inequalities
number theory unsolved
number theory

The powers $ 2^n$ and $ 5^n$ start with the same digit.Find the digit.

[hide]Suppose $ a10^{m_1} \leq 2^n < (a\plus{}1)10^{m_1}$, and $ a10^{m_2} \leq 5^n < (a\plus{}1)10^{m_2}$. Multiplying these inequalities, we see that $ a^2 10^{m_1 \plus{} m_2} \leq 10^n < (a \plus{} 1)^2 10^{m_1 \plus{} m_2}$. In other words, $ a^2 \leq 10^{n \minus{} m_1 \minus{} m_2} < (a \plus{} 1)^2$. For $ a$ between 1 and 9 inclusive, the only $ a$ that has a perfect power of 10 between $ a^2$ and $ (a \plus{} 1)^2$ is 3, so we must have that $ a \equal{} 3$. [/hide]


For how many positive integral values of $ a$ is it true that $ x \equal{} 2$
is the only positive integer solution of the system of inequalities

\[ \left\{
\begin{array}{c}
2x>3x\minus{}3\\
3x\minus{}a>\minus{}6
\end{array}
\right\}?\]

Eq1: it states that accept only values of x < 3.
Eq2: 3x > a-6. x=2 works only when a<12, and x=1 works when a<9.

Hence a={9,10,11} works, or three solutions.


suppose that u have an isolated spherical conductor where isolation has been used implying that the net external electric field is 0 .What will be the charge distribution of the sphere will the charge go to the surface or remain inside .....[/hide]

Sorry, I don't understand this, please make it more coherent :roll:

i mean to say that suppose you have a sphere(metallic) away from all kind of 'external' elctric field and charge it up then will the charge given to it come to the surface

[hide="i think"]i think and it should but just double checking may be i am goofing up my concepts[/hide]

Yes. If it didn't there would be an electric field in the conductor, and if an electric field appears in the conductor the free electrons will start moving until they are on the surface and the field is zero. And while the charges move you don't have electro[i]statics[/i], do you  :lol:

[quote="Djole"]   And while the charges move you don't have electro[i]statics[/i], do you  :lol:[/quote]
 :coolspeak:

[quote="color"]See [url=http://en.wikipedia.org/wiki/Faraday_cage]Faraday cage[/url][/quote]
that is waht i was studying when  this came out :wink:

The Faraday cage is an example of the conductor canceling an external field inside a cavity it encompasses; it works without being grounded. A 'reverse' Faraday cage that we've been discussing still can't cancel out the field itself.


He waits until someone else comes, and takes a ride with them :D

Simple...you gave it away.
He smashes the door down with a car...

But opening the trunk to see the spare tire will set off the booby trap  :D

[quote="nutz_for2.718281828"]A wealthy detective has just been tipped off that 4 of his cars are booby-trapped. He has 5 cars. They have colors red, white, blue, brown, and gray. One of them has a spare tire, but he cannot tell which. One of the cars has a black speck on it, but he can't tell which one has it. He is given these warnings:

The blue car is booby-trapped.
The one with the black speck is booby-trapped.
The one with the spare tire is safe.
The car in back of the specked car is booby-trapped.
Once you arrive in the garage, you can't get out except by smashing a wall with one of the cars.

He is in the garage now. How does he get out?[/quote]

It isn't the blue car for sure. He can't tell which car has a black speck, because the car is a dark color. If the car with the black speck was white, the detective could see it for sure. I might be wrong, but it is just my guess.

[quote="nutz_for2.718281828"]The one with the spare tire is safe.[/quote]This doesn't tell us anything, as he doesn't know which car has the spare tire.
[quote="nutz_for2.718281828"]The blue car is booby-trapped.
The one with the black speck is booby-trapped.
The car in back of the specked car is booby-trapped.[/quote]
This gives us information about (at most) three of the booby-trapped cars, but not the fourth. Thus, even if we know which car is specked and which one is behind it, we can't tell which car is safe.

There has to be some sort of trick, but I can't figure out what.

[size=142]OK, then.[/size]

[size=9][color=white]so what was the point of this topic then?[/color][/size]

Where did you get the problem? Do they have an answer?

[quote="nutz_for2.718281828"]A wealthy detective has just been tipped off that 4 of his cars are booby-trapped. He has 5 cars. They have colors red, white, blue, brown, and gray. One of them has a spare tire, but he cannot tell which. One of the cars has a black speck on it, but he can't tell which one has it. He is given these warnings:

The blue car is booby-trapped.
The one with the black speck is booby-trapped.
The one with the spare tire is safe.
The car in back of the specked car is booby-trapped.
Once you arrive in the garage, you can't get out except by smashing a wall with one of the cars.

He is in the garage now. How does he get out?[/quote]

He has to smash down the wall, but:

The north wall is booby-trapped.
The wall adjacent to the neon green wall is booby-trapped.
The wall that has a microfilm dot containing the proof of the Riemann Hypothesis is safe... :D

:wallbash:

Yeah, where did you get this problem?

[quote="moogra"][quote="nutz_for2.718281828"][hide="secret"]I don't know the answer. Maybe there isn't one.[/hide][/quote]

i don't think either. maybe if you gave us the order it would work[/quote]
it's nice how you admit that you dont think.

sell ur cars on EBay and buy some new ones  :blush:


In a convex quadrilateral $ ABCD$, lines $ AD$ and $ BC$ meet at $ E$. Let $ H$ and $ G$ be the midpoints of $ BD$ and $ AC$, respectively. Find the ratio of the area of triangle $ EHG$ to that of the quadrilateral $ ABCD$.

Let $ M, N$ be the midpoints of $ DC, AB$.
We have $ S_{MGNH}\equal{}S_{ABCD}\minus{}S_{BHMC}\minus{}S_{BNH}\minus{}S_{ANG}\minus{}S_{AGMD}$
$ \equal{}S_{ABCD}\minus{}(\frac{3}{4}S_{BDC}\plus{}\frac{1}{4}S_{ADB})\minus{}(\frac{3}{4}S_{ACD}\plus{}\frac{1}{4}S_{ABC})$
$ \equal{}S_{ABCD}\minus{}(\frac{1}{2}S_{BDC}\plus{}\frac{1}{4}S_{ABCD})\minus{}(\frac{1}{2}S_{ACD}\plus{}\frac{1}{4}S_{ABCD})$
$ \equal{}\frac{1}{2}(S_{ABCD}\minus{}S_{BDC}\minus{}S_{ACD})\equal{}\frac{1}{2}(S_{AIB}\minus{}S_{DIC})$
$ \Rightarrow S_{GHM}\equal{}\frac{1}{4}(S_{AIB}\minus{}S_{DIC})$
On the other side, $ S_{HME}\equal{}S_{HMC}\equal{}\frac{1}{4}S_{BDC}, S_{GME}\equal{}S_{GMD}\equal{}\frac{1}{4}S_{ADC}$
$ \Rightarrow S_{HGE}\equal{}\frac{1}{4}(S_{AIB}\minus{}S_{DIC}\plus{}S_{BDC}\plus{}S_{ADC})\equal{}\frac{1}{4}S_{ABCD}$

S(EGH)= S(PGH)+ S(EPH)+S(EPG)= S(PGH)+ S(PQH)+ S(PNG)=1/2 S(PQNM)=1/4 S(ABCD)


Due to the increase in enrollment in AoPS programs, there will be a number of changes in the formats of some of our programs beginning this Fall.  As these changes are made, I will post them here.

First, we are discontinuing audited enrollments.  Students who put in the time to work on the more challenging problems learn far more and the classes work together more both on the message board and in open chat times when there are challenging problems to discuss.

Second, we plan to cap enrollment in the Independent Study programs for the Fall.  We have not yet set final numbers for these caps, but I will post them here as we make that decision.  The Independent Study will also lack an audited option.

Additionally, Independent Study Office Hours will now be scheduled at regular times.  This will both facilitate an organized schedule and help us find more mentors as the enrollment of the program has grown by a factor of nearly 10 since it began in the Fall.  I am currently organizing our school year class schedules and I will post Office Hour times in this forum after they've been sorted out.

Another reason for getting rid of the audited option is that all of the database time and all of the questions we field (hundreds of hours of phone calls and emails) regarding the nuances of our programs eat time that could be better spent preparing materials.  For instance, the Intermediate Number Theory class is still yet to be completed, something we regret.  Hopefully this reorganization of the school will allow us to offer it as soon as is possible.

All of it sounds like it will help everyone in the long run, so that's good!

Just a question - will you announce when the amount of spots in the Intermediate Study/etc. is about to be gone? Because I'm seriously considering it and don't want to miss out if I wait a little too long and suddenly it's full.

Thanks and keep up the awesome work! We all greatly appreciate it!

[quote="WindSlicer"]
Just a question - will you announce when the amount of spots in the Intermediate Study/etc. is about to be gone? Because I'm seriously considering it and don't want to miss out if I wait a little too long and suddenly it's full.
[/quote]

Perhaps we will by email announcement.

It is also possible that we will loosen the caps depending on whether or not it looks like we can bring in enough mentors.  We will try to find highly qualified college students interested in this kind of part time work.  There are other factors that may change as well.


hey can anyone get me the free e-book of "challenge and thrill of pre - college mathematics ' by K.N. Ranganathan,V. Krishnamurthy, C. R. Pranesachar and B. J. Venkatachala, New Age International Publishers.

i also neeed

H. S. M. Coxeter and S. L. Greitzer,--- "Geometry Revisited"-- Mathematical Association of America.

hello
According to my knowledge I don't think there is any e-book of Challenge and Thrill  by Venkatachala.
It is one of the most common book U will definitely find it anywhere in Delhi.

hey can ne1 frm delhi tell frm which shop i can buy challenge and thrill of pre college maths
(except nai sarak)

Rijul saini, have you qualified the rmo?

Here is the Link for Geometry Revisited by Greitzer and Coxeter

http://rapidshare.com/files/66931249/Geometry_Revisited.rar

Also, There is no E-Book for "Pre college " by Venkatachala.
You can Easily Buy It From any Famous Book Store in New Delhi(only if you live in New Delhi)
A Friend Of Mine , Bought Pre College From Connaught Place.


1. What is the value of sin(pi/5)*sin(2pi/5)*sin(3pi/5)*sin(4pi/5)?

2. Compute (sin :^o:10)(sin :^o:50)(sin :^o:70)

1) I wrote about the explanation for this sort of problem earlier in this forum. Basically consider 1+x+x^2+x^3+x^4 as the product of its roots, substitute x=1, take the absolute value, and use the identity |1-e^(i*theta)|=2sin(theta/2). The answer is 5/16.

2) That's the same as cos(20)cos(40)cos(80). We evaluate it like this:

S=cos(20)cos(40)cos(80)
S*sin(20)=sin(20)cos(20)cos(40)cos(80)
=1/2sin(40)cos(40)cos(80)
=1/4sin(80)cos(80)
=1/8sin(160)
=1/8sin(20)

so S=1/8.

You can also do that with cos(12)cos(24)cos(48)cos(96) [That was in the final round of a team buzzer-style math contest once, and no one got it. I was very ashamed when I asked him how to do it after and the guy who wrote it told me.]


The basis of this question is a probability question which easily turns into a geometry one.

Point $A (5, 8)$ and $B (17,-1)$ lie on a circle with center $P$ such that $AB$ is a $72^{\circ}$ arc of the circle. A point is selected at random in the interior of this circle. Find the probability that this point lies in the interior of the circle inscribed in $\triangle ABP$

Suggestions?

Steps:
1. Determine location of point P with a system of equations.
2. Determine the area of circle P.
3. Determine the area of the incircle of ABP.
4. Divide the areas for the probability.

we can circumvent most of those steps

we don't really need the radius of the triangle or anything like that

the problem is basically, find (r/R)^2 for a 72-54-54 triangle

Can you elaborate on your method, Altheman? I'm not familiar with what you mean by (r/R)^2

Here is my solution:

The triangle formed from the arc has side lengths of 15, 12.7598, and 12.7598. The first was obtained by the distance equation, and the latter by Law of Sines. Therefore, the area of the triangle is $\approx 77.42148$. The incircle's area is $\approx 45.87797$ (By $R = \frac{2S}{a+b+c}$ where is area) and the circle's area is $\approx 511.4875$.

Hence the probability is
$\approx .0897$

[b]Extension: How would one do the same problem, but instead of solving for the incircle of the triangle, solving for the inscribed circle of the sector?[/b]


I always seem to have trouble with these types of problems. Can anyone show me how to solve this problem step by step?

V = L * I * K / C

Solve for L

So far, I have this:

1 / L = I * K / C * V

I can't seem to get L by itself though.

Any help would be greatly appreciated. Thanks.

[quote="i_like_pie"]From what you have, you can simply take the reciprocal of both sides (in reality, this is just cross-multiplying) to get $ L \equal{} \frac {CV}{IK}$.

From the original equation, this is how I would go about solving for $ L$:

[list]1. Multiply both sides by $ C$ to get rid of the denominator on the RHS: $ CV \equal{} LIK$
2. Divide both sides by $ IK$ to get $ L$ by itself: $ \frac {CV}{IK} \equal{} L$[/list]
You generally want to leave $ L$ in the numerator (don't divide by $ L$).

Hope that helps. :)[/quote]

Thanks

However, if you were to cross multiply both sides wouldn't you just end up with 1 on one side and the L would end up back on the other side?

It would be 1 / L * L  / 1 = I * K / C * V * L / 1

Simplified to: L / L = I * K * L / C * V

Or: 1 = I * K * L / C * V since L / L equals 1.

Sorry, I'm completely confused by this and I can't find many of these types of practice problems.

He multiplied by $ \frac {LCV}{IK}$, not just $ L$
You could also see it as taking the reciprocal or "flipping" both sides.

So it would look like this
$ \frac {1}{L} \equal{} \frac {IK}{CV}$
$ \left(\frac {LCV}{IK}\right)\frac {1}{L} \equal{} \frac {IK}{CV}\left(\frac {LCV}{IK}\right)$
$ \frac {CV}{IK} \equal{} L$

Here's another one for you to try, $ A \equal{} \frac {bh}{2}$. Solve for $ h$.

[quote="aegis"]If you were to cross multiply both sides wouldn't you just end up with 1 on one side and the L would end up back on the other side?

It would be 1 / L * L  / 1 = I * K / C * V * L / 1

Simplified to: L / L = I * K * L / C * V

Or: 1 = I * K * L / C * V since L / L equals 1.

Sorry, I'm completely confused by this and I can't find many of these types of practice problems.[/quote]
You were on the right track, but you only did half the work. ;)

After that, you multiply both sides by $ \frac{CV}{IK}$, thus simplifying the RHS to $ L$, while the LHS becomes $ \frac{CV}{IK}$.


A tennis star plays at least one match every day, but no more than 20 matches total, over a period of 2 weeks. Show that during some consecutive run of days the tennis start must play exactly 7 matches.

The statement needs touching up. As it is, he can play 20 matches every day.

I added the word "total" to clear up some ambiguity.

[hide]
This is equivalent to showing that you cannot pick $15$ distinct integers in the set $\{0, 1, 2, \ldots, 20\}$ such that no two differ by $7$ (one of them has to be zero, but that doesn't even matter).

Consider the pairs

$\{0,7\}$, $\{1, 8\}$, $\ldots$, $\{13,20\}$.

There are $14$ of them, but we need to pick $15$ so by Pigeonhole we choose two in the same set $\Rightarrow$ some consecutive run of days has exactly $7$ games.[/hide]

Interesting. I did it without including 0. You then have 14 and 13, so the pigeonhole still works.

[hide] Let' write the total number of games played after each day. We notice a couple of things
- we start with at least 1, and the number has to increase every day
- the last number must be 20
- there are a total of 14 numbers (1 for each day)
- if we have two numbers that differ by 7, then we are done

However, there are only 13 numbers that don't differ by 7. [/hide]


floor function
number theory unsolved
number theory

1. Prove that if m and n are nonnegative integers and $ p$ is a prime number, then the largest power of p dividing $ \binom{m\plus{}n}{m}$ equals $ pc$, where c is the number of carries when m and n are added in base p.

2. Prove that the exponent of a prime p in $ \binom{n}{k}$ equals the number of positive integers j such that the fractional part of $ k/p^j$ is greater than the fractional part of $ n/p^j$.

3. Prove that $ \binom{n}{k}$ is divisible by $ n/gcd(n,k)$.

i'll just do 2. the others are similar to it or follow from it.

the exponent of $ p$ in $ \binom{n}{k}$ is $ \sum_{j \equal{} 1}^{\infty} \lfloor \frac {n}{p^j}\rfloor \minus{} \lfloor \frac {k}{p^j}\rfloor \minus{} \lfloor \frac {n \minus{} k}{p^j}\rfloor$. now let $ n \equal{} p^ja \plus{} b$, $ k \equal{} p^jc \plus{} d$, where $ b, d$ are integers in $ [0,p^j)$. then  $ \lfloor \frac {n}{p^j}\rfloor \minus{} \lfloor \frac {k}{p^j}\rfloor \minus{} \lfloor \frac {n \minus{} k}{p^j}\rfloor \equal{} a \minus{} c \minus{} (a \minus{} c) \equal{} 0$ if $ b\geq d$ (i.e., the fractional part of $ k/p^j$ is smaller than that of $ n/p^j$) and is $ a \minus{} c \minus{} (a \minus{} c \minus{} 1) \equal{} 1$ if $ b < d$ (i.e. the fractional part of $ k/p^j$ is bigger than that of $ n/p^j$). the conclusion follows.

[quote="shoki"]and for 3:See PEN A28
and for 1,i can't understand this part:
"where c is [u]the number of carries [/u]when m and n are added in base p"
could anyone explain it ?
thx in advance. :)
(sorry for my poor english)[/quote]

a "carry" is when, in the process of adding two numbers, the sum of two digits is at least the value of the base, and the excess quantity is 'carried over' to the next column to the left...

okay thx i think this might help:
$ v_p(n!)\equal{}k\equal{}\frac{n\minus{}(a_0\plus{}a_1\plus{}...\plus{}a_r)}{p\minus{}1}$
where $ n\equal{}(a_ra_{r\minus{}1}...a_0)_p$


What is this SET program that's posted under this forum? Could someone tell me a little about it and what type of math opportunities are involved? Thanks.

Well, I searched it with yahoo! and found a link:

[url]http://nmp.jpl.nasa.gov/workshop/space_test_reqs.html[/url]

I don't know if this is actually what the thing posted on the forum.

-interesting_move

They might be talking about Johns Hopkins University's [url=http://www.jhu.edu/cty/set/]Study of Exceptional Talent[/url].

Yeah, we're referring to the Study of Exceptional Talent that intrepia cited...

hmm...i'm not entirely certain why anyone would want to talk about SET anyways. there's a set listserv if set people want to talk amongst themselves, but as far as set stuff there isn't much to talk about...unless for some strange reason you're a middle-schooler who really wants to get in or something... (no idea why that would be) and even so, admissions is pretty self-sufficient and happens on it's own... ah well, whatever

now if you wanted to talk about the game SET on the other hand...

[quote="Andrea"]now if you wanted to talk about the game SET on the other hand...[/quote]

Hmm...set is a fun game.  It's also a standard among math clubs, if I'm not mistaken.  When we went to nationals, our team must have been the only one without a set deck.

[quote="Andrea"]hmm...i'm not entirely certain why anyone would want to talk about SET anyways. there's a set listserv if set people want to talk amongst themselves, but as far as set stuff there isn't much to talk about...unless for some strange reason you're a middle-schooler who really wants to get in or something... (no idea why that would be) and even so, admissions is pretty self-sufficient and happens on it's own... ah well, whatever[/quote]

There are many interesting things to talk about on this forum, which is a real service to young people interested in mathematics. One of the best reasons to talk about the Study of Exceptional Talent (SET) outside of SET's own channels of communication

http://www.jhu.edu/gifted/set/index.html

is simply that a lot of people don't know about SET, and a few of those people, at least, would qualify for the program if they knew about it. The upper age limit for inclusion in SET is quite low, but I would venture to guess that many of the self-selected people who participate in this forum could qualify for SET on the math side, if only they know

1) to take the SAT I university-entrance test at an age well in advance of university age, for example through a talent search program, and

2) to report their qualifying scores, if they get them, to the Johns Hopkins University offices that administer SET.

I know someone locally who probably could have qualified for SET on BOTH the math section of the SAT and the verbal section, as did two of his local friends, but he simply didn't know he was up against a deadline for taking the SAT at an unusually young age. I don't see a lot of publicity about SET anywhere but on the SET Web site, and I'm pretty sure simply taking the SAT through a regional talent search and getting a qualifying score by the age limit does NOT, by itself, get a test-taker into SET. The test-taker in a non-Johns Hopkins talent search still has to affirmatively report the score to SET headquarters to get into the program.

The other reason that SET ought to be interesting to math competition people like the people who hang out here is that many famous names in math competitions are members of SET, as can be ascertained by reading the Web sites of SET members

http://www.jhu.edu/gifted/set/members.html

available for public viewing.

I wish I'd known about this when I was younger.

[quote="mathfanatic"][quote="Tare"]Oh :oops: I thought at first this was about that card game SET...

My friend took SAT and "qualified" or something at this but he didn't tell me much, is this going to hlp us with our college entrance?[/quote]

Don't do SET because you want it to help your college app!  Do it if you're interested in participating in various programs for advanced students or if you need resources to learn on your own.[/quote]

I am currently working at the Center for Talented Youth @ JHU, SET is mainly a STUDY of smart kids hence, (Study of exceptional talent).

There are some perks (like a pretty mundane newsletter that are mainly bragging rights) and some counselor that can help you through various troubles, but eh.. I don't think it's particularly worth it.  You guys can go for it if you want.

If you have any questions about that, you can email me @ jpo1@math.jhu.edu

I learned about SET through CTY's Talent Search, which is much more well known.  Is there an SET or CTY topic on here?

This forum is the forum for discussing CTY or SET. This thread, as the title suggests, is a good place for discussing SET. A lot of students in the non-JHU CTY regions of the country never hear about SET, which is too bad.

[quote="dkushnir"]I looked at SET closer. When in comes to some kind of counceling, it is for a fee. There are enough councelors who want you to pay for their services without SET.
Classes through JH CTY - one can take them without SET, as long as one qualifies and qualifying scores are MUCH lower.
So far it looks only like the priviledge of being a member and getting a special certificate during the Awards ceremony.[/quote]

My son is a member, and my daughter qualified this year.

Granted, there is not much to SET. They are a small organization, and part of their objective is to study exceptional talent - hence the name. The people who work there are dedicated and definitely want to do what they can.  Our local group occasionally gets together. Through this network, my son has discovered a  lot of option. In fact, we found out about AoPS from an LA Set meeting about 2 years ago.

It takes little effort to join. You get free magazines. Why not?

[quote="fireemblem13"]i think there are some requirements to apply like SAT score. What if u took the SAT last year for Duke TIP and got a good score? Do i have to retake it?[/quote]

No, you can apply to SET directly, using your TIP scores. Check it out here: http://www.jhu.edu/gifted/set/index.html

i faxed them what i needed to but i still haven't gotten anything back. how long does it take?
i hope they didn't reject me :( but there was no reason to reject me. :P

The reply to an inquiry about SET will come by postal mail. If you have qualifying scores, they will accept wherever in the world you live.

SET is a great opportunity for kids.  I can assure you that CTY spends significant amount of time and effort to help kids in SET advance in whatever they wish.

If anyone is having issues with SET or CTY, send me an email and I could help with some follow ups.  (I was a staff @ CTY for several years).

I used to be a member of SET, but it seemed it wasn't really doing much, so when the "fill out this 26 page form to continue being a member of SET" came, I trashed it.  :lol:


Hi to all Koreans
I like to have and solve Korea MOs . Could someone take them for me to download ?
Thanks

Please help me . Doesn't anyone have second or first round of Korean MO ?
감사합니다  :lol:

[quote="amparvardi"]Have you seen contests section?

http://www.artofproblemsolving.com/Forum/resources.php?c=155&cid=105&
http://www.artofproblemsolving.com/Forum/resources.php?c=155&cid=81&
http://www.artofproblemsolving.com/Forum/resources.php?c=155&cid=182&

Also there are some problems in imomath site. You can see them: http://imomath.com/index.php?options=oth|Kor|Ne|Ne&ttn=South%20Korea&p=0[/quote]
I've looked up all above sites , but I want the problem of recent years . ( 1st and 2nd stage )
Thank U :wink:

If you can read Korean, look around here for recent problems and solutions:
[url]http://cafe.naver.com/xmo.cafe?iframe_url=/ArticleList.nhn%3Fsearch.clubid=13507301%26search.menuid=20%26search.boardtype=L[/url]

and here for old problems and solutions (see the last page):
[url]http://cafe.naver.com/xmo.cafe?iframe_url=/ArticleList.nhn%3Fsearch.clubid=13507301%26search.menuid=8%26search.boardtype=L[/url]

Thank U
I'm rookie in Korean but now learning that , so could you give the last link of problems .I'm confused in the website  :o


P and Q are points on the side BC of a quadrilateral ABCD such that P is between B and Q, and BP = PQ = QC. S and T are points on the side AD of ABCD such that S is between A and T, and AS = ST = TD. Both AC and BD lie inside ABCD. Prove that the area of ABCD is three times the area of the quadrilateral PQTS.

hmm..sorry, but the hint was not very helpful, thx anyway! :)

here is what i think could be one way of solving...but i'm kinda stuck half-way through...  :blush:

|PDT| = |PST| because they share a common vertex P and common base (DT
= TS as given). Let |PDT| = |PST| = a

|PSQ| = |QSB| because they share a common vertex Q and common base (PQ
= QB as given). Let |PSQ| = |QSB| = b

|PQST| = a + b

Since the qn states that |ABCD| is 3|PQST|, I have to prove that
|APTD| = |AQTS| = |QBSC|= |PQST| = a + b

How can I do so?

Let $K, L, M$ be the midpoints of the segments $AS, ST, TD$. The triangles $\triangle BKP, \triangle PLQ, \triangle QMC$ have equal bases $BP = PQ = QC$ and their areas are in the ratio of their altitudes $h_K, h_L, h_M$ from the vertices $K, L, M$. Since the vertices $K, L, M$ are equally spaced, $KL = LM$, the altitude $h_L$ of the middle triangle is the average of the other 2 altitudes $h_L = \frac{h_K + h_M}{2}$ and the area of the middle triangle $\triangle PLQ$ is the average of the other 2 triangle areas,

$\triangle_{PLQ} = \frac{\triangle_{BKP} + \triangle_{QMC}}{2}$

The segments $PS, QT$ are medians of the triangles $\triangle KPL, \triangle LQM$, hence, the following triangle areas are equal:

$\triangle_{KPS} = \triangle_{LPS},\ \ \triangle_{LQT} = \triangle_{MQT}$

The triangles $\triangle KBA, \triangle KPS, \triangle MQT, \triangle MCD$ have equal bases $KA = KS = MT = MC$ and their areas are in the ratio of their altitudes $h_B, h_P, h_Q, h_C$ from the vertices $B, P, Q, C$. Since the vertices $B, P, Q, C$, are equally spaced, $BP = PQ = QC$, the altitude $h_P$ is the average of the 2 adjacent altitudes $h_P = \frac{h_B + h_Q}{2}$ and the altitude $h_Q$ is also the average of the 2 adjacent altitudes $h_Q = \frac{h_P + h_C}{2}$. It follows that

$\frac{h_B + h_C}{2} = h_P - \frac{h_Q}{2} + h_Q - \frac{h_P}{2} = \frac{h_P + h_Q}{2}$

$\frac{\triangle_{KBA} + \triangle_{MCD}}{2} = \frac{\triangle_{KPS} + \triangle_{MQT}}{2}$

Finally, for the 3 quadrilateral areas, we get

$\square_{PQTS} = \triangle_{PLQ} + \triangle_{LPS} + \triangle_{LQT} = \triangle_{PLQ} + \triangle_{KPS} + \triangle_{MQT} =$

$= \frac{\triangle_{BKP} + \triangle_{QMC}}{2} + \frac{\triangle_{KPS} + \triangle_{MQT}}{2} + \frac{\triangle_{KBA} + \triangle_{MCD}}{2} =$

$= \frac{\triangle_{BKP} + \triangle_{KBA} + \triangle_{KPS}}{2} + \frac{\triangle_{QMC} + \triangle_{MQT} + \triangle_{MCD}}{2} =$

$= \frac{\square_{BPSA} + \square_{QCDT}}{2}$

$\frac{\square_{PQTS}}{\square_{ABCD}} = \frac{\square_{PQTS}}{\square_{BPSA} + \square_{PQTS} + \square_{QCDT}} = \frac{\square_{PQTS}}{3\square_{PQTS}} = \frac 1 3$


A related question: can I scan in my permission form with my solutions and submit as one .pdf?

[quote="mathwizarddude"]This may sound stupid, but I've solved all the problems with their solutions written on the papers, and I'm planning on mailing them. I also have the entry and permission form filled out, but haven't sent it yet. Am I allowed to put both the solutions and the entry & permission form in a single mail, or should it be a separate mail?[/quote]

Send them in together. I think it says on the rules.

@ Brut3Forc3: I don't know. I am mailing the solutions. Just to be on the safe side, send them separately

Please do not submit as one pdf.  The entry form should be mailed or faxed separately.

[quote]You may submit the Entry Form and the Under 13 Permission Form with your first set of solutions, [b]if you are mailing in your solutions[/b]. Otherwise, you can send the Entry and Permission forms separately.[/quote](emphasis added)

Other related question:

Would it be okay if I uploaded my solutions now and faxed in the permission form later (like later tonight?)
I already have the solutions in a pdf, except I don't have a fax machine at my house.

On the safe side I would mail the permission form first. (you need permission before you send your answers!!!)

The rules state that the permission can be mailed with the solutions (says so in the RULES)

You do [b]not[/b] need to submit the Entry/Permission form before sending in your answers.  If you are mailing your solutions, the form can be enclosed with the solutions.  If you are web-submitted your solutions, the form can be mailed or faxed separately, either before or after you upload your solutions.

Does PDF version matter?  It says "For best results, when creating your PDF, select PDF version 1.3 or below.", but my scanner seems to generate files with version 1.6.

If you mail your solution through postal, when would you expect them to receive it?
(I sent in my solution last Monday but there is still no tick mark next to Round 1 on My USAMTS page)


linear algebra
matrix
analytic geometry
modular arithmetic
pigeonhole principle
combinatorics unsolved
combinatorics

Here is what i have until now:

Suppose we place $n$ squares in $(a_1,b_1)$, $(a_2,b_2)$, $(a_3,b_3)$...,$(a_n,b_n)$. Then we must have:

$a_1,a_2,a_3...,a_n$ are a permutation of $1,2,3....,n$, and the same for the $b_i$'s. So we also would have:

$a_1+b_1$, $a_2+b_2$,   $a_n+b_n$ are distinct modulo n, so we get, suming all of them that: $1+2+3...+n=0$ (mod n), so $n$ is odd.

Now, if $n$ is odd I still dont know what to do

As far as I can understand, it asks for this: prove that there is no permutation $\sigma$ of the residues $\pmod n$ s.t. all $\sigma(i)+i$ are distinct and all $\sigma(i)-i$ are distinct. I suppose I'm misunderstanding it.. :? What if you take $n$ to be a prime $>3$, and take $\sigma(i)=2i\pmod n$? We have $\sigma(i)+i=3i$, which are distinct, and $\sigma(i)-i=i$.

Where on Earth is the mistake? :?

Yes Grobber, there is no mistake :( . Your proof show in fact that for any $n$, wich is not a multiple of $6$, we can find such configuration! Now i am wondering iff this is a necessary condition? I have already showed that $n$ is odd.

 ;) Iura, what was the original question?

To translate it back into the language used here, why is the following not a counterexample for $n=5$? :?

$\begin{array}{ccccc}1&0&0&0&0\\0&0&1&0&0\\0&0&0&0&1\\0&1&0&0&0\\0&0&0&1&0\end{array}$

Edit: I missed the post above :)

[quote="iura"]Given a square matrix $n \times n$ call a diagonal a set of vertices with coordinates $(i,j)$ with $(i+j)$ $mod$ $n=const$ or a set of vertices with coordinates $(i,j)$ with $(i-j)$ $mod$ $n=const$ (so that a diagonal consists of $n$ squares).
Prove one cannot select $n$ squares on the matrix such that no two lie on same line, column or diagonal[/quote]
I can't find it in ARO 1997  :?
There is another problem (I think it is a true source of this problem).
Some cells of matrix $n\times n$ are colored in a black color. Prove there are two lines, or two columns, or two diagonals containing the same number of black cells.

Yes Grobber! I have some cool things! and i feel the answer will be: all $n$ such that $n$  is odd, and if $n$ is a multiple of $3$, $9$ divides it.

I have the following: let $f_a$ be a permutation that does the job for $1,2,3...,a$.

define $f_{ab}(ma+r)=f_a(r)+af_b(m)$  it is easy to see, that this works for the numbers $1,2,3....,ab$.

Now, I can show, that if $3|n$, then $9|n$. so if we could construct a permutation for $n=9$ we will be done!

(the alst thing that says we will be done is clearly false jaja, But anyway, the thing is if we can do it for a and b then we can do it for ab, i dont now what the answer going to be, sorry for my nonsense!)

Myth, are the diagonals in the ARO problem defined in the same way as in Iura's Problem?

(i bet they do, otherwise the problem is quite trivial)

Anyway here is the ARO problem: clearly we have: a column and a row all filled with black squares, or a column and a row all filled with white squares.

taking for each arrengement its complemet, we may assume the case is the first.

Now, clearly, every diagonal intersects our black row and our black column, it is easy to see both diagonals trought our row and column intersection must be white, (the rest, not the intersection!) (because no diagonal is completely white since it intersects every row and every column, in particular our black ones) This implies there is no row or column with n-1 black squares on it (it intersects both white diagonals), so there must be rows and columns that are all white, contradiction because they always intersect our black initial row and column.

Actuallyt there is a gap, if both white diagonals intersect again at another white square, then the row and column trought this square must also be black (the rest), but then, the diagonal with n-2 white squares must pass trough the other 2 intersections of this new almost black row and column, but then there cant be a diagonal with n-3 black squres, since it intersects the black thing in at least 4 different black squares.

[quote="Leonardo"]I dont get why you can asume that there is allways a row and column filled completely by one color.  :([/quote] It is a simple game with Pigeonhole principle.


A liver can absorb the alcohol in ten gallons of pure alcohol before dying. Assuming that 2 gallons of a 50% alcohol solution is just as liver-killing as one gallon of pure alcohol, how many gallons of 14.3% alcohol wine would it take to kill a liver? Round to the nearest integer.

I expect all of you to do this without a calculator.

I expect all of you to do this without a calculator.[/quote]

[hide]Wow, I have a feeling I'm overlooking something...
Anyway...

Basically, what this means is, we're looking for how many gallons of 14.3% alcohol wine will give us 10 gallons of alcohol, as a gallon of alcohol mixed with other stuff is as liver-killing as a gallon of alcohol not mixed with anything.

with 7 gallons of this wine, we get approximately one gallon of alcohol.
Mutiply by 10 and the answer is 70.[/hide]

[hide]Oh I think the solution is 10:!4.3%=70 gallons[/hide]

Math is my life :!:

[hide]Treething I hope this is not being asked because you have been drinking  :lol: .I believe it is just inverse variation is it not? Therefore the answer is (100*10)/14.3 which is approximately 70[/hide]

[quote="mathgeniuse^ln(x)"][quote="binonunquineist"]:idea:  BAD QUESTION

If you assume that two gallons of 50% alcohol solution is just as liver-killing as one gallon of pure alcohol doesn't have to mean that a 1000 hundred fourty-thirds gallons of 14.3% alcohol solution is as liver-killing as a gallon of pure alcohol. It might be like an inverse variation, or a linear relation or something like that.[/quote]

Actually, it is just 7 gallons of 14.3% alcohol.  Not 1000 hundred fourty-thirds.[/quote]
Well, it's not [i]exactly[/i] 7 gallons... more like 6.99, which you get from 1000/143

The point is that it is a  :idea:  BAD QUESTION

[quote="binonunquineist"][quote="mathgeniuse^ln(x)"][hide]Let's just hope that none of us ever have to deal with this, okay.

Bionunquineist, I first thought you said a number larger than 1000.

Next time could you use a number instead of words.

$\frac{1000}{143}$.[/hide][/quote]

ok[/quote]
I'm gonna pull a Jose.
 :spam:

[quote="mathgeniuse^ln(x)"]I'm gonna pull a Klebian

You forgot to put the accent over the e.[/quote]

 :rotfl:

^ :rotfl:

Anyway, I got 70, as an answer.
By the way have you noticed Treething's replacement for Michael Jackson.

Yeah I noticed it too.

btw you guys need to learn to read carefully, it's not 6.99 or 7...

[quote="mathgeniuse^ln(x)"][hide]
I already typed the answer in my post previously.

Oh, the answer is 70.

So the answer really is $\frac{10000}{143}$

That's alot.[/hide][/quote]
Well, when you round to the nearest integer... :roll:

I meant to say

but the answer really is...

instead of

so the answer really is...

[hide]I think the answer is 70. 14.3% is really close to 1/7, so it'd take 7 gallons to be like 1 gallon of pure alcohol, and then 70 to be like ten.[/hide]

[quote="TRAN THAI HUNG"]:spider: It's just a small question why do we have to discuss it long........

Math is my life :spam:[/quote]

Dude...stop saying "Math is my life"!!

Also, we can discuss any problem as long as we want if we want. ;)

[quote="TRAN THAI HUNG"]:starwars: Math is interesting if we know how to use it  :D

Math  is my life![/quote]
Just put it in your signature....it will save you some time


Our task is finding all positive integers n for which exists such sequence of positive integers $ x_1,x_2,\dots x_{2n}$ for which if $ x_i \equal{} x_j$ then i=j, and $ \sum\limits_{i \equal{} 1}^{n} x_i \equal{} \sum\limits_{i \equal{} n \plus{} 1}^{2n} x_i$ and  $ \prod\limits_{i \equal{} 1}^{n} x_i \equal{} \prod\limits_{i \equal{} n \plus{} 1}^{2n} x_i$.

I started with some small n, but only n=1 brings any result.

Can someone help me with it?

Trivially, there are no solutions for $ n \equal{} 1$ (these should have $ x_1 \equal{} x_2$).

Let's see there are no solutions for $ n \equal{} 2$ as well. Suppose $ a,b,c,d$ are positive integers such that $ a \plus{} b \equal{} c \plus{} d$ and $ ab \equal{} cd$. Then, $ d \equal{} a \plus{} b \minus{} c$, and $ ab \equal{} ac \plus{} bc \minus{} c^2$. Thus $ ab \minus{} ac \equal{} bc \minus{} c^2 \rightarrow a(b \minus{} c) \equal{} c(b \minus{} c)$, so $ b \equal{} c$ or $ a \equal{} c$, which are not possible.

For $ n \equal{} 3$ we have solutions. To find a solution, we proceed this way. Take any $ 6$ distinct positive integers $ a,b,c,d,e,f$ such that $ abc \equal{} def$, for example $ 5,6,7$ and $ 1,14,15$. If we multiply one factor from each side by a certain positive integer $ k$, the equality still holds, but the sum of the factors changes. If we can find some $ k$ such that $ 5k \plus{} 6 \plus{} 7 \equal{} k \plus{} 14 \plus{} 15$, then we are done. We solve this equation for $ k$ and we get $ k \equal{} 4$. So $ (20,6,7,4,14,15)$ is a solution. Notice that $ \forall j \in \mathbb N$, $ (20j,6j,7j,4j,14j,15j)$ is a solution as well. Of course, I reckon there could be some cases where this method fails, for example if we multiplied $ 5$ and $ 15$ by $ k$ instead of $ 5$ and $ 1$ we would have got the equation $ 5k \plus{} 13 \equal{} 15k \plus{} 15$, which has no integer solutions... but luckily I managed to found one where it doesn't  :P

Now let's see $ n \equal{} 4$. Again, we begin with a set that fulfills the product condition but not the sum condition: take for instance $ (5,7,26,33,11,13,14,15)$. We want a natural number $ k$ such that $ 5k \plus{} 7 \plus{} 26 \plus{} 33 \equal{} 11k \plus{} 13 \plus{} 14 \plus{} 15$. The only solution to this equation is $ k \equal{} 4$. So, $ \forall j \in \mathbb N$, $ (20j,7j,26j,33j,44j,13j,14j,15j)$ is a solution.

Now we'll prove that if there are solutions for $ n \equal{} k_1$ and $ n \equal{} k_2$, there are also solutions for $ n \equal{} k_1 \plus{} k_2$. Indeed, if the numbers $ (x_1,x_2,...,x_{2k_1})$ satisfy $ x_1 \plus{} ... \plus{} x_{k_1} \equal{} x_{k_1 \plus{} 1} \plus{} ... \plus{} x_{2k_1}$ and $ x_1...x_{k_1} \equal{} x_{k_1 \plus{} 1}...x_{2k_1}$; and the numbers $ (y_1,y_2,...,y_{2k_2})$ satisfy the analogous conditions, we have that the set $ (x_1,...,x_{k_1},y_1,...,y_{k_2},x_{k_1 \plus{} 1},...,x_{2k_1},y_{k_2 \plus{} 1},...,y_{2k_2})$ also fulfills both conditions. The only problem that could arise while doing this is that some $ y_i$ could be equal to some $ x_j$. But from what we've seen before we know that we can multiply all the $ y_i$'s by some natural number $ j$ in such a way that each $ y_i$ is larger than each $ x_j$, and the conditions still hold.

So far we know there are solutions for $ n \equal{} 3$, $ n \equal{} 4$ and $ n \equal{} 3a \plus{} 4b$ for all nonnegative integers $ a$ and $ b$ (not both $ 0$). But every positive integer $ n \ge 6$ can be written this way, since
$ 3k \plus{} 6 \equal{} 3\times(k \plus{} 2)$,
$ 3k \plus{} 7 \equal{} 3\times(k \plus{} 1) \plus{} 4$ and
$ 3k \plus{} 8 \equal{} 3 \times k \plus{} 2\times4$ for all nonnegative integer $ k$.
So, for all $ n\ge6$ there are solutions.

The only case missing is $ n \equal{} 5$: we'll do the same thing once again. We begin with $ (6,17,19,35,143,7,13,22,51,95)$, and we need $ k$ such that $ 6k \plus{} 17 \plus{} 19 \plus{} 35 \plus{} 143 \equal{} 7k \plus{} 13 \plus{} 22 \plus{} 51 \plus{} 95$, this is, $ 6k \plus{} 214 \equal{} 7k \plus{} 181 \rightarrow k \equal{} 33$. So $ \forall j \in \mathbb N$, $ (198j,17j,19j,35j,143j,231j,13j,22j,51j,95j)$ is a solution.

The proof is complete, and the answer to the problem is: all positive integers $ n$ with $ n\ge3$.

well that's problem number 10 from Polish MO and the deadline is today. is there any moderator to erase the previous post and block this topic?


I am thinking of doing a Mock MATHCOUNTS competition. The level will be hard chapter\easier state level. If I can get at least 5-10 people with interest, i will do it. No date yet. Post if you're interested

EDIT:
The targets are here if you can't do them Sunday. [b]Questions 5,6,7,8 are in my next post because of attachment limit.[/b]
OK:
The competition will be over the weekend I guess because of demand that it be over a few days. I am posting the sprint for people who can't take it Saturday, otherwise take it Saturday. The target will be posted Sunday or if you can't do it before, do them asap.

*signs up* Just try and PM me when it starts because I'm getting busier and busier these days.

Yay! some peoples are interested! i think ill have it.

Target questions 5,6,7,8 are attached to this post

I'll sign up, if you will PM me about it to remind me.  And, post the date it will take place at least a week before it starts.  Thanks.  (PM me twice, once for the date and once the day before to remind me, please)

I'll sign up...
Hopefully I can make it.
btw, 7 est or cst?

I'll sign up! However, I'm not exactly sure if ill be able to do it because I'm usually busy on Saturdays.

[quote="b-flat"]Probably CST.

So, do we PM you, or post them.  (I assume PM)

I won't be hear on the 12th.  I'm in Tennessee.  I might be able to do it before we leave, though.  If you don't get answers from me, don't be surprised.[/quote]Actually im from DE so EST. When are you leaving on Saturday? I can pm you the problems before you leave.

[quote="biffanddoc"]Lets say Saturday the 12th. I'll post the sprint round at 7:00. Answers are due within 45 minutes. The target round will be start 7:50 and each problem set will be posted 7 minutes after that. Target answers are due 10 minutes after the last set is posted. Maybe a CD round through AIM?

EDIT: all times are EST[/quote]

AM or PM?

What a cruel joke. I have a piano thing elsewhere on Saturday the 12th at 7:00 PM. (Of course, I'm in CST, but it lasts two hours.)

I suggest you make each round over a course of a few days--that's how it's traditionally been done.

What about i post them at 7PM EST and u have to pm me the answers by the end of the day?

I thought you were talking about in the morning.  I could do it if you could post them by 8:00 am EST.  I don't know when I could send in answers because I probably won't have access to the internet until Sunday night.  Having it over several days would work for me.  Hope for the best.

i'll do it 2. do i pm my answer or post it?

What about doing it over several days?  I can only do it if it is posted by 9:00 and I don't have to turn it in until Sunday night.

b-flat if you can't do it saturday, do the sprint now

Sorry, didn't know it was available.  What time do I have to turn it in?

Thanks, should have them in by Friday night.  I'll only use 40 minutes!  This will be good practice.

if we can't do it tomorrow, can we do it sunday or must we do it today?

The competiton was supposed to have started 70 min ago. What happened?

I didn't realize you were going to edit your first post and put them there.

I realize I don't really have the time this weekend. I was on vacation so I have tons of homework to do.

He said he has tons of homework to do; the problem isn't just that he was on vacation.

I sent in my answers.
I didn't realize that you wanted us to start at a specific time, I just started when I wasn't busy.
There were some weird problems, but I included what I had trouble understanding in the PM.

I'll either do the target later today or tomorrow.

 :D


function
real analysis
real analysis theorems

What is the easiest way to show that a monotonous function $f: R \rightarrow R$ is differentiable a.e ?

Sorry, but I don't know that a monotounous function is differentiable. :?
Actually, I think this is not true. :P

It says something like this in Siretchi ? A monotonous function is not even continuous, so... ;)

[quote="cezar lupu"]A monotonous function is not even continuous, so... [/quote]
But ... (and here I'm talking about a monotone function on a bounded interval, to make things easier), a monotone function can have at most a countable number of discontinuities, with each discontinuity a jump discontinuity.

Suppose our function is nondecreasing on $[a,b].$ From the monotonicity, we get that the function has right and left hand limits at every point. Further more the left hand limit is always $\le$ the right hand limit. So the only possible discontinuities are upward jump discontinuities.

Given $n,$ suppose there are $m$ discontinuities with jumps of height at least $\frac1n.$ Then $\frac mn\le f(b)-b(a),$ so there are at most a finite number of such jumps. Take the union over $n$ and we get a countable number of discontinuities.

So such a function is continuous almost everywhere. That's not the proof that it's differentiable  almost everywhere. That is a well-known theorem and should be in most analysis books.

Anything you can say about monotone functions can also be said about functions of bounded variation.


check out this lecture at the university politehnica bucharest: http://riccati.pub.ro/Lecture%20Notes/TS-global/TSII-global.pdf (it's a pdf) :D

I wish I spoke that language.  Cause that would make it a lot more interesting.  Anyways some genius profesor did I nice job with making that .pdf cookies all aroung ( : : ) ( : : )

um...I only see a bunch or rectangles...and % signs...and the occasional dots. is that supposed to happen?

No, it's a pdf of some math lecture notes I believe. (500 some pages of lecture notes) But I don't get what's so interesting about it. I thought it was just some papers turned upside down and made into a pdf?

[quote="Valentin Vornicu"]Yes, it's an up-side-down pdf file. That is supposed to be a class for students to read :) How can you read it if it's upside down? (and nobody wants to print out 500 pages :) ).[/quote]

Interesting...well, Adobe Reader allows you to rotate the page, so two clicks and you're done.

I suspect when the pages were made using some PDF creator, the author just clicked 'rotate' a couple of times, maybe as a joke.

The author is the teacher that teaches that class :) I think he has no idea the pages are up-side-down :D It just seemed funny to me.  :lol:


inequalities
integration
trigonometry
calculus
calculus computations

Prove the inequality $ \int_{0}^{\frac{\pi}{2}} \frac{(\cos x)^{\sin x}}{(\cos x)^{\sin x}\plus{}(\sin x)^{\cos x}}dx<1$

__________________________________
$ \textbf{Azerbaijan Land of Fire }$   :rotfl:

Let $ \int_0^{\frac {\pi}{2}} \frac { (\cos x)^{\sin x} }{(\sin x)^{\cos x} \plus{} (\cos x)^{\sin x}} dx\equal{} I$

By the identity that $ \int_0^a f(x) dx \equal{} \int_0^a f(a \minus{} x) dx$, we get
$ I \equal{} \int_0^{\frac {\pi}{2}} \frac { (\sin x)^{\cos x} }{(\sin x)^{\cos x} \plus{} (\cos x)^{\sin x}} dx$

Adding, $ 2I \equal{} \int_0^{\frac {\pi}{2}} dx \equal{} \frac {\pi}{2}$
Thus, $ I \equal{} \frac {\pi}{4} < 1$
(QED)

CMIIW


[quote="catcurio"]Charlyn walks completely around the boundary of a square whose sides are each $5$ km long. From any point on her path she can see exactly $1$ km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?

$\text{(A)}\ 24 \qquad\text{(B)}\ 27\qquad\text{(C)}\ 39\qquad\text{(D)}\ 40 \qquad\text{(E)}\ 42$[/quote]

[hide]I got $\boxed{\text{(C)} 39}$
The total area of the square that is in her line of sight including the corners that are  not included is $7^2=49$. The region in the middle is $3\times3=9$ and the corners have area $\frac{4-\pi}{4}$ and four of these corners have area $4-\pi$. So $49-9-(4-\pi)=36+\pi\approx39.$[/hide][hide][/hide]

[hide]The area that she can see inside the square is 25-9=16 sq.km. The area outside the square consists of 4 rectangle of sides 5x1 and 4 quarter circles with a 1km radius, for a total of $20+\pi$ sq.km.

Thus the total area is $16+20+\pi\approx39$ square kilometers. [/hide]

[quote="jli"][quote="ryan22"]the answer is 30[/quote]

1)  please explain

2)  how is it 30 if there is no answer choice for 30[/quote]
Don't argue with it, you'll go nuts like I did with this mathcounts question. :roll:

[quote="Mike Tang"]It's just not 30. That guy only said that to be random.[/quote]

Hello, welcome to AoPS.  Couple of things you should know.  First, hide your solutions by highliting the solution and clicking the "Hide" button.  Second, don't double post.  And third, don't revive old topics.  For example, this topic is well over a year old so that classifies it as old.   :)


A biologist catches a random sample of 60 fish from a lake, tags them and releases them. Six months later she catches a random sample of 70 fish and finds 3 are tagged. She assumes 25% of the fish in the lake on the earlier date have died or moved away and that 40% of the fish on the later date have arrived (or been born) since. What does she estimate as the number of fish in the lake on the earlier date?

[hide="Answer"]Say that the previous number of fish was $x$, and she marked $60$, so we have $\frac{60}{x}$ are marked. When she takes her second sample, there are $\frac{3x}{4}$ of those fish remaining, so we can assume $45$ tagged fish. $28$ fish from the second sample are new, so we have $3$ out of $42$ are tagged, meaning that there were originally $42\cdot 15=630$. However, 25% of the fish had died, so we actually have $\frac{4}{3}\cdot 630=\boxed{840}$ fish on the earlier date.[/hide]

Agreed, except for the last step.  If 3/42 fish are now tagged then 3/42=60/x, so x= 60*42/3=42*20=[b]840[/b]

Oh, you're right, I forgot to factor back in the fish that had died.


[color=darkblue]Let $ x\plus{}y\plus{}z\equal{}2009$. Find $ S\equal{}\frac{{x^3  \plus{} y^3  \plus{} z^3  \minus{} 3xyz}}{{(x \minus{} y)^2  \plus{} (y \minus{} z)^2  \plus{} (z \minus{} x)^2 }}$[/color]

[hide="solution"]
Since

\begin{eqnarray*} x^3 + y^3 + z^3 - 3xyz & = & (x + y + z)\left(\dfrac{(x - y)^2 + (y - z)^2 + (z - x)^2}{2}\right) \\
x^3 + y^3 + z^3 - 3xyz & = & \left(\dfrac{1}{2}\right)(x + y + z)((x - y)^2 + (y - z)^2 + (z - x)^2) \end{eqnarray*}
We can now substitute into the equation, so that the variables cancel out.
\begin{eqnarray*} S & = & \dfrac{x^3 + y^3 + z^3 - 3xyz}{(x - y)^2 + (y - z)^2 + (z - x)^2} \\
S & = & \dfrac{(x + y + z)\left(\dfrac{(x - y)^2 + (y - z)^2 + (z - x)^2}{2}\right)}{(x - y)^2 + (y - z)^2 + (z - x)^2} \\
S & = & \left(\dfrac{1}{2}\right) \dfrac{(2009)((x - y)^2 + (y - z)^2 + (z - x)^2)}{(x - y)^2 + (y - z)^2 + (z - x)^2} \\
S & = & \left(\dfrac{1}{2}\right)(2009) \\
S & = & \boxed{\dfrac{2009}{2}} \end{eqnarray*}  :ninja:

Edit: I was in a hurry before, now it's more detailed. Thanks varunrocks.
[/hide]


At a party, there are 15 males, 6 of whom wear glasses. Nine people at the party wear glasses. Eight people at the party are neither male nor wear glasses. How many people are at the party?

If 6 males wear glasses, and 9 wear glass in total, that means 3 females wear glasses. There are 8 other females, who do not wear glasses, making a total of 11 females. $ 11 \plus{} 15 \equal{} \boxed{26}$.

since $ 9$ people wear glasses, and $ 6$ were already counted in the male part, there are $ 3$ pplz that wear glasses that aren't male
there are also $ 8$ people that aren't male and don't wear glasses
so, we have $ 15$ males, $ 3$ females with glasses, and $ 8$ females without glasses

so...
$ 15 \plus{} 3 \plus{} 8 \equal{} 26$

ans is $ \boxed {26}$

edit: beaten to it.....


[hide="solution"]
$7^2=49
8^2=64
9^2=81
10^2=100
\text{9 is the answer because 100 is over 95 and 81 is the only square under 95}$[/hide]

[quote="236factorial"][hide]$\sqrt{(8)(32)} = 2\cdot4{2\cdot2}$, bringing out perfect squares.
So we have $8\sqrt{4}$, 8*2, 16.

Or, 32*8 is 256, sqrt of that is 16 [/hide][/quote]

how'd u get that? did joejia change the problem after u solved the original one..........

[quote="math92"]Maybe he answered a problem in the wrong topic
I have no idea where that answer came from :huh:[/quote]

I think it's the wrong topic I posted on  :roll:

:rotfl:
i think i did that  once.

Extension:
What is the greatest positive integer $n$ such that $n^3$ is less than $3375$

[hide]
$n$ is the greatest integer less than $\sqrt[3]{3375}$. Therefore, $n$ is $14$.[/hide]

[quote="joejia"]What is the greatest positive integer $n$ such that $n^2$ is less than 95?[/quote]
$95=n^2\\n=9$

[hide]Since $9^2=81$ and $10^2=100$, the answer must be $9$[/hide]


Find the greatest real number $k$ such that, for any positive $a,b,c$ with $a^{2}>bc$, $(a^{2}-bc)^{2}>k(b^{2}-ca)(c^{2}-ab)$.

yes, k = 1!

now, opening, we prove that $a^{4}+c^{3}a+b^{3}c \geq 3a^{2}bc$ by AM-GM!

but the equality can't occur, because a^2>bc!

(if k>1, we easily find one counter-example in the AM-GM inequality!)

[quote="M.SH"]Let $f(a,b,c)=\frac{(a^2-bc)^2}{(b^2-ca)(c^2-ab)}$, where $a^2> bc$. we have
\[f(a,b,c)\geq f(a,\sqrt{bc},\sqrt{bc})\iff bc(\sqrt{bc}-a)^2\geq (b^2-ac)(c^2-ab)\\\iff b^3+c^3\geq2bc\sqrt{bc}\]
which is true. so it is enough to consider the case $b=c$. in this case, $a>b$ and
\[f(a,b,c)=f(a,b,b)=\left(1+\frac ab\right)^2>4\]
Now it is obvious that $k=4$ is the best constant. indeed, when $b=c\to a^{-}$, $f(a,b,c)$ approaches to 4.[/quote]
Very good:)

Without loss of generality, we can assume that abc=1.
Let $x = a^3$,$y = b^3$ and $z = c^3$. then $x>1$ and the inequality is equivalent to: $\frac{(x-1)^2}{x}>k(y-1)(z-1)$.
the case: $(y-1)(z-1) \leq 0$ is obvious. Lets assume that:  $(y-1)(z-1) > 0$, then:
$\frac{(x-1)^2}{x(y-1)(z-1)} = \frac{(x-1)^2}{x(yz-(y+z)+1)} \geq \frac{(x-1)^2}{x(\frac{1}{x}-\frac{2}{\sqrt{x}}+1)} = (\sqrt{x}+1)^2 > 4$

So the minimal value is 4.


Find the greatest prime factor of $ 11!\plus{}13!$

$ 11! \plus{} 13! \equal{} 11!(1 \plus{} 12\cdot 13) \equal{} 11!(157) \Longrightarrow \boxed{157}$.

[hide="...."]

[url=http://en.wikipedia.org/wiki/Sexy_prime]SEXAY PRIME!![/url][/hide]


Find all functions $ f: \mathbb N \to \mathbb R$ such that $ f(x)\equal{}f(x\plus{}22)$ for all $x \in \mathbb N$ and
\[f(x^2y)\equal{}(f(x))^2 \cdot f(y)\]
for all positive integers $x$ and $y$.


algebra polynomial geometry rectangle quadratics function calculus

Ok

here's the answer:
[hide]
You know that the equation for the area of a rectangle is:
Length*Width

so you get:
Length*width=96

then sub in the values:
(x+2)(x-2)=96

multiply(FOIL) out:
x^2-4=96

then solve
[/hide]

Is this one of your homework problems?

Eh, I know this is spam, but I used to use those books :D

Anyway, we know:

$ (x\plus{}2)(x\minus{}2)\equal{}96$

Expanding with difference of squares knowledge gives us

$ x^2\minus{}4\equal{}96$

Adding $ 4$ to both sides:

$ x^2\equal{}100$

Taking the square root from both sides:

$ x\equal{}\pm 10$

We can immediately discard the negative root because if $ x\equal{}\minus{}10 \implies l\equal{}\minus{}8, w\equal{}\minus{}12$ which is impossible.  Thus, $ x\equal{}\boxed{10}$.

y wouldn't -8 times -12 make sense?
-8 times -9=96

Rectangles cannot have negative lengths or widths.  :)

You also know that the length is four more than the width, so you could write;

(w + 4)w = 96

Of course, it is usually easier to deal with a quadratic that has no linear term.  We can still use the idea in reverse though

If we were given that the length was four more than the width and the area is $ 623$, so $ l\equal{}w\plus{}4$ and $ lw\equal{}623$, we could instead make the substitution $ w'\equal{}w\plus{}2$, so $ (w'\plus{}2)(w'\minus{}2)\equal{}623 \Rightarrow w'\equal{}\sqrt{627}$, which gives us a length of $ \sqrt{627}\plus{}2$ and a width of $ \sqrt{627}\minus{}2$.

Since the area of a rectangle is the length times the width, and the length is x+2 and the width is x-2, you multiply them to get the area, right? So you simply multiply (x+2)(x-2)by using the FOIL method. Front,Outer,Inner,Last. You get x^2 -2x+2x-4=96, or x^2 -4=96, which becomes x^2=100. Taking the positive root, we get [b]10[/b] as our answer.

[quote="JokoMoko"]I'm new here~
lol testing testing  :lol:

Ok...on to polynomials....
A rectangle's area is 96 (l=x+2, w=x-2)
What would the polynomial be and the starting out equation?
Please help! Thanks  :ninja:[/quote][hide]

Polynomials are "the Key to Algebra" because any continuous, differentiable function -- [i]continuous[/i] and [i]differentiable[/i] are concepts one learns in calculus, but for now you can assume that the functions you're going to see are continuous and differentiable -- can be written as a polynomial thanks to Taylor's theorem (another calculus concept).

Polynomials are sums of variable terms, where the exponents of each variable term are whole numbers.

For your problem: $ \mbox{Area} \equal{} \ell \times w \equal{} (x \plus{} 2)(x \minus{} 2) \equal{} x^2 \minus{} 4 \equal{} 96$.

The polynomial is $ x^2 \minus{} 4$.

This is called a polynomial of degree 2 in x -- "degree 2" because the greatest exponent in any term of the polynomial is 2.

It is worth noting that polynomials like $ (x \plus{} 2)(x \minus{} 2)$ are special, in that the binomial factors, except for the signs of the second terms of each, are identical. Whenever you see this pattern you can tell immediately that when expanded by FOIL multiplication the middle term will vanish. For this reason the polynomial resulting from such expansion is called [b]a difference of two squares[/b]. For example,

\[ (x \plus{} 2)(x \minus{} 2) \equal{} x^2 \underbrace{\minus{}2x \plus{} 2x}_{0} \minus{} 2^2 \equal{} x^2 \minus{} 4\]

This difference of two squares is the difference between $ x^2$ and $ 2^2$.

If the objective of this problem is to solve for $ x$ then

\[ x^2 \minus{} 4 \equal{} 96 \\ x^2 \minus{} 4 \minus{}96 \equal{} 96 \minus{} 96 \\ x^2 \minus{} 100 \equal{} 0 \\ (x \plus{}\sqrt{100})(x \minus{}\sqrt{100}) \equal{} 0 \\ (x \plus{} 10)(x \minus{}10) \equal{} 0\]

Whenever you have the product of two factors equal to zero that means one of the factors or the other is itself equal to zero. In this case,

\[ \mbox{If }x \plus{} 10 \equal{} 0\mbox{ then } x \equal{} \minus{}10\]

Since $ x \plus{} 2$ is actual physical dimension (length), the fact $ x \plus{} 2 < 0$ when $ x \equal{} \minus{}10$ means this solution, though mathematically possible, makes no sense.

\[ \mbox{Else if }x \minus{} 10 \equal{} 0\mbox{ then } x \equal{} 10\]

$ x \equal{} 10$ is a legitimate solution because in this case, $ \mbox{length} \equal{} x \plus{} 2 \equal{} 12$ and likewise, $ \mbox{ width} \equal{} x \minus{} 2 \equal{} 8$[/hide]

just take the algebraic formula (a+b)(a-b)=(a^2 - b^2).


I'm interested to see who's going on February 13th, especially given registration filled up quickly. Unfortunately, unlike the UMD competition, you can't see who else is registered, so I'm attempting to circumvent that here. :P

I [i]think[/i] our school always does it, so I think this year, we are, too.
But I've personally never been able to participate due to schedule conflicts... :(


function
search
limit
algebra proposed
algebra

$f : R \to R$ satisfies : $f(f(x)) +f(x) =x^4+3x^2+3$ , $\forall x \in R$
 Prove that : $f(x)$ is an even function (f(x)=f(-x))

If $f$ is also continuous, this is problem 4, RMO 1992, proposed by Gh. Eckstein, discussed before in this forum. Try the search function.

I searched with the keyword "x^4+3x^2+4 " and founded an "big" solution .
 Enescu , could you please give me a hint to solve this problem more easily ?

[quote="Jarod"]I searched with the keyword "x^4+3x^2+4 " and founded an "big" solution .[/quote]

I did the same and didn't find anything but this topic ! However, I found a solution in a Romanian book (RMO 1954-2003). The function is supposed to be continuous and the solution is still "big" enough, I'm afraid! :(  Here are the main ideas:

(1) if f(a)=f(b),  we show that a=b or a=-b;
(2) so f is injective on $[0,+\infty)$ and also on $(-\infty,0]$; being continuous too, it follows that f is monotonic on these two intervals;
(3) we show that ${\lim_{x\to\infty}f(x)=\lim_{x\to -\infty}f(x)=\infty}$ (it's not so easy as it looks!);
(4) since f is continuous and monotonic on $(-\infty,0]$ and on $[0,+\infty)$,  (3) yields to
$f((-\infty,0])=f([0,+\infty))=[f(0),\infty)$.
So for every x in  $[0,+\infty)$, there exists a t in $(-\infty,0]$ so that f(x)=f(t). From (1) it follows that f(x)=f(-x) for every x in $R$.

If $f$ is not continuous, I doubt the conclusion holds. Unfortunately, I couldn't find a discontinuous function to match our functional equation. I tried to slightly "discontinue"  one of its continuous solutions, $f(x)=x^2+1$, but it didn't work.

Use \lim_{n\to \infty} to create $\lim_{n\to \infty}$ ;)


Hey, I just got AoPS over Christmas break and I'm looking through it. Does anybody know what chapters to focus on, if I want to practice for a higher score on the AIME?

I'm really looking for something to just help me get another point or so to maybe qualify for USAMO (at the moment, not very good shot)

Here is a post from rrusczyk that lists chapters from AoPS volume 2 to study for AIME:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=70754&#p70754[/url].


If I'm not mistaken, I think there's something wrong with your ineq.

try : $ a\equal{}b\equal{}c\equal{}1$

OK , i edited it . NOW you can solve it.   :)

[hide="Hint 1"]Well know inequality[/hide]

[hide="Solution 1"]
CS on the two sequences $ \{1,1,1\}$ and $ \{a,b,c\}$[/hide]

[hide="Hint 2"]
Multiply out
[/hide]

[hide="Solution 2"]
Multiplying out gives us

$ \sum_{cyc} a^2 \ge \sum_{cyc} ab$

Which follows from AM-GM or rearrangement
[/hide]

If $ a\equal{}b\equal{}c\equal{}0$ it is true, otherwise $ a\plus{}b\plus{}c>0$ so we can normalize to $ a\plus{}b\plus{}c\equal{}3$. Then $ (a\minus{}1)^2\ge 0 \iff a^2\ge 1\minus{}2a$. Then $ 3\sum a^2\ge 3(1\plus{}1\plus{}1\minus{}2a\minus{}2b\minus{}2c)\equal{}9$ so we're done.

[quote="Altheman"]If $ a \equal{} b \equal{} c \equal{} 0$ it is true, otherwise $ a \plus{} b \plus{} c > 0$ so we can normalize to $ a \plus{} b \plus{} c \equal{} 3$. Then $ (a \minus{} 1)^2\ge 0 \iff a^2\ge 1 \minus{} 2a$. Then $ 3\sum a^2\ge 3(1 \plus{} 1 \plus{} 1 \minus{} 2a \minus{} 2b \minus{} 2c) \equal{} 9$ so we're done.[/quote]

They're not necessarily nonnegative

$ \parallel{}a| \plus{} |b| \plus{} |c\parallel{}\ge |a \plus{} b \plus{} c|$ then...

it is basically from the inequality $ \frac{1}{2}[(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2] \geq 0$ , expanding gives the desired result  :)

[quote="Elio (n)"]Prove that $ 3(a^{2} \plus{} b^{2} \plus{} c^{2}) > \equal{} (a \plus{} b \plus{} c)^{2}$  :)[/quote][hide]It is equivalent to $ 2a^2\plus{}2b^2\plus{}2c^2\ge2ab\plus{}2bc\plus{}2ca$, which is true since $ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\ge0$.[/hide]

[hide="Proof"]We can expand this inequality to get $ 3a^2\plus{}3b^2\plus{}3c^2\geq  a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2bc\plus{}2ac$. This simplifies to $ 2a^2\plus{}2b^2\plus{}2c^2\geq 2ab\plus{}2bc\plus{}2ac\ (*)$. We know that $ \dfrac{1}{2}[(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2]\geq 0$. This means that $ (*)$ is true.[/hide]


The Fibbonacci Sequence 1,1,2,3,5,8,13,21... starts with two 1s, and each term afterwards is the sum of two predeccessors. Which one of the ten digits is the last to appear in the units of a number in the Fibbonacci Sequence?
A, 0
B, 4
C, 6
D, 7
E, 9

I dont understand what this question is asking for? (my english is bad!)

Basically, you look at the last (right most) digit, and see which one of the answer choices you see latest.

...Write all these down? I don't know much about Fibonacci...

1,1,2,3,5,8,3,1,4,5,9,4,3,7,0

so the last one of 10 digits appearing in the units is 6.

Waiting for better method.

Unfortunately, that's the only thing I can think of too. Wasn't this like an ASHME or AMC problem?

[quote="Fraenkel"]oh so when it said the ten digits it meant which of 1,2,3,...9 was the last digit to be in the units.
lol. that makes it easier to understand.[/quote]

0 is included as a digit  :wink:


I just want to see who in this forum is from Don Mills CI. I know Yufei is in here and I think I've seen Lin Fei around somewhere. Anyone else?

I've seen Jenny Wang somewhere..

Are you from Don mills too?

The preceding message is posted my me  :D

[quote]Are you from Don mills too?[/quote]
Yes.

I wonder if Oleg's here :? .

Oleg doesn't believe in mathlinks.

He has a [url=http://oleg.moltenstudios.com/blog/]blog[/url] btw, in case you're interested.

[quote="billzhao"]
He has a [url=http://oleg.moltenstudios.com/blog/]blog[/url] btw, in case you're interested.[/quote]

Oh wow, he has a blog?!  Heheh, I'll start reading it, Oleg's a cool/amusing guy

me?  Mathcamps.  Waterloo Seminars, National Camp and Winter Camp.  Good guy.

lol Doubt he remembers me, even though we were on the same team last year in Waterloo.


calculus
integration
trigonometry
inequalities
function
limit
real analysis

Here are two nice integral inequalities:

A) Let $f$ be a decreasing function and $a>0$. Prove that

$\int_{0}^{\infty}f(x)\sin(ax)dx>0$.

B) Let $f$ be a decreasing function such that $f''(x)<0$, and let $a>0$.

Prove that

$\int_{0}^{\infty}f(x)\cos(ax)dx>0$.

Of course $f$ is a function for which the integrals make sense.

(a) Integrate by parts. (Since all you said is that $f$ is a function "for which the integral makes sense" I'll use a Stieljes integral.)

$\int_{0}^{\infty}f(x)\sin(ax)\,dx= \left.\frac{f(x)(1-\cos(ax))}a\right|_{0}^{\infty}+\int_{0}^{\infty}\frac{1-\cos(ax)}a\,(-df(x))$

At this point, we have to assume that $\lim_{x\to\infty}f(x)=0$ in order for the integral to converge. Making that assumption, we have

$\int_{0}^{\infty}f(x)\sin(ax)\,dx= \frac1a\int_{0}^{\infty}(1-\cos(ax))\,(-df(x))$

And this is nonnegative: $(1-\cos(ax))\ge 0$ everywhere and $(-df(x))$ is nonnegative because $f$ is a decreasing function.

In fact, we also have an upper bound for the integral. It is no greater than

$\frac2a\int_{0}^{\infty}(-df(x))=\frac{2f(0)}a.$

On part (b), I think you meant to say $f''(x)>0.$

"For which the integral makes sense" works in the sense of improper integrals. We need not require that $f\in L^{1},$ and in fact we can have $f$ decay arbitrarily slowly. The only restrictions we need are that $f$ is monotone nonincreasing, $f(0)<\infty,$ and $f(x)\to0$ as $x\to\infty.$

In fact, we don't even really need $f(0)<\infty$ (although not having that would ruin my upper bound inequality.) $f(x)=o(x^{-2})$ is sufficient for the positivitiy argument, although we need a little more than that to assure that the integral converges.

Kent,
You are right. For part b) $f''(x)>0$, and part b) is a simple application of part a).

I do have another method for solving part a) of the problem, not so elegant as yours, which is the following one:

$\int_{0}^{\infty}f(x) \sin(ax)dx=\sum_{k=0}^{\infty}\int_{\frac{2k\pi}{a}}^{\frac{(2k+1)\pi}{a}}f(x)\sin ax dx+\sum_{k=0}^{\infty}\int_{\frac{(2k+1)\pi}{a}}^{\frac{(2k+2)\pi}{a}}f(x)\ sin ax dx$

$>\sum_{k=0}^{\infty}f(\frac{(2k+1)\pi}{a})\int_{\frac{(2k)\pi}{a}}^{\frac{(2k+1)\pi}{a}}\sin ax dx+\sum_{k=0}^{\infty}f(\frac{(2k+1)\pi}{a})\int_{\frac{2k\pi}{a}}^{\frac{(2k+2)\pi}{a}}\sin ax dx$

$=\sum_{k=0}^{\infty}f(\frac{(2k+1)\pi}{a})\int_{\frac{2k\pi}{a}}^{\frac{(2k+2)\pi}{a}}\sin ax dx =0$.


geometry
circumcircle
inradius
geometry unsolved

Let M inside triangle ABC. The lines AM,BM,CM meet the circumcircle ABC at A',B',C',respectively.Let r,r' is the irradius triangle ABC,A'B'C',respectively
Prove that $4rr^{\prime}\leq R^{2}-OM^{2}$.

Denote R is radius of circumcircle triangle ABC

I have a question:if you take two equilateral triangles on the same circumcircle such that AA',BB',CC' concur, must they concur in O?I think not :)


[quote="jensen"]be nazare man samane farhat ghatan yeki az oonast. :)[/quote]

jaye shoma too team vaghean khali nist.

ta akhare hafteye dige team maloom mishe.

Be bachehaye team tabriiiik migam makhsoosan be hadikhanloo va takapooyi . omidvaram teame iran emsal gol bekare :)

[quote="mahmood"]be shayan o arman be tore khaas tabrik migam (team : sefidgaran , firoozi , haadi khaanloo , takaapooyi , faazeli , dashmiz).[/quote]

Ahmadi pour ozve team nashod???? :o

na oon dge olympiad nemide,daneshgahe alan,faghat dore tabestoono bood,badesh raft...

oWwwwwWwWW!!!!!!!!!!!!!! be takapoooyi va hadikhanlou va baghie doosta tabrik migam;) IMO2007  iran aval misheeee//khashi to chi migi inja?

man ham be nobeye khodam be onvane ye studente alameh tabatabyi be  m.reza takapooyi va  saeed hadikhanloo tabrik migam.
zemnan be sadra yazdanbod va  amir sharifzadeh bekhatere daryafte medale talaye olympiade computer tabrik migam.

[quote="sepehr ahmadzadeh"]sigary u ki hasty? pas chera man nemishnasamet?[/quote]
 :rotfl:


we know that

$ x \equal{} w\log_{a}70$
$ y \equal{} w\log_{b}70$
$ z \equal{} w\log_{c}70$

by base change formulas, you get:

$ y \equal{} \frac{x}{\log_{a}b}$
$ z \equal{} \frac{x}{\log_{a}c}$

therefore

$ \frac{1}{x} \plus{} \frac{\log_{a}b}{x} \plus{} \frac{\log_{a}c}{x} \equal{} \frac{1}{w}$

$ \frac{\log_{a}{abc}}{\log_{a} 70} \equal{} 1$

therefore

$ abc \equal{} 70$

cough....yeah...:ewpu:, someone finish it

[hide="I PHAIL TOO"]huh? no..
\[ \cfrac{1}{x} \plus{} \cfrac{\log_ab}{x} \plus{} \cfrac{\log_ac}{x} \equal{} \cfrac{1}{w}
\]

\[ \cfrac{1}{x} \plus{} \cfrac{\log_ab}{x} \plus{} \cfrac{\log_ac}{x} \equal{} \cfrac{\log_a70}{x}
\]

\[ 1 \plus{} \log_ab \plus{} \log_ac \equal{} \log_a70
\]

\[ \log_abc \equal{} \log_a70
\]

\[ bc \equal{} 70
\]
okay so $ \cfrac{1}{w} \equal{} \cfrac{\log_a70}{x} \equal{} \cfrac{\log_b70}{y} \equal{} \cfrac{\log_c70}{z}$

a bunch of trig stuff also gets us:

$ \cfrac{1}{x} \equal{} \cfrac{\log_ba}{y} \equal{} \cfrac{\log_ca}{z}$
$ \cfrac{1}{y} \equal{} \cfrac{\log_cb}{z}$
$ \cfrac{1}{z} \equal{} \cfrac{\log_bc}{y}$

So substituting like we did before, we see that:
\[ ac \equal{} 70
\]

\[ bc \equal{} 70
\]

okay WTF:
\[ a \equal{} b \equal{} c \equal{} \sqrt {70}
\]
so the answer is $ a \plus{} b \plus{} c \equal{} \boxed{3\sqrt {70}}$?

wait but you said this is an AIME problem.. that can't be right..[/hide]

no its a modified one...i've seen something like that

undefined here is where you went wrong:

you  said

[quote]
$ 1 \plus{} \log_ab \plus{} \log_ac \equal{} \log_a70$
[/quote]

then in the next step, you left out the $ 1$ and you wrote

$ \log_abc \equal{} \log_a70$, when in fact $ 1 \equal{} \log_aa$, therefore the sum should be

$ \log_aa \plus{} \log_ab \plus{} \log_ac \equal{} \log_aabc \equal{} \log_a70$

therefore

$ abc \equal{} 70$

um.. um.. algebra 2 was more than two years ago :D

it was like a few months ago for me. so in a year or 2 i will forget alg. 2.  :maybe:

I'll post my solution later this week. Logarithms was only a few weeks away for me xD

disclaimer: I, unlike your mother, do not have a solution.

edit: what's your point?

[quote="mathcrazed"]Given that $ a^x \equal{} b^y \equal{} c^z \equal{} 70^w$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {1}{w}$, show $ a \plus{} b \equal{} c$.

Err edit: lol. fyi i posted the wrong problem since it's impossible to find a+b+c. xD.

So #H34N1 was obviously bsing about having a solution =P OWNED.[/quote]

are you sure thats right?  Looking at the problem it seems that its symmetrical throughout so a+c=b and b+c=a if a+b=c...

furthermore, say x=1, y=1, z=1.  Then w=1/3 from the second equation.  Thus a^1=b^1=c^1=70^(1/3) so we can make a=b=c=70^(1/3) and then a+b=c is false.

I think you mistyped again.

btw please respond to my pm.

$ a<b<c$ and $ a,b,c$ are all integers. $ x,y,z,w$ are reals.

but then that makes (a,b,c) unique as (a,b,c) is (2,5,7) and you can find a+b+c=14...

(7,-2,-5) also works but that wouldnt make a<b<c...

but im probably wrong...i just skimmed it and i just used abc=70 and a,b,c are integers, nothing else...

i didin't get any further than abc=70 either.

would it make sense to replace the 70 in the first equation with abc?


i really dont get this...

Let n be any number from 1-1000. For how many values of n is (the sqroot of n) a rational number? Explain to me please

[hide="Hint"] The square root of an integer is either irrational or an integer (in other words, it is never a rational non-integer).  So this question is just asking for the number of squares between $ 1$ and $ 1000$. [/hide]

[quote="MrRubix"][quote="t0rajir0u"] So this question is just asking for the number of squares between $ 1$ and $ 1000$.[/quote]
Which happens to be 31:
1,4,9,16,25,36,49,.... 961[/quote]

In other words it's just $ \lfloor\sqrt{1000}\rfloor$


I believe that your inequality doesnot hold(just check $a_{1}=\dots=a_{n},x_{1}=\dots=x_{n}$,but the inequality $\sum_{i=1}^{n}$    $\frac{a_{i}}{1+x_{i}}\leq \frac{1}{1+x_{1}^{a_{1}}\cdots x_{n}^{a^{n}}}$ holds,but easy to prove( just use jenson :$f(x)=\frac{1}{1+e^{x}},f^{''}(x)\leq 0$


geometry
perpendicular bisector
geometry unsolved

Let $ABC$ be a triangle, and $M$ be the midpoint of its side $AB$. Let $H$ be the orthocenter of triangle $ABC$, and $K$ be the midpoint of the segment $HC$. Prove that the point of intersection of the angle bisectors of the angles $HAC$ and $HBC$ is a point on the line $MK$.

Let [b]AD[/b] and [b]BE[/b] be the altitudes in [b]ABC[/b]. It is simple that [b]ME[/b] = [b]MD[/b] and [b]KE[/b] = [b]KD[/b], so [b]KM[/b] is a perpendicular bisector of [b]ED[/b]. Now we have only to prove that [b]XE[/b] =[b] XD[/b] (where [b]X[/b] is a point of intersection of bisectors of angles [b]HAC[/b] and[b] HBC[/b]). But [b]AXB[/b] = [b]90[/b], so points [b]A[/b], [b]B[/b], [b]D[/b], [b]E[/b], [b]X[/b] lie on one circle and [b]EAX[/b] = [b]DAX[/b],  so [b]XE[/b] = [b]XD[/b] :D


Show that every positive integer is a sum of one or more numbers of the form (2^r)(3^s), where r and s are nonnegative
integers and no summand divides another.

we will prove it with induction:

for the first numbers its easy to check...

first denote $S$ as ${S=\{2^{r}3^{s}| r,s \in \mathbb{N}\cup{0}}\}$ and name every number in $S$ with $s_{i}$...

now assume that its true for numbers less than $n$ and we want to prove it for $n$...

$i)$if $n$ was an even number then we would have $n=2k$ so $k<n$ so from the hypothesis of the induction we get that
$k=\Sigma s_{i}$ so $n=\Sigma 2s_{i}$ works...

$ii)$ if $n$ was an odd number then we would have $n=2k+1$

in this case assume the greatest number $\alpha$ such that $3^\alpha \leq n$ and put $p=3^{\alpha}$ and also put $q=\frac{n-p}{2}$

now we have $p<n<3p$ so we get that $0<q<n$ so from the hypothesis of the induction we can write $q=\Sigma s_{i}$

now put $n=p+\Sigma 2s_{i}$...

now hence $p=3^{\alpha}$ so $p$ isnt divisible by any of the numbers $2s_{i}$ and also if there exists $i$ such that $p \mid 2s_{i}$

then we would have $2p\leq 2s_{i}$ which contradicting the fact that $n<3p$ so the form $p+\Sigma 2s_{i}$ is a desired
writing for $n$...


Fie $ABCD$ un tetraedru a.i. $DB\bot DC$ iar $DH\bot (ABC)$ ,unde $H$ este ortocentrul triunghiului $ABC$.Aratati ca $AD\bot DB$ si $AD\bot DC$

[hide="idee"]
eu nu prea stiu geometrie in spatiu, insa metoda asta m`a salvat si in clasa a 8a. :)
$\overrightarrow{DH}\cdot\overrightarrow{AB}=0$ (pentru ca $DH\bot AB$).
$\overrightarrow{DH}=\overrightarrow{DC}+\overrightarrow{CH}$, insa $CH\bot AB$, deci $\overrightarrow{CH}\cdot\overrightarrow{AB}=0$.
deci $\overrightarrow{DC}\cdot\overrightarrow{AB}=0$, adica $DC\bot AB$.
cum $DC\bot AB$ si $DC\bot BD$, rezulta ca $\overrightarrow{DC}\cdot(\overrightarrow{AB}+\overrightarrow{BD})=0$, adica $DC\bot AD$.
analog cealalta.
[/hide]

pai, e destul de simplu din moment ce problema este o proprietate a tetraedrului tridreptunghic.
$BH\perp AC$
$DH\perp AC$, deci $BD\perp AC$.
Dar $BD\perp DC$ deci $DC\perp AD$.
Analog pt cealalta relatie.

am si eu o intrebare..cum pot sa scriu problemele..k lumea..cu linie de fractie si alea alea....zicetimi va rog..  :(

@alex_tataran ...

Trebuie sa folosesti limbajul $LaTeX$ :) intra pe linkul [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php]asta[/url] ca sa vezi cum sa-l folosesti :)


rotation
trigonometry
geometry
geometric transformation
trig identities
Law of Cosines

[geogebra]304345e815d5f122829924ed270447e4285f6b9e[/geogebra]

There is a regular hexagon ABCDEF with side length 2.
Find the MINIMAL value of $ AZ \plus{} BZ \plus{} EZ$.

EDIT: Blah, that was stupid.

Help?

EDIT: Oops, not maximal; minimal. :blush:

[hide="Hint"]
Rotate $ E$ and $ Z$ clockwise $ 60^\circ$ around $ B$.
[hide="Hint"]
Look for something equal to $ AZ \plus{} BZ \plus{} EZ$. When is this minimal?
[hide="Giveaway"]
$ Z$ is known as the [url=http://en.wikipedia.org/wiki/Fermat_point]Fermat point[/url] (or Torricelli point) of $ \triangle ABE$.
[/hide]
[/hide]
[/hide]

Edit: I still misread it as minimal the first time. :P

Edit: Um, the answer isn't $ 6$. You can already minimize it more by letting $ Z$ be the projection from $ A$ onto $ \overline{EB}$.

Edit: [hide="I think the answer is"]$ 2\sqrt7$[/hide]

[hide]

Rotate points $ E$ and $ Z$ clockwise $ 60^\circ$ around $ B$. Call the new points $ E'$ and $ Z'$. $ EBE'$ is equilateral, so $ Z$ must be on line $ AE'$, as $ Z$ is the Fermat point. I think you can show that $ A$, $ Z$, $ Z'$, and $ E'$ are collinear by constructing a point $ X$ on the other side of $ AB$ such that $ ABX$ is equilateral. Then, show that $ \triangle{XBZ}\sim\triangle{ABZ'}$, which means $ \angle{AZ'B}\plus{}\angle{BZ'E'}\equal{}180^\circ$. Since $ BZZ'$ is equilateral, $ BZ\equal{}ZZ'$. Also, $ EZ\equal{}E'Z'$, so $ AZ\plus{}BZ\plus{}EZ\equal{}AE'$, and you can find that by using Law of Cosines to get $ 2\sqrt7$

[/hide]

Blah, sorry if I wasn't coherent.

[hide="In case you do not understand"]Note that $ \triangle ABE$ is a $ 30\minus{}60\minus{}90$ triangle. Since I screwed up the diagram on accident, rotate $ E$ and $ Z$ counterclockwise instead of clockwise $ 60^\circ$ around $ B$ to get $ E'$ and $ Z'$, respectively. From the rotation, we see that $ m\angle EBE'\equal{}60^\circ$ and $ E'B\equal{}EB$ so $ \triangle EBE'$ is equilateral. Similarly, $ \triangle ZBZ'$ is equilateral as well. Since rotation preserves lengths, $ EZ\equal{}E'Z'$ too. Thus $ AZ\plus{}BZ\plus{}EZ\equal{}AZ\plus{}ZZ'\plus{}Z'E'$. Since $ AE'$ remains constant no matter where $ Z$ is, it's clear that the minimum possible sum of $ AZ\plus{}ZZ'\plus{}Z'E'$ is $ AE'\equal{}\sqrt{AE^2\plus{}EE'^2}\equal{}\sqrt{(2\sqrt3)^2\plus{}4^2}\equal{}2\sqrt7$, which occurs when $ A,Z,Z',E'$ are collinear (when $ m\angle AZB\equal{}m\angle BZE\equal{}m\angle EZA\equal{}120^\circ$). $ Z$ is the Fermat point.[/hide]

Now I understand... I was never good at geometry :blush:

Thank you very much :D


real analysis
function
integration
limit
real analysis unsolved

Let $ \mu$ be the Lebesgue-Stieltjes measure on $ \mathbb{R}$ associated to the function:
$ F(x)=\{ \begin{array}{c l}
3-e^{-3x} & x\geq 0 \\
0 & x<0 \\
\end{array} $
Recall that this means that $ \mu$ is determined by the condition $ \mu(a,b]=F(b)-F(a)$.
With $ \lambda$ denoting the Lebesgue measure, find the Lebesgue decomposition of $ \mu$ relative to $ \lambda$.

The absolutely continuous part is $ F'$. The singular part is a point mass at $ 0$, where $ F$ is discontinuous.

Thanks, mlok.

Let $ A=\{0\}$ and $ B=\mathbb{R}-A$. Then $ A\cap B=\emptyset$ and $ \mathbb{R}=A\cup B$.
Let $ \mu_s(E)=\{ \begin{array}{c l}
1 & 0\in E \\
0 & 0\notin E \\
\end{array} $ and $ \mu_a(E)=\int_E F'(x)\lambda (x)=\int_{E\cap [0,+\infty)} 3e^{-3x}\lambda (x)
=\int_E \chi_{[0,\infty)}3e^{-3x}\lambda (x)$.
Then $ \mu_s(B)=0$ and $ \lambda(A)=0$ gives that $ \mu_s\perp\lambda$.
By RNT, $ \mu_a<<\lambda$ and $ \frac{d\mu_a}{d\lambda}=F'(x)$ $ \lambda-a.e.$ where $ F'(x)=\{ \begin{array}{c l}
3e^{-3x} & x>0 \\
0 & x<0 \\
- & x=0 \\
\end{array} $.
So $ \mu(A)=\lim_{n\rightarrow\infty}\mu(-\frac{1}{n},0]=\lim_{n\rightarrow\infty}[F(0)-F(-\frac{1}{n})]=
2$ and $ \mu(a,b]=F(b)-F(a)=\mu_a(a,b]$ for $ 0<a<b<+\infty$.
By HET, $ \mu$ is unique on such intervals, so $ \mu(E)=\mu_a(E)$ for all Borel-sets $ E\subseteq (0,+\infty)$.
Hence for any Borel-set $ E$,
$ \mu(E)=\mu(E\cap (-\infty,0))+\mu(E\cap A)+\mu(E\cap (0,+\infty))$
$ =\mu_a(E\cap (-\infty,0))+\mu_s(E\cap A)+\mu_a(E\cap (0,+\infty))$
$ =\mu_s(E\cap A)+\mu_a(E\cap B)$
$ =\mu_s(E)+\mu_a(E)$.


Find all positive Integers $ x,y$ such that $ (x\minus{}1)^5\plus{}x\equal{}y^3$.

$ x \minus{} 1 \equal{} a$  $ \implies$   $ a^5 \plus{} a \plus{} 1 \equal{} y^3$   $ \implies$
$ (a^2 \plus{} a \plus{} 1)(a^3 \minus{} a^2 \plus{} 1) \equal{} y^3$
$ (a^2 \plus{} a \plus{} 1,a^3 \minus{} a^2 \plus{} 1) \equal{} (a^2 \plus{} a \plus{} 1,a^2 \minus{} 2) \equal{} (a^2 \minus{} 2,a \plus{} 3) \equal{} (a \plus{} 3,3a \plus{} 2) \equal{} (a \plus{} 3,7) \equal{} 1 V 7$
$ (a \plus{} 3,7) \equal{} 1$ $ \implies$
$ i)a^3 \minus{} a^2 \plus{} 1 \equal{} r^3$   $ a^3 \minus{} 3a^2 \plus{} 3a \minus{} 1 < a^3 \minus{} a^2 \plus{} 1 < a^3$    $ \implies$  we get any solution.
$ a^{3} \minus{} 3a^{2} \plus{} 3a \minus{} 1\geq a^{3} \minus{} a^{2} \plus{} 1$ $ \implies$ $ 3a\geq 2a^{2} \plus{} 2$ we get any solution
$ a^{3}\leq a^{3} \minus{} a^{2} \plus{} 1$ $ \implies$  $ a\leq 1$
$ a \equal{} 1$ $ \implies$  $ x \equal{} 2 ,y^{3} \equal{} 3$ $ \implies$ we get any solution.
$ a \equal{}0$   $ x \equal{} 1 ,y^{3} \equal{} 1$  $ (1,1)$
$ ii)(a \plus{} 3,7) \equal{} 7$  $ \implies$   $ (a,7) \equal{} 1$ $ \implies$
$ a^3 \minus{} a^2 \plus{} 1 \equal{} 7r^3 ,a^2 \plus{} a \plus{} 1 \equal{} 49m^3$   or   $ a^3 \minus{} a^2 \plus{} 1 \equal{} 49r^3 , a^2 \plus{} a \plus{} 1 \equal{} 7m^3$

[quote="cnyd"]$ x \minus{} 1 \equal{} a$  $ \implies$   $ a^5 \plus{} a \plus{} 1 \equal{} y^3$   $ \implies$
$ (a^2 \minus{} a \plus{} 1)(a^3 \minus{} a^2 \plus{} 1) \equal{} y^3$:[/quote]
That is wrong. Because $ a^5\plus{}a\plus{}1\equal{}(a^2 \plus{} a \plus{} 1)(a^3 \minus{} a^2 \plus{} 1)$.


inequalities
AMC
AIME
AMC 10
function
MATHCOUNTS
geometry

Look at the topic title.

[b]EDIT:[/b]
Please post the person's signature, as they are subject to change.

Hmm hw due on monday, to do it on sunday night!

Absolute value inequalities are evil.  :ninja:

Lesse. It's currently 9:15. I "started" my english paper at noon. Wow. (true story)

GAH! I wish I didn't make any stupid mistakes!

that only applies to inequalities between people in our world

I'm doing a newscast project on Archimedes and his fondlings with the transcendental number pi.  :lol:

true story also

you signature is weird.  i dont get it.  is it supposed to be funny?

Uhhh... no he won't? But he knows exactly what he would do if he were to hypothetically kill me.

haha and they will also never learn basic cliches.

well too bad, b/c your mom is a muffin man..

[quote="kimmystar94"]your [b]mom[/b] is a muffin [b]man[/b]..[/quote]

Um....

--sig response--
Eh whatever. KILL KILL KILL!

oh.. don't worry..
i once lost after like three moves...

I once lost after four moves. Thank goodness I didn't lose after two moves.

--sig--

It's [i]awesome[/i]. It means "some awe". Very easy to remember. keep trying on that speech!

yeah. Found out a way to copy images to stick them into your avatar.

eh..
i still haven't improved on chess...

stop reading that stupid article on the dangers of weight loss!!

the article is about the dangers of weight gain...
smarty pants...

--------sig---------

I still haven't gotten better at chess...

yes..
why.. is a change of signature not good?

----sig----

i suck//

Okay, this is getting out of hand  :mad: .


geometry
inequalities
circumcircle
geometry unsolved

Let $ ABC$ be a triangle，AB≤AC，the $ excircle$ and $ incircleI$ ⊙Iof $ triangle ABC$ cut $ BC,AC,AB$ at $ D,E,F,M$ .
⊙O is the circumcircle of $ triangleABC$，$ L$ is the midpoint of $ OI$，$ LN$ vertical $ EF$ at $ N$，$ MN$ intersect $ AB,AC$ at $ P,Q$ .

Please Prove：SΔBPN≥SΔCQN，and find out when the equality holds .

[quote="lym"]Let $ ABC$ be a triangle with $ AB \le AC$, its incircle (centers at $ I$) touches $ BC,CA,AB$ at $ D,E,F,$ respectively,and excircle touches $ BC$ at $ M$. Denote by $ O$  the circumcenter of $ \bigtriangleup ABC$，Let $ N$ be the foot of perpendicular to $ EF$ through the midpoint of $ OI$. Line $ MN$ intersects $ AB,AC$ at $ P,Q$,respectively.

Show that $ S_{\bigtriangleup BPN} \ge S_{\bigtriangleup CQN}$,and find the conditions the equality holds. where $ S_{\bigtriangleup XYZ}$ denotes the area of triangle $ XYZ$ .[/quote]


Who took the exam? What did you think of it?

I thought the MC was sort of hard, and the free response was very reasonable.

i took the physics c exams.. but i did look thru the B free response after.. it was really easy.. think i'll take the test w/o taking the class next year... haha.. yep... cuz im so confident..


[quote="13375P34K43V312"][hide]Basically we just have $3a\equiv4\pmod{10}$ and 8 is the only thing that works here so 8.[/hide][/quote]
That's what I thought right away... :P

[quote="13375P34K43V312"][hide]Basically we just have $3a\equiv4\pmod{10}$ and 8 is the only thing that works here so 8.[/hide][/quote]
But $238+528+368=1134\neq1124$...so there are no solutions... :wink:

[quote="tridecalogism_lover"]another mathcounts prob...

+++++++++++++++++++++++++++
what units digit a will make the addition problem correct?

        23a
        52a
      +36a
     ________
       1124[/quote]
[hide]$230+520+360+3a=1124$

$3a=14$

$\texttt{no solution}$[/hide]

[quote="Rafaelloaa"]it's easy

[hide]8[/hide][/quote]
That's wrong.

Nice question, especially if one is rash and doesnt' check his work.


geometry
geometric transformation
rotation
linear algebra
matrix
analytic geometry
rotation matrix

A glass with the form $z=x^{2}+y^{2}$ and $z\leq 1$ is filled with water to a certain height $h$.
If the glass is tilted $45^{o}$, the water is just about to stream out. Determine $h$.

[hide="answer"]
$h=1/4$
[/hide]

I guess you should use the rotation matrix here,
http://en.wikipedia.org/wiki/Rotation_matrix

Since two of the quantities (z,y coordinates) are zero and the other is rotated 45 degrees, I dont this its so messy.

first of all thanks.

Rotation matrix, ok, could you specify how this would be done?

Is there any other way? Like Rotate and se where the water surface lies and dtermine the equation for this water surface and the rotate back and consider it as a problem of plane intersect paraboloid.....?


Summation
number theory
equation
IMO Shortlist
IMO Longlist

A given natural number $N$ is being decomposed in a sum of some consecutive integers.

[b]a.)[/b] Find all such decompositions for $N=500.$

[b]b.)[/b] How many such decompositions does the number $N=2^{\alpha }3^{\beta }5^{\gamma }$ (where $\alpha ,$ $\beta $ and $\gamma $ are natural numbers) have? Which of these decompositions contain natural summands only?

[b]c.)[/b] Determine the number of such decompositions (= decompositions in a sum of consecutive integers; these integers are not necessarily natural) for an arbitrary natural $N.$

[b]Note by Darij:[/b] The $0$ is not considered as a natural number.

If $N=2^{\alpha}M$, where $M$ is odd and the total number of divisors of $M$ is $d(M)$ then exists $d(M)$ such decompositions in sum of positive integers.
If $N=2^{\alpha} $ prod($p_i^{\alpha_i}$) , where each $p_i$ is odd , $N$ is written as sum of $b+1,b+2,...,a$ then we get $a-b$ and $a+b+1$ as divisors with different parity of $2N$ and having the product exactly $2N$. So:
$a=2^{\alpha}$prod($p_i^{\gamma_i}$) + 1/2 prod($p_i^{{\alpha_i}-{\gamma_i}}$) ,
where any ${{0\leq\gamma_i}\leq\alpha_i}$.
It is easy to check that each two of these possible values of $a$ are different. The desired answer is:
prod(${\alpha_i}+1$).

small latex hint: you can use \prod to make a product sign $\prod$, or for example \prod_{i=1}^n a_i for $\prod_{i=1}^n a_i $ ;)

...did you notice the difference between a bad programmer and a good one?
1st always believes that a 1Kb has 1000 bits, while 2nd thinks that 1Km has 1024 meters  :)

Notice the fact that for every [i]decomposition[/i] of $N$ which has negative numbers, it is basically an extension of a decomposition having only natural numbers, and vice-versa, that is for every [i]decomposition[/i] of $N$ having positive numbers, we can make a decomposition of $N$ having negative numbers.
So now we consider only the [i]decompositon[/i] of $N$ having positive numbers.
Let the decomposition be $$N = x + (x-1) + (x-2) \dots + (y+1) + y \implies N = \frac{x(x+1)}{2} - \frac{y(y+1}{2} \implies 2N = x^2 - y^2 + x - y = (x-y)(x+y+1)$$
We want that $N$ has to be factorised into  two factors  such that one of them is even, one of them is odd.
Hence, we can simply find a general answer and we are done!