inequalities
function
algebra
polynomial
inequalities unsolved

$ a,b,c$  be non-negative real numbers
Proof
1)
$ \frac{a^2}{b^2\minus{}bc\plus{}c^2} \plus{} \frac{b^2}{c^2\minus{}ca\plus{}a^2} \plus{} \frac{c^2}{a^2\minus{}ab\plus{}b^2} \geq 2$
2)
$ \frac{a}{\sqrt{b^2\minus{}bc\plus{}c^2}} \plus{} \frac{b}{\sqrt{c^2\minus{}ca\plus{}a^2}} \plus{} \frac{c}{\sqrt{a^2\minus{}ab\plus{}b^2}} \geq 2$
3)
$ \frac{1}{b^2\minus{}bc\plus{}c^2} \plus{} \frac{1}{c^2\minus{}ca\plus{}a^2} \plus{} \frac{1}{a^2\minus{}ab\plus{}b^2} \geq \frac{6}{ab\plus{}bc\plus{}ca}$

[quote="colorfuldreams"]$ a,b,c$  be non-negative real numbers

1)
$ \frac {a^2}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^2}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {c^2}{a^2 \minus{} ab \plus{} b^2} \geq 2$
[/quote]

$ \sum{\frac {a^2}{b^2 \minus{} bc \plus{} c^2}}$

$ \equal{} \sum{\frac {a^4}{a^2b^2 \minus{} a^2bc \plus{} c^2a^2}}$

$ \ge \frac {\left(\sum{a^2}\right)^2}{\sum{(a^2b^2 \minus{} a^2bc \plus{} c^2a^2)}}$

$ \equal{} \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \minus{} abc(a \plus{} b \plus{} c)},$

so we just need to prove that

$ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge 4(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \minus{} 2abc(a \plus{} b \plus{} c)$

$ \iff (a \plus{} b \plus{} c)(abc \plus{} \sum{a(a \minus{} b)(a \minus{} c)}) \ge 0,$

true.

equality holds if $ a\equal{}0,b\equal{}c$.

[quote="colorfuldreams"]$ a,b,c$  be non-negative real numbers
Proof
2)
$ \frac {a}{\sqrt {b^2 \minus{} bc \plus{} c^2}} \plus{} \frac {b}{\sqrt {c^2 \minus{} ca \plus{} a^2}} \plus{} \frac {c}{\sqrt {a^2 \minus{} ab \plus{} b^2}} \geq 2$
[/quote]

by holder, we have

$ \left(\sum{\frac {a}{\sqrt {b^2 \minus{} bc \plus{} c^2}}}\right)^2\sum{a(b^2\minus{}bc\plus{}c^2)} \ge (a\plus{}b\plus{}c)^3,$

so we just need to prove that

$ (a\plus{}b\plus{}c)^3 \ge 4\sum{a(b^2\minus{}bc\plus{}c^2)}$

$ \iff 15abc\plus{}\sum{a(a\minus{}b)(a\minus{}c)} \ge 0,$

true.

equality holds if $ a\equal{}0,b\equal{}c$.

[quote="colorfuldreams"]$ a,b,c$  be non-negative real numbers
Proof
3)
$ \frac {1}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {1}{a^2 \minus{} ab \plus{} b^2} \geq \frac {6}{ab \plus{} bc \plus{} ca}$[/quote]

try $ a\equal{}0,b\equal{}1,c\equal{}2$... :wink:

[quote="kuing"][quote="colorfuldreams"]$ a,b,c$  be non-negative real numbers
Proof
3)
$ \frac {1}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {1}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {1}{a^2 \minus{} ab \plus{} b^2} \geq \frac {6}{ab \plus{} bc \plus{} ca}$[/quote]

try $ a \equal{} 0,b \equal{} 1,c \equal{} 2$... :wink:[/quote]

thanks
I start learning inequalities one weaks ago.
But I fell it's hard for me. :|
I fell that inequalities is such as memory games :D

[quote="kuing"][quote="colorfuldreams"]
I fell that inequalities is such as memory games :D[/quote]
but I don't think so  :maybe:[/quote]

If I haven't seen this before
I can't do it by myself..
If I have seen before
then I can know how to solve it. :stretcher:

there are simpler solutions:
1. WLOG suppose $ a \ge b \ge c$
then
$ \frac{a^2}{b^2\minus{}bc\plus{}c^2} \ge  \frac{a^2}{b^2}$
$ \frac{b^2}{c^2\minus{}ca\plus{}a^2} \ge  \frac{b^2}{a^2}$
$ \frac{c^2}{a^2\minus{}ab\plus{}b^2} \ge  0$
hence
$ LHS \ge  \frac{a^2}{b^2}\plus{} \frac{b^2}{a^2} \ge 2$

2nd inequality is similar

[quote="Dumel"]there are simpler solutions:
1. WLOG suppose $ a \ge b \ge c$
then
$ \frac {a^2}{b^2 \minus{} bc \plus{} c^2} \ge \frac {a^2}{b^2}$
$ \frac {b^2}{c^2 \minus{} ca \plus{} a^2} \ge \frac {b^2}{a^2}$
$ \frac {c^2}{a^2 \minus{} ab \plus{} b^2} \ge 0$
hence
$ LHS \ge \frac {a^2}{b^2} \plus{} \frac {b^2}{a^2} \ge 2$

2nd inequality is similar[/quote]

well done :lol:

$ \frac {a}{\sqrt {b^{2} \minus{} bc \plus{} c^{2}}} \plus{} \frac {b}{\sqrt {c^{2} \minus{} ca \plus{} a^{2}}} \plus{} \frac {c}{\sqrt {a^{2} \minus{} ab \plus{} b^{2}}}\geq 2$

WLOG suppose $ a\ge b\ge c$

$ \frac {a}{\sqrt {b^{2} \minus{} bc \plus{} c^{2}}}\ge\frac {a}{b}$
$ \frac {b}{\sqrt {c^{2} \minus{} ca \plus{} a^{2}}}\ge\frac {b}{a}$
$ \frac {c}{\sqrt {a^{2} \minus{} ab \plus{} b^{2}}}\ge 0$

hence

$ LHS\ge\frac {a}{b} \plus{} \frac {b}{a}\ge 2$
 :D

[quote="kuing"][quote="colorfuldreams"]$ a,b,c$  be non-negative real numbers

1)
$ \frac {a^2}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^2}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {c^2}{a^2 \minus{} ab \plus{} b^2} \geq 2$
[/quote]

$ \sum{\frac {a^2}{b^2 \minus{} bc \plus{} c^2}}$

$ \equal{} \sum{\frac {a^4}{a^2b^2 \minus{} a^2bc \plus{} c^2a^2}}$

$ \ge \frac {\left(\sum{a^2}\right)^2}{\sum{(a^2b^2 \minus{} a^2bc \plus{} c^2a^2)}}$

$ \equal{} \frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \minus{} abc(a \plus{} b \plus{} c)},$

so we just need to prove that

$ (a^2 \plus{} b^2 \plus{} c^2)^2 \ge 4(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \minus{} 2abc(a \plus{} b \plus{} c)$

$ \iff (a \plus{} b \plus{} c)(abc \plus{} \sum{a(a \minus{} b)(a \minus{} c)}) \ge 0,$

true.

equality holds if $ a \equal{} 0,b \equal{} c$.[/quote]

I have one question. How did you get the identity:

$ (a^2 \plus{} b^2 \plus{} c^2)^2 \minus{} 4(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \plus{} 2abc(a \plus{} b \plus{} c) \equal{}
(a \plus{} b \plus{} c)(abc \plus{} \sum{a(a \minus{} b)(a \minus{} c)})$

You've seen it earlier, or what?

Thanks! :)

I've got two ways to find the identity.
First is to figure out the Reminder theorem and use some facts about the degree of the cyclic/symetric function of the homogeneous function. Second use some indentity that is very easy to prove and just factor $ a \plus{} b \plus{} c$ out.

The basic is Let $ P(x)$ is the polynomial function of the form $ a_nx^n \plus{} a_{n \minus{} 1}x^{n \minus{} 1} \plus{} ... \plus{} a_1x \plus{} a_0$. If we put $ x \equal{} c$ to the $ P(x)$ and it's identically zero, then $ x \minus{} c$ is a factor of the polynomial function $ P(x)$. So $ P(x)$ can write in the form $ P(x) \equal{} (x \minus{} c)R(x)$, where $ R(x)$ is a polynomial of degree $ n \minus{} 1$ (because $ x \minus{} c$'s degree is 1).

Let $ f(a,b,c) \equal{} \sum_{cyc}a^4 \minus{} 2(\sum_{cyc}a^2b^2) \plus{} 2abc(\sum_{cyc}a)$ be a polynomial function in $ a$ variable which is cyclic and symetric in three variable. Observe that to put $ a \equal{} \minus{} b \minus{} c$ we get $ f(a,b,c)$ is zero, so we get that $ a \plus{} b \plus{} c$ is one factor of $ f$. Since $ a \plus{} b \plus{} c$'s degree is 1, so there're exist polynomial which polynomial's degree is 3 (Observe that $ f$'s degree is 4). So by the fact that the multplication or the division of the cyclic polynomial is still the cyclic polynomial, we get that $ f(a,b,c) \equal{} (a \plus{} b \plus{} c)g(a,b,c)$ where $ g$ is cyclic and $ g$'s degree is 3. there are three possible way to figure out the polynomial $ g$ which is cyclic and degree 3.
Now, we can write the polynomial in the form $ f(a,b,c) \equal{} (a \plus{} b \plus{} c)[P(\sum_{cyc}a^3) \plus{} Q(\sum_{cyc}a^2(b \plus{} c)) \plus{} R(abc)]$ where $ P,Q,R$ is the constant. We have to find $ P,Q,R$ by expanding the L.H.S.,R.H.S. We'll get all of $ P,Q,R$ but the much better way is to choose $ a$ from R.H.S. from $ a \plus{} b \plus{} c$ multiplie with $ Pa^3$ and compare with L.H.S. We get $ Pa^4 \equal{} a^4$, yielded $ P \equal{} 1$ by the like manner, we get $ (P,Q,R) \equal{} (1, \minus{} 1,4)$

The second way is rewrite the polynomial in the form
\[ \sum_{cyc}a^4 \minus{} \sum_{cyc}a^2b^2 \minus{} [\sum_{cyc}a^2b^2 \minus{} abc(\sum_{cyc}a)] \plus{} abc(\sum_{cyc}a)\]

\[ \equal{} (a \plus{} b)^2(a \minus{} b)^2 \plus{} (a \plus{} c)(b \plus{} c)(a \minus{} c)(b \minus{} c) \minus{} c^2(a \minus{} b)^2 \minus{} ab(a \minus{} c)(b \minus{} c) \plus{} abc(\sum_{cyc}a)\]

\[ \equal{} [(a \plus{} b)^2 \minus{} c^2](a \minus{} b)^2 \plus{} [(a \plus{} c)(b \plus{} c) \minus{} ab](a \minus{} c)(b \minus{} c) \plus{} abc(a \plus{} b \plus{} c)\]

\[ \equal{} (a \plus{} b \plus{} c)(a \plus{} b \minus{} c)(a \minus{} b)^2 \plus{} c(a \plus{} b \plus{} c)(a \minus{} c)(b \minus{} c) \plus{} abc(a \plus{} b \plus{} c)\]

\[ \equal{} (a \plus{} b \plus{} c)(\sum_{cyc}a^3 \minus{} \sum_{sym}a^2b \plus{} 4abc)\]

\[ \equal{} (a \plus{} b \plus{} c)(\sum_{cyc}(a^3 \minus{} a^2b \minus{} a^2c \plus{} abc) \plus{} abc)\]

\[ \equal{} (a \plus{} b \plus{} c)(\sum_{cyc}a(a^2 \minus{} (b \plus{} c)a \plus{} bc) \plus{} abc)\]

\[ \equal{} (a \plus{} b \plus{} c)(\sum_{cyc}a(a \minus{} b)(a \minus{} c) \plus{} abc)\]
For more identities, see here http://reflections.awesomemath.org/2007_5.html
I bring it 2 identities from the SOS-Schur method archive. Enjoy! :)


Mrs. Turner gives a multiple-choice test where each question has four choices. If the test contains five questions and Mark randomly guesses all of the answers, what is the probability that he gets all five correct? Express your answer as a common fraction.

The probability that Mark guesses any one question correct is $ \frac{1}{4}$. Therefore, the probability of answering all five questions correctly is $ (\frac{1}{4})^5\equal{}\boxed{\frac{1}{1024}}$.


Steven has 100 flowers in his garden. Of these, 45 have red flowers and green stems, 27 have red flowers with yellow stems, 20 have blue flowers and green stems, and the rest have blue flowers with yellow stems. Steve picks a flower at random from the garden. Given that it is a blue flower, what is the probability that it has a yellow stem?

first we have to find out how many blue flowers with yellow stems there are

we add up all the other numbers: $ 45\plus{}27\plus{}20\equal{}92$

and we subtract that from $ 100$: $ 100\minus{}92\equal{}8$

now we know there are $ 8$ blue flowers with yellow stems

we now move onto the probability part
we know that Steve picked a blue flower, and that there are $ 20\plus{}8\equal{}28$ blue flowers

since there $ 8$ of those blue flowers have yellow stems, our probability is $ \frac{8}{28}\equal{}\boxed{\frac{2}{7}}$


7.	How would the model: 100(1.04)^x = V(t) change if your Grandmother told you that your money will never be worth 100 dollars or less?

Your model, as stated has nothing to do with money, so it wouldn't change.

You should define your variables, their units, and the context from which they originate for a more relevant answer.

I think he's referring to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259373]this[/url].


With $ a,b,c \geq 0$ such that $ a\plus{}b\plus{}c\equal{}1$. Prove that
      $ \frac{1}{3} \leq \frac{ab}{3ab\plus{}c}\plus{} \frac{bc}{3bc\plus{}a}\plus{} \frac{ca}{3ca\plus{}b} \leq \frac{1}{2}$

[quote="nguoivn"]Nice result, 2424  :) . By CS:
$ \sum\ \frac {ab}{3ab \plus{} c} \le \frac {1}{2}(\sum\ \frac {ab}{2ab \plus{} c^2} \plus{} \sum\ \frac {ab}{ab \plus{} bc \plus{} ca}) \equal{} RHS$
$ \sum\ \frac {ab}{3ab \plus{} c} \ge \frac {(ab \plus{} bc \plus{} ca)^2}{\sum\ a^2b^2 \plus{} 3abc} \ge LHS$[/quote]

My solution:
$ \sum\ \frac {ab}{3ab \plus{} c} \ge \frac {(ab \plus{} bc \plus{} ca)^2}{\sum\ a^2b^2 \plus{} 3abc} \ge LHS$
+)$ \sum\ \frac {ab}{3ab \plus{} c} \le \frac {1}{2}$
$ < \equal{} > \sum \frac {c}{3ab \plus{} c} \ge \frac {3}{2}$
We have:
$ \sum \frac {c}{3ab \plus{} c} \ge \sum \frac {c}{\frac {3(1 \minus{} c)^2}{4} \plus{} c} \ge \sum(\frac {1}{2} \plus{} \frac {4}{3}(3c \minus{} 1)) \equal{} \frac {3}{2}$  :)

[quote="caubetoanhoc94"][quote="nguoivn"]
$ \sum \frac {c}{\frac {3(1 \minus{} c)^2}{4} \plus{} c} \ge \sum(\frac {1}{2} \plus{} \frac {4}{3}(3c \minus{} 1)) \equal{} \frac {3}{2}$  :)[/quote][/quote]

hello, can you show me how to do this strategy more detailed? maybe a link or article? thank you very much.


Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA = 3, PB = 4,$ and $AB = 5$, what is $PD$?

$\textbf{(A)}\ 5 \qquad
\textbf{(B)}\ \sqrt{34} \qquad
\textbf{(C)}\ \sqrt{41} \qquad
\textbf{(D)}\ 2\sqrt{13} \qquad
\textbf{(E)}\ 8$

[hide]$\Delta PAB$ is right angled.

$PH = \frac{12}{5}$.

$5x = 16$ where $x$ is projection of $PB$ on $AB$. Hence $x = \frac{16}{5}$.

$PD^2 = 5^2 + \left(\frac{16}{5}\right)^2 + \left(\frac{12}{5}\right)^2 = 41$.

So $PD = \sqrt{41}$.[/hide]

[hide]Since AB=BC=CD=AD=5, PD certainly isn't 5. It isn't 8 either, so we have narrowed down our choices to 3.

PD, if PA was on the same "horizontal" plane as AD" would be $\sqrt{41}$, but since PA is not on the same "horizontal" plane, it is more than $\sqrt{41}$ The only possible answer is $2\sqrt{13}$.
[/hide]

[hide]Let $PE$ be the altitude $PE$ from $P$ to side $AB$. We need to find $PD=\sqrt{PE^2+EA^2+AD^2}$. We can find $PE$ by finding the area $[PAB]$ in two ways--clearly $[PAB]=(1/2)(3)(4)=6$, but also $[PAB]=(1/2)(5)(PE)$, so $PE=12/5$. Next we can find the length $EA=\sqrt{3^2-PE^2}=9/5$. Finally, $AD=AB=5$, so our answer is \[ PD = \sqrt{(12/5)^2+(9/5)^2+5^2}=\boxed {\sqrt{34}}. \][/hide]

Interesting how everyone got different answers. :P

[hide="I agree with chess64"]Call the altitude from $P$ to $\overline{AB}$ $\overline{PH}$. $\overline{AH}^2=9-\frac{144}{25}=\frac{81}{25}$, so $\overline{DH}^2=\frac{81}{25}+25=\frac{706}{25}$, so $\overline{PD}=\sqrt{\frac{144}{25}+\frac{706}{25}}=\sqrt{34}$, and our answer is $\boxed{B}$.[/hide]

I don't understand why you guys' approaches are so complicated... isn't $ PD$ just $ \sqrt{AD^2\plus{}AP^2}\equal{}\sqrt{25\plus{}9}\equal{}\sqrt{34}$?

why isn't that obvious? is there a case where angle PAD isn't a 90 degree angle?

Consider what happens when $ AP$ is the common line of the two perpendicular planes, and then rotate $ AD$ about $ A$. :wink:

How would $ AP$ be the common line, though?  Wouldn't the fact that $ PAB$ and $ ABCD$ are in different planes mean that $ AB$ was the common line?  And since it's a triangle, $ A$, $ B$, and $ P$ can't be collinear.

Sorry, I'm pretty bad at 3d geometry.

EDIT: Oh, that makes sense.  Thanks.

For the problem, we can find that those two lines are perpendicular.  I assumed shen's question was referring to [b]any[/b] two lines in two perpendicular planes with one point in common.

[quote="solafidefarms"]mathwizarddude: you're saying $ AD\perp AP$, which is not obvious to me. Just because they are on perpendicular planes doesn't imply any line in one plane is perpindicular to any line sharing a point in the other plane.[/quote]

Well, there's only one kind of line that does not follow this - the intersection of the two perpendicular planes, but in the problem the intersection of the two planes is AB, so AP and AD must be perpendicular...


let $ a,b,c$ three  positive reals such that $ abc\equal{}1$ .
prove that $ 2(a^2\plus{}b^2\plus{}c^2)\plus{}a\plus{}b\plus{}c>\equal{}6\plus{}ab\plus{}bc\plus{}ca$

[quote="greatestmaths"]let $ a,b,c$ three  positive reals such that $ abc \equal{} 1$ .
prove that $ 2(a^2 \plus{} b^2 \plus{} c^2) \plus{} a \plus{} b \plus{} c > \equal{} 6 \plus{} ab \plus{} bc \plus{} ca$[/quote]
[hide="Hidden by mod"]Bt AM-GM inequality, we have
$ a^2 \plus{} b^2 \plus{} c^2 \ge ab \plus{} bc \plus{} ca$
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} a \plus{} b \plus{} c \ge 6\sqrt [6]{a^3b^3c^3} \equal{} 6$[/hide]


Check out [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=120115]this topic.[/url] Note that mlok does know what he's doing, but doing the right thing doesn't work for a link with an underscore in the address. Unfortunately, that's quite a few wikipedia addresses.

Correction: it's the apostrophe rather than the underscore which breaks the code.

The link works (but generates an error page) if the apostrophe is removed but the underscore is not.

http://www.mathlinks.ro/Forum/viewtopic.php?t=109572

If you're lucky, removing the apostrophe will lead you to a page that redirects you to the one with '. But that's pretty improbable.

You can replace the apostrophe with %27.

[b]Very nice![/b] It must be some sort of code (ASCII probably) since I've noticed many times URLs that contain things like %xx instead of spaces.  [i]EDIT:[/i] It's ASCII in hex notation.

But visually speaking, there's no effect (try some [i]url/hide[/i] tags).


calculus
integration
limit
function
calculus computations

Let $I_{n}=\int_{0}^{1}\frac{e^{x}}{x^{n}}dx$

[hide="here is what i have so far"]

so by using by-parts I have

$nI_{n+1}=I_{n}-e+\frac{1}{n!}$

this is from the fact that  $\lim_{x\rightarrow 0}\frac{e^{x}}{x^{n}}=\frac{1}{n!}$ by repeated L'Hopitals rule

and it is plain to see that $I_{0}=e-1$
[/hide]

is there a way to express $I_{n}$ in terms of $n$?

EDIT: (woops)
oops sorry i just noticed that

$lim_{x\rightarrow 0}\frac{e^{x}}{x^{n}}\neq \frac{1}{n!}$ but is undefined.

silly me.  :blush:  :oops:

change the bound to $1\rightarrow 2$ instead

Could you restate once more what exactly you wish to evaluate?

$I_{n}=\int_{1}^{2}\frac{e^{x}}{x^{n}}dx$

find a way to express $I_{n}$ in tems of $n$ if possible.

Integration by parts works to give a nice recursive definition, but you're going to end up having to calculate $I_{1}$ which doesn't seem to be nice.

yes JBL, i ran into that same problem when trying to find $I_{1}$.

using that

$I_{n}=\frac{e^{2}}{2^{n}}-e+nI_{n+1}$

we have

$I_{1}=\frac{e^{2}}{2}-e+I_{2}$
and
$I_{2}=\frac{e^{2}}{4}-e+2I_{3}$
so

$I_{1}-I_{2}=\frac{e^{2}}{2}-e=\frac{e^{2}}{4}-2I_{3}$

using this to solve for $I_{3}$ i get

$I_{3}=\frac{4e-e^{2}}{8}$.

but when i check this with what my calculator gives I find that this is wrong.
where did i make a mistake?  :maybe:

[quote="maokid7"]$I_{1}-I_{2}=...=\frac{e^{2}}{4}-2I_{3}$[/quote]  And how did you get this equality?

well since
$I_{1}=\frac{e^{2}}{2}-e+I_{2}$ it is clear that
$I_{1}-I_{2}=\frac{e^{2}}{2}-e$

and now we have

$I_{2}=\frac{e^{2}}{4}-e+2I_{3}$  so when we use this value for $I_{2}$ and subtract it from $I_{1}$ we have

$\frac{e^{2}}{4}-2I_{3}$

and they are both equal to $I_{1}-I_{2}$

Whatever you're doing, it doesn't work.  I think when you're "substituting" you are disappearing an $I_{1}$ without cause.

o woops i found my mistake.

it was when i said that

$I_{1}-I_{2}=\frac{e^{4}}{4}-2I_{3}$ it should have been

$I_{1}-I_{2}=\frac{e^{4}}{4}+I_{2}-2I_{3}$ and this gets us nowhere.

 :blush:

Just to let you know, Maple v10 gives

I_k=-exp(4)*GAMMA(1-k,2)+exp(2)*GAMMA(1-k,1)+(-(-1)^k*(Ei(1,-1)-Ei(1,-2))+(-1)^k*exp(2)*Ei(1,1)-(-1)^k*exp(4)*Ei(1,2))/GAMMA(k)

and I_1=Ei(1,-1)-Ei(1,-2)

where Ei(a,b) is an Exponential Integral GAMMA(a,b) is the incomplete gamma function, and exp(x) = e^x.

-Ben


[quote="bugzpodder"]I should ask the question a lot of others are wondering about?
what is Cody Bowl?  can you provide some info/link?[/quote]

Here is a link to a previous thread: [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14487]Cody Bowl[/url]

Yeah, it's down.  IM "Mathjoker" on AIM (Minus the quotes) and he'll provide you with any information ya want.  He's really helpful (and usually online too).          All of you guys should sign up!  You don't have to organize a team or anything to be in the Cody Bowl.


calculus
integration
function
geometry
angle bisector
national olympiad

Time allowed 3 hours

1. Find the integral solutions of the equation $7(x+y)=3(x^2-xy+y^2)$
http://www.mathlinks.ro/Forum/viewtopic.php?t=49911

2. The function $f(x,y)$, defined on the set of all non-negative integers, satisfies
(i) $f(0,y)=y+1$
(ii) $f(x+1,0)=f(x,1)$
(iii) $f(x+1,y+1)=f(x,f(x+1,y))$
Find f(3,2005), f(4,2005)
http://www.mathlinks.ro/Forum/viewtopic.php?t=49912

3. In triangle ABC, the altitude, angle bisector and median from C divide the angle C into four equal angles. Find angle B.
http://www.mathlinks.ro/Forum/viewtopic.php?t=49913

4. Let x,y,z be positive real numbers such that $x+y+z=1$.
For positive integer n, define $S_n = x^n+y^n+z^n$
Furthermore, let $P=S_2 S_2005$ and $Q=S_3 S_2004$.
(a) Find the smallest possible value of Q.
(b) If $x,y,z$ are pairwise distinct, determine whether P or Q is larger.
http://www.mathlinks.ro/Forum/viewtopic.php?t=49914

5. Given finitely many points in a plane, it is known that the area of the triangle formed by any three points of the set is less than 1. Show that all points of the set lie inside or on boundary of a triangle with area less than 4.
http://www.mathlinks.ro/Forum/viewtopic.php?t=49915

6. Find $2^{2006}$ positive integers satisfying the following conditions.
(i) Each positive integer has $2^{2005}$ digits.
(ii) Each positive integer only has 7 or 8 in its digits.
(iii) Among any two chosen integers, at most half of their corresponding digits are the same.
http://www.mathlinks.ro/Forum/viewtopic.php?t=49916

[quote="Skyward_Sea"]The TST began so early...
I'm still looking forward to the China TST...I hpoe I could take part in it...[/quote]
We have one TST in August (this one), one in October, one in December, and the last one in March (APMO) because it's impossible for us to have intenstive TSTs.... so the TSTs may seem much earlier to be held than other countries.

[quote="mecrazywong"][quote="leepakhin"]I could only do x questions where x is between 3.5 and 4... :(
Everyone scored full mark in this test... :(
So bad... :([/quote]
It is not really that bad for your first TST.
btw, I think problems 1 and 4 are really too easy. Problem 2 is just too ugly in my mind...
Problem 6 is quite nice, but it'd better to replace the word "Find" by "Prove that there exist". This makes the question much clearer.[/quote]
I think for 6 the general case should be asked.

[quote="shobber"]How do you select people for CMO in HongKong? And do you take CMO as another TST?[/quote]
1. based on the tests in August, October and December
2. no.

It's really easier that China's so-called "2nd-Test" of ChNO, which is held every October.


Show that if $a$ and $b$ are positive integers, then

\[\left(a+\frac{1}{2}\right)^{n}+\left(b+\frac{1}{2}\right)^{n}\]

is an integer for only finitely many positive integers $n$.

[hide]If the sum is an integer, then the sum of $(2a+1)^{2}+(2b+1)^{2}$ must be divisible by $2^{n}$.
Hence, $(2a+1)^{2}+(2b+1)^{2}\equiv 0 \mod_{4}$ for all integer n greater than one, which always works.  But that's not possible for integer n greater than one since the sum of the squares will always equal $2 \equiv \mod_{4}$.  Hence, there's only one integer n ensure that the sum is an integer, namely, n=1.[/hide]

I think the exponents in the first expression should be $n$? Or I'm missing something? Besides, the second expression, the way it's written, is $\equiv 2\pmod{4}$. Please review what needs to be reviewed.

It's problem 30 :D

1. http://www.mathlinks.ro/Forum/viewtopic.php?highlight=newman&t=3833
2. http://www.mathlinks.ro/Forum/viewtopic.php?highlight=newman&t=26833
3. http://www.mathlinks.ro/Forum/viewtopic.php?highlight=newman&t=35430


What are the next 2 numbers in the following sequence?

-1, 0, 1, 0, 1, 2, 3, 8, 25, ..., ...

first i tried to see the relation between the two former.  it looked like the one before the former squared plus the former.

but since it repeats 0,1 twice with different results, it has to be like last 3 added.

i think you wrote it wrong.
-1,0,1,0,1,2,3,6,12,21...

can someone give an answer for this!  ive spent too much time and I still cant figure it out.  :blush:

This arbitrarily came to mind.

$(-1*0)+1=1$      Sum of $1$ after product of previous two numbers
$(1*0)+0=0$       Sum of $0$ after product of previous two numbers
$(0*1)+1=1$       Sum of $1$ after product of previous two numbers
$(1*2)+1=3$       Sum of $1$ after product of previous two numbers
$(2*3)+2=8$       Sum of $2$ after product of previous two numbers
$(3*8)+1=25$     Sum of $1$ after product of previous two numbers

However, this doesn't provide a definate pattern for the sums so i'm not sure.

okay, maybe i wrote it wrong...but i got it from a book...and i quintuple checked and it is the same sequence in the  book. :?

Could you check if the next number is 201 for me? Thanks.

it says 201. Good job 4everwise!!!!!!!! :omighty:

its the product of the 2 numbersa before it +1 i think but it doesnt work

Polynomial pattern inference is certainly beyond MOEMS topic level.

If you do not know the MOEMS curriculum well enough to make appropriate posts, please do not post in the MOEMS forum.  I have begun banning users who repeatedly post off-difficulty from particular forums.


pigeonhole principle
ceiling function
number theory proposed
number theory

Consider the poset such that $a$ is related to $b$ iff $a$ divides it. then there is an antichain of size $n+1$ or a chain of size $n+1$. since there is no antichain, because they say that if i pick $n+1$ elements there are always to related, then there is a chain of size $n+1$, and this chain gives the desired set...I think it already appeared in the forum, right?

[quote="harazi"]Allow me please to express my sincere doubts about the quality of the given problems.  :?[/quote]
he-he

[quote="perfect_radio"][quote="Pascual2005"]Consider the poset such that $a$ is related to $b$ iff $a$ divides it. then there is an antichain of size $n+1$ or a chain of size $n+1$. [/quote]

How can you prove this?[/quote]
Did you see topic's title? It is "dilworth particularization"!!!

Whats "dilworth particularization"?
In the fact we must prove that $r(n,n+1)\geq n^2+1$!!($r(k,l)$ are Ramsey's numbers).

Just use [url=http://en.wikipedia.org/wiki/Dilworth's_theorem]Dilworth's Theorem[/url] on the poset of the divisibility lattice. By Dilworth's Theorem the smallest partitioning of $X$ into chains has a number of elements equal to the cardinality of same antichain. As we are given no antichains have more than $n$ elements, it follows by pigeonhole some chain has at least $\left \lceil \frac{n^2+1}{n} \right \rceil = n+1$ elements, as desired.

[quote=Valentin Vornicu]Let $n\geq 1$ be an integer and let $X$ be a set of $n^2+1$ positive integers such that in any subset of $X$ with $n+1$ elements there exist two elements $x\neq y$ such that $x\mid y$. Prove that there exists a subset $\{x_1,x_2,\ldots, x_{n+1} \} \in X$ such that  $x_i \mid x_{i+1}$ for all $i=1,2,\ldots, n$.[/quote]

Consider the problem setup in graph theoretical language. A directed edge $x \rightarrow y$ is drawn iff $x \mid y$. Now from the given condition we have that the size of the maximal anti-chain of the graph is, clearly, at most $n$. So by Dilworth's theorem it follows that the minimum number of disjoint directed paths the graph can be split into is at most $n$. Consequently, one of them contains a chain of size $\left\lfloor \frac{n^2+1}{n} \right\rfloor\ = n+1.$

Suppose there is no $n+1$-size subset $\{x_1,x_2,\ldots, x_{n+1} \}$such that no $x_i|x_{i+1}$ for all $i$.

We claim that all the points of $X$ can be partitioned into $n$ subsets such that in each subset we can't find two $a,b$ such that $a|b$.

Proof:Proceed by induction on $n$.Suppose $X$ has no subset $\{x_1,x_2,...,x_{m+1}\}$ of size $m+1$ such that $x_i|x_{i+1}$.Consider all the subsets of $X$ with maximum length such that each subset is a $chain$; meaning if $\{x_1,x_3,...,x_r\}$ is a such subset then $x_i|x_{i+1}$
Now let $S$ be the set with all maximum number from each maximum chain.We note that in $S$ no two elements devide one another.
If $X-S$ has a $m$- length chain then this is a maximum chain in $X$ also since $X$ has no $m+1$- length chain.So maximum of this chain $a$ is in $S$ as well.So it's not in $X-S$.Contradiction.So $X-S$ has no chain with length $m$.By induction hypothesis, $X-S$ is union of  $m-1$ length set in which no $a|b$.And together with $S$ ,$X=X-S\cup S$ is union of $m$ such subset.
So X is union of $n$ subsets in each which no $a|b$.So by php one of them is of length $\left\lfloor \frac{n^2+1}{n} \right\rfloor\ = n+1.$   Contradiction.
So our initial assumption was wrong.

Wonderful application and a Direct Consequence of Dilworth's Theorem -

[size=150][b]Dilworth Theorem - [/b][/size] Consider graph $\mathcal{G}$  ,which is directed acyclic , then consider a set of vertices $\mathcal{V}$ of  $\mathcal{G}$ such that there doesn't exist any directed path between any $2$ vertices in $\mathcal{V}$. Then maximum possible size of $\mathcal{V}$ is same as minimum number of disjoint paths , in which $\mathcal{G}$ can be decomposed.

Now , suppose we represent $n^2+1$ numbers as vertices in $\mathcal{G}$ , and there is a directed edge from $V_i$ to $V_j$ if and only if $ i \mid j $ , so for any subgraph $\mathcal{S}$ of $\mathcal{G}$ , with $n+1$ vertices, there exist atleast one directed edge.   $\clubsuit$

Now we want a path of size atleast $n+1$ in $\mathcal{G}$ , suppose assume the opposite .
Then suppose minimum number of disjoint paths in which $\mathcal{G}$ can be decomposed is $k$ , then from our assumption $ nk  \ge n^2 + 1$ , or $k  \ge n +1$ .

Now we claim that there doesn't exist an directed cycle in  $\mathcal{G}$ , suppose it exist , let vertex encountered be $x_1 , x_2 , .... x_i$ , then from definition of  $\mathcal{G}$ , $x_1 \mid x_2 \mid x_3 ,........\mid x_i \mid x_1$ , but this leads to all numbers being equal , A contradiction.

Now from Dilworth's Theorem  ,we get a subgraph  $\mathcal{X}$ of $\mathcal{G}$ such that there is no directed edge between any $2$ vertices , having size atleast $k = n+1$.   A Contradiction to $\clubsuit$.

Hence we are done.

Let $G$ be the directed graph on $X$ such that an edge $x \rightarrow y$ is placed iff $x \mid y$. The problem condition implies that the maximum antichain of $G$ has size at most $n$. Thus, Dilworth's theorem yields that the minimum number of disjoint chains $G$ can be split into is at most $n$, so by Pigeonhole there is some chain of size at least $\left \lceil \frac{n^2+1}{n} \right \rceil = n+1$, as desired.


calculus
integration
limit
logarithms
inequalities
LaTeX
calculus computations

I'm studying for my calculus midterms and I am going through past midterms my professor has posted. He has not posted the solutions for these midterms, so on a couple of these questions, I am just looking for verification of my answer....I have hit a couple bumps on the way, so if anyone could show me the way to solve the problems I have not yet figured out, I'd really appreciate it....

[b]Questions:[/b]

Determine if the integral is convergent or divergent. Evaluate the integral if it is convergent.

$\int_{-1}^{1} \frac{x+1}{x^\frac{4}{3}} \, dx$

So far, I have broken this down to
$\int_{-1}^{1} x^{-\frac{1}{3}} \, dx + \int_{-1}^{1} x^{-\frac{4}{3}} \, dx$

I believe the first part of this converges, but the second part can't be evaluated at zero, so I don't know how to go about solving it....I thought I should break the second part into two separate integrals from -1 to 0 and 0 to 1 and use limits, but when I tried that, it did not work out for me....

[b]Verification[/b]

$\int_{2}^{\infty} \frac{x}{x^2 - 1} \, dx$

= $\frac{1}{2} [ \lim_{t\to\infty} \ln (t^2 -1) - \ln 3]$

$= \infty$
So this integral is divergent

Determine if the integral is convergent or divergent. Evaluate the integral if it is convergent.

$\int_{e}^{\infty} \frac{dx}{x{(\ln x)}^2} \, dx$

I have a feeling that this one can be proved conv or div using the comparison test
I am not sure but I think
$\int_{e}^{\infty} \frac{dx}{(\ln x)^2}$ is convergent so then multiplying it by $\frac{1}{x}$ must make the original integral convergent.

Is the answer that it converges to 1? That's my conclusion after quick calculations

$\int_0^1x^{-4/3}\,dx=\left.-3x^{-1/3}\right|_{0^+}^1=\infty.$

The integral $\int_{-1}^1x^{-4/3}\,dx$ is improper for two reasons: the left side of zero and the right side of zero. When an integral is improper for two or more reasons, we will only accept it as convergence if we can break it into pieces, each one improper for only one reason, and each separate piece converges. Since I've shown that one of the pieces diverges, this integral must diverge. (The other piece also diverges, to $-\infty.$)

General rule: $\int_0^ax^{-p}\,dx$ converges if $p<1$ and diverges if $p\ge 1.$

$\int_2^{\infty}\frac{x}{x^2-1}\,dx$

I wouldn't even bother finding the antiderivative. For large $x,\,\frac{x}{x^2-1}$ closely resembles $\frac1x,$ and that's an integral that diverges. Think of what I just said as the limit comparison test.

[quote] $\int_{e}^{\infty} \frac{dx}{(\ln x)^2}$ is convergent so then multiplying it by $\frac{1}{x}$ must make the original integral convergent.[/quote]
That's a wrong argument. Even though the integral does actually converge, if this were my test that I were grading, I would give you zero points for this answer.

You see, $\int_{e}^{\infty} \frac{dx}{(\ln x)^2}$ diverges, and with that, you have no argument.

You want to think about comparison? Remember that the logarithm goes to infinity slower than any power of $x.$ Hence for large $x,$

$\frac1x>\frac1{x(\ln x)^2}>\frac1{x^p}$ for some $p>1,$

But $\int_b^{\infty}\frac1x\,dx$ diverges and $\int_b^{\infty}\frac1{x^p}\,dx$ converges.

Our inequality signs point in the wrong direction; the comparison is useless and gives us no information.

In this case DO go ahead and find the exact antiderivative.

Once again, thanks Kent so much for your clear explanations. I am going to attempt the third one once more to make up for my disastrous first try, haha

$\int_{e}^{\infty} \frac{dx}{x{(\ln x)}^2} \, dx$

$u = \ln x$
$du = \frac{dx}{x}$
$so$

$\int_{e}^{\infty} \frac{dx}{x{(\ln x)}^2} \, dx$
$= \int_{1}^{\infty} \frac{du}{u^2}$
$= \left.-\frac{1}{u} \right|_{1}^{\infty}$ (How do I make the "evaluated from" signs in LaTeX?)
$= -0 - (-1)$
$= 1$

Did I follow though correctly?
Thanks again for your time

Yes. The integral converges and your computation is correct. And I edited in one possible version of your evaluation sign.


Para quien esta leyendo este post: La discusion es acerca de temarios en el area indicada por el titulo. Se pretende que si usted pone algo aqui, que sea referente al temario, discutiendolo (criticas completamente aceptadas), haciendo preguntas sobre el o proponiendo otros posibles resultados dentro del tema. Por favor, solo por mantener el orden, si su post no cae en esta categoria, pues inicie un nuevo topic para ello. Esta peticion es unicamente para conservar el orden para quienes estamos interesados en seguirle la pista al temario!!

Solo cuestion de notacion, las secciones, subsecciones y subsubsecciones

estan indicadas asi:
- Seccion
* Subseccion
** Subsubseccion

[b]Combinatoria[/b]
- Conseptos basicos de conteo:
  * Cardinalidad
  * Regla de Suma
  * Regla de Producto
- Permutationes y Combinationes
  * Bolas y Paredes
- Inclusion Exclusion
- Coeficientes Binomiales ${n \choose k}$
  * Triangulo de Pascal
  * Teorema del Binomio
  * Formula de Convolucion de Vandermonde
  * Formulas especiales
    ** $\binom{n+m}{k}=\sum_{j=1}^{\max(m,k)}\binom{n}{k-j}\binom{m}{j}$
    ** $\displaystyle \sum_{k=0}^{n} \binom{n+k}{k} = \binom{2n+1}{n+1}$
    ** $\displaystyle \sum_{k=0}^{n} \binom{n}{k}^2 = \binom{2n}{n}$
- Denumerantes e invariantes
- Funciones Generatrices
- Teoria de Grafos

[quote="manuel"]Demuestra que para un conjunto X, donde |X|=n , no hay mas de ( n escoge floor{n/2} ) subconjuntos tal que ninguno de estos es subconjunto de otro.[/quote]

Hable con Manuel y pues, ahora hago la aclaracion general. Si proponen un problema aqui, es con la idea de aclarar como algun tema puede ser de interes. Por ejemplo, si proponen un problema como este y dicen que se puede aplicar tal o cual teorema para resolverlo (Supongo que por eso Manuel propone el Lema de Sperner). Si lo que quieren es proponer un problema para que alguien lo intente, pues tenemos una seccion de olimpiadas en Espaniol (que podemos hacer mas activa :D ), y alli, son totalmente bienvenidos a proponer problemas, a ver si le llegamos aunque sea a los tobillos a lo activos que son los foros en Ingles :P .

Bueno, ciao!


Is $ x^4\plus{}11$ irreducible over $ \mathbb{Z}_{17}$?

What's the connection between irreducibility in $ \mathbb{Z}[x]$ and $ \mathbb{Z}_p[x]$?

[quote="1=2"]I may be just being stupid, but what does being irreducible over $ Z_{17}$ mean? I have tested $ x^4 \plus{} 11$ and have found that it cannot be factored over the integers.[/quote]

Similar example but with $ \mathbb{Z}_5$

Factor $ x^2\plus{}4$

Clearly this is irreducible over the integers, but

$ x^2\plus{}4 \equiv x^2 \minus{} 1 \equal{} (x\plus{}1)(x\minus{}1) \equal{} (x\plus{}1)(x\plus{}4)$

in $ \mathbb{Z}_5$

[quote="t0rajir0u"]Irreducible in $ \mathbb{Z}_p[x]$ implies irreducible in $ \mathbb{Z}[x]$, but not the other way around.  Can you apply some variant of the Sophie-Germaine identity?[/quote]

Hmm...I don't really see how it applies...because 11 isn't a quad residue mod 17 so it is always the product of a quad residue and a quad non-residue...and that's kind of important in the Sophie-Germain identity, $ a^4\plus{}4b^4$ works because $ 4$ is a square and $ b^4$ is a square...

Right.  So it turns out that the Sophie Germaine identity is the only way to factor a polynomial of the form $ x^4 \plus{} a$ into a product of two irreducible quadratics, so if it can't be done then you know the polynomial's irreducible.

[quote="t0rajir0u"]So it turns out that the Sophie Germaine identity is the only way to factor a polynomial of the form $ x^4 \plus{} a$ into a product of two irreducible quadratics[/quote]

Oh, that's interesting. How would you go about proving that? I tried it directly...$ (x^2\plus{}px\plus{}q)(x^2\plus{}rx\plus{}s)\equal{}x^4\plus{}a$ and then showing $ a\equal{}4b^4$ for some $ b$...but it wasn't pretty and I didn't get the result in the end anyway...

See the more general question [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2026937520&t=228729]here[/url].  And the statement I meant to write was "the only way... if $ \minus{}a$ is not a quadratic residue."


parameterization
calculus
integration
function
trigonometry
calculus computations

Hi everyone. I'm trying to solve these differential equations, but i have run into troubles after doing a few steps. Would any of you help me out or give me some kind of a hint. I greatly appreciate it. x*y''+y'=cosx-xsinx (the original equation)
(x^2)*y''+xy'=xcosx-(x^2)sinx
First i find the homogenous solution:
So, I let (x^2)*y''+xy'=0
m(m-1)+m=0
m^2=0
m=0,0
y(homogenous)= C1x^0+ C2(x^0)Lnx= C1+ C2Lnx
Then I find the Wronskian W= 1/x
I use variation of parameter to solve the equation:
u1= Integral[ -Lnx* (xcosx- (x^2)sinx]/ (1/x)] dx
u1= -Integral[ (x^2)*Lnx*cosx-(x^3)*Lnx*sinx]dx
I got stuck right here when trying to integrate (x^2)*Lnx*cosx-(x^3)*Lnx*sinx. First i separated these two functions and tried to integrate each of them at a time. I used integration by part to integrate the first one (x^2)*Lnx*cosx. I let u=(x^2)*Lnx, du= (2x*Lnx+ x)dx, dv= cosxdx, v= sinx. So the small integral= (x^2)*Lnx*sinx- Integral[sinx*( 2x*Lnx+x)]dx. From there i don't know what to do next. Please help me.

Another problem I have is to solve a system of differential equations
dx/dt=x+ y+ t
dy/dt=x- y- t
I put it in standard form:(differential operation)
(D-1)x- y= t
(D+1)y -x= -t
I got stuck there also.
If you could give me some hint or tell me what I did wrong in those two which made them hard to solve or unsolvable, I would appreciate your help very much.
                                                                   John Adam

You know, we could find a particular solution to the first just by guessing- matching $xy''$ to $-x\sin x$ and $y'$ to $\cos x$, $y=\sin x$ is a solution.

as for second one we can get

$\frac{d^2x}{dt^2}=\frac{dx}{dt}+\frac{dy}{dt}+1=2x+1$

$\frac{d^3x}{dt^3}=2\frac{dx}{dt}$

So you solve the third degree DE with auxialiary equation $m^3-2m=0$ $\implies m=0,\pm\sqrt{2}$

And obtain the general form to be $x=Ae^{\sqrt{2}}+Be^{-\sqrt{2}}+C$

after that u substitute into what u found above into $\frac{dy}{dt}=x-y-t$ and solve the first degree DE .. :)


For me, I lost a realationship with a friend because he began telling people some of my personel information that I trusted in him. Also, this same friend began lying to me, saying things like "I didnt tell anyone" and such to me. He did however tell me the truth in the end, but its not the same type of friendship anymore.

friendships change....i had a really close friend, and we used to tell each other everything...but her priorities changed...and she started hanging out with some bad kids...we are still friends, but more like acquaintances.

[quote="Nienna"]Everyone of us needs a friend, but sometimes it is difficult to keep friendship. I'm asking myself where is the line between tolerance and trying to understand on the one side, and useless investment in somebody on the other side.
In other words what couldn't you forgive and what would be enough for you to break any relation with your former friend?[/quote]

Forgiving was never a problem, as far as I'm concerned.. But in my case, other side is not as forgiving most of the time...  :maybe:

BTW, has anyone read Aristotle's opinion on friendship (in Nycomachean Ethycs)? It's quite interesting (we had a debate about it in our Philosophy class).

^Indeed; rarely does one find something so interesting and relevant to modern life in the moldy old archives of Aristotle's thought :roll:

Well i have no friend at all for years. I dont understand anyone and no one understand me.
Anyways I am happy in my skin. :)
Friedrich Nietzche says that love is blind and friendship closes the eyes.

Lets see...

Before Pre-school:
A bunch of kids I played with

Pre-school:
No friends :(  at first then 3 or 4 friends :)

Kindergarten:
Decent amount of friends :D

First Grade:
Everyone in my class was almost a friend

Second Grade:
Everyone wanted to be my friend (I think it was because I was smart :rotfl:)(back when being smart was really cool)

Third Grade:
Decent amount of friends (I switched from a private school to a gifted and talented one)

Fourth Grade:
ok amount of friends (I switched to another gifted and talented school)

5th Grade:
Awesome amount of friends

6th Grade:
A few friends then I became popular :D

7th Grade:
Picked out real friends and was in the process of changing my attitude on life (from wannabe not nerd to accepting smart nerd) :D

8th Grade:
I got MySpace :lol: . I knew who was and who wasn't going to be tolerated by me. Finally understood the kind of people I could get along with real well.

after getting a high score on a state math test all the girls in my math class wanted to be my friends  :D

Who got the high score? You or the girls?

[quote="mdk"]only after getting the high score?[/quote]

lol we took the test like first thing when i started middle school....there were little girls in the GT math program at my elementary school

Let's see:

1st grade
No friends

2nd grade
No friends

3rd grade
No friends

Etc...

I don't want to tell you how far it goes;otherwise you'll know what grade I am in.

Tried it. Almost never works, except for some of my Mathcounts and Math Circle teammates. I could only find friends when we were similar in talent.

Generally for me, the tie of friendship could only break when the friend doesn't consider me a friend. However, it doesn't mean that every friendship is broken when the friend doesn't consider me a friend any more. It can be a one-way friendship. I just meant that most friendships of mine are broken after the other friend decides that i'm not a friend anymore.
so a friend deciding that i'm not a friend anymore doesn't generally mean the friendship is over, but a friendship being over generally does mean that the friend decided that i'm not a friend anymore.
it's just like one of those "if" or "only if" but not "iff" statements.

see to get friends you have to be open to many subjects

i usually will try to find out about everything  so the odds are that we'll have something in common.
of course the best friends will have my interests in common
1st
every1 in class
2nd
every1 in class
3rd
every1 in class
4th
every1 in class
5th
every1 in class cept 4 like 3 kids
6ths
 like 1 /20 of the school and 1/5 of grade
7th
1/10 of school and 3/4 of grade
8th
1/10 of school and almost all of grade

i think part of my popularity is my last name
 knock knock
whos there
Tommy
Tommy Hu!!!!!!!!
HAHAHAHAHAHAHAHAHAHAHHHAHAHA
AHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA
OMFG AHHAHAHAHAHAHAHHAHAHAHAHAHA
LMAO HAHAHAHAHAHAHAHAHAHAHAHAHAHAHA
ROTFLOL HAHAHAHAHHAHAAHAHAHAHHAAHAH
FDSLOLHAHAHHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA
(FDSLOL= falling down stairs laughing out loud)

For me

Easy to be friends with:
Smart People
Funny People
Crazy (in a funny and stupid way) People
Sports Knowlegable (sp?) People
People who like my music

Hard to be friends with:
Mean People
Humorless People
Ignorant People
People who hate my music
People who pretend


induction
inequalities proposed
inequalities

Let $n$ denote a positive integer and let $a>0,a\not=1$ prove that:

$n(a^{2n+1}+1)>a+a^{2}+\ldots+a^{2n}$

I think even $a>-1$ is sufficient. It is a nice property by the way. :)

we can prove it using [u]induction on n.[/u]


a) There is a row of seven seats in a classroom. Bob does not want to sit next to Julie. How many ways are there to arrange the seating?

b) How many ways are there to arrange the seating if Garrett must sit by Greg (assume Bob still dislikes Julie)?

c) Suppose one student was absent. How many ways are there to arrange the seating now? (assuming that everybody can sit anywhere)

d) Supposed one student was absent. How many ways are there to arrange the seating if Bob and Julie still hate each other and if Garrett still must sit by Greg?

[hide="a"]Bob and julie cannot sit by each other. We could either a) split into cases and find all arrangements (tedious) or b) Use complementary counting. I choose b).

Seats:
[u]1[/u] [u]2[/u] [u]3[/u] [u]4[/u] [u]5[/u] [u]6[/u] [u]7[/u]
There are a total of 7! for the kids to arrange themselves.
If bob sits next to julie, then there are 6 spots they can sit together at (Bob at 1, julie at 2, bob at 2, julie at 3......bob at 6, julie at 7). There will be 2 ways they can arrange themselves in those six spotes (either bob july or julie bob), and there will be 5! for the other kids to sit once we have figured out where bob and julie sit.
Since we used complementary counting, we subtract that from 7!. so our answer is 7!-(6x2x5!) [/hide]

Ughhhhh.. I hate doing problems like b). It involves lots of casework bash which im really not in the mood to do...

[hide="c)"]If one person is gone, there are 6 people left. There will be 7x6! for the 6 kids to sit if there are no restrictions.[/hide]

d) is the same as b). They both involve lots of casework bash which I dont feel like doing.


Let $d(n)=\sum_{0<d|n}{1}$. Show that, for any natural $n>1$,

\[ \sum_{2 \leq i \leq n}{\frac{1}{i}} \leq \sum{\frac{d(i)}{n}} \leq \sum_{1 \leq i \leq n}{\frac{1}{i}} \]

We have $\sum{\frac{d(i)}{n}} =\frac 1n \sum_{j \le n} [n/j]$ Because $n/j-1<[n/j]\le n$
$\sum_{2 \leq i \leq n}{\frac{1}{i}} \leq \sum{\frac{d(i)}{n}} \leq \sum_{1 \leq i \leq n}{\frac{1}{i}}$


algebra
polynomial
Rational Root Theorem

First of all, I apologize if I posted this in the wrong forum, but I'm looking for some assistance finding and proving zeros in a couple polynomials.

1.) Prove that (1 + i) is a zero of f(x) = 3x^3 - 7x^2 + 8x -2

2.) Find all zeros of f(x) = x^3 - 5x^2 + 2x - 10

Thanks in advance for any help!

[hide="hint for 1"]we could use the rational root theorem and get $\pm\frac{1}{3}, \pm\frac{2}{3}, \pm1, or \pm2$ and use Descartes rule of signs to further reduce it. This will find a factor and then the other two imaginary follows[/hide]

The equation in $2$ factors over $\mathbb{C}$
[hide="into"]$f(x)=(x-5)(x-\sqrt{-2})(x+\sqrt{-2})$.[/hide]

The reason is that like many others, t0rajir0u likes nice solutions.

Hmm.  I could actually have been quicker than that.

[hide="You see..."] If $x^{2}-2x+2$ is a factor of the polynomial then the only factor left must be a linear factor $ax+b$, but multiplying we must have $a = 3, b =-1$ so we're already done.

Then the "proof" consists of a quick multiplication to verify the factorization. [/hide]


[color=darkblue]Let $ABCD$ be the trapezoid for which $\{\begin{array}{c}AB\parallel CD\\\ AB=BC\\\ AC=AD\end{array}\ .$ Prove that

$\{\begin{array}{c}M\in AB\ ,\ MA=MB\\\ N\in MC\cap BD\end{array}\|$ $\Longrightarrow$ $BN=2\cdot MN\ .$[/color]

There's something in your style that strongly reminds me of Virgil Nicula, young_boy  :wink:

Anyway, here's my solution:
[hide]First of all, notice that $\Delta DAC$ and $\Delta ABC$ are similar.
Call $G$ the midpoint of $AD$.
Clearly, because $\Delta DAC$ and $\Delta ABC$ are similar, we know that $\Delta GMC$ is also similar to these two triangles.

Therefore, $\angle ABC = \angle GMC = \angle DNC = \angle MNB$

We have now proven that $\Delta MNB$ and $\Delta MBC$ are similar, so we got what we wanted since $2\left|MB\right|=\left|BC\right|$.
[/hide]

Strange but i have the same feeling as Jan :lol:

[quote="Jan"]
First of all, notice that $\Delta DAC$ and $\Delta ABC$ are similar.
Call $G$ the midpoint of $AD$.
Clearly, because $\Delta DAC$ and $\Delta ABC$ are similar, we know that $\Delta GMC$ is also similar to these two triangles.
[/quote]
$\Longrightarrow GM=GC=\frac{1}{2}BD$
On the ortherhand, apply Thales's theorem we have:
$\frac{MN}{BN}=\frac{NC}{ND}=\frac{MC}{BD}=\frac{1}{2}\quad\mathbb{QED}$

What is Thalet's theorem? I could not find it on mathworld or google

It's typo. :D  [b]Thales[/b] not [b]Thalet[/b]  :blush:


Which number follows logically in this sequence:

4, 9, 50, 147, 605, 1352, 3757, ...

[hide]Factor the numbers:

4 = 1 * 2^2
9 = 1 * 3^2
50 = 2 * 5^2
147 = 3 * 7^2
605 = 5 * 11^2
1352 = 8 * 13^2
3757 = 13 * 17^2

The first number in the factorization is a Fibonacci number. The second number is a square of a prime. Following this pattern, I believe that the next number is

21 * 19^2 = 7581[/hide]

What can you say about the existance of cubes in that sequence?

did you used programming lang for this chesspro

yes or your brain's right part do the work

I figured it out myself. I think it was the left part of my brain though, since it's more analytical.

[quote="polymorphic"]why are there no cubes?[/quote]

I agree.  It's not as trivial as it first seems.  You don't need F_n=p_n (being the nth Fibb and prime respectively).  If F_n=q^3p, where q was some prime, you would also get a cube.


[quote="pinkmuskrat"]Oh wow. You sure you let your computer load all the way?. Look's like it's even more :diablo: than :spam: !!!!!!!!!![/quote]

It isn't just the computer. It shows up that way on my computer as well.

Same thing here. I don't know why. For anyone who is wondering, it can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1529118#1529118]here[/url], at post number 2.

I just rated the post a 6, and now it says the overall rating is 1. A screen shot is attached. I will cancel it, so it will go back to .5.

[quote="jmerry"]I'm one of the people who rated it. I just messed around a bit for testing: after I leave a rating, the post shows a normal rating. When I reload after that, the displayed rating drops by half a cube: from 1 to 0.5 normally, from 2 to 1.5 if I change my rating to 6.

I'm not aware of any other posts doing that.[/quote]
I also rated that post and it shows up as 1/2 a cube. Maybe when it's averaged it's 0.5??

No. It is impossible to give less than one cube, so the average should never be less than one.

It also clearly affects this post and not (most) others. I tried rating some posts in the test forum, and got normal behavior there.

actually, when the average is something point 5, its a half cube, take Dr. Vornicu for example, his average is 4 and a half.

Yeah, but since the lowest rating is 1, the average cannot be lower than it. You can have a 4.5, for example if there are ratings of 4 and 5.

Also, we are talking about a single post's rating, not the overall rating of a user (which comes from post ratings).

EDIT: I didn't mean ratings for single posts can't be a non integer. I meant the source of the rating.

[quote="PowerOfPi"]Also, we are talking about a single post's rating, not the overall rating of a user (which comes from post ratings).[/quote]

Individual post cube rating are averaged, too, so it is possible to get 1.5 cube, 2.5 cube, 3.5 cube, 4.5 cube, or 5.5 cube for a post.

as a serious problem, shouldn't this be moved to support? or should it remain is QSA?

Very serious  :wink:
And only admins can move it, but at the moment of an admin seeing it, there is technically no need to move it anymore (they may move it nonetheless).

[quote="louiethechewy123"]Is it still half-cube? because I see 1 cube  :huh:[/quote]

You look at the second post on the thread, cause the rest of the ratings for Gen8 is whole cube.
  :arrow:  :arrow:  :arrow:

I see a half cube on the second post.  :huh:


probability
ratio
real analysis
topology
geometry
limit
expected value

So, if we have a sequence of iid random variables which take values in $\{0, 1\}$ each with probability 1/2, then we know by the Central Limit Theorem or whatever that the ratio of 0s to 1s converges to 1:1.

The measure on this space of sequences is the Lebesgue measure on $[0, 1]$.

There are 2 possible things that can happen (with probability 0 of course) that are unusual:

(a) the ratio of 0s to 1s converges to $p: q$ where $p \ne q$.
(b) the ratio of 0s to 1s does not converge.

Let $A$ be the set of points in $[0, 1]$ which do (a) (that is, their binary representations do).
Let $B$ be the set of points that do (b).

How big are $A$ and $B$? I am looking for estimates of Hausdorff dimension. (They both look uncountable to me although I didn't really check carefully because I am supposed to be doing topology homework and it's due in 8 hours and I need to sleep too....)

I wanted to give a very stupid lower bound for the Hausdorf dimension when the ratio of 0s to 1s converges to $n: 1$, $n>1$. To do this I wanted to find the Hausdorf dimension of the set of real numbers of the form
$x=\sum_{k=1}^{\infty}\frac{2^{a_{k}}}{2^{nk}}$
where $0\leq a_{k}\leq n-1$ for every $k$. I think that the Hausdorff dimension of this set is $d=\frac{\ln(n)}{n \ln(2)}$, but I'm not really sure. If it is the case, this gives a (very rough) lower bound for a (very) particular case  :D

this is the first theorem of chapter10 of Falconer's book "Geometry of Fractal sets", with a beautiful proof. Here is the idea:
One wants to find a probability measure $\mu$ on $F$ such that
$\lim_{r \to 0}\frac{\mu(B(x,r) \cap F)}{r^{s}}= 0$ or $\infty$, depending wether $s$ is superior or inferior to a certain fixed number $d$. Of course, this will implies that $d$ is the Hausdorff dimension of $F$(classic). This probability measure is the more natural one  :)  That is, the measure of those real numbers of the form $x=\overline{0.b_{1}b_{2}\ldots b_{k}\ldots}$ is $p(b_{1})p(b_{2}) \ldots p(b_{k})$ where $b_{i}\in \{0;1\}$ and $p(0)=\lambda$, $p(1)=1-\lambda$.

PS: it's funny to know that one cannot (at least when Falconer wrote the book) exhibit a real number $x$ such that it's $b$-ary expansion is equi-distributed for every integer $b$. But the set of those numbers is of full measure :)

[quote="alekk"]this is the first theorem of chapter10 of Falconer's book "Geometry of Fractal sets", with a beautiful proof. [/quote]
What? I have that book, and it only has 8 chapters...

:rotfl:  sorry for that... It is in "Fractal geometry: mathematical foundations and applications ", his other well known book  :lol:

If you have read Falconer's book(s), you should be able to do much better than just rough estimates :roll:. Recall that the dimension of a self-similar Cantor set constructed by choosing finitely many disjoint subintervals $I_{j}$ on $[0,1]$ and repeating this construction in a self-similar manner on each interval $I_{j}$ is the unique number $d$ such that $\sum_{j}|I_{j}|^{d}=1$. Now, assume that the limit you are looking for is rational, say $p: q$, $p,q\in \mathbb N$, $p<q$. (the general case is a bit more technical but very similar). Choose a huge $n$ that is a multiple of $p+q$. Consider the set of dyadic intervals $I$ of length $2^{-n}$ such that among the first $n$ digits of any point $x\in I$, the ratio of $0$s to $1$s is exactly $p: q$. Construct a self-similar set based on these intervals. This will give you an estimate from below. To get a matching estimate from above, look at the number of intervals where the ratio is less than $p/q+\epsilon$ among the first $n$ digits. To say more would be the same as to post the entire solution, so I'll stop here. :)

this is amazing that such a simple approach gives exactly the right dimension! (I've only checked the lower estimate) Thank you very much!

So, I was trying to use a similar technique (as Falconer's in ch10 as alekk cited, not fedja's way) to estimate what happens when the limit does not converge. So I defined a probability measure on [0, 1] in the following way:
Let $n_{0}= 1, n_{1}= 1, n_{k}= 2^{k-1}$. Let $x = 0.x_{1}x_{2}\ldots \in [0, 1]$ be a typical number.
If $n_{2k}< m \le n_{2k+1}$, then $P\{x_{m}= 1\}= q$.
If $n_{2k+1}< m \le n_{2k+2}$, then $P\{x_{m}= 1\}= 1-q$.

As $m \to \infty$, the expected number of 1s in the first $m$ digits will oscillate between $\frac{1+q}{3}$ and $\frac{2-q}{3}$.

So, my impression at first was that this is enough to show that these kinds of sets have full dimension too, because we can make $q$ really close to 1/2. I mean, if we repeat Falconer's technique of checking how big the measure of balls are according to this measure and obtain a lower bound for the Hausdorff dimension, it seems to be bigger than the value obtained by $P\{x_{m}= 1\}= \frac{1+q}{3}$ for all $m$. Does this seem ok? I was a little dubious because of how weird this measure feels.

Yes, it is fine and the set where there is no limit is, indeed, of Hausdorff dimension $1$. One more interesting thing is that, given $\alpha<\beta$  you can even find the exact dimension of the set where $\liminf=\alpha$ and $\limsup=\beta$. When $\alpha$ and $\beta$ are both close to $1/2$, you'll get a number close to $1$.


Let $ A$ be a $ n\times n$ ,$ n\geq 2$, symmetric, invertible matrix with real positive elements. Denote by $ a_{ij}$ and $ b_{ij}$ be elements of $ A$ and $ A^{\minus{}1}$,respectively. Show that for $ k \not\equal{}m$ ,$ \sum _{i\equal{}0}^n a_{ki}b_{im}\equal{}0$

[hide]By $ A^{ \minus{} 1} \equal{} \frac1{\det(A)}\,\text{adj}(A)$ and

$ (A\,\text{adj}(A))_{km} \equal{} \sum_{i \equal{} 1}^n a_{ki}C_{mi} \equal{} \left\{\begin{array}{*{20}c} {\det A,} & {k \equal{} m} \\
{0,} & {k \ne m} \\
\end{array} \right. ,$

then we get the result.[/hide]


I know that this has been posted over a billion times :D but all of them are too old to revive. Where can I old competions of MC?
Not someting like those drills on mathcounts.saab.org or whatever it is.

Thanks,
Dhaivat Pandya (Uh huh, you read correctly that's my name :D )

You can buy them of the mathcounts website.

[quote="math154"]If you bother enough people on AoPS, you could probably eventually get all of them from 1983-2008...after all, you do have a whole summer :D[/quote]
How did you think everyone would have that many Nats problems in all?

I don't even have any :(

[quote="math154"]A lot of states are really really serious about MATHCOUNTS, so they go everywhere to get the problems (every state coach gets a copy of that year's nationals tests) and so basically the coaches ask other coaches for tests and stuff. And the coaches, obviously, give the tests to students. So many people have lots of tests.[/quote]
Wow.

I feel jealous :blush:  :(

It's OK, I only have a few.

EDIT: Most people only have a few or none.

EDIT2: tinytim, you have 1999 too.

Uh, and what happened to the website : [url]http://www.yashar.biz/mathcounts[/url]

We're talking 'bout nats.

Oh, I dunno what happened to it, but if you search MC tests on google you'll eventually find all of them that were available there. There's also this Minnesota state MC website and like Mr. Stutz middle school blablabla with lots of tests.

i have 91-08 nats sprint, target, team, cd. 01-07 state and chapter :D

How????

*Please answer, then you'll have 1000 posts. :D

actually, now this is my 1001st :D I did my 1000th in the How To Succeed in MC topic

To find the chapter and state from 2001-2008, just look on the web. You can find 2008 on the mathcounts website, and others on sites such as http://www.stutzfamily.com/mrstutz/mathcounts/index.html . It takes awhile to find all of 2001-2008 including cd and team, but it is manageable.

For nats, just search old mathcounts tests, and look through topics. You'll find people who say that they have old mc tests, and if you ask tons of different people, you will find some who give them. I don't have the full set (it goes back to 1984), but I have a lot.

P.S. don't ask me, my labtop with all of my tests recently died and while I have the paper form, I don't have a scanner.

I have like 1990-2005... But don't ask me for them, because they're written on and I'm too lazy to scan them.

[quote="perfect628"]I have 1983-2006 personally...See above for my reason not to give them to you.[/quote]
I have none... :(  :(  :(


Sinda purchased three blouses and two scarves for $ \$87.75$. The two scarves were the same price, but one of the blouses cost $ \$5.00$ more than each of the others. Five scarves would have cost $ \$64.75$. What was the cost of the most expensive blouse?

Ok, so we have $ 2s$ for the scarves, and $ 3b \plus{} 5$ for the blouses.  We know that $ 5s \equal{} 64.75$.  Thus, the cost of one scarf is $ 12.95$.

Then, you have $ 2(12.95) \plus{} 3b \plus{} 5 \equal{} 87.75$.

$ 3b \equal{} 56.85$

Thus, the least expensive blouse is $ 18.95$, so the answer is $ \boxed{23.95}$.


[quote="Kouichi Nakagawa"]1=I=1
2=II=2
3=III=3
4=IV=3
5=V=2
6=VI=[color=red]3[/color]
7=VII=4
8=VIII=5
9=IX=3
10=X=2
11=XI=3
12=XII=4[/quote]

if you cant find the pattern, he means the number of lines used in its roman numeral form is the number it equals

[quote="Kouichi Nakagawa"]1=I=1
2=II=2
3=III=3
4=IV=3
5=V=2
6=VI=[color=red]3[/color]
7=VII=4
8=VIII=5
9=IX=3
10=X=2
11=XI=3
12=XII=4[/quote]

Excellent!

Have A Great Day!

[quote="moogra"]Do you have to find the loopholes in every problem?[/quote]

yes. yes he does. apparently.

;]

[quote="Brut3Forc3"]Don't feed the trolls...[/quote]

[quote="mewto55555"][quote="Arrange your tan"]This is an arbitrary pattern.  Though it can continue somewhat, what happens with the "L" for instance.

This puzzle has the expectation that there is one and only one way to come up
with the code in response, but another person COULD legitimze some other method other than the one given.

This is a can-you-read-MY-mind puzzle.[/quote]

Then please demonstrate another valid solution  :D[/quote]

Well, I guess this pattern is screwed then.  And meggy-meg, don't spam.

I think it was a nice puzzle.

And Meggy-meg wasn't spamming, she was reminding us NOT to spam...

I don't really see how the puzzle doesn't work anymore with the L...

Well, for sure it doesn't work with $ C$ :P

Actually, it could. That's one stroke, which I guess could count as a line, so it would be 2.

Puzzles are meant to be not-overanalyzed.

the L wasn't even in the problem, so please. stop overanalyzing.


Prove this:
For any two fractions in the form $\frac pq$ where both are integers and $q$ is not 1, the product and sum cannot both be integers.

[hide="I dont think proofs are allowed in Getting Started"]
For coprime $p, q$, the product is an integer, so write the two fractions as $\frac{p}{q}, \frac{kq}{p}$ where $p \not | k$.  Then their sum can be written $\frac{p^2 + kq^2}{pq}$.  Note that if this is an integer, then $q | p^2$, which is impossible because they are coprime.  Hence no such fractions exist.  Q.E.D.
[/hide]


Given two circles $ w_1, w_2$ in the plane that have at most one common point. Let $ l_1, l_2$ be their common external tangents and $ l_3$ be their common internal tangent that meet $ l_1$ at $ P$ and $ l_2$ at $ Q$. Parallel tangents $ l_4$ to $ w_1$ and $ l_5$ to $ w_2$ meet $ l_1$ and $ l_2$ at points $ A, D$; $ B, C$ respectively. Prove that $ BP \parallel DQ$.

It is very nice, I will post my solution later I have in mind

[quote="plane geometry"]Denote E is the ontersection of BC and AD  Draw CK//QD K=CK∩EA
EK/ED=EC/EQ =>  ED*EC=EQ*EK=EO*EO1 =>  ⊿EKO∽⊿EO1Q  => ∠EOK=∠EQO1=∠PQO1
Notice ∠AO2B=∠COD=180-∠CO1D
∠CO1D=180-1/2∠QCD-1/2∠CDP=1/2(∠CQP+∠QPD)=∠QO1P  => ∠QO1P=180-∠AO2B=∠QO2P   =>   Q,O1,P,O2 are concyclic  =>  ∠PQO1=∠PO2O1   =>  OK//O2P  =>  K and P are corresponding points w.r.t E ,  we have already known C and B are corresponding points w.r.t E  =>  CK//BP  =>  QD//BP , which completes our proof

ps:It is 2:30am in china. [b]good night [/b]dear mathlinkers![/quote]

Perfect solution

[quote="April"]See: http://www.mathlinks.ro/viewtopic.php?p=462115#462115[/quote]
[hide]dear [b]April[/b]:
I think they are different [/hide]


inequalities
analytic geometry
absolute value

write an absolute value inequality for the pharse "all real numbers $x$ at  most 4 units away from 2" .

ok this is a stupid question but...
Why is the answer  $|x-2|\le{4}$  and not $|x+2|\le{4}$.

Thanks :)

The distance between two points is just the (absolute value of the) difference of their coordinates, right?

As a reality check, you can have two very large numbers $x,y$ which can be very close together, but then $|x+y|$ will be very large, which isn't what you want.

well suppose on a line segment there are points A, B, and C.

Point B is 2 cm away from Point A, and Point C is 5 cm away from Point A.

If i asked u to find the distance between points B and C you would do 5-2=3, not 5+2=7. The same concept applies with number lines.

The distance is the absolute value of the difference.

"Away from 2"

What do you do when you want to find distance?

PositionA - PositionB

Here, Position A is x
Position B is 2.

x-2


Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that \[ f(x^2 - y^2) = x f(x) - y f(y)  \] for all pairs of real numbers $x$ and $y$.

When replying to the problem, I ask that you make posts for solutions and submit comments, jokes, smilies, etc. separately. Furthermore, please do not "hide" any portion of the solution. Please use LaTeX for posting solutions. Thanks.

We establish a few initial conditions:

$f(0^2-0^2) = f(0) = 0$.

$f(x^2-(-x)^2) = f(0) = xf(x)+xf(-x) = 0 \Rightarrow f(x) = -f(-x)$ (1)

$f(x^2-0) = f(x^2) = xf(x)$ (2).

Now let $a,b$ be positive reals. Then

$f(a-b) = \sqrt{a}f(\sqrt{a})+\sqrt{b}f(\sqrt{b})$,

which by (2) equals

$f(a-b) = f(a)-f(b)$.
$f(a-b)+f(b) = f(a)$ (3).

Using (1), we can derive the equations
$f(a-b)+f(-a) = f(-b)$ (4)
$f(-a)+f(b) = f(b-a)$ (5).

The following casework is necessary because $a,b$ are positive reals while $x,y$ can be positive or negative.

From (3), we set $a = x+y$, $b = y$ to get $f(x)+f(y) = f(x+y)$ for $x,y \ge 0$.

From (4), we set $a = -y$, $b = -(x+y)$ to get $f(x)+f(y) = f(x+y)$ for $x,y \le 0$.

From (5), we set $a = -x$, $b = y$ OR $a = -y$, $b = x$ to get $f(x)+f(y) = f(x+y)$ for one of $x,y$ negative and the other positive.

Hence we have $f(x)+f(y) = f(x+y)$ for all real $x,y$.

So $f(x) = f(1)+f(1)+\cdots+f(1) = xf(1)$. Since $f(1)$ is constant, let $f(1) = k$. Then we have $f(x) = kx$ for all real $k$. QED.

[quote="paladin8"]
So $f(x) = f(1)+f(1)+\cdots+f(1) = xf(1)$. Since $f(1)$ is constant, let $f(1) = k$. Then we have $f(x) = kx$ for all real $k$. QED.[/quote]

so um you've proven it for integers x... what if x isn't an integer?

paladin8, a typo:  (1) should read $f(x)=-f(-x)$.

ThAzN1's objection also stands.  The Cauchy equation $f(x)+f(y) = f(x+y)$ which you have derived does not imply $f(x)=kx$ for all $x$, unless you can also establish that $f(x)$ is continuous, or monotonic, or bounded on some finite interval.

I'll prove that  $f(x)=xf(1)$ for every [b]real[/b] x.
Let $x\in \mathbb{R}$.
$f(x)=f(x\cdot 1)$
Looking for $a$ and $b$ such that $x=a+b$ and $1=a-b$ we find $a=\frac{x+1}{2},\ b=\frac{x-1}{2}$, thus
$f(x)=f((a+b)(a-b))=f(a^2-b^2)=af(a)-bf(b)=\frac{x+1}{2}f(\frac{x+1}{2})-\frac{x-1}{2}f(\frac{x-1}{2})$.
Since $f(t+u)=f(t)+f(u)$ (see Paladin8's solution) we have $f(t/2+t/2)=f(t/2)+f(t/2)$, hence $f(t/2)=f(t)/2$.
We get
$f(x)=\frac{x+1}{2}\cdot \frac{f(x)+f(1)}{2}-\frac{x-1}{2}\cdot \frac{f(x)-f(1)}{2}$,
so $f(x)=\frac{xf(1)+f(x)}{2}$, hence $f(x)=xf(1)$ for every [b]real[/b] x.

I guess this is more or less the same thing, but

After establishing that $f(x)$ is odd and the Cauchy thing... we need only consider $x \geq 0$.  Note that

$f(2x+1) = f((x+1)^2-x^2)$
$= (x+1)f(x+1) - xf(x) = f(x+1) + x(f(x+1)-f(x))$
$= f(x+1)+ xf(1)$

Subtracting yields

$f(x) = f(2x+1) - f(x+1) = xf(1)$

Replace $x$ with $-x$ and using $f(x) = - f(-x)$ you get the result...

we immediately obtain  $f(0)=0$
and we know $f(x-y)=f(x)-f(y)$
let $x+y=1$
then $f(x-y)=xf(x)-yf(y)$
$\Rightarrow f(x)-f(y)=xf(x)-yf(y)$
$\Rightarrow xf(y)=yf(x)$
$\Rightarrow f(x)=xf(1)$
let $f(1)=k$and $k\in R$
$\Rightarrow f(x)=kx$
it`s good but i am not safe

[quote="mathisfun1"]Writing the functional equation in the form $f(xx - yy) = xf(x) - yf(y)$, we see that all linear functionals from R to R would suffice, thus, since all functions of the form $f: \mathbb R \rightarrow \mathbb R, x \rightarrow kx$ are linear functionals, all such functions f would satisfy the functional equation. I do have a question, though: do there exist linear functionals from R to R that are not linear functions?[/quote]

Well you could have showed that simply by plugging in $f(x) = kx$. Since there are no restrictions on the problem (continuity, boundedness, differentiability, etc.) it's very hard to come up with an "argumentative" solution since functions can pretty much do whatever they want without restrictions and you can come up with some really goofy stuff (as you may have discovered in an Analysis course)

[hide="Not Sure if This Works"]
As everyone else before me as done, we can quickly notice that  $ f(0) = 0$.  Plugging in $ y = 0$, we have that $ f(x^2) = xf(x)$ and similarly, $ f(y^2) = yf(y)$.  Thus, $ f(x^2) - f(y^2) = xf(x) - yf(y) = f(x^2 - y^2)$.  Letting $ a = x^2$ and $ b = y^2$, we have that $ f(a - b) = f(a) - f(b)$.  Now, since the function maps any real number to another real number, we see that $ f$ is a continuous function.  Hence, we can take the derivative of it.  Letting $ a = b + \epsilon$, where $ \epsilon$ is an infinitely small number, we have that $ f(\epsilon) = f(b + \epsilon - b) = f(b + \epsilon) - f(b)\implies \frac {f(\epsilon)}{\epsilon} = \frac {f(b + \epsilon) - f(b)}{\epsilon} = f' (b)$.  Notice that $ \frac {f(\epsilon)}{\epsilon}$ is constant, so $ f'(b)$ is constant.  It follows that $ f(b)$ is linear for positive $ b$.  Letting $ f(x) = kx + j$, the given equation becomes $ k(x^2 - y^2) + j = f(x^2 - y^2) = x^2k + xj - y^2k - yj = k(x^2 - y^2) + j(x - y)$.  If $ x - y\ne 1$, then it follows that $ j = 0$.  Hence, $ f(x) = xk$.  Similarly, if $ x$ is negative, then $ f( - x) = - f(x) = - xk = x( - k)$.  [/hide]

EDIT: Oh, sorry, [b]MellowMelon[/b].  I have a slightly different idea now; does this work?
[hide="Hopefully Correct Solution"]
As before, we have that $ f(a) - f(b) = f(a - b)$.  Now, let $ \epsilon$ be the number "next to $ 0$" on the number line, as defined in the "proof" above.  Notice that $ (0, 0)$ and $ (\epsilon, f(\epsilon))$ are collinear since any two points are collinear.  We prove through induction that $ (a + \epsilon, f(a + \epsilon))$ is on this line.  By the inductive hypothesis, $ (a, f(a))$ is on the line.  Now, $ \frac {f(a + \epsilon) - f(a)}{\epsilon} = \frac {f(\epsilon)}{\epsilon} = \frac {f(a) - f(a-\epsilon)}{\epsilon}$, so the slope formed by $ (a, f(a))$ and $ (a + \epsilon, f(a + \epsilon))$ is the same as that formed by $ (a, f(a))$ and ${ (a - \epsilon, f(a - \epsilon)}$, which means that $ (a + \epsilon, f(a + \epsilon))$ lies on the line, as desired. [/hide]
Anyways, are we allowed to say that this is the well-known Cauchy function, so the function must be linear?

[quote="The QuattoMaster 6000"]Now, since the function maps any real number to another real number, we see that $ f$ is a continuous function.[/quote]
This is not always true. Take the function that's outputs $ 1$ given a rational and $ 0$ given an irratinoal.

[hide="Building off paladin8s solution"]
We try to relate $ f(1)$ and $ f(x)$ to our conjecture from Cauchy's functional equation that $ f(x) = xf(1)$.
$ f(1) = f(x + 1) - f(x)$
So we look for an expression involving $ f(x + 1)$. We know we have to go back to the original equation, since Cauchy's functional equation itself is insufficient. Hence,
\begin{eqnarray*} f((x + 1)^2 - 1) & = & (x + 1)f(x + 1) - f(1) \\
f(x^2 + 2x) & = & \\
f(x^2) + f(2x) & = & \\
(x + 1)f(x) + f(x) & = & (x + 1)f(x + 1) - f(1) \\
f(x) & = & (x + 1)f(1) - f(1) \\
& = & \boxed{xf(1)} \end{eqnarray*}
and our conjecture is satisfied.
[/hide]

[quote="The QuattoMaster 6000"]Anyways, are we allowed to say that this is the well-known Cauchy function, so the function must be linear?[/quote]That isn't necessarily true - you need continuity at one point or monotonicity, otherwise [url=http://en.wikipedia.org/wiki/Cauchy%27s_functional_equation]there are funny non-continuous solutions[/url]. Don't know if you're allowed to say that if you can prove continuity or monotonicity.

Let $P(x,y)$ denote the assertion.

$P(0,0) \rightarrow f(0)=0$
$P(x,-x) \rightarrow f(x)=-f(-x)$
$P(x,0) \rightarrow f(x^2)=xf(x)$

Putting this in the original equation and using the fact that $f$ is odd gives $f(x+y)=f(x)+f(y)$. Now putting $(x+1, x)$ in the original function and the new function gives $f(x)=cx$ as the only solution, which works.

We claim that the only solutions are $\boxed{f(x) = cx}$ for any constant $c$, which work. We will now show these are the only solutions.

Let $P(x, y)$ denote the given assertion. $P(x, x)$ gives us $f(0) = 0$, and $P(x, -x)$ gives us $f(x) + f(-x) = 0$, so $f$ is odd. On the other hand, $P(x, 0)$ gives us $f(x^2) = xf(x)$. So, we may now rewrite the functional equation as
$$f(x^2 - y^2) = f(x^2) - f(y^2).$$
Since $f$ is odd, we may write
$$f(a + b) = f(a) + f(b)$$
for all nonnegative $a$ and $b$. Since $f$ is odd, this implies that $f$ is additive, or $f(a + b) = f(a) + f(b)$, for any real $a$ and $b$. Now, we return to the equation $f(x^2) = xf(x)$; plugging in $x + 1$ for $x$ gives us $$f((x + 1)^2) = (x + 1)f(x + 1) = xf(x) + f(x) + xf(1) + f(1),$$
yet we also have
$$f((x + 1)^2) = f(x^2 + 2x + 1) = xf(x) + 2f(x) + f(1),$$
so $f(x) = xf(1)$. This implies the solution set.

The solutions are $f(x) =cx$ for real numbers $c$.

By setting $y=0$, we have $f(x^2) = xf(x)$, and furthermore $f(-y^2) = yf(y)$ by setting $x=0$, hence $f(-x^2) = -f(x^2)$. Substituting this back in, $$f(x^2-y^2) = f(x^2) - f(y^2),$$ so $f$ is additive. Now, set $a \in \mathbb Q$ and $b \in \mathbb R$. It follows that $$(a+b)f(a+b) = f(a^2+2ab+b^2) = af(a)+bf(b) + 2f(ab) = af(a)+bf(b)+bf(a)+af(b).$$ Canceling terms, we have $bf(a) = af(b)$, so $\frac{f(a)}a = \frac{f(b)}b$. As $f(x) = cx$ on $\mathbb Q$ by Cauchy's equation, $f(x) = cx$ on $\mathbb R$ too.

We let $P(0, 0)$ to get $f(0) = 0$.
We let $P(x, 0)$ to get $f(x^2) = xf(x)$.
We let $P(0, y)$ to get $f(-y^2) = -yf(y)$. Thus we can see that $-f(x) = f(-x)$.
We can rearrange the given equation to get $f(x^2 - y^2) = f(x^2) - f(y^2)$.
We let $P(\sqrt{x}, \sqrt{y})$ to get $f(x-y) = f(x) -f(y)$. Not that $(x, y)$ are positive.
We can rearrange to get $f(y) + f(x-y) = f(x)$. This is just Cauchy's functional equation.

We claim that $f$ has the property that either $f(x) > 0$ or $f(x) < 0$ over the positive reals.
Assume FTSOC that we do not have the given.

Assume that we have an outlier $x$ such that $f(x) < 0$ (we will do the arguement only this way and the other way follows exactly opposite) and this is the smallest positive $x$ with this property. We can choose a smaller $y$ such that $f(y) > 0$ and we get $f(x^2 - y^2) < 0$. This is against our assumption. Thus we can see that $f$ is linear. We can thus see that our solution is $\boxed{f(x) = cx}$. $\blacksquare$

AHHH I can't figure out the case when $x = 0$ for cauchy's bounded someone help

We claim the answer is $f(x)=kx$ where $k\in\mathbb{R}$, which clearly satisfies the given conditions. Let $P(x,y)$ denote the given assertion. From $P(x,-x)$ we have $f$ is odd, and from $P(x,0)$ we have $f(x^2)=xf(x)$. Hence, \[f(x^2-y^2)=f(x^2)-f(y^2)\] Letting $a=x^2-y^2$ and $b=y^2$ we see $f(a)+f(b)=f(a+b)$ for non-negative $b$ and all $a$. Multiplying both sides by $-1$ and using $f$ is odd, we get $f(-a)+f(-b)=f(-a-b)$ and letting $c=-a$ and $d=-b$ we see $f(c)+f(d)=f(c+d)$ for non-positive $d$ and any $c$ so $f$ is additive. Then, \begin{align*}f((x+1)^2)=(x+1)f(x+1)&\implies f(x^2)+f(x+1)+f(x)=(x+1)(f(x)+f(1))\\&\implies xf(x)+2f(x)+f(1)=xf(x)+f(x)+f(1)+xf(1)\\&\implies f(x)=xf(1)\end{align*} so $f(x)=kx$. $\square$

The only solutions are $f(x)=cx$ for all real $c.$

First, we show that this function satisfies the given equation. Denoting the given assertion as $P(x,y),$ we plug in certain values to derive useful statements: from $P(0,0),$ we have $f(0)=0,$ from $P(x,0),$ we have $f(x^2)=xf(x),$ and from $P(0,y),$ we have $f(-y^2)=-yf(y).$ From the latter two statements, we have $-f(x)=f(-x),$ thus showing that $f$ is an odd function. Therefore, we write

\[f(x^2-y^2)=f(x^2)-f(y^2).\]

Extending this to all nonnegative $a$ and $b,$ we have $f(a-b)=f(a)-f(b),$ and noting that $f$ is odd,

\[f(a+b)=f(a)+f(b),\]

for any real $a$ and $b,$ showing that $f$ is Cauchy's functional equation. Therefore, this function does satisfy the given equation.

Now, we show that this is the only function satisfying the given equation. Plugging $x+1$ into $f,$ we have

\begin{align*}
f((x+1)^2)=(x+1)f(x+1)=(x+1)(f(x)+f(1))=xf(x)+f(x)+xf(1)+f(1) \\
f((x+1)^2)=f(x^2+2x+1)=f(x^2)+f(2x)+f(1)=xf(x)+2f(x)+f(1) \\
xf(x)+f(x)+xf(1)+f(1)=xf(x)+2f(x)+f(1) \\
f(x)=xf(1).
\end{align*}

Therefore, this is the only function to satisfy the given equation.

Great problem! I found a two liner solution lol
Our preliminary linear solution checks yield $f(x)=kx$. We claim that these are the only solutions. Note that $$f(0)=0,f(x^2)=xf(x)\implies f(x^2-y^2)+f(y^2)=f(x^2),f(0)=xf(x)+xf(-x)\implies f(x)=-f(-x)$$ $$4f(x)=f(4x)=f((x+1)^2-(x-1)^2)=(x+1)f(x+1)-(x-1)f(x-1)=2xf(1)+2f(x)\iff f(x)=xf(1),$$ as claimed (k=f(1)). $\blacksquare$

We claim the answer is $f(x)=cx$ for all real $c$, which clearly works, so we know prove that these are the only solutions.

We can say that:
\[f(x^2)=xf(x),\]
so:
\[f(x^2-y^2)=f(x^2)-f(y^2).\]

Also, $f(y^2)=-f(-y^2)$, so:
\[f(x^2+(-)y^2)=f(x^2)+f(-y^2),\]
implying that $f$ is additive.

We can also say:
\[f((x+1)^2)=f(x^2)+2f(x)+f(1)=xf(x)+xf(1)+f(x)+f(1),\]
\[f(x)=xf(1),\]
as desired.

Plugging in $(x,y)=(0,0)$ gives us $f(0)=0$. Plugging in $(x,y)=(x,-x)$ gives us that $f$ is odd. Plugging in $(x,y)=(x,0)$ gives us $f(x^2)=xf(x)$.

Now, we can use that substitution to get

\[f(x^2-y^2)=f(x^2)-f(y^2)\]
\[\implies f(x^2+y^2)=f(x^2)+f(y^2)\]

so $f$ is additive. Then,

\[f((x+1)^2)=(x+1)f(x+1)\]
\[\implies f(x^2)+2f(x)+f(1)=(x+2)f(x)+f(1)=(x+1)f(x)+(x+1)f(1)\]

meaning $f(x)=f(1)x$, or $f$ is linear. Thus our answer is $\boxed{f(x)=cx, \ c \in \mathbb{R}}$

[u][b][color=#3D85C6]Claim[/color][/b][/u]The only solutions are $f(x) = cx$ , $c \in \mathbb{R}$

Let $P(x,y)$ denote the assertion.
From, $P(x,0)$ and $P(0,x)$ we have: $f(x^2) = xf(x)$ and $f(-x^2) = -xf(x)$
\newline
$\implies f$ is odd and f(0) = 0
From this we now have $f(x^2-y^2)+f(y^2) = f(x^2)$
\newline
This implies that $f$ is an additive function.
Now make the substitution $P(x+1,y)$, obtaining:
\begin{align*}
   f((x+1)^2) &= (x+1)f()x+1 \\
    \implies f(x^2) + 2f(x) + f(1) &= (x+1)f(x) + (x+1)f(1) \\
    \implies f(x) = xf(1)
\end{align*}
From this the conclusion is clear

Note that $f$ is an odd function, as
\[f(x^2-y^2)=xf(x)-yf(y)=-(yf(y)-xf(x))=-f(y^2-x^2).\]

Substituting $y=0$, we find $f(x^2)=xf(x)$. Hence $f$ must be additive as
\[f(x^2-y^2)=xf(x)-yf(y)=f(x^2)+f(-y^2).\]

Now if we plug in $\left(\frac{t+1}{2}, \frac{t-1}{2}\right)$, using the fact that $f\left(\frac{t-1}{2}\right) = f\left(\frac{t+1}{2}\right)-f(1)$, we get
\begin{align*}
f(t) &= \frac{t+1}{2} \cdot f\left(\frac{t+1}{2}\right) - \frac{t-1}{2} \cdot f\left(\frac{t-1}{2}\right) \\
&= \left(\frac{t+1}{2} - \frac{t-1}{2}\right) f\left(\frac{t+1}{2}\right) +f(1) \cdot \frac{t-1}{2} \\
&= \frac 12 \left(f(t+1)-f(1)+t \cdot f(1)\right) \\
&= \frac 12 \left(f(t)+t \cdot f(1)\right)
\end{align*}

Rearranging, we find $f(t)=t \cdot f(1) \implies \boxed{f(x)=cx, \quad c \in \mathbb{R}}$, with clearly works. $\blacksquare$

Taking $y=0$ gives $f(x^2)=xf(x)$. Swapping $x$ with $y$ and setting $y=0$ gives $f(-x^2)=-xf(x)$, which implies $f(x^2)=-f(-x^2)$, and thus $f$ is odd. Therefore, $f(0)=0$.

Since $f(y^2)=-f(-y^2)=yf(y)$, we may rewrite the given as
\[f(x^2-y^2)=f(x^2)-f(y^2)\Longleftrightarrow f(x^2+y^2)=f(x^2)+f(y^2)\]
therefore $f$ is additive. Plugging in $x=a+1$ to $f(x^2)=xf(x)$, so \[f(a^2)+2f(a) + f(a) = (a+1)f(a)+(a+1)f(1),\] thus $f(a)=af(1)$, and so $f$ is linear. This gives the solutions to be of the form of $kx$ for all $k$, which obviously work. $\square$

$f\equiv 0$ and $f \equiv kx$ work.

$P(0,0) \implies f(0) = 0$.

$P(x,0) \implies f(x^2) = xf(x)$ and $P(-x,0)\implies f(x^2) = -xf(-x)$. Thus we can combine these to get $f(-x) = -f(x)$ for all $x\neq 0$, that is $f$ is odd.

Now,
\begin{align*}
    &f(x^2 - y^2) = xf(x) - yf(y) = f(x^2) - f(y^2)\\
    \implies &f(x^2 (-y^2)) = f(x^2) + f(-y^2)\\
    \implies &f(a+b) = f(a) + f(b) \; \forall a\in\mathbb R^+, b\in\mathbb R^-
.\end{align*}

Now $f(x+y) = f(x) + f(y)$ is clearly true if either of them is $0$. Otherwise if $a\in\mathbb R^+$ and $b\in \mathbb R^-$, we then also have $f(a+b) = f(a) + f(b)$.

Now first we show that $f(a+b) = f(a) + f(b)$ for all $a,b\in \mathbb R^+$. Then we just use the odd property of the function to derive that $f(a+b) = f(a) + f(b)$ for all $a,b\in \mathbb R^-$. We will then combine all these information to conclude that $f$ is Cauchy. So first let's prove $f(a+b) = f(a) + f(b)$ for all $a,b\in \mathbb R^+$.

Pick some $c\in \mathbb R^+$ such that $c>b$.

So,
\begin{align*}
    f(a+b) &= f((a+c) + (b-c)) = f(a+c) + f(b-c)\\
    &=f(a+c)+f(b+(-c))\\
    &=f(a+c) + f(b) + f(-c)\\
    &=[f(a+c) + f(-c)] + f(b)\\
    &=f((a+c) + (-c)) + f(b)\\
    &=f(a) + f(b)
.\end{align*}

Thus we now conclude that $f$ is indeed Cauchy!

Now note that we have $f(x^2) = xf(x)$.

So, $f((a+b)^2) = (a+b)f(a+b)\implies af(a) + f(2ab) + bf(b) = af(a) + af(b) + bf(a) + bf(b)$ from which we get $f(ab) + f(ab) = af(b) + bf(a)$.

Now set $b=1$ to get $f(a) + f(a) = af(1) + f(a)\implies f(a) = af(1) = ka$ and we are done. :yoda:

plug in $y=0$ to get $f(x^2)=xf(x)$, and $x=0$ to get $f(-y^2)=-yf(y)$
we can easily get that $f(0)=0$, and $f(a)=-f(-a)$
plug these into the original equation for $f(x^2-y^2)=f(x^2)-f(y^2)$ and $f(a-b)=f(a)-f(b)$
we then get that $f(x)$ is linear with $f(0)=0$, so $f(x)=kx$

Plug in $(x,0)$ to get that $f(x^2)=xf(x)$
So $f(x^2-y^2)=f(x^2)-f(y^2)$
Plug in $(-x,0)$ to get that $f(x^2)=-xf(-x)=xf(x)$, so $f(x)=-f(-x)$ and $f$ is odd
So $f(x^2)-f(y^2)=f(x^2)+f(-y^2)=f(x^2-y^2)$, so $f$ is additive for a nonnegative and a nonpositive.
Let $x,y=a,b$ where $a \le b$
$f(x^2-y^2)+f(2y^2)=f(x^2+y^2)$ because $x^2-y^2$ is nonpositive but $2y^2$ is nonnegative.
$f(x^2+y^2)=f(2y^2)+f(x^2)-f(y^2)=f(2y^2)+f(-y^2)+f(x^2)=f(y^2)+f(x^2)$, so $f$ is additive for two nonnegatives. Because $f$ is even, $f$ is also additive for two nonpositives. So $f$ is additive.
Because $f(x^2)=xf(x)$, $af(a)+f(2a)+f(1)=f(a^2+2a+1)=(a+1)f(a+1)=af(a+1)+f(a+1)=(a+1)f(a)+(a+1)f(1)$
$f(a)+af(1)=f(2a)$
$f(2a)-f(a)=af(1)$
$f(a)=af(1)$, so the answer is $\boxed{f(x)=kx}$


inequalities
trigonometry
calculus
calculus computations

Here is a inequality I have recently done. I want to see if anyone has any other solution.

Prove that $ \forall x\in\left[0,\sqrt{\frac{\pi}{2}}\right]$ that $ \sin^2(x)\le\sin\left(x^2\right)$

[quote="IndoChina"]Guess it's a better way to start off by showing others what you have done.[/quote]

I'm sorry. I don't think I correctly communicated what I meant. I have already done his problem. I am posting it for fun. I'd like to see how many solutions people come up with.


ARML
geometry
rhombus
trigonometry
analytic geometry
algebra
polynomial

Time limit: 10 minutes per pair.

1.) If $ x \equal{} 2008(a \minus{} b)$, $ y \equal{} 2008(b \minus{} c)$, and $ z \equal{} 2008(c \minus{} a)$, compute the numerical value of $ \frac {x^2 \plus{} y^2 \plus{} z^2}{xy \plus{} yz \plus{} xz}$, given that $ xy \plus{} yz \plus{} xz \neq 0$.

2.) The diagonals of rhombus ABCD are 10 and 4. Rhombus CDEF is drawn in the same plane, with DE perpendicular to DA. Compute the area of rhombus CDEF.

[hide="Number 1"]
Every individual term will have a coefficient of $ 2008^2$, so we can just cancel out the $ 2008^2$ from the top and bottom.  We are left with $ \frac{(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2}{(a\minus{}b)(b\minus{}c)\plus{}(b\minus{}c)(c\minus{}a)\plus{}(a\minus{}b)(c\minus{}a)} \equal{}$ $ \frac{2(a^2\plus{}b^2\plus{}c^2 \minus{}ab\minus{}bc\minus{}ca)}{\minus{}a^2\minus{}b^2\minus{}c^2\plus{}ab\plus{}bc\plus{}ca} \equal{} \boxed{\minus{}2}$
[/hide]

[hide="Number 2"]
The area of a rhombus is given by $ s^2\sin \theta$, where $ s$ is the side length, and $ \theta$ is an internal angle.  We know $ s\equal{}\sqrt{29}$ so all we need to find is $ \theta$.

Let $ \angle ADC\equal{}\theta$ and so $ \angle CDE \equal{} 90\minus{}\theta$
Use Law of Cosines on triangle ADC to get $ \cos \theta \equal{} \frac{21}{29}$
Since $ \cos \theta \equal{} \sin (90\minus{}\theta)$, $ \sin (90\minus{}\theta) \equal{} \frac{21}{29}$

So plug into the area formula and get
$ \sqrt{29}^2\cdot\frac{21}{29} \equal{} \boxed{21}$
[/hide]

lol the first one all you had to do was see $ x\plus{}y\equal{}\minus{}z$

ARML '94: Regular dodecagon $ P_1, P_2, .... P_{10}$. is drawn in the coordinate plane with $ P_1\equal{}(1,0)$ and $ P_6\equal{}(3,0)$. If $ P_n\equal{}(x_n,y_n)$, compute:

$ \prod_{k\equal{}1}^10 (x_k\plus{}iy_k)$

Hm awfully similar to #3 on the iTest ToC Round 2. I'm assuming you meant $ \displaystyle\prod_{k \equal{} 1}^{10}(x_{k} \plus{} iy_{k})$.

[hide]
Note that for a regular decagon centered about the origin, it's vertices can be represented by the 10th roots of unity,
\[ \omega^{10} \equal{} 1 \implies \omega^{10} \minus{} 1 \equal{} 0
\]
In this case, we want to center it about $ (2,0)$, so apply a shift two units to the right as
\[ (\omega \minus{} 2)^{10} \minus{} 1 \equal{} 0
\]
The solutions to the above equation correspond to the vertices of the desired decagon. We wish to find to find the product of these vertices in the complex plane, so by Vieta, we have
\[ \displaystyle\prod_{k \equal{} 1}^{10}(x_{k} \plus{} iy_{k}) \equal{} 2^{10} \minus{} 1 \equal{} \boxed{1023}
\]

[/hide]

Edit: Decagon, not dodecagon  :blush:

I think you meant decagon, not dodecagon... (and 10th roots of unity, not 12th roots of unity... but the equation was right.)

Anyways...

[b]Problem 4:[/b]
The sides of a non-right isosceles triangle are $ \sin x$, $ \cos x$, and $ \tan x$.  Compute $ \sin x$.

[b]Problem 5:[/b]
A strictly increasing sequence of [i]positive[/i] integers contains the terms 305 and 1990.  If each term after the second is equal to the sum of the two previous terms, compute the least possible value for the first term.

Problem 4:
Since the problem says that the triangle is isosceles, two of the sides must be congruent.
Let's say that sinx=tanx.
This means that sinxcosx=sinx
cosx=1. But whenever cosx=1, sinx=0 and this cannot be true because then the triangle would be degenerate.
Therefore sinx=cosx when x=45. This means that sinx= sqrt(2)/2.
problem5:
we call the term before 305, x.
x,305,305+x,610+x,915+2x,1525+3x.
1525+3x=1990.
x=155.
Going backwards from 305 we get.
305,155,150,5
So 5 is the answer.

Actually, for problem 4, if $x=45^\circ$, then the triangle is a right triangle. Hence, we instead have

$\cos x=\tan x$

$\implies \cos^2x=\sin x \implies \sin^2x+\sin x-1=0$.

Using the Quadratic Formula, we find that

$\sin x=\frac{-1\pm\sqrt{5}}{2}$.

However, $\frac{-1-\sqrt{5}}{2}<-1$, so the answer is

$\sin x=\boxed{\frac{-1+\sqrt{5}}{2}}$.

Sorry, you are totally right. I never saw the part where it said the triangle could not be right.

I will restart this marathon.
So I guess I will post a problem:
A number is called reversable if it is reversed and is still a prime. N is a two-digit number. What is the maximum possible value of N is N,4N,5N are all reversible primes.
The person who solves this easy problem, please post a new problem.

It is not really a hard problem, just try playing around with a few two digit numbers.


trigonometry
geometry
cyclic quadrilateral
geometry solved

Let us have three concurrent circles w1, w2, and w3. Let the point of concurrence
be P. Let the intersection of w1 and w2 be A, the intersection of w2 and w3 be B, and the other
intersection be C. Select a point X on w1. Let the intersection of XA and w2 be Y . Let the
intersection of YB and XC be Z. Prove that:
(a) Z lies on w3.
(b) XY/XZ is a constant.
enjoy!

a) We can solve the first part by viewing the problem from a different perspective....
Let Z be the intersection of the line YB with w3 join ZC and XC its enough to proove that X,C,Z are collinear or <XCZ= \pi ...

In the cyclic quadrilateral XAPC <XAP+<XCP= \pi ....(#)
Angles XAP & PAY form a linear pair.....(*)....

Repeating (#) & (*) cyclicly on quadrilaterals AYBP and CZBP....

We get <XCP+<PCZ= (\pi-<XAP)+( \pi -<PAY)= \pi (From # & *)

Hence A,C,Z are collinear

b) In the given scenerio points A,B,C,P are fixed. Hence
<CPA,<APB,<CPB are constant. Let <CXA=x,<AYB=y,<CZB=z which are constant as well...(since they form linear pairs with the respective former angles).

Hence in triangle XYZ using the sine rule we have
XY/sinz = YZ/sinx = XZ/siny

Hence XY/XZ = sinz/siny is a constant.
We also have XY/YZ = sinz/sinx a constant as well.


Wouldn't it be more efficient if the square root of a number (when using radicals) included both the primary and nonprimary roots?  And if one sought the primary square root, simply put an absolute value sign around it?  Because this confuses me:

((-3)^2)^(1/2)
Combining the exponents, one gets (-3^1)=-3, but each part done individually, you get 3 as your answer.

Another example:  My calculator can't do (-512)^(2/3), but there is obviously a solution.

I realize that if the notation was changed, exponential graphs would be very, uh, weird.  But I think this would be something interesting to look into.

[hide]
Well, in your first question, it says $\sqrt{(-3)^2}$. After all, $(-3)^2=(3)^2$, and put it in then it's done...
[/hide]


Hello everyone, got a “sub-problems” teaser for ya. Be sure to show which fractions you used, don’t worry about showing how you calculated them. Enjoy!

1.	“Reading Rate Pt. 1”

	If you read 15 minutes per day every day and end up reading 12 books of 200 pages each in 1 year, what is your reading rate in pages per minute?

2.	“Reading Rate Pt. 2’

	If you increase your reading speed so that each page takes you 30 seconds less than it did before and you begin reading 20 minutes per day, how many 200-page books can you now read in a year?

Problem 1:

[hide]First you have 12 X 200 = 2400 pages in a year
2400 / 365 = 6.575342466 pages a day
since he reads only 15 minutes a day : 6.575342466 / 15
                                                       = 0.438356164 pages a minute

so it's roughly 0.44 pages a per minute[/hide]

Problem 2:

[hide]
So you can read $.44$ pages a minute.  Take the inverse, this means it takes you $2.27$ minutes to read a page.  Subtract $.5$ and you get $1.77$ minutes a page.  Take the inverse again and we get $.56$ pages per minute.  Multiply by $15$ and we get $8.47$ pages a day.  Multiply by $365$ (or $366$ if its a leap year) and get $3093$.  Divide by $200$ and we get $\boxed{15.46}$ books.
[/hide]


[quote="Ulanbek_Kyzylorda KTL"]Natural number n>1 is that decimal part of number 9997n consists only odd numbers. Prove that
 minimal possible value of n is 3335.[/quote]

If "decimal part" means the part right of decimal point : if $ n$ is a natural number, so is $ 9997n$ and so decimal part of $ 9997n$ is made of zeroes.

If "decimal part" means the decimal representation, then $ n\equal{}1$ $ \implies$ $ 9997n\equal{}9997$ made of only odd numbers.

[quote="Ulanbek_Kyzylorda KTL"](If "decimal part" means the decimal representation, then    made of only odd numbers.)
n>1[/quote]

Ok, and sorry for having forgotten $ n>1$
 :blush:


geometry
parallelogram
analytic geometry

ABDC is a parallelogram of consecutive vertices in the order written. Find the vertex A, and knowing that B (0,1,3), C (2,3,5) and D (-1,0,2).

hello, it must be $ \vec{AB}\equal{}\vec{DC}$ and from here we get $ 0\minus{}x_A\equal{}3$ and  $ 1\minus{}y_A\equal{}3$ and $ 3\minus{}z_A\equal{}3$ so the point $ A$ has the coordinates $ A(\minus{}3,\minus{}2,0)$.
Sonnhard.