trigonometry
geometry
inradius
circumcircle
inequalities
Cauchy Inequality
inequalities proposed

Let $ ABC$ be a triangle .Prove that :
$ \frac{2r}{R} \le \frac{8 abc} {(a\plus{}b)(a\plus{}c)(b\plus{}c)}$,where r and R are  its inradius and circumradius,respectively.

[quote="soruz"]Let $ ABC$ be a triangle .Prove that :
$ \frac {2r}{R} \le \frac {8 abc} {(a \plus{} b)(a \plus{} c)(b \plus{} c)}$,where r and R are  its inradius and circumradius,respectively.[/quote]
Let $ b\plus{}c\minus{}a\equal{}x,$ $ a\plus{}c\minus{}b\equal{}y$ and $ a\plus{}b\minus{}c\equal{}z.$ Then
$ \frac {2r}{R} \leq \frac {8 abc} {(a \plus{} b)(a \plus{} c)(b \plus{} c)}\Leftrightarrow$
$ \Leftrightarrow(a\plus{}b)(a\plus{}c)(b\plus{}c)(a\plus{}b\minus{}c)(a\plus{}c\minus{}b)(b\plus{}c\minus{}a)\leq8a^2b^2c^2\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(x^4y^2\plus{}x^4z^2\plus{}2x^3y^3\minus{}x^3y^2z\minus{}x^3z^2y\minus{}2x^2y^2z^2)\geq0.$

[quote="dduclam"][quote="soruz"]Let $ ABC$ be a triangle .Prove that :
$ \frac {2r}{R} \leq \frac {8 abc} {(a \plus{} b)(a \plus{} c)(b \plus{} c)}$,where r and R are  its inradius and circumradius,respectively.[/quote]
Note that $ R\geq 2r$ and $ (a \plus{} b)(a \plus{} c)(b \plus{} c) \geq 2\sqrt {ab}.2\sqrt {bc}.2\sqrt {ca}$ by AM-GM :wink:[/quote]
Are you sure that it helps? :wink:

[quote="soruz"]Let $ ABC$ be a triangle .Prove that :
$ \frac {2r}{R} \le \frac {8 abc} {(a \plus{} b)(a \plus{} c)(b \plus{} c)}$,where r and R are  its inradius and circumradius,respectively.[/quote]
We have:
$ \frac {4abc} {(a \plus{} b)(a \plus{} c)(b \plus{} c)} \equal{} \frac {8Rr}{p^2 \plus{} r^2 \plus{} 2Rr} \ge \frac {r}{R}$
$ \Leftrightarrow 8R^2 \ge p^2 \plus{} r^2 \plus{} 2Rr$
By Gerretsen inequality, we have:
$ p^2 \plus{} r^2 \plus{} 2Rr \le 4R^2 \plus{} 4Rr \plus{} 3r^2 \plus{} r^2 \plus{} 2Rr \equal{} 4R^2 \plus{} 6Rr \plus{} 4r^2 \le 8R^2$(By Cauchy inequality $ R \ge 2R$)

Let $ ABC$ be a triangle .Prove that :
$ \sqrt \frac{2r}{R} \le \frac{8 \sqrt {abc}} {(\sqrt a\plus{}\sqrt b)(\sqrt b\plus{}\sqrt c)(\sqrt c\plus{}\sqrt a)}$,where r and R are  its inradius and circumradius,respectively.

Let $ ABC$ be a triangle and $ t \in [0,1]$.Prove that :
$ (\frac{2r}{R})^t \le \frac{8 (abc)^t} {(a^t\plus{} b^t)(b^t\plus{}c^t)(c^t\plus{}a^t)}$,where r and R are  its inradius and circumradius,respectively.

[quote="soruz"]Let $ ABC$ be a triangle and $ t \in [0,1]$.Prove that :
$ (\frac {2r}{R})^t \le \frac {8 (abc)^t} {(a^t + b^t)(b^t + c^t)(c^t + a^t)}$,where r and R are  its inradius and circumradius,respectively.[/quote]
By Jensen inequality, we have:
$ a^t+b^t \le 2(\frac{a+b}{2})^t$
$ b^t+c^t \le 2(\frac{b+c}{2})^t$
$ c^t+a^t \le 2(\frac{c+a}{2})^t$
It follows that:
${ (a^t + b^t)(b^t + c^t)(c^t + a^t)} \le 8(\frac{(a+b)(b+c)(c+a)}{8})^t$
So we need to prove:
$ \frac{(abc)^t}{(\frac{(a+b)(b+c)(c+a)}{8})^t} \ge (\frac {2r}{R})^t$
Or:
$ \frac{8abc}{(a+b)(b+c)(c+a)} \ge \frac{2r}{R}$

[quote="soruz"]Let $ ABC$ be a triangle .Prove that :
$ \frac {2r}{R} \le \frac {8 abc} {(a \plus{} b)(a \plus{} c)(b \plus{} c)}$,where r and R are  its inradius and circumradius,respectively.[/quote]
Stronger Inequality: Let $ ABC$ be a triangle .Prove that : $ \frac{r}{2R} \leq \frac{abc}{\sqrt{2(a^2\plus{}b^2)(b^2\plus{}c^2)(c^2\plus{}a^2)}}$,where $ r$ and $ R$ are its inradius and circumradius,respectively.


[I posted this file in November 2004, but I think it got lost during the server crash around that time.  So I am reposting.]

Here is a collection of 31 olympiad-level problems that can be solved with probabilistic methods.  At the end of the document, I provide a brief hint for solving each problem.

If you are interesting in seeing the full solution to any problem, post the problem in the appropriate subforum, and hopefully someone will help you out.

Comments are welcome.  Enjoy!

Wow.  I can do about 5 of these.  Where should I post for solutions to problems such as number 13?

The solution to Problem 13 happens to be in a thread called [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6653]Challenge Problem 2[/url], back in the earliest days of AoPS!

For other problems, you might want to post them in the Olympiad section, such as in the Combinatorics Unsolved Problems forum.

Where can I learn more about this probabilistic method?

I recommend the book [url=http://www.cs.nyu.edu/faculty/spencer/nogabook/nogabook.html][i]The Probabilistic Method[/i][/url], by Noga Alon and Joel Spencer.  You can find the first chapter of their book at that link.

I added 7 problems to my list of problems solvable with the probabilistic method.  Below is the updated document.


how has AMSP changed your performance on math contests?

me i haven't really gotten far enough.  we find out tomorrow.  (in 19.5 hours)

and then again in 43.5 hours.

yay.

"Tomorrow" refers to the USAMO, correct?  You're already farther than I am (2 on the AIME :wallbash: )

I'll see if it helped next year (I'm hoping to actually make USAMO [and preferably get at least a seven :maybe: ]).


algebra
system of equations
algebra proposed

solve for positive reals a,b,c,d,e;such that these reals satisfy the
given system of equations
                            a+b=c^2,
                            b+c =d^2,
                            c+d=e^2,
                            d+e=a^2,
                            e+a=b^2;
 :roll:

very easy :P
simply use cauchy-schwartz ineq :wink:

vas2007, your posting is useless.

"Very easy" doesn't help and you showed ZERO steps.

$\mu=\min\{a,b,c,d,e\}>0$, then $\mu+\mu\leq \mu^{2}$ and $a,b,c,d,e\geq \mu\geq 2$.

what u all have done is absolute nonsense :P
 a,b,c,d,e are not positive integers, they are +ve reals :bruce:
  i don't understand what made u think like that  :censored:

Summing up all the ineqs. we have
                       [color=indigo]$2[a+b+c+d+e]=a^{2}+b^{2}+c^{2}+d^{2}+e^{2}$[/color]
                            [color=red]by Cauchy it's evident that [/color]
[color=olive][i]$a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\leq 20$[/i][/color]

         FROM THE GIVEN EQNS.,WE HAVE
                   [color=orange]$c^{2}\geq 2[ab]^{1/2}$[/color]
                      and similarly  other ineqs. and multiplying we get

                   [color=green]$abcde\geq 32$[/color]
         BY am-gm ineq we have $b^{2}+c^{2}+d^{2}+e^{2}+a^{2}\geq 20$

[quote="vas2007"]Summing up all the ineqs. we have
                       [color=indigo]2[a+b+c+d+e]=a^2+b^2+c^2+d^2+e^2[/color]
                            [color=red]by Cauchy it's evident that [/color]
[color=olive][i]a^2+b^2+c^2+d^2+e^2<=20[/i][/color]

         FROM THE GIVEN EQNS.,WE HAVE
                   [color=orange]c^2>=2(ab)^.5[/color]
                      and similarly  other ineqs. and multiplying we get

                   [color=green]abcde>=32[/color]
         BY am-gm ineq we have [color=darkred]sigma[/color] a^2>=20[/quote]
I think your solution is more complicated than mine.

[color=red]HIKARU UR SOLN. IS ABSOLUTE NONSENSE U'VE TAKEN $a,b,c,d,e; as+ve integers$ they're only +ve reals [/color]

[quote="hikaru123"]$a,b,c,d,e$ are positive integers so $a^{2}\ge2$
$a\ge2$, the same for the others.
we also have
$2(a+b+c+d+e)=a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\ge \frac{(a+b+c+d+e)^{2}}{5}$ (by Cauchy-Schwarz ineq)
$10\ge a+b+c+d+e$
Then there's only one solution $a=b=c=d=e=2$[/quote]

   Who told u that  $a,b,c,d,e$ $\in$ $Z^{+}$  :?:

[quote="vashist"]solve for positive reals a,b,c,d,e;such that these reals satisfy the
given system of equations
                          $a+b=c^{2}$,
                          $b+c =d^{2}$,
                          $c+d=e^{2}$,
                          $d+e=a^{2}$,
                          $e+a=b^{2}$;
 :roll:[/quote]

Leave it already, guys.
hikaru123 read the problem wrong. So what?
His argument was still correct, and with my above post
I completed the solution long before you came here shouting.

I'm so sorry about my madness. :wallbash:  You have a right solution, vas2007.


is computing the base numbers first, then adding, then transfering to a different base be the easiest way to do this

could u also explain what u write for numbers above 9 for something like base 16 (isn't a number then a letter or something?)

-jorian

[quote="13375P34K43V312"][hide]Convert to base 16:

1B+177=192.[/hide][/quote]

This solution requires some explaining.  Can anyone discover a fast way of converting these expressions to base $16$?

[hide="Hint"] For a concrete example, given the binary expression $1001_{2}$, we can write this as

$1 \cdot 2^{3}+0 \cdot 2^{2}+0 \cdot 2^{1}+1 \cdot 2^{0}= (1 \cdot 2^{1}+0 \cdot 2^{0}) \cdot 4^{1}+(0 \cdot 2^{1}+1 \cdot 2^{0}) \cdot 4^{0}= 21_{4}$ [/hide]

[hide="click me!!!"]Note that $4^{2}=16$, thus, each pair of digits can be condensed into one for base 16 (multiply the first digit of each pair by $4^{1}$, then add the second digit).

$123_{4}=1B_{16}$

Base 8 is a bit trickier.  Convert to base 2, then to base 16.

To convert to base two, simply expand each digit into 2.

$567_{8}=101110111_{2}$

Now, to convert to base 16, each group of 4 makes 1 digit ($2^{3}$ times first digit, $2^{2}$ times second, $2^{1}$ times third, $4^{0}$ the fourth).

$101110111_{2}=177_{16}$

$1B+177=192_{16}$

Is this good enough explanation?

NOTE:  it takes less time than it looks.[/hide]

Is there a way to use the fact that converting from binary to base 8 you would use partitions of 3 digits, and converting from binary to base 16, you use partitions of 4 digits?  You can't take $\frac{4}{3}$ of a partition very well, but it seems like that might lead to a way to do it directly...

Not really, no.  The partitions from binary to base $16$ are different from the partitions from binary to base $8$.

Yeah, I guess you have to go back to binary to figure out one digit of a partition anyway...


Prove that for all a,b,c,d>0 with a^2+b^2+c^2+d^2=1 we have
sum a^2/(bc+cd+bd)>=8/(3+48abcd).

Ok, here is my solution:
  Apply Cauchy to find that sum a^2/(bc+cd+bd)>=(a^2+b^2+c^2+d^2)^2/(sum a^2(bc+cd+db)). So we have to prove that sum a^2(bc+cd+db)<=6abcd+3/8. Observe that sum a^2(bc+cd+db)=sum (ab(c^2+d^2)+cd(a^2+b^2)). We prove now that 2ab(c^2+d^2)+2cd(a^2+b^2)<=4abcd+1/4. Adding up these inequalities the problem will be solved. So, let's prove that 2ab(c^2+d^2)+2cd(a^2+b^2)<=4abcd+1/4, which can be written 2(bc+ad)(ac+bd)<=1/4+4abcd. But 2(bc+ad)(ac+bd)<=(bc+ad)^2+(ac+bd)^2=(a^2+b^2)(c^2+d^2)+4abcd<=1/4+4abcd.

Congratulation, Harazi, your solution is beautiful.

Thank you very much.


We can set $P(x)-2^x=a(x-1)(x-2)\cdots\cdots(x-n)\ (a\neq 0)$.
Therfore $P(n+1)-2^{n+1}=an(n-1)(n-2)\cdots\cdots2\cdot1=an!=0$? :?

But $P(x)-2^x$ isn't really a polynomial, is it? :)


induction
number theory unsolved
number theory

The sequence $ (F_n)$ of Fibonacci numbers satisfies $ F_1 \equal{} 1, F_2 \equal{} 1$ and $ F_n \equal{} F_{n\minus{}1} \plus{}F_{n\minus{}2}$ for all $ n \ge 3$. Find all pairs of positive integers $ (m, n)$, such that $ F_m . F_n \equal{} mn$.

lemma:
we have $ F_n > n$ for $ n > 5$ ,and $ F_n > 2n$ for $ n > 7$
prof (by induction ).
wlog suppose that $ n \le 5$ .
if $ n \equal{} 1$ then $ m \equal{} 5$.
$ n \equal{} 2 ,F_m \equal{} 2m$ (no solution  from $ F_m > 2m$ for $ m > 7$).
$ n \equal{} 3,2F_m \equal{} 3m$ (no solution  from $ F_m > 2m$ for $ m > 7$).
$ n \equal{} 4,3F_m \equal{} 4m$ (no solution  from $ F_m > 2m$ for $ m > 7$).
$ n \equal{} 5$ then $ m \equal{} 5$ or $ 1$.


function
calculus
derivative
calculus computations

Prove that there doesn't exist function $ f(x)$ sutch that $ f'(x)=sgn(x)$.

say such a function $ f(x)$ exists
well the given cond.s imply
$ f'(x)=1,x>0$
$ f'(x)=-1,x<0$
$ f'(x)=0,x=0$
the first condition implies $ f(x)=x+c,x>0$
the second implies $ f(x)=-x+b,x<0$

well getting the left and right derivative we clearly see no matter what $ a,b$ are LHD $ \neq$ RHD

Assume such a function exists. Since the derivative of any differentiable function verifies the Darboux property, the contradiction follows.


[quote="M.A"]I'm sorry here is the correct problem
\[ 45 \plus{} (2 \plus{} \frac {c^2}{ab})(2 \plus{} \frac {b^2}{ac})(2 \plus{} \frac {a^2}{cb})\ge 8(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})
\]
[/quote]
This problem is also wrong.  :wink:

The inequality equalivents
\[ \Leftrightarrow \sum {\left( {a^2 \left( {ab \plus{} bc \plus{} ca} \right) \plus{} abc\left( {a \plus{} b \plus{} c} \right) \minus{} 8a^2 bc} \right)\left( {b \minus{} c} \right)^2 }  \ge 0
\]
Let $ a\equal{}b$ then the inequality is true only with the condition $ a^2  \plus{} c^2  \ge 4ac$.  :lol:

[quote="Honey_S"][quote="M.A"]I'm sorry here is the correct problem
\[ 45 \plus{} (2 \plus{} \frac {c^2}{ab})(2 \plus{} \frac {b^2}{ac})(2 \plus{} \frac {a^2}{cb})\ge 8(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})
\]
[/quote]
This problem is also wrong.  :wink:

The inequality equalivents
\[ \Leftrightarrow \sum {\left( {a^2 \left( {ab \plus{} bc \plus{} ca} \right) \plus{} abc\left( {a \plus{} b \plus{} c} \right) \minus{} 8a^2 bc} \right)\left( {b \minus{} c} \right)^2 } \ge 0
\]
[/quote]
I have obtained $ 45 \plus{} \left(2 \plus{} \frac {c^2}{ab}\right)\left(2 \plus{} \frac {b^2}{ac}\right)\left(2 \plus{} \frac {a^2}{cb}\right)\ge 8(a \plus{} b \plus{} c)\left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)\Leftrightarrow$
$ \Leftrightarrow\sum_{sym}(2a^4bc\plus{}a^3b^3\minus{}8a^3b^2c\plus{}5a^2b^2c^2)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2c((a\plus{}b)c^2\plus{}2(a\plus{}b)ab\minus{}5abc)\geq0,$ which is true. :maybe:


calculus
integration
calculus computations

How do you "show" that
\[\int_{0}^{1}\int_{0}^{\sqrt{x}}\frac{2xy}{1-y^{4}}dydx=\int_{0}^{1}\int_{y^{2}}^{1}\frac{2xy}{1-y^{4}}dxdy \]
I mean how is the integration limits changed like that?

Consider the region you are integrating over: compare the region $R_{1}$ satisfying $0 \le x \le 1$ and $0 \le y \le \sqrt{x}$ to the region $R_{2}$ satisfying $0 \le y\le 1$ and $y^{2}\le x \le 1$.


this is not the right forum to post calculus

this is trig, not calclulus, but this is borderline difficulty, i guess it should be in intermediate

This problem actually only uses the identity $\sin^2x+\cos^2x=1$, which is covered in AoPS book 1, although the problem itself looks like one in volume 2.

[quote="ch1n353ch3s54a1l"][quote="anonymous_poster"]x = 1.54(1 - cos Θ) + 2sin² Θ

find Θ, 0° < Θ < 360°, if x = 2

im not too bright, so i need help solving this problem. TIA[/quote]

[hide]$x=1.54(1-\cos \theta)+2\sin^2 \theta\\ 2=1.54(1-\cos \theta)+2\sin^2 \theta\\ 2-2\sin^2 \theta=1.54(1-\cos \theta)\\ 1-\sin^2 \theta=.77(1-\cos \theta)\\ \cos^2 \theta=.77(1-\cos \theta).$
If $y=\cos \theta$, then... yeah.
$\boxed{x\approx32.844}$[/hide][/quote]
im confused. where'd the y come from and u meant to put theta in place of x right?

The y was an introduced variable so that the equation, instead of looking like $\cos^2\theta=.77(1-\cos\theta)$, looks like the nicer $y^2=.77(1-y)$, which is clearly a quadratic equation.

And yes, the x was supposed to be a theta sign.

[quote="236factorial"]$y^2=.77(1-y)$, which is clearly a quadratic equation.[/quote]
im a bit slow, could u rearrange it or at least tell me how one goes around to solving it.

$y^2=.77-.77y$
$y^2+.77y-0.77=0$.

Then use the quadratic formula.

After you find y, you know that $\cos\theta=y$, so use that to find theta.


Hi, I'm stuck on a problem and I would appreciate it very much if you could help me.

You have the transformation $Tx = \alpha x+\beta$ from the real line to the real line, where $\alpha$ and $\beta$ are real numbers and $\alpha \neq 0$.
Then you have a mapping from sets $E$ onto sets $T(E)$. Denote by $\mu(\mu^{*})$ the Lebesgue measure (outer measure) on the real line.
The first part of the problem is to prove that for any set $E$: $\mu^{*}(T(E)) = \left| \alpha \right|\mu^{*}(E)$.

This I managed to prove, it's the second part that caused me trouble.
Prove that $E$ is Lebesgue-measurable if and only if $T(E)$ is Lebesgue-measurable
How do you prove this?

Note that by bijectivity $T(A\cap E) = T(A)\cap T(E)$ and $T(A-E) = T(A)-T(E)$. Now test if $T(E)$ is $\mu^{*}$-measurable with $T(A)$ (instead of a general set $A$) and use the formula you've already shown.

By the way, how did you do on the exam? I found it quite difficult.

Thanks a lot, I didn't do the exam so I guess I'll probably need help with more problems this summer. This seems to be quite a good forum, are there others that are good too?


Let $a,b,c\in\mathbb{R}\setminus\{0\}$ satysfies following conditions:
\[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} =4 \]
and
\[ \frac{a}{c}+\frac{c}{b}+\frac{b}{a} = 5. \]

What is product of that \[ \left(\frac{a}{b}\right)^3 + \left(\frac{b}{c}\right)^3 + \left(\frac{c}{a}\right)^3 = \]
  sum?


Let $ A\in M_n(K)$ (where $ K$ is a generic field) and $ A$ triangularizable e let $ C(A)\equal{}\{X\in M_n(K) | AX\equal{}XA \}$.

Find basis and dimension of $ C(A)$.

Let $ K[A]\equal{}\{p(A)|\ p(X) \in K[X] \}\equal{}Im(ev_A)$  where $ ev_A: K[X]\rightarrow M_n(K)$ that $ p(X) \rightarrow p(A)$.

Demostrate that:
1) $ K[A] \subseteq C(A)$
2) $ K[A]\equal{}C(C(A))\equal{}\{B\in M_n(K)| \forall X\in C(A)\ BX\equal{}XB \}$.

What is $ C(C(C(A)))$?

For the first question cf. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38036


Exuse me but I don't understand how you determine the true age of participants. Why somebody of age 40 years can't participate saying that he is 16 years old?

It's all based on trust man. What would people aged $40$ gain by competing against high schoolers and lying about their age? :) It sounds pretty ridiculous, doesn't it? :D

[quote="Myth"]Finally, we will know true age :)
I have another question: why do some strong young members not participate in ML Contest? Eh, Grobber?[/quote]That's my question too :)

About the age limit, it's basically designed to encourage young people to participate. If you are 40 and you want to participate, you can obivously lie about it. However I'm not sure if you'll feel to good about it later ;)

[quote="Myth"]Finally, we will know true age[/quote]
How?

As I understand, the participant don't have to prove his age.

I'm 18 now. And I use to think about participation in the MLC.

There are various ways to know your true age :)

[quote="Remike"]I'm 18 now. And I use to think about participation in the MLC.[/quote]You can safely participate. The rule is to prevent Myth or Pierre to win the contest and give younger people change to prove themselves :)

It's not a strict rule, however I have met all the winners until now (places 1,2,3) personally and I would figure it out if you were much more than  20 years older :)

In plus it would be indeed a shame that an older guy participates with younger kids and gets lower scores, especially if it's against regulations. That's all.

[quote="pbornsztein"][quote]
The rule is to prevent Myth or Pierre to win the contest[/quote]

So, why don't you just give a rule that stipulate Myth and I are not allowed to participate  :P

Pierre.[/quote]
Because you can officially change your name, get new documents and then participate.  :lol:

[quote="Myth"]
Because you can officially change your name, get new documents and then participate.  :lol:[/quote]Indeed. Myth is right, you can easily and legally change your name, however you cannot change your age (or if you can, let us know how :D).

I won't tell you how since I can't participate...

Pierre.

Hmm... Forever young Pierre...  :lol:  :lol:  :lol:

[quote="Valentin Vornicu"][quote="Remike"][quote="Valentin Vornicu"]It's not a strict rule, however I have met all the winners until now (places 1,2,3) personally and I would figure it out if you were much more than 20 years older[/quote]
But I'm not going to be a winner of the contest. And even if I am I don't think you will have an opportunity to meet me. We are living in different countries, far from each other.[/quote]I've met Billzhao which lives in a different country, much far from mine than yours :) So anything is possible :)

If you know you obey all the regulations, I suggest you stop thinking about others and start solving the problems :P[/quote]

You met Billzhao!? So it means you came to China just to meet the winner of the MLC and confirm that he is under 20?? :D

Valentin met Billzhao on IMO (he took participation in all last IMOs since 2001, I think).

[quote="billzhao"]Actually, IMO 2004 Greece was the first IMO that I attended. And I did meet Valentin there. Although I didn't recognize him when I first saw him (which was when all the scores were up and everybody was trying to predict the cutoffs).[/quote]

I think Myth was rather referring to Valentin, not you, Yufei.   :lol:

[quote="http://www.mathlinks.ro/Forum/viewtopic.php?t=3683"]The first rule is that anyone (from any country) is free to participate in the contest - all except those of you who are [b]20 YEARS [/b]old in the first day of the contest (that is for the 4th Edition - 11th of October 2004). [/quote]

[quote="Valentin Vornicu"][quote="N.T.TUAN"]I'm 29 year old,can I?[/quote]You can send in solution, but you cannot actually participate. It is for high-school students (or university freshmans at most).[/quote]

Thank you  :(


Let $ ABCD$ be a convex quadrangle, $ P$ the intersection of lines $ AB$ and $ CD$, $ Q$ the intersection of lines $ AD$ and $ BC$ and $ O$ the intersection of diagonals $ AC$ and $ BD$. Show that if $ \angle POQ\equal{} 90^\circ$ then $ PO$ is the bisector of $ \angle AOD$ and $ OQ$ is the bisector of $ \angle AOB$.

It is very easy and clearly if you use harmonis division.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=161310[/url]
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=168866[/url]

In this problem we have:
$ (OB,OC,OP,OQ) \equal{} \minus{} 1$
And $ \angle{POQ} \equal{} 90^0$ then $ PO$ is the bisector of $ \angle AOD$ and $ OQ$ is the bisector of $ \angle AOB$.

PS: If don't use harmonis division we can solve this problem by Menelaus theorem!

Yes, it´s very simple if you use these projective ideas.

First, consider that there are points $ E, F$ such that $ \{E\}\equal{}AC \cap PQ$ and $ \{F\}\equal{}BD \cap PQ$. By Menelaus theorem on $ \triangle PBQ$, line $ CE$ and Ceva theorem on $ \triangle PBQ$, cevians $ BF, QA, PC$, we discover that $ \displaystyle\frac{EP}{EQ}\equal{}\displaystyle\frac{FP}{FQ}$. So, $ E,F$ divide harmonically the segment $ PQ$.
If $ AC\parallel PQ$, we use Ceva theorem on $ \triangle PBQ$, cevians $ BF, QA, PC$ and we use $ \triangle BPQ \sim \triangle BAC$ to discover that $ PF\equal{}QF$, so, again, $ E_\infty,F$ divide harmonically the segment $ PQ$ (here, $ E_\infty$ is the point of infinity corresponding to the lines parallel to $ PQ$).
Hence, we always have $ E,F$ dividing harmonically the segment $ PQ$. So, the lines $ OE, OF$ divide harmonically the lines $ OP, OQ$. Suppose that $ OP \perp OQ$. Choose a line $ r$ parallel to $ OQ$ through $ P$. $ r$ intersects the lines $ OE,OF,OP,OQ$ in the points $ \bar{E}, \bar{F},P,Q_\infty$. And the points $ P,Q_\infty$ should divide harmonically the segment $ \bar{E} \bar{F}$. So, it’s clear that $ \bar{E}P\equal{}\bar{F}P$, which gives us $ \angle \bar{E}OP\equal{}\angle PO\bar{F}$, and that the lines $ OP,OQ$ bissect the lines $ OE,OF$.

i found something wrong in this question.
assume that the question is true. bisectors of angles (APC) and (AQC) are perpendicular. so m(ABC)+m(ADC)=180 and ABCD is circumscribed. also intersection of this bisectors is {O} and m(OAP)=m(OBP) hence triangles OAP and OBP is equal. this gives a contradiction because it must be AD=BC and AB and CD are parallel so they can't intersect.

[quote="anonymous1173"] bisectors of angles (APC) and (AQC) are perpendicular[/quote]

Please read this problem carefully!
$ OP,OQ$ are bisector of $ \angle{AOD},\angle{AOB}$

:blush:  :blush: I'm so sorry. thanks tdl :)


In AoPS Volume 2, they talk about the extended version of the binomial theorem. In other words, the binomial theorem applies to expressions $(x+y)^n$ where n isn't necessarily a positive integer (it only has to be a real number). How do you show that this works, since it's obviously impossible to base the proof on counting (like in the "special case" of the theorem)

I would imagine that most of the proofs utilize calculus of some sorts.

It's not calculus. we just define the coeeficients in a different way


function
limit
logarithms
search
algebra proposed
algebra

Describe the family of continuous solutions satisfying $ f: \mathbb{R}\to \mathbb{R}$
\[ f(ax) \equal{} bf(x)
\]
where $ a,b\in \mathbb{R}^ \plus{}$.  What if we drop the continuity requirement?

$ f$ is uniquely defined by its values on $ [1, a)$, which are arbitrary.  If we desire that $ f$ is continuous, we let its values on $ [1, a)$ be a continuous function $ g(x)$ such that $ \lim_{x \to a} g(x) \equal{} b$.  Otherwise, $ g(x)$ can be chosen arbitrarily.

Basically the right idea, except that $ f$ is not quite uniquely defined by its values on $ [1,a)$, but rather by its values on $ ( \minus{} a, \minus{} 1]\cup [1,a)$ (where it's understood that $ [1,a)$ means $ (a,1]$ if $ a < 1$).  Also, if $ a \equal{} 1\neq b$ the only solution is $ f\equiv 0$, and if $ a \equal{} b \equal{} 1$ then of course $ f$ is arbitrary.

But really I guess I was looking for the following slightly more explicit, though equivalent, expression:
\[ f(x) \equal{} \begin{cases}\phi_ \plus{} (\log x)x^r & x > 0 \\
0 & x \equal{} 0 \\
\phi_ \minus{} (\log |x|)|x|^r & x < 0 \end{cases}
\]
where $ \phi_ \minus{} ,\phi_ \plus{}$ are arbitrary periodic functions with period $ \log a$ and $ r \equal{} \frac {\log b}{\log a}$.

Here's the non-homogeneous generalization: What are the solutions of $ f(ax) \equal{} bf(x) \plus{} g(x)$ where $ g: \mathbb{R}\to \mathbb{R}$ is given, and is the restriction to $ \mathbb{R}$ of an entire function.

Well, sure (and my apologies for the first error).  When $ a \equal{} b$ I gave almost that example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=277003597&t=209029]here[/url].


Show that

\[ \frac{\pi}{2}\equal{}\prod_{k\equal{}1}^{\infty}\frac{4k^{2}}{\left(2k\plus{}1\right)^{2}}\]

$ \prod_{k\equal{}1}^{n}\frac{4k^{2}}{\left(2k\plus{}1\right)^{2}} \equal{} \dfrac{2^{2n}(n!)^2}{((2n\plus{}1)!!)^2} \longrightarrow 0 \ \ (n \rightarrow \infty)$

[b]similar infinite product[/b]
$ \prod_{k\equal{}1}^{\infty} \left( 1 \minus{} \frac{1}{\left(2k\plus{}1\right)^{2}} \right) \equal{}\prod_{k\equal{}1}^{\infty} \frac{2k}{2k\plus{}1} \cdot \frac{2k\plus{}2}{2k\plus{}1} \equal{} \frac{\pi}{4}$
$ \prod_{k\equal{}1}^{\infty} \left( 1 \plus{} \frac{1}{\left(2k\minus{}1\right)\left(2k\plus{}1\right)} \right) \equal{}\prod_{k\equal{}1}^{\infty} \frac{2k}{2k\minus{}1} \cdot \frac{2k}{2k\plus{}1} \equal{} \frac{\pi}{2}$

Thanks, I must have copied it down incorrectly.  How do you prove the products you posted equal $ \pi /4$ and $ \pi /2$, respectively?


how do i prove, from the definition of limit of a sequence, that $lim (( n^{2}-2 ) /(3n^{2}-3 )) = 1/3$? Thanks in advance.

What you want to prove is given any $\epsilon > 0,$ there exists a positive integer $N_{0}$ (which may depend on $\epsilon$) such that $|\frac{n^{2}-2}{3n^{2}-3}-\frac13| < \epsilon$ for all $n > N_{0}.$

In such problems, it pays to work backwards. Write $|\frac{n^{2}-2}{3n^{2}-3}-\frac13| < \epsilon$ and from there, find $n=N_{0}$ which guarantees that the condition noted above is satisfied.

thanks otem and fieryhydra for your replies... however, i just realized i posted the wrong question.. sorry my bad.. the actually question was 'prove $lim ((n^{2}-n)/(3n^{2}-2))=1/3$. But thank you for answering my previous question.. :)

It doesnt really change much though.. :)

$\left|{\frac{{n}^{2}-n}{3\,{n}^{2}-3}}-\frac13 \right| =\frac13\, \frac1{ n+1 \right}<\frac1n$

So you can chose $N$ such that $N>1/\epsilon$..

ottem, i think u used a wrong denominator, i get $\left|{\frac{{x^{2}-x}}{{3x^{2}-2}}-\frac{1}{3}}\right| = \left|{\frac{{3x^{2}-3x-\left({3x^{2}-2}\right)}}{{3\left({3x^{2}-2}\right)}}}\right| = \left|{\frac{{-3x+2}}{{3\left({3x^{2}-2}\right)}}}\right|$.

i know i must somehow get a $N$ such that for $x\geq N$,$\left|{\frac{{x^{2}-x}}{{3x^{2}-2}}-\frac{1}{3}}\right| < \varepsilon$. but i really have no idea how i should choose the N.

Couldn't you just divide by $n^{2}$ on both sides to get
$\lim_{x \to \infty}{\frac{n^{2}-n}{3n^{2}-2}}=\lim_{x \to \infty}{\frac{1-\frac{1}{n}}{3-\frac{2}{n^{2}}}=\lim_{x \to \infty}{\frac{1}{3}=\frac13}}$?
Or is that not a proof?

There was no mistake in my solution..
[hide="Full solution"]Let $N>\frac1\epsilon$. Then for $n\geq N$

$\left| a_{n}-a\right| =\left|{\frac{{n}^{2}-n}{3\,{n}^{2}-3}}-\frac13 \right| =\frac13\, \frac1{n+1}<\frac1n<\epsilon$
[/hide]

Ottem, he changed the denominator of the fraction.

Nebula, that "works," but it's not rigorous.  In particular, the question asks for the proof to be based on the (epsilon) definition of the limit of a sequence.

is it possible to find the limit using the delta-epsilon definition of a limit? or is this just used to show that a number $L$ is the limit?

My friend pointed out that
$\frac{{3n-2}}{{3\left({3n^{2}-2}\right)}}\le \frac{n}{{3n^{2}}}= \frac{1}{{3n}}$
so, that kinda solve the question. Thanks for all e replies. :)


Show that there exist four integers $a$,$b$,$c$,$d$ whose absolute are values greater than $1000000$ such that:
      $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$ +$\frac{1}{d}$ $=$ $\frac{1}{abcd}$

I think, it was the worst problem on RMO 2006

[quote="Rust"]Let $a=x,b=-x-1,c=1-x(x+1),d=x(x+1)[x(x+1)-1]-1.$ It give solution for all x.[/quote]

Hey,can you continue ? because I don't know where we use the condition the absolute values of a,b,c,d greater than 1000000


Euler
function
number theory theorems
number theory

Determine all value of $ n$ is a positive integer number such that there exist a prime divisor$ p$ of $ d(n)$ where $ d(n)$ is Euler function and $ a^{{d(n)}/{p}}$ $ \minus{}1$ is divisible by $ n$ for all  $ (a,n)\equal{}1$.

In other words, $ n$ such that the group of units $ \bmod n$ doesn't have an element of order $ \varphi(n)$.  The answer is well-known: all $ n$ except $ \{ 1, 2, 4, p^k, 2p^k \}$ for $ p$ an odd prime and $ k$ a positive integer.  See [url=http://en.wikipedia.org/wiki/Primitive_root_modulo_n]primitive root[/url].


[hide]Two cases:

If $x$ is odd: $ax^{2}+bx+c$ = the product of three odd integers (odd) + the product of two odd integers (odd) + an odd integer. The sum of 3 odd integers is $\boxed{\text{odd}}$.

If $x$ is even: $ax^{2}+bx+c$ = the product of two even integers and an odd integer (even) + the product of an even integer and an odd integer (even) + an odd integer. The sum of 2 even integers and 1 odd integer is $\boxed{\text{odd}}$.[/hide]

thats incorrect..the product of and even an odd is $\boxed{\text{even}}$: even + even + odd = $\boxed{\text{odd}}$

Thanks for the effort and for the correction.

how does the hint help? All integers are either odd or even.

The hint is a reminder of that fact and a suggestion that your solution should have something to do with that.

[quote="nobody89"]I am not sure... but tried.

[hide]
our function is equal to = x(ax+b)+c where a,b and c are odd.
this is obvious : odd*odd = odd
                      odd*even = even
                      even+odd = odd
                      odd+odd = even
using these facts;
1.st case, where X is odd:
odd*(odd*odd+odd)+odd = odd*(even)+odd = odd.
2.nd case, where X is even:
even*(odd*even+odd)+odd = even*(odd)+odd = odd.
 [/hide][/quote]

yes that is correct

[quote="Interval"]Let f(x) = ax^2 + bx + c, where a, b, and c are odd integers.  If x is an integer, show that f(x) must be an odd integer.

HINT GIVEN IN MATH BOOK:

x is either an even integer or an odd integer.[/quote]
[hide]
the first term will be either an even or odd integer.
the second term will be the same type, because an even integer squared is an even integer, and an odd integer squared is an odd integer.
the first and second terms add up to an even integer.
since the third term is odd, the three terms add up to an odd integer.
[/hide]


Find all injective functions $ f: S\rightarrow S$ such that $ |f(x) \minus{} x| \equal{} |f(y) \minus{} y|\ \forall x,y\in S$, where $ S$ is a finite set and $ |S|\equiv 1\pmod 2$.

You need to define subtraction on $ S$, or is $ S$ a subset of the reals?

Suppose that $ S$ is a finite subset of $ \mathbb{R}$.

Since $ f$ is an injective function from a finite set to itself, it is bijective. Assume $ |S| > 1$, since the problem is trivial otherwise.

Observe that $ f(x) \equal{} x \forall x \in S$ works. Suppose $ f(x)$ is not always fixed. Let $ a, a\plus{}r$ be two elements of $ S$ ($ r$ is a nonzero real) and suppose $ f(a) \equal{} a \plus{} r$. Then the condition implies that $ f(a\plus{}r) \equal{} a$ or $ f(a\plus{}r) \equal{} a \plus{} 2r$. Suppose the latter. Then either $ f(a \plus{} 2r) \equal{} a \plus{} r$ (impossible since $ f$ is injective) or $ f(a \plus{} 2r) \equal{} a \plus{} 3r$. We can show that $ f(a \plus{} kr) \equal{} a \plus{} (k\plus{}1)r$, contradicting $ S$ finite. Therefore $ f(a\plus{}r) \equal{} a$. This implies that for any distinct $ a,b \in S$ such that $ f(a) \equal{} b$, $ f(b) \equal{} a$. So we can pair off elements of $ S$. However, because $ |S|$ is odd, one will be left over. This must map to itself, and this contradicts the condition. So only $ f(x) \equal{} x$ works.

[quote="MellowMelon"]So we can pair off elements of $ S$. However, because $ |S|$ is odd, one will be left over. This must map to itself, and this contradicts the condition. So only $ f(x) \equal{} x$ works.[/quote]

What if there exist an element such that f(a) = a??

Then the $ |S|$ can still be odd.

If $ f(a) \equal{} b$ and $ a \neq b$, and $ f(c) \equal{} c$, then $ |f(a) \minus{} a| \equal{} |b \minus{} a| \neq 0 \equal{} |f(c) \minus{} c|$. The use of the condition is the basis of this method. ;)

[u]A Solution.[/u] Since $ |f(x)\minus{}x|\equal{}|f(y)\minus{}y|\ \forall x,y\in S$, $ f(x)\equal{}x\pm c$ for some $ c\in\mathbb{R}$.  Since $ f$ is injective, hence bijective, $ \sum_{x\in S}x\equal{}\sum_{x\in S}f(x)\equal{}\sum_{x\in S}\plus{}c\sum_{x\in S}\pm 1$.  But since $ |S|$ is odd, $ \sum_{x\in S}\pm 1\neq 0\implies c\equal{}0$.  Ergo, $ f(x)\equal{}x$ is the only possible function. $ \Box$


Jane is 25. Dick's age is d > 25. In n years both will have two-digit ages which are obtained by transposing digits (so if Jane will be 36, Dick will be 63). How many possible pairs (d, n) are there?


Let f  \in R[x] be a monic polynomial s.t :
    for all n \in  N* ,exist m  \in N* such that : f(m)=2^n
    Prove that :degf=1

Suppose that P with dP = n > 1.
and P(x) = x n + a_(n-1)*x n-1  + ... + a_0.
We suppose a weaker property which will be more easy to use in the following :
We suppose that there exists an integer N > 0 such that for each integer k > N there is a integer x_k \geq  0 such that P(x_k) = 2 k . (*)

Let Q(x) = P(x) - x n .

Lemma : We have lim x_k = +oo when k - > +oo.

Proof :
For k > N, the sequence (x_k) has pairwise distinct terms, since P(x_i) =/= P(x_j) for i =/= j). Thus, this sequence has no infinite and bounded subsequence, hence result.

We deduce that we may assume that a_(n-1) is in [0,n[ :
This follows from the fact that there exist an integer t such that nt  \leq  a_(n-1) < n(t+1). Thus if R(x) = P(x-t), then R is a monic polynomial with dR = n, it satisfies the property (*) (with y_k = x_k + t) and its coefficient of x n-1  is  b_(n-1) = a_(n-1) - nt which belongs to [0,n[.

- If Q is identically 0 :
Then P(x_(nN+1)) = (x_(nN+1)) n = 2 nN+1 , where x_(nN+1) is an integer.
Then n divides nN+1. A contradiction.

- if Q is not identically 0 :
a) If 0 < a_(n-1) < n :
Then, as x tends to +oo, we have limQ(x) = +oo and lim Q(x)/(nx n-1 ) = a_(n-1)/n in ]0,1[.
Thus, there exists A > 0 such that for x > A, we have 0 < Q(x) < n*x n-1 . (1)
From the lemma, we deduce that there exists an integer k  \geq  0 such that :
(i) P(x_(kn)) > (x_(kn)) n , and
(ii) Q(x_(kn)) < n*(x_(kn)) n-1  .

From (i), we deduce that : 1 = P(x_(kn))/2 kn  > ( x_(kn)/2 k ) n .
Thus 2 k  > x_(kn), that leads to 2 k   \geq  x_(kn) + 1.
Then :
Q(x_(kn)) = P(x_(kn)) - (x_(kn)) n  = (2 k ) n  - (x_(kn)) n   \geq  (x_(kn) + 1) n  - (x_(kn)) n   \geq  n*(x_(kn)) n-1 , which contradicts (ii).

b) If a_(n-1) = 0 :
Let a_k be the leading coefficient of Q. Then k < n-1 and a_k =/= 0.
- If a_k > 0, we still have (1) which leads to the same contradiction.
- If a_k < 0 : Then Q < 0 or lim Q(x) = -oo as x - > +oo, and lim Q(x)/x n-1  = 0.
Then there exists A > 0 such that if x > A then Q(x) < 0 and |Q(x)| < x n-1  .  (2)

From the lemma, there exists an integer N_1 > 0 such that, for each integer k  \geq  0 with kn > N_1 we have
(x_kn) n  - (x_(kn) - 1) n  = n(x_(kn)) n-1  + sum{0 \leq i \leq n-2} C(n,i)(x_(kn)) i (-1) n-1+i   \geq  (x_(kn)) n-1 .  (3)

From (2) and (3), we deduce that there exists an integer N_2 > 0 such that for each integer k > N_2 we have :
(i') P(x_(kn)) < (x_(kn)) n
(ii') |Q(x_(kn))| < (x_(kn)) n-1
(iii') (x_(kn)) n - (x_(kn) - 1) n  \geq  (x_(kn)) n-1.
From (i'), we deduce as above that 2 k  < x_(kn), and then 2 k   \leq  x_(kn) - 1.
Thus :
|Q(x_(kn))| = (x_(kn)) n - 2 kn  \geq  (x_(kn)) n - (x_(kn) - 1) n  \geq  (x_(kn)) n-1 from (iii'), which contradicts (ii').

Thus, in each case, we obtain a contradiction, and we are done....

Pierre.


algebra
polynomial
trigonometry
abstract algebra
calculus
derivative
function

My book read:

[i]If any term of the assumed solution, $ y_p$, disregarding multiplicative constants, is also a term of the homogeneous solution, $ y_h$, then the assumed solution must be modified by multiplying it by $ x^m$, where m is the smallest positive integer such that the product $ x^my_p$ has no terms in common with $ y_h$.[/i]

It doesn't explain why at all. So why is this necessary?

We're trying to solve $ L(y)\equal{}g(t),$ where $ L$ is a linear constant coefficient differential operator, and $ g$ is the right kind of nonhomogeneity - the right kind being itself a product of polynomials and exponentials. (Note: I'm considering $ \sin at$ and $ \cos at$ to be exponentials.) We started by finding the solution to the equivalent homogeneous problem $ L(y)\equal{}0.$ (In linear algebra terms, we're finding the kernel or null space of the linear operator $ L.$) Our task now is to find a single particular solution $ y_p$ such that $ L(y_p)\equal{}g,$ to which we may add the solutions of the homogeneous problem in order to find the whole solution set.

Suppose our problem is this:
\[ y''\minus{}2y'\plus{}y\equal{}e^t\]
We note that the characteristic polynomial equation for this linear operator is $ r^2\minus{}2r\plus{}1\equal{}0,$ which has $ r\equal{}1$ as a double root. We clearly have $ e^t$ as a solution to the homogeneous equation, but it's not enough - we need a two-dimensional solution space. By experience, we learn that in this situation, we can multiply this by $ t$ (the independent variable) to get another solution. So the solution to the homogeneous equation is $ y_h\equal{}c_1e^t\plus{}c_2te^t.$

Now we look for the particular solution. It won't do to guess $ y_p\equal{}Ae^t.$ Why not? Because $ L(Ae^t)\equal{}0$ and that can't be $ e^t.$ Similarly, it won't do to guess $ y_p\equal{}Ate^t,$ because $ L(Ate^t)$ also equals zero. We have to go at least to a $ t^2$ term.

Could we guess $ y_p\equal{}At^2e^t\plus{}Bte^t\plus{}Ce^t?$ We could, but we'd be doing more work than we have to. The $ Bte^t\plus{}Ce^t$ terms are going to be killed off by $ L.$ On the other hand, we were going to put those terms there in the general solution just because they're in the complementary solution.

So we guess $ y_p\equal{}At^2e^t.$ Put that in the original equation. (Algebra done off-stage.) We get that $ A\equal{}\frac12.$ Putting that together with the complementary solution gets us that the general solution is
\[ y\equal{}\frac12t^2e^t\plus{}c_1te^t\plus{}c_2e^t\]
Here's an alternate approach that winds up in the same place. Denote the differentiation operator by $ D$ and write the original equation as $ D^2y\minus{}2Dy\plus{}y\equal{}e^t.$ Then factor the left hand side (you can do that) so it reads $ (D\minus{}1)^2y\equal{}e^t.$ Now note that the right hand side satisfies $ (D\minus{}1)y\equal{}0.$ Multiply the whole equation by the operator $ (D\minus{}1),$ getting $ (D\minus{}1)^3y\equal{}0.$ We know that the solutions of this are of the form ($ At^2\plus{}Bt\plus{}C)e^t.$ Put that into the original equation to check it; we find we must have $ A\equal{}\frac12$ but $ B$ and $ C$ are unrestricted. So we got to the same place with this, using some of the same arithmetic.

[quote="Kent Merryfield"]Why not? Because $ L(Ae^t) \equal{} 0$ and that can't be $ e^t.$ Similarly, it won't do to guess $ y_p \equal{} Ate^t,$ because $ L(Ate^t)$ also equals zero. We have to go at least to a $ t^2$ term.
[/quote]

This is the part I'm having trouble understanding. Can you explain why exactly $ L(Ae^t) \equal{} 0$ and $ L(Ate^t)\equal{} 0$? I think my misunderstand lies with the fact that I don't really understand what $ L$ actually is.

$ L$ is what you see on the left hand side of the equation. $ L$ is the linear operator $ L(y)\equal{}y''\minus{}2y'\plus{}y.$ $ L$ annihilates (that is, sends to zero) precisely the elements of what you called the homogeneous solution. And you would agree that includes $ e^t$ and $ te^t,$ right?

[quote="Kent Merryfield"]$ L$ is what you see on the left hand side of the equation. $ L$ is the linear operator $ L(y) \equal{} y'' \minus{} 2y' \plus{} y.$ $ L$ annihilates (that is, sends to zero) precisely the elements of what you called the homogeneous solution. And you would agree that includes $ e^t$ and $ te^t,$ right?[/quote]

Ohhh, I get it now. My book switches back and forth, using $ L$ and $ \phi$, and I never really understood what $ L$ was.

[quote="Kent Merryfield"]We're trying to solve $ L(y) \equal{} g(t),$ where $ L$ is a linear constant coefficient differential operator, and $ g$ is the right kind of nonhomogeneity - the right kind being itself a product of polynomials and exponentials. (Note: I'm considering $ \sin at$ and $ \cos at$ to be exponentials.) [/quote]

I have some questions here - what if $ g$ is a function of the form $ \ln t$ or some other non-standard function such as the Dirac delta function (which I've just learnt about)? Also, there's something about the solution space being determined by the solution vectors $ {y_1 \choose y_1'}$ and $ {y_2 \choose y_2'}$ where $ y_1$ and $ y_2$ are solutions to $ L(y) \equal{} 0$; also it is given by the method of variation of parameters that $ y_p \equal{} u(x)y_1 \plus{} v(x)y_2$. Why is this so and how does that work?

Undetermined coefficients is a strongly limited method - it can only be applied in cases where $ g(t)$ is of the right form: linear combination of $ t^ne^{at},$ including complex $ a$ which gives you sines and cosines. Anything not of that form, such as $ g(t)\equal{}\ln t\ ?$ That simply outside the scope of the method, and there's no way to make it work.

The Dirac delta isn't actually a function, which doesn't stop us from using it. Mention of that opens up a whole other world that includes unit step functions and other piecewise defined function. The computational method of choice: Laplace transforms. Do you have a textbook with a chapter on those?

And variation of parameters is another method, not as limited as undetermined coefficients but not as computationally efficient, either. It's different enough that we shouldn't mix talking about it into an undetermined coefficients thread.

No I don't have a textbook on Laplace transforms, and it apparently won't be taught in my math course either, but I'll Google and see what I can get.

I've appropriately started a new thread for the method of variation of parameters:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1680852#1680852[/url]


Prove: If $ \sqrt []{aa'} \plus{} \sqrt []{bb'} \plus{} \sqrt []{cc'} \equal{} \sqrt []{a \plus{} b \plus{} c}.(a' \plus{} b' \plus{} c')$
$ (a,b,c,a',b',c' > 0)$
so :
       $ \frac {a}{a'} \equal{} \frac {b}{b'} \equal{} \frac {c}{c'}$

Looks to me like the equality case of Cauchy.

However,not HSB, so mods please move this.


probability
number theory
prime factorization

If $ n$ is a positive integer with a prime factorization of $ p^{a_1}_1 p^{a_2}_2 \ldots p^{a_k}_k$ What is the probability that a randomly chosen factor of $ n^2$ greater than $ n$ is not a multiple of $ n$?

I think "greater than $ n$" might present a problem here... since primes are random, right? :maybe:

I don't see the problem.  :)  The question you really need to answer here is "how many factors of $ n^2$ are greater than $ n$"?

[hide="Hint"] If $ d | n^2$ and $ d > n$, what can you say about $ \frac{n^2}{d}$? [/hide]

Hmm, that might be true.

On the other hand, there is still a problem.

A factor of $ n^2$ has the form $ p_1^{m_1}p_2^{m_2}\dots p_k^{m_k}$ where:
(1) For any $ i$ from 1 to $ k$, $ 0 \le m_i \le 2a_i$.

And if this factor has at least one $ m_i$ such that $ m_i < a_i$, then the corresponding factor will not be a multiple of $ n$.

The problem here is that we have to explicitly tell when such a factor is greater than $ n$ just by looking at its prime factors and its exponents, and not from its actual value.

[quote="Yongyi781"]
The problem here is that we have to explicitly tell when such a factor is greater than $ n$ just by looking at its prime factors and its exponents, and not from its actual value.[/quote]

Right, but on the other hand, we do know the number of divisors of $ n^2$ that are multiples of $ n$ (that would just be the number of divisors of $ n$), and clearly any multiple of $ n$ is greater than $ n$ (besides $ n$ itself). And from t0rajir0u's hint, we know the number of factors of $ n^2$ greater than $ n$.

number of divisors of a number are defined by $ d(n) \equal{} (a_1 \plus{} 1)(a_2 \plus{} 1)......(a_n \plus{} 1)$.

so, we can calculate no. of divisors of $ n^2$$ (d(n^2))$ and of $ n$ $ (d(n))$ and subtracting them wud give divisors greater than $ n$ which are multiples of $ n^2$. the multiples of $ n$ are $ n, 2n, 3n .....n^2$.
so, probability shud be $ 1 \minus{} \frac {[(d(n^2) \minus{} d(n)) \minus{} (n \minus{} 1)]}{d(n^2)\minus{}d(n)}$


geometry
function
area of a triangle
Heron\u0027s formula

Let $ a\in\mathbb{R}$ . Show that the area of the triangle with side lengths $ \sqrt{a^2\minus{}a\plus{}1},\ \sqrt{a^2\plus{}a\plus{}1},\ \sqrt{4a^2\plus{}3}$ doesn't depend on $ a$ .

Applying the formulae of area as function of semiperimeter we obtain A=3/16

[hide]$ a \equal{} 1 \implies$ the sides are equal to $ 1, \sqrt {3}, \sqrt {7}$
Applying itsmiko's form of Heron's formula, this is $ \frac{\sqrt {(1 \plus{} 3 \plus{} 7)^2 \minus{} 2(1 \plus{} 9 \plus{} 49)}}{4} \equal{} \frac {\sqrt {3}}{4}$
Which is the [b]square root[/b] of castigioni's answer.
likely castigioni forgot this square root.

Clearing out itsmiko's work for the general a,

let the sides be equal to d, e, f and let
$ d^2 \equal{} g \equal{} a^2 \minus{} a \plus{} 1, e^2 \equal{} h \equal{} a^2 \plus{} a \plus{} 1, f^2 \equal{} i \equal{} 4a^2 \plus{} 3$

then the area is $ \frac {\sqrt {(g \plus{} h \plus{} i)^2 \minus{} 2\left(g^2 \plus{} h^2 \plus{} i^2\right)}}{4}$
substituting...
$ \frac {\sqrt {\left(a^2 \minus{} a \plus{} 1 \plus{} a^2 \plus{} a \plus{} 1 \plus{} 4a^2 \plus{} 3\right)^2 \minus{} 2\left(a^4 \minus{} 2a^3 \plus{} 3a^2 \minus{} 2a \plus{} 1 \plus{} a^4 \plus{} 2a^3 \plus{} 3a^2 \plus{} 2a \plus{} 1 \plus{} 16a^4 \plus{} 24a^2 \plus{} 9\right)}}{4}$
$ \frac {\sqrt {\left(6a^2 \plus{} 5\right)^2 \minus{} 2\left(18a^4 \plus{} 30a^2 \plus{} 11\right)}}{4}$
$ \frac {\sqrt {36a^4 \plus{} 60a^2 \plus{} 25 \minus{} 36a^4 \minus{} 60a^2 \minus{} 22}}{4} \equal{} \frac {\sqrt {25 \minus{} 22}}{4} \equal{} \frac {\sqrt {3}}{4}$
This agrees with your (itsmiko's) solution except for the step before "sqrt or (1) / 4" where i'm sure that a calculation error occured.[/hide]

I think the answer should be $ \sqrt{3}$ and the reason for that. we have

$ S^2\equal{} 2\sum_{cyc}(ab)^2 \minus{} (a^4\plus{}b^4\plus{}c^4)$

Stevenmeow's right, since you can just plug in $ a\equal{}0$ to get a triangle with base $ \sqrt{3}$ and height $ \frac{1}{2}$.


Read the title.

And after you answer my question, try to solve this problem:
The pool is half full. If I drain it while filling it with a hose, when will it be drained completely if the drain drains the pool in 45 minutes and the hose fills it in 75 minutes? Note: express this in hours.

i need 75/2= 37.5 minutes to fill the pool right now. It takes 45/2=27.5 minutes for me to drain the pool right now. 37.5-27.5=10
75/10=7.5
7.5*37.5=266.25 minutes
i am probably wrong so tell me if i am.

the answer i got is:-

1 hour= 60 minutes
x hours=(45+75)minutes
x hours=120 minutes
so,2 hours  = 120 minutes

the answer  i got is this. :maybe: correct.

Is it 56.25 minutes..?  I'm not sure though.

Well it's definitely not 75+45. That's just a weird out of the blue step.

Ha Ha, nobody's answering the other question.

Anyway, it's pretty self-explanatory. In this forum, post and respond about middle school math, normal-classroom concepts.

@ jothanchou711 I just tried to solve it.  :blush:

it's 56.25 because:

Let x be number of minutes until pool is completely drained.

So
x((1/45)-(1/75))=1/2

Each minute, (1/45)-(1/75) of the pool is drained and we need half of the pool too be drained.

Then... with the calculator...
x=56.25

(thought I'd post this because no one posted how they got the answer)

[quote="tcjy"]Note: express this in hours.[/quote]

Knowing this, we do $ 56.25 \div 60$, which gives us the answer $ \boxed {.9375}$, or $ \boxed {\frac {15}{16}}$


Let be $ a,b,c>0$ such that $ abc \equal{} a \plus{} b \plus{} c$ . Prove that :

$ \frac {1}{a(b \plus{} c)} \plus{} \frac {1}{b(a \plus{} c)} \plus{} \frac {1}{c(a \plus{} b)}\ge \frac {\sqrt {3}}{2}\cdot \frac{1}{\max\{a,b,c\}}$

max{$ a,b,c$} $ \geq\ \frac {a \plus{} b \plus{} c}{3}$
And let $ x \equal{} \frac {1}{bc},...$ we obtain an easy result  :)


A computer printed out the (base 10) numerals $ 2^{2002}$ and $ 5^{2002}$. How many digits did it print out?

crap, i forgot about the extra 1... no wonder i kept getting 2004

Walk Around The River,

you can be more direct by cutting down the number of characters
used and the number of steps by involving both the numbers 2 and 5:

floor(2002*log2) + 1 + floor(2002*log5) + 1 =

602 + 1 + 1399 + 1 =

2003

Except the whole point of my solution was that it did not require a calculator, as yours would (unless you know the approximation of log2...)

@Ragnarok.  Yeah, when x is positive, floor(-x)=-floor(x)-1.  Always keep that in mind.

Then I recommend typing this, because yours is relatively harder to follow from the beginning:

floor(2002*log2) + 1 + floor(2002*log5) + 1 =

floor(2002*log2) + floor{2002[log(10/2)]} + 2 =

floor(2002*log2) + floor(2002(log10 - log2)) + 2 =

floor(2002*log2) + floor(2002(1 - log2)) + 2 =

floor(2002*log2) + floor(2002 - 2002log2)) + 2 =

floor(2002*log2) + floor(2002) - floor(2002log2) - 1 + 2 =

2002 - 1 + 2 =

2003


geometry
trigonometry
pigeonhole principle
area of a triangle
geometry solved

In a square of side 6 the points $A,B,C,D$ are given such that the distance between any two of the four points is at least 5. Prove that $A,B,C,D$ form a convex quadrilateral and its area is greater than 21.

[i]Laurentiu Panaitopol[/i]

Let's assume ABCD is concave. Then we can assume D is in the interior of triangle ABC. Then one of the angles ADB, BDC, CDA is >= 120 deg. We again assume BDC >= 120. Because DB and CD >=5 we have from cosint thm CD>= (5^2+5^2+5^2)^1/2 = 5*3^1/2 = 75^1/2 > 72^1/2 = 6*2^1/2 =the length of the diagonal, but we know that the longest segment that can be inside a square is the diagonal, so we got a contadiction, so ABCD is convex.

For the second part consider the triangles ABC and BCD. Each of them has an angle >= 60 and they have all their angles <=120 (we proved this above), so each one of them has an angle with its sine>= (3^1/2/)2 and all their sides are >=5, but if you compute 2*(the area of a triangle with angle 60 and sides 5) you get a little over 21. The sum of the areas of our 2 triangles is, however >= what we calculated > 21, so it's finished.

I would have enjoyed participating in that selection test (the problems weren't too hard), but unfortunately this year I completely messed up at the final round so.. :D

cool stuff,grobber!!
even i think this year's selection tests were easier ,4th one in particular.
do u have solutions to all of them.?in case it is so,i'll ask for some of the solutions thru private messages,u wudnt mind,wud u? :D
best regards

Sure, I don't mind, but I don't actually have that many solutions. I solved less than half of all the problems (in all the tests), but I could think about them.. :D

Hey I just want to tell my opinion about the 4th selection test.It wasn't that easy , first of all because the first problem could easily trick you into taking M=A , which costed you 4 points out of 7 (It said M included or equal to A , M<>A).

The second problem (this one) was probably the hardest geometry problem out of all the selection tests (except perhaps for the one in the first test , in which almost no-one dared to use computations).

And the combinatorics problem was clearly the most difficult out of all the tests , since no student at all managed to solve it or do any real progress with it (the highest mark was 2 , obtained by 2-3 students).

I think this was probably the most difficult selection test except the 2nd , which was really challenging. So , all in all , I don't think it was that easy. (P.S. I took part in the actual selection test)

oh i am soory abt this miscommunication. i meant the 5th selection test which was "easier " as compared to the other tests.4th one was difficult,no doubt
cheers

First of all we observe that no angle formed with 3 of the 4 points can be greater of equal with $120^\circ$, because otherwise if we suppose that WLOG $\angle ABC\geq 120^\circ$, then from $AB\geq 5$ and  $BC\geq 5$ we deduce $AC\geq 5\sqrt 3 >6\sqrt 2$ contradiction.

Therefore if the quadrilateral $ABCD$ is not convex, then one of the 4 points lies inside the triangle formed by the other 3. Suppose WLOG that $D\in$int$[ABC]$. But then one of the angles $\angle ADB$, $\angle BDC$ and $\angle CDA$ would be, by the Pigeonhole Principle, greater or equal than $120^\circ$, contradiction. Thus $ABCD$ is a convex quadrilateral.

Now because each angle of the triangle $ABC$ is smaller than $120^\circ$ and there is at least one angle, say $\angle ABC$, which is greater than $60^\circ$ it follows that \[ \sin \angle ABC \geq \frac {\sqrt 3}2 \Rightarrow \]
\[ \sigma[ABC]=\frac 12 AB\cdot BC\cdot \sin \angle ABC \geq \frac {\sqrt3}4 \cdot 25 > \frac {21}2 \Leftrightarrow 625 > 12 \cdot 49 = 588 \]

Analogously one can prove that $\displaystyle \sigma[ACD]\geq \frac {21}2$, and thus $\sigma[ABCD]>21$.

Prove convexity as done above.

Now, knowing that our quadrilateral is convex, we have that all sides and diagonals have length at least $5$ and at most $6 \sqrt 2.$ Let our quadrilateral be called $ABCD,$ and $x = |AC|$ with $5 \le x \le 6 \sqrt2.$ We claim that if we let $B', D'$ be so that $AB' = B'C = CD' = D'A = 5$, then $[ABCD] \ge [AB'CD'].$

Indeed, it suffices to show that the distance from $B$ to $AC$ is at least the distance from $B'$ to $AC.$ Let $\omega_1, \omega_2$ be the circles of radius $5$ centered at $A, C$ respectively. We know that $B$ is not in the interior of either $\omega_1$ or $\omega_2.$ Therefore, it's clear from drawing a diagram that the foot from $B$ to $AC$ must be not be on segment $AC$, and that it must lie at least $\frac{x}{2}$ away from both vertices. This means that $|AB^2 - AC^2| \ge |(\frac{3x}{2})^2 - (\frac{x}{2})^2| = |2x^2| \ge 50.$ However, we have that $|AB^2 - AC^2| \le |(6 \sqrt2)^2 - 5^2| = 47$, which is a contradiction.

$\square$


calculus
integration
geometry
function
conics
parabola
calculus computations

Consider the functions f and g, both in the IR+---> IR, and whose laws are $ \ f(x) \equal{} 2x \minus{} x^2$ and $ \ g (x) \equal{} x^n$ where n is a positive integer. The area between  f and g equal $ \dfrac {1}{2}$. It is correct to affirm that n is a divisor:

A) 30
B) 28
C) 27
D) 26
E) 24

First, start by solving:

$ 2x\minus{}x^2\equal{}x^n$

From this we can see that $ x_1 \equal{} 0$ and $ x_2 \equal{} 1$. Now that we have our integration boundaries, we calculate:

$ \int^1_0 \left(2x\minus{}x^2\minus{}x^n\right)\,dx \equal{} \left(x^2\minus{}\frac{x^3}{3}\minus{}\frac{x^{n\plus{}1}}{n\plus{}1}\right)\Big|^1_{x\equal{}0}\equal{}1\minus{}\frac{1}{3}\minus{}\frac{1}{n\plus{}1}\equal{}\frac{1}{2}$

And finally we calculate:

$ 1\minus{}\frac{1}{3}\minus{}\frac{1}{n\plus{}1}\equal{}\frac{1}{2}$

$ \frac{2}{3}\minus{}\frac{1}{2}\equal{}\frac{1}{n\plus{}1}$

$ \frac{1}{6}\equal{}\frac{1}{n\plus{}1} \Rightarrow n \equal{} 5$

So I believe the answer is "A) 30".

[quote="vitoriogauss"]The area of the region bounded above by the graph of f below the graph of g
[/quote]

Do you mean to say between $ f$ and $ g$???

All right. I thought that too. But how is 1 and 0?

you solved the equation: $ \ 2x\minus{}x^2\ \minus{}x^n \equal{}0$... I am in doubt here.

how to find 0 and 1

Just try the roots....

yes, between, sorry.

how it can be above f and below g?I mean f is parabola facing downwards and g is below it in the interval $ [0,1]$
${ x(2-x^{n-1}-x}=0$
$ x=0,1$
thank u

Sketch the graphs of $ x^n$ and $ 2x \minus{} x^2$, then observe they intersect each other only at $ (0,0)$ and $ (1,1)$.


Is there an algorithm to arrange numbers from 1 to n^2 in a nxn square so that the obtained square is magic? (ie the sum on all the columns and rows and the 2 diagonals is the same).

Sincerely,
                                        Al Jaber.

Yep. In fact, it was one of the practice problems my teacher gave us for a computer science lesson in two-dimensional arrays.

Here is one source (http://www.delphiforfun.org/Programs/magic_squares1.htm):
Start with 1 in the middle of the top row; then go up and left assigning numbers in increasing order to empty cells; if you fall off of the square imagine the same square as tiling the plane and continue; if a cell is occupied, move down instead and continue.

3x3 example:\[\begin{array}[b]{|c|c|c|}
\hline-&1&-\\
-&-&-\\
-&-&-\\\hline
-&1&-\\
-&-&-\\
2&-&-\\\hline
-&1&-\\
-&-&3\\
2&-&-\\\hline
-&1&-\\
-&4&3\\
2&-&-\\\hline
5&1&-\\
-&4&3\\
2&-&-\\\hline
5&1&-\\
-&4&3\\
2&-&6\\\hline
5&1&-\\
-&4&3\\
2&7&6\\\hline
5&1&-\\
8&4&3\\
2&7&6\\\hline
5&1&9\\
8&4&3\\
2&7&6\\\hline
\end{array}\]
And we're done! :)

There certainly is for odd n, but im not so sure for even n.

It involves writing the numbers in a diagonal square enclosing the one you want, and then 'folding' the corners over an bang! Do you want the details?

Sorry for being imprecise...n^2 must be odd :blush:
Thanks for the first method....but I'd also like to see the one with the folding diagonal square...

Sincerely,
                                            Al Jaber

Im in a real rush right now, sorry, ill be back in either 1 week or 2 weeks, depending on wether i get home in between. Till then!


superior algebra
superior algebra unsolved

If $ f(g)\equal{}g^m$, where $ f$ is a homomorphism from a group $ G$ to a group $ G$ (also) ,then if $ |G|$ and $ m$ is coprime ($ |G|$ is finite), prove or disprove that $ f$ is (also) an isomorphism. Thanks in advance.

Write $ 1 \equal{} ms \plus{} |G| t$.  Then consider the map $ g(x) \equal{} x^s$.  Show that $ g(x)$ is an inverse for $ f(x)$ and hence
that $ f(x)$ is a bijection.

Proving that $ f$ is a bijection is not enough; you need to prove that it is a homomorphism.  For general $ G$ it is not--this fails for $ m\equal{}2$ and $ G$ a nonabelian group of odd order, for example.


Let a hexagon ABCDEF. Prove that AD,BE,CF are concurrent if each of them bisects the hexagon. ( a line bisects the hexagon if this line divide the hexagon into two parts which have the same area)

[quote="treegoner"]Try to prove opposite sides are parallel.[/quote]
Dear treegoner,
I don't think so. Please see attach picture for example!
Best regards,
Bui Quang Tuan

Hi, everyone
Here is my solution. I wonder if someone find out a synthetic proof.

[quote="treegoner"]Sorry for the wrong direction. What I meant was triangles BDF and ACE have parallel corresponding sides.[/quote]
Thank you dear treegoner! Now we can make synthetic proof:
Denote $ S(XY...Z)$ as area of polygon $ XY...Z$
$ S(ABCDEF)/2\equal{}S(BCDE)\equal{}S(FCDE) \Leftrightarrow S(BCE)\equal{}S(FCE) \Leftrightarrow FB//CE.$
Similarly we can show: $ DF//AC$ and $ BD//EA.$
$ \triangle ACE$ and $ \triangle DFB$ are homothetic therefore $ AD$, $ CF$, $ EB$ are concurrent.
Best regards,
Bui Quang Tuan


The two digits in Jack's age are the same as the digits in Bill's age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages?

$ \textbf{(A)}\ 9 \qquad
\textbf{(B)}\ 18 \qquad
\textbf{(C)}\ 27 \qquad
\textbf{(D)}\ 36 \qquad
\textbf{(E)}\ 45$

Let Jack's age be $ J$ and Bill's age be $ B$.  We know that $ J \equal{} 10a \plus{} b$ and $ B \equal{} 10b \plus{} a$ for some positive integers $ a,b$.  From the second sentence, we can set up an equation:
$ J \plus{} 5 \equal{} 2(B \plus{} 5) \rightharpoonup 10a \plus{} b \plus{} 5 \equal{} 20b \plus{} 2a \plus{} 10 \rightharpoonup 8a \minus{} 19b \equal{} 5$
Notice that the difference in their current ages is given by:
$ 10a \plus{} b \minus{} (10b \plus{} a) \equal{} 9(a \minus{} b) \equal{} 9(a \minus{} b)$.
We know that $ a$ and $ b$ have to be within the set $ \{0,1,2,3,4,5,6,7,8,9\}$, so we
just test numbers for $ a$ and $ b$.  Quickly, we find that $ a \equal{} 3, b \equal{} 1$ works:
$ 8(3) \minus{} 19(1) \equal{} 24 \minus{} 19 \equal{} 5$, so we know that the difference in their ages is:
$ 9(3 \minus{} 1) \equal{} 9(2) \equal{} 18 \rightharpoonup \boxed {B}$.

$ 10a \plus{} b \plus{} 5 \equal{} 2(10b \plus{} a \plus{} 5)$
$ 10a \plus{} b \plus{} 5 \equal{} 20b \plus{} 2a \plus{} 10$
$ 8a \minus{} 19b \equal{} 5$
Since we know a and b must be digits, we know b is less than 4 since $ 8(9) \minus{} 5 \equal{} 67$. We quickly find that $ (a,b) \equal{} (3,1)$ Thus the difference is $ 31 \minus{} 13 \equal{} 18 \rightarrow \boxed B$
Edit: Ninja'd  :ninja:

$ J = 10a+b, B=10b+a \rightarrow a>b$

$ \Rightarrow J-B = 9(a-b) \Rightarrow J-B \equiv 0 \mod 9$

Also, $ \Rightarrow 10a+b+5=2(10b+a+5)$

So $ 10(2b+1-a)+2a-b-5=0$

$ \Rightarrow 2b+1-a=0, 2a-b-5=0 \rightarrow a=3,b=1$

$ 31-13=18 \longrightarrow \boxed{B}$


the inscrible circle in a triangle $ ABC$  hits  [$ AC$] in $ K$.prove that the line which join the midpoint of [$ AC$] to the center of the inscriptble  circle  intersect  [$ BK$] in its midpoint.

it's equivalent to show that the incenter is the nagel of the medial triangle or that I,G,N are collinear and GN=2 GI (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=139691]here[/url]).


For how many dates in 2004 do the three numbers form an arithmetic sequence? For example, 06/05/04 satisfies the condition because 6, 5 and 4 form an arithmetic sequence with a difference of –1 between each successive term. Notice 5 – 6 = –1 and then 4 – 5 = –1.

let's start with extremes.

12-8-4
and
2-3-4

only even months work.
six evens from 2-12
12-8-4
10-7-4
8-6-4
6-5-4
4-4-4
2-3-4
6 dates


What is the reason why some mods and administrators of AoPS may dislike the Missouri Forum?

Take part in the poll and defend your point.

There envious because they don't live in Missouri...

Obviously, everyone knows there's too much spam here.

Of course, everyone's jealous of living in Missouri.

1. Mark Twain was born there.
2. Missouri = approximately dead center of USA.
3. Missouri is bordered by eight states, so it already pwns atleast 8 other forums.
4. Missouri = :flex:

missouri needs to have a private arml forum

Yeah, because Missouri is that special  :D

I can't believe I'm saying this but you guys need to be more productive (yes I truly changed from my previous spamming ways..).  Hmm then again I also used to live in MO..

But IL rawks enough to spell it the cool way :P

Edit:  I feel so hypocritical.   :| Maybe MO just has [i]too[/i] much spam.

Missouri is always productive.

That's why MO placed 8th in NATS and is looking forward to doing better this year. :wink:

[quote="suma_milli"]No I mean as in forum spamminess that's why everyone dislikes the MO forum.[/quote]
That's exactly what I said before in several posts above you... :wink:


Hi everyone
I was recently confronted with the following problem:
$ x^4 \plus{} 3x^3 \minus{} 9x^2 \plus{} 2x \plus{} 3$

Using the calculator, i got the following roots:
$ \minus{}4.896077, \minus{}.4538893, 1, 1.349967$

According to the rational roots theorem, the only rational roots allowed are $ \pm 1, \pm 3$, so that means that theres 3 irrational roots. The problem is that I thought that irrational roots only came in pairs!! So how can one have an odd number?? I asked my teacher, and he was like wtf that's really weird, and then my math teacher asked other match teachers and they're also like wtf this is really weird. Any explanation would be awesome thanks!!

[i]Non-real[/i] roots come in pairs (if your polynomial has real coefficients).  Irrational roots can do pretty much anything.

Edit:  You aren't quite asking the right question yet.  I can construct a polynomial like $ (x \minus{} \sqrt{2})(x \minus{} \sqrt{3})...$ that can have any number of irrational roots, but what you're really asking about is irrational roots of polynomials [i]with rational coefficients.[/i]  So this boils down to showing that there exist [url=http://en.wikipedia.org/wiki/Irreducible_polynomial]irreducible polynomials[/url] with all real roots of any degree.

Without the realness constraint, this is a straightforward application of [url=http://en.wikipedia.org/wiki/Eisenstein%27s_criterion]Eisenstein's criterion[/url].  I am not in fact sure of how to prove this result with the realness constraint, but I am certain it is true because of the existence of some good [url=http://en.wikipedia.org/wiki/Chebyshev_polynomials]near-examples[/url].


Two circles have radii of $a$ and $b$, respectively, and are externally tangent. Their common internal tangent intersects their common external tangents at $E$ and $F$. Find $EF$ in terms of $a$ and $b$.

[quote="GlenthemanN"]how can two circles be externally tangent and internally at the same time?[/quote]

They are externally tangent; it is just that they have a common internal tangent.

[hide]
Drop a parallel line from the external tangent then use pythagoras. EF = length of that parallel line = $2\sqrt{ab}$
[/hide]


algebra
polynomial
vector
combinatorics proposed
combinatorics

Find the number of polynomial $ P(x)\in F_p$ with degree $ n$ (fixed) such that the equation $ P(x)\equal{}0$ has a solution in $ \{0,1,2,...,p\minus{}1\}.$ Here $ p$ is a prime number.

I'll count the number of [i]monic[/i] polnomials that do [i]not[/i] have a solution, for simplicity and to avoid annoying factors of $ (p\minus{}1)$.

Unless I've made some dreadful mistake, the answer seems to be $ \sum_{k\equal{}0}^{min\{n,p\}} {(\minus{}1)^k {p\choose k}} p^{n\minus{}k}$, by inclusion-exclusion. (if anyone wants details I'll make it more clear)

For $ n<p$, there is probably not a closed form, but for $ n\geq p$, this is $ p^{n\minus{}p} (p\minus{}1)^p$, which is the answer you'd guess with some kind of handwavy vector space argument.

Maybe it's not that handwavy, actually.  For any $ p$-tuple $ v\in ({\mathbb{F}_p}^*)^p$, the number of polynomials satisfying $ P(i)\equal{}v_i$ for all $ i$ is exactly $ p^{n\minus{}p}$, as this is just a set of $ p$ equations in $ n$ variables. (it is not hard to see that these equations are linearly independent)  Since these vectors represent the possible values that "good" $ P$ take on $ \mathbb{F}_p$, we count $ p^{n\minus{}p} (p\minus{}1)^p$ solutions.

Cool problem!


[hide]
$697\equiv4(\mod 7)$
So each year the day is 4 more than the previous.
3 years later, it recycles, so year $3$ is the answer[/hide]

[quote="dragongrl6"]Really? That's weird. :?
I checked the answer key (2002 Chapter Sprint)
and it said the answer was 12.[/quote]
[hide]thats what i got
697=17x41, it shares no factors with 12. so the next time a 12 day week will start the year on the first day, it would be the next multiple of 12.. i think [/hide]

[quote="dragongrl6"]This is probably an easy problem for you guys, but I'm not exactly sure how to get the answer.

The length of the year on the planet Mars is exactly 697 days. If Mars has a calendar with a 12-day week, and year 0 begins on the first day of the week, what is the next year which will begin on the first day of the week?  :alien:[/quote]
[hide]
Lets name our days Mon, Tues, Wed, Thurs, Fri,Sat,Sun,Mony,Tuesy,Wedy,Thursy,Friy
After a year, the year 1 starts on Tues, year 2 start on Wed, Thus year 11 starts on Friy. And year 12 starts once again on Mon[/hide]

[quote="Klebian"][hide]Wrong mod, biff.
$697\equiv1(\mod 12)$

Which means, every year, you go ahead a day. So it'd recycle in 12 years.[/hide][/quote]

Isn't it 697x = 0 (mod 12), which is x = 0 (mod 12), so for this year (year x=0) then x = 12 (after 12 years)

Huh?

All you need to do is say $697\equiv 1\bmod{12}$, so $12(697)\equiv 0\bmod{12}$.

[hide]Divide 12 from 697. That will give you 58 R1. That means every year it will gain 1 day. Since there are 12 days in a week, it will take 12 years.[/hide]


Where I can find Mosp homework 2006 solution?  :omighty:

I don't think there are any "official" solutions for that.

Ask Alex Zhai.  He knows the solutions to half of them off the top of his head, and can solve the other half on the spot.

But yeah, there aren't any official solutions, I don't think.  Some of the problems might come from other sources, though I'm not sure...

You could search to see if people have posted them before, or just post them up yourself.  Or you could send problems to me: I'm always happy to do more good problems.

Does anyone have the PDFs of the homework of the later years?


limit
logarithms
calculus
calculus computations

what is $ \boxed{\lim_{a\to\pi}\frac{a^{\pi}\minus{}\pi^{a}}{a^{a}\minus{}\pi^{\pi}}}$?

[i]pls [/i]don't use l'hospital... :D

hello, the limit is $ \frac{1\minus{}\ln(\pi)}{1\plus{}\ln(\pi)}$.
Sonnhard.

Yes Dr Sonnhard Graubner, but how did you evaluated the limit?

[quote="misan"]what is $ \boxed{\lim_{a\to\pi}\frac{a^{\pi}\minus{}\pi^{a}}{a^{a}\minus{}\pi^{\pi}}}$? [/quote]
[hide="Quite elementary solution"]
Let $ a: \equal{}x$, we have to compute:

$ \lim_{x\to\pi}\frac{x^{\pi}\minus{}\pi^{x}}{x^{x}\minus{}\pi^{\pi}}\equal{}\lim_{x\to\pi}\frac{x^{\pi}\minus{}\pi^{x}}{x\minus{}\pi}\cdot\frac{x\minus{}\pi}{x^{x}\minus{}\pi^{\pi}}$

Let $ A\equal{}\lim_{x\to\pi}\frac{x^{\pi}\minus{}\pi^{x}}{x\minus{}\pi}$ and $ B\equal{}\lim_{x\to\pi}\frac{x^{x}\minus{}\pi^{\pi}}{x\minus{}\pi}$ and:

$ A\equal{}\minus{}\lim_{x\to\pi}x^{\pi}\frac{\frac{\pi^{x}}{x^{\pi}}\minus{}1}{x\minus{}\pi}\equal{}\minus{}\lim_{x\to\pi}x^{\pi}\frac{e^{x\ln\pi\minus{}\pi\ln x}\minus{}1}{x\ln\pi\minus{}\pi\ln x}\cdot\frac{x\ln\pi\minus{}\pi\ln x}{x\minus{}\pi}$, so let compute now:

$ \lim_{x\to\pi}\frac{x\ln\pi\minus{}\pi\ln x}{x\minus{}\pi}\equal{}\lim_{x\to\pi}\left(\ln\pi\minus{}\pi\frac{ln x\minus{}\ln\pi}{x\minus{}\pi}\right)\equal{}$

$ \equal{}\lim_{x\to\pi}\left(\ln\pi\minus{}\frac{\ln\frac{x}{\pi}}{\frac{x}{\pi}\minus{}1}\right)\equal{}\ln\pi\minus{}1$, so:

$ A\equal{}\minus{}\pi^{\pi}\cdot 1\cdot (\ln\pi\minus{}1)\equal{}\pi^{\pi}(1\minus{}\ln\pi)$

And:

$ B\equal{}\lim_{x\to\pi}\frac{\pi^{\pi}\minus{}x^{x}}{\pi\minus{}x}\equal{}\lim_{x\to\pi}x^{x}\frac{\frac{\pi^{\pi}}{x^{x}}\minus{}1}{\pi\minus{}x}\equal{}\lim_{x\to\pi}x^{x}\frac{e^{\pi\ln\pi\minus{}x\ln x}\minus{}1}{\pi\ln\pi\minus{}x\ln x}\cdot\frac{\pi\ln\pi\minus{}x\ln x}{\pi\minus{}x}$, and:

$ \lim_{x\to\pi}\frac{\pi\ln\pi}{\pi\minus{}x}\equal{}\lim_{x\to\pi}\left(\ln\pi\plus{}x\frac{\ln\pi\minus{}\ln x}{\pi\minus{}x}\right)\equal{}$

$ \equal{}\lim_{x\to\pi}\left(\ln\pi\plus{}\frac{\ln\frac{\pi}{x}}{\frac{\pi}{x}\minus{}1}\right)\equal{}\ln\pi\plus{}1$, so:

$ B\equal{}\pi^{\pi}(1\plus{}\ln\pi)$, hence:

$ \lim_{x\to\pi}\frac{x^{\pi}\minus{}\pi^{x}}{x^{x}\minus{}\pi^{\pi}}\equal{}\frac{A}{B}\equal{}\frac{1\minus{}\ln\pi}{1\plus{}\ln\pi}$[/hide]

Nice solution LideR_M, the second question in this post is a easier one... [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=114142[/url]
misan, this is a problem-book question or you made it by yourself joining the two easier ones?

Or use substitution $ a\minus{}\pi\equal{}y$ consider limit when $ y\to 0$ and use Taylor


[hide="hint"]How many times do you write the digit 7 in the 1s place?  The tens place?  etc?[/hide]

[quote="Cartman"]If you write down all integers from 0 to 1000 000, how many times do you write down the integer 7?[/quote]
Try it out. You can see the pattern:

[hide="Answer"]10......1 seven
100....20 sevens
1000...300 sevens
...
...
...
1000000...[b]600000[/b] sevens[/hide]

Also, like b-flat said, look at each specific digit place.

[hide="Second Answer"]There are 10000 7s in the ones place, 10000 7s in the tens place, on and on to the 100000th place. Added up, it makes [b]600000[/b].[/hide]


The NC MathCounts team is as follows:

1.Calvin Deng
2. Remy Lee
3. Jeffrey An
4. Yu Wang

Coach: Rohan Lewis

Congratulations to all that went!

mathcounts was entertaining

(i.e. the awards ceremony because that's all I saw :/)

argh I should've been studying molgen then too :( oh well

Did you know that Ashley Reiter the WOOT instructor is from NC and was on the NC mathcounts team in 1987

calvin has gotten into the habit of congratulating himself


function algebra domain logarithms vector calculus integration

There is a method to plot a Function Equation without solving the equation?

I would be glad if someone give me some site,pdf or ideas of the method.

The question is far too broad. Is there a particular type of equation you are interested in? Do you have an example?

There need to be a specify method to a specific functional equation?!

So give only a basic Idea to make a plot of $ f(x y) \equal{} f(x) \plus{} f(y)$

I know that this equation have a solution, but since it is simple, there could be a easy way to evaluate it numerically without solving the equation.

If you have better examples, I would be glad to see them.

What exactly do you mean by "plotting a functional equation"? :? The equation you wrote has not one but infinitely many solutions some of which are quite impossible to plot.

The functions for which we have any hope of doing anything numerical must have the property that we can fairly represent what they do using only a finite sampling of points. This - the idea that such a finite sampling tells us something - leads us to make assumptions that our functions are "nice" - usually a statement about a certain number of continuous derivatives. And making any assumption like that would rule out the wilder, ungraphable solutions that fedja mentions.

With any functional equation the first step should be exploration of specific value. Let's try $ y\equal{}0.$ Then the equation becomes $ f(0)\equal{}f(x)\plus{}f(0)$ for all $ x.$ That would give $ f(x)\equal{}0$ for all $ x,$ which is, well uninteresting.

The only way out of that would be to assume that $ 0$ is not in the domain of $ f.$ So let's go ahead and make that assumption. Since we also plan to eventually impose some sort of continuity or smoothness condition, omitting any point breaks up the domain. So we'll restrict further: only assume that $ f(x)$ is defined for $ x>0.$

Now try $ y\equal{}1.$ Then $ f(x)\equal{}f(x)\plus{}f(1).$ This implies that $ f(1)\equal{}0.$

Now assume that $ f$ is differentiable. Hold $ y$ constant and differentiate with respect to $ x:$

$ f'(xy)\cdot y \equal{} f'(x).$

Let $ x\equal{}1$ in that expression:

$ yf'(y)\equal{}f'(1),$ or $ f'(y)\equal{}\frac{f'(1)}y.$

We can let $ f'(1)$ be any number. But (assuming that we have already developed the logarithm), we have that $ f(x)\equal{}c\ln x$ for some constant $ c.$

(There are solutions with no continuity properties at all - if you're curious, try searching for the words "Hamel basis" on this site.)

This argument used nothing numerical at all. We could, of course, make some arbitrary assignment for $ f(2)$ and use that to compute $ f(4),f(8), f(1/2),$ and so on.

Also, how about $ f(x)\plus{}f(x\plus{}1)\plus{}f(x\plus{}\sqrt 2)\equal{}0$? ;) Here there are many continuous and even analytic solutions as well but how can one possibly plot any of them without solving the equation? :? Unlike the previous example, in this case we cannot even hope to compute infinitely many new points from finitely many given ones.

I think I didn't express myself clear.

To fedja first.

The equation $ y' \equal{} y$ has infinitely many solutions, so is impossible to plot, using your proper argument.

In plotting a general solution for a differential equation we can make something similar to a Contour Line. We can also use vectors to express the Set of all solutions for a DE. Etc... However, for more than 1 constant of integration, the plot can become very difficult and another’s methods to plot it will be necessary. Etc.

But this isn't the point I desire, since for numerical analysis we expect to have finite solutions not infinitely ones.

To Kent.

When I said numerically is to use numerical analysis to plot the functional equation, or in another words, I only want an approximation of what would be the exactly plot.

There is a plenty of methods to approximate a ordinary differential equation. All of them if I’m not wrong, transform the ODE in a recursive equation, and them make a approximation of the real curve.

But I have never seemed any methods to approximate a functional equation.

Your attempt is a rather good. To transform the functional in a differential equation and them plot, supposing it is a ”very nice function”.

But what I want is, there is a way to transform a FE in a RE without need of a DE?

Thanks to the time.


calculus
geometry
geometric transformation
rotation
trigonometry
absolute value

I hope this isn't calculus:

Why is e^(theta*i) equal to cis(theta)=cos(theta)+isin(theta)  where theta is in radians???

And why do the nth roots of 1 form a regular n-gon in the complex plane???

And why is abs(z)=a^2 + b^2 where abs is absolute value and z is a complex number???

Why is it called absolute value anyway if it's for a complex number???

Sorry...that's a lot of questions.

1. The derivation comes from looking at Taylor Series expansions of both, I believe (so it is calculus). I've also seen it proven with differential calculus.

2. Well, think about it. The nth roots of unity are of the form $ e^{2\pi ik/n}$, and $ 1$ is always a root of unity. Therefore, for each $ k$ from $ 1$ to $ n\minus{}1$ is a rotation of $ 1$ in the complex plane by equally spaced angles which all sum to $ 2\pi$; moreover, the roots of unity lie on the unit circle, so we get a regular polygon.

3. $ |x|$ for $ x\in\mathbb{R}$ is the positive distance from $ 0$ to $ x$ on the number line. For $ x\in\mathbb{C}$, $ x$ is on the complex plane, which is in some ways an extended number line. Therefore, to generalize the absolute value to a complex number $ x$, we want its positive distance from $ 0\plus{}0i$, which follows from the Pythagorean theorem.
$ |a\plus{}bi|\equal{}\sqrt{a^2\plus{}b^2}$. When $ b\equal{}0$, $ a\plus{}bi\in\mathbb{R}$, and $ |a\plus{}bi|\equal{}\sqrt{a^2\plus{}0^2}\equal{}|a|$.

For number 2, why are they in that form???
Why is that so unique???

Well, $ e^{2\pi i}\equal{}\cos2\pi\plus{}i\sin2\pi\equal{}1$, so for our $ n$th root of unity, we'll want it to reduce to that.

If we have $ z\equal{}e^{2\pi i/n}$, then $ z^n\equal{}e^{2\pi in/n}\equal{}e^{2\pi i}\equal{}1$, so we win!

But there are more solutions. Consider: $ e^{2\pi in}\equal{}\cos 2\pi/n\plus{}i\sin 2\pi/n$. Of course, sine and cosine are periodic over $ 2\pi$, so we can add or subtract as many $ 2\pi$ from $ 2\pi/n$ as we want.

$ 2\pi/n\plus{}2\pi \equal{} (2\pi\plus{}2\pi n)/n$. But we can factor the top to get $ 2\pi(n\plus{}1)/n$.

If we keep adding $ 2\pi$, we'll get $ 2\pi(n\plus{}2)/n$, etc. Eventually these just overlap and give the same solutions (which must happen because there can only be $ n$ solutions to $ z^n\equal{}1$).

Equivalently, subtracting $ 2\pi$ gives $ 2\pi(n\minus{}1)/n$, $ 2\pi(n\minus{}2)/n$, etc. If we let $ k$ be any integer from $ 0$ to $ n\minus{}1$, we encompass all of these solutions.

Thank You so much!!!
That last reply really made more since, especially the first proof!!!
Thanks, I really appreciate it!!!


group theory
abstract algebra
graph theory
superior algebra
superior algebra unsolved

I solved this problem with P. Hall Theorem but I 'd be so glad to see more solutions :

Suppose $G$ is group and $H$ is subgroup of $G$ with finite index. Prove that there exist $z_1,\dots,z_r\in G$ that $[G: H]=r$ and \[ \bigcup_{i=1}^{r}z_iH=\bigcup_{i=1}^{r}Hz_i=G \]

Beautiful problem, man :). At first, I thought you were referring to a group-theoretic result due to P. Hall, but then I realized that you were probably referring to Hall’s Marriage Theorem, right?

First let's notice that it can be reduced to the case when $G$ is finite. It is a very well-known result (which has been discussed several times on the forum; see these topics, for instance: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=22796[/url], [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=normal&t=19724[/url]) that there is a subgroup $N$ of $H$, normal and of finite index in $G$. Now, instead of working with $G$ and $H$, we work with $G/N$ and $H/N$ respectively. Everything is preserved: $[G/N: H/N]=[G: H]=r$, and for $z\in G$ the equivalence $z\in H\iff \bar z\in H/N$ holds, where $\bar z$ is the image of $z$ through the projection $G\to G/N$. In order to keep things simple, we just assume $G$ to be finite rather than introducing $N$.

Partition $G$ into the left cosets of $H$ and then into the right cosets of $H$. We have two partitions, each one into $r$ parts of equal size $|H|$. Construct a bipartite graph having bipartition $A,B$, where $A$ consists of the left cosets, $B$ consists of the right cosets, and two vetices are connected iff the respective sets (the left and right cosets representing the two vertices) have non-empty intersection. It's easy to check that this graph satisfies Hall's condition (every subset of size $1\le t\le r$ of $A$ has at least $t$ neighbours in $B$), so, by the Marriage Theorem, there is a perfect matching in the graph, or, in other words, we can number the left cosets $H_1,\ldots,H_r$ and the right cosets $H^1,\ldots,H^r$ s.t. $H_i\cap H^i\ne\emptyset,\ \forall i\in\overline{1,r}$. Now, for each $i\in\overline{1,r}$, just choose some $z_i\in H_i\cap H^i$.


geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

I have a question regarding connected sets.

Suppose two spheres in $R^{n}$ extouch each other at a single point. If $A$ is the set of point not outside the two spheres, is $A$ connected?

Also, I would be grateful if someone can tell me some general methods to show a set is connected besides using the definition  :D.

One method is to find a curve $\gamma_{ab}$ connecting any two points $a$ and $b$ in the set. In fact, it is enough to do this for a fixed point $b$. In your example, let $b$ be the common point of two spheres, and let $\gamma_{ab}$ be the line segment with endpoints $a$ and $b$.

Ok, now I see. So that means a set is path-connected iff it is connected?

Thank you.

Only one implication is true: path-connected $\implies$ connected.


parameterization
trigonometry
trig identities
Law of Cosines

The problem is under-determined: there are many points $ M$ such that angle $ AMB$ has measure $ \theta$.

Sorry about that, I meant to add another restriction. :blush:

[hide="Outline of solution"]Let's hold the parameter $ \theta$ and vary the parameter $ x$. Note that this results in an arc $ AB$ of some other circle with center $ P$, such that $ AOBP$ is a kite. Note that if we project $ M$ onto the opposite side of circle $ P$ to give $ M'$, the measure of $ \angle AM'P \equal{} 180 \minus{} \theta$, and thus the measure of $ \angle APB$ is $ 90 \minus{} \frac {1}{2} \theta$.

Note that $ AB \equal{} \sqrt {2} r \implies \frac {1}{2} AB \equal{} \frac {r}{\sqrt {2}}$. Let $ s$ be the radius of circle $ P$. Then, $ \frac {1}{2} AB \equal{} s \sin \left( \frac {90 \minus{} \frac {1}{2} \theta }{2} \right)$. Thus, $ s \equal{} \frac {2}{\sin \left( 45 \minus{} \frac {\theta}{4} \right) \sqrt {2}}$

Now, we know $ s$, and using LoC on $ \triangle MPB$, can find $ \angle MPB$ in terms of $ x$, $ r$, and $ \theta$.

We can also find $ PO$, since $ \angle APO \equal{} \frac {1}{2} \angle APB \equal{} 45 \minus{} \frac {\theta}{4}$ and $ \angle AOP \equal{} 45$, and we can thus determine $ \angle OAP\equal{} 90 \plus{} \frac{\theta}{4}$ and use law of cosines on $ \triangle OAP$.

With these, we can execute LoC on $ \triangle OMP$ to find $ OM$.[/hide]


find x,y integers $ 9x^2\minus{}39x\plus{}40\equal{}y^2$

hello, i have found
$ x\equal{}3,y\equal{}\minus{}2$ or $ x\equal{}3,y\equal{}2$
Sonnhard.

$ \Delta\equal{}81\plus{}36y^2$ Since the equation has an integer solution then it must be a perfect square

Let $ 81\equal{}c^2\minus{}(6y)^2 \Rightarrow c\minus{}6y\equal{}\plus{}\minus{}3^l$ and $ c\plus{}6y\equal{}\plus{}\minus{}3^q$ where $ l\plus{}q\equal{}4$ Hence $ 4y\equal{}3^{q\minus{}1}\plus{}3^{l\minus{}1}$ if $ q\minus{}1>0$ and $ l\minus{}1>0$ then $ 3\mid y$ whis is not true Since by the initial equation we have $ y^2\equiv 1 (\mod3)$ hence $ c\minus{}6y\equal{}\plus{}\minus{}3$ and $ c\plus{}6y\equal{}\plus{}\minus{}27$

conclusion $ (3,2);(3,\minus{}2)$ are the solutions  :)

$ (3x\minus{}6)^2 \equal{} 9x^2 \minus{} 36x \plus{} 36$, $ (3x\minus{}7)^2 \equal{} 9x^2 \minus{} 42 x \plus{} 49$. For $ x<0$ we have $ 9x^2 \minus{}39x \plus{} 40$ strictly between them, so no solution. For $ x\geq 0$ we have $ 9x^2 \minus{}39x \plus{} 40$ between them only if $ x\geq 3$. Accordingly $ x\equal{}3$ yields the only solutions.

It's a nice problem. Could anyone explain me why the only possible value of x is 3 ? And how do I know it ?

Thanks.

[quote="luiseduardo"]Could anyone explain me why the only possible value of x is 3 ? And how do I know it ?[/quote]

Well, in mavropnevma's solution he took two perfect squares and compared with $ 9x^{2}\minus{}39x\plus{}40$ accordingly the value of $ x$. So, some cases was deleted and the cases $ "x \in \{0,1,2,3 \}$, and only $ x\equal{}3$ worked.


A wooden cube $n$ units on a side is painted red on all six faces and then cut into $n^{3}$ unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is $n$?

$A.3$
$B.4$
$C.5$
$D.6$
$E.7$

[hide]
There are 6 faces on a cube. The total number of faces painted red will be $(6) (n^{2})$. The total surface area will be $\text{surface area for each cube}\times \text{number of cubes}= (6) (n^{3})$.
\begin{eqnarray*}\frac{6n^{2}}{6n^{3}}&=& \frac{1}{4}\\ \frac{1}{n}&=& \frac{1}{4}\\ &\implies& \boxed{n=4}\end{eqnarray*}
I divided by $n^{2}$ assuming $n \neq 0$.
[/hide]


Actually, if you have any experiance in biology (and judging by your post, you don't), then you'll realize that biology, chemistry, and math are all interconnected. But then again you think biology is worth anything, so I don't think you realize the importance of biology and chemistry in life.

[quote="pardesi"]
whom do u go to if u r ill???
to a mathematician :rotfl:  :rotfl:[/quote]

Haha. Here is a request [b]not[/b] to delete this topic; I want to show my bio teacher. We had a conversation along these lines before  :lol: .

[quote="Poincare"]why do you loser even like biology, it sucks, if you like biology you can't like math, period[/quote]

Who said that? Many Mathematics majors end up working as geneticists. Biology and Mathematics are actually more closely linked than you think. Using Mathematics we can model many biological scenarios and hence come up with solutions to biological problems.

[quote="pardesi"][quote="Poincare"]why do you loser even like biology, it sucks, if you like biology you can't like math, period[/quote]
whom do u go to if u r ill???
to a mathematician :rotfl:  :rotfl:[/quote]

LMAOWNED

[quote="7h3.D3m0n.117"][quote="BanishedTraitor"]Using Mathematics we can model many biological scenarios and hence come up with solutions to biological problems.[/quote]

Examples include the Hardy-Weinberg Equation and chi-square analysis for certain ascertaining genetic inheritance.  :)[/quote]

Exactly I mean eventually everything are related to some extent. People used to think that Biology and Chemistry might be incompatible until the need for chemical analysis became crucial. Also Physics and Biology and certainly Mathematics and Biology.

So yeah, Biology is not given the credit it deserves. It is a very diverse field. It is really more of an experimental Science at the moment (which might account for why Poincare doesn't like it) but nevertheless it is interesting especially when you do apply Chemistry/Physics/Mathematics to it. :D

[quote]7h3.D3m0n.117 wrote:
BanishedTraitor wrote:
Using Mathematics we can model many biological scenarios and hence come up with solutions to biological problems.

Examples include the Hardy-Weinberg Equation and chi-square analysis for certain ascertaining genetic inheritance.

Exactly I mean eventually everything are related to some extent. People used to think that Biology and Chemistry might be incompatible until the need for chemical analysis became crucial. Also Physics and Biology and certainly Mathematics and Biology.

So yeah, Biology is not given the credit it deserves. It is a very diverse field. It is really more of an experimental Science at the moment (which might account for why Poincare doesn't like it) but nevertheless it is interesting especially when you do apply Chemistry/Physics/Mathematics to it. [/quote]

Personally, I get loads of comments about biology being a subject that is meant only for those who like to learn things by rote, but then it has never bothered me as I always loved the subject.. Its just sad to think that in today's world, when cutting edge biotechnology is saving millions of lives, there are still people like the poster who believe that its a waste of time. In all honesty, its a waste of time replying to this post...  :!:

[quote="shruthibaskaran"]In all honesty, its a waste of time replying to this post...  :!:[/quote]

Then why are you doing that?? :rotfl:  :rotfl:  :rotfl:

well those have studied bio with interest only they can enjoy it
it may have been but now it is no parrot learning

after all,bandar kya jane nariyal ka swad
   i hope its right--i m no good in hindi


Four bugs are situated at each vertex of a unit square. Suddenly, each bug begins to chase its counterclockwise neighbor. It the bugs travel at $1$ unit per minute, how long will it take them to crash?

Please prove your answers!

Harder problem (more of a generalization):
$n$ bugs are situated at each vertex of a unit $n-gon$. Suddenly, each bug begins to chase its counterclockwise neighbor. It the bugs travel at 1 unit per minute, how long will it take them to crash?

[quote="tim1234133"][hide="The first problem"]
Not sure how rigorous this is, but...

As the bugs chase each other, they are eating up the side towards the neighbour that they are chasing. When they start, the side length is 1. When they crash in the middle, the side lengths have vanished, so they have each travelled 1 unit- and this would have taken 1 minute

This idea should also hold for any n-gon?[/hide][/quote]

That is the correct answer, but the n-gon problem is more difficult.

[quote="vishalarul"]For the $n-gon$ problem, I got:
[hide="Try it before you see the answer"]$\sin(\frac{180}{n})$[/hide]

[b]Also, please see my Problem Solving Marathon![/b] :P[/quote]

That doesn't work for a square.  :huh: And I thought the distance would increase as $n$ increases.

Sorry, I think I meant:
$\sin(\frac{360}{n})$

If the answer was $\sin\frac{360}{n}$, as n became very large (and the polygon tended towards a circle) the time taken would tend towards 0.

As lotrgreengrapes7926 said, the distance should increase as $n$ increases

[quote="vishalarul"]This is my final guess. If anyone finds a counterexample, I won't post the solution. If everyone thinks I'm right, then I will post the solution.

[hide="Answer :?: "]$\frac{\csc^{2}(\frac{180}{n})}{2}$[/hide][/quote]

I believe that's correct. Could you post a solution? :D

[hide="Solution"][b]Key Insight[/b]: The situation is rotationally symmetric in that there is no one "distinguished bug". If the starting position is an $n-gon$, then they will always maintain that configuration.

All calculuations are in degrees.

It's obvious that the center of the $n-gon$ doesn't move. Pick one of the bugs, and look at the line segment from the center of the $n-gon$ to the bug. This segment will rotate counterclockwise and (most importantly) shrink. When it has shrunk into 0, the bugs will crash. [i]From the point of view of this radial line[/i], the bug is always traveling at a $\frac{180}{n}$ angle. Since the bug travels at unit speed, its radial velocity component is just $1\cdot\cos(90-\frac{180}{n})=\sin(\frac{180}{n})$ units per minute, i.e., the radial line shrinks at this speed. Since the original length of the radial line was $\frac{1}{2\sin(\frac{180}{n})}$, it will just take
$\frac{1}{2\sin^{2}(\frac{180}{n})}$

$=\frac{\csc^{2}(\frac{180}{n})}{2}$ minutes for them to crash.[/hide]


Write $ 25^{2008}$ as a sum of five consecutive positive integers.

The five integers have an average of 25^2008/5, which also happens to be the middle number. Thus, the five integers are
$ 5(25^{2007}) \minus{} 2$,
$ 5(25^{2007}) \minus{} 1$,
$ 5(25^{2007})$,
$ 5(25^{2007}) \plus{} 1$, and
$ 5(25^{2007}) \plus{} 2$.

Yeah, and if they ask $ 25^{2008}$ as a product of 5 consecutive numbers, you can just take the fifth root of $ 25^{2008}$, you will get the middle number also.


I have a Mac OS X and it won't let me run BibDesk, why?

It keeps on saying: Drag this icon to your hard disk, once you are done, eject this icon.

I've done that like a million ltimes already!

How do I properly install it and open it?

Most of the things in the directions (configuring 10.3) I cannot do because the applications will not open.

I really want to use TeXnIC Center on vacation, but it's not turning out so well... :(

Please stop creating new threads for this single complaint.  There are several very dedicated users who read this forum and answer questions, but they are not your personal employees.  Sooner or later one of them will come along and help you, so be patient.  I am not a Mac user and so I can't help you, but it would probably be helpful if (on one of the threads you've already created, like this one) you listed in one place and in some detail what your system is and what exactly you've done so far and what exactly is happening (or not happening) as a result.  You also probably should try to read carefully through the documentation associated with the various programs you're trying to install.


At my school there are 6 8th grade teams. One of those teams is a special education team so none of us can be on it.

There are 9 friends I want to be on the same team as and 1 person I really don't want to be on the same team as.

a) What is the probability that those 9 friends will all be on my team and that one person will not?

b) What is the probability that those 9 friends WILL NOT be on my team and that one person will?

c) What is the probability exactly 4 of my friends will be on my team?

Assuming that each team can have any number of people:
[hide]By using the fact that the probability of someone being on your team is $\frac{1}{6}$, and the probability someone is not on your team is $\frac{5}{6}$, we have
a)$\frac{1^{9}}{6^{9}}*\frac{5}{6}=\frac{5}{6^{10}}$
b)$\frac{5^{9}}{6^{9}}*\frac{1}{6}=\frac{5^{9}}{6^{10}}$
c)$\frac{1^{4}}{6^{4}}*\frac{5^{5}}{6^{5}}=\frac{5^{5}}{6^{9}}$[/hide]

That's a nice solution, but they do have to even out the number of people on each team.

Oh Yeah, and you gorgot that you can't be on the special team, so there are only 5 reams left (assuming that "none of us" reffers to all the people talked about).

[quote="laughinghead505"] (assuming that "none of us" reffers to all the people talked about).[/quote]

yes it does.

And any number of people can be on a team. The teams usually have about 100 people.

Yes, but then everybody can be on the same team, which wouldn't work out too well.


Letters have replaced some of the digits in the addition expression below. Given that each digit 1 through 9 appears exactly once in the expression when the digits are revealed, which digit should replace the letter b?

           a 4 b
         +2 c 5
       _________
          d  1 e

(A) 3 (B) 6 (C) 7 (D) 8 (E) 9

[hide] It is clear that we have either $ c \equal{} 6$ or $ c \equal{} 7$. However, if $ c \equal{} 6$, by the given conditions, $ b \equal{} 8$ and $ e \equal{} 3$. Since each digit is used only once, $ a$ would have to be $ 7$. But combining this with the carry from the middle yields $ 7 \plus{} 2 \plus{} 1 \equal{} 10$, which is a contradiction. Hence, $ c \equal{} 7$. Once again, by utilizing the given conditions, we find that $ b \equal{} \boxed{3}$. ($ 643 \plus{} 275 \equal{} 918$). So the answer is A.[/hide]