$ 1$. Let $ ABC$ be a triangle with circum-circle $ \tau$. Let $ M$ be a point in the interior of triangle $ ABC$ which is also on the bisector of $ \angle A$. Let $ AM, BM, CM$ meet $ \tau$ in $ A_{1}, B_{1}, C_{1}$ respectively. Suppose $ P$ is the point of intersection of $ A_{1}C_{1}$ with $ AB$; and $ Q$ is the point of intersection of $ A_{1}B_{1}$ with $ AC$. Prove that $ PQ$ is parallel to $ BC$.

Strange but one actually prove more that $ M$ is collinear with $ P$ and $ Q$.Clearly $ \angle MC'C\equal{} \angle BB'M\equal{}\frac{\angle A}{2}$.Join $ PM,QM$.Observe that quadrilateral $ AC'PM$,$ AB'PM$ and $ AC'MB'$ are cyclic.The first because of the previous angle equality the last one because of $ \tau$.Thus $ \angle AC'M\equal{}\angle AMQ$,and $ \angle AMP\equal{} \angle AB'M$.Thus $ P,M,Q$ are collinear and since $ \angle APM\equal{}\angle AC'M\equal{}\angle ABC$ we have $ PQ \parallel BC$

As given $ M$ lies on angle bisector of $ \angle A$, $ \angle BAA' \equal{} \angle BB'A' \equal{} \angle CAA' \equal{} \angle CC'A' \equal{} \frac{\angle A}{2}$
Also $ \angle PC'M \equal{} \angle PAM \equal{} \frac{\angle A}{2}$ so $ PMAC'$ is a cyclic quadrilateral.
And thus $ \angle AC'M \equal{} \angle APM$. Because $ \angle AC'M \equal{} \angle AC'C \equal{} \angle ABC$
$ \equal{}> \angle APM \equal{} \angle ABC$ and hence $ PM \parallel BC$

Using similar arguments we can easily prove that $ MQ \parallel BC$
As $ M$ is the common point in the two segments $ PM$ and $ MQ$ both of which are parallel to $ BC$, thus $ M$ lies on $ PQ$ and $ PQ \parallel BC$.
Hence Proved. :)

I think that this is a nice application of Pascal theorem  :)  . If $ l$ is the tangent to the circumcircle by $ A'$ Just apply pascal theorem to the degenerated hexagon $ A'A'ABCC'$ and we get that $ M$ , $ P$ and the intersection of $ BC$ with $ l$ are collinear, but clearly $ l // BC$ therefore $ MP//BC$ . Using again pascal theorem for  $ A'A'ACBB'$ we get $ MQ//BC$ , hence $ P,M,Q$ are collinear and the line containing them is parallel to BC.


AMC
AIME
USA(J)MO
USAMO
AMC 10
AMC 12
search

Okay, I was wondering some things about these competitions.

1.) Why do they have an AMC10,[b] and[/b] an AMC12?

2.) How do they determine who goes to the AIME?

3.) How do they determine who goes to the USAMO, based on the scores on the AIME?

4.) What happens after USAMO?

5.) How do they determine who goes to the next place?

6.) How many people go to whatever is after the USAMO?

7.) How does it get from USAMO to IMO?

8.) What happens at the IMO?

Most of these questions are actually answered on the AMC website or on the tests themselves.

1) This question is common sense.
2) It's on the back of the test, and probably the website.
3) It's based on an index that's determined from year to year.
4) MO(S)P
5) On the website
7) MO(S)P
8) See the IMO or website, or the site of any country participating.

Tell me if I am right or not.

 I think that 10,000 people go to AIME, and 500 go to USAMO, and they determine the people who go, by the top scores.

AIME is top 1% from AMC 10 and top 5% from AMC 12 (approximations)

USAMO is usually around 500, since like 2005.

A general suggestion: do your own looking around for answers before posting questions. This applies to your topics in the math forums too; you want everyone to post problems in your topic rather than finding them yourself (we have years of message board posts with them and a search function).

In line with that, here are the answers to your questions through logical links from [url=http://www.unl.edu/amc/]this site[/url]. Try this yourself next time.

1. http://www.unl.edu/amc/f-miscellaneous/faq.shtml#amc1012 (AMC 10/12 #5 and #6)

2. http://www.unl.edu/amc/f-miscellaneous/faq.shtml#aime (AIME #7)

3. http://www.unl.edu/amc/f-miscellaneous/faq.shtml#usamo (USAMO #5)

4.
- http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml (The twelve top scoring...)
- http://www.unl.edu/amc/a-activities/a6-mosp/factoidsmosp.shtml (Participation Selection), or just look at the next entry on the list of contests on the left

5. http://www.unl.edu/amc/a-activities/a6-mosp/factoidsmosp.shtml (Participation Selection)

6. http://www.unl.edu/amc/a-activities/a6-mosp/factoidsmosp.shtml (Participation Selection)
Admittedly doesn't give the whole story, but you could also look at the list of participants at past MOPs:
- http://www.unl.edu/amc/a-activities/a6-mosp/a6-1-mosparchives/2008-ma/08mosppart.shtml
- http://www.unl.edu/amc/a-activities/a6-mosp/a6-1-mosparchives/2007-ma/07mosppart.shtml
Additionally, this is asked literally at least once every three months on the forums here.

7. http://www.unl.edu/amc/e-exams/e8-usamo/usamo.shtml (The twelve top scoring)... This is the only question that I couldn't fully answer from following obvious directions on the site. For those top twelve there is a test called the TST very similar in format to the USAMO/IMO; your score on that is added to your USAMO score to get a final index. The top 6 are selected.

8. http://www.unl.edu/amc/e-exams/e9-imo/imo.shtml

Your second post: http://www.unl.edu/amc/f-miscellaneous/faq.shtml#usamo (USAMO #5, the 7th item)


You can find quite a few proofs [url=http://en.wikipedia.org/wiki/Proof_that_the_sum_of_the_reciprocals_of_the_primes_diverges]here[/url].


calculus
integration
trigonometry
function
limit
complex analysis

Find the complex integral $I=\int_{|z|=1}\frac{\sin z}{\sin (z^{2})}\ dz$.

The function $\sin (z^{2})$ does not vanish on the boundary of the unit circle and has a pole of order 2 at 0 with residue 1.
$Res(f(z),0)=\lim_{z\to 0}\frac d{dz}\left(\frac{\sin z}{\sin z^{2}}\right)=1$ (if I did not make any mistake)
So our integral equals (by Residue Theorem) $2\pi i$

Needless to say, your answers are correct.
Laurent expansion.


inequalities
combinatorics unsolved
combinatorics

There are $ 25$ towns in a country. Find the smallest $ k$ for which one can set up two-direction flight routes connecting these towns so that the following conditions are satisfied:
1) from each town there are exactly $ k$ direct routes to $ k$ other towns;
2) if two towns are not connected by a direct route, then there is a town which has direct routes to these two towns.

$ k > 4$ because from a city we can reach at most $ k$ city directly and $ k(k \minus{} 1)$ with passing from exactly one city, but we must reach each of the cities with using at least one of the two ways (directly or passing from exactly one city) so $ k^2\geq 24$.
If $ k \equal{} 5$ we get the sum of degreess of all vertices $ 125$. Contradiction because it is an odd number.
$ k \equal{} 6$, Construct $ 5$ cycles with lenght $ 5$ and adjoin any two of them like in the Petersen Graph. Which is [url]http://en.wikipedia.org/wiki/Petersen_graph[/url]

A direct solution: :lol:

Consider an arbitrary town $ A$, the set of $ 24$ other towns is divided into to subsets: $ R(A)$ is the set of towns which has direct route to $ A$ and $ N(A)$ is the set of towns which does not have route to $ A$. We have $ |R(A)|\equal{}k$ and so $ |N(A)|\equal{}24\minus{}k$.

For every town in $ N(A)$, it must have a route to a town of $ R(A)$ (in order to have a common town to with $ A$), but the sum of the number of the route from any town in $ R(A)$ to any town of $ N(A)$ is $ (k\minus{}1)^2$ (we can not count $ A$ again), so the following inequality must hold:

$ (k\minus{}1)^2 \ge 24\minus{}k \Leftrightarrow k(k\minus{}1) \ge 23 \Leftrightarrow k \ge 6$

When $ k\equal{}6$, Petersen's graph yields a way of making a system of routes which satisfies the desired.

It is clear (by above solutions) that $k \ge 6$. A construction for $k = 6$ is as follows: let the vertices be labeled $0, 1, 2, 3, \dots, 24$. Connect $a, b$ by an edge if and only if $a + b \equiv 0, 1, 3, 7, 12, 18 \pmod {25}$. (The six vertices $a$ satisfying $2a \equiv 0, 1, 3, 7, 12, 18$ will have degree 5, but then draw three edges among them to make each vertex degree 6.) Observe that every (nonzero) residue mod 25 can be represented as $x - y$ for some $x, y \in \{0, 1, 3, 7, 12, 18\}$. Then the path $a \rightarrow x - a \rightarrow x - (y - a) = x - y + a$ shows that any $a$ can reach any other $b \neq a$ in two steps or fewer.


Can someone please help me out with this?
'Find the sum of the series
$\frac{4}{3}+\frac{9}{8}+\frac{16}{15}+...+\frac{n^{2}}{n^{2}-1}$'

Thanks for your time.
jimi

Um, that series diverges. The terms tend to $1$, when for an infinite series, the terms have to tend to $0$ for the series to converge (Though that condition alone doesn't guarantee convergence. Take the series $\sum_{i=1}^{\infty}\frac{1}{i}$. That series is well-known to diverge, even though its terms tend to $0$.).

EDIT:

Wait, is this an infinite series or do you just want the first $n$ terms?  I just immediately took it to be an infinite series. My apologies if it's the latter.

Well, first of all they want to have it in terms of n, so it shouldn't say diverge.

Second of all, notice that

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

You figure out what to do next.

Yeah my bad. :roll:  I fixed my (unhelpfu) hint.

do the partial fraction decomposition helps...

Not terribly different, you just compressed $\frac{1}{k+1}+\frac{1}{k}$.

What is a good book or good site to practice/learn the methods used for this problem? It appears that some of the steps posted were semi-trivial to the poster, whereas it would take me a lot of thought. Thanks.


Many thanks to webmaster for making Vietnamese box.
But i thinks you should send messages to everyone to tell about this. I think many people havent known about it, for example me. Today i just know it.

Hi :)

Glad you like it. I am still waiting to add some features to make it complete (including controls in Vietnamese) and after that I was planning to announce all members :)

But you can announce them quicklier :D

i'm glad to know you ! what's your name ??? Can i make you be friend ?? you like maths , physics , chemistry ??? like me about hobbies ??


modular arithmetic
quadratics
number theory
greatest common divisor
search
rotation
Asymptote

If x and y are both integer, not both zero,
find the minimum value of the absolute value of $ 5x^2\plus{}11xy\minus{}5y^2$

$ (x,y) \equal{} (1,0)$ gives $ 5$, because discriminant is non-square the given expression cannot be $ 0$; now suppose $ 5x^2 \plus{} 11xy \minus{} 5y^2 \equal{} z$ where $ |z| < 5$. Completing the square gives $ (10x \plus{} 11y)^2 \minus{} 221y^2 \equal{} 20z$; taking this modulo $ 13$ gives that $ (10x \plus{} 11y)^2 \equiv 7z \pmod{13}$, but $ \pm 7, 21, 28$ are not quadratic residues $ \mod 13$; similarly taking modulo $ 17$ gives that $ (10x \plus{} 11y)^2 \equiv 3z \pmod{17}$ but $ \pm 3,6,12$ are not quadratic residues. Hence $ |z| \neq \pm 1,2,3,4$, and $ 5$ is minimal.

Last year's PuMAC's [url=http://www.princeton.edu/~ahesterb/PUMaC08Problems/PUMaC%20Power%20Test%202008-9]power round[/url] gives a quick easy way to attack this also; in gist, let $ f(x,y)$ denote the given quadratic form, obviously the minimum absolute value occurs when $ \text{gcd}\,(x,y) \equal{} 1$; further observe $ f(x_1 \plus{} x_2,y_1 \plus{} y_2) \plus{} f(x_1 \minus{} x_2,y_1 \minus{} y_2) \equal{} 2(f(x_1,y_1) \plus{} f(x_2,y_2))$. One can then generate all $ f(x,y)$ with relatively prime $ x,y$ starting from $ f(0,1),f(1,0),f(1,1)$; using this "arithmetic sequence" rule tells us to search between values of $ f$ that switch from positive to negative and vice versa. Tracing along this "river" gives $ (5, \minus{} 11, \minus{} 7, \minus{} 5,7,11,5)$ where it then repeats.

Here is my solution:

[hide]We write the expression in the following manner.

$ 5((x\plus{}y)^2) \plus{} xy \minus{} 10y^2$
Then we arrange to receive.

$ 5((x\plus{}y)^2\minus{}2y^2) \plus{} xy$

We want to minimize $ (x\plus{}y)^2\minus{}2y^2$ and get rid of $ xy$, which implies on should be zero.  This happens when $ x\equal{}0$ and $ y\equal{}0$. We then get 0. But this is not allowed.  We then increase x to 1 and get 5.[/hide]

[b]cipos[/b], the equation in question is a rotated hyperbola. Let $ f(x,y) \equal{} 5x^2 \plus{} 11xy \minus{} 5y^2$.

The level curve $ f(x,y) \equal{} 0$ is just the union of the two asymptotes. The level curve $ f(x,y) \equal{} 1$ departs from this just slightly. What's to tell you that there won't be a lattice point somewhere way up near the asymptotes?

$ f$ close to 0 does imply that the point should be near the level curve $ f(x,y) \equal{} 0$, but the latter is two entire lines, and not just the origin -- in other words, there's absolutely no reason whatsoever to guess that the minimum value will occur near the origin.

In this case, the number theory approach is more useful.


[quote="Inspired By Nature"][quote="D.P.L"]Would someone be kind enough to tell me how hard the book is. Aops math book page says it's about getting started level. Whereelse amazon says it's for 12-14 years old?
thanx[/quote]
Its not fair that your posts seem to go unnoticed; WAKE UP PEOPLE!!! :mad: :furious:
I don't have personal experience with the book but I think from what i have heard, the Aops math book page is right. I think for the most part, it'll be quite easy with probably very little hard material. :)[/quote]

Will it be helpful in helping me to score a 8 on aime?

[quote="Inspired By Nature"]Its not fair that your posts seem to go unnoticed; [/quote]

First of all, there is no issue of fairness.  Second of all, it's totally reasonable that posts like this go unanswered because they're in the wrong forum (they should be in the Books and Journals forum) and because a search of the forum would lead to lots of past opinions of it.

Finally, the notion that your AIME score is going to be determined by which books you study from (as in, "Will it be helpful in helping me to score a 8 on aime?") is just plain silly.

[quote]First of all, there is no issue of fairness. Second of all, it's totally reasonable that posts like this go unanswered because they're in the wrong forum (they should be in the Books and Journals forum) and because a search of the forum would lead to lots of past opinions of it.
[/quote]

Where is the book and journal forum?

It's under Other Problem Solving Topics.

[quote="D.P.L"][quote]OF COURSE IT WILL HELP YOU GET AN 8 ON THE AIME

INTERPRET THAT AS YOU WILL[/quote]

[quote]No, it won't. It might help a bit on mathcounts, but if you have read AoPS 1 entirely, skip it.[/quote]

[hide=" :maybe: "]im confused[/hide][/quote]

It is a MathCounts level book and won't help you much with AIME.


calculus
function
real analysis
real analysis unsolved

Consider a function $f$ on $[0, 2]$ with these properties;
$\forall x\in[0, 2]\;|f(x)|\leq1,\;|f''(x)|\leq1$. Prove that $|f'(x)|\leq2$ on that interval and that, this bound is reachable.
PS Does anyone know the origin of this problem?

Littlewood stated (and proved) a qualitative result of this kind in his book [i]A Mathematician's Miscellany[/i]:
[quote]if $\lim_{x\to\infty}f(x)=0$ and $f''$ is bounded, then $\lim_{x\to\infty}f'(x)=0$. [/quote]
He remarks that the result was known earlier (he found it independently), but does not say who was the first to prove it.


Prove that for each natural number $ n>1$, we always find out a natural number $ z$ such that $ z$ can not be expressed into the form $ x^n\minus{}y!$ where $ x,y$ are natural numbers.

posted already:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=216111


[quote="Davron"] Given that 1/a+1/b+1/c=1/(a+b+c), then show that inequality holds...

0>=Ssym(a^2b)    ;)  maybe nice

Davron[/quote]
Your condition is equivalent with $\frac{(a+b)(b+c)(c+a)}{abc(a+b+c)}=0$ maybe this helps.

$a=1 b=-1 c=1$
 give $S=2$
I dont under stand this problem.
can you clarify it Davron?

Ok sure i will but i have no time now ok sorry i am gonna do that later ...

Davron


Find all pair integer numbers $ (x,y)$ x,y>0 satisfy equation: $ x^2\minus{}8xy\plus{}y^2\plus{}6\equal{}0$

$ x^2 \minus{} 8yx \plus{} y^2 \plus{} 6 \equal{} 0$

$ x \equal{} \frac {8y\pm 2\sqrt {15y^2 \minus{} 6}}{2} \equal{} 4y\pm\sqrt {15y^2 \minus{} 6}$

$ 15y^2 \minus{} 6 \equal{} z^2$, i.e. $ 15y^2 \minus{} z^2 \equal{} 6$.

$ (y,z) \equal{} (1,3)$ is a solution. now, if $ (y,z)$ is a solution, so is $ (4y \plus{} z,15y \plus{} 4z)$: $ 15(4y \plus{} z)^2 \minus{} (15y \plus{} 4z)^2 \equal{} 15y^2 \minus{} z^2 \equal{} 6$. so $ (y,z)\mapsto (4y\plus{}z,15y\plus{}z)$ gives infinitely many solutions to $ 15y^2\minus{}z^2\equal{}6$, which in turn gives solutions to the original equation. are these all solutions?

[quote="mrmohamed"][quote="duythuc_lqd"]Find all pair integer numbers $ (x,y)$ x,y>0 satisfy equation: $ x^2 \minus{} 8xy \plus{} y^2 \plus{} 6 \equal{} 0$[/quote]

Hello

 (x , y ) = ( 1 , 1 )   :wink:  :wink:
mrmohamed[/quote]

find *all* solutions. plus, if you'd been paying attention you would have seen my post...


[quote="Jongao"]:?:
This is a problem from the 2006-2007 MathCounts handbook (Workout 8, Question 3).
The problem is:
In MathCountia, the only currency is the 7-zorkmid and 11-zorkmid coins.  Some prices, like 42,43,and 44 zorkmids can be paid with exact change, while others, like 5 or 20 zorkmids, cannot.  How many prices cannot be paid with exact change?[/quote]

[hide]
First off, we can just list them all.  However, if we take this approach, we need to know when to stop.  The chicken mcnugget theorem tells us that 59 is the largest that can't be made from just 7 and 11 coins.  If we use complementary counting, we then find the number of multiples of 7 from 1-59 are there and the number of multiples of 11 from 1-59, then add.  (Make sure you see why we don't need to subtract for overcounting here)  There are 8+5=13 in this list, so 59-13=46.  However, we need to count the cases where we add multiples of 7 and 11.  We eventually find 4+4+3+2+2+1=16 more.  So the answer is 46-16=30?
That almost makes no sense.
[/hide]

[hide="general method"]basically consider a table of 7's and 11's, so down the column we write 0, 7, 14, 21, etc., and down the row we write 0, 11, 22, etc. Now in each row, note that the residue modulo 11 is the same, and each number modulo 11 is covered in each row starting from some number on. We just have to consider the first 11 rows and look at what is not covered (eh that's pretty annoying)[/hide]


1.propanoyl chloride and propanoic acid
2.Propanamide and propanoic acid

the second answer:
One can reduce both the compounds with suitable reducing agents.
for eg: using Li Al H4. then one will answer test for Amines the other wont. :!:
OR
Amides: use NaOH test: the dense white fumes of Ammonium are formed.
The first one: I dunno.

tell exactly what u will do and what u will see :)

Prepare a sample (which is easy since all compounds are liquids) and take an IR spectrum:

1. The acid will show the C=O band at about 1720 cm-1 (plus the very large -OH band centered at 3000 cm-1), while the acid chloride will show that band at 1810 cm-1.

2. The amide's C=O band will appear at about 1630 cm-1.

[quote="pardesi"]sorry we aren't into NMR spectroscopy :oops:
 i meant chemical methods[/quote]
Pardesi it is IR not NMR.

1.Sodium bicarbonate test:Effervescence-acid and acyl chlorides fumes in water giving irritating smelling gas HCl which gives white fumes with glass rod dipped in $ \ce{NH4OH}$
2.I think $ \ce{NaHCO3}$ itself is enough. Acid will get brisk effervescence whreas amides don't.

sorry da there for NMR :oops:  :rotfl:  :rotfl:
1)it's wrong and that was my basic question seems thsi test doesn't hold...i don't understand why
2)it's right

Yes pardesi that's what I said add water and we see that acyl chlorides fume in water and test it by glass rod dipped in $ \ce{NH_4OH}$

That's right. And for 2. just add $ \ce{NaHCO3}$.

sorry for not posting earlier, got my exams now.
can we use agno3 for the acid chloride?

no it seems u can't...but i don't know why..... :)

@Varun:have ur boards started :o

nope, but got my preps, and pracs, also IIT and AIEEE mock exams(which i doubt that ill be taking).
besides boards will start off soon. :D
when r ur boards?

i am in CBSE so from 1st March...which board do u belong to?

karnataka state board(altho i did my Xth from ICSE) rite now. its absolute crap.
you've got loads to mug. neways all the best for ur boards n stuff


vector
combinatorics unsolved
combinatorics

1. Let two fixed vector $ u, v\in \{0,1\}^n$ such that their hamming distance $ d(u,v)\equal{}m$, where $ m$ is a fixed natural number between $ 0$ and $ n$. What is the number of vertors $ w\in\{0,1\}^n$ such that $ d(u,w)\plus{}d(w,v)\equal{}m$.
2. Suppose that $ u<v$ (meaning $ u_i<v_i$) and such that their hamming distance $ d(u,v)\equal{}k$, where $ 0<k\leq n$. What is the number of vectors $ w\in\{0,1\}^n$ such that $ u\leq w\leq v$.
Please help me to solve them as specific as possible. Thanks very much.

No one can not help me??? Please help me. Thank you very much.

I have an answer.
1. $ u,v$ must have $ m$ different components, and the remainders are the same.
Since $ d(u,w),d(w,v)\leq m$, so $ w$ has  the same components to $ u$ and $ v$ in that $ n\minus{}m$ positions. In $ m$ positions, inspite $ w$ take the value $ 0$ or $ 1$ then we always have $ d(u,w)\plus{}d(w,v)\equal{}1$.(Explaination : if at the position $ i$ in that $ m$ positions, $ w(i)\equal{}u(i)$ , and the remaiders are the same, then $ w$ has $ m\minus{}1$ different positions to $ v$, so we have $ d(u,w)\equal{}1, d(w,v)\equal{}m\minus{}1$.).
Therefore, we have $ 2^m$ vectors $ w$ satisfies $ d(u,w)\plus{}d(w,v)\equal{}m$.
2. Similar to 1, we have $ 2^k$ vectors.


Okay, I understand that to pass a lesson and see more videos, you have to pass a certain amount of questions on that topic. And there was a point where I had more videos which disappeared, but that is not my question. The questions taht I am being asked are on the things where I am not allowed to watch the videos (Which I understand) but I am not being asked questions under the topics that I should be mastering, thus I can not move further, and when/if I eventually do, I will be in a larger hole because I would have missed all of the questions it is currently asking me, without having any knowledge of how to do them.
Please help!

what you have to do is:
1. watch the videos you have and master them
2. get questions right, leading 2 earning badges,
once you get more videos i think you'll be able to
be asked different/harder questions!!! ;)

I have investigated your account and it appears that you do have access to all the videos from topics which you should be primarily receiving problems from.


inequalities
trigonometry
LaTeX
inequalities unsolved

I'm very angry with myself but I can't prove this inequality ( It's a small part of the big problem and the solution did't prove it )
     The sequence of number [tex]S_n[/tex] is defined by:
          \[ S_n= \sum_{k=1}^n k\sin \frac k{n^3}\]
          Prove that:    $S_{n+1}<S_n $.

[Admin Edit: Fixed latex code].

Is it $ \sum^n_{k=1} k \times sin( \frac {k}{n^3}) > n+1$ ?
In that case, it is obviously false (just check $n=1$).

More generally, using that $sin(x) \leq x$ for all $x \geq 0$, it easily follows that
$S(n) \leq \frac {2n^2 + 3n + 1} {6n^2} = \frac 1 3 + \frac 1 {2n} + \frac 1 {6n^2}$.

Pierre.

Oh no, I didn't type n+1<S(n)
  The right one is:
   Prove that S(n+1)<S(n).
 I don't know how to use Latex, so sorry for the mistake.

I think that someone in this forum has proved it completely.Who know the answer, please help me.

We will use following inequality
(*) $x>\sin x> x-x^3/6$.

Due to (*) we have
1) $S_n>\sum_{k=1}^n (\frac{k^2}{n^3}-\frac{k^4}{6n^9}) \geq \frac{(n+1)(2n+1)}{6n^2}-\frac{1}{6n^4} =\underline{S}_n$.
2) $S_n<\sum_{k=1}^n \frac{k^2}{n^3} = \frac{(n+1)(2n+1)}{6n^3} = \overline{S}_n$.

Now it is sufficient to prove that $\underline{S}_n > \overline{S}_{n+1}$. Last inequality equivalents to
\[
\frac{(n+1)(2n+1)}{6n^2}-\frac{(n+2)(2n+3)}{6(n+1)^2} > \frac{1}{6n^4}.
\]
Actually, $LHS=\frac{1}{2n(n+1)}+\frac{2n+1}{6n^2(n+1)^2} >  \frac{1}{6n^4} = RHS$.


analytic geometry
geometry unsolved
geometry

Dear Mathlinkers,
let ABC be a triangle, L the de Longchamps point of ABC, A' the point of intersection of AL and BC, A'b, A'c the feet of the perpendiculars issue from A' onto AC, AB.
Prove synthetically that AL, BA'b and CA'c are concurrent.
Sincerely
Jean-Louis

We prove a stronger version of the problem.

[b]Proposition.[/b] $P$ is a point on the side $BC$ of $\triangle ABC.$ $X,Y$ are the projections of $P$ onto $AC,AB.$ Then $AP,BX,CY$ concur $\Longleftrightarrow$ either $P \equiv B$ or $P \equiv C,$ or $P$ is the cevian trace of the De Longchamps point $X_{20}$ on $BC.$

$AP,BX,CY$ concur, i.e. pedal triangle $\triangle PXY$ of $P$ WRT $\triangle ABC$ is perspective with $\triangle ABC$ $\Longleftrightarrow$ $P$ lies on the Darboux cubic $\mathcal{Q}$ of $\triangle ABC.$ Hence $P$ is the trace of $\mathcal{Q}$ on $BC.$ Barycentric equation of $\mathcal{Q}$ is given by

$\mathcal{Q} \equiv \sum_{\text{cyclic}}x(a^2S_A - S_BS_C)(b^2 z^2-c^2y^2) = 0 \ (\star)$

Thus, setting $x=0$ in the equation $(\star)$ gives the coordinates of $P$

$a^2(b^2S_B-S_CS_A)yz^2+a^2(c^2S_C-S_AS_B)zy^2=0$

$yz[z(b^2S_B-S_CS_A)-y(c^2S_C-S_AS_B)]=0$

Therefore, the lines $AP,BX,CY$ concur $\Longleftrightarrow$ either $yz=0,$ i.e. $P \equiv B \equiv (0:1:0)$ or $P \equiv C \equiv (0:0:1),$ or $P \equiv (0:b^2S_B-S_CS_A:c^2S_C-S_AS_B),$ which is the cevian trace of $X_{20}$ on $BC.$

P.S. For a derivation of $(\star)$ see the topic [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=334909]Locus[/url].

Lemma
Lat HA is height of ABC and N is it's midpoint .
Then angle NA'bA = HAB , angle AA'cN = CAH
I prove it here :
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=398778&hilit=Find+relation
Let A'A'b and A'A'c intersect line thru point A and || to BC at Y and X
Not hard to prove using Cheva theorem that if YA = AX then this problem done
Circle (A'YA) intersect A'bN at points P and A'b
Angle NPA = A'bA = CAH = AA'cN and angle PNA = A'cNA , so triangle PNA = A'cNA
A'cA = AP , angle PYA = NA'bA = NAA'c = AXA'c
TTriangle PYA = A'cAX , so YA = AX


I've tried to solve this problem but I got stuck, can someone please help explain it to me, thanks.

Simplify for all values of a for which the expression is defined

1  -   2
       ___
        a
______
4
__   -   1
a^2

no, its 1 then - 2 over a, sorry i'm kinda new at this

[quote="Julieny17"]sorry for being such a noob, hope this clears it up

1 .....    -      2
.......            ___

........            a[/quote]
$\frac{1-\frac{2}{a}}{\frac{4}{a^{2}}-1}$.

Thanks, you've helped me see where I went wrong  :)


Hello everyone,

Do you know if there is a program or a dictionary in which I will be able to add words (in english) and their meanings in my first language, if possible without any other words in it?

Thank you in advance,

S.

I don't think you can do that unless you make your own web.
http://dictionary1.classic.reference.com/help/faq/language/t54.html

There are things such as note cards that are designed precisely for this purpose ;)

Specifically, here is a free flash-card program in which you can add your own cards:
[url]http://www.mnemosyne-proj.org[/url].

Do it in the C++ or in any other programming languages. But using of map in the C++ makes it way easier.

I have written somewhat a dictionary described, but it's quite primitive.

http://www.definition-of.com/
is a site where you can add words defined by you, in any language.
Unfortunately, some  people used that site to add a variety of slang words.

There may be others in your native language, which I am not aware of.

Or you can use the wiktionary (en.wiktionary.org), the wikipedia dictionary. You can add definitions, pronunciations, expressions, quotes and so on.


Greetings! For those who don't know me very well, I'm Sasarai, a long time reader but not so much of a poster here on AoPS. Now, I've been wondering if anyone can give me some insight. You see, my family ordered 4 AoPS books for my birthday, but it doesn't seem like the books have ever arrived. We placed the order in late December, and I know that it said it'd take 2-4 weeks to deliver. Just so you know, I live in an igloo (see: living in Canada) that's right across the border from Detroit, MI. I don't mean to come off as impatient or rude, so please don't take it that way. I'm just curious to see if anyone has had a similar experiencing with mailing delay. I'm rather anxious to get the books into my hand as I'd really like to start reading them and doing the problems for fun (having completed AoPS 1 and 2, which I thought were absolutely wonderful). In any case, thank you for your time :)

-Sasa

Hello,

Please send an email to orders@artofproblemsolving.com explaining your issue. Include an order number if you have it. If not, please include your billing and shipping information so that the shipping department can look up your order.

As you are in Canada, shipments may take longer as they have to go through extra inspection through the postal system.

The same thing happened to me, but the problem was it got lost in the mail. AoPS sent me another set, so I wouldn't worry too much.

Thank you for the replies, guys, I appreciate it, and it's brought me some comfort as well. I suppose I shall continue and wait patiently :)


superior algebra
superior algebra unsolved

If $ p$ is a prime, prove that any transposition and any $ p$-cycle generates $ S_p$.

wlog the cycle is $ c\equal{}(0,1,2,...,p\minus{}1)$ and the transposition is $ t\equal{}(0,a)$ (otherwise just rename the elements $ S_n$ is acting on.) Now we have $ c t c^{\minus{}1}\equal{}(1,a\plus{}1), c^2 t c^{\minus{}2}\equal{}(2,a\plus{}2), ...$ (read mod $ p$ if necessary) so we can generate all elements of the form $ (k,a\plus{}k)$.
We have $ (k,a\plus{}k)(a\plus{}k,2a\plus{}k)(k,a\plus{}k)\equal{}(k,2a\plus{}k)$ and so on. Hence all transpositions of the form $ (k,ma\plus{}k)$ (read again mod $ p$) can be generated. Since $ a$ is relative prime to $ p$ we can choose an $ m\in\mathbb{N}$ s.t. $ ma\plus{}k\equiv 1 \mod p$.
Hence, we can generate the transposition $ (0,1)$. Conjugating by $ c$ yields all transpositions of the form $ (k,k\plus{}1)$ and these generate $ S_n$.

(In fact we have shown that a $ n$-cycle and the transposition $ (a,b)$ generate $ S_n$ if $ a\minus{}b$ is relative prime to $ n$.)

I think you have made a trivial error, barring which the proof is correct.

$ c^{ \minus{} 1}tc \equal{} (1,a \plus{} 1)$ and not $ ctc^{ \minus{} 1} \equal{} (1,a \plus{} 1)$.

$ (k,ma\plus{}k)$  possible to generate $ \Rightarrow (a,b) a \in \{0,1 \cdots ,p\minus{}1\} b \in \{1,\cdots ,p\minus{}1\}$ possible to generate.
Any disjoint cycle of a permutation  looks like $ (a \sigma(a) \cdots \sigma^{o(\sigma)\minus{}1}(a))$ where $ a$ is smallest among the members of the cycle.(i.e. $ a$ may be $ 0$, but not other members.

$ (a \sigma(a) \cdots \sigma^{o(\sigma)\minus{}1}(a))\equal{}(a \sigma(a))(a \sigma^2(a)) \cdots (a \sigma^{o(\sigma)\minus{}1}(a))$.
And each transposition in RHS is generated by $ c$,$ t$.


geometry
3D geometry
algebra
polynomial
trigonometry
function
complex numbers

[quote="Myth"][quote="spx2"]how do i find $sin(5^o)$[/quote]I think you cannot.[/quote]
No......Everyone can. It's just a cubic equation :lol: I think 3 hours will be enough to solve it :D
I will appreciate with the greatest sincere if anyone can work it out ;)

It is a cubic equation for which Cardano's formula doesn't work.

can't we use the formulas to arrive at $sin(10^o)$ then to $sin(20^o)$
then to $sin(40^o)$ and here we must make somethin else for sure,if i'm right.
continue please the solution...

No matter what you do, there's an angle trisection equation in there somewhere, and Cardano's formula generates cube roots of complex numbers in that case. There is no way to express the exact answer in terms of real radicals.

what do you mean by angle trisection stuff ?
i knew that cardano's formulas apply to finding 3-grade polynomials

An angle trisection equation is a cubic equation of the form $\cos 3t=a$, $4x^3-3x=a$ with $|a|<1$, or alternately any cubic with three real roots. The forms are all equivalent by real scaling and linear transformations of $x$.

The transformations to bring an equation into the standard form may involve taking square roots, but does not require anything more.

Since $\sin 5^\circ=\cos 85^\circ$ and $\cos 255^\circ$ is a known quantity, you can easily find $\sin 5^\circ$ by an angle trisection equation- but the equation will also have $\sin 125^\circ$ and $\sin 245^\circ$ as roots.

so can't i use expansion in taylor form ?

I don't exactly understand your question, but iff I understand it correct, you want to find the value by taylor series of $sin(x)$.
Well, this gives you some good aproximation, but doesn't give any chance for an exact value

[quote="spx2"]how do i find $sin(5^o)$[/quote]
[b] Nice question ! [/b]
Let us denote $ x=\cos{\frac{\pi}{36}}\; ,\; y=\sin{\frac{\pi}{36}} $ . If $a:= \cos{\frac{\pi}{12}} ,\;  $ then it  may be observed that $a=\frac{\sqrt{6}+\sqrt{2}}{4}\; . $  Because  $\cos{3\alpha}=4\cos^3{\alpha}-3\cos{\alpha}\; , $
 ($\alpha=\frac{\pi}{36} $) , we conclude that $ x $ is a root of equation
\[ f(z)\equiv 4z^3 -3z- a =0   \; ,\; \; z\in (0,1).\]
Further let $ f(z_j)=0 \; ,\; j\in \{1,2,3\}. $  An inspection of  the roots show us that $ -1<z_1<-0.5<z_2<0 $ and $ 0.5< z_3<1 . $  Therefore $ x= z_3 .  $
 [b]Method 1 ,  [to approximate $y$ ][/b] If you intend to use a PC, then try to find $ z_3 $ by means of an iterative method. For instance, by considering the equation $f(z)=0 \; ,\; z\in [0.5,1] , $ because $ f^{\prime}(z)\equiv 3(4z^2-1)> 0 $ on $ (0.5, 1] $ and $ f^{\prime\prime}(z)>0 $ you can use Newton method, that is starting with $ x_0=1 $ find the iterations $ x_1,x_2,...,x_p, ...  $  generated by
\[ x_{k+1}=F(x_k)\; ,\; \; \; F(x)=x-\frac{f(x)}{f^{\prime}(x)}=x-\frac{4x^3-3x-a}{3(4x^2-1)}.\]
For $p $ large you have $ x=z_3\approx x_p .$ It's clear that an  interesting question is to evaluate the absolute error $ |z_3-x_p | $ . According to some theoretical considerations it is known that there exists $ \varepsilon \in (0,1) $ and a positive constant $M $ such that
\[ |x-x_p|\le M\cdot {\varepsilon}^{2^p}\; \; ,\; \; p\in {\mathbb N}^* . \]
At the end $y=\sin{\frac{\pi}{36}}=\sqrt{1-x^2}\approx \sqrt{1-x_p^2}\; \; . $
 [b]Method 2 , [using Cardan formulae.] [/b] However it's possible to find explicitly the number   $ y  =   \sin{\frac{\pi}{36}} =\sqrt{1-z_3^2} .$ Define
\[
j=108(\sqrt{2}+\sqrt{6})\; \; \; ,\; \; \; \; D= \frac{27^2\cdot 32^2 (2+\sqrt{3})-1}{16}\; ,
\]
\[ u=\frac{j+\sqrt{D}}{2}\; \; \; ,\; \; \;
v=\frac{j-\sqrt{D}}{2}\; ,\; \; \; c=\frac{-1+i\sqrt{3}}{2}\; .
\]
\[
a_1= uc+vc^2\; \;  ,\; \; a_2=uc^2+vc\; \; ,\; \; a_3=u+v \; .
\]
Then the set $M: =\{a_1,a_2,a_3\} \subset {\mathbb R}\;\;  ,\; M= \{z_1,z_2,z_3\}. $ Therefore $    z_3=\max\{a_1,a_2,a_3\} $ and $ y=\sqrt{1-z_3^2}\; . $ Excuse for possible mistakes ! Alex

In general, if we know some trigonometric function of $x$, we know trig functions of $x/2$ and $x/3$ (up to some cases). So, since $36=2^23^2$, we may find $sin(\pi/36)$.


Let $a,b$ real positive numbers and let $\parallel A\parallel$  the numbers of element of set $A$. If  $S$ is the set of all solution of the equation:

     $|2X-a-b|+| \frac{X}{2}-\frac{ab}{a+b}| = (a-b)^{2}$

find $\parallel S\parallel$.

This solution not is completed , because if you take the case $\{-,-\}$ for example is necesary verificate that the value of x acomplish:

 $x < \frac{a+b}{2}$ and $x < \frac{2ab}{a+b}$

and this go to the inequality in terms of a and b and not all a,b are ok.

And the case $\{-,+\}$ is imposible becasuse:

   $(a+b) \geq \frac{4ab}{a+b}\forall a, b$

I arrive one necessary and sufficient condition on $a$ and $b$ in the case $\{+,+\}$  and the case $\{+,-\}$, and bot condition are excluding  one to another,but  in the case $\{-,-\}$   can't arrive to good result.

look easy, but ....not really

The problems in this question is that the numbers of element of S depend on the value of $a,b$.


Two infinite geometric series a + ar + ar^2 + ... and b + br + br^2 + ... have the same sum. Also, the second term of the second series is the first term of the first series. if r = 1/2 for the second series, find r for the first series.

Thanks..

can some one check this. my teacher said it was incorrect. thanks!

[hide]We know that $a/(1-r) = 2b$

We also know that $a=b/2$

This implies that $1/(1-r)=4$

We find that $r=3/4$[/hide]


Observing the masses of individual atoms of the elements cited below, calculate the mass of each of the following molecules:

Mass:
N = 14
S = 32
P = 31
C = 12
H = 1
F = 19
Zn = 65
Cl = 35,5
O = 16
B = 11
Fe = 56
Ag = 108
Ba = 137
Na = 23
BI = 209
Li = 7

a) NH4NO3
b) BaSO4
c) F3BNH3
d) NaH2PO4

[hide="HINT"]You just add up the mass of each individual atom in the molecule. Keep in mind the subscripts are how many of the atom there are. So H2O, there are 2 H's and 1 O.

[hide="#1 solution"]For example, in the first example, you add up N (14), 4 H's (4*1), another N (14), and 3 O's (16)
So you get 14 + 4*1 + 14 + 3*16 = 80[/hide][/hide]


function
calculus
derivative
algebra
polynomial
absolute value
real analysis

Okay so for the following function I have to:
a) Calculate an expression for the 5th derivative, g^(5)(x)
b) Find a bound for the absolute value of g^(5)(x) on the interval indicated.

The expression is g(x) = 1/x based at r=2 on (0.1, 3.9).

I've done a and this is what I got:
a) g'(x) = -x^-2
g''(x) = 2x^-3
g'''(x) = -6x^-4
g^(4)(x) = 24x^-5
g^(5)(x) = -120x^-6

However, I'm not entirely sure how to find part b so any assistance would be helpful!  Thanks!

How would you make the absolute value of $x^{-6}$ big?
This isn't that complicated; you don't usually have to go through the traditional Calc 1 method of finding maxima on this sort of problem, since you're usually dealing with reasonably simple functions.

(This is obviously related to Taylor polynomials- and in this special case, there's a reasonably simple exact expression for the actual error)