another problem i have tried long enuf  and havent come up with anything good

Problem code: ONEZERO

Certain positive integers have their decimal representation consisting only of ones and zeros, and having at least one digit one, e.g. 101. If a positive integer does not have such a property, one can try to multiply it by some positive integer to find out whether the product has this property.
Input

Number K of test cases
in each of the next K lines there is one integer n (1 <= n <= 20000)
Output

For each test case, your program should compute the smallest multiple of the number n consisting only of digits 1 and 0 (beginning with 1).
Example

Input:
3
17
11011
17
Output:
11101
11011
11101

So, we store powers of 10 modulo n. We then just need to keep checking if the first x powers have a subset adding to 0 mod n. As far as I'm aware something like this is NP-complete (not 100% sure, though), but the bounds are probably small enough to just go through all the possibilities (say, storing all possible sums in a boolean array. Of course, you would then have to go through again afterwards to find the actual answer.)

now since i have an ACC on this one(in .67 sec) here's the hint
consider the graph with vertices values from 1....n .now traverse this graph using a bfs and keep storing  the parent of every vertex... break from the bfs as and when u reach the vertex n.... contruct the string of 1's and 0's using the parent array..


These are from 2006 AMC 8..
I need solutions.

24. In the multiplication problem below, $ A,B,C,D$ are different digits. What is $ A\plus{}B?$

$ \ \ ABA$
$ \times \ CD$
$ \minus{}\minus{}\minus{}\minus{}$
$ CDCD$

I know that $ A$ has to be 1... Is there an easier way to do this?
[hide="Answer"]1[/hide]

25. Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. THe sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?

I had no clue on this question...
[hide="Answer"]14[/hide]

Maybe this should go in the AMC section.

Anyway, the second one:
[hide="Solution"]
Notice that 2 of the numbers are even, one is odd. For the 3 sums to be equal, they must all have the same parity. And there is only one even prime number, so the middle card must have the prime number 2. So now you know that 44 + x = 61 and 38 + y = 61. Solve:
x = 17
y = 23

So now you average them. 42/3=

$ \boxed{14}$

That was a pretty cool prolem, actually.[/hide]

24. Lets call ABA 101A+10B and CD 10C+D and CDCD 1010C+101D.

We are given that A=1, so (101+10B)(10C+D)=1010C+101D and 10B=0. B=0. 1+0=1. answer:1

[quote="alanchou"]For number 1, you can also divide $ CDCD$ by $ CD$, which is obviously 101, assuming that the letters are digits.[/quote]
slick.  :)

I originally bashed away with algebra, then with a little time left, I realized I could have done that and saved like 3 minutes.

Which should have been enough to catch at least one of my stupid mistakes...

[quote="alanchou"]For number 1, you can also divide $ CDCD$ by $ CD$, which is obviously 101, assuming that the letters are digits.[/quote]

Now that is cool...


I was just wondering, if you lay the piano, how long have you been playing, and what piece are you working on right now?

I've been playing for about 5.5 years, (6 in August), and am right now playing Hyden's 47th Sonata, according to my book.  For some reason, it's a different number for different books, which makes it really confusing.

6 years, playing Chopin's Fantasie Impromptu

Around 7ish years.  Haydn's Sonata in E Minor Hob. XVI/34; L.53.

 I just got a 1 at the Ohio OMEA competition for that song (best possible rating)! :D

By the way: How do you 'lay' the piano? :P

7 years, I'm playing the Grieg Piano Concert at the moment!

Started when I was 5.. so 11 years? but I haven't been playing steadily.
The most recent classical piece I played was Beethoven's Moonlight Sonata 3rd Movement.

Playing 11 years also :D  Currently playing Liszt's Hungarian Rhapsody #2

4 years, I started late.  Currently playing Chopin's 'Grand Valse Brillante'

Chopin is a genius.

6 or 7 years (I forget which  :D )
Currently playing Beethoven's Moonlight Sonata, trying to play Chopin's Fantaisie-Impromptu, Easy Living (jazz, by Ralph Rainger and Leo Robin), Macdowell's Hungarian, Debussy's Prelude 1....

Chopin is the best.

Played 7 years, currently working on Prokofieff's "Diabolical Suggestions"

And Chopin is Awesome. :)

I've been playing about 6 years. Playing a mozart song. Forgot the name and I'm too lazy to check. :P

I've played for 8-9 years...kind of quit now.  I started when I was like 5, and don't really play anymore.  Last piece I played was accompanying a friend in his cello competition...got $20 dollars  :D

[quote]Played 7 years, currently working on Prokofieff's "Diabolical Suggestions"[/quote]
It's great, isn't it.  Gotta love the minor ninths part.

Anyways, 11 years
Currently working on Brahms Rhapsody #1, Beethoven Sonata Op. 10 No. 3, Ginastera's Danza's Argentinas, and Chopin Etude no. 3 in E-major (not sure which op. number).
Anybody in here do Guild Auditions?

Fantaisie-Impromptu rocks! I learned it a year and a half ago and still play it often.

Playing for 9 years I think...

Currently working on Rhapsody in Blue and Scriabin's etude op 42 no 5

[quote="pteromys"]Playing on and off since late 1996, so a little over 9 years at the time of this writing...
Taking lessons for about 8 years.
Currently working on a sonata by Mozart (K. 284 #6?) and a polonaise by Chopin (Op. 44 #5?).

Eryaman: Was the $20 a share of the competition prize money, or did your friend just pay you from his pocket to accompany him?[/quote]No, my friend gave it to me for helping him, since he was desperate and we only had two weeks to prepare.

Level $1$ pieces that my piano teacher forces me to play despite my more advanced ability.

I practice Moonlight Sonata Movement $3$ and La Campanella in my spare time...

I also practice pieces that I wrote.

Gasp!
Moonlight Sonata Movement 3?!?!?!
That piece is insanely hard when played at full speed!
I haven't reached that level of playing yet, but I hope to soon.

I play Haydn's sonata in E minor and C major, third and second movement.

I have been playing for technically 9 years now but technically only like 4 of those years were actually with my new, much better teacher (she's also really strict), and I'm playing Chopin's Etude Op. 25 No. 1, Hexentanz/Witches Dance by MacDowell (it's easy and I hate that but I actually really like it), and one of Haydn's Sonatas.

I've been playing the piano for 4 years, but my actual training started from 2021. So, I've finished Moonlight Sonata 1st movement and I'm going to start the grade 6 practical practice. I attempted the grade 5 exam about 6 months ago or so, and I got 89%.

6 years, trying to learn every mozart piano sonata

[quote=Pi31][quote=iadzh]ive been playing piano for 6 years and im playing Waltz by Fredric Chopin (look it up on YouTube)[/quote]

But he wrote multiple Waltzes right[/quote]

ummm :| I don't know  ;)

Lots of Bach and a bit of Chopin and Grieg right now in my repertoire, played for 8 years

Been playing since i was 3. so... oh wow. 10 years. I can play Mozart and some Beethoven


function
linear algebra
matrix
calculus
derivative
linear algebra unsolved

Let $ f$ be a $ C^2$ function from an open convex subset of $ R^n$ to $ R^n$. $ f$ is said to be convex if $ f(tx_1 \plus{} (1 \minus{} t)x_2) \le tf(x_1) \plus{} (1 \minus{} t)f(x_2)$ for any $ x_1, x_2$ and any $ t \in [0 ,1]$
Prove that $ f$ is convex iff $ d^2_f$ is positive semidefinite

I'll use $ H(x)$ to stand for what you called $ d_f^2.$ It's the Hessian matrix - the symmetric $ n\times n$ matrix whose entries are $ \frac{\partial^2f}{\partial x_j\partial x_k}.$

Fix $ x_1$ and $ x_2.$ Define the function $ g$ on $ [0,1]$ by $ g(t)\equal{}f(tx_1\plus{}(1\minus{}t)x_2).$ Note that the definition of $ f$ convex given above is precisely the statement that $ g$ is an ordinary convex function for all choices of $ x_1$ and $ x_2.$ Use $ x(t)$ to mean $ tx_1\plus{}(1\minus{}t)x_2.$

I claim that the chain rule for differentiation shows that $ g\in C^2$ and

$ g''(t)\equal{}(x_1\minus{}x_2)^TH(x(t))(x_1\minus{}x_2).$

$ g$ is an ordinary convex function iff $ g''(t)\ge 0$ for all $ t$ (at least for $ C^2$ functions.)

Since we may choose $ x_1$ and $ x_2$ freely, we may arrange that any particular point $ x$ appear as $ x(t)$ for some choice of $ t,$ and we may choose to do that with $ x_1\minus{}x_2$ pointing in any arbitrary direction. Hence, for $ H\equal{}H(x(t)),$ since $ g''(t)\ge 0,$ we have that $ u^THu\ge 0$ for all choices of $ u.$ Hence $ f$ convex implies that $ H$ is positive semidefinite. Conversely, if $ H$ is positive semidefinite, then $ g''(t)\ge 0$ for all choices of $ x_1,x_2,$ and $ t,$ and hence $ f$ is convex.


For which integer $n$ , $11(14)^n+1$ is a prime number?

Let's denote $p=11 \cdot 14^n +1$. It is obvious that $p>11$.

If $n$ is even, then $p \equiv (-1) \cdot (-1)^n +1 \equiv (-1)^{n+1} +1 \equiv -1+1 \equiv 0 (mod3)$. So $3|p$, which means that $p=3$, but that is impossible, because $p>11$.
If $n$ is odd, then $p \equiv 1 \cdot (-1)^n +1 \equiv -1+1 \equiv 0 (mod5)$, which means that $5|p$ or $p=5$, but that is also impossible, as $p>11$.

So in both cases, $p$ cannot be a prime, so there is no solution for $n$ either.

Where did you get $\displaystyle p \equiv (-1) \cdot (-1)^n +1 \equiv (-1)^{n+1} +1 \equiv -1+1 \equiv 0 (mod3)$ from?

Same thing for n is odd case. How did you wind up with that equivalency?

Well it's using 11 and 14.  $11 \equiv 14 \equiv -1 \mod 3$ and similar for 5.  Nice solution by the way.  I'm not sure I would have been able to quickly see the use of those particular mods.

It's just a routine. Always(!) when you have this kind of problems, try putting different $n$-s in the expression and thus figuring out something particular for a particular $n$. For example, in our case for $p$, i just put in some $n$-s and then found out the divisibility. The only thing left to do then is to prove it.  ;)


In $\triangle ABC$, $D$ is on $BC$ such that $2BD=DC$. $F$ is on $AD$ such that $2AF=FD$. $BF$ intersects $AC$ at $E$. Find the ratio of $AE$ to $EC$.

[hide]its 1/6, ill give a solution later[/hide]

Can someone post a solution to this, please?


superior algebra
superior algebra unsolved

let $ G$ be a finitely presented group and $ p : G \to G$ a projection, i.e. a homomorphism with $ p^2\equal{}p$. is then $ im(p)$ finitely presented?

(it would be cool if this holds because this leads to a nice categorical definition of finitely presented)


[quote="elcric"]Could someone please explain the solution for the following problem?  Thanks in advance![/quote]
first, you might want to convet all the grams into mol. then decide which one is the limiting reactant, which one is excess based on the balanced equation.then, use the mole of the limiting reactant to figure out the mol of water.
mol of water * molar mass of water = gram of water produced.

Could you please show your work?  I followed your steps, but I need to make sure that everything I did is correct.

Thanks a lot!

I dont mean to sound rude or anything but you should post your working based on what i_t_f_k said and someone can comment on it otherwise its almost like other people doing your homework for you

[hide]i didnt have a calculator at hand so i might have gone wrong in many places but i calculated 2 moles of $ B_5 H_9$ reacts with 12 moles of $ O2$ which is the same ratio of moles as in the equation so 9 moles of water should be produced.[/hide]


integration
ratio
calculus
calculus computations

integrate secx^4 dx

tanx/(a+b*[(tan(x))^2] )

[(sinx)^5]/( (cosx)^4 )

hello, do you want to calculate $ \int\sec(x)^4\,dx$?
Sonnhard.

All the powers in the ratios are of the ratio not the angle.

[quote="Zed_darkx"]tanx/(a+b*[(tan(x))^2] ) [/quote]

multipley and divide with cos^2 x. that will do


Let $a_1=21$ and $a_2=90$, and for $n\geq 3$, let $a_n$ be the last two digits of $a_{n+1}+a_{n+2}$.
What is the remainder of $a_1^2+a_2^2+...+a_{2005}^2$ when it is divided by $8$ ?

[hide]
Notice that $x^2 \equiv (x-4)^2 \mod{8}$. So if $a_k+a_{k+1} \ge 100$, the new $a_{k+2} \equiv a_k+a_{k+1}-4 \mod{8}$ and thus the square is the same as if you just added and didn't take the last two digits.

So then we have the sequence ($\mod 8$): $5, 2, 7, 1, 0, 1, 1, 2, 3, 5, 0, 5, 5, 2,...$ which we find repeats every $12$ numbers with a sum of $0 \mod{8}$. Thus we take $2005 \equiv 1 \mod{12}$ so the sum is $5^2 \equiv 1 \mod{8}$.[/hide]


algebra
polynomial
absolute value
algebra unsolved

Prove that the polymomial: $z^n+\pi\cdot i\cdot z^{n-1}+\pi\cdot i\cdot z-1$ connot have roots of absolute value greater than 1.

Is it correct? I might have made some mistake in my computation but it seems that when $n=4$ and $z=ri$, we have
$z^n+\pi iz^{n-1}+\pi iz-1=0 \iff f(r)=r^4+\pi r^3-\pi z-1=0$
But $f(\pi)=-\pi^2-1$ while $f(4)>0$

Actually $f(\pi)=2\pi^4-\pi^2-1 > 180$. By the way.. this polynomial cannot have any real roots. Because, else we would have that $x_0^n-1=0$ and $x_0^{n-1}+x_0=0$ for some $x_0$ which have no common root. Bu I still don't know why it cannot have complex roots of absolute value greater than 1.
Can anyone help me with this problem?

The statement is not true. For $n=4$ solutions are $\pm i,-i\frac{\pi \pm \sqrt{\pi^2-4}}{2}$. One of them has the absolute value $>1$.

Thank you! It was a problem from my teatcher and I didn't think that it was wrong. I'll show it to him!


Find the sum of the coefficients in the expansion of $ (2x\minus{}3y\plus{}3)^{4}$.

[hide]Well you could calculate each term fairly easily, as it is only raised to the fourth power,  but the nicer and more general solution is to let x and y equal 1, as when you expand it that will just be the sum of the coefficients. So thats 16 I believe[/hide]

[hide]Intuitively, let $ f(x, y)\equal{}2x\minus{}3y\plus{}3$. Then if we raise $ f$ to the fourth power, we get another polynomial. Since the sum of the coefficients is taken when $ x\equal{}y\equal{}1$, the sum of the coefficients of $ (2x\minus{}3y\plus{}3)^{4}$ is $ (2\minus{}3\plus{}3)^{4}\equal{}16$.[/hide]


hi guys ...
can any body suggest me a good book on geometry & math
olympaid problems .... for undergrads....

thank u..

Coxeter and Greitzer's "Geometry Revisited" is good for geometry. and a good olympiad training book is Titu Andreescu's and Razvan Gelca's "Mathematical Olympiad Challenges"

In a more specific subject, "The Real projective Plane" by Coxeter is also good.


May be stupid... so apologies in advance.

In a kingdom a disease is spreading and the king orders that all people of the kingdom be tested for it.

A girl goes for the test and tests positive. She starts crying that she has tested positive for the disease and her life is over now.
The doctor comes and tells that the test results are only 99% accurate.

Question: What is the probability that the girl is actually infected with the disease?

[quote="blahblahblah"]I think you need to know what proportion of people have the disease to begin with.[/quote]

unfortunately thats  the only data given... i also dont know the answer...

if u what proprtion have the disease, then  how to solve? say x% have the disease to begin with...

[quote="dwx314"][hide]4 Cases:
The test is positive, She has the disease
The test is positive, She doesn't have the disease
The test is negative, She doesn't have the disease
The test is negative, She has the disease

We want cases 1,2
Case one: x/100*99/100 = 99x/10,000
Case two: (100-x)/100*1/100 = (100-x)/10,000

Add both cases:
(99x+(100-x))/10,000 = 98x-100

Divide:
99x/(98x-100)

[/hide][/quote]

I think that x<1, does this not make the final answer negative?


Does anybody know where to find MATHCOUNTS National Competitions?

Also, where can you find the 1999, 1998 etc. MATHCOUNTS Competitions (State and Chapter)?

Thanks!  :D

I don't really know how to prepare for MATHCOUNTS.

Trade with other people.  I got nearly everything from 1984-2008 that way.  Same for your other question.

[quote="JBoyd"]Only concentrate on the test from 1990 on.  Before that time the tests were in a different format.  The sprint was 40 questions and there were 5 sets of targets questions that you only got 4 minutes on.  The only part of the test that you might be able to use are the team rounds.[/quote]

I recommend 1991 on.  1990 is basically like a chapter level test.


geometry
3D geometry
number theory proposed
number theory

A natural number is called [i]good[/i] if and only if it is the sum of a perfect square and a perfect cube. Prove that if $ r,s$ are any positive integers, then

(a) there exist infinitely many positive integers $ n$ such that both $ r\plus{}n$ and $ s\plus{}n$ are good.

(b) there exist infinitely many positive integers $ n$ such that both $ rn$ and $ sn$ are good.

1) there are infinitely many solution to the equation $ a^3\plus{}b^2\equal{}c^3\plus{}d^2\plus{}x$ for any x. these are
    e.g.   $ a\equal{}t\plus{}1, b\equal{}(3t^2\plus{}3t\minus{}x)/2, c\equal{}t, d\equal{}(3t^2\plus{}3t\minus{}x)/2\plus{}1,$ for even x
    and similarly for odd x

2) $ n\equal{}r^2s^{6k\plus{}3}\plus{}r^{6k\plus{}3}s^2$


function
vector
real analysis
real analysis unsolved

Prove that any real function $ f$ that is increasing and concave is subadditive, i.e. if $ f: X\rightarrow R$, for $ X$ a vector space in $ \mathbb{R}$, then $ f(a\plus{}b)\leq f(a)\plus{}f(b)$ for all $ a,b\in X$.

Your question makes no sense since we dont know what kind of order in  "  The vector space X"

Well, the usual addition in \mathbb{R}, I mean $ X$ is a subspace of $ \mathbb{R}$...
I guess if you prove it for $ X\equal{}\mathbb{R}^{\plus{}}$ for example, the proof can be generalized to other vector spaces.

Does $ \mathbb{R}^\plus{}$ mean the set of positive real numbers?  That's not a vector subspace.

Anyway, for $ X \equal{} \mathbb{R}^\plus{}$, doesn't a function like $ f(x) \equal{} \sqrt{x} \minus{} 1$ show that the problem is incorrect?

alright, I must be tired. Sorry for being silly.
To be clear, if $ f: [0,\infty)\rightarrow \mathbb{R}$ is increasing, concave and $ f(0)\equal{}0$ then $ f$ is subadditive.
Can you prove it and/or find a generalization of this?

Then the assertion is true, but we don't need the increasing hypothesis.  By concavity, we have
\[ f(a) \ge \frac{a}{a \plus{} b} f(a \plus{} b).
\]
Similarly for $ f(b)$.  Then just add.


Let $ABC$ be a triangle and let $M,N,P$ be points on the sides $BC$, $CA$ and $AB$ respectively such that \[ \frac{AP}{PB} = \frac{BM}{MC} = \frac{CN}{AN}. \] Prove that triangle if $MNP$ is equilateral then triangle $ABC$ is equilateral.

Hey, this problem was given in local Olympiad in Constantza 2003. :o

I am sorry ! I think that here was $\mathrm {\ an\ acute\ absence\ of\ the\ imagination\ (inspiration)\ !}$

Let $x+y = 1$, where $AP = x * AB, PB = y * AB$.  Let $s = |PN|$.

Lets calculate $s^2$ directly using cosine law. It's $(xc)^2 + (yb)^2 - 2xybc \frac{c^2 + b^2 - a^2}{2bc}$

or $(xc)^2 + (yb)^2 + xy(a^2 - b^2 - c^2)$.

We get similar equations for PM, MN.  Adding them up, we get $3s^2 = (\sum a^2)(x^2 + y^2 - xy) = (\sum a^2)(1 - 3xy)$.

Now $[APN] = xy[ABC]$, whereby $[PMN] = [ABC](1-3xy) = \frac{ \sqrt{3} }{4} s^2$.

It follows that $\sum a^2 = 4\sqrt{3}[ABC]$.  Squaring both sides, we get $\sum a^4 + \sum a^2b^2 = 6\sum a^2b^2 - 3\sum a^4$.  It's equivalent to $\sum a^4 = \sum a^2b^2$, which occurs only when a=b=c.  Hence, ABC is equilateral.


AMC
USA(J)MO
USAMO
AMC 12
ARML
AIME
AMC 8

This is very strange, but I made a comparison, and if I did not miss anything only like 2 of USAMO winners got perfect on the AMC. How can that be?

The two are different.  AMC is a faster, easier test.  USAMO is a slower, harder test--and it requires proof-writing skills and I have a serious need of more of those.

You can do well on AMC and not so well on USAMO, and someone who did worse than you on AMC can do better on USAMO.  It depends on what you're good at.

but still shouldnt USAMO winners have perfect amc scores?

Not necessarily. Getting a perfect score on the AMC is difficult for anyone, simply because of the time constraints and the chance for a careless error. And like bubala said, there isnt a high correlation.

actually the correlation between scoring perfect on the AMC and doing well on the USAMO doesn't seem so bad to me.  Of course neither is a guarantee for the other.

I'd say that if the AMC12 were a 9 hour test like the USAMO, you can expect that the Winners, Honorable Mention, and a good deal of the rest of the qualifiers would generally all have perfect scores. If not, then a 140 or above is definitely expected.

If AMC were 100 minutes, there would be an explosion in additional perfects. It's simply that 75 minutes isn't a very long time, and you want to spend a decent amount of that time checking your work to avoid silly mistakes.

Also success on the AMC doesn't ensure future success on the AMC either. I got a perfect as a freshman on 12A (easy year but still) and I never managed to repeat the feat.

In fact, many of the prefect scoring AMC takers didn't even get to take the USAMO.

a friend of mine who got honorable mention on the USAMO got a 146.5 on the AMC12A.

it's not perfect, but it's as close you can get

[quote="darktreb"]Technically I'm 11th grade but I went to a 3 year boarding school after Jr. High so I'm going to college next year.[/quote]
He should have guessed on the problem he didn't answer  :D .  In terms of prestige, the difference between a 144 and 146.5 is much less than the difference between a 146.5 and a 150.

I agree with this!  :D  A 146.5 makes you look like a wuss, while a 150 stands out (especially on the AMC 12A)!

Well, actually, I might be closer, unless his/her USAMO score was 28. I had 150 AMC and 26 USAMO.

-Adam Hesterberg

I think Sherry Gong is the highest scoring AMC perfect scorer on the USAMO, since she was a winner.  But shouldn't be any suprise, since she got a silver medal on the IMO in eighth grade (but with Puerto Rico).

I suppose having trouble writing proofs would cost a student time, but I think a good problem solver should have no problem learning to write proofs within a year or so if he or she really wants to.

I agree.  In professional mathematics, speed and avoiding careless mistakes aren't nearly as important as pure problem solving (and unfortunately proof writing) skills.

[quote="D.P.L"]Huh how could someone who score 41 in USAMO and not get in to the black group :huh:[/quote]

Hmmm it is possible that someone who scored 41 on USAMO did not score 41 every time he took it. I don't know though, I'm just saying it might be a possibility.

hmm. I respectfully disagree. I'm pretty sure if you score 41, you have to score 41 every single time (both past and future)?

gobble!

wait a second are you saying that zhaizhai won't get a 42 this year

zhaizhai... classy.

What the dickens, man. We've had this conversation before. No reviving really old threads unnecessarily. Especially without looking at the DATES of the posts.

If I recall correctly, he turned 18 last month.

[quote="Phelpedo"]But...he's in 11th grade. That's cheating  :roll:[/quote]

Wait...he's in 12th grade, but anyway, how is that cheating?

Why is Iura even being discussed in this thread!?

DUDE IURA IS NOW MY NEW IDOL!!!

MY NON AMERICAN IDOL!

I'm not sure why this thread was revived after over 18 months of dormancy, but it's already been covered in excess, so it's done now.


did u like delete all the vowels? it is quite hard to read

Actually its easy to write.Sorry if u have felt any problem.
>tarik adnan

[quote="Moonmathpi496"]Here is a prob.Its from bangladesh math olympiad selection level(divitional)

if ABC is a triangle nd D nd F  are 2 points on AB nd E is a point on AC such that DE nd BC are parallel DC nd EF are parallel nd if AF=4 DB=15 DF=??
Frndz try 2 solve this ez one
nd plz send solu as text not image
>tarik adnan
i love math i love bangladesh[/quote]

 :mad:

please rewrite the problem in an appropriate way

[quote="Moonmathpi496"]Here is a prob.Its from bangladesh math olympiad selection level(divitional)

if ABC is a triangle nd D nd F  are 2 points on AB nd E is a point on AC such that DE nd BC are parallel DC nd EF are parallel nd if AF=4 DB=15 DF=??
Frndz try 2 solve this ez one
nd plz send solu as text not image
>tarik adnan
i love math i love bangladesh[/quote]
I think I can translate.
If ABC is a triangle and D and F are 2 points on AB and E is a point on AC such that DE and BC are parallel, DC and EF are parallel and if AF=4,DB=15. What does DF equal?
Hope that helped.

[hide]
let $DF = x$
$DE \parallel BC$, so $\triangle ADE \sim \triangle ABC$ $ \Longrightarrow \frac{CE}{EA}=\frac{BD}{DA}=\frac{BD}{DF+FA}=\frac{15}{x+4}$
$FE \parallel DC$, so $\triangle AFE \sim \triangle ADC$ $ \Longrightarrow \frac{CE}{EA}=\frac{DF}{FA}=\frac{x}{4}$
so $\frac{CE}{EA}= \frac{15}{x+4}= \frac{x}{4}$
$x=6$
[/hide]

Thank you for that translation.But i like to write like that because that saves time.And i dont like to waste time.
And thank you for the solution.
Here is another easy problem
find the last two digits of 3^999
its another bdmo problem
>moon
i love math i love bangladesh


Assuming it's degrees, you can pair them

$\sin1+\sin89+ \sin2+\sin88+ ... +\sin44+\sin46+ \sin45$

which is the same as...

$\sin1+\cos1+ \sin2+\cos2+ ... +\sin44+\cos44+ \frac{\sqrt{2}}{2}$

I'm not really sure where to go from there or even whether that helps...

[hide]
Hmm... doesn't seem to simplify that much, or maybe I did something wrong :oops:

$S = \sin 1 + \sin 2 + ... + \sin 88 + \sin 99$

$-2\sin (1) S = -2\sin 1(\sin 1 + \sin 2 + ... + \sin 88 + \sin 99$

Using the idendity $-2\sin(x)\sin(y)=\cos(x+y)-\cos(x-y)$

$-2\sin (1) S = \cos 90 - \cos 88 + \cos89 - \cos 87 + ... + \cos 4 - \cos 2 + \cos 3 - \cos 1 + \cos 2 - \cos 0$

The sum telescopes into

$-2\sin (1) S = \cos 90 + \cos 89 - \cos 1 - \cos 0$

$S = \displaystyle \frac{\cos 89 - \sin 89 - 1}{-2\sin 1}$

[/hide]

[quote="chess64"][hide="generalization"][u][b]Lemma.[/b][/u] $\sin \theta + \sin (\theta + k) + \sin (\theta+2k) + \cdots + \sin (\theta + nk) = \frac{\sin \left(\frac{1}{2}(n+1)k\right)\sin\left(\theta+\frac{nk}{2}\right)}{\sin \frac{k}{2}}$.

The proof is complicated, I might post it if you want. Basically, we want the imaginary part of the sum \[ \cos \theta+i\sin\theta + \cos (\theta+k)+i\sin (\theta+k)+\cdots+\cos(\theta+nk)+i\sin(\theta+nk).  \] We can simplify with de Moivre's and stuff.[/hide][/quote]
There's a similar formula, and I believe it's a bit easier to prove.
[hide]$\sin{\theta}+\sin{2\theta}+...+\sin{n\theta} =\frac{\sin \left(\frac{n+1}{2}\theta\right) \sin \left(\frac{n}{2}\theta\right)}{\sin\left(\frac{\theta}{2}\right)}$
[/hide]

[hide]Convert the expression into exponential form and it simplifies to

$\frac{\cos 89 + \cos 1 - 1}{2-2\cos1}$[/hide]

Ummm....this might help you out.  See attachment.

EDIT: I see this idea has already been posted, but I'll leave up the PDF anyway...

[quote="Iversonfan2005"]Find the value of $\sin1+\sin2+\sin3+......+\sin88+\sin89$[/quote]

[hide]
this is the imaginary part of $e^{\frac{n\pi i}{180}}$ where $n = 1\ldots89$
Let $w = e^{\frac{\pi i}{180}}$
then the equation is the real part of $w + w^2 + \ldots w^{89}$ which factors into:
\[ w\frac{w^{89} - 1}{w - 1} = \frac{i - w}{w - 1} = \frac{i(1 - \sin{1^{\circ}}) - \cos{1^{\circ}}}{\cos{1^{\circ}} - 1 + i\sin{1^{\circ}}} \]
and well, you can rationalize the denominator to get the same answer.
[/hide]


inequalities geometry ratio ARML trigonometry inradius circumcircle

By any chance, do we have an SC Math Team Website? (a real one, not the un-updated one mysmartmouth made) I know for the past years, we've had websites like this:
http://155.225.48.46/statemat/stuf2005/teampage.htm

This year though I don't know if we have one or not because I was trying to look for the list of All State Qualifers/ARML members but I couldn't find anything :(

Oh, and btw, tommorrow my lecture is going to cover the following topics if anyone wants to review stuff.

Introduction to Triangles
-Triangle Inequality
-Triangle Area Formulas (Heron's Formula)
-Altitude/Base area ratio, etc.
-Brief mention of Stewarts Theorem, Menelaus Theorem, and Ceva's Theorem

Similar Triangles
-side ratio/area ratio
-ARML practice problem set

Trig
-Law of Cosines
-Law of Sines/Extended Law of Sines
-ARML practice problem set

Angle Bisector Theorem
-ARML practice problem set

Miscellanious Theorems
-Shoelace Theorem
-Pick's Theorem

Introduction to Circles
-Circle Area/sectors
-ARML practice problem set

Power of Point
-ARML practice problem set

Inscribed Figures
-Inscribed Polygons
-Inradius
-Circumradius
-ARML practice problem set

Cyclic Quadrilaterals
-Ptolemy's Theorem
-Brahmagupta's Formula

Team ARML practice problem set

Power Round ARML practice problem

All in all, the entire lesson should last about 4.5 hours. There are 6 individual problem sets, which should take about 150 minutes. There's a team problem set which takes another 30 minutes, and the Power Round question will be about 20 minutes. The other 70 minutes I'll be lecturing about the topics themselves and presenting problems (the problems will be given before the individual problem set so that people feel comfortable doing them). And it's not like the Individual problems are graded or anything, in fact I let a few people have a copy of the individual problems to work on before hand. But the team problems are for a contest between three teams of 6 or 7 people along with the power round.

No, there is no site like that this year. Those sites used to be created by the previous organizer of the SC Math Team, Dave Troutman.

I think mysmartmouth's website would be a great "official" Team site if we added some new information to it. I'll see if I can get Mr. R to publish a list of the offical SC All-State Math team so that everyone can see it.


Before Rick open his gym locker, he must remember the combination.Two of the numbers of the three-term sequence are 17 and 24, but he has forgotten the third, and does not know the order of the numbers. He remembers that all the numbers were between 1 and 40. At ten seconds per try, at most how long will it take him to exhaust all possibilities?

it really isn't that tough. if the other number is 17, there are 3 possibilities. similarly, there are 3 possibilities if the other number is 24.

if the other number, $ x$, is different from 17 and 24, there are 3!=6 ways to order 17, 24, and $ x$. there are 40-2=38 possibilities for $ x$, so there are $ 6\cdot 38 \equal{} 228$ possibilities in this case. so 228+3+3=234 possibilities in total. so it would take 2340 seconds, or 39 minutes, to exhaust all the possibilities.

a more interesting question would be, what is the expected time it would take him to open the lock?

If all of the possibilities are equally likely, then the average time would be 2340sec/2. This is 1170 seconds or 19.5 minutes.


[  { 2 ^ ( x ^2 + 1 ) } ^ ( y ^4 + 1 ) / ( x + y ) ] ^ 0.5

$ x$ and $ y$ are positive integers with $ 0 < x < 100$ and $ 0 < y < 100$

at what maximum value of $ x$ and corresponding value of $ x \plus{} y$ the above expression is an integer?

If any one can also help me with the latex for the same.

$ [ { 2 ^{(x^2 \plus{} 1)} } ^ \frac {(y^4 \plus{} 1)} {(x \plus{} y)} ] ^ \frac {1} {2}$

Would be your latex'ed question. I'll solve this later as I got no time now.


Hello,

I always get numerous e-mails from AoPS that a new post has been posted in a topic. How do I stop these?

Profile > Preferences > Posting a message > Always notify me of replies
Turn that option off.

I've got it set to basically never email me for anything, since I'm on more often than I check email anyway.


Let $\mathcal{K}$ be a convex polygon in the plane. Show that for any triangle of the
minimum possible area containing $\mathcal {K}$, the midpoints of its sides lie on $\mathcal{K}$

This was proven in http://www.mathlinks.ro/Forum/viewtopic.php?t=35947 for any convex shape $\mathcal{K}$.

  darij


geometry
circumcircle
trapezoid
perpendicular bisector
geometry unsolved

Triangle $ABC$ is inscribed in circle $O$. Tangent $PD$ is drawn from $A$, $D$ is on ray $BC$, $P$ is on ray $DA$. Line $PU$ ($U \in BD$) intersects circle $O$ at $Q$, $T$, and intersect $AB$ and $AC$ at $R$ and $S$ respectively. Prove that if $QR=ST$, then $PQ=UT$.

I assume that P, Q, R, S, T, U follow on the line in this order. Since QR = ST, the circumcenter O of the triangle $\triangle ABC$ lies on the perpendicular bisector of the segment RS. Let (O') be the circumcircle of the triangle $\triangle ARS,$ the circumcenter O' also lies on the perpendicular bisector of RS. The circumcircles (O), (O') intersect at the common vertex A and at another point A'. $AA' \parallel RS$ (both perpendicular to OO'). Quadrilateral ARSA' is a cyclic trapezoid, isosceles. Hence $\angle BAA' \equiv \angle RAA' = \angle USA'.$ Quadrilateral ABCA' is cyclic, hence $\angle BAA' = \angle UCA'.$ It follows that $\angle USA' = \angle UCA'$ and the quadrilateral UCSA' is cyclic. Let a tangent to the circle (O) at A' meet the line PU at a point U'. Since $AA' \parallel SU',$ $\angle SCA' \equiv \angle ACA' = \angle SU'A'$ and the quadrilateral U'SCA' is also cyclic. But the circumcircle of the triangle $\triangle CSA'$ intersects the line $PU \equiv SU$ at the point S and at one other point, therefore $U' \equiv U$ are identical.  The points Q, T and the points A, A' are symmetrical with respect to OO' (the perpendicular bisector of RS). Since AP is a tangent to (O) at A and A'U a tangent to (O) at A', the points P, U are also symmetrical with respect to OO' hence PQ = UT.

use Butterfly theory and Pascal theory to solve this problem, easy


1)Prove that every non empty finite set has just as many subset with an even number of element as it has with an odd number of element.
2)Define lambert series with respect to Dirichlit series stating the generating function and its relationship with eulers product.
3)show that  multiplying a residue class $ {mod n}$ by $ {x}$ is in fact a permutation of the residue class.
4)show that if A is square free and has $ {k}$ primes then the number of divisors is $ {2^k}$.
5)what is the significance of diriclet series to möbius function with example.
6)prove that if $ {m and n}$ are coprime,$ {F(mn)\equal{}F(m)F(n)}$.for multiplicative function.
This question are from an old book.
The scoring is
No1-4points.
No2-2points.
No3-4points.
No4-1point.
No5-2points.
No6-4points.
Solve as many as you can if you can and maximise your point i got a total of 10 points showing i really need to improve in NT.

1) We establish a 1-1 correspondence between subsets with an even number of elements and subsets with an odd number of elements. Pick one element. To go from an even-numbered subset to an odd-numbered one (and vice versa), add/remove this element.

Alternatively, use the identity
\[ (1\minus{}1)^n\equal{}\binom{n}{0}\minus{}\binom{n}{1}\plus{}\binom{n}{2}\minus{} \cdots \plus{}(\minus{}1)^n \binom{n}{n}\equal{}0.\]

3) What do you mean by residue class? Do you mean that the numbers $ 0,1,2,3, \ldots ,n\minus{}1$ are a permutation of the numbers $ 0,1x,2x,3x, \ldots ,(n\minus{}1)x$? In this case, the statement is false, take $ n\equal{}4, x\equal{}2$ as a counterexample.

4) $ A\equal{}p_1p_2\cdots p_k$. Then the number of divisors is clearly $ 2^k$.

6) Isn't this the definition of multiplicative function? How does your book define "multiplicative function"?

3)$ {x}$ and $ {n}$ are coprime.
6)Prove that that property is true for eulers totient and möbius.
1)nice try make you proof more clear and obvius

I thought the proof for #1 was clear. Single out an element $ x$. Let $ A_e$ and $ A_o$ be the set of all subsets of even and odd size respectively that contain x, and $ B_e$ and $ B_o$ be the set of all subsets of even and odd size respectively that do not contain x.

Then there is a bijection from $ B_e$ to $ A_o$ and from $ B_o$ to $ A_e$, which is just to add x to every subset in $ B_e$ and $ B_o$. Therefore, $ |B_e| = |A_o|$ and $ |B_o| = |A_e|$ so since there are $ |A_e| + |B_e|$ subsets that are even and $ |A_o| + |B_o|$ subsets that are odd, we are done.

And $ \#3$ can be immediately solved simply by showing $ U(n)$, the set of all residues mod n, is a group under multiplication. For it is clearly closed, multiplication mod n is associative, and every element has an inverse mod n, so we are done.

For other question?
Are they that hard?
AOPSERS :rotfl:

I am not clear with the answer of 4th question from THinkflow.
d(n) is written as (a-1)(b-1).......(k-1) where a,b,... are powers of primes of n.
What if any one of the power is even .


If we have a function f: A -> B, the domain of f is A, but the range of f is not necessarily B, only a subset of B. How come so many people have studied how functions worked, but not as many have considered a "thing" g: A -> B where the range of g is B and the domain of g is a subset of A (but not necessarily A itself, as with surjective functions)? Is there a name for the type of "thing" g is?

Thanks for any insight you can provide.

That would be the inverse of an injective function.

There really isn't much reason to independently name every possible relation.

maybe they come up with this definition when they were studying the graphs of the functions...


Euler
function
floor function
LaTeX
number theory unsolved
number theory

Suppose that $n\in N$ and $S_{n}={\{1\leq k \leq 2n \mid{\{\frac{n}{k}}\}\geq \frac{1}{2}\}}$.
Prove that $\sum_{d\in{S_{n}}}\varphi(d)=n^{2}$.($\varphi$ is the Euler function and ${\{\frac{n}{k}}\}=\frac{n}{k}-[\frac{n}{k}]$ and $[\frac{n}{k}]$ is the floor of $\frac{n}{k}$).

[tex]\sum_{d \in S_n} \varphi ( d ) = \sum_{k = 1}^{2 n} ( \lfloor \frac{n}{k} + \frac{1}{2} \rfloor - \lfloor \frac{n}{k} \rfloor ) \varphi ( k )[/tex]. By using the [tex]\lfloor mx \rfloor = \sum_{j = 0}^{m - 1} \lfloor x + \frac{j}{m} \rfloor[/tex] identity, we get that [tex]\lfloor 2 \frac{n}{k} \rfloor = \lfloor
\frac{n}{k} \rfloor + \lfloor \frac{n}{k} + \frac{1}{2} \rfloor[/tex]. Thus [tex]\lfloor \frac{n}{k} + \frac{1}{2} \rfloor = \lfloor \frac{2 n}{k} \rfloor -
\lfloor \frac{n}{k} \rfloor[/tex] and [tex]\sum_{k = 1}^{2 n} ( \lfloor \frac{n}{k} + \frac{1}{2} \rfloor - \lfloor \frac{n}{k} \rfloor ) \varphi ( k ) = \sum_{k =
1}^{2 n} \lfloor \frac{2 n}{k} \rfloor \varphi ( k ) - 2 \sum_{k = 1}^{2 n} \lfloor \frac{n}{k} \rfloor \varphi ( k )[/tex] which is equal to (EDIT: Latex did some strange things here...) [tex]\sum_{k = 1}^{2 n} \lfloor \frac{2n}{k} \rfloor \varphi ( k ) - 2 \sum_{k = 1}^n \lfloor \frac{n}{k} \rfloor \varphi ( k )[/tex]. Now knowing that [tex]\sum_{d|n} \varphi ( d ) = n[/tex] and that for any arithmetic function the sum [tex]\sum_{d|n} f ( d ) = g ( n )[/tex] is equivalent to [tex]\sum_{j = 1}^n g ( j ) = \sum_{j = 1}^n f ( j ) \lfloor \frac{n}{j} \rfloor[/tex], we obtain the fact that [tex]\sum_{j = 1}^n \lfloor \frac{n}{k} \rfloor \varphi ( j ) = \sum_{j = 1}^n j = \frac{n \cdot ( n + 1 )}{2}[/tex]. Finaly  [tex]\sum_{k = 1}^{2 n} \lfloor \frac{2 n}{k} \rfloor \varphi ( k ) - 2 \sum_{k = 1}^n \lfloor \frac{n}{k} \rfloor \varphi ( k ) = \frac{( 2 n ) ( 2 n + 1 )}{2} - 2 \frac{n \cdot ( n + 1 )}{2} = n^2[/tex]

EDIT 2: For some strange reasons [tex]\varphi ( k )[/tex] disaperead twice  :huh:

Very nice solution. Congratulation Mr. cyberjoe ;)

I believe too , it's nice solution ,very good :oops:  :D  :lol:


The points $(2,7)$ and $(10,16)$ are given. Find the point that cannot be on the same circle as the given points.
A)$(22/3,13)$           B)$(13,10)$
C)$10,7)$                 D)$7/2,13$

What does it mean to be on the same circle? I thought you neded at least three points to determine one circle?

[hide="Hint"]
The only situation in which three points do not determine a circle is when they're in a line.  :wink: [/hide]


Find all 4-tuples of naturals such that each one divides sum of the other three.

Let $a,b,c,d$ the four naturals.
You can put :  $a=<b=<c=<d$.

but you have :
$d|a+b+c$ and thus $d=<a+b+c$

This implies that $d=a+b+c$

You have too :

$c|a+b+d$ and thus $c|2(a+b)$ ==> $c=<2(a+b)$ ==> $c =a+b$ or $c=2(a+b)$

And you continue like this to go to the solution ! :)

[quote="Labiliau"]
$d|a+b+c$ and thus $d=<a+b+c$

This implies that $d=a+b+c$
[/quote]
 :?:  :?

Yes, it was too quickly :

we have : $a=<b=<c=<d$
Thus :  $a+b+c=<3d$

like $d|a+b+c$, we have too : $d=<a+b+c$ and $a+b+c=kd$ with k integer.

Then :

$d=<a+b+c=<3d$ and $a+b+c=kd$

Finally, $a+b+c = d$ or $a+b+c =2d$ or $a+b+c=3d$ and you have to process with each case.

Let $a,b,c,d$ be such positive integers, and $S = a+b+c+d.$
Then, each of the integers divides $S$, so that $x = \frac S a , y \frac S b , z = \frac S c , t = \frac S d$ are positive integers.
Moreover they satisfy $\frac 1 x + \frac 1 y + \frac 1 z + \frac 1 t = 1.$ This last problem can be solved by standard method. The number of cases to check is not too big but quite boring to handle here.

Pierre.

What is the standard solution for that? :?

Assume that $a \leq b \leq c \leq d$ so that $1 \leq \frac 4 a$. Thus $a = 2,3,4.$
Then check cases  : if $a=2$ then $\frac 1 2 = \frac 1 b + \frac 1 c + \frac 1 d \leq \frac 3 b$ then $b =2,3,4,5,6$ and so on...

Pierre.

Ah! I see. Not very practile, is it? Is there any better method? Or can you reduce the number if cases you mut check?


Problem 1:[url]http://www.mathlinks.ro/viewtopic.php?t=207592[/url]
Problem 2:[url]http://www.mathlinks.ro/viewtopic.php?t=207590[/url]
Problem 3:[url]http://www.mathlinks.ro/viewtopic.php?t=208238[/url]
Problem 4:[url]http://www.mathlinks.ro/viewtopic.php?t=207595[/url]

Somebody should post these problems to Resources page. It was 2008 test.

Added [url=http://www.artofproblemsolving.com/Forum/resources.php?c=23&cid=285&year=2008]here[/url]. Thanks for letting us know.


If f(x) is a polynomial and all of its coefficients are whole numbers, then prove that if the square root of any prime number is a root of f(x), the negative square root of that prime number must also be a root.

See [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=134466]here[/url].  This is generalized by the Conjugate Root Theorem.


Let $ a$, $ b$ and $ c$ be positive real numbers. Prove that
\[ \sum{\dfrac{1}{b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca)}}\ge\sum{\dfrac{1}{a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}3bc\plus{}ca}}\]

[quote="ElChapin"]Let $ a$, $ b$ and $ c$ be positive real numbers. Prove that
\[ \sum{\dfrac{1}{b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca)}}\ge\sum{\dfrac{1}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} 3bc \plus{} ca}}
\]
[/quote]
$ \sum_{cyc}{\dfrac{1}{b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca)}}\ge\sum_{cyc}{\dfrac{1}{a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} 3bc \plus{} ca}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}\frac{(a\minus{}b)(a\minus{}c)}{(b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca))(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} 3bc \plus{} ca)}\geq0.$
Let $ a\geq b\geq c.$
Hence, it remains to prove that
$ \frac{1}{(b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca))(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} 3bc \plus{} ca)}\geq$
$ \geq\frac{1}{(a^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca))(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} 3ca)}.$
Let $ c^2 \plus{} 2(ab \plus{} bc \plus{} ca)\equal{}x$ and $ a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} ca\equal{}y.$
Hence, $ \frac{1}{(b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca))(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} 3bc \plus{} ca)}\geq$
$ \geq\frac{1}{(a^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca))(a^2 \plus{} b^2 \plus{} c^2 \plus{} ab \plus{} bc \plus{} 3ca)}\Leftrightarrow$
$ \Leftrightarrow(a^2\plus{}x)(2ac\plus{}y)\geq(b^2\plus{}x)(2bc\plus{}y),$ which is obvious.


calculus
integration
geometry
conics
parabola
calculus computations

What's the area between the parabola y=x^2+x+3 and the x-axis, fromx=-1 to x=4?

Given that $ \zeta(s)\equal{}\sum_{n\equal{}1}^\infty\frac{1}{n^s}$, derive that $ \zeta(s)\equal{}s\int_1^\infty\frac{[x]}{x^{s\plus{}1}}dx$ where $ [x]$ denotes the greatest integer less than $ x$.

(Extra credit: show this converges for all $ s>0$).


inequalities
Cauchy Inequality
inequalities proposed

Let $ a,b,c > 0$ such that $ a \plus{} b \plus{} c \equal{} \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}$
Prove that
$ \frac {a^3b \plus{} b^3c \plus{} c^3a}{a^2 \plus{} b^2 \plus{} c^2}\plus{} \frac { \sqrt {3}}{ \sqrt {a^2 \plus{} b^2 \plus{} c^2}} \geq 2$

[quote="2424"]Let $ a,b,c > 0$ such that $ a + b + c = \frac {a}{b} + \frac {b}{c} + \frac {c}{a}$
Prove that
$ \frac {a^3b + b^3c + c^3a}{a^2 + b^2 + c^2}$ + $ \frac {\sqrt {3}}{\sqrt {a^2 + b^2 + c^2}}$ $ \geq 2$[/quote]
You must have meant:
$ \frac {a^3b + b^3c + c^3a}{a^2 + b^2 + c^2} +\frac {\sqrt {3}}{\sqrt {a^2 + b^2 + c^2}} \geq 2$
.
Am I right?

Because I read the article states that this is an easy question so I was determined to give it a try, here is the answer

Using Cauchy Schwartz, we have

$ (a^3b\plus{}b^3c\plus{}c^3a)\left(\frac{a}{b} \plus{} \frac{b}{c} \plus{} \frac{c}{a} \right) \ge (a^2\plus{}b^2\plus{}c^2)^2$

It is deduced to prove that

$ \frac{a^2\plus{}b^2\plus{}c^2}{a\plus{}b\plus{}c} \plus{} \sqrt\frac{3}{a^2\plus{}b^2\plus{}c^2} \ge 2$

Using Cauchy inequality for the above estimation, we get the final result easily
Thus, our proof is completed.


An error of $ .02''$ is made in the measurement of a line $ 10''$ long, while an error of only $ .2''$ is made in a measurement of a line $ 100''$ long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:

$ \textbf{(A)}\ \text{greater by }.18 \qquad\textbf{(B)}\ \text{the same} \qquad\textbf{(C)}\ \text{less}$
$ \textbf{(D)}\ 10 \text{ times as great} \qquad\textbf{(E)}\ \text{correctly described by both (A) and (D)}$

[hide="Click for solution"]
Relative error equals $ \frac{\text{error}}{\text{measurement}}$.  Since $ \frac{.02}{10}\equal{}\frac{.2}{100}$, the relative errors are the same, or $ \boxed{\textbf{(B)}}$.
[/hide]


Suppose $*$ is an associative operation on set $S$. Define $x^n$ to mean $x*x*x*x*....*x$), $n$ times. (For example, $x^3=x*x*x$). Suppose further that an element $a$ of $s$ is such that all of $a, a^2, a^3, ..., a^9$ are different but $a^{10}=a^3$. Then there is some element $b$ belonging to $s$ such that $b=b^2$. One such element $b$ is:

(a) $a^4$
(b) $a^5$
(c) $a^7$
(d) $a^9$
(e) $a^{13}$

I don't quite understand this problem.

[quote="scorpius119"][hide]$a^{10}=a^3\Rightarrow a^{10}*a*a*a*a=a^3*a*a*a*a\Rightarrow a^{14}=a^7\Rightarrow \boxed{C}$.[/hide][/quote]

Okay sorry that was a little too easy  :blush:

Because it works?

More generally, $a^{10} = a^3$, so it follows that $a^{10} \ast a = a^3 \ast a$, and applying this operation repeatedly gives $a^k = a^{k+7}, k \ge 3$.  So we are looking for a $b = a^k$ such that $2k = k+7, k \ge 3$, which implies $k = 7$.  Using 4 $a$s just makes the solution very obvious and fast without ever needing to derive the rule.


We must prove that if a>b then f(a)<f(b) where
f(x)=(x^n-1)/(x^(n+1)-1), just study the variation of f
f'(x)= (x^(n-1)/(x^(n+1)-1)^2)Q(x)
Q(x)=-x^(n+1)-n+(n+1)x
Q'(x)=(n+1)(1-x^n)
Q"(x)=-n(n+1)x^(n-1)
=>Q'  \downto , Q'(0)=n+1>0, Q'(1)=0, Q'(x)<0 for x>1
Q is increasing on [0,1] decreasing on [1,+oo[, Q(1)=0
Q(x)  \leq 0
f is  \downto


For $ a,b,c$ are the sides of a triangle, prove that:
$ \sum_{cyc}\sqrt[3]{\frac{a^{2}+bc}{b^{2}+c^{2}}}\geq  \frac{9\sqrt[3]{abc}}{a+b+c}$

Wait. So why do we need $ a,b,c$ are sides of a triangle?
For easier to prove Shur??  :huh:

[quote="bibobeo"]Wait. So why do we need $ a,b,c$ are sides of a triangle?[/quote]
No! We don't need it!
[quote="bibobeo"]For easier to prove Shur??  :huh:[/quote]
Yes! :D  Your inequality is true for all positive $ a,$ $ b$ and $ c.$ :wink:

Uhm, I've tried with Ravi the first time I saw it but I've failed :(
It became so complex and I cannot simplify it

[quote=Qruni]Can we use Ravi's substitutuion to the inequality[/quote]

True. But this has no obvious advantage in simplifying.


function
floor function
algebra unsolved
algebra

Let $a,b,c,$ and $d$ be given real numbers. Find all functions $f: \mathbb R \to \mathbb R$ such that
\[f(ax+b) = c\cdot f(x)+d,\]
for all $x \in \mathbb R$.

[quote="Dtrung"]Let $ a,b,c,d$ be real numbers. Find all functions $ \large f : R \to R$ such that :
            $ f(ax + b) = c.f(x) + d$
for all $ \large x \in R$.[/quote]

It's a rather easy problem but you must be clever about all the different cases.
Here is the answer (demo is below in the second part of this post). Let's go

$ (a,b,c,d) = (0,b,1,0)$. Solution is $ f(x) = u$ for any real $ u$.
$ (a,b,c,d) = (0,b,1,d\neq 0)$. No solution
$ (a,b,c,d) = (0,b,c\neq 1,d)$. Solution is $ f(x) = \frac {d}{1 - c}$ constant.
$ (a,b,c,d) = (1,0,1,0)$. Solution is any function $ f(x)$.
$ (a,b,c,d) = (1,0,1,d\neq 0)$. No solution
$ (a,b,c,d) = (1,b\neq 0,1,d)$. Solution is $ f(x) = h(\{\frac {x}{b}\}) + \frac {dx}{b}$ for any function $ h(x)$ defined on $ [0,1)$.
$ (a,b,c,d) = (1,b\neq 0,c\neq 1,d)$. Solution is $ f(x) = c^{\lfloor\frac {x}{b}\rfloor }h(\{\frac {x}{b}\}) - \frac {d}{c - 1}$ for any function $ h(x)$ defined on $ [0,1)$.
$ (a,b,c,d) = ( - 1,b,c = 1,d\neq 0)$. No solution
$ (a,b,c,d) = ( - 1,b,c = 1,d = 0)$. Solution is $ f(x) = h(x - \frac {b}{2}) + h(\frac {b}{2} - x)$  for any function $ h(x)$
$ (a,b,c,d) = ( - 1,b,c = - 1,d)$. Solution is $ f(x) = h(\frac {b}{2} - x) - h(x - \frac {b}{2}) + \frac {d}{2}$  for any function $ h(x)$
$ (a,b,c,d) = ( - 1,b,c\notin\{ - 1, + 1\},d)$. Solution is $ f(x) = \frac {d}{1 - c}$ constant.
$ (a,b,c,d) = (a\notin\{ - 1,0, + 1\},b,1,d\neq 0)$. No solution
$ (a,b,c,d) = (a\notin\{ - 1,0, + 1\},b,1,0)$. Solution is $ f(x) = h(Sign(x + \frac {b}{a - 1})Sign(a)^{\lfloor\frac {\ln(|x + \frac {b}{a - 1}|)}{\ln(|a|)}\rfloor}{\{\frac {\ln(|x + \frac {b}{a - 1}|)}{\ln(|a|)}\})}$ for any function $ h(x)$ defined on $ ( - 1,1)$
$ (a,b,c,d) = (a\notin\{ - 1,0, + 1\},b,c\neq 1,d)$; Solution is $ f(x) = c^{\lfloor\frac {\ln(|x - \frac {b}{1 - a}|)}{\ln(|a|)}\rfloor}h(Sign(x - \frac {b}{1 - a})Sign(a)^{\lfloor\frac {\ln(|x - \frac {b}{1 - a}|)}{\ln(|a|)}\rfloor}{\{\frac {\ln(|x - \frac {b}{1 - a}|)}{\ln(|a|)}\}) + \frac {d}{1 - c}}$ for any function $ h(x)$ defined on $ ( - 1,1)$

Demos :

[u]Case 1[/u]: $ a = 0$, $ c = 1$ and $ d = 0$. The equation is $ f(b) = f(x)$ and the solutions are the constant functions $ f(x) = u$.

[u]Case 2[/u]: $ a = 0$, $ c = 1$ and $ d\neq0$. The equation is $ f(b) = f(x) + d$ and have no solution (since, for $ x = b$, we have $ 0 = d$)

[u]Case 3[/u]: $ a = 0$ and $ c\neq 1$. The equation is $ f(b) = cf(x) + d$ and the solution is $ f(x) = \frac {d}{1 - c}$

[u]Case 4[/u]: $ a = 1$, $ b = 0$, $ c = 1$ and $ d = 0$. The equation is $ f(x) = f(x)$ and any function $ f(x)$ is solution.

[u]Case 5[/u]: $ a = 1$, $ b = 0$, $ c = 1$ and $ d\neq 0$. The equation is $ 0 = d$ and no solution exist.

[u]Case 6[/u]: $ a = 1$, $ b\neq 0$, $ c = 1$. The equation is $ f(x + b) = f(x) + d$, so $ f(x + nb) = f(x) + nd$, so $ f(x) = f(b\{\frac {x}{b}\} + \lfloor\frac {x}{b}\rfloor b) = f(b\{\frac {x}{b}\}) + \lfloor\frac {x}{b}\rfloor d$
So $ f(x)$ is completely defined on $ \mathbb{R}$ as soon as $ f(x)$ is defined on $ [0,b)$ (or $ (b,0]$).
The general solution is then $ f(x) = g(b\{\frac {x}{b}\}) + \lfloor\frac {x}{b}\rfloor d$ for any function $ g(x)$ defined on $ [0,b)$ (or $ (b,0]$).
Considering $ h(x) = g(bx) - dx$, we have a prettier form for the solution : $ f(x) = h(\{\frac {x}{b}\}) + \frac {dx}{b}$ for any function $ h(x)$ defined on $ [0,1)$.
And it is easy to check that this necessary condition is sufficient.

[u]Case 7[/u]: $ a = 1$ and $ c\neq 1$. The equation is $ f(x + b) = cf(x) + d$ and so $ f(x + nb) = c^nf(x) + d\frac {c^n - 1}{c - 1}$ and so $ f(x) = f(b\{\frac {x}{b}\} + \lfloor\frac {x}{b}\rfloor b) = c^{\lfloor\frac {x}{b}\rfloor }f(b\{\frac {x}{b}\}) + d\frac {c^{\lfloor\frac {x}{b}\rfloor } - 1}{c - 1}$.
So $ f(x)$ is completely defined on $ \mathbb{R}$ as soon as $ f(x)$ is defined on $ [0,b)$ (or $ (b,0]$).
The general solution is then $ f(x) = c^{\lfloor\frac {x}{b}\rfloor }g(b\{\frac {x}{b}\}) + d\frac {c^{\lfloor\frac {x}{b}\rfloor } - 1}{c - 1}$ for any function $ g(x)$ defined on $ [0,b)$ (or $ (b,0]$).
Considering $ h(x) = g(bx) + \frac {d}{c - 1}$, we have a prettier form for the solution : $ f(x) = c^{\lfloor\frac {x}{b}\rfloor }h(\{\frac {x}{b}\}) - \frac {d}{c - 1}$ for any function $ h(x)$ defined on $ [0,1)$.
And it is easy to check that this necessary condition is sufficient.

[u]Case 8[/u]: $ a = - 1$,$ c = 1$ and $ d\neq 0$ ; The equation is $ f(b - x) = f(x) + d$ and so, with $ x = \frac {b}{2}$ : $ 0 = d$ and no solution exist.

[u]Case 9[/u]: $ a = - 1$,$ c = 1$ and $ d = 0$; The equation is $ f(b - x) = f(x)$. Let then $ g(x) = f(x + \frac {b/2})$. we have $ f(x) = g(x - \frac {b}{2})$ and the equation becomes $ g(\frac {b}{2} - x) = g(x - \frac {b}{2})$ and so $ g( - x) = g(x)$.
The solution is then $ f(x) = g(x - \frac {b}{2})$ for any even function $ g(x)$.
Considering that any even function can be written $ g(x) = h(x) + h( - x)$, we have another form for the solution :
$ f(x) = h(x - \frac {b}{2}) + h(\frac {b}{2} - x)$  for any function $ h(x)$
And it is easy to check that this necessary condition is sufficient.

[u]Case 10[/u]: $ a = - 1$ and $ c = - 1$; The equation is $ f(b - x) = d - f(x)$. Let then $ g(x) = f(\frac {b}{2} - x) - \frac {d}{2}$. We have $ f(x) = g(\frac {b}{2} - x) + \frac {d}{2}$ and the equation becomes $ g(x - \frac {b}{2}) = - g(\frac {b}{2} - x)$ and so $ g( - x) = - g(x)$.
The solution is then $ f(x) = g(\frac {b}{2} - x) + \frac {d}{2}$ for any odd function $ g(x)$.
Considering that any odd function can be written $ g(x) = h(x) - h( - x)$, we have another form for the solution :
$ f(x) = h(\frac {b}{2} - x) - h(x - \frac {b}{2}) + \frac {d}{2}$  for any function $ h(x)$
And it is easy to check that this necessary condition is sufficient.

[u]Case 11[/u]: $ a = - 1$ and $ |c|\neq 1$ The equation is $ f(b - x) = cf(x) + d$ and so $ f(x) = f(b - (b - x)) = cf(b - x) + d = c^2f(x) + cd + d$. Then :
$ f(x) = \frac {d}{1 - c}$ and the original equation becomes $ \frac {d}{1 - c} = \frac {cd}{1 - c} + d$ which is true.
So the only solution is $ f(x) = \frac {d}{1 - c}$.
And it is easy to check that this necessary condition is sufficient.

[u]Case 12[/u]: $ a\neq 0$, $ |a|\neq 1$, $ c = 1$ and $ d\neq 0$. The equation is $ f(ax + b) = f(x) + d$. let then $ x = \frac {b}{1 - a}$. We have $ 0 = d$ and no solution exist.

[u]Case 13[/u]: $ a\neq 0$, $ |a|\neq 1$, $ c = 1$ and $ d = 0$. The equation is $ f(ax + b) = f(x)$. let then $ k(x) = f(x - \frac {b}{a - 1})$. We have $ f(x) = k(x + \frac {b}{a - 1})$ and the equation is $ k(a(x + \frac {b}{a - 1})) = k(x + \frac {b}{a - 1})$ and so $ k(ax) = k(x)$.
This equation is well known and a general solution is :
$ k(0) = u$ and, for $ x\neq 0$, $ k(x) = h(Sign(x)Sign(a)^{\lfloor\frac {\ln(|x|)}{\ln(|a|)}\rfloor}{\{\frac {\ln(|x|)}{\ln(|a|)\}}}$ for any function $ h(x)$ defined on $ ( - 1,1)$
And so a general solution :
$ f(\frac {b}{1 - a}) = u$ (any real constant)
For $ x\neq\frac {b}{1 - a}$ : $ f(x) = h(Sign(x + \frac {b}{a - 1})Sign(a)^{\lfloor\frac {\ln(|x + \frac {b}{a - 1}|)}{\ln(|a|)}\rfloor}{\{\frac {\ln(|x + \frac {b}{a - 1}|)}{\ln(|a|)}\})}$ for any function $ h(x)$ defined on $ ( - 1,1)$
And it is easy to check that this necessary condition is sufficient.

[u]Case 14[/u]: $ a\neq 0$, $ |a|\neq 1$ and $ c\neq 1$. The equation is $ f(ax + b) = cf(x) + d$. Let then $ k(x) = f(x + \frac {b}{1 - a}) - \frac {d}{1 - c}$. We have $ f(x) = k(x - \frac {b}{1 - a}) + \frac {d}{1 - c}$ and the equation becomes : $ k(ax) = ck(x)$
This equation is well known and a general solution is :
$ k(0) = 0$ and, for $ x\neq 0$, $ k(x) = c^{\lfloor\frac {\ln(|x|)}{\ln(|a|)}\rfloor}h(Sign(x)Sign(a)^{\lfloor\frac {\ln(|x|)}{\ln(|a|)}\rfloor}{\{\frac {\ln(|x|)}{\ln(|a|)\}}}$ for any function $ h(x)$ defined on $ ( - 1,1)$
And so a general solution :
$ f(\frac {b}{1 - a}) = 0$
For $ x\neq\frac {b}{1 - a}$ : $ f(x) = c^{\lfloor\frac {\ln(|x - \frac {b}{1 - a}|)}{\ln(|a|)}\rfloor}h(Sign(x - \frac {b}{1 - a})Sign(a)^{\lfloor\frac {\ln(|x - \frac {b}{1 - a}|)}{\ln(|a|)}\rfloor}{\{\frac {\ln(|x - \frac {b}{1 - a}|)}{\ln(|a|)}\}) + \frac {d}{1 - c}}$ for any function $ h(x)$ defined on $ ( - 1,1)$
And it is easy to check that this necessary condition is sufficient.


integration
inequalities
algebra unsolved
algebra

Can anyone solve this?

Given Sn = sum from 1 to ((2^n )- 1) of K^((1/n)-1). Prove that: n<=Sn<=n+1

Proof:
Consider $C_{n}= \int_{1}^{m}x^{\frac{1}{n}-1}\,dx$. Then standard techniques give $C_{n}= n \cdot ( \sqrt[n] x-1)$.
It's lower sum is $a_{n}=\sum_{k=2}^{m}k^{\frac{1}{n}-1}$, it's upper sum is $A_{n}= \sum_{k=1}^{m-1}k^{\frac{1}{n}-1}$. So we have $a_{n}\leq C_{n}\leq A_{n}$.
Now set $m=2^{n}$, thus we get $C_{n}= n$ and $A_{n}= S_{n}$ and $a_{n}= S_{n}-1+m^{\frac{1}{n}-1}> S_{n}-1$.
Combining all the inequalities gives us $S_{n}\geq n$ and $S_{n}-1 \leq n$, so the result.

Edit: Rust was faster...


Prove or disprove: the only prime numbers of the form $ 100 \ldots 001$ are $ 11$ and $ 101$.

This is essentially as hard as discussing the existence or nonexistence of Fermat primes; in other words, this is open.  See [url=http://mathworld.wolfram.com/GeneralizedFermatNumber.html]generalized Fermat number[/url].


Suppose that $ m$ and $ n$ are odd integers such that $ m^{2} \minus{} n^{2} \plus{} 1$ divides $ n^{2} \minus{} 1$. Prove that $ m^{2} \minus{} n^{2} \plus{} 1$ is a perfect square.

This (difficult) problem already appeared here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=5935
:)

[quote="Rust"]Because (k,k+1)=1, it give k and k+1 are perfect square.[/quote]

Would you mind describing this in a little more detail? :blush:

Also, if $n=1$, then how can $m^{2}-n^{2}+1$ divide $n^{2}-1$?


abstract algebra
group theory
superior algebra
superior algebra theorems

Hi,

this has been bugging me for months now, I still am not really convinced  :(

let $M$ and $N$ be resp right and left $R$ modules with R a ring

click here if you wish to see definition of balanced product and tensor product
[hide]
now let us define a balanced product : $(Q,q)$ with $Q$ an abelian group is a balanced product of $M$ and $N$ if $q$ is a map from $M\times N$ to $Q$ such that :

$q(x,y+z)=q(x,y)+q(x,z)$
$q(x a,y)=q(x,a y)$
$q(x+z,y)=q(x,y)+q(z,y)$

Now a morphism from one balanced product $(F,f)$ to $(Q,q)$ is an abelian groupmorphism $k: F\rightarrow Q$

such that $k(f(x,y))=q(x,y)$

A tensor product finally is a balanced product $(P,f)$ such that for any other balanced product $(Q,q)$ there is a unique morphism from the first to the latter[/hide]

My questions now :

prove that
1) when you take  the abelian subgroup $W$ of $P$ generated by all elements $f(x,y)$, then $(W,f)$ is itself also a tensor product

2) prove that $W=P$

my teacher said that it had to be done in this order
Now there is a subtlety which occurred in class, it's not because $A$ is an isomorphic subgroup of $B$ that $A=B$ (you can construct weird infinite groups), so we used that while we shouldn't have.

Now when yu start constructing tensor products explicitly you can see this probably a lot easier, but the point is to see it as soon as possible, even before proving existence!

It's not a homework problem  :oops:  it's just something that I really wanna get to the bottom of. :roll:

1. What can I say? It's trivial. Check the definition of a tensor product.

2. The inclusion $W \to P$ is a homomorphism of balanced products. Now by 1. and the universal property of the tensor product, this must be an isomorrphism. Hence $W=P$.


inequalities
quadratics
calculus
calculus computations

Given a postive constant $ c,$ find the range of $ b$ such that $ (b \plus{} x^2)^2 > (1 \plus{} 4x^2)c^2$ for all real numbers $ x$.

[quote="FOURRIER"]Sorry, I didn't understand what do you mean by:

The range of $ b$[/quote]

The range of value of $ b$ for which the given inequality holds for any real numbers $ x$.

I can't edit. :(

It's inconveinient.

[hide]Note that $ b$ cannot be negative. After substitution $ t\equal{}x^2$ the inequality becomes $ b>\max_{t\ge 0}(c\sqrt{1\plus{}4t}\minus{}t)\equal{}\max\{c^2\plus{}1/4,c\}$.[/hide]


function
induction
algebra proposed
algebra

Let $ Z$ and $ R$ denote the sets of integers and real numbers, respectively.
Let $ f: Z \rightarrow R$ be a function satisfying:
(i) $ f(n) \ge 0$ for all $ n \in Z$
(ii) $ f(mn)\equal{}f(m)f(n)$ for all $ m,n \in Z$
(iii) $ f(m\plus{}n) \le max(f(m),f(n))$ for all $ m,n \in Z$

(a) Prove that $ f(n) \le 1$ for all $ n \in Z$
(b) Find a function $ f: Z \rightarrow R$ satisfying (i), (ii),(iii) and $ 0<f(2)<1$ and $ f(2007) \equal{} 1$

(a) m=1 implies $ f\equiv 0$ or f(1)=1. $ f\equiv 0$ satisfies all the conditions for part (a).
If f(1)=1; $ m\equal{}n\equal{}\minus{}1 \Rightarrow f(\minus{}1)\equal{}1 ( f \ge 0)$
$ m\equal{}\minus{}m,n\equal{}\minus{}n ; f(m)f(n)\equal{}f(\minus{}m)f(\minus{}n) ; m\equal{}1 \Rightarrow f(n)\equal{}f(\minus{}n)$
You can easily prove $ f(km) \le f(m) , k \ne 0$
therefore, $ m\equal{}1; 1\ge f(k)\equal{}f(\minus{}k) ,k\in Z\minus{}{0}$

$ m\equal{}n\equal{}0$ in original gives $ f(0)\equal{}0 or 1$

therefore, $ f(n) \le 1 \forall n\in Z$

(b) Use third equation repeatedly. $ f(2005\plus{}2)\equal{}1 \le f(2005) \Rightarrow f(2005)\equal{}1$
therefore, f(odd integer)=1

It can be easily shown that $ f(2^mp)\equal{}c^m , p\equal{}2k\plus{}1, k,m \in N_0,$ where $ 0<c<1$ and it is an even function.

I give my solution to this problem :
(a) Let [img]http://data.artofproblemsolving.com/aops20/latex/texer/cd3f48cd02f3a7f274b491773d9b2e83b48460b5.png[/img] the assertion of (ii), hence :
[img]http://data.artofproblemsolving.com/aops20/latex/texer/0ac38836f725120125ce4797e8cc29d7cb5c6778.png[/img].
-Case 1: [img]http://data.artofproblemsolving.com/aops20/latex/texer/aed73027fc054c58ff8bb34017cb8edba183c72b.png[/img], we can verify that [img]http://data.artofproblemsolving.com/aops20/latex/texer/a43d3e2405d10860e57153309a26f26e24bc1d7e.png[/img].
And then we are done with this case.
-Case 2: [img]http://data.artofproblemsolving.com/aops20/latex/texer/e6001daa49127576c49a5a0d31d344b7beaf500b.png[/img].
Let's prove by induction that :[img]http://data.artofproblemsolving.com/aops20/latex/texer/3aedbf1ee5709359f3871d2cee363e7a8dc32c88.png[/img].
- For n=0, we have f(0)=0 or f(0)=1 which satisfies the conditon.
- Now let n be a non-negative integer such that [img]http://data.artofproblemsolving.com/aops20/latex/texer/111ba6edf7d26c35ed073a86a34fd84a0f387f29.png[/img] and we are left to prove that : [img]http://data.artofproblemsolving.com/aops20/latex/texer/abc71045e6238e54a0eac4d1e5de31aa64630450.png[/img].
By (iii) we get : [img]http://data.artofproblemsolving.com/aops20/latex/texer/bed1b9b9e291a6b3f40b5a14ee838e71e2f60aba.png[/img].
And so we are done in this case too.
Now by symetry in (ii) we get : [img]http://data.artofproblemsolving.com/aops20/latex/texer/dcd20ad9dc32249946da5305bdcc3ed49990229e.png[/img], and it's obvious that f(-1)=1 and f(1)=1 and so f(n)=f(-n) which finish the proof.

(b) It's easy to check that for every odd integer f(n)=1,  now let k an even integer so there exists[img]http://data.artofproblemsolving.com/aops20/latex/texer/703393e38b35a93355d8763e765b34a14f4c3c64.png[/img].
Now and using (a) we find that for every even integer t , 0<f(t)<1.
And we can easy check that this fuction verify the conditions

Q.E.D ;)

(a) is trivial
(b):let f(0)=0,$f(n)=\frac{1}{2^{v_2(n)}}$
easy to verify it


Prove that sine squared x(1+cot squared x)=1
Prove that tan x (cot x + tan x)=sec squared x

cany anyone help me with this problem because im completely lost any help will be greatly appeciated.


induction
combinatorics proposed
combinatorics

Let $p,q \in \mathbb N^{\ast}$, $p,q \geq 2$. We say that a set $X$ has the property $\left( \mathcal S \right)$ if no matter how we choose $p$ subsets $B_i \subset X$, $i = \overline{1,n}$, not necessarily distinct, each with $q$ elements, there is a subset $Y \subset X$ with $p$ elements s.t. the intersection of $Y$ with each of the $B_i$'s has an element at most, $i=\overline{1,p}$. Prove that:

(a) if $p=4,q=3$ then any set composed of $9$ elements doesn't have $\left( \mathcal S \right)$;

(b) any set $X$ composed of $pq-q$ elements doesn't have the property $\left( \mathcal S \right)$;

(c) any set $X$ composed of $pq-q+1$ elements has the property $\left( \mathcal S \right)$.

[i]Dan Schwarz[/i]

(a) is a consequence of (b) and (b) is almost obvious: we partition the set into $p-1$ parts of size $q$, and take these to be $p-1$ of the sets, while the $p$'th set is just equal to one of the first $p-1$. No matter how we choose $p$ elements, some two of them will be in the same part in our partition, and we're done.

For (c), we can use induction on $p$. For $p=2,\ X$ has $q+1$ elements, and we choose two subsets $X_1,X_2$ of size $q$ of $X$. If $X_i$ don't cover $X$, then we choose one of our elements to be outside both $X_i$, and the second element arbitrarily, and if they do cover $X$, we take the first element in $X_1\setminus X_2$, and the second one in $X_2\setminus X_1$, which are clearly both non-empty.

Suppose now that (c) has been proven for values smaller than our current $p$, which we assume to be $\ge 3$. Let's count the pairs $(x,A)$, where $x\in X$ and $A$ is one of the $p$ sets which contains $x$. On the one hand, this count gives $pq$, because there are $p\ A$'s and each of them has $q$ elements. On the other hand, it's $\sum_{i=1}^{(p-1)q+1}n(x_i)$, where $(x_i)_i$ is an enumeration of the elements of $X$, and $n(x_i)$ is the number of sets among the $p$ chosen ones that contain $x_i$. Since $2((p-1)q+1)>2(p-1)q>pq$, there is at least one $x_i$ with $n(x_i)\le 1$. We take this $x_i$ to be one of the chosen elements of $X$, and eliminate the $q$ elements of one of the $p$ chosen subsets of $X$ as follows: if $x$ belongs to no such set, this set will be arbitrary, while if it does belong to such a set, we eliminate precisely this (unique, because $n(x_i)=1$) set. We are now in the same setting as before, except that now $p$ has been replaced by $p-1$, and the induction hypothesis gives us the extra $p-1$ elements of $X$ that we have to choose.

Nice proof, grobber, but not without a little flaw:

[quote="grobber"]We take this $ x_i$ to be one of the chosen elements of $ X$, and eliminate the $ q$ elements of one of the $ p$ chosen subsets of $ X$ as follows: if $ x$ belongs to no such set, this set will be arbitrary, while if it does belong to such a set, we eliminate precisely this (unique, because $ n(x_i) \equal{} 1$) set. We are now in the same setting as before, except that now $ p$ has been replaced by $ p \minus{} 1$, and the induction hypothesis gives us the extra $ p \minus{} 1$ elements of $ X$ that we have to choose.[/quote]

should be

[quote="grobber"]We take this $ x_i$ to be one of the chosen elements of $ X$, and eliminate $ q$ elements of $ X$ as follows:
- If $ x$ belongs to none of the $ p$ chosen subsets of $ X$, then we select an arbitrary of these $ p$ subsets and eliminate all of its elements but one (which we choose arbitrarily), and eliminate $ x$ afterwards.
- If $ x$ belongs to one of the $ p$ chosen subsets of $ X$, then we eliminate precisely the $ q$ elements of this (unique, because $ n(x_i) \equal{} 1$) set.
We are now in the same setting as before, except that now $ p$ has been replaced by $ p \minus{} 1$, and the induction hypothesis gives us the extra $ p \minus{} 1$ elements of $ X$ that we have to choose.[/quote]

(The mistake was that you eliminated $ q \plus{} 1$ rather than $ q$ elements in the first case.)

See also http://www.mathlinks.ro/viewtopic.php?t=299325 for further discussion.

  darij


Let's see what do you think about this one

ABDHPEJTMZY.....

[hide="HINT"]It is very easy, but a little math is required  :lol: [/hide]

[quote="probability1.01"]Without checking it seems to be [hide]$2^n \mod 26$[/hide][/quote]Almost. There is just one slight mistake!

Great, the explanation is simple: There are 26 letters, and 2 is a generator of $(\mathbf Z_{27}^*,\cdot)$, and it has exactly 26 elements  :D

You solved quickly


calculus
derivative
function
vector
inequalities
linear algebra
calculus computations

Prove the following theorem:

Suppose f is a real valued function which is defined and differentiable in a neighborhood of $a \in \mathbb{R}^{p}$, and suppose that $df(a) \not= 0$. Then the gradient df(a) points in the direction of greatest increase for f at a – that s, $D_{u}f(a)$ has its maximum value when the unit vector u is a positive scalar multiple of df(a).

Moderator's ruling: although it contains linear algebra elements, this belongs in the calculus section.

$f$ is differentiable at $\vec{a}$ iff there exist $\frac{\partial f}{\partial x_{i}},\ 1\le i\le n,$ such that as $\|\vec{b}\|\to0,$

$f(\vec{a}+\vec{b})=f(\vec{a})+\sum_{i=1}^{n}\frac{\partial f}{\partial x_{i}}(\vec{a})\,b_{i}+o(\|\vec{b}\|)$

We can write this in vector language as

$f(\vec{a}+\vec{b})=f(\vec{a})+\nabla f(\vec{a})\cdot \vec{b}+o(\|\vec{b}\|)$

Now let $\vec{b}=t\vec{u}$ for scalar $t$ and unit vector $\vec{u}.$ Take the expression above, subtract $f(\vec{a}),$ divide by $t,$ and let $t\to 0,$ getting

$D_{\vec{u}}f(\vec{a})=\nabla f(\vec{a})\cdot \vec{u}.$

At this point what we need is either the geometric (angle) interpretation of the dot product, or we need Cauchy's inequality (that is Cauchy-Schwarz for a finite sum.) By Cauchy:

$|\nabla f(\vec{a})\cdot \vec{u}|\le\|\nabla f(\vec{a})\|\,\|\vec{u}\|=\|\nabla f(\vec{a})\|$

with equality only if $\vec{u}$ is a scalar multiple of $\nabla f(\vec{a})$ Futhermore $\nabla f(\vec{a})\cdot \vec{u}=\|\nabla f(\vec{a})\|$ only if $\vec{u}$ is a positive scalar multiple of $\nabla f(\vec{a}).$

[Note: there is no important difference in the mathematical information contained in the notation between $\nabla f$, as I have been using it here, and $df,$ as you stated in your post.]


geometry rectangle counting distinguishability MATHCOUNTS rotation

In the Land of Eternal Darkness, there is no light, so no one can see. All paper money consists of rectangles with triangles cut off some, all or none of the four corners of the rectangles. Currency denominations are determined by shape. The dimensions of the rectangles are 2.5 units by 6 units and the triangles are right triangles with legs of length 1 unit. Citizens can recognize the value of the monoey by feeling which corners have or have not been cut off. Of course, they cannot tell the front from the back of these bills. How many different distinguishable denominations are possible for the currency in this land?

 :huh:

Land of Eternal Darkness, Mathcounts is so clever

You can have no corner cut off, one corner cut off, corners on one long side cut off, corners on one short side cut off, diagnol corners cut off, 3 corners cut off or all corners cut off. This makes 7. Sorry that I cant make image.

[hide="solution"]

Case 1-No corners are cut off.
There is clearly only one rectangle possible.

Case 2- One corner is cut off.
Once again only one possible rectangle, since you can always flip or rotate one rectangle to get them all.

Case 3- Two corners are cut off.
If they are connected to each other by a diagonal, that's one possible rectangle. If they are on the same, shorter side of the rectangle, that is another. If they are on the same, longer side of the rectangle, that is another. So 3 here.

Case 4- Three corners are cut off.
Basically the same as Case 2, only one possible.

Case 5- All four corners are cut off.
Basically the same as Case 1, only one rectangle.

Thus there are 1+1+3+1+1=7 possible.[/hide]

Echen, that is incorrect.

There are actually $ \boxed{8}$ because there are $ 2$ choices for each corner, whether you want to cut it or not.

---EDIT---
I was being really, really stupid; the answer is $ \boxed{7}$. I'm sorry, Echen  :oops: .

They are not all distinguishable, akhil.

Oh thanks.. Its really hard... I didn't get it cause it was so long..


1. How many different three-letter codes can be formed from the set of letters(a,b,c,d,e,f,g,i) if each letter may be used only once and the last letter in the code must be a vowel?

[quote="buzzer11"][quote="The QuattoMaster 6000"][quote="ckck"][hide="Can someone check this"]So you have three spaces for the code. Lets get rid of the last word first.You have three choices for the letter. Then go back to the first two. The first choice has 7 choices. Then  the second choice has 6 choices so the possible choices for the code is $3*7*6=126$.[/hide][/quote]
That solution is a much, much, much better solution than my long and painful solution.  Here is my awful solution:
[b]Case 1[/b]: There is only one vowel
There are $5$ consonants to start out with for the first vowel, and therefore there are $4$ consonants for the second one (because of the $4$ consonants are left).  Finally, there are $3$ vowel choices.  Therefore, there are $5*4*3=60$ combiniations.
[b]Case 2[/b]: There are two vowels.
There are $3$ vowels to start out with, then $5$ consonants, then $2$ vowels to end out with, so $3*5*2=30$.  Similarly, if the consonant is first, then you have $5*3*2=30$, therefore $30+30=60$ for Case 2.
[b]Case 3[/b]: There are 3 vowels
This one is simple, $3*2*1=6$.
Total=$60+60+6=126$[/quote]

Casework is usually not used for these types of problems.
For instance, if a problem said,

Using each letter in the alphabet five times, how many five digit sequences can be made?

This would be very torturous using casework.[/quote]
Yes, that is true.  However, I am just trying to show another way, to check his answer.

but the correct answer is supposed to be 168...
??

Oh!  You forgot the "h" when giving us the problem, so it would be 3*8*7=168.


Para quien esta leyendo este post: La discusion es acerca de temarios en el area indicada por el titulo. Se pretende que si usted pone algo aqui, que sea referente al temario, discutiendolo (criticas completamente aceptadas), haciendo preguntas sobre el o proponiendo otros posibles resultados dentro del tema. Por favor, solo por mantener el orden, si su post no cae en esta categoria, pues inicie un nuevo topic para ello. Esta peticion es unicamente para conservar el orden para quienes estamos interesados en seguirle la pista al temario!!

Solo cuestion de notacion, las secciones, subsecciones y subsubsecciones

estan indicadas asi:
- Seccion
* Subseccion
** Subsubseccion

[b]Algebra[/b]
- Simplificaciones con radicales
- Ecuaciones lineales y sistemas
- Ecuaciones cuadraticas y cubicas
- Polinomios
  * Numero de raices
  * Teorema del factor y teorema del reciduo
  * Identidades de Viete
- Desigualdades
  * Medias Armonica-Geometrica-Aritmetica.
  * Desigualdad de Cauchy Schwarz Buniakowsky.
  * Media Potential
  * Desigualdad de Jensen
  * Desigualdad de Holder
  * Desigualdades con $e$
- Interpolacion de Lagrange.
- Algo de Teoria de Grupos y Anillos podria ser util.

[quote="Leonardo"]-Polinomios Irreducibles
*Criterio de Einsenstein
*Lema de Gauss
-Numeros Algebraicos, Polinomios minimales
*Polinomios Ciclotomicos[/quote]

Mira, estos no los habia pensado, pero la verdad me parece que si podrian ser bastante utiles!


geometry
ratio
rectangle
power of a point

Circles with radii 6 and 12 are externally tangent to each other. Now, a new circle with radius 18 is drawn so that it is internally tangent to the first two circles. A chord is drawn in the larger circle such that the chord is externally tangent to both the smaller circles. Find the length of the chord.

I remember seeing this problem from some AMC...Am I right?

[quote="zolojetto"][quote="happyme"]Online
Joined: 10 Feb 2007
Posts: 267
Location: [u][b]Get some rest, you look tired[/b][/u].[/quote]

i love that movie, bourne ultimatum???[/quote]

It was in Supremacy too.

Um, I think its an AMC, I got it from a collection of problems from everywhere.

[hide="Solution"][img]http://img104.imageshack.us/img104/8046/circlesbi5.png[/img]
Referring to the image, you can see that there are two similar triangles. Setting up a ratio, we have
\[ \frac {12}{x} \equal{} \frac {18}{6} \Rightarrow x \equal{} 4
\]
Then, we can determine that the length of the line from the center of the largest circle to the chord is $ 4 \plus{} 6 \equal{} 10$.

Using the power of the point of the intersection of the chord and the line of length $ 10$, we have
\[ {\left(\frac {x}{2}\right)}^2 \equal{} (18 \plus{} 10)(18 \minus{} 10) \Rightarrow \boxed{x \equal{} 8\sqrt {14}}
\][/hide]

Is there an easier way to do this problem?

wow how do you make that picture, i tried paint but it didnt work, lol maybe im just aweful at paint!!!

Could you explain where the ratios come from (which segments they refer to?)

Especially the $ \frac {18}{6}$ one.

EDIT: Got those. However, how do we know that:

a) The line bisects the radius of the circle with radius length 12.
b) That that figure is a square (so that the length of the segment from the center to the chord is 4+6)

$ 18$ is the distance between the two centers, and $ 6$ is half of the radius of the circle with radius $ 12$. (Can you find this right triangle? You'll see there is a right triangle with legs $ 12$ and $ x$ that is similar to it.)

Do you see the two right triangles that share a common angle?

Oh, you mean the two triangles you used for your original ratios?

Yes. If you find out the side lengths, you'll see how I set up the ratio.

Yeah I figured that out (see my edit).

:D

[quote="happyme"]However, how do we know that:

a) The line bisects the radius of the circle with radius length 12.
b) That that figure is a square (so that the length of the segment from the center to the chord is 4+6)[/quote]

a) The lines form a rectangle, which has congruent opposite sides.
b) The figure is a square (but you don't need to know that). We're only interested in the two similar triangles.

[quote="happyme"]But we use the fact that the figure is a square to be able to add the 6 to get a length of 10.[/quote]
We use the fact that it is a [i]rectangle[/i].

Oh right, this is me being stupid.

Thanks!

No problem. Also, instead of Power of a Point, you could draw in a triangle with one leg being the radius of the radius-12 circle, another leg being half of the chord, and the hypotenuse (not drawn) being a radius of the largest circle.


algebra
polynomial
ratio
geometric sequence
algebra unsolved

Let $ P(x)$ be polynomial of degree $ n$ with rational coefficients and with $ n$ irrational roots,that forms geometric progression.
Find all possible values of $ n$?
Note:This problem is very similar to [url=http://www.mathlinks.ro/viewtopic.php?p=1157561#1157561]another one[/url],and as you can guess it is not easy problem too,furthermore,nobody was able to solve this problem at olympiad...

Here is the positive half of the solution:
Such polynomials exist for all even $ n\equal{}2k$.

Let $ q$ be a solution of $ q\plus{}\frac1q\equal{}1$.
Then $ q$ is irrational, and all powers $ q^a$ and $ q^{\minus{}a}$ with integer $ a\ge1$ are irrational.
Furthermore, $ q^a\plus{}q^{\minus{}a}$ is rational for all $ a\ge1$; this can be proved by an easy induction.

Now consider the sequences $ q,q^3,q^5,\ldots,q^{2k\minus{}1}$ and $ q^{\minus{}1},q^{\minus{}3},q^{\minus{}5},\ldots,q^{\minus{}2k\plus{}1}$.
Together they form a geometric progression of irrational numbers with ratio $ q^2$.

The polynomial with these $ n\equal{}2k$ roots can be written as a product of terms $ (x\minus{}q^a)(x\minus{}q^{\minus{}a})\equal{}x^2\minus{}(q^a\plus{}q^{\minus{}a})x\plus{}1$.
Since all these terms have rational coefficients, also the polynomial has rational coefficients.


[quote="jecruzs"]Hello, this problem was in the Colegio Espíritu Santo 2003 Annual Math competition.

I've tried using logarithms, but to no avail. I'd love some help.

[quote]
Solve within the set of real numbers R, the equation

$ 9^{x}+5\cdot4^{x}+1 = 2\cdot6^{x}+4\cdot2^{x}$

Explain.
[/quote]

I tried reducing it to
$ 3^{2x}+5\cdot2^{2x}+1 = 2^{x+1}\cdot3^{x}+2^{x+2}$, but after that I don't know what to do.[/quote]

Hello!
I managed to solve this problem.

[hide]
$ 9^{x}+5\cdot4^{x}+1 = 2\cdot6^{x}+4\cdot2^{x}$
$ (3^{x})^{2}+5\cdot (2^{x})^{2}= 2\cdot 2^{x}\cdot 3^{x}+4\cdot 2^{x}$

Let $ Y=2^{x}$, $ Z=3^{x}$. From the equation, we immediately see that

$ Z^{2}+5Y^{2}+1 = 2YZ+4Y$,

which reduces to
$ Z^{2}+Y^{2}+4Y^{2}+1 = 2YZ+4Y$
$ Z^{2}-2ZY+Y^{2}+4Y^{2}-4Y+1 = 0$
$ (Y^{2}-2YZ+Z^{2})+(4Y^{2}-4Y+1) = 0$
$ (Y-Z)^{2}+(2Y-1)^{2}= 0$

Since $ Y,Z\in\mathbb{R}$, both terms must be greater than or equal to $ 0$. Since their sum must be $ 0$, both terms must equal $ 0$. This implies that:
$ Y = Z \text{ and }2Y = 1$

The second implies that $ Y=1/2$, and the first implies that $ Z=1/2$

Hence,
$ (Y,Z) = (1/2, 1/2)$

Since $ Y=2^{x}$, $ Z=3^{x}$.,

$ \boxed{(x,y) = (\log_{2}1/2, \log_{3}1/2) = (-1,\log_{3}1/2) = (-1,-\log_{3}2)}$,

or if you can only use natural logarithms with a calculator,

$ \boxed{(x,y) = \left(-1,-\frac{\log 2}{\log 3}\right)}$[/hide]

EDIT: Added more steps to make the solution clearer.

Ah, wow. I don't think I would have ever solved it. Thank you for your response!!


abstract algebra
group theory
superior algebra
superior algebra unsolved

Let G be a finite cyclic group having a  unique composition series.  Prove that G must be a p-group for some prime p.

Also, show that S_3 has a unique composition series.

suppose that $G = \mathbb{Z}/n\mathbb{Z}$, and $p,q$ are two different primes such that $pq|n$: both $\frac{n}{p}G$ and $\frac{n}{q}G$ are maximal (they both have prime order) and normal (since $G$ is abelian), so each of them is the last factor of a composition serie (and the two resulting series are obviously different, and non-isomorphic).

as regarding the second question, the only normal subgroup of $S_{3}$ is $A_{3}$, and it has order $3$, so the only composition change can be $(e) \lhd A_{3}\lhd S_{3}$.


O problema din GM:
Sa se arate ca oricare ar fi a1,a2,..,an>0 are loc
(a1^n+n-1) (a2^n+n-1)… (an^n+n-1)≥(a1+a2+…+an)^n


The following polynomials in x, P(x), have exclusively integer
coefficients.  The number of non-zero terms of P(x) is n.
The coefficient of the highest degree of all of the polynomials
is set at 1.  If the remaining coefficients for the lower degreed
terms are allowed to vary (to be chosen) for each successive
degree polynomial for P(x) so that when P(x) is squared, then
you could get at least one polynomial with the minimum number
of non-zero coefficients for that desired degree.

Question:  Is there a determinable relationship/bound between
the number of non-zero terms, n, in P(x), and the number of
non-zero terms in the square of P(x), where P(x) has a minimum
number possible of non-zero terms?

Examples:

I. $ (x^2 \plus{} 2x \minus{} 2)^2 \equal{} x^4 \plus{} 4x^3 \minus{} 8x \plus{} 4$
II. $ (x^3 \minus{} 4x^2 \minus{} 8x \minus{} 32)^2 \equal{} x^6 \minus{} 8x^5 \plus{} 320x^2 \plus{} 512x \plus{} 1024$
III. $ (x^4 \plus{} 2x^3 \minus{} 2x^2 \plus{} 4x \plus{} 4)^2 \equal{} x^8 \plus{} 4x^7 \plus{} 28x^4 \plus{} 32x \plus{} 16$

Summary of the examples:

I. n = 3 => 4 terms
II. n = 4 => 5 terms
III. n = 5 => 5 terms

Maybe put into a less precise way:

What is the fewest number of terms that you can end up
with in a polynomial of any degree when you square one
that has only integer coefficients?

So, just some quick thoughts -- hopefully I'll have time to think a bit more seriously about this later.  The square of any polynomial with nonzero coefficients (integer, real, complex, or rational doesn't matter) must have a nonzero constant term, a nonzero coefficient on the $ x$ term, and the two highest coefficients must be nonzero as well.  (The coefficients for these terms are just products of the coefficients of the original polynomial, so nonzero.)  This means that for $ n > 3$, the answer is always at least 4.  However, we have an awful lot of flexibility in trying to kill off the middle terms.  It wouldn't surprise me if the sequence is bounded above by 5, for example.

Hopefully this afternoon I'll be able to hammer out some details to go along with that.

Edit: well, some computer experimenting actually suggests that this is a fairly difficult problem.  It also seems to me just as interesting to replace nonzero integers with nonzero reals and ask the same question.  The sequence is not bounded above by 5, as there is no monic fifth-degree polynomial with nonzero integer coefficients whose square has only five nonzero coefficients, assuming I haven't mucked anything up.

If $ P(x)$ has a degree $ D$, then $ P(x)^2$ has a degree $ 2D$.  There are then at most $ 2D\plus{}1$ terms in the expansion of $ P(x)^2$.

Multinomial theorem may help:
$ (a_1\plus{}a_2 \plus{} \ldots \plus{} a_m)^2 \equal{} \sum_{i\equal{}1}^m a_i^2 \plus{} 2 \sum_{1 \le i < j \le m} a_i a_j$

Then it remains to let $ a_i\equal{}b_ix^i$ and try to isolate the coefficients of each power of $ x$ in the expansion.  At that point, we'll have to examine the conditions under which coefficients turn out to be zero.

Quite a subtle problem, indeed. [url=http://matwbn.icm.edu.pl/ksiazki/aa/aa49/aa4916.pdf]Here[/url] it is shown that the sequence in question is unbounded but, of course, it would be great to obtain some better growth rate than double logarithm.


geometry
rectangle
real analysis
real analysis unsolved

Consider some discs of the same radius $r$ around each laticial point in the plane. For a given $N$ (large   enough), give an estimation of the minimal radius $ r=r(N)$ that these discs should have so that one cannot see the origin of the plane from the exterior of the square $NxN$ centered in the origin.

Any idea? I believe the answer should be something like $1/N$, but there are great chances to be wrong. I proved only that the value should be at least $\frac{1}{\sqrt{1+(N+1)^2}}$. Probably it's just non-sense...

I have given up. :blush: . Fedja, do you have a solution, a hint, something?

It is indeed of order $1/N$. Let $y=ax$ be a ray starting at the origin. Without loss of generality, we may assume that $0\leqslant a\leqslant 1$. Recall that for each positive integer $N$, there exists a positive integer $p\leqslant N$ such that $|ap-q|\leqslant \frac 1N$ for some integer $q$. Since $0\leqslant a\leqslant 1$, we have $0\leqslant q\leqslant N$, so the disk of radius $1/N$ around the point $(p,q)$ prevents the origin from being seen from the outside of the $2N\times 2N$ square centered at the origin in the direction $a$.

Thank you fedja. It was really idiot from my part.  :(

This is called "Polya's orchard problem".  Bibliography of the Polya/Szego book(s) of analysis problems may have a more precise reference for the source.


[quote="4everwise"]Can we discuss the round 1 USAMTS problems on this forum after the submitting date has passed?[/quote]

Not until we give the green light to do so.  Generally, this will be 3-4 days after the deadline has passed.


real analysis
calculus
integration
limit
function
advanced fields
advanced fields unsolved

Proof with Lebestgue-integration that

\[\lim_{k \to \infty} \int_{0}^{k} x^n(1-\frac{x}{k})^k dx = n!\]

voor n = 0,1,2,...

The limit is $\int_{0}^{+\infty}x^{n}e^{-x}dx=\Gamma(n+1)$

and use  $\Gamma(x+1)=x\Gamma(x)$

Note that $x^n\left(1-\frac xk\right)^k\chi_{[0,k]}(x)\le x^n e^{-x}$ on $[0,\infty).$

Therefore, we can use $x^ne^{-x}$ (which is integrable) as the dominating function in the Dominated Convergence Theorem to justify interchanging limit and integral.

Of course, interchanging limit and integral leads to the calculation in Moubinool's post.

It is convenient to regard all integrals as being over $[0,\infty),$ using the characteristic function of $[0,k]$ as appropriate.


There was a prisoner about to be executed, the king decided to give him a chance of living

The king said "I have 3 baskets, 100 black marbles and 100 white marbles. i want you to arrange these marbles in a way that which the probability of picking out a white marble is the greatest" name how many marbles of each color are in each basket

Ps: All marbles r used    :D  :D  :D  :D  :D

We will call the 3 baskets A, B, and C
A has 1 white marble
B has 1 white marble
C has 98 white marbles and 100 black marbles.
The probability of picking a white marble is 0.83

1 basket = 1 white marable
2 basket = 1 white marable
3 basket = 100 black and 98 white marables


1. Denote by $p_k$ the kth prime number. Show that $p_1p_2...p_n+1$ cannot be the perfect square of an integer.

No idea where to start from...

2. Prove that it is impossible for thre consecutive squares to sum to another perfect square.

[hide="Number1"]
well... the number certainly cannot be divided by any of the primes $p_1, p_2,...,p_k$, which means it is divisible by another prime. since $p_1=2$, the whole LHS becomes odd. Then use the mod4 to figure out.[/hide]

ive come up with a proof, somewhat. See above post edited.

[hide="Problem 1"]
The first and only even prime is $2$, so for any positive $n$, $p_1p_2...p_n+1$ will be congruent to $3\pmod{4}$. Since perfect squares are always congruent to $0$ or $1\pmod{4}$, $p_1p_2...p_n+1$ will never be a perfect square.
[/hide]

[hide="2"]$(x-1)^2+x^2+(x+1)^2 = 3x^2+2 \equiv 2 \mod{3}$. But all squares are $0,1 \mod{3}$ so it cannot be a square.[/hide]

[quote="xxreddevilzxx"]Couldnt observe that one. good job. Is there any way to do it with out using mods?[/quote]

Not that I can see offhand. Usually when you want to prove something is not a square, mods in 3 and 4 kill it pretty quickly.

Yeah, I kinda did #2 in amirhtlusa's way but not sure if it's correct.

Never thought you could do this with mods... Interesting.... Hmm...  :D

Because all squares are $0$ or $1$ $\bmod 3$

Well ,taking the route of proof by contradiction, let's

assume that p 1 p 2... p n  + 1 = m 2  where m is an integer

=> p 1 p 2... p n = m 2 - 1 = (m-1)(m+1)

Now if m is even, (m-1)(m+1) is odd .But LHS wich has 2 as the only even factor, is even.

Also if m is odd, each of m-1 and m+1 is even,making the product  (m-1)(m+1) div. by 4.
But LHS is not div. by 4

Hence, proved :lol:

For 1, use the Sieve of Erastothenes. Since none of the primes less than $\sqrt{p_1p_2p_3\ldots p_n+1}$ divide $p_1p_2p_3\ldots p_n+1$, $p_1p_2p_3\ldots p_n+1$ must be prime.

Edit: Oh, never mind, I misread the problem :blush:  :blush:  :blush:

[quote="amirhtlusa"][quote="xxreddevilzxx"][quote="amirhtlusa"]let the three consecutive numbers be : $(n-1),n,(n+1)$ then:
 $(n-1)^2+n^2+(n+1)^2=3n^2+2$ now this is OBVIOUS why it cant be a square....

EDIT-NO, Paladin 8 proved it before me. :(  didn't see it first.[/quote]

could you expand on your reasoning? How is it OBVIOUS?[/quote]
yes, because:
$3x^2+2\equiv 2  (mod 3)$
(paladin 8, said it before ;) ,why did you ask again?)[/quote]

I was just checking.... of course i saw paladin's.


Is it just me or have the submit and preview buttons switched places? No big deal, but I've gotten so used to submit being on the right I keep accidentally hitting preview :D

yeah but that makes more sense for me. on the last forum i kept hitting preview

Oddly I do the same thing that both of you say, before I kept hitting preview and now I'm still hitting preview :P

This might actually not be a bad thing because then you get to breifly check your post with the smilies/BBCode and instantly notice if a BBCode doesn't work or you misspelled something or the diagram isn't working or something like that.

It's the style change.  Apparently [url=http://www.trushkin.net/index.php?style=classic]Vjacheslav Trushkin[/url] prefers his submit button on the left, whereas the old black-and-yellow style's creator preferred it on the right.


For $ a_i, x_i\geq0, i \equal{} 1,2,...,n,$ and $ p,q > 0,$ prove that
$ \left.(a_1^{p \plus{} q}x_1 \plus{} a_2^{p \plus{} q}x_2\cdots \plus{} a_n^{p \plus{} q}x_n)\cdot(x_1 \plus{} x_2 \plus{} \cdots \plus{} x_n)\geq (a_1^{p}x_1 \plus{} a_2^{p}x_2\cdots \plus{} a_n^{p}x_n)\cdot(a_1^{q}x_1 \plus{} a_2^{q}x_2\cdots \plus{} a_n^{q}x_n)\right.$

We have from Holder's inequality that $ (a_{1}^{p\plus{}q}x_{1}\plus{}a_{2}^{p\plus{}q}x_{2}\cdots\plus{}a_{n}^{p\plus{}q}x_{n})^{\frac{p}{p\plus{}q}}\dot(x_{1}\plus{}x_{2}\plus{}\cdots\plus{}x_{n})^{\frac{q}{p\plus{}q}}\geq(a_{1}^{p}x_{1}\plus{}a_{2}^{p}x_{2}\cdots\plus{}a_{n}^{p}x_{n})$ and $ (a_{1}^{p\plus{}q}x_{1}\plus{}a_{2}^{p\plus{}q}x_{2}\cdots\plus{}a_{n}^{p\plus{}q}x_{n})^{\frac{q}{p\plus{}q}}\dot(x_{1}\plus{}x_{2}\plus{}\cdots\plus{}x_{n})^{\frac{p}{p\plus{}q}}\geq(a_{1}^{q}x_{1}\plus{}a_{2}^{q}x_{2}\cdots\plus{}a_{n}^{q}x_{n})$. Multiplying these together then yields the desired result.

Well done, [b]bakerbakura[/b]! Thanks! By the way, my solution doesn't use any classical inequality.
I'm still waiting for new ideas.

Here's a proof using no classical inequalities:

$ \left.(a_{1}^{p\plus{}q}x_{1}\plus{}a_{2}^{p\plus{}q}x_{2}\cdots\plus{}a_{n}^{p\plus{}q}x_{n})\cdot(x_{1}\plus{}x_{2}\plus{}\cdots\plus{}x_{n})\minus{}(a_{1}^{p}x_{1}\plus{}a_{2}^{p}x_{2}\cdots\plus{}a_{n}^{p}x_{n})\cdot(a_{1}^{q}x_{1}\plus{}a_{2}^{q}x_{2}\cdots\plus{}a_{n}^{q}x_{n})\right.$
$ \equal{}\sum_{i\equal{}1}^n a_i^{p\plus{}q}x_i^2\plus{}\sum_{i\neq j}a_i^{p\plus{}q}x_i x_j\minus{}\sum_{i\equal{}1}^n a_i^{p\plus{}q}x_i^2\minus{}\sum_{i\neq j}a^p_i x_i a^q_{j} x_j$
$ \equal{}\sum_{i\neq j}a_i^{p\plus{}q}x_i x_j\minus{}a^p_i x_i a^q_{j} x_j\equal{}\sum_{i\neq j}a_j^{p\plus{}q}x_j x_i\minus{}a^p_j x_j a^q_{i} x_i$
$ \equal{}\frac{1}{2}\sum_{i\neq j}x_i x_j (a_i^p\minus{}a_j^p)(a_i^q\minus{}a_j^q)\geq0$ since $ p,q>0$.

This is the elementary solution to this inequality! Good job, [b]bakerbakura[/b]! We can obtain some interesting results using this inequality.

In fact, if we have the same $ x_i$ and $ a_i$ as above and four real numbers $ a,b,c,d$ such that $ a\plus{}b\equal{}c\plus{}d$, we have that if $ (a\minus{}b)^2\geq(c\minus{}d)^2$, then
$ (a_{1}^{a}x_{1}\plus{}a_{2}^{a}x_{2}\cdots\plus{}a_{n}^{a}x_{n})\cdot(a_{1}^{b}x_{1}\plus{}a_{2}^{b}x_{2}\cdots\plus{}a_{n}^{b}x_{n})\geq(a_{1}^{c}x_{1}\plus{}a_{2}^{c}x_{2}\cdots\plus{}a_{n}^{c}x_{n})\cdot(a_{1}^{d}x_{1}\plus{}a_{2}^{d}x_{2}\cdots\plus{}a_{n}^{d}x_{n})$

Since the inequality and conditions are symmetric in $ a$ and $ b$, we may assume $ a\geq b$. Similarly we may assume $ c\geq d$. Then we have that $ a\minus{}b\geq c\minus{}d\iff a\geq c$ since $ a\plus{}b\equal{}c\plus{}d$. Also, $ 2c\geq c\plus{}d\equal{}a\plus{}b\geq2b$.

Then $ (a_{1}^{a}x_{1}\plus{}a_{2}^{a}x_{2}\cdots\plus{}a_{n}^{a}x_{n})^{\frac{c\minus{}b}{a\minus{}b}}\cdot(a_{1}^{b}x_{1}\plus{}a_{2}^{b}x_{2}\cdots\plus{}a_{n}^{b}x_{n})^{\frac{a\minus{}c}{a\minus{}b}}\geq a_{1}^{c}x_{1}\plus{}a_{2}^{c}x_{2}\cdots\plus{}a_{n}^{c}x_{n}$ and
$ (a_{1}^{a}x_{1}\plus{}a_{2}^{a}x_{2}\cdots\plus{}a_{n}^{a}x_{n})^{\frac{a\minus{}c}{a\minus{}b}}\cdot(a_{1}^{b}x_{1}\plus{}a_{2}^{b}x_{2}\cdots\plus{}a_{n}^{b}x_{n})^{\frac{c\minus{}b}{a\minus{}b}}\geq a_{1}^{d}x_{1}\plus{}a_{2}^{d}x_{2}\cdots\plus{}a_{n}^{d}x_{n}$ by Holder's inequality. Multiplying these two then gives the desired result. The inequality can also be proved by multiplying out and using the results deduced in the first paragraph to write the difference as a sum of nonnegative terms.


The date for ARML 2009 is Friday May 29 and Saturday May 30.

As was the case in 2008, this is not an SAT date.

Yay! Maybe I'll be on the SoCal A Team this time. :D

(By the way, good job on the title of the topic...)

LOLOLOLOL edit the topic...

lol maybe I'll be on the Freshman Team at the age of 12... likely!

That's great!  It means I don't have to schedule my final exams around ARML, as I thought I would have to!

[quote="BOGTRO"] lol maybe I'll be on the Freshman Team at the age of 12... likely![/quote]

What exactly was the point of that comment?

Yay, I'm marking this on my calender now. Thanks!

[quote="Ignite168"]Ugh, weekend before finals again although next year my finals will be Monday, Wednesday, Thursday versus Monday, Tuesday, Wednesday, Thursday, Friday, and my hardest exam won't be Monday which will be much more manageable with ARML.[/quote]The exam schedule probably isn't going to be the same next year, dude.

Sigh...during Finals again. Can we appeal to move ARML a week sooner?  :P

A week earlier would be the Memorial Day weekend. Always.

I don't even know when my exams are next year...

anyone in the RI/Mass area who needs someone for ARML???

No one's going to know at this point in the year.

There is no Rhode Island team. The Western Massachusetts team is closest to you; you should either join them or try to form a new team.

Right at the last day for seniors next year.


What is the tens digit in the sum 7! + 8! + 9! + . . .
+ 2006! ?
(A) 1 (B) 3 (C) 4 (D) 6 (E) 9

Explain how you would do this

EDIT: Sorry about that, never looked for another  :blush:

There is already [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=433612#433612]a thread for this[/url].  Next time, check the Contests section before posting an AMC/AIME problem.