Find [b]p,q[/b] in prime number such that [b](5 ^p -2^p)(5^q-2^q)[/b] divisible by [b]pq[/b]. :harhar:  :gleam:

First of all if $p=3$ then easily $q=3$
Suppose that $p,q\ge5$ then $5^q\equiv 2^q(mod p),5^{p-1}\equiv 2^{p-1}(mod p)$(by Fermat's little theorem)
---------------------------------------
Lemma:If $x=f(p)$ is smallest integer s.t. $p|5^x-2^x$ and $p|5^m-2^m$ then $x|m$
Proof:from here we will always look modp.
Suppose that $x\ge 1\Rightarrow 5^x\equiv 2^x$ let  $m=ax+y,y<x$ then $5^{ax+y}\equiv 2^{ax+y}\Rightarrow 5^{ax}5^y\equiv 2^{ax}2^y\Rightarrow 2^{ax}5^y\equiv 2^{ax}2^y\Rightarrow 5^y\equiv 2^y$ contradiction.
So we must have $x|m$
--------------------------------------
So using Lemma we must have $x|q\Rightarrow x=q,q|p-1$
Similarly we have $5^p\equiv 2^p(mod q),5^{q-1}\equiv 2^{q-1}(mod q)\Rightarrow f(q)=x'=p,p|q-1$
so the eq. doesn't give solutions unless  $(p,q)=(3,3)$
I didn't consider some cases that were very easy.
May be I have got some mistakes? :?

If I remeber correctly, its from Bulgarian olympiads, 2001 perhaps.


ratio
trigonometry
calculus
integration
geometry
Pythagorean Theorem
trig identities

In $ \triangle ABC$, $ AB\equal{}5$, $ BC\equal{}7$, $ AC\equal{}9$ and $ D$ is on $ \overline{AC}$ with $ BD\equal{}5$.  Find the ratio $ AD: DC$.

uh

[hide]By the law of cosines $ \cos{\angle{BAC}}\equal{}\frac{19}{30}$, so the height of triangle $ BAD$ to $ AD$ is $ \frac{19}{6}$, then we find by the Pythagorean Theorem that $ AD\equal{}\frac{\sqrt{539}}{6}$, and $ DC$ is 9 minus that, then the ratio is whatever.[/hide]

I think you made a mistake somewhere, because there is a nice integral non-trig solution. :)

[hide]would it involve heron's formula?

i get the area is $ \frac{21\sqrt{11}}{4}$. draw a perpendicular from B to AC. let the point of intersection be E and the length be x. plug the numbers into the triangle area formula and get $ \frac{9x}{2}\equal{}\frac{21\sqrt{11}}{4}$. solve. i get the height is $ \frac{7\sqrt{11}}{6}$. use the py-thag theorom(one kid on our team calls it that), we get $ CE \equal{} 35/6$. subtract that from 9 and get $ AE \equal{} 19/6$. therefore, $ AD \equal{} 19/3$. then we get $ DC \equal{} 8/3$ so $ AD/DC \equal{} 19/8$

i probably messed up somewhere using the formulas[/hide]

[size=75][color=darkred]Hide your answers.
-nebula42[/color][/size]

blah you could also just use Stewart's Theorem...

[hide][quote="archimedes1"]Correct, abacadaea.
There is another solution which doesn't involve areas, and only uses the Pythagorean Theorem.  I think it's a little faster since you don't have to compute the areas.  [/hide][/quote]

Label $ AD \equal{} x$ and $ DC \equal{} 9\minus{}x$ and the height as y. Um $ \left(\frac{x}{2}\right)^{2}\plus{}y^{2}\equal{} 25$ and $ \left(9\minus{}x\plus{}\frac{x}{2}\right)^{2}\plus{}y^{2}\equal{} 49$ and just solve from there. [/hide]

[quote="abacadaea"]whats up with the

[size=75][color=brown]Hide your answers.
-nebula42 [/color]
[/size]
underneath my post[/quote]

A mod edited your post cuz you hadn't hidden your answer.

Obviously the mod edited your post to include the hide...


I found the following improved result
$a^5+b^5+c^5\leq \frac{1}{\sqrt{27}}$.

I have such idea:
Let us denote $x = \frac{ab}{c}\, y = \frac{ac}{b}\, z = \frac{bc}{a}$.
Then $ a^2 = xy, b^2 = xz, c^2 = yz $
Condition of the problem is equalent to $x^2 + y^2 + z^2 = 1$
We have to find $max((xy)^{5/2} + (xz)^{5/2} + (yz)^{5/2})$.

$ 1 = (\sum x^2)^3\ = \sum x^6\ + 6x^2y^2z^2 + 3\sum x^4y^2 + y^4x^2 $ > $4\sum x^4y^2 + y^4x^2\geq 8\sum x^3y^3 $
(by Schur Inequality)
And it is easy to see $ 1/3 = 1/3(\sum x^2)^2\geq \sum x^2y^2.$

$ 1/24$ > $(\sum x^3y^3)(\sum x^2y^2)\geq (\sum (xy)^{5/2})^2 $

$(xy)^{5/2} + (xz)^{5/2} + (yz)^{5/2}$ < $ 1/\sqrt{24}  $.

I will think about Vasc proposition.

I am thinking how to solve it but it seems not to much easy.

Your approach, Prowler is good, but how to improve it?

[quote="Vasc"]$a^5+b^5+c^5\leq \frac{1}{\sqrt{27}}$.[/quote]

Let $a = qr$, $b=pr$, $c=pq$. Then the constraint becomes $p^4+q^4+r^4=1$, and we are required to prove $p^5q^5 + q^5r^5 + q^5p^5 \leq 1/\sqrt{27}$. Homogenizing and squaring, the ineq becomes equivalent to
\[
\left( \frac{p^5q^5 + q^5r^5 + q^5p^5}{3} \right)^2 \leq \left( \frac{p^4 + q^4 + r^4}{3} \right)^5
\]
this certainly looks a lot nicer, but I can't yet find any nice way of proving it.

But expanding (with Maple), the above becomes
\[
(20,0,0) + 10(16,4,0) + 20(12,8,0) + 20(12,4,4) + 30(8,8,4) \geq 27(10,10,0) + 54(10,5,5)
\]
where $(m,n,l)$ denotes $\sum_\mathrm{sym} p^mq^nr^l$.

Weighted AM-GM gives:
\[
(20,0,0) + 3(16,4,0) + 20(12,4,4) + 30(8,8,4) \geq 54(10 \frac{4}{27}, 6\frac{4}{27}, 3\frac{19}{27})
\]

so the ineq reduces to
\[
7(16,4,0) + 20(12,8,0) + 54(10 \frac{4}{27}, 6\frac{4}{27}, 3\frac{19}{27}) \geq 27(10,10,0) + 54(10,5,5)
\]
which readily follows from Muirhead. (hope I didn't make any calculation errors)

Could you post your solution Vasc, I failed to prove your statement.
For billzhao post I can say only that it is monstrous.
I don't know but I have a small doubt that knowing that
$ x^n + y^n + z^n = 1$
we can find maximum of $ (xy)^{n+1}+(yz)^{n+1}+(xz)^{n+1} $
but maybe I am wrong.

Of course, my solution is identical with Harazis solution.

[quote="harazi"]I think that I can solve your problem, prowler. Thus, let us prove that if $ x+y+z=3$ then $ (xy)^{\frac{4}{3}}+(yz)^{\frac{4}{3}}+(zx)^{\frac{4}{3}}\leq 3$. This reduces immediately to the Vasc's problem to which I have already given a solution: if $ a^3+b^3+c^3=3$ then $\sum {a^4b^4}\leq 3$. Since $\frac{4}{3}\geq\frac{n+1}{n}$ we also have the inequality $\sum { (xy)^{\frac{n+1}{n}}}\leq 3$, which solves your problem and darkmaster's improved problem. For $ n=2$ I do not think we have problems in finding the maximal value, I guess.[/quote]

Harazi, I don't understand you. Where can I find Vasc's problem to which you already gave a solution?

[u][b]The author of this posting is : manlio[/b][/u]
____________________________________________________________________

[quote="harazi"]if $ a^3+b^3+c^3=3$ then $\sum {a^4b^4}\leq 3$.[/quote]

I remember Harazi solution of Vasc problem. It is nice.

Use AM-GM to write $a^4b^4 \leq a^3b^3 \cdot \frac{a^3+b^3+1}{3}$

now put $x=a^3$, $y=b^3$, $z=c^3$ and use Schur.

I remember that someone posted the claim that if $x, y, z$ are positive reals that add to 3 then

\[
x^ry^r + y^rz^r + z^rx^r \leq \max(3, 3^{2r} / 2^{2r})
\]
can someone give proof to this inequality?

A proof is done in Gazeta Matematica A, 1/2004.
A similar proof is in Crux Math. 1/2004 (page 44-46), too.


Prove that in any tringle $ABC$ the following inequality holds:

                \[ 3\sqrt{3} r\leq\frac{3[(ab+bc+ca)-p^{2}]}{p}.  \]

 Remark. The inequality above is stronger than $p\geq 3\sqrt{3} r$ which is due to Mitrinovic. ;)

Cool, I'll use $s$ instead of $p$ for semiperimeter.  Let $x,y,z$ denote $s-a, s-b, s-c$ where $a,b,c$ are the sides of triangle $ABC$.  The inequality is equivalent to

$\frac{27xyz}{x+y+z} \leq \frac{9(\sum (x+y)(x+z) - (x+y+z)^2)^2}{(x+y+z)^2}$

$3xyz(x+y+z) \leq (xy+yz+zx)^2 \iff \sum x^2(y-z)^2 \geq 0$

Nice ThAzN1. My solution is almost the same with a few differences. :)


geometry
3D geometry
sphere
function
calculus
calculus computations

I have no clue how to do this problem either... :( ive been doing reviews and of the 11 review questions this is one of the 4 that has stumped me

The circumference of a sphere was measured to be 84 cm with a possible error of 0.7 cm. Use differentials to estimate the maximum error in the calculated surface area.

Estimate the relative error in the calculated surface area.

Hint: The circumference of a sphere of radius r is C = 2 \pi r, and its surface area is A = 4 \pi r^2. Eliminate r first!

Eliminate $r$ first, to represent $A$ as a function of $C$. That is, $A=C^{2}/\pi$.
Then find the differential of $A$ as a function of $C$, which is $dA=2C\,dC/\pi$.
Then use the numerical data.


Suppose that f: [a, $ \infty)\longrightarrow\mathbb{R}$ for some a $ \in\mathbb{R}$. Prove that f(x) $ \longrightarrow$ L as x $ \longrightarrow\infty$ if and only if f($ x_n)\longrightarrow$ L for any sequence $ x_n\in$ (a, $ \infty$) that converges to $ \infty$ as n$ \longrightarrow\infty$.

$ (\implies)$
$ \lim_{x\rightarrow\infty} f(x) \equal{} L$ $ \implies$ $ \forall$ $ \epsilon > 0\exists M > 0$ such that $ x > M\implies f(x)\in V_{\epsilon}(L)$ ... (1)

Consider a sequence $ x_n$ wich diverges to $ \plus{} \infty$

In this sequence, $ \forall$ $ N > 0$,$ \exists$ natural $ n$ such that $ x_m > N$ $ \forall$ $ m > n$...(2)

Let $ \epsilon > 0$ be given. Then from (1) we have $ M > 0$ such that all $ x$ larger than $ M$ satisfy limit condition. From (2) we have that $ \exists$ $ n$ such that all $ x_m$ with $ m > n$ are greater than $ M$. Hence, sequence satisfies condition.

I still am working on the converse. :)


The number 5 x 7^44 has a great many digits in its decimal representation. Without evaluating those digits, prove that some digit occurs at least four times in the number.

k dont we just have to show that:

$7^{44}\cdot{5}$ has atleast 31 digits..thus that means that atleast one digit showed up atleast 4 times...

[quote="SM4RT"]k dont we just have to show that:

$7^{4}4\cdot{5}$ has atleast 31 digits..thus that means that atleast one digit showed up atleast 4 times...[/quote]

I think you mean $7^{44}\cdot5$. Enclose both 4's in brackets, like {44}.

[hide="Going off on that idea,"]$\log(7^{44}\cdot5) = \log 5+44 \log 7 = 37.88$.

Hence, the number has 38 digits. By Pigeonhole, etc.[/hide]


Hey, All there in India.
Gave the NSEP. How was it?
I gave it. Hope adequate marks in it.
Please inform how was your paper.

The following question I was not able to solve. Please help.

A gas expands when heated whereas it gets heated when compressed. From these two statements we see that
(a) the behaviour of the gas is anomalous
(b) the expansion and the compression can be adiabatic and isothermal respectively
(c) the expansion and the compression can be isobaric and isothermal respectively
(d) both the expansion and the compression can be either isothermal or adiabatic.

[b]Only one option is correct[/b]
My choice was (c). [Know that you all must be thinking I'm fool :oops: ]

It wasw a lenghty paper - but quite easy! Most subjective problems from Irodov!

Anyways, I marked (c) too in this question - but I think the question is unclear and probably wrong!

Hey varun, who are you ? -where do you live?

Hi Varun  :D

I too have marked (c).
You know how I performed in the test.

Bye  :ninja:

Hi Anjani
Nice seeing you in AoPS.
You may post the other nice problems from the test.


function
integration
limit
real analysis
real analysis solved

Find all continous functions [tex]f:\mathbb{R}\to\mathbb{R}[/tex] such that [tex]\int_a^bf(t)dt=(b-a)f\left(\int_a^bf(t)dt\right),\;\forall a,b\in\mathbb{R}[/tex].

$F(x)=\int_{0}^{x}f(t)dt$

The given relation is

$\frac{F(b)-F(a)}{b-a}=f(F(b)-F(a))$ for any a,b real

Fixed a,  let's tend b to a we have

$F'(a)=f(0)$

$f(a)=f(0)$ constant

another (completely equivalent) way is by using the MVT:

$\int_a^b f(t)dt = f(\xi)(b-a) = (b-a)f\left(\int_a^b f(t)dt\right),$

where $\xi \in (a, b)$ and $a<b$. Therefore

$f(\xi) = f\left(\int_a^b f(t)dt\right)$.

Take $\lim_{a\rightarrow b^{-}}$ from both sides and game over.

BTW ... The "Maybe easy" question seems a bit harder than the "Not so easy" one. Actually how have you ordered your set of problems? Just joking  :) Good problems, though! Thanks!


Another very interesting problem, any assistance is always helpful. Thanks.

In Euclidean space R^2, consider the subset D = {(x,y) belonging to R^2 | precisely one of the numbers x and y is rational}. Is D a connected subset of R^2? Is its complement (R^2) \ D connected?

How do you show this is the case?

$D$ is disconnected by the line $y=x$. The complement is connected, even by piecewise linear curves. Or maybe not.  :maybe:


Dr.Surgeryandnomistakes has performed brain surgery for 25 years and never made any mistakes and did every surgeries successfully 25,239,529 times....

what's the probabilities that he would make another successful surgery on the next patient?

1 because he is doctor no mistakes... :?

[hide]Assuming he is as fast as  [img]http://www.artofproblemsolving.com/Forum/images/avatars/45961890041e4a6671adbd.jpg[/img] finishing  his  food (1.2 seconds)  he is working 24 hours per day .. 7 days a week .. 365 days a year.. for last 25 years...!!! :)

No way he is going to make an error ! [/hide]

Well, what if I replace his scalpel with a rusty chainsaw?

just because he haven't made any mistakes so far doesn't mean he won't make any mistakes in the future....

If someone were to answer this he would have to make a huge number of assumptions.

Mamat, you got it!!

 :D

this is a very common trick question

again just because you never made any mistakes in the past, doesn't mean you will still make no mistakes in the future,

A+MATH, what do u mean?

[quote="shinwoo"]this is a very common trick question[/quote]It is a very common trick question, just not in the way that you think it is.

A+MATH is right.  You can't answer this question without making some underlying assumptions.

You are making the (false) argument that if we ask the question "What is the probability of event E" occurring, and we have no other information, then it is correct to say that the probability is 1/2.

Here's a little problem to hopefully convince you why this is not the right thing to do.

My car is in the parking lot.
What is the probability that my car is white?
What is the probability that my car is white and has 4 doors?

By your reasoning, since you have no other information, you would say that the answer to each question is 1/2.

However, if this were so, then you have just "proven" that there are no 2-door white cars!

I would say that if you flipped a coin 1,000,000 times, and it came up heads every time, that there is an overwhelming likelihood that the coin is weighted to come up heads every time, so the probability that the 1,000,001th flip would come up heads is very very close to 1.  Surely you would not argue that if you watched someone flip a coin 1,000,000 times, and it came up heads every time, that the probability of a tail on the 1,000,001th flip is 1/2?

On the other hand, if you had irrefutable proof that the coin was fair, then you would be correct to say that the probability of the 1,000,001th flip being a tail is 1/2.  But here you have some additional information.

Hm..

then I think I made a mistake here.....

so I was supposed to give more information that would convince people that humans make mistakes... :D

For sure, very few people would undergo surgery if the base probability of a surgical mistake is 1/2.

[hide]he either makes it or he doesn't so it is $\frac{1}{2}$[/hide]

[hide]
1/2 because each case is independent.
[/hide]


function
trigonometry
calculus
integration
calculus computations

Please, solve
$x^{2}y'=3(x^{2}+y^{2})\arctan{\frac{y}{x}}+xy$
with the method one uses when we have
$M(x,y)dx+N(x,y)dy=0$
where $M$ and $N$ are homogeneous functions of same degree.
How do you do it without getting messy expressions?
This kind of expression will show up after substitution by multiplying each argument in $f(x,y)$ with $1/x$ giving $f(1,z=y/x)$:
\[\frac{dz}{f(1,z)-z}=\frac{dx}{x}\]

[url=http://en.wikipedia.org/wiki/Homogeneous_function]by homogeneous functions i mean this[/url]

I am studying for my exam right now, but can you try to use polar form

Here we go:
Let $z = \frac{y}{x}$, then $y' = z+x\frac{dz}{dx}$.
We have:
$x^{2}y' = 3(x^{2}+y^{2})\tan^{-1}(\frac{y}{x})+xy$
$\Rightarrow x^{2}(z+x\frac{dz}{dx})= 3(x^{2}+(zx)^{2})\tan^{-1}(z)+x^{2}z$
$\Rightarrow \frac{dz}{3(1+z^{2})}\tan^{-1}(z) = \frac{dx}{x}$

Let $z = \tan\theta$, then we have:
$\frac{d\theta}{3\theta}= \frac{dx}{x}$
Now it is a matter of integration and using backward substitution


A slightly modified quote from Sysiphus:

"I'm still interested, but -, this is proving to be a pestilence."

I agree with elyot, that the use of the words pestilence is slightly missplaced ( although it still works if you ask me ).

[b]So lets play a game.![/b]

You finish this sentance, in the best way possible:
[i]
[b]"I'm still interested, but -, this is proving to be a _______."[/b][/i]

Brownie points for funny things that are clean yet horrible.

I'll start:

I'm still interested, but -, this is proving to be a bad way to pick up chicks.

I'm still interested, but -, this is proving to be a hemorrhoidectomy.

[quote="Sisyphus"][i]misplaced?[/i]

I created the English language, son  :mad:[/quote]
Hey, I never said I was any good at english!  :D  :rotfl:

I'm still interested, but -, this is proving to be a lot of work for a bad cup of coffee.
[hide=" :D "]I got it from this show called becker[/hide]

I'm still interested, but, this is proving to be a bad idea and I still not talk to you anymore.

recent personal experience, although not in words, but it was clearly. girls always break my heart. *sniff*

[quote="J-I-M-B-O"]I'm still interested, but, this is proving to be a [b]bad idea and I still not talk to you anymore.[/b] [/quote]

You have been victicem of evil moster GRAMBLOR, THE GOD OF GRAMMARZ.  :lol:   :rotfl:

[quote="rzsolt"][quote="J-I-M-B-O"]I'm still interested, but, this is proving to be a [b]bad idea and I still not talk to you anymore.[/b] [/quote]

You have been [b]victicem[/b] of evil [b]moster[/b] GRAMBLOR, THE GOD OF GRAMMARZ.  :lol:   :rotfl:[/quote]
And you have been the victim SPELLICUS, THE GOD OF SPELLINGS.
lol

I'm still interested, but -, this is proving to be a pricy bazooka!

I'm still interested, but-, this is proving to be a bad batch of viagra


I'm preparing for AMC-10. I saw many of mock AMC's and good classes but I wonder if anyone wants to do any of AMC Math Chat. I never took any of AMC's yet so doing something like that would give me a good time to see what part of math I need more studying on.

If we get a lot of people, we might separate them into AMC-10 and AMC-12 chat, or just get bunch of people who would do for both. Anyway, if interested, please do this:

1. Your IM sn
2. In favor of: AMC-10 or AMC-12
3. Are you more likely to: teach or learn

1. jwec05
2. AMC-10 but willing to do both sides
3. Learn

1.  BaseballJoe2002
2.  AMC 12  but I ll do 10 since its not very different
3.  It depends on who goes but maybe I d teach...maybe...

1)Feiqi J1989

2)AMC12

3)both(teaching is a way of learning :) ), but depends on who goes

Schedule for AMC Math Chat:

Sunday at 7:00 EST

Time may change but not date.

If you think you want to see the thing and not able to do anything but need a transcript, mail me.

g89unit@gmail.com

Silverfalcon

Edited: I had wrong mail address.

whitehorseking88

AMC 12

LEARN - I need to!

tetrahedr0n

AMC 12 = (cant take the other lol)

learn, of course.

If you have problems that you want to talk about, please mail them so I can update on my website and then we can save the time typing the problems. I think it's better idea so, thanks!

Mail them to my mail.

Math chat's today! So be prompt.

So far I got:

joml88
beta
h_s_potter2002
white_horse_king88
tetrahedr0n

Not on here but:

MysticTerminator
theone853

Transcript who will receive:

MathFiend
yif man 12

So be there!

What time is the math chat, and what will the room be called?

It's tonight at 7 PM EST.  The room will most likely be called Math Chat.  If you need any help IM any of the ppl who are listed above.

No, its not too late to sign up.  In fact, I would say that you can just show up and it would be fine.

I have the transcript for those who care.  It's very long so I m not going to post it here.  Also, I didn't edit it all... I started on the first page and realized it would take too long.  So if anyone wants it ask me by pm.  I was gone for some of it so there is quite possibly some inappropriate stuff...hopefully not  ;) .

I am NOT doing anymore of this AMC Math chat.

After several minutes, it went up to MOP question and it was just not "me."

Anyone can take this idea or trash it or do anything they want to do it.

But, I'm OUT.

brb.

math is a subject that's very essential for everyone

and u shouldn't just prepare for amc

don't study math to just prepare for tests or competition

think it as a subject that u need to know for rest of ur life

once again

don't study math to just prepare for some tests and competition

there might be mop questions in amc or usamoc or mathcount problems in amc test

for the one last time

don't study math to just prepare for tests or competition

[quote]there might be mop questions in amc or usamoc or mathcount problems in amc test [/quote]

Unlikely questions from MOP will show up in the AMC 10 or 12.

Yes. I agree. When we have these math chats, it would be nice to stay in the fair range where the majority of people can fully comprehend the problem at hand. As for me, I can't do MOP problems.


modular arithmetic
number theory
number theory unsolved

Hello, I don't know if this problem belongs to number theory forum or algebra forum. It was given as an exercise in an abstract algebra class

Find all primes $p$ such that $x^{3}-2 \equiv 0 \pmod p$ has a solution.

I will move it to Number Theory as it fits better there I think.

At first, for all $p \equiv 2 \mod 3$ it has a solution as $x \mapsto x^{3}$ is injective, thus surjective.
Now by Chebotarev's theorem, for one third of the primes $p \equiv 1 \mod 3$ we have $2$ to be a cubic residue $\mod p$ and we can prove the following, too:
It has a solution iff $p$ can be written as $p=a^{2}+27b^{2}$.


For any three positive reals a, b, c, we have

$\frac{2a}{b+c}+\frac{6b}{c+a}+\frac{3c}{a+b} \ge 4$

$ (\frac{2a}{b+c}+\frac{6b}{c+a}+\frac{3c}{a+b})(2a(b+c) + 6b(c+a) + 3c(a+b)) \ge (2a + 6b + 3c)^2$

I am thinking that this should solve it, but I have no energy to open the brackets at this time.

Okay, lets try to show that

$\frac {2a}{b+c}+2+\frac {6b}{c+a}+6+\frac {3c}{a+b}+3\geq 15$

or that

$(a+b+c)\left(\frac {2}{b+c}+\frac {6}{c+a}+\frac {3}{a+b}\right)\geq 15$

But by Cauchy, the LHS is $\geq (1+\sqrt{3}+\sqrt{3/2})^2>15$.

Equality is unattainable.


[quote="xpmath"]Find the value of $ \cos^2x$  such that
$ 6(36^{\frac {\sin^2x}{4}}) \equal{} 6^{\sin^2x}$.[/quote]
[hide="Solution"]
We call $ \frac {\sin^2x}{4} \equal{} y$.  Thus, we obtain that
\[ 6(36^y) \equal{} 6^{4y}
\]

\[ 6(6^{2y}) \equal{} (6^{4y})
\]

\[ 6^{2y \plus{} 1} \equal{} 6^{4y}
\]

\[ 4y \equal{} 2y \plus{} 1
\]

\[ y \equal{} \frac {1}{2}
\]

\[ \sin^2 x \equal{} 2
\]

\[ \sin^2 x\le 1
\]

\[ \boxed{\text{No Solution}}
\]
[/hide]

EDIT: Okay, here is a revised solution:
[hide="Revised Solution"]
\[ y \equal{} \frac {\sin^2 x}{80}
\]

\[ 6(36^{20y}) \equal{} 6^y
\]

\[ 6(6^{40y}) \equal{} 6^y
\]

\[ 6^{40y \plus{} 1} \equal{} 6^y
\]

\[ 40y \plus{} 1 \equal{} y
\]

\[ y < 0
\]

\[ y \equal{} \frac {\sin^2 x}{80}\ge 0
\]

\[ \boxed{\text{No Solution}}
\]
[/hide]

Oops, typo on my part, I'll fix it up immediately.

[quote="xpmath"]Find the value of $ \cos^2x$  such that
$ 6(36^{\frac {\sin^2x}{4}}) \equal{} 6^{\sin^2x}$.[/quote]
[hide="solution"]
$ 6((6^2)^{\frac {\sin^2x}{4}}) \equal{} 6^{\sin^2x}$
$ 6^{\frac {2\sin^2x \plus{} 4}{4}} \equal{} 6^{\sin^2x}$
$ 2\sin^2x \plus{} 4 \equal{} 4\sin^2x$
$ 2\sin^2x \equal{} 4$
$ \sin^2x \equal{} 2$
$ \sin x \equal{} \sqrt {2}$
this is not possible.
[/hide]

Repaired Problem:
[quote="xpmath"]Find the value of $ \cos^2x$  such that
$ 6(36^{\frac {\sin^2x}{4}}) \equal{} 6^{\frac {\sin^2x}{8}}$.[/quote]
[hide="repaired solution"]
$ 6((6^2)^{\frac {\sin^2x}{4}}) \equal{} 6^{\frac {\sin^2x}{8}}$
$ 6^{\frac {2\sin^2x \plus{} 4}{4}} \equal{} 6^{\frac {\sin^2x}{8}}$
$ 4\sin^2x \plus{} 8 \equal{} \sin^2x$
$ 3\sin^2x \equal{} \minus{} 8$
This is not in the real world
$ \sin^2x\equal{}\frac{\minus{}8}{3}$
$ \sin x\equal{}\frac{\pm2i\sqrt{6}}{3}$
If i made a mistake somewhere or misinterpreted the problem, please tell me.
[/hide]

Apparently, I'm being an extreme idiot today, and forgot to put a 0. :oops:  :blush:
Really sorry about that.

could you check it one more time? I still am getting imaginary numbers

That was the right solution. As in no solution. You know, maybe I shouldn't have edited it at all, and just said no solution. :wallbash_red:

okay...here's what i had:
Rerepaired problem
[quote="xpmath"]Find the value of $ \cos^2x$  such that
$ 6(36^{\frac {\sin^2x}{4}}) \equal{} 6^{\frac {\sin^2x}{80}}$.[/quote]
[hide="rerepaired solution"]
$ 6(6^{(2)\frac {\sin^2x}{4}}) \equal{} 6^{\frac {\sin^2x}{80}}$
$ 6^{\frac {\sin^2x \plus{} 2}{2}} \equal{} 6^{\frac {\sin^2x}{80}}$
$ \sin^2x \plus{} 2 \equal{} \frac {\sin^2x}{40}$
$ \frac {39}{40}\sin^2x \equal{} \minus{} 2$
$ 39\sin^2x \equal{} \minus{} 80$
$ \sin^2x \equal{} \frac { \minus{} 80}{39}$ <--imaginary $ \sin x$
$ \frac { \minus{} 80}{39} \plus{} \cos^2x \equal{} 1$
$ \cos^2x \equal{} \frac {60}{13}$ <--impossible
[/hide]
I guess i got the right answer after all :D


integration
inequalities proposed
inequalities

Prove that:

$ 1\plus{}\frac{1}{\sqrt{2}}\plus{}\frac{1}{\sqrt{3}}\plus{}...\plus{}\frac{1}{\sqrt{n}}\geqslant \sqrt{n}$

$ n$ is natural number

[quote="arqady"][quote="Invariant"]Prove that:

$ 1 \plus{} \frac {1}{\sqrt {2}} \plus{} \frac {1}{\sqrt {3}} \plus{} ... \plus{} \frac {1}{\sqrt {n}}\geqslant \sqrt {n}$

$ n$ is natural number[/quote]
If $ n \equal{} 1$ it's true.
For $ n > 1$ we obtain:
$ 1 \plus{} \frac {1}{\sqrt {2}} \plus{} \frac {1}{\sqrt {3}} \plus{} ... \plus{} \frac {1}{\sqrt {n}} > \int_1^{n \plus{} 1}\frac {1}{\sqrt x}dx \equal{} 2\sqrt {n \plus{} 1} \minus{} 2 > \sqrt n.$[/quote]
Oh arqady,It is too hard
If $ n \equal{} 1$ it is obivius
Suppose the ineq is true with $ n$,we will prov with $ n \plus{} 1$
We have $ \frac {1}{\sqrt {n \plus{} 1}} \plus{} \sqrt n\ge \sqrt {n \plus{} 1}$<=>$ \sqrt {n^2 \plus{} n}\ge n$It is true
  :lol:
Beside,I think we can make it be stronger

[quote="Allnames"][quote="arqady"][quote="Invariant"]Prove that:

$ 1 \plus{} \frac {1}{\sqrt {2}} \plus{} \frac {1}{\sqrt {3}} \plus{} ... \plus{} \frac {1}{\sqrt {n}}\geqslant \sqrt {n}$

$ n$ is natural number[/quote]
If $ n \equal{} 1$ it's true.
For $ n > 1$ we obtain:
$ 1 \plus{} \frac {1}{\sqrt {2}} \plus{} \frac {1}{\sqrt {3}} \plus{} ... \plus{} \frac {1}{\sqrt {n}} > \int_1^{n \plus{} 1}\frac {1}{\sqrt x}dx \equal{} 2\sqrt {n \plus{} 1} \minus{} 2 > \sqrt n.$[/quote]
Oh arqady,It is too hard
If $ n \equal{} 1$ it is obivius
Suppose the ineq is true with $ n$,we will prov with $ n \plus{} 1$
We have $ \frac {1}{\sqrt {n \plus{} 1}} \plus{} \sqrt n\ge \sqrt {n \plus{} 1}$<=>$ \sqrt {n^2 \plus{} n}\ge n$It is true
  :lol:
[/quote]
But my estimation is stronger. :P


:lol: Really? How cool.  I did not think they broadcasted that.

[quote="goldendomer"][quote="LawOfSigns"]I didn't make top 4 in my state, but I want to watch the national countdown round.[/quote]

It will be aired on ESPN I think.[/quote]
Actually, due to lack of something or another (probably ratings), it will not be on ESPN this year.

Anyways, it might be on the Discovery Channel, last I heard.

If and when someone finds out what channel it is on post it.
 :)

There was nothing about MathCounts on [url]http://www.discovery.com[/url] whatsoever, so I'm assuming it's not going to be on Discovery Channel.

you could ask a friend's parent to record it for you.

[quote="LawOfSigns"]hmm...maybe the travel channel

Once they were the only channel broadcasting this one poker competition.[/quote]

And that means they'll automatically broadcast every obscure competition?

I don't think anything will be broadcast this year simply because the MC website has no mention of anything at all. There might be like a short 3 minute video that MC always makes but I think that'll be it. ESPN didn't broadcast anything on MC last year and MC didn't say anything either although they did the past 3 years. Your best bet is to email MC but there's nothing about any broadcast at all right now so I doubt it'll happen.

I heard that a student recorded it and would post it online.

Hmm, I've heard that Kevin won.
What place did Bobby come in?

Bobby came 13th. Awesome for a 6th grader!

[quote="13375P34K43V312"]It's on the Mathcounts website, you're a few days behind, my friend.[/quote]

May you post the link? thanks

just go to mathcounts.org and scroll down to national competition results and click on the top 57 thing

it's pretty straightforward...

Thanks Carl. I thought you meant TV program...

I wanted to watch TV program.

leet was answering zhangpeijin, uldivad9, i think...

i am 99% that it's not gonna be on tv...

Do they show it on TV during the actual thing, or after?

generally they show it a few weeks to a few months after

[quote="13375P34K43V312"]They only did it in 2003-2005, and they only show the countdown round.[/quote]

Well, that really is about the only thing they can show...

yeah, they can't really show the written.

umm...it'd be nice if someone did a documentary tho

[i want a name to put here but don't have one because i wanna be like people who are paranoid and don't give out their name lol jk about this whole thing]

[quote="Sly Si"]I think the last time they showed it, it was shown on the day of the National Spelling Bee.  The spelling bee was shown live, and during their lunch break, ESPN aired Mathcounts.[/quote]

I would think that a math competition in which viewers are potentially amazed with speed at which contestants solve problems is more interesting than a contest in which contestants are standing there struggling to spell something when the viewers can simply look it up in a dictionary and move on to the next channel, checking the results later - no offense to any spelling masters out there.

Yeah, but the spelling bee is much longer than the Countdown Round. Plus, words are formed out of the things everyone learned in kindergarten, letters. The appeal of a spelling bee is mainly how hard the words are. I think both are important, spelling bees have a lot of pressure and teach you how to study intsnsively, Mathcounts has the same thing. Mathcounts is useful if you are interested in math, the spelling bee if you are in the history of words

True, maybe my opinion is just so biased because I failed miserably at spelling bee... (something like 9th in the school -_-).  But, are you saying we did not learn our numbers in kindergarten?  The appeal of competitive mathematics, I believe, is strong, but not yet felt in American culture - how many people on the street have heard of MATHCOUNTS?  Compare that to how many hear of the spelling bee.  *cry*

Yeah, and a lot of people find math a boring subject, so they don't even bother to watch. I'd say about 80% of the people in my class puts math as their least favorite subject, which is a fair amount.

[quote="solafidefarms"]Please don't discuss the relative merits of the spelling bee vs. Mathcounts. Both appeal to viewers because it's astonishing what the students know. I personally find NSB more exciting, between my love of words and the number of people I know at NSB, but I know that other people like Mathcounts more. So let's keep this thread on-topic and not an argument. Thanks! :)[/quote]

In that post you both asked people not to compare the two and then went on to compare the two yourself.  Hmmm....

So.........while I obviously like MC more, I think that the "average" person can't do the Countdown Round questions at Nats, and if they could they can not do them nearly as fast as the competitors.  The TV is on to the next problem before the viewer has even read the previous one.  Whereas the spelling bee is slow and methodical and the "average" person at home can at least try to spell the words and is given time to try themselves.

I know my mother was really excited to watch the ESPN broadcast to see what MATHCOUNTS was and what I coached, but she only watched like 5 minutes before she said she felt stupid and couldn't keep up so she turned it off.

Whichever one you like better(cough cough MATHCOUNTS) the Bee is more accessible to a wider audience and therefore brings in more in commercial revenue and therefore gets to stay on TV :lol:

[quote="frost13"]
I know my mother was really excited to watch the ESPN broadcast to see what MATHCOUNTS was and what I coached, but she only watched like 5 minutes before she said she felt stupid and couldn't keep up so she turned it off.
[/quote]

I think this is the key point - people like to play along with the Spelling Bee and Jeopardy and Millionaire, even though the contestants are often way better than the viewers, because the skills required for success in these events (esp. the Spelling Bee) are pretty unimportant.  No one is going to have a better life simply through spelling better.

But these students in MATHCOUNTS are intimidating because they are masters of a very valuable skill - the non-mathy viewer is reminded watching MATHCOUNTS that they couldn't compete against these people in something that is important.  I don't care if a 12-year-old can out-spell me.  But if a 12-year-old is better at a highly employable skill than I am, that's depressing.   (No offense to your mom, frost13; my mom wouldn't have much tolerance for watching the video if I weren't the sportscaster.  And even then, I'll bet she flipped the channel more than once.)

[quote="rrusczyk"]And even then, I'll bet she flipped the channel more than once.[/quote]
 :P So true about the other things you said.

But the countdown just isn't exciting, unless you're in it (to the average person, it's just kids quickly saying numbers). The spelling competition is more nationally organized and viewed and shows the stories of individuals from each state.

I've always thought the spelling bee had more TV appeal because it requires a thought process that anyone can relate to even if they can't go through it.  If the average ESPN viewer sees the Countdown round, they're going to see the problem, and in most cases have no idea how to do it.  30 seconds later, the kid buzzes in and gets the correct answer.  Then sometimes it's explained to the viewer, but they still can't see what was going on in the kid's head.  In the spelling bee, though, it's easy to try to spell the word and at least start to go through the kid's thought process.

[quote="gdtea"]How about breaking real recorded TV program and inserting some educational pieces, such as displaying the problems and even with solutions...? In that way, we might even promote MATHCOUNTS more especially MATHCOUNTS problems are very practical.[/quote]
I had the same thought. I remember when our PBS channel broadcast a program for kids illustrating the Fibonacci sequence- and how it occurs in nature. It was appealing even to the non-math types. Maybe someone  can create a MATHCOUNTS docu-drama along the lines of "Spellbound" and explain the countdown problems in slow speed ! One can only hope!

I love all the comments, but I have one word to say, just one word

"business"


induction
number theory unsolved
number theory

Prove that for each $ n$, there exists a number $ A_n$, divisible by $ 2^n$, whose decimal representation contains $ n$ digits, each of them equal to $ 1$ or $ 2$.

$ A_1\equal{}2$ and $ A_{n\plus{}1}\equal{}x*10^n\plus{}A_n$, were $ x\equal{}1$ if $ A_n/2^n$ is odd, else $ x\equal{}2$. These sequence is unique.

it's easy :)
note that $ N\equal{}a_n...a_1\equal{}10^{n\minus{}1}a_n\plus{}...\plus{}10^0a_1$
suppose that $ 2^n | 10^{n\minus{}1}a_n\plus{}...\plus{}10^0a_1$
it implies $ 2^{n\minus{}1} | 10^{n\minus{}1}a_n\plus{}...\plus{}10^0a_1$ or $ 2^{n\minus{}1}| 10^{n\minus{}2}a_{n\minus{}1}\plus{}...\plus{}10^0a_1$
now we use induction:
suppose that the sequence is unique until the $ (n\minus{}1)\minus{}th$ term.i.e. if there exists a number like $ B$ with $ k$ digits ($ 1\le k \le n\minus{}1$) such that its digits are only $ 1$ or $ 2$ and $ 2^k | B$ then we have $ A_k\equal{}B$.
as mentioned above if the number $ N$ has $ n$ digits and its digits are only $ 1$ or $ 2$ and it is divisible by $ 2^n$ then we must have $ 2^{n\minus{}1} | N\minus{}10^{n\minus{}1}a_n$ but obviously this number ($ N\minus{}10^{n\minus{}1}a_n$) has $ n\minus{}1$ digits and its digits are either $ 1$ or $ 2$ and it is divisible by $ 2^{n\minus{}1}$.so we must have $ N\minus{}10^{n\minus{}1}a_n\equal{}A_{n\minus{}1}$.
now the rest is trivial.


$ m,n$ positive integer number . solve : $ n^{5}\plus{}n^{4}\equal{}7^{m}\minus{}1$

Here is my solution :
$ n^{5} \plus{} n^{4} \equal{} 7^{m} \minus{} 1$ so $ (n^{2} \plus{} n \plus{} 1)(n^{3} \minus{} n \plus{} 1) \equal{} 7^{m}$
Thus we have $ n^{2} \plus{} n \plus{} 1 \equal{} 7^{r}$ and $ n^{3} \minus{} n \plus{} 1 \equal{} 7^{s}$
now assume that $ n > 2$ so we have :
$ n^{2} > n \plus{} 2$ so we have $ n^{3} \minus{} n \plus{} 1 > n^{2} \plus{} n \plus{} 1$ and then $ s > r$ so $ (n^{2} \plus{} n \plus{} 1,n^{3} \minus{} n \plus{} 1) \equal{} 7^{r}$
now we have
$ 7^{r}|n^{3} \minus{} n \plus{} 1$ and $ 7^{r}|n^{2} \plus{} n \plus{} 1$ which gives $ 7^{r}|n \minus{} 2$
but $ n > 2$ so we have $ n \equal{} 7^{r}k \plus{} 2$ where $ k\geq 1$
now it's easy to show that $ n^{2} \plus{} n \plus{} 1 > 7^{k}$ .

so remain only two case :
1)$ n \equal{} 1$ in this case we haven't any solution.
2)$ n \equal{} 2$ then $ m \equal{} 1$ .
$ *$ the only solution is $ (m,n) \equal{} (1,2)$


function
calculus
derivative
real analysis
real analysis unsolved

I'm having trouble with this question and I hope this is the right place to post it.

The logistic map is the following iteration: $x_{n+1} = cx_{n}(1-x_{n})$ where  $c$ is a non-zero constant.  Note that $a_{c}=1-\frac 1c$ is a fixed point for this iteration.  For what value(s) of $c$ would $x_{n}$ converge to $a_{c}$ if $x_{0}$ is chosen to be sufficiently close to but not necessarily equal to $a_{c}$?  Is there any value of $c$ where the convergence will be faster than linear?  If so, for which value(s) of $c$ and what is the order of convergence?

General principles: suppose we have a differentiable function $f(x)$ and we define an iteration by $x_{n+1}=f(x_n).$ A fixed point $c$ is a solution to the equation $f(c)=c.$

Find $f'(c).$

If $|f'(c)|>1,$ then the fixed point is unstable; if $x_n$ is near $c,$ then $x_{n+1}$ will be further away. We will not have convergence to $c$ at all (unless by some piece of luck some $x_n$ happens to land exactly on $c.$)

If $|f'(x)|=1,$ we cannot settle the question of convergence so simply. Further analysis must be undertaken. But even if we do have convergence, it is likely to be very slow.

If $0<|f'(x)|<1,$ then we have a stable fixed point. For sufficiently close starting points, the iteration will converge. This will be a case of first order convergence.

If $|f'(x)|=0$ then we will have faster than first order convergence.

What does this say in your case?

Well I'm not sure if I've cought on or not but I get

$\mid f^{\prime}(x_{n})\mid>1$ but if I keep going I find that $\mid f^{\prime\prime\prime}(x_{n})\mid=0$

now does the fact that $c = 0$ would allow for a first order convergence help me at all

Sorry I don't think I'm catching on, any other help?

Second and third derivatives are mostly uninformative; stick with the first derivative here.

$f'(x_n)$ makes no sense as a notation. What we need to talk about is $f'(x),$ evaluated where $x$ is the fixed point.

If $f(x)=cx(1-x),$ then in general $f$ has two fixed points, two solutions to $f(x)=x.$

One of these solutions is at $x=0,$ the other is at $x=1-\frac1c.$

$f'(x)=c(1-2x).$

$f'(0)=c.$ Hence for $0\le c<1,$ the fixed point at $0$ is stable and $x_n\to0.$ This is an example of first order convergence.

If $c=1,$ then there's only one fixed point, at $0$. We still have $x_n\to 0$ but now the convergence is slow.

In general, $f'\left(1-\frac1c\right)=2-c.$

If $1<c<3$, then this fixed point is stable and $x_n\to1-\frac1c.$ This is first order convergence, except in the special case $c=2.$ In that case the first derivative of $f$ vanishes at the fixed point. Since the second derivative is nonzero, we have $x_n\to\frac12$ at a second order rate.

For $c=3,$ we do have $x_n\to \frac23,$ albeit slowly.

For $3<c<4,$ the sequence $x_n$ does not converge. Exactly how it does behave, and how that depends on $c,$ is well-studied and often simulated and pictured.


trigonometry
algebra
system of equations
algebra unsolved

We have a system of equations:
$2x+x^2y=y$
$2y+y^2z=z$
$2z+z^2x=x$

Firstly I check if $x$ or $y$ or $z$ can take value of $1$ or $-1$. They cannot, so from the first equation I try to get $y$, from the second one $z$ and from the last one $x$.

$y=\frac{2x}{1-x^2}$
$z=\frac{2y}{1-y^2}$
$x=\frac{2z}{1-z^2}$

We know, that if $\tan{\frac{\alpha}{2}}=t$ then
$\tan{\alpha}=\frac{2t}{1-t^2}$
And in this moment I have doubts.
Can I conclude from this, that:

$x=\tan{\frac{\alpha}{2}}$
${y=\tan{\alpha}}$
${z=\tan{2\alpha}}$
${x=\tan{4\alpha}}$
If I keep substituting I will have:
${x=\tan{\frac{\alpha}{2}}=\tan{4\alpha}}=\tan{32{\alpha}}\dots$, so $x=y=z=0$

Have I done it correctly???

You did right where you are doubted

but in the last line ... check it:

tan a/2 = tan 4a = ...

dose it mean that a=0 (I dont think so)

Jast find a in the first equation: tan a/2 = tan 4a

it mean  a =2k(PI)/7

The next equation mean

8a = 2k'(PI)/7

but we know

8a = 8 * 2k(PI)/7 = 2*(8k)*(PI)/7

and let k'=8k ... and so on

so a=2k(PI)/7

and x=tan(k*PI / 7)

so x = tan(PI/7) , tan(2*PI/7), ... , tan(6*PI/7) ,

and tan(0)=0


The number $5. 7^{34}$ has a great many digits in its decimal representation. Without evaluating those digits, prove that some digit occurs at least four times in the number.

[hide="Solution"] It suffices to show that $5 \cdot 7^{34}$ has at least $28$ digits; then the result follows by Pigeonhole.

Well, $7^{6}= 117649 > 10^{5}$, so

$5 \cdot 7^{34}> 5 \cdot 7^{4}\cdot 10^{25}= 12005 \cdot 10^{25}> 1.2 \cdot 10^{29}$

So the result follows. [/hide]

[quote="t0rajir0u"]It suffices to show that $5 \cdot 7^{34}$ has at least $28$ digits; then the result follows by Pigeonhole.[/quote]
Surely you would need it to have at least 31 digits to apply Pigeonhole?  (Unfortunately, it only has 30, so proving it has 31 would be difficult.)

Right, right, $0$ is also a digit.  This complicates matters.   :oops:

=270584780189760558344798304245

7,5,8,4 appear at least 4 times each

[yes i know this is a lame way to do it]

The problem stated "...without evaluating the digits."

If $5 * 7^{34}$ has >30 digits then by pigeonhole. If it has 30 digits, the only way without one digit being repeated 4 times is that each digit 0-9 occurs 3 times, but if that were the case, the number would be divisible by 3, contradiction.

Very nice, I think I would have posted something similar had I seen this earlier :oops:

Was that the intended solution? (Ashrafmod)


[asy]size(250);
import olympiad;
import math;
pair A,B,C,H,D,M,N,E,F,G,J;
H=(0,0);
A=(0,15);
B=(-8,0);
C=(8,0);
draw(H--C--A--B--H--A,blue);
D=foot(H,A,C);
M=midpoint(H--D);
draw(H--D,blue);
draw(A--M,blue);
draw(B--D,blue);
N=intersectionpoint(A--M,B--D);
dot(A--B--H--C--D,red);
dot(M--N,red);
label("$A$",A,N);label("$B$",B,S);label("$H$",H,S);label("$C$",C,S);label("$D$",D,NE);label("$M$",M,SE);label("$N$",N,NE);
draw(rightanglemark(A,H,B,20));draw(rightanglemark(H,D,A,20));
add(pathticks(B--H,2,0.5,0,25,red));add(pathticks(H--C,2,0.5,0,25,red));
add(pathticks(M--H,2,0.5,7,25,red));add(pathticks(M--D,2,0.5,7,25,red));[/asy]
$ \triangle ABC$ is an isoceles triangle with $ AB\equal{}AC$
Also, Given : $ HM \equal{} MD$
[b]Prove that :[/b]
$ \angle ANB \equal{} 90^\circ$

[quote="guru_bhai"][asy]size(250);
import olympiad;
import math;
pair A,B,C,H,D,M,N,E,F,G,J;
H=(0,0);
A=(0,15);
B=(-8,0);
C=(8,0);
draw(H--C--A--B--H--A,blue);
D=foot(H,A,C);
M=midpoint(H--D);
draw(H--D,blue);
draw(A--M,blue);
draw(B--D,blue);
N=intersectionpoint(A--M,B--D);
dot(A--B--H--C--D,red);
dot(M--N,red);
label("$A$",A,N);label("$B$",B,S);label("$H$",H,S);label("$C$",C,S);label("$D$",D,NE);label("$M$",M,SE);label("$N$",N,NE);
draw(rightanglemark(A,H,B,20));draw(rightanglemark(H,D,A,20));
add(pathticks(B--H,2,0.5,0,25,red));add(pathticks(H--C,2,0.5,0,25,red));
add(pathticks(M--H,2,0.5,7,25,red));add(pathticks(M--D,2,0.5,7,25,red));[/asy]
$ \triangle ABC$ is an isoceles triangle with $ AB \equal{} AC$
Also, Given : $ HM \equal{} MD$
[b]Prove that :[/b]
$ \angle ANB \equal{} 90^\circ$[/quote]
[hide="Solution"]
Let $ AN'\perp BD$ and $ N'$ be on $ BD$.  Let $ AN'$ meet $ HD$ at $ M'$.  It follows that $ \angle AN'B \equal{} 90 \equal{} \angle AHB$, so $ AN'HB$ is cyclic.  This implies that $ \angle DHC \equal{} 90 \minus{} \angle ACB \equal{} \angle HAC \equal{} \angle HAB \equal{} \angle HN'B$.  This means that $ \angle BHD \equal{} 180 \minus{} \angle DHC \equal{} 180 \minus{} \angle HN'B \equal{} \angle HN'D$.  This means that \[ \angle N'AH \equal{} \angle DBH \equal{} 180 \minus{} \angle BDH \minus{} \angle BHD \equal{} 180 \minus{} \angle HN'D \minus{} \angle N'DH \equal{} \angle N'HD\]  Also, $ \angle N'DH \equal{} \angle BN'H \minus{} \angle N'HD \equal{} \angle HAD \minus{} \angle HAN' \equal{} \angle N'AD$.  This means that the circumcircles of $ \triangle AN'H$ and $ \triangle AN'D$ are tangent to $ HD$.  Hence, $ AM'$ bisects $ HD$, so $ M \equal{} M'$, and it follows that $ N \equal{} N'$, so $ AN\perp HD$. [/hide]

[b][u]Proof.[/u][/b]

Denote $ E\in AC$ for which  $ BE\perp AC$ . Observe that $ DE \equal{} DC$ and $ \triangle\ BCE\ \sim\ \triangle\ AHD$ .

The line $ AM$ is the $ A$ - median in $ AHD$ and the line $ BD$ is the $ B$ - median in $ BCE$

These medians are omologously in the similarity $ AHD\sim BCE$ . Denote $ F\in BD\cap AM$ .

Therefore, $ \widehat {HAM}\equiv\widehat {CBD}$ , i.e. $ \widehat {HAF}\equiv\widehat {HBF}$ . Therefore, the quadrilateral $ ABHF$

is cyclically , i.e. $ \widehat {AHB}\equiv\widehat {AFB}$ . But $ HA\perp HB$ . Hence $ FA\perp FB$ , i.e. $ BD\perp AM$ .


Let ABC an acutangle triangle such  $ \angle BAC\equal{}60^{\circ}$  and  $ AC>  AB$
O  circumcentre , H  orthocentre
OH meet AB and AC resp. at P and Q
Prove that  OP=HQ

[color=darkblue][b]1. [/b]
$ AO\equal{}R$
$ AH\equal{}2R\cos A\equal{}R$
$ \Longrightarrow$ $ AO\equal{}AH$.

[b]2. [/b]
$ \angle{BOC}\equal{}2\angle{BAC}\equal{}120^\circ$
$ \angle{BHC}\equal{}180^\circ\minus{}\angle{BAC}\equal{}120^\circ$
$ \Longrightarrow$ $ BOHC$ is cyclic.

[b]3.[/b]
$ \angle{POB}\equal{}\angle HCB\equal{}90^\circ \minus{}\angle B$
$ \angle{PBO}\equal{}90^\circ\minus{}\angle{C}$
$ \angle{APO}\equal{}\angle POB\plus{}\angle PBO\equal{}180^\circ\minus{}\angle B\minus{}\angle C\equal{}60^\circ$
$ \Longrightarrow$ $ APQ$ equilateral.

[b]4.[/b]
Lines $ AO$ and $ AH$ are simmetric wrt bisector of $ \angle BAC$
$ \Longrightarrow$  $ OP\equal{}HQ$.[/color]


Stie cineva daca nationala de matematica va fi in februarie, ca si olimpiadele de fizica si chimie?

Eu zic ca nu are cum sa fie in februarie....programa ar fi prea mica

Si eu am fost uimit cand am auzit ca judetenele s-au mutat in decembrie :wink: .M-am interesat si am aflat ca numai la chimie si fizica nationala este in februarie :lol:

[quote="Vrajitoarea Andrei"]Si eu am fost uimit cand am auzit ca judetenele s-au mutat in decembrie :wink: [/quote]

Si judeteana de matematica s-a mutat in Decembrie ?  :huh:

[quote="turcas_c"][quote="Vrajitoarea Andrei"]Si eu am fost uimit cand am auzit ca judetenele s-au mutat in decembrie :wink: [/quote]Si judeteana de matematica s-a mutat in Decembrie ?  :huh:[/quote]Daca este adevarat, inseamna ca O.M. devine ridicola ...
Eu as propune ca etapa finala O.M. sa aiba loc in Octombrie pe materia din anul trecut ...

:)  Defapt , totul a fost un zvon , nu stiu cine l-a lansat .La matematica,cel putin, nu s-a schimbat nimic. Astazi m-am interesat si etapa locala la matematica e in Ianuarie , etapa judeteana in 1 Martie si nationala banuiesc tot in vacanta de pasti . :)

Catalin are dreptate. Etapa judeteana e in 1 martie iar etapa nationala va fi in vacanta de Paste, la Timisoara.


LaTeX
modular arithmetic
number theory
greatest common divisor
number theory unsolved

Prove that the equation \[ 6(6a^2 + 3b^2 + c^2) = 5n^2  \] has no solutions in integers except $a = b = c = n = 0$.

Let a = 3 ^ m * a1, b = 3 ^ l * b1, c = 3 ^ k * c1,  n = 3 ^ p * n1, where a1, b1, c1, n1 are not divisible by 3.
Then LS simplifies and factors to:
3 * 2 * ( 2a1 * 3 ^ (2m + 1) + b1 * 3 ^ (2l + 1) + c1 * 3 ^2k). Because 3 doesn't divide a1, b1, c1, the left side can be expressed as d * 3 ^ (2i + 1), i >= 0, d isn't divisible by 3.
RS is 5n1 * 3 ^ 2p and because 3 doesn't divide n1, we get RS = e * 3 ^ 2j, j >= 0, e isn't divisible by 3.
But LS = RS , so because 3 doesn't divide e or d, we must have 3 ^ 2p = 3 ^ (2i + 1), which is impossible, or d = e = 0.
So, d = e = 0, so a = b = c = n = 0 as required.

[quote="Micchi"]SORI!!! i got lost in your solution how did you get the simplification of the left side???[/quote]
Let a = 3 ^ m * a1, b = 3 ^ l * b1, c = 3 ^ k * c1, n = 3 ^ p * n1, where a1, b1, c1, n1 are not divisible by 3.
Then:
6(6*a^2+3*b^2+c^2)
=3 * 2 * (2 * 3 * a1 ^ 2 * 3 ^ 2m + 3 * b1 ^ 2 * 3  ^ 2l + c1 ^ 2 * 3 ^ 2k)
=3 * 2 * ( 2 * a1 ^ 2 * 3 ^ (2m + 1) + b1 ^ 2 * 3 ^ (2l + 1) + c1 ^ 2 * 3 ^2k)
Is that good enough?

Hey could you please use $\LaTeX$? It's not very difficult to learn and then maybe I'll at least be able to read your solution without getting a head ache :)

[quote="rem"] Because 3 doesn't divide a1, b1, c1, the left side can be expressed as d * 3 ^ (2i + 1), i >= 0, d isn't divisible by 3.
[/quote]

 :| I don't understand how this come from?? Can somebody explain this more clearly? Thanks!

Hmmm, I'm not sure about your solution REM.  You don't necessarily have an odd number of powers of 3 on the LHS.  My first inclination was to do an infinite descent on factors of 3 too, but it doesn't work immediately like that. What if for example that a and b have no factors of 3 and c has one factor of three, then there are two on the left which is fine.

Here I got some solution from the APMO website:

The left side is divisible by 3, so 3 must divide n. So $5n^{2}-36a^{2}-18b^{2}$ is divisible by 9, so 3 must divide c. We can now divide out the factor 9 to get: $4a^{2}+2b^{2}+6d^{2}=5m^{2}$. Now take  a, b, d, m to be the solution with the smallest m, and consider residues mod 16. Squares = 0, 1, 4, or 9 mod 16. Clearly m is even so $5m^{2}$ = 0 or 4 mod 16. Similarly, $4a^{2}$ = 0 or 4 mod 16. Hence $2b^{2}+6d^{2}$ = 0, 4 or 12 mod 16. But $2b^{2}$ = 0, 2 or 8 mod 16 and $6d^{2}$ = 0, 6 or 8 mod 16. Hence $2b^{2}+6d^{2}$ = 0, 2, 6, 8, 10 or 14 mod 16. So it must be 0. So b and d are both even. So a cannot be even, otherwise m/2, a/2, b/2, d/2 would be a solution with smaller m/2 < m.

So we can divide out the factor 4 and get: $a^{2}+2e^{2}+6f^{2}=5k^{2}$ with $a$ odd. Hence k is also odd. So $5k^{2}-a^{2}$ = 4 or 12 mod 16. But we have just seen that $2e^{2}+6 f^{2}$ cannot be 4 or 12 mod 16. So there are no solutions.

Doh, shouldn't have given up at mod 8.  I really dislike the diophantine equations where all you do is take a bunch of clever mods and it falls apart.

Well what can I say, I don't know latex :rotfl: ... But analize the power parity

First note that $6|n$, so for some $n_1$ we have $6a^2+3b^2+c^2=30n_1^2$. From here we have $3|c$, so
\[2a^2+b^2+3c_1^2=10n_1^2\]for some $c_1$, whence $b\equiv c_1\pmod2$. If both are odd, then taking modulo 8 we get
\[2a^2-2n_1^2\equiv4\pmod8\implies a^2-n_1^2\equiv2\pmod4,\]a contradiction. So both are even, and integers $b_1,c_2$ exist such that $a^2+2b_1^2+6c_2^2=5n_1^2$. Now $a,n_1$ have the same parity, and like before both cannot be odd or else
\[b_1^2-c_2^2\equiv b_1^2+3c_2^2\equiv2\pmod4.\]So both are even, and $a_1,n_2$ exist with
\[2a_1^2+b_1^2+3c_2^2=10n_2^2.\]By the method of infinite descent we must have $(a,b,c,n)=(0,0,0,0)$.

[quote="NuncChaos"][quote="math154"]If both are odd, then taking modulo 8 we get
\[2a^2-2n_1^2\equiv4\pmod8\implies a^2-n_1^2\equiv2\pmod4,\]a contradiction. [/quote]

Could someone explain this step more? I don't understand how the initial mod was obtained.[/quote]
If $ac\equiv bxc\mod n,a\equiv b\mod \frac n {gcd (n,c)}$

The L.H.S is $36a^2+18b^2+6c^2$ The right hand side is divisible by 5.
Now , $36a^2+18b^2+6c^2\equiv {a^2+3b^2+c^2} (mod 5) $
and $a^2+3b^2+c^2\equiv 0 (mod5)$.....................(1)
We know $x^2\equiv 0,1,-1 (mod 5)$
It is not possible to show that sum of them are divisible by 5 according to the equation (1), a contradiction
So the only answer is
$a = b = c = n = 0$

@above its not that simple if $a^2 \equiv 1$ and $c^2 \equiv -1 $and $b \equiv 0 $ then there is no problem mod $5$

Note that $6|n$ and $3|c$, so subbing $n=6n'$ and $c=3c'$ yields $b^2+2a^2+3c'^2 = 10n'^2$. Now we have $b^2+3c'^2\equiv 0$ or $\pm 2\pmod{8}$ (the latter is impossible), so $2|b$ and $2|c'$, and subbing $b=b'$ and $c=2c''$ yields $a^2+2b'^2+6c''^2=5n'^2$. This time we get $5n'^2-a^2\equiv 0$ or $\pm 2\pmod{8}$ (the latter is impossible) so $2|a$ and $2|n'$ and we win by infinite descent.


In tennis, duce is called when both players have 40 points. When this happens, the two players play a point. If one side wins the next two points, they win the game. If neither team wins the next two games, it goes back to duce. On average, how many times will it return to duce before there is a winner, given that each team has a 1/2 chance of winning?

well ,the chance that there will be another duce is $\frac12$. if $E$ is the expected value of the number of duces, then we get:

$E=\frac12\cdot 0+\frac12 (E+1)\implies E=1$, so it is expected that there will be another duce ...


trigonometry
inequalities
symmetry
geometry unsolved
geometry

Let $ ABC$ be an isosceles triangle $ AB\equal{}AC$ and let $ P$ be an interior point. The lines $ BP$ and $ CP$ intersect the opposite sides at the points $ D$ and $ E$ respectively. Find the locus of $ P$ if:

$ PD\plus{}DC\equal{}PE\plus{}EB$

At least all points that are on height from $ A$ to $ BC$ can be desired locus.

Most probably it is only solution but we have to prove that.

I hope that there is no mistake in this. Please remind me if you find it.

Let $ AF$ be height in triangle $ ABC$. Assume that point $ P$ lies inside of triangle $ AFC$.
Let $ K$ be an intersection point of $ EC$ with $ AF$ and $ R$ intersection of $ BP$ with $ AF$
Then obviously order of points is $ A \minus{} K \minus{} R \minus{} F$ so if we denote an intersection of $ BK$ with $ AC$ as $ L$  then obviously order of points will be $ A \minus{} L \minus{} D \minus{} C$.

Applying Ceva's theorem we have $ \frac {AE}{EB}\frac {BF}{FC}\frac {CL}{LA} \equal{} 1$ and since $ BF \equal{} CF$ this one is equivalent to:

$ \frac {AE}{EB} \equal{} \frac {AL}{LC}$ which is equivalent to: $ \frac {AE \plus{} EB}{EB} \equal{} \frac {AL \plus{} LC}{LC}$ and since $ AC \equal{} AL \plus{} LC \equal{} AB \equal{} AE \plus{} EB$ we can conclude that $ EB \equal{} LC$, and that triangles $ LCB$ and $ ECB$ are congruent by $ SAS$ and since order of points is $ A \minus{} L \minus{} D \minus{} C$ we can conclude $ DC < LC \equal{} BE$ hence $ DC < BE$.

 Equation $ \frac {AE}{EB} \equal{} \frac {AL}{LC}$ also implies that $ EL$ and $ BC$ are parallel.
Now, through $ D$ draw parallel $ p$ to $ BC$. Then points $ E$ and $ L$ will remain above line $ p$ which means that line $ p$ will cut segment $ EC$ at certain point $ H$ so $ PH < PE$ and by Tales theorem we have:

$ \frac {PD}{PH} \equal{} \frac {PC}{PB} \equal{} \frac {\sin PBC}{\sin PCB} < 1$.

Last inequality s true becouse:
$ \angle PBC \equal{} \angle CBL \minus{} \angle LBP \equal{} \angle BCE \minus{} \angle LBP < \angle BCE \equal{} \angle PCE$ and both of these are from $ (0,90)$.

Hence $ \frac {PD}{PH} < 1$ so $ PD < PH < PE$   so we have $ PD < PE$.

Now if we combine   $ PD < PE$ with $ DC < BE$ we conclude that $ PD \plus{} DC < PE \plus{} BE$ which contradicts condition of problem.

So point $ P$ can not be interior point of triangle $ \triangle AFC$.
In analogous way, $ P$ can not be interior point of triangle $ ABF$ so if there exist point $ P$ that satisfies conditions of problem then it must be on line $ AF$.

Let $ P$ be any point on line $ AF$ and let $ BP$ and $ CP$ intersect opposite sides at $ D$ and $ E$.

Then by Ceva's theorem, like in previous part of solution, we conclude that $ AE \equal{} AD$ and $ EB \equal{} CD$.
Since $ P$ is on height $ AF$ of an isosceles triangle $ ABC$ we have $ PD \equal{} PE$ so this combined with $ EB \equal{} CD$ gives us: $ PD \plus{} DC \equal{} PE \plus{} EB$.

Finally, we conclude that desired locus of point $ P$ is height from $ A$ to $ BC$ in triangle $ ABC$.

Because of symmetry, it is easy to show that any point lies on the line segment $ AA',$ as the $ A \minus{}$altitude of $ \bigtriangleup ABC,$ inwardly or outwardly to it, has the property as the problem states and then, it is enough to prove that the point $ P,$ for which $ PD \plus{} DC \equal{} PE \plus{} EB$ $ ,(1)$ lies on $ AA'.$

So, we suppose that $ P,$ doesn’t lie on $ AA'$ and we denote the points $ P'\equiv AA'\cap CP$ and $ D'\equiv AC\cap BP'.$

Because of $ P',$ lies on $ AA',$ we have $ P'E \plus{} EB \equal{} P'D' \plus{} D'C$ $ ,(2)$

$ (1),$ $ (2)$ $ \Longrightarrow$ $ PD \plus{} DC \plus{} P'E \plus{} EB \equal{} PP' \plus{} P'E \plus{} EB \plus{} P'D' \plus{} D'D \plus{} DC$ $ \Longrightarrow$ $ PD \equal{} PP' \plus{} P'D' \plus{} D'D$ $ ,(3)$

 which is true only when $ PP' \equal{} 0$ and $ D'D \equal{} 0,$ that is, when $ P\equiv P'$ and $ D\equiv D'.$

So, the point $ P,$ lies on the line segment of the $ A \minus{}$altitude of $ \bigtriangleup ABC,$ as the wanted locus and the solution is completed.

Kostas Vittas.


function
quadratics
algebra
polynomial
arithmetic series
arithmetic sequence
quadratic formula

Determine $ m > 0$ so that $ x^4 \minus{} (3m\plus{}2)x^2 \plus{} m^2 \equal{} 0$ has four real solutions forming an arithmetic series: i.e., that the solutions may be written $ a, a\plus{}b, a\plus{}2b,$ and $ a\plus{}3b$ for suitable $ a$ and $ b$.

A. 1
B. 3
C. 7
D. 12
E. None of these

I don't know how to describe my reasoning too well, but here goes nothing...

[hide]It is clear that this function is even.  Thus, its four roots are symmetric about 0.  Let the roots be $ \pm a$ and $ \pm b$, with $ b>a$.  Since they form an arithmetic sequence, it is clear that $ b\minus{}a\equal{}a\minus{}(\minus{}a)$, and thus, $ b\equal{}3a$.

Now, we see also, that the original equation is a quadratic in $ x^2$.  Let $ x^2\equal{}y$.  Then the equation $ y^2\minus{}(3m\plus{}2)y\plus{}m^2\equal{}0$ has roots $ a^2$ and $ b^2$, notice that $ b^2\equal{}9a^2$.  Now, the two roots of this quadratic are given by the quadratic formula:

$ y\equal{}\frac{3m\plus{}2 \pm \sqrt{5m^2\plus{}12m\plus{}4}}{2}$

Therefore...

$ \frac{3m\plus{}2 \plus{} \sqrt{5m^2\plus{}12m\plus{}4}}{2}\equal{}9\cdot \frac{3m\plus{}2 \minus{} \sqrt{5m^2\plus{}12m\plus{}4}}{2}$

Solving for m is not terribly difficult...

$ 10\sqrt{5m^2\plus{}12m\plus{}4}\equal{}8(3m\plus{}2)$

$ 5\sqrt{5m^2\plus{}12m\plus{}4}\equal{}4(3m\plus{}2)$

Squaring and rearranging, we get another quadratic...

$ 19m^2\minus{}108m\minus{}36\equal{}0$, which has the only positive root $ m\equal{}6$.  Thus, our answer is $ E)$, none of the above.[/hide]

[hide="Solution"]

As usual, I'll introduce two solutions to this problem, one solved by me and one that actual solution gives (which is probably often cleverer). So, here goes the actual solution:

Note that because the coefficient of $ x^3$ is 0, we can set the four roots to be $ 3a,a,\minus{}a,3a$. This is a nice choice because the thing is, then we get the polynomial as:

$ (x\plus{}3a)(x\plus{}a)(x\minus{}a)(x\minus{}3a) \equal{} (x^2\minus{}a^2)(x^2\minus{}9a^2) \equal{} x^4\minus{}10a^2 x^2 \plus{} 9a^4$. So, $ m \equal{} 3a^2$ by matching the constant term and $ 3m\plus{}2 \equal{} 10a^2$. It can be easily solved that $ m \equal{} 6$ from this. Again, very clever choice in the beginning makes this problem to be solved really fast.

My solution is bit longer but not very hard either. By using the notation that the problem gave, we know, by similar manner as the first solution, that $ a\plus{}a\plus{}b\plus{}a\plus{}2b\plus{}a\plus{}3b \equal{} 4a\plus{}6b \equal{} 0$ which tells us that $ a \equal{} \frac{\minus{}3b}{2}$.

So, the four roots, in terms of $ b$, are $ \frac{\minus{}3b}{2}, \frac{\minus{}b}{2}, \frac{b}{2}, \frac{3b}{2}$. Using the constant term, we know that $ \frac{9b^4}{16} \equal{} m^2 \rightarrow m \equal{} \frac{3b^2}{4}$. Now, we know that the coefficients of $ x^2$, which is $ \minus{}(3m\plus{}2)$ is the sum of roots taken by two.

Doing that, we get that $ \frac{\minus{}10b^2}{4} \equal{} \minus{}3(m\plus{}2)$. Substituting $ m \equal{} \frac{3b^2}{4},$ we get $ b^2 \equal{} 8$. Now, going back to $ m \equal{} \frac{3b^2}{4}$ again, we see that $ m \equal{} 6$ as desired. It is slightly longer but note that both solutions did not involve using Quadratic Formula. [/hide]


Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ f(x \plus{} f(y)) \equal{} f(x) \plus{} y.$

There are lots of bad (i.e. nonlinear, discontinuous) solutions to this functional equation. You need an extra condition, such as increasingness, continuity, or just restricting $ f$ to the rationals.

Here's an example: pick a Hamel basis of the reals over the rationals, and pick two special elements of that basis $ e_1, e_2$. Let $ f$ be an additive function that swaps $ e_1$ and $ e_2,$ and keeps all of the other basis elements the same. Then if $ x \equal{} x_1e_1 \plus{} x_2e_2 \plus{} x',$ $ x_1, x_2 \in \mathbb{Q},$ and similarly for $ y$, then $ f(x \plus{} f(y)) \equal{} f(x \plus{} y_2e_1 \plus{} y_1e_2 \plus{} y') \equal{} x_2e_1 \plus{} x_1e_2 \plus{} x' \plus{} y_1e_1 \plus{} y_2e_2 \plus{} y' \equal{} f(x) \plus{} y$.


What is the toughest AMC 10, and is there a correlation between AMC 10 difficulty and AMC 8 difficulty?

Answering your second question: A super-hard AMC 8 problem would be considered medium on an AMC 10, possibly medium-easy.

thats not what i was asking AIME15 when i said correlation, i was asking in terms of if there was an easy amc 8 one year, does that mean that the amc 10 that year will also be relatively easy, for example since this year the amc 8 was apparently very easy, will the 10 be easy as well

Ok, but is there any amc 10 that has stood out to anyone as unusually difficult or relatively difficult

[quote="chenhsi"]For your first question, I don't think there is a hardest AMC test. It varies from person to person, especially because in some years, there might be more geometry problems for example, which would mean that test would be easier for people good at geometry and harder for people bad at geometry.

Did that make sense?[/quote]

I agree. If someone suggested an AMC10 as "easy", others might find the same test hard. Even if both people are on the same "level' of mathematics.

The hardest AMC 10 for me was AMC 10A 2004...

so, as a general question, is there not a trend of tests getting harder (on the amc's) over the years?

I don't think they are getting harder. I also thought the hardest AMC 10 was 2004 A.

hmmm yea i just took the 2004A and got 138...
maybe ive improved...my last one was 141.5

What would i get on that if i got 135 on 2007 AMC 12 A?
i usually get 144 or 150 on AMC 10 practices


Let $ x,y,z>0$ such that $ x\plus{}y\plus{}z\equal{}1$. Prove that
\[ \frac 13<\frac{x^2}{\sqrt{x\plus{}y}}\plus{}\frac{y^2}{\sqrt{y\plus{}z}}\plus{}\frac{z^2}{\sqrt{z\plus{}x}}<1\]

[quote="nayel"]Let $ x,y,z > 0$ such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that
\[ \frac 13 < \frac {x^2}{\sqrt {x \plus{} y}} \plus{} \frac {y^2}{\sqrt {y \plus{} z}} \plus{} \frac {z^2}{\sqrt {z \plus{} x}} < 1
\]
[/quote]

$ \frac {x^2}{\sqrt {x \plus{} y}} \plus{} \frac {y^2}{\sqrt {y \plus{} z}} \plus{} \frac {z^2}{\sqrt {z \plus{} x}}> \frac {x^2}{\sqrt {x \plus{} y\plus{}z}} \plus{} \frac {y^2}{\sqrt {y \plus{} z\plus{}x}} \plus{} \frac {z^2}{\sqrt {z \plus{} x\plus{}y}}\equal{}x^{2}\plus{}y^{2}\plus{}z^{2}\geq \frac{1}{3}(x\plus{}y\plus{}z)^{2}\equal{}\frac{1}{3}$

$ \frac {x^2}{\sqrt {x \plus{} y}} \plus{} \frac {y^2}{\sqrt {y \plus{} z}} \plus{} \frac {z^2}{\sqrt {z \plus{} x}}<\frac {x^2}{\sqrt {x }} \plus{} \frac {y^2}{\sqrt {y }} \plus{} \frac {z^2}{\sqrt {z }}\equal{}x\sqrt{x}\plus{}y\sqrt{y}\plus{}z\sqrt{z}\leq\sqrt{x*x\plus{}y*y\plus{}z*z}\equal{}\sqrt{x^{2}\plus{}y^{2}\plus{}z^{2}}\leq\sqrt{x\plus{}y\plus{}z}\equal{}1$

[quote="zaya_yc"][quote="nayel"]Let $ x,y,z > 0$ such that $ x \plus{} y \plus{} z \equal{} 1$. Prove that
\[ \frac 13 < \frac {x^2}{\sqrt {x \plus{} y}} \plus{} \frac {y^2}{\sqrt {y \plus{} z}} \plus{} \frac {z^2}{\sqrt {z \plus{} x}} < 1
\]
[/quote]

$ \frac {x^2}{\sqrt {x \plus{} y}} \plus{} \frac {y^2}{\sqrt {y \plus{} z}} \plus{} \frac {z^2}{\sqrt {z \plus{} x}} > \frac {x^2}{\sqrt {x \plus{} y \plus{} z}} \plus{} \frac {y^2}{\sqrt {y \plus{} z \plus{} x}} \plus{} \frac {z^2}{\sqrt {z \plus{} x \plus{} y}} \equal{} x^{2} \plus{} y^{2} \plus{} z^{2}\geq \frac {1}{3}(x \plus{} y \plus{} z)^{2} \equal{} \frac {1}{3}$

$ \frac {x^2}{\sqrt {x \plus{} y}} \plus{} \frac {y^2}{\sqrt {y \plus{} z}} \plus{} \frac {z^2}{\sqrt {z \plus{} x}} < \frac {x^2}{\sqrt {x }} \plus{} \frac {y^2}{\sqrt {y }} \plus{} \frac {z^2}{\sqrt {z }} \equal{} x\sqrt {x} \plus{} y\sqrt {y} \plus{} z\sqrt {z}\leq\sqrt {x*x \plus{} y*y \plus{} z*z} \equal{} \sqrt {x^{2} \plus{} y^{2} \plus{} z^{2}}\leq\sqrt {x \plus{} y \plus{} z} \equal{} 1$[/quote]
Yup, your solution is almost correct except the last step as Altheman stated! Just consider x<1,y<1 and z<1 we will get the result(x+y+z=1)

Here's a stronger problem:

For $ x,y,z>0$ such that $ x\plus{}y\plus{}z\equal{}1$ prove that
\[ \frac {1}{2\sqrt 2} < \frac {x^2}{\sqrt {x \plus{} y}} \plus{} \frac {y^2}{\sqrt {y \plus{} z}} \plus{} \frac {z^2}{\sqrt {z \plus{} x}} < \sqrt{x^2\plus{}y^2\plus{}z^2}
\]
The right side was already proved. So I want to see a solution for the left part. :)

[quote="nayel"]Here's a stronger problem:

For $ x,y,z > 0$ such that $ x \plus{} y \plus{} z \equal{} 1$ prove that
\[ \frac {1}{2\sqrt 2} < \frac {x^2}{\sqrt {x \plus{} y}} \plus{} \frac {y^2}{\sqrt {y \plus{} z}} \plus{} \frac {z^2}{\sqrt {z \plus{} x}} < \sqrt {x^2 \plus{} y^2 \plus{} z^2}
\]
The right side was already proved. So I want to see a solution for the left part. :)[/quote]

$ (\sqrt {x \plus{} y} \plus{} \sqrt {y \plus{} z} \plus{} \sqrt {z \plus{} x})(\frac {x^{2}}{\sqrt {x \plus{} y}} \plus{} \frac {y^{2}}{\sqrt {y \plus{} z}} \plus{} \frac {z^{2}}{\sqrt {z \plus{} x}})\geq (x \plus{} y \plus{} z)^{2} \equal{} 1$ --->

$ \frac {x^{2}}{\sqrt {x \plus{} y}} \plus{} \frac {y^{2}}{\sqrt {y \plus{} z}} \plus{} \frac {z^{2}}{\sqrt {z \plus{} x}}\geq \frac {1}{\sqrt {x \plus{} y} \plus{} \sqrt {y \plus{} z} \plus{} \sqrt {z \plus{} x}} \equal{} \equal{} \frac {1}{1*\sqrt {x \plus{} y} \plus{} 1*\sqrt {y \plus{} z} \plus{} 1*\sqrt {z \plus{} x}}\geq\frac {1}{\sqrt {1^{2} \plus{} 1^{2} \plus{} 1^{2}}*\sqrt {(x \plus{} y) \plus{} (y \plus{} z) \plus{} (z \plus{} x)}} \equal{} \frac {1}{\sqrt {3}\sqrt {2(x \plus{} y \plus{} z)}} \equal{} \frac {1}{\sqrt {6}} > \frac {1}{2\sqrt {2}}$


Heys... this topic is just for evry1... to share thr solutions to problems of RMO 2008...
n tell us how many problems u solved!!

hello, keshav, there is already a thread under general discussion under indian community. check that.

Maybe this can be used for [b]academic[/b] discussion between the [b]Indians[/b] only?

Let me post the links for the problems for those who did not sit for RMO as I did in the General Discussion forum:
(Also Pardesi told me to do so through gmail chat)

[url=http://www.mathlinks.ro/viewtopic.php?p=1311906#1311906]Problem 1[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1311911#1311911]Problem 2[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1311915#1311915]Problem 3[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1311917#1311917]Problem 4[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1311919#1311919]Problem 5[/url]
[url=http://www.mathlinks.ro/viewtopic.php?p=1311921#1311921]Problem 6[/url]

Maybe it will be available in [url=http://www.mathlinks.ro/resources.php?c=78&cid=48]here[/url] soon. I've asked Valentin.


inequalities
algebra
polynomial
linear algebra
matrix
vector

Given a matrix $ A$ and a positive integer $ n$, show that there is a monic polynomial $ p$ of degree $ n$ minimizing $ \|p(A)\|_2$.
(Also true but much harder to prove is that this polynomial is unique, it's called the Chebyshev polynomial of $ A$.)

Clearly $ g: P^n\to\mathbb{R}^\plus{}: p\mapsto\|p(A)\|_2$ is continuous, but sadly $ P^n$ is not compact so Weierstrass' theorem can't be applied directly. I'm probably missing something obvious here?

I'm sorry if I'm missing something, but why isn't $ P^n$ compact?

compact = closed and bounded, while $ P^n$ is not bounded? Or by what would it be bounded?

Of couse, $ \|p(A)\|\to\infty$ if the coeffient vector grows huge, but how to put this formal?

Probably I am the one misunderstanding something, it was written here as a "routine exercice", but I fail to find it...

I am missing something, too, I guess: how can this polynomial be unique if, for $ n$ large enough, we can take various multiples of the characteristic polynomial as $ p$ and get $ p\left(A\right)\equal{}0$ ?

  darij

[quote="Kent Merryfield"]Darij - note the word "monic" in Peter's definition. That takes care of "various multiples."[/quote]

Not quite - you can multiply the characteristic polynomial with $ x\minus{}1$ or with $ x\minus{}2$ and get different monic polynomials both attaining $ 0$ at $ A$.

  darij

They probably assume $ n$ to be no larger than the degree of the minimal polynomial.

Let $ m$ be the minimal polynomial of $ A$.  If $ n \ge \deg m$, then as darij said, we can just choose a multiple of $ m$.  So we may assume that $ n < \deg m$.

Given a polynomial $ p$ of degree less than $ \deg m$, define the norm $ \left\Vert p \right\Vert$ by
\[ \left\Vert p \right\Vert \equal{} \left\Vert p(A) \right\Vert_2 .
\]
This is indeed a norm: The only thing interesting to check is that $ \left\Vert p \right\Vert \equal{} 0$ implies $ p \equal{} 0$, which follows because $ \deg p < \deg m$.

But all norms on a finite-dimensional space are equivalent.  So $ \left\Vert p \right\Vert$ is the same as any standard norm up to a constant factor.

From now on, assume that $ p$ is a monic polynomial of degree $ n$.  Then from the paragraph above, the set of $ p$ such that $ \left\Vert p \right\Vert \le \left\Vert x^n \right\Vert$ is a compact set.  So we can apply a compactness argument after all.


function
algebra
functional equation
algebra solved

Find all real functions f such that f(xy+x^2)=f(x)f(y)+f(f^2(x)).

>>Easy functional equation

That's for me  :)

Looonguish solution :

1. f = 0 is the only constant solution. Let's suppose f != 0 (and not constant).
2. x=0 : f(0) = 0 since f not constant and if f(a)=0, f(ay+a)=0 so a=0.
3. y=0 : f(x)=f(f(x))
4. x=y : f(2x) = f(x) + f(f(x)) = f(x) + f(x)
5. x=-y : 0 = f(x)f(-x) + f(f(x))
6. 4+5 : f(2x) = f(x)(f(x) - f(-x)) = f(2(-x))
7. x->-x,y->-y, f(x)f(y) = f(-x)f(-y)
8. 2+6+7 : f(-x) = f(x) for all x or f(-x) = -f(x) for all x
9. 4+6 : f(x) = -f(x)f(-x) so f(x) = f(x) (f(x) = -f(x) + 4 + 1 : contradiction) and f(-x)=f(x)
10. 3+4 : f(x) = f(f(x)), so f(x) = f(f(x)) for all x
11. xy+x = f(f(x)f(y) + f(f(x)) = f(f(x)f(y) + f(x)) = f(x)f(y) + f(x) for f(x)=x for all x.

f=Id only solution.

Now I see that the problem is even easier than I thought. Anyway, I will post my solution, too. First of all, if f(a)=0 and a<>0 then f(0)=f(ya+a^2) for all y, false. So f(x)<>0 for all x<>0. For any x<>0 we take y=(f^2(x)-x^2)/x and deduce that f((f^2(x)-x^2)/x)=0, so f^2(x)=x^2 for all x<>0. Put y=0 and x=1 and obtain f(0)=0.So, f(xy+x^2)=f(x)f(y)+f(x^2). Square this and use that f^2(x)=x^2. You will find that f(x)f(y)f(x^2)=x^3y. Change x and y and you obtain f(x^2)/x^2=f(y^2)/y^2. So f(x)/x is constant for positive x and from now on everything becomes trivial.


Whenever I try to attach anything to a post, it always says that I have used up my quota limit of 10MB, even though the attachment itself is only around 50KB or so. I don't think there's a problem, though, because other users have been able to upload files of a similar size. Have I attached too many files in the past, or something?

You've reached your [b]total[/b] quota of 10 MB.


There is a performance festival that lasts 7 days. There are 12 groups participating in the festival. Each day, some of the 12 groups are chosen to perform; the groups not performing that day watch all the performances. In the end, it is found that every group watched every other group's performance. Determine the minimum number of performances made.

I'm not entirely sure about this, but is it $22$?

It all depends on the number of days in which a single team performs. We may assume that a team which performs alone in one of the $7$ days does not perform again, in order to minimize the number of performances. Each team which does not perform alone must perform at least twice (so that it can be seen by the teams with which it performed in one of the days).

If there are no teams performing alone, then there are at least $24=2\cdot 12$ performances. If one team performs alone, there are at least $23=1+2\cdot 11$ performances. We can find an arrangement with $22$ performances when two teams perform alone (and this is the minimum we can get with $2$ teams which perform once); the schedules for the seven days are: $(1,2,3,4),(1,5,6,7),(2,5,8,9),(3,6,8,10),(4,7,9,10),(11),(12)$.

Now let's see what happens if we have more than $2$ teams performing once and we want to get $\le 21$ performances.

With $3$ teams performing once, the other $9$ teams must perform at least twice each, so we have at least $18$ performances. this means $21$ performances, and since we want less than $22$, it means that we have exactly $21$, and each of the $9$ teams which do not perform once performs exactly twice. There is at least a day when $5$ teams perform, and for each one of those teams, we must have at least one other day at our disposal, the days being different. This means that we need at least $6$ days, which we do not have.

Reasoning the same way as above we can conclude that in the cases left there are not enough days for the teams which perform more than once.


function
integration
limit
real analysis
real analysis unsolved

Let $ f$ be a continuous real valued function satisfying $ f(x) \geq 0$ for all x ,
and $ \int_{0}^{\plus{}\infty}{f(x)}<\plus{}\infty$
Prove that  :$ \frac{1}{n}\int_{0}^{n}{xf(x)} \to 0$ as $ n \to \plus{}\infty$
 :starwars:

Let $ F(x) \equal{} \int_{0}^{x} f(t) \, dt$. Then $ \frac{1}{R} \int_{0}^{R} x f(x) \, dx \equal{} F(R) \minus{} \frac{1}{R} \int_{0}^{R} F(x) \, dx$.

Now the result follows from the following theorem:

[b]Theorem.[/b] If $ f$ is continuous and $ \lim_{x\to\infty} f(x) \equal{} L$, then $ \lim_{x\to\infty} \frac{1}{x} \int_{0}^{x} f(t) \, dt \equal{} L$.

[i]proof.[/i] We may assume $ L \equal{} 0$. For any $ \epsilon > 0$, choose $ R > 0$ such that $ |f(x)| < \epsilon$ for $ x \geq R$. Then for $ x > R$, we have

$ \left| \frac{1}{x} \int_{0}^{x} f(t) \, dt \right| \leq \frac{1}{x} \int_{0}^{R} |f(t)| \, dt \plus{} \epsilon \left( 1 \minus{} \frac{R}{x} \right)$.

Taking $ x \to \infty$, we have $ \limsup_{x\to\infty} \left| \frac{1}{x} \int_{0}^{x} f(t) \, dt \right| \leq \epsilon$, from which the theorem follows. ////


One rod is laid perpendicularly on an equal rod.  Each rod is made up of a black and white halves(of equal length), the black half having twice the linear density of the white half.  The rods have a mass $M$, length $L$, and their white halves have linear density $\lambda$.  There is neither friction nor air resistance.

$(1)$ Calculate the initial acceleration that the two black halves have towards eachother due to gravitational force in terms of $M$,$L$,$\lambda$, and fundamental constants.

$(2)$ Calculate the time it takes for the two black halves to meet for the first time.

$(3)$ Calculate the work $W$ done by gravity from the blocks' initial positions to their first meeting

$(4)$ Calculate the period $T$ of the motion of the rods

A diagram would help, since the problem statement is kind of unclear.


let $ a,b,c$ is nonegative.prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{b \plus{} a} \plus{}\frac{25(a\plus{}b)(b\plus{}c)(c\plus{}a)}{(a\plus{}b\plus{}c)((a\plus{}b\plus{}c)^2
\plus{}ab\plus{}bc\plus{}ca)} \ge 7$
when equality hold?

[quote="tuantam1lan"]let $ a,b,c$ is nonegative.prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{b \plus{} a} \plus{} \frac {25(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \plus{} c)((a \plus{} b \plus{} c)^2 \plus{} ab \plus{} bc \plus{} ca)} \ge 7$
when equality hold?[/quote]

Nice  :P
Let:$ p\equal{}a\plus{}b\plus{}c\equal{}1;q\equal{}ab\plus{}bc\plus{}ca;r\equal{}abc$
Ineq
$ <\equal{}>\frac{1\minus{}2q\plus{}3r}{q\minus{}r}\plus{}\frac{25(q\minus{}r)}{1\plus{}q} \ge 7$
$ <\equal{}>\frac{1\plus{}q}{q\minus{}r}\plus{}\frac{25(q\minus{}r)}{1\plus{}q} \ge 10$
It's true by Am-Gm  :)

[quote="caubetoanhoc94"][quote="tuantam1lan"]let $ a,b,c$ is nonegative.prove that:
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{b \plus{} a} \plus{} \frac {25(a \plus{} b)(b \plus{} c)(c \plus{} a)}{(a \plus{} b \plus{} c)((a \plus{} b \plus{} c)^2 \plus{} ab \plus{} bc \plus{} ca)} \ge 7$
when equality hold?[/quote]

Nice  :P
Let:$ p \equal{} a \plus{} b \plus{} c \equal{} 1;q \equal{} ab \plus{} bc \plus{} ca;r \equal{} abc$
Ineq
$ < \equal{} > \frac {1 \minus{} 2q \plus{} 3r}{q \minus{} r} \plus{} \frac {25(q \minus{} r)}{1 \plus{} q} \ge 7$
$ < \equal{} > \frac {1 \plus{} q}{q \minus{} r} \plus{} \frac {25(q \minus{} r)}{1 \plus{} q} \ge 10$
It's true by Am-Gm  :)[/quote]
very good!


geometry
rectangle
algebra
polynomial
search
area of a triangle
Heron\u0027s formula

A convex pentagon inscribed in a circle has side lengths of 16,16,16,16, and 6.  Find the area of that pentagon

NOTE:please move this to intermediate

I accidentally posted this...sorry (so i had to make it up by posting a question so it wouldnt be spam)

[quote="igiul"]Sure.
[hide="Open house!"]$ A = 6\left(16+\frac{\sqrt{16^{2}-3^{2}}}{2}\right)$
$ ...=96+3\sqrt{247}\approx 143$[/hide][/quote]
Thats not correct..

It's right. There's more than one that fits that.

No Im 100% sure thats not right. (sorry if this sounds harsh im not trying to be)

First try explaing what you were doing in the first place..like explain your solution..

[quote="igiul"]Draw a house with a base of $ 6$. Vertical walls of $ 16$, then two 'roofs' that are $ 16$, and meet at a distance of $ 16+h$ from the base of the house. This is a Square, and a triangle. The rest is fairly straight forward.[/quote]
You didnt read "inscribed in circle" part...creating a unique pentagon

and a lot harder problem

[quote="igiul"]Sure.
[hide="Open house!"]$ A = 6\left(16+\frac{\sqrt{16^{2}-3^{2}}}{2}\right)$
$ ...=96+3\sqrt{247}\approx 143$[/hide][/quote]

No there are enough constraints so you cant adjust angles to be a rectangle at bottom and triangle at top.

The way I would approach it is to first use Ptolemys therom to find the diameter d.  You can actually find a exact value for d.  Then you divide d by 2 to get the radius r.  If yhou find the radius r, you will have 4 triangles of sides 16,r,r and 1 of 6,r,r.  Since you know all the side lengths you will be able to use Heron's formula.

However, the hard part is actually getting d.  Using Ptolemys theorm, you create a 4th degree polynomial so I used a caculator and found d somewehre around 24.  So r is somewhere around 12 (pluggin in actually value of course into ti-89).  Then you can use Heron's formula (with a caculator once again) and I get my answer to be about $ 325.4ish$) which amazingly gives a exact value of $ 123\sqrt{7}$
But this way is really bad because I dont think you are allowed to use calculators...and stuff so I think theres another way like extending sides or something like that...

$ f(x) = \pm\sqrt{16^{2}-(x+16-\sqrt{247})^{2}}$. That gets 3. I wasn't aware it had to hit all 5, gimme a few, ILL BE BACK!

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1713630974&t=23525[/url]

Basically, you apply ptolemy, solve the system, and use easy area formulas.


geometry
analytic geometry
calculus
integration
function

[b]Show that a convex region symmetric with respect to the origin O=(0,0) contains a lattice point other than O if it's area is > 4.[/b]

A region is called convex if the line segment connecting any two points of the region is contained in the region.

A point is called a lattice point if both of its coordinates are integers.

in spoiler

[hide]call our region R.

consider the square S defined by the points (-1,-1), (-1,1), (1,1), (1,-1) with area 4.  Bring all of the points in R onto S, and because the area of R is greater than 4 there must be two points [tex]p_1, p_2 [/tex] which are distinct and overlap.  Therefore [tex]\displaystyle p_1 \equiv p_2 \mod 2[/tex].  Because R is symmetric, [tex]\dislpaystyle -p_2 \in R[/tex] and because of convexity, [tex]\displaystyle \frac{p_1 - p_2}{2} \in R[/tex] but we know [tex]\displaystyle p_1 - p_2 \equiv 0 \mod 2[/tex] so [tex]\displaystyle \frac{p_1 - p_2}{2}[/tex] is a lattice point.[/hide]

You have to be very careful when you try to talk about overlapping points. Just because you have a set of points with area more than 4 going into an area of 4 doesn't mean two points must overlap. Consider a one dimensional case - take all the points on the line from 0 to n, and map each point x to x/n. Then you have all the points on the line 0 to 1 - but no two overlap.

He's talking about mapping each point [tex](x,y)[/tex] to the point [tex](x,y)\mod 2[/tex], which preserves area. Since area is preserved, there must be an overlap. The example you gave does not preserve distance.

I'm terribly sorry for not making it clear, I meant it so that each point would be translated in order to preserve distance, but it came out wrong.

zscool, be careful :-). As always, you provided a nice proof.

zabelman wrote:Ya, zs, it is a nice proof, but the only reason I followed it is cause I've seen that before... its known as Minkowski's theorem (at least that's the name I've seen).

zabelman, you are right. MathWorld called it "Minkowski Convex Body Theorem". MathWorld also pointed out it's related to "Blichfeldt's Theorem" stated as follows:
Quote:Any bounded planar region with positive area > A placed in any position of the unit square lattice can be translated so that the number of lattice points inside the region will be at least  A+1.

I saw something similar in Paul Zeitz's book (I don't have a copy here). Also, I remember that he proved for case when A=1. He didn't give a rigorous proof for A > 1. Any thoughts on how to make it more rigorous for A > 1?

[hide]It seems to me that we need to represent area as an integral of the characteristic function of a region.[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let G be a finite group. Suppose that G has a normal subgroup N whose order |N| and index [G : N] are
relatively prime. Show that N is the only subgroup of G having order |N|.

if there is another subgroup $H$ with $Card(N)$ elements, take $x \in H-N$. Then the projection of $x$ on $G/N$ has an order that divides $Card(G/N)$ and $N$. Thus $x \in N$. Contradiction.

That means that $N$ is a characteristic subgroup (and a Hall subgroup).

Can we show that $N = \{x^{[G: N]} | x \in G\}$?

A Hall subgroup is a subgroup N such that N and [G : N] are coprime.

(Thanks to Google ;))


calculus
integration
calculus computations

Find the volume of the intersection between $ 2$ infinite cylinders of radius $ r$. (With perpendicular axes)

Hint: WLOG you may assume that the cylinders are $ x^2\plus{}y^2\equal{}r^2$ and $ x^2\plus{}z^2\equal{}r^2$, and find corresponding integral using cylindrical coordinates.


calculus
integration
trigonometry
calculus computations

[quote="kunny"]Let $\alpha,\beta$ be the distinct positive roots of the equation of $2x=\tan x$.
Evaluate

\[\int_0^1 \sin \alpha x\sin \beta x\ dx\][/quote]

$\int\limits_0^1 {\sin \alpha x\sin \beta x\,dx}  = \frac{1}{2}\frac{{\sin (\alpha  - \beta )}}{{(\alpha  - \beta )}} - \frac{1}{2}\frac{{\sin (\alpha  + \beta )}}{{(\alpha  + \beta )}}$

we have $2\alpha  = \tan \alpha ,\,2\beta  = \tan \beta$

$\Rightarrow 2(\alpha  - \beta ) = \frac{{\sin (\alpha  - \beta )}}{{\cos \alpha \cos \beta }} \Rightarrow \frac{{\sin (\alpha  - \beta )}}{{2(\alpha  - \beta )}} = \cos \alpha \cos \beta$

and

$2(\alpha  + \beta ) = \frac{{\sin (\alpha  + \beta )}}{{\cos \alpha \cos \beta }} \Rightarrow \frac{{\sin (\alpha  + \beta )}}{{2(\alpha  + \beta )}} = \cos \alpha \cos \beta$

hence the answer zero.

Yes,that's right,vidyamanohar. This problem is easy for you,isn't it? :)


geometry
circumcircle
projective geometry
geometry unsolved

Given triangle $ ABC$, we construct two triangles $ BAD$ and $ CAE$ outward on sides $ AB,AC$ (the triangles can vary), such that the angles $ \angle DAB$ and $ \angle EAC$ are fixed and $ \angle DBA+\angle ECA=180^{\circ}$. Prove that the line $ DE$ passes through a fixed point.

[quote="nguyentrang"][color=darkred]Construct two triangles $ BAD$ and $ CAE$ outside of the triangle  $ ABC$ (these triangles can vary), such that $ m(\widehat{DAB})$ and $ m(\widehat{EAC})$ are fixed and $ m(\widehat{DBA})+m(\widehat{ECA})=180^{\circ}$. Prove that the line $ DE$ passes through a fixed point. [size=134][b]Nice and easy problem ![/b][/size][/color][/quote]
[color=darkblue][b]Proof.[/b] Denote the circumcircle $ w$ of the triangle $ ABC$ and the second intersections $ X$, $ Y$ of the circle $ w$ with the lines $ AD$, $ AE$ respectively.

Prove easily that the points $ X$, $ Y$ are fixed and the mobile point $ M\in DB\cap EC$ belongs to the circle $ w$, i.e. $ m(\widehat{BMC})=180^{\circ}-A$.

Apply the [b][u]Pascal's theorem[/u][/b] to the cyclical hexagon $ AXCMBY$ : $ \left\{\begin{array}{c}D\in AX\cap MB\\\ Z\in XC\cap BY\\\ E\in CM\cap YA\end{array}\right\|$ $ \implies$ $ Z\in DE$, where the point $ Z$ is fixed.[/color]


algebra
polynomial
inequalities
calculus
integration
algebra proposed

$ P(x)$ is polynomial with integer coefficients  Its degree is at least $ 1$. Prove that inequality is fulfilled by finitely many integers: $ |P(x) \minus{} P(0)| < |x|$

Note that $ P(x) \equal{} G(x) x \plus{} P(0)$ for all $ P(x) \in \mathbb{Z}[x]$ such that $ \deg P \geq 1$. So the statement is the same as saying any $ G(x) \in \mathbb{Z}[x]$ has finitely many integral point $ x$ such that $ |G(x)| \leq 1$. This holds since from some point, $ G(x)$ is increasing or decreasing to infinity.


calculus
integration
real analysis
real analysis unsolved

can you evaluate  :D

$ \int_{\minus{}\infty}^\infty\;\;\dfrac{\tanh\,x\cdot\text{sech}\,x}{x}\;\;\textbf dx$

Define $ f(z)\equal{}\frac{\sinh z}{z\, \cosh^2 z}$ and use the residue theorem.

$ I\equal{}2\pi i \sum_{k\equal{}0}^\infty \text{Res} \left(f(z)\;,\; z\equal{}i (2k\plus{}1) \frac{\pi}{2}\right)
\equal{}2\pi i \, \sum_{k\equal{}0}^\infty \frac{4 \, (\minus{}1)^k}{i \, (2k\plus{}1)^2\, \pi^2}\equal{}\frac{8}{\pi} \underbrace{\sum_{k\equal{}0}^\infty \frac{(\minus{}1)^k}{(2k\plus{}1)^2}}_{\equal{}G}$


Is there an enumeration of the rationals in $(0,1)$ such that $x_1+x_1x_2+x_1x_2x_3+...$ diverges?

Sure, but it's a bit hard to explain.

Let $(q_n)_{n\ge 1}$ be any enumeration of the rationals in $(0,1)$. We construct our enumeration along with an increasing sequence $(k_n)_{n\ge 1}$ of positive integers inductively, as follows:

Assume we've assigned values to $x_1,\ldots,x_{k_n}$. Let $x_{k_n+1}$ be $q_t$, where $t$ is the smallest index $k$ such that $q_k$ is not among $x_1,\ldots,x_{k_n}$. Now take $k_{n+1}$ large enough, and $x_{k_n+2},x_{k_n+3},\ldots,x_{k_{n+1}}$ so close to $1$ that $a_{k_n+1}+\ldots+a_{k_{n+1}}>1$, where $a_n=x_1\cdot\ldots\cdot x_n,\ \forall n\ge 1$. The series we're concerned with corresponding to this enumeration $(x_n)_{n\ge 1}$ does not satisfy Cauchy's criterion for convergence.


limit
induction
real analysis
real analysis unsolved

Let $\lim_{n\to \infty} (a_n+2a_{n+1}) = \alpha$ ($\alpha$ is a finite number) . Prove that : there exists $\lim_{n\to \infty}a_n$ ?

Consider $b_{n}=a_{n}-\frac{\alpha}{3}$. The problem is equivalent to show that $b_{n}+2b_{n+1} \rightarrow 0\Rightarrow b_{n} \rightarrow 0$ (indeed if $(b_{n})$ has limit $l$ then it must verify $l+2l=0$, meaning $l=0$).
But $\forall \epsilon, \exists N, \forall n \geq N, |b_{n}+2b_{n+1}| \leq \epsilon \Rightarrow |b_{n+1}| \leq \frac{\epsilon}{2}+|b_{n}|$. But then by induction $|b_{n+1}| \leq \epsilon (1-\frac{1}{2^{n+1-N}})+\frac{|b_{N}|}{2^{n+1-N}}$. Hence $\exists M, \forall n \geq M, |b_{n}| \leq 2\epsilon$, and we are done.


Let $ n\in N$ with $ n\geq 2$. Show that if $ n$ has exactly $ k$ distinct odd prime factors then $ 2^k|\phi(n)$

It is an easy problem
$ n \equal{} \prod_{i \equal{} 1}^r p_i^{a_i}$
$ \varphi (n) \equal{} \prod_{i \equal{} 1}^r p_i^{a_i \minus{} 1}(p_i \minus{} 1)$


Find generators for a Sylow p-subgroup of the symmetric group of 2p elements, where p is an odd prime.  Show that this is an abelian group of order p^2.

Hmm, let's see, can you think of two commuting $ p$-cycles?


Hi guys,

I found this little problem on some website ages go, and was just wondering what you think of it:

[img]http://img354.imageshack.us/img354/9052/cirlce8hk.gif[/img]

Anyone have any ideas?

You can find a square $OLMN,$ where $L,N$ are the contact points.

$r(\sqrt{2}+1)=R$

See the small square with sides $r$.Its diagonal is $r\sqrt{2}$.So $r\sqrt{2}+r=R$


geometry
perpendicular bisector
geometry proposed

[color=darkblue]Given are the isosceles triangle $ABC$ in the vertex $A$ with $A=100^{\circ}$ and
the points $M\in BC$, $N\in AB$ so that $MN\parallel AC$ and $\widehat {ACN}\equiv \widehat {MAN}$.
Ascertain $m(\widehat {ACN})\ .$[/color][hide="Here is the answer."]$30^{\circ}\ .$[/hide]

As the parallel $MN \parallel AC$ moves from AC to the vertex B, the angle $\angle ACN$ is increasing from 0 to $\angle ACB = 40^\circ$, while the angle $\angle MAN$ is decreasing from $\angle BAC = 100^\circ$ to 0. Hence, a unique position of this parallel exists, for which these 2 angles are equal, $\angle ACN = \angle MAN$.

Let the perpendicular bisector of CA meet BC at K. The triangle $\triangle AKC$ is isosceles with KA = KC, $\angle KAC = \angle KCA = 40^\circ$ and $\angle AKC = 100^\circ$. Then $\angle KAB = \angle BAC - \angle KAC = 60^\circ$. Let the perpendicular bisector of AK meet AB at N. The triangle $\triangle AKN$ is equilateral, hence KA = KN and $\angle AKN = 60^\circ$. Consequently, the triangle $\triangle KCN$ is isosceles with KN = KC and $\angle CKN = \angle CKA + \angle AKN = 160^\circ$, so that $\angle KCN = \angle KNC = 10^\circ$. Let the perpendicular bisector of KN meet BC at M. The triangle $\triangle MKN$ is isosceles with MK = MN and $\angle MNK = \angle MKN = 180^\circ - \angle CKN = 20^\circ$, so that $\angle KMN = 140^\circ$ and $\angle BMN = 180^\circ - \angle KMN = 40^\circ$. As a result, the triangle $\triangle BMN$ is isosceles with $\angle NMB = \angle NBM = 40^\circ$ and $MN \parallel CA$. But obviously, the perpendicular bisector of KN also passes through the opposite vertex A of the equilateral triangle $\triangle AKN$, so that $\angle MAN = \frac{\angle KAN}{2} \equiv \frac{\angle KAB}{2} = 30^\circ$ and $\angle ACN = \angle ACK - \angle NCK = 30^\circ$.


Hey, I was just wondering if anyone could possibly explain antimatter to me in layman's terms. I only have a beginning in 8th grade science (and its only early chemistry) so I dont understand all of the terminology presented in sites I find through google. What I understand of antimatter is that it has the same mass as its pair matter particle but the electrons are charged positively (or the opposite, im not sure if thats the same). When antimatter comes into contact with its pair opposite electron, the electrons cancel out and the result is pure energy. Thus $E=mc^2$ - and antimatter travels at the speed of light. I also know that when matter is created (it isnt created actually, except in labs w/ *i think* a particle accelerator and it creates extremely hot temperatures etc) an equal number of antimatter particles are formed.

Can anyone explain to me the specifics of antimatter and why it works, in simple terms?

Thanks

What exactly is antimatter? Well, antimatter wouldn't look any different to us if we could see a chunk of it flying through space. And, although this is highly unlikely, if a distant galaxy were made entirely of antimatter, we'd have absolutely no way of telling from the light it emitted. Every type of particle in the universe has a kind of evil twin, its antiparticle. The only difference between the two is that antiparticles have an opposite electrical charge than their regular-matter mates.

An antiproton is exactly like a proton, except that it has a negative charge instead of positive. An anti-electron has a positive charge, and thus is usually called a "positron." There's even an anti-neutron, which is also neutral, since the neutron has no charge. Antimatter isn't really dangerous in small amounts, as an antiparticle can only annihilate a similar amount of matter.


Let $ x;y;z$ be positive real numbers satisfying $ x\plus{}y\plus{}z\equal{}1$.
Prove that:
$ \sum\limits_{x,y,z}  \frac{x(1\minus{}x)^2}{1\plus{}x}  \geq  \sqrt{3xyz}$

[quote="chien than"]Let $ x;y;z$ be positive real numbers satisfying $ x \plus{} y \plus{} z \equal{} 1$.
Prove that:
$ \sum\limits_{x,y,z} \frac {x(1 \minus{} x)^2}{1 \plus{} x} \geq \sqrt {3xyz}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167965
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167949
Why you do it?


Given $ a, b, c$satisfy $ a\plus{}b\plus{}c\equal{}3$. Prove that:
$ \sqrt{a^2b\plus{}b^2c} \plus{} \sqrt{b^2c\plus{}c^2a}\plus{} \sqrt{c^2a\plus{}a^2b} \leq\ 3\sqrt{2}$
@can_hang,honey_s,mitdac123: toanthpt.net was locked.  :(

The first, we use Cauchy-swart ineq
Then with the sum that we have just had we let q=ab+bc+ac,r=abc (1)
Finally,the ineq is the not difficult with (1)  (we can find q<= 3)
The proof is complete

[quote="Allnames"][quote="can_hang2007"][quote="nguoivn"]I don't think your proof is true  :wink:  If Cauchy- Svac, the next inequality won't true.
My proof used Am-Gm and pqr.
@Allnames: Không phải bài nào dùng pqr cũng ra đâu em :P[/quote]
But actually, Cauchy Schwarz and pqr works here. Using my CYH technique, we obtain a very nice solution. :)
[hide="@nguoivn"]Cam on da hieu y cua anh[/hide][/hide][/quote]
Thank Can_hang my proof was used the technique which let $ \sum\ ab$ = $ \frac {p^{2} \minus{} q^{2}}{3}$ (#),the sum that i have  just said was $ (\sum\ ab)(\sum\ .....)$( after use C_S then use # we can find the ineq that we have to prove is $ q^{2}(4q \minus{} 1)^{2}$  $ > \equal{}$0
IT is true,so the proof is finished[/quote]
Can you post your full proof? Since I used a strange Cauchy Schwarz ;)


The state income tax where Kristin lives is levied at the rate of $ p\%$ of the first $ \$28000$ of annual income plus $ (p \plus{} 2)\%$ of any amount above $ \$28000$. Kristin noticed that the state income tax she paid amounted to $ (p \plus{} 0.25)\%$ of her annual income. What was her annual income?

$ \textbf{(A) }\$28000\qquad\textbf{(B) }\$32000\qquad\textbf{(C) }\$35000\qquad\textbf{(D) }\$42,000\qquad\textbf{(E) }\$56000$

this was pretty tricky...at first i was like wtf...then i was like oh yea...[hide]p cancels[/hide]