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IMO Shortlist

Does there exist functions $ f,g: \mathbb{R}\to\mathbb{R}$ such that  $ f(g(x)) \equal{} x^2$ and $ g(f(x)) \equal{} x^k$ for all real numbers $ x$

a) if $ k \equal{} 3$?

b) if $ k \equal{} 4$?

I only have a soln for a (I guess b was the tough part):
f(g(f(x)))=f(x^3)=(f(x))^2, so f(0), f(1), f(-1) are all roots of the eqn x^2-x=0. Then at least 2 of them must be equal. Let's assume f(0)=f(1). This means that g(f(0))=g(f(1)) so 0=1, false. The same happens if f(0)=f(-1) or f(1)=f(-1) because g(f(x))=x^3.

[color=blue]

For a) very easy

f(x) = x 2,  g(x) =x

f(g(x)) = f(x) =x 2

g(f(x))=g(x 2 )=x 2

[/color]

I think you got that wrong: in b) it's f(g(x))=x^2 and g(f(x))=x^4. Maybe the fact that it was written g(f(x))=(x^2)^2 confused you?

Just an idea (not the proof and the complete solutions)

I found in my drawers

If x > 1the following functions

f(x) =2^(2^sqrt(lnx/ln2))

g(x) = 2^((lnx/ln2)^2)

verify the conditions on the interval ]1;+ \infty [, i hope.

 :D

For b) Partial answer without proof

Sorry the exponent was not at the right place

for x>1

f(x) = 2^(2.sqtr(lnx/ln2))

g(x) = 2^((lnx/ln2)^2)

I hope now ot works

They satisfy the given relations indeed, but how did you come across them ? :?
Adn what about ]-oo, 1[ ?

I found this question in my note, but i don't understand the proof
some line missing It was  12 years ago (when i was student)

We have to find a clear nice proof.
Let's continous to think about it.

i have the solution of this again.... not by me of coz, again, troublesome to type it

myth 's correct

So, a solution, anyone ? Post it please ...

There is no solution, there is an answer

I don't know if it will work

but from

(1)  f(g(x)) = x^2

(2)  g(f(x)) = x^4,

the functions x^2 & x^4 are C^(oo) , (derivable as much as we want).

I wonder if from (1), (2) we can get a differentials equations

involving  only f and an other one with g, and solve them to get f,g

yes there is an answer but I remember thinking about that problem one year ago for half of hour and I just couldn't figure out the function ... so I guess that the question people are asking is just how in the world did you come up with those (not so easy to come up with) functions ?  :)

From (1) and (2) f(x^4)=f^2(x).
Let h(x)=f(e^x). Then h(4x)=h^2(x). Let H(x)=ln h(x). Then H(4x)=2H(x) - and what the simplest function satisfies with equation? Of course \sqrt x, and so we try H(x)=c*\sqrt x ==> h(x)=e^(c\sqrt x) ==> f(x)=e^(c\sqrt ln x). By the same reasons g(x)=e^(c'ln^2(x)). If we find c and c' we find f and g.

For the equation f(g(x)) =xx, let y=g(x), then f(y)=xx. Consider x=2^2^u, y=2^2^v. Then 2^2^v =g(2^2^u) => v=log_2 log_2 g(2^2^u) and f(2^2^v) = 2^2^(u+1) => u+1 = log_2 log_2 f(2^2^v). Let G(u) = log_2 log_2 g(2^2^u) and F(v) = log_2 log_2 f(2^2^v), then v=G(u), u+1=F(v). Hence u+1=F(G(u)). SImilarly, g(f(x))=x^4 leads to v=F(u), u+2=G(v), hence u+2=G(F(u)). To get such F and G, we try F(u)=au+b, G(u)=cu+d. Then u+1=a(cu+d)+b=acu+ad+b and u+z=c(au+b)+d=cau+bc+d. Hence we want ac=1, ad+b=1, bc+d=2, taking a=1/2, c=2, b=1, d=0, we get F(u)=u/2 +1, G(u)=2u satisfy F(G(u))=u+1, G(F(u))=u+2. Then g(x)=2^2^G(u)=2^2^2u=2^(2^u)^2=2^(log_2 x)^2 and f(y)=2^2^F(v) = 2^2^(v/2 +1) = 2^[2(2^v)^(1/2)]=2^[2 \sqrt log_2 y] Notice the f function is defined only if log_2 y>=0 or y>=1 In that case, we check f(g(x)=xx and g(f(x)=x^4.
To make this work for all x E R, we now extend f,g to the real number as follow:
define
f(x)
=2^[2 \sqrt log_2 y] if x>=1
=1/f(1/x) if 0<x<1
=0 if x=0
=f(|x|) if x<0
and
g(x)=2^(log_2 x)^2 if x>=1
=1/g(1/x) if 0<x<1
=0 if x=0
=g(|x|) if x<0
And this function satisfies the conditions

[quote="moubi"]I found this question in my note, but i don't understand the proof
some line missing It was  12 years ago (when i was student)[/quote]

That's strange ! This question was proposed for IMO 1997, and now you're saying you saw it twelve years ago :?

not all questions proposed at the IMO are original !!
for example the following question was used at a selection test in 2000 (Romania) and was submitted on the jury in '99 Romania.

prove that there are an infinity of positive integers a,b,c,d such that

a^3+b^3 | c^3 + d^3.

fortunately, the olympiad committe being formed by Romanians (that IMO was in Romania) they noticed the same problem could be found in an old (1984) Romanian book.

so it is indeed possible that Moubi solved that problem 12 years ago ! :D

The question you're talking about was a bit different Valentin :

Prove that for each positive n, there exist positive integers a, b, c and d such that
n = (a + b)/(c + d).

Am I mistaken or is this very tricky and difficult ? :D
I don't feel comfortable with the official solution.

"Your version" of the above problem is ... well ... hmm, original at least, but a bit trivial :)

yeah you are right Arne - although I don't agree with the problem being very difficult - the official answer was found by many Romanian contestants that year (almost all 6 who made it to the team) so i guess it's not that hard to come.

However, it is possible, although not probable, that some of the people who solved back then had knowledge of the shortlist 1999.

About the problem, hmm indeed you are right - i've just created a new problem .... :D let's solve it ... maybe put some more conditions like {0,1} \cap {a,b,c,d} = \emptyset

You know, I'm just an unexperienced Belgian guy ... so for me this is far from easy. :D
Indeed a was exaggerating a bit when I said this is "very" difficult.
But it's tricky and I don't like it :? (Did you solve it Valentin ?)

About your problem :D :
Putting c = ka, d = kb we get an infinite family of solutions.
But I guess this problem can't be solved properly.
There are too much possibilities, you've got four variables !

hmmm ... nope i didn't solve it back then - I was too busy preparing to go to Korea for the StarCraft BroodWar BattleNet Finals :D - so I didn't trained properly for the IMO TST's :D

StarCraft BroodWar the greatest strategy game ever - created by Blizzard. It's a RTS game ... :D

b)  Let
       g(x)= x^lnx     if x is bigger than or equal to 1
       g(x)= x^(-lnx)     if x is bigger than 0 and smaller than 1
       g(x)= (-x)^ln(-x)     if x is smaller than or equal to -1
       g(x)= (-x)^(-ln(-x))     if x is bigger than -1 and smaller than 0
       g(x)=0 if x=0
      and
       f(x)=y^2 ,where y is such that g(y)=x or g(-y)=x.
    these functions f and g are satisfying the given condition(it can proved by let y=f(x)^(1/2)...)
  I like this problem it is not very hard but it is nice :lol:

$f,g :R \rightarrow R$ үүд нь $f(g(x)) =x^2$ ба $g(f(x))=x^k$ байх функцууд байг.
а) Хэрэв $k=3$ бол ийм функцууд олдох уу?
б) Хэрэв $k=4$ бол яах вэ?


blogs
algebra
polynomial
number theory
modular arithmetic

What is the remainder when $x^{5}+2x^{4}-x^{3}+x+1$ is divided by $x^{2}+x+1$?

[hide="The usual trickiness"] Let $\omega$ be one of the roots of $x^{2}+x+1$.  Then $\omega^{3}= 1$, so

$\omega^{5}+2 \omega^{4}-\omega^{3}+\omega+1 = \omega^{2}+2 \omega-1+\omega+1 = \omega^{2}+3 \omega = 2 \omega-1$

So the remainder is $\boxed{2x-1}$. [/hide]

[quote="binomial_4eva"]can you explain this part please.how do u get from $\omega^{5}+2 \omega^{4}-\omega^{3}+\omega+1$ to $\omega^{2}+2 \omega-1+\omega+1$[/quote]

It was stated that $\omega^{3}= 1$, so $\omega^{5}= \omega^{3}\cdot\omega^{2}= \omega^{2}$, and so on.

Refer to my blog post on [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=125861]polynomial modular arithmetic[/url] for a solution that is

[hide="Alternate but equivalent"] Since $x^{3}\equiv 1 \bmod (x^{2}+x+1)$, we know that

$x^{5}+2x^{4}-x^{3}+x+1 \equiv x^{2}+2x-1+x+1 \equiv x^{2}+3x \equiv 2x-1 \bmod (x^{2}+x+1)$

This is phrased the same way as my solution but is actually very similar to spix's.  In fact, the rigorous way to produce the algebraic numbers is by the construction of modular polynomial rings. [/hide]


calculus
integration
number theory
relatively prime

[tex](\sqrt{a}+\sqrt{b})^2=c=a+b+2\sqrt{ab}[/tex]
So ab must be a perfect square. If a and b are both perfect squares then [tex](\sqrt{a}+\sqrt{b})[/tex] is an integer but [tex]\sqrt{c}[/tex] can never be integer. a and b can't be relatively prime,  let d be a common divisor of a, b
So we can factor d out from [tex]a+b+2\sqrt{ab}[/tex] Contradict definition of prime

Sqrt(5)+sqrt(0)=sqrt(5).. and 5 is prime...

[quote="beta"]$(\sqrt{a}+\sqrt{b})^2=c=a+b+2\sqrt{ab}$
So ab must be a perfect square. If a and b are both perfect squares then $(\sqrt{a}+\sqrt{b})$ is an integer but $\sqrt{c}$ can never be integer. a and b can't be relatively prime,  let d be a common divisor of a, b
So we can factor d out from $a+b+2\sqrt{ab}$ Contradict definition of prime[/quote]

You start off on a nice algebraic path, but what did you really discover?  a and b need not be perfect squares, but what must be?  How can that help you solve the problem?

We have to assume that $a$ and $b$ are both positive.

$a + b + 2\sqrt{ab} = c~.$

Since $c = 2$ works for the question, we can safely assume that $c \ge 3$; hence $c$ is odd. $ab$ must be a perfect square.

If $a$ and $b$ are both odd or both even, then $a + b$ is even. Then LS is even but RS is odd. Contradiction.

Suppose then that $a$ is even and $b$ is odd. Then, for $\sqrt{ab}$ to be integral, $a$ must be a multiple of 4. Hence, we let $a = 2^{2k}m$, with $m$ odd. Then the we must have

$c = 2^{2k}m + b + 2^{k + 1}\sqrt{mb}~.$

Suppose that $m$ and $b$ are not coprime, and we have $(m, b) = n$. Let $m = nx$ and $b = ny$. Then

LS = $2^{2k} nx + ny + 2^{k + 1}\sqrt{n^2xy} = n\left(2^{2k}x + y + 2^{k + 1}\sqrt{xy}\right)$, which isn't prime. Contradiction.

Then $m$ and $b$ must be coprime; for $\sqrt{mb}$ to be integral, both $m$ and $b$ must be perfect squares themselves. Then $\sqrt{a}$ and $\sqrt{b}$ are integral, so LS of the original equation is integral. But RS isn't. Contradiction.

QED.

[quote="h_s_potter2002"][quote="Scrambled"]:sqrt: a+ :sqrt: b <>  :sqrt: c when a,b, and c are integers and c is a prime number.[/quote]

I think I'm missing something, but can't this problem include many different solutions such as:

${\sqrt{100}+\sqrt{100}}  \not=  \sqrt{7}$
${10+10} \not= \sqrt{7}$
$20 \not= \sqrt{7}$

We can tell that $20$ is bigger than $\sqrt{7}$ becuase $20 > 7$ and $7 > \sqrt{7}$, so $20 > \sqrt{7}$.  If $a > b$ then $a \not= b$.[/quote]

The question asks you to PROVE :sqrt: a+ :sqrt: b <>  :sqrt: c[b] FOR ANY POSITIVE INTEGERS[/b] a, b, c, where c is a prime.

[quote="beta"]The question asks you to PROVE :sqrt: a+ :sqrt: b <>  :sqrt: c[b] FOR ANY POSITIVE INTEGERS[/b] a, b, c, where c is a prime.[/quote]

Okay. I didn't see the proove in the problem.


For $ x,y,z>0$, prove that $ \sum_{cyc} x^3(y^2\plus{}z^2)^2 \ge xyz\sum_{cyc}xy(x\plus{}y)^2$

[quote="The Nguyen"]For $ x,y,z > 0$, prove that $ \sum_{cyc} x^3(y^2 \plus{} z^2)^2 \ge xyz\sum_{cyc}xy(x \plus{} y)^2$[/quote]

 $ \Longleftrightarrow \sum_{cyc} y^2z^2(y\plus{}z)(x\minus{}y)(x\minus{}z)\geq 0$

[quote="The Nguyen"][quote="hedeng123"][quote="The Nguyen"]For $ x,y,z > 0$, prove that $ \sum_{cyc} x^3(y^2 \plus{} z^2)^2 \ge xyz\sum_{cyc}xy(x \plus{} y)^2$ (1)[/quote]

 $ \Longleftrightarrow \sum_{cyc} y^2z^2(y \plus{} z)(x \minus{} y)(x \minus{} z)\geq 0$                (2)[/quote]
1) Why (1) is equivalent  (2)?
2) Why (2) true?[/quote]

$ \sum_{cyc} x^3(y^2 \plus{} z^2)^2 \minus{} xyz\sum_{cyc}xy(x \plus{} y)^2 \equal{} \sum_{cyc} y^2z^2(y \plus{} z)(x \minus{} y)(x \minus{} z)$

assume $ x\geq y\geq z$

we have $ (x\minus{}y)(x\minus{}z)\geq 0$ and $ (z\minus{}x)(z\minus{}y)\geq 0$

$ \sum_{cyc} y^2z^2(y \plus{} z)(x \minus{} y)(x \minus{} z)\geq z^2x^2(z\plus{}x)(y\minus{}z)(y\minus{}x)\plus{}x^2z^2(x\plus{}z)(z\minus{}x)(z\minus{}$ $ y)\equal{}x^2z^2(y\minus{}z)^2(z\plus{}x)\geq 0$

[quote="The Nguyen"]For $ x,y,z > 0$, prove that $ \sum_{cyc} x^3(y^2 \plus{} z^2)^2 \ge xyz\sum_{cyc}xy(x \plus{} y)^2$[/quote]

$ \iff \sum_{cyc} {x^4(y^3 \plus{} z^3)} \plus{} xyz\sum_{cyc} {2x^2yz} \ge \sum_{cyc}{x^4(y^2z\plus{}yz^2)} \plus{} xyz\sum_{cyc}{x^2(y^2\plus{}z^2)}$

$ \iff \sum_{cyc} {x^4(y^3 \plus{} z^3\minus{}y^2z\minus{}yz^2)} \ge xyz\sum_{cyc}{(x^2(y^2\plus{}z^2)\minus{}2x^2yz)}$

$ \iff \sum_{cyc} {x^4(y\plus{}z)(y\minus{}z)^2} \ge xyz\sum_{cyc}{x^2(y\minus{}z)^2}$

$ \iff \sum_{cyc} {x^3(xy\minus{}yz\plus{}zx)(y\minus{}z)^2} \ge 0$

then use S.O.S Theorem ...

 :huh:


Prove:

$1+\cos\theta +\cos 2\theta+...+\cos n\theta=\frac{\sin\frac{n+1}{2}\theta\cos\frac{n}{2}\theta}{\sin\frac{\theta}{2}}$

"Just take the real part" is quite difficult in comparison to the first part of the proof.

Use $\sin \alpha\cos \beta=\frac{1}{2}\{\sin (\alpha+\beta)+\sin (\alpha-\beta)\}$,

$\sum_{k=0}^n \sin \frac{\theta}{2}\cos k\theta=\frac{1}{2}\sum_{k=0}^n \left\{\sin \left(k+\frac{1}{2}\right)\theta -\sin \left(k-\frac{1}{2}\right)\theta \right\}$

$=\frac{1}{2} \left\{\sin \frac{\theta}{2}+\sin \left(n+\frac{1}{2}\right)\theta\right\}$

Use $\sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}$

$=\sin \frac{n+1}{2}\theta \cos \frac{n}{2}\theta$

Divided by $\sin \frac{\theta}{2}\neq 0$ both sides,yielding the result.

Very nice.

Using the same method, you can prove the following identity also.

$\sin\theta+\sin 2\theta+...+\sin n\theta=\frac{\sin\frac{n+1}{2}\theta\sin\frac{n}{2}\theta}{\sin\frac{\theta}{2}}$

I used Gaussian pairing and got

$\sin\theta+\sin 2\theta+ \sin 3\theta+\dots+\sin n\theta=n\sin\left(\frac{n\theta+\theta}{2}\right)\cos\left(\frac{n\theta-\theta}{2}\right)$


For funtion $f: N^2\to N$ such that the following conditions:
i/$f(0,y)=y+1$
ii/$f(x+1,0)=f(x,1)$
iii/$f(x+1,y+1)=f(x,f(x+1,y))$
Determine $f(4,1985)$


How many 5-digit integers have their digits in descending order?

[quote="math154"]Strictly decreasing or otherwise?[/quote]

if abcde, a>b>c>d>e

Then it's simply $ \dbinom{10}5$, because everything is forced in a group of $ 5$ distinct digits.

[hide="clarification"]Let's first choose five digits. 10*9*8*7*6. Now we re-arrange the digits in all $ 5!$ ways. Only one of those ways will make the digits strictly descending. So we must divide by $ 5!$: $ \frac{10*9*8*7*6}{5!}\equal{}\binom{10}{5}$.[/hide]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=21600520&t=227632]here[/url] for a question on the number of positive integers with strictly [i]increasing[/i] digits.

Either way the solution remains elegant; we assume a string of length n listing 1->9 or 9->0:
"123456789" or "9876543210"
We may now remove any digit at random and still maintain strict order; we do so by choosing 1 from n, then 2, 3 and so on until it becomes clear that the answer is a sum of binomials (n k) with k incrementing--ie 2^n-1. take one away because the number zero does not have strictly decreasing digits just as "1" lacks strictly increasing digits.

a bit of a tangent, but the answer is 10 choose 5.  :icecream:

It's not $ \dbinom{10}5$ for strictly increasing digits.

This includes, for example, $ 01234$.

What it really is is $ \dbinom95$, because we just can't have $ 0$ at all.


If any point P in triangle ABC is equidistant from two sides AB and BC does that mean it lies on the angle bisector of vertex B?

Yes. If you draw the two distances from point P to AB and BC, you get two triangles congruent by H-L so the two angles must also be equal.


A block of mass 5.416 kg is released on a frictionless inclined track at a height of 4.82 m above the level of the surface. It slides down the track and makes a head-on elastic collision with a block of mass 10.832 kg, initially at rest. Calculate the height to which the block with mass 5.416 kg rises after rebounding from the collision.

The height is 4.82 m, so we can use $ v_{f}^{2}=v_{i}^{2}+2ad$ to determine the impact velocity.  Plugging in numbers, we get that $ v_{f}=9.72\mbox{m/s}$.  So $ p_{i}=5.416\mbox\frac{kg}\cdot9.72{m/s}=52.64\frac{kg \cdot m}{s}$.  Now since the mass of the first block is half that of the second block, it's velocity after impact should be twice that of the other one.  So 2$ 5.416\cdot v_{1}=52.64\rightarrow v_{1}=4.86\mbox{m/s}$ is the velocity of the smaller block.  For some reason, I can't figure out how to get the height though.  Sorry.  Hope what I gave you helps.

Yah i got that far too, but i wasn't sure if i can do energy conservation: $ \frac{1}{2}mv^2\equal{}mgh\plus{}\frac{1}{2}Mv_2^2$
actually, i know that doesn't work lol

Actually, it's not as useless as you think.    We recall that $ p_{i}\equal{}p_{T}\equal{}52.64$ (I'm omitting units for convenience).  Assuming that the larger block loses no energy or velocity, we can say that $ mgh\plus{}\frac{1}{2}Mv^{2}\equal{}52.64$.  Plugging in what we already know, $ 5.416\cdot 9.8\cdot h\plus{}\frac{1}{2}(10.832)(2.43)^{2}\equal{}52.64$ where $ mgh$ is potential energy of the first block.  So solving for $ h$, we get $ h\equal{}.389\mbox{m}$.

sorry, i'm a n00b at phys, but why can we assume the momentum is equal to the energy afterwards? (aren't they in different units too?)
i would think the KE of the bigger block and the GPE of the smaller one should be equal to the GPE of the small block at the beginning, since it is elastic and energy is conserved, but apparently i'm wrong somewhere in that thinking

Ah, it appears I plugged in the wrong numbers.  Momentum won't equal energy.  The initial energy in the system must equal the final energy.  The initial energy is $ mgh\equal{}5.416\cdot 9.8\cdot 4.82\equal{}255.83$.  So the energy equation must total this, so $ 5.416\cdot 9.8\cdot h\plus{}\frac{1}{2}(10.832)(2.43)^{2}\equal{}255.83\rightarrow h\equal{}4.22\mbox{m}$.  If I did the numbers right, that should be your answer.

sorry again, but where did that 2.43 come from?
why doesn't this work:
$ mgh\equal{}mgh_{unknown}\plus{}\frac{1}{2}Mv^2$
5.41*9.8*4.82=5.416*9.8*h+1/2*10.832*([b]4.86[/b])^2
h=2.41 m/s?

Because 2.43 is the velocity of the larger block.  We need the energy of the entire system, large block and small one.  So we're using the kinetic energy of the large block, coupled with the final potential energy of the small block.

hmm something's not right then cuz i punched it that answer, and i got it wrong on my hw ..
i was thinking maybe momentum conservation?
$ m_1v_1\equal{}m_2v_2\plus{}M_V$
5.146*9.71967=5.416*v_2+10.832*V
ugh... don't like phys..


Princeton college Harvard Yale MIT Stanford

Just out of curiosity...

What is # graduating undergraduate of top 5 schools going into the top 5 schools? i.e.,

Princeton -> Princeton, Harvard, Yale, CalTech, MIT, Stanford (in number)

Harvard -> Princeton, Harvard, Yale, CalTech, MIT, Stanford

Yale -> ...

etc.

It would be most advisable to make 7 by 7 table, but since I do not know how to make the table...

One thing that is misleading about this question is that many students with undergraduate degrees from the top colleges go immediately into very lucrative jobs in private enterprise and never think of going into graduate school. Sometimes it is more important to rate liberal arts colleges (colleges without their own graduate school) this way, by how many of their graduates get into a graduate degree program, because some of those graduates have much more trouble finding cool real-world jobs. But regardless of where you go as an undergraduate, whether or not to go to graduate school is a separate decision, and should only influence your decision about where to go to college insofar as you should avoid completely lousy undergraduate colleges at full list price.

Are you aware of the distinction between "graduate school" and "professional school"? In the United States, for instance, a law degree is not usually considered a "graduate" degree, but rather a "professional school" degree, and the same is true of a medical degree.

What question would this 5x5 or 7x7 table try to answer?
Certainly there is a correlation between the level of undergraduate and graduate institution, but this is obvious without any tables.
To some extent the top schools just permute their inputs and outputs, but it is not nearly a closed system.


calculus
integration
derivative
logarithms
calculus computations

For $x\geq 0,$ find the minimum value of $x$ for which $\int_0^x 2^t(2^t-3)(x-t)\ dt$ is minimized.

I am wondering if I am the right track, would the fundamental theroem of calculus be of any help?

(what i mean, in case i have the wrong name or w/e)

$\displaystyle I=\int_0^x f(t)dt.$
then
$\displaystyle \frac{d}{dx}I=f(x).$

Yes, you can use fundamental theorem of calculus. :)

[quote="maokid7"]so I have gotten this far.
$\displaystyle G(x)=\int_0^x2^t(2^t-3)(x-t)dt=$
$\int_0^x(x(2^t)^2-3x(2^t)-t(2^t)^2-3t(2^t))dt$ and now have
$G'(x)=-(x(2^x)^2+3x(2^x))+\frac{d}{dx}\int_0^x(x(2^t)^2-3x(2^t))dt$

I am not sure how you deal with the X in the second part of $G'(x)$? (i've only  taken up to calcBC...should i know this or is the procedure far beyond this class?)  :?[/quote]

Write $2^t(2^t-3)(x-t)$ as $x f(t)+g(t)$

and then $F(x) =x \int_{0}^{x} f(t) \, dt + \int_{0}^{x} g(t) \, dt$.

Now use product rule for the derivative .

 :)

well here is what i have.

$\displaystyle G(x)= x\int_0^xf(t)dt-3\int_0^xf(t)dt$
where $f(t)=2^t(2^t-3)$.
now we have
$\displaystyle G'(x)=\int_0^xf(t)dt+xf(x)-3f(x)=$
$\displaystyle\int_0^xf(t)dt+(x-3)f(x)$ to evaluate $\displaystyle\int_0^xf(t)dt$ we just use the reverse produce rule (more commonly known as by-parts)
after doing this we have

$\displaystyle G'(x)=(x-3)(2^x)(2^x-3)+\frac{(2^x)(2^x-3)}{\ln 2}-\frac{(2^x)^2\ln 2}{2}-\frac{4+\ln ^2{2}}{\ln 4}$.

then we find the value of $x$ such that $G'(x)=0$.
I think ill leave that for someone else to do...or ill do it when i have more time.  :P  :rotfl:


i searched this subject on google and this site's search, but i could only find 1980-1989 and 1995-2001( from http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm ). i also find 2002 from mit's site, but i couldn't find 2003 and 2004. if you have them please send me the site or attach it. thanx

It's not comming up for me, it says "error" when I click on the link :(

I think he means this link:

http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm

 :)

[quote="denizuzun"]sorry i didn't see the error... i will edit it now

but does anyone have 2003 and 2004 amc challenges??[/quote]

Yes, I have the AMCs, however, I didn't get them online, I'm afraid you'll have to buy them, or perhaps ask your math teacher or something.


Can somebody tell me when the math team practices are during this summer?  I'm going to be in Vestavia in early July and I'd like to stop by and give Tipton a copy of the new book.

Practice will be at VHHS this summer as follows:

Tuesday, June 6        6:30 - 8:30pm    first practice

Monday, June 12       6:30 - 8:30pm    answer questions
Tuesday, June 13       6:30 - 8:30pm    individual tests and ciphering

Monday, June 19       6:30 - 8:30pm    answer questions
Tuesday, June 20       6:30 - 8:30pm    individual tests and ciphering

Monday, June 26       6:30 - 8:30pm    answer questions
Tuesday, June 27       6:30 - 8:30pm    individual tests and ciphering

Wednesday, July 5    6:30 - 8:30pm    practice old tests
Thursday, July 6        6:30 - 8:30pm    practice old tests
Friday, July 7            6:30 - 8:30pm    practice old tests

Monday, July 10        6:30 - 8:30pm    practice old tests
Tuesday, July 11        6:30 - 8:30pm    practice old tests
Wednesday, July 12   6:30 - 8:30pm    practice old tests
Thursday, July 13      6:30 - 8:30pm    mini-tournament


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

[quote="LkNsngth"]if $ R$ is an integral domain contained in a local ring $ (S,Q)$, then show that there is a minimal prime of $ S$ contracting to $ 0$ in $ R$.

This is from exercise 10.7 of Eisenbud's Commutative Algebra with a view towards algebraic geometry. I've looked at the problem and I'm stumped as to where to start. Any input would be appreciated. Thanks  :)[/quote]

I don't see why we need $ S$ to be a local ring!  :(  $ R\minus{} \{0 \}$ is a multiplicative subset of $ S$ because $ R$ is an integral domain. So there's a prime ideal $ P$ of $ S$ with $ P \cap (R \minus{} \{0 \})\equal{}\emptyset.$

By Zorn's lemma $ P$ contains a minimal prime ideal $ P'$ of $ S$ and then it's clear that $ P' \cap R\equal{}\{0 \}.$

[quote="LydianRain"][quote="ysharifi"]So there's a prime ideal $ P$ of $ S$ with $ P \cap (R \minus{} \{0 \}) \equal{} \emptyset.$[/quote]

What result exactly are you using here?[/quote]

Let $ T\equal{}R\minus{} \{0 \}$ and consider $ C\equal{}\{I \lhd S: \ I \cap T \equal{} \emptyset \} \neq \emptyset,$ because $ (0) \in C,$ with inclusion. By Zorn's lemma $ C$ has a maximal element $ P.$

$ P$ then has to be a prime ideal of $ S$ because if $ ab \in P$ but $ a \notin P, \ b \notin P,$ then $ (P\plus{}Sa) \cap T \neq \emptyset, \ (P\plus{}Sb) \cap T \neq \emptyset.$ Then since $ T$ is closed

under multiplication and $ ab \in P,$ we'll get that $ P \cap T \neq \emptyset.$ Contradiction.

[quote="ysharifi"][quote="LydianRain"][quote="ysharifi"]I don't see why we need $ S$ to be a local ring!  :(  $ R \minus{} \{0 \}$ is a multiplicative subset of $ S$ because $ R$ is an integral domain. So there's a prime ideal $ P$ of $ S$ with $ P \cap (R \minus{} \{0 \}) \equal{} \emptyset.$

By Zorn's lemma $ P$ contains a minimal prime ideal $ P'$ of $ S$ and then it's clear that $ P' \cap R \equal{} \{0 \}.$[/quote]
Wait!  Are you sure we can use Zorn's lemma here?  What if our descending chain of prime ideals has intersection $ \emptyset$?[/quote]

That's not possible because every ideal contains $ 0$.  :)[/quote]

Ah, excuse me.  Never mind...   :roll:

In fact, it seems to me that the proof can be adapted to show that if $R$ is a (unitary commutative) ring with extension $S$ and $p$ is a prime ideal in $R$, there exists a prime ideal $q$ of S such that $q\cap R\subset p$.

Yes: Take any prime ideal (e.g. a maximal ideal) in the localization $S^{-1}B$ with $S=A\setminus\wp$.


Here are two functions:
\[ \begin{array}{ccc}
f(x) & \equal{} & 3x^2\minus{}2x\plus{} 4\\
g(x) & \equal{} & x^2\minus{}kx\minus{}6
\end{array}\]
If $ f(10) \minus{} g(10) \equal{} 10$, what is the value of $ k$?


topology
advanced fields
advanced fields unsolved

1. Let ~ be the equivalence relation $(x_{1}, x_{2})$ ~ $(y_{1}, y_{2})$ iff $x_{2}=y_{2}$, on $R^{2}$. Then $R^{2}/$~ is homeomorphic to $R$.

2. Let D be the decomposition of the plane into concentric circles about the origin. Prove that D is homeomorphic to $\{x \in R|x\ge0\}$.

3. Find a counterexample for:
"$E \subset X$ is disconnected iff there are disjoint open subsets $H$ and $K$ in $X$, each meeting $E$, such that $E \subset H \cup K$."

Thank you,

The homeomorphisms in 1 and 2 are quite natural: $(x,y)$  is sent into $y\in \mathbb R$; a circle of radius $r$ is sent into $r\in [0,\infty)$.

A counterexample for 3 is given by a three-point space $X=\{a,b,c\}$ with the topology generated by $\{a,c\}$ and $\{b,c\}$. The subset $E=\{a,b\}$ is disconnected, because $\{a\}$ and $\{b\}$ are open in $E$. Yet, one cannot find $H$ and $K$ as in the statement.


geometry
inradius
modular arithmetic
parameterization
ratio
Pythagorean Theorem

[size=150]Inradius of a triangle is 1 (unit).
It has integer sides.
Prove that it is a right triangle[/size]

The inradius of an triangle is equal to area/semiperimeter of the triangle.So if we get inradius =1 that means ,the semiperimeter and the area are numerically equal.Using area=(a*b*sin@)/2  we get a+b+c=a*b*sin@ .We got an integer in the left term witch means the right one is also an integer.That means sin@=1 ,and that means the angle measures 90 degrees...(we could also get sin@=0 but a point isnt a triangle right?)....hope my proof is corect,even that i can't find the "right triangle" from the problem

@Balint Florin
Nice method I must say.

by the way, I used semiperimeter=area thing. But instead, i converted the Rhs to the heron's formula.
So it was like:

         s=sqrt{s*(s-a)*(s-b)*(s-c)}
so,     s=(s-a)(s-b)(s-c)    on squaring both sides
As s=(a+b+c)/2 =>
=>     4(a+b+c)=(a+b-c)(a-b+c)(-a+b+c)

I couldnt simplify this further to get:
a^2=b^2+c^2 or anything of that form.

Thanks for the answer
 :)

Heron's formula was first thing i thinked of  but things weren't going to well

[quote="Balint Florin"]... we get a+b+c=a*b*sin@ .We got an integer in the left term witch means the right one is also an integer.That means sin@=1 ,and that means the angle measures 90 degrees...(we could also get sin@=0 but a point isnt a triangle right?) ...[/quote]

What if [b]@=30 degrees[/b]? Then [b]sin@=1/2[/b]. a and b are sides of a triangle. So there is always a possibility that at least one out of [b]a and b is divisible by 2[/b]. Thence, both sides of "[b]a+b+c=a*b*sin@[/b]" are integers in this case too.

Please think over it.

:blush:  ure right ,sin can also get 1/4 ,1/3  any value from 0 to 1(0<@<180) ... there was a and formula for triangles area when you know all three angles and one side  witch i dont remember ... maybe that would help here
  btw i still can't find a triangle with  integer sides and inradius 1 unit

[quote="thegreatindiankhali"][quote="bpgbcg"]A 3-4-5 right triangle has inradius 1.[/quote]

Yes I know this. But it would be better if you could prove that, there is no other triangle possible with integer sides to have inradius 1 unit except the 3-4-5 right triangle.[/quote]

For any inscribed circle within a right triangle,

$ r \equal{} ab/(a\plus{}b\plus{}c)$

where r is the radius of the circle (I'll leave the derivation to the individual, not hard).

Euler's formula states that for any right triangle with integer sides,
$ a \equal{} m^2\minus{}n^2$
$ b \equal{} 2mn$
$ c \equal{} m^2\plus{}n^2$
where (m,n) is a positive integer ordered pair, and m > n.

Substituting these expressions into the formula for the radius stated above, we can simplify the expression to become

$ r \equal{} n(m\minus{}n)$

It is now evident that if r=1, the only possible pair (m,n) is (2,1), which means (a,b,c) is (3,4,5), and that is also the only right triangle to have an inradius of 1.

Well, because $ a\plus{}b\plus{}c\equiv a\plus{}b\minus{}c\equiv a\minus{}b\plus{}c\equiv \minus{}a\plus{}b\plus{}c\pmod2$ and
\[ 4(a\plus{}b\plus{}c)\equal{}(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(\minus{}a\plus{}b\plus{}c)\]we can see that $ a\plus{}b\plus{}c$ must be even, so $ A\equal{}rs$ must be an integer.

Then this is a [url=http://mathworld.wolfram.com/HeronianTriangle.html]Heronian triangle[/url], so from the link we find that
\[ r\equal{}1\equal{}k(mn\minus{}k^2)\]and the rest should be easy. I'm not sure how easy the proof of the parametrization is though...

[quote="thegreatindiankhali"][quote="bpgbcg"]A 3-4-5 right triangle has inradius 1.[/quote]

Yes I know this. But it would be better if you could prove that, there is no other triangle possible with integer sides to have inradius 1 unit except the 3-4-5 right triangle.[/quote]
Sorry for the confusion.  I was not attempting to answer the problem; Instead I was responding to the poster above me about an example of of a triangle satisfying the desired property.

Hm, the proof of the parametrization for Heronian triangles (sides and area are rational) is fortunately pretty easy.
[hide="Proof"]With standard triangle notation, let $ c$ be the longest side, so we can drop an altitude of length $ h$ from $ \angle C$ which breaks $ c$ into segments $ d$ and $ e$ where $ a$ and $ d$ meet at $ \angle B$ and $ e$ and $ b$ meet at $ \angle A$. (There are two right triangles formed: one with legs $ d,h$ and hypotenuse $ a$ and the other with legs $ e,h$ and hypotenuse $ b$.) From the Pythagorean Theorem,
\begin{align*} h^2 = a^2 - d^2 & = b^2 - e^2 = (a + d)(a - d) = (b + e)(b - e) \\
d^2 - e^2 & = a^2 - b^2\implies d - e = \frac {a^2 - b^2}{d + e}. \end{align*}
But $ a,b$, and $ c = d + e$ are rational, so $ d - e$ is rational as well. Then $ ((d + e) + (d - e))/2 = d$ and $ ((d + e) - (d - e))/2 = e$ are rational too.

Thus we now have, from our Pythagorean Theorem relations above,
\[ a + d = m,\quad a - d = \frac {h^2}m,\quad b + e = n,\quad bee = \frac {h^2}n,\quad m,n\in\mathbb{Q}.\]
Finally,
\begin{align*} a & = \frac12((a + d) + (a - d)) = \frac1{2m}(m^2 + h^2) \\
b & = \frac12((b + e) + (b - e)) = \frac1{2n}(n^2 + h^2) \\
c & = d + e = \frac12((a + d) - (a - d) + (b + e) - (b - e)) = \frac1{2mn}(m + n)(mn - h^2). \end{align*}
Thus all Heronian triangles have sides in the ratio of $ x,y,z$ where
\begin{align*} x & = v(u^2 + w^2) \\
y & = u(v^2 + w^2) \\
z & = (u + v)(uv - w^2) \end{align*}
where $ u,v,w$ are integers such that $ uv > w^2$.[/hide]

An interesting corollary is that all solutions of
\[ pqr(p+q+r)=n^2\]can be expressed as
\[ p=w^2(u+v),\quad q=v(uv-w^2),\quad r=u(uv-w^2),\quad p+q+r=2uv(u+v),\]since letting $ a=p+q$, $ b=q+r$, and $ c=q+p$ this turns into
\[ (a+b+c)(a+b-c)(a-b+c)(-a+b+c)=16n^2,\]which is equivalent to the problem of Heronian triangles.


How can one make a single spaced row vector in a double spaced text?

My paper is double spaced in LaTeX, but row vectors such as $ \begin{bmatrix}X & Y & Z & W\end{bmatrix}$ look very ugly (see attached image). How do I make the brackets look like the ones in single space while keeping my paper in double spacing?

I've tried \begin{singlespace} and \end{singlespace}, but then it gives me a line break that I don't want. I've also tried \singlespacing with the setspace package, but it still gives me a line break before.

The easy answer is not to use double-spacing as it's not suitable for typeset documents - see [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=linespace]Double-spaced documents in LaTeX[/url].
I would argue with whoever says you must use double-spacing but assuming that you have no choice then put \singlespacing on the previous line as in something like this:
[code]This is line 1 \singlespacing
Line 2 where  $\mathbf{X}=\begin{bmatrix}X & Y & Z & W\end{bmatrix}$ \doublespacing\\
Line 3[/code]


function
algebra unsolved
algebra
functional equation

Consider  all  functions $ f: \mathbb{N} \to \mathbb{N}$  satisfying $ f(xf(y))=yf(x)$ for any $x,y \in \mathbb{N}$. Find the least possible value of $ f(2007)$.

$ x\equal{}1$ give $ f(f(y))\equal{}yf(1)$, therefore $ f(y)\not \equal{}f(z)$ if $ y\not \equal{}z$.
From $ x\equal{}y\equal{}1$ we get $ f(f(1))\equal{}f(1)\to f(1)\equal{}1$. Therefore $ f(f(y))\equal{}y$.
$ x\equal{}f(z)$ give $ f(f(z)f(y))\equal{}yz\to f(y)f(z)\equal{}f(f(f(y)f(z)))\equal{}f(yz).$ Therefore f is multiplicative.
Let p is prime.  If $ f(p)\equal{}mn$ is not prime, then $ p\equal{}f(f(p))\equal{}f(m)f(n)$ contradition.
Therefore $ f(p)\equal{}q,f(q)\equal{}p$ $ p$ - prime if and only if  $ q$ prime.
On the set of primes f defined $ f(p)\equal{}q,f(q)\equal{}p$, for some primes may be $ q\equal{}p$. By multiplicative condition we full defined $ f(n)$.
$ 2007\equal{}3^2*223$, therefore $ f(2007)\equal{}f(3)^2f(223)\equal{}p^2q$. If p=2, then $ q\ge 5$ and $ f(2007)\ge 20$.
If p=3,q=2 we get $ f(2007)\equal{}3^2*2\equal{}18$ minimfl possible value.

Nice solution..
But you consider that $ N\equal{}{1,2,3,...}$ ??
When I tried to solve it, I did'nt know if $ N\equal{}{1,2,3,...}$ or $ N\equal{}{0,1,2,3,...}$

But considering the first option, I proved that $ f$ is injective, as you proved.

Sorry for my poor English.

Many thanks.
Bye

If $ 0\in N$, then obviosly least mean is 0, For example $ f\equiv 0$.  :lol:

[quote="reza1370"]if x=1 then f(f(x))=x
in your solution f(3)=2 and f(223)=3  so 223=f(f(223))=f(3)=2 so 223=2[/quote]
If $ f(3)\equal{}2$, then $ f(2)\equal{}3$, therefore $ f(223)\ge 5$.
minimfl solution, when $ f(3)\equal{}3,f(223)\equal{}2$.


algebra
polynomial
calculus
integration
binomial theorem

prove that $ (3^4)^n*3^ 2 \plus{} (5^2)^n*5^ 1$ is a multiple of 14 where n is natural

Hmmm...

$ 9\plus{}5\equal{}14$, but $ 81\plus{}25\equal{}106$ which is not a multiple of 14...

But $ 729\plus{}125\equal{}854$ is a multiple of 14. $ 854\equal{}14 \cdot 61$.

What you really want to prove is $ a\plus{}b|a^{2k\plus{}1}\plus{}b^{2k\plus{}1}$.

This is true because if we view this a polynomial in $ a$, then a necessary and sufficient condition for $ (a\plus{}b)$ to be a factor is that $ \minus{}b$ is a root, which is true because:

$ (\minus{}b)^{2k\plus{}1}\plus{}b^{2k\plus{}1}\equal{}0$.

(Note that this doesn't work for even exponents, since $ (\minus{}b)^{2k}\equal{}b^{2k}$.

hello we have $ 81 \equiv 11 \mod 14$ and $ 25\equiv 11 \mod 14$ and from here we get
$ 81^n\cdot9\plus{}25^n\cdot5\equiv11^n\cdot9\plus{}11^n\cdot5\equiv11^n(9\plus{}5)\equiv11^n\cdot14\equiv0\mod14$
Sonnhard.

well gauss1181..you cant tell that $ a^x \plus{} b^x$ is always divisible by $ a \plus{} b$  . you can tell that when x is an odd natural number.maybe in this case it worked out but to be rigorous you have to specify . i think so. :)

if u have X raised to the power Y then break X into a sum of 14 or a multiple of 14 and a constant.
Expand it using binomial theorem and you should have 14 common from all of it and multiplied by some integeral sum in the form of a combinationary series. that should prove that is a multiple of 14.

[quote="gauss1181"]Well, $ 3^{4n}*3^2 \plus{} 5^{2n}*5 \equal{} 3^{4n \plus{} 2} \plus{} 5^{2n \plus{} 1}$ when simplified.

Now we also notice that we can write the base 3 exponential term so that it's base 9; since $ 3^2 \equal{} 9$, we have $ 9^{2n \plus{} 1} \plus{} 5^{2n \plus{} 1}$. $ 14 \equal{} 9 \plus{} 5$, and since a number $ a^x \plus{} b^x$ is always divisible by $ a \plus{} b$, the original expression is thus divisible by 14.

I'm not sure if I have to rigorously prove the $ \frac {9^{2n \plus{} 1} \plus{} 5^{2n \plus{} 1}}{9 \plus{} 5}$ part though...(I just suck at writing rigorous proofs :( ) I don't really know how that could be proved. All I know is that the "$ \frac {a^x \plus{} b^x}{a \plus{} b}$ is an integer part" is true...[/quote]

To prove $ (9 \plus{} 5)|(9^x \plus{} 5^x)$ for some odd $ x$, note the factorization

$ a^n \minus{} b^n \equal{} (a \minus{} b)(a^{n \minus{} 1} \plus{} a^{n \minus{} 2}b \plus{} ... \plus{} ab^{n \minus{} 2} \plus{} b^{n \minus{} 1})$.
Replacing $ b$ by $ \minus{} b$ with $ n \equal{} x$ we get $ a^x \minus{} ( \minus{} b)^x \equal{} a^x \plus{} b^x \equal{} (a \plus{} b)(a^{x \minus{} 1} \minus{} a^{x \minus{} 2}b \plus{} ... \minus{} ab^{x \minus{} 2} \plus{} b^{x \minus{} 1})$. Now substitute $ a \equal{} 9,b \equal{} 5$.


$a,b,m,n$ are integers,$k,p$ are odd numbers.
Prove or disprove :
If $p^m|a+b$ then $p^{m+n}|a^{kp^n}+b^{kp^n}$.

True.
It's enough to prove it for primes $p$ and $k=1$, the first one by factoring and the last one by looking $\mod p^{m+n}$.
It's also enough to prove it for $n=1$ then (so that $p^s | a+b \implies p^{s+1} | a^p+b^p$) because the general one will follow by induction.
Now $a=-b+t p^s$ by what is given, so $a^p = (-b+t p^s )^p = -b^p + \sum_{w=1}^p (-b)^{p-w} p^{ws} \binom{p}{w} = -b^p +p^{s+1} (and\ blah)$, using that $\binom{p}{w}$ is divisible by $p$ when $w \neq 0,p$.
Thus it's dead, dead, dead.


Does anyone know how to find the sum of the roots of a cubic function?

well, you can always use vieta's theorem.


abstract algebra
advanced fields
advanced fields unsolved

How do I show that in an abelian category if $f : A \to B$ and $\text{coker}(f) = 0$ then $\text{im}(f) : = \text{ker}(B \to \text{coke}(f)) = B$. I am sure that this result is true, but I need it as a lemma for a homework problem and cannot seem to figure it out.

Thank you for the help!

If the cokernel is zero, any map into it is zero. It should be clear from the definitions that the kernel of the zero map is (naturally isomorphic to) the domain.

yes... that is obvious. I guess that was not really my problem.

What I want to show is that if $f : A \to B$ and $\text{ker}(f) = \text{coker}(f) = 0$, then $f$ is an isomorphism. What I have been able to do is
\[A = \text{coker}(0 \to A) = \text{coim}(f) = \text{im}(f) = \text{ker}(B \to 0) = B \]
showing that $A$ is isomorphic to $B$, but I have not been able to show that $f$ is said isomorphism.

So, I need that $\text{im}(f)$ is isomorphic to $B$ and that the canonical map $\text{im}(f) \to B$ is said isomorphism. Then I can write $f$ as the composition of isomorphisms
\[A \to \text{coim}(f) \to \text{im}(f) \to B. \]

Again thank you for the help!


calculus
integration
derivative
function
probability
geometry
3D geometry

[quote="misan"]
i asked this question to a couple of 7th and 8th graders (11-13 yr olds), they failed to show it, but most of 9th and 10 graders did provide the answer. [/quote]

Wow! What kind of school do you teach in? I think teaching such a thing to children of that age is unheard of here in the US.

That's what I was thinking.  I'm not even sure undergraduates in the US would find it easy...  :|

Indeed. All the proposed solutions are easy to understand but none is straightforward with the usual tricks.
At the age of 15, no student in Belgium even know what is integration and they never heard the word Gaussian...

The same in Portugal. Actually, it would be almost a miracle if a 13-14 years old student in my country knows how to work with simple one variable equations and fractions ...
Unless, of course, Misan teaches in a school whose students are "gifted children".

well, i'm from india too (though not from bengaluru) and we don't learn hardcore integration till the 11th grade and some times even 12th.
as far as i know, only differentiation is common knowledge to most 14-15 year olds, but i very much doubt a couple of 15 year olds can solve this kind of integral problem.

[quote="misan"]

i asked this question to a couple of 7th and 8th graders (11-13 yr olds), they failed to show it, but most of 9th and 10 graders did provide the answer. here is the answer that i eventually posted.
[/quote]

I will be in bengaluru in June this year . pls tell me where I can meet these 9th graders.

Wow, 9th and 10th? Are you sure you don't mean 11th and 12th graders. And are you teachers? I too am from India (close to Benguluru) and would definitely like to meet your students.

Are you a teacher at a school or at some IIT-JEE/Olympiad coaching centre? I don't think calculus is there for MO and I definitely haven't heard of any 9th/10th'ers doing such good problems though I have seen young people do calculus.

[quote="pardesi"][quote="misan"]

that depends.

i asked this question to a couple of 7th and 8th graders (11-13 yr olds), they failed to show it, but most of 9th and 10 graders did provide the answer. here is the answer that i eventually posted.

l[/quote]
 :wow:  :wow:  where do u teach in india...i can beleive people from 8th-9th class solving difficult olympiad problems...but solving integrals is a real good thing???
and u gave this to 7th graders .... :rotfl:  :rotfl:  :huh:  they wouldn't be even knowing algebra completely that is function,relation and all forget about integration :huh:  :huh:[/quote]

 :D Very interesting where are such cute children... To know differentiation under the integral at the age of 11-13 years old they should study all elementary math in 5-6 years (go to school in 6 or 7 years old)... I think they should have only one subject in school - math... or they are all geniuses. :)))
Misan, how old are you?

This is great    :stretcher:
I  am from Chennai. I was jus thinking what i was doing in my 7th and 8th grade..
KUDOS to those two...

this is crazy. i am from bangalore and i havent come across any 13 or 14 year old who is capable of  deriving that formula. there is the probability of only one 8th grader who can do this. he got thru inmo this year. who did you ask misan?

Im Vignesh from Chennai.. That sum you gave was solved by 9th graders!! I'd change my name if that were true... Nowhere in India do ppl in class 11 and 12 know fully how to excersise Laplace transformations, chuck the 9th n 10th... Even a chuck of the olympiad gang in this very same india would not be able to solve that one.

Well, I am in 8th.It is not like I do olympiads very well but I am very good.
However, I have NEVER seen even my cousin who is in 12th solving these?
It is quite impossible to do so for 7th and 8th std.Not thinking tht since I do well, no one in 8th can do to such high levels.
But, practically it is NOT POSSIBLE, especially in India.

Sorry, not 7th and 8th.

Still, even if they r in 9th and 10th, it is not possible.

And it was a big funny thing tht u asked the 7th and 8th guys.

Crux: i am sorry if belgian 15 yr olds do not know integration. you will always find a handful of students,
mostly 15 and 16yr olds here within the indian subcontinent who educate themselves (outside of normal
schooling that is) using texts such as hardy's "pure mathematics" and david widder's "advanced calculus".
so maybe they are aware of a few things that many of you probably may not have heard of.

Carcul: probably the most sensible post i have read in this thread. yes, you are correct.... i am, shall i say,
 blessed to teach a group of gifted students. some of them have amazed me beyond belief. for
example, they have taught me a lot more about gamma functions after i used to think i was good at it.
to be frank, where normal teenagers play with sony playstations, these students opt to play with
ramanujan's formulas! it fascinates them.

hash_include: i myself didn't learn hardcore integration when i was in school. i did that outside of school,
on my own. Y R U surprised ? i was self educated using whittaker and watson's text and i read that when
i was 15 yrs old. sorry if that makes me look stupid, but i am not.  i am not claiming i am a genius of
any sort either. all i am saying is i and many others who have had a desire to learn chose to do so by
going beyond the norms and reading classic books on our own. it is not a big deal.

Cyber-Shreyas: no, i don't teach IIT or IISc students because most of them don't even have
a good foundation on differential or integral calculus to begin with. whatever they were taught in high
school apparently have misled them to think only inside the box.

VARUNARASU: what is crazy ? just because someone can do things that most can't doesn't mean
you have to be in disbelief. i have come across a 12 yr old who i  claim, without a doubt,
can explain differentiation or integration far better than most professors.

vignesh from chennai:   think carefully before you write, nowhere in my post did i claim
i teach laplace transforms in 8th grade. of-course, i knew laplace transform when i was 13, but that's
because.... well, let's not go there 'cause i don't want to brag. my olympiad gang (if i were
to teach one that is) can and will outperform yours any day, hands down! how you ask ? simply because
i will teach them the  fundamentals in such a manner that they will have no trouble
applying those to solve interesting and complicated problems. where you will use conventional
techniques found in most textbooks, they will hopefully opt to do it in different and quicker ways.

Aravind Srinivas L: if a student has the willingless to read and learn on his/her own, you have no
right to conclude anything based on your mild experience. do you teach ? have you
met anyone outside of your classmates who know calculus ? sorry to disappoint you, but i do, and that too,
on a daily basis. for example, it is not unusual to teach differentiation under integral sign in an introductory
differential calculus class, grade school or not. if you did not learn that in grade school, that's because
your instructor probably didn't even know what that was. sure, my instructor didn't know about it either,
but i sure learnt   about it when i read other texts at the library.

regardless of where it is, don't question the statements of other people. just because you and those
around you were not fortunate enough to be adept with advanced problem solving skills does not mean
you should say anything that comes to your mind. be appreciative of the fact that somewhere around
the world young minds are learning things at a much earlier age. that can either happen in school, or outside
of school, depending on how motivated the youngster is.

I'm not commenting on Misan's claim. I just want to know. Where are you a teacher? In a school? Or for students aspiring to get into the Maths Olympiad team or just a private maths learning home coaching.

P.S: Vignesh, maybe you should change your name.  :rotfl:

sorry. i dint mean to sound offending. but i second shreyas. where do you teach, misan??

I have to say that misan's questions (most of them) are extremely hard to solve but also interesting.

... and, those "gifted students," do they post here, in MathLinks?

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194627[/url]

nice sums!
Just a random doubt that's bugging me. MODS please dont mind this one post :D

Misan, did you give these sums to 7th stds as well? They solved this?  :D

well, misan hats off to you, thats all i can say.

Anyway, my math-loving cousin is now studying in 6th grade, so i REALLY want to know where this institute of yours is, to make him join.
even if commuting may be a problem, a teacher of your skills can not be left just like that, and can do wonders for our country.:)

NOTE: In case you are thinking i am being sarcastic here, i am NOT.

[quote="madness"]then let them all go to tc next year and win golds for india  :rotfl:   :D

Let all the present imotc guys quit next year ok? And i bet even Dang will be shocked at that  :rotfl:[/quote]

hmm. I'm not sure if you're referring to the olympiad in this post, but I just found it interesting that this forum is so amazed by children being able to assimilate material quickly. In the US, there's a sort of negative perception of that, especially in terms of the olympiad team, as most people agree it would be very easy to teach an eighth grader what a laplace transform is, but really kind of irrelevant (speaking in terms of the olympiad) -- in fact, the trainers discourage the olympiad team from spending much time on advanced mathematics, or really any mathematics at or above calculus level

[quote="MysticTerminator"]

hmm. I'm not sure if you're referring to the olympiad in this post, but I just found it interesting that this forum is so amazed by children being able to assimilate material quickly. In the US, there's a sort of negative perception of that, especially in terms of the olympiad team, as most people agree it would be very easy to teach an eighth grader what a laplace transform is, but really kind of irrelevant (speaking in terms of the olympiad) -- in fact, the trainers discourage the olympiad team from spending much time on advanced mathematics, or really any mathematics at or above calculus level[/quote]

Maths Olympiad is not the ultimate. Learning Mathematics is. IMO or any of its lower stages is a good test of your application and thinking skills, but is restricted to limited areas of maths.

Usually we see a lot of little kids working towards MO from classes 7/8 and learning a lot of stuff much beyond their age. However what surprised us was that calculus was not even in the olympiad portions... And for a eighth grader to truly understand this kind of stuff, thats great.

And no its not really frowned upon here. Many of the MO people do know a lot of stuff in other spheres of math as well, sometimes due self-interest sometimes as part of their preparation for other entrance exams.

What I wonder is to what extent misan teaches theory, or if it's all stated theorems and mind-numbingly boring calculations of integrals. If I had a teacher who insisted I spend my time reexpressing integrals in terms of other things, I would not abide.

Misan, your students teach you things about the Gamma function? So basically you've done basic complex analysis and more? How far do you go? Do these young people know integration theory, general topology and other "advanced" analysis-y undergrad topics? Do you teach more specialized topics? How do you justify testing your students on calculating integrals through grades 8 - 12? That is quite the outdated topic ...

[quote="anandghegde"]where do you teach sir? I live in Bangalore......I want to visit your institute once...... :D[/quote]

Thats what we've all been asking and till now no one's got a reply and nobody in b'lore have heard of this extrordinary coaching institute.

Hmm..Is there another Benguluru somewhere in India that we don't know of..  Or perhaps like Harry Potter we have to enter some secret door to enter the coaching institute. Anyway, I NV the students.

[quote="madness"]yes yes even i wish all the 8th and 9th imotc'ers go to his classes so that they can exceedingly develop their talents at a young age (read as chinna vayasu laye semmaya talenta cultivate pannalaam :D )[/quote]

Yes, its been quite long since we've got a gold in IMO and even longer since we've got a Top 10 ranking. I'm sure such a class can benefit lots of talented youngsters who are breeding in your country. Though it would be nice if they'd learn things other than integration.

atleast misan should encourage his present students to write the Maths Olympiad so that they can wash away all the places at the imotc :D :D

What puzzles me most about misan is not his ability to do some crazy formal computations that I would never be able to think of (and even not his claim that he teaches them to 7th graders) but his almost complete ignorance about such simple analytic things as convergence of series, etc. I cannot reconcile these two no matter how hard I try :?.

it seems that this guy can teach even the dogs to bark in the mathematical--or precisely--integral way :rotfl:

There's no doubt in what Misan said at all. Its bcos ' [b]He wants to believe[/b] ' :D

For all who don't get it, see his avatar :D :rotfl:

i pmed him asking his address but nooo reply :mad:

certainly u were not the ist one---his inbox is already 100% and can't recieve messages :D  :D

[quote="anandghegde"]If i knew there was such a resourceful and talented teacher in my own city, I would have gone to him for coaching :mad: (along with the 7th and 8th graders :P  :P )

I think all of you must have seen these

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179721

such a sum for 15 yr olds?  :wacko:

see this too

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194082
integrals for 8th mid term tests? :o

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=194974
no one has been able to solve this sum and he gives them as homeworks? :rotfl:  :P  :P[/quote]

Why not, just because you haven't heard of it nor because you are unable to solve them, it doesn't mean the extremely hyper-intelligent 10 year olds can't. Especially since they are taught gamma functions and the whole of integral calculus. See, thats a new breed he's training, they are Homo Mathematica.

Ofcourse, they aren't MO students, nor are they IIT aspirants. Nor are they interested in IISC or anything. I wonder, how they use their math knowledge. Or is it just

[hide]Chumma Jujubi[/hide]

[quote="Chronoz"] Homo Mathematica.
[/quote]
it seems as if the genus also has changed
in language of VMC
Geniii mathematica :rotfl:  :rotfl:  :rotfl:

btw that was shreyas who quoted it .....and not me :D

ahhhh-i deleted the end tag by mistake :wacko:    :oops_sign:

[quote="madness"][url=http://www.mathlinks.ro/Forum/viewtopic.php?t=207308]This one is the best[/url] :rotfl: :D :huh:[/quote]

 :rotfl:

btw, i am beginning to believe him  :|
he is such a talented fellow ......
and these days he ( and his students ofc :P ) seems to be doing some weird ramanujan-esque probs

but his student teasing him is just a bit too much......

[hide]yes
[hide] :D

 :P [/hide]
 [/hide]


topology
trigonometry
advanced fields
advanced fields unsolved

Show that the tautological line bundle on $ RP(1)$ is not a trivial bundle. In fact show that it is homeomorphic to the Mobius Strip.

The tautological line bundle $ p: E\to\mathbb{R}P(1)$ is a subspace of the trivial bundle $ E\subset\mathbb{R}P(1)\times\mathbb{R}^2$ consisting of pairs $ (l,v)$ with $ v\in l$ and $ p(l,v) \equal{} l$.

Can you see why it isn't trivial (note that it is locally trivial and that all trivial bundles have nowhere zero sections) and make the identification to the Mobius strip?

Can you please elaborate. I cant see how it gives the reqd result

It is nontrivial because $ w_1(\gamma) \neq 0$. It is homeomorphic to the Moebius strip because $ K_0(\mathbb{R}P^1) \equal{} \mathbb{Z}/2$.

Nontrivial: Two methods. Prove the Mobius strip part and clearly the Mobius strip is not isomorphic to the trivial bundle. Other:

Show: Every section $ s: B\to E$ has a zero. Consider $ S^1\to \mathbb{R}P(1)\to E$ by projection then $ s$. So $ x\mapsto ([x], t(x)x)$ where $ [x]$ is the equivalence class in $ \mathbb{R}P(1)$ and $ t(x)$ is continuous with $ t(\minus{}x)\equal{}\minus{}t(x)$. By $ S^1$ being connected the IVT gives that for some $ x_0\in S^1$ we have $ t(x_0)\equal{}0$, and thus $ s([x])\equal{}([x], 0)$.

Hint for identifying with the Mobius strip: think of how you write complex numbers. Identify with $ [0,\pi]\times \mathbb{R}$ by writing everything in the bundle as $ \left([\cos\theta, \sin\theta], t(\cos\theta, \sin\theta)\right)$


function
modular arithmetic
algebra solved
algebra
Functional Equations

Let $\mathbb N = \{0,1,2,\ldots\}$.  Determine all functions $f: \mathbb N \to \mathbb N$ such that
\[ xf(y) + yf(x) = (x+y) f(x^2+y^2)  \]
for all $x$ and $y$ in $\mathbb N$.

Greetings,
if $x = 0, \text{then} yf(0) = yf(y^2) \Rightarrow f(0) = f(n^2)$ when $n \in N$
if $x=y \neq 0$ then $2nf(n) = 2nf(2n^2) \Rightarrow f(n) = f(2n^2) \text{where } n \in N$
so that $f(n) = f(n^2) = f(2n^2) \text{ for } n \in N$
[Edit:Sorry, always the same mistake. :blush: ]

Excuse me, are you sure $f(x)$ is constant, Kalimdor?

Let $xf(y) + yf(x) = (x+y) f(x^2+y^2). \ y=0 \ give \ f(x^2)=f(0),$
$y=x>0 \ give \ f(x)=f(2x^2).$  Therefore $f(2^k)=f(0)$.
Let $g(x)=f(x)-f(0), x>0$ We have $g(z^2)=0,g(z)=g(2z^2)$. Let $x=2z^2,y=z^2 \Longrightarrow g(z)=g(2z^2)=3g(5z^4).$
If used fact, that g(z) is integer, we prove g(z)=0, because $g(z)=3g(5z^4)=9g(5^5z^16)=27g(5^{21}z^{64})=\dots.$
It give $3^k|g(z)$ for all k, therefore g(z)=0,f(x)=const. It may be proved visout use, that f(x) is integer.

Suppose $f$ is not a constant function. There must exist $x,y$ > 0 such that $f(y) > f(x)$. Choose $x,y > 0$ such that $f(y) - f(x)$ is minimal. Such $x, y$ exist by the well ordering principle. Then \[ f(x) = \frac{xf(x) + yf(x)}{x + y} < \frac{xf(y) + yf(x)}{x + y} < \frac{xf(y) + yf(y)}{x + y} = f(y). \] Hence $f(x) < f\left(x^2 + y^2\right) < f(y)$, which contradicts the minimality of $f(y) - f(x)$. Hence $f$ must be constant. It is obvious that all constant functions taking positive integer values work.

how do you know that such x and y exists?

Every non-empty set of positive integers must have a smallest element.

Wow, Arne's solution is very clever. Here is a less clever solution.

Substituting $x=0$ and $y=n>0$ into the equation gives that $f(0)=f(n^2)$ for all $n \in \mathbb{N}$. Now consider $x=k>0$ and $y=m^2$ where $m$ is selected such that $\gcd(k, m^2) = 1$ and $m^2 > |f(k) - f(0)|$. Now by the equation, it follows that $kf(0)+m^2f(k)=kf(m^2)+m^2f(k)=(m^2 + k)f(k^2 + m^4)$ which implies that $0 \equiv kf(0)+m^2f(k) \equiv (-m^2)f(0)+m^2f(k) \pmod{m^2 +k}$ which implies that $m^2+k$ divides $f(k)-f(0)$ since $\gcd(k+m^2, m^2)=\gcd(k, m^2) = 1$. However, $m^2 +k > m^2 > |f(k) - f(0)|$ which implies that $f(k)=f(0)$ for all $k>0$. Hence $f(x)=f(0)$ and is constant for all $x \in \mathbb{N}$.

Note that $f(m^2)=f(0)$ for all $m$; in particular $f(1)=f(0)$. Now pick $n \ge 1$, then we have $n+1 \mid f(n)-f(1)$. So for all $n>10^3(1+f(1))$, either $f(n)=f(1)$ or $f(n)>n$. Fix such an integer $n$ and define $a_k=2a_{k-1}^2$ for all $k \ge 1$, starting with $a_0=n$. Observe that $x=y=m \implies f(m)=f(2m^2)$ hence $f(n)=f(a_k)$ for all $k \ge 1$. However, if $f(n) \ne f(1)$ then $f(n)>a_k$, contradicting that $(a_k)_k$ is unbounded. Hence for all $n$ big, $f(n)=f(0)$. Again use $f(n)=f(a_k)$ for $a_k$ large so $f(n)=f(0)$ for all $n \in \mathbb{N}$.

I claim that the only solution is $f(x) = c$, for some $c \in \mathbb{N}$, which clearly works.

If $f$ is not constant, then by the well-ordering principle there exists an $x$ such that $f(x)$ is minimal, and a $y$ such that $f(y) \neq f(x)$ is minimal (the second smallest value of $f(x)$).

Substituting $x$ and $y$ into the functional equation, we get that
$$
xf(y) + yf(x) \equiv 0 \pmod{x + y},
$$
and indeed we get
$$
xf(y) - xf(x) \equiv 0 \pmod{x + y},
$$
and $xf(y) = xf(x) + k(x + y)$. So $f(y) = f(x) + k + \frac{ky}{x}$. Substituting this in, we get
\begin{align*}
	xf(x) + k(x + y) + yf(x) &= (x + y)f(x^2 + y^2)  \\
\implies \cancel{(x + y)}(f(x) + k) &= \cancel{(x + y)}f(x^2 + y^2),
\end{align*}
But then we have $f(x) + k = f(x^2 + y^2)$, and indeed $f(x^2 + y^2) < f(y)$, which contradicts the minimality of $f(y)$. Thus there is no such $y$, and $f$ must be constant.

[hide=Solution]
Setting $y=0$ we get that $f(x^2) = f(0)$ for all positive integers $x$.
\[x+y \mid xf(y) + yf(x) \Longrightarrow x+y \mid x(f(y)-f(x)).\]
Hence if $x$ and $y$ are relatively prime, $x+y\mid f(y)-f(x)$.
Let $y$ be a sufficiently large perfect square(say $z^2$) which is relatively prime to $x$.
\[x+z^2 \mid f(z^2)-f(x) \Longrightarrow x+z^2 \mid f(0)-f(x).\]
Since $x+z^2 > |f(0)-f(z)|$ we must have $f(x) = f(0)$. Hence $f$ is constant.
[/hide]


Putnam
logarithms
function
calculus
derivative
inequalities
analytic geometry

For which real $ k$ do we have $ cosh(x) \leq  e^{(k x^{2})}$ for all real $ x$?

This problem is Putnam B-1, 1980. We are to find $ k$ such that $ \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\leq 2kx$ for all real numbers $ x$, yielding $ k\geq \frac{1}{2}$.

My result was also $ K=\frac{1}{2}$ but why is the second the same expression?

yeah, and I don't understand how you got your answer. Can you clesrly show how you got 1/2?

hello, $ \cosh(x)=\frac{1}{2}\left(e^{x}+e^{-x}\right)$.
Sonnhard.

We knew that. One-line posts don't contribute much of anything to the discussion; we like solutions with explanation, not just answers.

Here's what kunny's argument is about:
Take the natural logarithm of both sides; we want $ \ln(\cosh x)\le kx^{2}$. Since both sides are even functions, it suffices to show this for $ k>0$.
The derivatives are $ \tanh x$ and $ 2kx$; if $ \tanh x\le 2kx$ for all $ x>0$, then we get the desired inequality by integration. Since $ \tanh$ is concave and has slope 1 at the origin, it is less than $ 2kx$ whenever $ 2k\ge 1$. Conversely, if $ 2k<1$, it is easy to show that $ \tanh x>2kx$ for $ x\in (0,\epsilon)$ for some $ \epsilon>0$, and then also $ \cosh x>e^{kx^{2}}$ for all $ x\in (0,\epsilon)$.
(The "for all real numbers x" should have been "for all $ x\ge 0$")

Working directly with the power series for $ \cosh x$ and $ e^{kx^{2}}$ is also reasonable; the latter dominates the former termwise for $ k\ge\frac12$.


Given a square of side 1, show that if 5 points are placed anywhere inside or on this square, then some pair of them must be at most sqrt(2)/2 units apart.

Connect the midpoints of opposite sides.  Then you have four regions.  Two points must lie in the same region (or on the borders of the same region), but the maximum distance between two points in the same region is $\frac{1}{\sqrt{2}}$
I don't think this one belongs in the olympiad section...

It's in easy with the pigeonhole principle. :D


Hello
i didnt solve this problem  :oops:
[img]http://img518.imageshack.us/img518/9400/fonksiyongzelsoruhm6fx0.jpg[/img]

Thank you i understand now..
i made a mistake i written positive  :oops:
Thank you again  :)

This might be the correct way.
[hide="Solution"]
$ g(50)\equal{}f(1)\plus{}f(2)...\plus{}f(50)$
The first ones are $ \frac{1}{6},\frac{1}{12},\frac{1}{20},\frac{1}{30}$ We see that there is aseries. Solving using a calculator yields your answer [/hide]

Yeah i solved ..
1/2-1/52 =25/104  :D
Thank you again  :)

[quote="MrRubix"]This might be the correct way.
[hide="Solution"]
$ g(50) \equal{} f(1) \plus{} f(2)... \plus{} f(50)$
The first ones are $ \frac {1}{6},\frac {1}{12},\frac {1}{20},\frac {1}{30}$ We see that there is aseries. Solving using a calculator yields your answer [/hide][/quote]

I don't really see a series. The denominator pattern doesn't work?
I think the best solution is to use partial fractions.
Then you would only need the first term and the last term since everything cancels out

[img]http://img84.imageshack.us/img84/4489/fonksiyongzelsoruhm6rz6.jpg[/img]
i think this way true  :(  am i right?  :(

The rest of the layout is correct. But I am not sure on the final calculation though. The answer I got was not a multiple choice option:[hide]$ \frac {1}{2} \equal{} \frac {26}{52}$
$ \frac {26}{52} \minus{} \frac {1}{52} \equal{} \frac {26 \minus{} 1}{52} \equal{} \boxed{\frac {25}{52}}$[/hide]

i am sorry  :oops:
i made mistake,   thank you again  :)

I also got 25/52.
I formed an equation based on cases. I first made a table with values of n, the value of f(n),and the sum of nth and the previous terms. Looking at the denominator first, factoring it yields 1/[(+1)(n+2)]. This leads to the conclusion that the final answer will have this denominator= 51*52. Then i looked at the sums, and found that they obeyed the sum formula such that the sum of f(1)-f(5) inclusive is 5*6/2 in the numerator and consequently 6*7 in the denominator. this leads to [(n)(n+1)/[2(n+1)(n+2)]= (n/2)/(n+2)


Let $ x,y,z$ be real numbers in the interval $ \left[ \frac{1}{2}, 2\right]$. Prove that
\[ \frac{x}{y\plus{}z}\plus{}\frac{y}{z\plus{}x}\plus{}\frac{z}{x\plus{}y} \ge \sqrt{\frac{x(y\plus{}z)}{y^2\plus{}6yz\plus{}z^2}}\plus{}\sqrt{\frac{y(z\plus{}x)}{z^2\plus{}6zx\plus{}x^2}} \plus{}\sqrt{\frac{z(x\plus{}y)}{x^2\plus{}6xy\plus{}y^2}}\]
:)

[quote="dduclam"][quote="can_hang2007"][quote="can_hang2007"]Let $ x,y,z$ be real numbers in the interval $ \left[ \frac {1}{2}, 2\right]$. Prove that
\[ \frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y} \ge \sqrt {\frac {x(y \plus{} z)}{y^2 \plus{} 6yz \plus{} z^2}} \plus{} \sqrt {\frac {y(z \plus{} x)}{z^2 \plus{} 6zx \plus{} x^2}} \plus{} \sqrt {\frac {z(x \plus{} y)}{x^2 \plus{} 6xy \plus{} y^2}}
\]
:)[/quote]
See my proof here: http://canhang2007.wordpress.com/2008/11/27/hard-inequality-vo-quoc-ba-can/
But actually I dont like this proof, so I am still waiting for another proof.[/quote]

Nice, Can. I think the following inequalities are also true

1/ $ \left(\frac a{b \plus{} c}\right)^2\ge 3\sum\frac {a(b \plus{} c)}{b^2 \plus{} 6bc \plus{} c^2}$ for all $ a,b,c\in\left[1,2\right]$

It's stronger then yours, but the condition is weaker.

2/ $ \frac a{b \plus{} c}\ge \frac1{2}\sum\sqrt {\frac {3a(b \plus{} c)}{b^2 \plus{} 4bc \plus{} c^2}}$ for all $ a,b,c\ge0$

3/ $ \left(\frac a{b \plus{} c}\right)^2\ge \frac {15}8\sum\frac {a(b \plus{} c)}{b^2 \plus{} 3bc \plus{} c^2}$ for all $ a,b,c\ge0$

Besides, we have a nice but very easy result : $ \sum\frac a{b \plus{} c}\ge\frac3{4}\sum\frac {a(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2}\ge\frac3{2}$

In fact, we have $ \sum\frac {a^k}{b \plus{} c}\ge\frac3{4}\sum\frac {a^k(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2}, \forall k\ge0$

 :)[/quote]
Hi, Lam. Do you have a different solution with mine? Can you post it? Since I really like this mixed root-fraction inequality of mine, and I would like to see a nicer proof. Thanks :)

[quote="dduclam"][quote="can_hang2007"][quote="can_hang2007"]Let $ x,y,z$ be real numbers in the interval $ \left[ \frac {1}{2}, 2\right]$. Prove that
\[ \frac {x}{y \plus{} z} \plus{} \frac {y}{z \plus{} x} \plus{} \frac {z}{x \plus{} y} \ge \sqrt {\frac {x(y \plus{} z)}{y^2 \plus{} 6yz \plus{} z^2}} \plus{} \sqrt {\frac {y(z \plus{} x)}{z^2 \plus{} 6zx \plus{} x^2}} \plus{} \sqrt {\frac {z(x \plus{} y)}{x^2 \plus{} 6xy \plus{} y^2}}
\]
:)[/quote]
See my proof here: http://canhang2007.wordpress.com/2008/11/27/hard-inequality-vo-quoc-ba-can/
But actually I dont like this proof, so I am still waiting for another proof.[/quote]

Nice, Can. I think the following inequalities are also true

1/ $ \left(\frac a{b \plus{} c}\right)^2\ge 3\sum\frac {a(b \plus{} c)}{b^2 \plus{} 6bc \plus{} c^2}$ for all $ a,b,c\in\left[1,2\right]$

It's stronger then yours, but the condition is weaker.

2/ $ \frac a{b \plus{} c}\ge \frac1{2}\sum\sqrt {\frac {3a(b \plus{} c)}{b^2 \plus{} 4bc \plus{} c^2}}$ for all $ a,b,c\ge0$

3/ $ \left(\frac a{b \plus{} c}\right)^2\ge \frac {15}8\sum\frac {a(b \plus{} c)}{b^2 \plus{} 3bc \plus{} c^2}$ for all $ a,b,c\ge0$

Besides, we have a nice but very easy result : $ \sum\frac a{b \plus{} c}\ge\frac3{4}\sum\frac {a(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2}\ge\frac3{2}$

In fact, we have $ \sum\frac {a^k}{b \plus{} c}\ge\frac3{4}\sum\frac {a^k(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2}, \forall k\ge0$

 :)[/quote]
1) I got an ugly solution by SOS. I will post it later.
2) and 3), see here: http://canhang2007.wordpress.com/2008/12/09/3-problems-duong-duc-lam-and-me/ :)
And the last inequality
\[ \sum \frac{a^k}{b\plus{}c} \ge \frac{3}{4} \sum \frac{a^k(b\plus{}c)}{b^2\plus{}bc\plus{}c^2}\]
is trivial because $ b^2\plus{}bc\plus{}c^2 \ge \frac{3}{4} (b\plus{}c)^2$.
:)

[quote="can_hang2007"][/quote][quote="dduclam"][quote="can_hang2007"][quote="can_hang2007"]Let $ x,y,z$ be real numbers in the interval $ \left[ \frac {1}{2}, 2\right]$. Prove that
\[ \frac {x}{y + z} + \frac {y}{z + x} + \frac {z}{x + y} \ge \sqrt {\frac {x(y + z)}{y^2 + 6yz + z^2}} + \sqrt {\frac {y(z + x)}{z^2 + 6zx + x^2}} + \sqrt {\frac {z(x + y)}{x^2 + 6xy + y^2}}
\]
:)[/quote]
See my proof here: http://canhang2007.wordpress.com/2008/11/27/hard-inequality-vo-quoc-ba-can/
But actually I dont like this proof, so I am still waiting for another proof.[/quote]

Nice, Can. I think the following inequalities are also true

1/ $ \left(\frac a{b + c}\right)^2\ge 3\sum\frac {a(b + c)}{b^2 + 6bc + c^2}$ for all $ a,b,c\in\left[1,2\right]$

It's stronger then yours, but the condition is weaker.

2/ $ \frac a{b + c}\ge \frac1{2}\sum\sqrt {\frac {3a(b + c)}{b^2 + 4bc + c^2}}$ for all $ a,b,c\ge0$

3/ $ \left(\frac a{b + c}\right)^2\ge \frac {15}8\sum\frac {a(b + c)}{b^2 + 3bc + c^2}$ for all $ a,b,c\ge0$

Besides, we have a nice but very easy result : $ \sum\frac a{b + c}\ge\frac3{4}\sum\frac {a(b + c)}{b^2 + bc + c^2}\ge\frac3{2}$

In fact, we have $ \sum\frac {a^k}{b + c}\ge\frac3{4}\sum\frac {a^k(b + c)}{b^2 + bc + c^2}, \forall k\ge0$

 :)[/quote][quote="can_hang2007"]
Hi, Lam. Do you have a different solution with mine? Can you post it? Since I really like this mixed root-fraction inequality of mine, and I would like to see a nicer proof. Thanks :)[/quote]

Yes, I do  :)  I have two solutions for your hard inequality, different and simple than yours.  :blush:

[quote="can_hang2007"]
1) I got an ugly solution by SOS. I will post it later.
2) and 3), see here: http://canhang2007.wordpress.com/2008/12/09/3-problems-duong-duc-lam-and-me/ :)
And the last inequality
\[ \sum \frac {a^k}{b + c} \ge \frac {3}{4} \sum \frac {a^k(b + c)}{b^2 + bc + c^2}
\]
is trivial because $ b^2 + bc + c^2 \ge \frac {3}{4} (b + c)^2$.
:)[/quote]

Yes, that all.  :lol:

[quote="dduclam"][/quote][quote="can_hang2007"][/quote][quote="dduclam"][quote="can_hang2007"][quote="can_hang2007"]Let $ x,y,z$ be real numbers in the interval $ \left[ \frac {1}{2}, 2\right]$. Prove that
\[ \frac {x}{y + z} + \frac {y}{z + x} + \frac {z}{x + y} \ge \sqrt {\frac {x(y + z)}{y^2 + 6yz + z^2}} + \sqrt {\frac {y(z + x)}{z^2 + 6zx + x^2}} + \sqrt {\frac {z(x + y)}{x^2 + 6xy + y^2}}
\]
:)[/quote]
See my proof here: http://canhang2007.wordpress.com/2008/11/27/hard-inequality-vo-quoc-ba-can/
But actually I dont like this proof, so I am still waiting for another proof.[/quote]

Nice, Can. I think the following inequalities are also true

1/ $ \left(\frac a{b + c}\right)^2\ge 3\sum\frac {a(b + c)}{b^2 + 6bc + c^2}$ for all $ a,b,c\in\left[1,2\right]$

It's stronger then yours, but the condition is weaker.

2/ $ \frac a{b + c}\ge \frac1{2}\sum\sqrt {\frac {3a(b + c)}{b^2 + 4bc + c^2}}$ for all $ a,b,c\ge0$

3/ $ \left(\frac a{b + c}\right)^2\ge \frac {15}8\sum\frac {a(b + c)}{b^2 + 3bc + c^2}$ for all $ a,b,c\ge0$

Besides, we have a nice but very easy result : $ \sum\frac a{b + c}\ge\frac3{4}\sum\frac {a(b + c)}{b^2 + bc + c^2}\ge\frac3{2}$

In fact, we have $ \sum\frac {a^k}{b + c}\ge\frac3{4}\sum\frac {a^k(b + c)}{b^2 + bc + c^2}, \forall k\ge0$

 :)[/quote][quote="can_hang2007"]
Hi, Lam. Do you have a different solution with mine? Can you post it? Since I really like this mixed root-fraction inequality of mine, and I would like to see a nicer proof. Thanks :)[/quote]

Yes, I do  :)  I have two solutions for your hard inequality, different and simple than yours.  :blush:

[quote="can_hang2007"]
1) I got an ugly solution by SOS. I will post it later.
2) and 3), see here: http://canhang2007.wordpress.com/2008/12/09/3-problems-duong-duc-lam-and-me/ :)
And the last inequality
\[ \sum \frac {a^k}{b + c} \ge \frac {3}{4} \sum \frac {a^k(b + c)}{b^2 + bc + c^2}
\]
is trivial because $ b^2 + bc + c^2 \ge \frac {3}{4} (b + c)^2$.
:)[/quote][quote="dduclam"]

Yes, that all.  :lol:[/quote]
So, post it, please. I am waiting for it.
P/s: The second proof of mine is to use
\[ \sum \frac {a}{b + c} \ge \sum \frac {(b + c)^2}{b^2 + 6bc + c^2}
\]
;)
This proof is based on my CYH technique. But I still like my first proof because of the way I found the inequality
\[ \sqrt {\frac {y^2 + z^2}{y^4 + 6y^2z^2 + z^4}} \le \frac {2(y + z)}{y^2 + 6yz + z^2}
\]

[quote="can_hang2007"]
So, post it, please. I am waiting for it.
P/s: The second proof of mine is to use
\[ \sum \frac {a}{b \plus{} c} \ge \sum \frac {(b \plus{} c)^2}{b^2 \plus{} 6bc \plus{} c^2}
\]
;)
This proof is based on my CYH technique. But I still like my first proof because of the way I found the inequality
\[ \sqrt {\frac {y^2 \plus{} z^2}{y^4 \plus{} 6y^2z^2 \plus{} z^4}} \le \frac {2(y \plus{} z)}{y^2 \plus{} 6yz \plus{} z^2}
\]
[/quote]

Yes, It's also my first proof :).  The second to use
\[ \sum \frac {a}{b \plus{} c} \ge\frac7{4}\plus{} \frac1{3}\sum \frac {a(b \plus{} c)\minus{}4bc}{b^2 \plus{} 6bc \plus{} c^2}
\]

A question was put out is: Find the maximum value of $ p$ such that the following inequality is true
\[ \sum\frac {a}{b \plus{} c} \ge \sum\sqrt {\frac {a(b \plus{} c)}{b^2 \plus{} 6bc \plus{} c^2}}
\]
for all $ a,b,c\in \left[\frac1{p}, p\right]$.

When $ p\in N$, I think the following Inequality holds
\[ \sum\frac {a}{b \plus{} c} \ge \sum\sqrt {\frac {a(b \plus{} c)}{b^2 \plus{} 6bc \plus{} c^2}}
\]
for all $ a,b,c\in \left[\frac1{5}, 5\right]$.  :)

[quote="dduclam"] A question was put out is: Find the maximum value of $ p$ such that the following inequality is true
\[ \sum\frac {a}{b \plus{} c} \ge \sum\sqrt {\frac {a(b \plus{} c)}{b^2 \plus{} 6bc \plus{} c^2}}
\]
for all $ a,b,c\in \left[\frac1{p}, p\right]$.

[/quote]

$ p_{max}\equal{}5.993964...$   :)

[quote="can_hang2007"][/quote][quote="dduclam"][/quote][quote="can_hang2007"][/quote][quote="dduclam"][quote="can_hang2007"][quote="can_hang2007"]Let $ x,y,z$ be real numbers in the interval $ \left[ \frac {1}{2}, 2\right]$. Prove that
\[ \frac {x}{y + z} + \frac {y}{z + x} + \frac {z}{x + y} \ge \sqrt {\frac {x(y + z)}{y^2 + 6yz + z^2}} + \sqrt {\frac {y(z + x)}{z^2 + 6zx + x^2}} + \sqrt {\frac {z(x + y)}{x^2 + 6xy + y^2}}
\]
[/quote]
See my proof here: http://canhang2007.wordpress.com/2008/11/27/hard-inequality-vo-quoc-ba-can/
But actually I dont like this proof, so I am still waiting for another proof.[/quote]

Nice, Can. I think the following inequalities are also true

1/ $ \left(\frac a{b + c}\right)^2\ge 3\sum\frac {a(b + c)}{b^2 + 6bc + c^2}$ for all $ a,b,c\in\left[1,2\right]$

It's stronger then yours, but the condition is weaker.

2/ $ \frac a{b + c}\ge \frac1{2}\sum\sqrt {\frac {3a(b + c)}{b^2 + 4bc + c^2}}$ for all $ a,b,c\ge0$

3/ $ \left(\frac a{b + c}\right)^2\ge \frac {15}8\sum\frac {a(b + c)}{b^2 + 3bc + c^2}$ for all $ a,b,c\ge0$

Besides, we have a nice but very easy result : $ \sum\frac a{b + c}\ge\frac3{4}\sum\frac {a(b + c)}{b^2 + bc + c^2}\ge\frac3{2}$

In fact, we have $ \sum\frac {a^k}{b + c}\ge\frac3{4}\sum\frac {a^k(b + c)}{b^2 + bc + c^2}, \forall k\ge0$

 :)[/quote][quote="can_hang2007"]
Hi, Lam. Do you have a different solution with mine? Can you post it? Since I really like this mixed root-fraction inequality of mine, and I would like to see a nicer proof. Thanks :)[/quote]

Yes, I do  :)  I have two solutions for your hard inequality, different and simple than yours.  :blush:

[quote="can_hang2007"]
1) I got an ugly solution by SOS. I will post it later.
2) and 3), see here: http://canhang2007.wordpress.com/2008/12/09/3-problems-duong-duc-lam-and-me/ :)
And the last inequality
\[ \sum \frac {a^k}{b + c} \ge \frac {3}{4} \sum \frac {a^k(b + c)}{b^2 + bc + c^2}
\]
is trivial because $ b^2 + bc + c^2 \ge \frac {3}{4} (b + c)^2$.
:)[/quote][quote="dduclam"]

Yes, that all.  :lol:[/quote][quote="can_hang2007"]
So, post it, please. I am waiting for it.
P/s: The second proof of mine is to use
\[ \sum \frac {a}{b + c} \ge \sum \frac {(b + c)^2}{b^2 + 6bc + c^2}
\]
;)
[b]This proof is based on my CYH technique[/b]. But I still like my first proof because of the way I found the inequality
\[ \sqrt {\frac {y^2 + z^2}{y^4 + 6y^2z^2 + z^4}} \le \frac {2(y + z)}{y^2 + 6yz + z^2}
\]
[/quote]
Dear Can, could you write something about this technique? :)


whoa thats really long picture...

the height of one non-hexagonal face is 36 inches, and width is 6 inches.

there are total 6 faces, so the total area is 36*6*6 square inches.

and one square foot is 144 square inches.

soooooo 36*6*6/144=9

answer : 9


Find the max possible value of k such that
$a^3+b^3+c^3+kabc\ge \frac{k}{27} +\frac{1}{9}$
forall $a,b,c\ge 0$ sastisfying $a+b+c=1$

[quote="bodan"]If we take $a=0,\; b=c=\frac{1}{2}$ we'll get $k\leq\frac{15}{4}.\;$ For $k=\frac{15}{4}$ we have $LHS=4(a^3+b^3+c^3)+15abc$ and $RHS=1.\;$  I got stuck here but it's probably true.[/quote]

$4(a^3+b^3+c^3)+15abc \geq 1$
$\rightarrow$$4(a^3+b^3+c^3)+15abc \geq (a+b+c)^3$
$\rightarrow$ $3(\sum a(a-b)(a-c)) \geq 0$ :)


geometry
analytic geometry
graphing lines
slope
vector
complex numbers

Given triangle ABC, D is the foot of the perpendicular from A to BC. Find the coordinates of D as a complex coordinate.

Given $ z_a,\ z_b\ and\ z_c$

$ z_d\ \equal{}\ ?$

(I need this to solve a bigger problem.)

[quote="mathemonster"]just do it as if in rectangular coordinates. (use pythagorean thm.)[/quote]

Could someone please do it and show me? I'd really appreciate it. (I've been trying this for a while now.  :blush: )

Ok let B have coordinates $ (x_1,y_1)$ C have coordinates $ (x_2,y_2)$ and A have coordinates $ (x_3,y_3).$  Then we have line BC (or BD w.e.) perpendicular to AD. Using point slope form we have line BC described by : $ y \minus{} y_2 \equal{} \frac {y_1 \minus{}y_2}{x_1}{x_2}(x \minus{} x_2)$ Line AD has slope equal to the negative reciprocal of the slope of BC so it is given by : $ y \minus{} y_3 \equal{} \frac{\minus{}(x_1\minus{}x_2)}{y_1\minus{}y_2}(x\minus{}x_3)$
Solve the two equations for x and y and we are done.

What problem is this from?

[quote="Poincare"]the most overwhelming thing in complex geometry is the complex numbers (NO!). unless its really really necessary, just use real stuff not imaginary.[/quote]

But complex numbers are really helpful in solving geometry problems (especially this one)!

Err, not really. Vectors are a lot easier. In fact, using the vectors approach makes the problem rather trivial. We can easily find the height vector's direction using a dot product, and we can find its magnitude by using the distance-of-a-point-from-a-line formula. Then we just add that to $ v_a$ and we're done.

I agree....solving with complex numbers it may take a wile....and you don't always aford to waste time instead of thinking on another problem...the complex numbers geometry, in my opinion...ar to be used if you have no chances to succeed with synthetic or vector geometry.


[hide]121, I think if I am wrong, then ignore me, if I am right, I need to know how to type squareroots and stuff[/hide]

Edit: me wrong, messed up, at least I gave N+1...

[hide]
Hopefully you can use a calculator :)

If you start computing it on your calculator, you see that when $n=3$, the sum is 1.
When $n=8$, the sum is 2.
The pattern is that when the sum is $x$, $n=(x+1)^2-1$
Apply that for 10.
$120=n$[/hide]

[hide]If you rationalize each denominator, the denominator is $-1$. Since the terms on the numerator cancel out when the term is rationalized, $\sqrt{n+1}-1=10$, $n+1=121$, and $\boxed{n=120}$. [/hide]

Note the title: rationalizing  ;)

[hide]The first term is $\frac{1}{\sqrt2+\sqrt1}=\frac{\sqrt2-\sqrt1}{1}=\sqrt2-1$

The second term, likewise, is $\sqrt3-\sqrt2$

So the terms will cancel until you get $\sqrt{n+1}-1=10$ because those are the first term and the last term.

Solving gives $n=120$[/hide]


How do you all think you did?

I can't remember my answers, since they wouldn't let me take the test booklets out, but my first few were E,D,C,B, and my last one was E.

I will be mad if I didn't get 25.

I really posted this to find out when the official answers will be put on the site.

The answers are on the AMC site.
http://www.unl.edu/amc/e-exams/e4-amc08/e4-1-8archive/2009-8a/2009-amc8answers.shtml

More discussion is being done [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=314200]here[/url]. I think I got a 25, unless I misbubbled.

For me, the hardest was #21 (the array of numbers). I spent about 5 minutes on it before I realized just letting them equal 1 would work. The second hardest was #25, but that was easy after I realized that the front and back surface areas are just one. The third worst problem was time. I finished with just under a minute left.

I didn't even finish. I spent too much time checking my answers. I probably got a 22 or 23, same as last year.  :(


On the first three MOP tests, CatalystOfNostalgia scored 10.5, 14.1, and 15.2. b-flat scored 12.8, 6.4, and 0.   If there are five more tests, and b-flat wants to beat CatalystOfNostalgia's MOP score by 30 percent, and CatalystOfNostalgia will average 14 points per test for the next five tests, how many more problems does b-flat have to solve over the next five tests (average per test) in order to achieve his goal? Assume that both MOPpers receive 4.9 MOP points for each problem solved and 0 otherwise.

Hmm. This problem sounds oddly personal. :)

[quote="BOGTRO"]Catalyst's total score is $ 10.5 + 14.1 + 15.2 + 14 \cdot 5 = 109.8$.

$ 1.3 \cdot 109.8 = 142.74$.

$ \lceil ing \frac {142.74}{4.9} \rceil ing = 30$.

So $ \frac {30}{5} = 6$.[/quote]I think you forgot that bflat started with a nonzero score.

On the first five MOP tests, CatalystOfNostalgia scored 10.5, 14.1, 15.2, 1.0, and -0.0. b-flat scored 12.8, 6.4, 0.1, 0.1, and 28. kyyuanmathcount scored 28, 28, 28, 28, and 28.5. aryth's score on each test was the harmonic mean of exactly two of the other three students. Find the mean of the possible average scores that aryth can achieve.

I am quite sure those scores are quite wrong. i believe the correct numbers should be 10.5, 14.1, 16.1, 12.6, ~16 for CatalystOfNostalgia, b-flat should be 12.8, 6.4, 1.8, 5, ~10, and kyyuanmathcount, ~10, ~6, ~4, 13.3, ~16.


calculus
integration
logarithms
calculus computations

can you solve :D

$ \int_{0}^{1}\;\frac{\ln x}{\sqrt{\;1\plus{}x\plus{}x^{2}\;}}\;\;\textbf dx$

Maybe using substitution $ x\equal{}t\minus{}\frac{b}{2a}$  :blush:


function
floor function
logarithms
algebra
functional equation
algebra unsolved

Let $a>0$ be a constant not equal to $1$. Find all functions $f: \mathbb R \to \mathbb R$ such that
\[f(ax)=f(x)+2f(-x),\]
for all real $x$.

[quote="The soul of rock"]Let $a=const >0$ and $a \neq 1$. Find all functions $f(x)$ such that
\[f(ax)=f(x)+2f(-x) \]
[/quote]

We have $f(0)=0$. Consider now that $f(x)$ is defined for $x>0$, then it will be defined, for $x<0$ by $f(x)=\frac{f(-ax)-f(-x)}{2}$ and the property is verified for any $x\geq 0$.

In order the property be verified for any $x<0$, we must have, for any $x>0$, $f(-ax)=f(-x)+2f(x)$, and so $\frac{f(a^{2}x)-f(ax)}{2}=\frac{f(ax)-f(x)}{2}+2f(x)$, and so $f(a^{2}x)=2f(ax)+3f(x)$

So the problem is equivalent to find a function $f(x)$ such that $f(a^{2}x)=2f(ax)+3f(x)$ $\forall x>0$, and to expand $f(0)=0$ and $f(x)=\frac{f(-ax)-f(-x)}{2}$  $\forall x<0$

For $x>0$, let $f(x)=g(\frac{ \ln(x)}{\ln(a)})$. We have then $g(\frac{\ln(x)}{\ln(a)}+2)=2g(\frac{\ln(x)}{\ln(a)}+1)+3g(\frac{\ln(x)}{\ln(a)})$ $\forall x>0$, and hence $g(x+2)=2g(x+1)+3g(x)$ $\forall x$
This last equation is rather well known and we have $g(x+n)=\frac{3^{n}-(-1)^{n}}{4}g(x+1)+\frac{3^{n}+3(-1)^{n}}{4}g(x)$

So the solution for $g(x)$ is :
$g(x)=\frac{3^{\lfloor x\rfloor}-(-1)^{\lfloor x\rfloor}}{4}h(x-\lfloor x\rfloor+1)+\frac{3^{\lfloor x\rfloor}+3(-1)^{\lfloor x\rfloor}}{4}h(x-\lfloor x\rfloor)$, with $h(x)$ any function defined on $[0,2)$

And so the solution for $f(x)$ is :

For $x>0$, $f(x)=\frac{3^{\lfloor \frac{ \ln(x)}{\ln(a)}\rfloor}-(-1)^{\lfloor \frac{ \ln(x)}{\ln(a)}\rfloor}}{4}h(\frac{ \ln(x)}{\ln(a)}-\lfloor \frac{ \ln(x)}{\ln(a)}\rfloor+1)$ $+\frac{3^{\lfloor \frac{ \ln(x)}{\ln(a)}\rfloor}+3(-1)^{\lfloor \frac{ \ln(x)}{\ln(a)}\rfloor}}{4}h(\frac{ \ln(x)}{\ln(a)}-\lfloor \frac{ \ln(x)}{\ln(a)}\rfloor)$, with $h(x)$ any function defined on $[0,2)$

For $x=0$, $f(0)=0$;

For $x<0$, $f(x)=\frac{f(-ax)-f(-x)}{2}$

Examples :
1) $h(x)=0\implies f(x)=0$

2) $h(x)=2$ $\implies$
For $x>0$, $f(x)=3^{\lfloor \frac{ \ln(x)}{\ln(a)}\rfloor}+(-1)^{\lfloor \frac{ \ln(x)}{\ln(a)}\rfloor}$
For $x=0$, $f(0)=0$;
For $x<0$, $f(x)=3^{\lfloor \frac{ \ln(x)}{\ln(a)}\rfloor}-(-1)^{\lfloor \frac{ \ln(x)}{\ln(a)}\rfloor}$

[quote="The soul of rock"]I need a solution base on cyclic function! Could you help me! Thanks![/quote]

I don't understand. Give some precisions.

My solution is, imho, the general one (all solutions has this form for some h(x)).

[quote="Just"]pco,

Could you please give some details about the solutions to the "rather well known equation". Thanks![/quote]

I'm sorry if I was not clear enough.
I say that functional equation $f(x+2)=af(x+1)+bf(x)$ ($a\neq0$ and $b\neq 0$) is rather simple and well known (according to me  :blush: ) :
If you define $f(x)$ as any function on $[0,2)$, then you can define it (thru $f(x+2)=af(x+1)+bf(x)$ with $x$ in $[0,1)$) on $[2,3)$ and then on $[3,4)$ and so on for any $x\geq 0$.
And $f(x+2)=af(x+1)+bf(x)$ ($a\neq0$ and $b\neq 0$) $\implies$ $f(x)=-\frac{a}{b}f(x+1)-\frac{1}{b}f(x+2)$ and so $f(x)$ can be defined on $(-\infty,0)$ in the same way.

In order to have a concrete expression of the solution, we can write $f(x+n)=u_{n}f(x+1)+v_{n}f(x)$ with $u_{n}$ and $v_{n}$ defined as :
$u_{n+1}=au_{n}+v_{n}$
$v_{n+1}=bu_{n}$
$u_{0}=0$, $u_{1}=1$, $v_{0}=1$, $v_{1}=0$,.

Or else :
$u_{n+1}=au_{n}+bu_{n-1}$, $u_{0}=0$, $u_{1}=1$ and $v_{n}=bu_{n-1}$

Once you have $u_{n}$ and $v_{n}$, you say :
$f(x)=h(x)$ $\forall x\in[0,2)$ with $h(x)$ being any function defined in $[0,2)$ and :
$f(x)=f( \lfloor x \rfloor+\{x\})=u_{ \lfloor x \rfloor }f(\{x\}+1)+v_{ \lfloor x \rfloor }f(\{x\})$ (with $\{x\}$ being the fractional part of $x$) and then :

$f(x)=u_{ \lfloor x \rfloor }h(\{x\}+1)+v_{ \lfloor x \rfloor }h(\{x\})$

[quote="Just"]Ok, got you. Thanks.

I was thinking of another method, what do you think?

$f(ax)=f(x)+2f(-x)$ and $f(-ax)=f(-x)+2f(x)$ hence $f(ax)+f(-ax)=3(f(x)+f(-x))$.

Letting $g(x)=f(x)+f(-x)$ we get $g(ax)=3g(x)$ whose solutions are (considering that $g(0)=0$) $kx^{\frac{\ln{3}}{\ln{a}}}$ (not sure they are the only ones).

Then, we try solving $f(x)+f(-x)=kx^{\frac{\ln{3}}{\ln{a}}}$.[/quote]

1) Yes, it could be a method and I think you'll obtain similar results.

2) Caution : $kx^{\frac{\ln{3}}{\ln{a}}}$ is absolutly not the only solution of $g(ax)=3g(x)$


There was an Intermediate Daily Contest for 13 weeks now, and i think most participants appricated it. but now the numbers of sent-in solutions reduced greatly. The question is do you still want a Daily Contest like this? I would like to hear, what you think about it ...

peeta

I think the last daily contest was a bit too hard, But I would be happy if there would be another one a bit easier.

i guess there are enough votes for a new trial next week for the Intermediate Daily Contest. I will try to find someone to moderate or moderate myself.
if there is anyone who disagrees with the "new" rules (look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=331006#p331006]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=335463#p335463]here[/url]) please pm me or post it in this thread ...
peeta

I think it would be best to have it during the winter and summer breaks.


number theory
greatest common divisor
number theory unsolved

Let $ S \equal{} \{ (a,b,c,d,e,f) | a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \plus{} e^2 \equal{} f^2, a,b,c,d,e,f \in N\}$. Then find $ GCD (abcdef | (a,b,c,d,e,f) \in S )$

$ (1,1,1,2,3,4)$ gives $ 24$, which is the answer.
Let the desired number be $ A$.
$ 3|A$, otherwise $ 2\equal{}1\mod3$, contradiction!.

If $ a,b,c,d,e,f$ has only one even, parity does not hold.
If $ a,b,c,d,e,f$ has greater-than-two evens, $ 8|abcdef$.
Suppose $ a,b,c,d,e,f$ has two evens, $ f$ has to be even, otherwise $ f^2\equal{}3\mod8$ or $ f^2\equal{}7\mod8$, contradiction!
Say $ 2|a$ and $ 2|f$, then $ f^2\equal{}a^2\plus{}4\mod8\Rightarrow 8|af$.
Hence, $ A\equal{}24$.


I've heard varying opinions, so let's see some numbers!

Math bowl was nice, but I think it'd be better if the competition started earlier to allow people who're in the individual finals to also do math bowl (quagga got screwed over for this reason).

I don't know how feasible this would be, but maybe if teams were grouped together and played sudden death elimination games of frisbee?  Nah, forget that.

I thought the Math Bowl was pretty cool.

[quote="K81o7"]I don't know how feasible this would be, but maybe if teams were grouped together and played sudden death elimination games of frisbee?  Nah, forget that.[/quote]
I have actually thought about related ideas before. For example, we could have several simultaneous fun competitions (speed chess/bughouse tournament, SET tournament, etc). Frisbee, though, is bad because we can't guarantee good weather (in fact, Nov/Dec in New Jersey means we can almost guarantee bad weather).

I think that math bowl is solid. It's not really my thing, but it was entertaining to watch at the end.

I enjoyed the math bowl, but that may be biased considering that I was on the winning team (I was the one that looked like an idiot when I missed the frisbee).  Eitherway, I think you should continue on the math bowl, relays are done everywhere, and while I enjoy math lectures I don't know if everybody there would agree with me, I know people that don't like going to the lectures at ARML. The math bowl was pretty unique, and I think its a fun way to set apart this competition from others

Catchphrase with math! :D

... okay, so that was a random suggestion that I doubt is even feasible. But let's throw it out anyways.

(for information about the game, try Catch Phrase (Game) on wikipedia, sorry the parentheses screw up the BB tags so I can't link to it)

[quote="ProtestanT"]The math bowl was awesome, but if time allows it'd be also cool for it to be double elimination like science bowl[/quote]
That may be possible. I'd also like it if we could do double elim. Double elimination would only add like 2 more rounds (I'd have to check -- since we don't have enough buzzers total in the first couple of rounds, it's not a trivial question).

[quote="CatalystOfNostalgia"]I wasn't at PUMaC this year, but a USEMO would be cool.  :D[/quote]
We were thinking of giving the contestants something like this to have fun with during lunch. We just didn't have time to write it because we were busy writing all of the other tests. Maybe we'll do it next year... ;)

I agree that Mathbowl gets huge originality points, although I'd personally love to do either Frisbee or Bughouse (or Soccer).

Are the questions for mathbowl going to be released, or will they be recycled in like 5 years?

[quote="solafidefarms"]One question was "A web comic..." and the answer was XKCD.[/quote]
Another question asked about $ e^{pi}\minus{}pi$.

A few more:
"First drawn in 1978..." - Mandelbrot Set
"If a rectangle's length is increased 30% and its width is decreased 30%, how many percent does the area change by? Indicate whether the area increases or decreases." - Decreases 9%
"To make this idealistic machine, all you need is an infinite tape and..." - Turing Machine
Something, then diagonal argument - Cantor

OK guys we decided that we're going to post a sample of the questions but not all of them. Please don't try to post all the ones you remember here as we may ask similar questions in the future. (Of course we won't repeat any word-for-word, but we might ask questions about the same topics.)

If you want to check it out, here's [url=http://www.princeton.edu/~nsavir/PUMaC/Math%20Bowl%20Sample%20Question%20Set.doc]the sample[/url].

I'd like to point out that some of the questions are not entirely accurate.

The only error on the samples was
Question 10: $ 0 \equal{} \{\} \neq \{\{\}\} \equal{} 1$.

[quote="K81o7"]Another possibility, though I guess it might take too long...

Guitar hero tournament![/quote]
dude
if you have this
im definitely going next year

Relating to having a cool lecture by a math professor.  The problem with this is that it is difficult to find a topic such that it will be interesting to everyone, and, more importantly, that the majority have not yet seen.

Anyway, math bowl was definitely good.

Also, chess tournaments are generally fun.

A Lot of people like relays. Love relays. I am one of them.

[quote="solafidefarms"]One question was "A web comic..." and the answer was XKCD.[/quote]

Uh.. Yeah. My team got beaten on that question in the Third Round (AAST 5)

but $ e^{\pi}\minus{}\pi$ was crazy

[quote="b-flat"][quote="BOGTRO"]but $ e^{\pi} \minus{} \pi$ was crazy[/quote]

Umm, same comic?[/quote]
Woah, Bryce, you read xkcd?

Ah I have to say this...

760 on SAT math at 8 years old, Australia...

Terence Tao!

How did it take [i]that[/i] long in the finals :wink:

haha yea i was wondering the same thing.

[quote="b-flat"]Relating to having a cool lecture by a math professor.  The problem with this is that it is difficult to find a topic such that it will be interesting to everyone, and, more importantly, that the majority have not yet seen.[/quote]

I don't buy that. Conway is a Princeton faculty member, and he has experience with lots of elementary, but super-creative things that an average person at PUMaC would probably understand most of it, and are still very interesting and not well-known.

For example, he wrote a paper containing a cool proof of the validity of Division by Three without AC, it's on arxiv.


trigonometry
inequalities
inequalities proposed

[quote="Minh Can"]Hi,

I thought about this question for a while wonder if a strategy can be used to unlock the trick. What about Lagrange Multiplier ? or even convexity ? I think the former may turn out effective.

Cheer. :([/quote]Please also post a problem ;)


Hover your mouse over hunter34's first LaTeX creation:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=944333#944333

It has extra
s in it, as I suppose linebreaks in posts are automaticall converted to
s upon submission.

Could something be done about this?


function
algebra
functional equation
algebra unsolved

find all continuous f so that
f(x+y) +f(x-y)=2.f(x).f(y) :?:

This is the well-known D'Alembert's functional equation, so it has certainly been posted before. (but the only related thread I could find is http://www.mathlinks.ro/Forum/viewtopic.php?t=26920 )

$f(x)=cosh(ax)$, a can be complex. For example $a=bi,b\in R$.


I seek to show that $e$ is irrational.

One has $e = \sum_{n=0}^{+ \infty} 1/n!$. I proceed by the absurdity by supposing that $e=p/q$ avec $p,q \geq 1$ and $p \wedge q = 1$. One has then: \[q!p = qq!e = \sum_{k=0}^q \frac{qq!}{k!} + q \sum_{k=q+1}^{ + \infty} \frac{q!}{k!}\].

The first sum is a natural entirety just as $q!p$. But I cannot show that the second sum is a real of $]0,1[$   :( .
Can you help me?


The inequality is equivalent to : $ \sum {[\frac{a}{3(a\plus{}b\plus{}c)}\minus{}\frac{a^2}{(2a\plus{}b)(2a\plus{}c)}]}\geq0$.

      $ \sum {\frac{a(a\minus{}b)(a\minus{}c)}{(2a\plus{}b)(2a\plus{}c)}}\geq0$

   Presume that $ a\geq b\geq c$

    Since  $ \frac{c(c\minus{}a)(c\minus{}b)}{(2c\plus{}a)(2c\plus{}b)}\geq0$, it sufices to show that $ \frac{a(a\minus{}b)(a\minus{}c)}{(2a\plus{}b)(2a\plus{}c)}\plus{}\frac{b(b\minus{}c)(b\minus{}a)}{2b\plus{}c)(2b\plus{}a)}\geq0$ (***)

   But (***) is equivalent to $ (a\minus{}b)^2[(a\plus{}b)(2ab\minus{}c^2)\plus{}c(a^2\plus{}b^2\plus{}5ab)]\geq0$ , wich is true.

Generally ,the inequality

$ \dfrac{a^2}{(ka + b)(ka + c)} + \dfrac{b^2}{(kb + a)(kb + c)} + \dfrac{c^2}{(kc + a)(kc + b)}\leq{\dfrac{3}{(k + 1)^2}}$

holds for all reals $ a,b,c > 0$ if and only  if $ k\geq2$.

[quote="kritikos"]Show that for all reals $ a,b,c > 0$

$ \dfrac{a^2}{(2a + b)(2a + c)} + \dfrac{b^2}{(2b + a)(2b + c)} + \dfrac{c^2}{(2c + a)(2c + b)}\leq{\dfrac{1}{3}}$[/quote]
It's the Tigran Sloyan's inequality. See  here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=66660

The inequality

$ \dfrac{a^2}{(ka + b)(ka + c)} + \dfrac{b^2}{(kb + a)(kb + c)} + \dfrac{c^2}{(kc + a)(kc + b)}\geq{\dfrac{3}{(k + 1)^2}}$

holds for all reals a,b,c>0 if and if $ 0\le{k}\le\frac {\sqrt3 + 1}{2}$.


We are given a sequence $c_{1},c_{2},c_{3},...$ of natural numbers. For any natural  numbers $m,n$ with $1 \leq m \leq \sum_{i=1}^{n}c_{i}$  we can choose natural numbers $a_{1},a_{2},...,a_{n}$, such that \[m= \sum_{i=1}^{n}\frac{c_{i}}{a_{i}}.\] For each $k$ find the maximum value of $c_{k}$.

[i]Edited by Myth[/i]

Why it is still unsolved?  :huh: Very nice problem.

It is clear $c_{1}\leq 2$. Note that $c_{n+1}\leq 2(c_{1}+c_{2}+...+c_{n})$, otherwise it is impossible to represent the number $m=c_{n+1}\leq \sum_{i=1}^{n+1}c_{i}$.
It follows that $c_{n}\leq 4\cdot 3^{n-1}$ for $n\geq 2$. Now it is easy to prove that the sequnce $c_{1}=2,c_{2}=4\cdot 3^{0},c_{3}=4\cdot 3^{1},c_{4}=2\cdot 3^{2},...$ satisfies condition. By induction. Indeed, it is clear how to represent numbers $c_{n}+1,...,c_{n}+(c_{1}+...+c_{n-1})$. Also it is clear how to represent $\frac{c_{n}}{2}+1,...,\frac{c_{n}}{2}+(c_{1}+...+c_{n-1})$. So we are only to represent $1,...,\frac{c_{n}}{2}=c_{1}+...+c_{n-1}$. Guess how ;) (don't forget about $\frac{c_{n}}{a_{n}}$).