Brazilian Undergrad MO
integration
inequalities
calculus
derivative
real analysis
real analysis unsolved

prove that for a continuous R-->R and crescent function:

$\int_{0}^{1}xf(x) \geq \frac{1}{2}\int_{0}^{1}f(x)$

This was for me an very easy problem for a contest level and i am not so good in math YET. I would like to know your solutions and your opinio about this problem that at least for me, didn´t required any hard trick to solve it.

Hi,
Your problem is equivalent to this one:

$\int_{0}^{1}xf(x)dx\geq \int_{0}^{1}xdx\int_{0}^{1}f(x)dx$
This is nothing else but Chebyshev`s inequality for monotonic nonnegative functions, $f$ is monotonic increasing , thus $f(x)\geq f(0)$ so you may assume that $f(0)\geq 0$ if not you replace $f$ by $f-f(0)$ so you can apply Chebyshev,s inequality.
Didi

yeah i didn´t noticed that. i just used integration by parts in my solution. Later i post it!

My idea was use integration by parts involving the dereivate of the function which i know that is positive, so:

$\int_{0}^{1}xf(x)dx = \int_{0}^{1}(\frac{x^{2}f(x)dx }{2})'-\int_{0}^{1}\frac{x^{2}f'(x)dx}{2}$
also:
$\int_{0}^{1}f(x)dx=\int_{0}^{1}( xf(x)dx)'-\int_{0}^{1}xf'(x)dx$
 dividing the last integral by two then subtract the terms:
$\int_{0}^{1}xf(x)dx-\frac{\int_{0}^{1}f(x)dx}{2}= \frac{f(x)}{2}-\int_{0}^{1}\frac{x^{2}f'(x)dx}{2}-\frac{f(x)}{2}+\int_{0}^{1}\frac{xf'(x)dx}{2}= \int_{0}^{1}\frac{f'(x)}{2}(x-x^{2}) \geq 0$
the last conclusion is obvious because f'(x) is higher then zero and $x-x^{2}$ is obviously higher then zero.

But you don't know if $f$ has derivative.

I think it can be assumed it it derivable, since
 if $f$ is a monotone increasing function on an interval of the form [a,b] then it is differentiable almost everywhere and since the Riemann integral does not depend on a set of Lebesque measure zero, without loosing generality we may assume that $f$ is differentiable.
Didi

You are of course right. But I've thought Brambila spoke about a derivative on whole interval.

SORRY, I FORGOT TO SAY THAT THE PROBLEM SayS THAT F is increasing and is integrable  for 0<=x<=1!
SORRY!!!!!!!

Let $f$ and $g$ be increasing and integrable functions. We'll prove that if $a < b$,
\[{\int_{a}^{b}f(x)g(x)\,dx \over b-a}\geq{\int_{a}^{b}f(x)\,dx\over b-a}\cdot{\int_{a}^{b}g(x)\,dx\over b-a}\]

First, notice that for every $x, y \in \left]a;b\right[$
\[(f(x)-f(y)(g(x)-g(y)) \geq 0\iff f(x)g(x)+f(y)g(y) \geq f(x)g(y)+f(y)g(x)\quad(*)\]

Integrating both sides in $\left]a;b\right[^{2}$:
\[\int_{a}^{b}\int_{a}^{b}(f(x)g(x)+f(y)g(y))\,dx\,dy \geq \int_{a}^{b}\int_{a}^{b}(f(x)g(y)+f(y)g(x))\,dx\,dy\]

But, if $\int_{a}^{b}f(x)g(x)\,dx = I$,
\[\int_{a}^{b}\int_{a}^{b}f(x)g(x)\,dx\,dy =\int_{a}^{b}\left(\int_{a}^{b}f(x)g(x)\,dx\right)\,dy = \int_{a}^{b}I\,dy = (b-a)\int_{a}^{b}f(x)g(x)\,dx\]
and, if $\int_{a}^{b}f(x)\,dx = A$ and $\int_{a}^{b}g(x)\,dx = B$,
\[\int_{a}^{b}\int_{a}^{b}f(x)g(y)\,dx\,dy = \int_{a}^{b}\left(\int_{a}^{b}f(x)\,dx\right)g(y)\,dy = \int_{a}^{b}Ag(y)\,dy = A\int_{a}^{b}g(y)\,dy = AB=\int_{a}^{b}f(x)\,dx\int_{a}^{b}g(x)\,dx\]

So $(*)$ reduces to
\[2(b-a)\int_{a}^{b}f(x)g(x)\,dx \geq 2\int_{a}^{b}f(x)\,dx\int_{a}^{b}g(x)\,dx\]
and the result follows.


1. What can you say about $ n$ if the value of $ \phi{(n)}$ is a prime number?

2. What can you say about $ n$ if the value of $ \phi{(n)}$ is the square of a prime number?

3. Solve $ \phi{(n)}=\frac{n}{6}$.

[hide]
1. If $ n$ is prime then $ \phi(n) = n-1$

2. If $ n$ is the square of a prime $ p$ then $ \phi(n) = \phi(p)\phi(p) = (p-1)(p-1)$.

[/hide]

You have it the other way around. Read the problems again.  :wink:


Thomas took three $ 100$-point quizzes, and the mean score for the three quizzes was $ 87$ points.  If the highest score is taken away, the mean of the remaining two scores would be $ 83.5$ points.  If the lowest score is taken away, the mean of the remaining two scores would be $ 89$ points.  What is the highest score?

x = highest
y = middle
z = lowest

x + y + z = 261
y + z = 167

We do not need the last equation, since subtracting the second from the first yields x = $ \boxed{94}$


What is the probability that you'll roll a sum of 14 when you roll two seven-sided dice labelled 1-7?

[hide]if you meant rolling 2 die, then the only way to get 14 is by getting 2 7's, so the probability is 1/49.[/hide]

[hide]
There are 49 sums with the numbers 1-7 2 times. So the answer is1/49[/hide]

[quote="math92"]do you mean with two rolls of the dice?[/hide][/quote]

Yeah, I trailed off and wrote you roll twice.  :blush:

[hide]There is only one possible way to get 14, and 49 total posibilities, so our answer is $\frac{1}{49}$ :idea: [/hide]

[hide]there is only one way to roll a sum of 14 (7 and a 7).  On two rolls of a 7 sided dice you would do 7^2 which is 49 so the final answer is $\boxed{\frac{1}{49}}$[/hide]


I don't even know if result is right or not:
 But is it true that all primes divide atleast
 one number of the type 10^l +1 ?
 where l is any integer? if so, can you prove it?

Have you tried $2,3,5$ ¿ ;)

A nice and easy problem:

Let $a$ be given and let $m$ be odd.
Then none of the primes dividing $a^{m}-1$ ever divides a number of type $a^{l}+1$.
Conversly, every prime never dividing any $a^{l}+1$ divides a number of type $a^{m}-1$ with $m$ odd.

By this, e.g. $37|10^{3}-1=999$ works.

And to get infinetely many such primes:
Prove that the numbers $a^{k}+1$ and $a^{l}+1$ are coprime if $k,l$ are coprime.


geometry
combinatorics unsolved
combinatorics

On a line we are given $ 2005$ points $ A_1,A_2,...,A_{2005}$. These points are painted black and white such that there is at least one point painted black as well as white. For each point $ A_i$ among them, let $ f(A_i)$ denote the difference between the number of the white points which are to the left of $ A_i$ and the number of the black points which are to the right of $ A_i$. Assume that there is a value $ k$ satisfies that there exist $ A_{i_1},A_{i_2},...,A_{i_{2m \plus{} 1}}$ such that $ f(A_{i_j}) \equal{} k$. Find every possible value of $ k$.

The second to last sentence doesn't make much sense. What is $ m$?

$ m$ is some possitive integer. $ i_{2m\plus{}1}$ just means that $ k$ appears in the sequence $ A_i$ for an odd number of times.

Okay, but if $ k$ appeared $ 4$ or $ 6$ times we could still select $ m \equal{} 1$ and three such $ A_i$. Are you intending for that sequence to cover all of the vertices for which $ f(A_i) \equal{} k$? (this is why I thought $ m$ wasn't arbitrary)

Oh, I see, so bad was I for the unclear statement of the problem.

You are right, Mellow Melon, the mentioned sequence covers all the $ 2005$ points  :blush:


Consider the set $R^2-IXI$. Where I is the set of all irrationals. Is this set connected or path-connected?

What do you mean by $IXI$? :? Maybe $I\times I$, or $I^2$? If this is what you mean, why not simply write $I^2$, just like you wrote $\mathbb R^2$?

Then of course it's path-connected (and thus connected): if $ (x,y),(z,t),\ x\in\mathbb Q,z\in\mathbb Q$ are points in our set, go from $ (x,y)$ to $ (x,t'),\ t'\in\mathbb Q$ vertically, then from $ (x,t')$ to $ (z,t')$ horizontally, and then from $ (z,t')$ to $ (z,t)$ vertically again. Something similar can be done when $ x \in \mathbb Q,t\in\mathbb Q$.


Let points $O$, $P$, $Q$ and $R$ have coordinates $O(0,0), P(0,1) Q(0,-2)$ and $R(5,0)$. There is a point $X(x,0)$ on the x-axis between $O$ and $R$ such that $m\angle PXR=2m\angle QXO$. For which value of x does this occur?

I think you want \angle ($\angle$) instead of $<$.

[hide]
Let $PXO = A, OXQ = B$.  Then $2B = 180 - A$.  Also, $\tan A = \frac 1 x$ and $\tan B = \frac 2 x$.
From (1), $\tan 2B = - \tan A \implies \frac{ \frac 4 x }{1 - \frac 4 {x^2} } = - \frac 1 x$.
$\frac{4x}{x^2 - 4} = - \frac 1 x \implies 5x^2 = 4 \implies x = \frac{2 \sqrt 5 } 5$
[/hide]


calculate.
$ \sqrt{\underbrace{44\cdots 4}_{\text{2n numbers}}+{\underbrace{11\cdots 1}_{\text{n+1 numbers}}-{\underbrace{66\cdots 6}_{\text{n numbers}}}}}$

Let $ a_n = 111\ldots1$ so that $ a_n$ has $ n$ digits. Observe that $ 10^n = 9a_n + 1$. The given quantity reduces to
\[ \sqrt{4 (10^n)(a_n) + 4 a_n + 10^n + a_n - 6a_n}\]
\[ \sqrt{4 ( 10^n )( a_n ) + 10^n - a_n}\]
\[ \sqrt{(4 a_n + 1) \cdot 10^n - a_n}\]
\[ \sqrt{(4 a_n + 1)(9 a_n + 1) - a_n}\]
\[ \sqrt{36 a_n^2 + 13a_n + 1 - a_n}\]
\[ \sqrt{36 a_n^2 + 12a_n + 1}\]
\[ \sqrt{(6a_n + 1)^2}\]
\[ 6a_n + 1\]

Basically, two thirds of $ 10^n$ rounded to the nearest integer.

Doesn't look complete... you can get the answer to something much simpler than $ \sqrt{4/9 \cdot 10^n \cdot (10^n \plus{} 1) \plus{} 1/9}$.


Here's a tricky problem:

Find 5 consecutive two-digit integers that can be combined into a 3-digit and a 7-digit number (e.g. 86, 94, and 72 can become 948267 or 869742 of 676892 etc.) that add up to a perfect square.

This is really hard; I can't figure it out.

I think the key is that they are consecutive, but one thing I dont get is what you mean by combined, is that just taking random digits from the numbers or what?

I guess then there would be a combination aspect?

Yeah, taking the digits and mashing them together.


$ x,y,z \geq 0,$prove that:

$ \frac{x}{\sqrt{x\plus{}y}}\plus{}\frac{y}{\sqrt{y\plus{}z}}\plus{}\frac{z}{\sqrt{z\plus{}x}} \leq \frac{5\sqrt{x\plus{}y\plus{}z}}{4}$

[quote="rachid"]$ x,y,z \geq 0,$prove that:

$ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}} \leq \frac {5\sqrt {x \plus{} y \plus{} z}}{4}$[/quote]
It is Jack Grafullhei ineq,it is old
But the proof of Can hang which be used CS is really nice
The ineq with 4 variables is still true

[quote="rachid"]$ x,y,z \geq 0,$prove that:

$ \frac {x}{\sqrt {x \plus{} y}} \plus{} \frac {y}{\sqrt {y \plus{} z}} \plus{} \frac {z}{\sqrt {z \plus{} x}} \leq \frac {5\sqrt {x \plus{} y \plus{} z}}{4}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1459
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88454
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=90923
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=152848

[quote=rachid]$ x,y,z \geq 0,$prove that:

$$ \frac{x}{\sqrt{x\plus{}y}}\plus{}\frac{y}{\sqrt{y\plus{}z}}\plus{}\frac{z}{\sqrt{z\plus{}x}} \leq \frac{5}{4}\sqrt{x\plus{}y\plus{}z}$$[/quote]
[url=https://math.stackexchange.com/questions/3280839/find-the-minimum-k-such-that-sum-cyc-fraca-sqrtab-leq-k-sqrtab]here[/url]
Let $a, b $ and $c$ be non-negative numbers such that $ab+ac+bc\neq 0$ and $a+b+c=1.$ [url=https://math.stackexchange.com/questions/2148909/if-abc-1-then-sum-limits-cyc-fraca-sqrt3ab-leq-frac3127]Prove that[/url]
$$\frac{a}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c}{\sqrt[3]{c+a}}\leq\frac{31}{27}$$


prove that n! squared is greater than n to the power n

I think $ n\geq 3$
$ n\equal{}3:$ ineq is right
Assume ineq is right to n
we have $ [(n\plus{}1)!]^2>(n\plus{}1)^{n\plus{}1}\Leftrightarrow (n!)^2>(n\plus{}1)^{n\minus{}1}\Leftrightarrow n^n>(n\plus{}1)^{n\minus{}1}\Leftrightarrow n^{\frac{n}{n\minus{}1}}>n\plus{}1$
that $ n^{\frac{n}{n\minus{}1}}\equal{}(1\plus{}n\minus{}1)^{\frac{n}{n\minus{}1}}\geq 1\plus{}n$
we are done


I really need some help with this

Assume you have a die that is made in a way so the probability to get "1" is 1/2, while the probability to get 2-3-4-5 and 6 is equal. X marks the nuber of eyes that you get when it is tossed.

a) Find the probability distribution!?
b) Find E(X)
c) Find Var(X)
------
Imagine that the die is tossed 8 times. Let Y be the number of times X shows "1 eye" on the die

a) Calculate: P(Y=y)
b) Calculate: E(Y)

Thank you for any help


AMC
USA(J)MO
USAMO
email
AIME
analytic geometry
AIME I

Preliminary List - April 10, 2009

Official emails with the official invitation will start going out on Friday, April 10, 2009. Paper postal invitations will be mailed shortly thereafter. Please see the reminder in Section V, page 5 of the AIME Teachers' Manual, no AIME results will be given out by phone (or otherwise) until after April 13. Please do not call or contact us now, we have much to do yet.

We expect to begin mailing AIME reports going out the week of April 13. Again, please do not call or email until after April 13.

SELECTION

Selection for the 2009 USAMO was as follows:

For each contestant the AMC computed the USAMO index as 10*AIME + AMC where AIME is the score on the 2009 AIME I or 2009 AIME II, and AMC is the score on the 2009 AMC 12 A, AMC 12 B, AMC 10 A, or AMC 10 B. If a contestant took contests on both the A-date and the B-date, we use the greater of the two scores.

Among all 2009 AMC 12 A and 2009 AMC 12 B contestants who qualified for and took the 2009 AIME I or 2009 AIME II the AMC selected the top 328 indices. This was an index of 201.0 or greater. The lowest AIME score among these indices was 7. Thus an AIME score of 7 becomes the "floor score".

Among all students in 10th grade or below, regardless of which contest they took on the A-date or the B-date or both, there were 282 contestants with an AIME I or AIME II score of 7 or greater not already in the first selection. Among these contestants, the highest 158 indices was an index of 215.0 or higher.

Together this gives 486 students invited to the USAMO. Among these contestants there were 12 states and U.S. territories not already represented, so the contestant with the highest index in each of those states and territories was added to the list. That makes a total of 498 students invited to the USAMO. This selection fulfills our selection process detailed in the USAMO Teachers' Manual.

The descriptive statistics for the AIME I and the AIME II are not significantly different, particularly at the upper end of the distribution. The statistics do not indicate separate consideration of the resulting indices.

After that process we carefully considered the contestants not already selected. Out of nearly 6000 eligible contestants, there were 15 students in 10th grade or below with qualifying AMC 12 scores who scored 8 or more on the AIME I or AIME II. We will make a special invitation to these students to the USAMO. This makes a grand total of 514 students invited to take the 2009 USAMO.

Are you guys planning on doing this selection method for AMC 12-takers above the floor again, or is it just for this year?

[quote="mathgeniuse^ln(x)"][quote="mathemonster"]So this year a 121 amc 12 and 8 AIME would make it?[/quote]
If only 121 were divisible by 1.5[/quote]
lol

[quote="dgreenb801"]Aw man, I missed it. I thought I got 121.5/11. How can this be?[/quote]
maybe u got 121.5/9

Maybe. I checked the answers though, and was pretty sure I got 11. Hmm, I dunno, maybe a scantron error because I found one during the test but thought I fixed it. Anyways, I'm a little disappointed  :( .

[quote="archimedes1"]So it looks like an AMC 10 score of 144 and an AIME score of 7 misses the cutoff by one point?[/quote]
Yes.  :P

Last year, I had an AMC 10 score of 145.5 and an AIME score of 7. That would have barely made it this year.

Um, so the "144/6" that 1=2 reported is false?

:( didn't qualify, as expected.

god!!!
i was already in my summer mood, not ready to practice any extremely challenging math.
and bam! "after careful consideration..."
120 amc 12/8 aime as a sophomore
time to whip out the old aops vol 2 again...

I'm a junior with 115.5/9. If I had made one more stupid mistake on either of the two tests, I would not have qualified.

wait, what do they mean by qualifying amc 12 scores and 8 aime? is it just all <10th graders with 8 aime who took amc 12?

ya, i guess so
but they had to qualify for the aime first
so that essentially means a 180 index with an aime floor of 8 for all sophomore and below amc 12 takers
its funny, i would have expected more than 15...

wait really??? i have a 105 amc 12 and an 8 aime and I didn't make it...
should i email them or what?

then again, maybe they just took the top 15
i'm not sure
you can ask if you want
or maybe you just didn't get your email yet

Hmm, I wasn't aware that 498 + 15 = 514.

well i didn't get my email and am not on the list either. i guess i'll sit patiently... ugh I wish I could forget..

Sweet!

Anyway, I agree with emmatheduck.

[quote]I wasn't aware that 498 + 15 = 514.[/quote]

If you really must know, several of us were working simultaneously on various aspects of compiling and posting the list, and one of us miscounted one set of lines in one spreadsheet tab, while another of us correctly counted the total and reported that.   In the pressure to post the list, we did not coordinate while doing the final proof-reading.  There are 514 names in the list.

Steve Dunbar
MAA Director, American Mathematics Competitions

AMCDirector, could you possibly explain why an underclassman who had an 8 on the AIME and took the AMC 12 didn't make it? Thanks for all your hard work!

Dang, missed the special 10th grade cutoff by 1.
But thank you for adding that rule in for this year, I hope it continues to be implemented. :)

simo14, you should probably email the AMC office with your real name and the scores you believe you have.  It's possible that you do not in fact have an 8 on the AIME.  It's also possible that you were left off the list by mistake, which I think has happened before.

Edit: As AMCDirector posted elsewhere, wait until Monday to do that.

thanks sly si. and btw great SMT this year!

oh at first i think you didn't take the 10... how do i delete this post?

The ultimate just-missed profile for this year: 144 AMC 10, 130.5 AMC 12, 7 AIME (as 10th grader or younger), missing all three cuts by the minimum. I wonder how close anyone came to that?
This does seem like a reasonably fair procedure; about as good as possible, given the framework in place and the test scores this year.

[quote="jmerry"]144 AMC 10, 130.5 AMC 12, 7 AIME [/quote]
I know a 126/124.5/7, which isn't too far away from that. I also know more conventional just-missed, like an 11th grade 120/8.

[quote="simo14"]@jmerry: had that person had an 8 on the aime, would they have qualified for the 3rd cut?[/quote]

In that case, he would have qualified for both (in reality the secondary cut).

As AMCDirector has said many, many times, [b]wait until April 13[/b]. Don't get worried until then.

Question: why did my post jump threads?  :maybe:

I moved it. It didn't seem to be in the correct place.

[quote="physisnu"]I qualified for 2008 USAMO, and participated in it.

this year I got  144.0 AMC 10A, 11 AIME I.

I'm from South Korea, and I'm an American citizen, which makes me eligible.

However, although I was on last year's list (with an idex of 230)

I'm not on this year's list (with an index of 254)

Can anybody tell me what happened??

I took the AIME I this year, so I don't think there is any reason my name should go on late....

Quite worried  :([/quote]

Contact Steve please. International qualifiers have to contact the AMC office, since there is no way for them to confirm the status from international papers.

Are you James? If so, please email me.


analytic geometry
function
calculus
derivative
graphing lines
slope

$AOB$ is a right angle. A material point is moving from $A\to O\to B$. Prove that the point will stop somewhere.

You mean that if we establish a suitable coordinate system , er have to prove that # \frac{dx}{dt} $ and $ \frac{dy}{dt} $ are zero at some time during the journey????

Also, what are the initial and final velocity?? (i'm thinking in terms of Rolle's theorm!!)

It is quiet simple the line function of the $AOB$ at point $O$ has no derivative.

So that means, it comes to rest there?

Yes.. this question is basically calculus as I observed. The critical point of a curve is where eithere derivative is zero or doesn't exist(which is the case here) !!!

does the point have to be moving along the lines connecting A to O and O to B?
if AOB were an inscribed right angle in a circle, what if the point were moving along the circle's circumference?

I think it is not, it basically moves along an arbitrary curve which passes throgh O

This question is basically an interesting application of Rolles theoerm (although I don't understand why O is required)The particle moves along a curve from A to B (at same level) Thus, the curve of this slope is zero somewhere. But how is it realted to speed??

Well, if it didn't have to move along the segments AO and OB, it wouldn't have to stop!

But since it does, it has to stop. It is clear to me intuitively, that if there is a sharp "break" in the trajectory, it is not possible to pass through there without stopping. But I was not quite able to give a precise or rigorous mathematical proof. Can you?

You I'll tell you actually it can move by $AOB$ without stopping if the angle $AOB$ is not acute because in this case point will have projection on the $OB$ while it moves on $AO$ and it can momentally turn the perpendicular componoent to 0, e.g. by absolute none-elastic collision, but keeping $OB$ component. But in case of acute angles and right angles it's impossible due to in first case the component is opposite to $\vec{OB}$ and the second case it equals 0.

Wait, wait, I thought there was a reason to it when you said that it is a "material point". That would mean it has some mass and it cannot have infinite acceleration. And thus, even if it had SOME velocity in projection to OB while still on segment AO, it would have to change in one moment by some amount. (or the other component would have to)

Actually you're right, if we assume that the velocity can't be changed in a moment.


modular arithmetic
analytic geometry
trigonometry
absolute value

Find the number of ordered pairs of real numbers $ (a,b)$such that $ (a\plus{}bi)^{2002}\equal{}a\minus{}bi$

[hide]hmm... lets see z^2002 = conj(z), looking at the magnitudes, we have either |z|= 0 which is 1 trivial case, or |z|=1.  So in the case |z|= 1, we know z*conj(z) = z^2003 =1, so z can be any of the 2003th roots of unity or 0.  So thats 2004 different ordered pairs[/hide]

[hide]
Re-write the equation as
$ r^{2002}\cdot (cis \theta)^{2002} \equal{} r\cdot cis ( \minus{} \theta)$
If $ r \equal{} 0$, then $ (a,b) \equal{} (0,0)$.
If $ r\neq 0$, then $ r^{2002}$=$ r$, so $ r \equal{} 1$. Use de Moivre Theorem,
$ cis (2002\theta) \equal{} cis( \minus{} \theta)$
$ 2002\theta \equiv \minus{} \theta \pmod {2\pi}, \theta \equiv 0 \pmod {2\pi}$. So $ (a,b) \equal{} (1,0)$.
Therefore, $ (a,b) \equal{} (0,0),(1,0)$.[/hide]

Let $ a\plus{}bi\equal{}z$
$ z^{2002}\equal{}\bar{z}$
$ z\equal{}0$ or $ (a,b)\equal{}(0,0)$ is a trivial solution. So assume $ z\ne 0$.
$ z^{2003}\equal{}|z|^2$
Taking | | both side implies $ |z|\equal{}1$.
Thus $ z^{2003}\equal{}1$ i.e. $ (a,b)$ are the coordinates of the 2003th roots of unity, namely
$ (a,b)\equal{}(\cos\frac{2n\pi}{2003},\sin\frac{2n\pi}{2003}), n\equal{}0,1,\ldots,2002$.

So, including (0,0), there are a total of 2004 pairs $ (a,b)$.

$ \sqrt[2003]{|z|^2}$ being real doesn't mean anything. $ |z|$ is always real, since it's a magnitude, so how wouldn't it be real? The conclusion that $ z$ is a real number doesn't follow.

$ 2002\theta \equiv \minus{}\theta \pmod{2\pi}$ does not imply that $ \theta$ is a multiple of $ 2\pi$. It implies that $ 2003\theta$ is a multiple of $ 2\pi$. That is very different.

[quote="packham"]Let $ a \plus{} bi \equal{} z$
$ z^{2002} \equal{} \bar{z}$
$ z \equal{} 0$ or $ (a,b) \equal{} (0,0)$ is a trivial solution. So assume $ z\ne 0$.
$ z^{2003} \equal{} |z|^2$
Taking | | both side implies $ |z| \equal{} 1$.
Thus $ z^{2003} \equal{} 1$ i.e. $ (a,b)$ are the coordinates of the 2003th roots of unity, namely
$ (a,b) \equal{} (\cos\frac {2n\pi}{2003},\sin\frac {2n\pi}{2003}), n \equal{} 0,1,\ldots,2002$.

So, including (0,0), there are a total of 2004 pairs $ (a,b)$.[/quote]

wait why does taking | | to both sides imply |z|=1?

Because of some properties of the absolute value function.

1. $ |z|$ is a positive real number.
2. $ r \equal{} |r|$ if $ r$ is a positive real number.
3. $ |z^n| \equal{} |z|^n$

So if $ z^{2003}\equal{} |z|^2$, then
$ |z^{2003}| \equal{} |(|z|^2)|$
$ |z|^{2003}\equal{} |(|z|)|^2$
$ |z|^{2003}\equal{} |z|^2$
$ |z|^{2001}\equal{} 1$ (since $ z \neq 0$ was assumed)
$ |z| \equal{} 1$

$ |z^{2002}|\equal{}|\overline{z}|\equal{}|z|\Longleftrightarrow |z| ^{2002}\equal{}|z|\Longleftrightarrow z\equal{}0$ or $ |z|^{2001}\equal{}1.$

[quote="MellowMelon"]
$ 2002\theta \equiv \minus{} \theta \pmod{2\pi}$ does not imply that $ \theta$ is a multiple of $ 2\pi$. It implies that $ 2003\theta$ is a multiple of $ 2\pi$. That is very different.[/quote]

mm...I see, it's really a critical mistake. :blush:


geometry
incenter
symmetry
ratio
circumcircle
geometry proposed

[b]Problem[/b] Let ABC be a triangle and let w be its incircle. Denote by D,E, F the inter-
sections of w with BC,CA,AB, respectively. Let T ∈ AD ∩ w, M ∈ BT ∩ w,
N ∈ CT ∩ w. Let p1 be a circle tangent to w at T, and p2 a circle tangent to
w at D, so that p1 and p2 intersect on chord (XY ). Prove that X, Y,M,N lie
on the same circle. :D

Proposed by Cosmin Pohoata, Bucharest, Romania

I am trying to prove that K, M, N are collinear, but still can't, anybody helps me???????? :blush:

$ t$ is the incircle tangent at $ T.$ Project the incircle $ (I)$ to a circle $ (I')$ and any line $ k$ through $ K \equiv EF \cap BC$ not intersecting the $ \triangle ABC$ to infinity. (You can even project to infinity the line $ AK.$) Projected lines $ E'F', B'C'$ become parallel $ \Longrightarrow$ projected $ \triangle A'B'C'$ becomes isosceles with incircle $ (I')$ $ \Longrightarrow$ projected tangent $ t'$ of $ (I')$ at $ T'$ and projected line $ M'N'$ become parallel to $ E'F' \parallel B'C'$ by symmetry $ \Longrightarrow$ the original lines $ EF, BC, t, MN$ are concurrent at $ K \in k.$ Etc.

Sorry, can you use some easier tool to solve this problem, now I just know Pole and Polar basically, so your solution......is really difficult for me to understand. Anyway, thanks for your help :)

How's this: The cross ratio $ G(K, D, B, C) \equal{} \minus{} 1$ is harmonic, the line pencil $ (TK, TD, TB, TC)$ is harmonic. Let $ TD, TK$ cut $ MN$ at $ P, Q,$ then the cross ratio $ G(Q, P, M, N) \equal{} \minus{} 1$ is also harmonic. $ TQ \equiv TK$ is tangent to the circumcircle $ (I)$ of $ \triangle TMN$ $ \Longrightarrow$ $ Q$ is center of its T-Apollonius circle $ \Longrightarrow$ $ TP \equiv TD$ is its T-symmedian $ \Longrightarrow$ tangents to $ (I)$ at $ M, N$ meet at $ S$ on the T-symmedian $ \Longrightarrow$ poles $ M, N, K$ WRT $ (I)$ of the concurrent lines $ MS, NS, TD$ are collinear (and $ Q \equiv K$ are identical).

Let $ \{ K \} \equal{} EF \cap BC$, since $ AD,BE,CF$ concur, $ (KDBC) \equal{} \minus{} 1$.

Let $ \{ L \} \equal{} EF \cap AD$, we also have $ (KLEF) \equal{} (KDBC) \equal{} \minus{} 1$. But $ KD$ is the tangent of $ (I)$ at $ D$ and $ \{ T \} \equal{} AD \cap (I) \minus{} \{D\}$ so $ K$ is the pole of $ AD$ wrt $ (I)$, i.e., $ KT$ is the tangent of $ (I)$ at $ T$.

On the other hand, $ T(TMDN) \equal{} T(KDBC) \equal{} \minus{} 1$ so $ TMDN$ is a harmonic quadrilateral, i.e., the intersection $ K$ of the tangents of $ (I)$ at $ T$ and $ D$ lies on $ MN$, which implies $ K,M,N$ are collinear.

Finally, $ \mathcal{P}_{K/(I)} \equal{} KD^2 \equal{} \mathcal{P}_{K/(I')} \equal{} \mathcal{P}_{K/(I'')}$, i.e. $ K$ is the radical centre of $ (I),(I'),(I'')$, so $ K\in XY$. This means: $ \overline{KM}.\overline{KN} \equal{} \overline{KX}.\overline{KY}$, so $ X,Y,M,N$ are concyclic, this ends the proof.

P/S: HAPPY LUNAR NEW YEAR OF VIETNAM  :lol:  :)  :wink:  :P


In convex hexagon $ ABCDEF$ all sides have equal length and

$ \angle A\plus{}\angle C\plus{}\angle E\equal{}\angle B\plus{}\angle D\plus{}\angle F$.

Prove that the diagonals $ AD,BE,CF$ are concurrent.

[quote="Carlez Tevos"]In convex hexagon $ ABCDEF$ all sides have equal length and

$ \angle A \plus{} \angle C \plus{} \angle E \equal{} \angle B \plus{} \angle D \plus{} \angle F$.

Prove that the diagonals $ AD,BE,CF$ are concurrent.[/quote]
It isn't difficult to show that $ \angle A\equal{}\angle D, \angle B\equal{}\angle E,\angle C\equal{}\angle F$
This problem is related to the prolem which is referred in International Olympiad 2005, Mexico.

If it really isn't difficult to show, please show it.

[quote="mr.danh"]See the topic: [url]http://www.mathlinks.ro/viewtopic.php?p=1129352#1129352[/url]
I hope you will find the solution soon.[/quote]

It's a different problem! Please show my problem.

Prove that ABCDEF is a quadrilateral with an inscribed circle. By definition, a tangential quadrilateral has its diagonals $ AB, BE, CF $ concurrent.

Umm.. I guess it's not necessarily a tangential quadrilateral.. It was just a quick thought.

Consider the following isometry: translate the entire hexagon until $C$ is mapped to $C'=A$, then rotate about $C'=A$ until the image of $D$ coincides with $F$. Notice now that, since $\angle B+\angle D+\angle F=2\pi$, we have $\angle EFE'=2\pi -\angle AFE - \angle AFE'=2\pi-\angle AFE-\angle CDE=\angle ABC$. Thus, $EE'=AC$. Also, by construction, $AE'=CE$. Thus, $CAE'E$ is a parallelogram, so $\angle AE'E=\angle ACE.$

We now seek to prove that $\angle BCD=\angle AFE$, and analogously will follow $\angle B=\angle E, \angle D=\angle A$.

In the following angle chase we use the fact that $\triangle ABC$ and $\triangle CDE$ are isosceles to establish $\angle BAC=\angle BCA$ and $\angle CED=\angle DCE$. We also know that the isometry preserves angles between corresponding sides, so $\angle AE'F=\angle C'E'D'=\angle CED$.  Since $\triangle EFE'\cong \triangle CBA$, $\angle EFE'=\angle BAC$.

Now notice that $F$ is the circumcenter of $AEE'$ since $FA=FE=DE=FE'$.

Thus, $\angle AE'E=\frac{1}{2}\angle AFE=\angle AE'F+\angle FE'E=\angle CED+\angle BAC=\angle ECD+\angle BCA=\angle BCD-\angle ACE=\angle BCD-\angle AE'E\iff$ $$\angle AFE=2\angle AE'E=\angle BCD.$$

Thus, opposite pairs of angles are equal. From here, it is easy to show that opposite sides are parallel.
Now, notice that since $AF=CD$, and from $\triangle BAC\cong \triangle EDC$ we must have $AC=FD$, $ACDF$ must be a parallelogram. It follows that the diagonals $AD,CF$ and $BE$ all bisect each other and are thus concurrent.


Square regions 5 cm by 5 cm are removed from each of the 4 corners of a 30 cm by 40 cm rectangle. The sides are folded up to create an open box. What is the volume, in cubic centimeters, of the interior of the open box?

Okay, the box will have dimensions $ (30\minus{}2(5)) \times (40\minus{}2(5)) \times 5$, so the volume will be $ 20 \times 30 \times 5$, or $ \boxed{3000}$.

That's not the question that shows up...
It shows "find out the number of hours in a 31 day month"


It's tonight, but I'm unable to attend. I have a question, though, and I was wondering if anyone is willing to ask it at the math jam for me: does psychology count as one of the topics for research? Really appreciated.

I couldn't copy/paste from the window, but
[quote="SWCmanager"]Intel and Siemens are very similar.  Main differences is that Siemens allows for teams w students in 9-12th grade.  We are also a regional competition.  And we do not accept behavioral or social science projects."[/quote]

Alright, thanks. Hmmm...My team mate wants to do Psychology while I want to do Graph Theory...Better for me I guess.


induction
combinatorics unsolved
combinatorics

Suppose that a permutation $ (a_1,a_2,...,a_n)$ of $ (1,2,...,n)$ satisfies
\[ \frac{a_k^2}{a_{k\plus{}1}}\leq k\plus{}2\]

for $ k\equal{}1,2,...,n\minus{}1$.

Prove that it must be the identity permutation.

I don't really see the point of this problem. If $ a_k \equal{} n$ for some $ k < n$, then $ a_{k \plus{} 1}\leq n \minus{} 1$ and $ k \plus{} 2\leq n \minus{} 1$, so that $ \frac {a_k^2}{a_{k \plus{} 1}}\leq k \plus{} 2$ is impossible (since $ \frac {a_k^2}{a_{k \plus{} 1}}\geq\frac {n^2}{n \minus{} 1} > n \plus{} 1\geq k \plus{} 2$). Hence, we must have $ a_n \equal{} n$. Now proceed by induction over $ n$.

  darij


Let a^2+b^2+c^2=1, prove that

$ \frac{1\minus{}c^2}{a\plus{}b}\plus{}\frac{1\minus{}b^2}{c\plus{}a}\plus{}\frac{1\minus{}a^2}{b\plus{}c} \geq a\plus{}b\plus{}c$

不難的
可以
$ 1\minus{}c^2 \equal{} a^2 \plus{} b^2$
$ 2(a^2\plus{}b^2)/(a\plus{}b) >\equal{} a\plus{}b$
因為
$ 2a^2\plus{}2b^2 >\equal{} (a\plus{}b)^2$
$ 2a^2\plus{}2b^2 >\equal{} a^2 \plus{} 2ab \plus{} b^2$
$ (a\minus{}b)^2 >\equal{} 0$
將3個式子相加就可以得出結果
 :lol:

[quote="colorfuldreams"]不難的
可以
$ 1 \minus{} c^2 \equal{} a^2 \plus{} b^2$
$ 2(a^2 \plus{} b^2)/(a \plus{} b) > \equal{} a \plus{} b$
因為
$ 2a^2 \plus{} 2b^2 > \equal{} (a \plus{} b)^2$
$ 2a^2 \plus{} 2b^2 > \equal{} a^2 \plus{} 2ab \plus{} b^2$
$ (a \minus{} b)^2 > \equal{} 0$
將3個式子相加就可以得出結果
 :lol:[/quote]

對啦!  :)

$ \frac{1\minus{}c^{2}}{a\plus{}b}\plus{}\frac{1\minus{}b^{2}}{c\plus{}a}\plus{}\frac{1\minus{}a^{2}}{b\plus{}c}\geq a\plus{}b\plus{}c$

is equivalent to proving

$ \frac{a^2\plus{}b^2}{a\plus{}b}\plus{}\frac{c^2\plus{}a^2}{c\plus{}a}\plus{}\frac{b^2\plus{}c^2}{b\plus{}c}\geq a\plus{}b\plus{}c$

Notice that by Q.M greater than or equal to A.M,
$ \frac{a^2\plus{}b^2}{a\plus{}b} \geq \frac{(a\plus{}b)^2}{2(a\plus{}b)} \equal{} \frac {a\plus{}b}{2}$

Adding the three terms we get the expression to be
$ \geq \frac{2(a\plus{}b\plus{}c)}{2} \equal{} a\plus{}b\plus{}c$


See above poll.

I am going to MathPath, and am 11 years old.

Post here if you want to!

I'm going and I'm 13.  (Honestly, why would someone post here if they weren't going?)

why wouold someonepost here at all if there werent a topic.....lol

Summary of 7 votes:

We have 1 Going, that is 11: BOGTRO

We have 4 Going, that are 13: avec_une_h, ???, ???, ???.

One not going that is 13: ???

One not going that is 12: ???

The majority is going and is 13.

Hope I'll see you guys there!

P.S. abacadaea omg make another post at the time of writing you have 1336!

1336 what? sorry. i am really slow. i didnt get your post script.

i am 13...

[quote="xpmath"]He meant that you had 1336 posts. Now you have 1337!

I'm not going because my parents don't want me to; I'm 12.[/quote]

Hear, hear to this entire post except I'm 13.


Demonstrati ca $\prod_{sym}(a^{2}+ab+b^{2})\geq\ (ab+bc+ca)^{3}$

Solutie foarte simpla $a^{2}+ab+b^{2}\geq \frac{3(a+b)^{2}}{4}$.Acuma $ (a+b)(b+c)(c+a)\geq\frac{8(a+b+c)(ab+bc+ca)}{9}$ si mai aplicam $(\sum a)^{2}\geq 3\sum ab$. :D

Cine poate da o solutie la aceasta frumoasa inegalitate?? :)


Find the probability of (At least a boy in a family of 5 children).
The answer is 5/6 or 31/32?
Please show the step.

For this, we can use a concept called complementary counting. Basically what you do is you count what you DONT want and then subtract it from the total number of outcomes to get what you DO want. In other words, it would look something like this:

$ \text{(Number Of Ways You Can Have At Least One Boy)\equal{}}  \text{(Total Number Of Ways To Have Children)} \minus{} \text{(Number Of Ways You Have No Boys)}$

We know that there are $ 2\cdot 2\cdot 2\cdot 2\cdot 2 \equal{} 32$ possible ways you can have 5 kids. How many of those ways results in no boys? Just 1--this occurs when everyone is a girl.

So now we have:

$ \text{(Number Of Ways You Can Have At Least One Boy) \equal{} (32) \minus{} (1)}$.

Hence:
$ \text{(Number Of Ways You Can Have At Least One Boy) \equal{} 31}$.

We already said that there are 32 possible ways you can have kids, and we found out that 31 out of those 32 times you will have a boy, so the probability is $ \boxed{\frac {31}{32}}$


Assume that $ A$ is a $ n\times n$ matrix such that $ \mathbf{rank}\{A\plus{}I\}\plus{}\mathbf{rank}\{A\minus{}I\}\equal{}n$

Prove that: $ A^2\equal{}I$

Here's a sketch: Let $ k \equal{} \text{rank}(A \plus{} I)$, by the rank-nullity theorem there is a set of $ n \minus{} k$ linearly independent vectors $ U^ \plus{}$ with all $ v^ \plus{} \in \text{span}(U^ \plus{} )$ satisfying $ (A \plus{} I)v^ \plus{} \equal{} 0$.  Likewise there is a set of $ k$ linearly independent vectors $ U^ \minus{}$ with all $ v^ \minus{} \in \text{span}(U^ \minus{} )$ satisfying $ (A \minus{} I)v^ \minus{} \equal{} 0$.  Furthermore, each nonzero $ v^ \plus{} \in \text{span}\left(U^ \plus{} \right)$ satisfies $ v^ \plus{} \not \in \text{span}(U^ \minus{} )$ (for, otherwise, $ (A \plus{} I)v^ \plus{} \equal{} (A \minus{} I)v^ \plus{} \plus{} 2v^ \plus{} \equal{} 2v^ \plus{} \neq 0$) and vice-verse.  In other words, $ \text{span}(U^ \plus{} )$ and $ \text{span}(U^ \minus{} )$ are disjoint except for the zero vector.  Furthermore, since the sum of their dimensions is $ n$, $ U^ \plus{} \cup U^ \minus{}$ spans the whole space.  Thus, for any vector $ v$,
\[ A^2 v \equal{} ((A \minus{} I)(A \plus{} I) \plus{} I)v \equal{} (A \minus{} I)(A \plus{} I)(v^ \plus{} \plus{} v^ \minus{} ) \plus{} v \equal{} \\ \equal{} (A \minus{} I)\left[(A \plus{} I)v^ \minus{} \right] \plus{} v \equal{} (A \minus{} I)\left[(A \minus{} I) \plus{} 2I\right]v^ \minus{} \plus{} v \equal{} \\ \equal{} (A \minus{} I)(2v^ \minus{} ) \plus{} v \equal{} v \\ \Rightarrow A^2 \equal{} I\]

This post, together with the one [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=308915]here[/url], allow us to state the following theorem:

A matrix $ A \in M_n$ satisfies $ A^2 \equal{} I_n$ iff $ rank(I \plus{} A) \plus{} rank(I\minus{}A) \equal{} n$.


I have a Windows XP platform and I want to create a diagram like this:

[code]beginfig (1)
    fill ((0,0)--(2,0)--(2,1)--(1,1)--(1,2)--(0,2)--cycle)
        transformed pagecoords withcolor green;
    draw ((2,0)..(2,1)..(1,1)..(1,2)..(0,2))
        transformed pagecoords;
    drawarrow ((0,0)--(2,2)) transformed pagecoords;
endfig;[/code]

using metapost.

How would I do it?

I think you misread the [url=http://en.wikipedia.org/wiki/MetaPost]Wikipedia example[/url] you were using which is a single file with 3 different diagrams. Yo can use just the one diagram provided you add the start [code]transform pagecoords;
pagecoords:=identity scaled 10mm shifted (100mm,150mm);[/code] and the end [code]bye[/code]
Save as foo.mp, go to Start, Run and type [i]cmd[/i], press OK.
Then navigate to where foo.mp is kept and then type [i]mpost foo.mp[/i]. The result will be foo.1 which you rename to foo.eps

Anyway, I found htis website called http://www.tlhiv.org/cgi-bin/MetaPostPreviewer/index.cgi

And you input your metapost in there, and it spits out an image. :)

Yeah that's what I usually do.

I think asymptote is easier to use, but it's kinda slow on this computer.


A rectangle has a perimeter of 48 feet. What is the greatest
possible number of square feet in the area of the rectangle?


[quote="qweretyq"]A good math problem:

How many different scores are possible on the AMC?
(6 points for correct answer, 2.5 for omit, 0 for wrong, 25 questions)[/quote]

It's the number of possible values of $6a+2.5b$, where $0 \le a+b \le 25$, right?

The < <  should be inclusive, you could get them all wrong or none wrong.


What is the units digit to $5^{2006}$?

The units digit of $n^3$ is 3, and n is an integer. What is the units digit of n?

[quote="236factorial"]What is the units digit to $5^{2006}$?
[/quote]
[hide]
5 becuase the units digit will always be five becuase 5*5 ends in a 5.[/hide]

[quote="236factorial"]

The units digit of $n^3$ is 3, and n is an integer. What is the units digit of n?[/quote]

i got 7... not sure though... just random picked a number and tried it... and it worked, i'm not sure if there's more than one answer... or maybe i'm missing out about something in the problem

Yes, that is correct. You can get it by looking at the repeating pattern of 1, 2, or 4 digits in the units place when each integer 0 to 9 is raised to successive powers, and see that $7$ is the only one that gives a $3$ in the units place when raised to the $4n+3$ power.


Let $A,B \in M_{2}(C)$ be such that ; $(AB)^{2}=A^{2}B^{2}$. prove that :
 $det(I+AB-BA)=1$
    :blush:

Since $\mathrm{tr}(AB)=\mathrm{tr}(BA)$, the matrix $I+AB-BA$ has trace $2$. So the problem can be restated as: show that the only eigenvalue of $I+AB-BA$ is $1$. Or, equivalently, show that $C=AB-BA$ has only zero eigenvalue. For this it suffices to prove that $\mathrm{tr}(C)=\mathrm{tr}(C^{2})=0$. It's clear that $\mathrm{tr}(C)=0$. I claim that all four terms in $C^{2}=ABAB-ABBA-BAAB+BABA$ have the same trace. Indeed, $\mathrm{tr}((A)BAB)=\mathrm{tr}(BAB(A))$ and $\mathrm{tr}((AB)BA)=\mathrm{tr}(BA(AB))$. Finally, use the given relation: $\mathrm{tr}(ABAB)=\mathrm{tr}(AABB)=\mathrm{tr}(ABBA)$.


inequalities
combinatorics unsolved
combinatorics

A graph has $n$ vertix.Prove that:
$\frac{n}{\alpha(G)}\leq \chi \leq n-\alpha(G) +1$

this is trivial.since  any monochrome set is an independent set, the first inequality follows immediately.next take a maximum independent set and color eah of the remaining vertices with a different color and finally color the independent set using another different color.


Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Denote
$ x\equal{}\sqrt{\frac{3a^2}{b^2\plus{}bc\plus{}c^2}}, \quad y\equal{}\sqrt{\frac{3b^2}{c^2\plus{}ca\plus{}a^2}},\quad z\equal{}\sqrt{\frac{3c^2}{a^2\plus{}ab\plus{}b^2}}.$
Prove that
a) $ x\plus{}y\plus{}z \ge xy\plus{}yz\plus{}zx;$
b) $ xy\plus{}yz\plus{}zx \ge 3;$
c) $ 4(x\plus{}y\plus{}z) \plus{}xyz \ge 13.$
:)

For more collection:
Prove that
$ \sum \sqrt{\frac{a^2}{b^2\plus{}bc\plus{}c^2}} \ge \sqrt{3}\sum \frac{a^2}{\sqrt{(a^2\plus{}ab\plus{}b^2)(a^2\plus{}ac\plus{}c^2)}},$
for any nonnegative real numbers $ a,b,c.$ :)

[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers, no two of which are zero. Denote
$ x = \sqrt {\frac {3a^2}{b^2 + bc + c^2}}, \quad y = \sqrt {\frac {3b^2}{c^2 + ca + a^2}},\quad z = \sqrt {\frac {3c^2}{a^2 + ab + b^2}}.$
Prove that
a) $ x + y + z \ge xy + yz + zx;$
b) $ xy + yz + zx \ge 3;$
c) $ 4(x + y + z) + xyz \ge 13.$
:)[/quote]
I prove the second (I feel that it is  the easiest).
It is equivalent to
\[ \sum \frac {ab}{\sqrt {(c^2 + cb + b^2)(a^2 + ac + c^2)}}\ge 1
\]

\[{ \Leftrightarrow (\sum ab\sqrt {a^2 + ab + b^2} )\ge \sqrt {(a^2 + ab + b^2)(a^2 + ac + c^2)(b^2 + bc + c^2)}}
\]
\[{ \Leftrightarrow
\sum a^3b^3+\sum a^2b^2(a^2+b^2)+2\sum a^2bc\sqrt{(a^2+ab+b^2)(a^2+ac+c^2)}}\ge (a^2+ab+b^2)(a^2+ac+c^2)(b^2+bc+c^2)
\]
Use
\[ x^2 + xy + y^2\ge \frac {3(x + y)^2}{4}
\]
And expand that inequality we will get that
\[ abc(\sum a^3 + 3abc - \sum ab(a + b))\ge 0
\]
Which is obvious.

I have work the first problem out.
Here is my solution.


Let $ C_1,C_2$ be two circles such that the center of $ C_1$ is located on $ C_2$. If $ M,N$ are the intersections of the circles, $ AB$ an arbitrary diameter of $ C_1$, $ A_1$ and $ B_1$ the intersections of $ AM$ and $ BM$ with $ C_2$ respectively, prove that $ A_1B_1$ is equal to the radius of $ C_1$.

I believe that you mean [b]"$ A_1B_1$ equals to the diameter of $ (C_2)$"[/b] instead of [b]"$ A_1B_1$ is equal to the radius of $ C_1$"[/b]
or it might be "$ A_1, B_1$ are the intersection of $ AM, BN$" instead of "$ A_1,B_1$ are intersections of $ AM, BM$"

In the second case, here is my solution.

It is well-known that the triangle $ NAA_1$ is similar to the triangle $ NC_1C_2$. Therefore the triangle $ NAA_1$ is isoscele (at $ A_1$) which means that $ A_1A \equal{} A_1N$.

We also have $ C_1A\equal{}C_1N$. So $ A_1C_1$ is the line bisector of $ AN$. Then $ A_1C_1$ is perpendicular to $ AN$ (1)

Because $ AB$ is a diameter of $ C_1$, $ BN$ is perpendicular to $ AN$ (2)

From (1) and (2), we deduce: $ BB_1 // A_1C_1$

Then $ A_1B_1 \equal{} C_1B \equal{} R_{(C_1)}$


Congratulations to Otis Chodosh, Sherry Gong, Men Young Lee, William Throwe, and Henry Tung, and alternate Ingmar Saberi!

Good luck in Singapore!

And better luck surviving on a long, full plane flight while chanting "World War Three!" every five minutes.

I lost

MIIIILD RECESSION! MIIIILD RECESSION!

I lost.

Titantic avoids Iceberg.

Titantic Explodes!

This thread seems to have deviated slightly from "physics" and degenerated into spamming...
oh well.

By the way, I won once this past week.  So that makes three times now, interluser.  Oooh, yeah!

[quote="AndersoN"]What wins and loses are you talking about?[/quote]to be honest, i was at physics camp and i still don't really understand the "i lost" meme.  where did that come from anyway?

edit: Time Vortex!  Noooooo!


[quote="math92"][quote="diophantient"][quote="math92"]Solve.

1.

$2^{(\frac{2}{3})x+1}-3 * 2^{(\frac{1}{3})x}-20=0$

2.

$2^{x}-3^{x}=0$[/quote]
[hide="1"]$2^{\frac{1}{3}x}=k$
$2k^{2}-3k-20=0$
$k=\frac{3 \pm 13}{4}$
$k=4$ (note that k can't be negative)
$2^{x}=64$
$x=6$[/hide]
[hide="2"]
$2^{x}=3^{x}$
$1.5^{x}=1$
$x=0$[/hide][/quote]

I'm not really sure what you did in your second step of the 2nd problem. Can somebody explain please?[/quote]
Basically, both sides of the equation were divided by $2^{x}$...


Show that $\sum_{n}^\infty [2\frac{( n+1}{n} )^{n}]$ is divergent. Use ratio test and comaprison test.

The terms go to $\infty$, rapidly. There isn't much more to say.

Sorry but I can't fix the latex.  That was suppose to be [(n + 1)/n ] raise to n.

Try this: $2\left(\frac{n+1}{n}\right)^{n}$. You want the parentheses outside the fraction.
Mouse over or copy for the code.

With this change, the terms tend to $2e$. Comparison with the divergent series $\sum_{n=1}^{\infty}1$ is sufficient to establish divergence.

The ratio test is inconclusive, since this diverges slower than any growing geometric series.


[hide] 0, lets say $A$ and $B$ want to shake hands.  For $A$, to shake hands with $B$, $B$ must be taller.  But for $B$ to shake hands with $A$, $A$ must be taller.  This is a paradox, since both are different heights (and no two people can possibly be taller than each other) and consequently no handshakes occur.[/hide]

[hide]i think no handshakes happen[/hide]

It's amazing how the world gets along.  :lol:

[hide]
0, because A can shake B's hand, but B can't  shake A's hand because he is taller[/hide]

[hide]the tallest person has nobody taller than him, and the next tallest has one. The next two, the next three, and so on. Therefore the equation is 0+1+2+3+4+5+6+7+8+9=45. 45 handshakes take place.[/hide]

Whoops. :blush: I didn't think of it that way...
 :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:  :blush:

[quote="xtrememath8"]Okay, there are 10 people in a room, all of different heights. If they shake hands only with people who are taller then them, how many handshakes take place in the room?[/quote]

[hide="Solution and Example"]This is an interesting problem:

I saw it at MATHCOUNTS once:

They shake hands with only the ones taller than them, so...

Example:

Bob and Sam - Bob is shorter, so he wants to shake hands with Samuel. Unfortunately, Samuel is taller than Bob, so he won't shake hands with Bob.

This is the basic logic involved in this problem. It is quite amusing.

Answer:

$\boxed{\boxed{\boxed{\text{0 handshakes take place.}}}}$[/hide]

[hide]They can never shake hands because they would both have to be taller than eachother :D [/hide]

Okay, I'm am dumb, but I am not that dumb. Simple logic people, the answer is zero!

That was not as hard as it looked. :D

[hide]0 [/hide]

The question is tricky  :roll:

0  because is I said so.  And because it's the answer. and because if person A is shorter than B, A wants to shake B's hand but B can't shake A's.

 :starwars:

Hmmdi...

[hide] If I shake hands with you, you have to be taller than me... AND, I have to be taller than you...

So - Impossible?   :ewpu: [/hide]

It is obvious that it's zero since when hand shaking, two peole are involved, a taller one and a shorter one.  However, a tall person cannot shake with a shorter person, so $\boxed{0}$.

Masoud Zargar

[quote="easyas3.14159..."]buzzer11, isnt there a limit on how many smilies to a post? :P jk.[/quote]

Yea...I have noticed that I can only put 2 smilies to a post......how did you put in so many?
Also....the Tough Handshake Problem isn't that hard if you understand it.

There can be no handshakes!  There will always be a taller and shorter person


function
real analysis
real analysis solved

Let $T:C(\mathbb{R}) \to C(\mathbb{R})$ be linear map such that $\phi_1=\phi_2$ on any interval $I$ implies $T(\phi_1)=T(\phi_2)$ on $I$. Prove that $T \phi(x)=f(x)\phi(x)$ for some $f(x) \in C(\mathbb{R})$

Is $I$ the unit interval, or a fixed interval, or is it true for any interval?

I'm sorry for being not accurate: $I$ is supposed to be any interval from $\mathbb{R}$

By linearity, $T(0)=0$. Let $f=T(1)$, where $1$ is the constant function $1(x)=1$.

For any continuous function $p(x)$ and any real $a$, we can decompose $p(x)=p(a)+g(x)+h(x)$ where $g(x)=0$ if $x\ge a$ and $h(x)=0$ if $x\le a$.
Then $T(p)(x)=p(a)f(x)+T(g)(x)+T(h)(x)$. Since $g(x)=0$ for $x\in [a,\infty)$, $T(g)(x)=0$ for $x\in [a,\infty)$. Similarly, $T(h)(x)=0$ for $x\in (-\infty,a]$.
Evaluating at $a$, we have $T(p)(a)=p(a)f(a)$. Since this was true for all $a$ and $p$, we are done.

This isn't very dependent on the real numbers.


Ya sure he is a nice guy, though we once had a fight over something here.

who is this guy/girl and what did (s)he do to u?

Yeah sure. You should let anybody view you blog.


[hide]r and s should be the largest primes below 15, and t should be the smallest.
$\frac{11+13}{2}=12$[/hide]

[quote="aznmonkey1992"]if r,s,and t are all distinct prime numbers and are lower than 15, whats the largest possible value of (r+s)/t?[/quote]
[hide]
prime numbers less than 15 are:
2,3,5,7,11,13

r+s must be as big as possible, so it's 11+13 = 24
t must be as small as possble, so its 2.

24/2 = 12
[/hide]

[quote="aznmonkey1992"]if r,s,and t are all distinct prime numbers and are lower than 15, whats the largest possible value of (r+s)/t?[/quote]

[hide]
(r+s)has to be highest. t has to be lowest. lowest prime =2. highest primes are 11 and 13. $\frac{11+13}{2}=12$[/hide]

[quote="aznmonkey1992"]if r,s,and t are all distinct prime numbers and are lower than 15, whats the largest possible value of (r+s)/t?[/quote]
not too hard...
solution[hide]first, obviously, get the top two prime #s, which are 11 and 13. then, get the lowest prime which is 2.(11+13)/2
                                                                 = 24/2=[b]12[/b][/hide]

[hide]
the numerator should have the highest primes under 15 which is 11 and 13. The bottome should have the lowest prime which is two.  (11+13)/2=$\boxed{12}$[/hide]

[quote="anirudh"]I thought joejia wrote this.???>[/quote]
is that relevant?whos joejja

[hide]t has to be as small as possible, and r+s has to be big. (13+11)/2 = 24/2. the answer is [color=indigo]12[/color].[/hide]

[quote="anirudh"]I thought joejia wrote this.???>[/quote]
it doesn't matter who wrote this.


Isaac attempts all six questions on an Olympiad paper in order. Each question is marked on a scale from $ 0$ to $ 10$. He never scores more in a later question than in any earlier question. How many different possible sequences of six marks can he achieve?

it's the same as the number of multisets of 6 elements using only elements from $ \{0,1,...,10\}$. this is because each set of scores forms a multiset and each multiset can be put in non-increasing order in a unique way. the number of multisets in question is the same as the number of solutions of $ x_0 \plus{} \cdots \plus{} x_{10} \equal{} 6$ in non-negative integers (the bijection being, $ x_0$ represents the number of 0's in the multiset, ..., $ x_{10}$ the number of 10's). by your typical "bars and shorts" method (or "balls an urns", or whatever you call it), this number is $ \binom{6 \plus{} 11 \minus{} 1}{6} \equal{} \binom{16}{6}$.


Since exactly right now,the latexed text,numbers appear to be like thin(not as dark as before).Is there a sudden problem in my comp or is it AoPS?

I'm getting the same thing. Strange thing is, it only affects certain strings of text. Look:

$AB = CD$

$BA = CD$

pheww.... and i thought it was my computer's problem... :lol:

I noticed the same thing.  I think it looks better this way.

I don't like it at all. It's harder to read. :(

I think this is just some weird thing that happened on our server last night.

$\text{This is a test}$

We are porting to a new server in the next 1-2 weeks, which should wipe this out entirely.

Ohh! I though it was my PC, and I edited one post like 6 times!!!

yeah, this font really annoys me. I'm glad that the AoPS staff is working on fixing it :) , hopefully it's not more than a week  :(

[quote="joml88"]I think I might have accidentally caused this  :blush:   Last night I was messing around with the verbatim command (\verb).  \verb prints out exactly what is typed so something like a_i is shown just as that (i.e. the i isn't shown as a subscript as in $a_i$).  It also makes changes the font to typewriter typeface.  After I tried \verb it seems like everything is being printed in the typewriter style (though, thankfully, it is being compiled properly).  Well, I deleted the \verb command I used last night since it wasn't what I wanted.  So I dunno...

I could be wrong and maybe this is just a really big coincidence that I did that last night and right after that the $\LaTeX$ is in tt font.

Oh, I imagine that some things still look fine because they are cached as imaged.  That is,  the image is pulled up from the cache instead of a new one being produced.

I informed Valentin of this, by the way.[/quote]Don't worry, I accidentally put verbatim command last year and it did nothing, so it can't be from your 'messing around' :)

:o  So it was you all this time joml,he was right infront of our eyes and we did nothing  :D

O.K i promise,no more spamming here  :roll:

Too many smilies in your post: 3.
You can have at most 2 smilies in this message. Please remove some smilies before submitting this message.

[quote="joml88"]Perhaps the timing was just a really big coincidence.  I dunno  :?[/quote]Don't worry, it wasn't your fault. It was just merely a coincidence  :)

I think that if you put everything in quotes then you're not going to get these weird fonts.  ;)

You don't really understand how the systems works, do you? ;)

This has nothing to do with syntax (quotes, brackets, whatever), but with the program that calculates the gray values of the images. Somehow that thing altered.
[ or regarding to the smaller \sum, with an old image which hasn't been deleted from the cache ]

Perhaps the font type has changed. Maybe the admins want to check the font type used because I see that the font type now is very similar to Garamond font, it used to be Times New Roman...

[quote="Peter VDD"]You will delete these ugly misshaped images after this, right? :P

Just asking as it didn't happen with $\sum a_i$ vs. $\sum$ (which is not the recent $\sum{}{}{}$, but an older thing from before \displaystyle was added.)

[ do you actually plan to ever delete those pre-displaystyle pics? ][/quote]Yeah, all the pictures will be regenerated in about a week or so. Again don't worry about it, it's a very temporary situation :)

The problem is still not completely solved by now......  :(  :?

Completely not solved is a more appropriotate word. :)

But I guess it won't take too long anymore now?

So when are we going to get the old LaTeX font back?

[quote="rrusczyk"]We hope to move to a new server this week, at which point the LaTeX problem will be solved.[/quote]
Any progress?  :?

[quote="rrusczyk"]We hope to move to a new server this week, at which point the LaTeX problem will be solved.[/quote]
any progress  :?

Edit: This is a doublepost. I don't know how it happened. So a moderator please delete it.

To Peter: I don't know that you can reponse in such a short time...... :P

Not within the last 2 minutes, no. :P [size=0][/quote][/size]

We will be on a new server soon (a matter of days now) - the LaTeX will then (hopefully!) be fixed.


Given a [tex](2m+1)\times(2n+1)[/tex] checkerboard in which the four corners are black squares, show that if one removes any one white and two black squares, the remaining board is coverable with dominoes (i.e., [tex]1\times2[/tex] rectangles).

I'm pretty sure my answer doesn't count as a proof but it might be a good start:

For a (2m+1)(2n+1) checkerboard colored in the usual fashion where the four corners are black squares we know that there are an odd number of squares and therefore one more black square than white (because the color alternates every square).  So removing two black squares and one white square leaves an even number of both black and white squares remaining.  Because one domino covers two adjacent squares one black and one white, every time one domino is placed on the board we are again left with an even number of black and white squares and can continue this process until the board is covered.

Well, I hope most of that is right.

Unfortunately, that only doesn't prove that you can't do it -- it doesn't prove that you can.

I remember once solving a related question.  Given a usual 8 by 8 chessboard, with one white square and one black square removed, prove that you can cover the remaining squares with dominos.  I will put an outline of its solution below, in case it is of use for this more general problem.

Spoiler: [hide]First, create a Hamiltonian cycle of all the squares on the chessboard.  (A Hamiltonian cycle means a path starting at one square, visiting all the rest exactly once, and then ending at the original square.)  Then imagine one white and one black square removed.  That divides the Hamiltonian cycle into two paths each with an even number of vertices.  But it's easy to cover such paths with dominos.[/hide]

this might be of some use:

http://www.cut-the-knot.org/do_you_know/chessboard.shtml

Below is a rough sketch of something cumbersome I think might work. Assume first of all that the board can be tiled if only one black square is removed. I think that can be proven fairly easily but I'm too lazy to think about it. BUT if it can be done:

[hide]Construct the correct tiling. Begin by removing only one of the black squares, and tile the board. Now, remove the domino over the white square. Thus there is an uncovered black square.

We rearrange the dominos in the following fashion: suppose the empty square is in square (a,b). Remove the domino covered (a+1, b) and place it over (a,b) and (a+1,b). Now there is a different black square uncovered, farther to the right than the first. Repeat this process until the uncovered square is in the column farthest to the right.

Now, by a roughly similar argument with some modifications, we can rearrange stuff until the empty square is in one of the corners. Then do basically the same thing with the second removed black square, with some modifications for specific cases, and you can get the created empty white square next to the corner where we put the empty black square, and put a domino over both of them.[/hide]

Uh, that's the opposite of rigorous, but I think it works once you fill in the blanks. I'm just too lazy to do that now. There's probably a better way in any case.

I'm BUMPing this because I really like the problem. Eugevir had an interesting idea, but I don't know if I would be able to make it rigorous. Anyway, there are other ways it can be done. Hint in spoiler: [hide]try induction [/hide]


ARML
geometry
3D geometry
search
USAMTS
AMC
USA(J)MO

[quote="Silverfalcon"]
I was thinking if AoPS people in Penn State site can meet in some place. It's my first year so I don't know which is a good place to meet (and time as well). If anyone has a good idea, we should meet and at least talk for little.
[/quote]

That's an excellent idea!

I think the best time would be some time after Individual, since by then we won't have all that tension.

[color=red][size=200]Mod Edit: Meeting Place Poll Closed
Meet At:

Bigler Avenue Swimming Pool
Friday 7:30 PM [/size][/color]

well, is anyone going a day early? we (the SC team) are, so we have all of Friday.
We will be wearing our navy blue SC shirts, and I'm the indian kid with the glasses, um, ok, so that's about 4 possibilities on our team :P

Yes, we (Upstate New York) will bew going a day early. I am the guy with the long hair.

I think everyone has the be there Friday night.  But yea, I'm the indian guy with long hair that used to wear glasses.

yeah, that's a cool idea, i'm going with upstate new york, and i'm the one with all the rubik's cubes.  oh, that could describe a lot of people.  i'm the girl with 4 rubik's cubes that are definitely amazing.

I will be the Canadian.

Search me on College Facebook.  Self-explanatory.

[quote="Silverfalcon"]I think Friday should be a good time.. What time of Friday do you guys prefer most? (and most importantly, where at?)[/quote]

good . :P

:|

Shobhit, stay focused.  We're talking about meeting at Penn State.  ARML is on June 2nd and 3rd.

I remember last year in the building where we have breakfast at there was some free time.  Perhaps we can meet up then and meet on the lower floor (the upper floor is the cafeteria).  I suppose we could meet up after the individual round...but that's really hard since they have everyone in an auditorium and usually everyone stays in the same seats until they leave after awards.

I think Friday night at around 11 should be good.

11?? isn't that late? well it's late for me but everyone else probably sleeps at 11 every day... :blush:

I thought hspotter was kidding...?  In either case, the doors to the dorms are locked at 11 PM so unless we're in the same dormitory that would be impossible.

huh, who sleeps at 11???  Honestly now...that's the craziest thing I've heard in a long time.

I wasn't kidding.  But I thought people stayed up past 3, playing ultimate or something.  But yea, if the dorms are going to be locked at 11, then we should meet at 8.

[quote="Kent Merryfield"][quote="paladin8"]Well it's going to be 100+ degrees in Las Vegas! Whee![/quote]
Somehow I don't think paladin8 understood when he wrote that that it would be 100 degrees [i]indoors[/i] during the individual and relay rounds[/quote]

Yeah... that was not cool. I mean, I can live with 100 degrees outside fine, but when we have to sit for 3 hours without air conditioning...

[quote="Philip_Leszczynski"]Something is very, very strange...

I don't recall being the author of this thread.[/quote]

Yes, when I closed the poll, it somehow made you the author of the thread, totally killing Silverfalcon's post.  I don't know if this is a glitch or a human error.  I would believe it is a glitch, and I will ask about it.

[quote="K81o7"]Guys at Penn State, I apologize for not being at the meeting...i could not find the pool (and the NYC team meeting went until 8).  Anyways, most of you probably saw me anyways for the tiebreaker round  :D[/quote]
Don't worry- only about 8 people arrived at the pool
and Congratulations!

Actually, I was at the pool.  But it was raining, and I couldn't find anyone.

Yeah ... I got there late (around 6) , then had to unpack, get something to eat, and then attend the lecture.

3 w3s there to3. Three h3d the number 3 written on h3s shirt to3 if anyone n3ticed.

P.S. The number of posts i have divided by 3 is 3

No, it is 4.

NC got in late, and we couldn't even find our dorms until after 7, so by the time I had been downtown to eat and gotten back, it was 8:30, so I played ultimate with some of the other NC people.

[quote="Boy Soprano II"]Kind of a pity not more people showed up.  It was fun anyway.  There were B-tone_party, Phillip_Leszczynski, perfectnumber628, frankthetank, Xantos C. Guin, solafidefarms, and myself, in addition to someone claiming to be no-one (by which I think he meant he wasn't registered).  I was fairly true to my appearance, ironically minus the umbrella (Karth absconded with it, thinking that I had left it behind some place).  Actually, there seemed to be a lot of AoPSers at the song contest preliminaries anyhow.[/quote]

I would've showed up but rain killed my spirit.

By the way, next year, we should meet up at Findlay Commons. That should be a good area because there are always a lot of people there and obviously, rain doesn't come there.

All the NC AoPSers were eating... that's why we won :D
Not really...

Personally, I'd prefer that intersection of the walkways between the dorms. The AoPSers ended up going back to a dorm for shelter anyway, so might as well meet nearby.

[quote="Boy Soprano II"]I second that.  Also, there is food there.  (Xantos C. Guin can tell you about our "dinner" that evening.)[/quote]

Our coach said that we had to have a group of 3 to go downtown and eat. Boy Soprano II, solafidefarms, and I went to the AoPS meet after our practice instead of dinner. solafidefarms didn't want to eat dinner, and everyone else on our team already ate, so Boy Soprano II and I decided to eat dinner at the commons. We were 30 minutes late for dinner, so we ate the chips and cookies laid out near the song preliminaries. That wasn't the most nutritious dinner, but it was edible and digestable. (unlike breakfast.)

Huh what was wrong with breakfast?  Until you've had the food they served us at our senior breakfast at my high school, I don't think you really know what bad food is :P  And plus, the muffins were good...and I assume the donuts were good although I didn't try any.


geometry
parameterization
vector
advanced fields
advanced fields unsolved

How would I prove that curves with constant curvature and no torsion must be circles?

Let $\kappa (s)=k$. From the Serret-Frenet equations we have:

$\dot{T}=kN$ and $\dot{N}=-kT$, so differentiate the first one (assuming we're in a natural parameter): $\ddot{T}=k\dot{N}$. Now we substitute in to get $\ddot{T}=k(-kT)$ or $\ddot{T}=-k^{2}T$ but we can now integrate to get $\dot{T}=-k^{2}x(s)+C$ or $\frac{d^{2}x}{ds^{2}}+k^{2}x(s)=C$. Now solve the (vector-valued) differential equation to get a circle.

without using DEQ you can also prove that the locus of the curvature centres is constant.

Let $ \alpha(s)$ your plane curve, s=natural parameter, for which $ \kappa(s) = \| \alpha''(s) \| \equiv k$.

Fix s, the curvature centre is given by $ C(s)= \alpha(s)+\frac 1{ k}\mathbf n(s)$ (n is the normal versor). First of all C(s) is a constaint point C (differentiate C(s) and use the frenet equation to see this). Then $ f(s)=\| \alpha(s)-C\| = \frac 1{k}\|\mathbf n(s) \|$ is constant since n(s) is a versor. So forall s, alpha(s) has a constant distance from a fixed point C and this distance is the radius R=1/k. Note: I didn't put the abs value for k because I'm used to define it as $ \|\mathbf t'(s) \| \geq 0$
bye :)


Solve equation
$ 1)2x \plus{} 1 \plus{} x\sqrt {x^2 \plus{} 2} \plus{} (x \plus{} 1)\sqrt {x^2 \plus{} 2x \plus{} 3} \equal{} 0$

$ 2)12\sqrt x \plus{} 2\sqrt {x \minus{} 1} \equal{} 3x \plus{} 9$

$ (1)$ [hide]Let $ y\equal{}x^{2}\plus{}2$.  Solve for $ y$, and then solve for $ x$.[/hide]
$ (2)$ [hide]Square both sides.  Subtract like terms.  Square again.  Solve for $ x$.[/hide]

hello, rewriting your equation $ (1)$ in the following form $ (x\plus{}1)\sqrt{x^2\plus{}2x\plus{}3}\equal{}\minus{}(2x\plus{}1\plus{}x\sqrt{x^2\plus{}2})$ and squaring both sides and simplifying we get
$ 4x^3\plus{}2x^2\plus{}4x\plus{}2\equal{}(4x^2\plus{}2x)\sqrt{x^2\plus{}2}$, squaring again and simplifying we get $ 16x^2\plus{}16x\plus{}4\equal{}0 \Leftrightarrow 4(2x\plus{}1)^2\equal{}0$ which gives us $ x\equal{}\minus{}\frac{1}{2}$, this fulfilles our equation.
Sonnhard.

hello, squaring both sides of your equation $ (2)$ we get $ 48\sqrt{x}\sqrt{x\minus{}1}\equal{}9x^2\minus{}93x\plus{}85$, squaring again, putting all terms to the left-hande side and factoring we get
$ (9x\minus{}25)(x\minus{}1)(9x^2\minus{}154x\plus{}289)\equal{}0$.
Only $ x_1\equal{}1$ and $ x_2\equal{}\frac{77}{9}\plus{}\frac{16}{9}\sqrt{13}$ are solutions of our equation.
Sonnhard.


http://books.google.com/books?id=AA-lOA1nPDcC&pg=PR8&dq=putnam+competition

If you go down to page 53 of the book, you'll see a problem about an ordered triplet of sets when their union is {1,2,..10} and their intersection is the empty set.  The first solution talked about a bijective proof between the number of matrices and the number of ordered triplets of subsets.  I understand the concept of bijection, but I don't get how matrices come into the picture and why they can be used to do bijective proofs; I haven't seen the concept of bijection enough to really see how it can be applied to specific problems.

Thanks for the help!

I think you're laboring under several misconceptions. This isn't really a matrix problem or a matrix solution - no linear algebra was used at all. And there's no theory of "how matrices are related to bijections" to learn. What there is to learn is principles of counting and ideas in counting. Note also that the book excerpt gives a second solution which involves Venn diagrams rather than matrices. To my eyes the two solutions are virtually identical. They're phrased in different language but they're the same mathematical idea.

Let's back up. This is a counting problem. We want to know the cardinality of a set $ E.$ We come up with the idea that this set is somehow the same as set $ F,$ in disguise, and we might have an easier time counting $ F.$ What do I mean by "really the same set in disguise"? For that we need a bijection. And it's likely that we'll check that it is a bijection constructively. That is: suppose we have an arbitrary element of $ E.$ Verify that we can unambiguously identify an element of $ F$ to connect it to. And reverse that: start with an arbitrary element of $ F$ and verify that we can unambiguously identify an element of $ E$ to connect it to.

The matrix involved is an [i]incidence matrix[/i] - an array of zeros and ones with a $ 1$ when something (a membership or a connection) is valid and a $ 0$ otherwise. Such matrices can be a very useful notation in a counting problem. They're seldom necessary; we could usually also count in some other way. But they might be the nicest way to represent things.

So everyone else doesn't need to follow the link:

The problem is to count the number of ordered triples of sets $ (A_1,A_2,A_3)$ such that $ A_1\cap A_2\cap A_3\equal{}\emptyset$ and $ A_1\cup A_2\cup A_3\equal{}\{1,2,3,\dots,10\}.$ Let the set of such things be the set $ E.$

The proposed solution talks about set $ F,$ which is a set of $ 10\times 3$ matrices of zeros and ones. Each of the three columns of this matrix stands for one of the sets $ A_1,A_2,A_3$ and each of the rows stands for one of the elements of $ \{1,2,3,\dots,10\}.$ We have a one if that element is in that set and a zero otherwise. No element can belong to all three sets, so there are no $ 111$ rows, and no element can belong to none of the sets, so there are no $ 000$ rows. But given an ordered triple of sets I know exactly how to build the matrix and given a matrix I know exactly what the three sets are - if I can say that, that's a bijection. Then we count $ F$ to get that there are $ 6^{10}$ elements.

Now to address the main issue. Being able to follow that solution isn't worth all that much by itself. In the whole structure of solving this problem, being able to follow and verify that solution is maybe 10% of the work and effort involved. 90% of the work and effort came in coming up with the idea of that solution. Following it and verifying it once you have it - for an idea like this, that's supposed to be quick. And it needs to be quick - you also needed to have the time to verify that the dead-end ideas you had before you came up with this one [i]don't[/i] work. Correct solutions tend to emerge at the end of a process of trying, testing, and winnowing ideas. You need ideas to try - lots of ideas. Coming up with an incidence matrix is just one possible idea among many. When will it work? The answer to that is that "it works when it works." That might sound like doublespeak, but it's really about the messy nature of problem solving. You have to just try things.


vector
function
Functional Analysis
real analysis
real analysis unsolved

I find this absolutely intractable, even though at first glance it appeared as a standard density problem: is it true that the vector space spanned by the functions of the form $f(x)=[a/x]-a[1/x]$ where $a$ belongs to $(0,1)$ is dense in $L^{2}((0,1))$?  :idea: Here $[\cdot]$ is the integer part.

IT IS INTRACTABLE! Do not try to solve this! Or you may try... but then you would prove RH! Indeed, Nyman has shown that this problem is actually equivalent to the Riemann Hypothesis! Oh, my God, and when I think I was happy I found a nice functional analysis problem...

weird,  was it listed in the problem book? :)

Biler, Piotr /  Nadzieja, Tadeusz /
Problems and examples in differential equations
1992


From the title, I assume you meant the product of all [i]odd[/i] positive integers.

[hide]
$1\cdot 3\cdot 5\cdot... 9999$

$=\frac{1\cdot 2\cdot 3\cdot...\cdot 10000}{2\cdot 4\cdot 6\cdot... \cdot 10000}$

$=\frac{10000!}{(2\cdot 1)(2\cdot 2)(2\cdot 3)...(2\cdot 5000)}$

$=\frac{10000!}{2^{5000}\cdot 5000!}$

$\boxed{\text{D}}$
[/hide]

[hide="Answer"]The product of the odd integers is equal to the product of all integers divided by the product of the even integers. The product of $n$ even integers is $2^n(n!)$, and we have $5000$ even integers in the first $10000$, so we have $\frac{10000!}{2^{5000}(5000!)}\Rightarrow \boxed{D}$.[/hide]


calculus
algebra
polynomial
blogs
analytic geometry
graphing lines
slope

This problem has been knocking around in my head for a while and I finally figured out a very nice solution.  I want to see some thoughts on this.

Describe all polynomials $ P(x)$ such that $ P(x\plus{}1) \minus{} 1 \equal{} P(x) \plus{} P'(x)$.

[hide]Let $ P(x) \equal{} a_n x^n \plus{} a_{n\minus{}1} x^{n\minus{}1} \plus{} \ldots \plus{} a_1 x \plus{} a_0$ with $ a_n \neq 0$. Suppose $ n > 2$. By the binomial theorem, the coefficient of $ x^{n\minus{}2}$ on the LHS is $ a_n {n \choose 2} \plus{} n a_{n\minus{}1} \plus{} a_{n\minus{}2}$. On the RHS, it is $ a_{n\minus{}2} \plus{} n a_{n\minus{}1}$. This implies that $ a_n \equal{} 0$, contradiction. So $ n \leq 2$ and $ P(x) \equal{} ax^2 \plus{} bx \plus{} c$. Substituting this in, we get $ ax^2 \plus{} (b \plus{} 2a)x \plus{} (a \plus{} b \plus{} c \minus{} 1) \equal{} ax^2 \plus{} (b \plus{} 2a)x \plus{} (b \plus{} c)$, or $ a \equal{} 1$ by cancelling. So all polynomials are $ x^2 \plus{} bx \plus{} c$ for any reals $ b,c$.[/hide]

Interesting method, but there's a slightly simpler solution.

Here's the solution I posted in miyomiyo's blog:

[quote="me"][hide="3"]
First I tried out some examples
For $ P(x) \equal{} c$, it obviously doesn't hold since $ c\ne c \plus{} 1$
For $ P(x) \equal{} cx \plus{} d$, we have $ cx \plus{} c \minus{} 1\ne cx \plus{} c$.
For $ P(x) \equal{} x^2$,
$ P(x \plus{} 1) \minus{} 1 \equal{} (x \plus{} 1)^2 \minus{} 1 \equal{} x^2 \plus{} 2x \plus{} 1 \minus{} 1 \equal{} x^2 \plus{} 2x \equal{} P(x) \plus{} P'(x)$.
For $ P(x) \equal{} x^2 \plus{} x$, it also holds. In fact, it holds for $ P(x) \equal{} x^2 \plus{} c_1x \plus{} c_2$ in general.
For $ P(x) \equal{} 2x^2$, we have $ 2x^2 \plus{} 4x \plus{} 2 \minus{} 1\ne 2x^2 \plus{} 4x$. It doesn't hold!
For $ P(x) \equal{} 2x^2 \plus{} x$, we have $ 2x^2 \plus{} 5x \plus{} 2\ne 2x^2 \plus{} 4x \plus{} 1$. It doesn't hold!
For $ P(x) \equal{} 3x^2$, we have $ 3x^2 \plus{} 6x \plus{} 3 \minus{} 1\ne 3x^2 \plus{} 6x$.
For $ P(x) \equal{} x^3$, we have $ x^3 \plus{} 3x^2 \plus{} 3x\ne x^3 \plus{} 3x^2$.
Conjecture It only holds for: quadratics with $ a_2 \equal{} 1$.
We have already shown that no zero-degree and fisrt-degree functions satisfy this. For quadratics, we get $ cx^2 \plus{} 2cx \plus{} c \minus{} 1 \equal{} cx^2 \plus{} 2cx\Leftrightarrow x \minus{} 1 \equal{} 0\Leftrightarrow x \equal{} 1$. We now prove by induction that in degree three or higher, it doesn't hold. We have already shown it for $ \deg [P(x)] \equal{} 3$, so now we assume it for $ \deg [P(x)] \equal{} n \minus{} 1 > 3$. The $ a_n(x \plus{} 1)^n$ term would contain a $ a_nx^{n \minus{} 2}$ term, which obviously could not be in the RHS unless $ a_n \equal{} a_{n \minus{} 1}$. We continue this logic all the way to $ a_0$ (the constant term of the polynomial) and we see that not all of these can hold. Thus, the quadratic specified is the only possibility and the conjecture holds. QED.

Note: You can see why my logic can only be applied if the degree is greater than or equal to $ 3$, right?[/hide]

Took almost half an hour of work... really nice problem.[/quote]
Miyomiyo's solution is a bit less elementary.

The High School forums are not the place for calculus.  You should at least define the formal derivative.

[hide="Solution (reworded)."] Note that $ (x \plus{} 1)^2 \minus{} 1 \equal{} x^2 \plus{} 2x$, so $ x^2$ is a solution (is this the motivation?).  Given any other solution $ P(x)$, the polynomial $ Q(x) \equal{} P(x) \minus{} x^2$ satisfies the homogeneous condition

$ Q(x \plus{} 1) \minus{} Q(x) \equal{} Q'(x)$;

in other words, the slope of the tangent line at $ (x, Q(x))$ is equal to the slope of the secant line from $ (x, Q(x))$ to $ (x \plus{} 1, Q(x \plus{} 1))$.  It follows that the second derivative is zero infinitely often, so is identically zero, and $ Q(x)$ can then be any linear polynomial.  The conclusion follows. [/hide]


Show that for each natural number $k$ there exist at most finitely many triples of mutually distinct primes $p,q,r$ such that$p|qr-k,q|pr-k,r|pq-k$

I think i can help you,and i also believe in the following statement:
  We have only  finite number of triples $ (p,q,r)$ of natural numbers with the property that:
                              $p|qr-k,q|rp-k,r|pq-k$
 with provided integer $k$.
  I'm expecting arguments about this problem.

Let take $p>q>r$  $>$ $2$ are primes.

Assume that we have :$p|qr-k,q|rp-k,r|pq-k$ so $pqr|pq+qr+rp-k$

But $-k<pq+qr+rp-k<pqr$,so $pq+qr+rp<=k$,which have only finite solutions in $N^3$.

Now if$r=2$,so $2pq|pq+2p+2q-k$ note two cases:

1.$q<k+2$ from $p<=2q-k$ imply the result.

2.$q=>k+2$ then $0<pq+2p+2q-k<2pq$,contradiction.


You are given $9$ points in a plane. Now draw $9$ straight lines using these points such that:
$a.$ Each line passes through exactly $3$ points
$b.$ Each line can pass through more than $3$ points.

I don't think this is possible in general. You can draw nine points in the plane so that no three of them are colinear (eg. the regular 9-gon). If no 3 points are colinear, then your 9-line construction is impossible.


If a, b, c > 0, a + b + c = 1,then:
$ ab\plus{}bc\plus{}ac \le \frac{1}{3 }$

We know that $ ab \plus{} bc \plus{} ca \leq a^2 \plus{} b^2 \plus{} c^2,$ since it is equivalent to $ 0 \leq \sum (a \minus{} b)^2$.
So
\[ 3(ab \plus{} bc \plus{} ca) \leq a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2bc \plus{} 2ca \equal{} (a \plus{} b \plus{} c)^2 \equal{} 1.\]

Another version is to take Mac-Laurin-Inequality:
\[ \sqrt{\frac{ab\plus{}bc\plus{}ca}{3}}\leq\frac{a\plus{}b\plus{}c}{3}\]
Squaring and plugging in the condition immediately gives the desired result.  :lol:

Variation (still easy):
Let $ a,b,c \in \mathbb{R}^ \plus{}$ with $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$. Show that
\[ a^2bc \plus{} ab^2c \plus{} abc^2 \leq \frac {4}{3}.\]

$ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{}2$, $ a^{2}bc \plus{} b^{2}ac \plus{} c^{2}ab \equal{} ab.ac \plus{} ab.bc \plus{} ac.bc\leq \frac {(ab \plus{} ac \plus{} bc)^{2}}{3}\leq \frac {4}{3}$


vector
linear algebra
matrix
linear algebra theorems

[color=darkblue]Let $\mathcal{A}$ be a nilpotent transformation on a vector space V.
Then V is a direct summ of cyclic subspaces under the $\mathcal{A}$.[/color]

Definition: Linear space V is cyclic under the $\mathcal{A}$ if $\exists a \in V : \\ a,a \mathcal{A}, a \mathcal{A}^{2}, \cdots , a\mathcal{A}^{n-1}$

are basis in V and n=dim(V).

P.S.: How one can name this theorem? In my old book Мальцев "Основы линейной алгебры" its called 'Первая теорема о разложение'.
 :blush:  4 bad english

Use the Jordan decomposition theorem to see that $ A$ is similar to a matrix with zeros everywhere but some ones above the diagonal. Now reinterpret this in terms of a corresponding basis of $ V$.

Privjet. :)


A particle P moves on a circle with centre O and radius b. At a certain instant the speed of P is v and its acceleration vector makes an angle a with PO. Find the magnitude of the acceleration vector at this instant.

Seems like some information is missing?

let the total accelartaion of the particle be $\vec a_{net}$
now it has a component $\frac{v^{2}}{b}= \vec a_{n}$ at that moment along $PO$
now $\vec a_{n}.\vec a_{net}=|\vec a_{n}\vec a_{net}| \cos a$
so $\cos a= \frac{\frac{v^{4}}{b^{2}}}{\frac{v^{2}}{b}|\vec a_{net}|}$
so $|\vec a_{net}|= \frac{v^{2}}{b \cos a}$

Thanks. Instead of a(net), I was thinking of the dot product with the position vector or something.


algebra
system of equations
algebra unsolved

Find all triples $ (m,n,p)$ of positive integers such that $ m\plus{}n\plus{}p\equal{}2007$ and the system of equations :
$ \frac{x}{y}\plus{}\frac{y}{x}\equal{}m, \frac{y}{z}\plus{}\frac{z}{y}\equal{}n , \frac{z}{x}\plus{}\frac{x}{z}\equal{}p$ has at least one solution in nonzero real numbers


[quote="silouan"]Έστω $E$ το εμβαδόν του τριγώνου και $R$ η ακτίνα του περ. κύκλου.
kai τ η ημιπερίμετρος

Νδο $4e^3RE\leq e^{2t}$[/quote]

Πράξεις και $e^{x-1}\geq x$

H τελευταία προκύπτει απο την ιδιαίτερα γνωστή $e^x \geq x+1$.

Ειπα να λύσω και καμια άσκηση τωρα το τριήμερο....... :D
 :)


Find all functions $f: (0, + \infty)\to(0, + \infty)$ such that for all $x,y\in (0, + \infty)$,
\[ f(f(x) + y) = x \cdot f(1 + xy).\]

[quote="cnyd"]$ \forall x,y\in\((0, + \infty))$
$ f(f(x) + y) = x.f(1 + xy)$
$ f: (0, + \infty))\rightarrow (0, + \infty))$[/quote]

Let $ P(x)$ be the assertion $ f(f(x)+y)=xf(1+xy)$

$ P(\frac{u}{f(2)},\frac{f(2)}{u})$ $ \implies$ $ f(\text{something})=u$ and $ f(x)$ is surjective.

Let then $ x>y>0$ :
If $ f(x)>f(y)$, the number $ \frac{f(x)-f(y)}{x-y}\in(0,+\infty)$ and so :

$ P(x,y\frac{f(x)-f(y)}{x-y})$ $ \implies$ $ f(\frac{xf(x)-yf(y)}{x-y})$ $ =xf(1+xy\frac{f(x)-f(y)}{x-y})$

$ P(y,x\frac{f(x)-f(y)}{x-y})$ $ \implies$ $ f(\frac{xf(x)-yf(y)}{x-y})$ $ =yf(1+xy\frac{f(x)-f(y)}{x-y})$

And so $ x=y$, hence contradiction.

So $ x>y>0$ $ \implies$ $ f(x)\leq f(y)$ and so $ f(x)$ is non increasing and, since surjective, continuous.

As a consequence (since non increasing surjective), $ \lim_{x\to +\infty}f(x)=0$

Let then $ z>0$ and $ x_0$ such that $ f(x_0z)<1$
Let then $ x>x_0$. We have $ f(xz)\leq f(x_0z)<1$ and so $ z(1-f(xz))>0$
$ P(x,z(1-f(xz)))$ $ \implies$ $ f(f(x)+z(1-f(xz)))=xf(1+xz(1-f(xz)))$
$ P(xz,1-f(xz))$ $ \implies$ $ f(1)=xzf(1+xz(1-f(xz)))$

And so $ f(f(x)+z(1-f(xz)))=\frac{f(1)}{z})$

Raise now $ x\to +\infty$. Since $ f(x)$ is continuous, LHS$ \to f(z)$ and we get $ f(z)=\frac{f(1)}{z}$
Putting back in the original equation, we get $ f(1)=1$

And so the solution is $ f(x)=\frac 1x$


determine with amathematical     prove[x] the greatest integer less than or equal to  x  where
1+1\sqrt2  +1\sqrt 3+....+1\sqrt 1000000

I'm assuming that $x=\sqrt1+\sqrt2+\sqrt3+\cdots+\sqrt{1000000}$?

i think it means reciprocals... that is, $x=\sum_{k=1}^{1000000}\frac{1}{\sqrt{k}}$

[hide="hint"]
\[\frac{2}{\sqrt{n}+\sqrt{n+1}}<\frac{1}{\sqrt{n}}<\frac{2}{\sqrt{n}+\sqrt{n-1}}\]
[/hide]