The circumference of a circle is arbitrarily divided into four arcs. The midpoints of the arcs are connected by segments. Show that two of these segments are perpendicular.

\[ \text{Angle formed by the joining chords}=\dfrac{\text{Sum of the arcs opposite to the angle}}{2} \]
Therefore in this problem it is $90^o$.


Can somebody post similar identities to:
$ a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc \equal{} (a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}bc\minus{}ac)$.

Maybe involving $ a^{5}\plus{}b^{5}\plus{}c^{5}$ or something similar.

Why don't you try Newton's Identities? Maybe it helps.
More, $ deg(a^{n})\equal{}n$ while $ deg(abc)\equal{}3$. Maybe we cannot generalize it  :maybe:

There was an interesting topic awhile back (I cannot find it) about generalizing the proof of $ a^{3}\plus{}b^{3}\plus{}c^{3}\ge 3abc$ (for positive $ a, b, c$) using such identities.  Virgil Nicuda, I think, provided such an identity for $ n \equal{} 5$, but it is not a factorization; rather, it is a sum of squares similar to

$ a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc \equal{}\frac{1}{2}(a\plus{}b\plus{}c)( (a\minus{}b)^{2}\plus{}(b\minus{}c)^{2}\plus{}(c\minus{}a)^{2})$

which is used to prove an inequality.  If someone could find this thread (because I can't :oops: ), that might be informative.

[quote="sylow_theory"]Does the identity I posted in the beginning have a name? I just love that identity.[/quote]

lol there is an interesting discussion on its name here

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=165625[/url]

[quote="tjhance"]
lol there is an interesting discussion on its name here

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=165625[/url][/quote]
That's kinda funny, nobody knows. I seen it in Titu Andresscu's preview book (I don't own it). And I seen in in Zeitz's book which I own. (I guess everyone try to take credit for it).

I think this is the thread t0raji0u is talking about

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1556383416&t=137976[/url]

[quote="sylow_theory"]Does the identity I posted in the beginning have a name? I just love that identity.[/quote]
It was posted and proven [url=http://www.mathlinks.ro/viewtopic.php?t=164348]here[/url].

As an answer to the first post, in the Romanian Magazine [url=http://recreatiimatematice.uv.ro/]Recreatii Matematice[/url] (issue 1/2006, page 34), [i]Marian Tetiva[/i] gives, as an exercise, the following identity:
$ \sum_{k \equal{} 1}^{n}a_{k}^{n}\minus{}n\prod_{k \equal{} 1}^{n}a_{k}\equal{}$ $ \sum_{k \equal{} 1}^{n\minus{}1}a_{n}^{n\minus{}k}\left(\binom{n}{k}\sum_{m \equal{} 1}^{n\minus{}1}(a_{m}\minus{}a_{n})^{k}\minus{}n\sum (a_{1}\minus{}a_{n})...(a_{k}\minus{}a_{n})\right)\plus{}$ $ \sum_{k \equal{} 1}^{n\minus{}1}(a_{k}\minus{}a_{n})^{n}$,
for any complex numbers $ a_{1}, a_{2}, ..., a_{n}$.
Notice that the sum without indices contains $ \binom{n\minus{}1}{k}$ terms.
This yields after a few simplifications, for example,
$ \left\{\begin{array}{lc}n \equal{} 2\Rightarrow & a^{2}\plus{}b^{2}\minus{}2ab \equal{} (a\minus{}b)^{2}\\ n \equal{} 3\Rightarrow & a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc \equal{} 3c(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}bc\minus{}ca)\plus{}(a\minus{}c)^{3}\plus{}(b\minus{}c)^{3}\end{array}\right.$.
The last identity shows yet another proof of AM-GM for three non-negative numbers, if we assume $ c \equal{}\min\{a, b, c\}$.


logarithms
calculus
calculus computations

Study the convergence of the serie \[ \sum_{n\equal{}1}^{\infty}\frac{\sqrt{p_{n}}\cdot\log p_{n}}{n^2},\]
  where $ p_{n}$ denotes the $ n\minus{}$th prime number.

$ p_n$ is asymptotic to $ n\log n,$ so the terms of your series look like $ n^{\minus{}3/2}$ times a collection of logarithms. This should converge by comparison to $ \sum n^{\minus{}r}$ for any $ r$ strictly between $ \minus{}\frac32$ and $ \minus{}1.$


If anyone of you think this is ( the $n = 4$ case) just the Lagrange's four square theorem, you're wrong. The four square theorem does not include the word [i]uniquely [/i]!!!

So any ideas now!??!???!??!

Note that
5 = 1+4
I think you can complete the proof from here.... :)

[quote="Rushil"]Find all positive integers $n$ that can be [i]uniquely[/i] expressed as a sum of five or fewer squares.[/quote]
Anyone for this interesting question?


Euler
induction
number theory unsolved
number theory

Let $ n>2$ be a integer
Prove that there are exist $ x,y$ be odd number such that
$ 2^n\equal{}7x^2\plus{}y^2$

We shall prove by induction.
Notice that for $ n\equal{}3$, $ x\equal{}y\equal{}1$ is a solution.
Suppose our statement is true for $ n$, and $ x,y$ is a solution for $ n\equal{}k$.
If $ \frac{x\plus{}y}{2}$ is odd, then $ \frac{x\minus{}y}{2}$ is even. We have $ 3x\plus{}\frac{x\minus{}y}{2}$ must be odd. So $ \frac{7x\minus{}y}{2}$ is odd.
Notice that $ 7(\frac{x\plus{}y}{2})^2 \plus{} (\frac{7x\minus{}y}{2})^2 \equal{} \frac{56x^2\plus{}8y^2}{4} \equal{} 2(7x^2\plus{}y^2) \equal{} 2^{n\plus{}1}$, so $ (\frac{x\plus{}y}{2},\frac{7x\minus{}y}{2})$ is a solution.
If $ \frac{x\plus{}y}{2}$ is even, then $ \frac{x\minus{}y}{2}$ is odd. We have $ 3x\plus{}\frac{x\plus{}y}{2}$ must be odd, so $ \frac{7x\plus{}y}{2}$ is odd.
Notice that $ 7(\frac{x\minus{}y}{2})^2 \plus{} (\frac{7x\plus{}y}{2})^2 \equal{} \frac{56x^2\plus{}8y^2}{4} \equal{} 2(7x^2\plus{}y^2) \equal{} 2^{n\plus{}1}$, so $ (\frac{x\minus{}y}{2},\frac{7x\plus{}y}{2})$ is a solution.
Thus, there exists a solution $ (x,y)$ for $ n\plus{}1$. By induction hypotesis, the solution exists for all natural number $ n$.


Each of the $ 39$ students in the eighth grade at Lincoln Middles School has one dog or one cat or both a dog and a cat.  Twenty students have a dog and $ 26$ students have a cat.  How many students have both a dog and cat?

$ \textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ 39\qquad \textbf{(E)}\ 46$

[hide]

When we add the number of students with dogs and the number of students with cats, we get $ 46$.  When we subtract it form the original number of students, we will find out how many have both, or $ \boxed{\textbf{(A)}\ 7}$[/hide]


limit
logarithms
ceiling function
calculus
calculus computations

For natural number $ n$, let denote $ k_n$ the number of digit of $ 3^n$. Find $ \lim_{n\to\infty} \frac {k_n}{n}$.

What do you mean the number of $ 3^n$? Do you mean $ 3^n$ itself? Or how many digits $ 3^n$ has? (implied by the title)

[quote="Fermat's Little Turtle"]Or how many digits $ 3^n$ has? (implied by the title)[/quote]

He meant this one. :)

And from basic logarithm skill we have $ k_{n}\equal{}\lceil n\log 3\rceil$

Furthermore, $ (n\minus{}1)\log 3\leq\lceil n\log 3\rceil\leq n\log 3$

Hence, by Squeezing Theorem, $ \lim_{n\rightarrow\infty}\frac{k_{n}}{n}\equal{}\log 3$

Note that in here $ \log$ represents logarithm base 10. :roll:


calculus
integration
limit
logarithms
real analysis
real analysis unsolved

Compute: $\displaystyle \lim_{n \to \infty} \sqrt[n]{ \int_{0}^{1} \arcsin^{n} x \cdot \ln (1+x) dx}$.

Asymptotic form of $\int_{a}^{b}g\left( x\right) e^{n\phi \left( x\right) }dx$ when $n\rightarrow \infty $, where $g\left( x\right) $, $\phi \left(x\right) $ are slowly-varying functions.  Without lose of generality, suppose $\phi \left( x\right) $ reaches its only maximum (not necessary extremum) at $b$.   $i.e.$ $\phi ^{\prime }\left( b\right) =\phi ^{\prime \prime }\left( b\right)=\cdots =\phi ^{\left( k-1\right) }\left( b\right) =0$, $\left( -1\right)^{k}\phi ^{\left( k\right) }\left( b\right) <0$, then $\int_{a}^{b}g\left(x\right) e^{n\phi \left( x\right) }dx$~$\frac{1}{k}\Gamma \left( \frac{1}{k}\right) \sqrt[k]{\frac{k!}{n\left\vert \phi ^{\left( k\right)}\left( b\right) \right\vert }}g\left( b\right) e^{n\phi \left( b\right) }$.
Note: if $\phi \left( x\right) $ has maximum at $c\in \left( a,b\right) $, we can break $\int_{a}^{b}$ into $\int_{a}^{c}+\int_{c}^{b}$

[quote="Kent Merryfield"]Clearly, the limit it $\frac{\pi}2.$

The integrand is increasing, so we have

$\int_0^1[\arcsin(x)]^n\ln(1+x)\,dx\le\int_0^1\left(\frac{\pi}2\right)^n\ln2\,dx = \left(\frac{\pi}2\right)^n\ln2.$

On the other hand, we can find small fixed positive numbers $\alpha, \beta, \gamma$ such that

$\int_0^1[\arcsin(x)]^n\ln(1+x)\,dx\ge\int_{1-\gamma}^1 \left(\frac{\pi}2-\alpha\right)^n\ln(2-\beta)\,dx = $
$\gamma\left(\frac{\pi}2-\alpha\right)^n\ln(2-\beta).$

Take the $n$-th root and take the limit as $n\to\infty$ and the quantity we are estimating lies between $\frac{\pi}2-\alpha$ and $\frac{\pi}2,$ with $\alpha$ arbitrarily small.

Just a special case of the $\infty$ norm being the limit of the $p$ norm as $p\to\infty.$[/quote] nice... ;)

but not [quote="Kent Merryfield"]Clearly, [/quote] for any people!  :blush:  :o

[quote="Lagrangia"]The 2nd one is $\displaystyle \frac{\pi}{4}$, at least this is what I got.. the result $\displaystyle \frac{\pi}{2}$ is if we compute:

$\displaystyle \lim_{n \to \infty} \sqrt[n]{ \int_{0}^{1} \arccos x \cdot \arctan^{n} x dx }$. :D

cheers! :D :D[/quote]
Unfortunately, you got the wrong number. It's still $\frac{\pi}{2}$

[quote="Lagrangia"]And .. what about:

$\displaystyle \lim_{n \to \infty} \sqrt[n]{ \int_{0}^{1} \arccos^{n} x \cdot \arctan{x} dx}$.

cheers! :D :D[/quote]
It must be $\frac{\pi}{2}$. Actually $\displaystyle \lim_{n \to \infty} \sqrt[n]{ \int_{0}^{1} f^{n}(x) \cdot g(x) dx}=\displaystyle \max_{[0,1]}f(x)$, if $f(x)$ has nonnegative lower bound on $[0,1]$ and $g(x)$ is bounded on $[0,1]$


geometry
perpendicular bisector
angle bisector
geometry unsolved

Let be given points $A,B$ on a circle and let $P$ be a variable point on that circle. Let point $M$ be determined by $P$ as the point that is either on segment $PA$ with $AM=MP+PB$ or on segment $PB$ with $AP+MP=PB$. Find the locus of points $M$.

I think it is supposed to be $AM=MP+PB \ \ (1)$ and $BM=MP+PA \ \ (2)$

$AM=MP+PB \Rightarrow $
$AP=AM+MP=2MP+PB \Rightarrow $
$AP>PB$

Similarly, from $(2)$ we get $AP<PB$.

Let $CD$ be the perpendicular diameter to $AB$.

Suppose that $P$ is such that $AP>PB$ so we use the formula $(1)\ \ AM=MP+PB$

On the extension of $AP$ we get the point $E$ such that $PE=PB$. Then $AM = MP+PE \Rightarrow  AM = ME\Rightarrow M$ is the midpoint of $AE$

Now suppose that $P\in\stackrel{\frown}{BD}$.
We'll prove that the point $D$ is on the perpendicular bisector of $BE$.
Indeed, the external angle bisector of the isosceles triangle $PBE$ is the line $PC$. But $PC\perp PD$, so the internal angle bisector of $\triangle PBE$ is the line $PD$. But $PBE$ is isosceles, so $PD$ is also the perpendicular bisector of its base $BE$.
On the other hand, $CD$ is the perpendicular bisector of $AB$, so we get $DA=DB=DE$.

So the point $E$ is moving on the circle $(D,DA)$
As a result, $DM\perp AM$, so $M$ lies on the circle with diameter $AD$
Similarly:
$P\in \stackrel{\frown}{BC}\Rightarrow M$ on the circle with diameter $AC$.
$P\in \stackrel{\frown}{AC}\Rightarrow M$ on the circle with diameter $BC$.
$P\in \stackrel{\frown}{AD}\Rightarrow M$ on the circle with diameter $BD$.

The locus consists of the red arcs.


Hi, a quick question

What is the best way to increase your SAT reading score? On practice tests, I usually get between 590 and 630on practice tests, and I do at least 1 reading passage a day.
My math is solid (700-800), and so is my writing (700-800).

UM READ!!!!! read novels, take AP Literature, etc

haha I am gonna have to start doing that, reading lots of novels . I am in 8th grade lolz but i will have to take this 2 years later. I can get a solid score in math also (700+) but currently am not very good at reading. 550 to 630 right now in reading lol

I do read, and I'm getting an A in AP Lit

haha, are AP classes realistically that hard? or just slightly harder than normal classes? I am curious since I will have to decide about that stuff next year.

AP classes can be very hard, or they can be bad jokes, hardly above the meager standards of regular classes. One clue is how well previous students in a course do on the AP exams.

oh gotcha. I am thinking of taking AP courses in subjects I am stronger in.

[quote="funcia"]where do you get these practice tests?[/quote]

I have about 6 sat books, and the sat online course.

[quote="kimby_102"][quote="neelnal"]My advice would be:

DON'T ANALYZE THE PASSAGES, TAKE THEM LITERALLY!!!

seriously, when i used to take the sat, i analyzed the life out of the passages and got a 600

then my teacher told me to take everything literally and i got 800!!![/quote]
what do you mean by taking literally? Do you mean skimming through the passages?[/quote]

What do you mean by taking the passages literally neelnal ?

I think he means to take the words for what they actually mean instead of looking into the text for stuff like symbolism and the sort. I don't know how well that idea actually works since there are usually some questions related to analysis.

Thanks for the advice. Maybe I'll finally be able to beat my 7th grade verbal score of 670...

woah ur good lol

I got 650 math 500 verbal and 540 writing as a 7th grader something like that. :(

[quote="K81o7"]Exactly what neelnal said.  Don't think too hard.  The answers are almost ALWAYS in the passage.

When I first did a cr practice, I thought too much about the passages when reading them, and then went on to the questions to find that they were vague.  Because I had formulated my own opinions about the subject, I got thrown off and selected the wrongs answers.  I ended up with a 610.

Well my brother told me not to really think while reading the passage.  I took that advice.  The practice after that one, I scored a 720.  Not bad: a 110 point improvement with just one general piece of advice.

Now, I actually look at the questions with specific line numbers first.  I then look up those specific lines in the passage and the lines around them (for context).  Those questions generally point to key ideas in the passage, so after answering those questions, I keep them in mind while skimming the passage.  Then when I get to the rest of the questions, I know better what the author is trying to say (and am more objective).[/quote]

Thanks for the advice K81o7. I will try what you said tomorrow.

wow that is really good advice. I might try it sometime this month when practiicing.  :roll:

A revision of my advice.  Sometimes, it isn't such a good idea to read the questions first (only do that if it's a short passage).  For long passages, maybe read the first paragraph first to get the main point of the passage, and then go on to follow the aforementioned strategy...

Yea. Don't get distracted at all. Just focus on the essay and don't try to interpret what you think the author means.

Collegeboard doesn't care what you think, so don't even try to interpret it your own way.

Usually, [b]don't[/b] read the questions first. Read the passage carefully first, then answer the questions and revert back quickly each time to the passage.


My email service, although it never bothered to tell me, only lets me send one attachment.  When i sent in solutions, I realized this (after i first emailed), so i sent a second email (first with .tex, second with .pdf) within minutes of each other.  Will this be a problem?  Do i need to write another email to solutions@usamts.org to explain it?

Thanks.

For this round, it's fine.

In the future, either (a) fix your email so you can send more than one attachment, or (b) don't send your stuff by email: use mail or fax instead.

We'll cross that bridge if we come to it.

[quote="tjhance"]I also have a concern involving attachments:

Earlier today I had problems with e-mail attachments not being received correctly on the other end. What will happen if my attachments in my e-mail for USAMTS cannot be opened?[/quote]

that just happened to me, i just got 3 emails saying " your message failed we will try to send it for 2 more days DO NOT RESEND YOUR MESSAGE" then i got another one saying "your message failed we will try for 1 more day" and lastly "your message permanently failed"

[quote="cognos599"]that just happened to me, i just got 3 emails saying " your message failed we will try to send it for 2 more days DO NOT RESEND YOUR MESSAGE" then i got another one saying "your message failed we will try for 1 more day" and lastly "your message permanently failed"[/quote]Send everything (including the error messages you received) to usamts@usamts.org so that we can look into it.  Be sure to include your USAMTS ID#.


If the remainder of the following quotient:
(x^100-x^99+x^98-x^97+....+x^2-x+1) / (x+1)
is written as A/(x+1), then find the value of A


A hoop of radius a and mass M hangs from a rough peg in a vertical plane .The angle of friction between the hoop and the peg is $\lambda$.
An insect of mass m starts at the bottom point of the hoop and slowly crawls upward.
Prove that the insect can reach the peg if $sin\lambda$>$\frac{m}{M+m}$.
If $sin\lambda$ <$\frac{m}{M+m}$ prove that the hoop will slip on the peg when the insect has travelled a distance

a[$\lambda$ +$sin^{-1}${$\frac{(M+m)(sin\lambda)}_{m}$}] along the hoop.

I've never heard of that before, either. But luckily there's google :): [url]http://dict.die.net/angle%20of%20friction/[/url]. So if there is an object on a plane, and you increase the angle of elevation of that plane, the angle of friction is the angle of elevation at which the object slips.

For the first part, I get:
$\sin{\lambda}> \frac{m}{\sqrt{(M+m)^{2}+m^{2}}}$
and
$\tan{\lambda}> \frac{m}{M+m}$

I'm not sure what's wrong. I found the center of mass of the hoop and the insect. So if the insect crawls an angle $\theta$ up the hoop, the center of mass rotates with it. Since the CM will rest on the vertical line through the point where the hoop rests on the peg, the line through that point and the CM is vertical. So let $\phi$ be the angle between this line and the segment connecting the center of the hoop and the rest point. This is also the angle at which the hoop rests on the peg, so $\phi$ is always less than $\lambda$. The maximum $\phi$ occurs when $\theta=90^\circ$, and here $\sin{\phi}= \frac{m}{\sqrt{(M+m)^{2}+m^{2}}}$ because one leg of the right triangle here is $a$ and the other is $\left(\frac{m}{M+m}\right) a$. And $\sin{\lambda}>\sin{\phi}$, which leads to the result above.


modular arithmetic
floor function
algorithm
arithmetic sequence

[quote="xenon_nightmare152"]How many pairs of positive integers (x,y)  satisfy 2x + 3y = 515? This type of problem shows up everywhere.[/quote]
[hide]Just start by finding a pair that has the smallest value of $x$ (using $y$ is also fine). If the coefficients are small, as in this case, we can just use trial and error, and we find that $(x,y)=(1,171)$ is the one we want. Then increasing the $x$ value by $3$ and decreasing the $y$ value by $2$ gives us the next highest solution (try to figure out why this works), and so our solutions are
$(1,171)$
$(4,169)$
$(7,167)$
...
...
...
$(256,1)$

Looking at the x-values, we can see that counting the number of solutions is the same as counting the number of terms in the arithmetic sequence
$1,4,7,\dots,256$, which has $\frac{256-1}{3}+1=\boxed{86}$ terms, our answer.[/hide]

[quote="mad_skillz_aops"]Is it-
[hide]86[/hide][/quote]

Please, remember to post your solution.

What I can only prove is that there're infinite integrer numbers that satisfied the equations

[quote="José"]
What I can only prove is that there're infinite integrer numbers that satisfied the equations[/quote]

[b]Positive.[/b]

[quote="José"][quote="mad_skillz_aops"]Is it-
[hide]86[/hide][/quote]

Please, remember to post your solution.

[/quote]
Sorry, but it looks like nebula42 did it the same way I did.

[quote="xenon_nightmare152"]How many pairs of positive integers (x,y)  satisfy 2x + 3y = 515? This type of problem shows up everywhere.[/quote]

Put it in mod 2:
$2x+3y = 515$
$y\equiv 1 \pmod{2}$
So y = $1, 3, 5, ....171$
That is $\boxed{86}$ numbers.

I guess I kind of did it similar to the mod method.  [hide="what I did"]It's obvious that $515-3y$ must be even for it to meet the conditions, so every odd value of $y$ such that $0<y\leq\lfloor\frac{515}{3}\rfloor$, or $0<y\leq171$, works.  Since 171 is the 86th odd number, there are $\boxed{86}$ pairs...[/hide]

[quote="srulikbd"]cann't we use the Extended Euclidean algorithm?[/quote]

Can you explain what that is, and how to use it?

Google loves you: http://en.wikipedia.org/wiki/Extended_Euclidean_algorithm


Let $ a,b,c$ be positive reals such that $ a\plus{}b\plus{}c\equal{}1$ . Prove that :

$ \sqrt{ab\plus{}c} \plus{} \sqrt{bc\plus{}a} \plus{} \sqrt{ac\plus{}b} >\equal{}1\plus{} \sqrt{ab} \plus{} \sqrt {bc} \plus{}\sqrt {ac}$

[quote="andrew_max"]Let $ a,b,c$ be positive reals such that $ a \plus{} b \plus{} c \equal{} 1$ . Prove that :

$ \sqrt {ab \plus{} c} \plus{} \sqrt {bc \plus{} a} \plus{} \sqrt {ac \plus{} b} > \equal{} 1 \plus{} \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ac}$[/quote]

It's not hard.  :P ;)

\[ \sum {\sqrt {ab \plus{} c} }  \equal{} \sum {\sqrt {ab \plus{} \left( {a \plus{} b \plus{} c} \right)c} }  \equal{} \sum {\sqrt {\left( {a \plus{} c} \right)\left( {b \plus{} c} \right)} }
\]
\[ \Rightarrow \sum {\sqrt {ab \plus{} c} }  \ge \sum {\left( {c \plus{} \sqrt {ab} } \right)}  \equal{} 1 \plus{} \sum {\sqrt {ab} }
\]

Done.


At a school, all 60 students play on at least one of three teams: Basketball, Soccer, and Mathletics. 8 students play all three sports, half the students play basketball, and the ratio of the size of the math team to the size of the basketball team to the size of the soccer team is $ 4: 3: 2$. How many students at the school play on exactly two teams?

Because we know that there are 30 people of the basketball team, we then know the number of players on the other teams using the ratio given: Math team: 40 people, and soccer team: 20 people. When we add these numbers all up we get 90 and see that it's bigger than 60, so we know that there are going to be people that are in two or three teams. We know that we counted the people that played in 2 teams one time more and those who played in 3 teams two times more, and we have 90=60+x+2y, with x=the number of people that plays in two teams and y be the number of people that plays in three teams. Except, we can plug 8 in as y since we know that there are 8 people that plays in 3 teams and we get x = 14.

[quote="james4l"]40:30:20. 60-(90)+x-2=0, so x=32.[/quote]
james4, you do not subtract by $2$, you must add $16$ instead.  Do you know why? [hide]We subtract $16$ because we overcount for the $8$ students who are on all three teams twice.[/hide].  Then you get:
$60-90+x-2=0$

$-14+x=0$

$x=\boxed{14}$

I'm not very good at formulas and what not, so here's what I did (my method takes longer):

60 students-  4:3:2   - M,B,S
B=30

Now we have ratio-   40:30:20  - but then the total would be 90.

*BUT* 8 play all three so now-   32:22:12-   our total is now 66.

That's six too many.     *THERFFORE* we can conclude the 8 we know play three, plus the extra 6 on our total= 14

OR 8+6=14

Hope I helped for those who aren't good at formulas!!! ~Anna

[quote="JSGandora"][quote="james4l"]40:30:20. 60-(90)+x-2=0, so x=32.[/quote]
james4, you do not subtract by $2$, you must [b]subtract[/b] $16$ instead.  Do you know why? [hide]We subtract $16$ because we overcount for the $8$ students who are on all three teams twice.[/hide]  Then you get:
$60-90+x-2=0$

$-14+x=0$

$x=\boxed{14}$[/quote]
EDIT: In my post, I wrote "...you must add $16$...".  It should be "..you must subtract $16$...".  Sorry for my mistake! It is corrected and bolded in the quoted text.


Joker ;)
(i) $f(x)=1$ for all $x$
(ii) ...
Problem: prove that $f$ is periodic ;)

Rather, it should perhaps be $f(x)\leq 1$ for all $x$.  Then the sequence $a_n=f(x+\frac{n}{42})$ (for some $x$) has recurrence $a_{n+13}=a_{n+7}+a_{n+6}-a_n$, with characteristic equation $0=t^{13}-t^7-t^6+1=(t^7-1)(t^6-1)$, and hence general term $a_n=cn+d+\sum_{i=1}^{11}c_ir_i^n$, where $r_1$, $r_2$, ...$r_{11}$ are the complex sixth and seventh roots of unity.  As the function, and hence sequence, is bounded, $c=0$, so $f(x+1)=a_{n+42}=d+\sum_{i=1}^{11}c_ir_i^{n+42}=d+\sum_{i=1}^{11}c_ir_i^n=f(x)$.

[quote="Amir.S"]Let $f: \mathbb R \to\mathbb R$ be a function such that $|f(x)| \leq 1$ for all $x \in \mathbb R$ and
\[f \left( x + \frac{13}{42} \right) + f(x) = f \left( x + \frac 17 \right) + f \left( x + \frac 16 \right), \quad \forall x \in \mathbb R.\]
Show that $f$ is a periodic function.[/quote][hide="solution"]for simplicity, denote $g(x):= f(\frac x{42})$. Then we have
$g(x)-g(x+6)=g(x+7)-g(x+13)$, iterating this $\implies g(x)-g(x+6)=g(x+42)-g(x+48)$.
We also have
$g(x)-g(x+7)=g(x+6)-g(x+13)$, iterating this $\implies g(x)-g(x+7)=g(x+42)-g(x+49)$.

So, putting $y=x-6$, we have $g(y)-g(y+1)=g(y+42)-g(y+43)$. Summing this up for $y,y+1,...,y+42$ yields
$g(y)-g(y+42)=g(y+42)-g(y+84)$.. Now suppose $g(y)\ne g(y-42)$. Then by iterating this, we see that $g$ and thus $f$ cannot be bounded, contradiction to $|f|\le 1$.

So $f$ has period $1$.[/hide]


algebra
polynomial
inequalities
absolute value
algebra proposed

Let $f \in \mathbb C[X]$ be a polynomial of degree $n \geq 0$.

Prove that for all reals $r \geq 1$, $f$ can be written as a sum of four polynomials from $\mathbb C[X]$, each of them having all roots of absolute value $r$.

What happens for $r < 1$?

How incredibily nice! Here's an attempt to prove it for all $r$.
 $\textbf{ Lemma 1}$: any polynomial $f$ is the sum of two polynomials all of whose roots lie in the open disc of radius $1$ centered at $0$.
   Proof: let $d$ be the degree of $f$. Take some large $M$ such that for all $|z|=r$ we have $\frac{|f(z)|}M<1$. Rouche's theorem assures in this case that the polynomials $MX^{d+1}+f(X)/2$ and $-MX^{d+1}+f(X)/2$ have all their roots of modulus smaller than $1$. They clearly add up to $f$.
 $\textbf{ Lemma 2}$ Any polynomial $f$ all of whose roots have modulus smaller than $1$ is the sum of two polynomials with all roots of modulus $1$.
  Proof: We may assume $f$ is monic, let $f(X)=(X-a_{1}) \cdots (X-a_{n})$. I claim that $f=P_{+}+P_{-}$ where $P_{\pm}=1/2(f\pm(1-X\overline{a_{1}}) \cdots (1-X\overline{a_{n}}))$ works. Indeed, suppose that $|(z-{a}_{1}) \cdots (z-a_{n})|=|1-z\overline{a_{1}}| \cdots |1-z\overline{a_{n}}|$. If $|z|>1$, then $|z-a_{i}|<|1-z\overline{a_{i}}|$ for all $i$, contradiction. In the other case, all inequalities are reversed, contradiction again. The conclusion follows.
   Finally, take some $r>0$ and let $f$ be a complex polynomial. Write $f(rX)=g_{1}(X)+\ldots+g_{4}(X)$ where all $g_{i}$ have roots on the unit circle. Then $f(X)=g_{1}(X/r)+\ldots+g_{4}(X/r)$ and all $g_{i}(X/r)$ have roots of modulus $1$.
   Hope this works...

I was sure you'd like it :) His (i.e. Marian Andronache's) proof is the same, but he applied the two lemmas in reverse order and only got the weaker result (with $r \geq 1$).


Hi, all.

 The second edition of PEN Problems Book will be weblished around early September.
 Next month, we are going to edit PEN Problems Book(ver. July 11, 2007):

 1. Find ALL typos in the problem statements.
 2. Eliminate ALL non-elementary (or non-appropriate) problems(e.g.  E38).

 We are going to check every pages in here at ML. If you find more mistakes,
 let us know by posting comments at each page.

 Currently, PEN TEAM has five people:

   Andrei Frimu (Moldova)
   Cosmin Pohoata (Romania)
   Yimin Ge (Austria)
   Hojoo Lee (Korea)
   Peter Vandendriessche (Belgium)

 WE WANT MORE STAFFS. If you want to be a member of PEN TEAM, then please
 let us know by sending an email to us at pen@problem-solving.be

 Mainly, you are required to proofread the solutions here or the-problems-of-the-biweek.
 Of course, you need to know very baby skills in TeX. But, we do NOT want experts in number
 theory. It is enough that you like number theory. That's all!

 All the best,
 PEN TEAM.

 PS. The draft of PEN Extended is available at http://projectpen.wordpress.com. Eventually,
 we will add it to the original PEN Problems Book.

Just a question: In PEN Extended, Year 2007 Problem 1, it says [quote]Let x, y be integers diﬀerent from −1 such that$ \frac{x^4 -1}{y + 1}+\frac{y^4-1}{x+1}$ is an integer. Prove that $ x^4y^{44} - 1$ is divisible by $ x + 1$.[/quote]
That seems not symmetric. Should it be $ x^4y^4-1$? (Sorry for wasting your time if it should be $ x^4y^{44}-1$, but I am not exactly up to date with number theory problems...  :blush: )

[quote="Brut3Forc3"]Just a question: In PEN Extended, Year 2007 Problem 1, it says [quote]Let x, y be integers diﬀerent from −1 such that$ \frac {x^4 - 1}{y + 1} + \frac {y^4 - 1}{x + 1}$ is an integer. Prove that $ x^4y^{44} - 1$ is divisible by $ x + 1$.[/quote]
That seems not symmetric. Should it be $ x^4y^4 - 1$? (Sorry for wasting your time if it should be $ x^4y^{44} - 1$, but I am not exactly up to date with number theory problems...  :blush: )[/quote]

Hmm, I think that someone from Vietnam needs to help us.

Yes ,it is $ x\plus{}1|x^4y^{44}\minus{}1$

[quote="TTsphn"]Yes ,it is $ x \plus{} 1|x^4y^{44} \minus{} 1$[/quote]

Thanks, TTsphn!

[quote="ideahitme"]TeXing work for second edition is almost done. The following problems will be deleted in the 1st edition
 of PEN Problems Book. The corresponding threads for those problems are deleted or locked!  They are

[quote] B02 B08
 E25 E26 E31 E38 E39 E41
 F13 F15
 G01 G07 G15 G16 G17
 H41 H53
 J07 J09 J10
 K30
 M08
 P02 P23
 Q11
 R09 R11
 S25 [/quote][/quote]

Plus, we have the following corrections :

[quote]A 33 [Corrected] Reference corrected : IMO Short List 1993
A 87 [Corrected] Reference corrected : Romania 2005

D 17 [Corrected] Reference corrected : AMM problem S9 [1979, 306] by M. S. Klamkin and A. Liu

G02 [Corrected] Correct reference: A.Mirotin Belarus 2002
G03 [Corrected] Correct reference V. Bernik Belarus 2001
G05 [Corrected] correct reference "CRUX(No. 7, Volume 25, 1999), Problem A240, Mohammed Aassila"
G06 [Corrected] Says: Putnam 1980 but tells that it might be wrong

G12 [Corrected] Reference  "CRUX(No. 1. Vol. 24, 1998), Problem 2305, Richard I. Hess".
G13 [Corrected] 'AMM, Problem 10630, Richard Strong'   Seems completely ugly
G14 [Corrected] reference wrong
G24 [Corrected] Source may not be ok

H16 [Corrected] wrong reference
H37 [Corrected] n>2 is required  correction suggested at ML
H55 [Corrected] Has a Typo

I14 [Corrected] typo

K19 [Corrected] Has a typo

N12 [Corrected]   correct statement

O02 [Corrected]  modify statement
O03 [Corrected] modifiy statement
O09 [Corrected]  modify source from India 1998 to India 2000.
O13 [Corrected]  modify statement
O33 [Corrected]  modify source from IMO Short List 1990 CZE3 to IMO Short List 1990 CZE2.
O35 [Corrected]  modify source from USA 1998 to Romania TST 1998
O39 [Corrected]  typo in the statement: modify 1985 with 1995.
O50 [Corrected]  reference
O53 [Corrected]  modify source to "Schur's theorem"

P01 [Corrected] modify second author from Irving Kaplanky to Irving Kaplansky.
P13 [Corrected]  additional reference: also used as Putnam 2005 A1.

Q06 [Corrected] modify source from [Twh, pp. 114] to Gauss, [Twh, pp. 114].

R01 [Corrected]  modify source from MM, Problem 1409, Gerald A. Heur to J. R. Arkinstall, Bull. of the Australian Math. Soc., 1980[/quote]

I mean let's think about the origin of problems. Nobody has a complete overview of all problems ever proposed to some competition or well-known journals as Gazeta Mathematica, [url=http://www.maa.org/pubs/mathmag.html]Mathematics Magazine,[/url] [url=http://www.maa.org/pubs/monthly.html]American Mathematical Monthly,[/url] [url=http://kvant.mirror1.mccme.ru/]Kvant,[/url] [url=http://journals.cms.math.ca/CRUX/]Crux Mathematicorum[/url] and [url=http://www.komal.hu/info/bemutatkozas.e.shtml]Kömal.[/url] And especially it is hard to really invent a genuinely new easy problem that has never been proposed before. So we mostly just use that source where we found the problem or the souce which is most popular. It would be desirable to mention the oldest known source but then again it might be some obscure source which only few people have access to.
The issue of re-posting problems seems also very widely spread on the AoPS/Mathlinks forum. But then again if you are new to the forum it is tedious to search for previous occurrences of the problem. Taking into consideration that people use different notation and it is not that easy to search within .tex-code does not make things easier. Experience members sometimes post the URLs of the previous occurrences which keeps the forum consistent but students may be less motivated to work on it as somebody already approached it. Not sure whether these hide tags are useful in this context.
And students even have less experience so it does not make such a big difference whether the problem has been used elsewhere, esp. for lower level local and national olympiads. It is also a question of effort one wants to put in organizing those competitions and it might be just easier to recycle some old problems to fill the paper.
The goal is to develop problem-solving expertise and skills and not solve unique problems but hopefully to approach very different challenges. Look at universities: many of them use the same text books and hence students get assigned similar problem sets.
The most challenging olympiads are closer to research as they can be more technical in their discussion. Often researchers extract a special case of previous or on-going research and pose a problem that can (also) be solved with "elementary" techniques. If students come up some special argument in the solution this might also enrich research in return, or researchers may take olympiad problems and generalize them which requires advanced and sophisticated machinery. Quite often it seems very appealing to use special case of advanced theorems. But what to do if students quote that fact? Assign zero or full credit to his "solution"?!
It seems to be the same at the IMO. Weaker countries lobby for easy problems that well-trained countries get no competitive advantage from. At IMO countries are assumed to state knowledge if a certain/similar problem has appeared before elsewhere. But then again not saying that a particular problem is known in their problem-solving community and culture benefits their students. Sam Nariman from Iran suggests that due to the strongly disparate score distribution of the lowest and highest quantile there should be a harder problem and an easier problem set at IMO depending on the nation's strength. To get a higher score might encourage many students but others feel inferior to participate in the second class paper which solves the easier problem set, which brings us back to political correctness. Should the choice of problem paper be enforced or voluntary? What about weaker nations that have a "stronger" students every few years? One different but nice approach is the [url=http://www.rmm.lbi.ro/]Romanian Master of Mathematics[/url] which invites the top 20 countries from the past IMO to participate. But it is not clear whether the host country like Romania is also required to satisfy this condition. Unless the set of hosts is restricted to very few nations which are always expected to place in the top 20 the organization might turn out to be difficult.

How hard are the problems in this book in comparison to certian competitions?


AMC
AIME
function
probability
AMC 12
search
AMC 10

[b][size=200]THE DEADLINE IS SUNDAY, JULY 6, 2008.[/size][/b]

I am granting a one week window to solve this test.

These are the instructions on page one of the test.

[quote]1. DO NOT OPEN THIS BOOKLET UNTIL YOUR PROCTOR TELLS
YOU.
2. This is a sample twenty-five question MCQ test. Each question has answer
choices (A), (B), (C), (D), and (E). Only one of these is correct.
3. There is a penalty for guessing. You will receive 6 points for a correct answer,
1.5 points for a blank answer, and 0 points for an incorrect answer. It is not
to your benefit to answer all of the questions.
4. No aids are permitted other than scratch paper, graph paper, rulers, and
erasers. NO CALCULATORS ARE PERMITTED.
5. Figures are not necessarily drawn to scale.
6. When your proctor gives the signal, begin working on the problems. You will
have 40 minutes to complete this test.
7. When you finish your exam, please go to http://mersenne.zapto.org/amc/amc
to submit your answers. Once your answers have been received, you will be
given an electronic participation certificate.
8. The statistics will be computed and you will be notified of your score.
Once you have read the above directions, you may turn the page. You have 75
minutes to finish this examination. While you are completing this test, a timer is
available online. Please refer to this timer. If you are unable to input your answers
within 75 minutes, there will be no reprieve and the score that you will receive
will be based only on the answers received by the server.
Good luck.[/quote]

It is to your advantage to read these.

75 minutes are granted for the test. Get this much time ready, a pencil and paper, a printer (optional), and your brain. Then head over to http://mersenne.zapto.org/amc/amc.

It should automatically authenticate you. Click on "test," follow the directions. The site has a timer. It will auto-submit after time is up.

When you start, you will get the questions.

Once again, the site is:

http://mersenne.zapto.org/amc/amc.

Good luck.

Note: EVERYONE is welcome to take this test.

Eh, what the heck, I got a 58.5 too. And the thing is, it would have been impossible for me to have gotten that (I answered 20, left 5 blank, to be safe).

Also, I believe there was a typo on one of the problems around 22 or so. It said closer to the center than any order, and I didn't know what that meant.

Okay I'm sorry people

The scoring thing was messed up...

I let my sister take it and she got two problems and also got a 58.5 :maybe:

Ignore the above post. My sister was using my computer.

And @xpmath: you didn't get a 58.5

what did i get then?

I know i did horrible, as in reeeally bad.

Urc - you did get a 58.5. I had a bad variable somewhere so for everyone it read your score.

I saw you spent like 5 minutes on it... :wink:

[b]Clarification - everything has been fixed[/b]

EDIT: As of 1:38 pm, the typo that xpmath pointed out has been fixed. @xp: I'll give you credit for #22

Yeah, I had to eat lunch and stuff and i didnt realize that till after i hit the button...lol. i would have done bad anyway, i didnt undertsnad a lot

Actually your score wasn't a failure in the grand scheme of things.

Hmm, according to the answers it says I put (and the fact that you gave me #22), my score isn't possible. Did you forget to change it because of #22?

i am not taking the test, but i briefly looked over the test, it seems quite challenging. good job with it!

i want to look over the test in greater detail afterwards and i can't look at it now because i exited after like 3 minutes. so will you be posting the questions some time?

#22 is fixed...

@cognos: I'll post them all in separate threads for solutions later.

It seems that the server at Mersenne is down.

Try coming back tomorrow or wait until #H34N1 fixes it.

OKay server is down.

I forgot

but ya

server is down

because upgrading to linux

Warning: file_put_contents(loadedtest.txt) [function.file-put-contents]: failed to open stream: Permission denied in /var/www/Site1/amc/amc/test1.php on line 17

Warning: file_put_contents(log.txt) [function.file-put-contents]: failed to open stream: Permission denied in /var/www/Site1/amc/amc/test1.php on line 19

Warning: fopen(tmp/urls.txt) [function.fopen]: failed to open stream: Permission denied in /var/www/Site1/amc/amc/generate_url.php on line 14
Could not open tmp/urls.txt! - Refresh link.

[color=blue]So then I put in random letters and submitted it. And then:[/color]

Warning: file_put_contents(submittedtest.txt) [function.file-put-contents]: failed to open stream: Permission denied in /var/www/Site1/amc/amc/posttest.php on line 14

Warning: file_put_contents(log.txt) [function.file-put-contents]: failed to open stream: Permission denied in /var/www/Site1/amc/amc/posttest.php on line 16

Warning: file_put_contents(scores.txt) [function.file-put-contents]: failed to open stream: Permission denied in /var/www/Site1/amc/amc/posttest.php on line 31

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[quote="georgiamathdude"]Is here a typo on number 16?[/quote]

Anyone see the irony here?

[quote="The Zuton Force"][b]Fatal error:[/b] Call to undefined function curl_init() in [b]/var/www/Site1/amc/amc/preload.php[/b] on line [b]54[/b][/quote]
I had nothing to do with the above error. :noo: I swear!

wow..
i completely failed
then again....i wuz listening 2 my ipod when i wuz taking it...
and i didn't exactly read each question completely.....
hmm.....

I'm interested in seeing the problems. Mind posting them?


In the following ,$i$ represents  $\sqrt{-1}$

1)If $(a+ib)^5=\alpha+i\beta$,then what is is $(b+ia)^5$ in therms of $\alpha$ and $\beta$(like $\alpha i+\beta$ or $\beta i+\alpha$,etc)

2)If $n$ is an odd integer,then  $(1+i)^{6n}+(1-i)^{6n}$  is equal to?

3)Let $z$ be a complex nuumber of constant modulus such that $z^2$ is purely imainary then the number of possible values of $z$ are?

4)If $3^{49}(x+iy)=[\frac{3}{2}+\frac{\sqrt{3}i}{2}]^{100}$ and $x=ky$,then what is the value of $k$?

5)The smallest positive integral value of $n$ for which  $(1+\sqrt{3}i)^{\frac{n}{2}}$ is real is?

[hide="hint for 5"]Write it in polar form and use de Moivre.[/hide]

what about the 1st one?Any short method?

Just use de Moivre again.

[hide="1"]
Let $a=\cos\theta$ and $b=\sin\theta$. Now $(a+bi)^5=\cos 5\theta+i\sin 5\theta$ and $(b+ai)^5=\sin 5\theta+i\cos 5\theta = \beta+i\alpha$ or something.
[/hide]

In the third one,if we take the complex no. as $x+iy$,the  constant modulus would be  $x^2+y^2=k^2$  or  $x^2+y^2=2k^2$?


limit
logarithms
inequalities
real analysis
real analysis unsolved

Let the real number $a\geq 1$. Prove that: $\lim_{n \rightarrow\infty}\left({2a^{\frac{1}{n}}-1}\right)^{n}=a^{2}$.

A Taylor expansion leads to the solution without any trouble.

:? I'm new at series and limits... What's a Taylor expansion?

You might want to check this : [url=http://mathworld.wolfram.com/TaylorSeries.html]Taylor expansion/series[/url] :)

Come on, such arsenal for this problem, since it follows from Bernoulli's inequality. ;)

I've already checked that :P! But I didn't understand so much. Can anyone write a solution? My book has two other solutions and  one includes Bernoulli inequality. But they are not explainded so well..  :?

Nice solution Moubinool :)

Anyway, I'd really be interested in seeing the one using Bernouilli's ineq.!

Hmm... How did you get from the limit as x goes to zero to the expression immediately following that?

Edit:  Nevermind; L'Hopital right?

[quote="julien_santini"]Approximations using Taylor expansions make (in my opinion) much more sense than using an inequality
"out of the blue" and being lucky enough so that it works. If the Taylor Series are not known, that's of course another story. By the way, Bernoulli's inequality is nothing else than a disguised short Taylor expansion.[/quote]
can u show me how to do it using Taylor? thanx a lot!

A standard method!

if the limit has the form $1^\infty$ we write

     $a^x = e^{xlna}$ and then we use de L' Hospital to find the limit on the exponent.This limit has the form $0.\infty$ and can be found with a new but well known triq.

Babis


AMC
AIME
geometry
geometric transformation
rotation

I know that this is kinda old, but, the older threads post solutions immediately. I wanted to ask for hints on how to solve this problem:

Two of the squares of a  checkerboard are painted yellow, and the rest are painted green. Two color schemes are equivalent if one can be obtained from the other by applying a rotation in the plane of the board. How many inequivalent color schemes are possible?

Firstly, I'm not sure if I understood the problem correctly. Say you have a 2x2 board. If you paint the 2 upper squares, then rotating it 4 times will give you 4 equivalent color schemes.

If you paint the 2 diagonal squares, rotating it twice will give you 2 equivalent color schemes.

Because we are looking for the inequivalent color schemes, do we have $ 4x2\equal{}8$ pairs of them?

[b]Rephrase 1:[/b] How many distinct coloring schemes are possible, up to rotation?

[b]Rephrase 2:[/b] Let S be the set of all coloring schemes. Let M be a subset of S such that every element in S\M (S exclude M) is the rotation of exactly one element of M. Find the number of coloring schemes in M.

[b]In a 2x2 checkerboard:[/b] There are 2. You have
XX
OO

XO
OX

And the rest are all rotations of one of the above.


1)What is the smallest possible integer which uses only the digits 0 and 1 and is divisible by 225? (Your answer must be exact)

2) In quadrilateral ABCD, AC is perpendicular to BC and AD is perpendicular to CD. AB=11. What is the sum of the squares of the lengths of the sides of this quadrilateral?

Better someone else to do the exercises, I just posted those for reference.

Solutions:

[hide="1"]Assuming you meant positive. A number is a multiple of 25 if and only if it ends in $ 25$, $ 50$, $ 75$, or $ 00$. By the problem conditions, the number must end in $ 00$. A number is a multiple of 9 if and only if the sum of its digits is a multiple of 9. We can only have $ 1$s and $ 0$s, and clearly $ 0$s aren't going to help. Since we want the smallest number, we just stick 9 $ 1$s in the number. This gives $ 11111111100$.[/hide]

[hide="2"]We have $ AD^{2}\plus{}CD^{2}\equal{} AC^{2}$ and $ AC^{2}\plus{}BC^{2}\equal{} AB^{2}\equal{} 11^{2}$. Then we want $ AD^{2}\plus{}CD^{2}\plus{}BC^{2}\plus{}AB^{2}$.
$ \equal{} AC^{2}\plus{}BC^{2}\plus{}AB^{2}$
$ \equal{} 2AB^{2}$
$ \equal{} 2(11^{2})$
$ \boxed{242}$[/hide]


If $ 2^n \plus{}1$ is prime number then prove $ n\equal{}2^k$

If $ n$ is odd then

$ 2^n \plus{} 1 \equal{} 2^n \plus{} 1^n \equal{} (2\plus{}1)(2^n\minus{}2^{n\minus{}1} \plus{}2^{n\minus{}2} \minus{} ...\pm 1) \equal{} 3(2^{n\minus{}1} \plus{} 2^{n\minus{}3}...\plus{}1)$ not prime

So let $ n\equal{}2m$, but then we have; $ 4^m \plus{} 1^m$ and the same argument applies

So we can continue like this foremver until we find $ n\equal{}2^k$

[hide]
[b]Lemma 1:[/b] For any positive integer $ x$ and odd positive integer $ n$, the number $ x^n \plus{} 1$ is prime if and only if $ x \plus{} 1$ is prime and $ n \equal{} 1$.
[i]Proof.[/i] If $ n$ is odd, then we have the factoring $ x^n \plus{} 1 \equal{} (x \plus{} 1)(x^{n \minus{} 1} \minus{} x^n \plus{} \dots\pm 1)$, which is prime if and only if $ x \plus{} 1$ is prime and $ x^{n \minus{} 1} \minus{} x^n \plus{} \dots\pm 1 \equal{} 1$, which is only true if $ n \equal{} 1$.

If $ 2^n \plus{} 1$ is prime and $ n$ is odd, then $ n \equal{} 1$ by Lemma 1, which is indeed in the form $ 2^k$ when $ k \equal{} 0$.

So suppose $ n$ is even. Let $ M$ be the largest power of 2 that divides $ n$ and let $ b \equal{} \frac nM$. Then, we have that $ (2^M)^b \plus{} 1$ is prime, so (on top of the fact that $ 2^M\plus{}1$ is prime) $ b$ must be $ 1$. The result follows directly.
[/hide]

[quote="enndb0x"]If $ 2^n \plus{} 1$ is prime number then prove $ n \equal{} 2^k$[/quote]
Another good and well known question is:
Prove that there are infinetely many numbers $ n$ so that $ n|2^n\plus{}1$ :wink:

Take $ n \equal{} 3^k$ for integers $ k \geq 0$, lifting the exponent, qed. (This was a problem on my blog that pythag011 took quite a while to solve :P)


inequalities
induction
inequalities proposed

It is not trivial, but the strongest inequality of this type is the following

$\displaystyle n^2(\frac{1}{a_1}+...+\frac{1}{a_n})\geq 4(n-1)(a^2_1+...+a_n^2)+n(n-2)^2$.

Whoa :o ,this inequality is too strong and too nice.Do you create it,Vasc?
I have try to use the method that I have used for the old one,but it doesn't work :cry: .
However,it is so great because I have a solution now :lol:(Hey this face looks too old) .
Everyone,try it too.Trust me,this one is very nice and fantastic. :)

I don't think "the strongest inequality of this type" is particularly difficult.

Observe that $\frac{n^2}{x} - 4(n-1)x -(n^2-4(n-1) + (x-1)(-n^2-(8n-1))) = \frac{4(n-1)(x-1)^2(2n+14+\frac{8}{n-1} - x)}{x} \geq 0$ for $0<x \leq n$, with equality iff $x = 1$.  It follows that

$\sum_{i=1}^n (\frac{n^2}{a_i}- 4(n-1)a_i^2) \geq \sum_{i=1}^n (n^2 - 4(n-1)+(a_i-1)(-n^2-(8n-1))) = n^3 - 4n^2 + 4n = n(n-2)^2$, with equality iff $a_1 = a_2 = ... = a_n = 1$.

A splendid problem!!!  I will prove the homogenuous inequality:
    $n\cdot (a_1+...+a_n)\cdot (\frac{1}{a_1}...+\frac{1}{a_n})\geq\ n(n-2)^2+\frac{ 4n^2(n-1)(a_1^2++a_n^2)}{(a_1++a_n)^2}$ by induction. The case n=2 is easy. So, suppose the inequality is true for n-1 and let us prove it for n. Now, we can homogenise it and it remains to show that $ \frac{1}{a_1}+...+\frac{1}{a_n}\geq 4n(n-1)(a_1^2+...+a_n^2)+(n-2)^2 $ if $ a_1+...+a_n=1 $. I will only prove how we can reduce this one to the case when n-1 variables are equal since after that its not very hard. So, let us take $ a_n$ be the greatest among the n variables and let us take $ f(a_1,,a_n)= LHS-RHS $. Let also A be the arithmetic mean of $ a_1,...,a_{n-1}$. I will prove that $ f(a_1,,a_n)\geq f(A,...,A,a_n) $. Of course, this comes down to $ \frac{1}{a_1}+...+\frac{1}{a_{n-1}}-\frac{(n-1)^2}{a_1+...+a_{n-1}}\geq 4n(n-1)(a_1^2+...+a_{n-1}^2)-4n $. (**) Now, we see that $ a_1++a_{n-1}\leq\frac{n-1}{n} $. Also, both LHS and RHS are nonnegative in (**) and it is very easy to see that we may in fact assume that $ a_1+...+a_{n-1}=\frac{n-1}{n} $ ( the LHS decreases and the RHS increases if we put $ a_i=kx_i $ with $ k\leq 1$ and $ x_1+...+x_{n-1}=\frac{n-1}{n} $) . Now, we use the inductive hyphotesys for $ a_1,...,a_{n-1}$ and we find that $\frac{1}{a_1}+...+\frac{1}{a_{n-1}}\geq \frac{4n^3(n-2)}{(n-1)^2}\cdot (a_1^2+...+a_{n-1}^2)+\frac{n(n-3)^2}{n-1} $ and its easy to see that this inequality implies (**) (here we need Cauchy-Schwarz to find that $ a_1^2+...+a_{n-1}^2\geq \frac{n-1}{n^2} $). Thus, the inductive step is reduced to a one variable inequality, which isnt extremely difficult. Super-nice problem!!!

I would like to see also Vasc and Anh Cuong solutions of this very hard and very nice inequality.

Thank you very much.

Dear manilo,I have used induction and it is similar to Harazi's one.I just want to add that my first idea for this one is using those formula:
[tex] n(a_1^2+...+a_n^2)-(a_1+...+a_n)^2 =\sum(a_i-a_j)^2 [/tex]
[tex] (\frac{1}{a_1}+..+\frac{1}{a_n})(a_1+...+a_n) -n^2 =\frac{ \sum(a_i-a_j)^2}{a_i \cdot a_j} [/tex]
Use them and the first inequaltiy that I posted is trivial. :)

Harazi solution is nice but very hard to find and last one variable inequality is difficult to solve for me.

Lipnowsky approach is interesting but if you try to prove that:

$ n^2 -4(n-1)x^2 -(n^2-4(n-1))x +x(x-1)(-n^2-8n+1) \geq 0$

you find

$ x^2(n^2+4n+3) -x(2n^2+4n+3) +n^2  \geq 0 $

which is true only if  $ x\geq 1$ and that doesn't match the condition of the problem     $ a_1+.....+a_n =n$

Do you think it is possible to find a more elementary solution or Harazi solution is the only ?

I cannot explain it, since it's wrong. Anyway, the solution is good, with all those typos present in the solution.

Yes, I verified now that your solution remains valid.
Super-solution ! :lol:

Vasc, can you please post the proof of

$ f(A,A,...,A,a_n) \geq 0 $  ?

Thank you very much.

Sorry, Manlio.
But it is the proof of Harazi.
 :)


geometry
3D geometry
tetrahedron
sphere
triangle inequality
angles
IMO Shortlist

Prove that the sum of the six angles subtended at an interior point of a tetrahedron by its six edges is greater than 540°.

I will give you a solution due to Klamkin to that problem :

Lemma.
A point O lies in the base ABC of a tetrahedron PABC.
Then, the sum of the angles formed by the line OP and the edges PA, PB, PC is less than the sum of the faces angles at vertex P.

Proof.
We will use the duality between trihedral angles and spherical triangles. First note that  it suffices to assume that PO is an interior ray of a trihedral angle whose edges are the rays PA, PB, PC. With $ \alpha , \beta , \gamma $ denoting the angles BPC, CPA, APB respectively, and $ x,y,z $ the angles OPA, OPB, OPC respectively, we want to prove that $x+y+z < \alpha + \beta + \gamma$.

Draw an unit sphere centered at P, intersecting the ray PO at O'. The points A',B',C' are defined in a similar way.
Clearly the angle OPB is equal to the angle O'PB', and we have similar equalities for the other angles.
Now, the intersection of the sphere with the trihedral angle is a spherical triangle A'B'C' which can be triangulated by great circle arcs meeting at O'. If all the faces angles of the trihedral angles are mesured in radians, then (the radius of the sphere being unity) the lengths of all connecting arcs are :
arc B'C' = $ \alpha $, arc C'A' = $ \beta$ and arc A'B' = $\gamma$,
arc O'A' = $ x$, arc O'B' = $y$ and arc O'C' = $z$.
Let arc O'B' and arc A'C' meet at B''.
Since all the spherical triangles we are concerned with here are convex, we have :
arc O'B'' + arc B'C' > $z$ and $ \gamma $ + arc A'B'' > $y$ + arc OB''
Adding theses two inequalities, we obtain $ \beta + \gamma > y + z$.
We deduce similar inequalities for the other angles. Summing leads to the desired result.

Now, we return to the original problem :
Let ABCD be the tetrahedron and P the interior point. let A' be the intersection of line AP with face BCD.
Let S be the sum of the angles APB, APC and APD.
Let S' be the sum of the angles BPC, CPD and DPB.
Let S'' be the sum of the angles A'PB, A'PC and A'PD.
Then :
Clearly S + S'' = 3$\pi$ and, using the lemma, we have S' > S''.
Thus S + S' > 3$ \pi $, and we are done.

Pierre.

Pierre, thanks but there is a point I don't quite understand:

[quote="pbornsztein"]If all the faces angles of the trihedral angles are mesured in radians, then (the radius of the sphere being unity) the lengths of all connecting arcs are :
arc B'C' = $ \alpha $, arc C'A' = $ \beta$ and arc A'B' = $\gamma$,
arc O'A' = $ x$, arc O'B' = $y$ and arc O'C' = $z$.
Let arc O'B' and arc A'C' meet at B''.
Since all the spherical triangles we are concerned with here are convex, we have :
arc O'B'' + arc B'C' > $z$ and $ \gamma $ + arc A'B'' > $y$ + arc OB''
Adding theses two inequalities, we obtain $ \beta + \gamma > y + z$.[/quote]

How? Maybe, there are some typos, or I am just blind?

Also, some very fundamental questions (sorry, I am a beginner in stereometry):

- When is a spherical triangle called convex?
- How to prove the triangle inequality for spherical triangles?

Thanks!
  Darij

1) I think that since we start from a tetrahedron and an interior point, the resulting spherical triangle is contained in an half-sphere. Thus, it is convex.

2) The great circles arcs are paths of minimal length between two points on the surface of a sphere. Thus :
arc O'C' < arc O'B'' + arc B''C', since O'C' is a great circle arc.
and
arc O'B'' + $y$ = arc O'B'' + arc O'B' = arc B'B'' < arc B'A' + arc A'B'' = $ \gamma$ + arc A'B'', since B'B'' is a great circle arc, which pass through O'.

Adding these two inequalities, we obtain :
arc O'C' + arc O'B'' + $y$ < arc O'B'' + arc B''C' + arc A'B'' + arc A'B'.
But arc A'B'' + arc B''C' = arc A'C' = $ \beta $, arc O'C' = $z$ and arc A'B' = $ \gamma$
Thus, eliminating arc O'B'', it leads to $z + y < \beta + \gamma$.

Pierre.

Thanks a lot, I have now understood your proof. And a polygon on the sphere is called convex if for any two points P and P' in its interior, the smaller of the two great circle arcs PP' completely lies inside the polygon, isn't it?

  Darij


A very bizarre weighted coin comes up heads with probability $ \frac12$, tails with probability $ \frac13$, and rests on its edge with probability $ \frac16$.  If it comes up heads, I win 1 dollar.  If it comes up tails, I win 3 dollars.  But if it lands on its edge, I lose 5 dollars.  What is the expected winnings from flipping this coin?  Express your answer as a dollar value, rounded to the nearest cent.

This problem is an excellent example of how one calculates the "house edge" of casino games. Las Vegas: the town that applied probability theory built!  :D

MrTeacher

[quote="yankeeprez"]I don't know if I am correct, but this is how I would go to solve the problem.

I multiply the probability of getting each event by the value of money it is worth, then I add them together.

So:

1/2(1)+1/3(3)+1/6(-5)                    It is -5 because you are losing that money.

=1/2+1+(-5/6)
=3/2-5/6
=9/6-5/6
=4/6
=2/3
2/3=.66
So that is 66.6 cents which rounds to [hide][b][u]67 Cents[/hide][/u][/b][/quote]

Good job! The only correction is that it says dollar value, so it would be 0.67


geometry
rectangle
FTW
Pythagorean Theorem

If the 9-inch by 12-inch piece of paper shown was first folded flat along the dotted line $ EF$, then folded flat along the dotted line $ GH$, how many inches closer would $ A$ be to $ C$.

[asy]draw((0,0)--(12,0)--(12,9)--(0,9)--cycle);
draw((0,3)--(12,3),dashed);
draw((8,0)--(8,9),dashed);
label("A",(0,0),SW);
label("H",(8,0),S);
label("D",(12,0),SE);
label("E",(0,3),W);
label("F",(12,3),E);
label("B",(0,9),NW);
label("G",(8,9),N);
label("D",(12,9),NE);
draw("3", (4,0)--(4,3), Bars, Arrows, PenMargins);
draw("4",(8,6)--(12,6),Bars,Arrows,PenMargins);
draw("6",(-0.5,3)--(-0.5,9),W,Bars,Arrows,PenMargins);
draw("8",(0,9.5)--(8,9.5),N,Bars,Arrows,PenMargins);[/asy]

Mods, Admins-- There's an error with this question. No C! Please fix it.

I'm guessing that $ E$ is supposed to be the intersection point of $ EF$ and $ GH$?

Maybe C is the intersection that you just mentioned.

Anyways moderators please fix these errors.

C is the first D so that the rectangle reads ABCD.  This is consistent with the answer that FTW gives.  A is 15 away from C at the start because of the pythagorean theorem, 9-12-15.  Afterwards A is 5 away from C, 3-4-5.  So the answer is 15-5=10


topology
function
inequalities
triangle inequality
real analysis
real analysis unsolved

Let $ (X,d)$ metric space , and $ F(X)$ set of all closed sets from $ X$. Let $ U(A,\epsilon): =\{x\in X;d(x,A)<\epsilon\},\epsilon >0$. Prove that function $ d'(A,B): =\inf \{\epsilon ;\epsilon >0, A\subseteq U(B,\epsilon), B\subseteq U(A,\epsilon)\},A,B\in F(X)$, is metric on $ F(X)$, so called Hausdorff metric. All the properties except the triangle inequality are obvious.

And the triangle inequality is obvious too.

This "metric" may assume infinite values unless $ X$ is bounded.


[quote="jelly119"]Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100.  S is given the value x+y and P is given the value xy.  They then have the following conversation.

[b]P:  I cannot determine the two numbers.
S:  I knew that.
P:  Now I can determine them.
S:  So can I.[/b]

Given that the above statements are true, what are the two numbers?[/quote]
what dyou mean by $\text{S is given the value x+y??}$ :?
And how on earth does the conversation help???!!! :wallbash_red:  :wallbash:

[quote="riddler"]
what dyou mean by $\text{S is given the value x+y??}$ :?
And how on earth does the conversation help???!!! :wallbash_red:  :wallbash:[/quote]

First of all, trivially, xy cannot be prime.  It also cannot be the square of a prime, for that would imply x = y

P's statement implies that xy cannot have exactly two distinct proper factors whose sum is less than 100.

Think about the rest three of these conversation....

[hide]Not to spoil the fun but you may ask my [url=http://artofproblemsolving.com/Forum/viewtopic.php?highlight=sanjay&t=9979]nephews[/url][/hide]

Ok here is how I work out

[hide="click to begin"]
First of all , both of the logicians know that $5\leq x+y\leq 99$ since $x$ has smallest value $2$ while $y$ has smallest value $3$ .So there are $95$ possible answer for $x+y$ now.

(i)$P$ first statement :arrow: $x$ and $y$ are not both prime (or else $P$ would have guessed it )

(ii)$S$ first statement :arrow:  $x+y$ cannot be even(=odd) or else $S$ wont be sure that $P$ doesnt know the answer since even number $can$ be made up by adding two prime numbers . So from here we can cancel all $even$ number in the set [5,99] . But , sum of two primes can also be in form $2+\text{odd prime}$ , so all $prime+2$ in the remaining set can be cancel as well . This left only $25$ possible answer for $x+y$ for now .

Up to here , $x+y$ can be $11,17,23,27,29,35,37,41,47,51,53,57,59,65,67,71,75,77,79,83,87,89,93,95,97$

(iii)$P$second statement :arrow: Indicate that for all the possible combo of sum of both $x$ and $y$ , there is only one which sum up to be an $odd$ . Why ? Because after he heard $S$'s statement that all $x+y$=even is not possible , he knows both $x$ and $y$ if and only if after he cancels all the even sum there is exactly $one$ odd sum .

(iv)$Q$second statement :arrow: He can determine $x$ and $y$ if all the combo of sum of the number he received have exactly one type of product such that it consist of only a single odd prime(not even with square!) .

Lets look at $11$ , all combo of sum here are (2,9) , (3,8) ,(4,7) , (6,5) .
For (2,9) ,the product is 18 .But 18 can be factor into 2x9 , 3x6 and notice that both of them add up to odd number so (after fixing himself in the situation of $P$) $S$ wont be able to determine the number.So (2,9) cannot be the pair .
For(3,8),the product is 24 .But 24 can be factor into 2x12,3x8,4x6 which you could see there are only one possible comeout to have odd sum which is 8+3=11.
For (4,7) , similarly the product are 2x14,4x7 where only one odd sum obtained.
Now ,since there are two possible outcome of odd sum for $11$ . That means $S$ wont be able to detemine the number .So $11$ is out .
(A little tip here is you only needa check for combo (4,k) or (8,k) or (16,k) ..... ($2^n,k$) . Because in this way , you will surely get another even factor if you find the product. And even factor is the only way for $P$ to cancel off all the combo except one) .

Ok so after checking all the integer set with those property , I conclude that (4,13) , (13,16) , (4,37) , (16,43) , (4,61) , (16,73) , (8,89) would be the pairs for $x,y$ .

That is why I said previously in my post that if we have to know at least one number given to $S$ or $P$ so that we can have the $ONLY$ answer .  :roll:
[/hide]

Well , correct me if theres something wrong there  :P  :P  :P  :P

[quote="shyong"]Weird , how come I found more than one solution ! :?
Could it be at least we should have to know what is the number of $S$ or $P$ . Because there are $>1$ pairs of numbers which follows the property above .

[hide](4,13) , (16,13) , (4,37) , (16,43) , (4,61) , (16,73) , (8,89) . If you put yourself as either one of the two person , then you would get your answer among these 7 pairs of number . But since we are third person here , so I guess there would appear more solutions to us :P [/hide][/quote]

There are more than one answers, but not all the answers you have given will work.

For example, (16,13) will not work, Here Sum = 29 and  although S can say first "I knew that" because 29 can not be written as sum of two primes. But S can NOT  find the numbers (even after P says he "now knows the numbers) because   the  numbers could be (16,13) OR (2,27) . P would know the answer if the product is 54 too (not only for  16*13) .. (as factors of 54  are  2*27, 3*18,6*9 and P would know it is 2*27).

(4,61) and  (16,73)  from your notes, would work, others will not.

Side note: As the question posed above, (That is if the restriction is  x+y<100)  there are more than  one answers,  (Namely (4,61) , (4,13) and (16,73)  [b]That's why my "nephews" question (Link in previous message) I restrict  both the  numbers  to be less than 50[/b], and then the solution is unique... (For fun one may even restirct the numbers to be less than 20  and still keep the problem interesting)

(Long ago I tried to prove that one of the  numbers  has to be of the form  $2^n$ but that's not correct too.. (40,937) works too!)

Dear Gyan ,

Sorry cause I cant see why $P$ would know the answer if the product i.e. is $54$ .
$54$ can be in 2x27,3x18,6x9 . But according to what I have used , first $P$ observes that it is not product of two distinct prime . So he wont know. Then according to $S$, I interpret what he said as the sum must not be even and it cannot be sum of two distinct prime . But here all 2x27 , 3x18 , 6x9 add up to become odd so in this case , $P$ wont be able to guess what number is it . So accordingly $xy$=$54$ cannot be true .  Do I interpret what $S$ says wrongly or do i miss something else ? :?

[quote]But here all 2x27 , 3x18 , 6x9 add up to become odd so in this case ,[/quote]
...But you see  21 (=3+18) and  15 (=6+9) , even though  odd, can be written as sum of two primes.. only 29 can not be written as sum of two primes...

Following  your own argument number II
[hide="given above"][quote](ii)S first statement Arrow x+y cannot be even(=odd) or else S wont be sure that P doesnt know the answer since even number can be made up by adding two prime numbers . So from here we can cancel all even number in the set [5,99] . But , sum of two primes can also be in form 2+\text{odd prime} , so all prime+2 in the remaining set can be cancel as well . This left only 25 possible answer for x+y for now .

Up to here , x+y can be 11,17,23,27,29,35,37,41,47,51,53,57,59,65,67,71,75,77,79,83,87,89,93,95,97[/quote]
Note that In the list above, not only all even numbers,  but 21 and 15 are also missing...:)  (and 29 is in the list)

[/hide]

[hide="More details"]  ...

Assuming P=54, and  S=29  You see the first [b]three[/b] statements.:

 P: I cannot determine the two numbers. === Okay... (Since there are three possibilities) .
S: I knew that. ==== Okay  (Since 29 is  can not be written as sum of 2 primes)
P: Now I can determine them. ===Okay  (Becuase  - We are going only from P's point of view - out of three possible sums (29,21,15) only 29 can not be wrrtten as sum of two primes - 21 (19+2) and 15 (2+13) can be written as sum of two primes and thus can be ruled out - from P's point of view. If the sum was 21 (or 15) , then S  could not make the above statment "I knew that"  .. right!

Similarly if P=16*13 and S=29  The above three statments would be same.
[hide="details"]
P: I cannot determine the two numbers. === Okay... (Since there are four possibilities) .
S: I knew that. ==== Okay  (Since 29 is  can not be written as sum of 2 primes)
P: Now I can determine them. ===Okay  (Becuase  - We are going only from P's point of view - out of four  possible sums (29,34,56,106) only 29 can not be wrrtten as sum of two primes - .. other than 29, all are even and thus can be written as sum of two primes .... right!

[/hide]
[/hide]
So knowing only S=29, and not knowing P, S can not say the last statement  "So can I" because from his point of view
P could be 54 or  208)

Hope this helps.
Let me know if I missed something...

[quote="Gyan"][quote]But here all 2x27 , 3x18 , 6x9 add up to become odd so in this case ,[/quote]
...But you see  21 (=3+18) and  15 (=6+9) , even though  odd, can be written as sum of two primes.. only 29 can not be written as sum of two primes...

Follow your own argument number II
[hide-given above]([quote]ii)S first statement Arrow x+y cannot be even(=odd) or else S wont be sure that P doesnt know the answer since even number can be made up by adding two prime numbers . So from here we can cancel all even number in the set [5,99] . But , sum of two primes can also be in form 2+\text{odd prime} , so all prime+2 in the remaining set can be cancel as well . This left only 25 possible answer for x+y for now .

Up to here , x+y can be 11,17,23,27,29,35,37,41,47,51,53,57,59,65,67,71,75,77,79,83,87,89,93,95,97[/quote]
Note that In the list above, not only all even numbers,  but 21 and 15 are also missing...:)

[/hide]

More details ...

Assuming P=54, and  S=29  You see the first [b]three[/b] statements.:

 P: I cannot determine the two numbers. === Okay... (Since there are three possibilities) .
S: I knew that. ==== Okay  (Since 29 is  can not be written as sum of 2 primes)
P: Now I can determine them. ===Okay  (Becuase  - We are going only from P's point of view - out of three possible sums (29,21,15) only 29 can not be wrrtten as sum of two primes - 21 (19+2) and 15 (2+13) can be written as sum of two primes and thus can be ruled out - from P's point of view. If the sum was 21 (or 15) , then S  could not make the above statment "I knew that"  .. right!

Similarly if P=16*13 and S=29  The above three statments would be same.
[hide="details"]
P: I cannot determine the two numbers. === Okay... (Since there are four possibilities) .
S: I knew that. ==== Okay  (Since 29 is  can not be written as sum of 2 primes)
P: Now I can determine them. ===Okay  (Becuase  - We are going only from P's point of view - out of four  possible sums (29,34,56,106) only 29 can not be wrrtten as sum of two primes - .. other than 29, all are even and thus can be written as sum of two primes .... right!

[/hide]

So knowing only S=29, and not knowing P, S can not say the last statement  "So can I" because from his point of view
P could be 54 or  208)

Hope this helps.[/quote]

Oh ya , I think I get it now  :D  I forget that sum of two primes not only gives us even sum but also odd sum as well when $2$ comes into consideration .I thought I already cancel all those $prime+2$ possibility but Im wrong .....  :oops:  Well , thanks Gyan , I appreciate your help . :D


algebra
polynomial
search
superior algebra
superior algebra unsolved

Find a condition on $ n\in \mathbb{N}$ s.t. there exists a monic irreducible polynomial of degree $ n$ in $ \mathbb{Z}[X]$ that is reducible modulo all prime.

The title seems to suggest that you are looking for conditions on $ n$ such that there exists a
monic, [b]irreducible[/b] polynomial of degree $ n$...

Posted before, the condition is: $ n$ has to be composite.

Thanks ZetaX, posted by harazi of course:)
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2140567689&t=34935


There will be a lecture by Prof. Leo Schneider on problem solving.  It will be on Saturday February 6, 2010 at Columbus State, from 10AM for at least 2 hours.  The lecture will be at AMC12/ARML level.

Prof. Schneider is from Math Dept. of John Carroll University, and is an officer of Pi Mu Epsilon, the Honor Mathematics Society.  Prof. Schneider is a seasoned speaker; he also is an author of AMC Contest Problem Book.  A few years ago, he was the Director of AMC contest for the State of Ohio; currently, he is a member of the writing team for ARML and NYSML.

This lecture is open to anyone interested.

Directions to the college and parking can be found by
Campus Map: http://www.cscc.edu/Tour/mapscroll.htm

Directions: http://www.cscc.edu/DOCS/visitor/directions.htm

Alan

Unfortunately I can't go due to a prior commitment :(
Is it possible for lecture notes/problems to post online as it would serve good review for the 12A exam :lol:

ooh, this sounds fun. Do we have to let you know ahead of time if we can go?

No, you don't have to let me know.
However, I prefer knowing you are coming or not, since you probably are not yet a member of Ohio ARML.

My e-mail: ayang@cscc.edu

Ah, sorry, I just realized I can't go. It's on the same day as Science Bowl.

Please let us know which room and building the lecture will be in ?

Due to the winter storm that's going to hit us Fiday and Saturday, the lecture has been postponed to 02/20 9also a Saturday).

It still will be held at 4th floor Davidson Hall (at the northeast corner of Mt. Vernon Ave. and Spring St.)  You can park in the big lot at the northwest corner of Mt. Vernon Ave. and Spring St., across from the Davison Hall.

Please come straight up to the 4th floor, I'll greet you there.

Directions to the college and parking can be found by
Campus Map: http://www.cscc.edu/map/EC.htm

Directions: http://www.cscc.edu/Visitor/directions.htm

Alan

Just making sure.

The new date is Februar 20, 2010.

Feb 20, 2010 will also be the date for Mathcounts Chapter Competition. Location is also at Columbus State Community College.

Any one interested is welcome!

It will be held at DH408, (4th floor Davidson Hall, at the northeast corner of Mt. Vernon Ave. and Spring St.) You can park in the big lot at the northwest corner of Mt. Vernon Ave. and Spring St., across from the Davison Hall.

Please come straight up to the 4th floor, I'll greet you there.

Directions to the college and parking can be found by
Campus Map: http://www.cscc.edu/map/EC.htm

Directions: http://www.cscc.edu/Visitor/directions.htm

Alan


inequalities
modular arithmetic
algorithm
induction
number theory
number theory unsolved

We have $2p-1$ integer numbers, where $p$ is a prime number. Prove that we can choose exactly $p$ numbers (from these $2p-1$ numbers) so that their sum is divisible by $p$.

This is the Erdos-Ginzburg-Zif Theorem (actually, the theorem is stated identically, only for arbitrary $n$, not merely primes $p$).

We can prove it by using the Cauchy-Davenport inequality, for example: if $A,B$  are subsets of $\mathbb Z_p$, then $|A+B|\ge\min(p,|A|+|B|-1)$. Partition the multiset of $2p-1$ residues $\pmod p$ in $p-1$ classes of size $2$ (call them $A_1,\ldots,A_{p-1}$) and one class of size $1$ (${x}$). Now apply the Cauchy-Davenport to get $|A_1+A_2+\ldots+A_{p-1}|\ge p$. This means that we can find $p-1$ elements of our multiset which add up to $-x\pmod p$. Adding $x$ to this yields the result.

A very instructive topic: [url]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=combinatorial+nullstellensatz&t=19496[/url]

You can find something useful in http://www.mathlinks.ro/Forum/viewtopic.php?t=14797

[quote="grobber"]

$|A_1+A_2+\ldots+A_{p-1}|\ge p$. This means that we can find $p-1$ elements of our multiset which add up to $-x\pmod p$.
[/quote]
Sorry, but more explanation ? If x is divisible to 0 then we have this problem : in 2p-2 numbers, we can chose p-1 number which sum is divisible to p ?

If all $2p-2$ numbers are 1, then it is impossible to choose $p-1$ numbers, whose sum is divisible by $p$.  :?
What does actually your question mean?

Hmm, may be i have lost something, but myth, your example is one of my problem. If we have 2p-2 numbers 1 and 1 number 0 , then if we choose a_2p-1 = 0, and follow grobber's solution, it is wrong.

We are to make one small correction/addition to grobber's solution. Note if we unable to divide $2p-1$ numbers into $p-1$ sets of 2 numbers and 1 sets of 1 number s.t. all 2 element sets contain distinct numbers (i.e. there are no sets of form $\{1,1\}$), then there are at least $p$ equal numbers, so we are done. Otherwise apply grobber's arguments.

[quote="Myth"]Note if we unable to divide $2p-1$ numbers into $p-1$ sets of 2 numbers and 1 sets of 1 number s.t. all 2 element sets contain distinct numbers (i.e. there are no sets of form $\{1,1\}$), then there are at least $p$ equal numbers[/quote]
I think it is not so trivial. Or I don't see something really obvious. Please write a simple proof if you have it.

Yes, guys, I'm sorry about this slip-up :).

Here's a maybe not so simple but cool way of showing that what Myth said works :):

On our multiset with $2p-1$ elements, define a partial order $\le$ by setting $x\le y\iff x=y$. By Dilworth's Lemma, if no chain has length $>p-1$, then the set can be expressed as the union of $p-1$ disjoint antichains, and from this partition we can easily extract a partition in one $1$-element set and $p-1$ $2$-element sets.

[quote="grobber"]define a partial order $\le$ by setting $x\le y\iff x=y$.[/quote]
Excuse me, but this is not a partial order. We have to show that if $x \leq y$ and $y \leq x$, then $x=y$ (as elements of our multiset). But this is not true. I suppose that the confusion apeared from more than one meaning of the sign "=".

[quote="Myth"]Are you kidding?  :?[/quote]
No! I still don't know an easy proof.

[quote="Remike"][quote="grobber"]define a partial order $\le$ by setting $x\le y\iff x=y$.[/quote]
We have to show that if $x \leq y$ and $y \leq x$, then $x=y$ (as elements of our multiset). But this is not true. I suppose that the confusion apeared from more than one meaning of the sign "=".[/quote]

You are right  :blush:. I don't know what the heck I was thinking about. Indeed, that's why I got confused  :blush:.

However, I still think the result is very intuitive and easy to prove, but I may be wrong yet again.  :blush:

Take out any element of the $2p-1$-element multiset. We're left with $2p-2$, and we want to pair them up so that no two in the same pair are equal. In order to do this, we divide them in two groups of $p-1$ as follows: put in one group all elements of the type appearing the most times among the $2p-2$, and then continue to add elements in their decreasing order of the number of appearances (if we add a $1$, for instance, we try to add as many $1$'s as possible, i.e. either until we finish the $1$'s or until we have $p-1$ elements in the first group).

If in the first group we only have elements of one type, then we are done, because the other $p-1$ elements are of different types from the type in the first group (assuming there are no $p$ equal elements). On the other hand, if in the first group there are at least two types of elements, there's at most one type from which we have elements in both groups of $p-1$ elements. We can pair the elements in this type from the second group with those of the first type in the first group, because by our construction there are more of these latter ones. After that, the elements we are left with in the second group are of types different from those of the elements in the first group, so we can pair them up however we want.

It may have been a bit hard to explain, but I think it's Ok and it can be considered simple.

This is pretty much what I explained, but with more words, you know :).

[quote="Myth"]I am shocked.
Write $2p-2$ numbers in a row grouping all equal numbers:
0, 0, ..., 0, 1, 1, ..., 1, ..., $p-1$, ..., $p-1$
Divide now it into two equal part of $p-1$ elements and write them one under another:
[code]0, 0, ..., 0, 1, 1, ..., a, a,
a, a,           ...         p-1[/code]
Take columns as 2-element sets.[/quote]
This is just the same what grobber has written in the previous post.

[quote="Myth"]And do you think I read what grobber wrote?  ;)[/quote]
I suppose sometimes it is very usefull to read what others write.

[quote="Myth"]"read" тут ?тоит в прошедшей форме, по?тому ? имел ввиду ?тот конкретный по?т. Я очень редко читаю длинные ?кучные объ??нени?. Я думаю, дл? форума ооочень полезно находить короткие пон?тные аргументы, идею которых ?хватываешь ? первых ?лов.[/quote]No Russian in here, Mikhail! Please edit your statement and translate it into English :) Thanks!

About the original problem, here is a solution that I think is nice :

Let $a_1, \cdots , a_{2p-1}$ be the given integers. Assume, for a contradiction, that no $p$ among them add up to $p$.
Then let $S = \sum (a_{i_1} + \cdots + a_{i_p})^{p-1}$ where the sum is over all the $p$-subsets $\{i_1, \cdots , i_p \} \subset \{1, \cdots , n \}.$
There are $t = C_{2p-1}^p $ such subsets. Note that $t = \frac {(2p-1)(2p-2) \cdots (p+1)} {(p-1)!} \neq 0 \mod [p].$
From Fermat's little theorem, we then have $S = 1 +  \cdots  + 1 = t \neq 0 \mod [p].$

In another hand, expanding each of the terms $(a_{i_1} + \cdots + a_{i_p})^{p-1}$ in $S$ will lead to a sum of numbers of the form $a_{i_1}^{e_{i_1}} \times \cdots \times a_{i_p}^{e_{i_{p}}}$, where the $e_i$'s are non-negative integers which add up to $p-1.$ For a given number of that form, say $a_1^{e_1} \cdots a_r^{e_r}$, where $r$ denotes the number of $a_i$'s which appear with a positive exponent : This number appear in the sum $S$ as many times as we may choose $p-r$ of the $a_i$'s among the $2p-1-r$ remaining (those which have an exponent equals to $0$). It follow that the coefficient of $a_1^{e_1} \cdots a_r^{e_r}$ in $S$ is $C_{2p-1-r}^{p-r} = \frac {(2p-1-r) \cdots p}{(p-r)!} = 0 \mod [p].$
Since each coefficient is divisible by $p$, it follows that $S = 0 \mod [p].$ A contradiction.

Pierre.

It is written in the mentioned topic devoted to Combinatorial Null...

[quote="hardsoul"]I have seen this problem before and the problem is valid for all naturals.
The most difficult part was to prove it for primes(then for composite numbers it can be proved  by an induction).[/quote]
Yes, we repeat it everytime we see this problem.  :lol:  :P

Prove that, among $32$ integer numbers, one can always find $16$ whose sum is divisible by $16$.

Write $m = pk$ for a prime $p$, define set addition by $A + B = \{a+b \pmod{p} | a\in A, b\in B\}$, and suppose the $2m-1$ given integers are $n_1, \dots, n_{2m-1}.$ Let $T_1$ be a set of $p$ elements composed from $n_1 + \{n_2, n_3 \} + \cdots + \{n_{2p-2}, n_{2p-1} \}$ with sum divisible by $p$, whose existence is guaranteed by the Cauchy-Davenport lemma where we have arranged for $n_{2l} \ne n_{2l+1} \pmod{p}$ (many algorithms exist - for one, take first and second most common residues of remaining items), or otherwise simply choosing $p$ of the same residue, which concurrently proves the theorem for primes - our base induction case. Form $T_{n+1}$ by composing a similar set addition with $2p-1$ of the $2m-1-pn$ elements in $\{n_1, n_2, \dots, n_{2m-1}\} - \{\cup_{i=1}^{n} T_i\}$ until we have $T_1, T_2, \dots, T_{2k-1}$, and now write $\sum_{d\in T_i} d = p \cdot S_i$ for $i = 1, 2, \dots, 2k-1$. By induction, however, we can choose $S_{a_1} + \cdots + S_{a_k}$ so that $k | \sum_{i=1}^{k} S_{a_i}$, so that $m = pk | p\sum_{i=1}^{k} S_{a_i} = \sum_{i=1}^{k} \sum_{j=1}^{p} \left[T_{a_{i}}\right]_j$.


Source: 2005 AMC-10B #8

An $8$-foot by $10$-foot floor is tiled with square tiles of size $1$ foot by $1$ foot. Each tile has a pattern consisting of four white quarter circles of radius $\frac1 2$ foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the floor are shaded?

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=606[/img]

$\text{(A)}\ 80-20\pi \qquad\text{(B)}\ 60-10\pi \qquad\text{(C)}\ 80-10\pi \qquad\text{(D)}\ 60+10\pi \qquad\text{(E)}\ 80+10\pi$

[hide]
Since the area of a tile is $1$, and the area of one of the white quarter-circles is $\frac{\pi}{16}$, so the area of the black region on a tile is $1-\frac{\pi}{4}$.  Multiply this by $80$ tiles to get $80-20\pi$.
[/hide]

[hide]The four white quarter circles make a circle with radius 1/2. So the area of white on each tile is $\frac{\pi}{4}$. The area of the square is 1, and the shaded area on each tile is $1- \frac{\pi}{4}$. There are 80 tiles, so the total shaded area is $80-20 \pi$. The answer is [color=green]A[/color]. [/hide]

[hide]We can first find the area of the shaded region in one tile, then multiply it by 80. The shaded region is the area of a tile minus the area of a circle with radius 1/2, so it is 1-pi/4. Multiply this by 80 and you get 80-20pi, answer A.[/hide]
Just wondering catcurio, how did you get the image to show up in the post?

You use the img tag (phpBB) and put the source of the image.

[hide]We know that the total are is $1 ft^2$ for one tile.  We can take the four parts of the circle and turn them into one circle and get a total unshaded area of $1/4 \pi$.  Then we can subtract that from the square and get $1ft - 1/4 \pi$.  After that, just multiply the whole thing by the number of tiles, (10*8 = 80 tiles) and we get 80 - 20 pi ft squared.[/hide]

[hide]The four circles form a whole circle with radius 1/2. Thus the shaded area equals $1-\frac{1}{2}^2\pi=1-\frac{1}{4}\pi$. Since there are 80 tiles, the total shaded area on the floor is $80(1-\frac{1}{4}\pi)=80-20\pi$. [/hide]


Let x,y,z be three poítive real numbers.Prove that:$ \frac{xy}{z}\plus{}\frac{yz}{x}\plus{}\frac{xz}{y}> 2\sqrt[3]{x^{3}\plus{}y^{3}\plus{}z^{3}}$

[quote="duythuc_lqd"]Let x,y,z be three poítive real numbers.Prove that:$ \frac {xy}{z} + \frac {yz}{x} + \frac {xz}{y} > 2\sqrt [3]{x^{3} + y^{3} + z^{3}}$[/quote]
Let $ xy=c,$ $ xz=b$ and $ yz=a.$ Hence, $ x=\sqrt{\frac{bc}{a},}$ $ y=\sqrt{\frac{ac}{b}}$ and $ z=\sqrt{\frac{ab}{c}.}$
Thus, $ \frac {xy}{z} + \frac {yz}{x} + \frac {xz}{y} > 2\sqrt [3]{x^{3} + y^{3} + z^{3}}\Leftrightarrow$
$ \Leftrightarrow a^2+b^2+c^2>2\sqrt[3]{a^3b^3+a^3c^3+b^3c^3}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6+3a^4b^2+3a^4c^2+2a^2b^2c^2-8a^3b^3)>0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6-a^4b^2-a^4c^2+a^2b^2c^2)+4\sum_{cyc}(a-b)^2a^2b^2+3a^2b^2c^2>0,$
which is true by Schur.

[quote="arqady"][quote="duythuc_lqd"]Let x,y,z be three poítive real numbers.Prove that:$ \frac {xy}{z} + \frac {yz}{x} + \frac {xz}{y} > 2\sqrt [3]{x^{3} + y^{3} + z^{3}}$[/quote]
Let $ xy = c,$ $ xz = b$ and $ yz = a.$ Hence, $ x = \sqrt {\frac {bc}{a},}$ $ y = \sqrt {\frac {ac}{b}}$ and $ z = \sqrt {\frac {ab}{c}.}$
Thus, $ \frac {xy}{z} + \frac {yz}{x} + \frac {xz}{y} > 2\sqrt [3]{x^{3} + y^{3} + z^{3}}\Leftrightarrow$
$ \Leftrightarrow a^2 + b^2 + c^2 > 2\sqrt [3]{a^3b^3 + a^3c^3 + b^3c^3}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6 + 3a^4b^2 + 3a^4c^2 + 2a^2b^2c^2 - 8a^3b^3) > 0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^6 - a^4b^2 - a^4c^2 + a^2b^2c^2) + 4\sum_{cyc}(a - b)^2a^2b^2 + 3a^2b^2c^2 > 0,$
which is true by Schur.[/quote]
Nice solution,it 's China MO 2007 :)

[quote="quykhtn-qa1"]
Nice solution,it 's China MO 2007 :)[/quote]
it is incorrect :maybe:


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14650[/img]

 :(

The incircle (I) touches BC at D. Draw diamneter DE of (I). F is symmetric to D wrt Ma. We knew that EF passes through A.
[u]Construct D :[/u] Let A' be symmetric to A wrt IMa. The line through A' and parallel to IMa cuts the circle whose diameter IMa at D. Easy to construct B,C.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14669[/img]

 :thumbup:

already posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=199145]here[/url]  :|


abstract algebra
superior algebra
superior algebra unsolved

Prove or disprove :

$ G/N\cong G$ iff $ N\equal{}\{e\}$.

I guess it's false , $ N$ is not necessary to be identity .

I guess in the group $ (\mathbb Z[x],\plus{})$ with $ N\equal{}\mathbb Z$ , it will  contradicts .

What do u see guys  :)

It is necessarily true when $ G$ is a finite group, though... ;) A cool way to see that a slight modification of your idea provides you with a counterexample is as follows: associate to every $ p \in \mathbb{Q}[x]$ its first derivative. This gives you a homomorphism $ \mathbf{d}$ from $ \mathbb{Q}[x]$ to itself. Now, the fundamental theorem of homomorphisms allows you conclude that $ \mathbb{Q}[x] / \ker \mathbf{d} \sim \mathbb{Q}[x]$. Can you see what the kernel of $ \mathbf{d}$ is?

Nice example .

Surely $ Ker d\equal{} \mathbb Q$

thx  :)


i have been trying to solve this all day.

Show that at least one member of the set
[i][b]{1, 3, 7, ..., (2^k) - 1, ..., (2^2000) - 1}[/b][/i]
is divisble by 2001.

Any help would be appreciated.

$2^{308}-1$ is divisible by $2001$

[hide] you know that $2^{28} \equiv 1 \bmod 29$
and $2^{22} \equiv 1 \bmod 23$
etc.. (notce that 3 , 23, 29 are prime, and   what is the product?
 And What is the LCM of 2,22,28 ?

In general Euler's theorem says $ 2^{\phi (2001)} \equiv 1 \bmod 2001 $
 [/hide]

If you are not familier with Ferma't Little theorem, and just want to use  a simple method, You can use what is called
"pigeon hole" principle.

Take 2001 numbers  ($ 2^0(=1),2^1,2^2,2^3,\dots 2^{2000}) $  Find out the remainder when you divide each number by $ 2001 $

Obviously this can not be  $ 0 $ (why? - 2001 can not divide into  any even numer) and is less than $ 2001 $  We have only $ 2000 $ possible remainders and we have $ 2001 $ (above) cases, so all can't be different, and we must have a case where:

$ 2^a $ and $ 2^b $  gives the same remainder when divided by $ 2001 $  (assume $ a>b $
so $ 2^a-2^b = 2^b(2^{a-b} -1) $ is divisible by $ 2001 $

Now $ 2^b $ can not have a common factor with $ 2001 $ so $ 2^{b-a} - 1 $ must be divisible by 2001, and of course $ b-a $ is less than 2001.
Have fun.

I'm moving this post from Getting Started to the Intermediate forum.

[quote="Gyan"]If you are not familier with Ferma't Little theorem...[/quote]

[b]Fermat's Little Theorem:[/b]

\[ a^p\equiv a (mod p), \]
where $ p $ is a prime number and $ a\epsilon N $.

\[ a^b-1\equiv 0 (mod p)\rightarrow a^b\equiv 1(mod p), \]
where $ p $ does not divide $ b $, however $ b|(p-1) $.
[b]Example:[/b] $ 2^6\equiv 1(mod 3) $.

\[ a^{p-1}-1\equiv 0(mod p)\rightarrow a^{p-1}\equiv 1 (mod p) \].

Masoud Zargar