[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ \sqrt {a \plus{} bc} \plus{} \sqrt {b \plus{} ca} \plus{} \sqrt {c \plus{} ab}\geq \ \sqrt {2}(ab \plus{} bc \plus{} ca)$
  :)[/quote]
Of course, It can be killed by AMGM inequality only,My friend,
Setting $ q \equal{} ab \plus{} bc \plus{} ca$ thus $ q\le {3}$
\[ \leftrightarrow 3 \plus{} q \plus{} 2\sum \sqrt {(a \plus{} c)(a \plus{} b)(b \plus{} c)(a \plus{} b)}\ge 2q^2
\]
But
\[ \sum (a \plus{} b)\sqrt {(a \plus{} c)(b \plus{} c)}\ge 3((a \plus{} b)(b \plus{} c)(c \plus{} a))^{\frac {2}{3}}
\]
And
\[ (a \plus{} b)(b \plus{} c)(c \plus{} a)\ge \frac {8q}{3}
\]
We can complete the proof since the condition $ p\le 3$
I think it is weak
Editted: I am wrong, sorryy. I will try later

[quote="Allnames"][quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a + b + c = 3$. Prove that: $ \sqrt {a + bc} + \sqrt {b + ca} + \sqrt {c + ab}\geq \ \sqrt {2}(ab + bc + ca)$
  :)[/quote]
Of course, It can be killed by AMGM inequality only,My friend,
Setting $ q = ab + bc + ca$ thus $ q\le {3}$
\[ \leftrightarrow 3 + q + 2\sum \sqrt {(a + c)(a + b)(b + c)(a + b)}\ge 2q^2
\]
But
\[ \sum (a + b)\sqrt {(a + c)(b + c)}\ge 3((a + b)(b + c)(c + a))^{\frac {2}{3}} And \[ (a + b)(b + c)(c + a)\ge \frac {8q}{3}
\]
We can complete the proof since the condition $ p\le 3$
I think it is weak[/quote]
Are you sure? $ a + b + c = 3$,then $ a + bc$ not equal to $ (a + b)(a + c)$.:wink:
Regards :oops:

[quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ \sqrt {a \plus{} bc} \plus{} \sqrt {b \plus{} ca} \plus{} \sqrt {c \plus{} ab}\geq \ \sqrt {2}(ab \plus{} bc \plus{} ca)$
  :)[/quote]
By Holder we obtain:
$ \left(\sum_{cyc}\sqrt {a \plus{} bc}\right)^2\sum_{cyc}\frac{1}{a\plus{}bc}\geq27.$
Thus, we need to prove that $ 27\geq2(ab\plus{}ac\plus{}bc)^2\sum_{cyc}\frac{1}{a\plus{}bc},$ which is obvious.

Another solution.
[hide]
By AMGM we have that
$ \sum{\sqrt{a\plus{}bc}\plus{}\sqrt{a\plus{}bc}\plus{}\dfrac{(a\plus{}bc)^2}{2\sqrt{2}}}\ge\sum{\dfrac{3(a\plus{}bc)}{\sqrt{2}}}$

So it suffices to prove that
$ \sum{\dfrac{3(a\plus{}bc)}{\sqrt{2}}\minus{}\dfrac{(a\plus{}bc)^2}{2}}\le 2\sqrt{2}(ab\plus{}bc\plus{}ca)$

Which is true because we have that
$ (a\plus{}b\plus{}c)^2\ge 3(ab\plus{}bc\plus{}ca)\Leftrightarrow a\plus{}b\plus{}c\ge ab\plus{}bc\plus{}ca\Leftrightarrow$

$ 2(a\plus{}b\plus{}c)^2\ge (ab\plus{}bc\plus{}ca)^2\plus{}(a\plus{}b\plus{}c)^2\Leftrightarrow$

$ 6(a\plus{}b\plus{}c)\ge \sum{b^2c^2\plus{}2abc\plus{}a^2\plus{}2bc}\Leftrightarrow$

$ 6\sum{(a\plus{}bc)}\ge \sum{(a\plus{}bc)^2\plus{}8bc}\Leftrightarrow$

$ \sum{\dfrac{3(a\plus{}bc)}{\sqrt{2}}\minus{}\dfrac{(a\plus{}bc)^2}{2}}\le 2\sqrt{2}(ab\plus{}bc\plus{}ca)$
[/hide]

:)

[quote="arqady"][quote="nguoivn"]Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that: $ \sqrt {a \plus{} bc} \plus{} \sqrt {b \plus{} ca} \plus{} \sqrt {c \plus{} ab}\geq \ \sqrt {2}(ab \plus{} bc \plus{} ca)$
  :)[/quote]
By Holder we obtain:
$ \left(\sum_{cyc}\sqrt {a \plus{} bc}\right)^2\sum_{cyc}\frac {1}{a \plus{} bc}\geq27.$
Thus, we need to prove that $ 27\geq2(ab \plus{} ac \plus{} bc)^2\sum_{cyc}\frac {1}{a \plus{} bc},$ which is obvious.[/quote]
Nice proofs, arqady and ElChapin  :lol: . And how'll you solve with this similar problem:
Given $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ \sqrt [3]{a \plus{} bc} \plus{} \sqrt [3]{b \plus{} ca} \plus{} \sqrt [3]{c \plus{} ab} \geq\ \sqrt [3]{2}(ab \plus{} bc \plus{} ca)$

PS: Both of them can be solved easily by Am-Gm :)

I have to proofs for your second inequality. It is nice and relatively easy.
Here is one of them, it is not the simplest of both, but is the first one, so I guess I like it the most.
[hide]
Case 1: $ ab + bc + ca\le 1$.
Suppose $ a = max\{a,b,c\}$, so we have that $ a\ge 1$.
It is also true that $ b + c < 1$. Indeed, suppose the contrary then $ 1\ge ab + bc + ca > (b + c)a\ge a$, $ \rightarrow\leftarrow$.
From here it follows that $ a\ge 2$, then $ \sum{\sqrt [3]{a + bc}} > \sqrt [3]{2} > \sqrt [3]{2}(ab + bc + ca)$.

Case 2: $ ab + bc + ca > 1$.
First, observe that inequality $ \sum{2(a + bc)}\ge\sum{(a + bc)^2}$ is true, it is equivalent to $ (a^2 + b^2 + c^2 - ab - bc - ca)(ab + bc + ca - 1)\ge 0$

Because of AMGM
$ \sum{\sqrt [3]{a + bc} + \sqrt [3]{\dfrac{(a + bc)^4}{2^3}} + \sqrt [3]{\dfrac{(a + bc)^4}{2^3}}\ge\sum{\dfrac{3(a + bc)}{\sqrt [3]{4}}}}$

So it suffices to prove that
$ \sum{3(a + bc)}\ge\sum{2bc + \sqrt [3]{4(a + bc)^4}}$

Again, by AMGM we have that
$ \frac {1}{3}\sum{(a + bc)^2 + (a + bc)^2 + 4}\ge\sum{\sqrt [3]{4(a + bc)^4}}$

So it is enough to prove that
$ \sum{3(a + bc)}\ge 4 + \sum{2bc + \frac {2}{3}(a + bc)^2}$

This last inequality holds because of
$ \sum{\frac {4}{3}a} = 4$, $ \sum{\frac {4}{3}(a + bc)}\ge\sum{\frac {2}{3}(a + bc)^2}$ and $ \sum{\frac {1}{3}a + \frac {5}{3}b}\ge\sum{2bc}$

This concludes the proof.
[/hide]

As I tried to prove this I grew up interest for this inequality. At this moment I'm thinking in a few things:
* I found another proof for the first inequality and  my second proof for the second inequality is almost the same, so ... How about the problem $ \sqrt [n]{a + bc}...$
* Can these inequalities can be made stronger (and how much stronger) with "simple and nice" methods... (I know this is very vague and uncertain)
* I think that both inequalities can easily be proved using CBS...

I will play around for a time and then post what happens...

:)

* In the same way as in my other proof it can be proven that if $ a, b, c > 0$, $ a \plus{} b \plus{} c \equal{} 3$ and $ 0 < n\le 1$
   $ \sum{(a \plus{} bc)^n}\ge \dfrac{2^n}{2}(3 \plus{} ab \plus{} bc \plus{} ca)$ if $ ab \plus{} bc \plus{} ca\ge 1$
and (just to say something...) $ \sum{(a \plus{} bc)^n} > \dfrac{2^n}{2}(1 \plus{} ab \plus{} bc \plus{} ca)$ if $ ab \plus{} bc \plus{} ca < 1$

* Another proof of these inequalities is possible using CBS at the begining, then we must use AMGM or somthing equivalent to finish the proofs.
:)

$ 27\geq2(ab\plus{}ac\plus{}bc)^{2}\sum_{cyc}\frac{1}{a\plus{}bc},$

Why is it obvious?

Thank you very much, Arqady?

[quote="manlio"]$ 27\geq2(ab \plus{} ac \plus{} bc)^{2}\sum_{cyc}\frac {1}{a \plus{} bc},$

Why is it obvious?

Thank you very much, Arqady?[/quote]
I don't remember. :D I am sorry. :(
But $ \sum_{cyc}(a\plus{}bc)(b\plus{}ac)\equal{}\sum_{cyc}(ab\plus{}a^2c\plus{}b^2c\plus{}c^2ab)\equal{}$
$ \equal{}\sum_{cyc}(ab\plus{}a^2b\plus{}a^2c\plus{}abc)\equal{}4(ab\plus{}ac\plus{}bc).$
Thus, we need  prove that
$ \sum_{cyc}(9abc\plus{}9a^2b^2c^2\plus{}27a^3bc\plus{}27a^2b^2)\geq8(ab\plus{}ac\plus{}bc)^3,$
which is equivalent to $ \sum_{sym}(3a^4b^2\minus{}a^3b^3\plus{}5a^4bc\minus{}6a^3b^2c\minus{}a^2b^2c^2),$ which is indeed obvios.


geometry
geometric transformation
reflection
parallelogram
trapezoid
geometry proposed

Let $ABC$ be an acute triangle inscribed in a circle centered at $O$. Let $M$ be a point on the small arc $AB$ of the triangle's circumcircle. The perpendicular dropped from $M$ on the ray $OA$ intersects the sides $AB$ and $AC$ at the points $K$ and $L$, respectively. Similarly, the perpendicular dropped from $M$ on the ray $OB$ intersects the sides $AB$ and $BC$ at $N$ and $P$, respectively. Assume that $KL=MN$. Find the size of the angle $\angle{MLP}$ in terms of the angles of the triangle $ABC$.

$\angle AKL = \angle BNP = \angle C,$ $\triangle MKN$ is isosceles, MN = MK = KL, K is the midpoint of ML. In addition, $\triangle ALK,\ \triangle PBN$ are both oppositely similar to $\triangle ABC.$ $\angle AMB = 180^\circ-\angle C,$ $\angle MAK+\angle MBN = \angle C.$ Let A' be reflection of A in K, AMA'L is a parallelogram, its diagonals cutting each other in half and $\angle MAK+\angle A'LK = \angle LA'K+\angle A'KN = \angle LKA = \angle C.$ It follows that $\angle MBN = \angle A'KN,$ $\triangle PBM \sim \triangle ALA'$ are similar, with corresponding points K, N, so that N is the midpoint of MP. Thus NP = MN = KL, KLPN is an isosceles trapezoid, $LP \parallel KN \equiv AB$ and $\angle MLP \equiv \angle KLP = \angle C.$

We may as well obtain a solution without any extra points, lines and etc.
Just observe that by angle chasing $KM=MN=KL$ and $\triangle AKL\sim \triangle PNB$ and $\triangle AKM \sim \triangle MNB$. Using the ratios between the segments, we  see that $ NP=KL$, so $\triangle MLP$ is isosceles. This implies the same result. :)


geometry
perimeter
trapezoid
exterior angle

In the attached picture, what is the perimeter and the area of the regular hexagon ABCDEF, if $\overline{BG}=\frac{2}{3}$ and $\angle{G}=90^{\circ}$?

[quote="Quevvy"][quote="vishalarul"][hide]The side length would be $\frac{4}{3}$, since it is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

Perimeter = $\frac{4}{3}\cdot 6 = 8$

Area = $\frac{aP}{2}= 4a$

The apothem is just $CG$, which is $\frac{2}{3}\sqrt{3}$:

Area = $\frac{aP}{2}= 4a = \frac{8}{3}\sqrt{3}$[/hide][/quote]
Why is it a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle?

What does Area = $\frac{aP}{2}= 4a$ mean? What does CG mean?

Sorry about so many questions... :P[/quote]

It's a 30-60-90 triangle since the hexagon's angle is 120: subtract 180 from that to get the 60. You already gave me the 90.

That is the area formula for any polygon: apothem times perimeter divided by 2.

CG is the length.

Or just divide them into six equilateral triangle and find the area of each and multiply that by 6. Or multiply the area of $\triangle{BCG}$ by 12 to find the total area of the hexagon .

My solution is slightly different but essentially the same
[hide]
[b]Perimeter:[/b] Since each exterior angle of a regular hexagon is 360/60=60, we have that Angle G=90 and Angle CBG=60.  Therefore, BCG is a 30-60-90 triangle.  Thus, BC=2(2/3)=4/3.  So, each of the six sides of the regular hexagon is 4/3.  Hence, the perimeter is 6(4/3)=$8$.
[b]Area:[/b] Draw FC to divide the regular hexagon into two congruent trapeziods.  Since CG is perpindicular to AB, we have that the area of trapezoid ABCF is (AB+FC)(CG)/2.  Since Trapezoid ABCF is isoceles, $FC=AB+2BG=4/3+2(2/3)=8/3$.  Now, CG is part of the 30-60-90 Triange: BCG, so $CG=2\sqrt{3}/3.$ Therefore, the area of Trapezoid ABCF is $((4/3+8/3)/2)*((2\sqrt{3})/3)=4\sqrt{3}/3$.  Therefore, the area of trapezoid FEDC is the same (because they are congruent).  Thus, the area of the other trapezoid is $4\sqrt{3}/3$.  Add them up to get $8\sqrt{3}/3$. [/hide]

[hide="solution"]Since

$\angle ABC+\angle CBG = 180^{\circ}$

and

$\angle ABC = 120^{\circ}$,

$\angle CBG = 60^{\circ}$.

Therefore,

$\angle BCG = 30^{\circ}$

and

$\overline{BC}=\frac{4}{3}$.

We want the perimeter of the triangle, which is

$P= 6 \times \overline{BC}= \boxed{8}$

For the area, we want

$A = \frac{3s^{2}\sqrt{3}}{2}$,

where s is the side length.

$A = \frac{3s^{2}\sqrt{3}}{2}= \boxed{\frac{8\sqrt{3}}{3}}$[/hide]


Let be given two polynomials $f,g$ not constant in ${\mathbb Q}[x]$ ie.. they have their coefficients in rational numbers. Assume that the set of all values of $f(x)$ at $x\in \mathbb Q$ equals to the set all values of $g(x)$ at $x\in \mathbb Q$. Show that there are $a,b\in\mathbb Q$ so that $f(x)\equiv g(ax+b)$ forall $x$.


trigonometry
function
linear algebra
matrix
complex numbers
calculus
calculus computations

Bah, I hate it when classes assume that we have learned things that we have not. Could someone please explain to me how to represent:

$ F_x\equal{}a(cos\omega t\plus{}sin\omega t)$ as the real part of a complex expression?

well, if a.t and $ \omega$ are real, than the number is real, so it is the real part of itself

$ \cos t \equal{} \textrm{Re}\left(e^{it}\right)$ and $ \sin t \equal{} \textrm{Im}\left(e^{it}\right)$ -- can you figure it out from there?

>>
well, if a.t and \omega are real, than the number is real, so it is the real part of itself

Yes, it is real, what I should have said is, can someone show me how to write it in the form of a complex exponential.

JBL:

I can write cos or sin by themselves as a complex exponential, it is having them together in the real part that is screwing me up. I'm sure there is some identity I can use here?

MAde some progress:

$ a(cos\omega t \plus{} sin\omega t)$=$ Asin(\omega t \plus{} \phi)$

but $ Asin(\omega t \plus{} \phi)$ = $ Asin\omega t cos\phi \plus{} Acos\omega t sin\phi$

so $ Acos\phi \equal{} Asin\phi \equal{} a$==>$ \phi \equal{} \pi/4$ and $ A \equal{} \sqrt2 a$

so $ a(cos\omega t \plus{} sin\omega t)$=$ \sqrt2 asin(\omega t \plus{} \pi/4)$

Now how to I make that a complex exponential?

The complex expression should be of the form $ c_1e^{c_2t}$, with $ c_1$ and $ c_2$ complex numbers. In this case, $ c_1$ will be neither purely real nor purely imaginary.

my idea is to write the number as a cosine instead of sine and then apply what JBL wrote

alternatively, use that $ cos(x\minus{}\pi/2)\equal{}sin(x)$

So then

$ \sqrt2 asin(\omega t \plus{} \pi/4) \equal{} \sqrt2 acos(\omega t \minus{} \pi/4)$?

and that could be written as the real part of $ \sqrt2 a e^{i(\omega t \minus{} \pi/4)}$

Is that correct?

Hey guys, last question - everything worked, but I am just wondering if someone could confirm both the factor of $ \sqrt2$ and the $ \minus{}\pi/4$ - those are the only things in my solution that I am not fully confident of.

The set of all simple harmonic motions of angular frequency $ \omega,$ which is to say the set of all solutions to the differential equation $ y''+\omega^2y=0,$ is a two-dimensional set of functions. Members of this set can be written in several useful forms; each of those forms has two arbitrary constants. I devote an early section of my notes on Fourier analysis to precisely this.

(1) Cosine-sine form: $ y= A\cos\omega t+B\sin\omega t.$

(2) Amplitude-phase form: $ y=R\cos(\omega t-\delta).$

(3) Complex form: $ y=Ce^{i\omega t}+C^*e^{-i\omega t}.$

(There's a fourth form, the "Hartley" or "cas" form, but I think I'll omit that from this post.)

Relationship between the cosine-sine and amplitude-phase forms:
\begin{align*}A&=R\cos\delta\\ B&=R\sin\delta\end{align*}
Note that this is algebraically the same as the conversion between rectangular and polar coordinates. We can reverse the equivalence, but it's not as clean. We write $ R=\sqrt{A^2+B^2}$ and $ \delta=\text{angle}\,(A,B),$ where the function $ \text{angle}$ has a quadrant-by-quadrant definition. In the first quadrant, $ \text{angle}\,(A,B)=\tan^{-1}\frac{B}{A}.$ (And I tacitly assumed $ y$ was real-valued in this paragraph.)

Relationship between the cosine-sine and complex forms:
\begin{align*}A&=C+C*\\B&=i(C-C^*)\end{align*}
or, writing it backwards (it's an invertible matrix transformation),
\begin{align*}C&=\frac{A-iB}2\\C^*&=\frac{A+iB}2\end{align*}

There are more formulas that we can make use of in the particular case that the function $ y$ is strictly real-valued. In that case:
\[ C^*=C\]
(along with $ A,B,R,$ and $ \delta$ being real.)
\begin{align*}A&=2\,\text{Re}\,C\\B&=-2\,\text{Im}\,C\end{align*}
and, to connect the complex form to the amplitude-phase form:
\begin{align*}R&=2|C|=2|C^*|\\ \delta&=-\text{arg}\,C=\text{arg}C^*\end{align*}


Rephrased problem: Prove that
\[ {n \choose 0}^2 \plus{} 2{n \choose 1}^2 \plus{} \ldots \plus{} (n\plus{}1){n \choose n}^2 \equal{} (n\plus{}2) {2n\minus{}1 \choose n\minus{}1}.\]

[hide]Let us suppose we have a house of representatives and senate each with $ n$ people. We need to choose an $ n$-person joint committee and either elect one member of the senate a leader or choose no leader at all.

On one hand, we can either select a leader from the senate ($ n$ ways) and then choose the remaining members ($ {2n\minus{}1 \choose n\minus{}1}$ ways), or if we don't have a leader at all we either select the first person from the senate and $ n\minus{}1$ others ($ {2n\minus{}1 \choose n\minus{}1}$ ways), or we don't select the first person and choose  $ n\minus{}1$ others that [i]aren't[/i] selected ($ {2n\minus{}1 \choose n\minus{}1}$ ways). Adding all of these up gives the right hand side.

On the other hand, let us choose from the house and senate separately. The number of people chosen in the senate must equal the number of people [i]not[/i] chosen in the house. If $ k$ people are selected from the senate ($ {n \choose k}$ ways), then $ k$ people are not selected from the house ($ {n \choose k}$ ways). Then we can select a leader in $ k$ ways or none at all (total $ k\plus{}1$ ways). Summing over all $ k$ gives the left hand side.[/hide]

consider n boys and n girls and a maths teacher. Make a delegation of n students + maths teacher with a condition that a boy or the maths teacher will be the leader
this can be done in two ways
(i) first select k boys and n-k girls + maths teacher and then one team leader from k+1
 now let k=0,1,2,...,n .
(ii)first a team leader from n boys then remaining n-1 students out of 2n-1 students or make maths teacher team leader and select n members out of 2n students


[quote="aznmonkey1992"]gertrude is 6 times as old as gunther
in 20 years, gertrude will only be 2 times as old
how old is gertrude now?[/quote]

i think you worded this incorrectly... its hard to understand...

[quote="aznmonkey1992"]gertrude is 6 times as old as gunther
in 20 years, gertrude will only be 2 times as old
how old is gertrude now?[/quote]

[hide="this time i remembered"]
x=gertrude
y=gunther

x=6y
x+20=2(y+20)

6y+20=2y+40
4y=20
y=5
therefore, x=30

[b]Gertrude is right now 30 years old[/b][/hide]

[quote="math92"][hide]
let G be getrude and g be gunther
6G=g
G+20=2g+20
solve the system of equations and........
i think i messed it up :?
[/hide][/quote]

It's okay. He worded it so it was hard to understand.

The wording isn't difficult. phi1.6188 had a correct solution, but he found the wrong answer (he found Gunther's age).

Also, another mistake of math92 was that he named G and g when it would have made lessc onfulsion if he had x and y.  :)

i'll give it another try:
[hide="click if you want a million dollars"]
let g be gertrude and y be gunther
g=6y
and
g+20=2(y+20) <--That's what i changed
y=g/6
so
g+20=2(g/6+20)
g+20=g/3+40
3g+60=g+120
2g=60
$\boxed{g=30}$
[/hide]

[hide]
6g=Gertrude, g=Gunther
6g+20=2(g+20)
6g=2g+20
4g=20
Guther is 5, so Gertrude is 30[/hide]

[hide][quote="aznmonkey1992"]gertrude is 6 times as old as gunther
in 20 years, gertrude will only be 2 times as old
how old is gertrude now?[/quote]

Let's consider the following:

Gertrude is 6 times older than Gunther, so:

Let:

x = Gunther's age

6x = Gertrude's age

Now, in 20 years, Gertrude will be only 2 times older:

2 (x + 20) = 6x + 20

It's quite obvious now :D

2x + 40 = 6x + 20

-4x = -20

x = 5 (Gunther)

6x = 30 (Gertrude)

Gertrude's age is 30. [/hide]

[hide]if gertrude is 30 and gunther is 5, (30+20)/2=5+20, so the answer is 30 :idea: [/hide]

[hide]a=gertrude
b=gunther

a=6b
a+20=2(b+20)
6b+20=2b+40
4b=20
b=5
6b=30

gertrude is 30[/hide]

[quote="aznmonkey1992"]gertrude is 6 times as old as gunther
in 20 years, [b]gertrude[/b] will only be [b]2 times as old [/b]how old is gertrude now?[/quote]

Why do you need the first sentence? In 20 years gertrude will be 2 times older...so if he is 20 now, 20 years later he will be 40, or 2 times older. I dont think this is what you meant, but if i saw this on a competition, this is what I would have interpreted.

i think he means:
In 20 years gertrude will be 2 times older than gunther


group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Verify the Galois correspondence explicitly for the splitting field of $x^3-2\in Q[x]$ and its Galois group.

Thanks a lot!

The decomposition field of $x^3-2$ is $\mathbb{Q}(\sqrt[3]{2},j)$
If $\sigma$ is in the Galois group then it is entirely determinated by the image of the generators.
$2=\sigma(2)=\sigma(\sqrt[3]{2}^3)=\sigma(\sqrt[3]{2})^3$ consequently $\sigma(\sqrt[3]{2})\in\{\sqrt[3]{2},j\sqrt[3]{2},j^2\sqrt[3]{2}\}$
Using same reasonment, $\sigma(j)\in\{j,j^2\}$.
Then you have $6$ different elements in the Galois group, it is isomorphic to $S_3$ and there is two subgroups.
First subgroup isomorphic to $\mathbb{Z}/3\mathbb{Z}$ wich is defined by $\sigma(j)=j$ for every element of its subgroup
Second subgroup isomorphic to $\mathbb{Z}/2\mathbb{Z}$ which is defined by $\sigma(\sqrt[3]{2})=\sqrt[3]{2}$
First subgroup gives the subfield $\mathbb{Q}(j)$ and second is $\mathbb{Q}(\sqrt[3]{2})$

Have you understood how it works ?

The Galois group can't be the cyclic group with $6$ elements! (I assumed that $j^3=1, j \neq 1$)
Indeed it's $S_3$, the permutations on $3$ elements.

Oups you're right, I'm going to edit my post

Well, $S_3$ has also $6$ subgroups (4 nontrivial).


group theory
superior algebra
superior algebra solved

Let G and H be 2 finite groups s.t. (|G|, |H|)=1. Show that Aut(GxH) is isomorphic to Aut(G)xAut(H).

Define $\varphi: Aut(G)\times Aut(H)\to Aut(G\times H)$ by $\varphi(u,v)(g,h)=(u(g),v(h))$, where $u\in Aut G$, $v\in Aut H$, $g\in G$ and $h\in H$. It is easy to see that $\varphi$ is an injective grup homomorphism. We need to prove that it is surjective. This means that for all $f\in Aut(G)\times Aut(H)$ we must find $u\in Aut(G)$ and $v\in Aut(H)$ such that $\varphi(u,v)=f$.

Consider the following sequence:
\[ G\stackrel{i_G}{\to}G\times H\stackrel f{\to}G\times H\stackrel{\pi_H}{\to}H,  \] where $i_G$ is the canonical inclusion $g\to (g,1)$ and $\pi_H$ is the canonical projection map $(g,h)\to h$. These are all group homomorpisms.
The composure $p=\pi_h\circ f\circ i_G$ is a group homomorphism from $G$ to $H$. Let $g\neq 1$ in $G$. Assume it has an order $n$. Then $n||G|$ but also $p(g)^n=1$ implying that the order of $p(g)$ is a divisor of $n$. But the order of $p(g)$ divides $|H|$. These and $gcd(|G|,|H|)=1$ imply $p(g)=1$. But $g$ was chosen arbitrarily, so $p$ is the trivial morphism. This means that $f(G\times\{1\})\subseteq G\times\{1\}$. Similarly $f^{-1}(G\times\{1\})\subseteq G\times\{1\}$. These yield $f(G\times\{1\})=G\times\{1\}$. Analogously $f(\{1\}\times H)=\{1\}\times H$.
Define $u=f|_{G\times\{1\}}^{G\times\{1\}}$ and $v=f|_{\{1\}\times H}^{\{1\}\times H}$. It is easy to prove that $u\in Aut(G)$, $v\in Aut(H)$ and $\varphi(u,v)=f$.


hi!
let ABC be a right triangle. Show that  $ R \geq (1\plus{}\sqrt{2})r$. When does the equality hold? :?:
God bless u!

[quote="kunny"]Let $ ABC$ be a right-angled triangle with $ AB \equal{} 1,\ BC \equal{} \cos \theta ,\ CA \equal{} \sin \theta \left(0 < \theta < \frac {\pi}{2}\right)$, W.L.O.G.
$ R \equal{} \frac 12,\ r \equal{} \frac {\cos \theta \plus{} \sin \theta \minus{} 1}{2}$, thus we have $ \frac {R}{r} \equal{} \frac {1}{\cos \theta \plus{} \sin \theta \minus{} 1} \equal{} \frac {1}{\sqrt {2}\sin \left(\theta \plus{} \frac {\pi}{4}\right)}\geq \frac {1}{\sqrt {2} \minus{} 1} \equal{} \sqrt {2} \plus{} 1$, yielding $ R\geq (1 \plus{} \sqrt {2})r$. The equality holds when $ \theta \equal{} \frac {\pi}{4}$, or $ ABC$ is right-angled isoceles triangle.[/quote]

Ey baba in aghaye kunny ham az hame chi sar dar miare ke,too forume iran chi mikhad dige??!! :rotfl:

Sorry, I made a typo. Here is the correction.

[quote="kunny"]Let $ ABC$ be a right-angled triangle with $ AB \equal{} 1,\ BC \equal{} \cos \theta ,\ CA \equal{} \sin \theta \left(0 < \theta < \frac {\pi}{2}\right)$, W.L.O.G.
$ R \equal{} \frac 12,\ r \equal{} \frac {\cos \theta \plus{} \sin \theta \minus{} 1}{2}$, thus we have $ \frac {R}{r} \equal{} \frac {1}{\cos \theta \plus{} \sin \theta \minus{} 1} \equal{} \frac {1}{\sqrt {2}\sin \left(\theta \plus{} \frac {\pi}{4}\right)\color{red}\minus{}1}\geq \frac {1}{\sqrt {2} \minus{} 1} \equal{} \sqrt {2} \plus{} 1$, yielding $ R\geq (1 \plus{} \sqrt {2})r$. The equality holds when $ \theta \equal{} \frac {\pi}{4}$, or $ ABC$ is right-angled isoceles triangle.[/quote]


calculus integration function limit logarithms search geometry

Integral test, just prove that's smaller than an integral that you know that converges.

The [b]Cauchy Condensation Test[/b] (CCT) might come in handy, too.

CCT. Let $ \{a_{n}\}_{n \in \mathbb{N}}$ be a decreasing sequence of positive real numbers.  Then, $ \sum_{n=1}^{\infty}a_{n}$ converges if and oly if $ \sum_{n=1}^{\infty}2^{n}a_{2^{n}}$ converges.

coquitao, I am very stupid to see how it works.
Can you please give some sketch proof?

Thank you for helping. :P

[quote="coquitao"]Gimme a sec...[/quote]

Thank you coquitao, :P  I look forward to your reply. :)

Preempting coquitao's reply:

The proof of the condensation test is based on the same principle as Oresme's classic proof that the harmonic series diverges; group the series into blocks of $ 2^{n}$.

For $ n\ge 1$,
$ 2^{n}a_{2^{n+1}}\le\sum_{k=2^{n}}^{2^{n+1}-1}a_{k}\le 2^{n}a_{2^{n}}$
The sum of the terms in the middle is the sum of the original series; the sums of the left hand side and of the right hand are constant multiples of the condensed series.

This method is not normally taught in calculus classes.

[quote="coquitao"]Gimme a sec...

By CCT all you to need to show is that

$ \sum_{n=1}^{\infty}2^{n}\left(\frac{1}{2^{n}}\right)^{s}$ converges when $ s > 1.$

Since,

$ 2^{n}\left(\frac{1}{2^{n}}\right)^{s}= 2^{n-ns}= 2^{n(1-s)}= (2^{(1-s)})^{n},$ it follows that this should be the case...

Remember the geometric series, right?[/quote]

Oh I see, :)

$ \sum_{n=1}^{\infty}(2^{(1-s)})^{n}= \frac{2^{(1-s)}}{1-2^{(1-s)}}$,

this series only converges if $ -1<2^{(1-s)}<1$, hence the series converges if $ s>1$.

coquitao, thank you for helping indeed. :P

[b]I wonder if anyone else has any [i]elementary proof [/i]of this question, please post it here.[/b] :!:
Your help is greatly appreciated, thank you for helping. :P

[quote="IrinaSharova"]$ \sum_{i=1}^{\infty}\frac{1}{i^{s}}$ where $ s\in\Re$[/quote]
This is essentially the same series as what I was taught as a "p-series" -- a lot of comparisons for convergence/divergence can be made based on the value of $ p\in\Re$ in the series $ \sum_{n=1}^{\infty}\frac{1}{n^{p}}$.  It is known that this series converges for $ p>1$ and diverges for $ p\leq1$.  This can be shown using the integral test.

The integral test basically says that if $ a_{n}= f(n)$ and $ \int_{k}^{\infty}f(x) dx$ exists and converges, then so does $ \sum_{n=k}^{\infty}a_{n}$, where $ k$ is some integer.  This can be roughly seen by seeing a definite integral as a "continuous" sum of all the terms over the involved interval, whereas a summation is a "discrete" sum of only terms corresponding to integers over the involved interval.  Thus, the summation will have fewer numbers to sum than the definite integral, and if the definite integral converges, the summation must converge -- one's just taken out a bunch of terms.  There may be other conditions for this test to work, but I don't really know them because I was never taught the integral test....  Am I wrong in any way?

to [b][u][i]prove[/i][/u][/b] the convergence/divergence of the zeta function $ \zeta{(s)}=\sum_{i=1}^\infty\;\frac{1}{i^{s}}$,
where $ s\in\mathbb{R^{+}}$, you can very simply and quickly apply the integral test
(this specific convergence/divergence of the zeta function using the integral test
is taught in high-school elementary calculus courses):

let $ \boxed{\textbf f(x)=\frac{1}{x^{s}}}$.
you will then need to investigate 3 distinct regions of interest:
(i) $ 0<s<1$
(ii) $ s=1$ (i.e. the harmonic series)
(iii) $ s>1$

we can easily see $ \textbf f(x)$ is a positive-valued, continuous, and decreasing function
for $ x \geq 1$. it therefore satisfies the conditions of the integral test right away.

now, $ \textbf I=\int_{1}^\infty\frac{1}{x^{s}}\;dx=\lim_{a\to\infty}\int_{1}^{a}\frac{1}{x^{s}}\;dx=\lim_{a\to\infty}\;-\frac{1}{(s-1)\; x^{s-1}}\; |_{1}^{a}$

$ =\lim_{a\to\infty}\frac{1}{s-1}\left(1-\frac{1}{a^{s-1}}\right)$

for (i), we have $ 0<s<1$, so we get $ \textbf I=\lim_{a\to\infty}\frac{1}{1-s}\left(a^{1-s}-1\right)\;\;\;\boxed{\rightarrow\infty}$
therefore, $ \zeta{(s)}=\sum_{i=1}^\infty\;\frac{1}{i^{s}}$ [b][u]diverges[/u][/b] for $ 0<s<1$

for (ii), i.e. $ s=1$, $ \textbf I=\int_{1}^\infty\frac{1}{x}\;dx=\lim_{a\to\infty}\int_{1}^{a}\frac{1}{x}\;dx$

$ =\lim_{a\to\infty}\ln x\; |_{1}^{a}=\lim_{a\to\infty}\ln a\;\;\;\boxed{\rightarrow\infty}$
therefore, $ \zeta{(s)}=\sum_{i=1}^\infty\;\frac{1}{i^{s}}$
[b][u]diverges[/u][/b] for $ s=1$ also (i.e. the harmonic series [b][u]diverges[/u][/b]).

for (iii), i.e. $ s>1$, we use the same integration technique as
we did to investigate (i).

we had, $ \textbf I=\int_{1}^\infty\frac{1}{x^{s}}\;dx=\lim_{a\to\infty}\int_{1}^{a}\frac{1}{x^{s}}\;dx=\lim_{a\to\infty}\;-\frac{1}{(s-1)\; x^{s-1}}\; |_{1}^{a}$

$ =\lim_{a\to\infty}\frac{1}{s-1}\left(1-\frac{1}{a^{s-1}}\right)$

now $ s>1$, so $ \textbf I=\lim_{a\to\infty}\frac{1}{s-1}\; (1)=\lim_{a\to\infty}\frac{1}{s-1}\;\;\;\boxed{\ll\infty}$
it therefore follows, $ \zeta{(s)}=\sum_{i=1}^\infty\;\frac{1}{i^{s}}$ [b][u]converges[/u][/b] for $ s>1$

$ \textbf{Q.E.D.}$

It's much more interesting when you treat of the case where $ s\in \mathbb{C}$.
You can prove that only when $ Re\,\,s>1$ does the series converge.

Zakary - the one key hypothesis for the integral test (the one hypothesis that you didn't mention) is that the function $ f(x)$ must be positive and [b]decreasing[/b]. (Or eventually decreasing - "eventually" is always good enough in comparison tests.) The Cauchy Condensation test has essentially the same hypothesis: the terms of the series must be decreasing.

Both the integral test and the condensation test are comparison test, meaning they can only be applied to positive-term series (or to series of absolute values, in search of absolute convergence.) Both the integral test and the condensation test are "two-sided" comparisons, capable of proving both convergence and divergence.

The best informal proof of the integral test: draw the graph of $ f(x),$ and draw a set of boxes of width $ 1$ and height $ f(n),$ whose areas represent the terms of the series. Draw the picture twice, once with the boxes fitting under the curve, once with then sticking out above. (Between the two pictures, the boxes have been shifted one unit sideways.)

Understand that picture and you understand not only the integral test but also how to use it to  provide sharp estimates for the "remainder," $ \sum_{k=n+1}^{\infty}f(k).$

Thank you everyone for replying, :P  I need sometime to read through and understand your replies.

[quote="IrinaSharova"]How to demonstrate that the series $ \sum_{i=1}^{\infty}\frac{1}{i^{s}}, s\in$ where $ s\in\Re$ converges when s>1.
Thank you for helping. :P[/quote]

For $ s>1,\ \sum_{i=1}^{\infty}\frac{1}{i^{s}}<1+\int_{1}^{n}\frac{1}{x^{s}}\ dx=1+\left[-\frac{x^{1-s}}{s-1}\right]$

$ =\frac{s}{s-1}-\frac{n^{1-s}}{s-1}<\frac{s}{s-1}$, which means that $ \sum_{i=1}^{\infty}\frac{1}{i^{s}}$ is an increasing sequence and is bounded upwards.


$ a,b,c,d  [0,1]$ , prove that there exist a $ x [0,1]$ satisfies $ \frac{1}{Ix\minus{}aI}\plus{}\frac{1}{Ix\minus{}bI}\plus{}\frac{1}{Ix\minus{}cI}\plus{}\frac{1}{Ix\minus{}dI}<40$

Pigeon hole method .. :D .. Let's suppose $ a_0 \equal{} 0 \leq a_1 \leq a_2 \leq ... \leq a_n \leq 1 \equal{} a_{n \plus{} 1}$ .
Let $ 0\leq m \leq n$ be such that $ a_{m \plus{} 1} \minus{} a_m \equal{} \max_{0\leq k\leq n} (a_{k \plus{} 1} \minus{} a_k)$ . Manifestly $ a_{m \plus{} 1} \minus{} a_m \geq \frac 1{n \plus{} 1}$
if  $ m \equal{} 0$ then  $ x \equal{} a_0 \equal{} 0$
if $ m \equal{} n$ then $ x \equal{} a_{n \plus{} 1} \equal{} 1$
if $ 0 < m < n$ then $ x \equal{} \frac {a_m \plus{} a_{m \plus{} 1}}2$
It's easy to see that $ \sum_{k \equal{} 1}^n \frac1{|a_k \minus{} x|} < 2n(n \plus{} 1)$ ...  :cool:


i hope none of you went to Berkeley today without being registered for the tryout, or else you will have learned an important if time-consuming lesson. Anyway, see below for an excerpt i sent to the ARML list this week about an extra tryout i was able to fit onto the calendar:

i have some good news about ARML tryouts - i have been able to add an extra tryout date for this year's team selection!  This brings the total this year to seven, up from four last year.  The extra tryout will be March 14 at 9am at the Stanford math building (same place two of the others have been).  There will also be a junior high contest in the same building, but we'll try to keep things moving as smoothly as possible.  If you're going to come to this one, you need to register by next Wednesday at http://mathleague.org/register.php (be sure to select the correct date from the drop down menu)..

 Since i haven't had time to advertise this as a contest, this tryout will only be offering sprint and target rounds.  No team events, no awards, just pay your 10 bucks and take the two tests so you can go to ARML.  i'm not expecting a big crowd - mainly i'm hoping to catch the students who otherwise wouldn't be able to make it to three tryouts.  i know there are a few of you who have only shown up at one or two tryouts, and this one and the one in April are your last chances to post three qualifying scores if you are interested in going to ARML this year.  i know it seems a silly hoop to have to jump through, but i have to have an objective way of determining who gets to be on what team.  None of you ever made it to national mathcounts without taking the easy tests at chapter first, so please don't expect me to make any exceptions at the high school level..

Except that doesn't even apply. Albert's argument is that the contests are not suited for ARML selection, because ARML is not a contest with all completely trivial problems. On the other hand, Chapter contests are suited for National Mathcounts because all the problems are on a low level and the aim (for the people who can place at state) is to have the people with the least mistakes and greatest accuracy.

[quote="farmertim"][quote="farmertim"]i know it seems a silly hoop to have to jump through, but i have to have an objective way of determining who gets to be on what team.  None of you ever made it to national mathcounts without taking the easy tests at chapter first, so please don't expect me to make any exceptions at the high school level..[/quote]

Yes i'm talking to you, Albert..[/quote]
that was completey unnecessary
[quote]
 (you're welcome)[/quote]
...thanks?


How many positive three-digit integers less than 400 satisfy each of the following characteristics simultaneously?

- Each of the three digits is prime.
- The sum of the three digits is prime.
- The number is prime.

Source: Go Figure! Math Challenge (http://www.mathcounts.org)

there are only 2*4*4 or 32 integers that satisfy the first condition.
To get a prime digit sum, we can have:
2+2+3 (3 ways)
2+2+7 (1 way, no 722 b/c over 400)
3+3+5 (2 ways)
3+3+7 (2 ways)
3+5+5 (1 way)

there are only 9 numbers left: 223. 232, 322, 227, 335, 353, 337, 373, and 355.
obvioiusly, evens other than 2 can't be prime, and numbers ending in 5 can't be prime, so that leaves 223, 227, 353, 337, and 373.
I don't know if all those numbers are prime, but at most 5 numbers satisfy all 3 of those conditions.

[hide="Da answer..."]223, 227, 353, 337, and 373 are all prime, so $\boxed{5}$[/hide]

[hide]okay
so you know that the digits have to be 2,3,5,or 7
so then you have to have either 2,2,odd
or odd odd odd
so for 2,2,odd there's
223,225,227
each can not be 522 or 252 so only 223,225, and 227.
225 is div by 5
223 and 227 can't be into 3 or 2 or 5
div each by 7 and 11 and 13
both are primes

with odd odd odd
none the same like aaa or 333...
335, 353, 337, 373, and 355[/hide]


Let $ a$, $ b$ and $ c$ be positive numbers satisfying $ ab \equal{} 11.25$; $ ac\equal{}20$; and $ bc \equal{} 36$. What is the value of $ c$?

We have $ \frac{ab}{ac}\equal{}\frac{b}{c}\equal{}\frac{9}{16} \implies b\equal{}\frac{9}{16}c$.

$ \implies \frac{9}{16}c^2\equal{}36 \implies c\equal{}\sqrt{36(\frac{16}{9})}\equal{}6(\frac43)\equal{}\boxed{8}$.


search
combinatorics proposed
combinatorics

How many ways are there to write either 0 or 1 in each cell of a 4 × 4 board so that the product of numbers in any two cells sharing an edge is always 0?

A search for "Korea 2007" (without quotes) leads to this: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=169651


[hide="Induction"]Base case of n=1 works. Now we prove that if it's true for n=k, it's true for n=k+1.

$ 1^2\plus{}2^2\plus{}3^3\plus{}\cdots\plus{}k^2\equal{}\frac{k(k\plus{}1)(2k\plus{}1)}{6}$

$ 1^2\plus{}2^2\plus{}3^3\plus{}\cdots\plus{}k^2\plus{}(k\plus{}1)^2\equal{}\frac{k(k\plus{}1)(2k\plus{}1)}{6}\plus{}k^2\plus{}2k\plus{}1$

\[ \equal{}\frac{2k^3\plus{}9k^2\plus{}13k\plus{}6}{6}\equal{}\frac{(k\plus{}1)(k\plus{}2)(2k\plus{}3)}{6}\]

Which is true.

$ \mathbb{Q.E.D.}$[/hide]

my favorite proof of it uses the hockey stick identity to create an essentially combinatorial proof. torajiru's post elaborates A Lot on that.


modular arithmetic
number theory unsolved
number theory

Find all solutions of $ a^2\plus{}b^2\equal{}n!$ in positive integers $ a,b,n$ with $ a \le b$ and $ n<14$.

It is well known that if $ p \equal{} 4k \plus{} 3$ is a prime divisor of $ a^2 \plus{} b^2$, then p divides both a and b.
Since 7 is of the form $ 4k \plus{} 3$, if $ n \ge 7$ then 7 would have to divide both a and b which would mean 49 divides n!, contradiction. Thus, $ n < 7$, in which case I think it's easy to brute-force the rest.
The solutions are (a,b,n)=(1,1,2), (12,24,6)

Here is a proof of that well known fact:
Suppose $ a^2 \plus{} b^2 \equiv 0 \pmod p$, where $ p \equal{} 4k \plus{} 3$, and p does not divide either a or b (it can't divide only one of them). Then, $ (\frac {a}{b})^2 \equiv \minus{} 1 \pmod p$, where $ m \equal{} \frac {a}{b} \equal{} ak$, where k is the inverse of b mod p (there exists an inverse since b and p are relatively prime). The inverse is the number k such that $ bk \equiv 1 \pmod p$. So $ m^2 \equiv \minus{} 1 \pmod p$. So we have $ 1 \equiv m^{p \minus{} 1} \equiv (m^2)^{\frac {p \minus{} 1}{2}} \equiv (m^2)^{2k \plus{} 1} \equiv ( \minus{} 1)^{2k \plus{} 1} \equiv \minus{} 1 \pmod p$, contradiction. Thus p must divide both a and b.


What is the product of the real roots to the equation

$x^2+18x+30=\sqrt{x^2+18x+45}$?

[hide="hint if you need it "]
Call $x^2+18$ $y$ and use the fact that $c$ is the product of the roots in this case[/hide]

This is old AIME 1983 problem. Clearly, posted before for about 5-6 times.

Locked.  :)


let $ a$ , $ b$ , $ c$ $ \in$ $ {\mathbb R}^{\plus{}}$ find minimum of :

$ \frac{(a\plus{}b\plus{}c)^3}{a^3\plus{}b^3\plus{}c^3\plus{}a^2(b\plus{}c)\plus{}b^2(c\plus{}a)\plus{}c^2(a\plus{}b)\plus{}3abc}$

[quote="y-a-s-s-i-n-e"]let $ a$ , $ b$ , $ c$ $ \in$ $ {\mathbb R}^{ \plus{} }$ find minimum of :

$ \frac {(a \plus{} b \plus{} c)^3}{a^3 \plus{} b^3 \plus{} c^3 \plus{} a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \plus{} 3abc}$[/quote]
If $ b\equal{}c\rightarrow0^\plus{}$ then
$ \frac {(a \plus{} b \plus{} c)^3}{a^3 \plus{} b^3 \plus{} c^3 \plus{} a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \plus{} 3abc}\rightarrow1.$
But $ \frac {(a \plus{} b \plus{} c)^3}{a^3 \plus{} b^3 \plus{} c^3 \plus{} a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \plus{} 3abc}\geq1\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(2x^2y\plus{}2x^2z\plus{}xyz)\geq0,$ which is true.
Id est, the infimum is $ 1$ and the minimum does not exist.

thank's arqady
if we put $ a\equal{}b\equal{}c$  we have

$ \frac {(a \plus{} b \plus{} c)^3}{a^3 \plus{} b^3 \plus{} c^3 \plus{} a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \plus{} 3abc}\equal{}\frac{27}{12}$

what is the diference betwin the minimum and the infimum

[quote="y-a-s-s-i-n-e"]thank's arqady
if we put $ a \equal{} b \equal{} c$  we have

$ \frac {(a \plus{} b \plus{} c)^3}{a^3 \plus{} b^3 \plus{} c^3 \plus{} a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \plus{} 3abc} \equal{} \frac {27}{12}$[/quote]
But $ \frac {27}{12}>1.$ :wink:

[quote="y-a-s-s-i-n-e"]
what is the diference betwin the minimum and the infimum[/quote]
The minimum holds but the infimum not necessarily holds.


At a graduation ceremony, names were read at the rate of five
names per minute.  If the names had been read at the rate of
seven names per minute, the ceremony would have taken
24 minutes less.  How many names were read?

let's make a equation.

let x be the number of names.

then the equation is...

$ \frac{x}{5}\equal{}\frac{x}{7}\plus{}24$.

multiply 35 at both side.

$ 7x\equal{}5x\plus{}840$

$ 2x\equal{}840, x\equal{}420$

answer : 420

[b]OR[/b]

We can solve for the amount of time it takes to read at $5$ names per minute, then solve for the number of names.

Let $t$ = The number of minutes it takes to read all the names at the rate of $5$ names per minute.

Using the fact that $\text{Work Rate}\times\text{Work Time}=\text{Work Done}$ and write and equation and solve for $t$.

The number of names read while reading at $5$ names per minute is $5t$.

The time it takes to read the names at $7$ names per minute is $t-24$ minutes.  That makes a total of $7(t-24)$ names read.  Since the number of names there are never changes, we can set the two expressions equal and solve for $t$:

$5t=7(t-24)$

$5t=7t-168$

$-2t=-168$

$t=84$

Therefore, the amount of time it takes to read all the names at $5$ names per minute is $84$ minutes.  Since the total number of names read is the rate at which the names are read times the times it takes, the total number of names is:

\[5t=5(84)=\boxed{420}\text{ names}\]

[quote=Helloman_1]FantasyLover --> :winner_first:[/quote]

Spamming -->  :mad:


real analysis
topology
real analysis unsolved

Okay, first of, this might be super easy, but it's something I do not know the answer to and would glady appreciate any input.

Can one partion any given bounded and closed interval in R into a union (possibly disjoint or whatever makes this possible) of generalized Cantor sets?

I believe, in general, Cantor sets have Lebesgue measure 0, hence any countable union of Cantor sets also have measure 0 (I think that is correct; my experience in measure theory has been limited).  Since any (nontrivial) interval has positive Lebesgue measure, this means that no interval can be decomposed into a countable union of Cantor sets.

There may be another argument not involving measure theory...?

You can construct cantor sets with Lebesgue measure [0,1), they all have the same topology and cardinality but not the same "size".


abstract algebra
superior algebra
superior algebra unsolved

Let $ E/K$ be a finite field extension. Show that if $ E$ is perfect so is $ K$.

Use the criterion for separable with $ \Omega^1_{L/K}$ and the sequence relating differential modules of a tower $ C/B/A$.

I have no idea what myan's answer means, but this is a very simple problem if you have the right definition of "perfect field":

A field $ F$ is called perfect if every finite extension of $ F$ is separable.


Column A: x
Column B: -y

2x-5y=-6
6y-x=6

Now given those two equations I should be able to figure out:

1. Is A bigger than B
2. Is B bigger than A
3. Are they the same
4. Not enough information

Now the answer is 3. They're equal. But why? Put together, the equations equal x+y=0. So they're equal. But in column B, it's a NEGATIVE y. So the equations prove that X and Y are equal, but in column B it's a NEGATIVE y, so how can they be equal when a negative is added to the y?

Given two equations:
$ 2x\minus{}5y\equal{}\minus{}6$ and
$ 6y\minus{}x\equal{}6$
We add twice the second one to the first to get:
$ (2x\minus{}5y\equal{}\minus{}6)$
$ \plus{}(\minus{}2x\plus{}12y\equal{}12)$
$ \equal{}(7y\equal{}6)$
Thus, $ y\equal{}\frac{6}{7}$
Now put this value in the second original equation to get:
$ (6)(\frac{6}{7})\minus{}x\equal{}6$
$ x\equal{}\frac{\minus{}6}{7}$
Since column B is $ \minus{}y$, $ x\equal{}\minus{}y\equal{}\frac{\minus{}6}{7}$

[quote="leroyjenkens"]
Now the answer is 3. They're equal. But why? Put together, the equations equal x+y=0. So they're equal. But in column B, it's a NEGATIVE y. So the equations prove that X and Y are equal, but in column B it's a NEGATIVE y, so how can they be equal when a negative is added to the y?[/quote]

How do you get $ x$ and $ y$ are equal when $ x \plus{} y \equal{} 0$? Try plugging in equal values of $ x$ and $ y$ and see if you get zero. You can get the fact that $ x\equal{}\minus{}y$ simply by subtracting $ y$ from both sides of the equation $ x \plus{} y \equal{} 0$.

Actually I just realized this math sheet is wrong. It says x+y=0 and says they're the same. What I meant to put was x-y=0. Ok thanks guys, sorry for the mistake.


Suppose $a,b,c\ge0$ and $ab+ac+bc=1$. Find the minimum value of :

\[ f(a,b,c) = \frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c} \]

This is pretty difficult... see for instance http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=31377  :)

From $ab+bc+ca=1$ we get
\[ f(a,b,c)= \frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}  \]
                  \[ =\frac{1}{a+b}+\frac{a+b+2c}{(b+c)(c+a)}  \]
                  \[ =\frac{1}{a+b}+\frac{a+b}{1+c^2}+\frac{2c}{1+c^2}  \]
let $t=a+b$ thus \[ \frac{1}{a+b}+\frac{a+b}{1+c^2}=\frac{1}{t}+\frac{t}{1+c^2}  \]
from AM-GM we get the minimum for \[ g(t)=\frac{1}{t}+\frac{t}{1+c^2}  \] is \[ \frac{2\sqrt{1+c^2}}{1+c^2}  \] thus we solve for $t$ get $t=\sqrt{1+c^2}$, and $g(t)$ in  $[\sqrt{1+c^2},+\infty)$ is an increasing function.

Can anyone solve the problem from here?


[hide="Solution"]$ x^2\minus{}20\plus{}25\minus{}4x\equal{}37$

$ x\equal{}\minus{}4,8$[/hide]

[quote="sri340"]Find all value(s) of x such that: x * 5 + 5 * x = 37, if a * b = a^2 – 4b[/quote]

$ x^2\minus{}20\plus{}25\minus{}4x\equal{}37$=$ x^2\minus{}4x\minus{}32\equal{}0.$ $ x\equal{}8, \minus{}4$

[quote="1337rox"]Case 1:  x=a,5=b. Case 2: x=b,5=a

So we have x*5+5*x=37. (x^2-20)+(25-4x)=37.-> x^2-4x+5=37 -> x^2-4x-32=0. factor: (x+8)(x-4)

$ {x \equal{} ( \minus{} 8,4)}$[/quote]

Ah, you factored wrong at the end.
$ x^2\minus{}4x\minus{}32\equal{}0$ factors into $ (x\minus{}8)(x\plus{}4)$, giving $ x\equal{}8,\minus{}4$


ARML
HMMT
floor function
function
modular arithmetic
calculus
integration

Here are some problems (some from ARML, some from elsewhere).

1. (ARML) Determine all positive primes $p$ such that $p^{1994}+p^{1995}$ is a perfect square.

2. (Mandelbrot) Given that $55555^2 = 3086358025$, calculate $55556^2$.

3.  (ARML) If, from left to right, the last seven digits of $n!$ are $8000000$, compute the value of $n$.

4. (ARML) A positive integer $N$ is self-descriptive if each digit of $N$ appears as often as the value of the digit. For example, 212 is a three-digit self-descriptive number.  Compute the sum of the digits of all the six-digit self-descriptive numbers.

5. (HMMT) What is the smallest integer $x$ for which $x^2$ ends in the same three digits as $x$ does?

6. (HMMT) Let $\lfloor x\rfloor$ denote the greatest integer less than or equal to $x$.  How many positive integers less than 2005 can be expressed in the form $\lfloor x\lfloor x\rfloor\rfloor$ for some positive real $x$?

7. (Mandelbrot) Find the last three digits of $9^{105}$.

8. (Mandelbrot) How many integers from 1 to 1992 inclusive have a base three representation that does not contain the digit 2?

9. (Engel) The four-digit number $aabb$ is a square.  Find it.

10. (from Engel's book) Is $4^{545}+545^4$ a prime?

[hide="1"]the answer is $3$[/hide]
[hide="7"]
$9^{105}=(10-1)^{105}$, using binomial theorem, we look at the last three which are $-\binom{105}{2}10^2 \cdot 1+\binom{105}{1}10 \cdot 1-\binom{105}{0}1 \cdot 1$, looking at it $\mod{1000}$, we have $000+050-001=49(\mod{1000})$ $\boxed{49}$[/hide]

leave others for other people, but if no one answers in 2 or 3 days, i'm solving almost all of them.

[quote="joml88"][quote="Iversonfan2005"]
$9^{105}=(10-1)^{105}$, using binomial theorem, we look at the last three which are $\binom{105}{2}10^2 \cdot 1-\binom{105}{1}10 \cdot 1+\binom{105}{0}1 \cdot 1$, looking at it $\mod{100}$, we have $000+050+001=\boxed{51}$[/hide][/quote]

[hide]You're a little bit off and made a few typos.  For example, you took it mod 1000 (not 100) and look back over the plus/minus signs.

You have all the major concepts down though :)  ... just need to be careful.[/hide][/quote]

hmmm...are you sure i got the sign changes wrong? :?

P.S. How is number 2 a Mandelbrot, it looks more like SAT to me ;)

[quote="joml88"]Yeah, I'm certain you got the signs wrong....and then you didn't simplify it right using mods now that I take another look at it :P

I checked using mathematica just in case ;)[/quote]

It's gotta be correct now :!:

I realized why I got the sign changes wrong, I forgot that I had switched it to $\binom{105}{2}$ instead of $\binom{105}{103}$ so I was getting the odd-even sign change thing wrong :blush:

number 3 is hard...I know through greatest integer function, $25\leq n\leq 29$ since it has $6$ zeroes on the end, but that's about all :(
[hide="2"]
$55556^2=(55555+1)^2=55555^2+111110+1=\boxed{3086469136}$[/hide]

[hide="2"]We are given $n^2$, where $n=55555$. We are asked to compute $(n+1)^2$. To do so, all we must do is add $2n+1$ to $n^2$. The answer is thus $\boxed{3086469136}$, unless i did my arithmetic wrong.[/hide]
[hide="3"]We must have $n!$ end in 6 zeros, implying $25\le n \le 29$. Dividing out $10^6$ from $25!$ and working $\mod{10}$, we find $\displaystyle25!/10^6 \equiv 4 \pmod{10}$. (This means that the last 7 digits of $25!$ are $4000000$. Working from this, it is easy to see that $\boxed{n=27}$.[/hide]
[hide="4"]The digits that can be in any given selfdescriptive number are $\{1,2,3\}$, $\{1,5\}$, $\{2,4\}$, and $\{6\}$. There are $\frac{6!}{1!\cdot2!\cdot3!}=60$ ways to rearrange $\{1,2,3\}$. The sum of the digits of all numbers of this form will be $(1^2+2^2+3^2)\cdot(60)=840$. Using similar logic, the sum of all six digit self descriptive numbers is $840+156+300+36=\boxed{1332}$.[/hide]
[hide="8"]Not entirely sure about this one, but.. $\displaystyle1992_{10}=2201210_3$. The greatest number that can be written in base 3 with just 1s or 0s is $\displaystyle1111111_3$. Note that if a number in base 3 does not contain a 2, then it must contain only zeros and/or ones, which reminds us of base 2. The answer is thus $\displaystyle1111111_2=\boxed{127}$.[/hide]

edit; oh shobhit already did 2...

[quote="arkmastermind"][hide="9"]$\boxed{7744}$...pretty much brute forced except i knew that $aabb$ must be the square of a multiple of 11..there arent too many to try out.[/hide][/quote]

What is bruteforce? trial and error? Because that's what I did, knowing that $\sqrt{11(100a+b)}$ would give me an answer, it needed to be in the form $11^2 \cdot k^2$, and I found that $8^2$ worked...not a great way though  :(

yeah pretty much trial and error. there werent that many cases..so i figured it was okay.

Number 9 can be done without _that_ much bruteforce; it can be reduced to only 4 cases, of which the second one works.

[hide]We know that $1100a+11b=c^2$. We take this equation mod 100 to find that $11b \in \{ 0,33,44,77,88\} \rightarrow b \in \{ 0,3,4,7,8\}$.

Now that we've reduced the possible $b$, we go back to the original equation. We observe $11(100a+b)=c^2$, thus 11|$100a+b$. This implies $100a+b\equiv 0 \pmod{11}\rightarrow a+b\equiv 0 \pmod 11$. a and b are single digits, thus the maximum of a+b is 18. Therefore, a+b=11, since a+b cannot equal 0.

Now we plug our b values into that equation to get $(a,b)\in\{ (8,3),(7,4),(4,7), (3,8)\}$, and then simply checking the results of each of these values, 7744 works.[/hide]

Billy

Let's see...5, 6, and 10 have gone unsolved so far.  These are some of the more fun ones (the harder ones of the set though :P )!  Does anyone want some hints?

Number 6: Let $x=a+\frac ba + c$, where $b<a$, $c<\frac 1a$, and $\lfloor x \rfloor = a$. The product is now $\lfloor a^2+b+ac$; we know that $a$ and $b$ are integral, thus we have $a^2+b+\lfloor ac\rfloor$. But $c<\frac 1a$, so $ac<1$, thus our equation is $a^2+b<2005$. The only restriction on the solutions of this equation is that $b<a$, thus we can now do lovely casework:

$a=1$: a=1, b=0  for 1(We don't have to worry about 0 since it isn't positive :D)
$a=2$: a=2, b<2 for 2
$a=3$: a=3, b<3 for 3
$\vdots$
(we continue until we find the largest integer for which $a(a+1)<2007$, namely 44)
$a=44$, $44^2=1936$, b<44 works, for 44
$a=45$, $45^2=2025$, no solutions.

Therefore there are 1+2+3+...+44=990 that can be represented in this manner.

oh yeah i did 5 a few days ago but didnt feel like writing out my solution...

basically you have $x\cdot(x-1)\equiv0\pmod{1000}$. from there its not too hard to find the smallest $x$ as $\boxed{376}$. (rem that either $x$ or $x-1$ will be divisible by 8 or 125 or both...)

i was working on ten too a few days ago, but i sorta used my calculator to help me see that the number was divisble by 73. its not too hard to show by hand that this is indeed the case,  but i suppose theres a more clever way to go about doing it..

Yes, there is definitely a better way to do #10 (why would you think of testing mod 73 anyway? :P ).

[hide="Hint for number 10"]
First try this problem:

Factor $n^4+4$.

[hide="More of a hint"]How about $x^4+4y^4$?[/hide][/hide]

lol, this problem was just posted in the intermediate forum so i'm going to close my mouth and not say anything about it  :lol:

Heh, well, it is a popular problem I imagine.  You can go ahead and post a solution Shobhit if you want.  Just try to see if you can do it by yourself this time :D

[hide="Another way of doing question 2"]
First,

$55556^2-55555^2 	\Longrightarrow (55556+55555)(55556-55555)	\Longrightarrow 111111$

$55556^2 \Longrightarrow  3086358025$ (Already given in question)

Then,

$3086358025+111111     \Longrightarrow          \boxed{\boxed{\boxed{\boxed{\boxed{3086469136}}}}}$

[/hide]

[quote=bassamali01][quote=Iamteehee]Here's a another way:
$55556^2=55556*55556=\boxed{3086469136}$[/quote]

This is very hard and long way and we can make mistakes in this because of these huge numbers.[/quote]

Implying you can't do that with any of the other methods?
Also, I was trolling,


On the contest page, many of the problems with a diagram say "External_Link". Where are the diagrams?

Their image probably disappeared or they were lost in the old album


limit
real analysis
ratio
real analysis unsolved

\[ \lim_{n\to \plus{}\infty} \frac{1}{\sqrt[n]{n!}}\]

[quote="zhoraster"]This is $ 0$. Try to prove that the (upper) limit is less that $ 1/2$, $ 1/3$, $ 1/4$ etc.[/quote]

can you say more details ? thanks

$ n!\ge 1\cdot 2\cdot 2\cdots 2 \equal{} 2^{n \minus{} 1}$ so $ \sqrt [n]{n!}\ge 2^{\frac {n \minus{} 1}n}$ and $ \lim\sqrt [n]{n!}\ge 2.$

$ n!\ge 1\cdot 2\cdot 3\cdot 3\cdots 3 \equal{} \frac29\cdot 3^n$, so $ \sqrt [n]{n!}\ge \sqrt [n]{\frac 29}\cdot 3$ and $ \lim\sqrt [n]{n!}\ge 3.$

$ n!\ge 1\cdot 2\cdot 3\cdot 4\cdot 4\cdots 4 \equal{} \frac6{64}\cdot 4^n$, so $ \sqrt [n]{n!}\ge \sqrt [n]{\frac 6{64}}\cdot 4$ and $ \lim\sqrt [n]{n!}\ge 4.$

And so on. That has to be the argument zhoraster was hinting at. (And those should really be liminf in the lines above rather than lim, but the idea is the same.)

Here's another proof: what is the radius of convergence of the power series $ \sum\frac {x^n}{n!}\ ?$ By the root test, this radius of convergence is $ \lim\sqrt [n]{n!}.$ By the ratio test, the radius of convergence is $ \infty.$ Since the root and ratio tests must give the same result whenever the ratio test is valid, we have that $ \lim\sqrt [n]{n!} \equal{} \infty.$

Here's the final version: what is the radius of convergence of the power series $ \sum_n\frac {n!x^n}{n^n}\ ?$ By the root test, this is $ \lim\frac {n}{\sqrt [n]{n!}}.$ By the ratio test, it is

$ \lim \frac {(n \plus{} 1)^{n \plus{} 1}}{n^n}\cdot \frac {n!}{(n \plus{} 1)!} \equal{} \lim \left(\frac {n \plus{} 1}n\right)^n \equal{} \lim\left(1 \plus{} \frac1n\right)^n \equal{} e.$

Since the radius of convergence is the same whichever way we measure it, we must have that $ \lim\frac {n}{\sqrt [n]{n!}} \equal{} e.$ This gives us the asymptotic formula $ \sqrt [n]{n!}\sim \frac ne,$ which can be considered a crude preliminary version of Stirling's formula.

$ \forall M \in Z^{ \plus{} },n > 2M,\frac {n!}{M^{n}} \equal{} \frac {(2M)!}{M^{2M}}\frac {n(n \minus{} 1)..(2M \plus{} 1)}{M^{n \minus{} 2M}} > \frac {(2M)!}{M^{2M}}2^{n \minus{} 2M}$
so it's easy to show $ \exists N,$when$ n > N,\frac {n!}{M^{n}} > 1$,it's equivalent to$ \frac {1}{\sqrt [n]{n!}} < \frac {1}{M}$
so $ \lim_{n\leftarrow \plus{} \infty}\frac {1}{\sqrt [n]{n!}} \equal{} 0$


A long roll of tissue paper is 200km in lenght. When it is rolled it is formed into a cylinder of inner radius 3cm and outer radius of 6cm. Find the thickness of the roll (approx).

[hide] If I understood the question well, the solution is:
We call thickness t
$V_{cilinder} = (36cm^{2} - 9cm^{2})h$ (With h unknown) Rolled off, $V_{rolled of}=20000000cm.t.h = V_{cilinder}$ $\longrightarrow 27cm^{2}=20000000 cm.t$ $\longrightarrow t=0,00000135cm^{2}=1,35 \mu m$ [/hide]

There was a problem almost exactly like this like 1 or 2 weeks back. Strangely, it was tissue paper also.

Hmm, I thought $\displaystyle V_{roll} = (36\pi - 9\pi)h = 27\pi h$.  That would make the thickness approximately 4.24 microns instead.  Or am I just missing something really obvious?

Oh damn, I feel so stupid right now...  :(
Of course I forgot multiplying with $\pi$ So the answer should indeed be $1,35µm.\pi = 4,241150082346220871924568563 \approx 4,24$ indeed...
Damn, perhaps I should learn my area formula's again  :)

[quote="riddler"]Yup thats right :D  :D[/quote]

the question mark icon means that you need help on this problem...just giving you a heads up.

for everyone else:

if this has a question icon, dont post the full solutions.  Give so advice or hints etc.

Oh my apologies, I didn't know that (I'm quite new...)
Fortunately my answer was wrong, so actually I didn't do anything wrong because I didn't give the solution...  :D


Contest Problem Book VIII, consisting of 350 problems and solutions from the first 8 years of the AMC 10, is now available in the AoPS Bookstore [url=http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=3013]here[/url].

Not worth fifty dollars -- through kids on AOPS you can find sites that have all the 10's

I agree with mathemonster, and it doesn't even have the AMC 12 or the AIME.

The point is that it's extremely useful for having as a portable version, the solutions are all right there, there are no missing diagrams, etc.

In addition:

1. The book includes a preface which has a history of the contests, including the process and philosophy of selecting the problems that go on the contest.  This is useful in understanding the contests and how to prepare for them using the book.

2.  All of the problems, solutions and figures were compiled from the original markup source by the Contest Chairman, ensuring that the problem statements are complete and correct.  In addition, the manuscripts were proofread by experts.  There should be no typos.

3.  Each year has a  table with item difficulties.

4.  The book has an index of problems by type and topic.

In short, you get the value-added features of a book.

Steve Dunbar,
MAA Director,  American Mathematics Competitions

haha yeah I ended up buying it for my birthday  :blush:

having some extra practice problems can never hurt. i got volume 7, and although it was somewhat expensive, it is useful to have a portable copy of all the problems with thorough solutions.


MATHCOUNTS
geometry
3D geometry
prism
LaTeX
calculus
integration

At long last, here are the results:
Winners (score, first Sprint question missed)
:winner_first: [size=200][color=blue]biffandoc (43, #11)[/color][/size]
:winner_second: [size=200][color=red]mathgeniuse^ln(x)  (43, #8)[/color][/size]
:winner_third: [size=167][color=green]robinhe  (43, #5)[/color][/size]
[hide="Statistics"]
Number of Participants: 15

Totals:
Max: 43
Min: 20
Mean: 35.2
Standard Deviation: 7.608266181

Sprint:
Max: 28 (noneoftheabove)
Min: 19
Mean: 24.66666667
Standard Deviation: 3.309438163

Target:
Max: 16 (biffandoc, mathgeniuse^ln(x), robinhe)
Min: 0
Mean: 10.53333333
Standard Deviation: 5.370111287[/hide]
[hide="Answers"]
Sprint Answers (Number of Correct Solvers)
1. 2006 (14)
2. 42 (12)
3. 40 (14)
4. 6 (15)
5. 4095 (13)
6. 12 (13)
7. 73 (15)
8. 24 (10)
9. 19 (13)
10. 0 (15)
11. $25\sqrt{3}$ (6)
12. 0 (10)
13. 6 (14)
14. 41 (13)
15. 141414 (13)
16. -3 (12)
17. -2006 (6)
18. 6 (14)
19. 40 (15)
20. 216 (13)
21. 21 (5)
22. 180 (13)
23. 8 (14)
24. 6 (14)
25. 84 (11)
26. 2 (13)
27. 10 (13)
28. 19 (12)
29. 3 (15)
30. 3 (10)

Target Answers (Number of Correct Solvers)
1. 775 (8)
2. 45 (13)
3. 10 (9)
4. 95 (11)
5. 26.5625 (12)
6. 168 (12)
7. $2\pi$ (5)
8. 1 (9)
[/hide]
[hide="Notes"]I apologize for the horrendous answer to #7 on Target. I intended the problem to have an integer answer but made errors in the calculations.

Thanks to my teacher Caroline Jones for putting the exam in Mathcounts format. Originally, the problems were closely spaced and had no answer blanks.

After Release 1, #6, #7, #19, #27, and #29 on Sprint all had incorrect answers. Fortunately, these were all caught by the Mathcounts students at my local middle school. Kudos to them.

solafidefarms mentioned the ambiguity of the definition of exterior angles in the original version of #22 on Sprint. Consequently, the problem was changed to involve clearer terms. He also noticed that just about every single one of my Target answers were in the wrong order. He also separated the calculation heavy problems on Target so that sufficient time would be given for each set. Finally, he proofread the questions and reworded several of them. Special thanks to him.

After Release 2, #11 and #30 on Sprint had incorrect answers, but both were noticed by some participant or another during the grading process.[/hide]

At long last, you may discuss the problems, which may be found here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76817

Whoops. Misread your post... thought it said to discuss at that link. Sorry.

How do you solve target #3?
If a prism has 20 vertices, how many rectangular faces does it have?

Are you, by any chance, thinking of a polyhedron?
Remember that prisms just have two polygons at the poles, and rectangular faces connecting them.

[quote="ch1n353ch3s54a1l"]How do you solve target #3?
If a prism has 20 vertices, how many rectangular faces does it have?[/quote]

This about this: in a prism, two congruent polygons are connected by lines. Those lines form rectangles.

Oh, and my score was a 40 (28/6).

YAY :wow:

I am so mad that I missed #11 and #12 though :wallbash_red:

What?! You should have had a tiebraker test.  :mad:  :mad:  :mad:

Oh well, I suppose I shouldn't be too mad.  I thought that $2^{12}=4098$.

How do you do #21? I had [hide]243[/hide]

Btw, what's standard deviation?

Basically standard deviation is the square root of the average of the sum of the squares of the difference between each element and the mean.
Theres a formula but Im too lazy...

To do #21,
You have 5 scoops of ice cream that can be any flavor.

There are 5 balls (scoops) and 2 walls (dividers)
Thus there are 7 total, and since the walls can be anywhere, its $\binom 72 =21$

The other 10 are given, so o_O

I didn't get #29 at all

Let S be the answer to this problem. What is the numerical value of 2S-3

EDIT: arrgh, just got it... i was like "which problem???" I thought it was talking about the problem above it, #28, after about 2 minutes of thinking about it so my answer was 38-3 = 35 UGH.

Could someone please explain biffanddoc's balls and walls with the ice cream or whatever? I Got [hide]243[/hide] but I now see how I missed it.

[quote="Fanatic"]
Let S be the answer to this problem. What is the numerical value of 2S-3
[/quote]

If S is the answer, then 2S-3=S, $S=3$

[quote="robinhe"]Btw, what's standard deviation?[/quote]

Have fun reading this:

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

and biffanddoc is right: it measures generally how far away the scores were from the mean. (For example, a set of {1, 2, 3} has a lower standard deviation than the set {-23, 0, 29}).

And I can't believe I missed target number 4! It was such a simple problem, but I was stupid and did it in my head and assumed 13*9 to be 91.  :mad:

And how do you do target number 7?

I'll answer that if you can tell me why 7 choose 2 works for #21 on the sprint.

j/k, here's how you do #7 (I got it wrong because I thought it was a triangle, but that was stupid mistake):

If you have a 4x4 square, a 45-45-90 right triangle could be drawn in with hypotenuse AB and legs 2 root 2 and 2 root 2.

Vertex C on that triangle is the right angle, and it's two away from the center of line AB. So if you draw like a semi-circle with a radius of two from the center of line AB and create triangles, they'll all be right angles (you're familiar with that property, right?). Everything outside that circle would be acute and everything inside would be obtuse. The area of that smie-circle is $2^2$ * PI/2 = 4PI/2 =

TWO PI

(Sorry for lack of latex!)

[quote="Fanatic"]I'll answer that if you can tell me why 7 choose 2 works for #21 on the sprint.

[/quote]
Say that Os are the scoops of ice cream and |s are the dividers between the flavors.

A possible arrangement is
||OOOOO (0,0,5)
O|O|OOO (1,1,3)
OOOOO|| (5,0,0)

The |s can go in any of the 7 spots
Thus $\binom 72=21$

You use the formula $mn-m-n$ to find the greatest number that cannot be obtained by adding an unlimited amount of them
$13\cdot9-13-9=107-13-9=85$

To find the smallest number which cannot be formed by adding integral numbers of $a$ and $b$, where $a$ and $b$ are integral and relatively prime( i.e. they do not have a common divisor), use the Chicken McNugget formula. The Chicken McNugget formula states that the largest number which cannot be made by adding $a$s and $b$s is $ab-a-b$.


Express as a common fraction: $ 3\plus{}\dfrac3{3\plus{}\frac13}$


[hide="Answer"]So we have the system $36+k=a^2$, $300+k=(a+b)^2 \Rightarrow 2ab+b^2=264$, and $596+k=(a+2b)^2 \Rightarrow 2ab+b^2=280$. This gives $b^2=16 \Rightarrow b=4$. Plugging in gives $a=31$ and so since $36+k=a^2=961 \Rightarrow k=\boxed{925}$[/hide]

How did you get $ 2ab \plus{} b^2 \equal{} 280$?
Shouldn't it be $ 2ab \plus{} 2b^2\equal{}280$?

Yeah, it was just a typo. The rest of the solution still holds.

[hide=Solution][quote=1989 AIME/7]If the integer $k$ is added to each of the numbers $36$, $300$, and $596$, one obtains the squares of three consecutive terms of an arithmetic series. Find $k$.[/quote]

[i]Solution:[/i] Let the difference of the arithmetic progression be $d$, and let the first term be $r$. Then $36+k=r^2$, $300+k=(r+d)^2$, and $596+k=(r+2d)^2$. Thus expanding the right side of each equation gives \begin{align}
36+k&=r^2, \\
300+k&=r^2+2rd+d^2, \\
\text{and }596+k&=r^2+4rd+4d^2.
\end{align}
Subtracting $(1)$ from $(2)$ gives $300-36=(r^2+2rd+d^2)-r^2$, so $264=d^2+2rd$.

Similarly, subtracting $(2)$ from $(3)$ gives $596-300=(r^2+4rd+4d^2)-(r^2+2rd+d^2)$, so $296=3d^2+2rd$.

Thus subtracting $264=d^2+2rd$ from $296=3d^2+2rd$ gives $296-264=(3d^2+2rd)-(d^2+2rd)$, so $32=2d^2$ and $d$ is either equal to $4$ or $-4$.

Since $596+k>300+k>36+k$, $(r+2d)^2>(r+d)^2>r^2$. Thus $d$ is equal to $4$, so $264=4^2+2r \cdot 4$. Simplifying gives $264=16+8r$, so $r=31$ and $k$ is equal to $31^2-36=\boxed{925}$.

Coincidentally, out of the two AIME problems I wrote out solutions to today, both had answer $925$.[/hide]


hi guys,

i've just arrived in athens (as a matter of fact I find myself browsing through the net in the airport right now). little chilli outhere, but I suppose it's just because it's morning.

more info soon :)

hope other people here will start posting. I also saw a sign in the airport with a flag of IMO 2004, however the office seemed completely deserted. hope it's not the case with the olympiad :) ( I would asume this since the Greeks have been partying a lot since they won the EuroCup :P )

Hey !
The French team is here too and ready for tomorrow  :D

Antony Igor Carlo Raf Vincent Pierre

I hope so! Because, otherwise do not forget that you will come back to France by swimming... :D

Pierre.

[quote="darij grinberg"]and yes, i'm in athens too...
and now i'm sitting in front of an internet terminal
together with peter, valentin, gabriel, misha and matthias
and we're attacking mathlinks :D

  dg[/quote] I think gabriel is more known as harazi on the site :P :)


[color=darkblue]Given are the function $f(x)=|x-1|e^{-\frac{1}{|x|}}$, where $x\not\in \{0,1\}$, a sequence $(a_{n})_{n\in N^{*}}$ and $L\in R$. Prove that

$a_{n}\rightarrow L\Longleftrightarrow (\forall )\epsilon \not\in \{0,1\}\ (\exists )\ n_{0}\in N$ so that $(\forall )\ n\in N\ ,\ n>n_{0}\ ,\ |a_{n}-L|<f(\epsilon )\ .$[/color]

It seems we should only to prove that there $f(x)>0$ and there exist $x_{0}$ such that $\limits_{x \to x_{0}}f(x)=0$ but it is not hard.(In fact $x_{0}=0,1$)


symmetry
analytic geometry
trigonometry
vector
function
abstract algebra

Suppose a Charge $ q$ is moving in a circular motion (the velocity is not constant in this case).

The Normal Acceleration is due a Magnetic Field.
The Tangential Acceleration is due a Eletric Field.

How can we descrive the Mag. Ele. fields?

By symmetry the magnetic field have $ x,y$ cordinates equal to zero and has the same value in each point in the space (the space where the particle is changing, the magnetic field in out of that is not important, so we can make this assumption) and is changing in time.

By symmetry the eletric field have $ z$ cordinate equal to zero. And the modulus is equal in all poins of the path.

The first thing to notice is: This is not a indefinied Accelerator. The magnetic field is dependent of the eletric field. Because the eletric field is due faraday law.

So we can think the movement of the particle will be periodic.

Using Farady and Ampere law

$ \text{E}2\pi r\equal{}\minus{}\frac{\partial\left(\text{B}\pi r^{2}\right)}{\partial t}\Rightarrow 2\text{E}(t)\plus{}r\text{B}'(t)\equal{}0$

$ \text{B}2\pi r\equal{}\frac{\partial\left(\epsilon\mu\text{E}\pi r^{2}\right)}{\partial t}\Rightarrow 2\text{B}(t)\equal{}r\epsilon\mu\text{E}'(t)$

The problem lies ahead. Solving this ODE

$ \text{E}(t)\equal{}k_{2}\cos\left(\frac{2 c t}{r}\right)\plus{}c k_{1}\sin\left(\frac{2 c t}{r}\right)$

$ \text{B}(t)\equal{}k_{1}\cos\left(\frac{2 c t}{r}\right)\minus{}\frac{k_{2}\sin\left(\frac{2 c t}{r}\right)}{c}$

How can I determine the constants above?

And How I determine the angular speed and acceleration of the particle?

Logically, assuming my results is correct. otherwise, how do I correct this?

please, give equations consistents and not only porlly advises or someting like that.

Thanks

i think it would be nice if u post the problem with the complete statement variables... so that we can work upon it and help u find ur mistakes(if any)  :)

I am not sure I understand the question.
We are given the equation of motion of charge $ q$, moving in a circle of radius $ r$ with constant angular acceleration $ \alpha$,
$ \vec{r}\equal{}r\cos\frac{\alpha t^{2}}{2}\widehat{i}\plus{}r\sin\frac{\alpha t^{2}}{2}\widehat{j}$
(As expected) tangential acceleration is due to $ \vec{E}$ and radial is due to $ \vec{B}$.
We have to find $ \vec{E}(t)$ and $ \vec{B}(t)$
Right??

[quote="shreyas_patankar"]I am not sure I understand the question.
We are given the equation of motion of charge $ q$, moving in a circle of radius $ r$ with constant angular acceleration $ \alpha$,
[/quote]

I never said that the accleration is constant!!

I said that the speed is not constant. or in another words. the vector equation of the particle would be

$ \mathbf{s}\equal{}\left(\begin{array}{lll}r\cos (\theta (t)) & r\sin (\theta (t)) & 0\end{array}\right)$ , $ \mathbf{v}\equal{}\frac{\partial\mathbf{s}}{\partial t}$ , $ \mathbf{a}\equal{}\frac{\partial\mathbf{v}}{\partial t}$

I used Partial derivatical cause $ \mathbf{s}$ have a constant $ r$ and to not make confusion thinking it is variable.

Like I said, The Tangential accleration is from eletric kind and the normal is magnetic.

So we can simply (working just with the modulus, since my first assumptions)

$ \text{E}q \equal{} m\frac{\mathbf{a}\cdot\mathbf{v}}{\|\mathbf{v}\|}$

$ q\|v\| B \equal{} m\frac{\|\mathbf{v}\times\mathbf{a}\|}{\|\mathbf{v}\|}$

The Left side is because $ B\cdot v \equal{} 0$

With this i have 2 equations and 3 functions. $ B,E,\theta$

I need one more equation.

I can use faraday, but not ampere law, since the angle is 90 degrees. i corrected this. but...

I just want a System of consistent equations to solve this problem for the 3 functions above, the modules of the ones


[quote="wesstar"]Given Q=4.58, S=14.5

[img]http://img255.imageshack.us/img255/1734/mathsamplekn6.png[/img]

I do use scientific calculator to calculate, and I manage to get 'n' approximately equal to 3.

However, what is the step and process to calculate the value 'n'?

Any experts?[/quote]

What is $ S$ ,$ Q$ and $ n$

@enndb0x:
[quote="wesstar"]Given Q=4.58, S=14.5 [/quote]
:wink:

Have you approximated S and Q?  $ \sqrt{21} \equal{} 4.58...$ and $ \sqrt{212}\equal{} 14.5..$

[hide]now basically bracket the expression and take -n outside. you'll have,s={1/[6Q+0.5]}^-n.simply substitute and get n[/hide]

Um, you can't just pull out the exponent blindly. Exponents are not distributive.

Is $ x^2\plus{}y^2$ the same as $ (x\plus{}y)^2$?

[quote="grn_trtle"]Um, you can't just pull out the exponent blindly. Exponents are not distributive.

Is $ x^2 \plus{} y^2$ the same as $ (x \plus{} y)^2$?[/quote]

I agree with grn_trtle , there's no radical form to solve this equations.Maybe in some special cases we can do something

 :@:grn_trtle

 $ (x\plus{}y)^{2}\equal{}x^2 \plus{}2xy \plus{}y^{2}$

Haha, that was a rhetorical question to illustrate that you can't distribute exponents, but thanks :lol:


If you know the answer, you shouldn't post it immediately.  Wait for others to respond (or not respond), and then post your solution.

[quote="tcjy"]what should the starting number have been?
start:
multiply by 4:
Add 8:
Divide by 3:
Result: 28[/quote]

$ \frac{4x\plus{}8}{3} \equal{} 28 \implies 4x \plus{} 8 \equal{} 84 \implies 4x \equal{} 76\implies \boxed{ x \equal{} 19}$


Should you shut down your computer when you aren't using it?

Energy/power/environmental concerns aside...

I always do.  Some people put it to sleep, which I guess saves energy, others just close it (laptop) or let the screensaver come on.  I think people should just turn it off.

I leave both my computers on 24/7 and barely notice a difference until the uptime reaches around a month. At that point they start slowing down. Servers should reboot approximately weekly, but normal computers can stay on much longer than that without trouble.

I rarely shut my laptop down, unless I have to take it somewhere and it won't last long enough when suspended. I also have a server (actually a computer from ~1998 with Linux on it) that I always keep on. The server has been running for 26 days without a restart, and that was due to a power outage in the area. It's a useful setup: I can always access files on my laptop from school by SSH-ing to the server and then to the laptop.

Generally, recent computers don't need to restart except for some installations and to clear up bad crashes. Power consumption really is the main issue here. For my laptop, battery life is a few hours actively running or a few weeks in sleep mode, so I just put it to sleep. The difference between sleep and shutdown power consumption is tiny.

I agree it takes some time to boot but I can deal with it.  I always keep my laptop plugged in, except when I'm going to another city, so I don't need to worry about battery life.  I still turn it off when I'm done with-maybe the power difference is small but it's just a few extra seconds of waiting.

[quote="serialk11r"]I put my computer on standby, because it takes very little time for it to start back up and it takes pretty much no more power than keeping it off.[/quote]

Yes, that is one of the advantages of Stand-By.  It's like a compromise between quick start-ups and power saving.

[quote="Aqua3993"]I put my laptop into sleep or hibernation whenever I am not using it. Rebooting only happens after I install or remove something and the OS requires me to restart the computer.

The biggest reason I leave it technically "on" is because Windows Vista takes FOOORRREEEVVEEER to boot.[/quote]

I use Gentoo Linux and therefore it's about 20 seconds from power on to desktop  :P . It also never requires me to restart for some update (except kernel updates, but those are rare and almost never necessary).

I still don't turn it off though, since it's about 3 seconds to go from suspend to desktop.

[quote="JRav"][quote="AIME15"]If you shut down your computer at night it'll acutaly take more energy than putting in standby (startup and shutdown take extra energy!)
[/quote]
But what saves more energy: keeping it off all night and starting and shutting it, or not starting and shutting it but leaving it on?[/quote]

If you leave it in standby (some weird mode of standby) it only takes 1 extra watt than keeping it on completely...and starts up instantly.  Turning on and off takes a while, and uses quite a bit of energy

There are different notions of standby -- using Vista terminology, the two options are Sleep and Hibernate. The former simply puts your computer into a low-power state where it doesn't do much. It is very fast to start up (since everything is in its correct state already). Hibernate, on the other hand, does something completely different. I believe it takes everything in memory and writes it onto disk and actually powers your computer off. This will definitely safe energy, but it has a longer start up time. I usually just let my computer Sleep at night, whereas my primary usage of Hibernate is to keep my laptop from losing tons of power when I am transporting it.

[quote="Poincare"]my brother (CS major) tells me that if you shut it down every time then that could kill your hard drive...[/quote]

I have never heard that before, but I guess it might be true for some computers.

As for vista Sleep vs. Hibernation: I like to let it sleep when I am off to the bathroom or dinner, but while I am at school, I make the laptop go into hibernation.

Although sleep is pretty simple, since it just involves putting comp in low-power (not NO power!) status, but hibernation can last forever, since it uses no energy. However, the downside is that the computer has to save its current status somewhere in the hard drive, so hibernation is like sacrificing space for time.

[quote="Poincare"]my brother (CS major) tells me that if you shut it down every time then that could kill your hard drive...[/quote]This was more true in the 80s and early 90s - not so much anymore. If your hard drive is gonna die, it'll die whether you keep your power on or consistently turn the computer on and off.

Usually, the rule of thumb is about 1 hour.
If you are not using it in an hour, you should probably turn it off.

[quote="Poincare"]hibernation uses SOME energy but shut down doesn't. But virtually hibernation uses no energy because if your laptop battery lasts just 20 minutes then it should last about 3 weeks in hibernation :o  :o[/quote]
No, when you hibernate your computer, it doesn't use any power (moreover, you can hibernate your computer and then remove the battery).

I can never figure out how to get out of Sleep mode on my mac...help!

Two suggestions:

1) Close it, then open it.
2) Press the power button.

As for me, I don't have a mac...

To get out of sleep, hit any key or click. It may take a few seconds.


geometry trigonometry function algebra polynomial

#2
[hide="Hint"]
The area of that triangle will be $(1/2)y^2 \sin C$ where $C$ is the angle between the two legs.  Now how do you maximize this quantity?[/hide]

Read it only if you want to find the answers .. skip it if you are trying ...

[hide]1. To multiple of 7 occurs in (7,14,...) But be careful, 7^2 comes in (49, 2*49 ...7*49) and 7^3 comses only once ..
Think about   it the till answer is clear ..:)
[hide][343/7] +[343/49] + [343/343] = 49+7+1 = 57[/hide]

2.  Area is   (1/2) b*c* sin A , [hide]here b=c=y and max value of b = y so answer is (1/2)y^2 (We do know that max value of sin of any angle is 1 :) )[/hide]
C.  Consider the function   P(x)*(x^2-4)+x  where P(x) is any polynomial)
[hide]The answer is obviously x
If you really want to see how it is done for more complex  given polynomials .[hide]. you assume  that the Polynomial is
P(x)*(x+2)*(x-2)+Ax+B
(x+2)*(x-2) is second degree so you  put the remainder   as Ax+B  (one degree in x)
When x=2, we get  2A+B = 2, when x=-2 we get -2A+B = -2 this gives A=1 and B=0 [/hide]

[/hide][/hide]

2.) [hide]In this triangle, the area will be $\frac{y^2}{2}\sin{C}$, where C is the angle between the legs. Since the maximum value of the sine of any angle is 1 (in other words, there is no angle of which the sine is greater than 1), the answer is $\frac{y^2}{2}$[/hide]

Gyan, I am not sure of what you mean when you say:

[quote]1. To multiple of 7 occurs in (7,14,...) But be careful, 7^2 comes in (49, 2*49 ...7*49) and 7^3 comses only once .. [/quote]

What do you mean when you say that $7^2$ comes in 49, 2*49....and $7^3$ comes in only once?

[quote]What do you mean when you say that 7^2 comes in 49, 2*49....and 7^3 comes in only once?[/quote]

[hide]343! = 1.2... [b]7 [/b].8.9...[b]14 [/b]...[b]21[/b] ...[b]28.[/b]........[b]343[/b]

So you  get 7^49 here... (along with other factors which do not contain 7, and thus can be ignored.)
But you forgot to count   another bunch of "additional " powers of 7

in the numbers     ....49 ...98 ...147 ...196 ............343.  so additional  7 factors here.
and finally 343 itself which is 7^3 and gives one more 7

If still not very clear .. samething in IOW..
[hide]IOW, if you ignore factors which do not contain 7 we see:

343! = (Bunch of numbers which  ontains no factors)* 7*14*21* ....241
       =  (Bucnh of numbers_etc)*7^49*(1.2.3.4.5.6.7..49)

        = 7^49*(7^7)(1.2.3.4.5.6.7)(Bunch of uninteresting numbers etc)
         =(7^49)*(7^7)*7*(Bunch of uninteresting numbers ..
[/hide]
Hope this helps.

.[/hide]

[quote]1.) The largest power of 7 which divides $343!$ is $7^x$. Find the value of $x$.[/quote] [hide]
$\text{Let }[x]\text{ be the 'greatest integer' function.}$

$343! = 1\cdot2\cdot3\cdot4\;\cdots\;343$

$\text{Clearly, every seventh number contains a factor of 7.}$
$.\quad\text{There are: }\left[\frac{343}{7}\right] = 49\text{ multiples of 7.}$

$\text{However, every }49^{th}\text{ number is a multiple of }7^2 = 49,$
$.\qquad\text{each of which contribute an extra 7.}$
$.\quad\text{There are: }\left[\frac{343}{49}\right] = 7\text{ of them.}$

$\text{And, every }343^{th}\text{ number is a multiple of }7^3 = 343,$
$.\qquad\text{each of which contribute yet another 7.}$
$.\quad\text{There are: }\left[\frac{343}{343}\right] = 1\text{ of them.}$

$\text{Hence, }343!\text{ contains: }49 + 7 + 1 = 57\text{ factors of 7.}$

$\text{Therefore: }343!\text{ contains }7^{57}\;\;\;\Longrightarrow\;\;\;x = 57$[/hide]


algebra
system of equations
algebra unsolved

For a real number $ x$ is $ [x]$ the next smaller integer to $ x$, that is the integer $ g$ with $ g\leqq<g+1$, and $ \{x\}=x-[x]$ is the “decimal part” of $ x$.
Determine all triples $ (a,b,c)$ of real numbers, which fulfil the following system of equations:
\[ \{a\}+[b]+\{c\}=2,9\]\[ \{b\}+[c]+\{a\}=5,3\]\[\{c\}+[a]+\{b\}=4,0\]

http://www.mathlinks.ro/viewtopic.php?p=1358623#p1358623
there is same problem...

Indonesia  TST 2009 Day 4 #1................

????????????


Aν $ 0\leq x,y,z\leq 1$ ν.δ.ο. ισχυει
$ 2(x^3\plus{}y^3\plus{}z^3)\leq x^2y \plus{}y^2z\plus{}z^2x \plus{}3$

[hide]Ισχυει οτι: $ a^2b \plus{} 1 \geq a^2 \plus{} b$, για $ a,b \leq 1$ με $ a \geq \minus{} 1$. Αυτο προκειπτει αμεσα απο την ανισοτητα της αναδιαταξης αφου οι δυαδες $ (a^2, 1), (b, 1)$ εχουν προφανος την ιδια διαταξη.

Ετσι για τους αριθμους $ x,y,z \in [0, 1]$ εχουμε:

$ RHS \equal{} x^2y \plus{} y^2z \plus{} z^2x \plus{} 1 \plus{} 1 \plus{} 1 \geq (x^2 \plus{} y^2 \plus{} z^2) \plus{} (x \plus{} y \plus{} z) \geq (x^3 \plus{} y^3 \plus{} z^3) \plus{} (x^3 \plus{} y^3 \plus{} z^3) \equal{} 2(x^3 \plus{} y^3 \plus{} z^3) \equal{} LHS$

αφου η συναρτηση $ f(x) \equal{} a^x$ ειναι φθινουσα στο $ (0, \plus{} \infty)$ οταν $ a \in [0, 1]$

ΥΓ: Με τη θεωρια αριθμων θα ασχοληθω αυριο γιατι με περιμενουν να παμε για καφε τωρα.[/hide] :D

wraios re nick! egw i lysi p skeftika einai
$ RHS\geq 2\sqrt{3(\sum x^2y)}$ kai $ LHS \leq 2 \sqrt{(\sum x^6y)(\sum \frac{1}{y})}\leq 2\sqrt{(\sum x^3y)(\sum \frac{1}{y})}$ apo B-C-S kai $ x^3\leq1$ etc
ara menei na apodeiksume
$ (\sum x^3y)(\sum \frac{1}{y}) \leq 3(\sum x^2y)$=> omogenis=> assume xyz=1
apomenws pernume
$ (\sum x^3y)(\sum \frac{1}{y}) =(\sum x^3y)(\sum xy) \leq  3(\sum x^2y)$
 :)
perimenw kai alles liseis !

Δεν ξερω αν μπορει να θεωρηθει διαφορετικη λυση, αλλα για να αποδειξουμε την ανισοτητα $ a^2b \plus{} 1 \geq a^2 \plus{} b$, η οποια κανει τη λυση μου τοσο απλη, δεν ειναι απαραιτητο να χρησημοποιησουμε ανισοτητα αναδιαταξης, αφου αν τα παμε ολα στο 1ο μελος λαμβανουμε: $ (a \plus{} 1)(a \minus{} 1)(b \minus{} 1) \geq 0$ το οποιο προφανος ισχυει. (Ετσι η ασκηση μπορει να λυθει και χωρις εξωσχολικες γνωσεις)


A collection of five positive integers has mean 4.4, unique
mode 3 and median 4. If an 8 is added to the collection, what
is the new median? Express your answer as a decimal to the
nearest tenth.

Five numbers with mean 4.4 means the sum of the numbers is $ 5\times 4.4\equal{}22$.  We know that 4 is smack in the middle and if the unique mode is 3, we know the other numbers are 3, 3, 4, __, __.  The other two numbers must sum to 12 since $ 22\minus{}3\minus{}3\minus{}4\equal{}12$.  They can't be the same since the mode is unique, and they must both be greater than 4 since otherwise, 4 wouldn't be the median.  So the other two numbers must be 5 and 7.  If we add 8, our set is 3, 3, 4, 5, 7, 8, which has mode $ \frac{4\plus{}5}{2}\equal{}\boxed{4.5}$.


Let $a,$ $b$ and $c$ are positive real numbers, such that $abc=1.$
Prove that $a^2+b^2+c^2+ab+ac+bc+6\geq4(a+b+c).$

This inequality is equivalent to

$(a + b - 2)^2 + (a + c - 2)^2 + (b + c - 2)^2 \; \geq \; 0$

which is valid for all real numbers a,b and c.


calculus
real analysis
real analysis theorems

[quote="Lucebike"]Hi,
the term is used everywhere and defined nowhere:
What does ist mean if you say that two norms are equivalent?
Thx,
Lucebike[/quote]

If I remember correctly, two norms, say $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent, if there exist strictly positive constants $C_*$ and $C^*$ such that for all $x$ in the normed space you have
$C_*\|x\|_1\leq\|x\|_2\leq C^*\|x\|_1$. I hope it helps.

By the way, unless you refer to some distinct type of norm, it should be in Analysis-Calculus, or in Advanced topics (I think!)

[quote="Lucebike"]Ok, that makes sense. Thank you very much!

I'm not quite familiar with this (great!) forum, so should I have placed this topic wrongly, I ask any mod to replace it.

Bye,
Lucebike[/quote]

Nevermind!, some efficient moderator already did it. Usually nobody bothers if you make a few posts in the wrong place when you are new in the forum (it always happens, even kids who place problems of grammar school in "Advanced Algebra" :D ). That is why we have moderators (well, and also to censor if someone gets dirty or rude, or something like that, but it does not happen that often).

Welcome to MathLinks (or AoPS, depending on what skin you use.) Best regards,


An example.

$x \in \mathbb{R}$ with $[\frac{x+3}{2}]=[\frac{x+12}{3}]$ where $[n]$ is the biggest integer less or equal to n. Find the interval of solutions for $x$.

I would like to solve this for me in as many ways as possible. I thought of something:

[hide]
If $[\frac{x+3}{2}]=[\frac{x+12}{3}]$ means that $\frac{x+3}{2}\ge \frac{x+12}{3}$ or $\frac{x+3}{2}\le \frac{x+12}{3}$ so $0 \le \frac{x+3}{2}-\frac{x+12}{3}< 1$ or $0 \le \frac{x+12}{3}-\frac{x+3}{2}<1$ right?

Would solving them separately give the result?
[/hide]

$[a]=[b]\iff (\exists n\in\mathbb{Z})(n\leqslant a\leqslant b<n+1\lor n\leqslant b\leqslant a<n+1)$

Thank you for that, but i'd like to know more about my approach, since my math teacher got mad and didn't manage to explain me what was wrong with  it :(

[quote="icx"]
$\frac{x+3}{2}\ge \frac{x+12}{3}$ or $\frac{x+3}{2}\le \frac{x+12}{3}$ [/quote]
I understand your teacher perfectly. If you have any $a$ and $b$ it is true that $a\le b$ or $a\ge b$ not oly for $a=\frac{x+3}{2}$ and $b=\frac{x+12}{3}$

[quote="digger"][quote="icx"]
$\frac{x+3}{2}\ge \frac{x+12}{3}$ or $\frac{x+3}{2}\le \frac{x+12}{3}$ [/quote]
I understand your teacher perfectly. This is always true.[/quote]

It is always true. Moreover, as i said [img]http://www.mathlinks.ro/Forum/latexrender/pictures/a3d4c43d20fce0aa2fa056900e1a1e7a.gif[/img] or [img]http://www.mathlinks.ro/Forum/latexrender/pictures/a05f0a2d574cf772ef1149c422217e8f.gif[/img] but why wouldn't this help me in solving the problem? I can find the interval i am looking for from these inequalities, can't i?

[quote="Farenhajt"]No, you can't.

$0<2.1-1.7<1$, but $[2.1]\neq[1.7]$[/quote]

We know $[x]=[n]$ $\Rightarrow$ $0 \le x-n < 1$ or $0 \le n-x <1$

We don't need the converse since in this problem we already know $[n]=[x]$ :)

Anyway, i remember my teacher saying almost the same thing as you did in your previous post, meaning there must be something fishy. I tried to replace $x$ with a number from the interval i found but it didn't work, so i suppose i should try her way since this one doesn't seem so sure :|

thank you

You DO need the converse, since the values you find by your method must work when you plug them in, which is exactly the converse statement:

IF $x=\mathrm{whatever}$, THEN $[\mathrm{blah}_{1}]=[\mathrm{blah}_{2}]$

[quote="icx"]We don't need the converse since in this problem we already know $[n]=[x]$ :)[/quote]
Suppose you are true. Now I take this truthfull conclusion
$[n]=[x] \Rightarrow$ I am sitting now on the chair.
What can you derive from this conclusion?  :D

You don't have to shout, now i understand, thanks


Can anybody make sense of this question?
5p^2q^2-7pq-6 &
4m^2-8mn+3n^2
Thanxx

do you probably mean
$5(pq)^2-7pq-6=0$
and
$4m^2-8mn+3n^2$
if so i'll give you an hint [hide]try with $pq=x$[/hide]

For the second one :
$4m^2-8mn+3n^2=0$. If $n=0$, then $m=0$. Suppose $n\neq 0$. Then the equation is equivalent to $(4m^2-8mn+3n^2)/n^2=0/n^2$, that is $4\left(\frac{m}{n}\right)^2-8\left(\frac{m}{n}\right)+3=0$ Put $x=m/n$ and you have to solve $4x^2-8x+3=0$. So $x\in\{1/2,3/2\}$. Finally the solutions are $(m,n)\in\{(\alpha,2\alpha),(\alpha,2\alpha/3)|\alpha\in\mathbb{R}\}$.

[hide="Solution to the first equation"]If it's $5(pq)^2-7pq-6=0$, you just call $pq$ $x$ to make it $5x^2-7x-6=0$ Then factor it into $\boxed{(5x+3)(x-2)}$[/hide]

[hide="Solution to the second equation"]You can actually just factor this off the bat easily. Since the middle term has $mn$, you know the $m$ and the $n$ are more "spreadout". You get $\boxed{(2m-n)(2m-3n)}$[/hide]

I'm not giving you solutions because they're not equal to anything, they're just expressions.


induction
integration
algebra unsolved
algebra

The sequence $ \{x_n\}$ satisfies $ x_1 \equal{} \frac {1}{2}, x_{n \plus{} 1} \equal{} x_n \plus{} \frac {x_n^2}{n^2}$.  Prove that $ x_{2001} < 1001$.

Prove $ x_n<\frac{n\plus{}1}2$ by induction.

Obviosly $x_n$ is monotonically increase.
If $x_n<C$ and $(C+\delta )^2(\frac{1}{n^2}+...+\frac{1}{(n+m-1)^2})<\delta$, then $x_{n+m}<C+\delta.$
Because $\frac{1}{k^2}<\int_{k-0.5}^{k+0.5}\frac{dx}{x^2}$, we get
if $x_n<C$ and \[(C+\delta)^2(\frac{1}{n-0.5}-\frac{1}{n+m-0.5})=(C+\delta)^2\frac{m}{(n-0.5)(n+m-0.5)}<\delta\] then $x_{n+m}<C+\delta.$
It gives, that $x_n$ is bounded. We can calculate $x_4=\frac{795}{768}$. Take $C+\delta =1.5$ it gives $x_{13}<1.5$. Then take $\delta =\frac{19-\sqrt{325}}{4}<0.25$. It gives $x_n<C+\delta \ \forall n$.
We prove $x_n<1.75  \ \forall n$.


calculus
derivative
algebra proposed
algebra

m,n are natural numbers,
Solve $m^3-n^3=mn+25$

so m>n consider
$f(x)=x^3-nx-n^3$ whose derivative is $3x^2-n$ for $m\geq\ n+1$ it is positive so f(x)is increasing so it takes the minimum value when $m=n+1$ so $2n^2+2n+1<26$ so $n<4$
$n=3, m=4$ for ${n \in\{1,2}$ there are no solutions for m.

Obviously $m>n$ . When $m=n+1$ , we can solve to get $m=4,n=3$

When $m-n\geq 2$ we have

$m^3-n^3=(m-n)(m^2+mn+n^2)\geq 2(3mn)=6mn$ which means $25\geq 5mn$ or $mn\leq 5$

For $mn=5$ , we only have $(m,n)=(5,1)$ but this does not satisfy .

For $mn=4$ , we have $(m,n)=(4,1)$ but this also doesnt satisfy

Similarly work out for case $mn=1,2,3$ and all does not satisfy the equation .

Hence we can only have $(m,n)=(4,3)$


Could you tell my some useful theorems about Newton Symbol.

${{n} \choose k}$

In my language Newton's Symbol is a number of possible choices of k elements from group composed of n elements

Here maybe : http://en.wikipedia.org/wiki/Newton

In fact, I'm not sure about what you call "newton symbol"..

In my language Newton's Symbol means that : ${{n} \choose k}$

Sorry, that You didn't know what I was talking about.
Anyway could you post some useful theorems, properties about that?

Oh! The binomial coeff. :)

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


abstract algebra
vector
group theory
superior algebra
superior algebra theorems

I tried to consider if $ Z^\infty$ is free abelian. Here I mean the group of countable sequences of integers under component-wise addition. I tried to look at the standard proof that each vector space has a Hamel basis, but obviously this does not seem to apply at all here.

Sorry if this is a really stupid question, but I can't figure it out  :maybe:

It is not, but that's not so easy. See http://www.mathlinks.ro/Forum/viewtopic.php?t=4924 and http://www.mathlinks.ro/Forum/viewtopic.php?t=89680 .

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=185366]This[/url] is also a proof. If this module were free, its dual would contain the free abelian group on its uncountable basis, and there would be nonzero linear functionals that are zero on the finitely supported sequences. As shown there, there are no such functionals, and the dual is actually countable.

"and there would be nonzero linear functionals that are zero on the finitely supported sequences" why?

Size. If there are no such functionals, values on the finitely supported sequences determine everything, and there are at most $ c$ such functionals. The dual of $ \bigoplus_{i\in \mathbb{R}}\mathbb{Z}$ has cardinality $ 2^c$.

By -oo-'s post, we can actually say that the dual of $ \mathbb{Z}^{\infty}$ is countable. The two things we're wondering about are two power sets away from each other.

thanks. so to show that $ \mathbb{Z}^\mathbb{N}$ is not free, we don't need that it's dual is free, but just what was asked in the other thread.


function
calculus
integration
number theory
prime numbers
number theory unsolved

Let $\pi(x)$ be the function that counts prime numbers less or equal than x. It is famous that $\frac{\pi(x)ln(n)}{x}$ converges tu 1. My question is: what do we know about $\pi(x)-\frac{x}{lnx}$. Does it maybe converges to 0?.

Bye, Uber

It can't converge to $0$ because $\pi(n)$ takes only integral values where the rest of the term can be very near to the half of an integer infinetely often.
But by the Riemann hypethesis (assumed as true) we get some sharper versions, see http://mathworld.wolfram.com/PrimeNumberTheorem.html , number (22) (this uses $Li(x)$ instead, but see (7) for some asymptotic).

Thank you ZetaX, I've read with interest your link. But I don't understand why it cannot be converges to zero (without using Riemann hypotesis)
Moreover, is it true that Li(n) is asympotic to n/ln(n)?
Bye, Uber


Okay, so here is the problem.

Ten tiles are numbered 1-10. A regular die with 6 sides numbered 1-6 is rolled and one of the ten tiles is flipped over. What is the probability that the the product of the two numbers is a perfect square?
(PLEASE NOTE: this may not be the EXACT wording of the problem, I sort of memmed the question.)

6 * 10 = 60 total combinations.

1, 2, 3, 4, 5, 6
1, 2, 3, 4, 5, 6, 7, 8, 9, 10

Perfect square pairs = (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), (2,8), (9,1), (4,1), (1,4) => 10 possible products.

What am I missing? The answer was 11/60. I got 10/60 = 1/6. (just one off!! fail)

Please help!

Hehe okay thanks AIME. 4, 9 = 2^2 * 3^3 = 36 yay so thats what i was missing.

Fail.

That's the one I got wrong on the AMC8 too...24 out of 25 :(.

And please don't click submit twice cliu.

Darn I clicked Submit twice. I'm too used to double-clicking.

That's ok...You're relatively new to the forum.

oh you can take 4 and 9 man I was stupid

I would hve gotten a  perfect score that was so easy

I wish there was a simpler way to solve that without listing them out.  I'm sure there is, but the AMC 8 solutions pamphlet doesn't specify one.

(Not that it is difficult, but what if there were 2008 pieces of paper and a 1043 sided die?)

Why are we reviving this post? This is something I posted over 3 months ago (when I was really new to AoPS).

Wow, I have improved a lot in 3 months.

[quote="AIME15"][quote="AdithyaGanesh"]Yes, but what if it was on the AIME or AMC 10/12?[/quote]

Indeed, a problem with such massive numbers would not appear because the goal of these exams is to test problem solving ability, not casework bashing (although the test-writers do seem to fail at this a lot :))[/quote]

That makes sense.  :)


Let V and W be vector spaces such that dim(V)=dim(W). Let T be a linear transformation from V to W. How do I show that there exists ordered bases a,b for V,W respectively, such that the matrix representation of T in the ordered bases a,b is a diagonal matrix?

It is essentially the same as the statement that each square matrix has a [url=http://en.wikipedia.org/wiki/Singular_value_decomposition]SVD[/url].  A proof of this can be found in most Linear Algebra texts, and some Numerical Methods texts.  More generally, the requirement $ \dim (V)\equal{}\dim (W)$ is unnecessary.

Singular values? This is Gaussian elimination, carried to the extreme. Choosing a basis on each side lets you use the full range of row and column operations. The entries in that diagonal matrix are all zeros and ones if you want.