In our LaTeX pages, you'll see that there are several places for 'Examples', and when you go to those pages you see a 'Coming Soon'.

If students are interested in submitting anything they have written for inclusion on that page, we'd be happy to review it and include it if suitable (with attribution to the author, of course!).  For students who are looking for something to write about for practice with LaTeX, you might consider convertiong portions of the [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]How to Write a Solution[/url] article to TeX.  If you do so, and choose one of the ones with a diagram, we can provide PDF files with the diagrams.

You can send your LaTeX source to us at aops@artofproblemsolving.com.

so you want the actual lesson changed into TeX?

Sure - you can take just one of the 11 or so sections, rather than taking on the whole thing at once.

What type of examples are you looking for?  Ones containing mainly math or ones containing words as well?  Also, should the examples be entire documents, or just sections from documents?  Should the examples use only ``standard'' macros or can it include ones that we define ourselves in the preamble?  Finally, should we post them here or e-mail them?

[quote="mathfanatic"]What type of examples are you looking for?  Ones containing mainly math or ones containing words as well?  [/quote]

I'm looking for full documents, not just snippets of math.  A solution to a USAMTS problem would be fine - a laundry list with explanations of useful math formulas would be fine.

[quote="mathfanatic"]Also, should the examples be entire documents, or just sections from documents?[/quote]

They should be full documents in the sense that I can take the TeX code you send me, open it in TeXnicCenter without adding any preamble or anything, and compile it without a problem.

  [quote="mathfanatic"]Should the examples use only ``standard'' macros or can it include ones that we define ourselves in the preamble? [/quote]

Both are fine.  If you include your own packages as part of a separate file, email me that file as well (I'll be putting up instructions in the next few days about how to make your own package if you haven't figured out how already).

[quote="mathfanatic"] Finally, should we post them here or e-mail them?[/quote]

Send 'em to me at aops@artofproblemsolving.com

[quote="confuted"]Would you like the examples to just be math and math related, or would a research paper on Maxwell's Equations be acceptable? I should be completing it within the next few days, perhaps tonight. It demonstrates the use of BibTeX as well, although perhaps not particularly thoroughly.[/quote]

Sounds great!

[quote="Spoon"]Confuted just showed me a beautiful report he texed completely, length of 16 pages! All about maxwell's equations... while the subject might be difficult to grasp, the display of TeX and it's power is there and I think it should be submitted as an example...

I won't give the link until his persmission is granted.[/quote]

I believe that's the one he was talking about two or three posts ago.

[quote="Spoon"]I won't give the link until his persmission is granted.[/quote]
The source files are on their way to Mr. Rusczyk; meanwhile, here is a link to the report: http://confuted.0catch.com/Maxwellsequations.pdf

If anyone finds any errors in that, I would of course appreciate knowing about them.

You can use one of my papers as an example. The source file can be found [url=http://www.albanyconsort.com/HaarMeasure/HaarMeasure.tex]here[/url], and a pdf can be found [url=http://www.albanyconsort.com/HaarMeasure/HaarMeasure.pdf]here[/url].


$ p^{n}\equal{}(x\plus{}y)(x^{2}\minus{}xy\plus{}y^{2})$-->$ x\plus{}y\equal{}p^{a},x^{2}\minus{}xy\plus{}y^{2}\equal{}p^{b}$ with $ 0\leq a,b\leq n$

[quote="Moonmathpi496"]Find all positive primes $ p$ for which there exist positive integers $ n, x, y$ such that $ p^{n}\equal{} x^{3}\plus{}y^{3}$.[/quote]
Let $ z\equal{}(x,y),x_{1}\equal{}x/z,y_{1}\equal{}y/z$, then $ z\equal{}p^{l},x_{1}^{3}\plus{}y_{1}^{3}\equal{}(x_{1}\plus{}y_{1})(x_{1}^{2}\minus{}x_{1}y_{1}\plus{}y_{1}^{2})\equal{}p^{k}*p^{l}$.
If k>0, l>0 then $ p^{2k}\minus{}p^{l}\equal{}3x_{1}y_{1}$. Therefore if $ p\not \equal{}3$ kl=0 (because $ (x_{1},y_{1})\equal{}1)$).
 If $ xy\not \equal{}0$ (else many solutions $ x\equal{}p^{k},y\equal{}0$), then k>0,l>0, therefore $ p\equal{}3$.
Therefore we can find only primitive solutions $ x_{1}\plus{}y_{1}\equal{}3^{k},x_{1}y_{1}\equal{}3^{2k\minus{}1}\minus{}3^{l\minus{}1}\to l\equal{}1,(x_{1},y_{1})\equal{}(2,1),\ (1,2)$.
Therefore all solutions are $ x\equal{}3^{k},y\equal{}2*3^{k}$ or $ x\equal{}2*3^{k},y\equal{}3^{k}$.

$ x\equal{}y\equal{}n\equal{}1,p\equal{}2$ is a solution.

For any other solution we can assume $ x\plus{}y\geq 3\,,\,gcd(x,y)\equal{}1$, since any joint prime-divisor must be equal to $ p$, i.e. we can reduce until $ x,y$ have no common divisor. Since $ x^{2}\minus{}xy\plus{}y^{2}>1$ we get $ p$ divides $ x\plus{}y$ and $ x^{2}\minus{}xy\plus{}y^{2}$. However, $ gcd(x\plus{}y,x^{2}\minus{}xy\plus{}y^{2})\equal{}gcd(x\plus{}y,3x^{2})\equal{}gcd(x\plus{}y,3)$.

Therefore, $ p\equal{}3$ and for example $ x\equal{}2,y\equal{}1,n\equal{}2$ does the job.

My solution:

It's not difficult to see that $ x\plus{}y\equal{}p^{a}$ and $ x^{2}\minus{}xy\plus{}y^{2}\equal{}p^{b}$. Substituting $ y\equal{}p^{a}\minus{}x$ in the latter, we get after some calculations, $ 3x^{2}\minus{}3p^{a}x\plus{}p^{2a}\minus{}p^{b}\equal{}0$. Which has solutions $ x\equal{}\frac{3p^{a}\pm\sqrt{12p^{b}\minus{}3p^{2a}}}{6}$. Thus \[ 12p^{b}\minus{}3p^{2a}\equal{}k^{2}\qquad(1)\]
for some $ k$. If $ 2a\le b$ we get $ (x\plus{}y)^{2}\le x^{2}\minus{}xy\plus{}y^{2}$, which is not possible. Thus $ 2a>b$ and we get $ p^{b}\mid k^{2}$. Let $ k^{2}\equal{}p^{b}q$. So Dividing by $ p^{b}$ we get $ 12\minus{}q\equal{}3p^{2a\minus{}b}$. Hence $ 3\mid q$. Checking for $ q\equal{}0,3,6,9$ we find that $ (p,q,2a\minus{}b)\equal{}(2,0,2),(3,3,1),(2,6,1),(p,9,0)$. But the last one is impossible as $ 2a\minus{}b>0$. And $ (2,6,1)$ implies $ k^{2}\equal{}3\cdot 4^{a}$, which is also impossible.

So we get the solutions $ p\equal{}2,3$.


A book company assesses shipping charges of $ \$3$ for the first
item in a package, $ \$2$ for the second item and $ \$1$ for each
additional item in a package. For example, the shipping charge
for one package of four items is $ \$3 \plus{} \$2 \plus{} \$1 \plus{} \$1 \equal{} \$7$. How
many dollars are saved by shipping 10 items in two packages of
five items each, rather than five packages of two items each?

Two packages of 5 is: $ 3\plus{}2\plus{}1\plus{}1\plus{}1\equal{}8\times 2\equal{}16$
Five packages of 2 is: $ 3\plus{}2\equal{}5\times 5\equal{}25$

The problem asks for how much is saved, so $ 25 \minus{} 16$ or $ \boxed{9}$.


function
Euler
algebra
polynomial
Vieta
quadratics
number theory

Let $ n>4$ be a positive integer such that $ n$ is composite (not a prime) and divides $ \varphi (n) \sigma (n) \plus{}1$, where $ \varphi (n)$ is the Euler's totient function of $ n$ and $ \sigma (n)$ is the sum of the positive divisors of $ n$. Prove that $ n$ has at least three distinct prime factors.

[quote="brianchung11"]Let $ n > 4$ be a positive integer such that $ n$ is composite (not a prime) and divides $ \varphi (n) \sigma (n) \plus{} 1$, where $ \varphi (n)$ is the Euler's totient function of $ n$ and $ \sigma (n)$ is the sum of the positive divisors of $ n$. Prove that $ n$ has at least three distinct prime factors.[/quote]
Because $ \varphi(n)\equiv 0 (\mod 2),\forall n > 4$ so we have $ \varphi(n).\sigma(n) \plus{} 1$ is an odd number . It follows that $ n$ must be an odd number .
It is easy to check that $ n$ must be a square free number .If not ,then there exist a prime p such that $ v_p(n) > 1$ .From the formulas of Euler function we have $ p|\varphi(n)$ ,therefore $ p$ can not be a divisor of number $ \varphi(n).\sigma (n) \plus{} 1$ . Thus $ n$ must be a square free number .
Suppose $ n$ has less than three prime divisor . And because $ n$ is not a prime so it must be product of two different primes .Let $ n \equal{} pq$ and $ p > q$ . Then
\[ \varphi(n) \equal{} (p \minus{} 1)(q \minus{} 1)
\]

\[ \sigma(n) \equal{} (p \plus{} 1)(q \plus{} 1)
\]
And from the condition we have : $ pq|(p \plus{} 1)(q \plus{} 1)(q \minus{} 1)(q \plus{} 1) \plus{} 1$
Thus $ p|q^2\minus{}2,q|p^2\minus{}2$ ,it gives contradiction .
Problem has been proved .

$ pq|(p \plus{} 1)(q \plus{} 1)(q \minus{} 1)(q \plus{} 1) \plus{} 1$<- i believe that you've made some mistakes here... and $ p|q^2 \minus{} 2$ instead of $ p|q^2$... i think the simplest way to tackle the case of $ n\equal{}pq$ is vieta jumping.

[hide="Solution"]
If $ n$ is a perfect prime power, then let $ n \equal{} p^l$ where $ l > 1$.  This means that $ p|\varphi (n)$, so $ p$ cannot divide $ \varphi (n)\sigma (n) \plus{} 1$, which is a contradiction since $ p^l$ divides it.  Thus, we have a contradiction, so $ n$ may not be a perfect prime power.

Assume that there are only two prime factors of $ n$, namely $ p$ and $ q$.  If $ p^r\parallel{}n$ and $ r > 1$, then we see that $ p|\varphi (n)$, so $ p$ may not divide $ \varphi (n)\sigma (n) \plus{} 1$, which is a contradiction as $ pq$ must divide $ \varphi (n)\sigma (n) \plus{} 1$.  Thus, $ r \equal{} 1$.  Similarly, $ q\parallel{}n$.  It follows that $ n \equal{} pq$.  Notice that $ \varphi (n) \equal{} (p \minus{} 1)(q \minus{} 1)$ and $ \sigma (n) \equal{} (p \plus{} 1)(q \plus{} 1)$, so the condition turns into $ \frac {(p \minus{} 1)(q \minus{} 1)(p \plus{} 1)(q \plus{} 1) \plus{} 1}{pq}$ must be an integer, so $ \frac {(p^2 \minus{} 1)(q^2 \minus{} 1) \plus{} 1}{pq}$ is an integer, so $ \frac {p^2 \plus{} q^2 \minus{} 2}{pq}$ is an integer.  Let this value be $ k$.  It follows that $ p^2 \minus{} p(kq) \plus{} (q^2 \minus{} 2) \equal{} 0$.  Now, allow $ (a, b)$ to be the solution with minimal $ a \plus{} b$ ($ a$ and $ b$ do not have to be prime anymore).  Let $ a\ge b$.  If we look at this equation as a quadratic in $ a$, then we may notice that $ (\frac {b^2 \minus{} 2}{a}, b)$ is a solution.  Since this is symmetric in $ a$ and $ b$, it follows that $ (b, \frac {b^2 \minus{} 2}{a})$ is a solution.  This means that $ a \plus{} b \le b \plus{} \frac {b^2 \minus{} 2}{a}$, so $ a^2 \le b^2 \minus{} 2\implies a < b$, which is a contradiction, so $ n$ has at least $ 3$ divisors.  [/hide]


Euler
floor function
number theory unsolved
number theory

Let $a_1,a_2 \dots a_n$ and $x_1, x_2 \dots x_n$ be integers and $r\geq 2$ be an integer. It is known that \[\sum_{j=0}^{n} a_j x_j^k =0 \qquad \text{for} \quad k=1,2, \dots r.\]
    Prove that
    \[\sum_{j=0}^{n} a_j x_j^m \equiv 0 \pmod m, \qquad \text{for all}\quad  m \in \{ r+1, r+2, \cdots, 2r+1 \}.\]

Put $S(k)=\sum_{i=0}^n a_{i}x_{i}^k$ and $P(x)=x(x-1)...(x-r)$
It is well-known that $P(x) \equiv 0 (\mod (r+1)!)$ for all $x \in Z$(1)
From the condition $S(1)=S(2)=...=S(r)=0$ so that $S(r+1)=\sum_{i=0}^n a_{i}P(x_{i})$
From (1) $\Longrightarrow S(r+1) \equiv 0 (\mod (r+1)!) \Longrightarrow S(r+1) \equiv 0 (mod (r+1))$
We investigate this problem with two cases:
      .If $m \in P$ then $\sum_{i=0}^n a_{i}x_{i}^m \equiv \sum_{i=0}^k a_{i}x_{i} \equiv 0 (\mod m)$
So that $\sum_{i=0}^n a_{i}x_{i}^m \equiv 0 (\mod m)$
      .If $m$ isn't a prime number $\Longrightarrow (r+1)! \equiv 0 (\mod m) (\forall m \in \{r+1,r+2,...,2r+1\})$
Now,$S(m)=a_{0}P(x_{m-(r+1)})+a_{1}P(x_{m-r})+...+a_{r}P(x_{m-1}) \equiv 0 (\mod (r+1)!)$
$\Longrightarrow S(m) \equiv 0 (\mod m)$ For all $m \in \{r+1,r+2,...,2r+1\}$ and $m$ isn't a prime number
From two cases,It follows that:$S(m) \equiv 0 (\mod m)$ for all $m \in \{r+1,r+2,...,2r+1\}$
So that the problem is proved. :)  :)
Remark:From this method we can also prove the problem (From Poland Mathematical Olympiad):
      Given $k \in N,k\geq 2$ and $a_{1},a_{2},...,a_{n} \in Z$satisfying:
                       $a_{1}+2^{i}a_{2}+...+n^{i}a_{n}=0$ for all $1 \leq i \leq (k-1)$
Prove that:$a_{1}+2^{k}a_{2}+...+n^{k}a_{n} \equiv 0 (\mod k!)$

[quote="{x}"]Very, very nice ! Can you just elaborate a little more on that equality

$S(m)=a_{0}P(x_{m-(r+1)})+a_{1}P(x_{m-r})+...+a_{r}P(x_{m-1}) (\mbox{ mod } (r+1)!)$[/quote]
Actually I don't understand how to arrive at this equality, namely because I don't know what $a_r$ is supposed to be if $r>n$ and so on... But I have to admit that I didn't really give thought to the identity..

[quote="shobber"]Let $a_1$, $a_2$, $\dots$, $a_n$ and $x_1$, $x_2$, $\dots$, $x_n$ be integers and $r \ge 2$ be an integer. It is known that $\sum_{j=1}^{n} a_j x_j^k =0$ for $k=1$, $2$, $\dots$, $r$. Prove that
\[ \sum_{j=1}^{n} a_j x_j^m \equiv 0 \mod m\]
holds for all $m \in \{ r+1$, $r+2$, $\dots$, $2r+1 \}$.[/quote]

[i]Proof.[/i] Fix $m=\prod p_i^{\alpha_i}$ and rewrite $\sum a_j x_j^m$ as  \[\sum_{j=1}^{n} a_j x_j^m=\sum_{j=2}^{n} a_jx_j^m-x_1^{m-k}\left(\sum_{j=2}^{n} a_jx_j^k\right)=\sum_{j=2}^{n}a_jx_j^k\left(x_j^{m-k}-x_1^{m-k}\right).\] It's clearly enough to prove that this sum is congruent to zero modulo $p_i^{\alpha_i}$ for every $i$.

Fix $i$ and suppose wlog that $p_i \not | x_1$ (if all the $x_j$'s are congruent to zero modulo $p_i$, just divide all of them by $p_i^{\alpha}$ for an appropriate $\alpha$). Moreover, let $k=m-t\varphi(m)>0$, $t$ as large as possible. We evidently have $p_i^{\alpha_i-1}| k$ and therefore $k \ge p_i^{\alpha_i-1} \ge a_i$.  We now prove that $a_jx_j^k\left( x_j^{m-k}-x_1^{m-k}\right) \equiv 0 \mod p_i^{\alpha_i}$ holds true for every $j$. If $p_i\not | x_j$, then we have $x_j^{m-k} = x_j^{t\varphi(m)} \equiv 1 \equiv x_1^{t\varphi(m)} \mod p_i^{\alpha_i}$. If $p_i |x_j$, then $p_i^{\alpha_i} | x_j^{\alpha_i} | x_j^k$, since we have shown that $k \ge a_i$. This completes the proof. :)

let $m=k\phi (m)+q$, where $k,q$ are integers with $k\ge 1$ and $0\le q\le \phi (m) $, $ \phi $ is Euler totient function.
Then we have $q\le min\{ \phi(m),m-\phi(m) \} \le \lfloor \frac{m}{2} \rfloor  $. :lol:


A certain right triangle has its area numerically equal to its perimeter. The length of each side is an even integer. What is the perimeter?
(1) 26 (2) 28 (3) 30 (4) 32 (5) None of the above

[hide="solution"]cant it be multiple numbers , cuz i got a 6-8-10 triangle, because 6+8=14+10=24, while 6x8/2=24[/hide]

Solution:
[hide]
I tried some small cases and got $ a\equal{}6$, $ b\equal{}8$, $ c\equal{}10$:
area = $ \tfrac{1}{2} \cdot 6 \cdot 8 \equal{}24$, perimeter = $ 6 \plus{} 8 \plus{} 10 \equal{} 24$,
so (5).
[/hide].


find all  natural number   $ n$  such that :  let $ S(n)$ be sum all digit  of number n satisfying following :
$ S( n ).(S(n)\plus{}1)\equal{}n$


I don't understand. But maybe this problem is same as http://www.mathlinks.ro/Forum/viewtopic.php?t=131843

[quote="carlosbr"][size=150][color=blue]TST PERU - IMO 2007 [Day 1- Q03][/color][/size]

Let $N$ be a natural number which can be expressed in the form $a^{2}+b^{2}+c^{2}$, where $a,b,c$ are integers divisible by 3.
Prove that $N$ can be expressed in the form $x^{2}+y^{2}+z^{2}$, where $x,y,z$
are integers and any of them are divisible by 3.

Spanish Communities[/quote]
i would guess that the problem is this way:

Let $N$ be a natural number which can be expressed in the form $a^{2}+b^{2}+c^{2}$, where $a,b,c$ are integers divisible by 3.
Prove that $N$ can be expressed in the form $x^{2}+y^{2}+z^{2}$, where $x,y,z$
are integers and any of them are NOT divisible by 3. (e.g. $3^{2}+3^{2}+3^{2}=1^{2}+1^{2}+5^{2}$ and $3^{2}+3^{2}+9^{2}=1^{2}+7^{2}+7^{2}$)

Naphthalin

It is my problem from 4th level of ARMO 2006.


I don't know, I can tell you that in the X=cos(Pi/4)+i*sin(Pi/4) thingy that X is equal to e^(i*(Pi/4)).  Remember e^(i*x)=cos(x)+i*sin(x).  Don't know if that helped you.  Try looking up complex exponentiation.  That might help.

I have no clue. I'm split between two definitions.

:(

Could anyone (admins?) could offer insight into what that notation means?

It was your question -- go back to where you found it and figure out what the notation means from there.  It looks like something roots-of-unity related, but I don't have any idea where 10 would have come from.

It was an early IMO longlist ('67 maybe?).

I'll most likely be able to answer this (if it's from 1967) from a book in about a week.  Please wait. :)

Masoud Zargar

So basically, $C_{10}=e^{i\frac{5\pi}{2}}$?

Masoud Zargar

Strange that you're fluent in Stirling numbers, boxedexe, yet you don't know some basic combinatorial notation: $C_{n}^{k}={n\choose k}$. How come?

Anyway, the identity doesn't hold hence it must be something else.

The combination of the numbers does not hold, so it is not that.  Besides, I do know the notation, and if you really think I don't, then what can I say?  :wink:

Masoud Zargar

to boxedexe:

why do you have a picture of yourself eating a piece of tin foil as your avatar?  just curious.

It's cookie...some people thought it was crack.  :o

EDIT: Could a mod please tell me why I have been notified by email that I have recieved a private message, but it is not there?  Thanks in advance.

[quote="boxedexe"]EDIT: Could a mod please tell me why I have been notified by email that I have recieved a private message, but it is not there?  Thanks in advance.[/quote]

Whoever sent it to you probably deleted it before you got to it.  (You can do this via your Outbox.)


algebra polynomial Vieta calculus integration

If a,b,c are positive integers such that   $ 0<\frac{a^{2}\plus{}b^{2}}{ab\plus{}1}$ is the square of an integer.

This is an [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=352683#352683]IMO problem[/url]. Definitely does not belong in HSB.

[hide="Solution"]
Let $ \frac{a^2\plus{}b^2}{ab\plus{}1}\equal{}k$

Suppose $ k$ is a nonsquare integer and suppose that the solution $ (a,b)$ that makes $ k$ a nonsquare is the smallest. Without loss of generality, take $ a\geq b$.

Consider the polynomial $ \frac{x^2\plus{}b^2}{bx\plus{}1}\equal{}k\implies x^2\minus{}bkx\plus{}b^2\minus{}k\equal{}0$. Clearly $ x\equal{}a$ is a root. By Vieta's formulas, the sum of the roots is $ bk$, so the other root is $ bk\minus{}a$.

Again, by Vieta's formulas, the product of the roots is $ b^2\minus{}k$, so the second root must be $ \frac{b^2\minus{}k}{a}$.

From $ bk\minus{}a$, we know that $ \frac{b^2\minus{}k}{a}$ is an integer. We also know that $ \frac{b^2\minus{}k}{a}\ne 0$; otherwise we would have $ b^2\equal{}k$, implying $ k$ is a square, which is a contradiction. Therefore, we have a second integral positive root to the quadratic.

$ a\geq b\implies a^2\geq b^2\implies a^2>a^2\minus{}k\geq b^2\minus{}k\implies a>\frac{b^2\minus{}k}{a}$, so the second root is strictly less than the first root, so by infinite descent we reach a contradiction.[/hide]


[quote="eminem"]$4^x+6^x=9^x$

x=?[/quote]

[hide="hint"]

Divide the equation by $9^x$ and see what you get .[/hide]

[quote="eminem"]$4^x+6^x=9^x$

x=?[/quote]

  [hide]The units digit of powers of 4 repeat as follows: 4, 6, 4, 6, 4, 6,4 ,6........

  The units digit of powers of 6 repeat as follows: 6, 6, 6, 6, 6, 6.........

  As we see, for an expression or number to equal $ 4^x + 6^x$  it must end in either, ($4+6=10$) a 0, or ($6+6=12$) a 2.

  However powers of 9 repeat as: 9, 1, 9, 1, 9, 1, 9, 1 , 9..........  (none of which is a 0 or 2)

  A contradiction, consequently no such value of x exists.  [/hide]

Of course it is possible,G-UNIT :lol:
Use what shyong said....

hmm,
why on earth is it coming wrong!
$[\frac{2}{3}]^{2x}+[\frac{2}{3}]^x=1$
So,$[\frac{2}{3}]^x=\frac{13}{9}$
So, no possible integral value exists....
Also because of Fermat's Last Theorem $a^x+b^x=c^x$ states that $x$ has no integral values if greater than $2$
Correct?

[quote="eminem"]$4^x+6^x=9^x$

x=?[/quote]

does this involve logs? :?

It can be done without them. Use shyong's hint.

riddler, how did you get to the second line of your latex?

[hide="almost complete solution"]
$4^x + 6^x = 9^x$
$\frac{4^x}{9^x} + \frac{6^x}{9^x} = 1$
$(\frac 49)^x + (\frac 69)^x = 1$
$((\frac 23)^2)^x + (\frac 23)^x = 1$
$(\frac 23)^{2x} + (\frac 23)^x = 1$
$((\frac 23)^x)^2 + (\frac 23)^x - 1 = 0$

substitute: $y = (\frac 23)^x$

$y^2 + y - 1 = 0$
$y = \frac{-1 \pm \sqrt{1+4}}{2}$
$y = \frac{-1 \pm \sqrt5}{2}$
$(\frac 23)^x = \frac{-1 \pm \sqrt5}{2}$
$x = \frac{log(\frac{-1 \pm \sqrt5}{2})}{log(\frac 23)}$
$x = \frac{log(-1 \pm \sqrt5)-log(2)}{log(2) - log(3)}$

using a calculator:

$x = 1.18681439...$ (the other answer is complex)
so we can change it to:
$x = \frac{log(\sqrt5-1)-log(2)}{log(2)-log(3)}$

Does anyone have a nifty way to simplify that?
[/hide]

what?? am I wrong it is obvious divide $9^x$ by both sides and then you get it

hamster 's is more understandable :blush:

The answer is 1.1868 something... I knew something existed, because i tried different values of x and knew it was somewhere between 1 and 1.5.

I guess that's the simplest form of the answer, unless someone wants to take the time  :D

for some reason i thought X was intended to be a positive integer, but i guess logs are fine

Hamster's method's perfect,u did need logs

hey riddler it suddenly struck me too , what about Fermat's last theorem? Isnt this violating it? or maybe i am wrong , i dunn know much about the theorem :D

oooops , sorry guys :blush: , that was stupid of me , i forgot that fermat's theorem is only for integral values! :D


Determine the number of ordered 17-tuples  of reals such that each of the quantities  is either 19 or 98.

Source: Gabriel D. Carroll, "Pre-Olympiad Problems".

I've got an answer ([hide]153[/hide]) but am not completely sure about it. Can anyone verify it?

Hi,

This is what I did, and I got the same answer you did.

[hide]You start with 500, and figure out that a17 must equal either 981 or 902.  The only way to get the 902 answer is if each "a" except for one is 19 more than the one before it (eg. a1=519, 19 more than the 500 starting point), and one is 98 more.  Since this one could be any of the 17 numbers, there are 17 different possibilities.  The way to get the 981 answer is if each a except for two is 19 more than the previous one, and two are 98 more.  Since the two could each be any of the 17, there are 136 possibilities with the two adding 98 (16+15+14...+1).  By adding these two answers together, since both are allowable, you get 153 possible sets.  You know each is different because each adds the 98's at different places.[/hide]


Given is an acute angled triangle $ ABC$. Determine all points $ P$ inside the triangle with
\[1\leq\frac{\angle APB}{\angle ACB},\frac{\angle BPC}{\angle BAC},\frac{\angle CPA}{\angle CBA}\leq2\]

I have solved this problem but I need to draw a diagram of it. can anybody help me :(

I'm not sure if this works or not.
Consider the circumcircles $ \omega_{1}$ of $ \triangle AOB$, $ \omega_{2}$ of $ \triangle BOC$, $ \omega_{3}$ of $ \triangle COA$ where $ O$ stands for the circumcenter of $ \triangle ABC$.
Then the point $ P$ inside the triangle must be on the same region with $ A,B,C$ considering over arc $ BOC,COA,AOB$ which containing $ O$, respectively.
This forces the point $ P$ in a way that $ P\equal{}O$ only.  :)

@above I think I did something similar:

[hide=Solution]
Let \(\Gamma_A, \Gamma_B, \Gamma_C\) be the circumcenters of triangles \(BOC, COA, AOB\), where \(O\) denotes the circumcenter.
Now, note that if \(P\) is inside \(\Gamma_A\), then \(\angle BPC > \angle BOC = 2A\), which is impossible. Similar results hold for \(\Gamma_B\) and \(\Gamma_C\). Let \(P(X)\) denote the statement "\(P\) is outside \(\Gamma_X\)" for \(X=A,B,C\). Then, the following statement must hold:
\[
(P\in\Gamma_A \vee P(A))\wedge
(P\in\Gamma_B \vee P(B))\wedge
(P\in\Gamma_C \vee P(C))
\]

So, in total, there are \(8\) cases to consider.

The first case is that \(P\) lies on all of \(\Gamma_A, \Gamma_B, \Gamma_C\). This case trivially reduces to \(P=O\).

The next three cases consider \(P\) lying on two of \(\Gamma_A, \Gamma_B, \Gamma_C\) and lying outside the third circle. Without loss of generality, it is sufficient to consider \(P\in\Gamma_B,\Gamma_C\) and \(P(A)\). The condition \(P\in\Gamma_B,\Gamma_C\) means either
\(P=O\) or \(P=A\), but both of these lie inside \(\Gamma_A\), which contradicts the third condition. Thus, this case is impossible.

The next three cases consider \(P\) lying outside two of \(\Gamma_A, \Gamma_B, \Gamma_C\) and lying on the third. Again, it suffices to consider the case \(P\in\Gamma_A \wedge P(B)\wedge P(C)\) due to symmetry. But this is impossible as well, because the first condition implies either \(\neg P(B)\) or \(\neg P(C)\).

The last case is \(P(A)\wedge P(B)\wedge P(C)\) which can obviously never take place.

Therefore the only possibility is \(P=O\).
[/hide]

Can someone confirm if this reasoning is correct?

[hide=two line solution by Gonzo17]
We have \(\angle APB \leq 2C\), \(\angle BPC \leq 2A\) and \(\angle CPA \leq 2B\). If any of the inequalities is strict, we can add them and get \(2\pi < 2\pi\), which is impossible. Hence, \(P=O\).
[/hide]


modular arithmetic
function
inequalities
floor function
Recursive Sequences

Let $p$ be an odd prime $p$ such that $2h \neq 1 \; \pmod{p}$ for all $h \in \mathbb{N}$ with $h< p-1$, and let $a$ be an even integer with $a \in] \tfrac{p}{2}, p [$. The sequence $\{a_n\}_{n \ge 0}$ is defined by $a_{0}=a$, $a_{n+1}=p -b_{n}$ \; $(n \ge 0)$, where $b_{n}$ is the greatest odd divisor of $a_n$. Show that the sequence $\{a_n\}_{n \ge 0}$ is periodic and find its minimal (positive) period.

That's a strange problem:

[quote="Peter"]Let $ p$ be an odd prime $ p$ such that $ 2h \neq 1 \; \pmod{p}$ for all $ h \in \mathbb{N}$ with $ h < p \minus{} 1$[/quote]
Take $ h\equal{}\frac{p\plus{}1}{2}$ so that $ 2h \equal{} 1 \; \pmod{p}$. Then we must have $ h\geq p\minus{}1$, implying that $ p\equal{}3$.
[quote="Peter"]and let $ a$ be an even integer with $ a \in] \tfrac{p}{2}, p [$.[/quote]
There is only one such integer: $ a\equal{}2$.
[quote="Peter"]The sequence $ \{a_n\}_{n \ge 0}$ is defined by $ a_{0} \equal{} a$, $ a_{n \plus{} 1} \equal{} p \minus{} b_{n}$ \; $ (n \ge 0)$, where $ b_{n}$ is the greatest odd divisor of $ a_n$. Show that the sequence $ \{a_n\}_{n \ge 0}$ is periodic and find its minimal (positive) period.[/quote]
This sequence is
$ 2,2,2,2,\dots$
and its period is 1.

$ a_n$ is even number from interval $ (p/2,p)$, number of distinct means is $ N\equal{}[\frac{p\plus{}1}{4}]$.
Because exist function f, suth that $ a_{n\plus{}1}\equal{}f(a_n)$, we get exist m and t, suth that $ m\plus{}t\le N, a_{i\plus{}t}\equal{}a_i \ \forall i>m$.
We can found $ a_n\equal{}g(a_{n\plus{}1})\equal{}2^{l_n}(p\minus{}a_{n\plus{}1})$, were $ l_n$ defined unique $ l_n\equal{}[log_2(\frac{p}{p\minus{}a_{n\plus{}1}})]$.
Therefore $ m\equal{}0$ and $ a_{i\plus{}t}\equal{}a_i \ \forall i$.

I'll try to look up what it should really be.

If $ a_{n\plus{}1}\equal{}f(a_n)$ and $ a_n\in S$, were $ S$ is finite set $ \#S\equal{}N$.
Then  in $ a_1,a_2,...,a_{N\plus{}1}$ exist at least 2 $ a_i\equal{}a_j \ (j>i)$, therefore $ a_{k\plus{}t}\equal{}a_k \ \forall k\ge i, \ t\equal{}j\minus{}i$.

[quote="Peter"]True. Maybe they mean $ 2^h\neq 1$?[/quote]
Yes. That makes more sense. And the length of the period in this case equals the number of even integers in the interval $ I \equal{} \left(\frac {p}{2},p\right)$, no matter what value of $ a$ was chosen. In other words, every even integer from the interval $ I$ appears in the period. The length of the period here can be also expressed as $ \frac {p\pm 1}{4}$ (whichever is integer).

Proof.

Inequality $ 2^h\neq 1$ for every $ h < p \minus{} 1$ simply means that the multiplicative order of $ 2$ modulo $ p$ equals $ p \minus{} 1$. Furthermore, in this case we have $ p\equiv 3, 5\pmod{8}$ and the smallest positive integer $ g$ such that $ 2^g\equiv \minus{} 1\pmod{p}$ is $ g \equal{} \frac {p \minus{} 1}{2}$.

The greatest odd divisor $ b(x)$ of an integer $ x$ can be expressed as $ b(x) \equal{} x\cdot 2^{ \minus{} k(x)}$ for some non-negative integer $ k(x)$. Since the number of integers in $ I$ is $ \frac {p \minus{} 1}{2}$, we trivially have the following estimate:
\[ \sum_{x\in I} k(x) \leq \left\lfloor \frac {(p \minus{} 1)/2 \plus{} 1}{2}\right\rfloor \plus{} \left\lfloor \frac {(p \minus{} 1)/2 \plus{} 1}{2^2}\right\rfloor \plus{} \dots \leq \frac {p \minus{} 1}{2}.
\]
Now suppose that $ a_m, a_{m \plus{} 1}, \dots, a_{m \plus{} t \minus{} 1}$ form the period, where $ t$ is the length of the period. Then all numbers $ a_m, a_{m \plus{} 1}, \dots, a_{m \plus{} t \minus{} 1}$ must be different (otherwise the period would be smaller). Furthermore, since $ a_{n \plus{} 1} \equal{} p \minus{} b(a_n) \equiv \minus{} a_n\cdot 2^{ \minus{} k(a_n)}\pmod{p}$ for every $ n$, we also have
\[ a_m \equal{} a_{m \plus{} t} \equiv a_m ( \minus{} 1)^t 2^{ \minus{} k(a_m) \minus{} k(a_{m \plus{} 1}) \minus{} \dots \minus{} k(a_{m \plus{} t \minus{} 1})}\pmod{p},
\]
implying that
\[ 2^{k(a_m) \plus{} k(a_{m \plus{} 1}) \plus{} \dots \plus{} k(a_{m \plus{} t \minus{} 1})} \equiv ( \minus{} 1)^t\pmod{p}
\]
which can take place only if $ k(a_m) \plus{} k(a_{m \plus{} 1}) \plus{} \dots \plus{} k(a_{m \plus{} t \minus{} 1}) \equal{} \frac {p \minus{} 1}{2}$ and the integers $ a_{m}, a_{m \plus{} 1}, \dots, a_{m \plus{} t \minus{} 1}$ represent all even integers from $ I$.
Q.E.D.


Hi all. I want to see all competitions from all countries online.
Here I'm sedning you Bulgarian Math Olympiad - 1960-1978.
Till 10-th of April I'll have all the problems from all the years.
I also would like to see these problems in the Resources section.
I wish you all the best.

--
B. Mirchev

P.S. I would like to thank to Mr. Ercole Suppa for inspiration to translate the problems I'm sending.

Thanks a lot Borislav!  :lol:  i have some questions for you:

1) Does the years 1960 and 1961 have Round 4 ?

2) In problem 2, 1960, round 3 says $ y\equal{}\frac{x^2\plus{}x\plus{}1}{x^\plus{}2x\plus{}1}$ , what is the correct power of x ?

Wow. Another great addition to AOPS's collection. Someone should upload it here.

http://www.mathlinks.ro/resources.php?c=27

to: tipe

1) I don't know - the problems are from a book it is all I've - probably not or they cannot be found.
2) x^2+2x+1 is the divisor of the fraction.


induction
combinatorics unsolved
combinatorics

Let $ G$ be a simple graph with $ 2 \cdot n$ vertices and $ n^{2}+1$ edges. Show that this graph $ G$ contains a $ K_{4}-\text{one edge}$, that is, two triangles with a common edge.

Induction works. Assume we have shown the result to be true for small $n$.

Suppose we are given $2(n+1)$ points with $(n+1)^2+1$ edges between them. By Turan's Theorem, there is a triangle. This means that we can choose two vertices $x,y$ of the triangle with degrees of the same parity. They are, of course, connected, since they form a side of the triangle. If the other $2n$ points have $n^2+1$ sides between them, then we're done by the induction hypothesis. If, on the other hand, they do not, there are at least $2n+2$ edges adjacent to either $x$ or $y$. One of these edges is between the two, and there can't be only $2n+1$ more because the degrees of $x,y$ have the same parity. This means that there are at least $2n+2$ edges adjacent to either $x$ or $y$ but not both. From here we easily conclude that $x,y$ have at least two common neighbours $u,v$, and $x,y,u,v$ is the "$K_4-$ one edge" :).

Strange...I'm quite sure to have seen that one before but I can't find it anymore  :?

Pierre.

Almost same as grobber' s solution:

With induction on $ n$. For $ n=2$ it is obviously.
Now let graph $ G$ have $ n^{2}+2n+2$ edges and $ 2n+2$ vertices. By
Turan' s theorem $ G$ contains $ K_{3}$. If some node from the
$ G'=G\setminus K_{3}$ is connected to two vertices form $ K_{3}$ we are
done. Now let us assume that every vertices from $ G'$ is connected
to at most one vertices from the set $ K_{3}$. So in the set $ G'$ we
have at least $ n^{2}$ edges and exactly $ 2n-1$ vertices.

Here is tricky part which I don't know if it is correct.

Let us add to $ G'$ artificial node which is connected with exactly one node from
$ G'$. Then new graph has $ n^{2}+1$ edges and $ 2n$ vertices. By
induction it has $ H=K_{4}$-edge and artificial node is not on it since it
has degree 1. So $ G'$ contains $ H$ and so does $ G$ and we are done.

I hope this with artificial node works.

Is this right? It seems a little strange.

We proceed using induction, with the base case of $n=2$ trivially true. Suppose that $n>3$ and some edge $v-w$ is not a part of a triangle. Then, $v$ and $w$ have no common neighbors, so $\deg v+\deg w\le 2n$. Removing $v$ and $w$ results in a graph with $2(n-1)$ vertices and at least $n^2+1-(2n-1)=(n-1)^2+1$ triangles, and we are done by the inductive hypothesis. Otherwise, every edge must be part of a triangle. If the problem is false, then each edge must in fact be a part of exactly one triangle. But as we have $n^2+1$ edges, we must then have $\frac{n^2+1}{3}$ triangles, a contradiction as that is not an integer for any $n$. Thus, there exists an edge that is a part of two triangles, as desired.

The case of $n=2$ is obvious, as any graph on $4$ vertices with $5$ edges is just two triangles with a common edge. We also check the case $n=3$ by hand. We now solve the problem by induction. It suffices to show that for any graph $G$ with $2n$ vertices and $n^2+1$ edges that we can delete some two vertices which removes at most $2n-1$ edges.

For any vertex $v$, let $d_v$ denote the degree of vertex $v$. Furthermore, for any two distinct vertices $v$ and $w$, let $e_{vw}$ be $1$ if $vw$ is an edge and $0$ otherwise. If we delete $v$ and $w$, then we remove at most $d_v+d_w-e_{vw}$ edges. Pick any $\{v,w\}\subset V$ uniformly at random. Then,
\[\mathcal{E}:=\mathbb{E}[d_v+d_w-e_{vw}]=\frac{4|E|}{|V|}-\frac{|E|}{\binom{|V|}{2}}.\]
We now compute
\[\mathcal{E}=\frac{2(n^2+1)}{n}-\frac{n^2+1}{n(2n-1)}=2n\left(1+\frac{1}{n^2}\right)\left(1-\frac{1}{2(2n-1)}\right)<2n\]
for $n\ge 4$. Thus, there are some $v$ and $w$ such that we remove at most $2n-1$ edges, as desired.

We will prove the problem by induction on $n$.Base cases can be checked by hand. For $n\ge 4$, consider any graph $G$ with $2n$ vertices and $n^2+1$ edges.By Turan's theorem this graph contains a triangle $ABC$.Let $V(G)$ denote the set of vertices of $G$.

For any vertex $v\in V(G)-\{A,B,C\}$ if $v$ is a neighbour of any 2 of $\{A,B,C\}$ then we will get a $K_{4}- \text{One edge}$.

Otherwise,assume any vertex in $V(G)-\{A,B,C\}$ is incident with atmost 1 of $\{A,B,C\}$.Hence we get ,
$$\deg(A)+\deg(B)+\deg(C)\le (2n-3)+6$$
+6 because each of $AB,BC,CA$ is counted twice.
So by php,
$$\min\{\deg(A)+\deg(B),\deg(B)+\deg(C),\deg(C)+\deg(A)\}\le \frac{2}{3}((2n-3)+6)$$
So WLOG assume $$\deg(A)+\deg(B)\le \frac{4}{3}n+2\le 2n-1$$
So $G'=G-\{A,B\}$ has $2n-2$ vetex and atleast $n^2+1-(2n-1)=(n-1)^2+1$.By induction hypothesis, $G'$ has a $K_{4}- \text{One edge}$.
We are done $\blacksquare$

Start off with $2n$ vertices.Consider vertices $A,B$ on an edge of a triangle(exists due to Turan's theorem) such that $d_{A}+d_{B}$ is even.If $d_{A}+d_{B} \ge 2n+2$,then there exists two common neighbours and we are done.
If $d_{A}+d_{B} \le 2n$,then on deleting vertices $A,B$ atmost $2n-1$ edges get deleted and we are done by induction.

We proceed with induction on $n$.

Assume the statement for $k$. Let graph $G$ have $(k+1)^2+1$ edges and $2k+2$ vertices. Choose two vertices $v$ and $u$ with $vu\in E(G)$. Let $G'$ be the graph resulting from removing $v$ and $u$ from $G$. In order for there to be no two triangles sharing a common edge, the number of edges incident to $v$ or $u$ (or both) is at most $2k+2$. There are two cases to consider:

[b]Case 1.[/b] There are at most $2k+1$ such edges, $|E(G')|\geq (k+1)^2+1-(2k+1)=k^2+1$, so $G'\subset G$ contains two triangles sharing a common edge.

[b]Case 2.[/b] There are exactly $2k+2$ such edges, $|E(G')|=(k+1)^2+1-(2k+2)=k^2$. Note that in order for there to be no two triangles sharing a common edge, there is exactly one vertex in $G'$ incident to both $v$ and $u$. Call this vertex $x$. Define
\begin{align*}
A&=\{a\in V(G')\mid av\in E(G),au\notin E(G)\}\\
B&=\{b\in V(G')\mid av\notin E(G),au\in E(G)\}
\end{align*}so $|A|+|B|=2k-1$. There are at most $|A||B|\leq k(k-1)$ edges between $A$ and $B$. No edge in $G'$ can be incident to $x$ (otherwise $G'$ contains a triangle sharing a common edge) so the among the remaining $\geq k$ edges, there must exist two adjacent edges with all endpoints in either $A$ or $B$. These edges, along with $v$ or $u$, determine two triangles sharing a common edge.

Thus the statement holds for $k+1$. $\square$

Perform induction, with base cases straightforward. The idea is to actually work backwards. Assume the inductive hypothesis and for sake of contradiction that the result is not true.

[b][color=#f00]Claim.[/b][/color] There exist a pair of vertices $u, v \in G$ such that deleting $u$ and $v$ removes at most $2n-1$ edges from the graph.

[i]Proof.[/i] The idea is to consider the sum $\deg u + \deg v - 1_{uv}$, where $1_{uv} = 1$ if $uv$ is an edge and $0$ otherwise. Across all $u, v$ this evaluates to $$\sum_{u, v \in G} (\deg u + \deg v) - (n^2+1) = 2(2n-1)(n^2+1),$$ so over all ${2n \choose 2}$ pairs of vertices, the average value of the sum is less than $2n$, hence the result.

Thus just remove these vertices and induct.


What's the coefficient of the $ m^3n^5$ term in the expansion of $ (m\plus{}n)^8$?

The Binomial Theorem states that the coefficient of the $ m^kn^{a\minus{}k}$ term in $ (m\plus{}n)^a$ is $ \binom{a}{k}$, so in our case that is simply $ \binom{8}{3}$, or $ \boxed{56}$.

This works because consider the product $ (m\plus{}n)(m\plus{}n)(m\plus{}n)(m\plus{}n)(m\plus{}n)(m\plus{}n)(m\plus{}n)(m\plus{}n)$.  From each binomial, we choose either an $ m$ or an $ n$.  Since we want $ 3$ $ m$'s, there are $ \binom{8}{3}$ ways to choose the binomials which we take the $ m$ from.


geometry rectangle algebra polynomial function vector

Let $t \in \mathbb{R}^+$. Prove that the unit square can be tiled with fnitely many rectangles similar to the $1 \times t$ rectangle if and only if $t$ is algebraic (i.e. the root of an irreducible polynomial $f \in \mathbb{Z}[x]$) and all of its algebraic conjugates have possitive real part. (The algebraic conjugates of $t$ are the roots of the minimal polynomial $f$ of $t$.)

So, for example, such a tiling cannot be done with $t=\sqrt{2}$ nor with $t=1+\sqrt{2}$, but it is possible produce a tiling with $t=2+\sqrt{2}$.

Sorry,Vess, I dont want to speak in anyones name, but I think this is too hard for an olympiad student :D

I have a question: why doesn't $1+\sqrt 2$ satisfy the conditions? Isn't its real part $1>0$? What exactly do you mean by "all its algebraic conjugates"?

P.S. I agree with DusT :D

Too hard? I'm not sure. After reading lhvietbao's magnificent solution (ref. http://www.mathlinks.ro/viewtopic.php?t=6043) to a problem of a similar nature, I decided to post this one.

To grobber: the algebraic conjugates of an algebraic number are the roots of its minimal polynomial. For example, the minimal polynomial of the algebraic number $\sqrt{2}$ is $x^2-2$, thus the algebraic conjugates of $\sqrt{2}$ are $\pm \sqrt{2}$, not both of which are positive (we have $-\sqrt{2}<0$.) Similarly, the number $1+\sqrt{2}$ has minimal polynomial $x^2-2x-1$ and the conjugates of $1 + \sqrt{2}$ are $1 + \sqrt{2}$ itself and $1 - \sqrt{2} < 0$. However, the conjugates of $2 + \sqrt{2}$ are $2  \pm \sqrt{2}$, which are both positive. Note that not all conjugates of  a positive $t \in \mathbb{R}^+$ are necessarily real.

I'm terribly sorry: when I read "real part" for some reason I was thinking about the rational part, i.e. the stuff outside the square root.I don't know if that notion really exists, but that's what I thought of :blush: :).

Well, lhvietbao is a IMO 2003 perfect scorer.
And there are not so many other students like him :).

vess, do you have a understandable solution??

It is much harder then (very hard too) problem solved by lhvietbao. At least, if the original solution is the simplest.

But one thing is easy to see with the approach used by lhvietbao: if a tiling exists, then $t$ must be algebraic.

By the way, I know more than one solution of the problem.

Well the idea is in fact the one used by Bao, but it is somehow an idea worked in some books and articles.

In our training program, and In a talk last year, Federico Ardila told us about this problem and also about the technique of redifining the area with other function.

I managed to prove then that $t$ is algebriac, and that its real conjugates are positive, but i am not done with the last part.

The idea is the following, if $t$ is not algebraic, then $Q(t)$ is isomorphic to $Q(-t)$, so we can define an aditive function such that f(t) is negative, and it will be in fact an automorphism from $R$ to $R$. We rthus have $f(1)=1$, so we get the area of our square is $1$ on one hand, and $\sum f(a)^2f(t)<0$ which gives a contradiction.

The secound part is similar: we have $Q(t)$ is isomorphic to $Q(t')$ where $t'$ is a real negative associate of $t$ over $Q$, then we can complete this isomorphism to an automorphism of $R$, which maps $t$ to $t'$. Henece we get the area of the square is again $1$, but the sum of areas is $\sum f(a)^2f(t)<0$, contradiction.

I am trying to use the idea that $Q(t)$ is isomorphic to $Q(t')$ for every associate of $t$, to construct an additive function, for example I think that viewing $R$ as a $Q(t)$ vector space may help, and then doing the switch from $Q(t)$ to $Q(t')$.

A few more words:

There is an article in Clay Math web site, where there is info about this problem. (by Federico)

This problem isnt anyway an olympiad problem, How on earth would a school student know about constructions of aditive functions?

However the technique is evry useful In olympiads, we can define the area of the rectangle whose vertices are $(a,b),(a,d)(c,b)(c,d)$ by: $(f(c)-f(a))(g(d)-g(b))$

And in this way, we get that if a ectangle is tiled, then the sum of the areas of the rectangles in the tiling is the area of the tiled one.

If we have $f$ and $g$ aditive, then the area is invariant under translations, and only depend on dimensions.

To illustrate an easier use of this technique, consider for example the well known property that if a rectangle is tiled with rectangles each having an integer side, then it has an integer side. To solve it, put one of the initial rectangle vertex at the origin and define $f(x)=g(x)=\{x\}$ (the fractionary part) It is easy to see that the area of every samll rectangle is cero, so the initial rectangle also has area cero, and therefore, one of its sides is an integer.


please give me feedback: on it's difficulty level (1-5), and did you enjoy it!?
thanks!
o yeah it's in latex! :D

Challenge Problem1:Find $ h^{\minus{}1}(\minus{}1)$ without finding $ h^{\minus{}1}(x)$ if $ h(x)\equal{}\frac {9^{2^{h(x)}}\minus{}2x} {5(5^{\minus{}1})\minus{}x}$ .Then, find $ h^{\minus{}1}(x)$ ,what can you conclude about h?

plz give me some feed back on difficulty level(1-5) and enjoyment, thanks!

[quote="esmalls"]Challenge Problem1:Find $ h^{ \minus{} 1}( \minus{} 1)$ without finding $ h^{ \minus{} 1}(x)$ if $ h(x) \equal{} \frac {9^{2^{h(x)}} \minus{} 2x} {5(5^{ \minus{} 1}) \minus{} x}$ .Then, find $ h^{ \minus{} 1}(x)$ ,what can you conclude about h?

plz give me some feed back on difficulty level(1-5) and enjoyment, thanks![/quote]

We suppose $ h(x)$ is some real function from $ \mathbb R\to\mathbb R$

In such case, no such function exists since, for example, $ h(0)$ cant be defined (equation $ y\equal{}\frac {9^{2^{y}}\minus{}0} {1\minus{}0}$ has no solution).

So, no $ h(x)$ $ \implies$ no $ h^{\minus{}1}(x)$ $ \implies$ no solution to this problem.

Except if $ h(x)$ is not defined over $ \mathbb R$. In such case, please give the precision.

[quote="pco"](equation $ y \equal{} \frac {9^{2^{y}} \minus{} 0} {1 \minus{} 0}$ has no solution).

So, no $ h(x)$ $ \implies$ no $ h^{ \minus{} 1}(x)$ $ \implies$ no solution to this problem.

Except if $ h(x)$ is not defined over $ \mathbb R$. In such case, please give the precision.[/quote]

Yes, as I'd stated in the previous message. Perhaps you were miss lead that if you can not solve for $ h(x)$ the same must hold true for its' inverse.I believe you can solve for $ h(x)$ only in terms of its' inverse for example find $ h(0)$ if $ h^{\minus{}1}(x)\equal{}\frac {\minus{}x\plus{}9^{2^{x}}} {\minus{}x\plus{}2}$ try to apply the same approach to this as I'd showed in the original problem.

And your solving for $ h^{\minus{}1}(\minus{}1)$ not $ h(x)$ ! I understand your point but indeed the function remains invertible and solvable.

No, he's saying plain out that this "function" is not a function over the reals. He's not saying anything about solvability. $ h(0)$ is undefined if its domain is the reals. You need to specify the function's domain.

In any case, what you'd probably intended...

[hide]Suppose $ h(x) = -1$. Then we have $ -1 = \frac{9^{2^{-1}} - 2x}{1 - x}$, so $ x - 1 = 3 - 2x = 5$, so $ x = 2$.

If $ h(x) = y$, we have $ y = \frac{9^{2^y} - 2x}{1 - x}$, so $ y - xy = 9^{2^y} - 2x$, so ${ x(2 - y) = 9^{2^y}} - y$, so $ h^{-1}(y) = x = \frac{9^{2^y} - y}{2 - y}$. [/hide]

[quote="esmalls"]
And your solving for $ h^{ \minus{} 1}( \minus{} 1)$ not $ h(x)$ ! I understand your point but indeed the function remains invertible and solvable.[/quote]                           Now, please try to solve the problem.

[quote="Zhero"]Suppose $ h(x) = - 1$. Then we have $ - 1 = \frac {9^{2^{ - 1}} - 2x}{1 - x}$, so $ x - 1 = 3 - 2x = 5$, so $ x = 2$.

If $ h(x) = y$, we have $ y = \frac {9^{2^y} - 2x}{1 - x}$, so $ y - xy = 9^{2^y} - 2x$, so ${ x(2 - y) = 9^{2^y}} - y$, so $ h^{ - 1}(y) = x = \frac {9^{2^y} - y}{2 - y}$. [/quote] Here's where you make the mistake...Answer this.$ h$ and $ h^{-1}(x)$ are only inverse functions of each other if and only if......

$ h \circ h^{\minus{}1} \equal{} h^{\minus{}1} \circ h \equal{} x$.

All I've shown was that if $ h(x) \equal{} y$ for any $ y$ in the image of $ h$, then $ x \equal{} \frac{9^{2^y}  \minus{} y}{2 \minus{} y}$. What more do you need from me? Do you want me to plug in $ h^{\minus{}1}$ into the original equation to check? Because that's not necessary.

Do you want me to show that $ h^{\minus{}1}$ is a function? How can I do that if $ h$ itself might not be a function?

[quote]I understand your point[/quote]
If you really understood the point, then you would have realized that we cannot solve for either $ h$ or $ h^{\minus{}1}$, [i]simply because $ h$ itself is not even a function[/i]. You need to specify a domain.

If I am indeed still wrong, I apologize for my many false assertions of correctness. In this case, please tell me the answer, because I do not know how to proceed further.


It says on my USAMTS profile I did not submit the entry form, however there is a checkmark next to the solns. Im pretty sure i sent my form in with the solns so what happened?? will mine still be graded?  :(  :(  :(

same for me

I uploaded my solutions on the website, but sent my entry form by mail.

All of the papers that we have received have been graded, entry form or no entry form.  We have a few late entry forms that have not yet been entered.


Given two positive integers a and n, to show that

(a-n)*a!/(a-n)!=(n+1)  \sum_{t=n}^{a-1} t!/(t-n)!

sorry, i don't know how to write summation signs (summation from t=n to a-1).

Rewritten,
Prove
$a,n\in\mathbb{N} \implies \frac{a!(a-n)}{(a-n)!}=(n+1)\sum_{t=n}^{a-1}{\frac{t!}{(t-n)!}}$
So this also requies the condition $a>n$, right?
I think you just forgot the $ signs  :)

[quote="cats..."]LT, try this, write the RHS as $(n+1)(n!)(\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\ldots+\binom{a-1}{a-n-1})$.  That should be $(n+1)!\binom{a}{n+1}$.[/quote]

Thanks, Steve.  By the way, do you or anyone know where I can find this identity?

\displaystyle \binom{n}{0}+\binom{n+1}{1}+...+\binom{a-1}{a-n-1}

you can see this in the Pascal triangle, all them form a diagonal line..but  the binomial of n in O = to n+1 in O and it sumed to the second by Pascal is n+2 in 1...and like that all the way...so finaly it is  \displaystyle \binom{a}{n+1}[/quote]

[quote="shyong"]How is it

$\binom{n}{0}+\binom{n+1}{1}+...+\binom{a-1}{a-n-1}$=$\binom{a}{n+1}$  :?[/quote]
I think it become easier if you write it in this way:
$\binom{n}{n}+\binom{n+1}{n}+...+\binom{a-1}{n}$=$\binom{a}{n+1}$ ;)


topology
function
geometry
rectangle
advanced fields
advanced fields solved

1. Construct an explicit deformation retraction of the torus with one point deleted onto a graph consisting of two circles intersecting in a point, namely, longitude and meridian circles of the torus. (Hatcher Exercise 0.1)

2. Construct an explicit deformation retraction of $ \mathbf{R}^n \minus{} \{0\}$ onto $ S^{n\minus{}1}$. (Hatcher Exercise 0.2)

Attempted Solutions:

1. I can kind of visualize the deformation retraction: draw lines going radially outward from the point removed; all the points move along those lines to either the meridian or the longitude. But I think some of these lines will intersect and I have no idea how to write that down as a formula (I assume that is what explicitly means).

2. I assume the retraction is x/norm(x). As in 1, I think we want to move all the points radially toward S^{n-1}. I don't know how to pace that points so that this will be continuous. Won't the points have to move "infinitely fast" near t = 0?

A deformation retract is a homotopy- which means you have to say how it moves. $ \frac{x}{\|x\|}$ is a retract, but not a deformation retract.

A way to create a deformation retract: let $ f(t,y)\equal{}y^t$ (or a similar homotopy between $ (0,\infty)$ and $ \{1\}$). Now, let the homotopy $ g(t,x)\equal{}f(t,|x|)\frac{x}{|x|}$.

[quote="jmerry"]A deformation retract is a homotopy- which means you have to say how it moves. $ \frac {x}{\|x\|}$ is a retract, but not a deformation retract.

A way to create a deformation retract: let $ f(t,y) \equal{} y^t$ (or a similar homotopy between $ (0,\infty)$ and $ \{1\}$). Now, let the homotopy $ g(t,x) \equal{} f(t,|x|)\frac {x}{|x|}$.[/quote]

Yes thanks that solves 2. Intuitively, it seems weird that there could be a homotopy between the identity on $ (0,\infty)$ and the constant map x to 1 since the range of the homotopy is $ (0,\infty)$ at all times greater than 0 but at 0 it suddenly becomes finite. Anyway, algebraic operations in $ \mathbb{R}^n$ are continuous and composition of continuous functions is continuous, so it must be true.

What about 1?

Interpret the torus as a rectangle with opposite sides identified. Cut out a point somewhere in the middle, and then shrink everything away from that point to the rim.

There's no need to quote the entire post immediately preceding yours, ever.

I see thanks. Can I move this thread to the "Solved Problems" thread or does a moderator need to do that?

$ \text{Unsolved Problems} \to \frame{\text{some miracle happens}} \to \text{Solved Problems}$

PS: No you can't  :wink:


Can someone explain how the dimensions of the nullspace and the column space of a matrix relate to the number of lead and free variables in the reduced row echelon form of the matrix.

Thanks

rank = dimension of row space = dimension of column space = number of pivot elements in the row echelon form that is row equivalent to the given matrix.

(I assume that's what you mean by "lead variables.")

The rank can also be thought of as the dimension of the range of the function given by multiplication by this matrix.

nullity = dimension of null space = number of free variables = number of columns without pivot elements in the row-equivalent row echelon form.

Suppose $A$ is an $m\times n$ matrix. ($m$ rows, $n$ columns.) Then $A$ defines a linear mapping from $\mathbb{R}^n$ into $\mathbb{R}^n.$

The null space is a subspace of $\mathbb{R}^n,$ which is the domain.

The range (column space) is a subpace of $\mathbb{R}^m,$ which is the codomain.

The basic dimension-counting theorem (proved by Guassian elimination, passing to row echelon form) is this:

(rank) + (nullity) = $n$ (the dimension of the domain.)

[Note: the use of $\mathbb{R}$ is for purposes of illustration; the same principles would hold for matrices over any field.]


analytic geometry
combinatorics unsolved
combinatorics

It is given a "rectangular lattice" with integer coordinates from $1$ to $a$, respectively $1$ to $b$, and $a$, $b$ $\geq 1$ positive integers.
Determine the total number of distinct lines generated by any $2$ such points as $F(a,b)$.
For example:
$F(a,1)=1$.
$F(a,2)=2+a^2$, $a$ $\geq 2$
$F(a,3)=3+2a^2-a$ mod $2$.
$F(a,b) \leq C_{ab}^2$.

I have a huge explicit formula for this... if someone is interested, maybe I'll post it...

Yes, I'm interested in seeing it since I couldn't find anything so far..  :(
Thanks in advance! :)


:blush: we make all the permutations to the number 1234567
whats theire sume

What the [insert place where Lucy lifes here] does the title mean¿


P3

a and b are positive integers. Prove that if (a+1)/b+b/a is an integer, then it's 3.

[hide] how about we rewrite it as (a/b) + (1/b) + (b/a). then in order for this to be an integer, b|1. this only happens if b=1, so then the expression becomes a + 1 + (1/a). (1/a) will only be an integer if a|1, so a also must be 1. hte expression then becomes 1+1+1, or 3.   :cry: [/hide]

No, that's just false.

From b|(a + 1) it does NOT follow that b|1 !

Well ... If (a, b) satisfies (a + 1)/b + b/a = k,
then so do the following pairs : (a, a(a + 1)/b) ; (b/a, b).

would this be right (please!?) ?

[hide] k. if (a+1)/b is an integer, then we have a (mod b) + 1 (mod b) = 0 (mod b), so a :equiv: -1 (mod b). so maybe for some integer k a=kb-1. then (a+1)/b becomes kb/b = k. and b/a is b/(kb-1). so the expr is k + b/(kb-1). now k is integer, k will always be integer, so we want kb-1|b. first of all, kb-1 :le: b. solve for b.....

b(k-1) :le: 1, so b :le: 1/(k-1). now for all positive integral k, 1/(k-1) is less than 1, which would make b not a positive integer. hence 1 = k-1 (hopefully!), and k=2. so from up above, b :le: 1. but b is positive integer, so b=1 (!?) then the expr becomes 2 + 1/1, or 3? [/hide]

is this right so far? i know where to go, just want to know if i should go on or stop.

Hmm ... that's wrong again, sorry. The fact that (a + 1)/b + b/a is an integer DOES NOT IMPLY THAT both (a + 1)/b AND b/a are integers. Do you see it ?

this may be totally wrong, but:

(a+1)/b +b/a = (a^2+a+b^2)/ab

Now if a=b, we have (2a^2+a)/a^2, and this is integral only if a = 1. So the previous expression is 2+1=3.

If a != b, we have (a^2 + b^2)/ab + 1/b, so 1/b will only be integral in the case b=1, and then once again (a^2 + 1)/a will be integral only if a=1, contradiction.

Thus a=b=1 and the expression is 3.

I was working along the same lines as Arne . . . don't have anything else, though.  Does a size comparison help?

Well grobber, I don't see where to go from there ...

[hide]

 (a+1)/b+b/a

=(a/b+b/a)+1/b>2

 (a+1)/b+b/a

=(a^2+a+b^2)/(ab)

a^2+a+b^2-kab=0, k>2

b^2-kab+a^2+a=0

consider the discrimant must be a perfect square

(ka)^2-4a^2-a=t^2

it should be obvious that a is a perfect square

let a=u^2

(ku)^2-4u^2-1=s^2

so we end up with some Pell's Equation

(k^2-4)u^2-s^2=1

:S

anyone know when a simple pell's equation have solutions?

[/hide]

I have not successfully solved a problem in weeks... this is not good esp for the TLE!!  good as dead

wow, they really do have no solutions (i verified to a couple of thousand using a program)??  what, must k^2-4 be prime or something?

I am currently trying to solve this one as hard as I can: I forgot my solution (I don't even know if it was OK) and the book where I can find the solution is somewhere else (I study in a different city from the one I live in and the book is in the city where I study, while I'm at home right now). However, I'll go there this evening and I'll take a look at the solution.

Wow! I've just figured out my old proof was wrong but I found a new one. I'm posting a hint right now and the soln a bit later (in spoiler, of course  ):

[hide]Don't try to prove that a=b=1. Those aren't the only solutions. In fact, there are infinitely many pairs; here's one example: (a, b)=(4, 10).[/hide]

Maybe this doesn't help much, but maybe it does...

Here's the soln:

[hide]

We get b|a^2+a=a(a+1) and a|b^2. Then a=p1^b1*..*pk^bk and b=p1^a1*..*pk^ak*q1^k1*..*qj^kj, where pi are the prime divisors of a, qi are SOME prime divisors of a+1, bi, ai>=1. Besides all that we get 2bi>=2ai>=bi.

b|(a^2+a+b^2)/a = a+1+b^2/a = a+1+p1^(2a1-b1)*..*pk^(2ak-bk)*q1^2k1*..*qj^2kj, so pi^ai|a+1+p1^(2a1-b1)*..*pk^(2ak-bk)*q1^2k1*..*qj^2kj for all i from 1 to k. If 2ai-bi > 0 for some i it means that pi|a+1, which is false because pi are the primes which divide a and (a,a+1) = 1. So we got the (very important) result: a = p1^2a1*..*pk^2ak; b = p1^a1*..*pk^ak*q1^k1*..*qj^kj. We replace this in ab|a^2+b^2+a to get p1^3a1*..*pk^3ak*q1^k1*..*qj^kj | p1^4a1*..*pk^4ak + p1^2a1*..*pk^2ak + p1^2a1*..*pk^2ak*q1^2k1*..*qj^2kj iff A = p1^a1*..*pk^ak|qi^2k1*..*qj^2kj+1 and B = q1^k1*..*qj^kj|p1^2a1*..*pk^2ak+1. The last 2 conditions are, in fact these:

A|B^2+1 and B|A^2+1. a=A^2 and B=AB, so (a^2+b^2+a)/ab = (A^4+A^2*B^2+A^2)/A^3*B = (A^2+B^2+1)/AB.

This is a well-known problem:

If AB|A^2+B^2+1 then (A^2+B^2+1)/AB must be 3. You can see that this ends the proof. I'll post the solution to this one later, but maybe I don't need to do that:    [url]

http://www.artofproblemsolving.com/phpBB/viewtopic.php?t=280[/url]

[/hide]

Ok here goes :

[hide]

Let gcd(a,b)=d , a=d*a_1, b=d*b_1 =>

d*a_1*b_1 | d*(a_1)^2+a_1 + d*(b_1)^2

This means that a_1 = d*x, if d>1 so :

d^2 *x * b_1 | d^3 * x^2 + dx + d* (b_1)^2 =>

d * x * b_1 | d^2 x^2 + x + (b_1)^2

This means that x=1 (otherwise x and b_1 share a common prime factor). Thus a=d and a|b therefore (a+1)/b is an integer, so (d+1)/(d * b_1) is an integer so d divides d+1 so d=1, contradiction.

Therefore we assume that d=1 from the start.

(a^2 + a + b^2)/(ab) is an integer => if p is a prime factor of a , p divides b^2 , a contradiction since gcd(a,b)=1. So a must be 1. => 2/b + b is an integer. Since b is a positive integer, b=1 or b=2. Regardless of the case , we get b/a + (a+1)/b = 3.

[/hide]

iandrei wrote:Ok here goes :
[hide]
Let gcd(a,b)=d , a=d*a_1, b=d*b_1 =>
d*a_1*b_1 | d*(a_1)^2+a_1 + d*(b_1)^2
This means that a_1 = d*x, if d>1 so :
d^2 *x * b_1 | d^3 * x^2 + dx + d* (b_1)^2 =>
d * x * b_1 | d^2 x^2 + x + (b_1)^2

This means that x=1 (otherwise x and b_1 share a common prime factor). Thus a=d and a|b therefore (a+1)/b is an integer, so (d+1)/(d * b_1) is an integer so d divides d+1 so d=1, contradiction.
[/hide]

How do you get a=d? Do you not mean a_1 = d? Since a_1 = dx.

Note that there are solutions where they do share a common factor, eg a=4, b=10 as stated before.

Yes sorry, my mistake. In that case, a=d^2 and b=d*b_1 and we get

d * b_1 | d^2 + b_1^2 + 1

Let k = (d^2 + b_1^2 + 1)/(d * b_1)

d^2 - k d^2 * b_1 + b_1^2 = - k

Then we continue in grobber's manner.

Sorry for the mistake


topology
function
advanced fields
advanced fields unsolved

We will call $ U$ be neighborhood of a set $ A\subset X$(An arbitrary topological space) if exist an open set $ T$ of $ X$ such that $ A\subset T\subset U$. Now give $ \Omega\subset X$ ia an open set and $ S\subset\Omega$ is compact set, prove that always exist an neighborhood $ U$ of $ S$ such that $ U\subset\Omega$.

I think when I wrote $ U\subset\Omega$ it mean $ U$ is really subset of $ \Omega$, $ U\not\equal{}\Omega$, sorry I don't remember some Tex formulas.

This contradicts common usage of the word "neighbourhood". Moreover, take $ X\equal{}\{x,y\}$ with the discrete topology, then $ S\equal{}\{x\}$ is a compact subset having no neighbourhoods at all.

To distinguish between proper inclusion and ordinary inclusion, use "\subset" and "\subseteq", or "\subsetneqq" and "\subset". owk

Ok thank you , I must rewrite my problem
We will call $ U$ be neighborhood of a set $ A\subset X$(An arbitrary topological space) if exist an open set $ T$ of $ X$ such that $ A\subset T\subseteq U$. Now give $ \Omega\subset X$ is an open set and $ S\subsetneqq\Omega$ is compact set, prove that always exist an neighborhood $ U$ of $ S$ such that $ U\subsetneqq\Omega$.
I think in your example and my fix problem then $ X \equal{}\{x,y\}$ is neighborhood of $ S \equal{}\{x\}$

New example: $ X\equal{}\{x,y\}$ with the trivial topology, whose open sets are $ \varnothing$ and $ X$. Then $ S\equal{}\{x\}$ is compact, but its only neighbourhood is $ X$. owk

ok thank, I think you are right but I think I don't false too maybe I don't know to use some concept, acutally I want to extend an problem from $ R^{n}$ as following
Give $ A\subset R^{n}$ then we call $ A_{\epsilon}\equal{}\{x|d(x,A)<\epsilon\}$ is neighbourhood of $ A$, forall compact set $ S\subset A\subset R^{n}$(an open set) the always exist a neighbourhood $ S_{\epsilon}\subsetneqq A$.
This problem is true and I think the general for an arbitrary topology space is true with my general definition for neighbourhood , but maybe I was false, sorry but Could you show me solution for the above problem?, please thank you!

The easiest proof of the statement for $ \mathbb R^{n}$ uses the fact that $ S$ has positive distance to the complement of $ A$ ($ d(s,\mathbb R^{n}\setminus A)$ is a continuous function of $ s\in S$, so has a minimum). Any $ \epsilon$ smaller than this distance will do. owk

But why do we need $ S$ is compact ? :maybe:

No $ \epsilon$-neighbourhood of $ y\equal{}\exp x$ is contained in $ \{(x,y)\in\mathbb R^{2}\mid y>0\}$. owk

oh I understand but I think in your proof we need $ A$ is bouned set for $ S$ is bouned therefore continuous function of $ s\in S$ has a minimum or infimum?
Thank you for all your helping  :) .


function
superior algebra
superior algebra unsolved

[quote="VolleyCandi04"]Show that the set A=(f_m,b : [b]R ->R[/b] | m /= 0 and f_m,b(x)=mx+b} of affine functions from R to R forms a group under composition of functions.

I have no idea what this question is even asking!![/quote]
You have to show that the set $ A\equal{}\{f_{m,b}(x)\equal{}mx\plus{}b, m\not\equal{}0\}$ together with the composition operator satisfy the four axioms of a group. That is
1.	Closure.	For all $ f, g$ in $ A$, the result of the composition $ f\circ g$ is also in $ A$
2.	Associativity.	For all $ f, g$ and $ h$ in $ A$, the equation $ (f \circ g) \circ h \equal{} f \circ (g \circ h)$ holds.
3.	Identity element.	There exists an element $ e$ in $ A$, such that for all elements $ f$ in $ A$, the equation $ e \circ f \equal{} f \circ e \equal{} f$ holds. In this case, $ e$ is the identity mapping $ Id$
4.	Inverse element.	For each $ f$ in $ A$, there exists an element $ g$ in $ A$ such that $ f \circ g \equal{} g \circ f \equal{} e$, where $ e$ is the identity element. In this case, the inverse element of $ f$ is the inverse composition $ f^{\minus{}1}$.


Ah yes, I remember this problem vaguely from mathcounts, I think. Then again, it could have been a problem similar, but not the same.

[hide]We square both sides to get  Subtract 2 from both sides to get

Now, we multiply  by  and get  Since , the answer is 18[/hide]

Of course, the answer would be extremely easy to get if you have a graphing calculator.

(a+b)^3 = a^3+b^3+3a*b(a+b)  substitute  a=x anb b=1/x  and get 27 = x^3+1/x^3 + 3*3  and what you want is18.

 Edit: (Typo removed in the above).

[quote="Gyan"](a+b)^3 = a^3+b^3+3a*b(a+b)  substitute  a=x anb b=1/x  and get 27 = x^3+1/x^3 + 3 and what you want is 24.[/quote]

you forgot to multiply (x + 1/x) by 3. so it would be: 27 = x^3+1/x^3 + 3 [b]* 3[/b], and the answer is 18.

Y - yes your are right... edited the original to remove confusion..

[quote="SnowStorm"]If x + 1/x = 3 then what is x3 + 1/x3 ?[/quote]

In fact, if x+1/x is an integer, then x^n + 1/x^n is necessarily an integer also.


Figures 0, 1, 2, and 3 consist of 1, 5, 13, and 25 nonoverlapping unit squares,
respectively. If the pattern were continued, how many nonoverlapping unit
squares would there be in figure 100?

(A) 10401 (B) 19801 (C) 20201 (D) 39801 (E) 40801

I need to learn how to find equations for patterns similar to this problems.

[quote="t0rajir0u"]It would help if we knew what the figures were.[/quote]
Are you sure it matters how they look like?

[hide]By taking a look at the sequence, we see that the second term is the first one plus four, the third term is the second one plus 8, the fourth one is the third one plus 12.

We are told that the third figure has 25 non overlapping squares which is the fourth term in the sequence.

We can see that $ F_3\equal{}25\equal{}4\sum_{k\equal{}1}^3 k\plus{}1$

WE generalize this with the formula $ F_n\equal{}4\sum_{k\equal{}1}^n k\plus{}1$. We can test the formula with the first couple of terms to make sure it is correct.

For figure #100, we evaluate

$ F_{100}\equal{}4\sum_{k\equal{}1}^{100} k \plus{}1\equal{}4 \frac{(100)(100\plus{}1)}{2} \plus{}1$
$ F_{100}\equal{}4(5050)\plus{}1\equal{}20200\plus{}1\equal{}20201$

So our answer should be $ C$[/hide]

[quote="t0rajir0u"]I mean that the figures might provide justification for the first differences being linear.  Clearly this is the implied assumption, but it is not necessarily true.[/quote]

The problem does state that the pattern continues, and the pattern is linear for the terms given, so I think the assumption should be correct


When the point $ (\minus{}3,\minus{}4)$ is reflected about the line $ y\equal{}\minus{}x$, what is the $ y$-coordinate of its image?

[quote="GameBot"]When the point $ ( \minus{} 3, \minus{} 4)$ is reflected about the line $ y \equal{} \minus{} x$, what is the $ y$-coordinate of its image?[/quote]

 [geogebra]82043b6a9b05fc6f0b09cc60b4c8bae6bda50496[/geogebra]

Congruent, so $ (3,4)$.


inequalities function calculus vector LaTeX

$y=\frac{1}{1+x}\ (x>0) $ is concave down and decreasing function,so by Jensen's theorem we have

$\frac{1}{3}\left(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\right)\geqq\frac{1}{1+\frac{a+b+c}{3}}\geqq\frac{1}{2}$.  we are done.

There's an even easier solution (read; no calculus)

By Cauchy or AM-HM, we have that $\sum \frac {1}{1+x}\geq \frac {9}{3+\sum x}$

But since $x+y+z=3$, we get the result.

Blahblahblah have use the next identity:\[
Having {\rm{ a,b}} \in {\rm{R}}_{\rm{ + }} ,rezults:\frac{{x^2 }}{a} + \frac{{y^2 }}{b} \ge \frac{{(x + y)^2 }}{{a + b}},\forall x,y \in R
\]After this we have:\[
\frac{{x_1^2 }}{{a_1 }} + \frac{{x_2^2 }}{{a_2 }} + \frac{{x_3^2 }}{{a_3 }} +  \cdot  \cdot  \cdot  + \frac{{x_n^2 }}{{a_n }} \ge \frac{{(x_1  + x_2  + ... + x_n )^2 }}{{a_1  + a_2  + ... + a_n }},\forall x_1 ,x_2 , \ldots ,x_n  \in R
\]This Lema was wrriten by prof.univ.TITU ANDREESCU(S.U.A).(Romania)
we can easy found the rezulte ofi Blahblahblah. :)

Actually, I just used the result that:

$\frac {1}{n}\sum \frac {1}{a}\geq n\frac {1}{\sum a}$, which is usually referred to as the arithmetic-harmonic means inequality.

For blahblahblah:you can use this Lema of Titu Andreescu to,and it can be used in many hard problems

Yeah, I know, it's an immediate consequence of Cauchy Schwarz applied to the vectors:

$\langle a_i\rangle = \langle \frac {x_i^2}{y_i} \rangle$ and
$\langle b_i\rangle = \langle y_i \rangle$

Yes, it's is easy and do you know any generalization of this [b]good[/b] inequality?

[quote="silviu"]A generalization of this good inequality is:
Having $a_{1},a_{2},...,a_{n}>0,x_{1},x_{2},...,x_{n} > 0$,and m>1.
We have:$\frac{x_{1}^m}{a_{1}^{m-1}} +\frac{x_{2}^m}{a_{2}^{m-1}}+...+\frac{x_{n}^m}{a_{n}^{m-1}}$ >= $ \frac { [x_{1}+x_{2}+...+x_{n}] ^{m}  } { [a_{1}+a_{2}+...+a_{n}]^{m-1} } $
(G.M.7-8/2002).[/quote]
Yeah, it is a cool inequality :) .I have proved it using Holder's inequality. Do you know another solution?

Sorry i dont have latex, this could get messy. If you want a neat solution to the original question, here you go:

[hide]9/(1/[1+a]+1/[1+b]+1/[1+c]) less than or equal 1+b+1+c+1+a by harmonic mean/arithmetic mean
=> 9/(1/[1+a]+1/[1+b]+1/[1+c]) less than or equal 6
=> 1/(1/[1+a]+1/[1+b]+1/[1+c]) less than or equal 2/3 an by simply inverting this we get the solution;
1/[1+a]+1/[1+b]+1/[1+c] greater than or equal 3/2[/hide]

Your solution is the same as blahblahblah's one. :D

Yeah i just realised that! No plagerism ment!


calculus integration limit function algebra polynomial geometry

Isn't this correct?  It's the natural consequence of some other integrals recently posted in here.
Let:
\[ P_n(z) \equal{} z_0 \plus{} z_1 z \plus{} z_2 z^2 \plus{} \cdots \plus{} z_n z^n
\]
Then show:
\[ \mathop\oint\limits_{|z| \equal{} R}\frac {dz}{\sqrt [n]{P_n(z)}} \equal{} \frac {2\pi i Exp[ \minus{} 2k\pi i/n]}{\sqrt [n]{z_n}}
\]
Where:
(1)  $ R$ contains all the zeros of $ P_n(z)$
(2) The integration is over an analytic branch of the root object
(3) $ \sqrt [n]{z_n}$ is the principal root of $ z_n$.

What determines the value $ k$?  I'm pretty sure I know, just asking others to think about it.  Personally I'd suggest numerically integrating it for say $ n \equal{} 3$ three different ways and compare with the expression above.  That's not too hard in Mathematica.  Just need to compute the principal branch cuts precisely over the contour and use the appropriate sheet in each interval.

[quote="shawtoos"]What determines the value $ k$?[/quote]

Maybe the same that determines the value of $ \sqrt [n]{P_n(z)}$ on the left hand side of the equation? You have not given any "canonical" way to say which of the $ n$ different roots has to be used here.

However, if you choose an [i]arbitrary[/i] analytic branch for $ \sqrt [n]{P_n(z)}$, and then define $ \sqrt [n]{z_n}$ accordingly (namely, by $ \sqrt [n]{z_n} = \lim_{w\to\infty}\frac {\sqrt [n]{P_n(w)}}{w}$), then you get $ k = 0$, if I am not mistaken.

Actually, it is already a nice exercise to prove that it is possible to choose an analytic branch of the function ${ \sqrt [n]{P_n(z)}}$ in the outside of a circle if all zeros of $ P_n$ lie inside that circle.

  darij

[quote="shawtoos"][quote="darij grinberg"]

Actually, it is already a nice exercise to prove that it is possible to choose an analytic branch of the function ${ \sqrt [n]{P_n(z)}}$ in the outside of a circle if all zeros of $ P_n$ lie inside that circle.
  darij[/quote]

Hi Darij.  Might you explain how one would begin to construct such a proof or at least in what direction I should begin?[/quote]

I have just been told that "circle" can be generalized to "convex closed set", so we have:

[color=blue][b]Lemma.[/b] If $ f$ is a polynomial of degree $ n$, and $ \Omega\subset\mathbb{C}$ is a convex closed set such that all the $ n$ zeros $ w_1$, $ w_2$, ..., $ w_n$ of $ f$ lie in $ \Omega$, then there is an analytic branch of $ \sqrt [n]{f\left(z\right)}$ in the exterior of $ \Omega$.[/color]

[i]Proof.[/i] In fact, we can WLOG assume that $ 0\in\Omega$ (else, do a parallel translation, i. e. consider the polynomial $ f\left(z+r\right)$ instead of $ f\left(z\right)$ for some $ r\in\Omega$). Since $ f$ is a polynomial of degree $ n$ with zeros $ w_1$, $ w_2$, ..., $ w_n$, we have $ f\left(z\right) = a\left(z - w_1\right)\left(z - w_2\right)...\left(z - w_n\right)$ for some $ a\in\mathbb{C}$ (by the Fundamental Theorem of Algebra). In other words, $ f\left(z\right) = az^n\left(1 - \frac {w_1}{z}\right)\left(1 - \frac {w_2}{z}\right)...\left(1 - \frac {w_n}{z}\right)$. Now, we obviously can find an analytic branch of $ \sqrt [n]{a}$ in the exterior of $ \Omega$ (just take any of the $ n$ possible values of $ \sqrt [n]{a}$) and an analytic branch of $ \sqrt [n]{z^n}$ (namely, $ z$), so it remains to find an analytic branch of $ \sqrt [n]{1 - \frac {w_i}{z}}$ for every $ i\in\left\{1,2,...,n\right\}$. But this is possible because the expression $ 1 - \frac {w_i}{z}$ for $ z$ lying outside of $ \Omega$ can never obtain a value in $ \mathbb{R}_{\leq 0}$ (since such a value would yield that $ z$ lies on the segment joining the points $ 0$ to $ w_i$, but this is impossible because $ 0\in\Omega$, $ w_i\in\Omega$ and $ z\not\in\Omega$, and $ \Omega$ is convex), and the $ n$-th root has an analytic branch in $ \mathbb{C}\setminus\mathbb{R}_{\leq 0}$. Thus we are done.

  darij

That's interesting Darij.  Thanks.  I'll work on it.:)

[quote="darij grinberg"]

I have just been told that "circle" can be generalized to "convex closed set", so we have:

[/quote]
it's very elementary fact .
More generally (Riemann theorem):
There exist analytic image between circle and $ D$ ($ C\minus{}D\neq 0$) where $ D$ Area in $ C$ and is one connected (I mean there exist $ id$ homotopic  )


Suppose $ \psi(abc) \equal{} ab^c$. Compute $ \psi(132) \plus{} \psi(321) \plus{} \psi(312)$.

It is what happens when someone presses "Ask on Forum".
GameBot will go post it on the forum.

$ 1(3)^2\plus{}3(2)^1\plus{}3(1)^2\equal{}9\plus{}6\plus{}3\equal{}18$.

I think that's what it should be anyways

it's possible their answer was 1162, but that's still incorrect

I don't see what else it could be though

[hide="possibly... "]they mixed up order of operatiions and got 9+6+9=24?[/hide]

Hmm, my guess is someone entered the answer wrong. I think I'll change the problem.

I'm pretty sure this is the correct answewer, but the computer marked me wrong:

$ 1\cdot 3^2 \plus{} 3\cdot 2^1 \plus{} 3\cdot 1^2 \equal{} 9 \plus{} 6 \plus{} 3 \equal{} 18$


[quote="ryan22"]solve this fraction thats complex

x + 3 + 2/x
/
1-4/x^2 this is a complex fraction[/quote]

$\frac{x+3+\frac{2}{x}}{1-\frac{4}{x^2}}$? Is this it?
And what do you mean by "this is a complex fraction"...

[quote="ryan22"]yes thats it i could not write it in latex because it said " dangerous latex".....yes how would i solve that complex fraction by finding the denominator??[/quote]

Lol.
In the whole thing, the common denominator on the denominator is $x^2$. In the numerator, the common denominator is $x$ (because they are the only denominators) and then you can solve from there...if solve=simplify. :dry:

ok i got it thanks, just brushing up on algebra

$\Pi 3.14159265358979323$


Consider an infinite sequence $\{a_n\}_{n \geq 1}$ of positive integers. For every $k$, do there necessarily exist $q,r$ such that $a_r \leq a_{q+r } \leq a_{2q+r} \leq \cdots \leq a_{kq+r}$?

Enjoy!  :D

--Vesselin

No, I don't think such subsequences necessarily exist for all $k$:

We can assign any value we want to $a_1$, then make $a_2>a_3>a_4>a_1$, and then, for every $k\ge 1$, we can make $a_{4^k+1}>\ldots>a_{4^{k+1}}$ and $a_{4^{k+1}}>a_{4^{k-1}+1}$. I think this should show that it doesn't even hold for $k=4$ (I think this works with powers of $3$ as well, but I wanted to make sure :)).

It does, however, hold for $k=3$. Assume WLOG that $a_1$ is minimal. We have $a_n\ge a_1,\ \forall n$, so in order to prevent arithmetic progressions to appear, we must have $a_{2n-1}<a_n,\ \forall n$, and this gives an infinite descent.


If $2x^2+2x+2=0$, then by algebra, $x^2+x+1$ must also equal zero. According to the rule:

$a=b$
$b=c$
then, $a=c$. and $a+b=x$
$\\
a+c=x
\\$
Then a+b must equal a+c

But $x^2+x+1 \neq 2x^2+2x+2$

[quote="anirudh"]But $x^2+x+1 \neq 2x^2+2x+2$[/quote]
You already said $2x^2+2x+2=0$ and $x^2+x+1=0$, so $0=0$.

[quote="anirudh"]But $x^2+x+1 \neq 2x^2+2x+2$[/quote]

Yes it is! $x^2+x+1=0$, and $2x^2+2x+2=0$, so $0=0$ like nat mc said.

OK, then I probably misconcepted the problem.

Sorry about the technicality, i am trinng to be pithy.


I have been struggling with the following problem, could anybody give me a hint? (Just a hint, please.)

Let G be a graph on vertex set V. Show that there is a partition $ V_1\cup V_2$ of V such that in each of $ G[V_1]$ and $ G[V_2]$ all vertices have even degree.

Thanks, J.

This is an unpublished theorem of Tibor Gallai, presented with solution as Problem $ 79$ in Bela Bollobas' book The Art of Mathematics (Coffee Time in Memphis) 2006.

To give you a hint from that, try induction on $ | V |$. Pick a vertex $ x$ of odd degree in $ G$, with $ N(x)$ the set of its neighbors. Take $ G \minus{} x$, but ,before applying the induction hypothesis, replace $ N(x)$ with its complement (remove all edges between vertices in $ N(x)$ and add those that were not present before).

Thank you. In fact, I found a solution a few hours after posting the problem here. But anyway, thank you for the Bollobas reference, it seems to be a good book.


[size=200][color=orange]Happy Diwali[/color] [color=white]and[/color] [color=green]Id to everyone[/color]!!!![/size]

While bursting crackers and visiting mosques, keep in mind the spirit of brotherhood and harmony that is the characteristic of Indian culture. Though events in history have undermined it, let us have progressive ideas and make great strides for the betterment of India!!

Sorry if I git a bit emotional above!!!  :lol:  The bottom line is ... Enjoy this Festival week!!!

[hide]Please correct me if I have forgotten some festival etc... Sorry!! [/hide]

How do you celebrate in India?  In the Bay Area, most people generally light many tea candles and line them around the house, make mithaees (not sure how to spell that), and since there are many Indian people here, we go to the temple, friends houses, and do pooja.

[quote="Rushil"]We clean and glow candles and lights around the house. Family gets together and prays at home or visits temples etc. We exchange gifts. We burst crackers through the night!!! At least that's how my family does it![/quote]

quite same here too :lol:  :lol:

have a nice time everyone!

since we arent allowed to light fireworks, we get little rose fireworks that sparkle on the ground.  too bad its on a school day this year!

[b]Aap sab ko shubh Deepawali aur Id mubarak [/b]:)

I love the things you get to eat on diwali and id :D

lol.  and with halloween the day before diwali.... :lol:


3.6 million square miles are to be divided so that 275 million people each have their own square plot of area K feet by K feet. What is K to the nearest whole foot?

Basically:
$K=\sqrt{\frac{(3.6)(10^{6})}{(275)(10^{6})}(5280^{2})}$

How would you do this? [b][u]I'm looking for the easy way that is accurate and fast.[/u][/b]

Note1: Notice the units: miles and feet!
Note2: If you simplify the radical so that you get as much as you can outside of the radical, you may be wrong because when you estimate the radical, you'll be off a little bit, [b]but when you multiply by the whole number, that "off a little bit" will be multiplied by that whole number, throwing off your answer.[/b]
Note3: [b]NO CALCULATORS![/b]

What I did was I simplified everything, but left it under the radical. Then I used a method that I know to get square roots very accurately (I'll post a link here later.) This way stinks because it's really tedious, but it's accurate.

The answer is:
[hide="Answer"]604[/hide]

[quote="fireninja0"]3.6 million square miles are to be divided so that 275 million people each have their own square plot of area K feet by K feet. What is K to the nearest whole foot?

Basically:
$K=\sqrt{\frac{(3.6)(10^{6})}{(275)(10^{6})}(5280^{2})}$

How would you do this? [b][u]I'm looking for the easy way that is accurate and fast.[/u][/b][/quote]
[hide]$K=\sqrt{\frac{3.6\cdot10^{6}\cdot5280^{2}}{275\cdot10^{6}}}$

$K=\sqrt{\frac{3.6\cdot5280^{2}}{275}}$

$K=5280\sqrt{\frac{36}{2750}}$

$K=31680\cdot\frac{2750}{\sqrt{2750}}$

$K=\frac{288\sqrt{110}}{5}$

$\sqrt{110}\approx10.5$

$K\approx604.8\approx\boxed{604}$[/hide]

[quote="espark52"][hide="just basically..."]$\sqrt{\frac{36000008*5280^{2}}{275000000}}$[/hide]
..and lol, u dont just give away the answer just like that before anyone even gets to answer  :rotfl:
i don know if there really is a faster way[/quote]
Simplify as much as you can before you do any multiplication and/or division. :wink:

[quote="i_like_pie"]
[hide]
$K\approx604.8\approx\boxed{604}$
[/hide]
[/quote]
^
^<< You rounded wrong?
$604.8\approx605$, not 604, so if you didn't know the answer, you would technically get it wrong. How would you find the square root of 110 quickly and [i]accurately[/i]? You can't use a calculator with this problem.
Remember when I said:
[quote]
Note2: If you simplify the radical so that you get as much as you can outside of the radical, you may be wrong because when you estimate the radical, you'll be off a little bit, but when you multiply by the whole number, that "off a little bit" will be multiplied by that whole number, throwing off your answer.
[/quote]
In this case, you would think that the difference between $10.5$ and the real value of $\sqrt{110}$ is so small, it would still be fine, but then, you take that discrepancy and then multiply it by $\frac{288}{5}$!
So $\mid\sqrt{110}-10.5\mid\cdot\frac{288}{5}$ would be how much you would be off by.
also, you wrote:
[quote]
$K=31680\cdot\frac{2750}{\sqrt{2750}}$
[/quote]
when I think you meant:
$K=31680\cdot\frac{\sqrt{2750}}{2750}$

Don't worry, I got that problem wrong by doing the exact same thing you did (only I rounded right) which is why I'm posting this question up. :wink:  :wink:

[quote="fireninja0"][quote="i_like_pie"]
[hide]
$K\approx604.8\approx\boxed{604}$
[/hide]
[/quote]
^
^<< You rounded wrong?
$604.8\approx605$, not 604, so if you didn't know the answer, you would technically get it wrong.[/quote]
Yes, but $\sqrt{110}$ is actually less than 10.5, and I thought that might make up for the error.

[quote="i_like_pie"]Yes, but $\sqrt{110}$ is actually less than 10.5, and I thought that might make up for the error.
[/quote]
But how much would the error change the answer? Your answer was 604.8, so would the error make it less than 604.5 so it would round to 604, or the error make it less, but be still bigger than 604.5 (like 604.56)? You might use t0rajir0u's way here...

[quote="t0rajir0u"][quote="fireninja0"]In this case, you would think that the difference between $10.5$ and the real value of $\sqrt{110}$ is so small, it would still be fine, but then, you take that discrepancy and then multiply it by $\frac{288}{5}$![/quote]

Would you like a way of estimating that error?  ;)

[hide="Hooooo"] There is a rational extension of the Binomial Theorem.  In this case, it states that

$(100+10)^{ 0.5}= 100^{0.5}+0.5 \cdot 100^{-0.5}\cdot 10+\frac{0.5(0.5-1)}{2}\cdot 100^{-1.5}\cdot 10^{2}+...$

We won't concern ourselves with terms beyond the third.  The sum of the first two terms is simply our predetermined $10.5$.  What is that third term?  Well, it's

$-\frac{1}{8}\cdot \frac{100}{100^{1.5}}=-\frac{1}{80}$

After multiplication by $\frac{288}{5}$, this will give you a much more accurate value. [/hide]

Edit:  [hide="Note"] The first three terms of the rational binomial expansion give

$\sqrt{110}\approx 10.4875$

Whereas its actual value is

$\sqrt{110}\approx 10.4880884817...$

Quite close, wouldn't you say?  :) [/hide][/quote]
Thanks for the help! c(^_^)"o
Only I don't understand the binomial theorem when the exponent is not a whole number c(>_<)"o...wah! I'll look it up later.
Thanks anyway, that's a cool way to find the difference.

Thanks for all your help, but I'm not satisfied! :wink:
[b]Does anybody have a way to look at the problem at a whole new angle?[/b]

Yup - take the square root by hand, up to two decimal places:

[code]
√110=10.48
  10:20x0
  1000:204x4
   18400:2088x8
[/code]

And now multiply $(57.6)(10.48)=603.648$, hence your answer is $604$.

Thanks Farenhajt. :lol:
I think we're all focusing too much on the $\sqrt{110}$ part. Does anyone have a total new approach to this problem? Like from the very beginning; something totally different?


Αν και είμαι άρρωστος βάζω μια ανισότητα for fun :lol:
Έστω $ a,b,c$ μη αρνητικοί πραγματικοί αριθμοί τέτοιοι ώστε $ a\plus{}b\plus{}c\equal{}3$. Να δείξετε ότι $ a\sqrt{1\plus{}b^3}\plus{}b\sqrt{1\plus{}c^3}\plus{}c\sqrt{1\plus{}a^3}\leq 5$

Γιώργο κατ'αρχάς περαστικά :) ....

Όμορφη ανισότητα....

$ \sqrt{1\plus{}a^3} \le \frac{a^2}{2}\plus{}1$

Άρα:

$ RHS \ge \frac{ab^2\plus{}bc^2\plus{}ca^2}{2}\plus{}a\plus{}b\plus{}c$.

Άρα αρκεί:

$ \frac{ab^2\plus{}bc^2\plus{}ca^2}{2}\plus{}3 \le 5 \Longleftrightarrow ab^2\plus{}bc^2\plus{}ca^2 \le 4 \Longleftrightarrow 4(a\plus{}b\plus{}c)^3 \ge 27(ab^2\plus{}bc^2\plus{}ca^2)$

που ισχύει.... :P

[quote="silouan"][quote="Dimitris X"] $ 4(a \plus{} b \plus{} c)^3 \ge 27(ab^2 \plus{} bc^2 \plus{} ca^2)$

που ισχύει.... :P[/quote]

Συμφωνώ ότι ισχύει, αλλά δεν είναι τόσο προφανής ώστε να το παραλείψεις... Εκτός και αν πάλι με πίασαν τα blind μου[/quote]

Πραγματι δεν ειναι προφανες, ενας τροπος να αποδειχθει ειναι με πολλαπλασιαστες Lagrange και αλλος ενας ειναι ο εξης:

Εστω $ y$ η διαμεσος των 3 αριθμων, τοτε ειναι $ y^2 \plus{} xz \leq xy \plus{} yz \Leftrightarrow (y \minus{} z)(y \minus{} x) \leq 0$, αρα:

$ xy^2 \plus{} yz^2 \plus{} zx^2 \equal{} x(y^2 \plus{} xz) \plus{} z^2y \leq xy(x \plus{} z) \plus{} z^2y \equal{} x^2y \plus{} z^2y \plus{} xyz \equal{} y(x^2 \plus{} z^2 \plus{} xz) \leq y(x^2 \plus{} z^2 \plus{} 2xz) \equal{} y(x \plus{} z)^2 \equal{} 4y\frac{x \plus{} z}{2}\frac{x \plus{} z}{2} \leq 4\frac{1}{27}(y \plus{} 2\frac{x \plus{} z}{2})^3 \equal{} \frac{4}{27}(x \plus{} y \plus{} z)^3$

Φυσικά και δεν είναι προφανές....
Απλώς έχω την εντύπωση ότι είχε συζητηθεί αρκετές φορές στα Mathlinks.....δεν βρίσκω τα Link όμως τώρα....αν κάποιος δεν μπορούσε να τη λύσει θα έβαζα τη απόδηξη.........
Νομίζω και θέτοντας όπου:
$ y\equal{}a\plus{}x,z\equal{}b\plus{}x$ ¨εχοντας θεωρήσει το $ x\equal{}min \{x,y,z\}$ βγένει επίσης......

se afto to link syzitithike ligo i anisotita http://www.mathlinks.ro/viewtopic.php?t=306625


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ H\lhd G$ and $ (|H|,[G: H])\equal{}1$, where $ G$ is a finite group. Prove that $ H$ is the only subgroup of $ G$ with order $ |H|$.

[quote="Abel"]If K is another subgroup of the same cardinality, HK is a subgroup of G (since H is normal). Hence, [G:H] is not co-prime to |K|=|H| which is a contradiction.[/quote]

your proof is not clear. Everything is fine and correct till your second sentence that you got a contradiction. it looks that there is no connection between your first and second sentence. Also cardinality of $ HK$ may not be equal to $ |H|^{2}$.

Yes, there is a connection. Have a closer look!

For subgroups H and K there is the identity $ |HK|\equal{}\frac{|H\parallel K|}{|H\cap K|}$. In our case this means, that $ |HK|$ is a divisor of $ |H|^{2}$. If K would be distinct from H, H would be a proper subgroup of HK. So $ |HK|$ would have nontrivial common divisors with $ |G: H|$. That is a contradiction to $ gcd(|H|, |G: H|)\equal{}1$ and to the fact, that $ |HK|$ divides $ |H|^{2}$.


If the second term in the expansion of (2.(x)^(4/3) + 3.(x)^(3/2))^n is 1344.x^(19/2) then the value(s) of n is/are :

A) 7
B) 14
C) 21
D) 42


Twelve friends met for dinner at Oscar's Overstuffed Oyster House, and each
ordered one meal. The portions were so large, there was enough food for 18
people. If they share, how many meals should they have ordered to have just
enough food for the 12 of them?

The ratio of people to meals is $ \frac{2}{3}$.  So you must find $ \frac{2}{3}$ of $ 12$, which is $ \boxed{8}$.

You could also set up a proportion.

$ \frac{12}{18} \equal{} \frac{x}{12}$

$ 144 \equal{} 18x$

$ \boxed{8} \equal{} x$


graph theory
combinatorics unsolved
combinatorics

A graph has $n$ vertices and $\frac{n^{2}-3n+4}{2}$ edges. There is an edge such that, after removing it, the graph becomes unconnected. Find the greatest possible length $k$ of a circuit in such a graph.

This question is silly.  It's a complete graph on $n$ vertices with $n-2$ edges missing.  The only way to disconnect a complete graph by removing $n-1$ vertices is if we remove all edges connected to a single vertex.  Thus, the longest circuit is just the longest circuit in a $K_{n-1}$.  If we mean "circuit with no duplicated vertices," this has length $n-1$.  If we mean "circuit with no duplicated edges," this is $\binom{n-1}{2}$ if $n$ is even and is $\binom{n-1}{2}-\frac{n-1}{2}$ if $n$ is odd.


2. In how many ways can the digits 1 through 9 be arranged into three 3-digit positive integers using each digit once such that the three integers add up to 1000?

Can you use the same digit in more than one of the three 3-digit numbers, as in
570+310+120=1000 ?
Otherwise, the answer is disappointing.

Sorry to revive but I need help on the same problem.  And KMST, it says "using each digit once".  Does anybody have a good solution?

Does anyone have any suggestions?  I am still stumped on this problem.

Approximately...0.
Where have I seen this before.. Hmmm.. "Counting and Probability Level 3" maybe. (which is supposedly independent work, but anyways,)

We can solve this using casework, with units, tens and hundreds digits.

First, if we use all nine digits one time each, the sum of all the digits will be 45.

There are only two possibilities the sum of the units digits could add to: either 10 or 20.
If the units digits add up to 10, then the sum of the tens digits is either 9 or 19, and the hundreds digits can add to only 8 or 9.

$10+9+9=28$

$10+19+8=37$

Therefore, no solutions are possible if the units digits sum to 10.

If the units digits sum to 20, then the tens digits may sum to either 8 or 18, and the hundreds can sum to either 8 or 9.

$20+8+9=37$

$20+18+8=46$

No solutions are possible if the units digits sum to 20 and so therefore, there are no solutions.

[hide="Solution"]In particular, we have the least number of possibilities for the hundreds digits, so it would be best to play around with possibilities for those and then see what we numbers we find. This is better than starting with the tens or units digits, because they could each be pretty much anything at this point.

In order to draw conclusions regarding the hundreds digits, we have to start with the lower digits because values carry over to higher digits which greatly complicates things. The sum of the units digits has to be 10 or 20, because the units digit of the sum is 0, and it is impossible for the three units digits to add up to 30 or more. This means that 1 or 2 carries over to the tens digits. The possibilities for the sum of the tens digits and numbers carried over is also 10 or 20, because it is, again, possible for those numbers to add up to 30 or more.

So, let the three hundreds digits be $a$, $b$, and $c$. In this case we know that:
$a + b + c + (1 or 2) = 10$
$a + b + c = 8 or 9$
The maximum value for any of these three numbers is $6$. Why? If one of those numbers, lets say, $a$ was equal to 7, then that would mean:
$7 + b + c = 8 or 9$
$b + c = 1 or 2$
It can't be one because that would imply that one of the two values was zero, which isn't allowed, but two isn't allowed either, because that would imply that there are two values that equal 1, when we have to use each digit exactly once.

So, this means that 7, 8, and 9 have to be amongst the ten and unit digits. However, they can't all be units digits or all tens digits, because then it would be impossible for the sum to be 1000.

Based on this information, we can conclude a very important point. The sum of the units digits has to be twenty. If the digits 7, 8, and 9 are all amongst the tens and units digits, but not all units digits or all tens digits, that means that at least one of those has to be a unit digit. Remember that the possible sums for the units digits are 10 and 20. If 9 is one of the units digits, then the smallest value for the sum of the units digits is 12, which is higher than ten. If 8 is one of the units digits, then the smallest value for the sum of the units digits is 11, which is also higher than ten. If 7 is one of the units digits, then the smallest value for the sum of the units digits is ten. This occurs if the units digits are 7, 1, and 2. However, this also doesn't work, because we know that 8 and 9 are both tens digits in this case. 8 + 9 + 1(carried over from the units digits) = 18. Which means, in order for the sum to be twenty, two must be a tens digit, but it has already been used as a units digit. This means that the sum of the units digits can not be anything other than 20.

Based on that point, you can draw another important conclusion. The sum of the tens digits and the value carried over from the units digits has to be twenty also. This is because the sum can't be ten. Let the tens digits be named $d$, $e$, and $f$. The sum of the tens digits and value carried over from the units digits' sum is:

$d + e + f + 2$
If this value was ten, that would imply:
$d + e + f = 8$
However, we know that at least one of the numbers 7, 8, or 9 must be a tens digit, so this is impossible. This means that two is carried over from the sum of the tens digits.

This knowledge tells us that 6 can't be a hundreds digit, because we now know:
$a + b + c + 2 = 10$
$a + b + c = 8$
And if one of those digits was 6, that would mean that two of the numbers were ones.

So, we have a lot of information right now and are close to being able to count the possible numbers. What we know:
*The sum of the hundreds digits is 8.
*The sum of the tens digits is 18.
*The sum of the units digits is 20.
*Six, seven, eight, and nine aren't hundreds digits.

The next important fact is that we know 1 and 2 aren't units digits or tens digits. They can't be units digits because then a sum of 20 would be impossible. Proving that they can't be tens digits is a bit trickier. I'll prove it by showing that it doesn't work if either of the values are tens digits. If 1 was a tens digit, that means 8 and 9 have to also be tens digit. However, with both 8 and 9 being tens digit, this means that the maximum possible sum for the units digits is 18 (7 + 6 + 5). If 2 was a tens digit, that means 7 and 9 have to also be tens digit. However, with both 7 and 9 being tens digits, this means that the maximum possible sum for the units digits is 19 (8 + 6 + 5). So, this means that 1 and 2 are hundreds digits, which means so is five, because they have to add up to 8.

Which means that the remaining digits that we need to figure out the locations of are 3, 4, 6, 7, 8, and 9. Let's take a look at the units digits now. The only possibilities for unit digit combinations are:

A. (3, 8, 9)
or
B. (4, 7, 9)

Which means the numbers used for the tens digits for each possibility are:
A. (4, 6, 7)
and
B. (3, 6, 8)

Neither of those two groups add up to 18, which means the answer to the problem is 0. There is probably an easier approach to this question, but I wasn't able to find one, and this one seems to work well. Do you understand it, because I wouldn't be surprised if I worded that strangely. [/hide]


let $ x_1,x_2,...x_n$ such that $ x_1 \ge x_2 \ge ... \ge x_n$ and $ \frac{x_1}{\sqrt{1}}\plus{}\frac{x_2}{\sqrt{2}}\plus{}...\plus{}\frac{x_n}{\sqrt{n}} \equal{}1$
 prove that $ x_1^2\plus{}x_2^2\plus{}...\plus{}x_n^2 \le 1$

[quote="dhvdochilinh"]let $ x_1,x_2,...x_n$ such that $ x_1 \ge x_2 \ge ... \ge x_n$ and $ \frac {x_1}{\sqrt {1}} \plus{} \frac {x_2}{\sqrt {2}} \plus{} ... \plus{} \frac {x_n}{\sqrt {n}} \equal{} 1$
 prove that $ x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2 \le 1$[/quote]

who can help me solve this problem?  :(

[quote="dhvdochilinh"]let $ x_1,x_2,...x_n$ such that $ x_1 \ge x_2 \ge ... \ge x_n$ and $ \frac {x_1}{\sqrt {1}} \plus{} \frac {x_2}{\sqrt {2}} \plus{} ... \plus{} \frac {x_n}{\sqrt {n}} \equal{} 1$
 prove that $ x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2 \le 1$[/quote]
It's wrong. Try $ x_1\equal{}1\plus{}\frac{1}{\sqrt n},$ $ x_2\equal{}...\equal{}x_{n\minus{}1}\equal{}0$ and $ x_n\equal{}\minus{}1.$ :wink:

I have seen this problem anywhere and as far I remember the condition is little different. Namely, all variables must be positive.
Anyway, the problem seems to be inconvenient but I remember that the solution is too easy and doesn't use any powerful inequality. :)

lasha, for positive numbers we can use the following reasoning.
$ \sum_{i \equal{} 1}^nx_i^2 \equal{} \sum_{i \equal{} 1}^ni\left(\frac {x_i}{\sqrt {i}}\right)^2\leq\left(\sum_{i \equal{} 1}^n\frac {x_i}{\sqrt i}\right)^2 \equal{} 1.$

[quote="dhvdochilinh"]let $ x_1,x_2,...x_n$ such that $ x_1 \ge x_2 \ge ... \ge x_n$ and $ \frac {x_1}{\sqrt {1}} \plus{} \frac {x_2}{\sqrt {2}} \plus{} ... \plus{} \frac {x_n}{\sqrt {n}} \equal{} 1$
 prove that $ x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2 \le 1$[/quote]
$ \boxed{x_1 \ge x_2 \ge ... \ge x_n \ge 0}$

Let : $ \boxed{c_k \equal{} \frac{1}{\sum_{i\equal{}1}^k\ \frac{1}{\sqrt{i}}}}$

$ \alpha_i \ge 0$ for i = $ 1..n$ and $ \sum_{i\equal{}1}^n\ \alpha_i \equal{} 1  ;$

We can assume :

$ \begin {cases} x_1 \equal{} \alpha_n*c_n \plus{} \alpha_{n\minus{}1}*c_{n\minus{}1} \plus{} ...... \plus{}\alpha_2*c_2 \plus{} \alpha_1*c_1 \\
x_2 \equal{} \alpha_n*c_n \plus{} \alpha_{n\minus{}1}*c_{n\minus{}1} \plus{} ...... \plus{}\alpha_2*c_2  \\
\dots  \dots\\
\dots  \dots\\
x_{n\minus{}1} \equal{} \alpha_n*c_n\plus{}\alpha_{n\minus{}1}*c_{n\minus{}1}\\
x_n \equal{} \alpha_n*c_n  \end {cases}$

By $ Mincovski$ :

$ \sqrt{x_1^2 \plus{} x_2^2 \plus{}  .... \plus{} x_n^2} \le \alpha_1*\sqrt{1}*c_1 \plus{} \alpha_2*\sqrt{2}*c_2 \plus{} ...... \plus{} \alpha_n*\sqrt{n}*c_n \le$ $ \alpha_1\plus{}\alpha_2 \plus{}...\plus{}\alpha_n \equal{}1$  ; ( $ \sqrt{k}*c_k \le 1$)

Equality will be for :  $ x_1\equal{}1 , x_2\equal{} ....\equal{}x_n\equal{}0 ;$


[quote="gr8sk8r"]ok, so there is a  house, and all four sides are facing north. then, a bear walks by, wut color is the bear?[/quote]
This house is at south pole, but not even a single bear lives there(I think this is true). Therefore, I don't have the answer.

well, thought that there were bears there, but now that you've mentioned it, i dont know.

This looks similar to a phisics problem. A bear fall in a hole, so what's the color of the bear?! :) We knew some information about the distance, speed and other things in order to calculate it's gravitational atraction, so we could determine where the bear lives :):). That was funny :clap:

since ther are no bears in the south pole therefore they should be some other animal,I like the Penguins

fine, i'll reword the problem. this time there's a house and all four edges are facing south, and a native bear walks by, what color is the bear?

[quote="math92"]white[/quote]

What?  The only way to have all four direction be north is to be as far south as possible, i.e. the South Pole.  And no bears live there at all!  I like the penguin answer.

[EDIT]  All right, [i]now[/i] white is right; it figures that gr8sk8er would reword the problem between the time I clicked 'Reply' and the time I clicked 'Submit'  :oops:  Note the  post times:  11:21 and 11:23 (depending on time zone, the hours may/will be different)  [/EDIT]

[quote="Danbert"][quote="math92"]white[/quote]

What?  The only way to have all four direction be north is to be as far south as possible, i.e. the South Pole.  And no bears live there at all!  I like the penguin answer.
[/quote]
I assumed bears lived in the south pole.