Prove:

$2AM \geq GM+RMS$

for two variables.

[quote="jhcreinhardt"]extention:

is this true for n variables?[/quote]

No:  for 5 or more variables, take the set $\{x, 1, 1, 1, 1\}$ which has AM equal to $\frac{x+4}{5}$ and QM equal to $\sqrt{\frac{x^{2}+4}{5}}$.  Then as $x$ grows, QM grows faster than 2AM -- even without the GM, the inequality fails.  It probably can also be made to fail with only 4 variables, but I believe it holds for 3 variables (which I suppose someone should prove or disprove).

perhaps a more elegant approach?  i havent made any progress on this (n=3)

Mathematica spent a few hours chugging away at it earlier and didn't make any progress.  (I had it normalize the largest of the three elements to 1.)  Potentially, everything could be squared out and turned into a polynomial in terms of $\sqrt[3]{x}$ and $\sqrt[3]{y}$ and then some other inequalities applied, but I don't really want to do that myself.

One other random note: my counterexample also shows that for any $r_{1}< r_{2}$ and $n$, there exists a set $S$ such that $n\cdot PM(r_{1}, S) < PM(r_{2}, S)$, where PM is the power-mean.

That's because it's supposed to be 3rd root, not 2nd. (In the 3 variable case that is...)

But perhaps another generalization is possible - that it holds $\frac{S_{p}(x,y)+S_{q}(x,y)}{2}\leqslant S_{{p+q}\over 2}(x,y)$ for two variables, where $S_{r}(x,y)$ is a power mean of the order $r$ and $S_{0}(x,y)$ is a geometric mean.

Speaking of which, what's about the relation between, say, $\sqrt{S_{p}(x,y)S_{q}(x,y)}$ and $S_{\sqrt{pq}}(x,y)$ - or for that matter, $S_{r}(S_{p}(x,y),S_{q}(x,y))$  and $S_{S_{r}(p,q)}(x,y)$?

[quote="lotrgreengrapes7926"]For 3 variables: counterexample: $(x,y,z)=(.01,.01,.1)$[/quote]

This is not, in fact, a counter-example.  We can scale everything up by 100 (because scaling affects all power-means proportionally) to 1, 1, 10, from which AM = 4, GM = $10^{\frac13}\approx 2.1544$ and QM = $\sqrt{34}\approx 5.8309$.

I might change Farenhajt's generalization slightly, to finding the relation between $S_{r}(S_{p}(x, y), S_{q}(x, y))$ and $S_{S_{s}(p,q)}(x, y)$.

Also, $(x, 1, 1, 1)$ is a counter-example for 4 variables for any $x > 1$.


modular arithmetic
geometry
3D geometry
number theory

Find all ordered triples of integers $(x,y,z)$ such that $x^3+y^3+z^3=3$

Post if you get an answer without guess and check.

I have a semi guess and check answer, but the math league question gave a little more information and I used it.
[hide]
The question also stated that there are 4 triples and (1,1,1) is one of them. They ask you to find the other 3.

I knew that if they were distinct, they could be permuted in $3!$ ways, but there were only $3$ other triples. So two of them had to be the same so $3!/2! = 3$.

So the triple was of the form (a, a, b) or $2a^3 + b^3 = 3$ or $b = (2a^3 - 3)^\frac{1}{3}$

Then you can set $2a^3 - 3 = k^3$ and after a few quick guesses for k we find the triple (4, 4, -5).[/hide]

Hmmm, at the time of this topic's posting, I was at school takin this thingy.  Might wanna hold off on discussion for a little bit  ;)

(unless there's a set time, but I didn't think there was)

Whatever, everyone's probably done with it now. Anyway, nobody cheats on math league :roll:

[hide]
$n^3\equiv 0,1,-1\pmod {7}$

If $x^3+y^3+z^3\equiv 3\pmod 7$ then $(x^3,y^3,z^3)\equiv (1,1,1)\pmod {7}$.

$x^3\equiv 1 \pmod {7}\implies x\equiv\{1,2,4\}$

$\Rightarrow x,y,z\in \{-6,-5,-3,1,2,4\}$

Let's not get too big; we can be pretty sure that the numbers won't be very large or else the sums will be huge.

Testing possibilities we find the solutions $(1,1,1),(4,4,-5),(4,-5,4),(-5,4,4)$.
[/hide]

Yeah, it's not a great solution.

The math league solutions paper made a table of sums.

My league mentioned that $(1, 1, 1)$ is a solution, find the other three.

Obvious hint:  two of the numbers are the same, and positive.  the third will be negative and have an ab. value one greater than the other two nos.  (can you see why?)

guess and check $\implies$ 3 permutations of $(-5, 4, 4)$

[quote="chess64"]
[/hide]

Yeah, it's not a great solution.

[/quote]

I beg to differ  :P

[quote="jhcreinhardt"]...the third will ... have an ab. value one greater than the other two nos.  (can you see why?)[/quote]

No

and it's http://www.mathleague.com/

Well, for me I tried to solve it with number theory, but at the end
I just brute forced it.......
How did u guys do? I think I got 6.

[quote="chess64"][quote="jhcreinhardt"]...the third will ... have an ab. value one greater than the other two nos.  (can you see why?)[/quote]

No[/quote]

I don't see why either.

I either got a 5 or a 6, i might've misread the average of 5 numbers one.

Don't remember if it was distinct integers or not.

3 is obviously very small.  When doubling the negative cube, you want it to be [i]slightly[/i] smaller than the next one.  Guess and check shows that they will probably have to be consecutive.  This is also using the fact that there are only three more solution, thus (a, a, b) will be used, not (a, b, c).

i got a 6 btw

i got a 5 because i missed this problem. i always make stupid mistakes on these tests. its really annoying when the first problems are extremely easy, yet i miss them because they have really trivial subtleties. does anyone else make a lot of mistakes on these math league tests?

It is also on the Cal Math League #6. I just wrote out the perfect cubes from 1-10. Then I instantly saw $(4,4,-5)$.

I got a 4, because for some reason it took me until there was about 1 minute left to see how I would need to do the circle problem, and then after I got $(4,4,-5)$ on this one, I didn't realize the other two would just be the rearranged ones until about 30 seconds after I turned it in... :wallbash:

So no, you're not the only one who makes stupid mistakes on these... :wallbash_red:

I got a 5 on the last VAML because I thought a 3 was a 4. stupid inequalities.
Anyway, The way you do this one is just educated guessing and checking.

or you can be smart and know that one of them is negative and 2 are positive

and you realize that two of them have the be the same value

and you realize the two with the same value must both be positive because the cube root of 3 is not an integer

and you realize that 128 can be written as the sum of 2 cubes and -125 can be written as a negative integer cubed

and you always check your answers before turning it in

of course, you always have a calculator to help you brute force it. :)

Rofl, I had to do english notecards, I couldn't afford more than 10 minutes.

Well, I did the circle one in less that 10 secs, and the rest were easy.  For the last one, I was really interested in proving that there were only 4 solutions, so I was thinking about it during the test. :P  Unfortunately, I didn't see the word different in the average problem, so I had the numbers 1,1,1,1,96.  Well, the lesson is to read every problem before and after you solve it.

Masoud Zargar


function
inequalities unsolved
inequalities

f(x) = (1+ \sqrt x) / (1-x) is convex.

PS It took me 10 minutes to get the "Post a reply" window :-(

I have a solution for it{hope it 's ok}
We have:
sum(1+ \sqrt a)/(1-a)=sum1/(1- \sqrt a)>=16/(4-sum( \sqrt a))(**)
But sum (\sqrt a)<= 2\sqrt (a+b+c+d)=2(*)
From(*) and (**) we have the result. :)

That's quite interesting. If my computations are right, then both solutions are wrong. I apologise if I'm mistaken. For Anh Cuong solution, proving that 16/(4-sum sqrt(a))>=8 comes down to sum sqrt(a)>=2.

You  are  right  harazi.  The  function  f(x) = (1+ \sqrt x) / (1-x)  is  convex  at  [1/9 , 1). This  way was  the  first  i  tried...

Something tells me that this problem is very hard. I will think about it, since it seems quite tough.

Not very difficult, but very very nice. It comes from the fact that 1/(1-a)>=1+4a^2 for any a<1. This is equivalnet to a(2a-1)^2>=0. Now add these inequalities.

Thanks  harazi!  You  have  no  idea  how much  helpfull  you  are!

[quote="pcoRO"][quote="sqing"]Thank arqady.
If $x\in(0,1)$,then $(2\sqrt{x}-1)^2\geq 0\iff \frac{1}{1-\sqrt{x}}\geq 4x+1,$
$\sum\frac{1+\sqrt{a}}{1-a}=\frac{1}{1-\sqrt{a}}\geq 4\sum a+4=8.$[/quote]
$(2\sqrt{x}-1)^2\geq 0\iff \frac{1}{1-\sqrt{x}}\geq 4x+1$?[/quote]
$\frac{1}{1-\sqrt{x}}\geq 4x+1\iff 1\geq 4x+1-4x\sqrt{x}-\sqrt{x}\iff(2\sqrt{x}-1)^2\geq 0$

The first inequality is absolutely easy after using x/(2-x)>=x^2, and sum 2/(2-x)=4+sum x/(2-x).


I've been on AoPS for a few months now, and I've decided I wanted to make a blog.  But I don't know how to make one.  Where's the button that shows how I can make a blog?

Also, my post count dropped by about 250 today.  WHY?

You go to your profile, and there's an option there where you can make a blog.

Steps are pretty self-explanatory.

A ton of Fun Factory threads in the Games forum were moved yesterday. So If you posted a lot in those threads, you probably lost a lot of posts.


A reaction is 50% complete in 30.0 minutes. How long after its start will the reaction be 75% complete if the reaction kinetics is

(a) first order?

(b) zero order?

(c) second order?

I think all you need are the equations:

$ [A] \equal{} [A]_{0}\minus{}kt$
$ [A] \equal{} [A]_{0}e^{\minus{}kt}$
$ \frac{1}{[A]}\equal{}\frac{1}{[A]_{0}}\plus{}kt$

Where $ k$ is a constant (not the same in the three equations).

So I think:

[hide="Answers"]

a) $ 60\text{minutes}$

b) $ 45\text{minutes}$

c) $ 90\text{minutes}$[/hide]


AwesomeMath
summer program
Mafia
MATHCOUNTS
email
number theory
complex numbers

Hmmmm...so why am I not at the dance but posting this...I suppose the dance is massive failing though :P
I'm going to China the day I return, but I think I'll pull an all nighter tomorrow anyways. That was random.

So I guess the best way to put my experiences here is...why the hell did I not come last year???? BLARGH
This was certainly the best camp I have ever been to (and I've been to a lot), and this was certainly one of the best experiences I ever had. Team contest was pretty unfair I must say, in a lot of ways, but I guess that's just because less and less MOP people are coming here :(

I really enjoyed Number Theory and Geometric Proofs, thanks to all the teachers, especially JNB, GABRIEL, AND ALEX ANDERSON (Al the man, Al the man, does whatever Al the man can)! I wish that we didn't do all that complex numbers though.

Honestly the food is a lot better than food served in a lot of college campuses, so I don't exactly have complaints. I think I gained like 10 pounds!!!

As Dr. Andreescu would say with a slight correction, this camp is not just great, it is AWESOME. I'm pretty sad it's over, but I had a lot of fun with yall.

Hello Jay, I assure you it is.
edit: OH SNAP 1337 POSTS

ahaha im not at awesomemath this year :P
i kinda wish i was though ahaha

Oh man I think a lot of people are pulling all-nighters tonight.

By the way, FightTheTide, ping-pong balls=awesome.

[quote="sooozyy"]serialk11r..

YOU DIDNT GO TO THE DANCE?!?
omg..WHY NOT?
just kidding.
it was okay [ehh] last year. but its the experience you get out of it.
anyways,

i totally agree with you that awesomemath is just purely AWESOME! :)[/quote]

THERE WAS A DANCE LAST YEAR??? THAT I MISSED????

[quote="serialk11r"] JNB, GABRIEL, AND ALEX ANDERSON[/quote]

too much win in one sentence; abort.

[quote="connaissance"]@pythag: Do you know Francis? He told me that you know him...

AMSP is over... sigh...

I wish that it could continue for another week... sigh...

Oh, and I have to admit, Number Theory was AWESOME!!!

BEST SUMMER CAMP EVER!!!!!!!!

but also my first ever...

I met sooooooooooooooooooooooooooooooo many people there... now i'm sad b/c i may never seen any of them ever again...[/quote]

Yeah I know francis...

Hmm which gabriel is it

Hmmm

it was all good until my mom couldnt find me at the airport:)

This camp was epic though... mod arith class was the best:p:o. it should win the most inside jokes award.
i think i left my coughdrops there... but=/.. :( nobody checks the fridge.
why didnt you go to the dance? we had great mafia games:)

hahaha i can imagine people playing mafia at a school dance

[quote="bpgbcg"][quote]THERE WAS A DANCE LAST YEAR??? THAT I MISSED????[/quote]
Yes, but apparently it failed (I wasn't at it last year either, so this is from what I heard).[/quote]

DID NOT FAIL LAST YEAR! :)
we had a pretty decent song selection and the girls had fun at least! hahaha
im not so sure about the guys (mostly because there only like 10 girls there) and there were like50guys standing around in circles outside of the girls circle ahahaha layers and layers of circles.

it was pretty good

and pythag011... hmm david im not so sure if you would've enjoyed it anyways...

You didn't miss much last year pythag - the dance was pretty much trivial. But I do wish you could have been there so some team would have been able to get better than 52% on my round of the team contest.

By the way, my guys from 6611 and 6614, you left so much stuff behind it's not even funny. I now have a pair of fake aviators, a Six Flags souvenir bottle, and the rest of the Whoppers that got stolen out of my room (and yes, these were stolen. Not ethernet cable stolen or wallet stolen, but actually stolen).

[quote="tonypr"][quote="iYOA"]lol i know who it was.

they randomly took a picture of me and then they were like

"why splash cold water on ur face to wake up"
so i go in the bathroom and then see the writing on my face

it got off easily though[/quote]

he he he I still have that photo...[/quote]
and i still have ur phone number on my dad's cell phone

i was deciding whether or not to call you when i was at the airport.

the bottle isnt mine. If u remember i threw away my bottle cause i was too lazy to carry it around so instead i wasted about 100 dollars on drinks =[
The aviaters are mine though... There broken though

Or they might be davids... who knows

The aviators were definitely KingJames's. They're probably in a dumpster now, along with gatorlover's watch and a certain laundry basket.

Fightthetide I just found out that you have the same bday as my mom...Pretty wierd...

Dang I'm going to Santa Cruz and not Dallas this year.

I just hope Santa Cruz is as epic as this year.

Dallas was extremely epic.. (I miss all of you!)

BTW, I left some yogurt in 5913. XD. And fightthetide, did you find any jackets? If you did, it's going to be daviiiids.

Just wondering, has anyone gotten the mathematica program?(I haven't)

I think any decent math competitor has at least one copy of Mathematica by now.

[quote="pythag011"]Hmm is wolfram research giving away mathematica at all the summer camps this year? They gave it at mop...[/quote]

Mr. Alan told us Wolfram sponsored AMSP this year. Not exactly sure why.

But at least this way I can get Mathematica for without paying through the nose or placing top 10 at ARML.
MOPpers should get it free. They deserve it.

what happens if our computer crashes and we lose the mathematica?

or is that even possible?

[quote="FightTheTide"]It's coming out later. Mr. Alan just sent us an email confirming email addresses.[/quote]

It's been like 2 weeks, and still nothing, has anyone else gotten it?

The email has been sent to the address you gave with your yearbook information.


function
logarithms
real analysis
real analysis unsolved

Let $ f: R \rightarrow R$ be a derivable function such that the expression:
\[ f'\left(x \plus{} n \ln\left(n \plus{} 1 \right)\right) \minus{} f'\left(x \plus{} n \ln n \plus{} 1\right)
\]
has a constant sign $ \forall x \in\mathbb{R}$ and $ \forall n \in\mathbb{N^{*}}$.Prove that $ f'$ is continuous on $ R$.

I'll write a proof assuming that the expressin has a constant sign $ \forall x \in \mathbb{R}$ and $ \forall n \in \mathbb{N^{*}}$.

I denote $ x_n = n\ln (n + 1) - n\ln n - 1 = \ln \left(1 + \frac {1}{n} \right)^n - 1$ and $ f^{'} = g$ ; it's obvious that $ x_n\leq 0$ and $ x_n \rightarrow 0$.

And now suppose that $ g(x + n\ln(n + 1)) - g(x + n\ln n + 1) < 0$ $ \forall x\in\mathbb{R}$, $ \forall n\in\mathbb{N^{*}}$, from which we can deduce that

$ g(x + x_n) < g(x)$, $ \forall x\in\mathbb{R}$ and $ \forall n\in\mathbb{N^{*}}$ $ \Rightarrow$ $ g(x + kx_n) < g(x)$, $ \forall k,n \in \mathbb{N^{*}}$ and $ \forall x \in \mathbb{R}$ ,$ (*)$

Now suppose that there exist $ a,b \in \mathbb{R}$ such that $ a < b$ but $ g(a) > g(b)$.Since $ g$ has the Darboux property, it follows that there exists $ \epsilon_0$ such that $ g(x) > g(b)$, $ \forall x \in ( - \epsilon_0 + a, \epsilon_0 + a)$.Because $ x_n \to 0$ we have that there is $ n_0 \in \mathbb{N}$ such that $ |x{}_n{}_0| < \frac {\epsilon_0 }{2}$.But there is $ k\in\mathbb{N^{*}}$ such that $ b + kx{}_{n{}_0} \in (a,a + \epsilon_0)$,therefore $ g(b + kx{}_{n{}_0}) > g(b)$.But on the other hand, from $ (*)$ we have that $ g(b + kx{}_{n{}_0}) < g(b)$, therefore , contradiction.So $ g$ is increasing.

$ g$ is monotone and has the Darboux property implies $ g$ continuous, since a function that has the Darboux property hasn't got points of discontinuity of first kind and a monotone function has only discontinuities of first kind.

You can work the case $ g(x + n\ln(x + 1)) - g(x + n\ln n + 1) > 0$ in the same manner.


I don't understand why BYC is 3y when BXY is 2y, please help.

.Let ABC be a triangle. Let X be a point on AB and Y be a point on AC such that AX=XY=YB=BC . If the angle ABC measures 132 degrees, find the number of degress in the measure of the angle BAC . a.  12
b.  13
c.  14
d.  9
e.  15

Solution:
[hide]
Let y be the measure of the angle BAC.
It is easy to verify that
measure of angle BXY= 2 y= measure of angle XBY and
measure of angle XYB= 180 - 4 y. This implies that
measure of angle BYC= 3 y = measure of angle BCY. Hence
measure of angle YBC= 180 - 6 y and therefore
measure of angle ABC= 180 - 6 y + 2 y = 180 -4 y = 132.
Solving for y = ( 180 -132)/4 = 12
[/hide]

If $ \angle XAY \equal{} y$ then $ \angle XYA \equal{} y$ since $ \triangle AXY$ is isosceles. $ \angle AXY \equal{} 180\minus{}2y$ since the number of degrees in a triangle is 180. $ \angle YXB \equal{} 2y$ since the number of degrees on a line is 180. It follows that $ \angle YBX \equal{} 2y$ and $ \angle XYB \equal{} 180\minus{}4y$ since $ \triangle XYB$ is isosceles. Since $ AYC$ is a line, $ \angle AYX \plus{} XYB \plus{} BYC \equal{} 180$. Thus, we have $ y \plus{} 180\minus{}4y \plus{} \angle BYC \equal{} 180$. $ \angle BYC \equal{} 3y$. Since $ \triangle YBC$ is isosceles, $ \angle YCB \equal{} 3y$ and $ \angle YBC \equal{} 180\minus{}6y$. Since $ \angle ABC \equal{} \angle ABY \plus{}\angle YBC \equal{} 132$, $ 2y \plus{}180\minus{}6y \equal{} 132$. Thus $ y\equal{}12$ and $ \angle BAC \equal{} \boxed {12}$


The sum of the last four digits of Tz's phone number is $30$. How many such four digit sequences are there??
[hide="how do u solve it"]The answer's $84$ but i don't know how to get it.[/hide]

Please show work.

You start with 9993 as the first number.
then:
9984
9975
9966
9957
9948
9939

9894
9885
9876
9867
9858
9849

Notice 99__ has 7 ways
and
98__ has 6 ways
So (7+6+5+4+3+2+1)=28
So 9___ has 21 ways

We assume the pattern will be (5+4+3+2+1)=21 for 8___

So the solution is 28+21+15+10+6+3+1=84
I'm not sure
I don't have pencil or paper near me.

Well, no offense but I was wondering about a quick slick way other than guess and check/trial and error. Maybe like a formula or something??

I always wanted to do that too.

I think Mathcounts loves giving pattern problems.
Once you've done all the past competitions (which I have,except for most of the nationals), you should be able to figure out what you're supposed to be looking for.
List a few numbers (like about 5) and you should be able to find the pattern.
Since Mathcounts won't make you list 84 numbers! :roll:

THIS IS NOT GUESS AND CHECK or TRIAL AND ERROR.
Its more like LIST SOME, FIND PATTERN, and SOLVE.

I doubt that there's a formula for this type of problem, but I could be wrong.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=phone+number++last+digits+sum&t=72777]You already posted this[/url]

[hide]
Not really a formula, but smart casework :coool:
There can be no 0s 1s or 2s
You just find the possible arrangements of the last 4 digits and find the number of permutations of each.
If you write the digits in decreasing order, you can be sure that you missed none.

If you start at the biggest and work your way down, it is a lot easier, namely because you know an arrangement is already counted if the digits are not in decreasing order
9993-4
9984-12
9975-12
9966-6
9885-12
9876-24
9777-4
8886-4
8877-6
The hyphen is how many permutations those numbers have
Sum them up and get 84
And just a note to why my method is easy and never misses a case: I got all those the first try and didn't have to change anything to make it sum to 84[/hide]


A nonlinear circuit element can be made out of a parallel plate capacitor and small balls, each of mass $m$, that can move between the plates.  The balls collide inelastically with the plates, dissipate all kinetic energy as thermal energy, and immediately release the charge they are carrying to the plate.  Almost instantaneously, the balls then pick up a small charge of magnitude $q$ from the plate; the balls are then repelled directly towards the other plate under electrostatic forces only.  Another collision happens, kinetic energy is dissipated, the balls give up the charge, collect a new charge, and the cycle repears.  There are $n_{0}$ balls per unit surface area of the plate.  The capacitor has a capacitance $C$.  The seperation $d$ between the plates is much larger than the radius $r$ of the balls.  A battery is connected to the plates in order to maintain a constant potential difference $V$.  Neglect edge effects and assume that magnetic forces and gravitational forces may be ignored.

(5)  a. Determine the time it takes for one ball to travel between the plates in terms of any or al of the following variables: $m$, $q$, $d$, and $V$.

(5)  b. Calculate the kinetic energy dissipated as thermal energy when one ball collides inelastically with a plate surface in terms of any or al of the following variables: $m$, $q$, $d$, and $V$.

(5)  c. Derive an expression for the current between the plates in terms of the permittivity of free space, $\varepsilon_{0}$, and any or all of the following variables: $m$, $q$, $n_{0}$, $C$, and $V$.

(5)  d. Derive an expression for the effective resistance of the device in terms of $\varepsilon_{0}$, and any or all of the following variables: $m$, $q$, $n_{0}$, $C$, and $V$.

(5)  e. Calculate the rate at which the kinetic energy of the balls is converted into thermal energy in terms of $\varepsilon_{0}$ and any or all of the following variables: $m$, $q$, $n_{0}$, $C$, and $V$.

This is similar to the first problem on the IPhO in South Korea in 2004. In the following I will assume that the balls are travelling only in one direction at a time.

a) The balls will have a constant acceleration of $a = qV/md$, so the time it takes them to travel from one plate to another is
\[t = d \sqrt{\frac{2m}{qV}. }\]
b) The kinetic energy of a ball is
\[E = \frac{mv^{2}}{2}= \frac{m (at)^{2}}{2}= qV. \]
c) The charge $n_{0}S q$ passes through the capacitor in time $t$, so the current is
\[I = \frac{n_{0}Sq}{d}\sqrt{\frac{qV}{2m}}= \frac{n_{0}Cq}{\varepsilon_{0}}\sqrt{\frac{qV}{2m}}. \]
d)
\[R = \frac{V}{I}= \frac{\varepsilon_{0}}{n_{0}Cq}\sqrt{\frac{2mV}{q}}. \]
e) In time $t$, $n_{0}S$ balls release an energy $qV$ each, so
\[P = \frac{n_{0}SqV}{t}= \frac{n_{0}qCV}{\varepsilon_{0}}\sqrt{\frac{qV}{2m}}. \]


I spent my flight back to Frankfurt with the following inequality, which is problem 18 in §1.2 of Vasile Cîrtoaje (Vasc)'s new book [i]Algebraic Inequalities - Old and new methods[/i] (Zalau: Gil, 2006). That problem asks for a proof that

$\frac{1}{a^{2}+ab}+\frac{1}{b^{2}+bc}+\frac{1}{c^{2}+cd}+\frac{1}{d^{2}+da}\geq\frac{4}{ac+bd}$,

where a, b, c, d are four positive reals. As I didn't come up with the proposed solution, I thought of improving the inequality. And indeed, the following is also true:

$\frac{1}{a^{2}+ab}+\frac{1}{b^{2}+bc}+\frac{1}{c^{2}+cd}+\frac{1}{d^{2}+da}\geq\frac{2}{\sqrt{abcd}}$.

This is none of the super-hard inequalities requiring SOS and tons of computations. There is a pretty nice proof!

What is strange is that, apparently, the corresponding 3-variables inequality:

$\frac{1}{a^{2}+ab}+\frac{1}{b^{2}+bc}+\frac{1}{c^{2}+ca}\geq\frac{3/2}{\left(abc\right)^{2/3}}$

for positive a, b, c, is harder! Try to prove it and see...

I leave the $\geq 5$-variables case to the braver MathLinkers. I suspect that the inequality will remain true for some time, while the proof will fail...

  Darij

Because of homogenity we may take $abcd=1$. Also take substitution $a=\frac{x}w,b=\frac{y}x,c=\frac{z}y,d=\frac{w}z$, Now inequality becomes $\frac{w^{2}}{x^{2}+yw}+\frac{x^{2}}{y^{2}+xz}+\frac{y^{2}}{z^{2}+yw}+\frac{z^{2}}{w^{2}+xz}\geq 2$. Using trivial $a+b\geq 2\sqrt{ab}$ we have $LHS \geq \frac{2w^{2}}{2x^{2}+y^{2}+w^{2}}+\frac{2x^{2}}{2y^{2}+x^{2}+z^{2}}+\frac{2y^{2}}{2z^{2}+y^{2}+w^{2}}+\frac{2z^{2}}{2w^{2}+x^{2}+z^{2}}$. Now to get rid of squares replace $x^{2},y^{2},z^{2},w^{2}$ by $a,b,c,d$ and inequality becomes $\frac{a^{2}}{2ab+a^{2}+ac}+\frac{b^{2}}{2bc+b^{2}+bd}+\frac{c^{2}}{2cd+ac+c^{2}}+\frac{d^{2}}{2ad+bd+d^{2}}\geq 1$.Now by Cauchy-Scwarc in Engel form we have $LHS\geq \frac{(a+b+c+d)^{2}}{(a+b+c+d)^{2}}=1$.

See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81588&highlight=michael

[quote="dule_00"]Because of homogenity we may take $abcd=1$.[/quote]
Asking for a silly question, why becuase of homogenity may we take $abcd=1$? What does homogenity mean?

[quote="suimong"]
Asking for a silly question, why becuase of homogenity may we take $abcd=1$? What does homogenity mean?[/quote]

If $a=ka',$ $b=kb',$ $c=kc'$ and $d=kd',$ where $k>0$ and $a'b'c'd'=1$
then $\frac{1}{a^{2}+ab}+\frac{1}{b^{2}+bc}+\frac{1}{c^{2}+cd}+\frac{1}{d^{2}+da}\geq\frac{2}{\sqrt{abcd}}\Leftrightarrow$
$\Leftrightarrow\frac{1}{k^{2}(a'^{2}+a'b')}+\frac{1}{k^{2}(b'^{2}+b'c')}+\frac{1}{k^{2}(c'^{2}+c'd')}+\frac{1}{k^{2}(d'^{2}+d'a')}\geq\frac{2}{k^{2}\sqrt{a'b'c'd'}}\Leftrightarrow$
$\Leftrightarrow\frac{1}{a'^{2}+a'b'}+\frac{1}{b'^{2}+b'c'}+\frac{1}{c'^{2}+c'd'}+\frac{1}{d'^{2}+d'a'}\geq2,$ where $a'b'c'd'=1.$ :wink:


If ABC is a triangle with the angles A, B, C less than 90 degrees, find the maximum of the expression:
\[ E \equal{} \frac {\cos A\cos B\cos C}{\sin ^{2} A\sin ^{2} B\sin ^{2} C} .
\]

To find extrem of $ E(A, B, C) \equal{} \frac {\cos A \cos B \cos C}{\sin^2 A \sin^2 B \sin^2 C}$ subject to $ G(A, B, C) \equal{} A \plus{} B \plus{} C \minus{} \pi \equal{} 0$ and $ 0 < A, B, C < \pi,$ define $ H(A, B, C) \equal{} E(A, B, C) \plus{} \lambda G(A, B, C)$ and put $ \frac {\partial H}{\partial A} \equal{} \frac {\partial H}{\partial B} \equal{} \frac {\partial H}{\partial C} \equal{} 0,$ yielding

$ \lambda \equal{} \frac {\sin A \cos B \cos C \sin^2 B \sin^2 C (\sin^2 A \plus{} 2 \cos^2 A)}{\sin^4 A \sin^4 B \sin^4 C} \equal{}$

[color=white]. .[/color] $ \equal{} \frac {\sin B \cos C \cos A \sin^2 C \sin^2 A (\sin^2 B \plus{} 2 \cos^2 B)}{\sin^4 A \sin^4 B \sin^4 C} \equal{}$

[color=white]. .[/color] $ \equal{} \frac {\sin C \cos A \cos B \sin^2 A \sin^2 B (\sin^2 C \plus{} 2 \cos^2 C)}{\sin^4 A \sin^4 B \sin^4 C}$

The 1st equation gives

$ \cos B \sin B (1 \plus{} \cos^2 A) \equal{} \cos A \sin A (1 \plus{} \cos^2 B)$

$ \sin (2B) (3 \plus{} \cos (2A)) \equal{} \sin (2A) (3 \plus{} \cos (2B))$

$ \sin (2(A \minus{} B)) \equal{} 3 (\sin (2B) \minus{} \sin (2A))$

$ \sin(A \minus{} B) \cos (A \minus{} B) \equal{} \minus{} 3 \cos (A \plus{} B) \sin (A \minus{} B)$

Either $ \sin (A \minus{} B) \equal{} 0,$ which means $ A \equal{} B,$ or $ \cos (A \minus{} B) \equal{} \minus{} 3 \cos (A \plus{} B),$ which reduces to $ \tan A \tan B \equal{} 2.$ Similarly, we get

either $ B \equal{} C$ or $ \tan B \tan C \equal{} 2,$
either $ C \equal{} A$ or $ \tan C \tan A \equal{} 2.$

Assuming $ B \neq C,\ C \neq A$ and dividing the alternate equations leads to $ \tan A \equal{} \tan B,\ A \equal{} B.$ Substituting $ A \equal{} B \equal{} \frac {\pi \minus{} C}{2},$ $ \tan A \equal{} \tan B \equal{} \frac {1}{\tan \frac {C}{2}}$ into either alternate equation yields

$ 2 \equal{} \frac {\tan C}{\tan \frac {C}{2}} \equal{} \frac {2}{1 \minus{} \tan^2 \frac {C}{2}},$

or $ \tan \frac {C}{2} \equal{} 0,\ C \equal{} 0,$ which is contradiction. Thus we also have $ B \equal{} C,$ the $ \triangle ABC$ is equilateral. Since $ E(\pi/3, \pi/3, \pi/3) \equal{} (2/3)^3 > 0,$ while on the acute triangle boundary $ A \equal{} B \plus{} C \equal{} \pi/2,$ etc., $ E(\pi/2, B, C) \equal{} 0$ and on the arbitrary triangle boundary $ A \equal{} 0, B \plus{} C \equal{} \pi,$ etc., $ E(A, B, C) \rightarrow \minus{} \infty,$ $ E(A, B, C)$ has absolute maximum for $ A \equal{} B \equal{} C \equal{} \pi / 3.$


An investment in a lease offers returns of 3713 dollars per month due at the
begining of each month for 4.5 years. What investment is justified if the
returns are deferred for two years and the interest required is 6.12%
compounded monthly?

[hide="answer"][/hide]$6.12/12=.51$. The discount at month $i$ is $1/1.0051^{i}$. If we call this ratio $r$, then the investment value is. $r^{24}3713(1-r^{54})/(1-r)$. I get this from the formular for the sum of a finite geometric series. Doing it out we get a value of $155 565.967$.[/hide]

What is the formula used for this particular word problem?

Can you break the formula down for me piece by piece?


AMC
USA(J)MO
USAMO
inequalities
Olimpiada de matematicas

HOLA ACA UN PROBLEMA QUE BUENO, NO SE SI ESTA BIEN,
----
SI $a,b,c > 0$ y $abc=1$, demostrar que
$(a+b-1)(b+c-1)(c+a-1)\geq 1$

[color=blue]Basta probar con [i]a[/i]=4 y [i]b[/i]=[i]c[/i]=1/2 para comprobar que no es correcto el enunciado.

¿Puedes revisar nuevamente donde encontraste este enunciado?[/color]

[quote="conejita"]Hola!!
Pienso que tal vez querias poner la sig. desigualdad:
http://www.mathlinks.ro/Forum/viewtopic.php?p=354109#p354109

Es bastante similar, con la diferencia de que esta si se cumple jeje[/quote]
bueno, jajaj si me confundi, aa conejita, tampoco es ese problema, es otro ya lo pondre, chaufas  :blush:

Bueno ya que la desigualdad anterior no se cumplia, pongo ahora si una que si cumple  :D
$a,b,c$ son los lados de un triangulo.
demostrar la siguiente desigualdad.
$(a+b+c)(1/a+1/b+1/c)\geq 6(a/(b+c)+b/(a+c)+c/(a+b))$,
---
[hide="SUGERENCIA"]BUENO MI SUGERENCIA, ES ALGO MMM BUENO SE PODRIA DECIR... LOCA PERO ES OPERAR COMO LOCO, JAJAJAJ  :rotfl:  SI SALE, PERO HACIENDO UN CAMBIO DE VARIABLE FAVORABLE, EN $a,b,c$, EN VARIABLES QUE CUMPLAN PARA TODOS LOS REALES.
$a=x+y$, $b=y+z$, $c=z+x$, para todo $x,y,z$ reales positivos.
 POSITIVOS[/hide][/hide]

Mi solucion es talachera. Usamos la transformacion de Ravi, reducimos lo mas que se pueda los terminos obtenidos y por ultimo una forma rapida de acabar es usando muirhead!!
 :lol:
Si alguien tiene una solucion bonita, xfavor que la postee!!

[color=blue]Una vez más, disculpen mi ignorancia. ¿De qué se trata la transformación de Ravi?[/color]

La sustitucion de Ravi es si $a,b,c$ son lados de un triangulo entonces existen $x,y,z \in \mathbb{R^{+}}$ tales que:
$a=x+y$
$b=y+z$
$c=z+x$

[color=green]Discúlpenne, pero soy nuevo y hay algunos términos que desconozco... Alguien podría explicarme qué significan \geq, \mathbb y otros términos semejantes?

Agradecería la aclaración...
[/color]

[color=blue]¡Así que ése era el nombre de la mencionada propiedad![/color]

Ya ps conejita pon tu solucion, jajaja aun es algo horrible pero pon la  :D .
chau.

Talvez si alguien pudiera demostrar la siguiente desigualdad:

$\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z}+9 \geq 6(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1})$; $x+y+z=1$ y $x,y,z \in (0,1)$


vector
linear algebra
matrix
combinatorics unsolved
combinatorics
Linear algebra in combinatorics

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

Parity High School has $n$ students. The school rules allow the students to form clubs, but dictate that each club must have an odd number of members, and that any two clubs must have an even number of members in common. What is the largest possible number of clubs?

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

Quite a few similar problems have been discussed before on the board.

Identify each club by its characteristic vector, i.e. consider it as a vector $\mathbf{v}_j = \left(x_1^j,x_2^j,\ldots,x_n^j\right)$, where $x_i^j$ is $1$ if student $i$ is in the $j$'th club, and $0$ otherwise. We have $AA^t=I_m$, where $A\in \mathcal M_{m,n}\left(\mathbb{Z}_2\right)$ is the matrix having the vectors as rows and $m$ is the number of clubs. This, in turn, shows that $m$ cannot be $>n$, since it would imply that the product of two matrices with rank $n<m$ has rank $m$, which is absurd. We thus get $m\leq n$, and the bound can obviously be achieved, by forming a club for each individual student.


I haven't had as much time as I would like to work on solutions; I've solved #'s 1 and 3, mostly done with 5, and have a solution for 4 that I think is wrong, but has some work that might get me some points. Would it be worth sending in my solutions even though I haven't been able to answer them all? It's possible that I'll get them all by the end of tomorrow, but I doubt it...

Yeah, you don't have to do all the problems. In 8th grade, I only did #1, 2, and 5. Number 5 I screwed up badly so I only got 1 point.

This year, I've done 1, 3, and 5. (I [i]think[/i] I have 5...) I'm working on #4 right now, and I'm almost done. I'm pretty sure I won't be able to do #2. It's so hard... :?

[quote="DPatrick"]If you don't have solutions to all the problems, that's fine.  Submit whatever you have.

Very important: only submit once.  Don't submit, say, 3 solutions now and a 4th solution tomorrow night.    And if emailing, submit as ONE file; don't submit a separate file for each problem.[/quote]

Eh..

The Latex template in the USAMTS site didn't work with my computer.. And I made my own form and apparently, I had problem with page numbers (Like since it was in same document, Problem 2 started in page 4).

So I sent in different PDF..  :? I didn't know if this wasn't allowed... I submitted all different PDF solutions attached in one email though..  :?  :?  :huh:  :huh:

[quote="DPatrick"]It's not ideal, but we'll allow it.  Next time, please submit only one file.[/quote]

Thanks.

I'm going to send my solutions by e-mail, but I'm kinda confused. Do I have to send just a PDF file, or do I have to send the PDF, TEX, and all the image files? And, I still have to send in the Entry and Permission forms. Do I have to do that by mail? Will there be a problem that I'll be sending part by mail and part by e-mail?

Also, what e-mail address should I send the answers to? What if I have more than 5 images to attach, but my e-mail provider only permits up to five attachments?

For the purposes of grading, we only need the output (PDF) file.  If you want your solution to be considered for publication, we need the source code: the .tex file and any images that are included.  Send to solutions@usamts.org.  If you have more images that your email provider can handle, zip them first and send them as a zip archive.

It's OK if the Entry/Permission forms arrive via a different method than your solutions.

Edit: I probably shouldn't say anything related to USAMTS for now... :|

My solution isn't very elegant, but it works! :D

[quote="JesusFreak197"]Hmm... I have a question about how I should write one of my solutions... it's a general question, but I don't want to ask it here, because it might give something away. Can I PM it to an admin?[/quote]If you even suspect that you might be "giving something away" by asking your question, then we wouldn't be able to answer it anyway.


modular arithmetic
number theory unsolved
number theory

[quote="N.T.TUAN"]
Note $\text{ord}_{3^{m+1}}(2)=2.3^{m}$.[/quote]

Why is this?

Also, it should be $-2$, not $2$,...

[quote="barasawala"]

Why is this?[/quote]
I think so, but I will prove later.
[quote]

Also, it should be $-2$, not $2$,...[/quote]
Because $3^{i-j}-1$ is even!  :wink:

[quote="N.T.TUAN"]

Because $3^{i-j}-1$ is even!  :wink:[/quote]

Okay. I'm sorry I failed to see this!

Another thing, I think $2.3^{m}|3^{j}(3^{i-j}-1)$ is not a contradiction!

[quote="barasawala"]Let $m\in Z^{+}$.
 Show that there are no $i,j\in Z$ such that $0\leq i<j\leq 3^{m}-1$ and
$3^{m+1}$ divides $(-2)^{3^{j}(3^{i-j}-1)}-1$[/quote]
Problem of you is wrong! $3^{j}(3^{i-j}-1)=3^{i}-3^{j}<0$, so $(-2)^{3^{j}(3^{i-j}-1)}-1\not\in\mathbb{Z}$ :D

[quote="barasawala"]Hey, I reversed $i, j$  :D  :blush:[/quote]
If so, What about if $i>j>m$ ? :D

[quote="barasawala"]Well, so what's wrong with that?[/quote]
If $i>j>m$ then $2.3^{m}|3^{j}(3^{i-j}-1)$, therefore $2^{3^{j}(3^{i-j}-1)}\equiv 1\pmod{3^{m+1}}$.


search
advanced fields
advanced fields unsolved

I was checking out arXiv.org on Linear Spaces and Finite Geometries and encountered the following article:

[url=http://arxiv.org/abs/0809.4144v1]The Continuum is Countable: Infinity is Unique[/url]

Is this one of those misinterpretations of 'infinity', much like the infamous '[url=http://www.mathlinks.ro/viewtopic.php?search_id=1303419311&t=30778]DoromShadmi[/url]' and his argument of $ 0.3333... < 1/3$ and Koch curves?

 :huuh:

EDIT: I'm not sure if this is the right forum to ask this but yea :P

Routine crankery. A good takedown is [url=http://scienceblogs.com/goodmath/2009/01/the_continuum_hypothesis_solve.php]here[/url].

I wouldn't classify it as "crankery". The author just makes a (rather bad, but understandable) mistake confusing the sets of finite and infinite paths in the tree but if, say, a 9th grader who just learned about the notion of cardinality had come up with something like this argument, I would consider this as a sign of cleverness, not wackiness.

A friend of mine (who is now a rather well-known mathematician) once told me that when he first learned about infinite sets, he was completely sure there were as many rational numbers as there were irrational ones just because between every 2 rationals there is an irrational and vice versa. This belief persisted until he learned the Cantor theorem, at which point he had to talk to his teacher for quite a while to figure out why his idea wasn't right.

Generally speaking, the "General mathematics" section of the arXiv is something like a junk yard for such things (just view it as an analogue of Fun Factory on AoPS). This doesn't mean that there are no good articles there at all but the "spam rate" is very high :(.

[quote="fedja"]Generally speaking, the "General mathematics" section of the arXiv is something like a junk yard for such things (just view it as an analogue of Fun Factory on AoPS). This doesn't mean that there are no good articles there at all but the "spam rate" is very high :(.[/quote]
Some say that GM really means "Garbage Mathematics"

Awesome link, jmerry, thx! The whole paper is quite deceiving and convincing, for the inexperienced like me  :blush: . And something I've learned from this: I have a better understanding about infinite sets now than I thought I did before :) I still enjoy reading the parody in DoronShadmi's post hehe.


Let a triangle ABC and $ X,Y,Z$ are three points on $ BC,CA,AB$ respectively such that $ AX\equal{}BY\equal{}CZ$ and $ \hat{BAX}\equal{}\hat{CBY}\equal{}\hat{ACZ}$. Prove that ABC is an equilateral triangle.

Hi! This problem is posted for third time...here I'm interested in it too.
The problem is not easy at all but it is very good.
Here you may take a look about the discussion about it:
http://www.mathlinks.ro/Forum/viewtopic.php?t=99674
It may be generalized. On 17-th of May I've a birthday. It means - on 19-th of May I'll post
the solution I've and moderators may put this problem on first page of mathlinks.ro...
 :wink:

[quote="borislav_mirchev"]
It may be generalized. On 17-th of May I've a birthday. It means - on 19-th of May I'll post
the solution I've.[/quote]
But why so late?  :maybe:

There are two reasons:
1. To give time to other mathlinkers to spend good time trying to solve this problem.
2. I'll say my second reason on 19-th of May - it is a good reason.

Take a look at this...http://www.mathlinks.ro/viewtopic.php?p=1131812#1131812


An employment agency employs a number of people.  Among these, 15 are American and 24 are women.  Only eight of the employees are neither American nor Swedish, and half of these are women.

a. Are there more American male employees than Swedish female employees?

b.  If it is known that at least one American man works for the company, could there be fewer than 30 employees in the agency?

sounds like a classic Inclusion-Exclusion Property problem

Try showing a solution for challenge problems. There is a good example of answers in spoiler text in this thread, but now let's talk through how to show a solution that reaches the answers.


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate ${ \int_0^{\frac{\pi}{8}}\frac{\cos  x}{\cos (x-\frac{\pi}{8}})}\ dx$.

Is it $ sin\left( {\frac {\pi}{8}\right) log(\cos(\frac {\pi}{8})) - \frac {\pi}{8}cos \left( \frac {\pi}{8}\right )}$?? :blush:

The magnitude is correct but the sign is wrong.

Oh.Yes. I am sorry.

Let $ u \equal{} x \minus{} \frac {\pi}{8}$

$ du \equal{} dx$ , $ u: \frac { \minus{} \pi}{8} \to 0$

Now the integral transforms as

$ \int_{\frac { \minus{} \pi}{8}}^{0} \frac {cos(u \plus{} \pi / 8)}{cos(u)}.du$

$ \equal{} \int_{\frac { \minus{} \pi}{8}}^{0} \frac {cos(u)cos(\pi / 8) \minus{} sin(u)sin(\pi / 8)}{cos(u)}.du$

$ \equal{} \frac {\pi}{8}cos(\pi / 8) \plus{} sin(\pi / 8)log[cos(\pi / 8)]$ :)

$ \int_{0}^{\frac {\pi}{8}}\frac {\cos x}{\cos\left(x \minus{} \frac {\pi}{8}\right)}dx$
$ \equal{} \int_{0}^{\frac {\pi}{8}}\frac {\cos x}{\cos{\frac {\pi}{8}x}\cos{x} \plus{} \sin{\frac {\pi}{8}x}\sin{x}}dx \equal{} \int_{0}^{\frac {\pi}{8}}\frac {\cos{x}}{a\cos{x} \plus{} b\sin{x}}dx$, where $ a \equal{} \cos{\frac {\pi}{8}}$ and $ b \equal{} \sin{\frac {\pi}{8}}$.

Now, the half angle substitution $ t \equal{} \tan{\frac {x}{2}}$ gives the answer, or thinking of $ I \equal{} \int_{0}^{\frac {\pi}{8}}\frac {\cos{x}}{a\cos{x} \plus{} b\sin{x}}dx$, $ J \equal{} \int_{0}^{\frac {\pi}{8}}\frac {\sin{x}}{a\cos{x} \plus{} b\sin{x}}dx$ is another good way. It is easily seen that $ aI \plus{} bJ \equal{} \frac {\pi}{8}$ and $ bI \minus{} aJ \equal{} \left[\ln{(a\cos{x} \plus{} b\sin{x})}\right]_{0}^{\frac {\pi}{8}} \equal{} 0(\because a^2 \plus{} b^2 \equal{} 1)$, we can find $ I$ by solving this simultaneous equations.

Yes, your solution is one of other solutions.

We have more two solutions at least.


geometry
geometric transformation
reflection
angle bisector
geometry unsolved

Reflect $ F$ in $ V_1, V_2$ into $ F_1, F_2.$ Draw directrices $ d_1 \perp FF_1, d_2 \perp FF_2.$ Construct 2 circles $ (A), (B)$ tangent to $ d_1, d_2$ and passing through $ F.$ $ AB$ is the common chord of the 2 parabolas.

don't see where are the directrices d1 and d2  :(

$ d_1 \perp FF_1$ through $ F_1,$ $ d_2 \perp FF_2$ through $ F_2.$

Circles $ (A), (B)$ do not have diameters $ FF_1, FF_2.$ Circle $ (A)$ is tangent to both $ d_1, d_2$ and goes through $ F.$ Same for circle $ (B).$
 [hide]Let $ D \equiv d_1 \cap d_2.$ Let $ l$ through $ D$ bisect $ \angle (d_1, d_2)$ containing $ F.$ Pick arbitrary point $ P \in l$ and let $ Q \in d_1$ be foot of perpendicular to $ d_1$ from $ P.$ Drav circle $ (P)$ with radius $ PQ.$ $ DF$ cuts this circle at $ U, V.$ Parallels to $ QU, QV$ through $ F$ cut $ d_1$ at $ X, Y.$ Perpendiculars to $ d_1$ at $ X, Y$ cut the angle bisector $ l$ at $ A, B.$ [/hide]

[quote="yetti"]...Perpendiculars to $ d_1$ at $ X, Y$ cut the angle bisector $ l$ at $ A, B.$[/quote]
You constructed perpendiculars through $ X, Y$ to the angle bisector $ l$ of $ \angle F_1DF_2.$ That's incorrect.


function algebra polynomial Vieta quadratics ARML

A straight line   touches to graf of the  function
$y=x^4+2x^3-3x^2-x+6$   in two various points.
Find the equation of the straight line.

by touch to you mean tangent?, if so, then that means there are 2 double roots when solving the system, so let the 2 solutions be a,b, then

$2(a+b)=-2$, $a^2+4ab+b^2=-3$ by vieta

$a+b=-1$, $ab=-2$

(easy to verify that these are real by making a quadratic with these roots and using discriminant)

then when intersecting with $y=mx+c$, we compare coefficients and use vieta to get:
$(-1-m)=-2ab(a+b)$, $6-c=(ab)^2$
then $m=3$, $c=2$, so $\boxed{y=3x+2}$

Nice solution, Altheman!
[quote="Altheman"]by touch to you mean tangent?[/quote]
My English is very bad.
To touch $\neq$ to tangent ? Thank you.

[quote="arqady"]To touch $\neq$ to tangent ? Thank you.[/quote]

There is no verb "to tangent" in English, just "to be tangent to."  One can use "to touch" there, but it is slightly ambiguous -- I personally would stick with "to be tangent to."

This is also not an easy question -- it was a number 8 at ARML a few years ago and there were very few perfect scores that year, I believe.

A harder problem from Sofia University 2005:
find all tangents to the graf of $x^4+2x^3-12x^2$ that have exactly [b]two common poitns with the graf.[/b]

It is not in fact so much harder, once you know the idea:  Consider any such tangent, $y = ax + b$.  Then the graph of $x^4 + 2x^3 - 12x^2 - ax - b$ has exactly two distinct roots, one for each intersection point, and one of these two is at least a double root.  Three real roots of a quartic means four real roots.  Thus it has either one triple root and another real root or two double roots.  From here it is essentially the same as arqady's question.


floor function
combinatorics unsolved
combinatorics

On sport games there was 1991 participants from which every participant knows at least n other participants(friendship is mutual).Determine the lowest possible n for which we can be sure that there are 6 participants between which any two participants know each other.

The Caro-Wei theorem states that $ \alpha(G) \geq \sum_{v \in V} \frac {1} {\deg_G(v) \plus{} 1}$, where $ \alpha(G)$ is the independence number of $ G$. To estimate $ \omega(G)$, the clique number of $ G$, i.e. the size of the largest complete subgraph of $ G$, we need to work with the complement graph $ G^*$ of $ G$, since $ \omega(G) \equal{} \alpha(G^*)$.
We have $ \deg_G(v) \geq n$ for all $ 1991$ vertices, so $ \deg_{G^*}(v) \leq 1990 \minus{} n$. Therefore $ \alpha(G^*) \geq \sum_{v \in V} \frac {1} {\deg_{G^*}(v) \plus{} 1} \geq \frac {1991} {1991 \minus{} n}$, and to be at least $ 6$ we need $ n \geq \left \lfloor \frac {5} {6} 1991 \right \rfloor \equal{} 1659$.
Alternatively, look up Zarankiewicz theorem.

aspava, actually yesterday your location was Izmir, Turkey. How come?

mavropnevma, you've shown that any such $ n$ is large enough, but not that this value of $ n$ is minimal.  By Turan, the graph on 1991 vertices with no $ K_6$ and maximal number of edges is $ K_{398, 398, 398, 398, 399}$.  This graph has $ 6\cdot 398^2 \plus{} 4\cdot 398 \cdot 399$ edges.  Thus, if the given graph has more than this number of edges, it contains a $ K_6$.  It follows that it suffices to have $ \frac{1991 n}{2} \geq 6\cdot 398^2 \plus{} 4\cdot 398 \cdot 399$, and this is achieved for all $ n \geq 1593$.  This also does not show that there is a 1592-regular graph on 1991 vertices with no $ K_6$, and I don't see an obvious way of modifying the Turan graph to find such a $ G$, so the answer could even be smaller.

There is such a simple modification. Label the vertices with ordered pairs $ (i,n)$ so that $ n\in \{1,2,\cdots,398\}$ for $ i\in\{1,2,3,4\}$ and $ n\in\{1,2,\cdots,399\}$ for $ i\equal{}5$. Connect each $ (i,n)$ to $ (i',n')$ whenever $ i\neq i'$, then remove the edges between $ (1,n)$ and $ (2,n)$ and the edges between $ (3,n)$ and $ (4,n)$. Each vertex has degree either $ 397\plus{}398\plus{}398\plus{}399\equal{}1592$ or $ 398\plus{}398\plus{}398\plus{}398\equal{}1592$. Since we only removed edges, we didn't add any $ K_6$ subgraphs.


How many ways are there to put 8 rooks on a chess board so that they do not attack each other

[hide=" solution"]

Not sure but: there must be a rook in each row and each column. This gives $ 8$ possibilities for the first row, $ 7$ for the second row, $ 6$ for the third row and so on, giving $ 8!\equal{}40320$ ways in all.

[/hide]

[hide="Click here to reveal hidden content"]

There are 8 ways to put a rook on the first row, 7 ways to put it on the second row, 6 ways to put it on the third row.... so there are 8!=8*7*6....*2 =[b]40320[/b] ways.[/hide]

Sorry,is this a joke?
$ 8!\equal{}40320$


How many odd perfect square factors does $2^4 * 3^6 * 5^{10} * 7^9$ have?

I know how to do this problem the long way, but is there a fast method to solve this problem?  Say in less than a minute.

[hide="I think..."] [b]120[/b]
Since the factors have to be odd, 2 can't be included, because 2*anything makes it even. Also the exponents in the prime factorization need to be even for the number to be a perfect square. So, the prime factorization can be
3^(0,2,4,6)*5^(0,2,4,6,8,10)*7^(0,2,4,6,8)
4*6*5
120[/hide]

I guess you could call that a fast method.  I just counted the ones with the 2nd power 4th power and so on.  Oh well, thanks.

[hide]
Odd means so powers of 2.
Then you have to have an even number of each other factor
4*6*5=120[/hide]


limit
induction
quadratics
geometric series
algebra proposed
algebra

[quote="nguyenvuthanhha"][i]A more beautiful solution :
... [/quote]

  :huh:

Alternately, call the sum $ S$. Then subtracting $ \pi^{\minus{}1}S$ will give reduce the degree of the numerators to linear (from quadratic), and repeating this with the new series will reduce the numerators to constants (ie a geometric series)


[quote]Fat [that] people eat accumulates.

When Fred eats[,] food gets thrown. (??)

I convinced her [that] children are noisy.
Until the police arrest[,] the drug dealers control the street.[/quote]

These are weird... :D

I think it could also be imperative:
Until the police arrest the drug dealers[,] control the street.

[hide=" for one of them"]The cotton [that] clothing is usually made from grows in Mississippi.[/hide]

Actually, some of these are grammatically incorrect. It is necessary to place commas behind introductory phrases.

Is it? The book claims they're grammatically correct.

I'm personally weirded out by "Prime number few" the books solution:

"Prime [people] number few"

I'm surprised nobody got this one:
The dog that I had really loved bones.

Basically, "I had a dog that really loved bones".

[quote="randomdragoon"]Fat people eat accumulates.
The cotton clothing is usually made from grows in Mississippi.
When Fred eats food gets thrown.
The horse raced past the barn fell.
Prime number few.
We painted the wall with cracks.
Mary gave the child the dog bit a plaster.
I convinced her children are noisy.
Until the police arrest the drug dealers control the street.
The dog that I had really loved bones.

Why are these grammatically correct sentences? How could they be improved to make them clearer?[/quote]

Fat people eat accumulates [of something; just as we can have "precipitates of Ca(OH)2," we can have "accumulates of [something]."  :D ]
The cotton [that] clothing is usually made from grows in Mississippi
When fred eats[,] food gets thrown.
The horse [that was] raced past the barn fell.
[The] prime number few.
We painted the wall [that was covered] with cracks.
Mary gave the child [that] the dog bit a plaster.
I convinced her [that] children are noisy.
Until the police arrest [them, ] the drug dealers control the street.
The dog that I had [was a dog that] really loved bones.


We need to prove this: $ \sqrt{(1\minus{}2a)(1\minus{}2b)(1\minus{}2c)}\le3\sqrt{3abc}$ ??

So let's try:
Let
x=c+b-a
y=a+c-b
z=a+b-c
then we have
   27(x+y)(y+z)(x+z)=>xyz
 which is obvious after multiplying. and about achieving equation(i doubt that it will achieve :huh: )
 IS something wrong?  :?:

[quote="behdad.math.math"]consider a,b,c>0 and a+b+c=1.prove that:
((1-2a)(1-2b)(1-2c))^1/2 <= 3((3abc)^1/2)[/quote]
This inequality, in fact, is weak! We can prove the stronger inequality:
Given a+b+c=1.Prove that: ((1-2a)(1-2b)(1-2c))^(1/2)=<(abc)^(1/2)(a,b,c>0)(1)
[hide]Proof:[/hide]
We have: LHS=((b+c-a)(a+b-c)(c+a-b))^(1/2)
W.L.O.G, assume that a=max(a,b,c). We will prove that: (b+c-a)(a+b-c)(c+a-b)=<abc(a,b,c>0)(2)
1st case: b+c-a<0.The inequality (2) is obviously true
2nd case: b+c-a>0.Then applying AM-GM we have: [(b+c-a)(a+b-c)(c+a-b)]^2=<(abc)^2, which means (a+b-c)(b+c-a)(c+a-b)=<abc.
From 2 cases above, we can state that (1) is always true.
The equality occurs when a=b=c=1/3.

[quote="dgreenb801"]I think that's wrong because x y or z may be negative.[/quote]

They can't:
Suppose x,y or z negative. It's obvious that only one of x,y,z can be negative(suppose otherwise and x,y-negative(which pair is in't necessary), but x+y=(b+c-a)+(a+c-b)=2c>0 which is contradiction). So we have at most one negative. Suppose x is negotive(so we have xyz<0), but

\sqrt{(xyz}\ = \sqrt{(1-2a)(1-2b)(1-2c)}\>0 contrudiction so x,y,z>0
  :wink: Am i wrong :huh:  :?:

Sorry for my poor English  :roll:


linear algebra
matrix
vector
symmetry
algebra
linear equation
linear algebra theorems

What is Cramer's Rule? How is it used in solving systems? What do the determinants in Cramer's method for solving systems represent?

P.S.: I don't know where to post this topic. Maybe the moderators will move it to another section.

i think this belongs to Advanced Algebra section.

You can find it in almost every linear algebra book.
You can also have a look at
http://mathworld.wolfram.com/CramersRule.html

I know this is in the wrong section, but I didn't want to begin a new thread.  Maybe this can be moved.

Cramer's Rule states the following:

$x_i=\frac{\det A_i(\mathbf{x})}{\det A}$

Here's an example:

$x_1+2x_2=5$
$3x_1+7x_2=17$

I will compute $x_1$.

$A=\begin{bmatrix}1 & 2 \\ 3 & 7\end{bmatrix}$

The meaning of $A_i$ is that you replace the "i"th column of A with the vector x, which in this case is <5,17>.

$A_1=\begin{bmatrix}5 & 2 \\ 17 & 7\end{bmatrix}$

So $\det A=7-6=1$ and $\det A_1(\mathbf{x})=35-34=1$.

$x_1=\frac{\det A_1(\mathbf{x})}{\det A}=1$

Maybe you can try to find $x_2$ with this method.  It is actually very simple and quick.  I can provide a proof of the rule as well if you need it.

Alex.

Cramer's Rule is: If we have a set of equations
\[ \\a_1x+b_1y+c_1z=k_1\\a_2x+b_2y+c_2z=k_2\\a_3x+b_3y+c_3z=k_3 \]
$x=\frac{\left|\begin{matrix}k_1&b_1&c_1\\k_2&b_2&c_2\\k_3&b_3&c_3\end{matrix}\right|}{\Delta}$  $y=\frac{\left|\begin{matrix}a_1&k_1&c_1\\a_2&k_2&c_2\\a_3&k_3&c_3\end{matrix}\right|}{\Delta}$  $z=\frac{\left|\begin{matrix}a_1&b_1&k_1\\a_2&b_2&k_2\\a_3&b_3&k_3\end{matrix}\right|}{\Delta}$  $\Delta=\left|\begin{matrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{matrix}\right|$

Of course, this also works for higher order determinants.

P.S.  I'm 14 years old, so I don't know if this is around what you wanted to know. :)

Masoud Zargar

What if our system has coeficients not in $R$(or $C$), but in a field $K$? If $K$ is commutative, it is clear, Cramer is still available. But what if it is not commutative?

Zero'th problem: fields are commutative for me ;)

First problem for noncommutative case:
A linear equation is not necessary of type $a_1 x_1 + a_2 x_2 +...=d$, but can contain more than one linear term of type $a_1 x_1 b_1$ with $x_1$.

Next problem: the definition of determinant looses it's symmetry of changing rows or columns, thus even interchanging the equations or variables could give us another determinant.

Third problem: even the division in the rule is not defined on which side to be taken.

Conclusion: by some random constants, e.g. from the quaternions, there should be a counterexample.

PS: moderators, please move this thread...


A patient takes one pill every 8 hours until the whole bottle of
30 pills is gone. How many hours elapsed between the time the
first pill was taken and the time the last pill was taken?

he took one so he has 29 left. he takes them in periods of 8 hours each so 29*8=232 hours


Find the sum of the first 12345 terms of the sequence 1/(4n^2-1)

well the sequence is as shown in my reply. it's $\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}...+\frac{1}{24689\cdot24691}$. since 3-1=2, 5-3=2, etc. this can be simplified to $\frac{1}{2}(1-\frac{1}{24691})$ which simplifies to the answer I got.

edit: yeah, telescope. :dry:

[quote="wly3298456"]Can you explain more of how you got the answer?[/quote]

He used what's called the "telescoping series." Unforunately, this is the term that doesn't appear on math books often.. I'm not sure why but the first time I saw was on Mathematical Circles. Anyway, note that $\frac{1}{4n^2-1}$ which has denominator that factors as $(2n-1)(2n+1)$. Moreover, we can see that $\frac{1}{4n^2-1} = \frac{1}{2} \left(\frac{1}{2n-1} - \frac{1}{2n+1} \right)$. Hence, the problem now becomes:

$\sum^{12345}_{n=1} \frac{1}{2} \left(\frac{1}{2n-1} - \frac{1}{2n+1} \right) = \frac{1}{2} \left(\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \cdots \right)$

As you see, the terms will get canceled (or telescoped) and that's how you'll reach the answer.


MIT
college
Putnam
real analysis
real analysis unsolved

Suppose $\frac{1}{1-x-y-z+6(xy+yz+zx)} = \sum_{r,s,t \geq 0} f(r,s,t) x^r y^s z^t$ for |x|, |y|, |z| sufficently small.  What is the radius of convergence of the power series $F(u) = \sum_{n \geq 0} f(n,n,n) u^n$?

Just out of curiosity, where is the MIT putnam site?

I found a paper containing this particular problem [url=http://www-math.mit.edu/~rstan/S34/]here[/url].

Yes, Grobber, this is my source.

Does anybody have any good ideas?


Prove that, for any prime $ p$, there exist integers$ x$, $ y$, $ z$, and $ w$ such that $ x^{2}\plus{}y^{2}\plus{}z^{2}\minus{}wp\equal{}0$ and $ 0<w<p$

http://www.mathlinks.ro/Forum/viewtopic.php?t=49271
The first thing it came to my mind was Arne's claim, but it is not nice at all
Zetax's approach seems quite interesting


algebra
polynomial
calculus
quadratics
algebra proposed

[quote="hoangvn.fix"]$ \frac {1}{x \minus{} 1} \plus{} \frac {1}{x \minus{} 2} \plus{} \frac {1}{x \minus{} 6} \plus{} \frac {1}{x \minus{} 7} \equal{} x^2 \minus{} 4x \minus{} 4$[/quote]

Reducing this equation gives a $ 6^{th}$ degree polynomial with $ 6$ real roots :

$ x_1\equal{}\minus{}0,588556838015879...$
$ x_2\equal{}0,814707811504444...$
$ x_3\equal{}1,88477221409876...$
$ x_4\equal{}4,72282255260248...$
$ x_5\equal{}6,10494973034297...$
$ x_6\equal{}7,06130452946723...$

I found no intelligent solution to find these values, except numeric calculus (I used Newton's method) and i would be very interested in your own clever method (you have one since you posted in "proposed and own" forum).

dear pco for this problem sorry for give you the wrong equation but it is not my fault ,the fault belongs to the crux because of mistake in typing the equation is
$ \frac{1}{x\minus{}1}\plus{}\frac{2}{x\minus{}2}\plus{}\frac{6}{x\minus{}6}\plus{}\frac{7}{x\minus{}7}\equal{}x^2\minus{}4x\minus{}4$
this equation is easy to solve after finding $ x\equal{}0 , x\equal{}4$ are 2 roots you can solve next by setting $ y\equal{}x^2\minus{}8x\plus{}7$ this is equivalent to a quadratic equation


i need some help with intergration... gettin on my nerves

basically i need to integrate [sec(^2)x]^3/2

i dunno if u can make that out, i cant do superscript..anyway its [sec squared ex]to the power 3/2

please help sumone

if i read it correctly
$\int (\sec^{2}x)^{3/2}\,dx=\int [(\sec x)^{2}]^{3/2}\,dx=\int sec^{3}x \,dx$

The problem turns out to one of classical integration by parts problem. When integrating by parts you will get a constant multiple of the integral you want. Try it and post back

[quote="chawdretto"]cheers mate, so do you mean that i choose u = secx and dv/dx = sec^2(x) ?[/quote]

yes, remember that $\sin^{2}x+\cos^{2}x=1$ and derive the formula for $\tan^{2}x$, you will need it in integration by parts.
Also know
 $\int \sec x \, dx=\ln |\sec x+\tan x|+C$

thanks for the promt replies everyone...i just wanted to know how you can get the formulas up? like get all the symbols so u can write out the whole equation properly

[url]http://en.wikipedia.org/wiki/LaTeX[/url]

That's the name of the type-setter we use, though it may take a while to learn it all. Plus, there's probably a thread on these forums about it.[/url]


In the Bohr model of the hydrogen atom, an electron in the lowest energy state follows a circular path at a distance of 5.29 multiplied by 10-11 m from the proton.
 What is the effective current associated with this orbiting electron?

note that the speed of the electron is $ 2.19\cdot 10^6$ m/s.

the ony formula relating the the current I is one involving area, but thats only for wires and stuff...

hello , can you explain waht you mean by 10-11 m from proton

and can you define [color=red]the effective[/color] current  :blush:

$ I \equal{} \frac {Q}{t}$ where $ t \equal{} \frac {2\pi r}{v}$, so the current associated with the electron is about 1.1*10^-3 A.

Furthermore, when you multiply this current by the area of the orbit you get the magnetic dipole moment which is about 9.3*10^-24 A*m^2 aka the Bohr magneton.


$ a,b,c\geq 0$
prove that:
$ \sum \frac {a}{b \plus{} c} \plus{} \frac {4(a \plus{} b)(b \plus{} c)(c \plus{} a)}{a^3 \plus{} b^3 \plus{} c^3} \geq 5$
I have an idea :
 +prove that $ f(a,b,c)\geq f(a,b,0)$after that,it's simple to prove that $ f(a,b,0)\geq 5$

can any body  tell me that my idea is true or wrong because I can't prove that $ f(a,b,c)\geq f(a,b,0)$
thanks :)

When $ a\equal{}b\equal{}c\equal{}1$ the inequality is false the way it is currently written.

Edit:NVM, thought the sign was reversed (tired :P)


Well, I know what sigmas are usually used for. The whole writing very long sums in a pretty, simple way. However, a long time ago, I learned that they use the sigma symbol to denote some sort of probability ratio. For example, sigma=something over something else. I haven't seen this for a long time, and it's been bothering me recently. Is anyone aware if this exists or is that lesson flawed?

...It has to do with some sort of probability outcome over something else. Meh. I'll continue to Google.

Hmm. But since I don't know a lot about them, are there any other reasons they're used for?
Oh, and how do you write them in LaTeX?

[code]$\sum _{x=0} ^{\infty} x^2$[/code]

produces $ \sum _{x\equal{}0} ^{\infty} x^2$.

Anyway, there is a similar formula in Statistics, where the arithmetic mean of a sequence can be calculated using a formula concerning sigmas. But that's probably not what you're looking for.

The lower case sigma
[code]
\sigma
[/code]
$ \sigma$
Is used to denote standard deviation, and $ \sigma^2$ is the variance. It is also common to denote permutations with either $ \sigma$ or $ \pi$. The last thing I can think of off the top of my head is that $ \sigma(n)$ denotes the sum of divisors function, that is $ \sigma(n) \equal{} \sum_{d|n}d$.

It is used in statistics. I forgot it all but I think I used it for binomial distribution?

We use $ \sigma$ when talking about [url=http://mathworld.wolfram.com/Sigma-Algebra.html]sigma algebras[/url].  $ \sigma_k(n)$ denotes the function $ \sum_{d > 0, d | n} d^k$.  When $ k \equal{} 0$, this is the divisor function (number of divisors of $ n$).

But the [url=http://mathworld.wolfram.com/StandardDeviation.html]standard deviation[/url] is probably what you were looking at.


Well in last topcoder a problem was given at DIV I.

that says

given some words... u have to find out the least length word of which the all the above words is a subsequence

oh i c, analysis is available now... thanks anyway!


quadratics
videos
inequalities
Putnam
algebra
quadratic formula
number theory

my calculator says no.  The factors are not real so there would be no base with a integer number.  However, i can't prove this.

Furthermore, the i parts are irrational.

I think you maybe are misunderstanding the question.

For example, the equation $x^2 -11x + 11 = 0$ has irrational solutions in base 10, but has the integer solution $x=2$ in base 3.

Well  .. yes.. kind of

[hide]

that is if base 1 or base 3 is allowed:

(All are in base 10 unless explicitly noted):

25 (in base 1) = 2*1^2+5 = 7  (base 10)

and 136  (in base 1)(  = 10 (base 10)

and you have (5,2)  as solution.

similary in base 3...

   x^2 - 11 x + 30 also has a solution in integers

Now if you have  an integer base >3 the solution, that is the roots  are  not even real... (why this is easy to show?

BTW normally when you see a "6", you assume that the base is > 6  [/hide]

Gyan there is no base 1 because otherwise we would violated the uniqueness of base representation of numbers. That way I can virtually write any number equal to another number with any number zeroes, or like 25 and 52 are the same in base 1!!!

demji, math does not mean calculator. The power of math is to NOT USE the calculator. So do your problems avoiding calculator as much as possible unless it becomes absolutely necessary to do faster computations.

[quote]Gyan there is no base 1 because otherwise we would violated the uniqueness of base representation of numbers.[/quote]

SO ? :D

Stirctly speaking even in base 10,  25 can be written as 25, 025, 0025, not to mention 25.0 or 24.999999......  And what about 50/2 or 100/4. Are you going to say "there is no such thing as  fraction"  just because it lacks "uniqueness"?  :D  (Of course one can even go further and say "no
"+' are allowed as 25 can be written as 15+10 or 24+1)

Sure,   it is  uncommon,   to put it mildly,   to use  base 1, (or for that matter   non_positive_integers) and many would agree that if you see a number such aw 126, most likely base >6.  But   as long as you define what you mean  (as I tried to do in the above post)  things  can have meaning.

Sorry to nit-pick here, but one should not get overly tied down by "rules". I am sure, you know, there are people who will  go and say "there is no such number as sqrt(2)  - just because it can not be written down as a  rational fraction - or  sqrt(-1)  - negative numbers do not have square-roots - .... and miss all the fun [i]and learning new things[/i]  in math.

Hope this helps. Regards.

Well thx flamer Gyan... you took things too personal. Just trying to say some suggestions, if you ain't like them, keep them to yourself.

I was just trying to help demji have more fun doing maths avoiding calculator, it's more fun, that's all.

And again, thx flamer for taking things too personal...and if you reply to this, you're flaming.

Hello, DPatrick!

[quote]Does the equation $x^2 - 25x + 163 = 0$ have an integer solution in any base?[/quote]
Let $b$ be the base.
The equation becomes: $x^2 - (2b+5)x + (b^2+6b+3) \;= \;0$

From the Quadratic Formula: $\displaystyle{x \;= \;\frac{(2b + 5) \pm \sqrt{(2b + 5)^2 - 4(b^2 + 6b+3)}}{2} }$

Then: $x \;= \;\frac{(2b+5) \pm \sqrt{13-4b}}{2} $

Since $13 - 4b \geq 0$, then $b \leq 3.$

But no base of 3 or less can have a "3", "5" or "6".

Therefore, the answer is: $NO.$

3x-   Honestly trying to help people learn some math, but it  looks likle I was wrong in your case.  Feel free to impress yourself.   Sorry can't help you.
Bye.

[quote="omicron"]What does the question mean by base? My math ain't that good :([/quote]
By base it means the base representation of numbers. You know a number like $982347$ means $9\cdot10^5+8\cdot10^4+2\cdot10^3+3\cdot10^2+4\cdot10+7$ and this number is written in base 10. Sorry, you will have to consult a book in Elementary Number Theory for more information, well sometimes an algebra book teaches number bases.

PS: I just impressed myself!!! :P  :cool:  :P
[Edit]: Nice solution Soroban. I just missed ur post for some unknown phenomenal reason. Maybe this thread is cursed or something  :D  :D

[quote="Gyan"]P.S Sorry  Have no intention to flame anyone but:
 [quote]math does not mean calculator. The power of math is to NOT USE the calculator.[/quote]

Says who? It is just your opinion, an axiom, or can you prove it? :D  :D
IMO,  just   as an astronomer does not mean "not using a telescope"   ... I would say more power to a calculator (or  a  pen, or a pencil , or  a paper, or a  ruler, or any other tool)  if it helps.[/quote]

Foul on the offense!  5 yard penalty for inapplicable parallel, because telescopes are used to make observations, while calculators are used to do that problem-solving on behalf of the student!  Second down, Gyan retains possession.

Seriously, though, calculators more often hinder than help the learning process.  The point of taking a math class is to learn, as the header says, "The Art of Problem Solving."  Button-punching is what we're supposed to do in our spare time (when I refer to spare time, I'm speaking hypothetically) on our video game consoles and PCs.  Honestly, while calculators can be useful, using them to multiply 2 and 3 (as high school math students have been seen doing) is a misapplication of their computing power.

[quote="RobinHood3000"]
Foul on the offense!  5 yard penalty for inapplicable parallel, because telescopes are used to make observations, while calculators are used to do that problem-solving on behalf of the student!  Second down, Gyan retains possession.

Seriously, though, calculators more often hinder than help the learning process.  The point of taking a math class is to learn, as the header says, "The Art of Problem Solving."  Button-punching is what we're supposed to do in our spare time (when I refer to spare time, I'm speaking hypothetically) on our video game consoles and PCs.  Honestly, while calculators can be useful, using them to multiply 2 and 3 (as high school math students have been seen doing) is a misapplication of their computing power.[/quote]

You're also incorrect.  Before setting out to prove a problem such as, say, Golbach's conjecture, it is helpful to know that the conjecture is true up to a certain point.  Similarly, one might assume that the converse of FLT is true, that a^(p-1) = 1 mod p iff p is prime.  However, there are 'pseudoprimes' which satisfy the converse but are not prime.

http://mathworld.wolfram.com/FermatPseudoprime.html

When I'm solving a tricky inequality, I have no qualms with using Maple to expand the whole thing into symmetric sums.  It's easy to make mistakes on such expansions, Maple guarantees that only user error will cause mistakes.  I mean, if I'm doing, say, a Putnam problem, I have a high degree of confidence that the problems will be correctly worded.  I do not have this confidence when doing problems which have a high degree of trickle-down in them.

Frankly, I think that Gyan is in general quite helpful, and that 3X is out of line in what he says.

When, might I ask, did I say I was taking sides?  Moreover, when did the name "3x.lich" even appear in my post?

If you're learning new material or doing something very complex (outside of a school curriculum, where the point is to make correctible mistakes), by all means, use a calculator as a crutch to help with learning.  The problem with calculator use/overuse among many students is that they fail to ever leave that crutch and continue to use it with topics they have neither trickle-down nor any substantial degree of complexity.  If you'd read my earlier post, you'd notice that I'd conceded that calculators can be useful, though if you'd prefer, I'd be more than happy to retract that concession.

Abracadabra, marimarihom!!! Evil begone!
Only Soroban did the problem. What's going on with all of you? So stuck up and keep on arguing about some useless issue that doesn't even get you to earn pennies. Move on and do more math problems, geezes.


calculus
integration
geometry
function
calculus computations

For constant numbers $ a,\ b$, let $ f(x)\equal{}e^{\minus{}x}(ax\plus{}b)$. Suppose that the curve $ y\equal{}f(x)$ passes through two points $ (t,\ 1),\ (t\plus{}1,\ 1).$
Find the area of the part bounded by the curve $ y\equal{}f(x)$ and the line $ y\equal{}1$.

From the fact that the function passes through points $ (t,1)$ and $ (t \plus{} 1,1)$ we get:

$ a \equal{} (e \minus{} 1)e^t$

$ b \equal{} (1 \minus{} t(e \minus{} 1))e^t$

so:

$ f_t(x) \equal{} e^{ \minus{} (x \minus{} t)}((e \minus{} 1)(x \minus{} t) \plus{} 1)$

We can see that for any $ t_1$ and $ t_2$ we have:

$ \int^a_b f_{t_1}(x)\,dx \equal{} \int^a_b f_{t_2}(x)\,dx$

So we can set $ t \equal{} 0$ and have:

$ f_0(x) \equal{} f(x) \equal{} e^{ \minus{} x}((e \minus{} 1)x \plus{} 1)$

Now we have to solve:

$ e^{ \minus{} x}((e \minus{} 1)x \plus{} 1) \equal{} 1$

to find the integration bounds, but that's easy because we know the function passes through points $ ((t \equal{} 0),1)$ and $ ((t \equal{} 0) \plus{} 1,1)$, so:

$ x_1 \equal{} 0$

$ x_2 \equal{} 1$

So, all we have:

$ A \equal{} \int^1_0 \left(e^{ \minus{} x}((e \minus{} 1)x \plus{} 1) \minus{} 1\right)\,dx$

and that's easily evaluated to be:

$ A \equal{} \frac {1}{e} \plus{} e \minus{} 3$

EDIT: Sorry about that, I forgot an integral part. Hope this is right.

You seem to have made a mistake in calculation. :(


Maria normally spends a half-hour driving to work. When her average speed is ten miles per hour slower than usual, the trip takes ten minutes longer. How many miles does she drive to work?

Use the formula d=rt. So first you do d=.5r, with .5 being the amount of time it takes (half an hour). Now, you also set the second bit of info into the equation: d=2/3(r-10). It takes 30+10, 40 minutes, or 2/3s of an hour to travel the distance if the rate is 10 less.

Now you just set the equations equal: 2/3(r-10)=.5r, and answer for r. You will get r=40.

Now plug this in, adyou find d=20.


geometry
parallelogram
circumcircle
perpendicular bisector
geometry unsolved

In a triangle ABC, let G the centroid of ABC, and L(lemoine point), its isogonal conjugate, A line trought L and parallel to BC meets AB and AC in A' and A'', define similarly B', B'', C' and C''. Prove that  A', A'', B', B'', C', C'' are concyclic.

The First Lemoine Circle
http://mathworld.wolfram.com/FirstLemoineCircle.html
you can download mathematica book to see solution!

Is in PDF format or ...?, I cant download

It is *.nb file you can read it by mathematica!

Don't bother with any download, it is too simple. AA'LA'' is a parallelogram, its diagonals AL, A'A'' cut each other in half. The A-symmedian AL is the locus of midpoints of antiparallels to $BC \parallel B''C$ with respect to the angle $\angle A$ (just like the A-median AG is the locus of midpoints of parallels), hence A'A'' is antiparallel to B''C', so that the quadrilateral A'B''C'A' is cyclic with a circumcircle (X) and $\angle A'B''C' = \angle ABC = \angle B.$ Similarly, CC'LC'' is a parallelogram and C'C'' antiparallel to AB with respect to the angle $\angle C,$ so that the triangles $\triangle LC''C' \sim \triangle CC'C'' \sim \triangle ABC$ are similar and $\angle L'C''C' \equiv \angle A'C''C' = \angle B.$ Therefore, $\angle A'C''C' = \angle B = \angle A'B''C',$ which means that C'' lies on the circumcircle (X) of the quadrilateral A'B''C'A''. In exactly the same way, $\angle B''B'A'' = \angle C = \angle B''C'A'',$ so that B' also lies on the circle (X).

Let O be the circumcenter of $\triangle ABC.$ The perpendicular bisector of A'A'' is parallel to AO, hence it is the midline of the triangle $\triangle ALO,$ so that it cuts KO at its midpoint. Similarly, the perpendicular bisectors of B'B'', C'C'' also cut KO at its midpoint, hence they concur at this midpoint, so that the midpoint of KO is the center of the circle (X).


Hi, i've got a couple of questions if someone could help me out with. Question is:
Consider the function f: N -> N, given for every x all natural numbers by:

f(x) = x + 3 (if x is odd) ; & x - 1 (if x is even)

a) What is the domain and codomain of the function
b) what is the range of this function
c) Is this function onto?

Thanks in advance.

I don't know if i'm right, but i think the domain and codomain are just N.
Not sure about range and if it's onto or not?

Ok i know it's not onto by testing it out with this:

f(1) = 4, f (3) = 6, f(5) = 8
f(2) = 1, f (4) = 3, f(6) = 5)

Obviously here f(x) = y never applies, hence it's not onto.
But how do i figure out if it's one-to-one or not?


I was looking at the IMO results for 2009, and only 3 people solved #6 (many people got 0s...). I personally can't do the problem (i'm not very good at olympiad problems), but for those of you who are more experienced: was this problem as difficult as the scores indicate? Does it get a place in IMO history with 2007/3?

If you look at imo-official.org you'll see the mean score on every problem

Some problems from last years:

09/6 - 0.168
08/6 - 0.260
07/6 - 0.152
07/3 - 0.304
06/6 - 0.187

So the hardest problem in the last few years was 07/6
But, if we take the percent of contestants who scored >0 we get

09/6 - 4.42%
08/6 - 9.9%
07/6 - 9.04%
07/3 - 15.96%
06/6 - 5.42%

So the problem on which most zeroes per number of contestants is 09/6

07/3 wasn't hardest of all times, just very hard for a #3

A lot of people I know solved it and didn't think it was that hard, but the problem's probably a lot harder when you try it on the actual exam since its a lot harder to [hide="Contains hints to solution"]stick with induction if it doesn't immediately work because 1. The problem's a #6, so I think a lot of people thought induction wouldn't work 2. There's a lot of pressure, and there isn't that much time.[/hide]

In my opinion, in general, IMO combinatorics is easier than the scores suggest, so though I think its probably a hard problem when you try it while taking the actual test, its not as hard as most other #6s.

[quote="Arachnotron"]So would it even be a "hard" problem in general? Or due to the reasons you gave, it was "hard" but the actual problem itself didn't require the creativity that a typical "hard" problem requires?[/quote]

It was hard if you had to solve the problem while under pressure, but if you were just trying the problem unofficially, it became a lot easier.

Are there any IMO problems that are hard in an unequivocal sense (i.e. just hard, not hard due to the time limit)?

Well pythag, every problem is easier with no time limit. And, probably you're good at combinatorics, so you say this. It's always better to take more persons and compare scores (and IMO does it well, so this problem is very hard)


calculus
integration
function
geometry
geometric transformation
reflection
calculus computations

We are to remark the inverse function of some function $ f(x)$.

But how do u get it as zero  ? Can u give the solution.

Let $ U$ be the region in the first quadrant bounded by the two axes and by the curve $ x^4\plus{}y^7\equal{}1.$

Let $ V$ be the region in the first quadrant bounded by the two axes and by the curve $ x^7\plus{}y^4\equal{}1.$

What happens if you mirror either of these two regions across the line $ x\equal{}y?$

And what does that tell you about $ \text{area}(V)\minus{}\text{area}(U)?$

[b]Solution.[/b]

If $ f(x)\equal{}\sqrt[7]{1\minus{}x^4}$, then $ f^{\minus{}1}(x)\equal{}\sqrt[4]{1\minus{}x^7}$. (Kunny was right on this)...Both graphs start at $ (0,1)$ and decrease to end at $ (0,1)$ and intersect at a point $ (A,A)$. (How would you find the numerical value of this point?) The region that is between the graphs is symmetric about the line $ y\equal{}x$. Any portion of this region that is above $ c<x<d<A$ for which $ f(x)>f^{\minus{}1}(x)$ for all $ c<x<d$ will be reflected, via reflection in the line $ y\equal{}x$, into a region of equal area for which $ f(x)<f^{\minus{}1}(x)$, and vice versa. (Actually, $ f(x)>f^{\minus{}1}(x)$ for $ x<A$.)

I have seen a problem like this $ \int_0^1(f\plus{}f^{\minus{}1})\equal{}0$ before under certain conditions, fedja talked about this long time ago, but don't remember the conditions.

Note that the $ f$ in the problem that started this topic is decreasing, which is why it made sense to take the difference. If you want to add, try an increasing function.

Suppose $ f$ is a continuous increasing function that maps $ [a,b]$ onto $ [a,b].$ Then
\[ \int_a^b(f\plus{}f^{\minus{}1})\equal{}b^2\minus{}a^2.\]

thank you guys , thats something i haven't thought about.


[b]Problem.[/b] Consider the prime numbers $ n_{1}<\hdots<n_{31}$. Prove that if $ 30| n_{1}^{4}\plus{}\hdots\plus{}n_{31}^{4}$, then among these numbers one can find three consecutive primes.

im just throwing out an idea here:
i think it has something to do with the fact that $ 30\equal{}2\cdot 3\cdot 5$

[hide="hint 1"]
The primes are $ 2, 3, 5$
[/hide]

[hide="hint 2"]
For all primes $ p\neq 2, 3, 5$, we have $ p^{4}\equiv 1\mod{30}$.
[/hide]


[size=150]Let ABC be a triangle,prove that:

$ \frac {b \plus{} c}{a} \plus{} \frac {x_b \plus{} x_c}{x_a} \geq \frac {x_b \plus{} x_c}{x_a}\frac {b \plus{} c}{a}.$
$ 1.x_a \equal{} m_a,x_b \equal{} m_b,x_c \equal{} m_c$holds;

$ 2.x_a \equal{} w_a,x_b \equal{} w_b,x_c \equal{} w_c$ holds.

$ 3.x_a \equal{} \frac {s(s \minus{} a)}{w_a},x_b \equal{} \frac {s(s \minus{} b)}{w_b},x_c \equal{} \frac {s(s \minus{} c)}{w_c}$ holds.

BQ[/size]

[size=150]$ y \equal{} {\frac {b \plus{} c}{a}} \plus{} {\frac {{\it w_b} \plus{} {\it w_c}}{{\it w_a}}} \minus{} {\frac {\left( {\it w_b} \plus{} {\it w_c} \right) \left( b \plus{} c \right) }{{\it w_a}\,a}} \minus{} 3\,{\frac {a}{b \plus{} c}} \minus{} 3\,{\frac {{\it w_a}}{{\it wb} \plus{} {\it w_c}}} \plus{} 3$
<=>
$ y \equal{} \frac {(mn \minus{} 3)( \minus{} mn \plus{} n \plus{} m}{mn}$
$ m \equal{} \frac {b \plus{} c}{a},n \equal{} \frac {w_b \plus{} w_c}{w_a}$

and $ \frac {1}{m} \plus{} \frac {1}{n} \geq 1$(Have been the result)
[/size]

Note:
yu dong wu prove this,once
BQ


calculus
derivative
logarithms
real analysis
real analysis unsolved

I'm trying to solve this problem:

Let a, b be constants.  consider the ODE:

y'' + (a/x) y' + (b/x^2) y = 0  on the half-line xE (0, infinity)

1.  make the substitution t = ln(x) and write the ODE with independent variable t

2.  seperate the problem into 3 cases of constant coefficient ODE's according to the values of a and b

3.  in each of the 3 cases, give the general solution to the original ODE (that is, in terms of x, not t)

-----------------------------------------------

I'm not sure about #1.  when i make the substitution, i get

y'' + (a/e^t) y' + (b/e^(2t)) y = 0

is this right?  or do i have to change y, y' and y'' too because of the substitution?

The substitution does affect the derivatives, essentially by the chain rule. If $ Y(x)$ is a solution to the old equation, we want $ Z(t)$ to be a solution to the new equation, where $ Z(\ln x)\equal{}Y(x)$. In order to transform the equation, write $ Y,Y',Y''$ in terms of derivatives of $ Z$.

Hint: if you do that correctly, you should get a constant coefficient linear ODE in terms of $ t.$ And that's really the point of part 2: that you should already be familiar with constant coefficient linear equations, and the three cases are something you should already know about.

Part 4 of this problem would be to find a method for solving a nonhomogeneous relative of this equation, in which the right hand side was some linear combination of terms of the form $ x^s,(\ln x)x^s,\dots,(\ln x)^nx^s, x^s\cos(q\ln x), x^s\sin(q\ln x),$ and other terms like those.


Show that there exists infinitely many positive integers which cannot be represented as
$ a^3 \plus{} b^5 \plus{} c^7 \plus{} d^9 \plus{} f^{11}$
where a,b,c,d, and f are positive integers.

Please try this problem. It has a very nice and simple solution.

Fix a number $ N$, $ N > 10^{10}$. We will find an upper bound on the number of numbers that can be represented like that. $ a$ can take on at most $ N^{1/3}$ values, b can take on at most $ N^{1/5}$ values, etc. The total number of quintuples is then at most $ N^{1/3}\cdot N^{1/5}\cdot N^{1/7}\cdot N^{1/9}\cdot N^{1/11} < N^{9/10}$, so we have that there must be at least $ N \minus{} N^{9/10}$ numbers below $ N$ which cannot be represented. Taking $ N$ approaching infinity, we have the desired result.


function
algebra
polynomial
functional equation
IMO Shortlist

Let $ f$ and $ g$ be two integer-valued functions defined on the set of all integers such that

[i](a)[/i] $ f(m \plus{} f(f(n))) \equal{} \minus{}f(f(m\plus{} 1) \minus{} n$ for all integers $ m$ and $ n;$
[i](b)[/i] $ g$ is a polynomial function with integer coefficients and g(n) = $ g(f(n))$ $ \forall n \in \mathbb{Z}.$

I think you forgot this line:

Determine f(1991) and the most general form of g.

For the RHS of the equation, it is $ \minus{}f(f(m\plus{}1))\minus{}n$ or $ \minus{}f(f(m\plus{}1)\minus{}n)$.

As we can choose $ n$ freely, so the function is surjective.

So there exists a constant integer $ k$ such that $ f(f(k))\equal{}1$, and we have $ f(m\plus{}1)\equal{}\minus{}f(f(m\plus{}1))\minus{}k$.

Again, by surjectivity, for all integer $ n$, we have $ n\equal{}\minus{}f(n)\minus{}k$, i.e. $ f(n)\equal{}\minus{}n\minus{}k$.

Substitute this into (a), $ \minus{}m\minus{}n\minus{}k\equal{}\minus{}m\minus{}1\minus{}n$, and $ k\equal{}1$.

So $ f(1991)\equal{}\minus{}1991\minus{}1\equal{}\minus{}1992$.


1)Prove that $ |1\plus{}z|\plus{}|1\plus{}z\plus{}z^2|\geq\ 1$ where $ z$ is a complex number.
2)If $ |z^2\plus{}1|\equal{}|z\minus{}1|$, find the extremes of $ |z|$, where $ z$ is a complex number.
3)If $ |z^2\plus{}1|\equal{}2|z\plus{}1|$, then find the maximal value of $ |z|$, where $ z$ is a complex number.

1)  Case a)  $ |z|\ge 1$

 Then  $ |z \plus{} 1| \plus{} |z^2 \plus{} z \plus{} 1|\ge | \minus{} 1 \minus{} z \plus{} z^2 \plus{} z \plus{} 1| \equal{} |z^2|\ge 1$

   Case b)    $ |z < 1$

  Then $ |z \plus{} 1| \plus{} |z^2 \plus{} z \plus{} 1|\ge |z \plus{} 1\parallel{}z| \plus{} |z^2 \plus{} z \plus{} 1|\ge| \minus{} z \minus{} z^2 \plus{} z^2 \plus{} z \plus{} 1| \equal{} 1$

  2)  $ \left|\frac {z^2 \plus{} 1}{z \minus{} 1}\right| \equal{} 1$  $ \Rightarrow$

   $ \frac {z^2 \plus{} 1}{z \minus{} 1}\cdot\frac {\overline{z}^2 \plus{} 1}{\overline{z} \minus{} 1} \equal{} 1$    $ \Rightarrow$

 $ (z \plus{} \overline{z})^2 \plus{} (z \plus{} \overline{z}) \plus{} |z|^4 \minus{} 3|z|^2 \equal{} 0$   $ \Rightarrow$

 $ \Delta\ge 0$  $ \Rightarrow$

 $ 4|z|^4 \minus{} 12|z|^2 \minus{} 1\le 0$  $ \Rightarrow$

  $ 0\le |z|^2\le \frac {3 \plus{} \sqrt {10}}{2}$   $ \Rightarrow$   $ 0\le|z|\le \sqrt {\frac {3 \plus{} \sqrt {10}}{2}}$

  3)  Similarly.


these are easy examples of mixing variables!

MV also works for 4 numbers of these forms.

For the first two inequalities, I think the following stronger results hold:
a) $(a+b+c)^{6}\ge 32(a^{2}+b^{2}+c^{2})(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$
b) $(a+b+c)^{9}\ge 256(a^{3}+b^{3}+c^{3})(a^{3}b^{3}+b^{3}c^{3}+c^{3}a^{3})$

(a) can be found at [u]http://www.mathlinks.ro/Forum/viewtopic.php?t=104719[/u]


AMC
AIME
function
AoPSwiki
algebra
polynomial
AIME I

Let $ C$ be the coefficient of $ x^2$ in the expansion of the product $ (1 \minus{} x)(1 \plus{} 2x)(1 \minus{} 3x)\cdots(1 \plus{} 14x)(1 \minus{} 15x)$. Find $ |C|$.

Supposing you had no clue how to do this elegantly, would you just multiply it out? I don't really understand aops wiki or kalva's solution. In particular for the aops wiki one, how and why do we want to break our product into different functions?[hide="Solution I am Asking About"]http://www.artofproblemsolving.com/Wiki/index.php/2004_AIME_I_Problems/Problem_7[/hide]

You could also think of it as taking the pairwise sums from the set $ \{\minus{}1,2,\minus{}3,4,\cdots,14,\minus{}15\}$. Then start with the pairs of $ \minus{}1$ and find the sum. Then move on to the pairs of $ 2$ and find the sum excluding the $ (\minus{}1,2)$ already accounted for and so on. You'll find a pattern as you go on but this method will be a bit computation heavy and you're more inclined to make a mistake along the way. However, if you don't see the solution shown in aopswiki, this might be the best way to go.

As for you other question. the reason we break the product into different functions is that we only really care about the $ x^{2}$ coefficient. So by constructing a polynomial $ Q(x)$ such that $ P(x)\equal{}1\minus{}8x\plus{}Cx^{2}\plus{}Q(x)$ and strategically plugging in values for $ x$, we can essentially eliminate all the undesired terms.

Just a note, the simplest solution is almost certainly Kalva's, although he doesn't explain it very well.  It is a one-step application of the common idea that "the square of the sum minus the sum of the squares equals twice the sum of the cross-terms"

EDIT: fixed.

oh that is really nice. On a related note, i think it should be "the square of the sum of the squares equals [b]twice[/b] the sum of the cross terms"

I'm don't understand the solutions in the first place I've seen so I'm having difficult following you guys :(

Tenoreoz

Why can we think of the problem as taking the pairwise sums from the set $ \{\minus{}1,2,\minus{}3,4,\cdots,14,\minus{}15\}$?
What do you mean by pairwise sums?

I didn't really make it clear in my post, but I don't even see how they constructed the polynomial.

E^(pi*i)=-1

"the square of the sum minus the sum of the squares equals twice the sum of the cross-terms"

What is the sum? Do you mean like... $ (x\plus{}y)^2\equal{}x^2\plus{}y^2\plus{}2xy$? How do we apply this to our problem?

The question asks for the sum of coefficients of all second degree terms when the polynomial is expanded. Such terms are obatained by choosing two $ x$'s and thirteen $ 1$'s. Thus, the requested sum will be the sum of pairwise products of the elements in $ \{c_i\} \equal{} \{ \minus{} 1, 2, \minus{} 3, ..., \minus{} 15\}$, which is the sequence of coefficients of the $ x$'s.
Now, observe that:
$ \left(\sum c_i\right)^2 \equal{} \sum c_i^2 \plus{} 2\sum_{sym}c_ic_j$
and then $ \sum_{sym}c_ic_j$, our desired sum, can easily calculated.


inequalities
inequalities proposed
three variable inequality

[quote="arqady"]Indeed, if it's true then, it's nice.[/quote]
This inequality alway true:
Let $ a,b,c$ be positive numbers such that: $ a\plus{}b\plus{}c\equal{}3$.Prove that:
$ \frac {a}{b \plus{} c^2} \plus{} \frac {b}{c \plus{} a^2} \plus{} \frac {c}{a \plus{} b^2} \plus{} \frac {ab \plus{} bc \plus{} ca}{72} \ge \frac {37}{24}$:)

[quote=quykhtn-qa1]Let $ a,b,c$ be positive numbers such that: $ a\plus{}b\plus{}c\equal{}3$.Prove that:
$ \frac{a}{b\plus{}c^2}\plus{}\frac{b}{c\plus{}a^2}\plus{}\frac{c}{a\plus{}b^2}\plus{}\frac{ab\plus{}bc\plus{}ca}{6} \ge 2$[/quote]
The following inequality a bit of stronger.
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=3.$ Prove that:

$$\frac{a}{b\plus{}c^2}\plus{}\frac{b}{c\plus{}a^2}\plus{}\frac{c}{a\plus{}b^2}\plus{}\frac{2(ab\plus{}bc\plus{}ca)}{9}\geq\frac{13}{6}$$


hello!
find the rest of division all $ x^{2n}$over$ (x\minus{}1)^2.(x\plus{}2)$

Solution
[hide]
It's easy to check that $ x^{2n} \minus{} 1 \minus{} 2nx \plus{} 2n$ is divisible by $ (x \minus{} 1)^2$.

Let's $ x^{2n} \minus{} 1 \minus{} 2nx \plus{} 2n \equal{} f(x)(x \minus{} 1)^2(x \plus{} 2) \plus{} r(x)$, and we know that $ r(x)$ is divisible by $ (x \minus{} 1)^2$.

So, $ r(x) \equal{} k(x \minus{} 1)^2$.

Putting $ x \equal{} \minus{} 2$, we will have:

$ 4^n \minus{} 1 \plus{} 6n \equal{} r( \minus{} 2) \equal{} 9k$.

So, $ r(x) \equal{} \frac {4^n \minus{} 1 \plus{} 6n}{9}(x \minus{} 1)^2$.

Thus, the answer is $ \frac {4^n \minus{} 1 \plus{} 6n}{9}(x \minus{} 1)^2 \plus{} 2nx \plus{} 1 \minus{} 2n$.
[/hide]

$ x^{2n} \equal{} P(x).(x \minus{} 1)^2.(x \plus{} 1) \plus{} a.x^2 \plus{} b.x \plus{} c$;
For $ x \equal{} 1\Rightarrow 1 \equal{} a \plus{} b \plus{} c$;
For $ x \equal{} \minus{} 1\Rightarrow 1 \equal{} a \minus{} b \plus{} c$;
$ \Rightarrow 0 \equal{} 1 \minus{} 1 \equal{} (a \plus{} b \plus{} c) \minus{} (a \minus{} b \plus{} c) \equal{} 2b \Rightarrow b \equal{} 0$;  and a+b+c=1;
$ \Rightarrow a \plus{} c \equal{} 1 \Rightarrow c \equal{} 1 \minus{} a \Rightarrow x^{2n} \equal{} P(x).(x \minus{} 1)^2.(x \plus{} 1) \plus{} a.x^2 \plus{} 1 \minus{} a$;
$ x^{2n} \minus{} 1 \equal{} P(x).(x \minus{} 1)^2.(x \plus{} 1) \plus{} a.(x^2 \minus{} 1)$;
$ (x^n \minus{} 1)(x^n \plus{} 1) \equal{} (x^2 \minus{} 1)[P(x).(x \minus{} 1) \plus{} a]$;
$ (x^2 \minus{} 1)(x^{n \minus{} 1} \minus{} x^{n \minus{} 2} \plus{} x^{n \minus{} 3} \minus{} \dots\plus{}(\minus{}1)^{n\minus{}1}).(x^{n \minus{} 1} \plus{} x^{n \minus{} 2} \plus{} \dots \plus{} x \plus{} 1) \equal{} \\
\equal{} (x^2 \minus{} 1)[(x \minus{} 1).P(x) \plus{} a]$;
$ (x^{n \minus{} 1} \minus{} x^{n \minus{} 2} \plus{} x^{n \minus{} 3} \minus{} \dots\plus{}(\minus{}1)^{n\minus{}1}).(x^{n \minus{} 1} \plus{} x^{n \minus{} 2} \plus{} \dots \plus{} x \plus{} 1) \equal{} [(x \minus{} 1).P(x) \plus{} a]$;
For x=1$ \Rightarrow a \equal{} \dots and \ c \equal{} \dots$.
If n=2k+1 then a=n, and for n=2k then a=0.


geometry
3D geometry
algebra
factorization
sum of cubes


Let $ a,b,c,d$ be positive integers such that $ a\plus{}b\plus{}c\plus{}d\equal{}38$
Find the Minimum and Maximum of $ P\equal{} \frac{a}{b^2}\plus{} \frac{c}{d}$

[quote="2424"]Let $ a,b,c,d$ be positive integers such that $ a \plus{} b \plus{} c \plus{} d \equal{} 38$
Find the Minimum and Maximum of $ P \equal{} \frac {a}{b^2} \plus{} \frac {c}{d}$[/quote]
Let's first find the max.
$ \frac{a}{b^2}$ is increasing in $ a$ and decreasing in $ b$. So if $ a\plus{}b \equal{} k$ then $ \frac{a}{b^2} \le k\minus{}1$. And likewise if $ c\plus{}d\equal{}j$ then $ \frac{c}{d} \le \frac{j\minus{}1}{1} \equal{} j\minus{}1$. So $ \frac {a}{b^2} \plus{} \frac {c}{d} \le 36$ with equality whenever $ b\equal{}d\equal{}1$.
The min:
$ \frac{a}{b^2}$ is increasing in $ a$ and decreasing in $ b$. So if $ a\plus{}b\equal{}k$ then $ \frac{a}{b^2} \ge \frac{1}{(k\minus{}1)^2}$. And likewise if $ c\plus{}d \equal{} j$ then $ \frac{c}{d} \ge \frac{1}{j\minus{}1}$. Let $ x \equal{} k\minus{}1$ and $ y \equal{} j\minus{}1$. Then $ x\plus{}y\equal{}36$ and we should find the min of $ \frac{1}{x^2} \plus{} \frac{1}{y}$. $ \frac{1}{x^2}\plus{}\frac{1}{y} \equal{} \frac{a^2 \plus{} 2a \plus{} 47 \plus{} a^{\minus{}1}}{36^2}$. Where $ a \equal{} \frac{y}{x}$. We should find the max of $ g(a) \equal{} a^2 \plus{} 2a \plus{} 47 \plus{} a^{\minus{}1}$. But $ g'(a) \equal{} \frac{a^3\plus{}2a^2\minus{}1}{a^2}$ and $ g''(a) \equal{} 1 \plus{} 2a^{\minus{}3} > 0$. Hence we should only find the two $ a$ such that $ g'(a)$ is close zero, and check which $ g(a)$ is smaller.


geometry
geometric transformation
homothety

Guys if you come across something new and nice during your IIT prep can you share them with us? You can post in this thread i guess. :lol:

Yay i get to post the first one. It is olympiad stuff.
Homothety: The connectors of corresponding points of two homothetic figures that is figures which are similar and have corresponding sides parallel to each other. Example to make this clear Consider a triangle and inside the triangle you have another small triangle which has sides parallel to the sides of the original triangle , then the two triangles are homothetic and the lines passing through the corresponding vertices of the two triangles intersect at a point called the homothetic center. Sorry guys i didn't give a lucid explanation because it takes me eternity to type all the stuff. I am adding a link though :D
[url]http://en.wikipedia.org/wiki/Homothetic_center[/url]
Ye all squeeze in time for olympiads and we should continue the legacy of our seniors. :rotfl:  :lol:


number theory
greatest common divisor
number theory proposed

Prove that there are not positive integers $ a,b$ satisfying $ a^2\plus{}b^2\plus{}26\equal{}5ab$.

If $ a,b$ both odd then $ X\equal{}a^2\plus{}b^2\plus{}26$ even and $ Y\equal{}5ab$ odd.
If one odd, second even, then $ X$ odd, $ Y$ even.
If both even, then $ X\equal{}2\mod 4$, $ 4|Y$.

[quote="SUPERMAN2"]Prove that there are not positive integers $ a,b$ satisfying $ a^2 \plus{} b^2 \plus{} 26 \equal{} 5ab$.[/quote]
$ 13 \nmid \text{gcd}(a,b) \text{ and } 13 \mid 26 \mid a^2\minus{}5ab\plus{}b^2 \mid (2a\minus{}5b)^2\minus{}21b^2$ but $ \left(\frac{21}{13}\right)\equal{}\left(\frac{\minus{}5}{13}\right)\equal{}\left(\frac{5}{13}\right)\equal{}\left(\frac{13}{5}\right)\equal{}\minus{}1$.[]


trigonometry
calculus
integration
calculus computations

hi, im having trouble getting the correct answer, can anyone check my work and help me out? thanks

1.)evaluate the definite integral

integral 0 to (pi/21) ((sec^12(7x))/(cot(7x)))dx => integral 0 to (pi/21) (1/cos^12(7x))*(sin(7x))/(cos(7x))

=> [cos(7x)]^-13 * sin(7x)

u=cos(7x) du=-7sin(7x)dx

=integral -7u^-13 = -7((u^(-12))/-12) = -7((cos(7x)^-12)/-12) and of course evaluate using the values given

2.)evaluate the indefinite integral

integra (1/((x^2)(sqrt(256-x^2)))

x=16sin(z)  dx=16cos(z)dz

integral (16cos(theta))/((256sin^2(z))(sqrt(256-256sin^2(z)) => integral 1/(256sin^2(z))

=> integral sin(z)^-2   => 256((sin(z)^-1)/(-1))

= 256((x^-1)/(-16))

since i dont know how to put theta on here z=theta

thanks again

On #1, you're right.

On #2, this line:
integral sin(z)^-2 => 256((sin(z)^-1)/(-1))
is wrong. $\frac1{256\sin^2 z}=\frac1{256}\cdot \csc^2 z$, and the integral of that is $\frac1{256}\cdot-\cot z$.

The "d(something)" should be considered part of the integral sign; never leave it off if there could be any confusion about the variable, and any substitution is possible confusion.

i see where i went wrong, thanks for you help.


Hello, how are you ? I think that you go to the wrong house, man. :gathering:  :spider:

[quote="cezar lupu"]Levi is perfectly right. i agree that this post should go to Intermediate or Basiscs. :D  ;)[/quote]

yes i made a mistake  sorry