Who will be the top-performers at IMO'2009?

Russia, Poland, Taiwan, Thailand, and Iran are all sending six fresh students.
The teams of China, USA, South Korea, Bulgaria, Vietnam are not known/public yet.

However, there are several students returning who made gold at IMO'2008:
37 István Tomon (Hungary)
36 Umut Varolgunes (Turkey)
35 Un Song Ri (North Korea)
35 Lisa Sauermann (Germany)
35 Makoto Soejima (Japan)
35 Melih Ucer (Turkey)
31 Luka Milićević (Serbia)

And there are lots of students who scored a strong silver with 25-30 points at IMO'2008:
30 Renan Henrique Finder (Brazil)
29 Dušan Milijančević  (Serbia)
29 Henrique Pondé de Oliveira Pinto  (Brazil)
28 Elena Mădălina Persu (Romania)
28 Jonathan Timur Schneider (Canada)
27 Teodor von Burg (Serbia)
27 Shiro Imamura (Japan)
27 Nazar Serdyuk (Ukraine)
26 Davi Lopes Alves de Medeiros (Brazil)
26 Fehmi Emre Kadan (Turkey)
26 Kazuhiro Hosaka (Japan)
25 Sviatlana Auchynnikava  (Belarus)
25 César Cuenca Lucero (Peru)
25 Cho Ho Lam (Hong Kong)

[quote="Agr_94_Math"]Beware of the Indian Team. There is a genius named Akashnil Dutta in the team. He is just in 10th grade. [/quote]
What age does 10th grade in India mean? Sixteen?

15 or 16. Depends on his birthdate. Also, can anyone tell whether any book is available with questions from math olympiads all over the world?
Thanks ina dvance.

Also Dongyi Wei (China) will be back!
Last year, he made gold with a perfect score.

http://www.imo-official.org/year_reg_team.aspx?code=CHN

Dear [b]test20[/b]
You forgot include our Yegor Klochkov in the list of gold medalists returning this year :D
I wish good luck to all participants :lol:
Hope Kazakhstan will be in top 20

[quote="Agr_94_Math"]Actually the Indian team is fresh this time too.
Beware of the Indian Team. There is a genius named Akashnil Dutta in the team. He is just in 10th grade. He is there in mathlinks too with username Akashnil.Most probably he will be winning a gold for India. There are others too who are very good like Gaurav patil who is also in mathlinks.
Anyway, I want India to be in the top 20 this time.
I think the winner is going to be CHina this time again.

All the best for all IMO contestants of 2009.[/quote]

I agree he is sure to win a Gold! :)  :lol:

i think that Teodor Von Burg will take gold because he was the first in Balkan Math Olympiad this year.

All students of Vietnam team this year are new.But I'm sure they will try their best and I hope one will become a top performer.(of course "they" not included me) :P

I hope Bo Lin (linboll) can become top-performer at IMO 2009, I believe he will achieve a perfect scores!

I think the winner is going to be China again  :)
I wish to our team (greece) the best wishes.
Dimitris Papadimitriou(aka bilot4)took gold at Balkan Mathematical Olympiad and i think he can achieve this and at the IMO...
Good luck guys :)
Write for your pleasure :P

[quote="KenHungKK"]Hope that Cho-ho Lam do good this year, and get a Gold medal. This is his last year before university. And hope that Hong Kong team, despite I am in it, can get into top 20.[/quote]
[quote="brianchung11"]Ya hope that Cho-Ho Lam rocks this year~~~ (in the 50th~~) he's the top among the Hong Kong team in the last two IMO~~[/quote]
You two share the same hope with me  :P

Of course there are the top-countries like China and Russia that always perform well.
It is no fun to predict that they will also do well at IMO'2009.

Here is my prediction for some countries that are not consistently at the top,
but that will perform extremely well at IMO'2009 and end up in the top-20:
* Serbia
* Turkey
* Thailand
* Brazil
* Peru

[quote="enndb0x"]How old are you Dimitris X ,aren't you going to IMO ?[/quote]

I'm sixteen years old...
Unfortunately no,i'm not going to IMO but i'll try my best the next(and my final) year to be there :) .
I wish you good luck too enndbox :wink:

[quote="nekojco"]this is my first time to see a math forum like this one... I'm still 14 so I'm not going to IMO, but I'm participating in the JBMO (junior balkan math olimpiade) so i hope I'll get in next year   :D[/quote]
from where are you???
Albania is bringing a 5th grade in JBMO.

[quote="enndb0x"]Who is he ? Is he the one who participated in $ Plus$'s magazine olympiad with [b]7th[/b] grade students?

[i] I won two times 2nd place in Plus's olympiad  :P [/i][/quote]

Yes he is!!!

Wrong.

[size=75]China: Gold medals: 101, Silver medals: 26, Bronze medals: 5, Honourable mentions: 0.

Sum: 132

Hungary: Gold medals: 74, Silver medals: 138, Bronze medals: 77, Honourable mentions: 4.

Sum: 293

USSR: Gold medals: 77, Silver medals: 67, Bronze medals: 45, Honourable mentions: 0.
Russia: Gold medals: 65, Silver medals: 28, Bronze medals: 9, Honourable mentions: 0.

Sum: 291

Romania: Gold medals: 66, Silver medals: 111, Bronze medals: 88, Honourable mentions: 2.

Sum: 265

Bulgaria: Gold medals: 50, Silver medals: 89, Bronze medals: 88, Honourable mentions: 1.

Sum: 228

USA: Gold medals: 80, Silver medals: 96, Bronze medals: 29, Honourable mentions: 1.

Sum: 206

UK: Gold medals: 34, Silver medals: 82, Bronze medals: 106, Honourable mentions: 11.

Sum: 233

Note that even former Yugoslavia has more medals than China:

Yugoslavia (1963-2002): Gold medals: 6, Silver medals: 46, Bronze medals: 96, Honourable mentions: 7.

Sum: 155
[/size]
It's worth reminding that China participated in 23 IMOs (so their avg is 5.7 medals), Hungary in 48 (avg 6.1 medals), Russia + USSR in 46 IMOs (avg 6.3 medals- remember that first 21 IMOs had 8-member teams participating), and Romania in 49 IMOs (avg 5.4 medals). US avg is 5.9 since they participated in 35 IMOs, and UK in 41 so their avg is 5.7.

And in the end, summing up all medals isn't showing much (summing weighted medals - multiplying G,S,B medals with pondered coefficients would be more indicative), but it is interesting.

[quote="b555"][quote="nik"]Did anyone count which countries have most IMO medalists ?[/quote]

thats  china[/quote]

Not China

Because China didn't attend IMO until 1986.

I think it is more worth comparing the numbers of gold medalists, since top countries such as China Russia will always have 6 medalists.

According to hsiljak, China has lower avg than Russia. That's because in 1986, China only sent two participants for trial, and this affected the avg.

[quote="nik"]These statistics are really interesting but I meant

which countries have most medalists in their team [b]this year[/b] :)[/quote]

lol I bet at least 20 countries are going to win 6 medals...

[quote="nicetry007"]If the problems on day 2 are similar in difficulty level to the ones on day 1, then we might have a lot of perfect scores unlike last year. I am curious as to what tomorrow's paper would be like. My guess is
(i)geometry
(ii)algebra(inequality/functional equation/polynomial)
(iii)combinatorics[/quote]

Guesses were here: http://www.mathlinks.ro/viewtopic.php?t=283543&start=40

I know what you mean now.... I think most people would understand "medalists" in another way...

I've no idea about the answer, but at least China is never the country you are looking for .,..

[quote="nik"]aaah, mathematicians find it hard undertanding anything except formulas   :roll:

:rotfl:

having medalists in their team means

1) Person who has won medal (in previous years)
2) Is contestant this year

is it very hard ?  :idea:[/quote]

[quote="nik"]aaah, mathematicians find it hard undertanding anything except formulas   :roll:

:rotfl:

having medalists in their team means

1) Person who has won medal (in previous years)
2) Is contestant this year

is it very hard ?  :idea:[/quote]

That was the question? Oh...  :lol:

Georgia: http://www.imo-official.org/year_reg_team.aspx?code=GEO

Thank you for making me go through all participating countries, since I've seen so many people I remember from my IMOs... ah, nostalgia...


A square table is positioned on an un-level surface so that all four legs of equal length are sitting firmly on the surface, What is the least number of degrees (the amount is greater than 0) that you can rotate the table and the table will still be stable?

How would you do this problem?

Weird question.  If the surface isn't level it could be anything, so the only thing you know about it is that the four spots the table is currently resting on are coplanar.

are you sure?
i think you may have forgotten to include some info......
?? :huh:

[quote="mr. math"]well i think i wrote the problem right. the only thing is that its a multiple choice problem. Here are the choices:

A) $45$ degrees
B) $90$ degrees
C) $360$ degrees
D) answer not given
E) not enough information

Note the answer is not E.[/quote]

Well, if you rotate it 90 degrees, you are guaranteed that it will be there, because each leg now takes care of the previous one.

45 is not necessarily true.

You can't really tell though, since you could move it 1/2 degree and it still fits.

well it says garaunteed...it's 90 because it asks for smallest, so you cross out 360.

-jorian

[quote="mathgeniuse^ln(x)"]

You can't really tell though, since you could move it 1/2 degree and it still fits.[/quote]

Hmm I'm skeptical of that statement. I think that might not always be the case, I haven't thought about it so much.
90 degrees seems like the most reasonable answer.

well, i think neal's saying that if there's just one inch or so of table on each of the legs, and you just move it half inch, it's still stable

to neal:remember, it says garaunteed, so you don't know if you're gonna have that 1 inch leverage

note:measurements aren't actually real, just part of what i'm saying

-jorian

actually, jhredsox, the problem itself never actually
states "guaranteed". however, it may be good to make
that assumption, else the answers could get very
unreasonable.

yes, on a mathcounts test they would be more clear (and it wouldn't be multiple choice :P )

-jorian

The answer is $90$ degrees. I checked.  The problem wasn't written well from the start though.

It also doesn't say where the legs are. If they are not on the corners, then 90 might not work.

What are you talking about?

The table is firmly in place now.  Good.  Lift it up and rotate it $90^{\circ}$.  It is exactly the same.  Put it back in place.  It should still be firmly in place because we assume it is perfectly square, so it has perfect rotational symmetry.

"a square table" somewhat implies the legs are at the corners and the center of gravity is in the middle.
-jorian

Well, sometimes these things tricks you by making you think that something is implied but is not really.
But, in this case, the legs are probably implied to be on the corners... :blush:

ya, but good point

i have a feeling this problem wasn't written exactly like where it was copied from

-jorian


[img]http://www.mathlinks.ro/viewtopic.php?mode=attach&id=14461[/img]

 :(

[hide="Answer."]A rectangle for wihich the lines $ l_1$ , $ l_2$ are diagonallines and $ P$ belongs to it.[/hide]


Are Theta Geometry and Circles and Polygons from past tests good practice for future geometry tournament competitions in our own state?

I have a holiday that I can practice some for an upcoming tournament in 14 days.  :D


superior algebra
superior algebra unsolved

Consider a group $ G$ acting transitively on a set $ X$ with prime cardinality. Show that any $ H \triangleleft G$ either acts trivially or transitively on $ X$.

All the stabilizers are conjugate, so the size of the intersection of a stabilizer with $ H$ must be constant.  This means that all of the orbits are the same size.


can u please explain how u did that? thx

[hide="Formula"]A zero is added whenever a number is multiplied by 10, or 2 and 5. Since there are always enough 2's, we look at the amount of 5's. For every 5, we add a zero. The number of zeros at the end of $n!=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{5^{k}}\right\rfloor.$[/hide]

It's a sigma (Greek letter). In mathematics, it is the sum symbol. $\sum_{k=1}^{n}k$ means you start with $k=1,$ do the indicated operations (in this case, just add $k$), and repeat the process until you get to $n.$

$\sum_{k=1}^{5}k+1=(1+1)+(2+1)+(3+1)+(4+1)+(5+1)=20$

well, for every multiple of five, there will be a five.  however, we haven't counted the second multiple of 5 in each multiple of 25.  or the third in each multiple of 125.  and so on.  thus, we count (multiples of 5)+(multiples of 25)+(multiples of 625)... to get the answer

[quote="robinhe"]Please don't explain this problem with such complexity in this forum.[/quote]
Sorry. :ninja:

Anyway, to find the number of zeros at the end of $n!$ without any formulas, you have to take every number $\leq n$ and see how many fives there are in that number's prime factorization. The sum  you get is the number of zeros at the end.

For example, take $20!$.

$20,\,15,\,10,\,5\Rightarrow1\mbox{ each}\Rightarrow4$

$1\ldots4,\,6\ldots9,\,11\ldots14,\,16\ldots19\Rightarrow0\mbox{ each}\Rightarrow0$

$\therefore20!$ ends in $4+0=\boxed{4}$ zeros.

ok i see what ure talking about now. But the part i don't understand now is in your original answer where u did:
920/625, 920/125... and so on...
I see the part where u find all the fives...so it would be like the fives in 915, 910, 905....
so the part i don't get is where did u get the denominators (625, 125....)?

Robinhe said we shouldn't discuss that here, so I can explain in PMs if you don't understand daermon's explanation.

[quote="xxxxx"][quote="robinhe"]Please don't explain this problem with such complexity in this forum.  You may explain it as so:

We want to find the number of 5's in $920!$ since $5\times 2$ gives a 0 and obviously there are more powers of $2$ than there are $5$.[/quote]

ok, i can understand that... but can u go on wit the explanation please? thx[/quote]

Mr. Frost's tips and tricks gives a great explanation:

[quote]Question :  How many zeros does 238! end in?

Solution : Since in order to end in zero, a number must be divisible by 10=5x2, each zero is the result of the number having a factor of both 5 and 2.  Since every other number is even there are plenty of 2’s as factors, so we need to see how many factors of 5 there are.  First we know 238! means factorial, which means 238x237x236x……1.  It would be impossible to do all that out, so we just look for factors of 5.  Since every 5th number has a factor of 5 we start with 238 divided by 5 = 47.6, so there are 47 numbers that have a factor of 5.  However, if a number has a factor of 25, it has 2 factors of 5.  Every 5th factor of 5 would then also be a factor of 25, so we can divide 47 by 5 to get 9.4, so there are 9 additional factors of 5.  We also know though that 53= 125, so we need to check for that.  So we divide 9 by 5 to get 1.8, so there is 1 more factor of 5.  Since 5 doesn’t go into 1, we are done.  There are 47+9+1= 57 factors of 5 in 238!, therefore it ends in 57 zeros!!
 	Work :  238 = 47.6, 47 9.4, 9 1.8,        47+9+1 = 57
[/quote]

Sorry if the formatting is a little bad.

Well...here's an image:

[img]http://i96.photobucket.com/albums/l167/adasarathy/frosttrick.gif[/img]

[quote="DaReaper"]What  pie says is very confusing the easy way is..

for any number n!...

first take the number and divide it by 5, (920/5) = 184

then take the number and divide it by 25, (920/25) = 36 (discard all decimals)

then divide by 125, (920/125) = 7

the by 625, (920/625) =  1

basically you divide n by the exponential outcomes of 5 (5,25,125,625...)[/quote]
1. Don't revive topics > 1 month old.

2. What you're saying is exactly what the formula does, if you think about it. :wink:

Nice job I_Like_Pie, that's a great formula.  However, people were trying to take your exemplary formula and explain it in their own terms.  No need to get angry.  Just live with it.


[quote="Tojiddin.S_90"]$2/gec a/bc+b/ca+c/ab For all a,b,c?$[/quote]
I don't understand . can you post your problem again? :blush:

[quote="Tojiddin.S_90"]$2\geq a/bc+b/ca+c/ab$ For all $a,b,c$?[/quote]

No. Counterexample: $a=b=c=1$

In fact no such inequality $k\geq \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ba}$ holds because the sum grows infinitely large when $a\mapsto \infty, b=c=1$


In a $ 40 \times 50$ array of control buttons, each button has two states: on and off. By touching a button, its state and the states of all buttons in the same row and in the same column are switched. Prove that the array of control buttons may be altered from the all-off state to the all-on state by touching buttons successively, and determine the least number of touches needed to do so.

I think that this is very similar to this problem:

http://www.mathlinks.ro/viewtopic.php?p=506217#506217

Generalization:

Consider an array of size $m\times{n}$. If both $m$, $n$ are odd then minimum number of steps = $min(m,n)$ obtained by touching each of the buttons along the side containing the $min(m,n)$ buttons.

If only one of the sides is odd say $m$, then minimum number of steps = $m$, obtained by touching each of the buttons along the side containing m buttons.

If both $m$ and $n$ are even, then one way to alter all lights to all-ON state is to touch each of the $m.n$ buttons EXACTLY ONCE sequentially going from one row/column to the next or in any other order.

We shall prove that minimum number of steps = $m.n$ when both $m$, $n$ are even.

To prove the above claim it suffices to prove that no button can be left untouched.

Assume that a button is left untouched, then this button will be turned to ON state only if there are a total of Odd number of touches along the row and column containing this button.

So either the row or column (say row) containing this button must have odd number of touches and the other (say column) must have an even number of touches. Let us label this row as $R$ and column as $C$.

Now, each button in $C$ needs to have Odd number of touches in total, so each row of the matrix obtained after excluding $C$ and $R$ must have an Odd number of touches.

This matrix has the dimensions of $m-1\times{n-1}$, so total number of touches in this matrix must be odd $-- (1)$

Now, each button in $R$ needs to have Odd number of touches in total, so each column of the matrix obtained after excluding $C$ and $R$ must have an even number of touches, so total number of touches in this matrix must be even $-- (2)$

From $(1)$ and $(2)$ we have a contradiction.


a_0,a_1,a_2,... and b_0,b_1,b_2,... are arithmetic progressions whose common differences share no factor. prove that for each integer n, there exists positive integers i,j such that  a_i - b_ j =n.

Am I wrong or is a direct consequence of Bezout's theorem?

Pierre.


Angle A of triangle ABC is 40 degrees.  The bisectors of the exterior angles of B and C meet at D.  What is the measure of angle BDC?

if angle A = 40 degrees, then B + C = 140, then .5B + .5C = 70.
Angle D = 180 - (.5B + .5C) = 180 - 70 = 110

Josh Frost

It says [i]exterior[/i] angles of B and C so the sum of that is 360-(180-40)=220. Bisect that, you get 110, the other angle is 180-110 (triangle has 180 degrees), therefore angle D=70.


topology
advanced fields
advanced fields unsolved

Let $ l_1,l_2,....l_k$ be distinct lines in $ R^n$ passing through the origin and
 $ A \equal{} \cup_{i \equal{} 1}^k l_i$

Compute the fundamental group of $ R^n \setminus A$

Let's consider more generally the problem with $ \mathbb{R}^n$ with $ k$ rays emanating from the origin removed (if you have $ k$ lines, simply replace $ k$ with $ 2k$). We suppose $ k > 0$, in which case we can write the resulting space as the product of $ S^{n\minus{}1}$ minus $ k$ points with $ (0, \infty)$. The latter space is contractible, so our original space is homotopy equivalent to $ S^{n\minus{}1}$ minus $ k$ points. Removing one point from $ S^n$ yields a space homeomorphic to $ \mathbb{R}^n$ by projection, and so removing $ k\minus{}1$ more points from this homeomorphism yields that $ S^{n\minus{}1}$ minus $ k$ points is homeomorphic to $ \mathbb{R}^{n\minus{}1}$ minus $ k\minus{}1$ points. This space in turn deformation retracts to a $ k\minus{}1$-fold wedge of $ S^{n\minus{}2}$. Hence, the original space is homotopy equivalent to $ \bigvee_{i\equal{}1}^{k\minus{}1} S^{n\minus{}2}$. In particular, this allows us to calculate the fundamental group: $ n > 3 \Rightarrow \pi_1 \equal{} 1, n \equal{} 3 \Rightarrow \pi_1$ is the $ k\minus{}1$ fold free product of $ \mathbb{Z}$, $ n < 3$ implies the space isn't connected and this isn't really terribly exciting anyway.


MATHCOUNTS
function
summer program
MathPath
geometry
3D geometry
AwesomeMath

I know since TI is a sponsor of MathCounts, they give free calculators to Nationals participants.  Does anyone know what it will be this year?  I really want a TI-89 Titanium.  I already have a TI-84 Silver...

wow, they do!!?? grrrr, i wish i made it

so do i. i guess when we see eachother, you can give me programs (my stupid computer for some reason won't accept it or something).

-jorian

O_o rich kids. I use a 20+ year old ancient calculator (no kidding) which I believe was very technologically advanced for the time because it even had logs/exponent functions and stuff like that. The buttons don't press down right, though :( .

For math competitions I borrow someone's TI-84+. The only reason I use it is because the interface is MUCH easier to work with than a TI-30A (which I would borrow from my sister if I couldn't get the TI-84+), because it doesn't have any programs so that part is useless.

Last year the competitors got TI-84 Silvers. Why don't you ask your coach  :roll: ?

[quote="Ubemaya"]O_o rich kids. I use a 20+ year old ancient calculator (no kidding) [/quote]

Me too. :P
Mine works just fine.

Are graphing calculators allowed? I borrowed my coach's 2000 competitions and in the booklet it says no programmable or graphing calculators, but at the chapter competition, there were people using graphing calculators.

They let you use Graphing from what I could tell (just see the instructions for the 2007 Chapters).

The calculator that I use is roughly the same size as a Graphing Calculator, but it can only perform basic functions such as adding, subtracting, multiplying, dividing, exponents, and square roots.

[quote=".:Logic:."]OMG You get a free calculator??- I need to get one anyways (borrowing a TI-83 now), this just makes me want to get to nationals even more! :lol:[/quote]

haha I think if you land within the top 4 at Nats then you get some savings bonds and laptops?

Well... I don't really expect to go there.. but I've been kinda hopefull about even going to nationals, ever since I took last year's state (last night) and got my state's exact cut-off :lol:

BTW which AoPSers here got in the top 20 at Nats? (last year)

[quote=".:Logic:."]Well... I don't really expect to go there.. but I've been kinda hopefull about even going to nationals, ever since I took last year's state (last night) and got my state's exact cut-off :lol:[/quote]

huh? i thought to advance into nats you had to place within the top 4?

[quote=".:Logic:."]Yes... exactly.

You said the laptops are top for the 4 AT NAIONALS....[/quote]

no i meant when you took last year's state you got exactly at the cutoff...what do you mean? i thought it wa the top 4 scorers to get into nationals

I meant the lowest score I would need to get last year to get into the nationals:
The score that #4 at state got in Ohio last year.

[quote="Ubemaya"][quote="jhredsox"]Not trying to stereotype, but didn't you go to MathPath last year?

jorian[/quote]

So what if I did? It's not like I understood anything...[/quote]

You had to pay some $\$$ for it

[quote="yhh3002"]they give you TI-83-gold edition ones[/quote]

Really?  Last year they gave out TI-84 Plus Silver Editions (I didn't make Nationals, my dad was a coach).

I have a TI-89, but pretty much any graphing calculator is good as long as you can program certain things to solve certain types of problems.

I have a TI84+Silver, but not as an award, just because I desperately need one at the 2006 School competition.

[quote="frost13"]I am 99% sure there is no such thing as TI83 GOLD.  They have given out TI84+Silver the last two years.  The history of what they've given has been discussed countless times here.  Just wait until they announce it or you get to Nationals.[/quote]

LOL, someone tried to prank call me once saying that I had won a TI-83 Gold Edition :roll:

Good luck everybody :)

[quote="happyme"]You get all the way to nationals, and all you care about is the free calculator you're getting?!

Morals man![/quote]
that was one of the reasons i even wanted to go through the suffering of studying in the first place

normally i never study...

You get a TI-84 + silver edition. Since I don't have a graphing calculator, it would be awesome to get it from nats  :wink:
And u get to see a rodeo,  :huh: which is really weird, and free trip to Texas :D and a trip to the zoo. Thats probably the fun part of Nationals where everyone would look up to.

Muahahahaha!!! I have a TI-89 Titanium!! Everybody at school makes fun of me because they all have TI-84s or Ti-83s, which are all basically the same. But I know they're just jealous. :D

[quote="cat810930"]I have a Ti-83 calculator, but I don't know how to use any functions besides the ones on a scientific calculator. I also don't have the cable to connect it to a computer, which makes it almost useless as a graphing calculator... :help:  :(  :oops:[/quote]

You should probably ask a teacher or someone who can actually press the buttons on the calculator and show you how to use it.

[quote="cat810930"]I have a Ti-83 calculator, but I don't know how to use any functions besides the ones on a scientific calculator. I also don't have the cable to connect it to a computer, which makes it almost useless as a graphing calculator... :help:  :(  :oops:[/quote]

Isn't there a guide or handbook that comes with it when  you buy it? Read some of the chapters and you should know everything about it.

[quote="cat810930"]My teacher isn't exactly good at using a calculator. One time I had to show her how to cube a fraction on a scientific calculator :wink: ... Reading the guide is probably a good idea, now that all my competitions are over :( , I'll have plenty of time to read the guide!
Thanks for the suggestions![/quote]

Yeah through my experiances most teachers are never that good at handling technology...

i have this really really cool brand new 3 year old casio scientific calculator. :roll:

Its pretty funny actually, the names of Ti Calcs. I have a friend who swears he has a TI-89 Plus (I mean, what the heck? I've never heard of such a thing)

umm...guys, this is turning into spam :wink:

please stop

 :) jorian

[quote="asianhottie8293"][quote="jhredsox"]Not trying to stereotype, but didn't you go to MathPath last year?

jorian[/quote]
HAHAHAHAHAHAHA they give u calcs at nats? 4 some reason i think that's rly hilarious.[/quote]
No joke, at nats last year I forgot to bring my calculator.
Thank god they gave us one, my team mates/coach didn't have a spare.
I would've been pwned utterly without a calc.

I've never used a graph thingy calculator :(

I got this thing called: fx-82MS

[quote="jli"][quote="asianhottie8293"][quote="jhredsox"]Not trying to stereotype, but didn't you go to MathPath last year?

jorian[/quote]

hey, which camp is better, mathpath, or awesomemath? HAHAHAHAHAHAHA they give u calcs at nats? 4 some reason i think that's rly hilarious.[/quote]

i think awesomemath

mathpath was interesting...  i went there 2 years ago and we talked about kinda random topics (math history, rubiks cubes, spherical geometry etc.)

maybe its different now[/quote]

I thought MathPath was better, but maybe it was because I didn't go to AwesomeMath... :harhar: Seriously, though, it wasn't that random, and we got to meet a bunch of famous math people :)

Forgot your calculator at nats :rotfl:

I'm going to lock this thread, because we now have a definitive answer, and the discussion is getting way off-topic.

Source: [url]http://www.mathcounts.org/webarticles/articlefiles/815-07_Awards.pdf[/url]

[quote="MATHCOUNTS"]All 228 National Competitors and 57 Coaches receive:
· TI 84+ Silver Edition Calculators
· Commemorative competition hats
· All-expense-paid trip to the Lockheed Martin MATHCOUNTS National Competition – 2007 in Fort Worth.[/quote]


$ 2)$ If $ x$ and $ y$ are positive integers such that $ x\plus{}x\plus{}x\plus{}x\plus{}x\plus{}x\plus{}x\plus{}x \equal{} y\plus{}y\plus{}y\plus{}y\minus{}y$, find the minimum possible value for $ x\plus{}y$.

I wasn't really sure what to do on this one. I just wrote down $ 8x \equal{} 3y$, and then couldn't think of what to do to get the right answer.

[hide="2"]$ x\plus{}x\plus{}x\plus{}x\plus{}x\plus{}x\plus{}x\plus{}x\equal{}y\plus{}y\plus{}y\plus{}y\minus{}y$

$ 8x\equal{}3y$

$ (3,8)$

$ x\plus{}y\equal{}11$[/hide]

I'm pretty sure there is a typo in the answer or in the problem.  (Maybe that last minus sign is supposed o be a plus sign?  Then we'd get 13.)

[quote="Bicameral"]The answer key says 13. The choices are 10, 13, 8, 5, and NOTA. Typo or something?[/quote]

Original probably was supposed to be $ x\plus{}x\plus{}x\plus{}x\plus{}x\plus{}x\plus{}x\plus{}x\equal{}y\plus{}y\plus{}y\plus{}y\plus{}y$ for an answer of 13- simple typo of a plus or minus sign...


When a satellite is a distance R from the center of Earth, the force due to gravity on the satellite is F. What is the force due to gravity on the satellite when its distance from the center of Earth is 3R?

(1) F/9    (3) F
(2) F/3    (4) 9F

[quote="Interval"]Can you show me how to use the formula you provided?[/quote]
The SI unit of force is Newtons, which is defined as $\frac{kg\cdot m}{s^{2}}$ ("kilogram meters per second squared") People didn't want to write that out so they changed it to Newtons (after Isaac Newton). The formula I used is just the definition of a force unit. Since meters is on the top of the fraction, it is a direct variation ($y=kx$). I doubt you would ever need this formula on a competition but since force varies directly (it should be given) with distance, you divide it by 3.

[quote="Quevvy"][hide]Knowing that force is measured in Newtons (and $N=\frac{kg\cdot m}{s^{2}}$) so force varies directly with distance so then since you are 3 times farther away, it will be 3 times less, or (2) F/3[/hide][/quote]

I would have thought it would follow the inverse square law so F/9... :huh:

im not a science expert, but i think it is direct.

Actually, applying Newton's Law of Universal Gravitation would yield that it's $\frac{F}{9}$. Here:

For an object on the surface of the earth, the gravitational force felt by the earth and the object is:

$F = \frac{GM_{E}M_{S}}{R^{2}}$, where $R$ is the radius of the Earth, $M_{E}$ is the mass of the Earth, and $M_{S}$ is the mass of the satellite.

For an object $3R$ from the center of the Earth, again applying Newton, the force is:

$F = \frac{GM_{E}M_{S}}{(3R)^{2}}= \frac{GM_{E}M_{S}}{9R^{2}}= \frac{F}{9}$.


Given: assume the number of items sold (n) depends linearly on the price charged for that item (x).  From this, you will show that profit (p) depends on price charged.

A college bookstore has been selling tshirts for a year.  During the fall term, 100 were sold at $5 each.  During the Spring term, the price was raised to $7 and only 80 were sold.

a. write an equation showing the dependence of price charged on the number of shirts sold.  what does this equation tell you in general?  graph.

b. using the relationship in part A, how many tshirts would be sold during a term at $3.50 each?  What should the cost of a shirt be in order to sell 45 in a term?
c. suppose it cost the bookstore $3 to buy each shirt.  calculate the total profit where they may charge $5 $6, $7, $8, $9 and $10.  Then calculate average rate of change in total profit between charging $5 \& $6, $6 \& $7, $7 \& $8, $8 \& $9, $9 \& $10. Use this data to explain why profit does not depend linearly on price.  Explain in general why profit should realistically depend quadratically on price.

d. generate a formular describing how P depends on x (profit = number sold * price per shirt).  Graph.  Find P intercept; what does this value mean?

e. calculate breakeven points.

f. find price that will maximize profit.  what will the profit be?

g. explain why this information is important to the bookstore.

AoPS is not a forum that does homework for you.  You'd have better luck getting a response if you give us your solution to check or ask about a specific concept that you do not get.

Most of these seem pretty straightfoward.

a, graph your points and connect them, then find the equation of the line.
b, read your graph
c, make a table with the prices, profit per shirt sold, number of shirts sold and total profit

The rest is just reading your new graph. This stuff shouldn't be completely new, so just look at some of the work you've done before, if you get stuck.


function
group theory
abstract algebra
algebra unsolved
algebra

[quote="keira_khtn"]A function f is called [b]Darboux[/b] if X is an interval implying f(X) is also an interval.
Find all Darboux funtion f s.t.:
$f(x+y)=f(x+f(y))$[/quote]
Let $P(x,y)$ be the assertion $f(x+y)=f(x+f(y))$

Let $A=\{f(x)-x$ $\forall x\in\mathbb R\}$
Notation : in the following, the notation $|u,v|$ indicates an interval $(u,v)$ or $[u,v)$ or $(u,v]$ or $[u,v]$ (where $u,v$ may be numbers or $\pm \infty$)

1) $A$ is an additive subgroup of $\mathbb R$
=======================
$A$ is not empty.
$P(x-y,y)$ $\implies$ $f(x)=f(x-y+f(y))$ $\implies$ $(f(x)-x)-(fy)-y)$ $=f(x-y+f(y))-(x-y+f(y))$
So $a,b\in A$ $\implies$ $a-b\in A$
So $a-a=0\in A$
So $0-b=-b\in A$
So $a-(-b)=a+b\in A$
Q.E.D.

2) If $A=\{0\}$, then $f(x)=x$ $\forall x$
========================
This is just the definition of $A$ and $f(x)=x$ $\forall x$ is indeed a solution.
Q.E.D.

3) If $A\ne \{0\}$ is dense in $\mathbb R$, then $f(x)=c$ constant
======================================
$f(\mathbb R)$ is an interval since $f(x)$ is Darboux.
If this interval contains two distinct numbers $a<b$, then it contains $[a,b]$
$P(0,x)$ $\implies$ $f(f(x))=f(x)$ and so $f(x)=x$ $\forall x\in[a,b]$

Since $A$ is dense in $\mathbb R$, let then $u$ such that $0<f(u)-u<b-a$
$a\in[a,b]$ and so $f(a)=a$
$a+f(u)-u\in[a,b]$ and so $f(a+f(u)-u)=a+f(u)-u\ne f(a)$
But $P(a-u,u)$ $\implies$ $f(a)=f(a+f(u)-u)$ and so contradiction.

So no such distinct $a,b$ exist and $f(\mathbb R)=\{c\}$ which indeed is a solution.
Q.E.D.

4) If $A\ne\{0\}$ is not dense in $\mathbb R$, then no solution
===================================
If $A\ne \{0\}$ and not dense in $\mathbb R$, then $\exists \Delta>0$ such that $A=\Delta\mathbb Z$ (remember that $A$ is an additive subgroup of $\mathbb R$)

$\Delta\in A$ $\implies$ $\exists t$ such that $f(t)-t=\Delta$ and then $P(x-t,t)$ $\implies$ $f(x)=f(x+\Delta)$ $\forall x$

Let $f(\mathbb R)=|u,v|$
$P(0,x)$ $\implies$ $f(f(x))=f(x)$ and so $f(x)=x$ $\forall x\in|u,v|$

If $v-u>\Delta$, then :
$\exists a\in|u,v|$ such that$ a+\Delta\in|u,v|$ too
$a\in|u,v|$ $\implies$ $f(a)=a$
$a+\Delta\in|u,v|$ $\implies$ $f(a+\Delta)=a+\Delta$
But $f(a)=f(a+\Delta)$
and so contradiction.

So $v-u \le \Delta$

If $v-u<\Delta$, let $x\in(v,u+\Delta)$
$f(x)\in|u,v|$ and so $f(x)-x\in (0,-\Delta)$ which is impossible since $f(x)-x\in A=\Delta\mathbb Z$

So $v-u=\Delta$

But then $f\left(\left[u+\frac{2\Delta}3,u+\frac{4\Delta}3\right]\right)$ $=\left[u+\frac{2\Delta}3,u+\Delta\right)$ $\cup\{f(u+\Delta)\}$ $\cup\left(u,u+\frac{\Delta}3\right)$ is not an interval
And so no solution.
Q.E.D

5) Synthesis of solutions
=================
The only solutions are :
$f(x)=x$ $\forall x$
$f(x)=c$ $\forall x$ and for any $c\in\mathbb R$


search
modular arithmetic
number theory unsolved
number theory

Solve in the set of natural numbers the following equation $ 3^{x}\plus{}4^{y}\equal{}5^{z}$.

We search for n=1 and for n=2, we have one solution, n=2

for n>2 of course n-2>0
3^n+4^n=(3^2)*3^(n-2)+(4^2)*4^(n-2)<(5^(n-2))*(3^2+4^2)=5^n
so n=2 is the answer

But $ x,y,z$ don't have to be equal (unless you prove it first :) )

[hide="@ MagdalenaW"]
też z Lublina jestem ;)
[/hide]

Looking $ \pmod{4}$ we see that $ 2|x$. Let $ x\equal{}2x_1$ and the equation becomes $ 9^{x_1} \plus{}4^y\equal{}5^{z}$.
Looking $ \pmod{3}$ we see that $ 2|z$. Let $ z\equal{}2z_1$ and the equation becomes
$ (3^{x_1})^2 \plus{}4^y\equal{}(5^{z_1})^2$
$ 2^{2y}\equal{}\left(5^{z_1}\minus{}3^{x_1} \right) \cdot \left( 5^{z_1}\plus{}3^{x_1} \right)\equal{}2^a \cdot 2^b$
which gives a system
$ \begin{cases} 5^{z_1}\equal{}2^{a\minus{}1} \\ 3^{x_1} \equal{}2^{b\minus{}1} \end{cases}$
so the answer is $ (x,y,z)\equal{}(2,2,2)$

[quote="micf"]
which gives a system
$ \begin{cases} 5^{z_1} \equal{} 2^{a \minus{} 1} \\
3^{x_1} \equal{} 2^{b \minus{} 1} \end{cases}$
[/quote]
$ (x,y,z)\equal{}(2,2,2)$ doesn't satisfy your system...
I think it should be
$ \begin{cases} 5^{z_1} \equal{} 2^{a \minus{} 1} \plus{}2^{b \minus{} 1}\\
3^{x_1} \equal{} 2^{b \minus{} 1}\minus{}2^{a \minus{} 1} \end{cases}$


Let a,b,c be sides of a triangle ABC prove that
$ \sqrt{a\plus{}b\minus{}c}\plus{}\sqrt{b\plus{}c\minus{}a}\plus{}\sqrt{c\plus{}a\minus{}b}\leq\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$

AM-GM
$ \sqrt{a\plus{}b\minus{}c}\plus{}\sqrt{b\plus{}c\minus{}a}\leq 2b$

see here http://www.mathlinks.ro/Forum/viewtopic.php?p=451720#451720 .


Given a right-angled triangle with hypotenuse $ c$, altitude to the hypotenuse - $ h_{c}$ and the other to side are $ a$ and $ b$. It is known that $ a$, $ b$ and $ h_{c}$ are naturals and also $ \gcd\left(a,b,c\right)\equal{}1$. Prove that the hypotenuse $ c$ is a perfect square of a natural number.

please check my solution, i'm not sure with it.

[hide]It's well-known that there exist integers $ m,n$ such that $ a\equal{}m^2\minus{}n^2,b\equal{}2mn,c\equal{}m^2\plus{}n^2$. So $ h_c\equal{}\frac{ab}c\equal{}\frac{(m^2\minus{}n^2)(2mn)}{m^2\plus{}n^2}$. But we have $ \gcd(mn,m^2\plus{}n^2)\equal{}1$ and $ \gcd(m^2\minus{}n^2,m^2\plus{}n^2)\le2$. So $ m^2\plus{}n^2\in\{1,2,4\}$. If $ m^2\plus{}n^2\in\{1,4\}$, we are done. If $ m^2\plus{}n^2\equal{}2$, then $ m\equal{}n\equal{}1$, which implies $ a\equal{}m^2\minus{}n^2\equal{}0$, a contradiction. Q.E.D.[/hide]

Ohhh I''m very sorry. It's $ \gcd\left(a,b,h_{c}\right)\equal{}1$ not $ \gcd\left(a,b,c\right)\equal{}1$.

we know that there exists integers $ d,r,s$ such that $ \gcd(r,s)\equal{}1$ and $ 2 \nmid r\minus{}s$ and
$ a\equal{}dr^2\minus{}ds^2$ , $ b\equal{}2drs$ , $ c\equal{}d(r^2\plus{}s^2)$
but we have $ h_{c}\equal{}\frac{ab}{c}\equal{}\frac{(r^{2}\minus{}s^{2})(2rs)d}{r^{2}\plus{}s^{2}}$
since $ \gcd(a,b)\equal{}d$ and $ (h_c,(a,b))\equal{}1$ it implies $ \gcd(d,h_c)\equal{}1$
but we have $ \gcd(r^2\plus{}s^2,2rs(r^2\minus{}s^2)\equal{}1$ and it implies $ r^2\plus{}s^2|d$
if $ r^2\plus{}s^2 \neq d$ then we'll have a prime for which $ q | \frac{d}{r^2\plus{}s^2} \implies q|c$ so we get a prime  $ q |d$ and $ q |c$
which is a clear contradiction so we must have $ r^2\plus{}s^2\equal{}d$ which implies $ c\equal{}d^2$
we r done!


calculus integration limit trigonometry function geometry ratio

Let $ p$ be a positive constant number, find $ \lim_{a\to\infty} \int_0^p |x\sin (ax^2)|\ dx.$

First, I applied $ u\equal{}x^2$

$ \frac{1}{2}\lim_{a\to\infty} \int_0^{p^2}|\sin(au)|du$

Since $ |\sin(au)| \leq 1$, the upper bound of that statement is $ \frac{p^2}{2}$

The function has a period of $ \frac{\pi}{a}$. This means that the interval can be subdivided into a number of intervals of width $ \frac{\pi}{a}$. If you sum all these intervals and take the limit as $ a$ goes to infinity, it forms a Riemann sum equivalent to the integral. Since each of these intervals is equivalent, we need only look at one.

The area under the curve in one of these intervals is $ \int_0^{\frac{\pi}{a}}\sin(au)du \equal{} \frac{1\minus{}\cos{\frac{\pi}{a}}}{a}$
The area of the box surrounding the interval (upper bound) is $ \frac{\pi}{a}$

The Ratio of function:upperbound is $ \frac{1\minus{}\cos{\frac{\pi}{a}}}{\pi}$

Taking $ \lim_{a\to\infty}\frac{1\minus{}\cos{\frac{\pi}{a}}}{\pi} \equal{} 0$, we see that this ratio approaches 0

Taking this ratio and summing it over the entire set of intervals, the value of this given problem is $ 0 * \frac{p^2}{2} \equal{} 0$

Seems right

I made a computational error in mine

$ \int_0^{\frac {\pi}{a}} \sin(au) du \equal{} \frac {2}{\pi}$

Then, moving a little later in my steps

$ \frac {2}{\pi}\cdot\frac {p^2}{2} \equal{} \boxed{\frac {p^2}{\pi}}$

[quote="Gib Z"]Let $ u \equal{} ax^2$, so $ du \equal{} 2ax$

$ I \equal{} \int^p_0 |x \sin (ax^2)| dx \equal{} \frac {1}{2a} \int^{ap^2}_0 | \sin u| du$ since 2ax is positive.

Since $ \int^{k\pi}_0 |\sin u| du \equal{} k \int^{\pi}_0 \sin u du \equal{} 2k$,

$ I \equal{} \frac {1}{2a} \frac {ap^2}{\pi} \int^{\pi}_0 \sin u du \equal{} \frac {2ap^2}{\pi} \frac {1}{2a} \equal{} \frac {p^2}{\pi}$.

$ \lim_{a\to \infty} \frac {p^2}{\pi} \equal{} \frac {p^2}{\pi}$.

I hope im correct.[/quote]
This proof is not correct as  $ \frac {ap^2}{\pi}$is not an integer.

Ahh Well I didn't think I would have to bother to justify that step, too much of a bother :P

If you really want it, just let a go through a sequence as such - define [tex]a_n = $ \frac{n\pi}{p^2}$.

As n--> infinity, so does a_n, and we get our fulfillment that $ \frac{ap^2}{\pi}$ is an integer, namely n.


probability
real analysis
real analysis unsolved

Let $ a \in Z_{ \plus{} }^n, a \equal{} (a_1,a_2,...,a_n) and |a| \equal{} a_1 \plus{} ... \plus{} a_n$.
How many elements has $ A \equal{} \{ a \in Z_{ \plus{} }^n / |a|\leq k\}$ ?? $ n,k\geq0$

$ n^{n\minus{}k}1_{n\ge k}$

[hide="probability style"]
$ A_{n,k}\equal{}\{a\in Z^n_\plus{}: |a| \equal{} k\}$

$ a\equal{}(a_1, ... , a_n ) \to n \le |a|$ for $ j \ge 1,$

$ A_{n,n\minus{}j} \equal{} \emptyset,  A_{n.n} \equal{} \{ (1,1,...,1) \}$

$ |A_{n, n\plus{}j}| \equal{}$ number of ways to put j items in n boxes $ \cdot |A_{n,n}|\equal{} n^j$ [/hide]

$ a\equal{}(a_1,...,a_n)$ and $ a_i\equal{}0$ is corect. also$ a_i$ can be 1,2,...,k. I think you are wrong.


Let $f: \mathbb{R}\rightarrow\mathbb{R}$ a  function such that $f(f(x))=x^3.$ Prove that there exists distinct $a,b,c\in\mathbb{R}$ such that $f(a)+f(b)+f(c)=f(a)f(b)f(c).$

[b] HINT[/b]

[hide]It is easy to prove that $f$ is one to one and that $f^3 (x) = f(x^3)$ . But then the set of  ${ f(0) , f(1) , f(-1)}$  is the same set as of set  $-1 , 0, 1$ and since these values are different , their sum and their product is equal to $0$.
 I hope I have not missed something ![/hide]

  [u]babis[/u]

yes stergiu is right just take $a=f(1),b=f(-1),c=f(0)$


hi guys, i have two questions.
sigma(an) converges and sigma(bn) diverges. is sigma(an+bn) divergent?
and
estimate 5^(1/3) correct to four decimal places.
thanks.

Hint for the first one
$b_n =(a_n+b_n)-a_n$

Now if $a_n+b_n$ converge, and it is given that $a_n$ converges...

The simplest way to do the second one is to start with some fairly good approximation and  make a few steps in the Newton method. If you start with $x_0=1.7$, then one step is already good enough, so a possible answer is $\frac {7413}{4335}=\frac{2471}{1445}$. This seems to be the best you can do with pen and paper fairly quickly (I hope, I haven't made a stupid mistake when counting). Of course, if a computer or calculator is allowed, the problem becomes very easy.


Select a number a and b between 0 and 1 independently and at random, and let c be their sum. Let A, B, and C be the results when a, b, and c respectively are rounded to the nearest integer. What is the probability that A +B = C?

C will either be 0 or 1.  If C is 0 then A and B both have to be 0.  If C is 1 then one of A or B has to equal 1 and the other has to equal 0.

It's equally likely for any of them to equal 0 or 1.

So the probability that they are all 0 is $ \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \equal{} \frac{1}{8}$

The probability A = 1, B=0, and C=1 is also $ \frac{1}{8}$

And the probability A=0, B=1, and C=1 is $ \frac{1}{8}$

So the probability that A+B=C is $ \frac{1}{8} \plus{} \frac{1}{8} \plus{} \frac{1}{8} \equal{} \frac{3}{8}$

Truffles, C can be 2 as well. for example, a=0.9, b=0.9 This means c=1.8 so A=1, B=1 C =2 (nearest integer)

Whoops sorry I actually misread the question.  I thought c was another number picked from 0 to 1.


While reading  the news coverage of the Southern California fires, I have been thinking about the good folks at AoPS, as well as all the others affected there.

The San Diego public TV station KPBS regularly updates a [url=http://maps.google.com/maps/ms?hl=en&ie=UTF8&msa=0&msid=114250687465160386813.00043d08ac31fe3357571&ll=32.990236,-116.930237&spn=0.946815,1.842957&z=9&om=1]googlemaps mashup[/url] showing the locations of the fires and evaculation areas.  Apparently, West Alpine (just west of AoPS location in Alpine) was ordered evacuated on Monday.

We understand that you're all pretty busy with pressing issues, and I'm sure that many AoPSers will join me in sending good thoughts your way.

[quote="sophia"]We understand that you're all pretty busy with pressing issues, and I'm sure that many AoPSers will join me in sending good thoughts your way.[/quote]Thanks a lot we appreciate that.

Yes, I hope that those fires don't get near you or any other AoPSers.  (Really I mean everyone, but thats like impossible seeing that this is the worst fire in California ever.)

Yep, wish you guy's luck in sorting out some of your problems. AoPS ftw!


AMC AIME USA(J)MO USAMO geometry parallelogram FTW

what did you guys get on the aimes???

please say which grade you are in. greatly appreciated.

9th; 8. a poll would be more effective, in my opinion.

6 in 7th grade....same score as 6th grade.

*suicide*

9th grade with a 9.

The funny thing is that in 8th grade, I didn't even qualify. So let's say I got a 0 in 8th grade.

By the property of arithmetic progressions, in 10th grade, I will get 18.

9th Grade, got an 8,
Pretty happy, got 6 last year, so, I improved  :lol:

8th grade, 8.

I got 3 last year so lots of improvements I guess :D

I'll be happier if I made USAMO too :D

9th grade = 7.

Last year I didn't qualify (-1) so by 12th grade I should get 8*4-1= 31 on the AIME.  :D

I guess I improved a lot, but making USAMO would have been cooler.

9th grade: 7
would have gotten 9 if not for stupid arithmetic errors  :wallbash_red:

8th, 9 should be 12 w/o stupid mistakes...:(

I made even more stupid mistakes --- 4 mistakes. Only got 8 with an expectation of 12...

9th, i got a 6.  Hmm...Everyone who posted is doing better than I am... :|

5th and 6th, 5. I epically failed computation this year.

Lets see...

grade 4,5,6
score 0,5,5
so the pattern should be
score 0,5,5,10,10,15,15
grade 4,5,6,7,8,9,10

Yay?

Dude, stop saying that you epicly phailed. It makes me feel bad. Additionally, what makes it worse is that you are a 6th grader and you got a 5 on the AIME...and you also got a 5 in fifth grade. That is beast. And I'm an 8th grader. :mad: And I got a 6.

fourth/fifth/sixth/seventh
0/4/7/[b]10[/b].
AIME0/AIME1/AIME2/[b]AIME1[/b]
Guess: eighth=13 but more likely 12 and ninth=15 (I wish)

I spot something fishy.

Jongao, I feel your pain.

9th, 6 but would be a 9 if I didn't make bad mistakes...

I'm in 9th grade! I got an 8 on the AIME, though I should have gotten number 14 and the one with the 41x+y=2009 triangles.

On the triangles one, The answer was 25 choose 2 times two. But though 25*24=600, I put 625 instead. :(

I hope I make USAMO -crosses fingers- >.<

8th grade, my first AIME, I got an 8 (i used cheap way for the parallelogram one).

I could've gotten the triangles one and #14 the sum of t^n ones, but I didn't try hard enough.

I didn't expect to understand the problems though :D.

BTW: Good job everyone. Remember that >0 is good on AIME :).

I'm in 9th grade and got an 8. Progress from a 7 that I got last year. Probably my first year for USAMO.

Hmm, who put scores >=13, please post up (remember this is grade<=9)

As far as I know pythag011 got an 11 and that's one of the highest (if not the highest) of the 9th and belows.

Well are there THREE people? Hard to believe.

We'd all like to know who you are, if you really did get >=13 and <=9th grade. That's quite amazing.

I doubt 3 people actually scored that high.  Some people obviously *misremembered* or purposefully inflated their scores.

7th, i got a 6 in the AIME.

This year was my first year going AMC 10, AMC 12, and AIME. I don't think it'll be the first year for USAMO though.

I got a 4 as an eighth grader. I got a 4 as a seventh grader as well. Hopefully this incredible streak of good fortune will continue through my high school years.

4 is a really, really high score, right?

4 is a really good score for a middle schooler.
ok, i really don't believe that 8 people got >=13..

[quote="abcak"]I am in 7th grade, got a 4 or a 5 (don't really know which one since I don't have the score, I think I might have guessed one right though), really underpreformed the old tests i did where i was able to get at least 8.   :(  I wasted lots of time on really easy problems[/quote]

Tests from the 1990s and 80s are generally much easier. I personally average around 2-3 points lower on practicing recent ones than old ones (though I've only taken around two old ones, so not much of a sample size).

[quote]4 is a really good score for a middle schooler. [/quote]

um, somehow that doesn't make me feel better.

Maybe, but the tests I took were fairly recent; almost all of them were 2000 or later

I'm in 8th grade, and got an 11. Sadly that was today as practice, and I have to go take AIME II on Wednesday as the real thing.

Last year was my first year doing AMC's and I got a 6.

To the people who got $ x$ as practice: be aware that on the real thing, you will likely get an average of $ x\minus{}3$. For example, if you got a 12 on practice, you're likely to get a 9 if you had taken it officially.

For example, I could've gotten a 12, but I really got a 9 due to stupid mistakes and time pressure.

A rule like that will never work more than around a third of the time. I averaged 11 on practices, but I got a 4, having solved 12. I know people who got a much higher score than they averaged. It all depends on the person.

sadly, my average on practices is a 9-10 and i only got a 6 on the real thing. :(
I really wasted too much time on some of the problems that I got right (and were trivial), so I had no time to do some other problems that were trivial (like #11, shoelace method ftw!)
And I had stupid mistakes on #1 (ARRGH), 9 (stupid casework), and 14 (arithmetic mistakes)
And, sadly, I had no idea why I couldn't do #5 (just so easy :o SAS congruence and angle bisector)
However, this was my first time doing the AIME for real, so...meh?

lol i got a 15 so where i do vote on the poll since the highest is 14???
(JUST KIDDING  :D )

Err..
Actually I don`t qualified to the AIME because of lot of stupid mistakes (actually I could do 24 out of 25, but I only got 114).
But I just tried to do the AIME II and assume I were in the exam, so I will got 7.  :D
Actually I`m in grade 9 and I`m just trying to studying for high school level mathematics olympiad.

9th grade with 11 on aime#1.
ROFL I MISSED NUMBER 1
 :rotfl:

Something makes me doubt 12 people under 9th grade scored 13+.

[b]People, it is respectful not to intentionally screw up poll results.[/b]

TYTIA.

9th grade with 8 on the AIME II. Should have gotten 9 though because I did 34*2=74...


Do there exist a polynomial $p(x)$ of degree $2$ such that, for any positive integer $n$, the equation $p^{n}(x)=0$ has $2^{n}$ real roots.

excuse me, all of the roots must be distinct! that is works?

$p(x)=2x^2-1$.Writing $x=cos y\Rightarrow p_n(x)=cos2^n.y\Rightarrow ...$


In a triangle ABC with AB > AC we consider  median AD. Let CE the altitude in triangle

CAD (CE is perpedicular to AD). If angles ACE and  B are equal , prove that triangle ABC is right at A, that is angle

A = 90.

Cute one :)

The perpendicular through $A$ to the median makes an angle equal to $\angle B$ with $AC$, so this perpendicular is tangent in $A$ to the circumcircle of $ABC$. This shows that the circumcircle of $ABC$ lies on the median $AD$, and the conclusion follows immediately.

This is the official solution
<< Let F a point on line CE such that EF = CE ( C, F are symmetric with respect  to E). Then (angles) AFC = ACF = B. This means that  AFBC is inscriptible , so (angles)
    CAB = CFB = 90 , since  FB // ED ( from triangle CFB) and (angles) CFB = CED = 90.

      Babis


hi guys

do you happen to know how amc 12 is scored?
its 25 questions, each worth 6 marks right?
and since its multiple choice, so how can u get .5 in ur score?

thank you!

Correct answers are worth 6 points.
Incorrect answers are worth 0 points.
Unanswered questions are worth 1.5 points.

That particular scoring system is new this year- it was 6, 2.5, 0 for all previous years since the AHSME became the AMC 10 and AMC 12. There are several old threads in the forum discussing the change.


algorithm
function
factorial
floor function
algebra
number theory
prime factorization

Hi everyone i think i have a 10sec algorithm for finding the numbers of zeros in $ {n!}$.
But first i need anybody to tell me how many zeros are in $ {1000!}$ if it tallies then i'll post the algorithm

as for the question, i believe the answer is 249.

It is a basic counting problem, as one considers the floor function of a number factorial (1000) in this case divided by 5 and then add that to 1000 divided by 5^2, etc.

This is also called Legendre's formula (at least one of them).

Are you referring to the total number of zeros or the number of trailing zeros?

Yes, I had the same question, since I believe Legendre's formula only gives the number of zeros at the end.

Trailing zero's.
249 is correct.
By my work(through GOD) today i have a faster way.
The algorithm is just recording quotients of divison of $ {n}$ by $ {5}$.
Up to the point where the quotient is 1.
Sum it all up(quotients).
E.g $ {n}$ $ {1000}$.
The quotients are $ \frac{1000}{5}\equal{}200$.
$ \frac{200}{5}\equal{}40$.
$ \frac{40}{5}\equal{}8$.
$ \frac{8}{5}\equal{}1$.
The sum of the quotients as you see is 249.
 Similarly for 10000! Factorial the number of trailing zeros is 2499.
Since i've not seen it used anywhere i'll like if its called "10 secs counter"(1sc).
You can PM me for a proof outline.

Essentially, you're still doing the same thing as theax did.
Just out of curiosity, what would be the result with your if $ n\equal{}1002$ or if $ 5\nmid n$?

21 7 15's method is exactly the same as what theax did.

Start with $ n$.

$ [\frac{n}{5}]$ is the required first part.

$ [\frac{n}{25}]\equal{}[\frac{[\frac{n}{5}]}{5}]$. One of these is theax's, one is 21 7 15's. They are equivalent. The fact that one includes the decimal part of the previous division makes no difference, as no extra 5's will divide into it.

Sorry to disappoint you, but that's known fact.

yfl20673, it's important to notice the idea behind the algorithm, we need to count how many $ 5$'s are there in factorization of $ n!$ because they will give $ 10$'s (thus $ 0$'s at the end) when multiplied by $ 2$. We can do it like this:

$ \left\lfloor\frac{1002}{5}\right\rfloor\equal{}200$

$ \left\lfloor\frac{1002}{25}\right\rfloor\equal{}40$

$ \left\lfloor\frac{1002}{125}\right\rfloor\equal{}8$

$ \left\lfloor\frac{1002}{625}\right\rfloor\equal{}1$

Or the shorter way:

$ \left\lfloor\frac{1002}{5}\right\rfloor\equal{}200$

$ \left\lfloor\frac{200}{5}\right\rfloor\equal{}40$

$ \left\lfloor\frac{40}{5}\right\rfloor\equal{}8$

$ \left\lfloor\frac{8}{5}\right\rfloor\equal{}1$

But it's always good to find out things and then prove them yourself (like you did).  :lol:

http://www.artofproblemsolving.com/Wiki/index.php/Legendre%27s_Formula

is the page where the theorem is cited. It's interesting that in that aops wiki page, the formula goes more in depth -  $ \frac {n \minus{} S_p(n)}{p \minus{} 1}$ is equal to the exponent of $ p$ - where $ p$ is a prime - in the prime factorization of $ n$. $ S_p(n)$ is the sum of the digits of $ n$ when written in base $ p$.

It seems the I assumed you meant trailing zeros  :blush: but luckily for me, my assumption was right.

It's also interesting to note that Legendre discovered many many formulas, so I assume when you cite legendre's formula on say a contest, you probably need to say which one you're going to use.

And as people already noted, $ \left\lfloor \frac {n}{25} \equal{} \frac {\lfloor\frac {n}{5}\rfloor}{5}\right\rfloor$.

Flame, I understand the algorithm, I was just making sure that 21 7 15 truncates the decimal from the quotients. The examples he's used were all divisible by 5, but I suppose after looking at the final line, he's done it without showing the floor function.


Discuss.

I got a 129 :( I missed two problems due to stupid mistakes (the tangent circles problem and the 100...64 (#25) problem)

And I overkilled time on several problems due to crappy solving methods...so I ran out of time and didn't do #23 and #24...even though they're so freakin' easy

I hope I make USAMO by beasting the AIME...what score might I need on the AIME for USAMO qualification?

This thread is another duplicate - merging and locking.


Find the sum of the $n+1$ binomial coefficients for $(a-b)^n$

[quote="eryaman"][hide]Plugging in $1$ for both $a$ and $b$, we get $0$.[/hide][/quote][hide="counterexample"]Plugging in $a$ and $b\ne{a}$, we don't get $0$.

\[(a-b)^n=\sum_{k=0}^n{C_n^k{a}^{n-k}b^k(-1)^k,}\]so we're looking for the sum of the coefficients, or \[\sum_{k=0}^n{C_n^k(-1)^k}.\][/hide]

[quote="joml88"]But that IS zero, Antonio.  Eryaman's method works but it should be explained some.  If you actually expanded that out you would want to sum up the coefficients. But if you let a and b be 1 once it is expanded that is the same thing as summing the coefficients.  Thus, just let a=b=1 before expanding it out and you get your answer.[/quote]I see now, thanks  :blush:.

[quote="eryaman"][hide]Plugging in $1$ for both $a$ and $b$, we get $0$.[/hide][/quote]

Hmm, that IS nice!  :)

Thanks for the explanation, joml88!


arithmetic sequence
Diophantine equation
number theory unsolved
number theory

Find infinitely many triples (a,b,c) of positive integers such that a,b,c are arithmetic progression and such that ab+1 , bc+1 , ca +1 are perfect square

But there no solution in that topic!Here is sln:
Take (x,y) be a root of Pell-Equation:$x^2-3y^2=1$(It have infinitly many roots!) and choose:$a=2y-x,b=2y,c=2y+x$ for (x,y) larges sufficiently will work(check it!)

your solution is a little unnatural. what make you know that the equation x^2-3*y^2=1 is relative to this problem
Does there exist any triples (a,b,c) are diffirent from a=2y-x, b=2y , c=2y+x
And if (x,y) is a root of the equation x^2 -3 * k^2 * y^2=1 and triple (a,b,c) satisfies a=2ky-x , b=2ky , c=2ky+x then
ab +1 , bc +1 ,ca +1 is also pefect square
 ( k is a natural number )


Does anyone know if homeschoolers can be on a team, and if so, under what circumstances?

The handbook only says that "Homeschoolers may compete in accordance with all other rules".

Thanks.

"in accordance with all other rules".
Under the same circumstances as a schol, a school being defined as the body of homeschoolers in a given area?

Just brainstorming.

I'm homeschooled

I'm homeschooled, and i do mathcounts through my math club, which draws from all over the state. If you're wondering about what geographical areas you can draw from, i'm pretty sure you can go to a chapter competition if you don't live in the chapter area, but you can't go to a state competition in another state.

OK, I got a reply:

Homeschoolers who live in the same chapter or in adjacent chapters may combine to form a team. (Note: This policy may change next year so please be sure to read the rules next year). If you choose to combine with students who live in adjacent chapters, you should contact your Chapter Coordinator to make him or her aware of the situation.

Thanks for the help.

EDIT: If you are of mathcounts age, homeschooled, in NJ, and interested in forming a team (IF STILL POSSIBLE), please contact me ASAP.


group theory
abstract algebra
superior algebra
superior algebra unsolved

If N and M are normal subgroups whose intersection is {e} prove that nm=mn. The problem is from herstein. I am not able to find out exactly how to use the hypothesis that the two groups have only 'e' as a common element.

[hide]The contrapositive of the statement to be proven is:

"If there exists an $n_{1}\in N$ and $m_{1}\in M$ such that $n_{1}m_{1}\neq m_{1}n_{1}$, then there exists an element in the intersection of normal subgroups M and N that is not the identity {e}."

If $n_{1}m_{1}\neq m_{1}n_{1}$, then $n_{1}m_{1}n_{1}^{-1}m_{1}^{-1}\neq e$. So it reduces to prove that $n_{1}m_{1}n_{1}^{-1}m_{1}^{-1}$ is in the intersection of normal subgroups M and N. This will prove the contrapositive statement, and hence the original statement.[/hide]


let $ a,b,c \in \mathbb{R}^*$ such that $ \frac{kabc}{a\plus{}b\plus{}c}\leq (a\plus{}b)^2\plus{}(a\plus{}b\plus{}4c)^2$
Find the maximum of $ k$

[quote="tobeno_1"]let $ a,b,c \in \mathbb{R}^*$ such that $ \frac {kabc}{a \plus{} b \plus{} c}\leq (a \plus{} b)^2 \plus{} (a \plus{} b \plus{} 4c)^2$
Find the maximum of $ k$[/quote]
$ 0$ because if $ k>0$ then for $ ab\neq0$ and for all $ \epsilon>0$ it exists $ c$ for which $ \frac {kabc}{a \plus{} b \plus{} c}>\epsilon.$ :wink:


I have a questions for homework which i have no idea how to approach

 Find the distance between the indicated point A and PLANE P
$A=(1,1,2)$ and $P: 3X1+X2-5X3 = 2$

Thnx

So we have

Theorem: The distance d from the point (x0,y0,z0) to a plane C: ax+by+cz+d=0 is

d=|ax0+by0+cz0+d|/sqrt(a^2+b^2+c^2)

If you want the proff, ask me...

(you can prove this on your own using the normal vector (a,b,c) )


geometry
incenter
analytic geometry
geometric transformation
circumcircle
trigonometry
ratio

Let the inscribed circle and circumscribed circle of an acute triangle $ABC$ be $I$, $O$ respectively. And $I$ meets $BC$, $CA$, and $AB$ at $P$, $Q$, and $R$ respectively. Let the circles that tangents $O$ and $BC$ at $P$, $CA$ at $Q$, $AB$ at $R$ be $O_{1}$, $O_{2}$, and $O_{3}$ respectively. Finally, let $O_{1}$, $O_{2}$, and $O_{3}$ meets $O$ at $P'$, $Q'$, and $R'$ respectively.
(1) $AP'$, $BQ'$, $CR'$ concur at a point $X$.
(2) $PP'$, $QQ'$, $RR'$ concur at a point $Y$.
(3) $X$, $Y$, $I$, $O$ are linear.

Maybe (2) is easier than (1).

You can check that $PP', QQ', RR'$ are concurrent at the incenter of similitude of the incircle and excircle.

You can check the barycentric coordinate of $X$ is $(\frac{a}{p-a},...)$

Here's my solution for $(1)$.

Let $AR=AQ=x, BR=BP=y, CP=CA=z$.

Using Ceva's Theorem of Angle, ETS $\frac{\angle BAP'}{\angle CAP'}\frac{\angle CBQ'}{\angle ABQ'}\frac{\angle ACR'}{\angle BCR'}=1$.

We'll prove this one.

It's clear that $\angle BAP' = \angle BCP' , \angle CAP' = \angle CBP'$.

Therefore, $\frac{\angle BAP'}{\angle CAP'}= \frac{\angle BCP'}{\angle CBP'}= \frac{BP'}{CP'}$.

so we get $\frac{\angle BAP'}{\angle CAP'}\frac{\angle CBQ'}{\angle ABQ'}\frac{\angle ACR'}{\angle BCR'}= \frac{BP'}{CP'}\frac{CQ'}{AQ'}\frac{AR'}{BR'}$.

Let's prove $\frac{BP'}{CP'}\frac{CQ'}{AQ'}\frac{AR'}{BR'}=1$.

OBSERVE. Define $A'$ by the intersection of line $PP'$ and circumcircle. Then, $A'$ is mid point of arc $BAC$.

(Think about Translation circle $O_{1}$ $\rightarrow$ circle $O$ which maps $P$ to $A'$ )

From OBSERVE, we get $P'A'$ is anglebisector of $\triangle P'CB$.

so $\frac{BP'}{CP'}=\frac{y}{z}$.

Using similar ideas, we get $\frac{BP'}{CP'}\frac{CQ'}{AQ'}\frac{AR'}{BR'}=\frac{y}{z}\frac{z}{x}\frac{x}{y}=1$

$Q.E.D.$

[quote="Sung-yoon Kim"]Let the inscribed circle and circumscribed circle of an acute triangle $ABC$ be $i=C(I,r)$, $c=C(O,R)$ respectively and  $i$ meets $BC$, $CA$ and $AB$ at $P$, $Q$ and $R$ respectively. Let the circles that tangents to $c$ and $BC$ at $P$, to $c$ and $CA$ at $Q$  to $c$ and $AB$ at $R$ be $c_{1}=C(O_{1},r_{1})$, $c_{2}=C(O_{2},r_{2})$ and $c_{3}=C(O_{3},r_{3})$ respectively. Finally, let $c_{1}$, $c_{2}$ and $c_{3}$ meets $c$ at $P'$, $Q'$ and $R'$ respectively.
$1.\blacktriangleright\ AP'$, $BQ'$, $CR'$ concur at a point $X$.
$2.\blacktriangleright\ PP'$, $QQ'$, $RR'$ concur at a point $Y$.
$3.\blacktriangleright\ X$, $Y$, $I$, $O$ are linear.[/quote]
[color=darkblue][size=117][b]Very nice problem ![/b][/size]

$1.\blacktriangleright$ Define the diameter $[A_{1}A_{2}]$ of the circumcircle so that $A_{1}\in AI$. From the well-known property $A_{2}\in PP'$ results $\frac{P'B}{P'C}=\frac{PB}{PC}\ .$
Denote the intersection $S_{a}\in BC\cap AP'$. Therefore, $A_{2}\in PP'$ $\Longrightarrow$ $\frac{S_{a}B}{S_{a}C}=\frac{P'B}{P'C}\cdot \frac{\sin \widehat{BP'S_{a}}}{\sin\widehat{CP'S_{a}}}=\frac{PB}{PC}\cdot \frac{\sin C}{\sin B}=\frac{c(p-b)}{b(p-c)}$, i.e. $\frac{S_{a}B}{S_{a}C}=\frac{c(p-b)}{b(p-c)}$.
Analogously denote the points $S_{b}$, $S_{c}$. Thus, the lines $AS_{a}$, $BS_{b}$, $CS_{c}$ are concurrently, i.e. the lines $AP'$, $BQ'$, $CR'$ are concurrently.

$2.\blacktriangleright$ Denote the intersection $Y\in OI\cap PP'$ and apply the Menelaus' theorem to the transversal $\overline{P'PY}$ in the triangle $OIO_{1}$ : $\frac{P'O_{1}}{P'O}\cdot\frac{YO}{YI}\cdot \frac{PI}{PO_{1}}=1$ $\Longrightarrow$ $\frac{r_{1}}{R}\cdot \frac{YO}{YI}\cdot\frac{r}{r_{1}}=1$ $\Longrightarrow$ $\frac{YO}{YI}=\frac{R}{r}$- symmetrical value w.r.t. the triangle $ABC\ .$ Therefore, the lines $PP'$, $QQ'$, $RR'$ are concurrently in a point $Y\OI$ for which $\frac{YO}{YI}=\frac{R}{r}\ .$ [b]Otherwise,[/b] $A_{2}\in PP'\Longrightarrow \frac{YO}{YI}=\frac{OA_{2}}{IP}$, i.e. $\frac{YO}{YI}=\frac{R}{r}\ .$
[b]Remark.[/b] The lines $P'A_{1}$, $RQ$, $BC$ are concurrently in the conjugate point $T$ of the point $P$ w.r.t. the points $B$, $C$ and $TI\perp AP$.
The intersection of the lines $AA_{2}$ and $BC$ is the conjugate point of the point $D\in BC\cap AI$ w.r.t. the points $B$, $C$.

$3.\blacktriangleright$ Above proved that $Y\in OI$ and $\frac{YO}{YI}=\frac{R}{r}$. Denote $X\in AP'\cap BQ'\cap CR'\ .$ The above relation $\frac{S_{a}B}{S_{a}C}=\frac{c(p-b)}{b(p-c)}$, i.e. $\frac{S_{a}B}{\frac{c}{p-c}}=\frac{S_{a}C}{\frac{b}{p-b}}$ means that
the point $X$ has the  barycentrical coordinates $\left(\frac{a}{p-a},\frac{b}{p-b},\frac{c}{p-c}\right)$ (see the previous [b]Treegoner's message[/b]). Therefore,
$X\in OI$ $\Longleftrightarrow$ $\left|\begin{array}{ccc}\frac{a}{p-a}& \frac{b}{p-b}& \frac{c}{p-c}\\\ a & b & c\\ a\cos A & b\cos B & c\cos C\end{array}\right|=0$ $\Longleftrightarrow$ $\left|\begin{array}{ccc}1 & 1 & 1\\\ (p-b)(p-c) & (p-c)(p-a) & (p-a)(p-b)\\ \cos A & \cos B & \cos C\end{array}\right|=0$ (prove easily).

 :!: [b]I didn't succed to ascertain the ratio [/b]$\frac{XO}{XI}\ .$ I think that it is nastely ![/color]

nice solutions! :)

Indeed, (2) can be proved by another way. As Woojin said, the intersection of the line $PP'$ and the circumcircle is the midpoint of the arc $BAC$. Let the midpoint $M_{a}$, and define $M_{b}, M_{c}$ similarly. Draw a tangent line of the circumcircle at $M_{a}, M_{b}, M_{c}$, then the three tangents makes a triangle, which is congruent to $ABC$. $M_{a}$ corresponds to $P$, so lines $M_{a}P, M_{b}Q, M_{c}R$ meets at the origin of the congruency, $Y$. Moreover, as the incircle of the big circle, which is actually the circumcircle of $ABC$, corresponds to the incircle of $ABC$, and so are the centers. So $Y,I,O$ are linear. :)

Another proof for part (3) for the part $ X,I,O$ are collinear.
Denote $ A''$ is the midpoint of arc $ BC$ which doesn't contain $ A$,$ C''$ is the midpoint of AB which doesn't contain C.
As showed in previous posts,$ A'$ is the midpoint of arc $ BAC$ and $ C'$ is the midpoint of arc $ BCA$.
Let $ P'C'' \cap R'A'' \equal{} \{Z \}$.Apply Pascal's theorem to the hexagon $ A''R'C'C''P'A'$ we have $ Z,O,Y$ are collinear.Apply Pascal's theorem to the hexagon $ A''R'CC''P'A$ we have Z,I,X are collinear.But $ Y,I,O$ are collinear (the previous part) so $ X$ lie on the line $ OI.$
[img]http://img38.imageshack.us/img38/4140/kakagpt.jpg[/img]


i need software that can build a grafic of  this :
Z=x^4+y^4+18xy-9x^2-9y2+1

you should try mathematica.

Try this: http://www.mathgv.com/ , I dont know how it works though :P

I think you can do it with excel too.

[url=http://www.gnuplot.info]gnuplot[/url] is a free program that can graph functions. You can use the tutorial [url=http://www.cs.uni.edu/Help/gnuplot/]here[/url] to get started.


geometry
geometric transformation
reflection
geometry unsolved

Let triangle $ ABC$ inscribed circle $ (O)$. $ PB,PC$ be the tangents of $ (O)$. $ M$ be the midpoint of $ BC$. Prove that $ AP$ respective with $ AM$ through the bisector of $ \hat{BAC}$. :lol:

See figure below. We have division $ (AHGP),(KIJP)$ is harnomic.Applying menelaus's theorem for $ \triangle PGJ$ we get $ A,G,M$ to be collinear=> QED :)
 Sorry for my bad english

Let's call $ D$ the second intersection of $ AP$ with the circle $ O$. $ BP$ and $ CP$ being tangent to the circle, we have the angles equalities: $ \measuredangle BAD = \measuredangle DBP$ and $ \measuredangle CAD = \measuredangle DCP$, so $ \triangle ABP \sim  \triangle BDP$ and $ \triangle ACP = \triangle CDP$, from their similarities getting: $ \frac {AB}{BD}=\frac{AP}{BP}=\frac{BP}{DP}$ and $ \frac{AC}{CD}=\frac{AP}{CP}=\frac{CP}{DP}$.
From the first double equality we shall get an equivalent equality by squaring the first term and equaling to the product of the next two terms: $ \left ( \frac{AB}{BD} \right )^2 = \frac{AP}{BP} \cdot \frac{BP}{DP}$, that is, $ \left( \frac{AB}{BD} \right)^2=\frac{AP}{DP} \ ( \ 1 \ )$. Similarly we shall get $ \left( \frac{AC}{DC} \right)^2=\frac{AP}{DP} \( \ 2 \)$, from (1) and (2) getting $ \frac{AB}{BD}=\frac{AC}{DC}$ or $ AB \cdot CD = AC \cdot BD \ ( \ 3 \ )$.
Applying Ptolemy's for $ ABDC$: $ AB\cdot CD+AC\cdot BD = AD\cdot BC$ which, with (3) becomes $ 2\cdot AC\cdot BD=AD\cdot BC$ or $ AC \cdot BD = \frac{BC}{2} \cdot AD$, or $ \frac{AC}{AD}=\frac{CM}{BD} \ ( \ 4 \ )$.
From (4) with $ \angle ACB = \angle ADB$ we get $ \triangle ACM \sim  \triangle ADB$, i.e. $ \angle CAM = \angle BAD$, or $ AD$ is the reflection of $ AM$ in the bisector of the angle $ \angle BAC$, $ M$ being the midpoint of $ BC$, done.

Best regards,
sunken rock


The line joining the midpoints of the diagonals of a trapezoid has length 3. If the longer base is 97, what is the shorter base?

[hide="hint"]i think you can use similarity on this one
[/hide]

I used similar triangles. But I reached a contradiction. Could someone show me their work?

Call the distance from the intersection of the diagonals to the longer base $ d$.
Call the height of the trapezoid $ h$.  The shorter base has length $ b$.
With similar triangles, $ \frac{h\minus{}d}{d}\equal{}\frac{h}{d}\minus{}1\equal{}\frac{b}{97}$.

The distance from the longer base to the segment that joins the midpoints is $ \frac{h}{2}$.  Use similar triangles to find that $ \frac{d\minus{}\frac{h}{2}}{d}\equal{}\frac{3}{97}$.
Simplify to $ \frac{94}{97}\equal{}\frac{h}{2d}$.
$ \frac{h}{d}\equal{}\frac{188}{97}$.
$ \frac{h}{d}\minus{}1\equal{}\frac{91}{97}$, so the shorter base has length $ 91$.

Maybe right?

I don't know if you still need a solution, but...

[hide]Since the two bases are the same distance away from each other, the segment connecting the midpoints is parallel to the bases. Parallel lines = similar triangles. Let the trapezoid be ABCD, where AB is the shorter side, CD is the longer side. Let the midpoints of AC and BD be X and Y, respectively, and let the intersection of AC and BD be E. K, so we know that: triangle BAE ~ triangle DEC ~ triangle YEX by AA similarity. We can write:

EX/EC = EX/(EX+XC) = 3/97
AE + EX = XC

Solving the first one, we get XC = 94/3*EX so EC = 97/3*EX
Plugging that result into the second one gets us: AE = 91/3*EX

From similar triangles again, we have:

AE/EC = (97/3*EX)/(91/3*EX) = 91/97 = AB/DC = AB/97. AB is then 91. :)[/hide]

[hide="Very simple"]The segment in question is the difference of midlines of the two triangles whose bases are the paralel sides of the trapezoid. Hence $ {a\minus{}b\over 2}\equal{}3\implies b\equal{}a\minus{}6\equal{}91$[/hide]


Let $ a_1a_2a_3...a_n$ be a natural number, where $ a_1$, $ a_2$...are digits. What must this number be, so that $ 12\cdot2a_1a_2a_3...a_n1 \equal{} 21\cdot1a_1a_2a_3...a_n2$?

[hide]
Call the number x, and the number of digits n.
The LHS of the equation becomes $ 24\cdot 10^(n\plus{}1) \plus{}120x\plus{}12$.
The RHS becomes $ 21\cdot 10^(n\plus{}1) \plus{}210x\plus{}42$.
Equate them and get $ 3\cdot 10^(n\plus{}1)\equal{}90x\plus{}30$.
Simplify. $ 10^n\plus{}1\equal{}30x\plus{}10$.
Finally. $ x\equal{}3333333333....$ with n digits.

[/hide]


I use GeoGebra to make all of my USAMTS diagrams: i export my picture and then use the command \includegraphics for my pdf... but this is a huge pain to size the picture correctly and then have it come up in high resolution on the pdf. Is there a better way to get GeoGebra diagrams into a pdf file??

Thanks!

Same question here, as well as another: Are we allowed to use general pictures (for example, a regular hexagon) from a resource off the web in our solutions?

Any image that you have permission to use, and whose source your cite, you may use.

Does anyone have a good way for high quality images?


The equation [b]sin³x . cos x - senx.cos³x = 1 / 4[/b] , at range [0;2pi] . How many and which solution the equation have?

I wait, thank you.

[hide="Easter Eggs"]$ \sin^3 x\cos x \minus{} \cos^3 x\sin x \equal{} \frac {1}{4}$
$ \Leftrightarrow \sin x\cos x(\sin^2 x \minus{} \cos^2 x) \equal{} \frac {1}{4}$
Now $ \cos 2x \equal{} \cos^2 x \minus{} \sin^2 x$ and $ \sin 2x \equal{} 2 \sin x\cos x$
$ \Leftrightarrow \sin 2x\cos2x \equal{} \minus{} \frac {1}{2}$
$ \Leftrightarrow \sin 4x \equal{} \minus{} 1$
$ \Leftrightarrow 4x \equal{} \frac {3\pi}{2}, \frac {7\pi}{2}, \frac {11\pi}{2}, \frac {15\pi}{2}$
$ \therefore x \equal{} \frac {3\pi}{8}, \frac{7\pi}{8}, \frac{11\pi}{8}, \frac{15\pi}{8}$[/hide]

EDIT: That was silly

hello, i have also
$ x_1\equal{}\frac{3\pi}{8}$
$ x_2\equal{}\frac{7\pi}{8}$
$ x_3\equal{}\frac{11\pi}{8}$
$ x_4\equal{}\frac{15\pi}{8}$
Sonnhard.

check your question.

[url=http://www.banishrosacea.com]rosacea treatment[/url]

Complex_Ninja , thank you.
But im really nb in trigonometry and i really don't understand this pass

$ \sin 2x\cos2x \equal{}\minus{}\frac{1}{2}$

how you put $ \sin 2x \equal{} 2\sin x\cos x$ and $ \cos 2x \equal{}\cos^{2}x\minus{}\sin^{2}x$ into the  $ \sin x\cos x(\sin^{2}x\minus{}\cos^{2}x) \equal{}\frac{1}{4}$ ;  how appears an $ \minus{}\frac {1}{2}$

I thought in $ \frac{1}{2}\sin2x (\minus{}\cos2x) \equal{} \frac{1}{4}$ ; 	but I still do not know how to proceed.

thank you.

[quote="Likon"]Complex_Ninja , thank you.
But im really nb in trigonometry and i really don't understand this pass

$ \sin 2x\cos2x \equal{} \minus{} \frac {1}{2}$

how you put $ \sin 2x \equal{} 2\sin x\cos x$ and $ \cos 2x \equal{} \cos^{2}x \minus{} \sin^{2}x$ into the  $ \sin x\cos x(\sin^{2}x \minus{} \cos^{2}x) \equal{} \frac {1}{4}$ ;  how appears an $ \minus{} \frac {1}{2}$

I thought in $ \frac {1}{2}\sin2x ( \minus{} \cos2x) \equal{} \frac {1}{4}$ ; 	but I still do not know how to proceed.

thank you.[/quote]

Here is the order
$ \sin 2x \equal{} 2\sin x\cos x$ or $ \sin x\cos x \equal{} \frac{\sin 2x}{2}$
and $ \cos2x \equal{} \cos^2 x \minus{} \sin^2 x \Leftrightarrow \minus{}\cos 2x \equal{}  \minus{}(\cos^2 x \minus{} \sin^2 x) \equal{} \sin^2 x \minus{} cos^2 x$

Now subbing the above we have $ \frac{\sin 2x}{2}(\minus{}\cos 2x) \equal{} \frac{1}{4} \Leftrightarrow \sin 2x\cos2x \equal{}\minus{}\frac{1}{2}$

Well for the second part we have the identity $ \sin 2y \equal{} 2\sin y\cos y$ now sub $ y \equal{} 2x$ to get $ \sin 4x \equal{} 2\sin 2x\cos 2x$


calculus
integration
logarithms
calculus computations

Verify that $ \int\sqrt{a^2\plus{}u^2}{du} \equal{}  \frac{u}{2}\sqrt{a^2\plus{}u^2}\plus{}\frac{u^2}{2}\ln{(u\plus{}\sqrt{a^2\plus{}u^2})}\plus{}c$

Excuse me, but are we supposed to do? Aren't you able to verify that by yourself?

Diffentiate the right side and then simp. I found it as a good clac 1 test.

hello, a better way is to substitute $ u\equal{}a\sinh(t)$ and to compute the integral.
Sonnhard.


April 12, 2005

[b]Ehlers joins Wolf in introducing interest-free loan bill[/b]

Congressmen tout proposal to increase math, science education

WASHINGTON - Concerned by the lack of U.S. students who choose to pursue degrees and careers in math, science, engineering and technology, Congressman Vernon J. Ehlers Tuesday joined with fellow Congressman Frank Wolf in introducing legislation that would pay up to 10,000 dollars in student-loan interest for students who agree to commit at least five years to teaching or working in related fields.

The Math and Science Incentive Act of 2005 was introduced in the U.S. House of Representatives Tuesday by Wolf, R-Va., with Ehlers, R-Mich., as the legislation’s primary co-sponsor. House Science Committee Chairman Sherwood Boehlert, R-N.Y., also co-sponsored the House legislation, while Sen. John Warner, R-Va., said he planned to introduce companion legislation in the U.S. Senate.

The bill would establish a new program through the U.S. Department of Education, under which the government would pay the interest on a student’s loans in exchange for five years in a job related to science, technology, engineering or math (STEM), including teaching in those fields at any level. The government would pay the interest on the loan starting at the beginning of the service requirement up to a maximum of 10,000 dollars per person.

[url=http://www.house.gov/apps/list/press/mi03_ehlers/041205STEMLoans.html]More info[/url]

They better pick a different name.  With the name STEM, they'll have the Schiavo people all over them for supporting stem-cell research.

Of course, I like the idea of putting a focus on math, science, etc., but think if there's going to be a pile of money thrown at the problem, I'd aim at younger people.

Well at least there exist [b]student loans[/b] in US.

In Romania it's so hard to get a loan, not even to speak about interest free: the conditions you have to pass usually mean that you have to earn so much money (they want to make absolutely [b]sure[/b] you'll pay back) that you don't need a loan after all :) For students it is impossible to get a loan.

[quote="rrusczyk"]They better pick a different name.  With the name STEM, they'll have the Schiavo people all over them for supporting stem-cell research.  [/quote]

STEM is an acronym the NSF has been using for several years now in most of its education-related grant programs.  I doubt the acronym will start causing problems now, as legislators are likely familiar with it.


Um, everyone;
I have Intermediate-Pre Olympiad level competitions coming up in 1-2 weeks and even MATHCOUNTS level problems are hard for me. I know that nothing I do will make a significant improvement in 2 weeks but can you recommend something FREE that I can do to increase my performance? As you can see, I'm PHAILING on even in MATHOCOUNTS forum.

Thanks!

What are your strengths and weaknesses? Strengthen your weaknesses these 2 weeks so that you can maximize your performance. If one of those weaknesses is counting/probability, I suggest playing around with Alcumus.


[quote="jgao"]I ve got some questions :?: :
What are bases?
How do you prove that points are collinear?
How do you prove lines are concurrent?
Thanx :)[/quote]
all these questions are answered in Aops vol.1 and vol.2. :)

Take a look at this:
http://www.artofproblemsolving.com/Wiki/index.php/Number_bases

for bases.

The other  two I don't know...

[quote="jgao"]I ve got some questions :?: :

What are bases?
How do you prove that points are collinear?
How do you prove lines are concurrent?

Thanx :)[/quote]
Which bases? The binary or the geometry kind.
Also to prove collinearness the points need to have the same slope.

[quote="jgao"]How do you prove lines are concurrent?[/quote]

Ceva's Theorem. :D
http://mathworld.wolfram.com/CevasTheorem.html

To show that 3 lines are concurrent, you can use Ceva's.
Or you can turn it into a collinearity problem.
for example, if you want to prove that altitudes of a triangle are concurrent, draw two altitudes, and from the intersection(let that be A), draw a segment to the other vertex(say that its B) and then draw a perpendicular to the third side(call the foot C). Then you just need to show that A,B and C are collinear.

To show collinearity, you can also show that the angle between the three points is 180.
And you can use menelaus' theorem in some cases:[url]http://mathworld.wolfram.com/MenelausTheorem.html[/url]

[quote="vishalarul"]To show that $3$ points are collinear, do one of these 2:
(1) Show that they are on the same line
(2) Let the $3$ points be $A$, $B$, and $C$, and let $B$ be between $A$ and $C$. Suppose, to the contrary that $B$ is not on the line, and that the $3$ points determine a triangle. Show that $AB+BC = AC$, violating the Triangle Inequality. So the temporary assumption was false. The statement is proven.

That's all I know... :D[/quote]

But how could you could you prove that they're on the same line. :?:

Choose an arbitrary point $D$. Prove that $\angle ABD+\angle CBD=180^\circ$.


Let be given two sequences of real numbers $a_1,a_2,a_3,...$ and $b_1,b_2,b_3,....$ satisfying:
$a_1=\alpha$
$b_1=\beta$
$a_{n+1}=\alpha a_n-\beta b_n$
$b_{n+1}=\beta a_n+\alpha b_n$
How many pairs of real numbers $(\alpha,\beta)$ satisfying $a_{1997}=\beta,b_{1997}=\alpha$
Note:don't forget the subject:[b]"use trigonomatric"[/b]

$\gamma^2 = \alpha^2 + \beta^2$
$a_1 = \gamma \cos \theta$
$b_1 = \gamma \sin \theta$
$a_n = \gamma^n \cos (n \theta)$
$b_n = \gamma^n \sin (n \theta)$

$a_{1997} = \beta, b_{1997} = \alpha \implies \gamma = 1$

Now we only need to find solutions to $\cos (1997 \theta) = \sin \theta$ and $\sin (1997\theta) = \cos \theta$.  Given $\cos (\frac{\pi}{2} - \theta) = \sin \theta$ we can reason:

$\frac{\pi}{2} - 1997 \theta + 2\pi k= \theta$
$1998 \theta = 2\pi k + \frac{\pi}{2}$
$\theta = \frac{\frac{\pi}{2} + 2\pi k }{1998}, k = 0, 1, 2, ... 1997$ so there are $\boxed{1998}$ solutions?


AoPSwiki
number theory
prime factorization
prime numbers
AoPS Books

I'm not quite sure how to go about doing this problem. Can anyone please help?  :)

Determine the five smallest natural numbers that have exactly 18 positive factors.

Thanks

well, you know that the prime factorization of any number is in the form $ n \equal{} p_1^{n_1} p_2^{n_2}...p_k^{n_k}$

And the number of distinct divisors is $ \tau (n) \equal{} (n_1 \plus{} 1)(n_2 \plus{} 1)...(n_k \plus{} 1)$.

In this particular problem, $ \tau (n) \equal{} 18$. First find sets of divisors that would make the product small.

[hide="Solution I think..."]Well the way we find the number of factors a number has, we have to prime factor it to the form $ a^xb^yc^z...$ where a, b, and c are prime and x, y, and z are greater than 0. The number of factors that this has is $ (x\plus{}1)(y\plus{}1)(z\plus{}1)...$ since x, y, and z can be zero also. Thus since 2 and 3 are the smallest prime numbers, we have $ 2^{17}, 2^{8}3^{1}, 2^{5}3^{2}, 2^{2}3^{5}, 2^{1}3^{8}$[/hide]

[quote="math154"][quote="Smartguy"][quote="crazypianist1116"][hide="Solution I think..."]Well the way we find the number of factors a number has, we have to prime factor it to the form $ a^xb^yc^z...$ where a, b, and c are prime and x, y, and z are greater than 0. The number of factors that this has is $ (x \plus{} 1)(y \plus{} 1)(z \plus{} 1)...$ since x, y, and z can be zero also. Thus since 2 and 3 are the smallest prime numbers, we have $ 2^{17}, 2^{8}3^{1}, 2^{5}3^{2}, 2^{2}3^{5}, 2^{1}3^{8}$[/hide][/quote]

actually that is incorrect consider the following:
[hide]$ a^2b^1c^1$[/hide][/quote]

$ (2 \plus{} 1)(1 \plus{} 1)(1 \plus{} 1) \equal{} 12$ :wink:

Of course, you might have meant $ a^2b^2c^1$, but I'm not sure.[/quote]

yes that is what i meant

I think I actually get the way to do it now.. thanks guys =D

hey little math kid, do you know what GBML is?

Yeah.. and as you pointed out, this problem's from the last team round

um im in 8th so this seems confusing to me so what grade do you think somehting like this is taught? or is it not taught in school?

I haven't seen this taught in school.. I learned this from books and my own research. Recommend the AOPSWiki for starters, or Wikipedia :P

Also, AOPS books are nice to use.

thanks, that's what i thought. What AoPS books would you think covered this?

Introduction to Number Theory, AoPS 1. Maybe if you actually look at the bookstore it will be obvious. :wink:

You can find find the answer to this problem in AoPS- Art of Problem Solving Volume 1 (not the way to find this exact problem but information that can be applied to solve problems like this one)

Before I start, let me just say that if you have a number, the amount of factors is 1 plus the exponents of its prime factorization (souds weird)..for ex,

 $ 48\equal{}2^4\cdot3^1$ It's exponents are 4 and 1. You add 1 to each to get 5 and 2. The you multiply 5 and to to get 10. So 48 has 10 factors.

This should be applied to answer this problem.

So to have 18 factors, the possibilities are:

$ a^{17}$
$ a^8b^1$
$ a^5b^2$
$ a^2b^2c^1$

As you can see,$ a^2b^2c^1$ will get you the smallest numbers.

Let's start with: $ a^2b^2c^1$
1. $ 2^23^34^1\equal{}144$
2. $ 3^24^22^1\equal{}288$
3. $ 4^22^23^1\equal{}192$
4. $ 2^23^25^1\equal{}180$
5. $ 2^23^26^1\equal{}216$

So the numbers are:

[b]144,180,192,216,288[/b]

I realize that the same method is given as the solution to that problem. I was more or less wondering about any other methods out there

[quote="Nandubetax"]You can find find the answer to this problem in AoPS- Art of Problem Solving Volume 1 (not the way to find this exact problem but information that can be applied to solve problems like this one)

Before I start, let me just say that if you have a number, the amount of factors is 1 plus the exponents of its prime factorization (souds weird)..for ex,

 $ 48 \equal{} 2^4\cdot3^1$ It's exponents are 4 and 1. You add 1 to each to get 5 and 2. The you multiply 5 and to to get 10. So 48 has 10 factors.

This should be applied to answer this problem.

So to have 18 factors, the possibilities are:

$ a^{17}$
$ a^8b^1$
$ a^5b^2$
$ a^2b^2c^1$

As you can see,$ a^2b^2c^1$ will get you the smallest numbers.

Let's start with: $ a^2b^2c^1$
1. $ 2^23^34^1 \equal{} 144$
2. $ 3^24^22^1 \equal{} 288$
3. $ 4^22^23^1 \equal{} 192$
4. $ 2^23^25^1 \equal{} 180$
5. $ 2^23^26^1 \equal{} 216$

So the numbers are:

[b]144,180,192,216,288[/b][/quote]

$ a$, $ b$, and $ c$ must be prime.


I can see/read the posts in the Canada community forum, yet I am not Canadian. (I live in Texas, very far away) Was I accidently added to the forum, or is there something wrong?

There is nothing wrong. The Canadian forum is open to the public  :D. Although we don't encourage non-Canadians to post on it.

[quote="billzhao"]Although we don't encourage non-Canadians to post on it.[/quote]Unless they have reasons to do so ;)


Blah, I'm hoping to get 100% on the math test we took today. There was one question that I wasn't sure of though. I remember it quite well.

There are 2 crates. One crate has 3 oranges and 3 apples while the other crate has 4 oranges and 6 apples. Given that the a person randomly chose a crate and chose an orange from it, what is the probability that the person got it from the first crate?

How I did it was I first found the odds.

The odds of the orange being chosen from crate 1 is 3/6 (3 oranges out of 6 total fruits) to 4/10 (crate 2).

Thus, the probability is (1/2)/(1/2+4/10) = 5/9

I don't think the problem is worded clearly.  What do you mean when you say he chose a crate "randomly"?  Wouldn't the answer be $\frac{1}{2}$?

[quote="Flipp"]There are 2 crates. One crate has 3 oranges and 3 apples while the other crate has 4 oranges and 6 apples. Given that the a person randomly chose a crate and chose an orange from it, what is the probability that the person got it from the first crate?[/quote]

You need to use conditional probability. $P(A|B)=\frac{P(A\cap B)}{P(B)}$. In this case, $P(\text {person got it from first crate}|\text {chose an orange})$ is equal to $\frac{P(\text {person got an orange, from first crate})}{\text {chose an orange}}$ or $\frac{3/6}{(1/2)(3/6)+(1/2)(4/6)}=\frac{6}{7}$, I believe.

@ matt: You are given that an orange was selected. Since it is more likely to select an orange from the first crate, there is a higher probability that the orange was taken from the first crate.
An easy way to see this effect: Consider two crates each with 100 fruit. In the first one, there is only one orange, but in the second, there are 50 of each. If you pick a piece of fruit from a random crate and it turns out it is an orange, from which crate do you think the orange came?
I do agree that there is some ambiguity when they say the person "randomly chooses" a crate. They should specify that there is an equal chance of choosing each crate.

@ chess64: First, P(orange from first crate) in the numerator should be (1/2)(3/6). 3/6 is the probability that an orange is chosen, given that the first crate is already selected.
Then the second term in the denominator should be (1/2)(4/10), since there are 10 total fruit in the second crate.
With these corrections, you get the same fraction as Flipp, who is correct.


να βρεθεί το $ \int\frac {xdx}{(x \plus{} 6)^2(x^2 \plus{} x \plus{} \frac {13}{2})}$

Ανάλυσε σε απλά κλάσματα της μορφής $ \frac{a}{x\plus{}6}\plus{}\frac{b}{(x\plus{}6)^2}\plus{}\frac{cx\plus{}d}{x^2\plus{}x\plus{}\frac{13}{2}}$


How do you do these block path walking problems with holes in them for number of ways from a to b?

hard to explain without a picture, but I'll try...

OK. first, in corner A, put a 1, since there is only 1 way to get to that vertex. Similarly, all of the vertices on the bottom edge and all the vertices on the right edge have only 1 way to get there, so label them with a 1. Now, we do a lot of addition. For any vertex V, we can reach it by going up from the one below it (X) or right from teh one left of it (Y). So, the number of ways to reach V is number of ways to reach X + number of ways to reach Y. Do this addition a lot and build up all the numbers until you finally get a number in vertex B.

Not for the version you've got there, because the hole is so big.  For smaller holes, though, it is possible.  I just posted a few questions in the Intermediate forum inspired by this.

for joel's simple hole.
the number of combinations to the hole times the number of combinations from the hole to the end.

Uh, if you were trying to solve the problems in the intermediate forum , can you please post them there, and more importantly, in spoiler? :) thanks.


calculus
integration
trigonometry
logarithms
function
real analysis
real analysis unsolved

sos, that is not correct... i will use your own solution box with some edits
to show where it failed... :( the answer i get, on paper, unfortunately
seems to involve what is called a [b]generalized [/b][i]hypergeometric function[/i]...

my answer:
$ \textbf I\;\; =\;\;\int_{0}^{1}\;\;\frac{\cot^{-1}\,\left(\sqrt{\;1+x^{2}\;}\right)}{\sqrt{\;1+x^{2}\;}}\;\;\textbf dx\;\; =\;\;\int_{0}^{\frac{\pi}{4}}\;\;\frac{\tan^{-1}(\cos\theta)}{\cos\theta}\;\;\textbf d\theta$
$ \;\; =\;\;\boxed{-\;\textbf K\;+\;\frac{\pi}{2}\cdot\left(\sinh^{-1}\,1\right)\;+\;\frac{1}{6}\cdot\left(\;_{3}\mathcal{HG}_{2}\;\left(0.5,1,1;1.5,1.5;0.1\dot 1\right)\;\right)}$
(i hope the notation i have used above is correct... :|)
[taken from whittaker and watson]
(where of-course, $ \textbf K$ is catalan's constant)

[quote="sos440"]\begin{eqnarray*}\int_{0}^{1}\int_{0}^{1}\frac{dx\, dy}{1+x^{2}+y^{2}}& = & 2\;\int_{0}^{1}\int_{0}^{x}\frac{dy\, dx}{1+x^{2}+y^{2}}\\ & = & 2\;\int_{0}^{1}\int_{0}^{\frac{\pi }{4}}\frac{x\sec^{2}\theta\, d\theta\,dx }{1+x^{2}+x^{2}\tan^{2}\theta }\\ & = & 2\;\int_{0}^{1}\int_{0}^{\frac{\pi }{4}}\frac{x\sec^{2}\theta\, d\theta\,dx }{1+x^{2}\sec^{2}\theta }\\ & = &\int_{0}^{\frac{\pi }{4}}\int_{0}^{1}\frac{2x\sec^{2}\theta\, dx\,d\theta}{1+x^{2}\sec^{2}\theta }\\ & = &\int_{0}^{\frac{\pi }{4}}\;\ln\left(1+x^{2}\sec^{2}\theta\right)\bigg|_{0}^{1}\, d\theta\\ & = &\int_{0}^{\frac{\pi}{4}}\ln\left(1+\sec^{2}\theta\right)\, d\theta\\ &\neq & 2\;\int_{0}^{\frac{\pi}{4}}\ln\tan\theta\, d\theta\;\;\;\quad\text{(LoL!)}\end{eqnarray*}[/quote]

[quote="misan"]can you solve :D

$ \int_{0}^{1}\;\int_{0}^{1}\;\;\frac {\textbf dx\;\;\textbf dy}{1 \plus{} x^{2} \plus{} y^{2}}$[/quote]
$ \int_0^1 {\int_0^1 {\frac {{dx\,dy}}{{1 \plus{} x^2 \plus{} y^2 }}} } \equal{} 2\int_0^1 {\int_0^x {\frac {{dy\,dx}}{{1 \plus{} x^2 \plus{} y^2 }}} } \equal{} 2\int_0^{\pi /4} {\int_0^{\sec \theta } {\frac {r}{{1 \plus{} r^2 }}\,dr} \,d\theta } .$

This is not good.

What is your solution misan?


algebra
polynomial
absolute value
algebra proposed
Sum

... it is also a power sum. But this is incredibly difficult, so let us satiate with the following special case:

Let $a_i; x_i; b_j; y_j \in \mathbb{N}$, $1 \leq i \leq k, 1 \leq j \leq l$, are positive integers such that, for every $n \in \mathbb{N}$, we have
\[
b_1y_1^n + \cdots + b_l y_l^n \mid a_1x_1^n + \cdots + a_kx_k^n.
\]
Prove that there exist integers $c_i; z_i$, $1 \leq i \leq m$, such that
\[
a_1x_1^n + \cdots + a_kx_k^n \equiv \big(b_1y_1^n + \cdots b_l y_l^n \big)\big( c_1z_1^n + \cdots + c_mz_m^n \big)
\]
identically.

[i]Hint: [/i] Again, the advantage we gain in this special case is not due to the fact that the poles of the power sums are integers, but because we are once again in the [i]dominant root case.[/i]  ;)

--Vesselin

Okay, here is my solution. Of course, we may assume WLOG that $y_1 > y_2 > \cdots > y_l$. If the number
\[
\frac{a_1x_1^n + \cdots + a_kx_k^n}{b_1y_1^n + \cdots + b_ly_l^n}
\]
is an integer, then so is the number
\[
\frac{a_1x_1^n + \cdots + a_kx_k^n}{1 + \frac{b_2}{b_1}\big( \frac{y_2}{y_1} \big)^n + \cdots + \frac{b_l}{b_1}\big( \frac{y_l}{y_1} \big)^n } = b_1y_1^n  \frac{a_1x_1^n + \cdots + a_kx_k^n}{b_1y_1^n + \cdots + b_ly_l^n}
\]

Denote $a(n) = a_1x_1^n + \cdots + a_kx_k^n, b(n)= -\Big( \frac{b_2}{b_1}\big( \frac{y_2}{y_1} \big)^n + \cdots + \frac{b_l}{b_1}\big( \frac{y_l}{y_1} \big)^n \Big)$. Then, the poles $\frac{y_i}{y_1}$ of the power sum $b(n)$ are rational numbers with absolute value less than $1$, and $\frac{a(n)}{1 - b(n)} \in \mathbb{Z}$ for all $n$. The goal is then to prove that $\frac{a(n)}{1 - b(n)}$ is a power sum as well, i.e. that it satisfies some linear recurrence relation (with rational coefficients and with no multiple poles, of course).

We have, for all $\ell \in \mathbb{N}$, the following expansion:
\[ (*) \quad
\frac{a(n)}{1-b(n)} = \Big( a(n) + a(n)b(n) + a(n)b(n)^2  + \cdots + a(n)b(n)^{\ell} \Big) + \frac{a(n)b(n)^{\ell+1}}{1 - b(n)} = C_{\ell}(n) + R_{\ell}(n).
\]
Evidently, $C_{\ell}(n) := \sum_{0 \leq i \leq \ell} a(n)b(n)^i$ is a power sum with [i]rational[/i] poles, so it satisfies some linear recurrence relation with integer coefficients. That is, there exists an integer polynomial $P_{\ell} \in \mathbb{Z}[x]$ such that, if $E$ denots the 'shift' $E: f(n) \to f(n+1)$, then the operator $P_{\ell}(E)$ [i] annihilates[/i] $C_{\ell}(n)$. Also, because $\frac{a(n)}{1 - b(n)} \in \mathbb{Z}$ and $P \in \mathbb{Z}[x]$, we have $P_{\ell}(E)\big( \frac{a(n)}{1  - b(n)} \big) \in \mathbb{Z}$. By (*), this integer equals
\[ (**) \quad
P_{\ell}(E)\big( \frac{a(n)}{1  - b(n)} \big) = P_{\ell} (E) \big( \frac{a(n)b(n)^{\ell + 1}}{1 - b(n)} \big).
\]
Now comes the crucial point: because all poles of $b(n)$ have absolute value [b]strictly[/b] less than $1$, by choosing $\ell$ large enough we may ensure that all poles of $a(n)b(n)^{\ell + 1}$ also have absolute value strictly less than $1$. Choose $\ell$ sufficiently large so that this holds, and fix it. Then, letting $n \to \infty$, we obtain that $a(n)b(n)^{\ell+1} \to 0$, and hence $\frac{a(n)b(n)^{\ell + 1}}{1 - b(n)} \to 0$. As $\ell$ is fixed, we certainly have that
\[
\Big| P_{\ell}(E) \Big( \frac{a(n)b(n)^{\ell + 1}}{1 - b(n)}  \Big) \Big| < 1
\]
for $n$ large enough. But this was an integer, so it must be equal to $0$. By (**), we obtain that all sufficiently large $n$ satisfy
\[
P_{\ell}(E) \Big( \frac{a(n)}{1  - b(n)}   \Big) = 0,
\]
i.e. the operator $P_{\ell}(E)$ annihilates $c(n) := \frac{a(n)}{1 - b(n)}$. Hence, $c(n)$ is also a power sum. QED

--Vesselin

P.S. I hope you like this series of problems  ;)

I adore them, vess, but they are too difficult for many of us. Anyway, thanks for the great solutions given.


algebra
polynomial
number theory open
number theory

Is is true that $ x^n\equiv a$ ($ mod$  $ p$) $ \Longleftrightarrow$ $ a^{\frac {p \minus{} 1}{(p \minus{} 1,n)}}\equiv 1(mod p)$ and is there a proof for it?

Yes.  You can use the fact that $ (\mathbb{Z}/p\mathbb{Z})^{*}$ is cyclic of order $ p\minus{}1$ or you can factor the polynomial $ t^p \minus{} t$ over $ \mathbb{F}_p$.

Can you send me a link for this kind of  theory because I'm not very familiar with it.


A square-shaped hole is cut through the center of a cube and is perpendicular to two opposite faces. Find the sum of the number of vertices, the number of faces, and the number of edges of the remaining solid. (Note: any point where three or more faces meet is a vertex.)

It is not difficult to visualize this: the number of faces is $ 6\plus{}4\equal{}10$, the number of vertices is $ 8\plus{}8\equal{}16$, and the number of edges is $ 12\plus{}12\equal{}24$, so the sum is $ 10\plus{}16\plus{}24\equal{}\boxed{50}$. This could also be thought of as a cube and a rectangular prism so we just double to get $ 2(6\plus{}8\plus{}12)\equal{}52$, but then subtract the $ 2$ faces that are just empty holes.