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inequalities unsolved

it is the same thing with proving that 243xyztv<=(s-x)(s-y)(s-z)(s-t)(s-v) where s=x+y+z+t+v and x,y,z,t,v are the side lenghts of a pentagon and that i think it is not so hard but yet i have no time now to try it.

if my memory is correct, it's a special case of the famous mongolia inequality, you can search for it.

Yep Soarer :)
And, here it is : http://www.mathlinks.ro/Forum/viewtopic.php?t=26103

Thank you, mathmanman.
Wow!  I was close to that this inequality is more likely incorrect, than it is true!

[i]Errare humanum est..[/i] (I think this is well known and not considered as a non-english speaking, correct me if I'm wrong and I'll immediately translate this :) )
Well, arqady, maybe you could find a really nice solution for the inequality? ;)

Thx. In pdf given in your link there was a nice proof for my ineq using just Jensens ineq.

a=c=e=1,b=d=0
\[ \frac{a+b}{2} \cdot \frac{b+c}{2} \cdot \frac{c+d}{2} \cdot \frac{d+e}{2} \cdot \frac{e+a}{2}=\frac{1}{2^4} \]
\[ \frac{a+b+c}{3} \cdot \frac{b+c+d}{3} \cdot \frac{c+d+e}{3} \cdot \frac{d+e+a}{3} \cdot \frac{e+a+b}{3} =\frac{2^4}{3^5} \]
but we have:$\frac{1}{2^4}>\frac{2^4}{3^5}$
so it is wrong.
If you think,$a \neq 0$
Let $a \to 0$

[quote="pavel kozlov"][quote="zhaobin"]
but we have:$\frac{1}{2^4}>\frac{2^4}{3^5}$
[/quote]
But it's again isn't true:
$\frac{1}{16}<\frac{16}{243}$, since $243<256$ ;)[/quote]
Sorry.

[quote="Big Bolzano"]$\frac{a+b}{2} \cdot \frac{b+c}{2} \cdot \frac{c+d}{2} \cdot \frac{d+e}{2} \cdot \frac{e+a}{2} \leqslant \frac{a+b+c}{3} \cdot \frac{b+c+d}{3} \cdot \frac{c+d+e}{3} \cdot \frac{d+e+a}{3} \cdot \frac{e+a+b}{3}$

$a,b,c,d,e, \in R_+$[/quote]
A.Khrabrov  has found the beautiful proof for this inequality.

With iterating and some tweaking (for odd $n$, as for even $n$ it's straightforward), it seems to me that this nice trick can be used to show more generally for $n\ge 5$

$\prod_{cyc}\frac{a_1+a_2}{2}\le\prod_{cyc}\frac{s-a_1-a_2}{n-2}$ where $s=a_1+\cdots+a_n$.

I wonder if [quote="cefer"]the weaker one
$\frac{a_1+a_2}{2}.\frac{a_2+a_3}{2}.....\frac{a_n+a_1}{2}\leq \frac{a_1+a_2+a_3}{2\sqrt{2}}.\frac{a_2+a_3+a_4}{2\sqrt{2}}..... \frac{a_n+a_1+a_2}{2\sqrt{2}}$
from St.Petersburg math olimpiad-2000[/quote] is sharp. Or can we prove for $n\ge 6$
$\frac{a_1+a_2}{2}.\frac{a_2+a_3}{2}.....\frac{a_n+a_1}{2}\leq \frac{a_1+a_2+a_3}3.\frac{a_2+a_3+a_4}3..... \frac{a_n+a_1+a_2}3$ ?  :?

[quote="silouan"]Of, course. This is the famous Mongolian inequality....[/quote]
don't forget $a_i$ is ordered

This one seems in its behaviour quite similar to Shapiro's ineq. $\sum_{cyc}\frac{a_i}{a_{i+1}+a_{i+2}}\ge c\frac n2$
with a constant $c\approx.989$ (see http://mathworld.wolfram.com/ShapirosCyclicSumConstant.html)

With [i]unordered [/i]$a_i$'s, I just found (and certainly others before me) that
$\frac{a_1+a_2}{2}.\frac{a_2+a_3}{2}.....\frac{a_n+a_1}{2}\leq \frac{a_1+a_2+a_3}3.\frac{a_2+a_3+a_4}3..... \frac{a_n+a_1+a_2}3$ (*)
is false for all even $n\ge4$ and all odd $n\ge 9$:
For even $n=2k$, just put $a_{2i-1}=0,a_{2i}=2,$ yielding $1\le (\frac89)^k$ false.
For odd $n=2k+1$ ($k\ge4$), put $a_1=a_2=1$ and $a_{2i-1}=0,a_{2i}=2 \ (i\ge2)$, yielding $\frac14\le \frac13(\frac89)^{k-1}$ which is false for $k\ge4$.

So only the case $n=7$ of (*) remains open. I would tend to think it's true :D.

I guess a weaker form with a constant slightly smaller than $3$ on the RHS (like the St.Petersburg one) should be valid for all $n$ but that must be a very very tough proof. :huh: Well, if it holds with $\sqrt{8},$ maybe even an 'elementary' proof (I mean one without functional analysis, but it might well be rather lengthy) is possible for even $n$, as the ineq would be sharp.
By the way, has somebody out there a proof of the St.Petersburg one?

[quote="spanferkel"]
So only the case $n=7$ of (*) remains open. I would tend to think it's true :D.
[/quote]
I think you are not right :
$a_1=a_3=a_5=0,a_2=a_4=a_6=a_7=1,$  id est,  $(0,1,0,1,0,1,1).$
Hence, $(\frac{1}{2})^6\leq\frac{2^5}{3^7}.$ This  is not true.   ;)

[quote="cefer"]Here is the weaker one
Let $n\geq 3$ is positive integer.For all positive numbers $a_1,a_2,....,a_n$ show that
$\frac{a_1+a_2}{2}.\frac{a_2+a_3}{2}.....\frac{a_n+a_1}{2}\leq \frac{a_1+a_2+a_3}{2\sqrt{2}}.\frac{a_2+a_3+a_4}{2\sqrt{2}}.....\frac{a_n+a_1+a_2}{2\sqrt{2}}$
P.S It is from St.Petersburg math olimpiad-2000[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=793443#p793443

[quote="cefer"]Here is the weaker one
Let $n\geq 3$ is positive integer.For all positive numbers $a_1,a_2,....,a_n$ show that
$\frac{a_1+a_2}{2}.\frac{a_2+a_3}{2}.....\frac{a_n+a_1}{2}\leq \frac{a_1+a_2+a_3}{2\sqrt{2}}.\frac{a_2+a_3+a_4}{2\sqrt{2}}.....\frac{a_n+a_1+a_2}{2\sqrt{2}}$
P.S It is from St.Petersburg math olimpiad-2000[/quote]
Let $n\geq 3$ is positive integer.For all positive numbers $a_1,a_2,....,a_n$ show that $$\frac{(a_1+a_2+a_3)(a_2+a_3+a_4)\cdots(a_n+a_1+a_2)}{(a_1+a_2)(a_2+a_3)\cdots (a_n+a_1)}>2^{\frac{n}{2}}.$$
[size=50]p/6246593744[/size]

[quote=arqady][quote="mathmanman"]
Well, arqady, maybe you could find a really nice solution for the inequality? ;)[/quote]
 I shall certainly try to make it.   :P  :D[/quote]

Hmm, one friend (Maybe with the help of software?) noticed that \[\begin{aligned}7776\!\left(\mathrm{RHS}-\mathrm{LHS}\right)&=32\prod_{\textup{cyc}}{\!\left(a+b+c\right)}-243\prod_{\textup{cyc}}{\!\left(a+b\right)}\text{,}\\&=\sum_{\textup{cyc}}{\!\left(a-c\right)^{\mspace{-1mu}2}\mspace{-4.5mu}\left[11ade+13\!\left(d^2a+e^2c+e^2d\right)+14ce\!\left(a+d\right)+15d\!\left(ac+de\right)\right]}+\sum_{\textup{cyc}}{\!\left(a+b-c-d\right)^{\mspace{-1mu}2}\mspace{-4.5mu}\left[abc+3ade+14cde+17be\!\left(a+d\right)+18ace+19e^2\mspace{-4mu}\left(b+c\right)\right]}\text{,}\\&\geq0\textnormal{,}\end{aligned}\] which is evidently true for all $a, b, c, d, e\in\mathbb{R_+}$.
Also, he conjectured and noted that $$\frac{a+b+c}{3}\cdot\frac{b+c+d}{3}\cdot\frac{c+d+e}{3}\cdot\frac{d+e+a}{3}\cdot\frac{e+a+b}{3}\leqslant\frac{a^3+b^3+c^3+d^3+e^3}{5}\cdot\frac{ac+bd+ce+da+eb}{5}$$ holds for all \(a, b, c, d, e\in\mathbb{R_+}\).

Two similar unsolved problems: \[(a + b)(b + c)(c + d)(d + a)\ge(a + b + c + d)(bcd + cda + dab + abc)\mathtt{,}\quad\forall{a,b,c,d}\in\mathbb{R}\] and \[(a^2 + b^2 + c^2 + d^2 + e^2)(ab + bc + cd + de + ea)\ge(a + b + c + d + e)(cde + dea + eab + abc + bcd)\mathtt{,}\quad\forall{a,b,c,d}\in\mathbb{R_+}\textnormal{.}\]

[quote=arqady][quote="Big Bolzano"]$\frac{a+b}{2} \cdot \frac{b+c}{2} \cdot \frac{c+d}{2} \cdot \frac{d+e}{2} \cdot \frac{e+a}{2} \leqslant \frac{a+b+c}{3} \cdot \frac{b+c+d}{3} \cdot \frac{c+d+e}{3} \cdot \frac{d+e+a}{3} \cdot \frac{e+a+b}{3}$

$a,b,c,d,e, \in R_+$[/quote]
Let $a=b=c=2,d=e=1.$ Then
$\frac{a+b}{2} \cdot \frac{b+c}{2} \cdot \frac{c+d}{2} \cdot \frac{d+e}{2} \cdot \frac{e+a}{2}=2\cdot2\cdot1.5\cdot1\cdot1.5=9$ and
$\frac{a+b+c}{3} \cdot \frac{b+c+d}{3} \cdot \frac{c+d+e}{3} \cdot \frac{d+e+a}{3} \cdot \frac{e+a+b}{3}=$
$=2\cdot\frac{5}{3}\cdot\frac{4}{3}\cdot\frac{4}{3}\cdot\frac{5}{3}=\frac{200}{81}.$ ;)[/quote]

Why did you assume the variables to be a constant like 1 or 2....which meathod is this ....i have never seen this before ??

[quote=Rhapsodies_pro]Two similar unsolved problems: \[(a + b)(b + c)(c + d)(d + a)\ge(a + b + c + d)(bcd + cda + dab + abc)\mathtt{,}\quad\forall{a,b,c,d}\in\mathbb{R}\][/quote]

After full expanding, the inequality becomes $(ac-bd)^2 \ge 0$, which is obviously true.


geometry
circumcircle
trapezoid
symmetry
geometric transformation
reflection
cyclic quadrilateral

An elegant proof is needed please!

1.Prove that in a circle arcs of equal lenth subtend equal angles at any point on the circle.

2.Prove that the angle subtended by an arc at the center of the circle is twice the angle subtended by it anywhere on the circle.

PLEASE NOTE: Donot use the result of 2. to prove 1 and vise versa.
i.e. do not deduce one from the other.

$\textbf{Lemma 1:}$ In convex cyclic quadrilateral $ABCD$, $\angle{ABC}$ and $\angle{CDA}$ are supplementary.

$\emph{Proof.}$ Draw in the circumcenter O.  Then

\begin{eqnarray*}
m\angle{ABC} +  m\angle{CDA} = m\angle{ABO} + m\angle{OBC} + m\angle{CDO} + m\angle{ODA} \\ =  m\angle{BAO} + m\angle{BCO} + m\angle{OCD} + m\angle{OAD} \\
= m\angle{BCD} + m\angle{DAB}
\end{eqnarray*}

But, because the angles in any quadrilateral add up to $360^{\circ}$, it must be that $m\angle{ABC} + m\angle{CDA} = 180^{\circ}$.

$\textbf{Lemma 2:}$ For any convex cyclic quadrilateral $ABCD$, $\angle{ABD} \cong \angle{ACD}$.

$\emph{Proof.}$ This follows immediately if we consider a point $E$ on the arc connecting $A$ and $D$ that does not contain $B$ or $C$, since both $\angle{ABD}$ and $\angle{ACD}$ are supplementary to $\angle{DEA}$ by Lemma 1.

$\textbf{Proof of 1:}$ By Lemma 2, it is sufficient to prove the claim for two non-overlapping arcs. Call the arcs $\omega_1$ and $\omega_2$. Let $A$ and $B$ be the endpoints of $\omega_1$, and let $C$ and $D$ be the endpoints of $\omega_2$ in a manner so that $ABCD$ is an isosceles trapezoid with $\overline{BC} \parallel \overline{AD}$. The perpendicular bisector of line segments $\overline{BC}$ and $\overline{AD}$ intersects the circumcircle of $ABCD$ at two points; call them $E$ and $F$. By symmetry, the angles subtended by $\omega_1$ and $\omega_2$ from $E$ are congruent; likewise for $F$. By Lemma 2, however, we may move $E$ and $F$ about on the circumcircle provided that we do not move onto $\omega_1$ or $\omega_2$.

Remark: The claim as stated is not exactly true; if we take a point that lies on one of the arcs but not the other, then the angles subtended by the two arcs are supplementary. This is a consequence of applying Lemma 1 to either $E$ or $F$.

$\textbf{Proof of 2:}$ Let $A'$ be the point diametrically opposite $C$. By identity 1, $\angle{A'BC} \cong \angle{CDA'}$ (alternatively, we could reflect $D$ over diameter $\overline{A'D}$ and apply Lemma 2), thus, $\angle{A'BC}$ is a right angle. By Lemma 2, $\angle{BAC} \cong \angle{BA'C}$. Now, let $x = m\angle{BA'C} = m\angle{BA'O}$. But now $m\angle{A'BO} = m\angle{BA'O} = x$. Subtracting, $m\angle{CBO} = 90^{\circ} - x = m\angle{BCO}$. But, because the angles in $\triangle{BCO}$ add to $180^{\circ}$, it must be that $m\angle{BOC} = 2x$, establishing our claim.

Remark: The above solution is for the case where $A$ and $A'$ are on the same side of $\overline{BC}$, but the other case is solved in a similar manner.


integration
algebra
partial fractions
calculus
calculus computations

I am solving a differential equation :

$y''-ty=0$ with $y(0)=1$ and $y'(0)=0$

and I get to a point where I need to find the inverse laplace transform of :

$\frac{\int (s*e^{\frac{s^3}{3}})ds}{e^\frac{s^3}{3}}$

Could anyone help me! Or is it something that cannot be tranformed ?

EDIT:

Well I have one  more problem if it comes to talking about inverse Laplace tranforms:

let $F(s) = \frac{1}{s^2(s^2+a^2)}$

what is $L^{-1}(F(s))$

For your second question: partial fractions!

As for your first question: the differential equation is not constant coefficient, and transform methods (Laplace, Fourier, etc.) work much more cleanly with constant coefficient linear equations. At that, I can't see how you would get to that particular formula.

Yes the second one is easy but I am still stuck with the first problem. I tried to solve the problem using the power series method but still i get expressions wich I cannot find the inverse Laplace tranform. If anyone has some patience and time could you please try and solve the problem using Laplace tranforms  because I am out of ideas!

Let $L$ be the Laplace tranform operator.

$L(\frac{1}{a^2}\int_{0}^{at}x*\sin(at-x)dx) = \frac{1}{a^2}L(at)L(\sin(at))=\frac{1}{a^2}*\frac{a}{s^2}*\frac{a}{(s^2+a^2)}=\frac{1}{s^2*(s^2+a^2)}$

but the correct answer is : $\frac{a}{s^2*(s^2+a^2)}$

Could anyone please show me where the mistake is !


What is the last digit of the product of the odd numbers less than 1000? What is the second-last digit?

[hide="last digit"]
$ 1*3*5\equiv5\pmod{10}$. Multiplied by any odd number, that remains $ 5\pmod{10}$. So the last digit is $ 5$.[/hide]

well, we know that the second to last digit must be 2 or 7, otherwise the big number wouldn't be divisible by 25

[hide="penultimate digit"]$ 1*3*5*7*9\equiv45\pmod{100}$, and $ 45^2\equiv25\pmod{100}$, and $ 25*45\equiv25\pmod{100}$, so the last two digits are $ 25$.[/hide]

Not necessarily true
First of all, i used [url]http://www.alpertron.com.ar/BIGCALC.HTM[/url] and brute forced it :D
secondly, taking the product mod 4 would help
[hide]
The product mod 4 is $ (1 \cdot 3)^{25} \equiv 3 \pmod{25}$, and we already know that the product is $ 0 \pmod{25}$.
Combining these 2 mods tells us that the product is $ 75 \pmod{100}$, which means that the penultimate digit is actually $ \boxed{7}$ (and the last digit is 5)
[/hide]
[hide="what is wrong with darmani's approach"]
$ 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \equiv 45 \pmod{100}$ but this is not true for
$ 11 \cdot 13 \cdot 15 \cdot 17 \cdot 19$. In fact, this product is $ 35 \pmod{100}$, meaning that the product of the first 10 odd numbers is really $ 75 \pmod{100}$
as are the next 4 groups of 10 odd numbers.
the answer is then $ 75^5 \equiv 75 \pmod{100}$
[/hide]


Compute:

$ \sin{\frac{\pi}{8}}\cdot\sin{\frac{3\pi}{8}}\cdot\sin{\frac{\pi}{12}}\cdot\sin{\frac{5\pi}{12}}$

[hide]Notice that
$ \cos(A \minus{} B) \minus{} \cos(A \plus{} B) \equal{} 2 \sin A \sin B$
Therefore,
$ \sin{\frac {\pi}{8}}\cdot\sin{\frac {3\pi}{8}}\cdot\sin{\frac {\pi}{12}}\cdot\sin{\frac {5\pi}{12}}$
$ \equal{} \frac {1}{2}(\cos\frac {\pi}{4} \minus{} cos\frac {\pi}{2})\frac {1}{2}(\cos\frac {\pi}{3} \minus{} cos\frac {\pi}{2})$
$ \equal{} \frac {1}{4}\times\frac {\sqrt {2}}{2}\times\frac {1}{2}$
$ \equal{} \frac {\sqrt {2}}{16}$[/hide]

EDIT: Latex error


中国2005西部竞赛试题

Sorry but your pdf is very ungly . Could you improve it

Indeed.. it is unreadable :(
So, please, try and make it look better ;)

The file is in Chinese.
The problems are already in the Olympiad Section, so there is no need for the PDF.


A liar will always speak falsely while a truther will always tell the truth.

An island comprises entirely of Liars and Truthers..

After  occurance of a bank theft in the island, six suspects A, B, C, D, E and F were summoned to the local police station for questening.

It was known from reliable eyewitness accounts (Truthers)   that precisely two persons committed the said robbery

When the query relating to the identity of the perpetrator of the crime was posited before each of the six suspects, the following statements were received by way of responses:

A's Statements:
1. At most one of B and E is guilty.
 2. D is guilty.

B's Statements:
1. Both A and C are guilty.
2. E is not guilty.

C's Statements::
 1. At least one of D and E  is guilty.
 2. B  is not guilty.

D's Statements:
1. Exactly one of A and B  is guilty.
2. C  is guilty.

E's Statements:
1. Neither C nor D is guilty.
2. A  is guilty.

F's Statements:
1. D is guilty.
2. A is not guilty.

Which two of the six suspects actually perpetrated  the said crime?

[hide="Solution"]

From A's statements, we know that:
If A is a liar, then both B and E are guilty.
If A is a truth-teller, then D is guilty.
Therefore at least one of B, D, or E must be guilty.

If B is a truth-teller, than A, C, and one of B, D, or E must be guilty.  However, only two are guilty, so B is a liar and E is guilty.

This means that C is a truth-teller, because his first statement is true.  Then B is not guilty.

Returning to A, because B is not guilty, his first statement is true and therefore he is a truth-teller and D is guilty.

Therefore D and E are guilty.[/hide]

Wow ... I think there might be a little too much information here, because I only used the first three people's statements.

[quote="rjt"][hide]
If A is guilty: F is a liar, D is not guilty, A is a liar, B and E are guilty. But now ewe have three guilty people! Contradiction, so A is not guilty.

If B is guilty: C is a liar, D and E are innocent, A is a liar, B and E are guilty, contradiction. B is not guilty.

If C is guilty: D is a truth-teller, either A or B is guilty, but they can't be as we've already shown. Contradiction. C is not guilty.

If F is guilty: B is a liar (as only two people can be guilty, and so A and B can't be guilty also). Therefore E is guilty, A is a truth-teller, D is guilty. But we now have three guilty people, contradiction.So F is not guilty.

So, D and E must be guilty.
[/hide][/quote]

[hide]You didnt have to do that much.  once you showed a and b were telling the truth, they both said that person "X" did it, so you know its D and E[/hide]


:D We have:
$[\sum(x)]^2-2\sum xy =b \Leftrightarrow 2\sum xy =a^2-b$
$\sum x^3-3xyz=(\sum x)(\sum x^2-\sum xy)\Leftrightarrow 3xyz=\frac{2c-3ab+a^3}{2}$
$\sum x^4 =(\sum x^2)^2-2\sum x^2y^2=b^2-2[(\sum xy)^2-2xyz\sum x]$
Now we get result!!! :rotfl:  :lol:


Show that for all positive a,b,c
$ \sqrt {\frac {a^3}{a \plus{} b}} \plus{} \sqrt {\frac {b^3}{b \plus{} c}} \plus{} \sqrt {\frac {c^3}{a \plus{} c}} \ge \frac {\sqrt {2}}{2}(a\plus{}b\plus{}c)$

[quote="dgreenb801"]Show that for all positive a,b,c
$ \sqrt {\frac {a^3}{a \plus{} b}} \plus{} \sqrt {\frac {b^3}{b \plus{} c}} \plus{} \sqrt {\frac {c^3}{a \plus{} c}} \ge \frac {\sqrt {2}}{2}(a \plus{} b \plus{} c)$[/quote]
[hide]By Holder we obtain:
$ \left(\sum_{cyc}\sqrt {\frac {a^3}{a \plus{} b}}\right)^2\sum_{cyc}(a\plus{}b)\geq(a\plus{}b\plus{}c)^3.$
It gives $ \sum_{cyc}\sqrt {\frac {a^3}{a \plus{} b}}\geq \frac {\sqrt {2}}{2}(a \plus{} b \plus{} c).$[/hide]

The following is also true:
\[ \sqrt {\frac {a^3}{a \plus{} b}} \plus{} \sqrt {\frac {b^3}{b \plus{} c}} \plus{} \sqrt {\frac {c^3}{a \plus{} c}} \le\sqrt{\frac 32(a^2\plus{}b^2\plus{}c^2)}\]

[quote="dgreenb801"]Show that for all positive a,b,c
$ \sqrt {\frac {a^3}{a + b}} + \sqrt {\frac {b^3}{b + c}} + \sqrt {\frac {c^3}{a + c}} \ge \frac {\sqrt {2}}{2}(a + b + c)$[/quote]

$ \iff$ $ \sum \frac {a}{\sqrt {1 + \frac {b}{a}} \geq \frac {\sqrt {2}}{2}$

by convexity of $ x\rightarrow \frac {1}{\sqrt {1 + x}}$ :

$ \sum af(\frac {b}{a})\geq f(a + b + c) = f(1) = \frac {\sqrt {2}}{2})$

Yes! That was the solution I had in mind. I wanted to make a point of using Jensen's with $ af(\frac {b}{a}) \plus{} bf(\frac {c}{b}) \plus{} cf(\frac {a}{c}) \ge (a \plus{} b \plus{} c)f(1)$.

the inequality is homogenous , so we can assume that $ a\plus{}b\plus{}c\equal{}1$ ;)

[quote="dgreenb801"]Show that for all positive a,b,c
$ \sqrt {\frac {a^3}{a \plus{} b}} \plus{} \sqrt {\frac {b^3}{b \plus{} c}} \plus{} \sqrt {\frac {c^3}{a \plus{} c}} \ge \frac {\sqrt {2}}{2}(a \plus{} b \plus{} c)$[/quote]
[hide="Solution"]
Just by C.S:

$ (\sum \sqrt {\frac {a^3}{a \plus{} b}})(\sum \sqrt {a^2 \plus{} ab}) \geq (a \plus{} b \plus{} c)^2,$

and:

$ \sum \sqrt {a^2 \plus{} ab} \equal{} \sum \sqrt {a} \sqrt {a \plus{} b} \leq\sqrt {2}(a \plus{} b \plus{} c),$

$ \Leftrightarrow \sum \sqrt {\frac {a^3}{a \plus{} b}} \geq \frac {(a \plus{} b \plus{} c)^2}{\sum \sqrt {a^2 \plus{} ab}} \geq \frac {(a \plus{} b \plus{} c) ^2}{\sqrt {2}(a \plus{} b \plus{} c)} \equal{} \frac {\sqrt {2}}{2}(a \plus{} b \plus{} c)$[/hide]

[quote="nayel"]The following is also true:
\[ \sqrt {\frac {a^3}{a \plus{} b}} \plus{} \sqrt {\frac {b^3}{b \plus{} c}} \plus{} \sqrt {\frac {c^3}{a \plus{} c}} \le\sqrt {\frac 32(a^2 \plus{} b^2 \plus{} c^2)}
\]
[/quote]

Nice one !  :)
Use Cauchy Swartz and the Vasc's inequality , namely : $ \sum\sqrt{\frac{a}{a\plus{}b}}\leq\frac{3}{\sqrt{2}}$

Thanks, but it's not actually mine. I also proved it with Cauchy-Schwarz a long time ago while solving another problem [url=http://www.mathlinks.ro/viewtopic.php?p=986999#986999]here[/url]. :)


Given $ a, b, c \geq\ 0$. Prove that:
$ (a^4\plus{}b^4\plus{}c^4)(a^3\plus{}b^3\plus{}c^3)(a\plus{}b\plus{}c) \geq\ 9abc(a^5\plus{}b^5\plus{}c^5)$

@tran300491: Nice to make friend with u.
My name is Tran Quoc Anh (you can call me is nhocnhoc)
My YM: langtukhongnha1012. We can talk more on YM  :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0$. Prove that:
$ (a^4 \plus{} b^4 \plus{} c^4)(a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c) \geq\ 9abc(a^5 \plus{} b^5 \plus{} c^5)$
[/quote]
SOS kills it easy enough. :wink:

@arquady: Yes, my solution is also SOS :D
Ineq became:
$ \Leftrightarrow  \frac{a^3\plus{}b^3\plus{}c^3}{3abc}  \geq\ \frac{3(a^5\plus{}b^5\plus{}c^5)}{(a\plus{}b\plus{}c)(a^4\plus{}b^4\plus{}c^4)}$
$ Sa \equal{} \frac{a\plus{}b\plus{}c}{2abc} \plus{} \frac{a^3\plus{}2bc(b\plus{}c)\minus{}(b^2\plus{}bc\plus{}c^2)(a\plus{}b\plus{}c)}{(a\plus{}b\plus{}c)(a^4\plus{}b^4\plus{}c^4)}$
We easily to see $ Sa,Sb \geq\ 0$
We only need to prove that: $ Sb\plus{}Sc \geq\ 0$
We have:
$ Sb\plus{}Sc \geq\ \frac{a\plus{}b\plus{}c}{abc} \minus{} \frac{2abc\plus{}2(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)}{(a^4\plus{}b^4\plus{}c^4)(a\plus{}b\plus{}c)}
\geq\ \frac{a\plus{}b\plus{}c}{abc} \minus{}\frac{20(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)}{9(a^4\plus{}b^4\plus{}c^4)} > 0$
Because $ (a\plus{}b\plus{}c)^5 \geq\ 81abc(a^2\plus{}b^2\plus{}c^2)$
$ \Rightarrow$ q.e.d


calculus ratio limit integration LaTeX factorial function

Find the limit as n-> :inf: of (2^n)/(n!).

Now the answer is obviously 0, but how would one show this rigorously (ie in such a way that one would receive full credit were this problem given on the calculus AP exam)? Thanks.

take the limit as x-> :inf: of f(x+1)/f(x), and if this is less than 1, it approaches 0. I can't remember the names of stuff well, but i think thats something like the ratio test.

edit:fixed a dumb typo

Ah, ok, you're all suggesting that I prove that the series converges first and THEN by that, say that the seq tends to 0. Sounds good. Thanks.

Just use the ratio test: each term after 2 is multiplied by something less than 2/3, so it's worse than (2/3)^n, so it's 0.

Oh.  Well, whatever :).  The ratio test works, too, maybe?  What is the ratio test?

I think that was both ratio and comparison, actually. Blah, names.

Hmm... Is testing values ok too?

n=0 --> 0
n=1 --> 1
n=2 --> 1.414
n=3 --> 1.817

Since its increasing, it will eventually be infinite... Is that good enough?

Intuitively, that gives you the answer, but its not formal. How do you know it will keep increasing?

And even if you knew it was increasing, how do you know it is increasing to infinity?  For example, (1 + 1/n)n is also increasing, but not to infinity.

Are questions like these -- of this difficulty-- likely to show up on the BC exam? If they are, then I am in big trouble because I have no clue how to approach them.

They don't ask you to <i>prove</i> anything on the BC exam, if I recall correctly. However, there are open-ended questions which ask for you to elaborate on how you get your answers. They ask you to show your work and explain your results.

[quote="rlee"]How about this one? Prove:

[tex]\lim_{n\to\infty}\sqrt[n]{n!}=\infty[/tex][/quote]

We just show that [tex]\sqrt[n]{n!}[/tex] get arbitrarily large for sufficiently large n.

say we want [tex]\sqrt[n]{n!} > k[/tex]  for some positive k.

then this is equivalent to [tex]n! > k^n[/tex]. We show [tex]\displaystyle\frac{n!}{k^n}[/tex] eventually becomes greater than 1, for large enough n.
[tex]\exists m[/tex] st. [tex]m\in\mathbb{Z}[/tex] and [tex]m > k[/tex].
Define [tex]\displaystyle c = \frac{m!}{k^m}[/tex]. c is a positive constant.

Consider terms of the form [tex]\frac{p}{k}[/tex] for integral p greater than m. Each of these terms is greater than 1, and as p gets larger,  they increase without bound, as the sequence of terms is in arithmetic progression. The product of all these terms increases without bound as we add more terms.

However [tex]\displaystyle\lim_{n\to\infty}\frac{n!}{k^n} = \lim_{n\to\infty}c \prod_{p=m+1}^n\frac{p}{k} = \infty > 1.[/tex]

The claim follows.

This board is helping me learn how to use LaTeX. Yay.

As for LaTeX, it is better to use \times or \cdot for multiplication rather than *.

Interesting. I'm too tired to think about that too much, so I'll look at it more in depth later.

I actually figured out a different method. Anyone want to try it using concepts only up to Calculus II (equivalent of BC)?

Well, it's pretty easy to see that n! :ge: (n/2)n/2.  Namely, in the product of n!, just compare the ceiling(n/2) largest terms with n/2.  Thus the nth root of n! is at least :sqrt:(n/2).

[hide]Ratio test: If the absolute value of f(n+1)/f(n) < 1, the series converges. So, under those rules, the series would converge:

{[2^(n+1)]/[(n+1)!]}/{[2^(n)]/[(n)!]} = (2)/(n+1)

As n approaches infinity, (2)/(n+1) approaches 0, so your series would converge.[/hide]

It's not a series. It's a limit.

Edit: Also, even assuming that the ratio test works, you would have proved that the limit is a constant value. You're trying to prove that it's  :inf: .

For the original question on this thread, we can prove that n! :ge: n 2n-2.  We get that by comparing each of the middle n - 2 terms of n! with 2.

Oops, sorry. I left out "if you know the series converges, than the limit obviously has to go to 0."

Oh, sorry. Thought you were talking about the other problem.

Convergence of a series is implicative of the limit approaching 0, it is a MUST for all functions.

Proving that a series will converge is sufficient for showing that the limit goes to zero.

Also, why wouldn't we just use the squeeze theorem for all of the problems aforementioned?

If you wish to prove that limit goes to zero, and it's something with a factorial in it, just squeeze it with another KNOWN function that also goes to zero, thus by the squeeze theorem, approaches zero.

[quote="zabelman"]here's a pretty slick solution:

Since [tex]\displaystyle \sum_{k=0}^{\infty} \frac{2^k}{k!}[/tex] converges to [tex]e^2[/tex], the terms in the infinite sum must tend toward 0.[/quote]

via ratio test:

[tex]\begin{eqnarray*}
\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n} &=& \lim_{n\rightarrow\infty} \frac{2^{n+1}\cdot n!}{(n+1)!\cdot 2^n}\\
&=&\lim_{n\rightarrow\infty} \frac{2\cdot 2^n\cdot n!}{(n+1)\cdot n!\cdot 2^n}\\
&=&\lim_{n\rightarrow\infty} \frac{2\cdot 2^n\cdot n!}{(n+1)\cdot n!\cdot 2^n}\\
&=&\lim_{n\rightarrow\infty} \frac{2}{n+1}\\
&=&0
\end{eqnarray*}[/tex]

btw, how does one do absolute value in latex?


By p, do you mean primes? In that case, I think I have a solution, but it probably isn't correct.
[hide="Solution"]
We can make both sides exponents of 2. After crossing out logs, we get:
$2^{x^{2}-x-1}=\frac{2^{2^{x}}}{x^{2}+2^{x}}$
We can multiply both sides by $x^{2}+2^{x}$ and expand. After simplifying, we can divide both sides by $2^{x^{2}-1}$ and subtract 1 from both sides. We can look at this equation modulo x. The left side of the equation becomes 0 and the right side reduces to $2^{2^{x}}-1$. After taking the logarithim base two of each side, we find that $0=2^{x}$. There are no solutions to this congruence modulo $x-1$, so the original equation has no prime x that satisfy the equation.[/hide]

[quote="ashrafmod"]solve the equathon on the set of  p .n

$x^{2}-x-1=2^{x}-\log_{2}\ (x^{2}+2^{x})$[/quote]
If $2^{x}>x^{2}$, then $log_{2}(x^{2}+2^{x})<log_{2}(2*2^{x})=x+1$, therefore $2^{x}-log_{2}(x^{2}+2^{x})>x^{2}-x-1$. Analogy consider case $2^{x}<x^{2}$.
Therefore it is equavalent to $2^{x}=x^{2}$. It give 2 solutions. Positive is x=2.

Dear always-so-untidy Rust, what about $x=4$?  :roll:

Yes. $2^{x}=x^{2}$ had 2 positive solutions x=2 and x=4 and 1 negative solutions.


I am enrolled in the Intermediate Algebra class.  The first class was yesterday, and I cannot find the message board.  It doesn't appear under the Online School section of the forum like it's supposed to.  Has the message board not been created yet, or is there something wrong?


Suppose that Bill and Chris each have a whole number of dollars.
Given that $ \frac{5}{6}$ the value of Chris's money is the same amount as $ \frac{3}{4}$ the value of Bill's money, what is the smallest whole number of
dollars they could have together?

[quote="GameBot"]Suppose that Bill and Chris each have a whole number of dollars.
Given that $ \frac {5}{6}$ the value of Chris's money is the same amount as $ \frac {3}{4}$ the value of Bill's money, what is the smallest whole number of
dollars they could have together?[/quote]

B=bill's money and C=chris' money

$ \frac56\cdot C=\frac34\cdot B$
Divide both sides by $ \frac34\cdot C$, you get
$ \frac{B}{C}=\frac{10}9$
The least is Bill 10 and Chris 9. the answer is 19


geometry
rectangle
combinatorics unsolved
combinatorics

1.If r is a mXn rectangle formed from unit squares, find the number of unit squares containing a segment of one diagnol of R.

2.Find the number of ways in which nine digits 1,2,3,....9 can be arranged in a 3X3 square array so that the determinant of this array is an odd number.Each number is to be used only once.

3.Given $ n^{2}$ different real numbers and having arranged them into an nXn array, we mark the greatest one in each column and the smallest one in each row. Find the number of arrangements in which exactly 2n numbers are marked.

no one is answering this post....
if u guys feel this is too easy can u atleast tell me how to get started on these problems?

[hide="Hint for third"]
It is obvious that there should be no number in the required array that is maximum in its column and minimum in its row. Prove that there may be at most one such number in any array.
[/hide]

1 is a rather well-known problem. The answer is $m + n - \gcd(m,n)$; I'll let you figure out how. Hint: Let $d = \gcd(m,n)$; let $m = dx, n = dy$. Find out what happens in an $x \times y$ board.


oh sorry i forgot the condition $abc = 1$

Well, indeed it is simple if you don't try tu use smart ideas. Just work! It follows immediately that $ (a^2b+b^2c+c^2a)(b^2a+c^2b+a^2c)=3+a^3+b^3+c^3+a^3b^3+b^3c^3+c^3a^3 $. Thus, it is enough to prove that
 $\sqrt{3+x+y+z+xy+yz+zx}\geq 1+\sqrt[3]{2+x+y+z+xy+yz+zx} $ for any $ xyz=1$. Since $ x+y+z+xy+yz+zx\geq 6$ it is enough to prove that $\sqrt{3+x}\geq 1+\sqrt[3]{2+x} $ for any $ x\geq 6$. Put $ 2+x=u^3 $ and this will reduce to a trivial inequality.

Just want to add that it is Korean MO 2001 problem.


floor function
algorithm
continued fraction

Could anyone kindly explian how to get the smallest solution to

$x^2-dy^2=1$

via continued fraction of d? I tried mathworld, but it wasnt to clear :(

$x^2 + dy^2 = 1$ isn't too interesting, so I think you meant $x^2 - dy^2 = 1$.  :?  :D
For this latter one, all the positive solutions are of the form $x = h_n, y = k_n$, where $\frac{h_n}{k_n}$ are the convergents to $\sqrt{d}$, so you can just check these convergents until you find the first one that satisfies the equation. (Not all these convergents will be solutions.)

I have a spreadsheet that generates them with small integer arithmetic; with arbitrary precision integers you could generate the exact convergents and easily pick out which one is the solution.

Let $a_{-1}=0,a_0=1,b_{-1}=1,b_0=0$. Also let $c_{-1}=-d$ and $c_0=1$.

For $n\ge0$, define
$k_n=\mathrm{sign}(c_{n-1})\sqrt{c_nc_{n-1}+d}$
(The $c_n$ have alternating sign; this is just a convenience to avoid having to carry an alternating sign variable. Also, this always produces an integer.)
$q_n=\lfloor\frac{\sqrt{d}}{|c_n|}-\frac{k_n}{c_n}\rfloor= \lfloor\frac{\lfloor\sqrt{d}\rfloor+|k_n|}{|c_n|}$
$c_{n+1}=c_nq_n^2+2k_nq_n+c_{n-1}$

Also for $n\ge0$, let
$a_{n+1}=q_na_n+a_{n-1}$
$b_{n+1}=q_nb_n+b_{n-1}$

Then $a_n^2-d\cdot b_n^2=c_n$, and the $\frac{a_n}{b_n}$ are the convergents of the continued fraction for $\sqrt{d}$.
There are other, easier to understand, algorithms, but I like the way that this one uses only integer variables. Even the square roots in the formula are integers.

The only large integers that appear here are the $a_n$ and $b_n$; the other sequences are uniformly bounded by $2\sqrt{d}$ except for $c_{-1}$.

Try it for $d=4861$; the minimal solution there is $a\approx 1.153\cdot10^{109}$, $b\approx 1.654\cdot10^{107}$.

Ok i tried it a bit, but thing dont seem to go together too well :(
For the $k_n$ bit, firstly for $k_0$ the number under the squre root is $\sqrt{-8}$
Even ignoring the minus sign, it still isnt an integer like you say it should be? Any clue???

That's my fault- wrong formula. That should be $k_n=\mathrm{sign}(c_{n-1})\sqrt{c_nc_{n-1}+d}$.

$k_0$ should always be 0.

I'll edit my previous post.


calculus
derivative
trigonometry
real analysis
real analysis unsolved

How do you prove that the Taylor series at 0 of sin x converges for all x?

That depends on what we have already proved. Let us suppose that we have proved Taylor's Theorem with the Lagrange or mean value form of the remainder: for sufficiently smooth $f,$

$f(x)=\sum_{k=0}^{n}\frac{f^{[k]}(a)}{k!}(x-a)^{k}+\frac{f^{[n+1]}(\xi)}{(n+1)!}(x-a)^{n+1}$

for some $\xi$ between $a$ and $x.$

Assuming we now that all of the derivatives of $\sin x$ of any order are uniformly bounded by $1,$ we can show that that remainder term is no larger than $\frac{|x|^{n+1}}{(n+1)!}$ and for any fixed $x,$ that tends to zero as $n\to \infty.$

The convergence is uniform on any interval $[-M,M]$ but is not uniform on $\mathbb{R}.$


[color=purple]Given triangle ABC in plane has area S.Prove that :
 $a^{2}+b^{2}+c^{2}\ge 4S(\sec\frac{A}{2}+\sec\frac{B}{2}+\sec\frac{C}{2}-\sqrt{3})+[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]$
 Good luck  :lol: [/color]

It is wrong the sign must be $\le$, and write clearly reference it is from Bottema's book.

[color=purple] Ok ,  :lol:  and where is the proof ? [/color]


integration
calculus
calculus computations

:blush: $\int_{0}^{+\infty}\frac{dx}{1+x^\alpha sin^{2}x}.$
    Find $\alpha$ to converge or diverge  intergration

Already posted by little_cat in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=129552]this thread[/url]. Please, continue the discussion there.


calculus
integration
logarithms
function
geometry
trigonometry
calculus computations

[quote="kunny"][color=darkred]Evaluate $ \int_{0}^{1}e^{\sqrt{e^{x}}}\ dx+2\cdot\int_{e}^{e^{\sqrt{e}}}\ln (\ln x)\ dx$.[/color][/quote]
[color=darkblue]For a increasing bijective and continue function $ f: [a,b]\rightarrow [c,d]$ is truly the well-known relation $ \boxed{\int_{a}^{b}f(x)\ dx+\int_{c}^{d}f^{-1}(x)\ dx = bd-ac}$.

[b]Proof.[/b] Using the substitution $ t = f^{-1}(x)$, i.e. $ x = f(t)$ and $ dx = f'(t)dt$ obtain $ \int_{c}^{d}f^{-1}(x)\ dx$=$ \int_{a}^{b}tf'(t)\ dt = bd-ac-\int_{a}^{b}f(t)\ dt$ a.s.o.

Can prove otherwise ([b]with areas[/b]) this remarkable identity using the simple property "[b]the graphs $ \mathcal G_{f^{-1}}$ and $ \mathcal G_{f}$ are symetrically w.r.t. the line $ y = x$"[/b]. Indeed,

denote the points $ \left\|\begin{array}{c}A(a,0)\ ,\ B(b,0)\ ;\ M(a,c)\ ,\ N(b,d)\\\\ C(0,c)\ ,\ D(0,d)\ ;\ M'(c,a)\ ,\ N'(d,b)\\\\ P(c,0)\ ,\ Q(d,0)\end{array}\right\|$. From the areas-relations $ [OBND]=[OAMC]+[A\widehat{MN}B]+[C\widehat{MN}D]$

obtain required identity because $ [OBND]=bd$, $ [OAMC]=ac$, $ [A\widehat{MN}B]=\int_{a}^{b}f(x)\ dx$ and $ [C\widehat{MN}D]=[P\widehat{M'N'}Q]=\int_{c}^{d}f^{-1}(x)\ dx\ .$

[b]Particular case.[/b] For the function $ f: [0,1]\rightarrow\left[e,e^{\sqrt e}\right]$ obtain  $ \int_{0}^{1}e^{\sqrt{e^{x}}}\ dx+2\cdot\int_{e}^{e^{\sqrt{e}}}\ln (\ln x)\ dx = e^{\sqrt e}$ because $ f^{-1}(x) = 2\cdot\ln (\ln x)\ .$[/color]

Sorry I made a typo, I should have multiply 2 by the second definite integral. :oops:

Your answer is correct,　Virgil Nicula. :)

I didn't know the solution related to the area of functions $ f(x)$ and $ f^{\minus{}1}(x)$.

Now this reminds of me Young's inequality.

My solution isn't same yours. We will wait other's solution, if no one posted it, I will post my solution later.

kunny

[quote="kunny"]Evaluate $ \int_{0}^{1}e^{\sqrt{e^{x}}}\ dx\plus{}2\int_{e}^{e^{\sqrt{e}}}\ln (\ln x)\ dx$.[/quote]

$ \text{(i)}\;\;\;\int_{0}^{1}e^{\sqrt{e^{x}}}\ dx\equal{}2\int_{1}^{\sqrt e}\;\frac{e^{u}}{u}\;du\;\;\;\quad (\text{putting}\;\;\;u^{2}\equal{}e^{x})$

$ \text{(ii)}\;\;\;2\int_{e}^{e^{\sqrt{e}}}\ln (\ln x)\ dx\equal{}2\int_{1}^{\sqrt e}\;e^{u}\cdot\ln u\;du\;\;\;\quad (\text{putting}\;\;\;u\equal{}\ln x)$

add the two to get $ \text{I}\;\equal{}\;2\int_{1}^{\sqrt e}\;e^{u}\left(\frac{1}{u}\plus{}\ln u\right)\;du$

now, $ \frac{d}{du}\left(e^{u}\cdot\ln u\right)\;\equal{}\;e^{u}\left(\frac{1}{u}\plus{}\ln u\right)$

and so, $ \text{I}\;\equal{}\;2\;e^{u}\cdot\ln u\bigg|_{1}^{\sqrt e}\;\equal{}\;{e^{\sqrt{e}}}$

Here is my solution, but this is same as misan's.

Let $ \sqrt{e^{x}}\equal{}t$, $ \int_{0}^{1}e^{\sqrt{e^{x}}}\ dx\equal{}2\int_{1}^{\sqrt{e}}\frac{1}{t}e^{t}\ dt$

$ \equal{}2\left(\left[e^{t}\ln t\right]_{1}^{\sqrt{e}}\minus{}\int_{1}^{\sqrt{e}}e^{t}\ln t\ dt\right)$

$ \Longleftrightarrow\int_{0}^{1}e^{\sqrt{e^{x}}}\ dx\plus{}2\int_{1}^{\sqrt{e}}e^{t}\ln t\ dt\equal{}e^{\sqrt{e}}$.

Let $ e^{t}\equal{}u$, $ \int_{1}^{\sqrt{e}}e^{t}\ln t\ dt\equal{}\int_{e}^{e^{\sqrt{e}}}\ln(\ln u)\ du$.

$ \therefore\int_{0}^{1}e^{\sqrt{e^{x}}}\ dx\plus{}2\int_{e}^{e^{\sqrt{e}}}e^{x}\ln x\ dx\equal{}e^{\sqrt{e}}$.

[color=darkred]For a increasing bijective and continue function $ f: [a,b]\rightarrow [c,d]$ is truly the well-known relation $ \boxed{\int_{a}^{b}f(x)\ dx+\int_{c}^{d}f^{-1}(x)\ dx = bd-ac}$.[/color]

[color=darkblue][b]Proof.[/b] Using the substitution $ t = f^{-1}(x)$, i.e. $ x = f(t)$ and $ dx = f'(t)dt$ obtain $ \int_{c}^{d}f^{-1}(x)\ dx$=$ \int_{a}^{b}tf'(t)\ dt = bd-ac-\int_{a}^{b}f(t)\ dt$ a.s.o.

Can prove otherwise ([b]with areas[/b]) this remarkable identity using the simple property "[b]the graphs $ \mathcal G_{f^{-1}}$ and $ \mathcal G_{f}$ are symetrically w.r.t. the line $ y = x$"[/b]. Indeed,

denote the points $ \left\|\begin{array}{c}A(a,0)\ ,\ B(b,0)\ ;\ M(a,c)\ ,\ N(b,d)\\ \\ C(0,c)\ ,\ D(0,d)\ ;\ M'(c,a)\ ,\ N'(d,b)\\ \\ P(c,0)\ ,\ Q(d,0)\end{array}\right\|$. From the areas-relation $ [OBND] = [OAMC]+[A\widehat{MN}B]+[C\widehat{MN}D]$

obtain required identity because $ [OBND] = bd$, $ [OAMC] = ac$, $ [A\widehat{MN}B] =\int_{a}^{b}f(x)\ dx$ and $ [C\widehat{MN}D] = [P\widehat{M'N'}Q] =\int_{c}^{d}f^{-1}(x)\ dx\ .$

[b]Another some particular identities.[/b]

$ 1\blacktriangleright$ $ \int_{0}^{\frac{\pi}{2}}\arctan (\sin x)\ dx+\int_{0}^{\frac{\pi}{4}}\arcsin (\tan x)\ dx =\frac{\pi^{2}}{8}\ .$

$ 2\blacktriangleright$ $ \int_{0}^{1}e^{-x^{2}}\ dx+\int_{1}^{\frac{1}{e}}\sqrt{-\ln x}\ dx =\frac{1}{e}\ .$[/color]


modular arithmetic
number theory proposed
number theory

Let $ p$ be an odd prime number.Show that if $ q$ is a prime divisor of $ x^{p\minus{}1}\plus{}x^{p\minus{}2}\plus{}...\plus{}1$ then either $ q\equal{}p$ or $ p/q\minus{}1$

Let $ o_n(a)$ denote the order of $ a$ modulo $ n$.

We have $ q|x^{p\minus{}1}\plus{}x^{p\minus{}2}\plus{}\ldots\plus{}1\implies q|x^p\minus{}1\implies o_q(x)|p\implies o_q(x)\in\{1,p\}$.

If $ o_q(x)\equal{}p$, then $ p|\varphi(q)\implies p|q\minus{}1$.

If $ o_q(x)\equal{}1$, then $ x\equiv1\pmod{q}$. But $ 0\equiv x^{p\minus{}1}\plus{}x^{p\minus{}2}\plus{}\ldots\plus{}1\equiv 1\plus{}1\plus{}\ldots\plus{}1\equiv p\pmod{q}$. So $ q\equal{}p$.

Q.E.D.

[quote="Johan Gunardi"]Let $ o_n(a)$ denote the order of $ a$ modulo $ n$.

We have $ q|x^{p \minus{} 1} \plus{} x^{p \minus{} 2} \plus{} \ldots \plus{} 1\implies q|x^p \minus{} 1\implies o_q(x)|p\implies o_q(x)\in\{1,p\}$.

If $ o_q(x) \equal{} p$, then $ p|\varphi(q)\implies p|q \minus{} 1$.

If $ o_q(x) \equal{} 1$, then $ x\equiv1\pmod{q}$. But $ 0\equiv x^{p \minus{} 1} \plus{} x^{p \minus{} 2} \plus{} \ldots \plus{} 1\equiv 1 \plus{} 1 \plus{} \ldots \plus{} 1\equiv p\pmod{q}$. So $ q \equal{} p$.

Q.E.D.[/quote]
I have solution the same with slution by you  :oops:


Does anybody know the date and location of this year's Balkan Olympiad?
As i know, it will be held in Macedonia. But i don't know the city.

Does anybody know anything about Balkan 2008?

(apart from the location  :P )

Also as I am informed it will be held in first week of May.
Location can be also Ohrid.


geometry
rectangle
rotation
function
complex analysis

Find a conformal map that takes
$\Omega = \{z \in \mathbb{C}\; | \; |(1-i)z+(1+i)\bar{z}| < 2 \, , \, (1+i)z+(1-i)\bar{z}> 0\}$
onto
$\Omega' = \{z \in \mathbb{C}\; | \; |z-1-i| < 2 \, , \,-i(z-\bar{z}) > 2\}$

The first set is a rectangle in the complex plane. My idea is to rotate it to a rectangle with axes-parallel sides. Then apply exponential function to make it into a sector of a circle ring. Then enlarge the sector with a power function so the new area is bounded by two genuine circles and a straight line. But I don't really know what to do with this.

Ideas?

Looks hard. $\Omega'$ is a half-disk, which can be conformally mapped onto a half-plane without much difficulty. However, it is not easy to transform a rectangle into a half-plane. The map that does it can be expressed in terms of an elliptic integral.

This problem is actually very easy, it is a half-strip not a rectangle. I confused $|x+y| < 2$ with $|x|+|y| < 2$. What a time-waster! I have spent a lot of time on this problem trying a lot of conformal functions. I don't have time for such mess-ups!


geometry
geometric transformation
homothety
geometry unsolved

$k_1, k_2, k_3$ are three circles. $k_2$ and $k_3$ touch externally at $P$, $k_3$ and $k_1$ touch externally at $Q$, and $k_1$ and $k_2$ touch externally at $R$. The line $PQ$ meets $k_1$ again at $S$, the line $PR$ meets $k_1$ again at $T$. The line $RS$ meets $k_2$ again at $U$, and the line $QT$ meets $k_3$ again at $V$. Show that $P, U, V$ are collinear.

Let $r_1, r_2, r_3$ be the radii of the circles $k_1, k_2, k_3$. The tangency points Q, R are the internal homothety centers of the circle pairs $(k_3, k_1), (k_2, k_1)$. Therefore, $\dfrac{QP}{QS} = \dfrac{QV}{QT} = \dfrac{r_3}{r_1}$ and $\dfrac{RP}{RT} = \dfrac{RU}{RS} = \dfrac{r_2}{r_1}$. The vertical angles $\angle PQV = \angle SQT$ are equal and the vertical angles $\angle URP = \angle SRT$ are also equal. Thus the triangles $\triangle PQV \sim \triangle SQT$ are centrally similar by SAS with the homothety center Q and the lines $PV \parallel ST$ are parallel. The triangles $\triangle URP \sim \triangle SRT$ are also similar by SAS with the homothety center R and the lines $PU \parallel ST$ are also parallel. Since only one line parallel to the line ST passes through the point P, the lines $PU \equiv PV$ are identical and the points P, U, V are collinear.

:)   :)  :)  :)  I am very glad to see YETTY again  :)  :)  :)  :)  :)

[b]A little  easier.[/b]

 Consinder the comon tangent $\el l = KRL$  at R. Then $\angle TSR = \angle TRL (or \angle TRK) = \angle KRP(or \angle LRP) = \angle RUP$ , so $PU // ST$ . In similar  way we take  $PV // ST$  and  Yetti's nice argument is on work!

 [u]Babis[/u]

Thank you for the returning welcome :blush:. I am still very busy, just finished one task in a difficult project, and I tought that I would just slip in quietly for a day or two, like a fish...

There is a little bit more in the problem, not only are the points P, U, V collinear, but the 5 points $P, U, V, K_2, K_3$ are collinear ($K_2, K_3$ being the centers of the circles $k_2, k_3$), which implies that the triangles $\triangle URP \sim \triangle SRT,\ \triangle PQV \sim \triangle SQT$ have right angles at the vertices R, Q.

The tangency points P, Q are the internal homothety centers of the circle pairs $(k_2, k_3),\ (k_3, k_1)$, which are collinear with the external homothety center H of the circle pair $(k_1, k_2)$ lying on the center line $K_1K_2$. Therefore, not only are the triangles $\triangle URP \sim \triangle SRT$ centrally similar, with the homothety center R and (negative) homothety coefficient $-\dfrac{r_2}{r_1}$, the triangles $\triangle PHU \sim \triangle SHT$ are also centrally similar, with homothety center H and (positive) homothety coefficient $+\dfrac{r_2}{r_1}$. Since the circle centers $K_2, K_1$ are the corresponding points of both the triangle pair $\triangle URP \sim \triangle SRT$ and of the triangle pair $\triangle PHU \sim \triangle SHT$. From the first homothety, the corresponding points $K_2, K_1$ have to be both inside or both outside of the triangles $\triangle URP \sim \triangle SRT$ or on the corresponding lines PU, ST.  From the second homothety, the corresponding points $K_2, K_1$ have to be both inside or both outside of the triangles $\triangle PHU \sim \triangle SHT$ or on the corresponding lines PU, ST.  Clearly, the only condition, which can be satisfied simultaneously, is that the points $K_2, K_1$ lie on the correspondind lines PU, ST. Similarly, we can show that the center $K_3$ of the circle $k_3$ lies on the line $PV \equiv PU$.

Really , very nice and interesting results !

 Babis

[hide="A very nice solution!"]Let's consider to homotheties:
$J_Q^{\alpha_1}(k_1)=k_3$
$J_R^{\alpha_2}(k_1)=k_2$

$J_Q^{\alpha_1}(T)=V \quad J_Q^{\alpha_1}(S)=P \Rightarrow l_{ST}\parallel l_{PV}$
$J_R^{\alpha_2}(S)=U \quad J_R^{\alpha_2}(T)=P \Rightarrow l_{ST}\parallel l_{PU}$

Hence $l_{PV} \parallel l_{PU}$, so $P, U, V$ are collineral. [/hide]

There is another solution by INVERSION.That's really nice but a little bit longer


I found this result accidentally, so I don't know whether it's hard or easy  :D . I'm sorry if this problem has been posted in this forum.
[b]Problem[/b]: Given triangle $ ABC$ and an arbitrary point $ M$. Denote $ O_a,O_b,O_c$ are the centers of the circles $ (BMC),(CMA),(AMB)$, respectively; $ l_a,l_b,l_c$ are the Euler lines of the triangle $ BMC,CMA,AMB$, respectively. Prove that $ AO_a,BO_b,CO_c$ are concurrent iff $ l_a,l_b,l_c$ are concurrent.
For example:
+ Let $ M \equiv O$, we have $ AO_a,BO_b,CO_c$ are concurrent at the Kosnita point and $ l_a,l_b,l_c$ are concurrent at $ O$
+ Let $ M \equiv I$, we have $ AO_a,BO_b,CO_c$ are concurrent at $ I$, and $ l_a,l_b,l_c$ are concurrent at the Schiffer point.
+ If $ \Delta ABC$ is equilateral, $ AO_a,BO_b,CO_c$ are concurrent at the centroid for all $ M$, and we can prove that, for all $ M$, $ l_a,l_b,l_c$ are concurrent at a point lies on the line connecting $ M$ and the orthocenter of triangle $ O_aO_bO_c$
Hope you'll interested in this problem  :wink:

Sorry for spamming but ... no one interested ?  :(


geometry
geometric transformation
rotation
probability
3D geometry
counting
distinguishability

1.	How many ways are there for seven identical dogs and 4 identical cats to stand in a line so that no two cats stand next to each other?

2.	Four distict circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect?

3.	Using the digits 1, 3, 5, 7, and 9 to form two numbers whose product is minimal, what are the two numbers?

4.	In a galaxy far away, a certain planet has a radius of 19000 km, It has a moon orbiting it in a circular path, which is exactly 8000 km away from the surface of the planet. One day on the moon is equal to one rotation of the moon, and the moon makes one rotation every 375pi km that it travels around its planet. How many days does it take for the moon to revolve exactly once around its planet?

5.	What is the remainder when (2009)(2010)(2011)…….(4015)(4016) is divided by the product of the first 2008 positive integers?

6.	A number is first rounded to two decimal places, and the answer is rounded to one decimal place. If the final number is 1.5, what is the probability that the original number is between 1.45 and 1.55 inclusive?

7.	The years 1900 to 2008 are written down consecutively. Then plus and minus signs are places alternately between the digits : 1+9-0+0-1+9-0+1-1+9-0+2-….-2+0-0+7-2+0-0+8, if the result is N. find N.

8.	A solid cone has a radius of 7cam and a height of 12cm. If we drill a hole with radius 2cm from the center of the base all the way to the other side, then what will be the volume of the remaining solid?

9.	Cupid celebrated his 25th birthday on a Sunday, February 14, Which of the 45th, 55th, 65th, 75th, 85th, 95th, 105th, 115th, and 125th birthdays is/are also on a Sunday?

[hide="5"]Every number in the product is twice one of the first 2008 integers, so the remainder is $ 0$.[/hide]

w8, for number 3 they are different right?
[hide="if so"]
You find the two smallest numbers
w8 can you use the numbers twice?
[hide="if so"]
$ \ 13\cdot\ 15\equal{}185$
[/hide]
If not then $ \ 13\times\ 35\equal{}455$
[/hide]

Cgyao, I think you're supposed to use [b]all[/b] of them.

[hide="3"]$ 1\cdot3579 \equal{} \boxed{3579}$[/hide]

EDIT: Can we use decimals?

[hide="1)"]
Line up the dogs first. Then "insert" the cats. 9 spots for the first cat, 8 for the second, 7 for the third, and 6 for the last. Since the cats are indistinguishable, we divide by $ 4!\equal{}24.$ So the number of ways to arrange them is \[ \frac{9\times8\times7\times6}{24}\equal{}\boxed{126}.\][/hide]

[hide="2)"]
I don't see any way other than testing. The desired picture is shown below.
 [geogebra]4b0f708cc57a255c337b6e15121fb5f7b895a536[/geogebra]
I believe the answer is $ \boxed{12}.$[/hide]

[hide="3)"]
It seems relatively obvious that we want one of the factors to be $ 1,$ which is possible in this case so the product is $ \boxed{3579}.$ Verify for yourself that there is no smaller product.[/hide]

[hide="4)"]
Imagine the orbital ring of the moon. It has a radius of $ 19000\plus{}8000\equal{}27000$ km. The circumference of this circle is $ 54000\pi.$ We're basically being asked "how many $ 375\pi$ fit into $ 54000\pi$" which is \[ \frac{54000\pi}{375\pi}\equal{}\boxed{144}.\] Therefore, it will take $ 144$ days for one revolution around the planet.[/hide]

[hide="5)"]
We can actually say that the value of \[ \frac{(2009)(2010)(2011)\dots(4016)}{(1)(2)(3)\dots(2008)}\] can be represented as \[ \binom{4016}{2008}.\] This must obviously be an integer. [hide="why?"]
If we divided it by $ 2008\plus{}1\equal{}2009$ we would get a Catalan number, which are all integers, so the binomial is an integer.[/hide]
Therefore, the remainder is $ \boxed{0}.$[/hide]

[hide="6)"]
[hide="Cheap"]
There are an infinite number of possibilities, so the probability is either $ 1$ or $ 0.$ Testing any random number in that range we find it works, so the probability is $ \boxed{1}.$[/hide]
[hide="Somewhat logical"]
If the number is finally rounded up, the tenths place must be a $ 4.$ In order for it to be rounded up at that place, the hundredths place must be $ \ge5.$
If the number is finally rounded down, the tenths place must be a $ 5.$ In order for it to be rounded down at that place, the hundredths place must be $ \le4.$ (I think there might be a mistake in the question, shouldn't it be "in the range $ 1.45\le n<1.55$"?)[/hide][/hide]

[hide="7)"]
Notice that the sum $ \minus{}1\plus{}9$ is always present (except for the first one). So part of the sum is \[ 1\plus{}9\plus{}99(\minus{}1\plus{}9)\plus{}9(\minus{}2\plus{}0)\equal{}10\plus{}792\minus{}18\equal{}784.\]
Now notice that the sum of the tens and units digits have the same form: $ \minus{}0\plus{}0, \minus{}0\plus{}1, \minus{}0\plus{}2,\dots,\minus{}1\plus{}0, \minus{}1\plus{}1, \minus{}1\plus{}2, \dots.$ The final part of the sum is \[ 45\plus{}35\plus{}25\plus{}15\plus{}5\minus{}5\minus{}15\minus{}25\minus{}35\minus{}45\plus{}36\equal{}36.\] Finally, the actually sum is \[ 784\plus{}36\equal{}\boxed{820}.\]Sorry if that was a little vague.[/hide]

[hide="8)"]
After the drilling, the cone on the top is similar to the original big cone with volume ratio $ \frac{8}{343}$ or dimension ratio $ \frac{2}{7}.$ That means the height of the small cone is $ \frac{2}{7}\cdot12\equal{}\frac{24}{7}$ implying the height of the cylindrical hole is \[ 12\minus{}\frac{24}{7}\equal{}\frac{60}{7}.\] Since the cylinder has a radius of $ 2,$ its volume is \[ 2^2\cdot\frac{60}{7}\pi\equal{}\frac{240\pi}{7}.\]
The volume of the cone is \[ \frac{7^2\cdot12\pi}{3}\equal{}196\pi.\]
Finally, the remaining volume of the solid is \[ 196\pi\minus{}\frac{240\pi}{7}\equal{}\boxed{\frac{1132\pi}{7}}.\][/hide]

I don't think you gave enough information for 9). I think you have to specify what year because there are leap years involved. Of course, if you don't want leap years to be involved, [hide="then..."]
In order for his birthday to be a Sunday, \[ 365(\text{year number})\equiv0\pmod{7}.\] 7 doesn't divide 365. That means \[ (n\text{th birthday})\minus{}25\equiv0\pmod{7}.\] The one it works for is his $ \boxed{95\text{th birthday}}.$[/hide]

[quote="RoFlLoLcOpT"]

I don't think you gave enough information for 9). I think you have to specify what year because there are leap years involved. Of course, if you don't want leap years to be involved, [hide="then..."]
In order for his birthday to be a Sunday,
\[ 365(\text{year number})\equiv0\pmod{7}.\]
7 doesn't divide 365. That means
\[ (n\text{th birthday}) \minus{} 25\equiv0\pmod{7}.\]
The one it works for is his $ \boxed{95\text{th birthday}}.$[/hide][/quote]
If it involved leap years, wouldn't you have to use mod 4?

But for some centuries, 00 is not a leap year, so we cannot assume that. We need orangefronted to tell us what he meant.

Uh actually if the year is divisible by 4 then it is a leap year and the divisibility rule for 4 states that if the last to digits are divisible by 4 then the whole number is... seing as 0 and 4 are divisible by 4 then all years that end in 0,4,8,12... are leap years

Uh actually no.  If a year is 0 mod 100 but nonzero mod 400, it is not a leap year :iwnk:

[quote="Utter Loser"]I believe that you have to assume that 00,04,08..... and so on, are leap years.[/quote]
We cannot simply assume this (see AIME's post).
I think we would also have to know what year Cupid was born in to solve this problem...

Hmmm....I guess you got a point there.

I just got the problem in one of the math contests. So I don;t have the correct answer though.

So the math contest problem was exactly how you typed it? Are you sure you typed it right then? I kinda doubt the math contest problem is unsolvable.

We have two cases:
Case 1: Cupid was born on a leap year
Case 2: Cupid wasn't born on a leap year

I think the problem applies to case 2...if he was born on a leap year, I think the problem would say that...would't it?
Case 1 is harder to solve...

It doesn't matter if he's born on a leap year or not.  The problem is unsolvable unless we are given the birth year.

@orangefronted: What math competition was this?


Given some nonnegative numbers. Prove the following inequality:
\[{a^{2}+b^{2}\over 2a+2b+c}+{b^{2}+c^{2}\over 2b+2c+a}+{c^{2}+a^{2}\over 2c+2a+b}\geq{2\sqrt{3(a^{2}+b^{2}+c^{2})}\over 5}. \]

Nice, ductrung. There is one more solution!


ya..same rules as last time.
no hurtful comments
scale from 1-10
COME ON, LETS START!

6.  One-third of the time, it says k00lpe.  This is a word I am pretty biased against.

AoPS wouldn't take in the 40.1 KB original.
7.2.

[quote="Richard_Min"]By having an animated gif[/quote]
6. Mine is 0/0=undefined.

0/10 right? Because it does not exist

And Undefined means it doesn't have a set value.

Lalalala

(my rating for you is the amount of characters in the series of 'la' above)

8.5

It's stuck in my head. I'm not at all sure if that's good or bad.

nice 8 i have been trying to find a good avatar so mine might change alot.

7, the brain has colors that I disagree with

AMAZING AVATAR OMGUS YEAH OMGUS YEAH HOLY COW!!!!
100/10

isnt rate the avatar still going around?

So technically, this is NOT a revival?

btw, 7

A revival is not necessary.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=262307&start=460]Here's the main thread.[/url]

Locked.


[quote="buzzer11"]ugh i still have a month.[/quote]

lmao owned

omfg!
bah i have 20 days left including a week of finals
that sucks
boohoo

I have 11 days left, not including final exams.

today was my last full day of school next three days are half days with finals but juniors and seniors are exempt from finals

[quote="1=2"][quote="Ignite168"]So today (Mon May 21) was the last day of classes for me, wut about u guys?[/quote]

A couple years... or until I can argue dad into letting me have a break, and if I do, it would be a month long.[/quote]

Pfffhhh.... You wacky homeschoolers :P .

Pff... you lucky guys.. for me school ends on the 1st of June but starting from the 25th June there's one week of the Bacalaureat and another month later the entry exams at the university. :ninja:

For me, school ends on June 26th or something like that.

last friday the sems were over
last class was on may 1, that too a special one
last regular class was on 25th april

13 days for me!!! counting finals AND the day I miss for ARML!

lmao June 19th but i have AP bio meetings that year so i can't go the last two days (w00t)

[quote="buzzer11"]ugh i still have a month.[/quote]
OUCH
anyways my last today is tommorow (wednesday the 23rd woot woot)


[quote="rem"][quote="CT_14"]Let p is prime number and $r_1,r_2,...,r_m$ are intege
with $1\leq r_1\leq r_2\leq r_3\ ...\leq r_m\leq p-1$
satisfy:$r^m_i\equiv 1(mod\ p)$
prove that: $\forall {x}\in Z$ then $x^{m-1}\equiv {(x-r_1)(...)(x-r_m)} (mod\ p)$[/quote]
I probably misunderstood the problem, but if $x=r_1$ we will have $r_1^{m-1}\equiv{0} (mod\ p)$ :?[/quote]

I afraid you're wrong, Because from $r^m_i\equiv 1(mod\ p)$ you don't have $r_1^{m-1}\equiv{0} (mod\ p)$

[quote="cats..."]right, but that contradicts what you're trying to prove...[/quote]
yes, that's what I mean. We subst x=r_1 and we do not get the result of the problem. :(


conics
parabola
symmetry
analytic geometry

[b]
If segment of parabola is given then how we obtain its focus , axis , vertex, ....et :| c[/b]

first you must find the equation of the parabola.  Get it in standard form (ax^2+bx+c).  Use the equation -b/(2a) to find the axis of symmetry in the form of x=-b/(2a).  To find the vertex, plug in the x-value found for the axis of symmetry into the equation of the parabola (ax^2+bx+c).  That will give you the y-coordinate  of the vertex, you have already found the x-coordinate.

dear it is the queston of construction
if i give you the portion of parabola means graph of parabola given not complete (small part of it )


Hi, i need help with this question
im kinda of stuck on it as there is not enuf info, but here folos...
A triangle has two altitudes h & k. h can't equal k. Determine the upper and lower bounds of the length of the third altitude in terms of h and k.
Do i use triangular inequalties, pythagoras or orthocentre? i dunno where to start.
 :blush:  :blush:
Thnx :D

i think this has been posted before...perhaps you can try to use the search function.

Let third altitude be "b", here we have:

1/h + 1/k > 1/b
(h+k)/hk>1/b
b(h+k)/hk>1
b>hk/(h+k)


geometry
inequalities
inequalities proposed

[quote="chien than"]Let $ a;b;c$ be the sides of a triangle with areas $ S$
Prove that:
$ a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}\frac{a^{3}}{b\plus{}c}\minus{}\frac{b^{3}}{a\plus{}c}\minus{}\frac{c^{3}}{a\plus{}b}\geq 2\sqrt{3}S$[/quote]
Very nice inequality, chien than!  :lol:
It's equivalent to the following inequality:
$ \sum_{cyc}\frac{a}{h_{a}\plus{}r_{a}}\geq\sqrt3.$ :wink:
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167949

[quote="nntien"]OK!
We can check following inequality:

$ \frac{a^{2}(p\minus{}a)}{b\plus{}c}\plus{}\frac{b^{2}(p\minus{}b)}{a\plus{}c}\plus{}\frac{c^{2}(p\minus{}c)}{a\plus{}b}\geq\sqrt{3}S$

Indeed, we have:
$ \frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{a\plus{}c}\plus{}\frac{c^{2}}{a\plus{}b}\geq p$, and
$ (p\minus{}a)\plus{}(p\minus{}b)\plus{}(p\minus{}c) \equal{} p$ and we can check easily:
$ xy\plus{}yz\plus{}zx\geq\sqrt{3(x\plus{}y\plus{}x)xyz}$ (Newton's inequality)

Solve above inequality, now![/quote]

I don't think it works  :(  as you cannot apply Chebychev in this way

$ \frac{a^{2}(p\minus{}a)}{b\plus{}c}\plus{}\frac{b^{2}(p\minus{}b)}{a\plus{}c}\plus{}\frac{c^{2}(p\minus{}c)}{a\plus{}b}\geq\frac{1}{3}(\frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{a\plus{}c}\plus{}\frac{c^{2}}{a\plus{}b})((p\minus{}a)\plus{}(p\minus{}b)\plus{}(p\minus{}c))$

[color=purple]Hi all ! This inequality due to mttit my friend not of chienthan  :mad: !
[hide] Chu' may` dung` post bai` linh tinh kieu nay` nua nhe' [/hide][/color]

To evarist, このフォーラムでベトナム語を使わないでください。  :wink:

Japanese Communities Moderator

kunny

[quote="kunny"] このフォーラムでベトナム語を使わないでください。  :wink:

[/quote]

What does it mean, kunny ? Maybe " don't post Vietnamese to this forum "  :D

[quote]I don't think it works  :(  as you cannot apply Chebychev in this way

$ \frac{a^{2}(p\minus{}a)}{b\plus{}c}\plus{}\frac{b^{2}(p\minus{}b)}{a\plus{}c}\plus{}\frac{c^{2}(p\minus{}c)}{a\plus{}b}\geq\frac{1}{3}(\frac{a^{2}}{b\plus{}c}\plus{}\frac{b^{2}}{a\plus{}c}\plus{}\frac{c^{2}}{a\plus{}b})((p\minus{}a)\plus{}(p\minus{}b)\plus{}(p\minus{}c))$[/quote]

I think that you can't apply the Trebysev's ineq, it's false!

This problem is simple, but we haven't any solution full and exactly!

By the way, the following inequality is wrong: $ \sum_{cyc}\frac {a}{l_{a} \plus{} r_{a}}\geq\sqrt3,$
where $ l_a,$ $ l_b$ and $ l_c$ are bisectors of the triangle. :wink:

From: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167949
[quote="chien than"][quote="nmtvn"]New trigonometrical inequality[/quote]
To Mods: Please lock this topic, this is a problem from M&Y, issue 9, and we can't discuss it now[/quote]

[quote="nmtvn"][b]chien than[/b]
Where's the solution for your problem? Don't you have it, do you?[/quote]

nmtvn: Haha this is total failure. Are you stupid or something? If you are going to cheat, you should at least ask somebody who told you not to cheat.

[b]Altheman[/b], who's stupid? You or me? Do you wnown that [b]chien than[/b] posted this topic? You don't known, do you? So, you are stupid, impolite and idiot! Why do you ask chien than? Why did he post this problem?

Hello. I am just having a little fun. Your response indicates that you missed my point.

Anyway,

This topic was started at: Tue Sep 25, 2007 10:57 am
He told you not to talk about it: Tue Sep 25, 2007 11:27 am

He made this topic before he told you not to discuss the problem in that other topic. That should resolve your question.


number theory
relatively prime
number theory unsolved

Let $\phi(n)$ denote the number of totatives of n (meaning the number of integers less than n that are relatively prime to n). Prove that if n is an integer greater than 1, then there exist infinitely many integers m such that

$\frac{\phi(n)}{n} = \frac{\phi(m)}{m}$

The powers of $n$ suffice. $\frac{\phi(k)}{k}$ depends only on the set of primes dividing $k.$

Where in number theory can this be found?

It is a number theory fundamental.
There is a formula $\varphi(n)=n\cdot\left(1-\frac{1}{p_1}\right)\left(1-\frac{1}{p_2}\right)...\left(1-\frac{1}{p_k}\right)$, where $p_1, p_2,...,p_k$ are all distinct prime factors of $n$.

Do you know of any very good text sources I could get for Number Theory?

I am not so familiar with english math literature.

I was told you could help me with this:

[quote]
spoudyal
PostPosted: Fri Apr 08, 2005 9:21 pm    Post subject: ANYTHING
Is there any really good book on number theory you know of?
[/quote]

[quote]
Myth
PostPosted: Fri Apr 08, 2005 9:24 pm    Post subject:
Ask James.
[/quote]

So any idea?

Since I didn't learn number theory from a book, I don't have any recommendations. Also, if you want to get my attention- you can send me a private message instead of posting in a forum I don't read very often.


observe that if $m \in \mathbb{Z}$ then $m^{3}\equiv m (\mod 6)$

so let`s write $a=M_{6}+\alpha_{1}$, $b=M_{6}+\alpha_{2}$, $c=M_{6}+\alpha_{3}$.

now the condition $(x^{2}+y^{2}+z^{2}= M_{6}\Longleftrightarrow (\alpha_{1}-\alpha_{2})^{2}+(\alpha_{2}-\alpha_{3})^{2}+(\alpha_{3}-\alpha_{1})^{2}= M_{6}$

after we study some cases we find the unique solution $\alpha_{1}=\alpha_{2}=\alpha_{3}$ , therefore $x_{n}+y^{n}+z^{n}= M_{6}$ is true.


Well I think this problem is a real challenge:
Enjoy thinking on it:
(darij style :) )
Here by construction we mean ,Ruler and compass construction.
[color=blue]Problem:Let[/color]$ABCD$[color=blue]a given convex(in the link that darij has posted, the problem is solved both for convex or concave)  quadrilateral .
Construct a square In the $ABCD$ such that the vertices of the square be on the sides of the equilateral .[/color]

This one was discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=19372 , and Grobber posted a sketch of a solution.

Note that a polygon with 4 vertices is called "quadrilateral", not "equilateral". "Equilateral" is an adjective and means that something has equal sidelengths: An "equilateral triangle" is a triangle ABC such that BC = CA = AB.  :)

  Darij

:oops_sign:  :oops_sign:
Sorry,I dont know how why on the earth i have wrote equilateral :blush:
I think ,i am getting a little  :fool: these days


[quote="Γιώργος"]Να βρείτε το κέντρο ενός κύκλου παραπλεύρως χρησιμοποιώντας μόνο διαβήτη![/quote]

παραπλεύρως????τεσπα το θέμα υπάρχει παντου στο forum.

Χρησιμοποιοντας διαβητη η ακομα ενα κερμα... :wink:


can you tell me with which program i can write correct the maths vales

Do you mean $\LaTeX$? Look at this [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]subforum[/url] and read some of the sticky threads and announcements.

yess exactely what i need because i couldn't write the maths symbols and formulas


linear algebra
matrix
geometry
3D geometry
sphere
topology
advanced fields

Help me?

Schwarz lemma and Negative Curvature

Definition of Gaussian curvature and  example ?

Definition of  Zariski cotangent space and example?

Definition of  Zariski tangent space and example?

Gaussian curvature at a point $P$ is defined to be $k_{1}k_{2}$ where $k_{1}$ is the maximum of the normal curvature through $P$ and $k_{2}$ is the minimum normal curvature through $P$. If you go through some fancy derivations you will find that this is equal to $\frac{L}{g}$ where $L=\text{det}(L_{ij})$ the determinant of the second fundamental form in matrix form and $g=\text{det}(g_{ij})$ the determinant of the first fundamental form in matrix form.

The normal curvature on a sphere of radius $r$ is $\frac{1}{r}$ for any point on the sphere, so the Gaussian curvature is $\frac{1}{r^{2}}$ for all points.

The Gaussian curvature for a cylinder of radius $r$ is $0$ since it has maximum normal curvature of $\frac{1}{r}$ along a tangent to a circular slice (if you can visualize that) and minimum of $0$ perpendicular to the last one.

Schwarz lemma and negative curvature

 I don't know anything about it?

Please recommend me some good books!


What does this notation mean:

http://upload.wikimedia.org/math/3/f/5/3f5c2191d5128f36ec6b726c77df3b06.png

Thanks.

It's called a combination. http://www.artofproblemsolving.com/Wiki/index.php/Combination

It's read "$ n$ choose $ k$", and represents the number of ways to choose $ k$ objects from $ n$ objects without regard to order.

This is how it's defined: Suppose we have $ n$ objects. Consider a row of $ n$ spaces. We can place one of $ n$ objects in the first space and have $ n\minus{}1$ left over, place one of $ n\minus{}1$ in the next space, and so on, so there are $ n!$ ways to distribute these objects (that is, $ n!$ permutations). If we divide out $ (n\minus{}k)!$ of the objects, we get the permutations of the first $ k$ objects. Now, for each sub-permutation, there are $ k!$ equivalent permutations that are being overcounted, so we divide those out as well. This gives us the following expression:
\[ \binom nk \equal{} \frac{n!}{(n\minus{}k)!k!}\]


Anyone loves it?
I have seen 1, 2, 3 and part of 4. My favourite team is the "team guido" in 1. Bill and Joe are brilliant. Can't forget how the made the final 3......
What about your feelings towards this show?

I like the Amazing Race.  Its interesting to watch. :lol:

I love it's premise, and secretly hope to someday apply and get accepted. It seems like it would be a great experience.

But that's the one show that I can never figure out what time it's on, I mean I catch it here and there but I never note when it's on when I'm watching it, and then the next week I can't remember what station/time/day.

I never really watch it full time but I've seen several episodes of it.

I too would really enjoy being on it. As I'm sure everyone does, I think I'd be rather good ;). Maybe after I retire (after college) I'll apply.

[quote="lulu"]:blush: who can tell me what is an Amazing mactch?Thank you[/quote]
This is a live show. 11 teams (2 members each) are selected to compete in a World wide race. They are given clues at the beginning and are told what they're going to do and where they are going. They set off from USA and go to Europe of South America and then Africa, then Asia and Australia or New Zealand, finally complete a circle back to USA. The first team to arrive will recieve 1,000,000 dollars.

You can try googling the Amazing Race to find more detailed informations.


Counting with Combinations Part 4

We solve a problem using constructive counting and casework, and get a simple combination as an answer.  This makes us wonder -- is there a simple solution using combinations?  We find two more solutions, and an interesting relationship regarding combinations.  Make sure you watch the whole lesson -- the solutions are more interesting than the answer!

Why isn't the answer just 11 choose 4?  Aren't there 11 vertices, of which you are choosing 4?

There are 6 points on the bottom so if all four points are on the bottom, we would be overcounting and so is the top. Also another reason why it is not $\binom{11}{4}$ is that we have to [b]trace[/b] a quadrilateral on the line.


[url=http://www.unidata.ucar.edu/staff/russ/mathcounts/diaz.html]Here's a website[/url] that you should read over, do what it says, and print it out.

If you know that stuff, it'll help you in chapter, state, national, and anything basically.

I've taken the liberty of correcting your post, tweak it, and make it a Sticky since it's so important. I'm definitely going to pass this out to all my more hopeful pupils...(a.k.a. my future members of MATHCOUNTS Team)

my mathcounts coach gave me this 2 years back before state. its great for beggining students, but dont most of you guys already know this stuff?

Yeah, but not all people are as intellectually advanced as we for we are all math 3l33ts. :twisted:

Well, that's just what you should remember, not what you should know right???

...What's the difference between the two?

having the knowledge is different from remembering stuff isn't it???

Like for example: some stuff that you should know aren't on the bible... and that bible just states the basic formulas you should remember...

martin, youre messed up

im trying to make this mc forum more simplistic for beginning peeps and such. and this is what they should know

That's not very nice...I still don't get what he's talking about though...do you mean "memorizing the formulas" and "knowing how to use them"?

sigh. just forget it. The point is you should not limit yourself to only those stuff.

lol, i know that

but for the beginning mcers its nice to start with

PRIMES FROM 2 to 109!! boy, that's a lot of numbers.

AND know factors for all non-prime numbers, hard work.

[quote="MysticTerminator"]hmm....i would say, pretend you're a girl. and then when they're least expecting it, jump out of the shadows, and say you're a guy. and then they'll start shrieking. and then you can pretend you're a girl some more. etc.[/quote]
I was thinking about that but it's just freaky when a boy does that...I won't mind if a girl does that though :)
[quote="mathfanatic"][quote="Tare"]
Let's just say all of us will know when we meet at nats... :wink:
[/quote]

I have to wonder how they'll all "know" that you're a boy.  Will you take off your shirt and write "Art of Problem Solving" on your sixpac?[/quote]
Well since I know Syntax Error's face, I'm guessing others are around him somewhere :)

I don't have a six-pac :(...it'll be evn freaki[b]er[/b] if I do that, and it'll be the freaki[b]est[/b] if a girl does that :lol:

You know, we might want to set up a system in which we'll know who's in AoPS, like have a red flag or something :lol:

[quote="MysticTerminator"]I've got a great idea! How 'bout just saying to each other "i'm in AoPS. are you?"[/quote]

LOL! Sometimes the low-technology means of communication work just fine.

Not expecting to be at Nats THIS year . . .

That...would be...er, inefficient to say the least :lol:

Think about it, 50 states, four team+top 4 people from each state, 50*(4*4+4)=1000 people...

How about us all chewing gum? If we keep on blowing bubbles it'll be pretty obvious, plus the extra sugar will help us a lot.

blah if you're a mc mod at least go to their website and read some rules. i mean look at my masterful studliness. ok, nm. there are 57 competing districts, top 4 from each district go: 228 students, (also coaches/parents/etc., but I don't think you meant them)

blah. no comment on my masterful studliness?

ugh. well if you know what a "stud" is, then youve got studliness down

See, I'm afraid I got the wrong definition...none of them describes a human except for one which doesn't work since we're talking about MATHCOUNTS...

From the (late) "Free Online Contests for Middle Schoolers" thread/sticky:
[quote="mathfanatic"]Hey you!  Yeah, you!

If you're in middle school, then try the Dositey.com Math Challenge at [url]http://www.dositey.com/problems/mathchall58.htm[/url].

Or try the ABACUS Internet Math Challenge at [url]http://www.gcschool.org/pages/program/Abacus.html[/url].  There are also years worth of past problems, so you'll have plenty of material to study for other stuff, too.

Enjoy!

[Note: cross-posted from News on the AMC board, since this is most pertinent to middle schoolers.][/quote]

[quote="MathFiend"]another challenge site is: http://www.olemiss.edu/mathed/contest/contests.htm

they give out prizes...TIs[/quote]

[quote="mathfanatic"]That was also on [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=849]News[/url].  Ole Miss POW is fun.

Here's another one you may enjoy:
"The Maine Page" - problem of the week site geared towards high school students (although middle schoolers can participate, too, presumably) - [url]http://www.theproblemsite.com/maml/default.asp[/url]

mathfanatic[/quote]

Also see the Mathcounts Toolbox on the mathcounts website it has a lot of that stuff and more.

Sorry, I had to get rid of at least one spoiler so I merged two together, you can make the thread again if you want to.

That's all right.  I see what you mean.  I'm trying to condense things as well.

Anyway, you can have the "contests for middle schoolers" and AMC can have "contests for high schoolers" threads.  Or maybe they should all just go in the Other Contests and Programs part...

Mmm, I think they should both be in the same forum, just to keep things organized.


Paul has a large bag of candies, each of which is one of 5 possible different flavors: apple, banana, cherry, dutch chocolate, and elderberry.

Assuming he can fit up to 10 candies in his mouth at once, how many different flavors can he make? Note that 1 apple and 1 banana is the same flavor as 2 apples and 2 bananas (justa a larger amount), but 1 apple and 2 bananas is not the same as 2 apples and 1 banana. Also note that he has more than 10 candies of each flavor.

Generalize the number of flavors and how many can be eaten at one time.

i assigned each flavor a digit 1 2 3 4 5 and 0 stands for no candy (0 is used to express less than 10 candies as a 10 digit combination)

so there are 6 choices, choose 10, repeating allowed
((6 + 10 - 1)!)/(10!(6 - 1)!) <-thats the formula right?

but if you notice, any flavor consisting of 5 or less candies can be at least doubled (12=1122=111222=11112222=1111122222), so remove them all
remove this many flavors: 6 choices, choose 5, ((6 + 5 - 1)!)/(5!(6 - 1)!)

now that all 5 or less candy combos are removed since none of them are unique, it leaves only the unique flavors.

((6 + 10 - 1)!)/(10!(6 - 1)!)-((6 + 5 - 1)!)/(5!(6 - 1)!) = 2751 flavors

is that right?


geometry
parallelogram
geometry proposed

Let $ ABCD$ be a convex cuadrilateral. Let $ P$, $ Q$, $ R$ and $ S$ be the midpoints of the segments $ AB$, $ BC$, $ CD$ and $ DA$ respectively.

Prove that $ AC\cap BD\equal{}PR\cap QS$ if and only if $ ABCD$ is a parallelogram.

:)

$ PR\cap QS \equal{} AC\cap BD$ implies $ AC\cap BD$ is the center of gravity of $ ABCD$. But the center of gravity is the midpoint of the segment joining the midpoints of $ AC$ and $ BD$. Hence $ AC\cap BD$ must be the midpoint of $ AC$ and $ BD$, and the conclusion follows. The opposite direction is trivial.

I would like to see a more geometric approach. :)

I don't know how much geometric is this considerated but, a solution is possible using this:

Let $ T\equal{}PR\cap QS$. Let $ E$ and $ F$ be points on $ (AB)$ and $ (CD)$, such that $ E$, $ F$ and $ T$ are collinear.

Prove that \[ \frac{(BPC)}{(DPA)}\equal{}\frac{EB\cdot CF}{AE\cdot FD}\]

:)

[quote="nayel"]I would like to see a more geometric approach. :)[/quote]
$ \textbf{Lemma: }$ (Geometry Revisited theorem 3.12),
"The segment joining the midpoints of pairs of opposite sides of a quadrangle and the segment joining the midpoints of the diagonals are concurrent and bisects one another"

Let,$ X,Y$ be the midpoints of the diagonals.
As the segment $ BD$ goes through $ X,O$ it goes through $ Y$.
So,$ BD \cap AC \equal{} Y \equal{} O$. So, the parallelogram $ PXRY$ degenerates in a segment $ PR$.
Which proves that, $ O \equal{} X \equal{} Y$ is the midpoint of the diagonals.So, $ ABCD$ is a parallelogram.

[and opposite direction is simple]:)


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let  p and  q be two primesand let  G a group of order   $ p^2q^2$.

  a) Show that  if p=q  then G  is not simple
 b) Show that  if p is distinct from  q  and G  is smple then G  is of order  36
 c) Show that  if  G  is a  group of order 36 then  G  is  not simple.

Is b) missing a condition like "... and $ G$ is simple..."?

oh  yeah  sorry!  i  have edited the original  post.  thanks

[hide="a"]
If $ p \equal{} q$, then $ |G| \equal{} p^4$, and it is a simple lemma that a group of prime-power order has a nontrivial center.
[/hide]

[hide="b"]
There must be exactly $ p^2$ of $ q$-Sylow subgroups, so we have $ p \equal{} \pm 1 \mod{q}$ as well. This means that $ p \equal{} q \minus{} 1$, which requires $ p \equal{} 2$, $ q \equal{} 3$.
[/hide]

Might as well finish this off (also, if you don't know the result about the center of a p-group being non-trivial, just examine the class equation).

Suppose there is a simple group of order 36. Then there are 4 Sylow 3-subgroups all conjugate to each other. In particular G acts on the set of Sylow 3-subgoups. Represent this action by $ \chi: G\to S_4$. But we assumed no normal subgroups, so $ ker \chi \equal{} 1$, but this is impossible since then $ 36 \equal{} |G| < 24$. Thus no such group can exist.

(Note this trick works well when you have a small number of Sylow conjugates, and a relatively larger group order).

c) If H is a proper subgroup of G and |G| doesn't divide $ [G: H]!$, then H contains a nontrivial normal subgroup of G.

Now, if $ |G| \equal{} 36$, let H be any 3-Sylow subgroup of G. The purported result follows now from the above criterion. Voilà!


When an object is in translational equilibrium, prove that torque acting on it is the same, about any point.

Nobody tried this? Here's a [hide="hint"]Vector product is distributive (BTW, why is it so?)[/hide]


counting
distinguishability
geometry
geometric transformation
rotation

In how many distinguishable ways can four identical red chips and two identical white chips be arranged in a circle?

[hide]The way 6 chips can be arranged is 6!. But we have 4 red, so we divide by 4!, and we have 2 white, so we divide by 2!.
Answer is 15. I think.[/hide]

[hide="I think..."]There are 3 ways:
white chips together
white chips seperated by one red chip on one side, 3 chips on the other
white chps opposite each other (seperated by two red chips on each side)[/hide]

Well, there has to be a sort of 'fixed point' (one of the chips) that you don't count in the combinations on the circle because rotating a circle would give a different combination but not a distinguishable way. I don't get it though...doing that I still get [hide]15[/hide]

[quote="biffanddoc"][hide]
The white chips can either
have no reds separating them
have 1 red separating them
have 2 reds separating them

All of the other ways are just rotations of those.
[/hide][/quote]
[hide]I agree. It says DISTINGUISIBLE ways.[/hide]


Suppose $ x$ is a solution to $ x^2 \plus{} 1 \equal{} 7x$. What is the sum of $ x$ and it's reciprocal?

The question asks for the sum of x and its reciprocal.
The sum of any number N and its reciprocal is $ N \plus{} \frac{1}{N} \equal{} \frac{N^2 \plus{} 1}{N}$
So, we want to find $ \frac{x^2 \plus{} 1}{x}$.
Using the handy equation that was given, we find that by dividing both sides by x, the equation becomes:
$ \frac{x^2\plus{}1}{x} \equal{} 7$, and therefore the sum of x and its reciprocal is $ \boxed{7}$.


[quote="ashrafmod"]detremain all triangle ABC for which
$\cos a \cos b+\sin a \sin b\sin c=1$[/quote]
[hide="solution"]$\cos{(a-b)}=\cos a\cos b+\sin a\sin b\geq\cos a \cos b+\sin a \sin b\sin c=1$, therefore $a=b=\frac{\pi}{4}$ and $c=\frac{\pi}{2}$.[/hide]


How many pairs of prime numbers have a sum of 40?

Listing primes less then 20,

2  -> +38 (not prime)
3  -> +37 (prime)
5  -> +35 (not prime)
7  -> +33 (not prime)
11 -> +29 (prime)
13 -> +27 (not prime)
17 -> +23 (prime)
19 -> +21 (not prime)

So, 3.

Except you only need to test: 3,7,11,13,17,19 since the rest clearly won't work since they're factors of 40.

Does the order only matter if it says ordered pair?

The easiest way is just to list out all the possibilities

2+38   X
3+37   O
5+35   X
7+43   O
11+29 O
13+27 X
17+23 O
19+21 X

... and it loops around.
There are 4 pairs of prime numbers that add up to 40.

[quote="dragon96"]The easiest way is just to list out all the possibilities

7+43   O
[/quote]

They add up to 40, not 50. It really should be 7 and 33, like BOGTRO said, which is not prime. So, 3 ways.


pigeonhole principle
search
combinatorics unsolved
combinatorics

Given a graph (simple graph, no self-loops) on $ n$ nodes such that each node has degree at least $ \frac{n}{2}$, show that it has a Hamilton Circuit.

Thanks,

If you're trying to work out a proof for yourself, here's a couple hints that might get you started:

1) First, see if you can prove the (much weaker) statement that $ G$ is connected.

2)  Then, try to prove the following two claims, which essentially amount to an inductive construction of the circuit:

A) If $ G$ has a cycle of length $ k$ and $ k < n$,  then $ G$ has a path of length $ k \plus{} 1$.

B) If $ G$ has a path of length $ k$, it either has a path of length $ k \plus{} 1$ or a cycle of length $ k$.  In particular, if $ k \equal{} n$ it must have a cycle of length $ n$.

3) For both 1) and 2), the pigeonhole principle is your friend.

If you're careful enough in your argument for (2), you'll actually end up proving the following generalization of Dirac's theorem due to Ore: If for a graph $ G$ on $ n$ vertices every pair of non-adjacent vertices $ x$ and $ y$ satisfies $ deg(x)\plus{}deg(y) \geq n$, then $ G$ has a Hamiltonian cycle.

[quote="kevinatcausa"]If you're trying to work out a proof for yourself, here's a couple hints that might get you started:

1) First, see if you can prove the (much weaker) statement that $ G$ is connected.

2)  Then, try to prove the following two claims, which essentially amount to an inductive construction of the circuit:

A) If $ G$ has a cycle of length $ k$ and $ k < n$,  then $ G$ has a path of length $ k \plus{} 1$.

B) If $ G$ has a path of length $ k$, it either has a path of length $ k \plus{} 1$ or a cycle of length $ k$.  In particular, if $ k \equal{} n$ it must have a cycle of length $ n$.

3) For both 1) and 2), the pigeonhole principle is your friend.

If you're careful enough in your argument for (2), you'll actually end up proving the following generalization of Dirac's theorem due to Ore: If for a graph $ G$ on $ n$ vertices every pair of non-adjacent vertices $ x$ and $ y$ satisfies $ deg(x) \plus{} deg(y) \geq n$, then $ G$ has a Hamiltonian cycle.[/quote]

hmm...exactly my proof from here:

http://www.mathlinks.ro/viewtopic.php?search_id=1724477919&t=184544


Electrons that "paint" the picture in a TV tube undergo constant acceleration over a distance of $3.8$ cm. If they reach a final speed of $1.2$ x $10^{7}$ m/s, what are (a) the electrons acclerations and (b) the time spent accelerating?

I'm assuming that they begin at rest.  All you need to do is solve the system of equations: \[\begin{cases}1.2\times 10^{7}=at \\ 3.8\times 10^{-2}=\frac{1}{2}at^{2}\end{cases}\]


Prove or disprove the following statement:
There exists $ a,b,y \in N$ that satisfy the condition:
$ 2a^{2}b^{2}(a^{2}\plus{}b^{2})\equal{}y^{3}$
 :wink:

[quote="ghjk"]Of course it's obviously right when (a-b)^2=2ab since this equation always has at least one root :)[/quote]

Unfortunately, you're wrong : $ (a\minus{}b)^2\equal{}2ab$ has no solution for $ a,b\in\mathbb{N}$

$ d \equal{} gcd(a,b)$ then
$ a \equal{} dm,b \equal{} dn$ where $ \gcd(m,n) \equal{} 1$
So $ 2a^2b^2(a^2 \plus{} b^2) \equal{} 2d^6m^2n^2(m^2 \plus{} n^2)$
It follow that
$ (mn)^2(m^2 \plus{} n^2) \equal{} 4k^3$
But $ \gcd(mn,m^2 \plus{} n^2) \equal{} 1$ and $ 4\not |a^2 \plus{} b^2$ so we must have :
$ m^2 \plus{} n^2 \equal{} z^3,(mn)^2 \equal{} 4x^3$
Exist t such that:
$ mn \equal{} 2t^3$
But from $ \gcd(m,n) \equal{} 1$ so exist $ p,q\in N$ such that :
$ (m,n) \equal{} (2p^3,q^3)$
So $ 4p^6 \plus{} q^6 \equal{} z^3$
But I think it has no solution on N as Fermat last theory .

Well, I may be dumb, but I don't quite see the connection between that equation and Fermat's Last Theorem  :wink: .

[quote="ghjk"][quote="pco"][quote="ghjk"]Of course it's obviously right when (a-b)^2=2ab since this equation always has at least one root :)[/quote]

Unfortunately, you're wrong : $ (a \minus{} b)^2 \equal{} 2ab$ has no solution for $ a,b\in\mathbb{N}$[/quote]
In fact, if a,b belongs to N*, it has no root! But in this case, a=b=0 is a root of this equation! :)[/quote]

The fact that $ 0\in\mathbb{N}$ is a decision which differ between countries (in mine, $ 0\in\mathbb{N}$).

So, on this forum, it is considered that $ \mathbb{N}$ does not contain $ 0$ (US choice, for example).

To be unambiguous, we could use the notations $ \mathbb{Z}^{\plus{}}, \mathbb{Z}^{\plus{}} \cup \{0 \}$.  :)

TTsphn:  I got to the same point as you, and then tried factoring over $ \mathbb{Z}[i]$, but it wasn't helpful.   :|


So, is anybody in the closed beta testing right now? I just enrolled for it, but I'm thinking that the open beta testing might open before I even get response, anyway. Discuss Mythos here.

Am I even supposed to know about it??
I have no idea what it is...

http://www.mythos.com/

Google is your friend.

1 d0n'7 kn0w wh47 17 15, 4nd 1 d0n'7 c4r3 wh47 17 15.


find all integer positives x,y so that
\[8^{x}=2^{x}+y\left({2y-1}\right)\left({2y-2}\right)\]

[hide]
We know that x can't be too big because the left side grows waaaay too quickly.  After much work, I found these solutions (x,y)
0,0
0,1
2,3
[/hide]

It's not enough.  $y$ can grow as large as necessary so that the right side is just as big as the left side.  You need to show that $8^{x}-2^{x}$ takes on the form $y(2y-1)(2y-2)$ at only the points you described.


geometry
function
calculus
integration
trapezoid
national olympiad

Today and tomorrow.

This is allowed as long as we don't discuss the content of the questions, right?

How did everyone find it? Manage okay? Where's everyone from? Discuss.

lol....it was okaii i spose. But definitely harder than the past papers.

[quote="shobber"]Can someone please post the problems? I really want to see them......[/quote]

The questions aren't allowed back to us till March and I don't think it's allowed, but anyway..

Ed- Well, I still think the '04 one was easy.. Today's could've been a lot worse.

Sen- Yeah, you probably should, but if you get a few marks out it won't be too much of a difference..

...I only answered 2 with complete solutions for day 2.  What about everyone else?  And how much of the paper does one have to solve for a gold certificate?

Around what's the cutoff for silver? In the past couple of years?

i think it's like around 34....not too sure

hmm think I got 28 or so..what's the cutoff for bronze around?

Ougoah- Oh, you'll get a bronze. Silver.. you've got a chance by what Edward said. Thirties?

Out of curiosity, what school are you at?

[quote="ougoah"]I'm from Fort St.  guessing you are from Ruse, since Sen is.  hope I can get [i]something[/i]...the 2001 paper was easier[/quote]

Yeah I'm there now. 2001 paper.. did you get that from school? I've got the ones up to '95, and then 2003 and 2004. 2003 was hard, 2004 was relatively easy. Were you the only one from Fort Street to do it?

I received the 2001 paper from Geoff Ball.  and yes I was the only one..it would have made 2 had my friend remembered to reply to the invitation.

cant expect too much relli..

i hope to get 20 overall

just pass the cutoff for bronze hopefulli

wens results gonna come out?

I'm thinking about 20 too..

How can someone manage to forget to reply to the invitation.

Results.. you said a few weeks. Ask someone who did it last year or something.

APMO is March 14 so it's probably not going to be long

lol....the AMO gets us into the April camp rite???.....if we're like recieving bronze, silver or gold awards

April camp.. That's like the Melbourne one. Top 12 overall plus 12 juniors who have a good chance later on. So, I'd say around Silver.

Ougoah- That's what I thought too. Wasn't it the 17th?

btw vulgarfraction have you received the AoPS books?  If so, how suitable are they for preparing for the westpac competition?

dun worry.....westpac will be easy...if u can do the AMO...:P

I didn't order them. Postage here is over thirty after conversion. I'm going to get a few from Amazon.

AMC's not hard unless you're aiming for a medal. But there's the new format and all.

is Engell among those in your list?  or Zeitz?  can you tell me your list? :)

I want to get the Engel one and [i]Geometry Revisited[/i] by Greitzer and Coxeter. ^^

AMO 2006

1. Find all positive integers $m$ and $n$ such that $1 + 5 \times 2^m = n^2$

2. f is a function defined for positive integers with positive integral values

(i) $f(ab) = f(a)f(b)$
(ii) $f(a) < f(b) if a < b$
(iii) $f(3) \geq 7$

Find the minimum value for $f(3)$.

3. $PRUS$ is a trapezium such that $\angle PSR = 2\angle QSU$ and $\angle SPU = 2 \angle UPR$. Let $Q$ and $T$ be on $PR$ and $SU$ respectively such that $SQ$ and $PU$ bisect $\angle PSR$ and $\angle SPU$ respectively. Let $PT$ meet $SQ$ at $E$. The line through $E$ parallel to $SR$ meets $PU$ in $F$ and the line through $E$ parallel to $PU$ meets $SR$ in $G$. $FG$ meets $PR$ and $SU$ in $K$ and $L$ respectively. Prove that $KF = FG = GL$

4. There are $n$ points on a circle, with each line segment connecting the two is either red or blue. $P{}_iP{}_j$ is red iff $P_i_+_1 P_j_+_1$ is blue, for distinct $i, j$ in set ${1, 2, ..., n}$

(a) for what $n$ is this possible?
(b) prove that one from get from any point to any other point through a maximum of 3 steps when a 'step' is from a point to another point through a red segment joining them.

5. In a square $ABCD$, $E$ is a point on diagonal $BD$. $O_1$ and $O_2$ are the circumcentres of $\triangle ABE$ and $\triangle ADE$ respectively. Prove that $AO{}_1EO{}_1$ is a square.

6. For any positive integer $n$, $a(n)$ is the product of the digits in $n$.

(a) Prove that $n \geq a(n)$
(b) Find all $n$ for which $n^2-17n+56 = a(n)$

7. Of a sequence $S_0$, another sequence $S_1$ exists called the offspring of $S_0$ in that each term $a_i$ in $S_1$ is the number of terms to the right of $a_i$ in $S_0$ larger than $a_i$.

eg. $S_0 =$ {$8, 2, 3, 5, 7, 0, 1, 2$} has offspring $S_1 =$ {$7, 2, 3, 3, 3, 0, 0, 0$}

$S_1$ is the offspring of any sequence $S_0$, and $S_2$ is the offspring of $S_1$, and so on.

Prove that $S_j = S_j_+_1$ for some $j$.

8. Alice and Bob play a guessing game. Alice picks a positive integer $a$ and tells Bob that $a \leq 2006$. In each turn, Bob gives Alice a positive integer $b$ and Alice tells Bob whether or not $a+b$ is a prime.

Prove that Bob can guess $a$ in less that 2006 turns.

Could you explain the 7th question? I'm not sure I understand it :?

take for example the first element of the set. count the number of elements to the right that are less that it (8). there are 7 items => first element of the oppspring set is a seven.

number of elements to the right of the second element less than it (<2) is 2 (0 and 1)
=> second element of offspring is a two

etc.


number theory
greatest common divisor
algorithm

1. If n is a 3-digit number with digits a,b,c in that order from left to right (n=abc), let

f(n)=a+b+c+ab+ac+bc+abc.

For example, f(234)=2+3+4+6+8+12+24=59

How many 3 digit positive integers n satisfy f(n)=n

2. Let a(n) denote the n-digit positive integer all of whose digits are 1.  For example a(3)=111.  What is the greatest common divisor of a_(45) and a_(140)

for the second problem observe that $ a_n$ = $ 10^n \minus{}1/9$

So $ a_{45}$ = $ 10^{45} \minus{}1/9$

and $ a_{140}$ = $ 10^{140} \minus{}1/9$

So let us first find the gcd of 9$ a_{45}$ and 9$ a_{140}$

By euclid's algorithm u get the gcd of 9$ a_{45}$ and 9$ a_{140}$ is $ 10^5 \minus{} 1$

So the gcd of $ a_{45}$ and $ a_{140}$ is $ (10^5 \minus{} 1)/9$ which is 11111.

Befuddlers

For #1, note that $ f(n) \plus{} 1 \equal{} (a \plus{} 1)(b \plus{} 1)(c \plus{} 1) \equal{} 100a \plus{} 10b \plus{} c \plus{} 1$
b=9 and c=9 gives a= anything
b=9 and c=8 \ b=8 and c=9 gives no solution
b=8 and c=8 gives no solution
Otherwise,
$ (b \plus{} 1)(c \plus{} 1) \le 72 \implies 100a \plus{} 10b \plus{} c \plus{} 1 \le 72(a \plus{} 1) \implies$
$ 28a \plus{} 10b \plus{} c \le 71 \implies a \equal{} 1$ or $ a \equal{} 2$.
$ a \equal{} 1 \implies 2(b \plus{} 1)(c \plus{} 1) \equal{} 101 \plus{} 10b \plus{} c \implies (2b \plus{} 1)(c \minus{} 4) \equal{} 95 \implies b \equal{} 9$, $ c \equal{} 9$
$ a \equal{} 2 \implies 3bc \plus{} 2c \minus{} 7b \equal{} 198 \implies (3b \plus{} 2)(3c \minus{} 1) \equal{} 580 \implies b \equal{} 9$, $ c \equal{} 9$
so 9 solutions.

[quote="befuddlers"]for the second problem observe that $ a_n$ = $ (10^n \minus{} 1)/9$

So $ a_{45}$ = $ (10^{45} \minus{} 1)/9$

and $ a_{140}$ = $ (10^{140} \minus{} 1)/9$

So let us first find the gcd of 9$ a_{45}$ and 9$ a_{140}$

By euclid's algorithm u get the gcd of 9$ a_{45}$ and 9$ a_{140}$ is $ 10^5 \minus{} 1$

So the gcd of $ a_{45}$ and $ a_{140}$ is $ (10^5 \minus{} 1)/9$ which is 11111.

Befuddlers[/quote]
bufuddlers,

I don't know about the correctness of your whole solution,
but you missed three sets of grouping symbols, such as
parentheses, that I placed in the highlighted "quote box"
above.

[quote="the future"]anybody?

thanks[/quote]
the future,

you bumped interest in your post after 2 hours 11 minutes
since it was first posted.  Don't go back in desperation
looking for hand-outs on your (?) problem (or perhaps someone
else really created it and it is part of homework, a project,
or challenge exam).  If it is good enough/interesting enough,
users will decide they want to invest time on it, and/or they
are working on it behind the scenes.

Don't make your desperation (which is what it is) an emergency
AND unfairness to the order/hierarchy of other math problems
that come along.  Don't bump up your problem(s).

alright my bad...too much CC maybe lol?

it was pretty far down (like below half way) because of all those AMC 12 problems that were posted just after I posted, so I bumped it up.

and nah, its from the 2008 Leigh University Math Exam, its actually got solutions but it was kinda confusing

besides i posted solutions to like 3-4 of the AMC 12 probelms that were posted so its not like im just here to have people answer my questions.


If $t_k=\frac{k(k+1)}{2}$ and $a_n=\frac{t_1+t_2+\ldots+t_n}{1^2+2^2+\ldots n^2}$

Find $a_{2007}$
I know these problems are a little easy but they don't really fit in middle school forum. I'm posting them for a very specific reason.


function
trigonometry
inequalities unsolved
inequalities

Find minimum and maximum of funtion $ f(x) \equal{}sin^{6}x\plus{}cos^{6}x\plus{}msinxcosx$, with $ m$ is a constan.

Let $ t = \sin x\cos x = \frac {1}{2}\sin 2x$, we have $ - \frac {1}{2}\leq t \leq \frac {1}{2}$.
$ f(x) = \sin ^ 6 x + \cos ^ 6 x + m\sin x\cos x$

$ = (\sin ^ 2 x + \cos ^ 2x)^3 - 3\sin ^ 2x\cos ^ 2x (\sin ^ 2x + \cos ^ 2x) + m\sin x \cos x$
$ = - 3t^2 + mt + 1$

$ = - 3\left(t - \frac {m}{6}\right)^2 + \frac {1}{12}m^2 + 1\equiv g(t)\ \left( - \frac {1}{2}\leq t\leq \frac {1}{2}\right)$.

We are to consider the graph of the qudratic function $ y = g(t)$ and classify the position of the axis $ t = \frac {m}{6}$ as follows.

$ 1^\circ\ \frac {m}{6} < - \frac {1}{2}\Longleftrightarrow m < - 3$

$ f_{max} = g\left( - \frac {1}{2}\right) = - \frac {1}{2}m + \frac {1}{4}$.

$ f_{min} = g\left(\frac {1}{2}\right) = \frac {1}{2}m + \frac {1}{4}$.

$ 2^\circ\ - \frac {1}{2}\leq\frac {m}{6} < \frac {1}{2}\Longleftrightarrow - 3\leq m < 3$

$ f_{max} = g\left(\frac {m}{6}\right) = \frac {1}{12} m^2 + 1$.

$ f_{min} = Min.\left\{g\left( - \frac {1}{2}\right),\ g\left( - \frac {1}{2}\right)\right\}$

$ = Min.\left\{ - \frac {1}{2}m + \frac {1}{4},\ - \frac {1}{2}m + \frac {1}{4}\right\}$

$ = \left\{\begin{array}{ll} \frac {1}{2}m + \frac {1}{4}\ (m < 0) & \quad \\
- \frac {1}{2}m + \frac {1}{4}\ (m\geq 0) & \quad \end{array} \right.$

$ 3^\circ \ \frac {1}{2}\leq \frac {m}{6}\Longleftrightarrow m\geq 3$

$ f_{max} = g\left(\frac {1}{2}\right) = \frac {1}{2}m + \frac {1}{4}$.

$ f_{min} = g\left( - \frac {1}{2}\right) = - \frac {1}{2}m + \frac {1}{4}$.

Therefore summarizing the above results,

$ f_{max}=\left\{
\begin{array}{ll}
-\frac{1}{2}m+\frac{1}{4}\ (m<-3)  &\quad  \\
\frac{1}{12}m^2+1\ (-3\leq m<3) &\quad　\\
\frac{1}{2}m+\frac{1}{4}\ (m\geq 3) &\quad
\end{array}
\right.$

$ f_{min} = \left\{\begin{array}{ll} \frac {1}{2}m + \frac {1}{4}\ (m < 0) & \quad \\
- \frac {1}{2}m + \frac {1}{4}\ (m\geq 0) & \quad \end{array} \right.$

Thank you very much. Now I'm buzy, I will read it later. :o