{array}{c}\{x,y,z\}\subset (1,\infty )\\\\ x+y+z+2=xyz\end{array}\right\|\Longrightarrow \frac{3+x+y+z}{6}\ge\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ }\ .$

Like a remark, with the substitutions: $x=\frac{a+b}{c}$, $y=\frac{b+c}{a}$, $z=\frac{c+a}{b}$ respecting the condition : $x+y+z+2=xyz$

the inequality is $\Longleftrightarrow \sum{\frac{a+b}{c}}+3\geq 6\sum{\frac{a}{b+c}}$ $(\star)$ , with $a,b,c$ sides of a triangle (because the other condition : $x,y,z > 1$ )

but $(\star) \Longleftrightarrow (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 6\sum{\frac{a}{b+c}}$, which is the almost famous now inequality given in Vietnam TST 2006.

Put $a=x-1$ ,etc and it is equivalent to
http://www.mathlinks.ro/Forum/viewtopic.php?t=122198

 :wink:


geometry
parallelogram
number theory
greatest common divisor
geometry unsolved

An acute triangle $ABC$ is given. Let $C'$ be the foot of the altitude on $AB$, and let $D,E$ be different points of segments $CC'$. Denote by $F,G$ be orthogonal projections of the point $D$ onto the sides $AC,BC$. Prove that if $DGEF$ is a parallelogram, then $ABC$ is isosceles.

$\triangle CDF$ and $\triangle CDG$ can be inscribed in the same semicircle because of the equal hypotenuses. But the perpendicular distances from $F$ and $G$ to segment $CC'$ are equal, so these triangles are congruent. If $\angle GDC=\angle FCD$, $\angle C$ would be a right angle. But $\triangle ABC$ is acute, therefore, $\angle GCD=\angle FCD$. We know that $CC'$ is the altitude so $\angle A=\angle B$.


The positive integers $ A$, $ B$, $ A \minus{} B$, and $ A \plus{} B$ are all prime numbers. The sum of these four primes is

$ \textbf{(A)}\ \text{even} \qquad \textbf{(B)}\ \text{divisible by }3 \qquad \textbf{(C)}\ \text{divisible by }5 \qquad \textbf{(D)}\ \text{divisible by }7 \\ \textbf{(E)}\ \text{prime}$

A cannot be 2, since there is no prime below it for A-B to equal. Therefore, A is an odd prime. B must be 2, because otherwise A-B and A+B would both be even and not prime. The only 3 consecutive odd numbers that are prime are 3, 5, and 7, so A = 5. These four primes sum up to 17, so the answer is E.
EDIT: Oops, messed up a bit in writing. Fixed.

[quote="Brut3Forc3"]A cannot be 2, since there is no prime below it for A-B to equal. Therefore, A is an odd prime. B must be 2, because otherwise A-B and A+B would both be even and not prime. The only 3 consecutive odd numbers that are prime are 3, 5, and 7, so B = 5. These four primes sum up to 17, so the answer is E.[/quote]

Not true, there may be more triple primes.
But you know that they are of the form a, a-2, and a+2.  Add those up  and you get 3a, so the answer is B

EDIT: Oh snaps, forgot to include 2.  Ehehe.  Still though, I don't think it will always be prime.

[quote="Brut3Forc3"][quote="CircleSquared"]
But you know that they are of the form a, a-2, and a+2.  Add those up  and you get 3a, so the answer is B[/quote]
First of all, you would have to add 2 to that sum, so it wouldn't be divisible by 3.
Also, there can't be any more triple primes because of any 3 consecutive odd numbers, there will be one that is divisible by 3.[/quote]

i see.  i knew i was missing something  :rotfl:

[quote="alexk"][quote]B must be 2, because otherwise A-B and A+B would both be even and not prime. The only 3 consecutive odd numbers that are prime are 3, 5, and 7, so B = 5.[/quote]
You said that B must be 2, and then you said that it is 5.
Did you mean that A is 5? that's the only thing that fits.[/quote][/quote]
Yup. Sorry.  :oops:


IN THE NAME OF GOD

Find all points (4, y) that are 10 units from the point (–2, –1).

THANKS
SALAH

Hi salahcool.

The points which are $ 10$ units from $ (\minus{}2, \minus{}1)$ lie on the circle, centre $ (\minus{}2, \minus{}1)$ and radius $ 10$, that is on the circle whose equation is $ (x\plus{}2)^{2} \plus{} (y\plus{}1)^{2} \equal{} 100$. The line $ x\equal{}4$ cuts this circle where $ (4\plus{}2)^{2} \plus{} (y\plus{}1)^{2} \equal{} 100$, that is where $ (y\plus{}1)^{2} \equal{} 64$, so $ y\equal{}\minus{}9$ or $ 7$ and so the points are $ (4, \minus{}9)$ and $ (4, 7)$.

hello, the distance between the two points is given by
$ \sqrt{(4\plus{}2)^2\plus{}(y\plus{}1)^2}$ and this must be $ 10$, hence we have
$ \sqrt{(4\plus{}2)^2\plus{}(y\plus{}1)^2}\equal{}10$ from here we get
$ y_1\equal{}\minus{}9$ or $ y\equal{}7$
Sonnhard.

ok,thank you very much

Well done

Dr Sonnhard Graubner

AndrewTom


Prove that if $ x \plus{} y$ is a square-free number then the equation $ x^{2009} \plus{} y^{2009} \equal{} z^{2009}$ has no solutions in natural numbers.


analytic geometry
vector
linear algebra
matrix

Does there exist a linear map T : R^6-------->R^2 such that 2rank(T) = nul(T)?

Yes. rank(T)+null(T)=dimm(R^6)=6
3rank(T)=6 so rank(T)=2

can you please give me an example ?

Thanks

Wow Thanks for superfast reply.

$ x_{1}$ has six components? i.e (1,2,3,4,5,6)

No, $ x_1$ is just the "first coordinate" of a vector $ x \equal{} (x_1,x_2,x_3,x_4,x_5,x_6) \in \mathbb{R}^6$. So, if you define $ y \equal{} (x_1,x_2) \in \mathbb{R}^2$, the linear mapping is $ T(x) \equal{} y$.

Written as a matrix, mavropnevma's example is $ \begin{bmatrix}1&0&0&0&0&0\\0&1&0&0&0&0\end{bmatrix}.$

But $ \begin{bmatrix}3&0&\minus{}7&21&9\\1&4&32&\sqrt{2}&11\end{bmatrix}$ would also work.

And if you write down just any $ 2\times 6$ matrix (which is exactly what I just did) it would be a staggeringly unlikely coincidence for it to have any rank other than $ 2.$

Thanks Kent Merryfield.

where do i plug this matrix and what do i get?

What do you mean by that? Those matrices are simply representations of your linear transformation (you are familiar with matrix representations of linear transformations, right?). Basically, for this case just write down any 2x6 matrix such that (for example) its 2x2 submatrix made up of rows 1,2 and columns 1,2 is invertible.


How do you put a strikethrough, cross, or line through an equation (to cancel it, in a fraction)? I have tried \not{equation}, but it only works on single digits, and then, not well. \cancel{equation} does not work in the AoPS Classroom or Forum. \sout{equation} does not work either. How would I do this? I have seen it done in the Classroom before, so I know it can be done. Please help! Soon! Thank you!

Since the cancel package is not available on the forum you need to be creative eg using \not\! as in \[ \frac{32}{64}
\equal{}\frac{{2}\times\not\!4\times\not\!4}{4\times\not\!4\times\not\!4}
\equal{}\frac{2}{4}
\equal{}\color{black}\boxed{\color{blue} \frac{1}{2}}\]
For cancelling out more complicated stuff you will just have to use lots of \not\!
\[ \frac{x^2\minus{}1}{x\minus{}1}\equal{}\frac{(x\plus{}1)(\not\!x\not\!\minus{}\not\!1)}{\not\!x\not\!\minus{}\not\!1}\equal{}x\plus{}1\]
It's not ideal but will have to do.


this is kind of a brain teaser. a easy one.

$(x-a)(x-b)(x-c)...(x-z)=?$

 :D

[quote="Zenith Xan"][hide="Click to reveal hidden text"]It has $(x-x)$, so $\boxed{0}$.[/hide][/quote]

Simple and nice!!!!!!!!!!!!


We introduce the following operators mapping classes of algebra to classes of algebras (all of the same type):
$ A\in S(K)$ iff is a subalgebra of some member of K
$ A\in H(K)$ iff is a homomorphic image of some member of K
$ A\in P(K)$ iff is a direct product of family of algebras in K

We know that:$ SH\subset HS, PS\subset SP, PH\subset HP$
Show:
$ HP \neq PH$
$ SP \neq PS$

So I don't know much about universal algebra, but hopefully these examples are easily modified to your purposes:

Firstly notice that HP contains central products of groups, but PH has very few indecomposable members.

Take K to be the class of dihedral groups G of order 8.  Denote their central involution by z.

H(K) is the class containing { 1, 2, 2 x 2, G }

The only directly indecomposable members of P(H(K)) are already in H(K)

P(K) contains G x G, and H(P(K)) contains $ G \times G / \langle (z,z) \rangle$, a directly indecomposable extra-special group of order 32, sometimes denoted $ 2^{2\cdot2\plus{}1}$ or $ G \circ G$.

Secondly notice that SP contains subdirect products, but PS has very few indecomposable members.

Take K to be the class of nonabelian groups G of order 6.

The only directly indecomposable members of P(S(K)) are already in S(K) and so have order at most 6.

However P(K) contains a group isomorphic to $ \langle (1,2,3), (1,2) \rangle \times \langle (4,5,6), (4,5) \rangle$, and so S(P(K)) contains a group isomorphic to the indecomposable group $ \langle (1,2,3), (4,5,6), (1,2)(4,5) \rangle$.


[quote="JSteinhardt"]A slightly nicer way:

1*3*5*...*199 = 200!/2*(100)!

...[/quote]

Not that it changes the answer any, but $2\cdot 6\cdot 8\cdots 200\neq 2\cdot 100!$, it equals $2^{100}\cdot 100!$...

[quote="technox"]Question: In the above solution it is mentioned:
[quote]  There are 33 multiples of 3, 11 multiples of 9, 4 multiples of 27 and 1 multiple of 81 for a total of 49 factors of 3. [/quote]

Aren't the 11 multiples of 9 (or the 4 multiples of 27 and so on...) counted with the 33 multiples of 3 (since they are multiples of this number as well) and, thus, they should not be re-counded;[/quote]'

We count the multiples of 3 and then add 1 more for each multiple of 9 (so we have counted the multiples of nine twice and the multiples of ONLY three once). This is repeated with 27 and 81.

[quote="bpms"][quote="ktm"][quote="E^(pi*i)=-1"][hide]There are 33 multiples of 3, 11 multiples of 9, 4 multiples of 27 and 1 multiple of 81 for a total of 49 factors of 3.[/hide][/quote]

Aren't there only 3 multiples of 27? I got 48 as my answer.[/quote]
27/81/135/189[/quote]

Oh, I read the question as the first 100 positive integers... woops.

[quote=solafidefarms]Let $P$ be the product of the first 100 positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k$.[/quote]

this is inclusion-exclusion principle.

[quote=Andrew_maybe]there's a formula for finding the highest power of a some number k such that it divides n![/quote]

Yes, but we are talking about the odd integers thus it doesn't work

[hide=Solution]
The first odds are $1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199$. So, we want to see how many $3'$s are in the first $200$ integers.
$$\left\lfloor \frac{200}{3}\right\rfloor - \left\lfloor \frac{200}{6}\right\rfloor +\left\lfloor\frac{200}{9}\right\rfloor - \left\lfloor \frac{200}{18}\right\rfloor +\left\lfloor \frac{200}{27}\right\rfloor - \left\lfloor \frac{200}{54}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor - \left\lfloor \frac{200}{162}\right\rfloor= 66 - 33 + 22 - 11 + 7 - 3 + 2 - 1 = 49.$$

aime p8

There are $33$ multiples of $3$, $11$ multiples of $9$, $4$ multiples of $27$, $1$ multiple of $81$, $0$ multiples of $243$, so answer is $33+11+4+1=\boxed{049}$.


Find
\[\int_{C}{\frac{1}{{z+2}}dz}\]
where C:
\[z = 2e^{i\theta },-\pi \leqslant \theta \leqslant 0 \]
Does anyone knows how to solve this complex integral when we have a singularity at the begining of the curve?

thanks!


linear algebra
matrix
superior algebra
superior algebra unsolved

Following up from http://www.artofproblemsolving.com/Forum/topic-36105.html where it was shown by Leva1980 that $\mathrm{SL}_{n}(\mathbb{Z})$ is generated by only two elements, $S, A_{1,2}$.
Show that there exist an integer $K_{n}$, s.t. every  matrix of $\mathrm{SL}_{n}(\mathbb{Z})$ is the product of at most $K_{n}$ elements belonging to $\{S, A_{1,2}\}$.

P.S.: I don't ask to find the smallest possible value for $K_{n}$ but that would be a very nice surprise (although I think it is extremelly difficult).

It seems to me that if you only allow finite products, you will only get a finite subset of $SL_{n}$

amfulger,
I think what I wrote is true, even if it seems a bit conter-intuitive. It's a nice result :wink:

As amfulger said, with at most $K_{n}$ factors from the set $\{S, A_{12}\}$ you can get at most $2^{K_{n}+1}-1$ different products. Since $SL_{n}(\mathbb Z)$ is infinite (the matrices $I+k N_{12}, k \in \mathbb Z$ are pw. different), you cannnot generate $SL_{n}(\mathbb Z)$ in this way.

Did you mean what we understand by what you wrote? :)

LiMa

Amfulger, LiMa,
You are absolutely right. I jumped to quickly from "boundly generation against a set of elementary matrices".
I should have written:

Show that there exist an integer $K_{n}$, s.t. every  matrix of $\mathrm{SL}_{n}(\mathbb{Z})$ is the product of at most $K_{n}$ elementary matrices, where an elementary matrix is a power of $I_{n}+E_{i,j}$ ($i\not=j$).
Sorry for the inconvenience, still a nice problem I hope.

read a proof of the fact that these matrices generate $SL_{n}(\mathbb{Z})$ (I think it has been done on the forum) and you will immediately get such a $K_{n}$, but maybe not a tight bound :)


logarithms
modular arithmetic
Recursive Sequences

For a given positive integer $k$ denote the square of the sum of its digits by $f_{1}(k)$ and let $f_{n+1}(k)=f_{1}(f_{n}(k))$. Determine the value of $f_{1991}(2^{1990})$.

It is clear that $ f_1(k) \leq (9(\log_{10}k\plus{}1))^2$ and, in particular, $ f_1(k)<k$ for all $ k\geq 400$. It is easy to find all limit cycles of $ f_1(k)$ - there are three of them:
$ 1,1,1,\dots$
$ 81,81,81,\dots$
$ 169,256,169,\dots$
It is clear that $ f_{1991}(2^{1990})$ ends up in one of these numbers. Since $ f_1(k)\equiv k^2\pmod{9},$ we have
\[ f_{1991}(2^{1990})\equiv 2^{1990\cdot 2^{1991}}\equiv 4\pmod{9},\]
implying that $ f_{1991}(2^{1990})\equal{}256.$


1. Divide 2008 numbers from 1 to 2008 into 8 groups with each group containing 251 numbers. Let $ x_i$
be the median of the ith group, where $ 1\le i\le8$. Let $ S \equal{} \sum_{i \equal{} 1}^8 x_i$.
Find the maximum and minimum of S. Justify your answer.

2. Let $ x \equal{} (\sqrt5 \plus{} 2)^{2009}$. Find $ x(x \minus{} \lfloor{x}\rfloor)$.

3. Three boys A, B, and C sit around a round table in that order. A has a ball in his hand. Starting
from A the boy having the ball passes the ball to either of the other two boys. After 5 passes the
ball goes back to A. How many different ways can the ball be passed?

[hide="my sketch for #3"]$ P(n)\equal{}(1\minus{}P(n\minus{}1))/2,P(0)\equal{}1$ with $ 2^5P(5)$ as the answer. Is this correct? (note tha P(n) is treated as the prob. of getting the ball back to A at the nth pass)[/hide]

[quote="mathwizarddude"]
2. Let $ x \equal{} (\sqrt5 \plus{} 2)^{2009}$. Find $ x(x \minus{} \lfloor{x}\rfloor)$.
[/quote]

[hide]
Let $ x' \equal{} (\sqrt 5 \minus{} 2)^{2009}$. We will prove that $ x' \equal{} x \minus{} \lfloor x\rfloor$. First, note that $ x' < 1$ since $ \sqrt 5 \minus{} 2 < 1$. Next, we will prove that $ x \minus{} x'$ is an integer.

[b]Lemma 1.[/b] If $ a \minus{} \frac 1a$ is an integer, then $ a^n \minus{} \frac 1{a^n}$ is an integer for all odd integers $ n\geq 3$.

[i]Proof:[/i] We use double induction on $ n$. The base case for $ n \equal{} 1$ is given. For $ n \equal{} 3$, just observe that $ a^3 \minus{} \frac 1{a^3} \equal{} \left(a \minus{} \frac 1a\right)^3 \plus{} 3\left(a \minus{} \frac 1a\right)$. The RHS is the sum of two integers by the induction hypothesis, implying that the LHS must be an integer. Also, we find that $ a^2 \plus{} \frac 1{a^2}$ is also an integer since
\[ a^2 \plus{} \frac 1{a^2} \equal{} \left(a \minus{} \frac 1a\right)^2 \plus{} 2.\]
Finally, the identity
\[ a^n \minus{} \frac 1{a^n} \equal{} \left(a^{n \minus{} 2} \minus{} \frac 1{a^{n \minus{} 2}}\right)\left(a^2 \plus{} \frac 1{a^2}\right) \minus{} \left(a^{n \minus{} 4} \minus{} \frac 1{a^{n \minus{} 4}}\right)\]
finishes the induction step, and the proof is complete.$ \;\;\blacksquare$

Now, let $ a \equal{} \sqrt 5 \plus{} 2$. It follows that $ \frac 1a \equal{} \sqrt 5 \minus{} 2$. Since $ a \minus{} \frac 1a \equal{} (\sqrt 5 \plus{} 2) \minus{} (\sqrt 5 \minus{} 2) \equal{} 4$, by Lemma 1, $ a^{2009} \minus{} \frac 1{a^{2009}}$ is an integer as well, and that proves that $ x \minus{} x'$ is an integer, and therefore $ x' \equal{} x \minus{} \lfloor x\rfloor$.

Finally, the computation $ (\sqrt 5 \plus{} 2)^{2009}(\sqrt 5 \minus{} 2)^{2009} \equal{} 1^{2009} \equal{} 1$.
[/hide]

1. The minimum value of the first median is 126.  In an attempt to be rigorous, we will prove that $ x_{I\plus{}1}\minus{}x_I \geq 126$.  This can be done in general with induction (not for 126); we proceed with the specific case.  After the numbers which have already been used preceding $ x_I$ are eliminated, we now have the choice to use all of the consecutive numbers greater than $ x_I$, of which there are 126.  Therefore, our minimum can be attained trivially from this fact.  We add numbers consecutively from the beginning and take the remaining 125 from the end of the list of numbers from {1-2008}.  If we do this continually, we end up with the greatest median as 126*8=2008-8*125, as desired.  We can do basically the same thing for the maximum value.

3.  Yeah, your method is relatively effective (and correct).  Alternately you can notice that Q(n)+Q(n-1)=$ 2^(n\minus{}1)$, where we denote Q(i) to equal the number of ways to end at person A after i moves.  After n-1 moves, you are either at A, B, or C.  Q(n) counts the cases where you are at B and C, because to get to A you must go through one of them.  Therefore, this recursion becomes elementary.  (And now, after a quick check, you can see that our forms are equivalent).


calculus
integration
real analysis
real analysis unsolved

Find
1. $ \int \sqrt {\frac {x^{3}}{a^{3} \minus{} x^{3}}}\,dx$
2. $ \int \frac {3x^{4} \plus{} 12x^{3} \plus{} 6}{x^{5} \plus{} 5x^{4} \plus{} 16x \plus{} 12}\,dx$
First one may be easy, but the second is good enough.
Selected from sosmath.com.

hello, your first integral leads to elliptic integrals of the first and second kind.
See here: http://www.wolframalpha.com
Sonnhard.


[quote="Treething"]Deep's Throat Mints come in three flavors, Wintermint, Whitemint, and Spearmint. If Whitemint lozenges are 60% sugar, Wintermint lozenges are 90% sugar, and Spearmint lozenges are 40% sugar, how many arrangements of mints can be in a pack of Deep's Throat Mints if each one has seven mints and it consists of 69% (rounded to the nearest integer) sugar? Assume that each mint is the same volume and the mints are arranged in a roll like Lifesavers or Mentos are.[/quote]

[hide]3 Spearmint, 4 Wintermint

19 arrangements

5 Whitemint, 2 Wintermint

12 arrangements

19+12=31?[/hide]