Max was walking past Benny’s house one day when he noticed 5 pieces of coal, a carrot and a scarf lying on Benny’s lawn. He thought that was odd so he knocks on Benny’s door and asks, “Why did you put those bits and pieces on your lawn, Benny?”

“I didn’t put them there,” says Benny,“ in fact, nobody put them on my lawn. But there is a perfectly logical reason for their being there.”

What is that reason?

"That reason" means I put down one scenario, not more than scenario for the answer (just as in another thread with different words requiring a list of solutions).

That reason, for my educated guess, is a snowman.  The carrot
was in the nose position, the scarf was around the neck area,
and two of the five pieces of coal made up the eyes.  The remaining three coal pieces could have made up the mouth area.

The best brainteasers have a very elegant solution where no other solution seems to suffice. The snowman is a really good answer: He made a snowman over the winter, and it melted.


Humphrey the magical pigeon flies at a constant speed of 60mph. He begins his flight from the head of car, which travels at 28mph headed east. 40min later, Humphrey reaches the head of car B, traveling on the same road west at 52mph. Humphrey then flies back to car A. since he is magical, he wastes no time in turning and continues back and forth in this manner until he is tragically crushed between the two cars. How many miles was Humphrey's final flight?

(No, the answer is not zero. We are talking about the other final flight, not THE final flight where instant death occurs.)

[hide] 56 minutes, counting the original 40, unless my calculator is screwed up, which is always possible, since it's not my calculator [/hide]

What I want to know is this: If Humphrey the magical pigeon is really so magical and talented, how is it that he fails to notice he is about to be squashed between two cars at the end of his life? Couldn't he just magically teleport himself out of there or regenerate or something? This tragic question leaves me wondering about the sad role of stupid pigeons in today's society.

[hide]is this not just a variant on the well-known problem or am i missing something critical?[/hide]

[hide] I believe it's just the variant. [/hide]

For some reason, I seem to have forgotten the solution to this classic fly-between-trains-headed-towards-each-other problem. Could someone post a quick solution?

find the time that he flies, multliply by his rate, done.

Hmm, please clarify. My answer appears to be different.

[color=olive]It is a variant of a well-known problem.  On the other hand, no one has yet posted a correct answer.  Jen may have done something right, but her answer is in minutes and the question definitely asks for a distance, so that might be a problem.[/color]

The famous anecdote is that someone gave this problem to John von Neumann to see how quickly he would get it. He then gave the answer instantly, and the person who gave it to him said ``I should have known that you would find the trick,'' and then von Neumann replied ``What trick? I summed the infinite series.'' Apparently he was being serious about that.

Whoops.

How about the same thing in miles?

hehe
this is funny
when the dude put up his solution to this problem, he specifically said "leave ur answer as just 56, becusae if u wrote 56 min by accidnet, u would be wrong."

First of all, we must find the initial distance between the 2 cars: if the pigeon and the second car meet after 40 mins, by taking their speeds into account, we get the initial distance between the 2 cars to be (40/60)(60+52)=2/3*112 miles so it's 224/3 miles. The time the pigeon flies is the time needed by the 2 cars to meet. Let this time be x minutes. We then get x/60=224/3*(28+52)=224/240, so 224/x=4, so x=224/4=56 mintues of flight. We then multiply the speed of the pigeon with this time to get a distance of 56/60*60=56 miles.

This is a puzzle-type question. I found numerous variants on websites with puzzle-type questions.

Wow! I'm so proud of myself: "numerous" is such a long word! :D


If $ r$ and $ s$ are the roots of $ x^2\minus{}px\plus{}q\equal{}0$, then $ r^2\plus{}s^2$ equals:

$ \textbf{(A)}\ p^2\plus{}2q \qquad
\textbf{(B)}\ p^2\minus{}2q \qquad
\textbf{(C)}\ p^2\plus{}q^2 \qquad
\textbf{(D)}\ p^2\minus{}q^2 \qquad
\textbf{(E)}\ p^2$

[quote="AIME15"]If $ r$ and $ s$ are the roots of $ x^2 \minus{} px \plus{} q \equal{} 0$, then $ r^2 \plus{} s^2$ equals:

$ \textbf{(A)}\ p^2 \plus{} 2q \qquad \textbf{(B)}\ p^2 \minus{} 2q \qquad \textbf{(C)}\ p^2 \plus{} q^2 \qquad \textbf{(D)}\ p^2 \minus{} q^2 \qquad \textbf{(E)}\ p^2$[/quote]
[hide]$ r^{2}\plus{}s^{2}\equal{}(r\plus{}s)^{2}\minus{}2rs$
$ r\plus{}s\equal{}p$
$ rs\equal{}q$
$ p^{2}\minus{}2q$ so B[/hide]


Rewrite the following in Cartesian form. Do not simplify your answer: $ r\equal{}1\plus{}2r\sin\theta$

I replaced $ \sin\theta$ with $ y$ to get $ r\equal{}1\plus{}2y$, but then I didn't know where to go from there...

In general, $ (x,y)\equal{}(r\cos\theta,r\sin\theta)$, so $ r\equal{}\sqrt{x^2\plus{}y^2}$.

[quote="math154"]In general, $ (x,y) \equal{} (r\cos\theta,r\sin\theta)$, so $ r \equal{} \sqrt {x^2 \plus{} y^2}$.[/quote]

So then, $ r\equal{}1\plus{}2y$
$ \sqrt {x^2 \plus{} y^2}\equal{}1\plus{}2y$
$ x^2\plus{}y^2\equal{}1\plus{}4y\plus{}4y^2$
$ x^2\equal{}3y^2\plus{}4y\plus{}1$
$ x^2\equal{}(3y\plus{}1)(y\plus{}1)$
I believe I understand it now, but in the original problem, it says, "do not simplify your answer" and I don't know how much I can simplify...

They probably wanted you to stop right after making those substitutions for $ r$ and $ r \sin \theta$.

Directions like that are usually given for one or both of these reasons:
(1.) The instructor just want to see that you are familiar with the substitutions, and doesn't want you to waste time trying to figure out the appropriate form to simplify it to.
(2.) The instructor does not want to have to check if your final answer is equivalent or not, and moreover, if you make the substitution correctly but then make an error simplifying it, the instructor should not have to decide how much partial credit to give.

Thank you soooo much! I appreciate the help!


trigonometry
modular arithmetic
calculus
calculus computations

Investigate convergence of series: $ \sum\limits_{ n\equal{}1}^{ \infty} \frac{ (\minus{}1)^n \sin nx}{ n}$ and $ \sum\limits_{ n\equal{}1}^{ \infty} \frac{ (\minus{}1)^n \cos nx}{ n}$

These are the imaginary and real parts of $ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^ne^{inx}}n.$

Note that for $ |z|<1,$ we have $ \sum_{n\equal{}1}^{\infty}\frac{(\minus{}1)^nz^n}n\equal{}\minus{}\log(1\plus{}z).$

By a summation by parts argument (Abel-Dirichlet test), this series converges for all $ |z|\equal{}1,z\ne \minus{}1.$ Which is to say that the original series converge whenever $ x\not\equiv \pi\pmod{2\pi}.$

But if $ x\equiv\pi\pmod{2\pi},$ then the sine series is identically zero. Hence the sine series converges for all real $ x,$ while the convergence of the cosine series has the restriction already noted.

Except in the cases where all the terms are zero, this convergence is conditional, not absolute.

Another not so different way is:

Break
$ ( \minus{} 1)^n {\sin nx\over n} \minus{} ( \minus{} 1)^n {\sin (n \plus{} 1)x\over n \plus{} 1} \equal{} ( \minus{} 1)^n\left({\sin nx\over n} \minus{} {\sin (n \plus{} 1)x\over n}\right) \plus{} ( \minus{} 1)^n\left({\sin (n \plus{} 1)x\over n} \minus{} {\sin (n \plus{} 1)x\over n \plus{} 1}\right) \doteq c_1 \plus{} c_2$

The series with $ c_2$ converges absolutely.

$ c_1 \equal{} {( \minus{} 1)^n \over 2n}( \minus{} \sin {x\over2} \cos(nx \plus{} {x\over2}))$ and $ \sum {\cos (nx \plus{} a)\over n}$ converges
for any $ x\ne0$ by using Abel "summation by parts". For $ x \equal{} 0$ $ c_1\equiv0.$  Note that the sign of the terms of the "grouped" series (both $ c_1$ and $ c_2$) is the same: minus.

The procedure for the second series is the same.


Diophantine equation
number theory proposed
number theory

Given an integer $k$. Prove that there are infinite many triple of integers $(a,b,c)$ such that $bc-k, ca-k, ab-k$ are all perfect square.

Case $k=0$: trivial

Case $k \neq 0,-1$:
Take $a=k,b=k+1$. Then $ab-k=k(k+1)-k=k^{2}$ is a perfect square. Now the idea is to set $c=kn^{2}+1$, giving that $ac-k=k(kn^{2}+1)-k=(kn)^{2}$. So we are left with making $bc-k=(k+1)(kn^{2}+1)-k=(k^{2}+k)n^{2}+1$ a perfect square $x^{2}$, too.
But this is simply the pell equation $x^{2}-(k^{2}+k)n^{2}=1$, and since $k^{2}+k$ is positive and not a square (it lies between $k^{2}$ and $(k+1)^{2}$), this equation has infinetely many solutions.

So we are left with $k=-1$. But here just take $a=1,b=3$, and $ab+1=4$ will be a perfect square. So we are left to find a $c$ with $c+1=x^{2}$, $3c+1=y^{2}$, giving the Pell equation $y^{2}-3x^{2}=-2$ which has also infinetely many solutions, because there exists the solution $x=y=1$.


The length of a rectangle is four times its width. The perimeter is
100 cm. What is the number of square centimeters in the area of
the rectangle?

$ 4x\plus{}x\equal{}50 \implies x\equal{}10, 40 \times 10\equal{}\boxed{400}$.


Which do you prefer?

Anything, though mostly a pencil of sorts.  :lol:

Just not a crayon. They're too thick, there's no way to fit all the things necessary to solve a decent problem on a decent piece of paper.

Plus, I don't like them.

I prefer mechanical pencils, but regular pencils are alright.
Pens are also okay but... crayons? Ha definitely not.

Mechanical pencils (actually to be specific, a PhD pencil). Great for spinning (the PhD is fatter than most pencils are the spin is a bit awkward but easier than normal sized pencils).

I like pencils.  Generally, I find that mechanical pencils break.

i use a pen.
they glide smoother..
and d3m0n, do people ask you how you twirl your pencil??
everyone at my school asks me how i do it..
frankly, it gets annoying and everyone tries to do it now.
my science teacher nearly banned the twirling of pencils...
the fatter ones are easier to spin, but they spin slower..
i prefer spinning faster...

and back to my pen theory..
i don't like flipping every other second to erase my mistakes..
i cross them out... :D

[quote="kimmystar94"]and d3m0n, do people ask you how you twirl your pencil??
everyone at my school asks me how i do it..
[/quote]

Of course. Maybe if I feel like it I'll make a video on how I spin. Frankly, I find the speed greatly increased with fatter pencils. The only pitfall is the loudness of the impact if I accidently mess up and my pencil falls to the ground.

Actually for spinning I use a rod about 3 feet in length. If I mess up I usually hit the cat. Then I get scratched. So not to injure myself I spin the rod better. I tried a pen, but I got bored and threw it at the blackboard. Same with the pencil.

On the correct side note, I use anything that writes, except for some crayons. (Somehow turning in a turquoise font for homework is unacceptable).

I'll take anything.  Generally, I like using a pen in math contests for authenticity issues (I don't think that I'd get sabotaged by any graders, but hey, anything can happen).  But for scratch work, I don't care.  But generally, I like using pens more than pencils.

Generally, I use the same mechanical pencil throughout the day. I think the crayon would effect the grade you get for math because the teacher couldn't read the numbers. And the regular pencil keeps breaking, which is a pain towards most people (generally me, others, and the teacher) in always consistent in sharpening it again. And, every classroom has an [i]electric[/i] sharpener.

not my science classroom...

yes. i do not prefer crayon..
i have a really thick obese pen that has thirteen colors..
but three of them are black b/c they're different thicknesses...

hehe..
but two of the blacks ran out... and my pink is running out, too...

My teachers don't allow pen, because, he says, that if you make a mistake, he won't let you have a big ink blot in the middle of the paper and the answer right next to it.

[quote="kimmystar94"]
and d3m0n, do people ask you how you twirl your pencil??
everyone at my school asks me how i do it..
[/quote]

i've been trying to learn, but I can't!  :(
i like mechanical pencils
i like regular pencils but only when they are super sharp
i think that when i write in pen, i write a little messy
crayons? ha! ya right. they're too thick

I'm normal pencils all the way. Pens are rather messy (for me). Mechanicals prohibit how hard you are allowed to write. Crayons suck for obvious reasons.

Pencil spinning is fun... I am okay at it.. I do it most when I am doing math  :lol:

Pencils.

Though I do occasionally use pens, especially if they are in reach and I have no sharp pencils. And I think I only own two mechanical pencils.

If all I had was a crayon, I'd use it, but they don't seem very practical on a regular basis.

[quote="7h3.D3m0n.117"][quote="Cyrazeno"]I do it most when I am doing math  :lol:[/quote]

I spin my pencil...anytime it's in my hand. It's interesting; I can't seem to stop spinning it subconsciously.[/quote]

yup.. same with me...
and people stare and can't concentrate on their own work..

If we talk math hw, then pencil. Otherwise, I use pens (you don't need to sharpen them).

I hate using pens, and hate even more mechanical pencils. I like writing with short pencils...And pens and mechanical pencils can't get shorter.

I'm an expert with using mechanical pencils, especially cheap ones.

I particularly like them because I find them to be significantly more efficient and effective than regular pencils, even the cheaper ones which I most commonly use on a regular basis.
Their marks/strokes remain very consistent and readable throughout, unlike its counterpart in my view. In addition, if you use them well and know how to fix them when they're jammed, and how to save extra lead, they can last for a long time (the majority of the cheaper ones I use throughout last through a whole busy school year).
I admit it's fun to twirl them every now and then.  :P Most of my classmates and teachers don't even notice.

Generally I use pens when I take notes or write essays because I write much faster with then. Plus the AP Exams require essays to be written in pen.

I have one curious question: Are there any other reasons for mechanical pencils to be banned from the SAT other than their distracting ability to spin?   :?: :huh:

[quote="Patterns_34"]I have one curious question: Are there any other reasons for mechanical pencils to be banned from the SAT other than their distracting ability to spin?   :?: :huh:[/quote]

I think it's the habit of jamming during unfortunate times. It's far more reliable to use two regular pencils than the possibility of a mechanical breaking.

I forgot the AP essays needed to be in pen  :ewpu: . I did some of my recent AP practice bio essays with my mechanical pencil. Thanks for reminding me before I take the real thing in May  :) .

Like Patterns, a mechanical pencil provides a uniform width throughout your writing. I find that regular pencils wear out and the marks are far more illegible when your tip wears out.

Mechanical pencils. Duh! they are amazig (at least to me)

[quote="Patterns_34"]I have one curious question: Are there any other reasons for mechanical pencils to be banned from the SAT other than their distracting ability to spin?   :?: :huh:[/quote]

I use them on the SAT...
As long as it's #2 lead, it's fine...
None of my proctors noticed, and nor do they care...
I like them better than normal pencils...


A bag contains two red beads and two green beads. You reach into
the bag and pull out a bead, replacing it with a red bead
regardless of the color you pulled out.  What is the probability that all beads in the bag
are red after three such
replacements? Express your answer as a common fraction.

We can divide this into three cases, since you must draw two green beads and a red one in any order:

Case 1: RGG
There is a $ \frac{1}{2}$ chance of drawing a red first, $ \frac{1}{2}$ for a

green afterwards, and a $ \frac{1}{4}$ for a the last green.  There is a

$ \frac{1}{2} \times \frac{1}{2} \times \frac{1}{4}= \frac{1}{16}$ of

drawing this combo.

Case 2: GRG

$ \frac{1}{2}$ for drawing a green first, $ \frac{3}{4}$ for the red, and

$ \frac{1}{4}$ for the last green, so the chance of drawing this combo is

$ \frac{1}{2} \times \frac{3}{4} \times {1}{4}= \frac{3}{32}$.

Case 3: GGR

$ \frac{1}{2}$ for drawing a green first, $ \frac{1}{4}$ for drawing a green

second, and $ \frac{1}{1}$ for drawing a red last, so the chance is $ \frac{1}{2}
\times \frac{1}{4} \times 1= \frac{1}{8}$.

Adding, the answer is $ \boxed{\frac{9}{32}}$.

I understand doing it through cases. However, is there a way to solve it from example? I found 7 ways of choosing the marbles and 3 out of the 7 ways ended up with 4 reds.

I have a question. Wouldn't the complement be the probability of having a green last? This gives 1/2... so exactly what did I do incorrectly?

How strange that the first question asked about this problem is only weeks away from being ten years old.

Alternate solution:  rather than having the four balls in a bag, put them into four boxes, labeled A, B, C, D.  Let the A and B boxes contain red marbles and the C and D contain green marbles.  So rather than picking a marble at random, you pick a box, and replace its marble with a red marble always.

Then the only way to get all red marbles is to pick boxes C and D at some point.  You can do this in four ways:

CCD, CDD, ACD, BCD,

where you can scramble the letters.  The first two each have three scramblings and the last two each have six.  This gives $3+3+6+6=18$ orderings that work, out of $4^3$ possibilities, so the answer is $18/64=9/32.$

[hide=Solution]The beads will all be red at the end of the third draw precisely when two green beads are chosen in the three draws. If the first bead drawn is green, then there will be one green and three red beads in the bag before the second draw. So the probability that green beads are drawn in the first two draws is$$
\frac{1}{2}\cdot\frac{1}{4}= \frac{1}{8}.
$$The probability that a green bead is chosen, then a red bead, and then a green bead, is$$
\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{1}{4} = \frac{3}{32}.
$$Finally, the probability that a red bead is chosen then two green beads is$$
\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{4} = \frac{1}{16}.
$$The sum of these probabilities is$$
\frac{1}{8}+ \frac{3}{32}+ \frac{1}{16}= \boxed{\frac{9}{32}}.
$$[/hide]

Wait I dont get it

if you are replacing the balls then how can you end up with 2 balls out of the bag?

i don't get it either how do you have all red in the bag after you replace them?

It says that no matter what color you pick out of the bag it will be replace by a red bead so like if for the first round i pick out red and replace with red then second round i pick out green and replace with red and do the same for third i will have all reds


Compute $ 1\minus{}2\plus{}3\minus{}4\plus{}...\minus{}98\plus{}99$ .

We see that there are $ 49$ pairs of $ \minus{}1$ from $ 1\minus{}2$, $ 3\minus{}4$, $ 5\minus{}6$, and so on.
There is a leftover $ 99$ at the end, so the sum is $ 49(\minus{}1)\plus{}99\equal{}\minus{}49\plus{}99\equal{}50$.


inequalities
calculus
three variable inequality
algebra
romania

Ok, i post my solution
First,we put $f(a,b,c)=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-(a^2+b^2+c^2)$
WLOG $a \leq b \leq c$.We will prove that $f(a,b,c) \geq f(\frac{a+b}{2},\frac{a+b}{2},c)$(1)
In fact:(1) $\Longleftrightarrow \frac{1}{a^2}+\frac{1}{b^2}-\frac{8}{(a+b)^2}+\frac{(a+b)^2}{2}-(a^2+b^2) \geq 0$
               $\Longleftrightarrow (a-b)^2(\frac{(a+b)^2+2ab}{a^2b^2(a+b)^2}-\frac{1}{2}) \geq 0$ (2)
Because $a \leq b \leq c \Longrightarrow a+b \leq 2 \Longrightarrow ab \leq 1$
So that:$a^2b^2(a+b)^2 <2(a+b)^2+4ab \Longrightarrow \frac{(a+b)^2+2ab}{a^2b^2(a+b)^2}-\frac{1}{2} >0 \Longrightarrow$(2) true
Hence $f(a,b,c) \geq f(\frac{a+b}{2},\frac{a+b}{2},c)$ (when $a \leq b \leq c$)
From this,we need only to prove the case:
    Give $t,c >0$ satisfy $2t+c=3$.Prove that:
              $\frac{2}{t^2}+\frac{1}{c^2} \geq 2t^2+c^2$(3)
Solution:
(3) $\Longleftrightarrow \frac{2}{t^2}+\frac{1}{(3-2t)^2} \geq 2t^2+(3-2t)^2$
     $\Longleftrightarrow (t-1)^2(12t^4-36t^3+33t^2-6t-9) \leq 0$
We're easy to check that $12t^4-36t^3+33t^2-6t-9 \leq 0$ when $t \in (0,\frac{3}{2})$
So that the problem is proved :D  :D

hi, i`m new here :)

the ineq. is equivalent to :

$(a+b+c)^{4}(\sum a^{2}b^{2})\geq 81\sum a^{4}b^{2}c^{2}$.

we calculate the left side and then AM-GM and Muirhead (or AM-GM only) for the resulting sums, grouped convenably.
it is a lot to write, but it is ok

$\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2\Leftrightarrow\sum_{cyc}\left(\frac{1}{a^2}-a^2+4(a-1)\right)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(a-1)^2(1+2a-a^2)}{a^2}\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(2a-b-c)^2((a+b+c)^2+6a(a+b+c)-9a^2)}{a^2}\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}\frac{(2a-b-c)^2(-2a^2+b^2+c^2+8ab+8ac+2bc)}{a^2}\geq0.$
Let $\frac{-2a^2+b^2+c^2+8ab+8ac+2bc}{a^2}=P_a,$
$\frac{-2b^2+a^2+c^2+8bc+8ab+2ac}{b^2}=P_b,$
$\frac{-2c^2+a^2+b^2+8ac+8bc+2ab}{c^2}=P_c.$
Hance, remain to prove that $\sum_{cyc}(2a-b-c)^2P_a\geq0.$
But $\sum_{cyc}(2a-b-c)^2P_a=\sum_{cyc}(a-b+a-c)^2P_a=$
$=2\sum_{cyc}(a-b)(a-c)P_a+\sum_{cyc}(a-b)^2P_a+\sum_{cyc}(a-c)^2P_a=$
$=2\sum_{cyc}(a-b)(a-c)P_a+\sum_{cyc}(a-b)(a-c+c-b)P_a+\sum(c-a)(c-b+b-a)P_a=$
$=4\sum_{cyc}(a-b)(a-c)P_a+\sum_{cyc}(b-a)(b-c)P_a+\sum_{cyc}(c-a)(c-b)P_a=$
$=\sum_{cyc}(a-b)(a-c)(4P_a+P_b+P_c).$
Let $a\geq b\geq c.$ Then $4P_c+P_a+P_b\geq4P_b+P_a+P_c\Leftrightarrow P_c\geq P_b.$ This is obviously true.
And $4P_a+P_b+P_c=-12+\frac{4b^2}{a^2}+\frac{a^2}{b^2}+\frac{4c^2}{a^2}+\frac{a^2}{c^2}+\frac{32b}{a}+\frac{8a}{b}+...\geq0.$
Similarly $4P_b+P_a+P_c\geq0$ and $4P_c+P_a+P_b\geq0.$ And our inequality is proved. :)

[quote="maky"]
the ineq. is equivalent to :

$(a+b+c)^{4}(\sum a^{2}b^{2})\geq 81\sum a^{4}b^{2}c^{2}$.

we calculate the left side and then AM-GM and Muirhead (or AM-GM only) for the resulting sums, grouped convenably.
it is a lot to write, but it is ok[/quote]

 $(a+b+c)^4(a^2b^2+a^2c^2+b^2c^2)\geq81(a^2+b^2+c^2)a^2b^2c^2\Leftrightarrow$
$\Leftrightarrow\sum_{sym}(a^6b^2+4a^5b^3+4a^5b^2c+3a^4b^4+12a^4b^3c+10a^3b^3c^2-34a^4b^2c^2)\geq0.$
AM-GM only. ;)

I got that the inequality is equivalent to $(6,2,0)+4(5,3,0)+4(5,2,1)+3(4,4,0)+12(4,3,1)+10(3,3,2) \geq 34(4,2,2)$. (arqady's post above seems to confirm my result)

How do I use AM-GM only? I couldn't cancel the last of the ten $(3,3,2)$ with AM-GM, I have to use Schur to complete the solution: $(4,4,0)+(3,3,2) \geq 2 (4,3,1)$.

By the way, why did they give an inequality which can be killed by dumb455-ing??? I find this just plain useless. Anyone who knows this "tehnique" (you can hardly call it a tehnique, but I couldn't find another word) should be able to do it.

Also, who proposed this?

The form of the inequality is nice.
Then, I think you can find at least two nice solutions to this inequality. :P

Give at least one mr Cirtoaje . :D
BTW, my solution is similar to Arqady's one . Arquady thank you for teaching me  the trick with the first derivative to the factorisation. :D

I think that this one might interest you too : http://www.mathlinks.ro/Forum/viewtopic.php?t=70815 .

Very nice  :D Thank you mr Vasile .You are the constructor of this nice problem ??

[quote="ciprian"]Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2.  \][/quote]
In my solution I have proved that
 \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2.  \]
<=>
\[ (b-c)^2S_a+(a-c)^2S_b+(a-b)^2S_c\geq 0  \]
wher $S_a=\frac{2}{b^2}+\frac{2}{c^2}+\frac{6}{bc}-\frac{1}{a^2}-\frac{27}{(a+b+c)^2}$
$S_b$ and $S_c$ definding similary.
And with leting $a\geq b\geq c$ I proved that $S_a,S_b\geq 0$ and $S_b+S_c\geq 0$,and how we know from sos the ineq is done ;).

[quote="Vasc"]A nice method: Use the first two lines from Arqady's solution, then considere two cases:
    [i]Case[/i] $c\leq b \leq a \leq 1+\sqrt{3}$;
    [i]Case [/i] $a>1+\sqrt{3}$, when $b+c<1/2$, $bc<1/16$ and
$\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}>\frac 1{b^2}+\frac 1{c^2}>\frac 2{bc}>32>(a+b+c)^2>a^2+b^2+c^2$.[/quote]
Vasc, you meant to:
    [i]Case[/i] $c\leq b \leq a \leq 1+\sqrt{2}$;
    [i]Case [/i] $a>1+\sqrt{2}$, when $b+c<3/5$, $bc<9/100$ and
$\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}>\frac 1{b^2}+\frac 1{c^2}\geq\frac 2{bc}>9=(a+b+c)^2>a^2+b^2+c^2$.
OK? ;)

Now that you've literally killed the problem, what do you think about this:

[quote]Let $a,b,c$ be positive real numbers such that $a+b+c=3$. For what $t \in \mathbb R$ is it true that we always have \[ \frac 1{a^t}+\frac 1{b^t}+\frac 1{c^t} \geq a^t + b^t + c^t \, ?  \][/quote]

[quote=Vasc]Since $\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq \frac 1{ab}+\frac 1{bc} +\frac 1{ca}$,
it suffices to show that
$\frac 1{ab}+\frac 1{bc} +\frac 1{ca} \geq a^2+b^2+c^2$.  :P
This is my third solution.[/quote]

Applying AM-GM $\sum_{cyc} \frac{1}{a^2}\geq \sum_{cyc} \frac{1}{ab}=\frac{a+b+c}{abc}=\frac{3}{abc}=\frac{27}{3abc(a+b+c)}\geq \frac{27}{(ab+bc+ca)^2}$

We need to show that $\frac{27}{(ab+bc+ca)^2}\geq a^2+b^2+c^2\Leftrightarrow (ab+bc+ca)^2(a^2+b^2+c^2)\leq 27$

By AM-GM: $(ab+bc+ca)^2(a^2+b^2+c^2)=(\sum_{cyc} ab)(\sum_{cyc} ab)(\sum_{cyc} a^2)\leq \left [ \frac{(a+b+c)^2}{3} \right ]^3$

The equality holds at $a=b=c=1$.

[quote=bel.jad5][quote=Sohil_Doshi]By Chebyshev's inequality, $3(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) \geq (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2$

Now we essentially have to prove that $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2 \geq 3(a^2+b^2+c^2)$

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

By AM GM $ab+bc+ca \geq 3(abc)^{2/3}$

$(ab+bc+ca)^3 \geq 27(abc)^2$

$(ab+bc+ca)^3 \geq 3(abc)^2(a^2+b^2+c^2)(ab+bc+ca)$

$27(abc)^2 \geq 3(abc)^2(a^2+b^2+c^2)(ab+bc+ca)$

$(a^2+b^2+c^2)(ab+bc+ca) \leq 9$

By Chebyshev's inequality, $(a^2+b^2+c^2)(bc+ca+ab) \leq 3abc(a+b+c)$

$(a^2+b^2+c^2)(bc+ca+ab) \leq 9abc$

So if we can prove that $abc \leq 1$ then we are done

By AM GM, $\frac{a+b+c}{3} \geq \sqrt[3]{abc}$

Since $a+b+c=3$ this implies that $abc \leq 1$. $\blacksquare$[/quote]

Your application of [b]Chebyshev's inequality[/b] is not right.[/quote]

Which application and I am not able to see why it is not right. How is it not right?

We can use AM-GM to get
\[\dfrac 1{a^2}+\dfrac 1{b^2}+\dfrac 1{c^2}\geq\dfrac 1{ab}+\dfrac 1{bc}+\dfrac 1{ca}=\dfrac{a+b+c}{abc}=\dfrac 3{abc}\]
However, we note that using the inequality
\[(x+y+z)^2\geq 3(xy+yz+xz)\]
for $x=bc,y=ac,z=ab$, we get that
\[(ab+bc+ca)^2\geq 3(a^2bc+ab^2c+abc^2)=3abc(a+b+c)=9abc\]
Thus, we can substitute to get
\[\dfrac 1{ab}+\dfrac 1{bc}+\dfrac 1{ca}\geq\dfrac{3}{abc}\geq\dfrac{27}{(ab+bc+ca)^2}\]
However, by AM-GM, we have
\[27(ab+bc+ca)(ab+bc+ca)(a^2+b^2+c^2)\leq (a^2+b^2+c^2+ab+bc+ca+ab+bc+ca)^3=((a+b+c)^2)^3=(a+b+c)^6\]
Thus, we get that
\[\dfrac 1{a^2}+\dfrac 1{b^2}+\dfrac 1{c^2}\geq \dfrac 1{ab}+\dfrac 1{bc}+\dfrac 1{ca}\geq\dfrac{27}{(ab+bc+ca)^2}\geq\dfrac{729(a^2+b^2+c^2)}{(a+b+c)^6}=a^2+b^2+c^2\]
where we used $a+b+c=3$.

[quote=Sohil_Doshi]
$(a^2+b^2+c^2)(bc+ca+ab) \leq 3abc(a+b+c)$
[/quote]
This step is wrong - inequality is the other way simply by AM-GM
\[bc+ca+ab\geq 3(abc)^{2/3}\]
and
\[\dfrac 23a^2+\dfrac 16b^2+\dfrac 16c^2\geq 6\sqrt[6]{\dfrac{a^8b^2c^2}{6^{6}}}=(abc)^{1/3}a\tag{*}\]
because adding the symmetric versions gives
\[a^2+b^2+c^2\geq (abc)^{1/3}(a+b+c)\]
Now, multiplying the two inequalities yields the conclusion.

[quote=Sohil_Doshi]Thank you for the explanation. I have edited that part of the proof and used a different method of proving the inequality $(a^2+b^2+c^2)(ab+bc+ca) \leq 9$.[/quote]

The inequality mentioned is incorrect. Try the counter-example $(x,y,z)=(\frac 32,\frac 34,\frac 34)$.

[quote=ciprian]Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \][/quote]

Let $q=ab+bc+ca$  ,  $ r=abc$,  $ \rightarrow r \le 1$ ,  $ q^2 \ge 9r$

Write the inequality in this form

$$LHS-RHS=\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}+2(ab+bc+ca) -9=\frac{q^2-6r}{r^2}+2q-9 \ge \frac{q^2-6(\frac{q^2}{9})}{ (\frac{q^2}{9})^2}+2q-9= \frac{81(q-3)^2(2q+3)}{q^2}\ge 0$$

[quote=ciprian]Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \][/quote]

$$\text{LHS-RHS}=\,{\frac {[ \left( ab+ca+bc \right) ^{2}+3\,abc]\cdot {[{{a}^{2} \left( b
-c \right) ^{2}+{b}^{2} \left( c-a \right) ^{2}+{c}^{2} \left( a-b
 \right) ^{2}}]}}{2{a}^{2}{b}^{2}{c}^{2} \left( ab+ca+bc \right) ^{2}}}+{\frac { \left( 2\,ab+2\,ca+2\,bc+3 \right)  \left( ab+ca+bc-3
 \right) ^{2}}{ \left( ab+ca+bc \right) ^{2}}}
\geqq 0$$
I have another but ugly :D

[quote=ciprian]Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \][/quote]

Equivalent to prove

$$  \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}+2(ab+bc+ca) \geq 9 $$

We write

$$  \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2}+2(ab+bc+ca) \ge \frac 1{ab}+\frac 1{bc}+\frac 1{ca}+2(ab+bc+ca) =\frac 3{abc}+2(ab+bc+ca) \ge \frac {27}{(ab+bc+ca)^2}+2(ab+bc+ca) \ge 3\sqrt[3]{ \frac{27}{(ab+bc+ca)^2} \cdot (ab+bc+ca)^2    }=9 $$

[quote=ciprian]Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq a^2+b^2+c^2. \][/quote]

I like this one.
So our ineq is equivalent to
$$\sum_{\text{cyc}} \frac{1}{a^2}-a^2 \ge 0$$
Now remember that $x+\frac{1}{x} \ge 2$ so $\frac{1}{x^2}-x^2 \ge 2 \left(\frac{1}{x}-x \right)$ and now let $f(x)=\frac{1}{x}-x$, then $f''(x)=\frac{2}{x^3}>0$ so $f$ is convex so by jensen $f(a)+f(b)+f(c) \ge 0$ and now replacing all of this stuff we get
$$\sum_{\text{cyc}} \frac{1}{a^2}-a^2 \ge 2 \sum_{\text{cyc}} \frac{1}{a}-a \ge 0$$
Thus we are done :D

Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\geq \frac{32}{81}(a^2+4b^2+c^2)$$
[url=https://artofproblemsolving.com/community/c6h2814149p24843123]h[/url]

Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that
$$\frac {a}{b^3}+\frac {b^3}{c^5}+\frac {c^5}{a} \geq a^2+b^2+c^2$$
$$\frac 1{a^2}+\frac 1{b^2}+\frac 1{c^2} \geq \frac 3{2} + \frac 1{2} (a^3+b^3+c^3)$$
$$\frac{a}{b^3}+\frac{b}{c^3}+\frac{c}{a^3}\ge a^3+b^3+c^3$$
Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$$\frac {a}{b^2}+\frac {b^3}{c^5}+\frac {c^5}{a} \geq a^2+b^2+c^2$$
$$\sqrt{\frac {a(1+b)}{b(1+c)}}+\sqrt{\frac {b(1+c)}{c(1+a)}}+\sqrt{\frac {c(1+a)}{a(1+b)}}\leq a+b+c$$
$$a^3+b^3+c^3-a^2b-b^2c-c^2a \geq \frac {479}{100} (a+b+c-3)$$
$$a^3+b^3+c^3-a^2b-b^2c-c^2a \geq \frac {84}{25} (ab+bc+ca-3)$$
Let $a,b,c$ be positive real numbers such that $a^2+b^2+c^2=3$. Prove that
$$\frac {1+a}{1+b^3}+\frac {1+b^3}{1+c^5}+\frac {1+c^5}{1+a} \geq a+b+c$$
$$\frac {1+a}{1+b^3}+\frac {1+b}{1+c^3}+\frac {1+c}{1+a^3} \geq a+b+c$$
[quote=sqing]Let $a,b,c$ be positive real numbers such that $a+b+c=3$. Prove that: \[ \frac 1{a^3}+\frac 1{b^3}+\frac 1{c^3} \geq a^3+b^3+c^3. \]
How to prove?[/quote]
[url=https://artofproblemsolving.com/community/c6h1908420p13065498]h[/url] [url=https://artofproblemsolving.com/community/c6h1701625p10933595]h[/url]
Let $a,b,c>0, \  a+b+c+abc=4$. [url=https://artofproblemsolving.com/community/c6h1957917p13531601]Prove that[/url]
$$\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\ge a^3+b^3+c^3$$

ij realized this problem was posted before I was born


An object moves along a line so that its position at time t is given by $ s(t) \equal{} t^{3} \minus{}6t^{2} \minus{}15t$. Determine the interval(s), if any, over which the object is moving left and slowing down.

My problem with this question is what is meant by "moving left". Seems a bit ambiguous to me, but I'm assuming it means that velocity must be negative, and for slowing down to occur, we need acceleration to be positive. However, if this isn't true, then I've screwed up the entire question automatically. Also, I got an answer of $ (2,5)$ as the interval but now I'm thinking there is no solution, from interpreting the graph.

Sorry for such a silly question  :blush:

That's the interpretation I'd use, and your answer sounds right. If you think it feels ambiguous, add a note about the interpretation.


geometry
analytic geometry
graphing lines
slope

Let $ A(2,2)$ and $ B(7,7)$ be points in the plane. Define $ R$ as the region in the first quadrant consisting of those points $ C$ such that $ \triangle ABC$ is an acute triangle. What is the closest integer to the area of the region $ R$?

$ \textbf{(A)}\ 25 \qquad \textbf{(B)}\ 39 \qquad \textbf{(C)}\ 51 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80$

[hide]consider the locus of points $C$ such that ABC=90, and BAC=90, these are the lines $y=14-x$, and $y=4-x$

consider if you move the point between these lines, then those 2 angles are acute

(to find these lines, consider that they are perpendicular to $y=x$, which has the given points, then do the negative reciprocal and point slope form)

the locus of points $C$ such that ACB=90 is a circle centered at the midpoint of A and B (think of inscribed angles)

if we move the point out of that circle, then ACB is acute (because the extra arc that it intercepts makes it less)

so R is between the 2 lines, outside the circle, and in quadrant 1, looking at the graph, we get 2 right triangles minus the area of the circle, this is

$\frac{1}{2}(14^2)-\frac{1}{2}(7^2)- \left(\frac{5}{\sqrt{2}}\right)^2\pi=50.73...$

so the answer is $\boxed{51}$[/hide]

EDIT: yikes, my wording was almost unreadable...

I didn't read Altheman's solution but here is how I think.

[hide="Imagine.."]

Imagine that $AB$ is one leg of the right triangle. Since $AB = \sqrt{50}$, the largest we can form is to have point $C$ such that $AB = BC = \sqrt 50$. We can place $C$ in two ways and each $C$ form two different large right triangles. Each has area of $\sqrt{50} \cdot \sqrt{50} \cdot 0.5$ and so the both of them is approximately 50. The closest choice is C, 51. [/hide]

I think I made some mistakes but it came out OK in drawing..

altheman i think you mean 14^2/2 - 4^2/2 - 25/2 (pi)


analytic geometry
function
linear algebra
matrix
topology
calculus
derivative

Let $ I \equal{} ( \minus{} 1,1)^n$ be the n-dimensional open cube.

According to my book, we have the following definition:

$ S \subset R^n$ is a k-dimensional smooth surface in $ R^n$ if $ \forall x \in S$ there is a neighborhood U of x and a diffeomorphism $ \varphi : U \to I$ under which the image of the set $ S \cap U$ is the subset of I defined by the condition that the last n-k coordinates be zero.

Here is a provided example of the definition in use:

Let $ F^{_i } (x^1 ,...,x^n )(i \equal{} 1,...,n \minus{} k)$ be a system of smooth functions of rank n-k; ie. the matrix A which has its ith row as the Jacobi matrix of $ F^{_i }$has rank n-k on $ R^n$.

We shall show that the relations

$ \begin{array}{l}
 F^{_1 }    (x^1 ,...,x^n ) \equal{} 0 \\
 .......................... \\
 F^{_{n \minus{} k} }    (x^1 ,...,x^n ) \equal{} 0 \\
 \end{array}$

define a k-dimensional surface S in $ R^n$.

Suppose that the last n-k columns of A are independent at a point b in S. By the inverse function theorem, the function $ (t^1 ,...,t^n ) \equal{} P(x^1 ,...,x^n )$ defined by the transformation
$ \begin{array}{l}
 t^i  \equal{} x^i   (i \equal{} 1,..,k) \\
 t^i  \equal{} F^{i \minus{} k} (x^1 ,...,x^n )  (i \equal{} k \plus{} 1,..,n) \\
 \end{array}$

is a diffeomorphism of a neighborhood at the point b.
In the new coordinates $ t^1 ,...,t^n$the original system will have the form $ t^{k \plus{} 1}  \equal{} ... \equal{} t^n  \equal{} 0$; thus S is a k-dimensional smooth surface in $ R^n$.

----

However I don't see how P can be understood as a function which is onto I (a surjection onto I) - or any other open cube/interval for that matter. This is frustrating because there are about 3 other examples like this where the author neglects to mention what the neighborhood U of x should be or why the suggested diffeomorphism is onto. This leaves me with no understanding of what is going on.

Any help showing how surjectivity is proven in this example, for instance, would be really appreciated.

In the book I learned this from, "submanifold" was not used in the smooth context without an additional identifying word; either "immersed" or "embedded". Also, what you quote as the definition here was instead part of a later cluster of theorems; it's an alternate characterization of embedded submanifolds.

I can't find the book right now, but the essential definition is this:
An immersion is a smooth map from one manifold $ M$ to another $ M'$ with injective derivative (as a map from $ T_pM$ to $ T_pM'$) at each point of $ M$. An immersed submanifold is the image of an immersion, with the topology and smooth structure induced by that map.
An embedding is an immersion which is also a topological embedding; the image of an open set is open in the subspace topology. An embedded submanifold is the image of an embedding.
Since the Wikipedia page lists the book I used as one of the references, I'm pretty sure this is the right definition.

Some examples of immersions that aren't embeddings:
Map the line into the torus $ \mathbb{R}^2/\mathbb{Z}^2$ by $ f(t)\equal{}(t,\sqrt{2}t)$. It wraps around infinitely many times without ever meeting itself, and the image is dense in the torus.
The "figure eight" in the plane: curl an open line segment around so that both ends meet the middle orthogonally.

The "real" meaning of your version of the definition: we can choose local coordinates so that this piece of the embedded submanifold is precisely one of the coordinate hyperplanes. The exact neighborhoods used don't matter.

Let's see if I understand

So instead of the open n-cube I mentioned in the definition, we can use a neighborhood of a point which belongs to the k-dimensional plane in $ R^n$ as long as $ S \cap U$ gets mapped [i]onto [/i]the intersection of the k-dimensional plane with that neighborhood. Correct?

And, just out of curiosity, (if what I said is right) are all such neighborhoods diffeomorphic to the open n-cube I (ie. in such a way that their intersections with the k-dimensional plane get mapped bijectively onto the intersection of I with the k-dimensional plane)?

I'm just trying to understand why the author chose to define submanifolds in terms of the open n-cube if such a bijection might not be possible in general. In local coordinates, the diffeomorphism given in the provided example doesn't take S to one of the coordinate hyperplanes (it takes it to a proper subset of a hyperplane). How does it still satisfy the definition?

Yes, they are diffeomorphic- as long as you don't put in any holes. Any two star-shaped (the segment from the origin to $ x$ is in the set) open sets centered at the origin are diffeomorphic, and we can choose the diffeomorphism to affect only distance from the origin and not angle.
In practice, you just cut it off in whatever shape you feel like. It doesn't have to be a big neighborhood.

As for the example- it does hit everything, at least in the proper neighborhood. The defining conditions of the hyperplane are exactly the same as the defining conditions of the subset.

I really think this works better as a theorem than as a definition. For one thing, it doesn't have the framework to talk about immersions.

Thanks jmerry.

This is really curious since the chapter is the introductory chapter on surfaces in R(n) and nothing about embeddings or star shaped surfaces has been mentioned. That definition is pretty much all I know so far. The author just sort of gave the definition in terms of cubes and then proceeded to give examples in terms of various open sets which might not even look like cubes without any justification for the switch.

Well, "star-shaped" is merely one of several ways to avoid holes. It's very much an overkill condition, and it's not suited to anything of less than full dimension. You can't just use an arbitrary open set; a ball and a solid donut are not homeomorphic.

More practically than trying to reshape into a cube, you can just cut a small cube out of the middle of an arbitrary open set.

The exposition I worked with started with an abstract coordinate-independent definition (following a quarter of topological manifolds). The theorems that let us say that surfaces defined by an equation are manifolds in their own right were several chapters in.

After sleep it all makes sense now. This is the statement which eluded me:

[quote]More practically than trying to reshape into a cube, you can just cut a small cube out of the middle of an arbitrary open set.
[/quote]

If the center of the cube is a point in the k-dimensional plane (ergo the cube is a neighborhood of the point) then the cube is (of course) diffeomorphic to the standard cube in such a way that their respective intersections with the plane are in one-one correspondence. If I'm not wrong (I haven't actually written it out) the diffeomorphism is even obvious: simple stretching and translation.

I wish the author had mentioned it though. Incidentally, in the second volume of the book he reintroduces k-dimensional surfaces and requires that for smooth surfaces, the charts are C1 homeomorphisms of rank k from R(k). He also shows that isn't any different from the previous definition but claims the previous one is more "intuitive".


define $f_{1}(x)=x^{2}-1$
$f_{i}(x)=f_{1}(f_{i-1}(x))$

find all $f_{j}(x)$'s produced this way which have multiple roots.

no one can do this?  :maybe:
i think a modartor should move this to Olampiad forum.

I think this is close to IMO 1976 # 2

[hide]
$f_{n+1}(x)=(f_{n}(x))^{2}-1$
$f_{n+2}(x)=(f_{n}(x))^{4}-2(f_{n}(x))^{2}=(f_{n}(x))^{2}((f_{n}(x))^{2}-2)$
So it follows that $f_{n+2}(x)$ has multiple roots(which are the roots $f_{n}(x)$). With this, we see that all $f_{n}(x)$ with $n>2$ have multiple roots, since they can be expressed as a product of $(f_{n}(x))^{2}$ and some other stuff. But also note that $f_{2}(x)=x^{2}(x^{2}-2)$ has multiple roots at 0. So all $f_{n}(x)$ with $n\ne1$ have multiple roots.
[/hide]
p.s. w00t 1000th post!


algebra
polynomial
algorithm
absolute value
algebra unsolved

Let $ a,b,c$ be integers with absolute value no more than $ 10$. The polynomial $ f(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$ satisfies the property $ |f(2 \plus{} \sqrt {3})| < 0.0001$. Determine if $ 2 \plus{} \sqrt {3}$ is a root of $ f$.

$ 2 \plus{} \sqrt {3}$ has minimal polynomial $ x^2 \minus{} 4x \plus{} 1$.  The division algorithm gives (we are assigning different variables now)

$ f(x) \equal{} (x^2 \minus{} 4x \plus{} 1)(x \plus{} a) \plus{} (bx \plus{} c) \equal{} x^3 \plus{} (a \minus{} 4)x^2 \plus{} (1 \plus{} b \minus{} 4a)x \plus{} (a \plus{} c)$

We are given that

$ \left| (2b \plus{} c) \plus{} b \sqrt {3} \right| < \frac {1}{10000} \Leftrightarrow$
$ \left| (2b \plus{} c)^2 \minus{} 3b^2 \right| < \frac {|(2b \plus{} c) \minus{} b \sqrt {3}|}{10000}$

Now, the bounds on the coefficients gives $ \minus{} 6 \le a \le 14$, giving $ \minus{} 35 \le b \le 65$ and $ \minus{} 24 \le c \le 16$.  The RHS is therefore clearly less than $ 1$, which implies $ (2b \plus{} c)^2 \minus{} 3b^2 \equal{} 0$ (since it cannot equal $ 1$).  This occurs only when $ b \equal{} c \equal{} 0$, hence $ 2 \plus{} \sqrt {3}$ is a root of $ f$ as desired.

I believe the bound on the coefficients can be loosened considerably.

Edit:  The condition on $ f$ can be strengthened to $ | f(2 \plus{} \sqrt {3}) | < \frac {1}{34}$.  And unless I am horribly off, the coefficients can actually (alternately) be bounded by $ 2994$.  For a more interesting problem, perhaps they should've been bounded by $ 2007$ :)

Is the fact that $ x^2 \minus{} 4x \plus{} 1$ the minimal polynomial for $ x \equal{} 2 \plus{} \sqrt {3}$ relevant?

We can apply the division algorithm to establish
\[ (x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} (x \plus{} a \plus{} 4)(x^2 \minus{} 4x \plus{} 1) \plus{} (4a \plus{} b \plus{} 15)x \plus{} ( \minus{} a \plus{} c \minus{} 4)
\]
Plugging in $ r \equal{} 2 \plus{} \sqrt {3}$ we get $ ur \plus{} v$ where $ u \equal{} 4a \plus{} b \plus{} 15$ and $ v \equal{} \minus{} a \plus{} c \minus{} 4$.

I agree that this is a strange problem. Perhaps the proposer had a different solution in mind.

I mention the minimal polynomial as a way to describe a general technique if $ 2 \plus{} \sqrt{3}$ had been replaced by some other algebraic number.


if you like soccer,come here and express your opinons about  leagues, teams,or  players

European football leagues are my favorites!!!  :rotfl:

In my opinion, I think that Liverpool should have the right to defend their title as European champions.

Me,too.Liverpoor is a great team with a great deal of passion,they have many good players and they won the champion in the champion league this season,that is enough to prove their power.
It is reported that Liverpoor's captain Gerrard may leave next season,many clubs are interested in him,including RealMadrid,will he go to Spain after Owen?

I used to watch European soccer, and I was fond of German's league...since Chinese TV broadcast it a lot.

Manchester U. just bought a korean player, which makes me very happy.

i personally never particularily liked the British league, nor the German for that matter, I always enjoyed Italian and Spanish soccer though, even though i usually dislike Italian soccer at international competitions, but their clubs are really good and there are many fantastic games even among less reknowned teams.
btw, thanks for not including MLS in this thread becuase thats not really soccer

I used to like the Italian league since Batistuta was there.( hope i spelled correctly).

Real Madrid is probably the best club team, I think.

Red Devil you are korean?

*EDIT

Shoulda know... Red Devil hehe...

Long live 2002!

I like this idea of buying the rights to a player, but then loaning him off to another team so he can gain in stature and make a name for himself while he plays for the other team.

[quote="themonster"][quote="math92"]I like David Beckham and Zidane until he retired.  I like Manchester United and Real Madrid.[/quote]
Zidane retired?  I must be really out of it.  I should keep up with European soccer more; it's pretty awesome.[/quote]He only retired from the National Team (France's Team) not from the club :)

Real has some cool players, but because each season new players arrive, they don't make a good team (in the real sense of the word team). Look at the last winners of the Champions League: FC Porto and Liverpool. I bet no one would have bet on these teams two years ago to defeat names like Real, Barca, Chelsea, Milan (both of them :P), Juve, Bayer, Manchester, etc. However both these teams had really good middlefielders (Deco for Porto and Gerrard for Liverpool) and a good morale and team spirit, which ultimately brought them the Cup :)

Hmmmm, I think it was Liverpool's dancing keeper that propelled them to victory in penalty kicks..heehee.

[quote="lulu"]i hear that Realmadrid will go to US this summer,who know it well please tell us more about it,thank you[/quote]

I think some other European soccer teams are coming like Chelsea. I'll double check just to be sure. ;)

RealMadrid will come to Beijing China and have a friendship match on July 23rd with the local team,which  now is ranking the 4th place in China's league.Would you like to guess the result of the match?
I guess it would be 4-1

We were talking about Manchester United this noon... :?

We all know that Real have no world class defencer except Samuel,but I got the news that Real have sold him to InterMilan,I don't know why.Do anybody know more about this news?

They sold him because they want to buy Robinho. ;)

[quote="cezar lupu"]hey,$ A.C.  MILAN $ IS THE BEST TEAM IN THE WORLD. :lol:  :lol:  :lol:[/quote]
They got European champion last year. Really cool.
Edit: My post number now is 1234.  :rotfl:

I heared that Robinho won't go to Realmadrid,Santos doesn't let him go,and Robinho is a great young striker,but Realmadrid has so many world class strikers,what Real need is a world class defencer.
And i also heard that Vieri has left Inter for ACMilan and Crespo has got back to Chelsea.
Figo has got to Liverpool.

Figo? The guy in the Real Madrid? He's gonna go!?

it seems to me that Real arguably have the best players in the world on their squad, but they don't seem to use them effectively. they lost the league to barcelona and finished in a close second with ronaldo, zidane, owen, and beckham?!?!?! it would be interesting to see if real can ever get ronaldinho. then, they would be unbeatable! after all figo did move from barca to real. that said, i think that ac milan and arsenal are very good as well. with henry's speed and shevchenko's tactics, they're very fun to watch :lol:


linear algebra
matrix
limit
integration
function
abstract algebra
algebra

The 15th Annual Vojtech Jarnik International
Mathematical Competition

Second announcement

1. In order to recognize the achievement and to increase the
interest in mathematics, the 15th Annual Vojtech Jarnik
International Mathematical Competition will be held at the
University of Ostrava on Wednesday, 6th April 2005. The
competition is intended for university students of
mathematics.

2. This competition is divided into two categories. Category I
is intended for the students of the first and second year and
Category II is intended for the students of the third, fourth
and fifth year.

3. The competition is free of charge. It is sponsored by the
University of Ostrava. The prizes for the winners are
CZK 7000, CZK 3000 and CZK 2000 for the student on the first,
second and third place, respectively, in each category.

4. The competition is controlled by the Competition Committee.
It consists of delegates from participating universities.
Each university sends one representative to the Committee.
The Competition Committee is responsible for:

a) selection and evaluation of competition problems which
will be solved in each category,
b) correction and evaluation of students' solutions and
c) determination of the list of the winners.

A majority of votes is necessary for the acceptance of the
winner. In the case of a tie, the chairman of the Committee
casts the deciding vote.

5. The first meeting of the Committee will take place on
5th April 2005 at 9 a.m. The members of the Committee will
elect their chairman and will select competition problems for
each category.

6. There will be four mathematical problems in each category.
Students will have four hours to solve these problems. Their
solutions will be evaluated by the Committee then. The total
score of a student will be equal to the sum of the points for
the four mathematical problems.

7. Each participating university shall announce its delegate's
name and his e-mail as soon as possible to
vjaimc@osu.cz

8. Each participating university should send at least one
problem for each category along with a solution. Please do
not forget to send it before 5th March 2005 in a TeX file
in English. The e-mail address is
vjaimc@osu.cz

9. You will help us if you send each problem in a separate file.
Please specify the name of your university and the category
for which the problem is proposed explicitely in the file.
If you wish, you can add a short note explaining the origin
of the problem (e.g., the name of the person who proposed the
problem).

Each participating university agrees not to reveal the
competition problems to any of its students before the
beginning of the competition.

9. The competition will take place on 6th April 2005 from
9 a.m. to 1 p.m. The results will be announced the next
day afternoon (at about 3 p.m.).

10. There is no limit on the number of students from one
university, but two students are usually sent to each
category.

11. The official language is English. The competition problems
will be in the English language too. Students must solve
these problems in English.

Programme of the competition

Tuesday, 5th April 2005
9:00 -- 11:30 meeting of the Committee (in the room
no. C107, street 30. Dubna 22)
11:30 -- 13:00 lunch
13:00 -- ? meeting of the Committee (in the the room
no. C107, street 30. Dubna 22)

Wednesday, 6th April 2005
8:30 opening ceremony (in the assembly hall,
Ceskobratrska 16)
9:00 -- 13:00 the competition (in the Faculty of Science,
30.dubna 22)
13:00 -- ? evaluation of students' solutions by the
Committee

Thursday, 7th April 2005
15:00 (?) closing ceremony

Organizational instructions

1. The arrival of the participants is expected on 4th April
2005 evening or on 6th April 2005 morning. Their departure
is expected on 7 th April 2005 evening or on 8th April 2005
morning.

2. The stay of the participants and of the members of the
Competition Committee is to be paid by their mother
university.

3. Accommodation for participants will be booked in hotels and
hostels. The possibilities of accommodation are as follows:

a) Hotel Jindrich (***). Ten minutes from the Faculty of
Science of the University of Ostrava, where the
competition will take place. Prices: CZK 830 for a
double room, CZK 1300 for a triple room, CZK 1300 for
apartments. The price does not include breakfast
(CZK 70). The deadline for booking is 17th March 2005.

b) Hotel Polsky dum (***). Ten minutes from the Faculty of
Science. Prices: CZK 800 for a single room, CZK 1300 for
a double room, CZK 1600 for apartments. (If ten or more
rooms are booked, a discount of 10 % is allowed.) The
price includes breakfast. The deadline for booking is
17th March 2005.

c) Hotel Maria (***). By the Faculty of Science. Prices
for foreigners: CZK 1000 for a single room, CZK 1300 for
a double room (single person), CZK 1400 for a double room
(two persons), CZK 1700 for a triple room. Prices for
the Czechs and Slovaks: CZK 750 for a single room,
CZK 850 for a double room (single person), CZK 950 for a
double room (two persons), CZK 1200 for a triple room.
The price includes breakfast.

d) Hostel NH. Twenty minutes from the Faculty of Science.
Price: CZK 150 for a bed.

e) Hotel DMH. Forty minutes from the Faculty of Science by
tram. Prices: CZK 290 for a double room, CZK 350 for a
double room (de luxe), CZK 370 for a triple room. The
deadline for booking: as soon as possible.

Important dates

1. Applying to enter the competition: as soon as possible, the
deadline is 1st March 2005.

2. Recommending a member of the Competition Committee: as soon as
possible, the deadline is 5th March 2005.

3. Sending competition problems: no later than 5th March 2005.

4. Sending the number of participants and their need of
accommodation: no later than 5th March 2005.

5. Sending the names* of participants in each category:
till 15th March 2005.

6. Reporting the exact time of arrival and departure of
participants: till 20th March 2005.

* Note: Please pay a special attention to the names in order
that we do not distort them. Specify explicitely the first
name(s) and the family name. We prefer that you always specify
the family name after the first name(s), i.e., first name(s) first,
then the family name.

If a name contains accented letters, then use the standard ASCII
text to describe the accents. Most easily, you can use the TeX
sequences to describe the accents. (Alternatively, you can use
words to describe the accents, e.g., the latin small letter z with
dot above -- \.z -- a Polish letter, and so on.)

Exchange rates

The exchange rate list as of 14th December 2004:

1 USD = 23 CZK
1 EUR = 31 CZK
100 SKK = 78 CZK
100 HUF = 12 CZK
10 PLN = 73 CZK

Further information about this Competition can be found at

http://www.osu.cz/fpr/kma/dokumenty/vjaimc/vjaimc.html

Hi folk

Do not forget this math contest

We don't but there's really nothing to talk about yet :)

[quote="Moubinool"]The constest is finished 6th a APRIL

Could you post the problems ?
thx[/quote]

Here are the problems for the second category (as far as I remeber):

1. For square matrix M define $exp(M)=I+\frac{1}{1!}M+ \frac{1}{2!}M^2+\frac{1}{3!}M^3+...$.
Construct $2\times2$ matrices A and B such that $exp(A+B)\not= exp(A)exp(B)$

2. Given a real $n\times n$ matrix $[a_{ij}]$ such that $a_{ii}=0$ for $i=1,2,...,n$ .
Prove that there is a set $J\subset\{1,2,...,n\}$ such that $\displaystyle\sum_{i\in J, j\not\in J}{a_{ij}} + \displaystyle\sum_{i\not\in J, j\in J}{a_{ij}} \ge \frac{1}{2}\displaystyle\sum_{i,j=1}^n{a_{ij}}$

3. Compute $\displaystyle\lim_{n\to\infty} \frac{(2n+1)!}{(n!)^2} \displaystyle\int^1_0 \int^1_0 (xy(1-x)(1-y))^n f(x,y) dx dy$  where f is a continuous function on $[0,1] \times [0,1]$.
(I'm not quite sure about this problem, but it was something like that)

4. R is a finite ring with the following property: for every $a, b \in R$ there exists $c \in R$ (depending on $a$ and $b$) such that $a^2+b^2=c^2$.
Prove that for every $a, b, c \in R$ there is $d \in R$ such that  $2abc=d^2$.

Looks like quite easy problems... I want to be polite here... ;)

First category problems are:

1. Let $S=\{ z \mid |z|=1, z\neq -1\}$ and $f(z)=\frac {Im(z)}{1+Re(z)}$ be function form $S$ to $R$. Proove that $f$ is bijection and find $f^{-1}$.

2. Let $f: A^3 \rightarrow A$ such that:
  a) for all $x,y \in A$ we have $f(x,y,y)=f(y,y,x)=x$;
  b) for all $x_i,y_i,z_i \in A$ (i=1,2,3)  we have
\[f(f(x_1,x_2,x_3),f(y_1,y_2,y_3),f(z_1,z_2,z_3))=f(f(x_1,y_1,z_1),f(x_2,y_2,z_2),f(x_3,y_3,z_3)).\]
For fixed $a \in A$ define $x+y$ by $x+y=f(x,a,y)$. Proove that $(A,+)$ is Abelian group.

3. Find all real numbers $\lambda$ such there exists real polynomial $p(x)$ where
$\frac {p(1)+p(3)+\hdots+p(2n-1)}n=\lambda p(n)$ holds for every natural $n$.

4. Let $x_2>0$ and $x_{n+1}=-1+\sqrt[n]{1+nx_n}$ for $n\geq 2$. Find $\lim x_n$ and $\lim nx_n$.

[quote="Maja"]

3. Compute $\displaystyle\lim_{n\to\infty} \frac{(2n+1)!}{(n!)^2} \displaystyle\int^1_0 \int^1_0 (xy(1-x)(1-y))^n f(x,y) dx dy$  where f is a continuous function on $[0,1] \times [0,1]$.
(I'm not quite sure about this problem, but it was something like that)
[/quote]

I'm afraid there is a mistake in this text. Could anybody write it in a correct way? I'm just curious if i could solve it :lol:

It should be \[\lim_{n\to\infty} \left(\frac{(2n+1)!}{(n!)^2}\right)^2 \int^1_0 \int^1_0 (xy(1-x)(1-y))^n f(x,y) dx dy,\] and, clearly, answer is $f\left(\frac 12,\frac 12\right)$.

Problem 4 from first category is not hard. i think that both limits are equal with zero.is there anyone with other
 opinion.

Hello Everybody!

Can somebody solve the 3. problem (first category)?
Moubinool, Maja, Myth, n0vakovic, Owca, cezar lupu please help me! It's very important.

Thank you all!

Valentin, I'm interrested. How can someone participate at this contest? ;)

Solution of problem 2.:

We have to show that:
1) $ (\exists e \in A)(\forall x \in A)\ e \plus{} x \equal{} x \equal{} x \plus{} e$
Let $ e \equal{} a$. Indeed, $ e \plus{} x \equal{} a \plus{} x \equal{} f(a, a, x) \equal{} x \equal{} f(x, a, a) \equal{} x \plus{} a \equal{} x \plus{} e$.

2) $ (\forall x \in A)(\exists y \in A)\ x \plus{} y \equal{} e$
For arbitrary fixed $ x \in A$ let $ y \equal{} f(a, x, a)$.
Hence $ x \plus{} y \equal{} x \plus{} f(a, x, a) \equal{} f(x, a, f(a, x, a)) \equal{}  f(f(a, a, x), f(a, x, x), f(a, x, a)) \equal{}$
$ \equal{} f(f(a, a, a), f(a, x, x), f(x, x, a)) \equal{} f(a, a, a) \equal{} a \equal{} e$.

3) $ (\forall x, y, z \in A)\ (x \plus{} y) \plus{} z \equal{} x \plus{} (y \plus{} z)$
$ (x \plus{} y) \plus{} z \equal{} f(x, a, y) \plus{} z \equal{} f(f(x, a, y), a, z) \equal{} f(f(x, a, y), f(a, a, a), f(a, a, z)) \equal{} f(f(x, a, a), f(a, a, a), f(y, a, z)) \equal{} f(x, a, f(y, a, z)) \equal{} x \plus{} f(y, a, z) \equal{} x \plus{} (y \plus{} z)$

4) $ (\forall x, y \in A)\ x \plus{} y \equal{} y \plus{} x$
$ x \plus{} y \equal{} f(x, a, y) \equal{} f(f(x, a, a), f(x, x, a), f(y, x, x)) \equal{} f(f(x, x, y), f(a, x, x), f(a, a, x)) \equal{} f(y, a, x) \equal{} y \plus{} x$

Therefore $ (A, \plus{})$ is Abelian group.
Why is in formulas so much blanks? Does anybody know how to avoid it?


logarithms
calculus
calculus computations

Let $\Gamma_{n}=\frac{n!}{n^{n+\frac{1}{2}}e^{-n}}$ for $n\geq 1$.  Prove that the infinite product:  $\prod_{1}^\infty \frac{\Gamma_{n+1}}{\Gamma_{n}}$ converges.

The actual value to which the product converges is well-known.  Do you know it?

$\Gamma_{n}\geq (\frac{2}{3}e)^{\frac{3}{2}})\ (n=2,\ 3,\ \cdots)$

Stirling will come. :D

your sum is equal to:
$\frac{1!2!3!...(n+1)!}{e^{-\frac{n(n+1)}{2}}\sqrt{n!}1^{1}2^{2}3^{3}...n^{n}}$
aproximatly valua of this $1!2!3!...(n+1)!$ expression is here
 http://www.mathlinks.ro/Forum/viewtopic.php?t=147678

here $\sqrt{n!}$ use stirling formula.

and here

$1^{1}2^{2}3^{3}...n^{n}$
the first step:

show by Stoltz theorem
$\sum^{n}k\ln{k}=\frac{1}{2}(n^{2}+n+\frac{1}{6})\ln{n}-\frac{1}{4}n^{2}+C_{1}+o(1)$
therefore
$1^{1}2^{2}3^{3}...n^{n}=e^{\frac{1}{2}(n^{2}+n+\frac{1}{6})\ln{n}-\frac{1}{4}n^{2}+C_{1}+o(1)}$
.
now you have everything to solve this problem (find exact aproximatly value) good luck  :wink:

I am not sure if I follow Kunny's approach.  Bounding the product above by $K \prod_{1}^{\infty}\Gamma_{n+1}$, with $K$ constant, is not helpful since that product diverges.

Also, I am not sure of Extremal's first equivalence, but after that his solution looks good -- but I think it may be disingenuous.  That is, both solutions mention using Stirling's approximation, but if you know Stirling's approximation, you can solve the problem immediately.  Indeed, we have:

$\prod_{1}^{\infty}\frac{\Gamma_{n+1}}{\Gamma_{n}}= \frac{\Gamma_{2}}{\Gamma_{1}}\times \frac{\Gamma_{3}}{\Gamma_{2}}\times \frac{\Gamma_{4}}{\Gamma_{3}}\cdots$.  So the product, if it converges, will equal $\frac{\Gamma_{\infty}}{\Gamma_{1}}=\frac{\sqrt{2\pi}}{e}$ by Stirling's approximation since the product telescopes.  (Note I should have started the product at $0$ with the understanding that $\Gamma_{0}=1$, then the product would converge to $\sqrt{2 \pi}$ exactly.

But with all of that said, does anyone have a solution that does not use Stirling's approximation?

Convergence is very simple to prove, because $\ln\frac{\Gamma_{n+1}}{\Gamma_{n}}=1-\left(n+\frac12\right)\ln\left(1+\frac1n\right)=O\left(\frac1{n^{2}}\right)$.


function
algebra
domain
summer program
PROMYS
number theory

I know how to solved the Pell's equations $ x^2 \minus{} by^2 \equal{} 1$ and $ x^2 \minus{} by^2 \equal{} \minus{} 1$, but I am interested in the more general

$ x^2 \minus{} by^2 \equal{} c$

Now I know if we have one solution $ (p,q)$ then we can find infinitely many solutions with $ (p \minus{} q\sqrt {b})(r \minus{} s\sqrt {b})^n$ where $ (r,s)$ is the first solution to $ x^2 \minus{} by^2 \equal{} 1$.

However, http://mathworld.wolfram.com/PellEquation.html says that this does not necessarily generate all solutions.

My question is, how do we know when we have all the "primitive" solutions $ (p,q)$ that will generate all the solutions? (Also, is it possible that there would be infinite primitive solutions?)

I couldn't find this answer searching anywhere.

The Mathworld article tells you what happens when $ c < \sqrt{b}$, so I assume you are asking about the (significantly harder) other case.  In the special case that $ \mathbb{Z}[\sqrt{b}]$ has prime factorization, the problem of finding the primitive solutions is equivalent (I think) to the problem of finding the prime factorization of $ c$ in $ \mathbb{Z}[\sqrt{b}]$.  (I believe this technique still works when $ \mathbb{Z}[ \frac{1 \plus{} \sqrt{b}}{2} ]$ - which is the appropriate ring of integers half the time - has prime factorization.)  See [url=http://mathworld.wolfram.com/QuadraticField.html]here[/url].

Unfortunately, for most values of $ b$ this is not the case.  The corresponding rings are much harder to work with.

Thank you, t0rajir0u, but how do I use this prime factorization to find the solutions?

It will be easiest to demonstrate using the case $ b \equal{} \minus{}1$ (where the unit solutions are trivial and everything is a primitive solution).  Then we are finding solutions to $ x^2 \plus{} y^2 \equal{} c$.  These solutions depend entirely on the factorization of $ c$ into primes congruent to $ 1 \bmod 4$ and primes congruent to $ 3 \bmod 4$ (which in turn determines its factorization over $ \mathbb{Z}[\imath]$, the [url=http://en.wikipedia.org/wiki/Gaussian_integer]Gaussian integers[/url].)

[b]Proposition.[/b]  $ c$ is representable as a sum of two squares if and only if each prime congruent to $ 3 \bmod 4$ divides it an even number of times.

The proof is left as an exercise; it relies on proving that primes congruent to $ 1 \bmod 4$ are a sum of two squares in exactly one way.  It will again be easiest to demonstrate how this leads to values of $ x$ and $ y$ by example.  When $ c \equal{} 65$ we have

$ 65 \equal{} 5 \cdot 13 \equal{} (1^2 \plus{} 2^2)(2^2 \plus{} 3^2) \equal{} (1 \cdot 2 \pm 2 \cdot 3)^2 \plus{} (1 \cdot 3 \mp 2 \cdot 2)^2 \equal{} 8^2 \plus{} 1^2 \equal{} 4^2 \plus{} 7^2$

where we have used the lemma $ (a^2 \plus{} b^2)(c^2 \plus{} d^2) \equal{} (ac \pm bd)^2 \plus{} (ad \mp bc)^2$ (which is in turn a consequence of permuting the factors of $ (a \plus{} bi)(a \minus{} bi)(c \plus{} di)(c \minus{} di)$.)  In other words, permuting the complex factors of $ c$ into two groups such that one group is the conjugate of the other group gives you the set of solutions.  (A product of three primes is expressible in more ways than a product of two primes, and so forth.  See [url=http://mathworld.wolfram.com/SumofSquaresFunction.html]sum of squares function[/url].)

If you are familiar with algebraic number theory, you can find a complete consideration in one of my posts in http://www.mathlinks.ro/viewtopic.php?t=212193 .

Ok so is this example right?

For $ a^2 \minus{} 2b^2 \equal{} 7$ the prime factorization of $ 7$ is

$ 7 \equal{} (3 \plus{} \sqrt {2})(3 \minus{} \sqrt {2})$

So $ (3,2)$ is my solution.

But if I have $ a^2 \minus{} 2b^2 \equal{} 161$ then

$ 161 \equal{} 7\cdot 23 \equal{} (3 \plus{} \sqrt {2})(3 \minus{} \sqrt {2})(5 \plus{} \sqrt {2})(5 \minus{} \sqrt {2})$
$ \equal{} (17 \plus{} 8\sqrt {2})(17 \minus{} 8\sqrt {2})$
$ \equal{} (13 \plus{} 2\sqrt {2})(13 \minus{} 2\sqrt {2})$

So I get $ (17,8)$ and $ (13,2)$.

I'm pretty sure that unique prime factorizations do not always exist for all $ x$ in the given set of numbers  in the form $ a \plus{} b\sqrt {x}, a,b\in \mathbb{Q}$.

You mean the ring of integers, not the field, but yes (it does exist when $ x \equal{} 2$).  The negative values of $ x$ for which this is true are the [url=http://mathworld.wolfram.com/HeegnerNumber.html]Heegner numbers[/url].  The question for positive $ x$ is open; the ones for which $ \mathbb{Q}[\sqrt{x}]$ is a [url=http://en.wikipedia.org/wiki/Euclidean_domain]Euclidean domain[/url] are given [url=http://mathworld.wolfram.com/QuadraticField.html]here[/url].  See the [url=http://en.wikipedia.org/wiki/Class_number_problem]class number problem[/url].

The counterexample that appears on the PROMYS 2006 shirt is $ 2 \cdot 3 \equal{} (1 \plus{} \sqrt{\minus{}5})(1 \minus{} \sqrt{\minus{}5})$ in $ \mathbb{Z}[ \sqrt{\minus{}5} ]$.

tjhance, the solution of $ a^2 \minus{} 2b^2 \equal{} 7$ is $ (3,1)$ not $ (3,2)$.

So for the general case $ a^2 \minus{}xb^2 \equal{} c$ we prime factorize $ c$ over $ \mathbb{Z}[\sqrt{x}]$ and then figure out how to write it as two (conjugate) factors, and then the factors are in the form $ a\plus{}b\sqrt{n}$ where $ (a,b)$ is the solution?

Also how can we determine prime factorizations of $ c$, and whether they are unique?


I currently use \twocolumn to create a two column page in a document and use \onecolumn when I am done.  These commands, though, automatically start a new page.  How can I create a page in a document that has two columns but does not automatically create a new page, i.e. the whole page is not in two columns?  Thanks

droovy

The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=multicol]multicol package[/url] will solve this problem.


Ummm... you might want to understand what a problem is saying before you post it....

As for limits, just look it up on wikipedia. Yes, its that simple (my explanation would be biased in some way)

The answer by the way is [hide]18[/hide]
off the top of my head

$\lim_{x\to\infty}\left(\frac{3x-5}{x}\right)\left(6+\frac{2}{x}\right)=\lim_{x\to\infty}(3-\frac{5}{x})(6+\frac{2}{x})=3.6=18$

see http://mathworld.wolfram.com/Limit.html  :wink:

The definitions at Wikipedia and Mathworld are really confusing and I was wishing that someone could explain it to me in simple terms.

Would you need to find out different values and then see what they approach?

[quote="Quevvy"]i think this is called a limit and I'm not sure what forum to put it in...
Here is the problem:
$\lim_{x\to\infty}\left(\frac{3x-5}{x}\right)\left(6+\frac{2}{x}\right)$
How would you do this, and what is the definition of a limit?[/quote]
Limits are when something converges to a number.  For example, $\lim_{x\to\infty}$ is when x [b]converges [/b]to infinity, but is not quite there yet (but it is so close that you can say it is).  For example, if you wanted to caculate x/x when x's limit is 0, then it is 1.  However, if you wanted to calculate x/x when x [b]is[/b] 0, then it is undefined.  Now, for your problem:
[hide] $\frac{3x-5}{x}=3-\frac{5}{x}$.  Now, since x goes to 0, we can pretty much say that a/x is 0 since x is so big.  Therefore, 5/x=0.  This implies that $\frac{3x-5}{x}=3$.  Now, for the second part, we have that 6+2/x=6 since 2/x is almost 0.  Therefore, you have that $(\frac{3x-5}{x})(6+\frac{2}{x})=3*6=18$.  [/hide]

with that definition, you see that as x grows bigger, the fractional part: 2/x gets smallers and tends to 0
then you are left with only a 6 to deal with.
similarly the -5/x also tends to 0 so you are left with a 3
then just multiply to get 18!!
limits are useful in this problem because they easily get rid of dumb fractions that hinder you......
without it, well good luck!

Thank you The QuattoMaster 6000 and davidlizeng, for explaining how limits work. The definitions at Mathworld and Wikipedia were [i]really [/i] confusing and now I understand. But I still have one question, in the sign, it looks like x goes to infinity but you people said that it goes to zero. Which one is it?

Oh, sorry for making that confusing.  In this case $\lim_{x\to\infty}$, x goes to infinity.  I was taking another example: $\lim_{x\to0}$.  In the new example, x converges to 0.  The notation says that for $\lim_{x\to\(y}$, x converges to y.

quatto, you got it all wrong....
when i say tend to 0 it means this:
take the fraction 2/x
as [b]x tends to infinity[/b]
the [b]fraction tends to 0[/b]
basically, the bigger the x, the smaller the fraction, till the point where x is soooooo large that the fraction is sooooo small that it basically is 0.

The best way to learn about limits is by the formal definition, like the one given at wikipedia. Using definitions lie "x tends to infinity, fraction tends to zero" etc. doesn't do much good (well at least for me) but consider what a limit really is: a bound on a quantity that isn't obtained. It suffices to prove that every number arbitrarily close (and in the correct direction) can be attained. That's all a limit is.


Έστω $ p,q,r$ θετικοί πραγματικοί αριθμοί με $ n\in N$. Αποδείξτε ότι εαν $ pqr\equal{}1$, τότε $ \frac{1}{p^{n}\plus{}q^{n}\plus{}1}\plus{}\frac{1}{q^{n}\plus{}r^{n}\plus{}1}\plus{}\frac{1}{r^{n}\plus{}p^{n}\plus{}1}\leq 1$

Θέτουμε  $ p^n\equal{}a^3$, $ q^n\equal{}b^3$, $ r^n\equal{}c^3$, οπότε αρκεί να δ.ο.

$ \frac{1}{a^3\plus{}b^3\plus{}1}\plus{}\frac{1}{b^3\plus{}c^3\plus{}1}\plus{}\frac{1}{c^3\plus{}a^3\plus{}1} \le 1$ , με abc =1 .

Είναι  $ a^3\plus{}b^3 \ge ab(a\plus{}b)$  <=>

        $ a^3\plus{}b^3\plus{}1 \ge ab(a\plus{}b)\plus{}1$  <=>

        $ \frac{1}{a^3\plus{}b^3\plus{}1} \le \frac{1}{ab(a\plus{}b)\plus{}1}$ <=>

        $ \frac{1}{a^3\plus{}b^3\plus{}1} \le \frac{abc}{ab(a\plus{}b)\plus{}abc}$  <=>

        $ \frac{1}{a^3\plus{}b^3\plus{}1} \le \frac{c}{a\plus{}b\plus{}c}$

Εργαζόμαστε κυκλικά προσθέτουμε και είμαστε έτοιμοι.


geometry
3D geometry
pyramid
Pythagorean Theorem

A pyramid has a rectangular base 10 cm by 25 cm and four triangular faces; two opposing faces both with altitude 13 cm and the other two faces with altitudes of 20 cm and 15 cm.  What is the volume of the pyramid?

formula for V (pyramid) = 1/3*Bh (u can latexify if u wanna, but im not doin it :D )

first, to find the height, use the pyth. theo.

find the apothem of the base (5)

the height of the triangle is 13.

it is a 5-12-13 triangle, thus the height is 12

then simply multiply, 1/3*250*12

                            250*4 = 1000

this concludes my 32nd post.

meh.

I do not understand which face should be the hypotenuse, and I cannot visualize it. Any suggestions?

formula for V (pyramid) = $ 1/3*Bh $

first, to find the height, use the $ Pythagorean $ $ theorem $

the $ apothem $ of the base is $ 5 $

the height of the triangle is $ 13 $

it is a $ 5-12-13 $ triangle, and thus the height is $ 12 $

then simply multiply, $ 1/3*250*12$

$ 250*4 = 1000 $

$ \boxed {1000} $ is your answer.

$ LATEXIFIED $


Let $ a>1$ and the sequence $ (x_n)_{n \ge 0}$ such that:

$ x_0\equal{}0, x_1\equal{}1, (a\plus{}1)^{x_{n\plus{}2}}\equal{}a^{x_{n\plus{}1}}\plus{}(\sqrt{a\plus{}1})^{x_n}$, $ \forall n \ge 0$

Prove that the sequence is convergent and determine its limit

Do you have a solution to this? I don't see how a limit might even be determined algebraically.

I don't have a solution to this. The only thing I can deduce from this problem is that if $ (x_n)$ is convergent then the limit of this sequence is $ 1$. It remains to prove that $ (x_n)$ is convergent.

Sorry, you're right. The limit isn't $ 1$. But this limit can be determined solving the equation:

$ 1\equal{}(\frac {a} {a\plus{}1})^x\plus{}(\frac {\sqrt{a\plus{}1}} {a\plus{}1})^x$

which has a unique solution


Starting with "lii", add a letter and then scramble it to make a new word.
If you were this new word, you would have even more difficulty solving this riddle :lol:

(No proper names, slang, etc.)
(Also, it's not roman-numeral related)

[hide]ilia?

I know it's a word (from Scrabble)[/hide]

[hide]An illiterate person would have a more difficult time solving
the puzzle, but....[/hide]

I don't know what word construction would fulfill this, if true.

[hide="I think you got it!"]If you add the words "a letter" to iil, you get illiterate.[/hide]

[quote="box of rocks"]That's what the creator of the puzzle of MAY have intended, BUT he/she can't have it both ways.

The instructions are not ambiguous; adding a letter cannot be be translated into attaching the letters themselves of the phrase "a letter" to those of "lii" and then later forming a new word.

Having to rely on this makes it a flawed puzzle.[/quote]
With your logic, pretty much EVERY puzzle here(and everywhere else) are flawed puzzles.

junggi, you were told to stop trolling these boards, weren't you?

If not yet, I am doing so now.  You are not adding anything; in fact you are disrupting the flow of them.  When I make a point,
it's made with a purpose.  You are harrassing on here to throw off the topic.

[quote="box of rocks"]junggi, you were told to stop trolling these boards, weren't you?
[/quote]
By who? you? well why should I listen to you, when you never listen to me?
And I respectfully disagree with you that I am "trolling." I am just pointing out that your logic is not very reasonable, as most puzzles have flaws
[quote="box of rocks"]
If not yet, I am doing so now.
[/quote]
By trolling?
[quote="box of rocks"]
You are not adding anything; in fact you are disrupting the flow of them.
[/quote]
And what would you say about YOUR post that I was responding to? Did that post add anything to the discussion of the puzzle? I'd have to say no.
[quote="box of rocks"]
When I make a point, it's made with a purpose.
[/quote]
I know(but your point isn't that good), and same goes for me.
[quote="box of rocks"]
 You are harrassing on here to throw off the topic.[/quote]
and you're not?   :huh:

[quote="box of rocks"]That's what the creator of the puzzle of MAY have intended, BUT he/she can't have it both ways.

The instructions are not ambiguous; adding a letter cannot be be translated into attaching the letters themselves of the phrase "a letter" to those of "lii" and then later forming a new word.

Having to rely on this makes it a flawed puzzle.[/quote]

Actually, box of rocks, this is the basis of very many puzzles, like the following:

"railroad crossing - look out for cars"
can you spell that without any r's?

[hide="answer"]
t-h-a-t
[/hide]

problems like this encourage people to think about the problems from different angles

Unfortunately for the puzzle, ilia is a word. And it would be very hard for you to solve this puzzle if you were a bone!


algebra
polynomial
calculus
derivative
calculus computations

Suppose $ f(x) \equal{} (2x \plus{} 1)^{10}(3x \minus{}1)^7$. Find a formula for $ f'(x)$. Decide on a reasonable way to simplify your result and find a formula for $ f''(x)$.

[hide="what I did"]Using the product rule:
$ f'(x) \equal{} (2x\plus{}1)^{10} \cdot 3 \cdot 7(3x\minus{}1)^6 \plus{} (3x\minus{}1)^7 \cdot 2 \cdot 10(2x\plus{}1)^9$
I tried to simplify it:
$ f'(x) \equal{} (3x\minus{}1)^6 (7 \plus{} (3x\minus{}1)) ...$
Help please?[/hide]

Offhand, I'd say factor out $ (2x \plus{} 1)^{9} (3x \minus{} 1)^{6}$ and then write the rest as another factor.  Generally, if $ (x \minus{} r)^k$ divides a polynomial $ f(x)$ then $ (x \minus{} r)^{k\minus{}1}$ divides $ f'(x)$.

I got $ f'(x) \equal{} (2x\plus{}1)^9 (3x\minus{}1)^6 (102x \plus{} 1)$ how can I apply the product rule to this?

When you're taking the derivative of three factors, you have to apply the product rule twice.  That is,

\begin{eqnarray*}
\frac{d}{dx} fgh &=& \frac{d}{dx} \left( f \cdot (gh) \right) \\
 &=& f' (gh) + f \cdot \frac{d}{dx} \left( gh) \right) \\
 &=& f' gh + f \cdot (g'h + gh') \\
 &=& f'gh + fg'h + fgh'
\end{eqnarray*}

This expression is actually analogous to the product rule for two factors, and is easily generalized.  That is, we take all the possible products of $ n$ factors where exactly one of them is differentiated.  That's why bookworm_vn has three terms in his expression.

In such questions, using Maple/Mathematica is a good way

To check your answer, maybe.  As far as concepts and skills go, doing it by hand at least once is necessary.


Solve for integer solution:
$ x^2\plus{}1\equal{}y^3$

[quote="bibobeo"]Solve for integer solution:
$ x^2 \plus{} 1 \equal{} y^3$[/quote]

where's your solution? i would like one...

This is the [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell curve[/url] with $ n \equal{} \minus{}1$.

[hide="Solution"] Write $ (x \plus{} i)(x \minus{} i) \equal{} y^3$ over the Gaussian integers.  Then $ y$ must be the norm of a Gaussian integer, so let $ y \equal{} (a \plus{} bi)(a \minus{} bi)$, hence

$ x \plus{} i \equal{} (a \plus{} bi)^3 \Leftrightarrow$
$ x \plus{} i \equal{} a^3 \plus{} 3a^2 bi \minus{} 3ab^2 \minus{} b^3 i \Leftrightarrow$
$ x \equal{} a^3 \minus{} 3ab^2, 1 \equal{} 3a^2 b \minus{} b^3.$

Since $ b | 1$ we have $ b \equal{} \pm 1$ and $ \pm 1 \equal{} 3a^2 \minus{} 1 \implies a \equal{} 0$, so $ (x, y) \equal{} \boxed{(0, 1)}$ is the only integer solution. [/hide]

I know little about Moedel curve,and Gaussian integer...does there exist fundamental solution??

[hide="my idea"]
the clear solution is $ (x,y) \equal{} (0,1)$.Let $ (x,y)\ne (0,1)$ in following.
$ (y \minus{} 1)(y^2 \plus{} y \plus{} 1) \equal{} x^2$ and $ (y \minus{} 1,y^2 \plus{} y \plus{} 1) \equal{} (y \minus{} 1,2y \plus{} 1) \equal{} (y \minus{} 1,3) \equal{} 1$ or $ 3$.
Consider in two cases.

1)When $ (y \minus{} 1,y^2 \plus{} y \plus{} 1) \equal{} 1$
We can let $ y \minus{} 1 \equal{} m^2,y^2 \plus{} y \plus{} 1 \equal{} n^2$($ m,n$ are non-negative integers.),and we have
$ (2m^2 \plus{} 3 \minus{} 2n)(2m^2 \plus{} 3 \plus{} 2n) \equal{} \minus{} 3$,easily see that there are no such $ (m,n)$

2)When $ (y \minus{} 1,y^2 \plus{} y \plus{} 1) \equal{} 3$
We can let $ y \minus{} 1 \equal{} 3m^2,y^2 \plus{} y \plus{} 1 \equal{} 3n^2$,we have

$ 3m^4 \plus{} 3m^2 \plus{} 1 \equal{} n^2$ i.e. $ (2n)^2 \minus{} 3(2m^2 \plus{} 1)^2 \equal{} 1$

How can I see this kind of Pell's equation doesn't have solutions?

[/hide]

Gaussian integers aren't that much more complicated than the solutions to Pell's equations, especially since what you've got isn't actually a Pell's equation but just another form of an elliptic curve.

[quote="t0rajir0u"]This is the [url=http://mathworld.wolfram.com/MordellCurve.html]Mordell curve[/url] with $ n \equal{} \minus{} 1$.

[hide="Solution"] Write $ (x \plus{} i)(x \minus{} i) \equal{} y^3$ over the Gaussian integers.  Then $ y$ must be the norm of a Gaussian integer, so let $ y \equal{} (a \plus{} bi)(a \minus{} bi)$, hence

$ x \plus{} i \equal{} (a \plus{} bi)^3 \Leftrightarrow$
$ x \plus{} i \equal{} a^3 \plus{} 3a^2 bi \minus{} 3ab^2 \minus{} b^3 i \Leftrightarrow$
$ x \equal{} a^3 \minus{} 3ab^2, 1 \equal{} 3a^2 b \minus{} b^3.$

Since $ b | 1$ we have $ b \equal{} \pm 1$ and $ \pm 1 \equal{} 3a^2 \minus{} 1 \implies a \equal{} 0$, so $ (x, y) \equal{} \boxed{(0, 1)}$ is the only integer solution. [/hide][/quote]

3 things: 1. Why does $ y$ have to be the norm of a Gaussian integer?
2. Wouldn't $ y\equal{}\sqrt{a^2\plus{}b^2}$ instead of $ y\equal{}a^2\plus{}b^2$ if it is the norm of a Gaussian integer?
3. Why does $ x\plus{}i$ have to equal $ (a\plus{}bi)^3$?

1.  The more elementary way to think about this is that every prime divisor of $ x^2 \plus{} 1$ must be congruent to $ 1 \bmod 4$, hence this is also true of $ y$.  In terms of the Gaussian integers, what's going on is that unless each prime factor of $ y$ (over the Gaussian integers) is paired with its conjugate, the prime factors of $ y^3$ won't be paired with their conjugates, but the prime factors of $ (x \plus{} i)(x \minus{} i)$ are always paired with their conjugates.

2.  Number theorists define $ N(a \plus{} b \sqrt{d}) \equal{} a^2 \minus{} db^2$ instead of its square root; from a number-theoretic (rather than geometric) point of view, this is the more interesting quantity.

3.  $ \gcd(x \plus{} i, x \minus{} i) \equal{} \gcd(x \plus{} i, 2i)$, and for $ x \neq 1, i, \minus{}1, \minus{}i$ this implies $ \gcd(x \plus{} i, x \minus{} i) \equal{} 1$.  Again, one can think of this as pairing up conjugate prime factors over the Gaussian integers.

Thank you torajirou,I would read texts about Gaussian integer and eliptic curve.


geometry inradius circumcircle trigonometry

The radius of the inscribed circle of a triangle is $r$, the radius of its circumcircle is $R$, and one of the angles is $\alpha$. Given that \[ r = 4R\cos \alpha\sin^2 \frac{\alpha}{2} \]
show that the triangle is isosceles.


$ \underbrace {\frac{{\gamma _{{\rm{lg}}} (1 \plus{} \cos \theta )}}{{2\sqrt
{\gamma _{{\rm{lg}}}^{{\rm{dis}}} } }}}_y \equal{} \underbrace {\sqrt
{\gamma _{{\rm{sg}}}^{{\rm{pol}}} }}_m\underbrace {\sqrt
{\frac{{\gamma _{{\rm{lg}}}^{{\rm{pol}}} }}{{\gamma
_{{\rm{lg}}}^{{\rm{dis}}} }}}}_x \plus{} \underbrace {\sqrt {\gamma
_{{\rm{sg}}}^{{\rm{dis}}} }}_c$

How do I make the braces on the same level?

Search here: http://en.wikibooks.org/wiki/LaTeX

If it isn't here, then I don't know where to look.

\vphantom can solve the problem. Just write
[code]
\newcommand{\vf}{\vphantom{\frac A{\frac BC}}}
$$
\underbrace{\vf XYZ}\underbrace{\frac A{\frac BC}}
$$
[/code]
to get
$ \underbrace{\vphantom{\frac A{\frac BC}} XYZ}\underbrace{\frac A{\frac BC}}$

The first formula has the greatest height so you add it using \vphantom to each of the other formulae:
[code]\newcommand{\formulaA}{\frac{{\gamma _{{\text{lg}}} (1 + \cos \theta )}}{{2\sqrt{\gamma _{{\text{lg}}}^{{\text{dis}}} } }}}
\newcommand{\formulaB}{\sqrt{\gamma _{{\text{sg}}}^{{\text{pol}}} }}
\newcommand{\formulaC}{\sqrt{\frac{{\gamma _{{\text{lg}}}^{{\text{pol}}} }}{{\gamma_{{\text{lg}}}^{{\text{dis}}} }}}}
\newcommand{\formulaD}{\sqrt {\gamma_{{\text{sg}}}^{{\text{dis}}} }}
$\underbrace{\formulaA}_{y}=\underbrace{\formulaB\vphantom{\formulaA}}_{m}\underbrace{\formulaC\vphantom{\formulaA}}_{x}+\underbrace{\formulaD\vphantom{\formulaA}}_{c}$[/code]
which looks like this
$ \underbrace{\frac {{\gamma _{{\text{lg}}} (1 \plus{} \cos \theta )}}{{2\sqrt {\gamma _{{\text{lg}}}^{{\text{dis}}} } }}}_{y} \equal{} \underbrace{\sqrt {\gamma _{{\text{sg}}}^{{\text{pol}}} }\vphantom{\frac {{\gamma _{{\text{lg}}} (1 \plus{} \cos \theta )}}{{2\sqrt {\gamma _{{\text{lg}}}^{{\text{dis}}} } }}}}_{m}\underbrace{\sqrt {\frac {{\gamma _{{\text{lg}}}^{{\text{pol}}} }}{{\gamma_{{\text{lg}}}^{{\text{dis}}} }}}\vphantom{\frac {{\gamma _{{\text{lg}}} (1 \plus{} \cos \theta )}}{{2\sqrt {\gamma _{{\text{lg}}}^{{\text{dis}}} } }}}}_{x} \plus{} \underbrace{\sqrt {\gamma_{{\text{sg}}}^{{\text{dis}}} }\vphantom{\frac {{\gamma _{{\text{lg}}} (1 \plus{} \cos \theta )}}{{2\sqrt {\gamma _{{\text{lg}}}^{{\text{dis}}} } }}}}_{c}$
(The code for this image is far too complicated and only for this forum but is necessary as \newcommand is not allowed here.)

Note the use of \text rather than the deprecated \rm

PS This does the same as fedja's post which I didn't see when I posted :)

Just something worth mentioning I think: you don't actually need the \newcommand to do this.


search
inequalities unsolved
inequalities

Let $x,y,z$ be real numbers satisfy: $x+y+z=0$ and $xyz=2$.
Find the maximum values of :
$\displaystyle P=\frac {x^2}{y}+\frac {y^2}{z}+\frac {z^2}{x}$.

I think Namdung posted a very similar problem. Try to search for it. I forgot the name of the topic, but it must have been something with homogenuous.


AMC
AIME
quadratics
algebra
quadratic formula

Find the sum of all positive integers $n$ for which $n^{2}-19n+99$ is a perfect square.

[hide]
$n^{2}-19n+99=k^{2}$
$n^{2}-19n+99-k^{2}=0$
By the quadratic formula:
$n=\frac{19\pm\sqrt{19^{2}-4(99-k^{2})}}{2}=\frac{19\pm\sqrt{4k^{2}-35}}{2}$
For $n$ to be an integer, the number in the radical in must be a perfect square and the numerator must be even. The second requirement is taken care of because the radical would be odd and so is 19.

For the first requirement,
$4k^{2}-35=n^{2}$
$(2k)^{2}-35=n^{2}$
$(2k)^{2}-n^{2}=35$
$(2k+n)(2k-n)=35$
The factors can either be $7,5$ (in which case $k=3,n=1$) or $35,1$ (where $k=9,n=17$)

We then plug the possible values of $k$ into the quadratic formula equation at the top, we get possible values of $n$ to be $1,9,10,18$, and the sum is $\boxed{38}$
[/hide]


[i]Solving Mathematical Problems[/i], exercise 4.4

Find all positive reals $ x,y,z$ and all positive integers $ p,q,r$ such that

$ x^p \plus{} y^q \equal{} y^r \plus{} z^p \equal{} z^q \plus{} x^r$

I think that it's obvious there are infinitely many of them, so we can't find them all. The equality will always hold for any reals and integers satisfying the condition:
$ (x\equal{}y\equal{}z)\wedge (p\equal{}q\equal{}r)$ :wink:

Another family of solution : $ x\equal{}y\equal{}z\equal{}1$.
Is there yet another?

I don't think there is a nice solution to that, at least not for reals. For example, putting x=y implies p=q and we land in the end in an equation:
$ 2x^{p}\minus{}x^{r}\equal{}z^{p}$ and we have infinitely many combinations here... :o

I've had the same problem with that one some months ago, but actually there's a mistake in the statement, as one can read on [url=http://terrytao.wordpress.com/books/solving-mathematical-problems/]Tao's blog[/url] :
[quote="Terence Tao"]In page 66, problem 44, there is a $ \equal{}2$ missing at the end of the string of equations, thus $ x^p\plus{}y^q\equal{}y^r\plus{}z^p\equal{}z^q\plus{}x^r\equal{}2.$[/quote]
Now you should be able to destroy this problem. :)


let $ a, b, c \in [0, 1]$. Prove that $ \frac{a}{b\plus{}c\plus{}1}\plus{}\frac{b}{c\plus{}a\plus{}1}\plus{}\frac{c}{a\plus{}b\plus{}1}\plus{}(1\minus{}a)(1\minus{}b)(1\minus{}c) \le 1$

[quote="aktyw19"]let $ a, b, c \in [0, 1]$. Prove that $ \frac {a}{b \plus{} c \plus{} 1} \plus{} \frac {b}{c \plus{} a \plus{} 1} \plus{} \frac {c}{a \plus{} b \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \le 1$[/quote]

It's old problem and easy  :)

[quote="aktyw19"]let $ a, b, c \in [0, 1]$. Prove that $ \frac {a}{b \plus{} c \plus{} 1} \plus{} \frac {b}{c \plus{} a \plus{} 1} \plus{} \frac {c}{a \plus{} b \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \le 1$[/quote]

Tell me, we can kill this excircise by method " Look at the end point "
we have:
suppose a=max{a;b;c}
we have:
calling T is left position of inequality
we have :
$ T\leq \frac{a\plus{}b\plus{}c}{b\plus{}c\plus{}1}\plus{}(1\minus{}a)(1\minus{}b)(1\minus{}c)$
comment $ T\leq max{f(0);f(1)}$
But f(1)=0
$ f(0)\equal{}\frac{b\plus{}c}{b\plus{}c\plus{}1}\plus{}(1\minus{}b)(1\minus{}c)\minus{}1\equal{}\frac{bc(b\plus{}c)\minus{}(b^2\plus{}c^2)\minus{}bc\minus{}1}{1\plus{}b\plus{}c}\leq \frac{(b^2\plus{}c^2)(\frac{b\plus{}c}{2}\minus{}1)\minus{}bc\minus{}1}{1\plus{}b\plus{}c}$
=>$ f(0)<0$=>It's nice :P  :P  :P  :P  :P  :P  :P  :P  :P  :P


1. How are the AMC tests graded? How much points per problem? How many problems?
2. What is the time limit for the tests?
3. What would you recommend for sumone who just wants to make USAMOs?
4. What would you recommend for preparing for the tests?

Please answer these questions for AMC-10, AMC-12, and/or USAMO.

Questions 1 and 2 can be answered by looking on the AMC website at:

http://www.unl.edu/amc

Questions 3 and 4 can be answered by doing a search on these forums. They have been asked many times.

Can people just reply anyway?
The tests are updated annually and changed annually.
I want the current opinions.

[quote="paladin8"]1. 6 right, 2.5 blank, 0 wrong
2. 75 min
3. practice aimes
4. practice math[/quote]

next year would be 1.5 blank i think


Find all $ (a_1,a_2,...,a_n)$ such that

                                          $ \sum_{k \equal{} 1}^{n}{|a_1x_k \plus{} a_2x_{k \plus{} 1} \plus{} ... \plus{} a_nx_{k \plus{} n \minus{} 1}|} \equal{}m \sum_{k \equal{} 1}^{n}{|x_k|}$

for all reals $ x_i$. Where $ x_i \equal{} x_{n \plus{} i}$ and $ m>0$ a parameter.

 :)

First, take $ x_1\equal{}1$ and $ x_i\equal{}0$ if $ i\neq1$. This gives
$ |a_1|\plus{}|a_2|\plus{}\cdots |a_n|\equal{}m$
If, on the other hand, we take $ x_1\equal{}\cdots x_n\equal{}1$, we get
$ n\cdot |a_1\plus{}a_2\plus{}\cdots a_n|\equal{}m\cdot n$ or
$ |a_1\plus{}a_2\plus{}\cdots a_n|\equal{}m$
Combining these results, we see that [i]all[/i] $ a_i$ are either nonnegative or nonpositive. Suppose $ a_i\geq0,\forall i$. (Other case is similar)
Now let us compare $ a_r$ and $ a_s$. Take $ x_r\equal{}1,x_s\equal{}\minus{}1,x_i\equal{}0(i\neq r,s)$. Then we have
$ |a_r\minus{}a_s|\plus{}|a_{r\plus{}1}\minus{}a_{s\plus{}1}|\plus{}\cdots\plus{}|a_{r\minus{}1}\minus{}a_{s\minus{}1}|\equal{}2m$
Now remark that
$ 2m\equal{}|a_r\minus{}a_s|\plus{}|a_{r\plus{}1}\minus{}a_{s\plus{}1}|\plus{}|a_{r\plus{}2}\minus{}a_{s\plus{}2}|\plus{}\cdots\plus{}|a_{r\minus{}1}\minus{}a_{s\minus{}1}|$
$ \leq(|a_r|\plus{}|a_s|)\plus{}(|a_{r\plus{}1}|\plus{}|a_{s\plus{}1}|)\plus{}\cdots\plus{}(|a_{r\minus{}1}|\plus{}|a_{s\minus{}1}|)\equal{}2m$
Therefore, $ a_r\plus{}a_s\equal{}|a_r\minus{}a_s|$
This means that $ \min\left\{a_r,a_s\right\}\equal{}0$
Comparing all the numbers $ a_i$ in this way, we must conclude that only one $ a_i$ can differ from zero, and moreover one $ a_i$ must differ from zero, for else $ m\equal{}0$. That unique $ a_i$ must therefore be equal to $ m$ (in our case, $ \minus{}m$ in the other case).
Answer: $ (\pm m,0,0\cdots,0),(0,\pm m,0,\cdots,0),\cdots(0,0,\cdots,0,\pm m)$

That's esentially my solution too. We only differ slightly on how to prove that there is only one $ a_i$ different from $ 0$.

The original problem I solve was for $ n \equal{} 3$ and $ m \equal{} 1$, but this method really simplifies it, and the general form isn't a hard problem anymore.

 :)


What are the requirements for entering Clay Mathematical Institute or MIT? And What should i do in this few years in order to enter these 2 universities. Im still in high school this year, so i am jz curious about knowing what should i do....

You should probably start by doing enough research to know that CMI is not a university that you can attend.  (As its name suggests, it's a research institute; most of its efforts are directed at those who already have obtained a doctorate, although it does sponsor activities aimed at talented younger people and the general public.)  For example, it might be a good idea to go to the websites of the institutions in question and look around -- university admissions officies typically have lots of information available for potential applicants.

Since you seem to be functioning basically at the level of "a few names of places you've heard before," it might also be wise for you to talk to whatever passes for a college counselor at your school and/or to acquire one of the many college guides available.  (I think I used Fiske when I was a junior and senior in high school, and it was inoffensive.)


Prove that in any acute triangle $ ABC$,the following inequality holds
$ \frac{tanA}{tanB}\plus{}\frac{tanB}{tanC}\plus{}\frac{tanC}{tanA} \geq \frac{sin2A}{sin2B}\plus{}\frac{sin2B}{sin2C}\plus{}\frac{sin2C}{sin2A}$

@mods: Move this to temp, please (MY active problem).


pigeonhole principle
number theory unsolved
number theory

There are [i]2s-1[/i] positive integers.Prove that we can choose [i]s [/i]of them such their sum is divisible by [i]s[/i].[b][i]s=2^n[/i][/b].Thank you!

You want $ EGZ_2$ and lemma 1 from http://www.mathlinks.ro/viewtopic.php?p=389030 .


probability
logarithms
integration
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Take a stick and break it at a location selected with uniform density along its length. Throw away the left-hand piece and break the right-hand one at a location selected with uniform density along its length. Continue forever. What is the probability that one of the discarded left-hand pieces is more than half as long as the original stick?

Nice problem! The answer appears to be $\log 2$.

A possible proof could go as follows: Let $X_{1},X_{2},\ldots$ random variables with values
in $[0,1]$ such that $X_{n}$ is distributed uniformly on $[0,X_{n-1}]$. We have to calculate
$p : = \sum_{n}P(X_{n}\ge 1/2)$.

Using the recursion
$P(X_{n+1}\ge c) = \int_{0}^{1-c}dy P(X_{n}\ge c/(1-y)) = \int_{c}^{1}dz \frac c{z^{2}}P(X_{n}\ge z)$
it is easy to prove inductively that
$P(X_{n}\ge c) = 1-c \sum_{k < n}\frac{(-\log c)^{k}}{k!}= c \sum_{k \ge n}\frac{(-\log c)^{k}}{k!}$
and thus interchanging the sums over $n$ and $k$ we get
$p = c (-\log c) \sum_{k \ge 0}\frac{(-\log c)^{k}}{k!}=-\log c$
where $c = 1/2$.

I hope I didn't do any miscalculations! Since the result is so nice, I wonder if there is a more direct proof?

for another solution, see the bottom of my second post in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=900774#900774]this[/url] thread.


Here is an easy problem from MOP,1997 .The solution using Familiar geometrical theorems is easy .But i want  a solution using Transformations....It is given as an exercise after the transformation chapter in kedlaya's notes,.
Consider a triangle $ \Delta \mathrm{ABC}$ with $ \mathrm{AB} \equal{}\mathrm{ AC}$, and points $ \mathrm{M}$ and $ \mathrm{N }$on $ \mathrm{AB}$ and $ \mathrm{AC}$ respectively
 The lines BN and CM intersect at P . Prove that M N and BC are parallel if and only if $ \angle APN \equal{}\angle APN$.

I don't know what you mean by "transformations" but a simple way to do it would be to construct $ MN' \parallel BC$ with $ N$ on $ AC$ and let $ P' \equal{} BN' \cap CM$. If $ N \neq N'$ we have either $ \angle APM > \angle AP'M$ $ \equal{} \angle AP'N' \equal{} \angle ABP'\plus{}\angle BAP'$ $ > \angle ABP\plus{}\angle BAP \equal{} \angle APN$ or with all the signs reversed.


For each non-empty subset of {1, 2, 3, 4, 5, 6, 7}, arrange the numbers in decreasing order with alternating signs and take the sum. For example, the subset {5} we get 5. For {6, 3, 1} we get 6 - 3 + 1 = 4. Find the sum of all the resulting numbers.

[quote="t0rajir0u"]This has been posted before.  Search for it.[/quote]

Sorry.


How many arrangements of the letters in MISSISSIPPI have no consecutive Is?

I was thinking that I should try the principle of inclusion and exclusion...

11!/(4!2!4!)-10!/(3!2!4!)+9!/(2!2!4!)-8!/(2!4!)

The 11!/(4!2!4!) is the number of ways to arrange the letters of Mississippi.  If we group two I's together 10!/(3!2!4!) would be the number of ways to spell MSSSSIPPI(I,I).  Etc... But my answer was far off!  Someone tell me why.

Good question - you have 2 problems.

Problem 1: Take a close look at your denominators.  Are they correct?

Next - I presume your second term below is counting (once you've fixed the denominator) the number of arrangements with and II in there.  What form of overcounting have you left out?

Must be very careful using PIE - think of it as a way to deal with overcounting.  There are more ways to overcount than those that are addressed with simple PIE.  When you see what you've left out, you'll see what I mean.

Finally - can you find a slicker way to do this problem using distributions?

(One last note - thanks for the problem - I think I'll add it to my Intermediate Counting class problems).

Actually, thank you for the problem.  That is from the Intermediate Counting problem set Problem set 3, Question 8.    :D

Did I forget to count the possibility of (II) and (II)?

What's wrong with my denominators?  I divided the number of letters by the number of repeated letters so that I don't get the same word twice. Is it wrong because once I put the (II), I can't count that as a single I?

Hah!  Sure enough, it's right there on that problem set.

The II and II is the second snafu I was referring to.  How many times does it get counted in your expression?

As for your denominators, consider when you count: MSSSSIPPI(I,I), that (I,I) that you're treating as a block is a different letter from I or from I (the other two Is).  Therefore, what should your denominator be?

a neat way to think about it is to consider what will happen when we replace, say, the four s's by four distinct letters (that dont already occur in mississippi). there are 4! arrangements of these, so the number of permutations of mississippi, say, x, is multiplied by 4!. if we replace the two p's in the same fashion, we have 2! as many permutations, replacing the four i's, 4! as many. so then we have eleven different letters replacing mississippi, and there are 11! arrangements of these. so 4!4!2!x=11!, and x=11!/4!4!2!.