How will the AMC office report USAMTS AIME Qualifiers to teachers? Do they just send the AIME qualifiers in an email? Do they come with AMC results?

We get a list of USAMTS qualifiers from the USAMTS office.  We send an AIME to the school, or contest manager for that school.  For the vast majority of cases, we send that AIME with the AMC 10/12 School reports, along with all our other AIMEs.  In the isolated cases that a school did not participate in the AMC contests but has a USAMTS AIME qualifier, we get in touch with the school and /or teacher directly.  In a few cases, we may ask the qualifier to take the AIME at a nearby college, University or school which did participate in one of the AMC contests.

If you have questions, contact the AMC office by one of the means printed on all of our literature.

-- Steve Dunbar
MAA Director, American Mathematics Competitions

I was one of the perfect scorers in this years USAMTS (Alex Paskov) and I was wondering if that qualified me for for the AIME? The first AIME assessment is tomorrow but my school didnt recieve a test for me, so am i just taking the second "giving" of the AIME on March 16, or will something else be arranged. (I called the company that runs the AMC and they said they were not notified of USAMTS being a qualifier for the AIME).
Thanks!

Anything above a 68/75 will qualify you for AIME, and your AMC12 index is calculated by roughly 4/3 times your USAMTS score.

Are you sure the index is roughly 4/3 times your USAMTS score? I heard that it was the percentage you got right (out of 75). But if you are really sure that it is 4/3 times the USAMTS score, it's definitely better than using percentage!

uh, 4/3*75=100 so it's literally the percentage you get right...

[quote=Eulersbestfriend271]Whoops, I was thinking about 4/3 times the percentage! You're right, but why is 100 the max index from USAMTS (its pretty low)?[/quote]

Because it's the minimum cutoff (I know, it's dumb, it should be DHR or something...)

EDIT: Wait lol the AMC12B just proved that this isn't even true anymore, so why can't they make it DHR?

[quote=qwerty137][quote=Eulersbestfriend271]Whoops, I was thinking about 4/3 times the percentage! You're right, but why is 100 the max index from USAMTS (its pretty low)?[/quote]

Well, if the max index from USAMTS is 150 instead, it would encourage cheating (USAMTS runs on the honor system and is a month long). They have to set a reasonable limit while ensuring that not much cheating occurs, since most people who can get 68/75 or more can get higher than 100 on the AMC12.[/quote]

It's not just cheating, it would be unfair if that many people had 150 AMC12 indices walking into AIME (since unlimited time on USAMTS is much better than 75 minutes).

So basically just take the AMCs if you're seriously aiming for AIME/USAMO...

They could make the index 5/3*USAMTS score, that would be fair. It would create a cap of 125 which would be motivating and more rewarding but not unfair to people who focus on AMC.


[quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove or disprove that:
$ [\sum\ a^4 \plus{} \sum\ ab(a^2 \plus{} b^2) \plus{} 9abc].(ab \plus{} bc \plus{} ca) \leq\ 54$[/quote]

Try $ a\equal{}0, b\equal{}c\equal{}1.$  :wink:

But I think the following is trues with the same condition

$ [\sum\ a^4 \plus{} \sum\ ab(a^2 \plus{} b^2) \plus{} 6abc].(ab \plus{} bc \plus{} ca) \leq\ 45$

[quote="dduclam"][quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove or disprove that:
$ [\sum\ a^4 \plus{} \sum\ ab(a^2 \plus{} b^2) \plus{} 9abc].(ab \plus{} bc \plus{} ca) \leq\ 54$[/quote]

Try $ a \equal{} 0, b \equal{} c \equal{} 1.$  :wink:

[/quote]
But $ a\plus{}b\plus{}c\equal{}3$, my brother  :maybe:

[quote="nguoivn"][quote="dduclam"][quote="nguoivn"]Given $ a, b, c \geq\ 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove or disprove that:
$ [\sum\ a^4 \plus{} \sum\ ab(a^2 \plus{} b^2) \plus{} 9abc].(ab \plus{} bc \plus{} ca) \leq\ 54$[/quote]

Try $ a \equal{} 0, b \equal{} c \equal{} 1.$  :wink:

[/quote]
But $ a \plus{} b \plus{} c \equal{} 3$, my brother  :maybe:[/quote]

Yes, try $ a\equal{}0,b\equal{}c\equal{}3/2$. :)

Sorry for my bad calculus. You are right  :(


limit
topology
calculus
calculus computations

Let $ \{E_n\}$ be a sequence of closed, nonempty, bounded sets in a complete metric space $ X$. Also assume that $ E_{n \plus{} 1}\subset E_n$. Prove that $ \bigcap_{n \equal{} 0}^{\infty}E_n$ is nonempty.

I know this is true for compact sets, but am not quite sure how to proceed with the given conditions. I am pretty sure though that the fact that $ X$ is complete is necessary.

This is false as stated. Here's the example: Let $ X$ be $ \mathbb{R}$ with the metric $ d(x,y)\equal{}\frac{|x\minus{}y|}{1\plus{}|x\minus{}y|}.$ Every set in $ X,$ including $ X$ itself, is bounded in this metric, and the space is complete. Let $ E_n\equal{}[n,\infty).$

There are two ways to escape from this counterexample. One is to replace the word "bounded" in the definition with "totally bounded." The other is to add the hypothesis that $ \lim\text{ diam}(E_n)\equal{}0.$

The "totally bounded" version is then just the same theorem as the version that uses compactness; that diminishes the interest. So assume the second version - the hypothesis that the diameters tend to zero.

To use completeness, you'll need a Cauchy sequence. Do you see how to find one?

Thanks the problem as stated in the book did say that $ \displaystyle\lim_{n\to\infty}\text{diam}(E_n) \equal{} 0$, but I didn't think that was necessary.

I'm not sure is this is correct but here is what I have so far.
[hide]
Let $ \{p_n\}$ be a sequence in $ X$ such that $ p_n\in E_n$. Then from the assumption about $ \{E_n\}$ we have that $ p_n\in E_j$ for all $ j\le{n}$. Now let $ \epsilon > 0$ so since $ \lim\text{diam}(E_n) \equal{} 0$ we have that $ d(p_n,p_m) < \epsilon$ for $ n,m$ large enough. Finally since $ X$ is complete we have that $ \{p_n\}$ converges to a point, $ p$, in $ X$. Also $ p$ is a limit point of this sequence and since each $ E_n$ is closed, $ p$ is in the set.[/hide]

Also what does totally bounded mean?

Your proof looks pretty much like the standard proof of this. (I would have actually used the words "Cauchy sequence" with respect to the $ p_n,$ but you did prove and use that anyway.)

A set $ E$ is totally bounded in a metric space if for any $ r>0,$ one can cover $ E$ with finitely many balls of radius $ r,$ where the number of balls needed may depend on $ r.$ It turns out that this is a characterization of compactness: $ E$ is compact if and only if it is both closed and totally bounded.

The need for the concept is revealed by my counterexample above. Every metric space is topologically equivalent to a bounded metric space, so "bounded" by itself doesn't mean what you might hope it would mean. In $ \mathbb{R}$ with that funny metric, $ [n,\infty)$ is a bounded set but not a totally bounded set.

The hypothesis of the diameters of the $ E_n$ tending to zero is the more natural one for this problem anyway; I probably shouldn't even have mentioned total boundedness. Note that you can strengthen your conclusion: not only is the intersection of the $ E_n$ nonempty, but it consists of precisely one point.


people have forgot this post and me too :?

[quote="minhkhoa"]Let a,b,c be three sides of a triangle and a+b+c=2p(p is a constant)(p <= 1). Find the smallest value of k such that
a 2 +b 2 +c 2 +2abc < k[/quote]

a 2 +b 2 +c 2=4p 2 -2ab-2bc-2ca
a 2 +b 2 +c 2 +2abc =4p 2 -2ab-2bc-2ca+2abc
a,b,c<1,so 0>(1-a)(1-b)(1-c)=1-2p+ab+bc+ca-abc
we can get abc-ab-ca-bc>1-2p

and now a 2 +b 2 +c 2 +2abc =4p 2 -2ab-2bc-2ca+2abc>4p 2 +2-2p

Oh, you don't understand my problem
I mean that :find k such that
$a^2+b^2+c^2+2abc< k$
and $k$ must be the best constan

Well,it is enough to prove that:
[tex] 2(a+b+c)(a^2+b^2+c^2)+4abc \leq (a+b+c)^3 [/tex]
[tex] \Leftrightarrow a^3+b^3+c^3 \leq a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2) +2abc [/tex]
[tex] \Leftrightarrow (b+c)(b-c)^2 \leq a(b+c)^2 +a^2(b+c-a) [/tex]
which is right since [tex] a(b+c)^2 \geq (b+c)(b-c)^2 [/tex] and [tex] a^2(b+c-a) \geq 0 [/tex]
We can have the equality [tex] a=0,b=c [/tex] so [tex] 2p^2 [/tex] is the best. :)
An easier  problem:
Find $ k $ when [tex] p=\frac{9}{5} [/tex] :)


calculus
inequalities
4-variable inequality

For all non-nagative real numbers a,b,c,d, we put S = a+b+c+d,T= ab+ac+ad+bc+bd+cd, R=abc+bcd+cda+dab, H= abcd
Prove that: S/4 >= (T/6)1/2 >= (R/4)1/3 >= H1/4.
I've solution, and you?

The fist ineq S/4>=T/6^1/2 is very simple.After some calculs you will have to prove that 3/2* \sum a^2>= \sum ab which is true.

     I hope I will come back with the solution for the rest of the ineq!

[quote="hxtung"]For all non-nagative real numbers $ a,b,c,d$, prove that

$ \frac {a \plus{} b \plus{} c \plus{} d}{4}\geq\sqrt {\frac {ab \plus{} bc \plus{} cd \plus{} da \plus{} ac \plus{} bd}{6}}\geq\sqrt [3]{\frac {bcd \plus{} cda \plus{} dab \plus{} abc}{4}}$.[/quote]Denote $ f(k) \equal{} \frac {k}{4}(ab \plus{} bc \plus{} cd \plus{} da) \plus{} \frac {1 \minus{} k}{2}(ac \plus{} bd)$, then

$ f(k)\leq\left(\frac {a \plus{} b \plus{} c \plus{} d}{4}\right)^2$ holds if and only if $ \frac {1}{2}\leq k \leq 1$;

$ f(k)\geq\left(\frac {bcd \plus{} cda \plus{} dab \plus{} abc}{4}\right)^{\frac {2}{3}}$ holds if and only if $ \frac {1}{3}\leq k \leq \frac {5}{6}$.

[quote="Charlie"]It's well known as Symmetric Mean Inequality.[/quote]

Is there a Symetric Mean Inequality for n unknowns?


What is the difference between the greatest whole number factor of 121 and the least whole number factor of 6?

The greatest factor of 121 is 121 and the least factor of 6 is 1, so 121-1=120.

[hide="answer"]1.greatest factor of 121=121

2.smallest factor or 6 is 1

3.121-1=120[/hide]

[b][i][u]rate my post it is greatly appreciated [=[/u][/i][/b]

please be fair though [=

 :clap:  :clap:  :clap:  :clap:


topology
real analysis
real analysis theorems

Prove that any non empty open set $ S$ in $ \mathbb{R}^{1}$ contains both rational and irrational numbers.
Thanks in advance.

Sorry for double-posting, but I would also like to know how to solve the following:

Prove that every nonempty closed set on $ mathbb{R}^{1}$ is the intersection of a countable cover of open sets.

First, any open interval contains both rational and irrational numbers.

Second, every open set in $ \mathbb{R}$ is the countable union of disjoint open intervals.  For $ x \in S$ construct $ a \equal{} \inf(a < x : (a, x) \in S), b \equal{} \sup(b > x : (x, b) \in S)$, where $ a < x < b$ by construction.  Then $ I_x \equal{} (a, b) \in S$ and each interval contains a rational number, so the set of such intervals for any $ x \in S$ is countable.

Third, it's essentially the same result as the above (take complements).

alrite thank you sir, i was finding it hard to prove the countability condition.


Duke college AMC USA(J)MO USAMO MATHCOUNTS ARML

How is the Duke (blehh...) math meet going to work?
Still teams of 15?
If it's by school and nothing else, my school will get schooled, since we're writing-intensive, not math.

You don't have to do it by school; creating one for a region is also ok, especially for most schools.  The first time I went, we had a team from Winston-Salem that consisted of about 4 different schools.  Last year, I also went with a team from Winston-Salem, but picked up a member from Raleigh and another member from Greenville (this was because we didn't have enough people in Winston-Salem due to All County Band).

However, if your school is capable of sending a team, then I would go ahead and go with it; the bigger problem I have found is getting people to go.

[quote="not_trig"]What? They're teams of 6, last time I checked. And it's by school, but there's individual stuff too. Hopefully my team won't be so sucky this year because we will probably actually have 2 (!) USAMO qualifiers.[/quote]

Hmm seems like you are assuming I will make USAMO??

must be someone else.

blah

[quote="not_trig"]dude I am indeed assuming you will make USAMO... you made it last year, didn't you?....[/quote]

yeah but that was last year  :maybe:

ummm dude you don't make much sense

you are supposed to improve

which means that making USAMO should become easier

dude why am I bothering? you're clearly just being modest, w/e

[quote="b-flat"]this was because we didn't have enough people in Winston-Salem due to All County Band[/quote]

Ahem. Of course.

Dude I am supposed to improve??

I MOVED BACK TWO PLACES AT [b][u]REGIONAL MATHCOUNTS[/u][/b]

Well then again the top 5 scores were

45
44
44
44
44

...
...

stupid pirate pete and pirate paul

...
...

[quote="mintybear"]can anyone explain what the duke math meet is? and what other school or team competitions there are? i just moved to NC and want to make a math team at my school, except i don't know anything about math here....[/quote]

The duke math meet is an arml style competition in November that takes teams of 6 from schools.
http://www.math.duke.edu/dumu/mathmeet/

The main event in North Carolina is here:
http://courses.ncssm.edu/goebel/STATECON/state.htm
There are qualifying tests and then the state test and then ARML.

so, guys - apparently TJ A is coming :O

it will be exciting, no?

Hello North Carolinians,

LET US GET PUMPED FOR DUKE MATH MEET, SHALL WE? LET'S BEAT THOSE GEORGIANS AND SOUTH CAROLINIANS AND TJ-ERS (because those are pretty much the only Virginians.)

also hopefully I will not fail. That would be extremely (!) helpful.

Thank you for your time.

So NC got thoroughly owned this year. Not only did no NC team place in the top 4, and no NC individual make it to the tiebreaker round, but there were several egregious typos on individual problem 4.

Maybe next year...?

LOL YOU GUYS HAVE FELT THE WRATH OF TJ A.

ALSO

LOL I AM DA MIEN XD

AND THE FIRST LADY BAKES COOKIES.

sure ok... darn I got this later because I remembered I COULD USE EQUATIONS NOT SERIES

so the octahedral graph is pretty much a K6 (6 vertices with all vertex pairs connected by exactly 1 edge) without the long diagonals

so - call the players A, B. adj = adjacent to other player (i.e. the other player can win now), win = win, opp = opposite other player (so that the other player must move adjacent)

1. A moves adj
2a. B moves opp (1/4 chance) -> 3a. A moves adj (recurse)
2b. B moves adj (1/2 chance)
-> 3b. A wins (1/4)
-> 3c. A moves adj (1/2) (recurse)
-> 3d. A moves opp (1/4) -> 4d. B moves adj (recurse2)

So let the probability that A wins after moving adj be $ p$, and the probability A wins after $ B$ moves adj be $ q$.

So, $ p \equal{} \frac {p}{4} \plus{} \frac {q}{2}$ and $ q \equal{} \frac {1}{4} \plus{} \frac {p}{2} \plus{} \frac {q}{4}$.

Solving, $ p \equal{} \frac {2}{5}$.

EDIT: @ hamster/kyl: LOL I AM XD
agh so I think NCSSM tied for 5th... bleargh? dude I think we should've gotten together an "NC ARML A" team. Then we would've pwned TJ. but... we are all going with schools yeah? and there aren't many super-stacked schools.

A problem with a similar type of solution: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10180

According to the results on the duke math meet website, I appear to have gotten a 2  :o

It seems like they messed up the grading somehow because I know I got 1,2,3,5,8 correct (they have me with 3 and 5 correct)...

Another guy on my team also seems to have his score incorrectly posted (it puts Winston-Salem A down several ranks on the list).

Finally, I think that Duke needs to give some type of award to an NC school, because otherwise, I doubt that a lot of NC schools will continue to send teams if we just get owned by out of state teams every year.  Maybe they could give an award to the top team in the state (or maybe top 2)?

Yeah an NC team would destroy TJ but that pretty thorougly defeats the purpose of the Duke Math Meet. And I might have gotten #7 if I realized an octahedron is not a dodecahedron. And hello I wrote the correct answer to #4, unlike some of the TJ people. So yeah that's about it.

... Congrats Hamster and TJ. We shall meet again on our terms at Princeton.

what was the correct answer to number 4? I got $ 16\pi$ =/

eh i thought that problem didnt count...

and where can you see the results?

Actually, that problem wasn't messed up, and it was actually harder than #7 according to the tiebreaking criteria. But, yes, it did not count.

[quote="33286"][quote="1076207"]You can ask your coach to log on to the [url=http://www.math.duke.edu/dumu/mathmeet/]duke site[/url] and look at the results. There's only individuals and team totals, not team or proof round scores.[/quote]

There are team and power round scores.[/quote]

Where? I've looked and haven't seen them.

If you click on a team, you can see that team's scores (power round, team round, each relay question, and individual scores). I'm not certain where the list of teams was; it may have just been the team results. I can't check because I'm not a coach.

[quote="n^4+4"]Haha life was so much more awesome when one NC school could beat TJ A ^^[/quote]

east, enloe '04-'06? or so? man. We are all so spread out now >.>

maybe the younger generation can continue this legacy eventually :P (or dude maybe I could convince all these good sophomores to come to NCSSM and then we could maybe win)


Let $f(n)=\frac{2^n+2}{n}=\frac{2(2^{n-1}+1)}{n}$ is integer number.
Known, that $2^{p-1}=1(mod p^2)\Longrightarrow p=1093 \ or \ p=3511 \ or \ p>3000000000$.
Therefore $n=p_1*p_2*\dots*p_k$, were $p_1<p_2<\dots <p_k$ are distinct prims (because (n-1,n)=1).
If k=1 we have n=2. Let $d_i \ or \ d(p)$  is minimal with satisfyed $2^{d_i}=-1(mod \ p_i)$. If $p=7(mod \ 8)$, then d(p) is not define, because has not d: $2^d=-1(mod \ p)$. Therefore $p_i\not =7(mod \ 8)$.
We have $d_i|(n-1)$ and $\frac{n-1}{d_i}$ is odd for all i (d(2)=1). Therefore $ord_2(d_1)=ord_2(d_2)=\dots =ord_2(d_k)$.
Because n<1998 and k>1, one of prime divisors is 2 or 3 or 5 or 11 or 13 or 17 or 19 or 29 or 37 or 41 or 43 and behaps one prime divisor p>43. We have
$d(2)=1,d(3)=1,d(5)=2,d(11)=5,d(13)=6,d(17)=4,d(19)=9,d(29)=14,d(37)=18,d(41)=10,d(43)=7$.
If k=2 and n is odd we have $d_1|(p_2-1),d_2|(p_1-1)$. If k=3 and $p_1=2$ then $d_2|(2p_3-1),d_3|(2p_2-1)$. It give easy property to chek:
If 43|n then 86|n (because (d(43) is odd) and 7|(n-1). It give n=86*11=946.
If 41|n then n=11(mod 20). It give n=41*11 or 41*31. Because d(11) is odd, d(31) is not defined in this case have not solution.
If 37|n then n=19(mod 36). It give n=37*19 or n=37*55  - have not solution.
If 29|n then n=15(mod 28). It give n=29*15 or n=29*43 - have not solution.
If 19|n then 38|n and n=1(mod 9). It give n=38(5+9k)=38*5,38*14,38*23,38*32,38*41,38*50 -have not solution.
If 17|n then n=5(mod 8). It give n=17*(5+8k) - have not solution.
If 13|n then n=7(mod 12). It give n=n=13*(7+12k) - have not solution.
If 11|n then 22|n and n=1(mod 5). It give n=22(3+10k). We have solution n=66.
If 5|n then n=3(mod 4). It give n=5(3+4k). It give that n have prime divisor p=3(mod 4). Then d(p) is odd or not define. It give contradition with $ord_2(d(p))=ord_2(d(5))=1$.
If 3|n then 6|n. If $\frac n6 =p$ is prime we have contradition with $d(p)|5=2p_2-1$. Else it have prime divisor p<43, and we consider all cases. Therefore, if f(n) is integer and n<2000, then $n=2 \ or \ n=6 \ or \ n=66 \ or \ n=946$.

[quote="pavel kozlov"][quote="shobber"]Find an integer $n$, where $100 \leq n \leq 1997$, such that
\[ \frac{2^n+2}{n}  \]
is also an integer.[/quote]
Find all such integers or at least one?[/quote]

It seems just one......

[quote="Rust"]$ 2^{p \minus{} 1} \equal{} 1(mod p^2)\Longrightarrow p \equal{} 1093 \ or \ p \equal{} 3511 \ or \ p > 3000000000$[/quote]
I didn't understand why are we using this. Shouldn't be $ 2^{p \minus{} 1} \equal{}\minus{} 1(mod p^2)$???Explain it please!

[quote="Rust"]Let $ f(n) \equal{} \frac {2^n \plus{} 2}{n} \equal{} \frac {2(2^{n \minus{} 1} \plus{} 1)}{n}$ is integer number.
Known, that $ 2^{p \minus{} 1} \equal{} 1(mod p^2)\Longrightarrow p \equal{} 1093 \ or \ p \equal{} 3511 \ or \ p > 3000000000$.
Therefore $ n \equal{} p_1*p_2*\dots*p_k$, were $ p_1 < p_2 < \dots < p_k$ are distinct prims (because (n-1,n)=1).
.[/quote]

why  $ n \equal{} p_1*p_2*\dots*p_k$ and not $ n \equal{} p_1^{a_1}*p_2^{a_2}*\dots*p_k^{a_k}$ !!!!!!!

The number $n = 946 = 2 \cdot 11 \cdot 43$ works.

The way you construct is: we try to look for examples of the form $n = 2pq$, where $p$ and $q$ are distinct odd primes.  This amounts to $p \mid 2^{2q-1}+1$ and $q \mid 2^{2p-1}+1$.  Looking at orders, it would suffices to have $q = 4p-1$ and $\left( \frac2q \right) = -1$, but the latter follows anyways by quadratic reciprocity (since $q \equiv 3 \pmod 8$).  In the first equality, this means $p \mid 2^{8p-3}+1 \implies p \mid 2^5+1=33$.  So we get the solutions $(p,q)=(3,11)$ and $(p,q)=(11,43)$.  These gives $n = 66$ and $n = 946$ as candidates; the latter works.

What is the motivation for your solution? Thanks!

Could someone please tell me the motivation for v_Enhance's solution too? Thanks!

@v_Enhance could you please elaborate on how you got $p \mid 2^{2q-1}+1$, since I got $p \mid 2^{4q-2}-1$ but this does not imply your result. Also, could you explain the latter part when you say: [quote] Looking at orders, it would suffices to have $q = 4p-1$ and $\left( \frac2q \right) = -1$,[/quote]. Thanks in advance.

For the first bit: if $p \mid 2^{n-1}+1$, then $0 \equiv 2^{n-1}+1 \equiv 2^{2pq-1}+1 \equiv 2^{2q-1}+1 \pmod p$, since $2^p \equiv 2 \pmod p$ by Fermat.

For the second part, I meant the second divisibility only when I said "it suffices to have" (that is unclear on my part, sorry). Since if $q = 4p-1$, then $2p-1 = \tfrac12(q-1)$, and $(2/q) \equiv 2^{\frac12(q-1)} \pmod q$.

Hey, this kind of reminded me of IMO 1990 P3... anyways does anyone know of a nice elegant solution to this one? The way I did it was [i]very[/i] bashy... first tried $pq$, then $pqr$, somehow stumbled across $2pq$, figured $p \equiv q \equiv 3 \pmod{4}$, took $p$ to be the smaller one, took the case when $p=7$, got no solutions, then tried $p=11$ to first try $q=23$ because $q \equiv 3 \pmod{5}$ and finally got $q=43$ to work (if you're wondering how I got these relations, it's just basic orders and $gcd$ bashing)... the problem with this method, I feel, aside from being unnecessarily long, is that it's very easy to just give up during a real exam, especially after 20-25 minutes of trying out numbers with no progress... so does anyone have a different solution, or even a way to shorten this approach?

This problem is such a troll; the number $n=946$ works. This is clear as $2^{946} \equiv 2^6 \equiv -2 \pmod {11}$ and $2^{946} \equiv 2^{22} \equiv -2 \pmod {43}$, as desired.

[b][color=#f00]Remarks: [/color][/b]The way I constructed $n$ is similar to the above solutions. First we can note that $n$ has to be even, for obvious reasons. It would be pretty unexpected if a lot of different primes divided $n$, since then we get a lot of different conditions on $n$. So we try $n=2p$, $n=2p^2$, then $n=2pq$ to gradually add more conditions. The first one yields $n=6$, too small, while the second one yields no value of $n$ less than or equal to $1997$. Now we try $n=2pq$, and using simple orders we find that $q=4p-1$ satisfies $p \mid 2^{2q-1}+1$. We can guess that the primes $p,q$ must be of that form, and trial and error shows one that works.

[quote=Mathloops][quote=mathlogician]This problem is such a troll; the number $n=946$ works. This is clear as $2^{946} \equiv 2^6 \equiv -2 \pmod {11}$ and $2^{946} \equiv 2^{22} \equiv -2 \pmod {43}$, as desired.

[b][color=#f00]Remarks: [/color][/b]The way I constructed $n$ is similar to the above solutions. First we can note that $n$ has to be even, for obvious reasons. It would be pretty unexpected if a lot of different primes divided $n$, since then we get a lot of different conditions on $n$. So we try $n=2p$, $n=2p^2$, then $n=2pq$ to gradually add more conditions. The first one yields $n=6$, too small, while the second one yields no value of $n$ less than or equal to $1997$. Now we try $n=2pq$, from which simple orders tells us that $q=4p-1$, the end.[/quote]

Hello, how do you get q=4p-1?[/quote]

I misphrased it, sorry. We want to solve both conditions $p \mid 2^{2q-1}+1$ and $q \mid 2^{2p-1}+1$. We can derive this by seeing that $2$ has order $4p-2$ mod $q$, so if we assume that $2$ is a primitive root we get $q=4p-1$. I didn't know how to show the second condition works, but there's only a small amount of primes to check so I just checked all the pairs of primes $(p,q) = (p,4p-1)$ and found one that worked.

[quote=tumbleweed][hide=Two questions:](i) Must we prove that the odd case does not work?  (The official solution did not do that, I think.)
(ii) We know that $pq$ works, but how do we prove that $pqr, pqrs, pqrst,$ and so on do not?  Again, I don't think the solution said anything about that.[/hide][/quote]

I don't understand the question, the problem asks to find an $n$ and we found an $n$, it's not necessary to find all $n$.

This is official solution, but I don't know... how did they get p-1l4pq-2 and q-1l4pq-2?

$946=2 \cdot 11 \cdot 43$ clearly works. Also, $6$ and $66$ work but unluckily doesn’t lie in the desired range. Now, we illustrate the process of construction of this number. First of all, we show some results which reduces the search space of such $n$.

	[b][color=blue]Lemma 1:[/color][/b] If $x^\gamma \equiv -1 \pmod p$ and $\gamma$ is a power of 2, then $\text{ord}_p(x)=2\gamma$.

	[i]Proof:[/i] Let $\text{ord}_p(x)=d$. Now $d \mid 2\gamma$ as $x^{2\gamma} \equiv 1 \pmod p$. Also, $d \nmid \gamma$ because if the contrary were true, then $x^\gamma \equiv 1 \pmod p$ which is a contradiction. Combining $d \mid 2\gamma$ and $d \nmid \gamma$ we get $d=2\gamma$. $\square$

	We will now show another lemma with the above result.

	[b][color=blue]Lemma 2:[/color][/b] If $p \mid a^\beta +b^\beta$ where $\beta =2^k$ for some $k \in \mathbb{N}$ and $p$ is some prime then $p \equiv 1 \pmod{2^{k+1}}$.

	[i]Proof:[/i] From the divisibility condition, we have
	\[a^\beta +b^\beta \equiv 0 \pmod p \implies \left(\frac{a}{b}\right)^{\beta} \equiv -1 \pmod p \]
	Applying Lemma 1, we get to know $2^{k+1} \mid p-1$. $\square$

	Coming back to the problem, the following claim discards the possibility of odd $n$.

	[b][color=red]Claim:[/color][/b] Any odd $n>1$ does not work.

	[i]Proof:[/i] Assume that $n$ is odd. The proof will be by infinite descent on $\nu_2(n-1)$ to show that $n-1=0$. As $n$ is odd, $n-1$ will be even. Pick a prime $q$ dividing $n$, which gives $q \mid 2^{n-1}+1$. By [i]Fermat's Christmas Theorem[/i], we have that $q \equiv 1 \pmod 4$ which in fact gives $n \equiv 1 \pmod 4 \iff 4 \mid n-1$. Notice that $2^{n-1}+1=2^{4a}+1$ for some integer $a$. By Lemma 2, we have that $8 \mid q-1$ which gives $8 \mid n-1$. Upon simple induction, we can show that $2^\alpha \mid n-1$ for arbitrarily large $\alpha \in \mathbb{N}$. This forces $n-1=0$ as desired. $\square$

	Now $n$ can only be even and it is easy to see that $\nu_2(n)=1$. We will show that we do have a solution if $n=2pq$ where $p$ and $q$ are primes.

	If $n=2pq$ then we have $pq \mid 2^{2pq-1}+1$. This can be simplified like this
	\[p \mid 2^{2pq-1}+1 \implies p \mid 2^{2q-1}+1\]
	Similarly, we get $q \mid 2^{2p-1} +1$. Another wild try could be to consider the case to consider the case when $p=q$ which gives that both primes must be equal to 5. Sadly this doesn't work. So, from here on assume $p>q$.

	$p \mid 2^{2p-1}+1 \implies 2^{2p} \equiv -2$. So we want $-2$ to be a quadratic residue. We now wish $\displaystyle
	\left(\frac{-2}{p}\right)=1$ where we used the [i]Legendre's Symbol[/i]. Now
	\begin{align*}
		1=\left( \frac{-2}{p} \right) &= \left(\frac{-1}{p}\right) \times \left(\frac{2}{p}\right) \\
		&= (-1)^{\frac{p-1}{2}} \cdot (-1)^{\frac{p^2-1}{8}}
	\end{align*}
	Set $\displaystyle \frac{p-1}{2} \cdot {\frac{p^2-1}{8}}=2f$ for some integer $f$. We can easily get that this is possible iff  $p \equiv 1,3 \pmod 8$. Now we can finally go for hit and trial. This was lengthy on paper as I personally tried too much for $p \equiv 1 \pmod 8$. If we try  $p\equiv 3 \pmod 8$ then we get $(p,q)=(11,3)$ and $(11,43)$ where the value of $q$ can be found out very easily from $q \mid 2^{2p-1}$ when you put $p=11$. We are finally done. $\blacksquare$
[hide=Remark]
One can easily show that $6$ is a solution when $n=2p$ for $p$ being a prime. Also, $6^3$ posts!
[/hide]

[quote=v_Enhance]The problem only asks to construct a single $n$, you can show (or guess) that $n$ should be even. If $n = 2p$ then the only solution is $n = 6$, so you do the next best thing and guess $n = 2pq$. Everything past that point is routine application of orders. :)[/quote]

How did would one prove that $n$ is even?

[hide=Experimental]
Clearly $4\nmid n$; henceforth take cases on the prime factorization of $n$.
[list=1]
    [*] If $n=p$, the assertion reads $p\mid(2^p+2)$, which is absurd by FLT.
    [*] If $n=2p$, the assertion reads $2p\mid(2^{2p}+2)$, i.e., $2^{4p-2}\equiv1\pmod{p}$. Taking the exponent (mod $p-1$) gives $2^2\equiv1\pmod{p}$, which yields the solution $n=6$.
    [*] If $n=pq$, the assertion reads $pq\mid(2^{pq}+2)$, i.e., $2^{2pq-2}\equiv1\pmod{p,q}$. Taking the exponent (mod $p-1,q-1$) gives $2^{2p-2}\equiv1\pmod{q}$ and $2^{2q-2}\equiv1\pmod{p}$. Assuming $2$ is a primitive root (mod $p,q$), we entail $(p-1)\mid(2q-2)$ and $(q-1)\mid(2p-2)$, meaning one of $p-1$ and $q-1$ is twice the other, say, $\tfrac{p-1}{q-1}=2$. Yet that makes $2^{p-1}\equiv1\pmod{q}$ and so $2^{pq-1}\equiv1\pmod{q}$, bad.
    [*] If $n=2pq$, the assertion reads $2pq\mid(2^{2pq}+2)$, i.e., $2^{4pq-2}\equiv1\pmod{p,q}$. Arguing analogously as above, it suffices to have $(p-1)\mid(4q-2)$ and $(q-1)\mid(4p-2)$ provided, of course, that $(p-1)\nmid(2q-1)$ and $(q-1)\nmid(2p-1)$. Evidently we need $p,q\equiv3\pmod{4}$, so we construct the following chart:
        \begin{tabular}{c|c|c}
            $p\equiv3\pmod{4}$ & $p-1$ & $2p-1$ \\ \hline
            3 & 2 & 5 \\
            7 & 6 & 13 \\
            11 & 10 & 21 \\
            19 & 18 & 37 \\
            23 & 22 & 45 \\
            31 & 30 & 61 \\
            43 & 42 & 85
        \end{tabular}
    We seek two rows where both second entries divide twice the other row's third entry, which we find is true for $p,q=11,43$. So $n=2\cdot11\cdot43=\boxed{946}$ works.

$946$ works since it is $2 \cdot 11 \cdot 43$ and letting $11=p$ and $43=q$ we see $\tfrac{p-1}{2} \mid 2q-1$ and $\tfrac{q-1}{2} \mid 2p-1$ so $\tfrac{p-1}{2} \mid 2q(p-1)+2q-1=2pq-1$ and we can check $2$ is a primitive root $\pmod p$ and $2pq-1$ is odd so we get $2^{2pq-1} \equiv -1 \pmod p$ so $p \mid 2^{2pq}+2$ and similarly for $q.$

Suppose $n$ is of the form $2pq$. From FLT, we have
\[pq \mid 2^{\frac{(p-1)(q-1)}{2}}-1\]
and we require
\[2pq \mid 2^{2pq}+2 \implies pq \mid 2^{2pq-1}+1 \mid 2^{4pq-2}-1.\]
(Note that we must check afterwards that $pq \nmid 2^{2pq-1}-1$.) Thus it suffices to find a solution to
\[9 \cdot \frac{(p-1)(q-1)}{2} = 4pq-2\]
\[(p-9)(q-9)=68\]
\[p=11, \quad q=43,\]
and since $11 \cdot 43 \nmid 2^{945}-1$, we get our possible value of $n$ as $n = 2 \cdot 11 \cdot 43 = \boxed{946}.$

It is clear that $4 \nmid n$, so we consider some formatting of the desired number. Working case by case, we see that $n=p$, $n=2p$, and $n=pq$ all do not produce valid solutions in the desired range. Hence, we move to $n=2pq$:

We require

\[2pq \mid 2^{2pq}+2 \implies pq \mid 2^{2pq-1}+1 \mid 2^{4pq-2}-1,\]

as long as $pq \nmid 2^{2pq-1}-1$. This essentially reads

\[2^{4pq-2}\equiv1\pmod{p,q},\]

implying $(p-1)\mid(4q-2)$ and $(q-1)\mid(4p-2)$, if  $(p-1)\nmid(2q-1)$ and $(q-1)\nmid(2p-1)$. It is clear that $p,q \equiv 3 \pmod{4}$ so some experimenting shows us our answer of $2 \cdot 11 \cdot 43 = \boxed{946}$, which works.

Groupsolved with [b]cursed_tangent1434[/b].
Our answer is $2 \cdot 11 \cdot 43 = 946$
\[2^{946} + 2 = 2(2^{945} + 1)\] Clearly this is divisible by $2$.
\[2^{945} + 1 \equiv 2^{5} + 1 \equiv 33 \equiv 0\pmod{11}\]
So it is divisible by $11$.
\[2^{945} + 1 \equiv 2^{21} + 1 \equiv 2097153 \equiv 0\pmod{43}\] so $946$ works.


group theory
abstract algebra
superior algebra
superior algebra solved

how can i prove that if a group G of finite order has all its sylow p groups normal in G, then G is soluble?

E.g., use the following theorems:

When $U,V$ are normal subgroups of $G$ with $U \cap V = \emptyset$, then $UV$ is a normal subgroup again and isomorphic to $U \times V$.

$A\times B$ is solvable iff $A$ and $B$ are solvable.

Every $p$-group is solvable.

Edit: I didn't see http://www.mathlinks.ro/Forum/viewtopic.php?t=62786 !
Why are you posting it again¿


AMC
AIME
AwesomeMath
summer program
Mathcamp
PROMYS
Ross Mathematics Program

Well, I was looking into some great summer math programs, however they all seemed far to advanced/hard for me.  I really don't know how to judge my math knowledge/ability...  I just got into extracurricular math this summer, but I've improved a lot.  So while I know I don't have amazing mathematical ability I feel like my knowledge is more of a obstacle than my ability.  (I'm around 100-115 AMC 12 and 2-5 AIME)

So I was wondering if you guys know of any awesome programs that someone like me would a) have a shot of getting into and b) be able to succeed at.

 Thanks!

AwesomeMath Summer Program is a great program for developing [b]contest math[/b] skills.  Given your scores, I'm sure you can get into it, and there are classes from AMC 10/12 level to Olympiad classes.  I went this year - it's a lot of fun and I learned a lot also.

Mathcamp is another good summer program; however, it's harder to qualify for and it focuses on exposing students to new advanced math topics, not necessarily for contest math.

Honors Summer Math Camp is for students of all backgrounds who are interested in math.  You don't have to be a "genius" to get in.  Rather, they're looking to develop a passion in students for mathematical discovery
[url]http://www.txstate.edu/mathworks/camps/hsmc.html[/url]

Well I actually haven't taken the AMC or AIME yet, because I didn't even know about them until this summer; those are my scores from "practice tests". Do you still think I would be able to "get into" Awesome Math?  Anyways, thinks for the suggestions, Awesome Math seems to be just the program I'm looking for!

This year, there were people with much lower scores than yours who qualified for AwesomeMath.  Just put a little bit of time into the qualifying test.

[quote="mathemonster"]if u can get in PROMYS is ILL[/quote]

I can say the same for Ross. :)
How was PROMYS, by the way?

mathematically it was pretty good.
socially, it was just OK.
how was ross?

Mathematically, Ross was intense. Lots of problems, lots of sets, and you really get pushed through a LOT of NT. It really depends on how much you want to pick up along the way. Socially, I can't complain, lots of chess, frisbee, cards, running around. Its not like we went into boxes and did sets, most people kind of socialized constantly, fitting in work where they could. :) But at the same time it WAS a lot of work, and most people did it.

right, i need advice too and it seems like this is the right topic :)

i would like to develop my math skills (like make it through to the second round of BMO-british maths olympiad- or maybe even higher) but, obviously, i can't apply to AwesomeMath anymore, as the deadline has passed.

however, i applied for SUMaC (Stanford University Mathematics Camp), i got all the questions and i got in.

so...i was thinking about applying to MathCamp, Ross and PROMYS (in this particular order of priorities and preferances). i am really into number theory, i mean i reaaaally love it, but i would like to get more problem solving practice and improve my performance in maths contests(like i said, especially BMO) as well.

soooo, what would you recommend between SUMaC, MathCamp, Ross and PROMYS for the particular 'goals' or 'expectations' that i have listed above? :)

thanks a loot :)

I would say both, although I haven't got quite a lot of experience with the rigorous, hardcore type :)

[quote="buzzer11"]I can't say anything about the other camps, but Ross and Promys is not oriented toward contest problems. Do you love number theory contest problems, or hardcore, rigorous number theory?[/quote]
I would say PROMYS Ross $\leq$ USAMO number theory

SUMaC is also not oriented toward contest problems.  Did you get into Program I or II?  SUMaC doesn't officially focus on number theory, but Program I has a fair amount of number-theoretic content.  Program I does abstract algebra (mostly group theory), which is a crucial tool for doing "real" number theory (i.e. not contest-math NT).  One of the perennial research projects in Program I studies cryptography, which is essentially all number theory; another does ruler-and-compass constructibility, which is really a problem in algebraic number theory.  At SUMaC, you won't do much of the sort of number theory that gets used in contests, but you'll probably (at least in Program I) get a taste of the "real" stuff, and build a solid foundation for studying it.

i got into program I.
@sly si : thanks for all the details. btw, did you ever go to SUMaC? :)

[quote="modularmarc101"][quote="modularmarc101"]How good is CTY compared to Awesome Math? Cuz I almost went to cty last summer but decided to wait another year, but now I heard that awesome math is really good too.[/quote]

Can someone answer this for me?[/quote]

With all respect to the CTY programs (I am a member of CTY/SET, so no rivalries or anything here), I think you'll definitely find a more rigorous curriculum at AwesomeMath, if you're looking for contest math that is.  And I think if you were looking for "higher level" mathematics, PROMYS, SUMaC, Ross, MathCamp or HCSSiM would probably all be better programs than CTY.  Their admission tests/standards are much higher (tests with real math problems vs SAT or other standardized testing).  I think at these math-focused camps, you'll probably find a peer group that's more similar to you intellectually and will probably work better with.

Note: I haven't actually been to a CTY camp, but have spoken with kids who have gone and read a lot of their syllabuses


Slove equation : $ 10x\plus{}15y\plus{}4z\plus{}t\plus{}2u\equal{}35$
 Here $ (x,y,z,t,u) \in \mathbb{N}^{5}$
 Please for answer

[quote="Nguyen thanh Giang"]Slove equation : $ 10x \plus{} 15y \plus{} 4z \plus{} t \plus{} 2u \equal{} 35$
 Here $ (x,y,z,t,u) \in \mathbb{N}^{5}$
 Please for answer[/quote]

I consider $ 0\notin\mathbb N$. So $ 10(x\minus{}1)\plus{}15(y\minus{}1)\plus{}4(z\minus{}1)\plus{}(t\minus{}1)\plus{}2(u\minus{}1)\equal{}3$ and so $ x\equal{}y\equal{}z\equal{}1$  and two possibilities for $ (t\minus{}1,u\minus{}1)$ : $ (3,0)$ or $ (1,1)$

Hence two solutions : $ (1,1,1,4,1)$ and $ (1,1,1,2,2)$

Obviously $ x \equal{} y \equal{} 1$ so $ 4z \plus{} t \plus{} 2u \equal{} 10$ and $ z\leq 2$.

If $ z \equal{} 1 \implies t \equal{} 4$ and $ u \equal{} 1$  and $ z\equal{}1 \implies t\equal{}u\equal{}2$

$ z\equal{}2$ there's no solution  $ 2u \plus{}t >2$

"Sloven":
   1. A man or boy habitually negligent of neatness and order; the male equivalent of slattern, or slut; untidy, dirty.
   2. low, base, lewd.
   3. (obsolete) immoral woman.


Let $n\geq 2$ be positive integers.Given the sequences $x_1,\ x_2,\ \cdots x_n$ and $y_1,\ y_2,\ \cdots y_n$ such that $x_1\geq x_2\geq \cdots \geq x_n$ and $y_1\geq y_2\geq \cdots \geq y_n.$ For the sequence $z_1,\ z_2,\ \cdots z_n$ which can be obtained by rearranging $y_1,\ y_2\ \cdots y_n,$ prove that we have $\sum_{j=1}^n (x_j-y_j)^2\leq \sum_{j=1}^n (x_j-z_j)^2.$

this was given in some past IMO as well, I think..... ;)

see http://www.mathlinks.ro/Forum/viewtopic.php?t=61050 .

  darij


How many ways are there to put 5 balls in 2 boxes if the balls are not distinguishable and neither are the boxes?

Since the balls aren't distinguishable, and neither are the boxes, we just count the different ways:

(0,5)
(1,4)
(2,3)

And so the answer is $ \boxed{3}$.

There are $\binom{6}{1}=6$ (balls and urns). However, the boxes are indistinguishable, so $\frac{6}{2}=\boxed{3}$


What is the formula for finding the number of diagonals in a convex n-gon? Sorry if this question has been asked before!

Never mind! I just remembered the formula! :P  :oops:

hello, here you can find the answer:
http://mathforum.org/t2t/message.taco?thread=3829&message=3
Sonnhard.

The answer is $ \frac{n(n\minus{}3)}{2}$.  See if you can derive it.

HINT: How many diagonals can you draw from just one vertex?


$ ABCD$ is a convex quadrilateral with $ AC\equal{}BD$. Equilateral trinagles are constructed externally on the sides of the quadrilateral with centroids $ P$, $ Q$, $ R$, $ S$ respectively on $ AB$, $ BC$, $ CD$, $ DA$. Prove that $ PR$ is perpendicular to $ QS$.


What is the for experienced by an electron when it is an electric field of magnitude $ 1.00$x$ 10^5NC^{\minus{}1}$ which is pointing from East to West/

I got the magnitude but what would the direction be??

Since an electron is negatively charged and the field is positive, the electron should move east I think due to attraction (right?).

The direction is the same, but the sense is opposite (West to East) since the electron has a negative charge.


[quote="zhou_qiang"]A farmer use four straight fences, with respective lenghts 1, 4, 7 and 8 units to form a quadrilateral. What is the maximum area of the quadrilateral can enclose?[/quote]

[hide]By Brahmagupta's Formula, the area is maximized when the quadrilateral is cyclic:
$ \begin{eqnarray*}A = \sqrt{(s-a)(s-b)(s-c)(s-d)}&=& \sqrt{(10-1)(10-4)(10-7)(10-8)}\\ &=& \boxed{18}$
Such a quadrilateral is possible:
$ AB=1$
$ BC=8$
$ CD=4$
$ DA=7$

[asy]import cse5; pair A=MP("A",D((0,0)),SW), C=MP("C",D((8.0622577483,0)),E), O=MP("O",D((A+C)/2)), B=MP("B",D(IP(CR(A,1),CR(O,8.0622577483/2))),NW), D=MP("D",D(OP(CR(A,7),CR(O,8.0622577483/2)))); D(A--B--C--D--cycle); D(CR(O,8.0622577483/2)); MP("$1$",(A+B)/2,W); MP("$8$",(B+C)/2,N); MP("$4$",(C+D)/2); MP("$7$",(D+A)/2);[/asy][/hide]
Looks like the farmer wants to get the most out of his money. However, if I were him, I'd glue the four fences, and create a circle with area $ 100/\pi$, much bigger than the answer given above. :P

[hide="Hint"]
There is a theorem (I can't remember the name) that says that, given the sides lengths of a quadrilateral, the maximum area comes when the quadrilateral is cyclic.[/hide]
[hide="Hint 2"]
Brahmagupta's Formula
[/hide]

EDIT: whoops. I didn't notice vishalarul's hidden content. I wouldn't have given these hints otherwise.


MATHCOUNTS
summer program
MathPath
\/closed

Hmm, I've been wondering that too. I'm assuming it's one of the Mathcounts directors, because people with one post don't generally get selected as moderators.

[quote="Ubemaya"]1337, I'm ashamed of you.

levans... Larry Evans... ?

Actually I'm not sure, I don't think he has anything to do with MC.[/quote]

oh, clever, clever...
He runs his own school in California I think, but I don't see why he would be a mod with one post...

Larry Evans is currently a middle-school teacher in Ojai, California (west of LA).  He has also been on the staff at the MathPath summer program.

We will have more to say about Larry and his role here at AoPS in a few weeks.


quadratics
AoPSwiki
algebra
special factorizations

What exactly is "completing the square"?  my prof lightly covered this subject and I don't fully understand how to do it.

Completing the square is a method for solving a quadratic equation without having to find clever factorizations.
Here's an explanation of it from AoPSWiki: http://www.artofproblemsolving.com/Wiki/index.php/Completing_the_square


abstract algebra
superior algebra
superior algebra solved

hello
im learning for some time now on my own abstract algebra
i dont see to improve in my problem solving skillz at all
i just find it verry hard to solve problems
i find myself staying 15minutes-3h to understand a single problem
does solving problems include understanding solutions to problem one
cannot solve on his own ?
i have to be motivated by something...i mean...if one cannot solve
by himself problems what satisfaction does he get?
i would like it if ppl here could make up a list of problems...group theory
problems
i know for the moment stuff like:
group,monoid,semigroup,ring,first isomorphism th for groups,second isomorp for
groups,third isomor...(2 and 3 i havent made any applications with them yet),
permutations,groups of permutations,problems concerning Z_n,
order of perm,Lagrange th,Cayley th(i have never applied this ?!?!
it says that G~Sn or something like that in some cond but i have never
applied it)
[maybe i forgot some stuff but i will soon post some more if i know something
more]

[for moderators]
I hope that if allot of problems are posted here...i try to solve them and then
give some results for them here...one moderator can retrieve them at a latter
time when all are done with them and can make separate problems out of them
and post them into the solved problem section

.
I am still learning.
If i am not welcome with this post i understand and i will deleted it at one.

[quote="spx2"]
im learning for some time now on my own abstract algebra
i dont see to improve in my problem solving skillz at all
i just find it verry hard to solve problems
[/quote]

Did you try making notes, while reading your abstract algebra book ? ( I ask since I had the same difficulty as you have, and solved it exactly be making notes)
The benefit of making them is that they force one to organize knowledge and clear our own contradictions.
Obviously, it's harder to write down something we do not fully understand, than read it and just skip to the next section.
Therefore by making notes, you force yourself, to understand everything..
Moreoever there is a way of writting proofs that does also clear a lot of misunderstandings, see: http://www.mathmlconference.com/2002/presentations/lamport/lamport.pdf
Rewritting each proof in your book, in the way presented in the above link should make everything perfectly clear.
Also there are probably better textbooks than the one you use (I heard you complain about it in a previous post), maybe you'll like the one at:
http://www.math.miami.edu/~ec/book/

As for the problems let's begin with : Prove that every group of order $\leq 5$ is abelian.

Thank you {x}!

Now about the problem...i read from my book from a previous
problem that it was prooved that groups of order 4 can be
either isomorphic to $Z_4$ or $\left\{e,x,y,xy(=yx) \right\}$(the last
one is Klein) wich are both
abelian with operation $\cdot$
(in some other book i read that Klein is $Z_2\times Z_2$).

if the group has order 3 then it cannot be but $Z_3$
if there were other groups with 3 elements then they should look
like {e,x,y} so they are different from $Z_3$ with $x^2\neq y$.
Now to proove that there are no more such 3-element groups:
let (we multiply by x on left)$x\times |x^2\neq y \Rightarrow e\neq xy$ so
$y\neq x^{-1}$ but for this to be group $x$ must have an inverse
and if it is not $y$ then there is no other element that could be $x$ inverse...

also there is another reason for wich this group of order 3 different
from $Z_3$ cant exist.It should also contain

so the proof is complete.

Now for groups of order 2 its obvious that $\left\{e,x\right\}$
is abelian.

[quote="spx2"][for moderators]
I hope that if allot of problems are posted here...i try to solve them and then
give some results for them here...one moderator can retrieve them at a latter
time when all are done with them and can make separate problems out of them
and post them into the solved problem section
.
I am still learning.
If i am not welcome with this post i understand and i will deleted it at one.[/quote]Create a weblog and post your problems there. For now the (yet unwritten) regulations state that each topic must discuss one problem only.

Actually for groups of order $2,3,5$ it suffice to state that they are cyclic.
Generalizing a bit, you can prove that every group of order $p$ with $p$ a prime, is cyclic, thus abelian.


Hello all!

I'm hoping you all can double check my answers.  They just don't seem right.  I'm not sure where I'm going wrong.

here's the questions:

Suppose you have just had a 10 by 15 foot in ground swimming pool installed in your backyard.  You have 3,000.00 left from the building project that you would like to spend on surrounding the pool with a patio, equally wide on all sides.  Write an expression in terms of x for the lenght and withdth of the patio.

Overall length:[hide="Clickable text"]15 x squared?[/hide]
Overal width:[hide="Clickable text"]10 x squared?[/hide]

You have talked to several local suppliers about options for building the patio and must choose between the following:

[u]Option            Material                             Price                                              [/u]
 A                       Poured Cement                 5.00 per sq. ft.
 B                       Brick                                  7.50 per sq. ft. plus a 30.00 delivery fee
 C                       Outdoor Carpet                4.50 per sq. ft. plus 10.86 per foot of the
                                                                   pool's perimeter to install edging.

Find the area of the swimming pool: [hide="Clickable text"]150 feet[/hide]

Write an algebraic expression for the total area of the region containing both the pool and the patio: [hide="Clickable text"]15x squared * 10x squared + 150?[/hide]

Write and algebraic expression for the area of the patio (not including the area of the pool): [hide="Clickable text"]15 x squared * 10x squared ? [/hide]

Find the perimeter of the swimming pool [hide="Clickable text"]50 feet[/hide]

For each patio materiual option, write an algebraic expression for the total cost of installing the patio based on its area and the given price information:
[hide="Clickable text"] and now I'm lost - I know at this point something is wrong in my equations and I can't come upt with an expression for the total cost.  I mean, it could be as simple as multiplying it by the materials, but that doesn't seem right)[/hide]

Help!!!!

I'm not sure whether if I interpreted your problem right, but here's my idea.

To start with, let's look at the patio again.

Let the width of the patio path be x, as shown. Then the width of the patio is the width of the pool plus twice the width of the patio path, or 10+2x. The length similarly is 15+2x.

So the area of the patio with the pool is $(10+2x)(15+2x)=150+30x+20x+4x^2=150+50x+4x^2$.

Without the pool, we take 150 away, or $50x+4x^2$.

Help...how did you get the 30x when multiplying your equation out.  I don't follow there.

Attached is a graphic of what the pool looks like, yes you are absolutely right about your interpretation.

When multiplying binomials, we can distibute or use F.O.I.L.   That means we do the first terms in each set of parentheses, then the outter terms, then the inners, and finally the last terms.

We do the first terms. $\displaystyle{\110\cdot\115=150}$

Then we do the outter terms. $\displaystyle{\110\cdot\12x=20x}$

Then the inner terms. $\displaystyle{\12x\cdot\115=30x}$[i] This is the part you were wondering about [/i]

And the last terms. $\displaystyle{\12x\cdot\12x=4x^2}$

Hey!  Thanks!  I guess cause I was doing FOIL in my head (and it's still a little new to me) I missed that Outter/Inner step.  It screwed me up that you put it as the second in your equation, and I couldn't see past the tree to see the forest!

HA!

Thanks for steping me through it!


I've attached a sample portion of my table below.  What I would like to do is center the headings and the data columns, but keep the data right-justified.  Right now my headings are fine, but the data and data columns are both right-justified.

[code]			\begin{tabular}{c r r} \hline
				Year&\multicolumn{1}{c}{BY Capital Costs}&\multicolumn{1}{c}{Discounted Costs}\\
				&\multicolumn{1}{c}{(\$000)}& \multicolumn{1}{c}{to 98/99 (\$000)}\\ [0.5ex]
				\hline
98/99	&	277,608 	&	277,608 	\\
99/00	&	473,842 	&	451,278 	\\
00/01	&	407,709 	&	369,804 	\\
01/02	&	388,159 	&	335,307 	\\
09/10	&	42,241 	&	24,697 	\\ [0.5ex]
						\hline
			\end{tabular}
[/code][code][/code]

[quote="ouadad"]I've attached a sample portion of my table below.  What I would like to do is center the headings and the data columns, but keep the data right-justified.  Right now my headings are fine, but the data and data columns are both right-justified.[/quote]
This sounds contradictory. You say
[i]What I would like to do is ... keep the data right-justified[/i]
then
[i]but the data and data columns are both right-justified[/i]
You'll need to explain in more detail.

Yes, I think there are clearly some typos that render it incomprehensible.  I wonder also if some of the words "column" are supposed to be "row"?

$ \begin{tabular}{crr}\hline
            Year&\multicolumn{1}{c}{BY Capital Costs}&\multicolumn{1}{c}{Discounted Costs}\\
            &\multicolumn{1}{c}{(\$000)}& \multicolumn{1}{c}{to 98/99 (\$000)}\\ [0.5ex]
            \hline
98/99   &   277,608    &   277,608    \\
99/00   &   473,842    &   451,278    \\
00/01   &   407,709    &   369,804    \\
01/02   &   388,159    &   335,307    \\
09/10   &   42,241    &   24,697    \\ [0.5ex]
                  \hline
         \end{tabular}$

Looks pretty good to me.

Yes, sorry if my message wasn't clear.

... and thank you.


induction
number theory unsolved
number theory

An integer sequence $ \left(a_1,a_2,a_3,...\right)$ is defined by the starting values $ a_1\equal{}1$ and $ a_2\equal{}2$ and by the recurrent equation $ a_n\equal{}3\cdot a_{n\minus{}1}\plus{}5\cdot a_{n\minus{}2}$ for all $ n\in\left\{3,4,5,...\right\}$.

Find out whether there exists an integer $ j>1$ such that $ a_j$ divides the product $ a_{j\plus{}1}\cdot a_{j\plus{}2}$.

TTsphn, i think your idea isn't nice  :) .

[quote="kisiu111"]I can't understand your resolve[/quote]
He meant to find general expression of this sequence.
You can find it using TTsphn's hint, or look for Andrei Jorza lecture about Integer sequence.I don't remember where i found it,but i could post his lectur here. :wink: Anyway you can always use google. :wink:

Do you mean $ \gcd(a_{n\plus{}1},a_{n})\equal{}1$
[hide="hint"]
$ \gcd(a_{k},a_{k\minus{}1})\equal{}\gcd(a_{k\minus{}1},5a_{k\minus{}2})$
[/hide]

[quote="kisiu111"]Can you explain, from where you have that equality? And how it help in this problem?[/quote]
Yes
First you must prove that
$ \gcd(x_{n},5)\equal{}1,\forall n$
Then induction on n
The equality I think very easy because
$ a_{n\plus{}1}\equal{}3a_{n}\plus{}5a_{n\minus{}1}$

I don't know how improve that gcd(a(n),a(n-1))=gcd(a(n-1);5*a(n-2)). And I know that gcd(a(n);a(n-2))=1 but I don't know how improve that too

Yes
This is lemma of Eclide
$ \gcd(a,b) \equal{} \gcd(a \plus{} tb,b)$
Prove it quite easy.
For you problem
$ a_{n \plus{} 2} \equal{} 3a_{n \plus{} 1} \plus{} 5a_n$

Can you improve for me that gcd(a(n);a(n-2))=1?

[quote="TTsphn"]Do you mean $ \gcd(a_{n \plus{} 1},a_{n}) \equal{} 1$
[hide="hint"]
$ \gcd(a_{k},a_{k \minus{} 1}) \equal{} \gcd(a_{k \minus{} 1},5a_{k \minus{} 2})$
[/hide][/quote]
No you prove $ (a_n,5a_{n\minus{}1})\equal{}1$

Please improve for me that $ gcd(a_{n},a_{n \minus{} 2}) \equal{} 1$ I don't know how improve only that

This is a problem from an ongoing math contest   :!:  :!:  :!:   (this year's Polish Mathematical Olympiad)

The deadline for this problem was 8 October 2007. I am moving it back. It can be discussed now.

Shame on Kisiu111 for posting this problem before the deadline. Of course he didn't mention the source "59th Polish MO, 1st Round, problem 3"; it was inserted by myself (and I also edited the problem statement).


Of the 50 students in the drama club, 36 take mathematics, 27 take physics and 20 students take both mathematics and physics. How many drama club students take neither mathematics nor physics?

We use the [url=http://www.artofproblemsolving.com/Wiki/index.php/PIE]Principle of Inclusion-Exclusion[/url] to come up with the following equation:

\begin{align*}
36+27-20+x&=50
\\ 43+x&=50
\\ x&=7
\end{align*}

Thus, there are $ \boxed{7}$ students that satisfy the requirements.


For any positive real numbers $a,b$ and $c$, \[\sqrt[6]{\frac{b^{2}+c^{2}}{a^{2}+bc}}+\sqrt[6]{\frac{c^{2}+a^{2}}{b^{2}+ca}}+\sqrt[6]{\frac{a^{2}+b^{2}}{c^{2}+ab}}\le \frac{a+b+c}{\sqrt[3]{abc}}\] There is a nice solution to the above. Actually, it is true that \[\sqrt[3]{\frac{b^{2}+c^{2}}{a^{2}+bc}}+\sqrt[3]{\frac{c^{2}+a^{2}}{b^{2}+ca}}+\sqrt[3]{\frac{a^{2}+b^{2}}{c^{2}+ab}}\le \frac{a+b+c}{\sqrt[3]{abc}}\] But my proof is too ugly. I hope somebody gives a nice solution.

[quote="hungkhtn"]Yes, I really like two inequalities "six degree" of yours. It's very nice, so I will try  :lol:[/quote]

Thanks for your interest, hungkhtn. I'm still working on a better proof for the second one.

[quote="ductrung"]For any positive real numbers $ a,b$ and $ c$,
\[ \sqrt [3]{\frac {b^{2} \plus{} c^{2}}{a^{2} \plus{} bc}} \plus{} \sqrt [3]{\frac {c^{2} \plus{} a^{2}}{b^{2} \plus{} ca}} \plus{} \sqrt [3]{\frac {a^{2} \plus{} b^{2}}{c^{2} \plus{} ab}}\le \frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}\]
[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-42-p-h-duc.html

[quote="nguoivn"]I think the stronger also trues  :maybe:
\[ (a^2 \plus{} b^2 \plus{} c^2) \sum \frac {1}{a \plus{} b} \le \frac {(a \plus{} b \plus{} c)^3}{6\sqrt [3]{a^2b^2c^2}}\]
[/quote]

try $ a \equal{} \frac {1}{3},b \equal{} \frac {1}{4},c \equal{} 1$

$ (a^2 \plus{} b^2 \plus{} c^2) \sum \frac {1}{a \plus{} b} \le \frac {(a \plus{} b \plus{} c)^3}{k\sqrt [3]{a^2b^2c^2}}$

$ k_{max} \equal{} 5.1934...$

:wink:

Sorry, indeed it's not trues  :oops:  . Thank you for your check, hedeng123.


quadratics
algebra
special factorizations

solve by completing the square

2m2+10m-3=0

I dont know if your aware of the formula to find the roots of a quadratic equation
say ax^2 + bx +c =0
the the roots can be computed by using
p,q =-b+-sq root (b^2 -4ac)/2a
where p and q are the roots

I hope its clear

I'd like to add something. Completing the square is used to convert a formula in the form of ax + bx + c into the form of a(x+k)^2+h. Sorry if i got the variables mixed up. I just can't remember at the moment. This form of the equation has many uses.

Example of how this is done:

0 = x^2+8x+12
0 = (x^2+8x+[]) +12 - []
0 = (x+4)^2 +12 -16
0 = (x+4)^2 -4

There! Now the equation is in the form that we need!


algebra
polynomial
logarithms
number theory unsolved
number theory

if $ p_n$ denotes the n-th prime number. proove that $ p_n < n^2$ for all natural numbers grater than $ 1$

i know that it is true asimpthotically and that Legendre's conjecture is still unprooven but can we proove this regardless?
any useful links on a topic of legendre's conjecture or any conected problem are also welcomed
Tnx for ur help

[color=green]I've found a better one

$ p_n \leq n(n \minus{} 1) \plus{} 1$

And something even more important is that: if this is no true, then Goldbach's conjecture is not true either.

How did I get this?
Well... I started supposing that Goldbach was true.

Then, let $ p_1;p_2;...;p_n$ be the first $ n$ primes.

There are $ \binom{n \minus{} 1}{2}$ sums of the form $ p_i \plus{} p_j$ with $ i > 1$ different from $ j > 1$
and $ n \minus{} 1$ sums of the form $ p_i \plus{} p_i$ with $ i > 1$

so there are exactly $ \binom{n \minus{} 1}{2} \plus{} n \minus{} 1$ sums of odd primes and we add $ 4 \equal{} 2 \plus{} 2$ we have in total
$ \binom{n \minus{} 1}{2} \plus{} n$ even sums of the n first primes.

The group of these sums contains the numers: $ 4;6;8;10;...;p_n \plus{} 3$ (all the even numbers, because of Goldbach)
These are $ \frac {p_n \plus{} 3 \minus{} 4}{2} \plus{} 1 \equal{} \frac {p_n \minus{} 1}{2} \plus{} 1$ numbers

so $ \frac {p_n \minus{} 1}{2} \plus{} 1 \leq \binom{n \minus{} 1}{2} \plus{} n$

$ p_n \minus{} 1 \plus{} 2 \leq (n \minus{} 1).(n \minus{} 2) \plus{} 2.n$

$ p_n \leq (n \minus{} 1).(n \minus{} 2) \plus{} 2.n \minus{} 1$

$ p_n \leq n.(n \minus{} 1) \plus{} 1$

And that's all :P [/color]

nice :lol:  gr8 idea
i think its currently prooven that every integer grater than 4 is a sum of at most six primes. adopting your method to this we can proove that $ p_n$ is always smaller than some polinomial of degree 6.
not quite $ n^2$ but unleast it's polinomial unlike $ 2^n$ which you get from bertrand postulate

Any statement of this form asymptotically weaker than the Prime Number Theorem only needs to be proven for small $ p$ and then one only needs an estimate on the error of the approximation $ p_n \sim n \ln n$.

yes i know that but i don't ever get to see the proof of that. also shouldnt than be simple to prove $ p_n<n^2$?

An estimate is easier to prove than an inequality.  In this case, though, the fact that Goldbach's conjecture is true for small primes and that PNT is true is enough; at some point, we should be able to show $ p_n < 2n \ln n$ and then we're done.

can u write a proof or unleast a sketch of a proof.
also do u know where to find a proof for that estimate?

Notice that $p_1=2 > n^2$,
If $n \geq 2$, then we have
$n < \sqrt{2} +1$ which gives,
$(n+1)^2 > 2n^2$.
Now, we show that $p_n < n^2$ by induction on $n$.
Notice that for $n+1$, via Bertrand's postulate, we have
$p_{n+1} < 2p_n < 2n^2 < (n+1)^2$
Since it is true for $n=2$, it follows that it is also true for all $n \geq 2$.

[quote=Tintarn][quote=SultansOfSwing]$2n^2 < (n+1)^2$[/quote]
Are you sure?[/quote]

It's not true for any n>2


What is the greatest possible number of digits in the product of a
4-digit whole number and a 3-digit whole number?

To find this answer, simply add the two number of digits up, to get $ \boxed{7}$.

You can check by multiplying the largest 4-digit number by the largest 3-digit number. $ 9999 \times 999$ gives you a 7-digit number.

And a fast way to see that $ 9999\times 999$ is going to have 7 digits without actually doing ALL the work,

$ 9999(999)\equal{}(10000\minus{}1)(999)\equal{}9990000\minus{}999$ clearly going to have 7 digits right? :)


inequalities
LaTeX
algebra
polynomial
inequalities open

Find the maximum of:

$\dfrac{ab}{a^2+b^2+4c^2}+\dfrac{bc}{b^2+c^2+4a^2}+\dfrac{ca}{c^2+a^2+4b^2}$

I wait a detail solution?

Ok. It's not difficult by use old way : sum of square. We can solve by this way because the maximim can be attended when $a=b=c$ ( We don't care for other case). THen it's not hard to prove the maximum takes then $a=b$!

   But in my first inequality, this's not true because the equal can't get if $a=b=c$. It's reason why I think It's difficult.

  Am I true, Vasc?

  I hope that your inequality has a nice solution other way sum of square !!!  :) If that, It's so beautiful.

Indeed, your inequality becomes equality for $a=1$ and $b=c$, but not for $a=b=c$. I verified this using a computer.
With regard to my proposed inequality, I am waiting for any solution.

Ok!
It's nice. I've check it this moring. I found a generel inequality with complete solution (not use computing) as follow:

If $k\le 3$ then:

$\dfrac{ab}{a^2+b^2+kc^2}+\dfrac{bc}{b^2+c^2+ka^2}+\dfrac{ca}{c^2+a^2+kb^2} \le \dfrac{3}{k+2}$

The equal holds if:
$k<>3$ only $a=b=c$
$k=3$ then $a=b=c$ or $b=c=\dfrac{3}{2}a$

And $k=3$ is the best constant for this.

How you prove yor general inequalty..

Bomb

Yeah... It's use a not new way. But the most interesting of this problem is the best constant is $3$...

[quote="hungkhtn"]For my first problem? Can you prove that the maximum attacked when 2 of 3 number equa, Vasc?[/quote]
Yes. For positive $k$, the expresssion
$\frac{ab}{a^2+b^2+kc^2}+\frac{bc}{b^2+c^2+ka^2}+\frac{ca}{c^2+a^2+kb^2}$
is maximal when $a \leq b=c$ (a<=b=c).

Why did you post this ineq at the solved problems? For variety? :rotfl:

My first inequality is express in general case:

Find maximum of
$S=\sum\frac{ab}{a^2+b^2+kc^2}$
With constant $k\ge 0$
And that time I don't have solution. May be It can't be attacked. :(

This inequality can be solved for $k \geq 1$ by (n-1)Equal Variable Principle.
Replace first $a,b,c$ by $\sqrt a, \sqrt b, \sqrt c$, then use (n-1)EVP for $p=0$:
For $abc=const$ and $a+b+c=const$, if $g(x)=f(1/x)$ is convex, then the sum $f(a)+f(b)+f(c)$ is minimal when $a \leq b=c$.
http://www.mathlinks.ro/Forum/viewtopic.php?p=227570#p227570.
In our case, $f(x)=-\frac{1}{\sqrt{x}(1+x)}$. ;)

[quote="Vasc"]I think the maximum is  0.543257.....
But the following inequality is very interesting

$\frac{ab}{a^2+b^2+3c^2}+\frac{bc}{b^2+c^2+3a^2}+\frac{ca}{c^2+a^2+3b^2} \leq \frac{3}{5}$,

because equality occurs for $a=b=c$, and also for $b=c=\frac{3a}{2}$ or any other permutation.[/quote]

Vasc could you post the solution of this ineq?

I posted a solution above. :)
According to this solution, it suffices to consider $b=c=1$.

Ok. Maybe It can be solved withiut $n-1EP$? I hope that...

Can Anybody solve it without $n-1EV$? I don't like problems solved by one way.

[quote="Vasc"]I think the maximum is  0.543257.....
But the following inequality is very interesting

$\frac{ab}{a^2+b^2+3c^2}+\frac{bc}{b^2+c^2+3a^2}+\frac{ca}{c^2+a^2+3b^2} \leq \frac{3}{5}$,

because equality occurs for $a=b=c$, and also for $b=c=\frac{3a}{2}$ or any other permutation.[/quote]

hung, which ineq are you referring to?
For vasc's inequality, I guess I have a solution.
Let $5z^2=a^2+b^2+3c^2$ and so on. WLOG assume $a \ge b \ge c$, then $x \ge y \ge z$.
Now we have to prove
$\sum x^2y^2 \sqrt{(4x^2-y^2-z^2)(4y^2-x^2-z^2)} \le 6x^2y^2z^2$ (Hope it's correct)
In fact, $\sum x^2y^2 \sqrt{(4x^2-y^2-z^2)(4y^2-x^2-z^2)} = \sum \sqrt{(4x^2y^4-y^6-z^2y^4)(4y^2x^4-x^6-z^2x^4)} \le \frac{4(x^2y^4+y^2x^4)-x^6-y^6-z^2y^4-z^2x^4}{2}$
If we sum them, $L.H.S. \le \frac{3(\sum_{sym} x^2y^4) - 2(x^6+y^6+z^6)}{2}$
In fact we can write R.H.S. - L.H.S. = $\sum (x^2+y^2-2z^2)(x^2-y^2)^2 \ge 0$, which is equivalent to $(x^2-y^2)(y^2-z^2)(x^2-z^2) \ge 0$. As $x \ge y \ge z$, this is true.

I think it's quite strange. Can anyone check it for me?

Ok. The problem if k=3 isn't difficult with me. But for $k >3$ may solution is not vaild. Dear Soarer, can you prove this if $k=4$ or more in my first post?

[quote="Soarer"]
In fact we can write R.H.S. - L.H.S. = $\sum (x^2+y^2-2z^2)(x^2-y^2)^2 \ge 0$, which is equivalent to $(x^2-y^2)(y^2-z^2)(x^2-z^2) \ge 0$. As $x \ge y \ge z$, this is true.
I think it's quite strange. Can anyone check it for me?[/quote]

These inequalities are not equivalent, because the first is symmetric, but the second is not.

Ok. IF there's no try, I will post my solution for $k\le 3$. But tomorrow I will take a school test then, wait me 2 day after I complete my test and translate solution from Vietnamese to English...

For $k\ge 3$, It's still unsolved.

hungkhtn,i think 20 days are gone now ;)
can you post the solution for $k le 3$ now?

and I wonder if your solution is:
let $f(a,b,c)=\frac{ab}{a^2+b^2+kc^2}+\frac{bc}{b^2+c^2+ka^2}+\frac{ca}{c^2+a^2+kb^2}$
$a \le b \le c$
then try to prove $f(a,b,c) \leq f(a,\sqrt{\frac{b^2+c^2}2},\sqrt{\frac{b^2+c^2}2})$
or  $f(a,b,c) \leq f(a,\sqrt{bc},\sqrt{bc})$ :)

Sorry every one. I haven't Latex verision new , my verision is 4.0 and It can't be created PDF from DVI? Then I cann't send it here...  :(

[quote=Vasc]I think the maximum is  0.543257.....
But the following inequality is very interesting

$\dfrac{ab}{a^2+b^2+3c^2}+\dfrac{bc}{b^2+c^2+3a^2}+\dfrac{ca}{c^2+a^2+3b^2} \leq \dfrac{3}{5}$,

because equality occurs for $a=b=c$, and also for $b=c=\dfrac{3a}{2}$ or any other permutation.[/quote]
See [url=https://artofproblemsolving.com/community/c6h402647p2243406]here.[/url]


[quote="The Rules, shamelessly stolen from iLord"]Guess who will post next.

Keep track of points in each post.

You get 1 point for guessing.

The person who is guessed get's 2 points.

You get 5 additional points for getting your guess right.

No double posting.

No guessing yourself. [/quote]

[hide="bluecarneal"]1 bluecarneal: 2
2 ernie: 1[/hide]

ernie

[hide]2 bluecarneal: 3
1 ernie: 8[/hide]

[hide="bluecarneal"]1 bluecarneal: 10
2 ernie: 9[/hide]

ernie

1 bluecarneal: 11
2 ernie:11

tell me if i score someone wrong.

You forgot to give me five points for guessing right.

[hide]1 bluecarneal: 18
2 ernie: 17[/hide]

I guess bluecarneal.  :D

[hide="ernie"]
1 bluecarneal:18
2 ernie: 19
3 jjfun1: 1[/hide]

[hide="jjfun1"]
1 ernie: 19
2 bluecarneal: 18
3 jjfun1: 3
4 math_explorer: 1[/hide]

[hide="jjfun1"]1 ernie: 19
2 bluecarneal: 18
3 jjfun1: 5
4 math_explorer: 1
5 FlyAgaric: 1[/hide]

[hide="james4l"]1 ernie: 20
2 bluecarneal: 18
3 jjfun1: 5
4 james4l 2
5 math_explorer: 1
5 FlyAgaric: 1[/hide]

[hide="ernie"]1 ernie: 22
2 bluecarneal: 19
3 jjfun1: 5
4 james4l 2
5 math_explorer: 1
5 FlyAgaric: 1[/hide]

[hide="jjfun1"]1 ernie: 22
2 bluecarneal: 19
3 jjfun1: 7
4 james4l: 2
5 math_explorer: 2
6 FlyAgaric: 1[/hide]

[hide="jjfun1"]1 ernie: 22
2 bluecarneal: 19
3 jjfun1: 9
4 james4l: 2
5 math_explorer: 2
6 FlyAgaric: 2[/hide]

[hide="jjfun1"]
1 ernie: 22
2 bluecarneal: 19
3 jjfun1: 11
4 james4l: 2
5 math_explorer: 2
6 FlyAgaric: 2
7 westiepaw: 1
[/hide]

Sooner or later, [i]someone[/i] has to be right!

[hide="bluecarneal"] 1 ernie: 22
2 bluecarneal: 20
3 jjfun1: 11
4 james4l: 2
5 math_explorer: 2
6 FlyAgaric: 2
7 westiepaw: 1 [/hide]

[hide="k00lperson"]
1 ernie: 22
2 bluecarneal: 20
3 jjfun1: 11
4 james4l: 2
5 math_explorer: 2
6 FlyAgaric: 2
7 westiepaw: 1
8 k00lperson: 1
9: Teh meewhee!!!: 1[/hide]

dang.
[hide="wrightboy"]
1. ernie: 152
2. k00lperson: 101
3. jjfun1: 62
4. PhireKaLk6781: 61
5. bluecarneal: 32
6. wrightboy10: 25
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. abcak: 1[/hide]

[hide="wrightboy10"]1. ernie: 152
2. k00lperson: 101
3. jjfun1: 62
3. PhireKaLk6781: 62
5. bluecarneal: 32
6. wrightboy10: 27
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. abcak: 1[/hide]

[hide="Phire"]
1. ernie: 152
2. k00lperson: 101
3. PhireKaLk6781: 64
4. jjfun1: 62
5. bluecarneal: 32
6. wrightboy10: 27
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. abcak: 1[/hide]

[hide="PhireKaLk6781"]
1. ernie: 152
2. k00lperson: 101
3. PhireKaLk6781: 66
4. jjfun1: 62
5. bluecarneal: 32
6. wrightboy10: 28
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. abcak: 1
[/hide]

[hide="Phire"]
1. ernie: 152
2. k00lperson: 101
3. PhireKaLk6781: 68
4. jjfun1: 62
5. bluecarneal: 32
6. wrightboy10: 28
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. abcak: 1 [/hide]

[hide="k00lperson"]
1. ernie: 152
2. k00lperson: 103
3. PhireKaLk6781: 68
4. jjfun1: 62
5. bluecarneal: 32
6. wrightboy10: 29
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. abcak: 1
[/hide]

[hide="k00lperson"]1. ernie: 152
2. k00lperson: 103
3. PhireKaLk6781: 68
4. jjfun1: 62
5. bluecarneal: 32
6. wrightboy10: 29
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. Sabien5 1
17. abcak: 1 [/hide]

[hide="wrightboy"]
1. ernie: 152
2. k00lperson: 105
3. PhireKaLk6781: 68
4. jjfun1: 62
5. bluecarneal: 32
6. wrightboy10: 31
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. Sabien5 6
17. abcak: 1[/hide]
haha, good job sabien :)
Just make sure to add 2 points to the person you guess though!

Rawr, I'm back. Did anyone ever beat my score in the previous one?

[hide="wrightboy"]
1. ernie: 152
2. k00lperson: 105
3. PhireKaLk6781: 68
4. jjfun1: 62
5. wrightboy10: 33
6. bluecarneal: 32
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. Sabien5 6
17. abcak: 1
17. chenhsi: 1[/hide]

[quote="chenhsi"]Rawr, I'm back. Did anyone ever beat my score in the previous one?[/quote]

I got within 200. :P

[hide="ernie"]
1. ernie: 154
2. k00lperson: 105
3. PhireKaLk6781: 68
4. jjfun1: 62
5. wrightboy10: 33
6. bluecarneal: 32
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. Sabien5 6
17. abcak: 2
18. chenhsi: 1[/hide]

[hide="abcak"]
1. ernie: 154
2. k00lperson: 105
3. PhireKaLk6781: 68
4. jjfun1: 62
5. wrightboy10: 33
6. bluecarneal: 32
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 11
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. Sabien5 6
17. abcak: 4
18. chenhsi: 2[/hide]

[hide="abcak"]1. ernie: 154
2. k00lperson: 105
3. PhireKaLk6781: 68
4. jjfun1: 62
5. wrightboy10: 33
6. bluecarneal: 32
7. math_explorer: 16
8. ksun48: 15
9. 15!!!: 14
10. KingSmasher3: 13
11. Meewhee009: 6
12. westiepaw: 5
12. james4l: 4
12. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. Sabien5 6
17. abcak: 4
18. chenhsi: 2[/hide]

congrats.  But I think I deserved 2 points back there

[hide="jjfun1"]1. ernie: 154
2. k00lperson: 105
3. PhireKaLk6781: 68
4. jjfun1: 63
5. wrightboy10: 33
6. bluecarneal: 32
7. KingSmasher3: 18
8. math_explorer: 16
9. ksun48: 15
10. 15!!!: 14
11. abcak: 7
12. Meewhee009: 6
13. westiepaw: 5
14. james4l: 4
14. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. Sabien5 6
18. chenhsi: 2[/hide]

[hide="abcak"]1. ernie: 154
2. k00lperson: 105
3. PhireKaLk6781: 68
4. jjfun1: 63
5. wrightboy10: 33
6. bluecarneal: 32
7. KingSmasher3: 18
8. math_explorer: 16
9. ksun48: 15
10. 15!!!: 14
11. abcak: 9
12. Meewhee009: 6
13. westiepaw: 5
14. james4l: 4
14. FlyAgaric: 4
15. cf249: 3
15. pacman2812: 3
16. Sabien5 6
18. chenhsi: 3[/hide][/quote]


AMC USA(J)MO USAMO trigonometry geometry circumcircle

Given a triangle $ABC$ such that $cot A$, $cotB$ and $cotC$ are respectively terms of an arithmetic progression. Prove that $\measuredangle GAC = \measuredangle GBA$, where $G$ is the centroid of the triangle.

Thanks. I'm not very good with definitions of terms and stuff.
I do not remember them since I have never found them useful in school or in the USAMO or AIME.

[color=darkblue]$1.\blacktriangleright$ From the well-known relation $4S=(b^{2}+c^{2}-a^{2})\cdot\tan A$ a.s.o. obtain easily $2\cot B=\cot A+\cot C\Longleftrightarrow a^{2}+c^{2}=2b^{2}\ .$
$2.\blacktriangleright\widehat{GAC}\equiv\widehat{GBA}\Longleftrightarrow$ $NA^{2}=NG\cdot NB$, where $N$- the middlepoint of the side $[AC]$ $\Longleftrightarrow$ $3b^{2}=4m^{2}_{b}$ $\Longleftrightarrow$ $b^{2}=a^{2}+c^{2}\ .$
[u]Otherewise[/u], $\widehat{GAC}\equiv\widehat{GBA}$ $\Longleftrightarrow$ $bm_{a}=cm_{b}$ $\Longleftrightarrow$ $2c^{4}-(3b^{2}-2a^{2})\cdot c^{2}+a^{2}b^{2}-2b^{4}=0$ $\Longleftrightarrow$ $(c^{2}+a^{2}-2b^{2})(2c^{2}+b^{2})=0$ $\Longleftrightarrow$ $a^{2}+c^{2}=2b^{2}\ .$

[b]Remark.[/b] $a^{2}+c^{2}=b^{2}$ $\Longleftrightarrow$ $OH\perp BN\ ,$  where : $O$- the circumcenter, $H$- the orthocenter.[/color]


inequalities
number theory unsolved
number theory

Let $ (p_{n})$ be the sequence of increasing primes. Do there exists constants $ A$ and $ B$ so that:
$ p_{n}\leq An+B$

It's well-known that we can establish an arbitrarily long sequence of composite numbers

$ n!+2, n!+3, n!+4, ... n!+n$

So the primes cannot have linear growth rate.

As $ p_{n}\sim n \cdot \log(n)$, such bounds cannot exist.
(Chebyshevs theorem gives non-asymptotic bounds)

@t0rajir0u:
The ordered sequence $ (x_{n})$ of numbers $ x$ with $ 2^{2k}\leq x \leq 2^{2k+1}$ for some $ k$ has arbitrary big gaps, but still has a linear growth rate, namely $ x_{n}\leq 4n$.


I got helpful replies in the topic 'Cyclic Inequalities', but that's not what I'm looking for.
Let me rephrase... Cyclic inequalities in which each term consists of homogenous algebraic expressions seem to hold equality when a=b=c.
example: (a/(b+c)) + (b/(c+a)) + (c/(a+b)) has the extremum value 3/2.   IF a,b,c are positive reals
OR
ab + bc + ca has the extremum value 12 PROVIDED THAT a+b+c = 6.  IF a,b,c are positive reals

These are simple examples.
NOW could anyone could explain (maybe with proof), or does no explanation exist?

Please let me know if there is any exception (homogenous algebraic expressions). I'm a complete beginner when it comes to Olympiad Math.

(a/(b+c)) + (b/(c+a)) + (c/(a+b) >= 3/2 is Nesbitt's Inequality

http://en.wikipedia.org/wiki/Nesbitt%27s_inequality

see also : http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=303607


Sorry if this is too basic a question, but the nth term divergence test does work for alternating series and other series with some negative terms, doesn't it?

Yes. The "test" is very simple: if the terms don't go to zero, the series diverges. Cancellation of terms with opposite signs can't help.

$\sum_{n=0}^{\infty}\frac{(-1)^nn}{\sqrt{n^2+1}}$ diverges because the terms don't tend to zero.

This is a consequence of the Cauchy criterion for convergence:

$\sum_{k=1}^{\infty}a_k$ converges if and only if

$\forall\,\epsilon>0,\,\exists\,N$ such that if $n>m>N,$ then $\left|\sum_{k=m+1} ^na_k\right|<\epsilon.$

That implies (and this is a one-way implication whose converse is false) that for $n>N,\,|a_n|<\epsilon,$ which is the same as saying that $a_n$ tends to zero as $n$ tends to infinity.


Let $ a_n \in R$ and let $ b_n \equal{} \frac{1}{n}\sum_{i\equal{}1}^{n} a_{i}$. Show thar, if $ a_n \rightarrow a$ as $ n \rightarrow \infty$, then $ b_n \rightarrow a$ also.

As $ a_{n} \to a$ ,  $ \lim_{n \to \infty}\left [ \sum_{i\equal{}1}^{n} a_{i} \le na \right ]$ Assuming $ a_{n} \le a ,     \forall n$ Making,  $ \lim_{n \to \infty} \left[ \frac{1}{n} \sum_{i\equal{}1}^{n} a_{i} \right ] \le a$.

Thus $ b_{n} \to a$. :blush:  :blush: Am I right??

[quote="DineshManohar"]As $ a_{n} \to a$ ,  $ \lim_{n \to \infty}\left [ \sum_{i \equal{} 1}^{n} a_{i} \le na \right ]$ Assuming $ a_{n} \le a , \forall n$ Making,  $ \lim_{n \to \infty} \left[ \frac {1}{n} \sum_{i \equal{} 1}^{n} a_{i} \right ] \le a$.

Thus $ b_{n} \to a$. :blush:  :blush: Am I right??[/quote]

I am sorry, but it looks like nonsense. Does anybody have any other ideas?


Two points $C,D$ lie in the semicircle with a diameter of $AB$. Suppose that $AC=2\sqrt{5},AD=8,\tan \angle CAD=\frac{1}{2}$.

Find the length of $AB$.

Let $\vartheta  = \angle CAD$. We have:

$\sin \vartheta  = \frac{{\tan \vartheta }}{{\sqrt {1 + \tan ^2 \vartheta } }} = \frac{1}{{\sqrt 5 }},{\rm{   }}\cos \vartheta  = \frac{2}{{\sqrt 5 }}$

From cosinus theorem follows:

$CD^2  = AC^2  + AD^2  - 2AC \cdot AD \cdot \cos \vartheta  = 20$

and therefore:

$R = \frac{{CD}}{{2\sin \vartheta }} = \frac{{2\sqrt 5 }}{{2\frac{1}{{\sqrt 5 }}}} = 5$

Best regards
Leon

Thank you for your reply,Leon.

Consequently you will have an answer $AB=2R=10$.


hey guys i have solved this but want to verify the answer

my answer is

no soln will satisfy this eqn

http://www.mathlinks.ro/resources.php?c=137&cid=39&year=2005
MathLinks :: Poland Finals 2005

problem 1

plz reply and tell whether it is correct or not

i will post up my soln here if u want.............

Actually we get infinitely many solutions.


[quote="ashrafmod"]\[x^{y}+y^{z}+z^{t}=x^{2005}\][/quote]
[hide="Hint"]\[\Longleftrightarrow x^{z-t}\left(x^{2005-z}-x^{y-z}-1\right)=1\]
\[\Longleftrightarrow \begin{cases}z=t\\ x^{y-z}\left(x^{2005-y\right)=1\end{cases}}\]
\[\Longleftrightarrow\begin{cases}z=t\\ x^{y-z}=1\\ x^{2005-y}-1=2\end{cases}\;\text{or}\; \begin{cases}z=t\\ x^{y-z}=2\\ x^{2005-y}-1=1\end{cases}\][/hide]

In the first case: They are integers so $x=1$ or $y=z$. If the former then $x^{2005-y}=3$, impossible because $x=1$. Therefore $y=z=t$, so we have $x^{2007-y}=3$ so $x=3$ and $y=z=t=2004$.
Second case: $x^{y-z}=2$  so $x=2$ and $y=z+1$. $x^{2005-y}=2$ so $y=2004$ and $z=t=2003$.


calculus integration logarithms limit trigonometry function geometry

Needless to say, another marathon thread. But, I want to say a few things about this marathon:

1) This marathon does NOT replace shadysaysurspammed's calc marathon.
2) This marathon should ONLY contain easy calculus problems, problems that only a first-year calculus student is expected to solve. Problems should be about the level of AP Calculus BC but not beyond!
3) This marathon is neither a contest, nor competition, nor 'show-off' galore, please. Problems can be challenging but not overly challenging.
4) No homework problems, we are not here to do your homework.

OK, I write like an old man but anyways let's start with the fun:

[color=blue][size=150]Problem #1[/size][/color]

Evaluate $\int_{-2006}^{2006}\frac{x^{1003}\ dx}{1+x^2}$

[EDIT]Deleted "pls read rules" from the title. Everyone seems to catch on with the word easy :) .

I wanted my marathon to be easy as well being a beginner :D

 hope the level here adheres to what you want 10000thuser.

For problem 2:
  Let f(x)=G(x^2). Obviously, G'(x)=ln(x), so f'(x)=2x*G'(x^2)=2xln(x^2)=4xln(x)
  so f'(3)=12ln(3)

[quote="kunny"][b]Problem 3[/b]

Find $\lim_{x\rightarrow 0} \frac{\sin (2\sin {1003x})}{x}$[/quote]

[hide]$\lim_{x\rightarrow 0} \frac{\sin (2\sin {1003x})}{x}=\lim_{x\rightarrow 0} \frac{\sin (2\sin {1003x})}{2\sin(1003x)}\cdot \frac{2\sin{1003x}}{x}=2\cdot\lim_{x\to 0}{\frac{\sin{1003x}}{1003x}\cdot 1003}=\boxed{2006}$[/hide]

I'm at my science(informatics) class now and I don't have any problem to post :)

Sorry, edited. I was in a hurry because my theacher was comming  :P

Good, Slizzel. :)

[b]Problem 4[/b] 1976 Yokohama National University entrance exam/Education

For three distinct constant numbers $a,b,c,$ let $f(x)=(x-a)(x-b)(x-c).$

Find the value of $\frac{a^2}{f'(a)}+\frac{b^2}{f'(b)}+\frac{c^2}{f'(c)}.$

Thanks. :) , but you are wrong in calculation of $f'(x).$ :(

Easy calculus Marathon.
[b]Problem 28.[/b]
Find:
\[ \int \frac{dx}{\sin x}  \]
This problem is easy. I posted it here just because there’s a very nice solution for it.

[hide="problem 27"]
let $x=r \cos \theta$ and $y= r \sin \theta$, then $(x,y) \rightarrow 0$ (together) implies that $r \rightarrow 0$ since $\sin \theta$ and $\cos \theta$ won't approach zero together.
so the limit is $\lim_{r \rightarrow 0} \frac{\sin r}{r} = 1$
[/hide]

Please numbered your problem Lovasz :) . I would say it is as problem 28,

[quote="Lovasz"]
Find:
\[ \int \frac{dx}{\sin x}  \]
[/quote]

[hide]
let $\sin x =\frac{2 \tan\frac{x}{2}}{1+\left( \tan\frac{x}{2} \right)^2}$ with parametric formula.

Now, let $\tan\frac{x}{2}=t \implies x=2 \tan^{-1} t$ $dx=\frac{2}{1+t^2}dt$

So $\int \frac{dx}{\sin x}=\int \frac{1+t^2}{2t}*\frac{2}{1+t^2}dt=\int \frac{dt}{t}=\ln \left| \tan\frac{x}{2} \right |$[/hide]

[hide="Alternate Solution 28"]
$\int\csc x\,dx=\int\frac{\csc^{2}x-\csc x\cot x}{\csc x-\cot x}\,dx=\ln\left|\csc x-\cot x\right|$
[/hide]

There is another similar problem, which is really easy if you know a good trick:

[size=150]Problem 29[/size]

Find $\int\frac{1}{1+\sin x}\,dx$.

$\int \frac{1}{1+\sin x}\ dx=\int \frac{1}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}\ dx$

$=2\int \frac{1}{\left(1+\tan {\frac{x}{2}}\right)^2}\left(1+\tan {\frac{x}{2}}\right)'\ dx$

$=-\frac{2}{1+\tan \frac{x}{2}}+C.$

[color=blue]Alternative solution[/color]:

$\left(\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\right)'=-\frac{2}{1+\sin x}$

and you can see....

Or alternatively, put $t=\tan\frac{x}{2}$.

You may post the next problem kunny.

[b]Problem 30[/b]

Evaluate $\int_0^1 \ln (x+\sqrt{1+x^2})\ dx$

Solution

[hide]Just integrate by parts taking the first func. as the integrand and second as 1.

The answer turns out to be ln(1+sqrt.2) + (1-sqrt.2)[/hide]

[size=150]Problem 31[/size]

Evaluate

Integral [x^(3m) + x^(2m) + x^(m)] [2x^(2m) + 3x^(m) + 6]^(1/m)  dx
(from 0 to 1)

The expression looks huge in this form. But the problem is quite simple.

[quote="Abishek"]$\int_0^1\left(x^{3m} + x^{2m} + x^{m}\right)\cdot \left(2x^{2m} + 3x^{m} + 6\right)^{\frac1m}\, dx$[/quote]

Just a translation into $\LaTeX$.

[b]Problem 28,[/b]
\[ \int \frac{dx}{\sin x}=\int \frac{\sin x}{1-\cos^2 x}dx  \]
This is the proof I wanted to say.

Hint for problem 31.

Guess the problem got tough. Here is a hint.

[hide]Multiply and divide the integrand by x.
Let the x in the num.r be consumed into the II term in the integrand and the x in denm.r be consumed into th first term.[/hide]


calculus
integration
algebra
polynomial
trigonometry
calculus computations

Let $f(t)$ be the cubic polynomial for $t$ such that $\cos 3x=f(\cos x)$ holds for all real number $x$.
Evaluate

\[\int_0^1 \{f(t)\}^2 \sqrt{1-t^2}dt\]

Expanding $\cos 3x$ we have, $\cos 3x = 4\cos^3 x - 3\cos x \Rightarrow f(t) = 4t^3-3t$, so the integral becomes $\int_0^1 \left(4t^3-3t\right)^2\sqrt{1-t^2} \ dt = \frac{\pi}{8}$.


AMC
AMC 12
USA(J)MO
USAMO
articles
AMC 10

So say you solved #1-24 and you have very little time left and you decide to guess on #25, what would you guess?

If this happens to me on AMC12 I'll go with the letter that wins this poll :D

Yeah, I probably wouldn't guess if I were you. If you don't guess you have an EV of 2.5 points; if you do guess you have an EV of (1/5)(6) = 1.2 points.

Well clearly that 2.5 point will not make any difference, at least for a 10th grader.

Man, if you're sure about #1-24, you would be stupid to not guess on #25. The amount of awesomeness you are if you get a perfect score greatly outweighs the really minor dent in your USAMO index a loss of 2.5 points would bring.

Hmm.  It will pretty interesting just to see what kind of distribution the poll gives :D (don't say anything to bias the results!...like what you chose or what the poll is currently at) ;)

I was reading this article a stats teacher wrote how b is the most common answer in multiple choices, occurring 22% of the time. Can't find it, though.

[quote="dakyru"]I wouldn't just randomly pick a letter.  You should look at the answer choices and pick the one that feels right.   :)[/quote]Agreed. That's why I voted B!

:rotfl:  :rotfl:

Hmm... well, when you don't know the answer on a fill-in-the-blank, it's always $2$, so I'll say $B$, since that corresponds with $2$. :D

Haha I wonder why less people are voting for E

[quote="Philip_Leszczynski"][quote="dakyru"]I wouldn't just randomly pick a letter.  You should look at the answer choices and pick the one that feels right.   :)[/quote]

I disagree. The correct choice on the AMC 12 for #25 is often the one that looks the most artificial and the one you would first eliminate. The people who make the AMC are [i]very[/i] good at this.

If you don't believe me, look at AMC 12 2001 #25.[/quote]

yeah, but sometimes it's not ... I would just pick randomly (roll random int from 1 to 5 on calculator or whatever)

unless you really can eliminate an answer choice (I don't just mean that it [i]looks[/i] bad, but you can actually give some argument why it's wrong) - then roll from 1 to 4

I put C.  Dunno why, it's just a habit for me on random guesses to guess C.  :P

I look for answers that look similar, and then pick one of those :D It works pretty well.

Well doesn't look like the poll's gonna change, so I'm going to guess C for #25 (or whichever problem that is left to guess on if I solve 24 problems).

[quote="white_horse_king88"]A all the way - cuz we all know we want to Ace that test. :D[/quote]

Hahaha A is just out of the race, but D is catching up... hmm maybe I'll change my mind but there are only 6 days left.

The payoff of not answering is more than the expected payoff of guessing (which is 1.2).

However, I tend to pck the number in the middle, or the letter which has not been used much.

[quote="K81o7"]The payoff of not answering is more than the expected payoff of guessing (which is 1.2).
[/quote]

Well again we only talking about a guess on the only problem you have left, and that additionally 2.5 points will not do any good anyway!

[quote="dasherm"]Yeah, but a score of 146.5 makes you look like a wimp.   :P[/quote]

True.  But if you can't do it and have to make a comletely random guess, it's better to leave it blank.

In the past with the AMC10, I've been able to do all of the problems, so I mark everything: along the way, I make a few careless msitakes and end up with a 138.  So this year, with the AMC12, I'll try taking both just for fun, and they'll take the higher score (apparently they are doing that...).

[quote="beta"][quote="K81o7"]
True.  But if you can't do it and have to make a comletely random guess, it's better to leave it blank.
[/quote]

Why?? What difference will the 2.5 points make for freshmen and sophomores?[/quote]

True.  If I have a shot at a 150, it's probably best to take it.
But how did you know I was a freshman/sophomore? (I am a sophomore)

 :ninja:  :ninja:  :ninja:   stalker

You implied in your second-to-last post that you're still eligible for the AMC-10.

[quote="beta"][quote="hockeyadi23"][quote="dasherm"]Yeah, but a score of 146.5 makes you look like a wimp.   :P[/quote]

At least on the AMC 10, yeah. ;)[/quote]

On the AMC12 too! Anyway looks like the answer could be D based on the poll so far.[/quote]

But at least if it's the 12A the 2.5 points matter for USAMO qualification (not much, but it does).

Yeah, but, uh, how many people who are scoring a 144, 146.5 or 150 on the AMC 12 are really worrying that much about qualifying for the USAMO?  It's definitely worth it to go for a 150...it's just not sensible to be [i]that[/i] overly cautious at that point.

Valid point. It passed my mind how rare 150s are on the AMC12s. Unlike AMC10s.


The numbers $x_1 , x_2 , ... , x_n$ are such that $x_1 + x_2 + ... + x_n=0$ and $x_1^2 + x_2^2 + ... + x_n^2 = 1$. Prove that there are two numbers among them whose product is no greater than $\frac{-1}{n}$.

Did you enter the question correctly? The problem as it stands seems intractable; however, modifying the problem slightly yields a solution. The modified problem statement:

The numbers $x_1, x_2, ..., x_n$ satisfy the equations $x_1+x_2+...+x_n=0$ and ${x_1}^2+{x_2}^2+...+{x_n}^2=1$. Prove that there exist $x_i, \ x_j \ (i \ne j)$ such that $x_ix_j \le \frac{-1}{n(n-1)}$.

To solve the new problem, note that

$(x_1+x_2+...+x_n)^2 = {x_1}^2+{x_2}^2+...+{x_n}^2+2x_1x_2+2x_1x_3+...+2x_1x_n+2x_2x_3+...+2x_{n-1}x_n = 1 + 2x_1x_2+2x_1x_3+...+2x_1x_n+2x_2x_3+...+2x_{n-1}x_n = 0$.

The expression after the 1 contains $\frac{n(n-1)}{2}$ terms and includes all possible products of elements in the set of numbers, so we then have:

$2x_1x_2+2x_1x_3+...+2x_1x_n+2x_2x_3+...+2x_{n-1}x_n = -1$
$x_1x_2+x_1x_3+...+x_1x_n+x_2x_3+...+x_{n-1}x_n = \frac{-1}{2}$
$\frac{x_1x_2+x_1x_3+...+x_1x_n+x_2x_3+...+x_{n-1}x_n}{\frac{n(n-1)}{2}} = \frac{-1}{n(n-1)}$

Because the arithmetic mean of the products equals $\frac{-1}{n(n-1)}$, at least one of the products must fall below that mean, so  there exist indeed $x_i, \ x_j \ (i \ne j)$ such that $x_ix_j \le \frac{-1}{n(n-1)}$.

The original problem, I admit, I have no idea how to solve.


conics
ellipse
geometry unsolved
geometry

It's equivalent to [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=209148]this[/url]:

In this case the perspector P is the isotomic conjugate of the anticomplement of C. Let A,B,C the intersection of the tangent. AP,BP,CP meet BC,CA,AB on D,E,F. Call D',E',F' the symmetric of D,E,F w.r.t. C. Then D,E,F,D',E',F' stay on the ellipse.

i've the G (centroid) of triangle defined by the 3 tangents and C center of ellipse
tell me where's de P perspector
[b]tell me the datum  G  C  and P[/b]  ???????

 :(  :(  :(

...call K the simmetric of C w.r.t. G, call J the simmetric of G w.r.t. K, the J is called the anticomplement of C, P is the isotomic conjugate of J. If you don't know what is the isotomic conjugate see [url=http://faculty.evansville.edu/ck6/encyclopedia/glossary.html]here[/url]

If $ a$ is tangent to the ellipse with center $ O,$ then parallel to $ a$ equidistant from $ O$ is also an ellipse tangent. Therefore, you can draw 3 more ellipse tangents, but draw only 2 more (to avoid confusion). The 5 lines have 10 pairwise intersections, but 2 are at infinity (because 2 pairs of lines are parallel). From the remaining 8 finite intersections, pick 5 intersections $ A, B, C, D, E$, such that the pentagon $ ABCDE$ is convex (again, to avoid confusion). Use Brianchon theorem to find the ellipse tangency points with $ AB, BC, CD, DE, EA.$

Diagonals $ BD, CE$ intersect at $ U.$ $ AU$ intersects $ CD$ at $ P.$
Diagonals $ CE, DA$ intersect at $ V.$ $ BV$ intersects $ DE$ at $ Q.$
Diagonals $ DA, EB$ intersect at $ X.$ $ CX$ intersects $ EA$ at $ R.$
Diagonals $ EB, AC$ intersect at $ Y.$ $ DY$ intersects $ AB$ at $ S.$
Diagonals $ AC, BD$ intersect at $ Z.$ $ EZ$ intersects $ BC$ at $ T.$

$ P, Q, R, S, T$ are the ellipse tangency points with $ CD, DE, EA, AB, BC.$

P is the isotomic conjugate of J

what ellipse is this not tangent to the 3 tangents lines ?

 :(


Given a quadratic polynomial $f(x)$, a fixed $\delta > 0\in \mathbb{R}$, and a fixed $n > 2 \in \mathbb{N}$, show that for all real $x$,

$ \sum_{j=0}^{n}{(-1)^{j}{n\choose j}f(x+\delta j)}= 0$

[quote="t0rajir0u"]A solution in three movements!

[hide="I:  Allegro"] Let $f(x) = ax^{2}+bx+c$.  We consider the summation for the constant term only.  It's obviously zero; consider $(1-1)^{n}$.  This is simple enough; nothing unexpected, as we walk sprightly along. [/hide] [hide="II:  Adagio"] Let's slow down a little and consider the linear term.  Now,

$\sum_{j=0}^{n}(-1)^{j}{ n \choose j }b(x+\delta j) = bx \left( \sum_{j=0}^{n}(-1)^{j}{ n \choose j}\right)+b \left( \sum_{j=0}^{n}(-1)^{j}{ n \choose j}j \right)$

As before, the first term is zero.  What happens to the second?  Well,

${ n \choose j}j = \frac{n! j}{j! (n-j)!}= n \frac{(n-1)!}{(j-1)!(n-j)!}= n{ n-1 \choose j-1 }$

A multiple of a summation that we determined previously to be $0$.  [/hide] [hide="III:  Presto"] We now consider the sum

$\sum_{j=0}^{n}(-1)^{j}{ n \choose j}a(x^{2}+2 \delta j+\delta^{2}j^{2})$

The first two terms are zero, as before.  What of the third?  Well, we rewrite it as

$a \delta^{2}\sum_{j=0}^{n}(-1)^{j}{n-1 \choose j-1}(j-1+1)$

We split again, and the summations are still zero.  Presto! [/hide][/quote]

beautiful.


Is there a time frome on when we can get our solutions graded?

If we give correct and accurate answers, how can I be graded down?

[quote="pakagawa"]Is there a time frome on when we can get our solutions graded?[/quote]
We expect that the grading will be complete and the scores posted by the end of the month.

[quote="pakagawa"]If we give correct and accurate answers, how can I be graded down?[/quote]
If you give a fully correct and accurate answer (which means showing your work, not just listing the final answer), you will get full credit.


I think there is a lag when posting replies.
Yesterday, I posed a message, and when I looked again on the forum, it was not there.
I used Opera and I know that when you go back, it doesn't refresh, but I'm  didn't go back, so the page refreshed, i made sure it did.
When I posed the reply again, the previous post was shown  :?

I guess there was a temporary bug.

At least, today when I logged in and clicked on "View posts since last visit", I read "No posts met your search criteria", what was actually wrong - there were several messages posted that night -, so I guess there were some temporary problems.

  dg

Very strange anyway  :?
My messages seemed like flooding :) 2 same messages one after the other

well it's more of a phpBB2.0.6 modded problem, so I can't help it too much. maybe I will try to upgrade to 2.0.8, but I think it's best to wait for the soon-coming 2.2.0 :D

I think we're still affected by what could be called "the curse of the Beta version", huh? :D:D.

These problems come and go. I've had quite a few of this sort, but you get used to it :).

don't worry, mathlinks will meet the princess and finally leap out of beta stage :)

Ok so the beta version is over, now it's the curse of the 1.0.0 version :D:D
well the site's speed will be a little bit low these couple of days, because ML takes a lot of RAM,  some RAM needs to be bought.

Until then, please give me some feedback in this thread about the site's current speed (and other possible errors).

It doesn't seem to be significantly slower. It's so nice to have Mathlinks back! I was getting bored :).

Yeah, it's a curse. Especially the speed is a curse. The slow server is definitively inappropriate for such a big math forum like this one. Maybe, we (= the German forum members, for instance me) can help financially?

  Darij

for now we have aquired 256 Mb of RAM, but we really need to upgrade the proc. speed (it's a PIII at 500).

I'll keep you guys up to date with everything.

there is also now a page generation time at the bottom of each page. please post it, for this particular page and for the portal and forum.

thanks.

The speed has got much better today!

For instance, for this page here, I get:
  [i]Page generation time: 11.6778s (PHP: 95% - SQL: 5%) - SQL queries: 21 - GZIP disabled - Debug on[/i]

For [url=http://mathlinks.ro/viewtopic.php?p=16917#16917]another one[/url], I see
  [i]Page generation time: 9.7897s (PHP: 89% - SQL: 11%) - SQL queries: 21 - GZIP disabled - Debug on[/i]

  Darij

it should normally be under 3 seconds. it's the server all right- I could not access the database earlier because there were over 200 connections (there are 3 other sites hosted on the same server).

please other people post your page generation times (located at the bottom of each page).

I got

Page generation time: 19.9932s (PHP: 98% - SQL: 2%) - SQL queries: 21 - GZIP disabled - Debug on

for this particular page;

For the link called "another one" in darij's last message on this topic I got

Page generation time: 13.2094s (PHP: 97% - SQL: 3%) - SQL queries: 24 - GZIP disabled - Debug on

I bet the server is on Windows, on Linux it should be better  :P