If $a,b,c,$ are integer numbers and ${a,b} \vdots\ {c-1}$ then prove that for every $a,b,c$ there exist integers $x,y,z$ which satisfy ${x^a}+{y^b}={z^c}$.And find all $(x,y,z)$ triples.

Do you mean $a,b \leq c-1$ or $a,b | c-1$ or $c-1 | a,b$ ?
The $\vdots$ sign is not clear ... :roll:

EDIT: Corrected a typpo ;)

$a,b\vdots {c-1}$ means $a$ and $b$ is divisible by $c-1$.

[quote="gopherhole112"]hmm, it doesnt seem true to me. are you sure it works? I tried moving one of the terms over and factoring but that didnt help. I would say the proof should start with something about Fermat's last theorem.[/quote]

I think u are right ;)

I can't find all (x,y,z)
but I can find infinitly.
suppose that $kab=lc-1$where $k,l$are postive integer.
then for any positive integer t
$(2^{kb+tcb},2^{ka+tca},2^{lc+tab})$ is a solution.
more,we can only use the condition $(ab-1,c)=1$to complete the proof.


[quote="moogra"]#1. The Pope has it but he does not use it.
Your father has it but your mother uses it.
Nuns do not need it.
Arnold Schwarzenegger has a big one,
Michael J. Fox's is quite small.
What is it?

#2. It is something you do in a bed at night.
Both a man and a woman do it.
Some people make noise when they do it.
There are even pills you can take to help you do it.
What is it?[/quote]
[hide="#1"]A car[/hide]

[hide="1"]Name[/hide]

[hide="2"]sleep[/hide]

there's also another thing for number 2....but I think you can figure it out :)

[quote="mustafa"]Same with the first, that's the point of these riddles I think.[/quote]

yeah, you all got them right
[hide="#1"]A [b]Last[/b] Name[/hide]
[hide="#2"][color=blue][b]Sleeping[/b][/color][/hide]
The first and second one are meant to sound wrong.

[quote="mustafa"]Same with the first, that's the point of these riddles I think.[/quote]
There's a double (sick) meaning for the first one????  :huh: But I can't think of any.....Please don't tell me if it is very sick...

Ohhhh ok....ewwwww
I am not programmed to think that sick!!  :lol:
Sorry I asked... :|

good for you. I posted this in another forum, and I got that for many solutions...

Yeah. I can think of "sensitive" answers for both of those. :rotfl:

it's really just the sleep one, the other one is "nasty" but theres a higher chance of thinking of the right answer

Actually, for the first one, I first thought of the clean answer:
[hide]Last name[/hide]

[quote="Padmavathi"]Oh!! Ya I can totally relate...all my friends said those answers when I asked them!  :D[/quote]

Said what  :roll:  the mum uses it but father doesn't...i wonder what

[quote="nutz_for2.718281828"]For #1, I immediately thought money... :rotfl:[/quote]

yet that's not right :) try again?

[quote="D3m0n Shad0w"][hide="#2"]Snoring[/hide]

lol[/quote]

i dont think anyone's interested in pills which help people snore louder :D

Yeah, the dirty answers are pretty obvious, but I actually got the second one a lot faster than the first one.

Yes! My 200th post!! :D

[quote="anjali"]i,too,think it's MONEY :?:[/quote]

uhhh the answers were posted already go check them

How can you have a big money?

I now agree with [hide="a"]last name.[/hide]


Mr. Wong has 10 grandchildren. Assuming that the gender of each child is determined independently and with equal likelihood of male and female, what is that probability that Mr. Wong has more grandsons than granddaughters or more granddaughters than grandsons?

Use complementary counting:

The probability that she has the same number of granddaughters as grandsons is $ \frac {\binom{10}{5}}{2^{10}} \equal{} \frac {63}{256}$. $ 1 \minus{} \frac {63}{256} \equal{} \boxed{\frac {193}{256}}$


Prove that:
$\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+3 \geq 2(a+b+c)$
where a;b;c are positive real numbers satisfying $abc=1$

$\sum a^{6}b^{6}+\sum a^{4}b^{4}c^{4}\geq 2\sum a^{6}b^{3}c^{3}$
after homogenizing and substituting $a^{3},$ etc. for $a$ resp.

[quote="andy_tok"]let a=x/y, b=y/z, b=z/x, assume x>=y>=z[/quote]

You cannot assume that. The inequality in x, y, z is not symmetric!

Anyway, the inequality in question is http://www.mathlinks.ro/Forum/viewtopic.php?t=19076 , applied to the numbers $\frac{1}{a}$, $\frac{1}{b}$, $\frac{1}{c}$ instead of a, b, c (since abc = 1 yields $\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}=\frac{1}{abc}=\frac11=1$, and $\frac{1}{b}\cdot\frac{1}{c}=\frac{a}{abc}=\frac{a}{1}=a$ and similarly $\frac{1}{c}\cdot\frac{1}{a}=b$ and $\frac{1}{a}\cdot\frac{1}{b}=c$).

  darij


Your formula provides the sequence
\[1,1,2,2,2,3,3,3,3,... \]
Which is a little bit off  :wink:

[quote="Phelpedo"]Find a formula for the $n$th term of the sequence $1,2,2,3,3,3,4,4,4,4...$[/quote]

Um, cosinator is right.  You thought that he had floor symbols not ceiling symbols, but cosinator had ceiling symbols.

So the answer is $a_{n}=\lceil \frac{-1+\sqrt{8n+1}}{2}\rceil$.  And that comes from the fact that, all triangular numbers when multiplied by 8, and then added 1, they become a perfect square, so you see that 1 goes with 1, 3 goes with 2, 6 goes with 3, etc.


Let ABC be an acute-angled triangle. Let D and E be points on BC and AC, respectively, such that AD is perpendicular to BC and BE is perpendicular to AC. Pet P be the point where ray AD meets the semicircle constructed outwardly on BC and Q the point where ray BE meets the semicircle constructed outwardly on AC. Prove that PC = QC.

Verify that $ CD\cdot BD \equal{} PD^2$ implies $ PC^2 \equal{} PD^2 \plus{} CD^2 \equal{} CD(BD \plus{} CD) \equal{} CD\cdot BC$.

Again $ CE\cdot AE \equal{} QE^2\implies QC^2 \equal{} QE^2 \plus{} CE^2 \equal{} CE(AE \plus{} CE) \equal{} CE\cdot AC$.

But since $ ABDE$ is cyclic we have $ \angle CDE \equal{} \pi \minus{} \angle BDE \equal{} A$, which implies $ \triangle DEC\sim \triangle ABC$, and thus $ CE\cdot AC \equal{} CD\cdot BC$. Therefore we conclude that $ PC^2 \equal{} QC^2\implies PC \equal{} QC$.

Btw, what is this doing in the Algebra forum?

tooooooo ez
$ PC^2\equal{}BC.CD\equal{}CE.CA\equal{}QC^2$
here used similar triangle and a cyclic quad :D


The ratio of the measures of two complementary angles is 7:11. What is the measure, in degrees, of the larger angle?

We can say $ 7x \plus{} 11x \equal{} 90 \implies 18x \equal{} 90 \implies x \equal{} 5$. Thus, $ 11 \times 5 \equal{} \boxed{55}$.


Question (1)
$ 6$ Cups are placed on a table, $ 3$ facing up and $ 3$ facing down. If you flip $ 2$ cups at a time, is it possible to get all $ 6$ cups facing up or all $ 6$ cups facing down?

yes? no ?  why and how?

question (2)
Each person in a room shakes everyone else's hand once. A total of 28 handshakes take place. How many people are in the room?

[hide="#1"]i dont beleive it is possible to get all of the cups facing down (unless you are totally 1337!) because an odd number minus and even number is always odd. So there is always an odd number facing down/up therefore one can never have an even number of cups facing up/down and cannot have all 6(even) cups facing one direction at the same time![/hide]

[hide="#2"]ok so if there are n people in the room on person will shake hands with n-1 people, the next will shake hands with n-2 people so we have $ \frac{(n\minus{}1)(n\minus{}1\plus{}1)}{2} \equal{} \frac{n(n \minus{} 1)}{2} \equal{} 28$ therefore we have n = 8 so there are eight people at the party![/hide]


inequalities
logarithms
function
inequalities proposed

Find the minimum value of $ a$ such that $ \log_{10} xy\leqq (\log_{10} a)\sqrt{(\log_{10} x)^2\plus{}(\log_{10} y)^2}$ always holds for any real numbers $ x>1,\ y>1$.

hello, we have $ p + q\le \log_{10} a\sqrt {p^2 + q^2}$ with $ t = \log_{10} a$ we get
$ p + q\le t\sqrt {p^2 + q^2}$ and $ t > 1$. This is equivalent with
$ 0\lep^2(t^2 - 1) + q^2(t^2 - 1) - 2pq$
$ 0\le \frac {p}{q}(t^2 - 1) + \frac {q}{p}(t^2 - 1) - 2$
$ 0\le u^2(t^2 - 1) - 2u + t^2 - 1$ with $ \frac {p}{q} = u$
te function $ f(u) = u^2(t^2 - 1) - 2u + t^2 - 1$ has a global minimum at the point $ u = \frac {1}{t^2 - 1}$ so
it must be $ f\left(\frac {1}{t^2 - 1}\right) = t^2 - 1 - \frac {1}{t^2 - 1} \geq 0$ this gives us
$ t \geq \sqrt {2}$ or $ log_{10} a \geq \sqrt {2}$ or $ a \geq 10^{\sqrt {2}}$.
The equal sign holds if and only if $ a=10^{\sqrt{2}}$ .
Sonnhard.

[quote="Potla"][quote="kunny"]Find the minimum value of $ a$ such that $ \log_{10} xy\leqq (\log_{10} a)\sqrt {(\log_{10} x)^2 \plus{} (\log_{10} y)^2}$ always holds for any real numbers $ x > 1,\ y > 1$.[/quote]
Let $ p \equal{} \log_{10} x; q \equal{} \log_{10} y$. We have  from RMS-AM inequality or CS that:
$ \sqrt {\frac {p^2 \plus{} q^2}{2}}\geq \frac {p \plus{} q}{2}$
$ \therefore \sqrt {p^2 \plus{} q^2}\geq \frac {1}{\sqrt {2}}(p \plus{} q)$
$ \implies \sqrt {2}\sqrt {(\log_{10} x)^2 \plus{} (\log_{10} y)^2}\geq \log_{10} xy$
So $ \log_{10} a \equal{} \sqrt {2}\Longrightarrow a \equal{} 10^{\sqrt {2}}$
I am not sure if this is correct.  :blush:
[hide](Edited to fix a mistake)[/hide][/quote]

Very Good! :lol:  Being added the equality case, the solution is worthwhile to full marks.

Thanks for your comment, kunny-san. Sorry for my incomplete solution.  :blush:
Equality holds if $ p\equal{}q\implies \log_{10} x\equal{}\log_{10} y\implies x\equal{}y$

Perfect!

[quote="Potla"]Thanks for your comment, kunny-san. Sorry for my incomplete solution.  :blush:
Equality holds if $ p \equal{} q\implies \log_{10} x \equal{} \log_{10} y\implies x \equal{} y>1.$[/quote]

To Dr Sonnhard Graubner:

Could you edit the solution?

You seemed to be hasty.


$1+2sinx = 2sinx \sqrt{1-sin^{2}x)}$
square both sides
we get $1+4sinx =-4sin^{4}x$

how would i use the quartic formula -> arc sin to find x?

I think the problem writer intended the solver to use $\sqrt{1-sin^{2}x}=\pm cosx$.

I highly suggest you don't even consider using the quartic formula unless you have lots of time to blow. If you really do, plug in 0 for a, 0 for b, 1 for c, and 1/4 for d in the formulae below:

[img]http://images.planetmath.org:8080/cache/objects/1525/l2h/img5.png[/img]

For all [b]practical[/b] purposes, the quartic formula doesn't exist.

Of course, the original problem read

$1+2 \sin x = 2 \sin x \cos x$

Edit #2:  Sorry, the factoring does not work.  :oops:

Here are the other 3 roots:

[img]http://images.planetmath.org:8080/cache/objects/1525/l2h/img8.png[/img]
[img]http://images.planetmath.org:8080/cache/objects/1525/l2h/img11.png[/img]
[img]http://images.planetmath.org:8080/cache/objects/1525/l2h/img14.png[/img]

Edit:   Deleted the first few lines, no longer necessary ...

:) BTW this is from now locked thread: :)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121061]link[/url]

Anyway a post of mine may be helpful:

[quote] Getting back to the topic:

To solve $x^{4}+x+(1/4)=0$

[hide](or similar equation), one can use numerical methods (or use calculator like TI-86/89 etc)
You can also (Method is standard and can be found in good math book )

You rewrite it as $(x^{2}+px+q)(x^{2}-px+r)$
Giving: $(r+q = p^{2}$
$r-q = 1/p$
$qr = 1/4$
This gives $(p^{6}-p^{2}-1 = 0$  A cubic equation in $p^{2}$
(which I assume you know how to solve)

The solution is $p^{2}= \frac{(108+12 \sqrt{69})^{1/3}+(108-12 \sqrt{69})^{1/3}}{6}\approx 1.3247$
 Now you have two quadratic equations-ready to be solved.[/hide][/quote]


I just started looking for tests a few days ago, and I have traded a couple times, but I have nothing before 2006 except for the 1984 State. I would appreciate it if I could have, maybe, the full 2001-2002 Nationals? By "2 years" do you mean school, chapter, state, nationals? It doesn't matter which years for me. Thanks. If you want to trade, I don't have many things. Thanks again for your offer!

[b]Current List[/b]
PowerOfPi - 2001, 2002 Nationals
eggylv999 - 1990, 1991 State

EDIT: sorry, 06-07 Nationals(sprint, target, team, solutions)
i found I already have the 08 Nationals test

Could I please have the 2007 and 1997 Narionals?

meh so apparently I barely have anything at all...

2007 nationals and 2008 nationals please, thanks :lol:

EDIT: mathcountstrading@gmail.com

2008 and 2009 national please
if not please send me 2007 national

Thanks a million.

Could I please have a copy of the 2009 Mathconts Nationals test (sprint, target, and team) and a copy of the 2008, Mathcounts Nationals test (sprint, target and team)? Thanks a lot  :lol:
e mail:twc7188368788@earthlink.net

95 chapter and 92 state please! igoroogenflagenstein@gmail.com

Everyone, I would appreciate it if from now on you included your e-mail in your post as Mewto55555 and Thunder365 did.

I still have to e-mail:
venkataraman (you don't even have an e-mail?)
Thunder365
tiger21
draops
Mewto55555

2009 Nationals (Sprint, Target, Team, Countdown) if you have it. Thanks so much!  :lol: If you don't mind, can you check my profile to see my email.. I don't really want to post it here.  :blush:
I might edit this post and add more tests once I check my other computer.. then I can figure out which tests I don't have.

Cool.

2009 Nationals
1999 State

mathnerd22@gmail.com

2001 and 2000 states. tyvm. i dont want to put my email up but u can find it on my profile. thhxx

[quote="akhil0422"]Izzy- I have 1999 State but not 09 Nats.[/quote]

All right, then instead of 09 Nats, please send me 1998 State.

I would like 1989 state and national, please. My email is chiuerh@yahoo.com.

Thanks!

91 state and 93 chapter, please. dragonlaird7@yahoo.com

Can I have the nats for free as I gave you free tests (and a lot of other people too) or do I still need to trade something?  :blush:    :blush:    :blush:

I will send them to you. What is your email?

Akhil, do you know when we should be expecting the tests by?

I would like to have 1990 and 1991 Chapters. Thanks.
isabellagandy@yahoo.com

Can I have the 1996 state and nationals? If you have them, can you please send them to nickchow99@gmail.com? Thank you so much.

Can I have 1987 State and 1987 Nationals.

Thanks!  My email is nchen07@gmail.com

Be cautious with multiples - I know some people that would definitely make multiple accounts for this.

To akhil: The answer to the question that you asked through PM is no.

TO ANYONE:
2009 Nationals (sprint, target, team, answers) would make me [u][b]forever grateful[/b][/u] to you. I made nationals last year but not this year, and was pretty bummed, and I would like to take it to see what I get compared to other people... Please and thank you even if you just read this!!!
NearlyHeadlessNickS@gmail.com

[quote="akhil0422"]I have recently had people private messaging me for free tests. As this form of communication, e-mailing, and posting in the trading forum are too confusing for me, I am starting a new thread wherein various people shall post their names and only [b]2[/b] years of MATHCOUNTS and/or AMC tests that you want (in which you include which rounds you want also). The deadline for this is 1 week from now, or [b]June 5[/b].

However, let it be known that I have every right to refuse giving tests to you, but would not do unless I have a reason accepted by someone who's judgement I trust (the reason for which is multis, people with non-genuine interest in math, etc.).

I hope this shall be a good service to all.  :)

EDIT - About me previous multis point, why is there are cgyao15 and a cyao15? And they both got the "spinning-ball" avatars around the same time...[/quote]

You can tell if they're multis created for this reason by their join date...

I would like 1998-1999 Nationals please. My email is remy1140@gmail.com Thanks.

Thanks for the tests! :lol:  Ur awesome!

[quote="remy1140"][quote="PowerOfPi"]My email is powerofpi@yahoo.com

@ anyone who wants 2009 Nats: I have them (they are scanned) and I can give them to you, although I would appreciate it if you could give me something too.[/quote]

How did you get those problems? Did your coach give you after the competition?[/quote]
I traded.

Am I allowed to request for *parts* of tests? For example, the sprint round from the 2002 Nationals. If so, can I have:

1990 State Countdown
1991 State Countdown
1993 State Countdown
1996 State Countdown
1997 State Countdown
2000 State Countdown
2000 National Answers
2001 National Solutions

Note: This is based on the fact that an average test will have 4-5 *parts* (Sprint, Target, Team, Answers, maybe Solutions/Masters), so 2 tests * 4 parts = 8 parts.

My email is mathfreak123@live.com

Thanks a ton!

I want as many as possible:
email: varunkmohan@gmail.com :roll:

I want 2009 nationals plz

yao1fan@yahoo.com

[quote="fermatsmymom"]Can you give me 2000, and 2001 ty.[/quote]

you have to tell state, national, chapter, or school

Hi, Could I have 1999 and 2000 state and nationals sprint round and answers?
My email is mrsaturns@gmail.com  If I am missing something, could you people post it here or pm me? Thanks.

Can I have the 2009 national mathcounts test?
and..any test from the 1990s

I am thinking to prepone this to end at 7:00 PM EST today as of Monday, June 1st as this thread already has way too many posts and I cannot cope with all of them.
As to when I'll distribute the tests, the answer is as soon as I can.
I will make a list to who I still have to e-mail later. Please tell me if I have messed up somewhere.

Akhilo: does 8:37 p.m. EST violate your timing?

I would like to have 2008 and 2007 Nats plz.

My e-mail is: alex960531@gmail.com

Thank You!

Fine, alex you're the last one. No more requests.

Please..can I be the last one?

I just want the 1994 and 1995 nats tests..please?


I was just curious... who here is also taking the AIMEs... I'm paying no attention to the AIMEs since this is my last year of Mathcounts... but how about you all? Responses would be nice...

I am too, and I'm acting just like you are. However, since MA's state comp. is three days before the AIME...how do you spell cram? (OK, maybe that's a bad idea) :D

I'm only considering the AIME as a practice this year. There's always next year for AIME. For now, I'm practicing MathCounts.

And, since you're in Mathcounts, the year after that, the year after that, and the year after that as well.  So, I'm really not worried about the AIME, though it would be nice to get more than 1 right.  ;)

is seventh graders allowed to take the AIMEs?

Yes, anyone who scores high enough on the AMC 10 or AMC 12 can take the AIME.

I am taking the alternate AIME on March 22, since the other one happens to take place on spring break for us. However, I won't be able to prepare for it because it is two days after State Mathcounts, and I need to prepare for Mathcounts.

nope. i dont think so. dwx314 actually has been bragging about it to me at school. i hope u know that it is very annoying,  :( but i really don't care. i take practice ones at home. i also voted wrong. i though u were talking about the amcs. SORRY!!!!   :?

So for the AIME takers, what did you get on the AMC 10/12 A/B s.
On the AMC 10A, I won the school with 19 right, 3 wrong, 3 blank, and a 121.5 score (just made the AIMES!) I couldn't take the Bs since we had Winter Break.... or else I'd take the 12B as well... :(

Now since it's the tenth, you can repost in this string.. and thanks solafidefarms for pming me about it... so the AIMES are a test that you have to qualify for (120+AMC 10, 100+AMC 12) out of 150 to take, and it's really hard (esp for us middleschoolers). It takes three hours to do the 15 questions w/o a calc and only like 6 ppl get perfects each year. See the AMC forum for details.

Question to the moderators. If the poll isn't active (sorry I can't see it), can you re-activate it for a few days? THanks...


calculus
integration
trigonometry
function
calculus computations

Evaluate $ \int_0^{n\pi} e^x\sin ^ 4 x\ dx\ (n\equal{}1,\ 2,\ \cdots).$

If we make the substitution: $ \sin^4 x\equal{}\frac{(e^{ix}\minus{}e^{\minus{}ix})^4}{16}$, then the integral is relatively easy (as long as we are comfortable with integrating function with a complex input...it behaves as it should...).

I am getting $ \frac{24}{85}*(e^{n\pi}\minus{}1)$...

Nice!

I'd like to see a well-motivated solution for this...

Let $ I\equal{}\int_0^{n\pi} e^x\cos^4 x\ dx,\ J\equal{}\int_0^{n\pi} e^x \sin ^ 4 x\ dx.$

[quote="kunny"]Evaluate $ \int_0^{n\pi} e^x\sin ^ 4 x\ dx\ (n \equal{} 1,\ 2,\ \cdots).$[/quote]

Let $ \textbf I_{a,k}\;\equal{}\; \int\; e^{ax}\sin ^ k \,x\; \textbf dx \;\equal{}\; \sin ^ k \,x\cdot \frac {e^{ax}}{a } \;\minus{}\; \int \; k\; \sin ^{k\minus{}1 }\,x \;\cos \, x \cdot \frac {e^{ax}}{a }\; \textbf dx$

integrating by parts again gives

$ \textbf I_{a,k}\;\equal{}\; \sin ^ k \,x\cdot \frac {e^{ax}}{a } \;\minus{}\; \dfrac{k}{a} \left[ \sin ^{k\minus{}1 }\,x \;\cos \, x \cdot \frac {e^{ax}}{a } \;\minus{}\; \int \;\left[ (k\minus{}1)\; \sin ^{k\minus{}2 }\,x \;\cos^2 \, x \;\plus{}\; \sin ^{k\minus{}1 }\,x \;(\minus{}\;\sin\, x) \right] \cdot \frac {e^{ax}}{a }\; \textbf dx \right]$

$ \equal{}\; \sin ^ k \,x\cdot \frac {e^{ax}}{a } \;\minus{}\; \dfrac{k}{a^2}  \sin ^{k\minus{}1 }\,x \;\cos \, x \cdot e^{ax}$
$ \;\plus{}\; \dfrac{ k(k\minus{}1) }{a^2} \; \int  \sin ^{k\minus{}2 }\,x\;(1\minus{}\sin^2\, x) \cdot e^{ax} \; \textbf dx \;\minus{}\; \dfrac{k}{a^2} \; \int \sin ^ k \,x\cdot e^{ax} \; \textbf dx$

rearranging gives

$ \equal{}\; \dfrac { e^{ax}\sin ^{k\minus{}1 }\,x  \left( a\;\sin\, x \;\minus{}\; k\; \cos\, x\right) } { k^2 \plus{} a^2 } \;\plus{}\; \dfrac {k(k\minus{}1) } { k^2 \plus{} a^2 } \cdot \textbf I_{a,k\minus{}2}$

put $ a\equal{}1,\;\;\; k\equal{}4$, to get

$ \textbf I_{1,4}\;\equal{}\; \dfrac { e^{x}\sin ^{3 }\,x  \left( \sin\, x \;\minus{}\; 4\; \cos\, x\right) } { 17 } \;\plus{}\; \dfrac {12 } { 17 } \;\textbf I_{1,2}$

$ \equal{}\; \dfrac { e^{x}\sin ^{3 }\,x  \left( \sin\, x \;\minus{}\; 4\; \cos\, x\right) } { 17 } \;\plus{}\; \dfrac {12 } { 17 } \; \left\{ \dfrac { e^{x}\sin \,x  \left( \sin\, x \;\minus{}\; 2\; \cos\, x\right) } { 5 } \;\plus{}\; \dfrac {2 } { 5 } \; \textbf I_{1,0} \right\}$

$ \equal{}\; \dfrac { e^{x}\sin ^{3 }\,x  \left( \sin\, x \;\minus{}\; 4\; \cos\, x\right) } { 17 } \;\plus{}\; \dfrac {12 } { 17 } \; \left\{ \dfrac { e^{x}\sin \,x  \left( \sin\, x \;\minus{}\; 2\; \cos\, x\right) } { 5 } \;\plus{}\; \dfrac {2 } { 5 } \; e^x \right\}$

now insert the limits

$ \textbf I_{1,4}\;\bigg|_0^{n\pi}$

$ \;\equal{}\; \dfrac { e^{x}\sin ^{3 }\,x  \left( \sin\, x \minus{} 4\; \cos\, x\right) } { 17}\;\bigg|_0^{n\pi} \plus{} \dfrac { 12\;e^{x}\sin \,x  \left( \sin\, x\minus{} 2\; \cos\, x\right) } { 85 } \;\bigg|_0^{n\pi} \plus{} \dfrac {24 } { 85 } \; e^x\;\bigg|_0^{n\pi}$

the [b]1st [/b] $ 2$  terms will yield $ 0$, so all we are left with is

$ \textbf I_{1,4}\;\bigg|_0^{n\pi} \;\equal{}\; \dfrac {24 } { 85 } \; e^x\;\bigg|_0^{n\pi} \;\equal{}\; \boxed {\dfrac {24 } { 85 } \; \left( e^{n\pi} \;\minus{}\; 1 \right) }$

Needless to say, your solution is correct. :)

[quote="kunny"]Let $ I \equal{} \int_0^{n\pi} e^x\cos^4 x\ dx,\ J \equal{} \int_0^{n\pi} e^x \sin ^ 4 x\ dx.$[/quote]
$ I\plus{}J \equal{} \int_0^{n\pi} e^x(\cos^4 x\plus{}sin^4x) dx$
$ I\minus{}J \equal{} \int_0^{n\pi} e^x(\cos^4 x\minus{}sin^4x) dx$
Then we find I and J.
Is this way ok?

[quote="kunny"] [color=darkred]Evaluate $ \int_0^{n\pi} e^x\sin ^ 4 x\ \mathrm {dx}$ , where $ n\in\mathcal N^*\ .$[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] $ \int_0^{n\pi} e^x\sin ^ 4 x\ \mathrm {dx}\equal{}$ $ \sum_{k\equal{}0}^{n\minus{}1}\int_{k\pi}^{(k\plus{}1)\pi}e^x\sin^4x\ \mathrm {dx}\equal{}$ $ \sum_{k\equal{}0}^{n\minus{}1}\int_{0}^{\pi}e^{x\plus{}k\pi}\sin^4x\ \mathrm {dx}\equal{}$
$ I\cdot \sum_{k\equal{}0}^{n\minus{}1}e^{k\pi}$ , where $ I\equal{}\int_{0}^{\pi}e^x\sin^4x\ \mathrm {dx}$ . Denote $ J\equal{}\int_{0}^{\pi}e^x\cos^4x\ \mathrm {dx}$ .
Prove easily that $ \sin^4x\plus{}\cos^4x\equal{}\frac 14\cdot (3\plus{}\cos 4x)$ and $ \cos^4x\minus{}\sin^4x\equal{}\cos 2x$ .
Therefore, $ I\plus{}J\equal{}\frac 14\cdot\int_0^{\pi}e^x(3\plus{}\cos 4x)\ \mathrm {dx}$ and $ J\minus{}I\equal{}\int_0^{\pi}e^x\cos 2x\ \mathrm {dx}$ a.s.o.[/color]


function
geometry
summer program
MathPath
\/closed

Hi all,

I just wanted to let you know about the existence of a [url=http://www.mathlinks.ro/Forum/album.php]Photo Album[/url] on the site. You can now share, upload and download pictures from the site. Each user will have his/her own gallery (full implementation of this function is going to be available very soon).

Now you can share the pictures you have taken along with your friends on the site at different math competitions.  Furthermore, you can upload geometry figures and then use them in the hide tags, and much, much more :)

I hope you will be enjoy this new feature we are providing for you at AoPS - MathLinks forums :)

Say cheese!

Very cool!

Forum's getting better every day :) heh

Wow, you've been hard at work, Valentin! Many thanks!

I was waiting for something like this  :)  thanks Valentin

Wow, this feature will really help the AoPS Flash/Art Contest!!  :D

Btw, do we have an upload limit in the album, like we do when we attach images?

[quote="NoSoupForYou"]Btw, do we have an upload limit in the album, like we do when we attach images?[/quote]There is a limit for uploading files into the album, but we might increase that limit if required ...

[quote="Valentin Vornicu"]There is a limit for uploading files into the album, but we might increase that limit if required ...[/quote]

how many pictures (384x512, or something) can we upload? :D

Great idea Valentin! I see that windslicer is the first to have taken advantage of the pesonal galleries.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=4[/img]
<img src="http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=4">

Why don't either of those work?   :?

because files need to end in .jpg or .gif or .png in order to work. so [img] doesn't work on that type of url's.

could someone provide more specific directions for uploading images for anti-computer geeks like me? :D

Go to the Album.  Now, if you want to put your pictures in a certain Public Category click on that category.  Then click Add Picture.   Where it says "Upload a pic from your machine" click the "Browse" button right beside it and find the picture you want.

If you want to put it in Your Personal Gallery go to the Album, click on Your Personal Gallery, click Add Picture and same thing as last time.

Is that enough help or are are you still unclear on something else?

Valentin, a suggestion: maybe make a Misc. category?

[quote="manuel"]Valentin i have some imo 2004 pictures, in which category can i post them.[/quote]

You can put them in your personal album.

Also, maybe you should make a category for the AoPS Flash/Art Contest!  :D

[quote="manuel"]I was trying to upload a pic but it says that the dimension size is too large, how do I fix this?[/quote] :oops_sign: my bad ... fixed. You can now upload files up to 1000 kB in size, and 1600x1200 in dimensions :)

[quote="Valentin Vornicu"][quote="manuel"]Valentin i have some imo 2004 pictures, in which category can i post them.[/quote]I am making now an IMO 2004 category.

About the [ img] bug, you guys are right. I am going to fix it, but only after I get back from the IMO (and probably after MathCamp).[/quote]

Sounds good, take your time! And have a blast at the IMO/MathCamp

valentin
can you put the pictures of the moderators in the photo album ? :D
to be exact, darij pictures in real action (i mean last imo's)

the photo album is the greatest thing since AOPS!!!!!

um...

I posted a picture and it says "No Thumbnail"... what is that? and how do i fix it to normal (preview mode ) pictures like others do

[quote="towersfreak2006"]Hmm, I'm not sure, but I think it only allows you to modify the picture name/description. Am I missing something? :?[/quote]This feature is not bug-proof yet, so there might be stuff missing :D I will check on it later  :oops:

[quote="Valentin Vornicu"][quote="towersfreak2006"]Is it possible for users to remove their pictures from the photo album? My "test" wasn't very successful... :oops:[/quote]It should be possible. There's a little edit/test button there :)[/quote]

But wheni go there, it says i CANNOT delete mypics

[quote="bubka"]Where did the photo album go?[/quote]

There was a server change and not all features were (properly) re-implemented. As far as I know this is Valentin's task and as you know he takes a break.  :)

I want to put in a photo from my hard disk, fredbel style. How do I do it?

is the photo album supposed to be up now? it isn't working for me.


Pythagorean Theorem
geometry
geometry proposed

Let circle $ (O)$ with $ AB$ is a diameter, let $ (d)$ is tangent of circle at point $ B$. Let $ C$ is a point lie on circle and let $ H$ is perpendicular foot of point $ C$ on  $ AB$. Let $ D$ is a point lie on $ CH$ ray pass point $ H$ such that $ HD \equal{} AC$. Two lines pass point $ D$ and contact with circle $ (O)$ meet line $ (d)$ at $ K$ and $ G$ respectively, with $ K$ lie on segment $ BG$. Prove that $ KG \equal{} AB$.

Hello!
Let $ R$ be the radius of the circle $ \left(O\right)$.
$ DE\equal{}DF$, $ GE\equal{}GB$ and $ KF\equal{}KB$.

$ DE^{2}\equal{}DO^{2}\minus{}OE^{2}\equal{}HD^{2}\plus{}OH^{2}\minus{}R^{2}\equal{}AC^{2}\minus{}\left(R^{2}\minus{}OH^{2}\right)\equal{}$
$ \equal{}AC^{2}\minus{}CH^{2}\equal{}AH^{2}\Rightarrow DE\equal{}AH$.
So $ DE\equal{}DF\equal{}AH$.

$ T\in GK$ such that $ DT\bot GK$ $ \Rightarrow DT\equal{}HB$ and $ BT\equal{}HD\equal{}AC$.

$ GD^{2}\equal{}GT^{2}\plus{}DT^{2}\Rightarrow\left(GE\minus{}DE\right)^{2}\equal{}\left(GB\minus{}BT\right)^{2}\plus{}HB^{2}\Rightarrow$
$ \Rightarrow\left(GB\minus{}AH\right)^{2}\equal{}\left(GB\minus{}AC\right)^{2}\plus{}HB^{2}\Rightarrow$
$ \Rightarrow GB^{2}\minus{}2\cdot GB\cdot AH\plus{}AH^{2}\equal{}GB^{2}\minus{}2\cdot GB\cdot AC\plus{}AC^{2}\plus{}HB^{2}\Rightarrow$
$ \Rightarrow 2\cdot GB\cdot AC\minus{}2\cdot GB\cdot AH\equal{}HB^{2}\plus{}AC^{2}\minus{}AH^{2}\Rightarrow$
$ \Rightarrow 2\cdot GB\cdot\left(AC\minus{}AH\right)\equal{}HB^{2}\plus{}HC^{2}\Rightarrow GB\equal{}\frac{BC^{2}}{2\cdot\left(AC\minus{}AH\right)}\Rightarrow$
$ \Rightarrow GB\equal{}\frac{2\cdot R\cdot HB}{2\cdot\left(AC\minus{}AH\right)}\Rightarrow GB\equal{}\frac{R\cdot HB}{AC\minus{}AH}$  (*)

$ DK^{2}\equal{}DT^{2}\plus{}KT^{2}\Rightarrow\left(DF\plus{}KF\right)^{2}\equal{}HB^{2}\plus{}\left(BT\minus{}KB\right)^{2}\Rightarrow$
$ \Rightarrow\left(AH\plus{}KB\right)^{2}\equal{}HB^{2}\plus{}\left(AC\minus{}KB\right)^{2}\Rightarrow$
$ \Rightarrow AH^{2}\plus{}2\cdot AH\cdot KB\plus{}KB^{2}\equal{}HB^{2}\plus{}AC^{2}\minus{}2\cdot AC\cdot KB\plus{}KB^{2}\Rightarrow$
$ \Rightarrow 2\cdot AC\cdot KB\plus{}2\cdot AH\cdot KB\equal{}HB^{2}\plus{}AC^{2}\minus{}AH^{2}\Rightarrow$
$ \Rightarrow 2\cdot KB\cdot\left(AC\plus{}AH\right)\equal{}HB^{2}\plus{}HC^{2}\Rightarrow KB\equal{}\frac{BC^{2}}{2\cdot\left(AC\plus{}AH\right)}\Rightarrow$
$ \Rightarrow KB\equal{}\frac{2\cdot R\cdot HB}{2\cdot\left(AC\plus{}AH\right)}\Rightarrow KB\equal{}\frac{R\cdot HB}{AC\plus{}AH}$  (**)

(*), (**) $ \Rightarrow GK\equal{}GB\minus{}KB\equal{}\frac{R\cdot HB}{AC\minus{}AH}\minus{}\frac{R\cdot HB}{AC\plus{}AH}\equal{}$
$ \equal{}R\cdot HB\cdot\left(\frac{1}{AC\minus{}AH}\minus{}\frac{1}{AC\plus{}AH}\right)\equal{}R\cdot HB\cdot\frac{2\cdot AH}{AC^{2}\minus{}AH^{2}}\equal{}$
$ \equal{}2\cdot R\cdot\frac{AH\cdot HB}{CH^{2}}\equal{}2\cdot R\equal{}AB\Rightarrow GK\equal{}AB$.

Best regards, Petrisor Neagoe :)

Thank you very much. I have a solution for this problem but it's very long same your solution. Who can solve this problem by other way? Thank you so much.  :P

[b]Problem:[/b] In $ \triangle ABC,$ let $ D$ the tangency point of the B-excircle $ (I_b,r_b)$ with $ BC$ and $ Q$ the antipode of $ D$ WRT $ (I_b).$ Perpendicular from $ A$ to $ DQ$ cuts $ DQ$ and $ (I_b)$ at $ P,E,$ respectively such that $ E,A$ lie on different sides of $ DQ.$ If $ EQ \equal{} AP,$ then $ a \equal{} 2r_b.$

In the right $ \triangle EQD$ we have $ EQ^2 \equal{} PQ \cdot DQ \equal{} 2r_b(2r_b \minus{} h_a)  \ \ (1)$

By Pythagorean theorem in $ \triangle API_b$ we obtain

$ AP^2 \equal{} {AI_b}^2 \minus{} {PI_b}^2 \equal{} {r_b}^2 \plus{} (s \minus{} c)^2 \minus{} (r_b \minus{} h_a)^2$

$AP^2 \equal{} (s \minus{} c)^2 \plus{} 2h_ar_b \minus{} {h_a}^2 \ \ (2)$

Since $ EQ \equal{} AP,$ from the equations $ (1)$ and $ (2)$ we obtain

$ 2r_b(2r_b \minus{} h_a) \equal{} (s \minus{} c)^2 \plus{} h_a(2r_b \minus{} h_a)  \Longrightarrow \ 2r_b \minus{} h_a \equal{} s \minus{} c$

Using the identity $ 2r_b \equal{} \frac {a \cdot h_a}{s \minus{} b},$ the above expression simplifies as

$ h_a \left (\frac {a \plus{} b \minus{} s}{s \minus{} b} \right) \equal{} s \minus{} c  \Longrightarrow  h_a \equal{} s \minus{} b \equal{} \frac {2r_b(s \minus{} b)}{a}  \Longrightarrow  a \equal{} 2r_b$


Evaluate: $ \displaystyle\sum_{i\equal{}1}^n\frac{1}{i^2}$ .
Also, establish the bound that $ \displaystyle\sum_{i\equal{}1}^n\frac{1}{i^2}< \frac{7}{4}$.


Five of the 10 digits 0,1,2,3,4,5,6,7,8,9 are randomly chosen (without replacement) and ordered to create a 5-digit number (that can start with 0).

a)What is the probability that the resulting 5-digit arrangement is an odd number exceeding 10000?
- I know that it being odd takes out half of them... but the greater than 10000 confuses me and I don't know where to go with it???

b)What is the probability that the 5-digit arrangement does not contain 0, is odd, and has odd and even digits alternating?
- This one from the get go has me completely confused??? Any help?

[hide="a)"]
[color=red]Edited[/color]: It must end in a 1,3,5,7,9 and begin with any digit except 0.
Chance of this: $\frac{9}{10}$.
Chance of the first and last digits being odd: $\frac{5}{9}\cdot \frac{4}{9}= \frac{20}{81}$
Chance of the first digit being even and the last being odd: $\frac{4}{9}\cdot \frac{5}{9}=\frac{20}{81}$

Total: $\frac{40}{81}$ [/hide]

[hide="2"]Now you have 9 numbers, 5 odd, and 4 even. If the number has odd and even digits alternating, and the last number must be odd, then the 1st, 3rd, and 5th numbers all have to be odd. The answer is $5*4*4*3*3$, or 720, because there are 5 possibilities for the first number, 4 possibilities for the second, etc.

edit: I forgot that it asked for probability, It would be $1/42, 720/(10*9*8*7*6)$[/hide]

[hide="for a,"]We need an odd number that does not start with 0. The probability that 0 is n front is 1/10 and that the number is odd is 1/2. We multiple 9/10 and 1/2: 9/20[/hide]

I'm confused with b.

a possible answer for b would be something like 12345

[hide]I thought the answer would be 1/42 = 5/10*4/9*4/8*3/7*3/6[/hide]


My violin teacher said that I'm pretty good technically, but that I need to make my playing more musical. Does anyone have any tips on how I can add more emotion to my playing? It's not that I don't like violin- I love it. If it helps, I play violin and have been playing 5 years, since 3rd grade. I'm turning 13 in July. I practice 3 hr.s a day.

Emotion isn't something you can add to music at will.  It's not something you can practice either, it just comes with time.  For me, no matter how much I would try, I could never seem to get emotion into my piano pieces.  Then not so long ago (around the time I turned 16) I just started playing the same songs but suddenly I could make myself cry doing so.  And not because they sounded terrible either :roll:

So yea, it comes with time.

Well, maybe when you play, try signing the music to yourself, either out loud or in your brian (out loud works better). This will cause you to be more "into" the music, and cause you to play with more emotion.

Well... I play used to play the piano. When I needed to express emotion when I was playing the piano, I would find that I would make my body up from my waist move around randomly. Try to go with the music. When the music becomes really loud and you have to mash the keys, then a common emotion for this is to scrunch up and get your face pretty close to the keys and bash up the piano.  :wink:

When I started playing cello, I played the Bach Suite No. 1 Prelude which is a very relaxing song. But then I started to notice that I had added a bit of improvisation to it. I started playing bottom notes a bit longer and started using a  lot of vibrato.

NOTE: There are 2 ways of expressing emotion. It's the style in which you play the piece and physically.

Not to be rude or anything, but I really don't think that is something anyone should consciously make an effort to do.

Playing musically, beautifully, is what separates the amateurs from the professionals. There are ways to evoke emotion through playing. Obviously dynamics are one thing. Contour is another. I think one of the best ways to learn would be to listen to professionals playing your pieces. Figure out what it is that you want to express from those recordings and find out how the professional does it. It may be as simple as playing faster or slower in a certain part of the piece, or as difficult as playing two sections of a piece that are written exactly the same differently (this can actually be very powerful; if you play the first A section of a rondo with subtleness and the last two with "broadness", the audience will notice).

[quote="jonathanchou711"]Well... I play used to play the piano. When I needed to express emotion when I was playing the piano, I would find that I would make my body up from my waist move around randomly. Try to go with the music. When the music becomes really loud and you have to mash the keys, then a common emotion for this is to scrunch up and get your face pretty close to the keys and bash up the piano.[/quote]

Sure this may help you become the emotion you are trying to convey, but you can convey an emotion in your playing without actually showing it through your body language. Lang Lang is horrible with this...

Putting emotion into pieces is something teachers can't teach you. They can only help you with it. What I like to do is listen to different artists and take parts which I like from them and put them into my music. As with the "moving your body" part, my first teacher strongly disliked moving your body, so I basically didn't do any of that. I still managed to play with emotion. My new teacher uses contrary motion, which is the hands doing half circles in opposite directions, to play legato, which my old teacher would never do. It's okay to move your body according to the music, but don't forget that the emotion comes first.

My only suggestion is to listen to alot of music. Especially the songs you play. Listen to multiple recordings of the same song, and try to emulate the artist.

Possibly, your first task could be to find a piece or two...or more, that you are really into.  It doesn't really matter what type of piece it is.  It could be country, or it could be Schindler's List for all I know.  But if you play pieces that you really like, you are definitely more likely to have more fun with your pieces.

I realized that when my teacher assigned me music in the past, if I really hated her selection, I would not do as well at competitions with them.

Then, a year ago or so, I somehow found something.  I chose a few of my pieces that I really wanted to play, and went out and played them better than I would have if I had not been so into them before I played them.

[quote="Dojo"]My only suggestion is to listen to alot of music.[/quote]
Agreed.  Along the way, find a song that you really like.
[quote="Dojo"]try to emulate the artist[/quote]
If you like watching a certain musician, find out what makes you like them so much, and see if you can do something with that. No matter how small it is, it will make a difference.

About body movement, I　don't think you should consciously go about moving your body to show emotion, it'll probably look forced and unnatural, not what you want to achieve.  Consciously thinking something like "this is a fast passage, I should scrunch up my face here" probably isn't going to work...  Ideally, it'll come naturally as you invest emotion into your playing.  I don't think there's a right or wrong about it though, some people do it a lot, some people a little, some people not at all, though a lot of people don't agree with this (see opinions about Lang Lang), everyone thinks something different is the most aesthetically pleasing.

I've had this problem too. I could never get that much emotion out, even though I loved to play. Then one day my teacher gave me this really beautiful, sad piece. And it just kind of happened. It has been appearing in my other pieces now, too. Music is like a language to me now, and in order to communicate that wonder, the emotions will flow out. Wow, that was not very helpful. But I DO find it helps to be emotional before you start playing. :wink:

I tell myself a story when I play a piece. Like a sad little girl learns how to fly or something, and then I can really pour emotions into the piece.

do you remember that random guy that gets all emotional about everything.
try acting like him/her and play with emotion and swing around like in those good players and for some reason, people say you are playing emotionally

"That random guy"? That's the most vague thing I've ever heard.

Personally, I really don't like people who just move around just for show, because it looks like they're just showing off. I've even seen some violinists practically dance, which is pretty stupid.
Sure, some musicians will make movements when they really get into a piece, but that is not going to give emotion.
Heifetz is known for rarely showing facial expressions while playing, and he is known as one of the greatest violinists of the 20th century!

To play musically, you have to get into the piece. But with some people it is a quality they just have, it is hard to learn.

I was fond of listening music, but now I have left listening music, since music is prohibited in Islam.


trigonometry
quadratics
probability
geometry
3D geometry
logarithms
AMC

Like some of us know in the Getting Started forum is a topic named: Problem Solving Marathon. Who start's a new one on Intermediade Topics?!  :D. Even if didn't post there iI found this full of fun!!
Sorry if I disturbed with this subject   ;)

We can start it in this post. I'll post the first question.  It's a classical question ;) :
[b][size=150]
Problem 1[/size][/b]

Source AHSME

If $\theta$ is a constant such that $0<\theta<\pi$ and $x+\frac 1x=2\cos \theta$, then for each positive integer $n$, find $x^n+\frac{1}{x^n}$ in terms of $n$ and $\theta$.

[quote="Slizzel"]Solution
[hide]Let $x=\cos\theta+\sin\theta$ then $x^n+\frac{1}{x^n}=\cos n\theta+i\sin n\theta+\cos(-n)\theta+isin(-n)\theta=2\cos n\theta$ i think :D.[/hide]

[size=150]Problem 2[/size]

Source:year 1988, a red book :)

Let $a\in R^*$ ant $x,y,z\in R^*$ such that $x+y+z=a$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{a}$. Show that exsists two numbers among $x, y$ and $z$ whose sum is null.[/quote]

I'm assuming that none of the four variables are 0.

[hide="Solution"]
Let us assume that x + y = 0 would make this work. It should be obvious that this doesn't result in any loss of correctness.

x + y = 0
z = a.

So 1/z = 1/a.

1/x + 1/y = 0.

1/x + 1/y = (x + y)/xy. This does not equal 0 unless x + y = 0, and since none of the variables are equal to 0, xy cannot equal 0. Thus, x + y must equal 0.
[/hide]

Umm, I don't know where you people get these questions, so I'm afraid I can't post one :blush: .

You can get them from anywhere you like...just find something and post it.

**EDIT** the whole thing turned out wrong
i need a tutorial on how to work the "math writing"

To the 1st problem by joml

anyway, since i dont know how to put this, i'll just explain it real quick and you can do the rest.
take the 1st identity given, multiply both sides by x and solve using the quadratic formula.  you'l get a root that looks like

 cos :theta: +/-  isin :theta:
Plug that into the equation with x^n's, and simplify using DeMoivres.  :lol:

Treething used hide tags.  The most basic way to use them would be to just type [ hide ] ... [ /hide ] around the text you want hidden (don't put any spaces in there like I did though.  Doing this should produce:

[hide]
Hidden text here :D [/hide]

If you want to change what the link says do something like this:

[ Hide=Solution ] .... [ /Hide]

and you get:

[hide="Solution"]Solution!!![/hide]

Ok, next problem:
[size=150]
Problem 3[/size]

Evaluate:

\[\sqrt{6+\left(1+\sqrt{3+\left(1+\sqrt{3+2\sqrt{2}}\right)^2}\right)^2}.\]

(Express in the form: $a+\sqrt{b}$.)

[hide]$\displaystyle \sqrt{6+\left(1+\sqrt{3+\left(1+\sqrt{3+2\sqrt{2}}\right)^2}\right)^2}$----1

$\sqrt{3+2\sqrt{2}} = \sqrt{2} + 1$

then 1 becomes $\displaystyle \sqrt{6+\left(1+\sqrt{3+\left(2+\sqrt{2} \right)^2}\right)^2}$

$\sqrt{3+\left(2+\sqrt{2}\right)^2} = \sqrt{9+4\sqrt{2}} = 2\sqrt{2} + 1$

then 1 becomes $\displaystyle \sqrt{6+\left(2\sqrt{2} + 2\right)^2}$
$= \sqrt{6+8+4+8\sqrt{2}} = \sqrt{18+8\sqrt{2}} = 4+\sqrt{2}$[/hide]

[size=150]Problem 4[/size]

You and a friend frequently meet up to go run, in between 2:00 and 3:00. Whoever arrives first stays for 15 minutes, and then leave if the other person does not arrive. ASsuming the time of arrival is random, what is the probability taht you and your friend dont run today and tomorrow. give you answers in the decimal to the hundreths

Source: NMC2005... variation

[quote="joml88"]Ok, once you answer a problem please post another problem for someone else to serve.  That keeps this thread going nicely :) .[/quote]

sorry. had to find the problem.

[hide]
Not sure how to explain this, but draw a square graph of 2:00 to 3:00 on each axis, and try to find the area of possible times that they will meet. You end up with a probability of $\frac{7}{16}$. Sorry for the bad explanation; I'm too lazy to draw the graph.

EDIT: So the probability that they do meet is $\frac{9}{16}$ and the probability that they don't meet two days in a row is $\frac{81}{256}$.[/hide]

Next Problem:

[size=150]Problem 5
[/size]
If $x+\frac{1}{x} = 1$, find $x^9+\frac{1}{x^9}$.

Source: My own, but based on some questions I've seen

[quote="xxreddevilzxx"][quote="paladin8"][hide]
Not sure how to explain this, but draw a square graph of 2:00 to 3:00 on each axis, and try to find the area of possible times that they will meet. You end up with a probability of $\frac{7}{16}$. Sorry for the bad explanation; I'm too lazy to draw the graph.[/hide]

Next Problem:

If $x+\frac{1}{x} = 1$, find $x^9+\frac{1}{x^9}$.

Source: My own, but based on some questions I've seen[/quote]

you didnt read the question right.[/quote]

Heh. Whoops. I'll edit it.

[quote="Rushil"]does $\Delta $ refer to area?? Also what do you mean by that $m$ and $N$ lie on a straight line. A figure would help!!![/quote]
Yes. $\triangle$ is the area. $M$, $N$, $B$ are collinear means $BN \cap AC = M$.
And I won't give you the figure until you "figure" it out yourself.

BTW. This is problem from China, 1983.

Yes it is.
BTW, please edit you figure, it is tooooooooooo huge.

RIDDLER needs to fix that picture... :mad:

The picture is too large, so make it smaller ;)

Yeah, indeed... :(

If anyone has a solution, please post it, or we won't be able to continue the marathon...

A hint: The three points are collinear so you can try to use Menelaus. Let the ratio be $r$ then use area ratios to complete a quadratic equation of $r$ with the Menelaus theorem.

[quote="K81o7"]Nobody else seems to be going at this, so i will...
Let $AC$ intersect $BD$ at $E$.  Let $AN$ intersect $BD$ at $F$.  Note that if $CN:ND$=$CM:MA$ then $BN || AD$.  So we have a trapezoid.  Let the area of the entire quadrilateral be 7.  Then $\triangle{ABC}=1$, $\triangle{ACD}=6$, $\triangle{ABD}=3$ and $\triangle{BCD}=4$.  Let $CN:CD=n$.  Then, $\triangle{ACN}=6n$, and $\triangle{ADN}=6-6n$.  Thus, $\triangle{AMN}=6n(1-n)$ and $\triangle{ABM}=1-n$.  Finally, $\triangle{BCN}=4n$.  Therefore, the area of $ABCD=7=6-6n+6n-6n^2+1-n+4n=-6n^2+3n+7 \implies 6n^2-3n=0 \implies n=0, \frac{1}{2}$.  The n=0 comes when $ABCD$ is a specific trapezoid, and otherwise $n=\frac{1}{2}$, which makes $MN$ the midsegment of $\triangle{ACD}$, and that implies what was wanted.

Is this Menelaus?[/quote]

Give the next problem

The citizens of towns A and B wish to build a road network and a trading market that connects both towns to the market and also connects them to a river. If the towns were points on a map and the river was a straight line, how would you construct, with straight-edge and compass, the road network and the location of the market (which is also a point) that used minimal length of the road, and prove that your network uses minimal length. The towns do not have to be directly connected to the market or river, the road network must just pass through the towns, market and river.

To be honest I think this marathon is redundant with the one posted here as a sticky:

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=33505


function
algebra
functional equation
algebra unsolved

Find all functions $f : \mathbb{R} \longrightarrow \mathbb{R}$ that satisfy
\[ (x - y)f(x + y) - (x + y)f(x - y) = 4xy(x^2 - y^2)  \]

Perform the substitution
$a=x+y$
$b=x-y$
To get the equivalent functional equation
$b f(a) - a f(b) = ab(a^2 - b^2)$
For $a,b \neq 0$, we can divide both sides by $ab$.
$f(a)/a - f(b)/b = a^2-b^2$
Let $g(a)=f(a)/a$, $g$ is defined on $\mathbb{R} - {0}$
$g(a) - g(b) = a^2 - b^2$
Suppose $b=1$
$g(a) = a^2 - 1 + g(1) = a^2 + c$, where $c$ is an arbitrary constant.
Therefore $f(a)=a^3 + c a$
This satisfies the functional equation for all $a \in \mathbb{R}$


I am not sure if this is the right place to ask,but anyway.Does anyone remember that during the football competition some other activities were conducted as well,namely I am talking about writing of fun inscriptions on a whatman. One of them was proving that "girls=evil"(by adding other "variables"-money and time),but I completely forgot the other inscriptions like mentioned one. So I suspect some of you has these inscriptions and hope you'll share them with me.

I took some photos of it to remember it. There not very sharp but I think most things are readable.

'' girls = evil " ....This  is complitely, complitely  wrong !   :mad:   :mad:

Of course it is absolutely wrong... It's just a male macho chauvinistic joke... don't take it too seriously  :wink:

[quote="hsiljak"][quote="Erken"]One of them was proving that "girls=evil"(by adding other "variables"-money and time)[/quote]

Hmmm... This isn't compeletely true, is it? girls = |evil|, if I recall correctly (at least that's how I remember the proof-using the fact that money is the root of all evil  :rotfl: )[/quote]
yup,my mistake...
money is root of all evil,girl is $ money\cdot time$ and time is money,therefore,$ girl\equal{}|evil|$.
big thanks for providing these nice photos,Mat(h)s...


LaTeX
number theory proposed
number theory

try to have the solution :D  :D  :D
Let $S$ be the sequence $n_1,n_2,....,n_k$ of positive integers such that \sum_{i=1}^k x_i\=m $<$ 2*k. prove that for each integer q with 1 <= q <=m ,there is a sequence $n_{i_1},n_{i_2},..,n_{i_u}$ ,where 1 <= $\i_{1}$<$\i_{2}$<...<$\i_{u}$ <= k, we have the sum of their equal to q.
i have the long solution and i want to have short solution.

[quote="nhat"]try to have the solution :D  :D  :D
Let $S$ be the sequence $n_{1},n_{2},....,n_{k}$ of positive integers such that {\sum_{i=1}^k x_i}=m $<$ 2*k. prove that for each integer q with 1 <= q <=m ,there is a sequence $n_{i_1},n_{i_2},..,n_{i_u}$ ,where 1 <= $\i_{1}$<$\i_{2}$<...<$\i_{u}$ <= k, we have the sum of their equal to q.
i have the long solution and i want to have short solution.[/quote]
sorry for my stupid to write it in the latex  when i post the problem


Problem 1: If you have a $ \frac{1}{10}$ chance of getting into any college you apply to, and you apply to 10 colleges, what is the chance that you will get into [b]at least[/b] one of the colleges?

Problem 2: If you have a $ \frac{1}{7}$ chance of getting into any college you apply to, how many colleges must you apply to to make your chances of getting into at least one college over $ \frac{3}{4}$?

I know these problems might be confusing, but if you wikipedia "bernoulli trials", it should help.
-MrRubix

Well, I'm not entirely sure bernoulli trials are necessary here.

I'll assume the probabilities are independent.
[hide="Problem 1"]
The probability that you don't get into any ONE of the colleges you apply for is $ \frac {9}{10}$
The probability that you don't get into ANY of them is $ \left(\frac {9}{10}\right)^{10}$
The probability that you get into at least one of them is then the complement of this, or $ 1 \minus{} \left(\frac {9}{10}\right)^{10} \equal{} 0.651\ldots$
[/hide]
[hide="Problem 2"]
Same idea.
We want $ 1 \minus{} \left(\frac {6}{7}\right)^{x} > \frac {3}{4}$
Solving for x, we get $ x > 8.99311$ so you need to apply to at least nine colleges.
[/hide]


This is a problem I'm having a lot of trouble with from 2005 National Sprint:

Consider the sequence $ a_1\equal{}1, a_2\equal{}13, a_3\equal{}135, ...$ of positive integers. The $ k$th term $ a_k$ is defined by appending the digits of the $ k$th odd integer to the preceding term $ a_{k\minus{}1}$. For example, since the $ 5$th term is 13579, then the $ 6$th term is $ a_6\equal{}1357911$. Note that $ a_3$ is the first term of the sequence divisible by 9. What is the value of $ m$ such that $ a_m$ is the $ 23$rd term of the sequence that is divisible by 9.

[hide="hint"] $ a_n\pmod{9}$ is congruent to the sum of the digits of $ a_n$, (mod 9). However, the sum of the digits of $ a_n$ is just $ n^2\pmod{9}$. If $ n^2\equiv0\pmod{9}$, what do you know about $ n$?[/hide]

EDIT: Hm I apparently forgot to type something.

Ohh, so the answer would be $ (22 \times 3) \plus{} 1 \plus{} 2 \equal{} \boxed{69}$, right?

Thanks!

Yeah, the answer's 69. Just note that the $ n$th term divisible by $ 3$ is $ 3n$, so it's just $ 23\cdot{3}\equal{}69$.

Guys, this is Mathcounts, we don't need fancy closed forms for summation.

1
13
135
1357
13579
1357911

Wow, terms 3 and 6 are divisible by 9. That is the first terms divisible by 9 is 3x1, second is 3x2, so the 23rd is 69.

[quote="isabella2296"]This is a problem I'm having a lot of trouble with from 2005 National Sprint:

Consider the sequence $ a_1 \equal{} 1, a_2 \equal{} 13, a_3 \equal{} 135, ...$ of positive integers. The $ k$th term $ a_k$ is defined by appending the digits of the $ k$th odd integer to the preceding term $ a_{k \minus{} 1}$. For example, since the $ 5$th term is 13579, then the $ 6$th term is $ a_6 \equal{} 1357911$. Note that $ a_3$ is the first term of the sequence divisible by 9. What is the value of $ m$ such that $ a_m$ is the $ 23$rd term of the sequence that is divisible by 9.[/quote]

Wow, Mathcounts tricks a lot of people with complicated, wordy problems.  The trick is, all sequence terms where n is a multiple of 3 have a number divisible by 9.  3 times 23 is 69, simple as that(No modifiers needed[I don't even know what a modifier is :lol: ])


How I can change only name of function [b]chapter[/b]? please help

You need to redefine \chaptername.

For example, if you were writing in French, you might do

\renewcommand{\chaptername}{Chapitre}


trigonometry
geometry
similar triangles
angle bisector
geometry solved

Let C1 and C2 be 2 circles. Their intersection is {M, N}. The tangents at M to the circles intersect C1 at A and C2 at B. C and D are the pts where AN intersects C2 for the second time and where BN intersects C1 for the second time, respectively. Prove that AC=BD.

It's not a hard one, but I like it.
Cheers! (as many people tend to say here on this forum :D )

Yes I am interested but I don't find the solution.
You said it was easy but it is not easy for me I think :(
Or maybe I'm just overlooking something which should be obvious.
So Grobber, can you post your solution (I suppose you have one) ?

AM 2 =AN.AC
BM 2 =BN.BD. we want to prove AC/BD =1.so it suffices to that [AM 2 .BN]=[BM 2 .AN]
applying rule of cosines to trianglea ANM and BNM and noting that <BNM=<ANM,we see it;s equivalent to proving MN 2 =AN.BN
,which can be proved easily by the sine rule to the same triangles
QED (hope the proof is correct,and i guess Grobber has an ace up his sleeve,inversion in this also :D )
cheers

Here's my proof:
Triangles MAC and MDB are similar because angle MCN=angle MBN and angle MDN=angle MAN. All 3-letter combos are angles from now on. AMN= ACM (AM tangent to C2), so ANM= AMC. BMN= BAM (BM tangent to C1) so BNM= DMB= AMC= ANM= Pi-DNM. By using sine law in C1 and the fact that sinX=sin(Pi-X) we get DM=AM. Since triangles MAC and MDB are similar (in this order of the vertices) and DM=AM we get the conclusion that they are equal, so AC=BD.

If O' and O" are the centers of the circles C_1 and C_2, it's a known property that all triangles having a vertex at M and the opposite side passing through N are similar to the triangle O'MO", hence triangles AMC and DMB are similar.

From ANMD cyclic, <ADN = <AMN, but <AMN = <MBN ( AM tangent to C_2), hence <ADN = <NBM, and AD || BM, that is, M is the midpoint of the arc AND and MD = AM; similarly MC = MB, hence the similar triangles AMC and DMB are congruent and AC = BD.

[color=darkblue][b]Remarks![/b]
[i]If T is the symmetrical of M about N, then ATBM is cyclic, and some additional angle chasing shows that AB is the A-symmedian in triangle MAT, hence ATBM is harmonic. See easily that MT is the angle bisector of <ANB, well known property of a harmonic quad.

The line joining the midpoints of AB and CD, being parallel to MN, is hence perpendicular onto O'O".

From the proof of the problem it has been noticed that AD || MB and BC || AM, hence AD and BC intersect on the M-median of the triangle AMB.[/i][/color]

Best regards,
sunken rock

by simple angle chasing we can conclude MD=MA  MB=MC
then we have triangle[b]AMC[/b]  triangle[b]DMB[/b] are congruent

Dear Mathlinkers,
1. By a particular case of Reim's theorem, AD//MB
2. In consequence, MAD is M-isoceles
3. Mutatis mutandis, MBC is M-isoceles
4. <DMB=<AMC
4. According to the c.a.c. theorem applied to MAC and MDB, AC=BD.
Sincerely
Jean-Louis


Euler
geometry
geometric transformation
homothety
circumcircle
cyclic quadrilateral
geometry proposed

Here's a not so hard question which looks like a difficult one:

Let ABCD be a cyclic quadrilateral and let Ea, Eb, Ec, Ed the centers of the Euler circles of the triangles BCD, ACD, ABD, ABC respectively. Prove that AEa, BEb, CEc, DEd are concurrent lines.

What is  Euler circle? I think russian equivalent has other name.

It's the 9 point - circle: the one which contains the middle pts of the sides, the feet of the heights and the middle pts of HA, HB, HC, where H is the orthocenter.

Good problem!

Let O be center of the circle, vec.OA=a, vec.OB=b, ...

If M is the mass-center of BCD then OM=(b+c+d)/3. Point Ea is a result of action of homothety with center M and coefficients -1/2 to point O ==> OEa=(b+c+d)/2.

Line AEa includes points k*OA+(1-k)OEa (k is arbitrary real number). Let k=1/3 then k*OA+(1-k)OEa=(a+b+c+d)/3 - it is a common point for all lines AEa, BEb, ...

Do you know that Euler circle is tangent to inscribed circle and three escribed circles? So its is 13 point-circle!

Myth said:

Do you know that Euler circle is tangent to inscribed circle and three escribed circles? So its is 13 point-circle

 I think grobber posted to prove that! (Euler circle  tangent to inscribed circle and three escribed circles)

cheers!

Lagrangia!
I don't understand your insinuation again!

Do you read the problem?

:? :?
myth yes I read the problem.. and I said that in a post there was a proof of the fact that: Euler circle is tangent to inscribed circle and three exscribed circles.
hope you understood now!!

cheers!!

Thx Myth! That's exactly my proof. This could, however, be nasty without complex numbers.

Yes, I know it is tangent to those 4 circles, but Euler's circle is also known as "the 9 point-circle".  :D

Dear Mathlinkers,
is there a synthetic proof of this problem?
Sincerely
Jean-Louis

Let $H_{a},\ H_{b}$ be the orthocenters of the triangles $\vartriangle BCD,\ \vartriangle ACD$ and we denote as $M,$ the midpoint of the side-segment $CD$ of the given cyclic quadrilateral $ABCD$.

From $AH_{b}\parallel = 2OM\parallel = BH_{a}$ as well, where $O$ is the circumcenter of $ABCD$, we conclude that $AB\parallel = H_{a}H_{b}\parallel = 2E_{a}E_{b}\ \ \ ,(1)$

From $(1),$ based on the [b]Thales theorem[/b], we conclude that $AP = 2PE_{a}\ \ \ ,(2)$ and $BP = 2PE_{b}\ \ \ ,(3)$ where $P\equiv AE_{a}\cap BE_{b}.$

By the same way, we can say that any two of the segment-lines $AE_{a},\ BE_{b},\ CE_{c},\ DE_{d},$ intersect each other at one point which divides them in to the same ratio.

Hence, we conclude that the four lines as the problem states, are concurrent at point $P\equiv AE_{a}\cap BE_{b}$ and the proof is completed.

Kostas Vittas.


algebra
polynomial
induction
algebra proposed

Determine all positive integers $ n$ such that $ g|f$, where:

$ f\equal{}X^n\plus{}X\plus{}1$ and $ g\equal{}X^3\minus{}X^2\plus{}1.$

[quote="moldovan"]Determine all positive integers $ n$ such that $ g|f$, where:

$ f \equal{} X^n \plus{} X \plus{} 1$ and $ g \equal{} X^3 \minus{} X^2 \plus{} 1.$[/quote]

Let the reminder of the polynomial division of $ x^n$ by $ x^3\minus{}x^2\plus{}1$ be $ a_nx^2\plus{}b_nx\plus{}c_n$

We get the induction rules : $ a_{n\plus{}1}\equal{}a_n\plus{}b_n$ and $ b_{n\plus{}1}\equal{}c_n$ and $ c_{n\plus{}1}\equal{}\minus{}a_n$

So we can write this reminder as $ a_nx^2\minus{}a_{n\minus{}2}x\minus{}a_{n\minus{}1}$ with sequence $ a_n$ defined as $ a_0\equal{}0$, $ a_1\equal{}0$, $ a_2\equal{}1$ and $ a_{n\plus{}3}\equal{}a_{n\plus{}2}\minus{}a_n$

Then $ g|f$ $ \iff$ $ (a_{n\minus{}2},a_{n\minus{}1},a_n)\equal{}(1,1,0)$ $ \iff$ $ (a_{n\minus{}5},a_{n\minus{}4},a_{n\minus{}3})\equal{}(0,0,1)$ $ \equal{}(a_0,a_1,a_2)$

So we have a first solution for $ n\equal{}5$ and we can have other solutions only if $ \{a_n\}$ is a periodic sequence.

But $ a_n\equal{}a.r_0^n\plus{}b.r_1^n\plus{}c.r_2^n$ with $ r_i$ roots of $ x^3\minus{}x^2\plus{}1\equal{}0$ and $ a_n$ periodic with period $ p$ would mean $ a_{n\plus{}p}\equal{}a_n$ $ \forall n$ and so $ a.r_0^n(r_0^p\minus{}1)\plus{}b.r_1^n(r_1^p\minus{}1)\plus{}c.r_2^n(r_2^p\minus{}1)\equal{}0$ $ \forall n$ and this implies $ a\equal{}b\equal{}c\equal{}0$ if $ r_i$ are pairwise distincts and different from roots of $ 1$.

So our sequence is not periodic and the only solution is $ n\equal{}5$


1. a) Determine if the set {1,2, ..., 96} can be partitioned into 32 sets of equal size and equal sum.

    b) Determine if the set {1, 2, ..., 99} can be partitioned into 33 sets of equal size and equal sum.

[hide="a)"]No, because the sum of all the numbers is not a multiple of 32.[/hide]

b)

[hide]The sets $ {100\minus{}3n,53\minus{}3n,6n\minus{}3}$ for $ n\equal{}1,2,...,17$.
and the sets $ {3n\minus{}2,3n\plus{}50,102\minus{}6n}$ for $ n\equal{}1,2,...,16$.

It is easy to check that there are no common terms for if you look at the terms 0(mod 3) in the first set, we have 3(mod 6), while in the second set it is 0(mod 6).

1(mod 3), we have no terms overlap and for 2(mod 3) the first set goes from 97 to 49, while the first set goes from 1 to 46.

So these are allowed.[/hide]


inequalities
Cauchy Inequality
inequalities proposed

[quote="Vasc"]

If $a,b,c,d$ are non-negative real numbers such that $a+b+c+d=4$, then
${\frac{a^2}{b^2+3}+\frac{b^2}{c^2+3}+\frac{c^2}{d^2+3}}+\frac{d^2}{a^2+3}\geq 1$.[/quote]

Manlio, please stop calling all the inequalities the same.

And a solution for this great problem:
   Using the old trick with Cauchy inequality and homogenizing it suffices to prove that
   $(a^2+b^2+...)( a^2+b^2+...-3((a+b+c+d)/4)^2)$ is greater than $ a^2b^2+b^2c^2+...$. This follows from the fact that $((a+b+c+d)/4)^2$ is at most $(a^2+...)/4) and $ (x+y+z+t)^2$ is at least $4(xy+yz+zt+tx)$

You  are right, Harazi. I have little fantasy about naming Vasc inequalities.

However, nice solution ;)

Harazi, please explain how did you reduce inequality to $(a^2+\dots)(a^2+\dots-3(...))\geq a^2b^2+\dots$.

What should I explain? I just applied the Cauchy-Schwarz inequality in the very well known form and then I homogenized the inequality, replacing 1 by the arithmetic mean of a,b ,c,d.

I try to understand it.but I failed.
does it mean to prove $3a^2+3b^2+3c^2+3d^2+a^2 b^2+b^2 c^2 +c^2 d^2+d^2 a^2 \leq 16$
but how to do about it?anyone?

[quote="harazi"]Please, apply correctly the Cauchy-Schwarz inequality.[/quote]
can you point where I was wrong? ;)  ;)

[quote="manlio"]Harazi refers to this application of Cauchy
${\displaystyle[a^2(b^2+3) +b^2(c^2+3)+c^2(d^2+3)] (\frac{a^2}{b^2+3}+\frac{b^2}{c^2+3}+\frac{c^2}{d^2+3}}+\frac{d^2}{a^2+3})\geq (a^2+b^2+c^2)^2$[/quote]
and complete it please,thanks :)

Either I am  loosing my minds, or something strange is happening. What is so difficult to understand? I applied the same method for hundreds of times.

This is Harazi solution.

From Cauchy we see that it is sufficient to prove that

$(a^2+b^2+c^2+d^2)^2 \geq 3(a^2+b^2+c^2+d^2)+a^2b^2+b^2c^2+c^2d^2+d^2a^2$

which can be rewritten as

$(a^2+b^2+c^2+d^2)(a^2+b^2+c^2+d^2-3) \geq a^2b^2+b^2c^2+c^2d^2+d^2a^2$
Now you must homogeneize to have Harazi form

$(a^2+b^2+c^2+d^2)(a^2+b^2+c^2+d^2-3(\frac{a+b+c+d}{4})^2) \geq a^2b^2+b^2c^2+c^2d^2+d^2a^2$

which follows from

$ a^2+b^2+c^2+d^2  \geq 4$

and

$(x+y+z+t)^2 \geq 4(xy+yz+zt+tx)$

with $x=a^2$ and similar.

I have only rewritten Harazi proof. ;)

In fact, the following stronger inequality holds:
If $a,b,c,d$ are non-negative real numbers such that $a+b+c+d=4$, then

${\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}}+\frac{d}{a+3}\geq 1$.

In addition, I conjecture the following statement:
If $a,b,c,d,e$ are non-negative real numbers such that $a+b+c+d+e=5$, then

${\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{d+1}}+\frac{d}{e+1}+\frac{e}{a+1}\geq \frac{5}{2}$.

Can somebody to prove this one?

[quote="Vasc"]In fact, the following stronger inequality holds:
If $a,b,c,d$ are non-negative real numbers such that $a+b+c+d=4$, then

${\frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}}+\frac{d}{a+3}\geq 1$.
[/quote]

I am sorry if this question is stupid: How to prove that this inequality is stronger than the original one?  :?

A relative answer.

(1) $a+b+c+d \ge 4$ implies ${\displaystyle \frac{a^2}{b^2+3}+\frac{b^2}{c^2+3}+\frac{c^2}{d^2+3}}+\frac{d^2}{a^2+3}\geq 1$ .

(2) $a+b+c+d \ge 4$ implies ${\displaystyle \frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}}+\frac{d}{a+3}\geq 1$ . But (2) is equivalent to
(2') $a^2+b^2+c^2+d^2 \ge 4$ implies ${\displaystyle \frac{a^2}{b^2+3}+\frac{b^2}{c^2+3}+\frac{c^2}{d^2+3}}+\frac{d^2}{a^2+3}\geq 1$ .
I said (2) is stronger than (1) because $a+b+c+d \ge 4$ implies $a^2+b^2+c^2+d^2 \ge 4$ .
Thus, we can use (2') to prove (1).

More interesting is to prove that $a+b+c+d = 4$ implies ${{\frac{a^2}{b^2+3}+\frac{b^2}{c^2+3}+\frac{c^2}{d^2+3}}+\frac{d^2}{a^2+3}\geq \frac{a}{b+3}+\frac{b}{c+3}+\frac{c}{d+3}}+\frac{d}{a+3}$ .
Let us start with $n=2$. Thus, prove or disprove the statement:

If $a,b$ are positive numbers such that $a+b=2$, then
$\frac{a^2}{b^2+1}+\frac{b^2}{a^2+1} \ge  \frac{a}{b+1}+\frac{b}{a+1}$.

O.K.
Let us continue with $n=3$. Thus, prove or disprove the statement:

If $a,b,c$ are positive numbers such that $a+b+c=3$, then
$\frac{a^2}{b^2+1}+\frac{b^2}{c^2+1}+\frac{c^2}{a^2+1} \ge  \frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}$.

I think a simpler case is for $abc=1$.


[img]http://i198.photobucket.com/albums/aa269/ConanKudo5/Geo.jpg[/img]
AB.BC=2AD.CD
Prove that:
$8BD^{2}\le 9AC^{2}$

[hide]Let $h_{B}=d(B,AC)$ and $h_{D}=d(D,AC)$

$\frac{(ABC)}{(ADC)}= \frac{AB\cdot BC \cdot \sin(ABC)}{AD \cdot DC \cdot \sin(ADC)}= \frac{AB\cdot BC}{AD \cdot DC}= 2 \Rightarrow$

$\frac{AC \cdot h_{B}}{AC \cdot h_{D}}=2\Rightarrow h_{B}= 2h_{D}\ \ (1)$

Now let $F\in AC \cap BD$. From $(1)$ we get $BF=2DF$.

Set $DF=a$. Then $BF \cdot DF=2a^{2}$

Let $AC$ be a chord passing through $F$.
$AF\cdot FC =2a^{2}$, so we have a minimum sum $AF+FC$ when $AF=FC=a\sqrt 2\Rightarrow AC_{min}=a\sqrt 8$

So we have $AC\geq a\sqrt{8}$

But $a=DF =\frac{BD}{3}\Rightarrow AC\geq \frac{BD}{3}\sqrt 8\Rightarrow \boxed{9AC^{2}\geq 8BD}$[/hide]


algebra
polynomial
search
Gauss
algebra unsolved

Prove that polynomial $x^{1383}+2003$ is not farctoriseable to two polynomials with rational coefficients.

Wow,What a nice thing .
I didnt know Eisenstein,so after your post i start to search for it.
Correct me if this is not the Eisenstein lemma:
Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$ and $p$ a prime number.
then if

$p\mid a_k,0\le k\le n-1$ and $p$dont$\mid a_n$ ,$p^2$dont $\mid a_0$ then
$P(x)$ is not fractorisable to to polynomail with integer cooefficients.

:idea:  :idea:  :idea:  :idea: Wait ,Eisenstein is just for integers but here problem askes for the rational coofficients.

Next well known theorem (special version of a well known lemma by Gauss):
A polynomial with integer coefficients and leading coefficient $1$ can be factored into a product of rational polynomials iff this already can be done into integer polynomials.

Gauss lemma,Its nice.
Zetax,I searched fo this too an i find something in a book .
It is exactly like your lemma but without the leading cooffiecient to be $1$.

So the problem is done.
Thanks i learned two nice lemma(indeed nice) :heli:
If any one knows a good problem related to these lemma's please post it.


13 lines pass through point A, yet no three lines are concurrent. How interesting.

[hide="2, assuming I understand correctly"]Ignoring the collinear business, we get ${10\choose2}=55$ lines.

There are 4 points that are collinear. Assuming that they are not, we have 6 different lines. Now that they are, all the 6 different ones merge into one line, causing 5 less possibilities. So the total number of lines is 55-5=50. [/hide]

[quote="236factorial"]13 lines pass through point A, yet no three lines are concurrent. How interesting.

[/quote]

I think it meant $besides$ points $A$ and $B$.....

The answer to the second one you have given is wrong(According to the book....)

[quote="shadysaysurspammed"]1)In a plane there are $37$ straight lines,of which $13$ pass through the point $A$ and $11$ pass through the point $B$.  Besides,no three lines pass through one point  ,  no line passes through both points $A$ and $B$  ,  and no two parallel.Find the number of  points intersection of  the straight lines.

2)There are $10$ points in a plane,no three of which are in the same straight line,excepting $4$ points,which are collinear.Find the  $\text{number of straight lines obtained from the pairs of these points}$.[/quote]

[hide="1"]
the 13 and the 11 lines intersect in a total of $13 \cdot 11 = 143$ ways. The 13 points intersect at one point, the 11 points intersect at one point. the other 13 lines not associated with either point each hit every line associated with A and B as well as them selves since they are all skew. so we have $13 \cdot 24 = 312$ and $\binom{13}{2} = 78$ Add it all together: $143 + 2 + 169 + 78 = 535$ [/hide]

[quote="shadysaysurspammed"]The answer to the second one you have given is wrong(According to the book....)[/quote]

Arithmetic is a killer ... ${10 \choose 2} \neq 55$


Let $ O$ be a point in the plane of a convex quadrilateral $ A_1 A_2 A_3 A_4$ and let $ B_1, B_2, B_3, B_4$ be points such that $ OB_i$ is parallel and equal in length to $ A_i A_{i\plus{}1}$ for $ i\equal{}1,2,3,4$ (where $ A_5\equal{}A_1$). Show that the area of the quadrilateral $ B_1 B_2 B_3 B_4$ is twice that of the quadrilateral $ A_1 A_2 A_3 A_4$.


I had (and have) a rating of 4 (yes, only 4). Then I noticed someone rated one of my post a 6, but my rating did not change. Why?

It's simple math. Your rating is the average of all your ratings. Do the math yourself to see how adding 1 rating to hundred others influences your rating.


Let $ a$ , $ b$ and $ c$ be positive real numbers. Prove that :  $ \sqrt{\frac{a^{2}\plus{}bc}{b\plus{}c}}\plus{}\sqrt{\frac{b^{2}\plus{}ca}{c\plus{}a}}\plus{}\sqrt{\frac{c^{2}\plus{}ab}{a\plus{}b}} \geq \sqrt{3(a\plus{}b\plus{}c)}$


analytic geometry
geometry
function
integration
calculus
calculus computations

a) Compute the area of the region that lies inside the polar curve  r = sin(theta)  – cos(theta)  but outside the polar curve
r^2 = sin (2*theta) .

b) Compute the length of the portion of the polar curve  r = 2 sin^2 (theta / 2) that lies outside the polar curve r = 2 + sin(theta).

a) Draw a picture.  Then you can see where each function is relative to the other, and where they intersect.  Then just use
$ \frac{1}{2}\int_{\theta_{1}}^{\theta_{2}}f(\theta)^{2}\ d\theta$


rotation
integration
calculus
calculus computations

The region bounded by $y = e^{ - x} ,y = 1,x = 1$ is rotated about x=1,
Find the volume


geometry
IMO Shortlist
geometric inequality
hexagon

The lengths of the sides of a convex hexagon $ ABCDEF$ satisfy $ AB \equal{} BC$, $ CD \equal{} DE$, $ EF \equal{} FA$. Prove that:
\[ \frac {BC}{BE} \plus{} \frac {DE}{DA} \plus{} \frac {FA}{FC} \geq \frac {3}{2}.
\]

just an idea!

have you thought of a way to make a connection between the following inequality that is always true for a,b,c >0

a/(b+c)+b/(c+a)+c/(a+b)>=3/2

perhaps this is a good idea perhaps this ain't such a good one! perhaps it can never be applied to this! :D it is you who diciede!

tschuess!

[quote="galois"]well this is an old problem lying since quite sum time but i think now i have a solution using lagrangia's idea.i will post it later[/quote]

well i think i met this problem be4 too and i used ptolemy then it simplifies to lagrangia's inequality..

Yeah... that inequality works.  I'll just post the quick solution since no one else has bothered.

In quadrilateral ABCE we have by Ptolemy's inequality

$CE \cdot AB + AE \cdot BC \ge AC \cdot BE$
$\implies BC \cdot (CE + AE) \ge AC \cdot BE$
$\implies \frac{BC}{BE} \ge \frac{AC}{CE+AE}$

Adding the analogous inequalities,

$\frac{BC}{BE} + \frac{DE}{DA} + \frac{FA}{FC} \ge \frac{AC}{CE+AE} + \frac{CE}{EA+CA} + \frac{EA}{AC+EC} \ge \frac{3}{2}$

Applying Ptolemy's inequality on $ACDE$ we have $(DE)(AC)+(DC)(AE)\ge (DA)(CE)$. Since $DE=DC$, this becomes $DE(AC+AE)\ge (DA)(CE)$ or $\frac{DE}{DA}\ge\frac{CE}{AC+AE}$. Similarly we have \[\begin{cases}\frac{DE}{DA}\ge\frac{CE}{AC+AE} \\ \frac{FA}{FC}\ge \frac{EA}{CE+CA}\\\frac{BC}{BE}\ge\frac{EA}{EA+EC}\end{cases}\] After adding the three inequalities, the result immediately follows from Nesbitt.

This is a nice application of [b]Ptolemy's[/b] inequality.Try another which is from the same.

Prove that in an acute-angled triangle ABC with orthocentre H, we have:

$AH+BH+CH\le2h_{max}$

where $h_{max}$ denotes the largest altitude of triangle ABC .

[quote=biomathematics][quote=WJ.JamshiD]:( What is Ptolemy's inequality?[/quote]

The Ptolemy's  Inequality states that given a quadrilateral $ ABCD $ the following is true :-

$ AB. CD + AD. BC \ge AC.BD $, with equality iff $ ABCD $ is cyclic.[/quote]

thanks you :)

Relatively straightforward, only needed Ptolemy's, and considering that I just made like 10 posts on inequalities Nesbitt's was a very easy application to apply and see.


[quote="cincodemayo5590"]Hey, I was wondering with what methods an explicit formula for a sequence can be derived from its recursive definition. I already know how to use the characteristic polynomial for a linear recursion and how to use finite differences, but I was wondering what methods should be used for non-homogeneous recusions (things such as $a_{n+1}=a_{n}+n^{2}$). Is there some sort of algorithm or methodology to it?[/quote]

oh come on...that should be easy...and non-homogeneous is unrelated to what you are talking about...that means different degrees, so like a_n=a_{n-1}^2, etc

also sam...just look in the precalc book...the difference equations section gives an algorithm for these problems...

Generating functions are always useful.  Simply sum both sides of the recurrence, $a_{n}= f(a_{n-1})$ and multiply by $a^{n}$, then find a relation to the generating function $A(x)$.

Some stuff of the form $a_{n+1}= a_{n}+f(n)$ can be written in the form $a_{n+1}-g(n+1) = a_{n}-g(n)$ or similar, and you can solve for $b_{n}= a_{n}-g(n)$.


Prove that for all real numbers $a, b, x, y$ satisfying
$x + y = a + b$. and $x^4 + y^4 = a^4 + b^4$ then $x^n + y^n = a^n + b^n$for all$n$ betwen $N$.


function
algebra
polynomial
advanced fields
advanced fields unsolved

Give a Taylor expansion of order 3 around 0 for the function
defined implicitly by x-->y=g(y) with g(0)=1 and
f(x,y)=1-ye^x +xe^y = 0

[quote="Moubinool"]Give a Taylor expansion of order $3$ around $0$ for the function
defined implicitly by $y=g(x)$ with $g(0)=1$ and
$f(x,y)=1-ye^x +xe^y = 0$[/quote]

 Let $y=g(x)=1+xQ(x)+o(x^3)$
 Where $Q(x)$ is of degree 2 and is what we have to compute!!
 To continue, let $e^x=P(x)+o(x^3)$ be the Taylor expansion of $e^x$ around $0$; and $e^y=e^{(1+xQ(x)+o(x^3)} = e(1+P(xQ(x))+o(x^3)$

 Substitute all this stuff into the equation

 $f(x,y)=1-ye^x +xe^y = 0$, cancel all the $o(x^3)$ term, we will get a polynomial identity from which we can compute $Q(x)$

  I don't know any shorter solution :P


Find $f(x) : \mathbb{R}\to \mathbb{R}$;$g(x) : \mathbb{R}\to \mathbb{R}$ such that:
${f(x-y)=f(x)g(y)-f(y)g(x)}$
and $g(x-y)=g(x)g(y)+f(x)f(y)$
with $x,y\in R$

[color=red][I edited the title (the original one was the new source)/source of this one for you, but this is the (first and) last time. -mathmanman][/color]

Posted before. Solutions $f,g\equiv 0$ and $cosax,sinax$.


10th grade
Number N,Z,D,Q,R, prime number
Absolute value |x| definition and properties
Equations AxB=0, A/B=0, inequations
Functions  monotony, maximum, minimum,
ax+b, x^2, 1/x, \sqrt x, sinx, cosx
No derivatives
Statistic mean, median, mode,"frquence f_i=n_i/n "
Triangles similar, isometrics
Vectors Chasles relation
Analytic geometry coordinates vectors
Equation of line (AB)
System of equation ax+by=c, a'x+b'y=c'

[quote="grobber"]Are these high school grades? If so, don't you study integrals in high school?[/quote]

integrals are not in the curriculum

Actually, if you do the IB program in Canada, you have to study integrals.

Meh, Canada's math curriculum sucks. I wish I went to one of those Math/Science academies in the States.

hmmm....I must be the only 16yr old guy who'd say that.

where in ontario are you? i lived in ottawa for 7 yrs.. now i go to one of those math/science academies in the States.. lol...
my math program for high school:

9th grade - Adv. Analysis II (some weird honors precal thing)
10th grade - Analytical BC+ calc
11th grade - multivariable calc
12th grade - linear algebra stuffs

Burlington, its close to Hamilton, if that doesnt ring a bell then its about a half hour to an hour away from Toronto

errr... Toronto rings the bell.. lol.. never heard of the other two..

The only reason I though you might have heard of Hamilton was if you watched the CFL, which I don't, but I watch enough TSN to know all the teams, especially the local ones.  Burlington's only known as the rich city (3rd richest in Canada)

CFL? TSN? Major apologies for my lack of knowledge  :blush: .. lol..

The curriculum for gr. 10 pre-IB is the following:

Radicals
Exponents and Polynomials
Solving Equations
Co-ordinate Geometry
The Straight Line
The Parabola
Congruent and Similar Triangles
Trigonometry

And that's it!  :D

canada won't let me or my family in their country

Thats better than when they let a family in, then decide they wont let the parents stay but leave the kids. This is a true story, it happened to a classmate of mine in gr.7, he was from Hungary.

is this post about canada, or about high school?

[quote="Ani"]well obviously they had a good reason for doing so. we don't just throw people out for no reason :D[/quote]of course! however, they usually don't let people [b]in[/b] without any reason  :?

I have no clue what you are talking about... :?

I mean they don't kick out people without any reason, but many times they don't let people in (ie give visas) for no reason.

grade 5:Algebra I(but then I swithched to Algebra II halfway through the year)
grade summer of 5-6: Geometry
grade 6:Pre-Calc

woah, thats major acceleration, but wait this was about Canada, right?

[quote="Ani"]Oh, I see, you mean they are strict about who they accept in their country. Yeah, that's true ;)[/quote]

Not really, we were the token white family. All the other people were other foreign ethnicities.

[quote=MathNinja234]Canadas system is trash
Also if you in ontario do not vote liberal
they want grade 13[/quote]

I don't know.

Honestly, the most reasonable party now seems to be NDP and Green.

I am a leftist, and I want to cheer for a leftist candidate, but the liberal one is just not it this year.

[quote=AlienGirl05][quote=pezzeepenguin69][quote=bugzpodder]Grade 9: 2d geometry, solving systems of equations, cartesian geometry
Grade 10: quadratics, factoring, more geometry
Grade 11: log, functions, conic sections
Grade 12: three courses: data management (combinatorics), Algebra and Geometry (2d geometry + vectors) , and Calculus (limits/derivatives)[/quote]

I feel like Canada's curriculum is much better than the US. It looks like they actually teach you stuff that you'll actually use.[/quote]

According to my best friend, the Canadian system sucks.

Sadly, Canada's school funding is not that great.

I don't know if I am $100\%$ correct, but we don't have enough chairs for children to sit on, janitors are one of the top-payed school workers, and there is more funding for playgrounds than for the actual education that children receive.

The material they learn is also far-behind.

While my best friend's class is learning the Pythagorean theorem, I am learning precalculus on AoPS.

In fact, I believe that students only learn it when they are $16$ or $17$ years old.

If I am not mistaken, even calculus is an OPTIONAL course to take in university.[/quote]

Great. I don't think I'm ever going to use the fundamental theorem, Euler's Theorem, or Squeeze Theorem ever. It should be a choice whether you learn Calculus.

I believe we mostly learn those things to prepare ourselves for whether we will get jobs involving those subjects or not.

However, I am not quite sure why, for example, if someone wants to become a historian, they learn history at special schools designed for aspiring historians, instead of making all students learn history.

Of course we should know general history of how we got here, but not the history of the most random things ever.

I likely won't need to know things such as how snap peas were created, or how some dude named Joe Dabadoo made a farm $300$ years ago.

[quote=pezzeepenguin69]
Great. I don't think I'm ever going to use the fundamental theorem, Euler's Theorem, or Squeeze Theorem ever. It should be a choice whether you learn Calculus.[/quote]

(i think you mean Euler's method) Calculus is super important in helping understand easier branches of math and the sciences such as physics. It's arguably the most important branch of math. All the geometry volume formulas can only be derived through calculus.

[quote=ValorKid12345]UMM... I don not know about you but this is kind of against what TRT was made for. Listen to BOGTRO[/quote]

This thread was made 14 years before BOGTRO lmao

[quote=pezzeepenguin69]I don't even know what the US is teaching us now for math. They are literally teaching Non-deterministic Finite Automata, Naive Bayes, Linearity of Expectation, Poison Distributions, Chernoff Bounds, and Markov Chains. I don't know one thing I'll use, but at least they're somewhat interesting.[/quote]
you literally only have to take up to precalculus to graduate

[quote=pezzeepenguin69]I don't even know what the US is teaching us now for math. They are literally teaching Non-deterministic Finite Automata, Naive Bayes, Linearity of Expectation, Poison Distributions, Chernoff Bounds, and Markov Chains. I don't know one thing I'll use, but at least they're somewhat interesting.[/quote]

In the US? The majority of those topics are usually seen in undergraduate courses, not in high school.

[quote=pezzeepenguin69][quote=bugzpodder]Grade 9: 2d geometry, solving systems of equations, cartesian geometry
Grade 10: quadratics, factoring, more geometry
Grade 11: log, functions, conic sections
Grade 12: three courses: data management (combinatorics), Algebra and Geometry (2d geometry + vectors) , and Calculus (limits/derivatives)[/quote]

I feel like Canada's curriculum is much better than the US. It looks like they actually teach you stuff that you'll actually use.[/quote]

bro they don't even teach integrals

[quote=pezzeepenguin69][quote=bugzpodder]Grade 9: 2d geometry, solving systems of equations, cartesian geometry
Grade 10: quadratics, factoring, more geometry
Grade 11: log, functions, conic sections
Grade 12: three courses: data management (combinatorics), Algebra and Geometry (2d geometry + vectors) , and Calculus (limits/derivatives)[/quote]

I feel like Canada's curriculum is much better than the US. It looks like they actually teach you stuff that you'll actually use.[/quote]

they used to. Unfortunatly this is no longer true

My best friend is learning how to solve inequalities in LINEAR equations at the same age when I was learning precalculus. Recently, I reconnected with friends of mine who are $16$ years old, and they are barely beginning to learn precalculus!

Linear Algebra is a part of precalc at least in the states.

I think she means linear equations, not linear algebra

My intention was to say "linear equations" instead of "linear algebra", my apologies.


What value of $ k$ would make the shaded triangle's area $ 12\frac12\%$ of the rectangle's area? Express your answer as a common fraction.
[asy]size(70);
draw((0,0)--(0,8)--(9,8)--(9,0)--cycle);
fill((0,0)--(3,8)--(4,0)--cycle,gray(.7));
draw((0,0)--(3,8)--(4,0)--cycle);
label("8",(0,4),W);
label("$k$",(2,0),S);
label("9",(4.5,8),N);[/asy]

$ 8\times9\times\frac {12.5}{100} \equal{} \frac {1}{2}\times k \times 8 \iff 9 \equal{} 4k \iff k \equal{} \boxed{\frac{9}{4}}$.

I didn't really understand that solution so I will give you my very own. As you know, 12.5 percent of something is also 1/8th of it.

So we know that the shaded are is 1/8th of 72, or 9.

Now we recall the formula for finding the area of a triangle which is A=bh/2.
So : 9=8b/2.

The height has to be 8 because. well you should see why in the picture.

base (or K) obviously equals 9/4


linear algebra
matrix
superior algebra
superior algebra solved

If A is a 2x2 matrix with integer entries such thar
|Tr(A)|>2, then [tex]A^n\neq I_2[/tex], for all n in N.
Generalize.
p.s. I think that the conclusion remain the same if we change 2 with 3. :)

Suppose A is a kxk matrix with complex elements such that |TrA|>k. Then we cannot have A^n=I_k for no n. Suppose we have A^n=I_k. Then the proper values of A have modulus 1. Let them be r_1,...,r_k. Then k<|TrA|=|r_1+...+r_k|<=|r_1|+...+|r_k|=k, false.


geometry
incenter
circumcircle
geometry proposed

Let $\Omega$ be the Brocard point, $I$ the incenter, $O$ the circumcenter of triangle $ABC$.

Prove the $O\Omega\geq OI$

I think guys who like geometry problems don't want to solve this one,
but guys who don't like them(I mean geometry problems :) ) even don't want to try. :(


Putnam
symmetry
calculus
calculus computations


Find the total resistance. All the resistors are identical with R=10 ohm.

[url=http://imageshack.us][img]http://img19.imageshack.us/img19/3579/hsj1bbtmp.jpg[/img][/url]

Points symmetric about diagonal $ AB$ are equi-potential.

Hence the circuit resembles the one given below. (Do not laugh at my diagram, please. Each "curve" is a resistor of 10 ohm.)

Hence net resistance,
$ R_{net} \equal{} \left(\frac {R}{2} \plus{} \frac {R}{4} \plus{} \frac {R}{6}\right)\times2 \equal{} \frac {11R}{6} \equal{} \frac {110}{6} \text{ ohm}$

are don tu kasa resolve kelas circuit reply

hello
Here is my solution.
The circuit can be seen in this manner too.
I hope u will easily calculate it by ur own.
Thank u

[offtopic]Hehe, this reminds me of [url=http://xkcd.com/356/]this[/url]  :D [/offtopic]

[quote="integration"]Kabi, and every1, I'm really, really confused on how you solve this nightmarish circuit.  I looked into my book, and the only tool that would seem to work is Kirchoff's junction and loop rule.  But your equations don't even look remotely close to Kirchoff's rule!!  :huh:

Can you elaborate on your simplifications/reasonment?  Otherwise, can you guide me on what I need to read to solve those circuits? (link, .... or any theory you can come up...)  :blush:[/quote]
Well i m really poor in explaining things therefore i haven't given any explanation with my solution.
Let me try once .
See  Kouichi Nakagawa's circuit diagram .There is symmetry in this figure .If u just join $ AP$ the upper part is similar to lower part.Now when a charge comes to point $ A$ whether it took path $ AB$ or $ AE$ its going to face a similar circumstances .Now come to point $ F$ whether it go through $ FG$ or $ FJ$ effects on the charge is same.
So we can consider a charge coming at $ E$ will directly go in path $ EFJ$ without going to $ FG$.
The same thing is happening at $ J$ .
Now see these two fig.
@Rajiv U have done mistake as i m getting $ \frac{13 R}{7}$.
thank u

Look at the diagram in the attachment below this post, the lines of the same colour have the same potential, by symmetry, as also from Kouichi Nakagawa's drawing too.

[quote="Kouichi Nakagawa"]
[img]http://alt1.artofproblemsolving.com/Forum/latexrender/pictures/8/2/e/82e712af832ee41f02dbbd21058ebd38775e8f7c.png[/img]

[/quote]

Hence, the circuit is a simple series parallel combination like the one I've drawn in my first post.

Hence, $ R_{net} \equal{} \left(\frac {R}{2} \plus{} \frac {R}{4} \plus{} \frac {R}{6}\right)\times2 \equal{} \frac {11R}{6} \equal{} \frac {110}{6} \text{ ohm}$.

I may be wrong, but so far I haven't found any fault with my reasoning. Are the points not at the same potential?

your picture is wrong you forgot to put 6 resistors in the middle not 5!!!
and your solution is wrong I can't explain you but it is
130/7 is correct answer.
Can anybody solve the problem if there are n resistors connected in that way.

[quote="Matematika"]your picture is wrong you forgot to put 6 resistors in the middle not 5!!![/quote]

Thanks for pointing it out. The picture may be wrong, but I considered $ 6$ resistors when calculating.

[quote="Matematika"]...and your solution is wrong...[/quote]

In your opinion...

[quote="Matematika"]...I can't explain you....[/quote]

If you can't, how can you say I'm wrong?

Rajiv
 You can't speak here (rather write, he he he ) of equipotential lines; it's not electrostatic but electrodynamics. You can have symmetric points having the same potential, as for instance, in your green line the points in the "edge" and not that in the middle (or in the diagonal).

Its not my opinion this problem has been on z-trening programming competion
and computer is generationg different solution so its not my opinion

[quote="Immanuel Bonfils"]
You can't write of equipotential lines

[/quote]

We can't? Lets see...
Consider the following circuit:C1

We can see that the net resistance is $ \left(\frac{1}{R\plus{}R}\plus{}\frac{1}{R\plus{}R}\right)^{\minus{}1}\equal{}R$ and by the method of equipotential is $ 2\times \left(\frac{1}{R}\plus{}\frac{1}{R}\right)^{\minus{}1}\equal{}R$.

Now a more complicated circuit: C2

By Kirchoff's Law, $ a\equal{}b$ and $ R_{net}\equal{}\frac{3R}{2}$ and by the method of equipotentials, $ R_{net}\equal{}2\times\left(\frac{R}{2}\plus{}\frac{R}{4}\right)\equal{}\frac{3R}{2}$.

[quote="Rajiv"][quote="Immanuel Bonfils"]
You can't write of equipotential lines

[/quote]

We can't? Lets see...
Consider the following circuit:C1

We can see that the net resistance is $ \left(\frac {1}{R \plus{} R} \plus{} \frac {1}{R \plus{} R}\right)^{ \minus{} 1} \equal{} R$ and by the method of equipotential is $ 2\times \left(\frac {1}{R} \plus{} \frac {1}{R}\right)^{ \minus{} 1} \equal{} R$.

Now a more complicated circuit: C2

By Kirchoff's Law, $ a \equal{} b$ and $ R_{net} \equal{} \frac {3R}{2}$ and by the method of equipotentials, $ R_{net} \equal{} 2\times\left(\frac {R}{2} \plus{} \frac {R}{4}\right) \equal{} \frac {3R}{2}$.[/quote]
he meant that the nomenclatureis wrong. not all points on the line are equipotential, only those in picture are.

It is not that  do not know Kirchoff's Law, I too get $ \frac {13R}{7}$ with KL, but I just wanted to know if what I did in the 1st post works-and after some reasoning by myself, I found it doesn't.


linear algebra
matrix
number theory
relatively prime
superior algebra
superior algebra solved

For any integer n > 2, show there there are at least 2
elements in U(n) that satisfy x^2 =  1

U(n) is a group with operation" multiplication in modulo n"

Is U(n) the group of n x n unitary matrices? If it is, then the answer is easy:
The only condition that has to hold is $A\cdot\overline{A}^{T}=1$. If you take $A_{1}$ as the identity matrix and $A_{2}=-A_{1}$ then you have two such matrices (even für n=1 and n=2).
In fact you have $2^{n}$ such matrices at least.

I think Mambo is talking about U(n)="the set of all positive integers less than n and relatively prime to n". This is a group under multiplication modulo n indeed, and of course has two elements with square equal to one: 1 and n-1.


function
induction
algebra unsolved
algebra

Find all function $f:X\to (0,1)$ such that \[(x+2)f(x)\cdot f(x+1)=x\quad\forall x \in X,\] where $X$ is set of positive real numbers.

[i]Edited by Myth[/i]


function
integration
real analysis
limit
calculus
real analysis solved

1. This one was given an the Romanian NMO 1995, final round, 12'th grade (problem 4):

Let M be the set of all continuous functions f:[0, Pi]->R s.t.                         \int {0->Pi} f(x)*sinx dx = \int {0->Pi} f(x)*cosx dx = 1
Find min {f is in M}  \int {0->Pi} (f(x))^2 dx

2. This one is from a Romanian contest (Gheorghe Titeica; I think I've mentioned it here b4):

Find lim {n->00}  \int {1->1996} sin(x^n) dx

Maybe you'll find them kind of easy (especially the second one), but it wouldn't be the first time I'm posting problems which are too easy... :D  :D

2/
Change variable x= y^(1/n) => dx=1/n.y^(1/n -1)dy
An = \int {1->1996} sin(x^n)dx
An=1/n \int {1 to 1996^n}(siny)y^(1/n-1)dy
integrat by part
An= (cos1)/n - cos(1996^n)/1996^(n-1) +
1/n(1-1/n) \int_{1 to 1996^n}(cosy)y^(1/n)/ydy
with Henri Lebesgue dominated convergence theorem

http://mathworld.wolfram.com/LebesguesDominatedConvergenceTheorem.html

\lim An = 0

Henri Lebesgue studied at cole Normale Suprieure. He taught in the Lyce at Nancy from 1899 to 1902. Lebesgue formulated the theory of measure in 1901 and the following year he gave the definition of the Lebesgue integral that generalises the notion of the Riemann integral.

oubinol

Thx for the solution. It's the same thing I did. About Lebesgue: do you think Lebesgue's criterion for integrability is accpeted in contests (I mean without proof)?. A teacher of mine told me not to take my chances because he's pretty sure that it's not accepted. That's too bad since it can be pretty useful sometimes.

by all means be sure that it is accepted at the National Olympiad. of course I can't say that for the city and district olympiads :)

anyway don't worry about it, there won't be any problems which can be solved by that at the olympiad :D

does anyone know the answer to the first problem?

Give me some time to remember the solution. I sent Moubinool the soln by PM (if I can remember correctly). Here's something I do remember: the answer is 4/Pi (maybe you can work on that).

I won't post it completely, but this should be enough:

It's obvious that  \int {0->Pi} f(x)[a*sinx+(1-a)*cosx] dx = 1 for all real a. We can apply Cauchy ineq to this (for the 2 functions f(x) and g(x)=a*sinx+(1-a)*cosx). After Cauchy we can see that in order to get as high an estimation for  \int {0->Pi} (f(x))^2dx we have to get as small a value of  \int {0->Pi} (g(x))^2 dx as possible. Now try to choose a in [0,1] such that this is as small as possible.

Hint: a=1/2


Suppose a, b, c are the lengths of the sides of a triangle. Prove that

ab(a - b) + bc(b - c) + ca(c - a) >= 0.

a=y+z...etc
 :Sigma:  xy^3 :ge:xyz(x+y+z)
From Cauchy we have
 :Sigma: xy^3(x+y+z) :ge: xyz(x+y+z)^2

I'll explain this a bit ...

If a, b, c are the sides of a triangle we can substitute as follows :
a = y + z
b = z + x
c = x + y
and the conditions a + b > c, b + c > a, c + a > b reduce to
x, y, z > 0.
This way you can simplify problems concerning sides of a triangle :).

The inequality - to - prove is now equivalent to

xy + yz + zx >= xyz(x + y + z)

Now Cauchy (again) gives

(xy + yz + zx)(z + x + y) >= xyz(x + y + z)

as Maverick noticed. Dividing by x + y + z gives the result.

But the fact that

xy + yz + zx >= xyz(x + y + z) can also be proved as follows :

Notice that SymSum xy >= SymSum xyz since (3, 1, 0) majorizes (2, 1, 1).
This is a direct consequence of Muirhead's theorem.

(This is IMO1983/6)

Btw, Maverick, how did you get those "sophisticated" >= and (sigma) ... signs ?
I don't manage to use them on this forum.
The "\geq" does not work as far as I know.

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group theory
abstract algebra
superior algebra
superior algebra unsolved

If $ G$ and $ H$ are two normal subgroups with finite index, prove or disprove that $ G \cap H$ is a subgroup with finite index.
Thanks in advance.

Correct me if I'm wrong, but I think this will suffice:

[u][b]Lemma:[/b][/u] If $ A \leq B \leq G$, then $ [G : A] = [G : B][B : A]$.

[u][b]Proof:[/b][/u] Let us call $ L_G(A)$ and $ L_G(B)$ the sets of left cosets of $ A$ and $ B$ in $ G$. For each left coset $ gB$ we choose a unique element $ g$. Now define
\[ \Phi : L_G(B) \times L_B(A) \mapsto L_G(A) : (gB,bA) \mapsto gbA\]
One can easiley check that $ \Phi$ is a bijection, so the statement of the lemma follows.

Now, if $ A$ and $ B$ are normal subsets of $ G$, then $ A \cap B$ is a normal subset of $ A$. Furthermore, $ A \slash (A \cap B) \cong AB \slash B \leq G \slash B$. It follows that $ [A : A \cap B]$ is finite. Now, using the lemma, we obtain that
\[ [G : A%Error. "capB" is a bad command.
] = [G : A][A : A \cap B].\]
This is a finite number.

Note that the fact that they are normal is completely irrelevant.


Find all real-valued continuous functions defined on the interval $ (\minus{}1, 1)$ such that $ (1 \minus{} x^2) f(\frac{2x}{1 \plus{} x^2}) \equal{} (1 \plus{} x^2)^2 f(x)$ for all $ x$.

Let $ x\equal{}tht,$ then $ \frac 12 f(th(2t)dth(2t)\equal{}f(th(t))dth(t)$. Therefore $ f(x)\equal{}\frac{dt}{dx}\equal{}ch^2t\equal{}\frac{1}{1\minus{}x^2}.$