inequalities
trigonometry
geometry
perimeter
geometry solved

Let ABC -- triangle such that  AB<BC<CA. l -- bisector from B to AC,
h -- altitude from C to AB. Prove, that  l \leq h.

[i]unknown source[/i]

So we have c<a<b.
l=(2ac/(a+c))*cos(B/2)=(2ac/(a+c)) \sqrt (p(p-b)/ac)=2 \sqrt (a*c*p*(p-b))/(a+c).
h=2S/c.

l \leq h <=>c* \sqrt (ac)/(a+c) \leq  \sqrt (p-a)(p-c) <=>
a*c 3 \leq (p-a)(p-c)*(a+c) 2 .(1)

Denote p-a=A and so on . We have c<aA>B.
(1) comes to (B+C)(A+B) 3 \leq AC(A+2B+C) 2.
I will use a instead of A and so on.
(b+c)(a 3+3a 2b+3ab 2 +b 3) \leq ac(a 2 +4b 2 +c 2 +2ac+4ab+4bc).
This comes by computations to a 3b+3a 2b 2 +3ab 3+b 4+b 3c \leq ac 3+abbc+aabc+4abcc+2aacc.
This is b(a+b) 3+b 3c \leq ac(cc+ab+4bc+2ac)+ab 2 c.

We have c>a>b, we have bbbc<abbc, so it suffices to show
b(a+b) 3 \leq ac(c 2 +ab+4bc+2ac).
Since c>a we have
ac(cc+ab+4bc+2ac)>a 3 (3a+5b).

So it suffices to show b(a+b) 3 \leq a 3(3a+5b). This come by computations to
3a 2 b 2 +3ab 3+b 4 \leq 3a 4+4a 3b.
Since a>b it suffices to show
a 2 b 2 +ab 3 \leq a 4+a 3b <=> (a+b)(a 3-b 3) \geq 0, which is true since a>b.

----------------------------------------------------------------------------------
Sorry about this monster. I know what you will say :"Ah ce scarbos"  :D .
I just really wanted to finish the proof of the first problem I sove in 2004  :D .

[quote="Fedor Bakharev"]Let $ABC$ be a triangle such that  $AB\leq BC\leq CA$,  $w_{b}$ is the angle-bisector from $B$ to $CA$, $h_{c}$ is the altitude from $C$ to $AB$. Prove that  $w_{b}\leq h_{c}$.[/quote]
Let we show a refinement of  the original inequality :

$c\leq a\leq b\Longrightarrow h_{c}\geq r_{a}\geq w_{b}$,

where $r_{a}$ is the radius of excircle.
[hide="Proof."]
$h_{c}-r_{a}=\frac{2\Delta}{c}-\frac{\Delta}{s-a}$

$=\frac{\Delta[2(s-a)-c]}{c(s-a)}$

$=\frac{\Delta(b-a)}{c(s-a)}\geq0$;

$r_{a}-w_{b}=\frac{\Delta}{s-a}-\frac{2ca}{c+a}\cos{\frac{B}{2}}$

$=\sqrt{\frac{s(s-b)(s-c)}{s-a}}-\frac{2\sqrt{cas(s-b)}}{c+a}$

$=\sqrt{s(s-b)}\left(\sqrt{\frac{s-c}{s-a}}-\frac{2\sqrt{ca}}{c+a}\right)$

$=\frac{(a-c)[a^{2}+5ca+ab+c(c+a-b)]\sqrt{s(s-b)}}{2(c+a)^{2}\sqrt{(s-c)(s-a)}+4(s-a)(c+a)\sqrt{ca}}\geq0$,

where $s$ is the semi-perimeter and $\Delta$ is the area of $ABC$. [/hide]


Find all real positive solutions (if any) to
\begin{align*}
x^3+y^3+z^3 &= x+y+z, \mbox{ and}  \\
x^2+y^2+z^2 &= xyz.
\end{align*}

Well as we have $x,y,z>0$ we have also \[ \sqrt[3]{\frac{x^3+y^3+z^3}{3}}\geq \frac{x+y+z}{3} \] so from the first equation \[ \sqrt[3]{\frac{x+y+z}{3}}\geq \frac{x+y+z}{3} \] and as $x+y+z>0$ we obtained $x+y+z \leq 3$.
Now we have also \[ \sqrt{\frac{x^2+y^2+z^2}{3}}\geq \sqrt[3]{xyz} \] and thus we may get $xyz \geq 27$ but this is contradiction cos' we would have \[ 1 \geq \frac{x+y+z}{3} \geq \sqrt[3]{xyz} \geq 3 \] this means that our system of equation has no solutions in positive real numbers  :)

Multiply the first equation by the square of the second, you get
$ (x^3\plus{}y^3\plus{}z^3)(x^2\plus{}y^2\plus{}z^2)^2\equal{}(x\plus{}y\plus{}z)x^2y^2z^2$
But the left side contains every term on the right side and more, hence there is no solution.

$x+y+z=x^3+y^3+z^3 \geq 3xyz=3(x^2+y^2+z^2) \geq (x+y+z)^2$

$\implies x+y+z \leq 1 \implies 0 \leq x,y,z \leq 1 \implies x^3 \leq x, y^3 \leq y, z^3 \leq z \implies x^3+y^3+z^3 \leq x+y+z$.

Equality occurs only at $x=y=z=1 \implies x+y+z=3$, contradiction.

Jastrzab's is the most classy solution....

Maths is the doctor of science..
Sayantan...

[quote="shobber"]Find all real positive solutions (if any) to \[
\begin{align*}
x^3+y^3+z^3 &= x+y+z, \mbox{ and}  \\
x^2+y^2+z^2 &= xyz.
\end{align*}. \][/quote]
Dividing the first equation by the second we get:
$\frac{\sum x}{\sum x^2}=\sum \frac{x^2}{yz}\ge \frac{\left(\sum x\right)^2}{\sum xy}$ [b][Cauchy-Schwarz][/b]
$\Rightarrow \frac{\sum x}{\sum x^2}\ge \frac{\left(\sum x\right)^2}{\sum xy}\Rightarrow \sum xy\ge \left(\sum x\right)\left(\sum x^2\right)$
$\ge \left(\sum x\right)\left(\sum xy\right)\Rightarrow \sum x\le 1$. But
$\sum x=\sum x^3$, so that $xyz\le \tfrac{1}{3}\left(\sum x^3\right)\le \tfrac{1}{3}$ [b][AM-GM][/b]
However, $xyz=\sum x^2\ge 3\sqrt[3]{x^2y^2z^2}\Rightarrow xyz\ge 27$, absurd!
Thus, $\exists$ no solutions in $x,y,z>0$ to the system. $\Box$


$p$ is a positive number. For given $p$ find the greatest $C$ for which the following inequality stands:
$x^{2}+y^{2}+pxy \geq C(x+y)^{2}$
where $x$ and $y$ are nonnegative numbers.

Take $x=y=1$ then $2+p\geq 4C\iff \frac12+\frac{p}{4}\geq C$
Now $C=\frac12+\frac{p}{4}$ is sufficient; indeed, \[x^{2}+y^{2}+pxy \geq \left(\frac12+\frac{p}{4}\right)(x+y)^{2}\iff (x^{2}+y^{2})\left(\frac12-\frac{p}{4}\right)+(\frac{p}{2}-1)xy\geq 0\\ \iff \left(\frac12-\frac{p}{4}\right)(x-y)^{2}\geq 0\]
which is true for $p\leq 2$.


Let  n>1  be an integer, and let  S  be a subset of  {1,2,...,n} such that  S  neither contains two coprime elements, nor does it contain two elements, one of which divides the other. What is the maximum possible number of elements of  S  ?


algebra
polynomial
vector
linear algebra
linear algebra unsolved

Let $A$ be a $C$-commutative algebra of dimension $n$ over $C$ and assume that 0 and 1 are its only idempotents. Prove that $A$ is generated by a finite number of nilpotent elements.

Is there something missing, or wrong, or have I gone crazy? :? What if this algebra is $\mathbb C$ itself, for instance? It's true that $0$ and $1$ are the only idempotent elements, but there's only one nilpotent element, $0$, and it certainly doesn't generate $\mathbb C$..

I'd also like to know what "to generate" means here, in the context of algebras. For instance, do you assume that $1$ and $X$ generate $K[X]$, because all elements of $K[X]$ are polynomials in $1$ and $X$, or do you mean that the elements have to generate the algebra as a vector space?

Oh, I am terribly sorry, I forgot to add what I understand by generates. Actually, we say that the family $x_1,..,x_k$ generates the algebra $A$ if every element of $A$ is a linear combination of $1_A$ and some of $x_1,...,x_k$. Again, I apologise.

As usual, identify $a\in \mathcal A$ with the endomorphism $x\mapsto ax$ represented in a basis of $\mathcal A$ over $\mathbb C$. What we get is an embedding of $\mathcal A$ as a subalgebra of $\mathcal M_n(\mathbb C)$. What we must prove can be rephrased thus:

Given a subalgebra $\mathcal A$ of $\mathcal M_n(\mathbb C)$ with no idempotent elements except for $0,1$ ($1=I_n$ here; in general, things like $A-\lambda,\ A\in\mathcal A$ mean $A-\lambda I_n$), prove that $\mathcal A$ is generated by $1$ and several nilpotent elements.

In order to prove this, it suffices to show that for every element $A\in\mathcal A$ there is a scalar $\lambda$ such that $A-\lambda$ is nilpotent. Every $A\in\mathcal A$ can be written uniquely as $A=D_a+N_a$, where $D_a,N_a$ are diagonalizable and nilpotent respectively, and they commute with one another. Moreover, they're polynomials in $A$, so they belong to $\mathcal A$. Suppose now that $D_a$ is not a scalar matrix. It must then have $k\ge 2$ distinct eigenvalues $\lambda_1,\ldots,\lambda_k$, and it can be written as $D_a=\sum_{i=1}^k\lambda_iP_i$, where $P_i$ is the projection onto the space of $\lambda_i$-eigenvectors associated to $A$. $P_i$ can be written as some scalar times $\prod_{j\ne i}(D_a-\lambda_j)$, so it too belongs to $\mathcal A$, but then $P_i$ would be an idempotent element of $\mathcal A$ different from $0,1$, so we have a contradiction. This means that $D_a=\lambda$, and we have $A-\lambda=N_a$, which is nilpotent, and this is exactly what we wanted.


Hi I'm new to the forum & looking for help displaying the algebraic logic to get from one equivelant formula to next.

1) Starting: Chrg = Pmt + Cont + Woff
2) Chrg/Chrg = Pmt/Chrg + Cont/Chrg + Woff/Chrg
3) ??
4) End: Chrg/Chrg = Pmt/(Chrg-Cont) + Woff/(Chrg -Cont)

Need help on showing the logic needed between step 2 & 4?

Thanx in advance

What are these 'symbols'  :maybe:
 Can anyone put this in LaTeX

I think each "string" of 4 letters is a different variable. (i.e. $ Chrg\equal{}a, Pmt\equal{}b\cdots$)


trigonometry
geometry
circumcircle
symmetry
geometry unsolved

the inscribed circle of ABC is tangent to each side at D,E,F, AD meets the inscribed circle at X,take DY=AX in AD,YB and YC meets the circle at P,Q.prove:FP,EQ,AD are parallel.

[quote="thanhnam2902"][color=darkblue]I have another nice problem.

The inscribed circle of $ \triangle ABC$ is tangent to each side $ BC, CA, AB$ at $ D,E,F$ respectively. Let $ AD$ meets the inscribed circle at $ X$, take $ DY \equal{} AX$ with $ Y \in AD$
Let $ BY$ meet $ AC$ at $ P$, $ CY$ meet $ AB$ at $ Q$. Prove that $ FP, EQ, AD$ are concurrent.[/color]

[color=darkblue]What do you think? Is it true?[/color][/quote]

This is obviously true by apply Pappus's theorem to 6 points BQFEPC (because $ AD,BE,CF$ are concurrent at Gergonne point).And you don't need the statement $ DY\equal{}AX$ in this problem.

[quote="livetolove212"]Draw the line which passes through $ E$ and parallel to $ AD$, it cuts $ (I)$ at $ Q$, similarly $ FP\parallel AD$. We will prove $ DY \equal{} AX$ and complete the proof.
Applying the law of sine we have:
$ \frac {YD}{DC} \equal{} \frac {YQ}{QC}\cdot\frac {\sin \angle YDQ}{\sin \angle QDC} \equal{} \frac {AE}{EC}\cdot\frac {\sin \angle AEX}{\sin \angle AXE}$
$ \Leftrightarrow YD \equal{} AE\cdot\frac {AX}{AE} \equal{} AX$
We are done![/quote]

typo: $ \frac {YD}{DC} \equal{} \frac {YQ}{QC} \cdot \frac {\sin \angle QDC}{\sin \angle YDQ}$

Let te parallel to $ AD$ through $ E$ cut $ (I)$ at $ Q$. Let $ CQ$ intersect $ AD$ and $ (I)$ at $ Y$ and $ Q'$, respectively. Let the tangent line to $ (I)$ at $ Q$ intersect $ AD$ and $ AC$ at $ Y'$ and $ M$, respectively. We will see that $ AX \equal{} DY$, which solves the problem.

Projecting from $ C$ we get $ (E,Q,D,Q') \equal{} (E,Q',D,Q)$. This implies $ (E,Q,D,Q') \equal{} \minus{}1$.

Now, consider the projection from $ Q$ which maps the incircle into line $ AD$. We get $ \minus{}1 \equal{} (E,Q,D,Q') \equal{} (\infty ,Y',D,Y) \equal{} DY/DY'$. This implies $ DY \equal{} Y'D$.

As $ QME$ is isosceles, $ Y'MA$ is isosceles too. By simmetry, we get $ Y'D \equal{} AX$. Then $ DY \equal{} AX$, which is what we wanted to prove.

Dear Mithril,
are you sure of some path of your proof?
For example:  (E,Q,D,Q')=(E,Q',D,Q)?
If you have some time can you explain again your third line...
Sincerely
Jean-Louis

Dear Mithril,
your idea is very nice and helpfull.
I come up with your notation and some explications:
the quadrilateral EQDQ' is harmonic; in consequence, the tangent to the incircle at Q' meet the tangent à Q on DE at a point N.
Let X be the meetpoint of QQ' and DE.
(E, D, X, N) is harmonic , so is the pencil with origin at Q…
the rest follows
Sincerely
Jean-Louis

Dear Mithril,
your idea conduct to a converse wich solve the initial problem of Bye.
Sincerely
Jean-Louis

$ (E,Q,D,Q') \equal{} (E,Q',D,Q)$ because if you project from $ C$, mapping the incircle into itself, $ E$ and $ D$ are fixed points, $ Q$ goes to $ Q'$ and $ Q'$ goes to $ Q$. This means $ EQDQ'$ is harmonic.

Then, project from $ Q$ and send $ E$ to the infinity, and $ Q$, $ D$, $ Q'$ to $ Y'$, $ D$ and $ Y$, respectively.

[quote="jayme"]
(E, D, X, N) is harmonic , so is the pencil with origin at Q…
[/quote]
What do you mean here? I don't think you need to project to $ ED$ before.

Based on the same idea as in [b][size=100]mithril's[/size][/b] solution, let us to try an alternative approach.

[b][size=100][color=DarkBlue]LEMMA. - A triangle $ \bigtriangleup ABC$ is given and let $ D$ be, the point of intersection of its circumcircle $ (O),$ from the $ A$-symmedian. We draw a line through vertex $ B$ and parallel to the line segment $ CD,$ which intersects $ AD$ at point $ E,$ the circle $ (O)$ at point $ F,$ and the tangent line of $ (O)$ at vertex $ C,$ at point $ Z.$ Prove that $ BE \equal{} FZ.$[/color][/size][/b]

[b][size=100]PROOF OF THE LEMMA.[/size][/b] - We draw the tangent line of $ (O)$ at point $ D,$ which intersects the line segment $ CZ,$ at point so be it $ K$ and we denote the points $ P\equiv BC\cap KD$ and $ Q\equiv BC\cap AD.$

The tangent line of $ (O)$ at vertex $ B$ of $ \bigtriangleup ABC,$ passes through the point $ A'\equiv AD\cap CK$ as well.

$ \bullet$ Because of the line segment $ BC,$ as the polar of $ A'$ with respect to $ (O),$ passes through the point $ P,$ we have that the polar of $ P$ with respect to $ (O),$ passes through the point $ A'.$

So, we conclude that the line segment $ A'D,$ connecting the point $ A'$ with the point $ D,$ as the tangency point of $ (O),$ from the line through $ P,$ is the polar of $ P$ with respect to $ (O)$ and then, we have that the points $ P,\ B,\ Q,\ C,$ are in harmonic conjugation.

That is, the pencil $ D.PBQC$ is also harmonic and because of $ BE\parallel DC,$ we conclude that $ BE \equal{} BE'$ $ ,(1)$ where $ E'\equiv DP\cap BE.$

But, it is easy to show that $ BE' \equal{} FZ$ $ ,(2)$ because of the symmetry with respect to the line segment $ KO.$

From $ (1),\ (2)$ $ \Longrightarrow$ $ BE \equal{} FZ$ and the proof of the [b][size=100]Lemma[/size][/b] is completed.

$ \bullet$ Return now in the proposed problem, we draw the line through the point $ F$ and parallel to $ AD,$ which intersects the incircle $ (I)$ of the given triangle $ \bigtriangleup ABC,$ at point so be it $ P$ and we denote the points $ Y\equiv AD\cap BP$ and $ B'\equiv (I)\cap BP.$

The line segment $ B'B,$ is the $ B'$-symmedian of $ \bigtriangleup B'DF$ and because of $ AD\parallel FP,$ where $ P\equiv B'B\cap (I),$ based on the above [b][size=100]Lemma[/size][/b], we conclude that $ DY \equal{} AX.$

By the same way, we can prove that the line segment $ CQ,$ where $ Q$ is the point of intersection of $ (I),$ from the line through the point $ E$ and parallel to $ AD,$ intersects $ AD$ at the same point $ Y$ and the proof of the proposed problem is completed.

Kostas Vittas.


Let $ a,b,c \ge 0$ such that $ a\plus{}b\plus{}c\equal{}3$. Prove that
\[ \frac{a^2}{2a^2\minus{}3a\plus{}3}\plus{}\frac{b^2}{2b^2\minus{}3b\plus{}3}\plus{}\frac{c^2}{2c^2\minus{}3c\plus{}3} \le \frac{3}{2}\]
:)

[quote="can_hang2007"]Let $ a,b,c \ge 0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \frac {a^2}{2a^2 \minus{} 3a \plus{} 3} \plus{} \frac {b^2}{2b^2 \minus{} 3b \plus{} 3} \plus{} \frac {c^2}{2c^2 \minus{} 3c \plus{} 3} \le \frac {3}{2}
\]
:)[/quote]
I think the following more general statement holds:

Let $ a_1,a_2,...a_n$ be nonnegative real numbers such that $ a_1\plus{}a_2\plus{}...\plus{}a_n\equal{}n$. If $ p\le \frac n{2(n\minus{}1)}$ and $ q\ge 0$, then

$ \frac {a_1\minus{}1}{(a_1\minus{}p)^2\plus{}q}\plus{} \frac {a_2\minus{}1}{(a_2\minus{}p)^2\plus{}q}\plus{}...\plus{} \frac {a_n\minus{}1}{(a_n\minus{}p)^2\plus{}q} \le 0$.

In the case $ p\equal{}\frac n{2(n\minus{}1)}$, equality holds for $ a_1\equal{}a_2\equal{}...\equal{}a_n$, and again for $ a_1\equal{}0$ and $ a_2\equal{}...\equal{}a_n$, or any cyclic permutation.

Yes, it can be rewrite:
$ \sum\ \frac{(2b\plus{}2c\minus{}a)^2}{15a^2\plus{}(2b\plus{}2c\minus{}a)^2} \geq\ \frac{9}{8}$
Using Cauchy Schwarts:
$ LHS \geq\ \frac{[\sum\ (2b\plus{}2c\minus{}a)^2]^2}{\sum\ (2b\plus{}2c\minus{}a)^2[15a^2\plus{}(2b\plus{}2c\minus{}a)^2]} \geq\ \frac{9}{8}$
(obviously with Schur).
It also canhang's proof for my similar problem:
$ \sum\ \frac{a^2}{2a^2\plus{}(b\plus{}c\minus{}a)^2} \leq\ 1$
Of course, with my problem, we can prove easily with Am-Gm  :)
Notice that with 2 ineqs above, equality occurs when $ a\equal{}b\equal{}c$ or $ a\equal{}b;c\equal{}0$

[quote="Vasc"][quote="can_hang2007"]Let $ a,b,c \ge 0$ such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \frac {a^2}{2a^2 \minus{} 3a \plus{} 3} \plus{} \frac {b^2}{2b^2 \minus{} 3b \plus{} 3} \plus{} \frac {c^2}{2c^2 \minus{} 3c \plus{} 3} \le \frac {3}{2}
\]
:)[/quote]
I think the following more general statement holds:

Let $ a_1,a_2,...a_n$ be nonnegative real numbers such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} n$. If $ p\le \frac n{2(n \minus{} 1)}$ and $ q\ge 0$, then

$ \frac {a_1 \minus{} 1}{(a_1 \minus{} p)^2 \plus{} q} \plus{} \frac {a_2 \minus{} 1}{(a_2 \minus{} p)^2 \plus{} q} \plus{} ... \plus{} \frac {a_n \minus{} 1}{(a_n \minus{} p)^2 \plus{} q} \le 0$.

In the case $ p \equal{} \frac n{2(n \minus{} 1)}$, equality holds for $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n$, and again for $ a_1 \equal{} 0$ and $ a_2 \equal{} ... \equal{} a_n$, or any cyclic permutation.[/quote]
Nice, VasC. Let me tell about the source of my inequality. Indeed, it is just my inspiration from your inequality: If $ a,b,c \ge 0,a\plus{}b\plus{}c\equal{}3$, then
\[ a^{\frac{3}{2a}}\plus{}b^{\frac{3}{2b}}\plus{}c^{\frac{3}{2c}} \le 3 \qquad (1)\]
In case $ a,b,c$ are the side lengths of a triangle, we can use Bernoulli's Inequality:
\[ \left( \frac{1}{a}\right)^{\frac{3}{2a}} \ge 1\plus{}\frac{3}{2a}\left( \frac{1}{a} \minus{}1\right) \equal{}\frac{2a^2\minus{}3a\plus{}3}{2a^2}\]
or
\[ a^{\frac{3}{2a}} \le \frac{2a^2}{2a^2\minus{}3a\plus{}3}\]
Thus, we only need to prove
\[ \frac{a^2}{2a^2\minus{}3a\plus{}3} \plus{}\frac{b^2}{2b^2\minus{}3b\plus{}3} \plus{}\frac{c^2}{2c^2\minus{}3c\plus{}3} \le \frac{3}{2} \qquad (2)\]
After working on (2), I see that it holds for all $ a,b,c \ge 0$. That's how I came with this inequality.  :blush:  :)

@nguoivn: Em co nghi den cach khac ko dung Cauchy Schwarz ko? Vi thuc su anh post bai nay de gioi thieu ky thuat khac. :)

Nice your idea Can_Hang. :lol:

[quote="Vasc"] Let $ a_1,a_2,...a_n$ be nonnegative real numbers such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} n$. If $ p\le \frac n{2(n \minus{} 1)}$ and $ q\ge 0$, then

$ \frac {a_1 \minus{} 1}{(a_1 \minus{} p)^2 \plus{} q} \plus{} \frac {a_2 \minus{} 1}{(a_2 \minus{} p)^2 \plus{} q} \plus{} ... \plus{} \frac {a_n \minus{} 1}{(a_n \minus{} p)^2 \plus{} q} \le 0$.

In the case $ p \equal{} \frac n{2(n \minus{} 1)}$, equality holds for $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n$, and again for $ a_1 \equal{} 0$ and $ a_2 \equal{} ... \equal{} a_n$, or any cyclic permutation.[/quote]
Actually, the condition $ q\ge \frac {n(3n \minus{} 4)}{4(n \minus{} 1)^2}$ is needed, too. We can reformulate this statement as follows:

Let $ a_1,a_2,...a_n \ge \frac { \minus{} n}{n \minus{} 2}$ such that $ a_1 \plus{} a_2 \plus{} ... \plus{} a_n \equal{} n$. If $ r\ge \frac {n(3n \minus{} 4)}{(n \minus{} 2)^2}$ , then

$ \frac {1 \minus{} a_1 }{r \plus{} a_1^2} \plus{} \frac {1 \minus{} a_2 }{r \plus{} a_2^2} \plus{} ... \plus{} \frac {1 \minus{} a_n }{r \plus{} a_n^2} \ge 0$.

@Can_hang: Ko biet y' tuong? nay` co' dc ko anh :
Assume that $ c \equal{}$ min$ {a, b, c}$.  => $ 3\geq\ a \plus{} b \geq\ 2 ; \frac {9}{4} \geq\ ab \geq\ 0.$
We can prove:
$ \frac {a^2}{2a^2 \minus{} 3a \plus{} 3} \plus{} \frac {b^2}{2b^2 \minus{} 3b \plus{} 3} \leq\ \frac {a \plus{} b}{2}$  (let $ a \plus{} b \equal{} S$; $ ab \equal{} P$...)
We only need to prove:
$ \frac {a \plus{} b}{2} \plus{} \frac {c^2}{2c^2 \minus{} 3c \plus{} 3} \le \frac {3}{2}$
<=> $ \frac {c(c \minus{} 1)(2c \minus{} 3)}{2c^2 \minus{} 3c \plus{} 3} \geq\ 0$ (always trues with $ c\equal{}$ min $ {a, b, c}$.)
 :maybe:

[quote="nguoivn"]@Can_hang: Ko biet y' tuong? nay` co' dc ko anh :
Assume that $ c \equal{}$ min$ {a, b, c}$.  => $ 3\geq\ a \plus{} b \geq\ 2 ; \frac {9}{4} \geq\ ab \geq\ 0.$
We can prove:
$ \frac {a^2}{2a^2 \minus{} 3a \plus{} 3} \plus{} \frac {b^2}{2b^2 \minus{} 3b \plus{} 3} \leq\ \frac {a \plus{} b}{2}$  (let $ a \plus{} b \equal{} S$; $ ab \equal{} P$...)
[/quote]
This inequality does not hold, nguoivn. Try $ a\equal{}1.1,b\equal{}c\equal{}0.95$ ;)


trigonometry
ratio
number theory
relatively prime
AMC

Triangle $ABC$ has a right angle at $C,$ $AC = 3$ and $BC = 4$. Triangle $ABD$ has a right angle at $A$ and $AD = 12$. Points $C$ and $D$ are on opposite sides of $\overline{AB}$. The line through $D$ parallel to $\overline{AC}$ meets $\overline{CB}$ extended at $E$. If

\[ \frac{DE}{DB} = \frac{m}{n} \]

where $m$ and $n$ are relatively prime positive integers, then $m+n =$

$\text{(A)} \ 25 \qquad \text{(B)} \ 128 \qquad \text{(C)} \ 153 \qquad \text{(D)} \ 243 \qquad \text{(E)} \ 256$

edit: better picture

first use Pythagorean to find all lengths (the 5 and the 13).
draw $AG$ parallel to $CE$ and $BF$ parallel to $AC$ and $DE$.  Then we see $GE = 3$ and $AF = 4$ because of the rectangles.  Label $DG = x$ and $FG = y$ then we get $x^2 + (y+4)^2 = 12^2$ and $(x+3)^2 + y^2 = 13^2$ from the right triangles.  At this point bust out your TI and graph the two equation in terms of $y$ (ignore the negative radical since lengths must be positive).  Then you get $x = 9.6$ and $y = 3.2$.  so bust out your TI again and find $(9.6+3)/13$ in fraction and adding top and bottom we get $128$.  answer is $B$.

[quote="K81o7"]If you say ${\angle{ABC}=\theta}$ and extend $CA$ past $A$, then the angle there is also $\theta$, and then really it's just ratios and pythagorean theorem.[/quote]

Now that I think of it, that's probably better because there are no systems of equations of powers of two.  But with TI that really is no problem at all...


Water weighs 62.4 pounds per cubic foot. What is the weight of the water in a full, right, cylindrical tank with a 7-foot diameter and height of 2 feet? Use 22/7 as an approximation for π. Express your answer to the nearest whole number

$ \mathrm{weight} \equal{} \left (\frac{22}{7}  \right )\left (\frac{7}{2}  \right )^{2}2 \hspace{1mm}\mathrm{ft}^{3}\times \frac{62.4 \hspace{1mm} \mathrm{pounds}}{\mathrm{ft}^{3}}\equal{}\boxed{4805 \hspace{1mm} \mathrm{pounds}}$


Solve the equation

x*((y')^2+1)=y^4

Thanx

hello, have someone solved this equation?
Sonnhard.


Γεια σε ολους. μια καλη (θεωρημα): αν p πρωτος της μορφης 4κ+1 να δειξετε οτι υπαρχουν 0<a<b ωστε p=a^2+b^2 οι οποιοι ειναι και μοναδικοι.        :jump:            και γενικα σαν δευτερη ασκηση οτι αν p ειναι ενας ακεραιος >0 ωστε να μπορει να γραφει ως αθροισμα δυο τετραγωνων με δυο διαφορετικους τροπους πχ    p=a^2+b^2=c^2+d^2 ,  τοτε p = συνθετος  :ewpu:

Όσον αφορά την πρώτη άσκηση, είναι γνωστή ως θεώρημα του Fermat (και αυτή...  :wink:  ) και αποδεικνύεται με αρκετούς τρόπους. Αν κάποιος τώρα θέλει να γράψω μια απόδειξη,ας μου το πει.  :wink:

Η δεύτερη προκύπτει απευθείας από το θεώρημα,αφού αν $p$ πρώτος της μορφής $4k+1$ τότε γράφεται με μοναδικό τρόπο ως άθροισμα δύο τετραγώνων,ενώ αν είναι της μορφής $4k+3$ τότε απλά δεν γράφεται ως άθροισμα δύο τετραγώνων  :P .   Αν τώρα $p=2$ τότε προφανώς ο $p$ δεν γράφεται με δύο διαφορετικούς τρόπους ως άθροισμα δύο τετραγώνων. Άρα, ο $p$ είναι σύνθετος.

Y.Γ. Καλύτερα να μην μου πει κάποιος ότι θέλει απόδειξη στο πρώτο θεώρημα γιατί είναι μεγάλη και βαριέμαι...  :P   :lol: .

Stelios...


function
topology
real analysis
real analysis unsolved

Can anyone help me with this problem?

Let $S$ be a subset of the metric space $E$ with the property that each point of $S^{c}$ (complement of S) is a cluster point of $S$. Let $E'$ be a complete metric space and $f: S \to E'$ a uniformly continuous function. Prove that $f$ can be extended to a continuous function from $E$ into $E'$ in one and only one way, and that this extended function is also uniformly continuous.

I don't get the initial definition: how can some cluster point of $S$ also be in $S^{c}$?

Can anyone give some pointers?

[quote="teacupontheridge"] I don't get the initial definition: how can some cluster point of $S$ also be in $S^{c}$?
[/quote]
Let $(\mathbb{R}, d)$ be a metric space with $d$ as the usual distance function, i.e. if $x, y \in \mathbb{R},$ then $d(x,y) = |x-y|.$ Now, let $S =$the set of all rational numbers $\subset \mathbb{R}.$ Then, $S^{c}=$ the set of all irrationals. Clearly, each point of $S^{c}$ is a cluster point of $S.$

More importantly, a cluster point of $S$ need not necessarily be in $S.$

Thanks, but can anyone give pointers for showing that $f$ can be extended to a continuous function from $E$ into $E'$ in one and only one way?

hmmm... i think there's a thm as such:  $f: X\to Y$ is continuous at $x$ if and only if for every sequence $\{x_{n}\}$ of $X$ such that $x_{n}\to x$, $f(x_{n})\to f(x)$.

this should help q a bit.


function
limit
real analysis
real analysis unsolved

find all function $f:R->R$ such that:
for every real series $\sum u_n$, if $\sum u_n$ converges then $\sum f(u_n)$ converges

This has been posted before. Only functions $f(x)$ which are locally equal to $ax$ near 0 for some $a$ work.

I remember I posted it a long time ago but I've never seen a proof for it ..

But if you have a proof for it, then you must be happy, isn't it? Here are the main steps:
  1) Prove that it is odd on a neighbourhood of 0. If not, consider a sequence of the form $ a_0, -a_0, a_0, -a_0,...,a_1,-a_1,a_1,-a_1,...$ where $a_n$ is decreasing to 0 and verifies $ |f(a_n)+f(-a_n)|>0$ and take each group $a_i,-a_i,a_i,-a_i,...$ such that $a_i, -a_i$ appears $N_i$ times and $N_i |f(a_i)+f(-a_i)|>1$.
 2) Prove that f is continuous in 0. This is trivial.
  3) Prove that f is additive on a neighbourhood of 0. If not, consider sequences of the form $ a_0+b_0,-a_0,-b_0,a_0+b_0,-a_0,-b_0,... $ of the same type as in 1), use the result in 1) and conclude.

harazi ?

Why is the continuity in 0 trivial ?

Okay, I found out for the continuity.

I have additiveness on a neighboorhood of zero and continuity in zero.

The only thing that I miss is to prove that f is continuous on some neighboorhood of zero, and then it's over.

I know there is a theorem which states that if f is additive on R and contiuous in one point, then it is linear on all R, but a can't figure out how to modify this theorem so that it works on the neighborhood...

If $x$ is in the neighborhood, then $f_x(t)=f(x+t)=f(x)+f(t)$ is continuous at $t=0$, so $f$ is continuous at $x$.

Actually, even if it is ``trivial'', I do have a problem for the continuity in zero  :blush:  :mad:.

Could anyone give a bit more details ?

If $f$ is not continuous in 0 you can find a decreasing sequence $a_n$ which tends to 0 such that $|f(a_n)-f(0)|\geq \epsilon$.But since $f(0)=0$ you have $|f(a_n)|\geq \epsilon$.Now choose the serie with the following terms:$a_1,-a_2,....,a_{2n-1},-a_{2n},...$and conclude.

Sorry in advance for the stupid question:
[quote="jmerry"]Only functions $f(x)$ which are locally equal to $ax$ near 0 for some $a$ work.[/quote]
Do you mean $\lim_{x \rightarrow 0} \frac{f ( x )}{x} = a$ ?

no, it means here that there exists some $\epsilon > 0$ such that $f(x)=ax$ on $]- \epsilon; + \epsilon[$

$\epsilon$-neighborhood ;) ? Ok, right ! Thanks.


ratio MATHCOUNTS AMC AIME USA(J)MO USAMO geometry

What's your dumbest stupid mistake?

Mine would have to be getting the correct answer but changing the answer when I transferred it from my scrap paper to my answer sheet.

My most dumbest stupid mistake (that i can remember) was question 10 the amc 10a. i still can't believe i got the wrong answer!!!.  :mad:

my dumbest mistake happened in a previous competition back when I was in middle school.

I bubbled in the right answer for the wrong problem! :blush:

Chapter CountDown Last Year

Final Round

Sudden Victory

Question: 5-x=8

Answer: x= -8

Last year my friends and I were quizzing each other before State Mathcounts.
Chris asks me what the ratio of the shortest diagonal in a regular hexagon to the longest diagonal is.
I automatically say sqrt3/2.
In Sudden Victory in the countdown the question is what is the ratio of the shortest diagonal in a regular hexagon to the longest diagonal.
I buzz in and say sqrt3/3.

Ouch JMI, sorry about that. I didn't have the luxury of going to a school that offered MathCounts and hadn't heard about it until 5 months ago.

The question asks for A:B, I gave B:A, where A and B are correct.

I forgot the 3s in the expansion of (a+b)^3.

At my school, we have someone who makes stupid mistakes all the time. We now call the act of making such mistakes "pulling a (person's name here)".

[quote="mcalderbank"]On a similar note, I said on a number of occasions 16*5 = 90[/quote]

Wow, I do that all the time..

Once in a test I found one solution to an equation, went on to prove there were no more by assuming that x/x = 0, and concluded by saying there was no solutions at all..

Or when I publicly said on this forum that some particular number ending with 5 was a prime..

Hmm...does state MC count?

Number 1...10 - 2 = 6

Went from like 3rd to 10th place...

AHSME 1999 #4 (paraphrased)
Find the sum of all primes less than 100 which leave a remainder of 1 when divided by 4 and a remainder of 4 when divided by 5.

My thought process: Okay, that's 9 mod 20. 9 is not prime, 29 is prime, 49 is not prime, 69 is prime, 89 is prime. 29+69+89=187. That answer was available.

I checked it later, and went through the same thoughts to verify my answer.

I got the other 29 questions right.

24/3=3.  Gah!

And 27-8=15.

Both mistakes cost me dearly.  :(

ok well there's probably a stupider one, but the previous one is the last stupid mistake i remember... :blush:  :D  ;)

MathCounts.  When I read my answer out loud at countdown, I had the right answer on paper, but read the wrong thing

In 8th grade:
21 --> 4 because the bottom line on the 2 went too far and crossed the 1, then on the next line I wrote a 4.

More recently: on my calculus tests, there are calculator and non-calculator parts. On the calculator part, one problem asked to find a definite integral of the function ln(xe^x) (before we had learned how to integrate ln(u)), and i spent about 10 minutes trying to find the antiderivative bfore giving up. That was the only problem on the test I got wrong.

Trying to eat a whole jar of pickles in one day...not really

Saying that 91 was the largest integer prime less than 100 :( forgot 97, and $7*13$ :rotfl:

saying that pi/6 rad is 60 degrees (ok its not that bad)

but it was a costly mistake, on this year's AIME I

School math test (thankfully not state mc):

6+2=7

:cursing::wallbash:

My best one (on AIME, though I caught it): 0*n=n

Got 20th instead of 8th because of:

I thought an angle equaled 49 when it was complementary to it (misread)

Countdown round:
Not paying attention and saying the same wrong answer as the one my opponent said 20 seconds earlier  :oops: . Didn't even know until my dad told me, which was after the awards ceremony.

Luckily, it didn't cost me much.

3*3 = 6, I argued with someone for 2 minutes over that once  :blush:

i think i put that 0.75=25% in a school math test  :blush:

my dumbest mistake is when i do my homework fast and misread a $\times$ sign for a $+$ sign

last geometry test-I did 16*4=16 for like 10 problems.

My dumbest mistake as of now would be tied between misbubbling two problems(got them right, bubbled them wrong), reading a 90 element subset as a three element one (wonder how i pulled that one off), and thinking .012>.123, all of which were made on the AIME this year, so my score right now is just below the predicted cutoff for USAMO.


inequalities
trigonometry
inequalities proposed

[quote="Virgil Nicula"]Let $ABC$ be a triangle. Define $2p=a+b+c$, the circumradius $R$, the inradius $r$,
the A-exinradius $r_a$ and the lengths $m_a$, $l_a$, $h_a$ of the median, angle-bisector and altitude respectively from the vertex $A$.

\begin{eqnarray} & \Longrightarrow & \left(A-90^{\circ}\right)\cdot\left[m_a-R(1+\cos a)\right]\ge 0\\\ & \Longrightarrow & \frac{a(p-a)}{\sqrt{2Rr}}\le \frac{b+c}{2}\cos \frac A2\le m_a\\\ & \Longrightarrow & \sqrt{p(p-a)}\le m_a\ \ \wedge\ \ \frac{b^2+c^2}{4R}\le m_a\\ & \Longrightarrow & \sqrt{p(p-a)}\le \frac{a(p-a)}{\sqrt{2Rr}}\Longleftrightarrow (a-b)(a-c)\le 0\\ & \Longrightarrow & \boxed {(4R+r)\sqrt{\frac{2r}{R}}\le m_a+m_b+m_c}\le 4R+r\\ & \Longrightarrow & 3\le \boxed {\frac 12\cdot \sum \frac{b+c}{a}\le \sum \frac{m_a}{h_a}}\le 1+\frac Rr\\ & \Longrightarrow & 3\le \boxed {\frac{a^3+b^3+c^3}{abc}\le \sum \frac{m_a}{r_a}}\\\ & \Longrightarrow & \boxed {3r\le \frac{am_a+bm_b+cm_c}{a+b+c}}\le R+r\\ & \Longrightarrow & \frac{a+b+c}{2R}\le \boxed {\frac{3}{2R}\cdot \frac{a^2+b^2+c^2}{a+b+c}\le \frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}}\\ & \Longrightarrow & \frac pR\le \boxed {\frac{p}{\sqrt{2Rr}}\le \frac{m_a}{a}+\frac{m_b}{b}+\frac{m_c}{c}} \end{eqnarray}[/quote][b][u]Remark 1.[/u][/b] The  non-enclosed inequalities are evidently or well-known and I shall prove them further.

[b][u]Remark 2.[/u][/b] Prove easily that $\frac{a(p-a)}{\sqrt{2Rr}}=(1+\cos A)\sqrt{2Rr}\le m_a$.

[b][u]Remark 3.[/u][/b] We can prove the inequality $(0)$ with two remarkable identities :
$1^{\circ} \blacktriangleright\ \boxed {(p-b)(p-c)(b-c)^2+p(p-a)(b+c-2a)^2=4\left[bcm^2_a-2ap(p-a)^2\right]}\Longrightarrow$
$\boxed {m_a\ge\frac{a(p-a)}{\sqrt{2Rr}}}$ (with equality if and only if $a=b=c$).

$2^{\circ} \blacktriangleright\ \boxed {16S^2+\left(b^2-c^2\right)^2=4a^2m^2_a}\Longleftrightarrow a^2(4Rm_a)^2=4(4RS)^2+4R^2\left(b^2-c^2\right)^2\Longleftrightarrow$
$a^2(4Rm_a)^2=4(abc)^2+4R^2\left(b^2-c^2\right)^2\Longleftrightarrow (4Rm_a)^2=4b^2c^2+4R^2\left(\frac{b^2-c^2}{a}\right)^2\Longleftrightarrow$
$(4Rm_a)^2=\left(b^2+c^2\right)^2-\left(b^2-c^2\right)^2+4R^2\left(\frac{b^2-c^2}{a}\right)^2\Longleftrightarrow$
$(4aRm_a)^2=\left[a\left(b^2+c^2\right)\right]^2+\left(4R^2-a^2\right)\left(b^2-c^2\right)^2\Longrightarrow \boxed {4Rm_a\ge b^2+c^2}$

(with equality if and only if $b=c\ \vee \ A=90^{\circ}$).

$3^{\circ} \blacktriangleright$ From the identity $(1^{\circ})$ we obtain $4bcm^2_a\ge 8ap(p-a)^2+p(p-a)(b+c-2a)^2\Longleftrightarrow$

$\frac{4bcm^2_a}{p(p-a)}\ge (b+c)^2\Longleftrightarrow \boxed {m_a\ge\frac{b+c}{2}\cos \frac A2}$ (with equality if and only if $b=c$).

[b][u]Remark 4.[/u][/b] $\boxed {\frac{a(p-a)}{\sqrt{2Rr}}\le \frac{b+c}{2}\cos \frac A2}\Longleftrightarrow 2a(p-a)\le (b+c)\sqrt{2Rr}\cos \frac A2\Longleftrightarrow$

$2a(p-a)\le (b+c)\sqrt{\frac{2Rrp(p-a)}{bc}}\Longleftrightarrow 2a(p-a)\ge (b+c)\sqrt{\frac{2Rpra(p-a)}{abc}}\Longleftrightarrow$

$\Longleftrightarrow 2a(p-a)\le (b+c)\sqrt{\frac{a(p-a)}{2}}\Longleftrightarrow$ $2\sqrt{a(b+c-a)}\le b+c\Longleftrightarrow (b+c-2a)^2\ge 0$

(with equality if and only if $b+c=2a$).

[b][u]Remark 5.[/u][/b] $\boxed {1\ge\cos \frac{B-C}{2}\ge \sqrt{\frac{2r}{R}}}\Longleftrightarrow 2\sin \frac{B+C}{2}\cos \frac{B-C}{2}\ge 2\sqrt{\frac{2r}{R}}\cos \frac A2\Longleftrightarrow$

$\sin B+\sin C\ge \sqrt{\frac{8rp(p-a)}{Rbc}}\Longleftrightarrow b+c\ge \sqrt{\frac{32Rpr(p-a)}{bc}}\Longleftrightarrow$

$b+c\ge \sqrt{\frac{32Rpra(p-a)}{4Rpr}}\Longleftrightarrow b+c\ge 2\sqrt{a(b+c-a)}\Longleftrightarrow (b+c-2a)^2\ge 0\ .$

(with equality if and only if $b+c=2a$).

[b]Another (geometrical) proof.[/b] Denote $w=C(O,R)$, $A_1\in (AI\cap w$, $A_2\in (A_1O\cap w$, i.e. $A_1A_2=2R$, $A_2A\perp A_1A$. Therefore, $\cos \frac{B-C}{2}=$ $\frac{A_1A}{A_1A_2}=$ $\frac{IA+IA_1}{2R}\ge$ $\frac{2\sqrt{IA\cdot IA_1}}{2R}=$ $\frac{\sqrt{2Rr}}{R}=$ ${\sqrt{\frac{2r}{R}}\ .}$

Ineq:
$\Large \frac{1}{2}\sum \frac{b+c}{a}\le \sum \frac{m_a}{h_a}$ is real hardy.
Can you post your solution,thank :P .

Maybe later. The all above non-enclosed inequalities have a common spring: the  primitive inequality $(0)$.

is it true that
$\sum \frac{1}{\sqrt{p-a}} \sum \sqrt{p-a} \geq \sum \frac{1}{a} \sum a$
because if it is we have a proof of three by using $m_a \geq \sqrt{p(p-a)}$


There might be some sections in the forum where this is allowed

Also, that was almost three years ago. Mathlinks was different then.

[quote="shobber"][quote="José"]Looking at some quotes about IMO 2003, I see guests posting....... how's that?[/quote]
What's wrong with that? Why can't guests have some place to post?[/quote]

Nothing wrong. But if you want to post, why don't you create an account?????

And why in 2003 you could and now you can not?

[quote="amcavoy"]Maybe it was like that so people could post without going through the account-making process and then people started to spam too much so they changed the rule.[/quote]

Yes, that would be a good reason then


geometry
3D geometry
summer program
MathPath

[quote="espark52"]a 4x4x4 in cube originally built from 1x1x1 in cubes is cut into exactly 29 cubes with integer edge leghths and no material left over. how any of the 29 cubes are 2x2x2 cubes?[/quote]
[hide="hint"]systems[/hide]

[hide]Um....... I think the answer is 5. [/hide]

[hide]Let $x$ be the number of 1x1x1 cubes and $y$ be the number of 2x2x2 cubes. Then we have

$x+y=29$
$x+8y=64$

This gives $y=5$ and $x=24$, so the answer is $5$.[/hide]

I figured out a solution for the 4x4x4 all by myself at MathPath!!!!

Of course, it only worked about 1/4 of the time cuz of the parity case which I couldn't figure out...

Another way to think about it is this:
[hide]If you had all 2x2x2 cubes, you'd have 8 cubes right?
So then for every 2x2x2 cube that you take off, you make 8 1x1x1 cubes so you add 7 every time. Start with 8 cubes (8 2x2x2's), then 15 cubes (7 2x2x2's), 22 cubes (6 2x2x2's) and finally 29 cubes (5 2x2x2's and 24 1x1x1's)[/hide]


Search here: http://mirror.hmc.edu/ctan/info/symbols/comprehensive/symbols-letter.pdf

If you can't find it here, it probably doesn't exist.

With the fancybox package, you can type something like \shadowbox{hello}.  I don't know if that's what you were looking for.


inequalities
LaTeX
inequalities unsolved

[quote="long14893"]prove that for every x1,x2,...,xn are positive real , we have:
$ \sum cos^2(x_i \minus{} x_j) \geq \frac {n(n \minus{} 2)}{4}$
plz help :([/quote]
Hello....no one's gonna help me ??? :((

could you explain the question properly?  :)

and use late3x preferably
for subscripts use eg: u_i in dollar signs = $ u_i$

sr ... but I think t used Latex :-/ ?
about my question....what makes u hard to understand it :-/?
For example , if n=3 , we need to prove:
$ cos^2(x_1\minus{}x_2) \plus{} cos^2(x_2\minus{}x_3) \plus{} cos^2(x_3\minus{}x_1)^2 \geq 3/4$
plz help me :| i really need the solution of this question >"<

[quote="long14893"][quote="HTA"][quote="long14893"]prove that for every x1,x2,...,xn are positive real , we have:
$ \sum cos^2(x_i \minus{} x_j) \geq \frac {n(n \minus{} 2)}{4}$
plz help :([/quote]
 It should be

$ \sum_{1\leq i < j \leq n } cos^2(x_j \minus{} x_i) \geq \frac {n(n \minus{} 2)}{4}$ instead of $ \sum cos^2(x_i \minus{} x_j) \geq \frac {n(n \minus{} 2)}{4}$  :)[/quote]
yes ... that's exactly what I wanted to say .... cuz I forget how to use Latex to write like that :D

[hide] làm ra chưa ... nản quá :(( tưởng nó tốt bụng =)) [/hide][/quote]
ok , here is my solution :
Firstly ,  notice that :
$ \sum_{i,j \equal{} 1}^{n} cos^2(x_i \minus{} x_j) \equal{} 2 \sum_{1 \leq i,j\leq n}^{n}cos^2(x_i \minus{} x_j) \plus{} n$
Hence , we need to prove
$ \sum_{i,j \equal{} 1}^{n} cos^2(x_i \minus{} x_j) \minus{} n \geq \frac {n(n \minus{} 2)}{2}$
$ \sum_{i,j \equal{} 1}^{n}(\frac {1 \plus{} cos(2x_i \minus{} 2x_j)}{2}) \minus{} n \geq \frac {n(n \minus{} 2)}{2}$
$ \sum_{i,j \equal{} 1}^{n}(1 \plus{} cos(2x_i \minus{} 2x_j)) \minus{} 2n \geq n^2 \minus{} 2n$
$ \sum_{i,j \equal{} 1}^{n}cos(2x_i \minus{} 2x_j) \geq 0$
$ (\sum_{i \equal{} 1}^{n}cos2x_i)^2 \plus{} (\sum_{j \equal{} 1}^{n}sin2x_j)^2 \geq 0$
which is true :)


pigeonhole principle
combinatorics unsolved
combinatorics

Let $ A$ be a subset of $ \{ 0,1,2,...,1997 \}$ containing more than $ 1000$ elements. Prove that either $ A$ contains a power of $ 2$ (that is, a number of the form $ 2^k$ with $ k\equal{}0,1,2,...)$ or there exist two distinct elements $ a,b \in A$ such that $ a\plus{}b$ is a power of $ 2$.

[quote="moldovan"]Let $ A$ be a subset of $ \{ 0,1,2,...,1997 \}$ containing more than $ 1000$ elements. Prove that either $ A$ contains a power of $ 2$ (that is, a number of the form $ 2^k$ with $ k \equal{} 0,1,2,...)$ or there exist two distinct elements $ a,b \in A$ such that $ a \plus{} b$ is a power of $ 2$.[/quote]

Assume that we can find a counterexample $ A$ , which should be a subset of $ \{0,1,2,...,1997\}$ \ $ \{2,4,8,...,1024\}$

Then divide $ \{0,1,2,...,1997\}$ \ $ \{2,4,8,...,1024\}$ into $ \{1997,51\}, \{1996,52\}, \{1995,53 \},...,\{1025,1023 \},$
$ \{50,14\}, \{49,15\}, \{48\}, \{47,17\}, ..., \{33,31\},$
$ \{13,3\}, \{12\},\{11,5\},...,\{9,7\}, \{1\}, \{0\}$ 998 groups.

So by pigeonhole principle, we can find two distinct elements of $ A$ in one group and so sum of the two elements is a power of $ 2$.


Show that

$ n\left(1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}\right) \geq \left(n \plus{} 1\right)\left(\frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n \plus{} 1}\right)$

for all natural numbers  $ \,\,n \geq 1$

See i have made progress, i just dont know how to finnish it off. This is what i did.

$ n \plus{} n\left( \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}\right) \geq \left(n \plus{} 1\right)\left(\frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}\right) \plus{} \frac {n \plus{} 1}{n \plus{} 1}$

$ n \plus{} n\left( \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}\right) \geq n\left(\frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}\right) \plus{} 1\left(\frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}\right) \plus{} 1$

(Subtracting $ n\left( \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}\right)$ from both sides)

$ n \geq 1 \plus{} \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{n}$

How do i prove this? What do i do to finnish my proof?

To get to:

$ n\geq 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$

You can express $ n$ as the sum of $ n$ one's: $ n=\underbrace{1+\cdots+1}_{n}$.

And for each $ n\ge k\ge 1$ : $ 1\ge \dfrac{1}{k}$,(equality in $ k=1$) you add all of those up and get the result you want.

In other words

$ \sum_{k=1}^n 1 \ge \sum_{k=1}^n \dfrac{1}{k}$

Example: $ n=2$,

\begin{eqnarray*}
1&\ge& 1\\
1 &>& \dfrac{1}{2}
\end{eqnarray*}

$ 1+1\ge 1+\dfrac{1}{2}\implies 2\ge 1+\dfrac{1}{2}$

Since the opposite is not true, the equation you have written is. If $ n \equal{} 2$, then we would need $ 1 \plus{} \frac {1}{2}$ to be greater than or equal to $ 2$. But it is not. Finally, since the integers increase by $ 1$ on the LHS, and on the RHS, the fractions increase by $ \frac {1}{n}$, we are actually trying to prove $ n$ is greater than or equal to $ \frac {1}{n}$ for every natural number $ n$. But this is always true.

I hope you understood my proof.

Thanks.

But if you writing a proof for a contest what would you write to finnish it off?

ie is it satisfactory to state that this is obviously true?

No; prove by induction. $ 1\ge 1/1$; $ 1>1/k$ for all $ k$ (which is what you add to both sides when you go from $ k\minus{}1$ to $ k$), hence the inequality is true.

I'm sorry; I'm not very good at writing proofs. I thought you didn't know how to solve it and that it was a casual question, so I just posted a proof for it.

Another way is induction:

1. It's true for $ n \equal{} 1$.

2. Assume it's true for $ n$
$ n\ge 1 \plus{} \dfrac{1}{2} \plus{} \cdots \plus{} \dfrac{1}{n}$  $ (1)$

3. Prove it's true for $ n \plus{} 1$:

$ n \plus{} 1\ge 1 \plus{} \dfrac{1}{2} \plus{} \cdots \plus{} \dfrac{1}{n} \plus{} \dfrac{1}{n \plus{} 1}$ $ (2)$

Subtract $ 1$ from $ 2$.

$ 1\ge \dfrac{1}{n \plus{} 1}$

$ n \plus{} 1\ge 1$ Which is true for all natural numbers.

Edit: oops didn't read Temperal's post

I considered induction and tried it but i just didnt think properly ;(

Thanks heaps  :lol:

Most of the time induction gets pretty messy!!!!!!!!! :(

[quote="tonypr"]
Subtract $ 1$ from $ 2$.
[/quote]

Nup, you can't do that !!

Fortunately it's fixable. Just add $ 1 \ge \frac{1}{1\plus{}n}$ to $ (1)$ to get $ (2)$


Why A contains at least two elements denoted by red line ?

Thanks

[quote="leshik"]If $ A$ contains just one point(for example $ a$) then our sequence $ x_n$ converges to $ a$ and we are done :)[/quote]

Thanks.

Although I might had thought of it , but I didn't.

When read books of math , we need to discuss and trigger off mutually.

in (i), why $ f(\ell) \neq \ell$ ? I see. if $ f(\ell) \equal{} \ell$, we will get contradition.

Then in (ii) , why $ [\ell_{\_}\plus{}f(\ell_{\_})]/2\in A$ ?


inequalities
AMC
USA(J)MO
USAMO
inequalities proposed

this is a very famous inequality from USAMO 1971 (I believe?)

let the LHS be f(a,b,c,d,e). Since it is convex in a,b,c,d,e, the maximum occurs on the boundary, that is, when $ a,b,c,d,e \in \{p,q\}$. Testing these cases yields the desired result.

EDIT: oops, usamo 1977.

[url]http://web.archive.org/web/20040308184647/www.kalva.demon.co.uk/usa/usa77.html[/url]


There are 30 people in a senate. Each pair of senatorsare either friends or enemiesof each other. Each senator has 6 enemies. Every three senators form a committee. Find the total number of committees whose members are either all friends or enemies of each other

Every triple of senators falls into one categories: they are either "good," if all three are friends or if all three are enemies, or they are "bad" if some pair are friends and some other pair are enemies.  For every "bad" triple, we have exactly two ordered triples of the form $(x, y, z)$ where $x, y$ are friends and $y, z$ are enemies.  For each "good" triple, we have no such ordered triples.  Thus the number of such ordered triples is exactly twice the number of "bad" triples.

How many triples are there?  We have 30 choices for $y$, and after fixing $y$ we have $6$ choices for $z$ and 23 choices for $x$.  Thus we have $30\cdot6\cdot 23 = 4140$ ordered triples and so 2070 "bad" triples.  The total number of triples of senators is $\binom{30}{3}= 4060$, and since any triple is either good or bad, there are $4060-2070 = 1990$ good triples.

More generally, for $n+1$ senators, each with $m$ friends, the number of triples which are the vertices of a $K_{3}$ in the friendship graph or its complement is $\binom{n+1}{3}-\frac{1}{2}(n+1)\cdot m \cdot (n-m)$.

In the case that $n$ is even and $m$ is odd, this gives a half-integer, which signifies the fact that no such graph exists in this circumstance.


For questions 1 and 2 below, find the average rate of change of f between 1 and x, f(x) - f(1)/(x - 1), where x does NOT = 0 for each function f.  Be sure to simplify.

(1) f(x) = -2x

(2) f(x) = sqrt{x + 3}

[quote="Interval"]For questions 1 and 2 below, find the average rate of change of f between 1 and x, f(x) - f(1)/(x - 1), where x does NOT = 0 for each function f.  Be sure to simplify.

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

(2) $f(x) = \sqrt{x+3}$[/quote]

This should probably be in the HSB forum.

All you need to do is plug in the function into the rate of change formula you've given, and simplify it.

(#1) $\frac{ (-2x)-(-2[1])}{x-1}= \frac{-2x+2}{x-1}= \frac{-2(x-1)}{x-1}=-2$, for $x \neq 1$ since that would make the fraction undefined.

(#2)  $\frac{ \sqrt{x+3}-\sqrt{ (1)+3}}{x-1}= \frac{ \sqrt{x-3}-2}{x-1}$, and you can't really simplify that any further.


Find all pair of prime numbers $ p,q$ such that: $ p^2\minus{}p\plus{}1\equal{}q^3$

$ p(p \minus{} 1) \equal{} (q \minus{} 1)(q^2 \plus{} q \plus{} 1)$ ,
$ i)$ $ q\equiv1(modp)$ $ \Longrightarrow$  $ q \minus{} 1 \equal{} kp$ and $ q^2 \plus{} q \plus{} 1 \equal{} \frac {p \minus{} 1}{k}$ but $ \frac {p \minus{} 1}{k} < pk$  so $ q^2 \plus{} q \plus{} 1 < q \minus{} 1$ is a contradiction.
$ ii)$ $ q^2 \plus{} q \plus{} 1\equiv0(modp)$ $ \Longrightarrow$ $ q^2 \plus{} q \plus{} 1 \equal{} k.p$ and $ q \minus{} 1 \equal{} \frac {p \minus{} 1}{k}$ $ \Longrightarrow$ $ p^2 \minus{} (k^3 \minus{} 3k \plus{} 2)p \plus{} 3k^2 \minus{} 3k \plus{} 1 \equal{} 0$ hence $ (k^3 \minus{} 3k \plus{} 2)^2 \minus{} 4(3k^2 \minus{} 3k \plus{} 1) \equal{} t^2$ where $ t$ is a positive integer. $ k^4 \minus{} 6k^2 \plus{} 4k \minus{} 3 \equal{} m^2$ ,       $ (m \equal{} \frac {t}{k})$, if $ m > 3$ so $ (k^2 \minus{} 3)^2 < k^4 \minus{} 6k^2 \plus{} 4k \minus{} 3 < (k^2 \minus{} 2)^2$ is a contradiction. If $ k \equal{} 1$ or $ k \equal{} 2$ so $ m^2 < 0$ , $ k \equal{} 3$ $ \Longrightarrow$ $ m \equal{} 6$ and $ p \equal{} 19$ , $ q \equal{} 17$ is only solution.
$ Remark$: If $ q$ is not a prime number, the problem is also true($ p,q >1$).The only solution is $ p \equal{} 19$, $ q \equal{} 17$.

I think $ p\equal{}19$ and $ q\equal{}7$  : :wink:


trigonometry
inequalities
inequalities unsolved

[quote="tkhtn"]$cos^2(A)+cos^2(B)+cos^2(C)=-1$?  :rotfl:[/quote]
there may be a small mistake,I think nttu may mean $cos^2(A)+cos^2(B)+cos^2(C)+2\cos A \cos B \cos C=1$
 :)

I think the first inequality can be solved by Cauchy.If anyone can do it, please post here

To nttu:For the second inequality, can you post all your solution in details?

The second ineq is very old and easy!
Please read ineq agains.

Please solve the first ineq by Cauchy,Bunhiacopki,Schwarz,AM-GM or Chebusev


Let $a_{1},a_{2},...$ be positive numbers such that $a_{n+1}=a_{n}^{2}-2(n=1,2,...)$. Prove that $a_{n}\geq 2$ for all $n\geq 1$.

[hide="complete proof"]
If $a_{i}\geq 2$, then $a_{i}^{2}\geq4$ and $a_{i}^{2}-2 = a_{i+1}\geq 2$.

So, we just have to show that $a_{1}\geq 2$. Suppose $a_{1}< 2$. Then $a_{1}^{2}-a_{1}-2 = (a_{1}-2)(a_{1}-1)$
Suppose $1<a_{1}<2$
Then, $a_{1}^{2}-a_{1}-2 < 0 \implies a_{1}^{2}-2<a_{1}$, so $a$ will decrease until $a_{i}\leq 1$.

Suppose $0 < a_{i}\leq 1$. Then $a_{i}^{2}\leq 1$ and $a_{i}^{2}-1 \leq-1 < 0$. Thus, if $a_{1}< 2$, the sequence will eventually become negative, so $a_{1}\geq 2$ as wanted.
[/hide]

[quote="tjhance"]this is kind of a lame question, but what does it mean by "bounded below 1"?[/quote]
ie. $a_{n}>1\forall n$  :wink:

so basically he was saying that it wasn't proven that $a_{i}$ will ever go below 1?

Because if $a_{i}<1$ then $a_{i+1}<0$, contradiction!

[hide]
If $a_{1}\geq 2$, then $a_{n}$ will always be $\geq 2$ by induction (as $a_{k}\geq 2\Rightarrow a_{k+1}=a_{k}^{2}-2\geq 2$).

Suppose that $a_{1}<2$ but $a_{n}$ is positive for all $n$. It follows by induction that $a_{n}<2$ for all $n$, so we can write
\[a_{n}=2\cos \alpha_{n}\]
where $0<\alpha_{n}<\frac{\pi}{2}$. But we have
\[2\cos \alpha_{n+1}=a_{n+1}=a_{n}^{2}-2=4\cos^{2}\alpha_{n}-2=2\cos 2\alpha_{n}\]
so $\alpha_{n+1}=2\alpha_{n}$. Let $\theta=\alpha_{1}$; then it follows that
\[a_{n}=2\cos 2^{n-1}\theta\]
for all $n$. Since $\theta>0$ and $2^{n-1}$ gets arbitrarily large for large enough $n$, there will be some $M$ such that $2^{M-1}\theta \geq \frac{\pi}{2}$. Consider the least such $M$. Then because $2^{M-2}\theta<\frac{\pi}{2}$, $2^{M-1}\theta <\pi$. It follows that
\[\frac{\pi}{2}\leq 2^{M-1}\theta<\pi\Rightarrow a_{M}\leq 0\]
but this is a contradiction!
[/hide]

Me too  :wink:
[quote="scorpius119"][hide]
If $a_{1}\geq 2$, then $a_{n}$ will always be $\geq 2$ by induction (as $a_{k}\geq 2\Rightarrow a_{k+1}=a_{k}^{2}-2\geq 2$).

Suppose that $a_{1}<2$ but $a_{n}$ is positive for all $n$. It follows by induction that $a_{n}<2$ for all $n$, so we can write
\[a_{n}=2\cos \alpha_{n}\]
where $0<\alpha_{n}<\frac{\pi}{2}$. But we have
\[2\cos \alpha_{n+1}=a_{n+1}=a_{n}^{2}-2=4\cos^{2}\alpha_{n}-2=2\cos 2\alpha_{n}\]
so $\alpha_{n+1}=2\alpha_{n}$. Let $\theta=\alpha_{1}$; then it follows that
\[a_{n}=2\cos 2^{n-1}\theta \]
for all $n$. Since $\theta>0$ and $2^{n-1}$ gets arbitrarily large for large enough $n$, there will be some $M$ such that $2^{M-1}\theta \geq \frac{\pi}{2}$. Consider the least such $M$. Then because $2^{M-2}\theta<\frac{\pi}{2}$, $2^{M-1}\theta <\pi$. It follows that
\[\frac{\pi}{2}\leq 2^{M-1}\theta<\pi\Rightarrow a_{M}\leq 0 \]
but this is a contradiction!
[/hide][/quote]
This is solution of my friends:
If $1<a_{1}<2$, by induction $1<a_{n}<2\forall n\geq 1$. Therefore $a_{n}>a_{n+1}\forall n\geq 1$. Now from above we have $(a_{n})$ convergen. Contradiction! (Because if $L=\lim_{n\to\infty}a_{n}$ then $L=-1$ or $2$).

[quote="JBL"][quote="Hamster1800"]Then, $a_{1}^{2}-a_{1}-2 < 0 \implies a_{1}^{2}-2<a_{1}$, so $a$ will decrease until $a_{i}\leq 1$.[/quote]
"Decreasing" is what you proved.  You didn't prove "decreasing and not bounded below by 1," which is what you used.[/quote]
Right. My bad. Well how about this.

[hide="fixing my proof"]
Let $d_{i}= a_{i+1}-a_{i}$.

Suppose $1 < a_{i}< 2$

$a_{i+1}= a_{i}^{2}-2$

So, $d_{i}= a_{i}^{2}-a_{i}-2$.

(Now I slip into calculus as it's probably the easiest way)

Differentiate $d_{i}$ with regard to $a_{i}$.

$\frac{d}{a_{i}}d_{i}= 2a_{i}-1$.

Thus, as $a_{i}$ decreases to 1, so does the rate at which it decreases.

This works, but I believe we can also do this:

Suppose $a_{i}\leq \sqrt3$. Then $a_{i+1}= a_{i}^{2}-2 \leq 3-2 = 1$. Thus, we just need $a_{i}\leq \sqrt{3}$.

Suppose $a_{i}\leq \sqrt{\sqrt3+2}$. Then $a_{a+1}= a_{i}^{2}-2 \leq \sqrt3+2-2 = \sqrt3$. Therefore we only need $a_{i}\leq \sqrt{\sqrt3+2}$.

I'm not sure exactly how to word this, but we can continue this arbitrarily far and the limiting value is two, so we're done.
[/hide]


Does there exist an infinite $\sigma$-algebra which has only countably many members?

Let $\mathcal{F}\subset 2^{X}$ be an infinite $\sigma$-algebra.

Assume that $\{A_{n}\}_{n \geq 1}\subset \mathcal{F}$, $A_{k}$ nonempty, $A_{1}\varsupsetneq A_{2}\varsupsetneq \ldots$. Then $\{B_{n}\}_{n \geq 1}$, where $B_{k}= A_{k}\setminus A_{k-1}, k \geq 1$, is a countable family of pairwise disjoint sets from $\mathcal{F}$. It implies that $\mathcal{F}$ is not countable.

So it is enough to show that in every infinite $\sigma$-algebra there is a decresing sequence of nonempty sets.

Let $A_{1}= X$. Assume that one defined $A_{1}, A_{2}, \ldots, A_{k}$, $A_{1}\varsupsetneq A_{2}\varsupsetneq \ldots \varsupsetneq A_{k}\neq \emptyset$. Because $\mathcal{F}$ is infinite one can find a nonempty set $A \subset \mathcal{F}$ such that $A \neq A_{i}, i = 1, \ldots k$. Then $A \cap A_{k}\neq \emptyset$ or $A^{c}\cap A_{k}\neq \emptyset$. Let $A_{k+1}$ be that nonempty product of set.

It is easy to proof that $\{A_{n}\}_{n \geq 1}$ is decresing sequence of nonempty sets from $\mathcal{F}$.

But is it true that $A_{k}\neq A_{k+1}$?

Suppose $S$ is a countably infinite $\sigma$-algebra in $X$.

Let $A(x) =$ the intersection of all elements of $S$ containing $x$. Then $A(x)\in S$.

We can show that those $A(x)$ consist a partition of $X$, and this partition must be infinite.

So the power set of the partition is uncountable and there is an injection from this partition into $S$. So $S$ is uncountable, contradiction.

[quote="cyqwhu"]I am sorry.
But can you explain more specifically why the power set is uncountable?[/quote]
I think it may be obvious,because the partition is infinite. :)

[quote="cyqwhu"][quote="zhaobin"]
I think it may be obvious,because the partition is infinite. :)[/quote]

Then why the partition is infinite?[/quote]
 :wink:
Assume the contrary,the partition is infinite assume it is $A(x_{1}),A(x_{2}),\cdots,A(x_{n})$
Notice the $A(x_{i})$ is a subset of $X$,and for each element in $\sigma$-algebra,if it contains a element of  $A(x_{i})$,then it must contain all element of $A(x_{i})$.So you can treat all the elements of  $A(x_{i})$ is the only one element when considering that $\sigma$-algebra,which implies the numbers of the elements of that $\sigma$-algebra will not exceed $2^{n}$,contradiction.Do you understand?


What does the notation $ \mathbb{R} \rightarrow \mathbb{R}$ mean?  How would I use this info to solve the problem:

Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ yf(x)\equal{}xf(y)$ for all real numbers $ x$ and $ y$.

Thanks!

$ f: A\to B$ means that the domain is $ A$ and the codomain is $ B$. The range is always in the codomain, but they're not necessarily the same.

Let $ f(1)\equal{}a$ and plug in $ y\equal{}1$
$ f(x)\equal{}ax$

Hence, $ f(x)$ must be a linear function f the form $ f(x)\equal{}ax$ for some x, checking with the original equaiton, we see that it works.

$ \mathbb{R} \rightarrow \mathbb{R}$ means the domain is the reals, and the range is the reals.

[quote="rofler"]Let $ f(1) \equal{} a$ and plug in $ y \equal{} 1$
$ f(x) \equal{} ax$

Hence, $ f(x)$ must be a linear function f the form $ f(x) \equal{} ax$ for some x, checking with the original equaiton, we see that it works.

$ \mathbb{R} \rightarrow \mathbb{R}$ means the domain is the reals, [b]and the range is the reals.[/b][/quote]

Not necessarily. Not all values of the codomain must be achievable.

[quote="rofler"]

$ \mathbb{R} \rightarrow \mathbb{R}$ means the domain is the reals, and the range is the reals.[/quote]

Let me try for an analogy:

Mathematically incorrect language:  You start off in one place of
the reals.  The arrow points in the direction of an intention to land
in another place of the reals, but you could have limited seating
when you get there compared to what you had before. Or you find
out that you could have the door closed on you altogether and not
get to land in the other place of the reals at all.


geometry
rectangle
function
conics
parabola
analytic geometry

ABC is a right triangle with AB perpedicular to BC, AB= 50 cm and BC = 120 cm.  find the area of the largest rectange that can be inscribed in ABC with one of its corners at B.

[hide]1500cm².
Here's how I found it: The heigth of the rectangle in function of the length, which we will call $x$, is given by: $\frac{-5}{12}x+50$
So the area of the rectangle is given by: $\frac{-5}{12}x^{2}+50x$
This is a parabola with maximum on \[\frac{-b}{2a} \rightarrow \frac{-50}{\frac{-10}{12}} = 60cm\]
So, we by filling in that value in the equation that gives the area, we get $1500cm^{2}$[/hide]

i know this is not for here, but a little knowledge of calc. would help here.
define coordiante system where, the Center of coordinate axis are at $B$, the find the equation of the hypathoneous. you get $y= -\frac{5}{12}x+50$
now the area of the rectangle for any x. is:
$A=x*y=x(-\frac{5}{12}x+50)$
$A'=-\frac{5}{6}x+50=0\Rightarrow x=60$ is the max
thus the rectangle with max area is has length of $60$ on $BC$ and height of $25$ on AB.$\Rightarrow A=1500$
note there are many other ways todo this. you can just ignore this if you don't understand it.
just wanted to help :)


LaTeX
combinatorics unsolved
combinatorics

please put a comment on it because it is important for me to know that how it is solved.
I myself proved that for the integers like 2^k this is not possible but I couldent prove that it is possible for the rest of the integers.and I know that the answer is this.
so I want you to think about the unsolved part.
thank you

I wonder why nobody answers to me. :mad:

Maybe the question is somewhat incomprehensible? Using LaTeX would help...

[quote="P.H"]find all the integers like "n" so that there are n real numbers like A1,A2and...An  that for each i which is between 1 and "n" the some of Ai,........to A2i is equal with 0.
I think this is a difficault problem because I put so much effort on it but I couldent solve it completely.so think on it. :lol:[/quote]
may be i can help you,here is your problem with $Latex$:
find all the integers like $n$ such that there are $n$ real numbers like $A_1,A_2,...,A_n$ that for each $i$ that $1 \leq i \leq n$,\[ \sum_{j=i}^{2i}=0 \].
is it $2i$ or $2i-1$ ?

i think the answer $n$ can be any natural numbers except $2^m,m \in N$ . :D


[quote="NoSoupForYou"]I took the Intermediate Independent Study A class last summer, and I believe I received an offer for a "free subject class." My question is: can I get a free problem series class instead?[/quote]

Sure.  Send us an email at classes@artofproblemsolving.com.

Sorry for reviving, but may I ask what are "free subject classes?" Are they being given out right now? Why were they given out before?

Back when they offered Independent Study (9 years ago or so) a subject class of your choice was included in Independent Study classes. They stopped offering Independent Study a long time ago though

Independent Study courses require a [i]lot[/i] of instructors, and we have so many students now that they wouldn't be practical.

We still encourage you to independently study on your own though and ask questions on our message boards!


conics trigonometry probability algebra polynomial calculus function

I'm working with some students on Algebra II this summer to get them ready for next year. They're all honors, but not at a gifted level or anything, so the curriculum can't be anything too crazy...We're meeting four times (for 2.5 hours each time) over the summer. The problem arises when i have to decide what topics to cover during those 4 sessions. Right now...
1st) systems of equations/sequences and series
2nd) Logs/exponents
3rd) (?)
4th) Conics
I'm trying to decide amongst Basic Triginometry (which their class covered much of already), Probability, or miscellanious topics (such as inverse,direct variation) and do a little practice test taking...
Just curious what topic(s) i should cover. If i'm missing one important one, please fill me in...
Thanks in advance

Matices

Polynomials

Recursions and Special Sequences?

Binomial Theorem

If you do probability, don't forget to make them draw ridiculously large tree diagrams that take up more than one page

I think probability, binomial theorem, etc. is good. Since in Pre-Calculus, polynomials are covered and trigonometry and complex numbers are deeply covered, at least at my school.

Wait, are they going into Algebra II or Pre-Calc next year?

They're just getting a little jump on the material that they'll learn next year in Alg. II
thanks for the imput btw!

Personally, the hardest part of Algebra II was matricies and conics. :D

What topics concerning matrices should i cover, if i decide to teach them?

it would be good if you could talk to a teacher of the class, or get the book that your student is using...also i don't see how you are going to teach a whole years worth of material in just 10 hours...

some other topics that i did were:

solving linear systems (cramer, matricies, elimination, row reduction)

trig (common values, identities, solving equations)

word problems (clock things, d=rt, work, etc)

polynomials (factoring, grouping, synthetic division, descartes, bounding)

rational functions (graphing, asymptotes, horiz/vert+oblique, poly division, finding holes)

matricies (determinants, row operations, add/sub/mult, inverses)

if i can think of more...i will get back to you


What is the least positive integer by which you could multiply 180 to get a product that is a perfect cube?

If $ n$ is an integer, then $ 180n$ must end in a zero.  So $ 180n\equal{}x^{3}$, where $ x^{3}$ ends in a zero as well.  This can be accomplished only if $ x$ ends in a zero.  If $ x\equal{}10$, we get $ x^{3}\equal{}1000$, but that's not divisible by 180.  If $ x\equal{}20$, then its cube is 8000, but that's not divisible by 180.  If $ x\equal{}30$, its cube is 27000, which when divided by 180, gives $ n\equal{}150$.

I hope I'm correct.

Another way that may prove more useful in harder problems:

First we know $ 180n \equal{} x^3$, just as JRav wrote.
Since 180=2*2*3*3*5, we know that
$ 2^2\cdot 3^2\cdot 5\cdot n \equal{} x^3$

In order to minimize $ n$, we must have multiply 180 by no factors other than those required.
In order for $ 180n$ to be a perfect cube, all prime factors of $ n$ must be raised to the third power.
This means that
$ n \equal{} 2^3\cdot 3^3 \cdot 5^3$
$ n \equal{} 30^3$
$ n \equal{} 27000$

Another way I found was that a perfect cube needs to have factors that are divisible by 3. The prime factorization of 180 is 2*2*3*3*5, meaning $2^2$ * $3^2$ * 5. In order for these to be a multiple of three, we multiply it by 2*3*$5^2$, making it a perfect cube because it would then be a multiple of three. Multiplying these out, we get the answer, 150


Hi all! :)

One choose a seven digits password at random.
Find the following probabilities :
1) All the digits are distincts
2) The product of those digits is divisible by 2.
3) The digits form :
a/ a strictly increasing sequence.
b/ an increasing sequence.

Enjoy! ;)

1) [hide]There are 10 choices for the first digit, 9 choices for the second such that the second is not the same as the first, 8 for third, and so on down to 4 choices for the 7th. (10*9*8*7*6*5*4)/10^7=189/3125
[/hide]
2) [hide]We must have at least 1 of the 7 digits a multiple of 2. There are 5 such digits to choose from. We may count the number of passwords whose product is not a multiple of 2 and subtract this from the total to find the number whose product is a multiple of 2. There are 5^7 passwords with a product not divisible by 2, since there are 5 digits not multiples of 2 to choose from and 7 positions possible. The probability of getting one of these is 5^7/10^7=1/2^7=1/128, so the probability that you get one whose product is divisible by 2 is 1-1/128=127/128.[/hide]

3a) [hide]I'm guessing that a strictly increasing sequence means that the common difference is not 0, but there is a common difference. I shall have the solution for the no-common-difference also.

If we have a common difference of 1, we can have the first term of the sequence any digit from 0 to 3 inclusive, and since each other member is equal to 1 more then the previous member, there are a total of 4 possible successful passwords. Therefore there is 4/10^7=1/2500000 chance of these. (You cannot have a common difference of any other positive real as those differences do not allow for our limited stock of digits)

If we are strictly increasing, we can have as the first term any digit from 0 to 3 inclusive. There is only 1 sequence for first digit=3 where all increase. There are 7 such for 2. There are 29 such for 1.   There are (I think) 67 such for 0. There are a sum of 100 such, giving 100/10^7=1/100000[/hide]

3b) [hide]Assuming one can have a common difference of 0: There are 10 with difference 0 and 4 with difference 1, for 14/10^7=7/1000000

With variable difference between terms, there are 1 for first digit 9, 7 for 8, 63 for first of 7, 189 for 6, 441 for 5, 903 for 4, 1695 for 3, 2937 for 2, 4789 for 1, and 7920 for 0. At least I hope so, because I've wasted a lot of time otherwise...There are thus 18945 passwords that fulfill our conditions. 18945/10^7=3789/2000000[/hide]

Did I get some of it right at least? :( I probably messed up somewhere...

I think I have got different answers for 3(a) and (b) .

3(a) [hide]
For strictly increasing sequence we just have to select sets of $7$ different numbers. They can be arranged in only one way in strictly increasing order.
Thus the probability is $\frac{\binom{10}{7}}{10^7}$
[/hide]

3(b) [hide]
For increasing sequence, we can divide the job as selecting 7 different numbers, 6 different numbers, 5 different numbers,…,1 number.
Now they have to be arranged in increasing order where numbers can be repeated to get a total of $7$ digits.
So, probability of obtaining such a sequence is
\[ \frac{\binom{7}{7}\binom{10}{7}+\binom{7}{6}\binom{10}{6}+\cdot\cdot\cdot+ \binom{7}{1}\binom{10}{1}}{10^7} \]
[/hide]

Someone can give a better a solution? :oops:

Yes another approach

3(a)

[hide]The number of such number is also given by the number of solution of

$a+b+c+d+e+f+g \leq 9$ (seventh digit is $\leq 9$) with $0 < a,b,c,d,e,f,g \leq 9$
Or $a'+b'+c'+d'+e'+f'+g'+h = 2$ with $0 \leq a',b',c',d',e',f',g',h$
This is a "combinaison with repetitions" an the number of solutions is $\binom{9}{7} = 36$
[/hide]
3(b)

[hide]Same thing, $a+b+c+d+e+f+g \leq 9$ with $0 < a, 0 \leq b,c,d,e,f,g \leq 9$

This gives $\binom{17}{7} = 6435$ numbers[/hide]

And the  winner is... Tµtµ!  :first:

(at least, according to my own results :P)

Apart from that, solafidefarms, your other results are good! :)

Varun, I think you misunderstood the questions...

Well well well, after I tried to search a bit more about this problem..
I've come to the conclusion that you were right for 3/a/ Varun!

And, tµtµ, your idea (combinations with repetitions) was good, but, you've made a little error ;). The 3/b/ should be : ${16} \choose {7}$.

But, once again, this is just my work..

ps: sorry for double-posting..

What is the difference between increasing and strictly increasing? :?

Read solafidefarms' post for an explanation ;)


The question is:
Find all continuous functions satisfying:
$ f(x + y) = f(x) + f(y) + xy(x + y) ; x, y \in \mathbb R.$

[hide="What I did is:"]
Let $ y = x$.
Then,
$ f(2x) = 2f(x) + x^2(2x)$
$ 2f'(2x) = 2f'(x) + 3x^2$
$ 2f''(2x) = f''(x) + 6x$
$ 4f'''(2x) = f'''(x) + 6$

Therefore:
$ 4f'''(0) = f'''(0) + 6$
$ f'''(0) = 2$
Now integrating from $ 0$ to $ x$, I got $ f(x) = \frac {x^3}{3} + cx^2$
The correct answer is ${ \color{red}{f(x) = \frac {x^3}{3} + cx}}$ where $ \color{red}c = f'(0)$
[b]Where did I possibly go wrong???[/b] :huh:

[/hide]

[quote="hsbhatt"]See http://goiit.com/posts/list/algebra-functional-equation-91610.htm#425429[/quote]
Thank you. As stated by other members of [hide="forum"] I accidentally posted the same thing thrice![/hide], solving it by $ Cauchy's$ $ Eq^n$ is also a quick and convenient trick.

Thank You for the assistance.

-$ Rajiv$ $ Patki.$


x,y,z reale astfel incat x+y+z=pi/4

demonstrati ca (sinx+cosx)(siny+cosy)(sinz+cosz)=cos2x+cos2y+cos2z

oups... scuze pt eroare

demonstrati ca sqrt2 * (sinx+cosx)(siny+cosy)(sinz+cosz)=cos2x+cos2y+cos2z

si mai pun una aici:

demonstrati ca orice numar natural se poate scrie ca o suma de termeni distincti ai sirului lui Fibbonacci .

a doua iese direct prin inductie
fie $F_{k}\leq n < F_{k+1}$. atunci exista inegalitatile :
$0 \leq n-F_{k}< F_{k+1}-F_{k}= F_{k-1}$.
insa numarul $n-F_{k}$ se scrie ca o astfel de suma, si toti termenii sunt distincti, si mai mici decat $F_{k-1}$. adaugand $F_{k}$ la suma, proprietatea se pastreaza.
verficarea pentru cazurile mici e imediata.

parca era o teorema problema 2, zeckendorf sau ceva in genu.

cat despre prima problema , iese cu complexe si cu calcule multe, si anume:
luam $z_{1}=\cos{x}+i \sin{x}$ etc, si avem $z_{1}z_{2}z_{3}={\frac{\sqrt{2}}{2}+i \frac{\sqrt{2}}{2}}$, si tre sa demonstram ca $\frac{\sqrt{2}}{8z_{1}z_{2}z_{3}}\cdot \prod{(z_{1}^{2}+1-i(z_{1}^{2}-1)))}= \sum{\frac{z_{1}^{4}+1}{2z_{1}}}$ , asta in caz ca nu ai ce face :)

E mai usor deoarece $2x+2y+2z=\frac{\pi}{2}$ avem $\cos 2z=\sin{2(x+y)}=2\sin{(x+y)}\cos{(x+y)}$ acum $\cos 2x+\cos 2y=2\cos{(x+y)}\cos{(x-y)}$ de unde $\cos 2x+\cos 2y+\cos 2z=2\cos{(x+y)}(\cos{(x-y)}+cos{\left(\frac{\pi}{2}-x-y\right)}) =4 cos{(x+y)}\cos{(y+z)}cos{(z+x)}=4\prod_{cyc}\cos\left(\frac{\pi}{4}-x\right)= \sqrt{2}\prod_{cyc}(\cos x+\sin x)$ :lol:

[b][u]Lemma (well-known or prove easily).[/u][/b]

$\{\begin{array}{c}\{\alpha ,\beta ,\gamma \}\subset\mathrm R\\\ \alpha+\beta+\gamma=\pi\end{array}\|\Longrightarrow\{\begin{array}{c}\sin 2\alpha+\sin 2\beta+\sin 2\gamma=4\sin\alpha\sin\beta\sin\gamma\\\\ \cos 2\alpha+\cos 2\beta+\cos 2\gamma =-4\cos\alpha\cos\beta\cos\gamma-1\end{array}$

[b][u]Particular cases.[/u][/b]

$\{\begin{array}{c}x+y+z=\frac{\pi}{4}\\\ \alpha : =\frac{\pi}{4}+x\ ,\ \beta : =\frac{\pi}{4}\ ,\ \gamma : =\frac{\pi}{4}+z\\\ (\alpha+\beta+\gamma =\pi )\end{array}\|\Longrightarrow\{\begin{array}{c}\sum \cos 2x=\sqrt 2\cdot\prod (\cos x+\sin x)\\\\ \sum\sin 2x=1+\sqrt 2\cdot\prod (\cos x-\sin x)\end{array}$


For positive real numbers $ a_1, ... , a_n, b_1, ... , b_n$ ($ n \geq 3$) with the $ b_i$'s pairwise distinct, denote $ S \equal{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n$ and $ T \equal{} b_1b_2 ... b_n$.
Let $ f(x) \equal{} (x \minus{} b_1)(x \minus{} b_2) ... (x \minus{} b_n)\sum_{j \equal{} 1}^n\frac {a_j}{x \minus{} b_j}$.
Find the number of distinct real zeroes of the polynomial $ f(x)$.

It is obviosly. Consider \[ f(b_j)\equal{}a_j\prod_{i\not \equal{}j}(b_i\minus{}b_j)\]
If $ b_1<b_2<...<b_n$, then because $ a_j>0 \forall j$ $ f(b_j)(\minus{}1)^{n\minus{}j}>0$, therefore they had n-1 roots $ b_1<r_1<b_2<r_2<...<r_{n\minus{}1}<b_n$.


Hi !!

$ a,b,c\,\geq\,0$  Prove that :

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

By Cauchy,

\[ (\text{LHS})(\sum\limits_{\text{cyc}}(b\plus{}c)(a^2\plus{}bc)) \ge (a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)^2\]

So it suffices to prove
\[ (a^2\plus{}b^2\plus{}c^2\plus{}ab\plus{}bc\plus{}ca)^2 \ge 2(a\plus{}b\plus{}c)(a^2b\plus{}b^2c\plus{}c^2a\plus{}ab^2\plus{}bc^2\plus{}ca^2)\]

Expand it out, it reduces to showing
\[ a^4\plus{}b^4\plus{}c^4 \ge a^2b^2\plus{}b^2c^2\plus{}c^2a^2\]
which follows by AM-GM.

we have to prove :
$ \sum(\frac {a^2 \plus{} bc}{b \plus{} c} \minus{} b)\,\ge\,0$

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

$ \iff$        $ \sum((a \plus{} b)^2(a \plus{} c)(a \minus{} b))\,\ge\,0$

$ \iff$        $ \sum(a^4) \minus{} \sum(a^2b^2)\,\ge\,0$

wish is true  :wink:

Let Add :a+b+c  to those parts of inequality and then set: a+b=x,b+c=y and c+a=z now we must prove:  Sum((xy)/z) >= x+y+z ,which it clearly true.

\[ \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b}\,\geq\,a \plus{} b \plus{} c \longleftrightarrow \sum_{cyclic}x(a \minus{} b)(a \minus{} c) \geq 0
\]where $ x \equal{} \frac {1}{b \plus{} c},y \equal{} \frac {1}{c \plus{} a},z \equal{} \frac {1}{a \plus{} b}$
WLOG $ a \ge b \ge c$. $ \longrightarrow x \ge y$
from Vornicu-Schur; $ \sum_{cyclic}x(a \minus{} b)(a \minus{} c) \geq 0$
thus,
\[ \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b}\,\geq\,a \plus{} b \plus{} c
\]

The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that  $ ab\plus{}ac\plus{}bc\neq0.$ Prove that:
\[ \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b}\geq\frac{2(a^2\plus{}b^2\plus{}c^2)\plus{}ab\plus{}ac\plus{}bc}{a\plus{}b\plus{}c}\]

[quote="arqady"]The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that  $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b}\geq\frac {2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} ac \plus{} bc}{a \plus{} b \plus{} c}
\]
[/quote]
The inequality is equivalent to

$ \sum \frac{a^3\plus{}abc}{b\plus{}c} \geq \sum a^2,$

$ \Leftrightarrow \sum \frac{a(a\minus{}b)(a\minus{}c)}{b\plus{}c} \geq 0,$

WLOG we can assume that $ a \geq b \geq c,$

we have $ \frac{c(c\minus{}a)(c\minus{}b)}{b\plus{}c} \geq 0,$

and $ \frac{a(a\minus{}b)(a\minus{}c)}{b\plus{}c} \geq \frac{b(a\minus{}b)(b\minus{}c)}{a\plus{}c}$ because it's equivalent to $ (a\minus{}b)(a^2\plus{}b^2\minus{}ab\minus{}c^2) \geq 0,$

and the proof is completed.

[quote="rachid"][quote="arqady"]The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that  $ ab \plus{} ac \plus{} bc\neq0.$ Prove that:
\[ \frac {a^2 \plus{} bc}{b \plus{} c} \plus{} \frac {b^2 \plus{} ca}{c \plus{} a} \plus{} \frac {c^2 \plus{} ab}{a \plus{} b}\geq\frac {2(a^2 \plus{} b^2 \plus{} c^2) \plus{} ab \plus{} ac \plus{} bc}{a \plus{} b \plus{} c}
\]
[/quote]
The inequality is equivalent to

$ \sum \frac {a^3 \plus{} abc}{b \plus{} c} \geq \sum a^2,$

$ \Leftrightarrow \sum \frac {a(a \minus{} b)(a \minus{} c)}{b \plus{} c} \geq 0,$

WLOG we can assume that $ a \geq b \geq c,$

we have $ \frac {c(c \minus{} a)(c \minus{} b)}{b \plus{} c} \geq 0,$

and $ \frac {a(a \minus{} b)(a \minus{} c)}{b \plus{} c} \geq \frac {b(a \minus{} b)(b \minus{} c)}{a \plus{} c}$ because it's equivalent to $ (a \minus{} b)(a^2 \plus{} b^2 \minus{} ab \minus{} c^2) \geq 0,$

and the proof is completed.[/quote]

Very nice proof. How about the even stronger one

$ \frac {a^2 \plus{} bc}{b \plus{} c}\plus{}\frac {b^2 \plus{} ca}{c \plus{} a}\plus{}\frac {c^2 \plus{} ab}{a \plus{} b}\geq\frac {2(a^2 \plus{} b^2 \plus{} c^2) \plus{}ab\plus{}ac\plus{}bc}{a\plus{}b\plus{}c} \plus{} \frac {27}{8} \frac {(ab\plus{}bc\plus{}ca) \sum (a^2(a \minus{} b)(a \minus{} c))} {(a \plus{} b \plus{} c)^4}$


Find $\sum_{k=0}^\infty \frac{\tan(k \sqrt2)}{2^{k}}$, or prove that it diverges. (The argument is in radians.)

$\tan{k\sqrt{2}}$ is in RADIANS???!!! :what?:

Whoa! Scary. But I'm pretty sure that there's a not-too-hard way to do this...

I've gotten it down to:

[hide]
$\frac{1}{i}\sum_{k=0}^{\infty}\left(\left(\frac{e^{\sqrt{2}i}}{2}\right)^{k}-\left(\frac{e^{-\sqrt{2}i}}{2}\right)^{k}\right)\left(\frac{1}{e^{ik\sqrt{2}}+e^{-ik\sqrt{2}}}\right)$
[/hide]

And now I'm stuck.  It's the factor on the right that gets me.  If it weren't there, I could separate it into geometric sequences.

Sorry to bump, but it would be neat if someone could solve this.


Prove that $\forall n\in\mathbb{N}$:
(1) $\prod_{k=1}^n{(n+k)}=2^n\prod_{k=1}^n{(2k-1)}$
(2) $\left(\frac 23\right)^n<\frac4{n(n+1)}$
by using mathematical induction.

Well, I don't think they're pre-olympiad level.

[quote="tµtµ"]This is very easy and probably would rather go in Getting Started ?
[/quote]
I have posted it in rather getting started or intermediate but someone moved it to this area.

[quote="tµtµ"]This is very easy and probably would rather go in Getting Started ?[/quote]
No proof in Getting Started ;)
See the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=32837]guideline[/url] (especially post #2).

So I think this is appropriate for Intermediate or Pre-olympiad.

[quote="tµtµ"]This is very easy and probably would rather go in Getting Started ?

Anyway note that $1/(n+1) * (2n+1)(2n+2) = 2(2n+1)$ and prove that $2/3 * \frac 4{n(n+1)} <\frac 4{(n+2)(n+1)}$[/quote]This explains (2) is not true $\forall n\in\mathbb{N}$ :) Counterexample: let $n=1$.


modular arithmetic
number theory
relatively prime

Find the number of solutions of $x^{2}\equiv x\pmod{m}$ for any positive integer $m$.

EDIT: Ok, I really have no clue about anything in NT I guess.

[hide]
\[x^{2}\equiv x\Rightarrow m|x(x-1)\]
[i]Let $k$ be the number of distinct prime factors of $m$.[/i] Since $x$ and $x-1$ are relatively prime, for each prime $p$ dividing $m$, either all factors of $p$ divide $x$ or all divide $x-1$. Since there are two choices for each prime, there are $\boxed{2^{k}}$ solutions, and the Chinese Remainder Theorem guarantees that they all work.
[/hide]


THE EPIC SPACE WAR:

II:

REVENGE OF THE CRONOSIANS: (What? I had to think of a name)

After the great battle, where many heroes like EggyLv.999, Spaceguy524 and others were defeated, the survivors fled to Titan, the largest moon of Saturn. (Also known as Cronos) However, on titan, there are strange beings that feed on the air particles. These are known as Cronosians and feared by all.

Signups available:

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Extra Game Mod:

Please copy and paste list then type your username:

Like this:

1. My username here!
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Extra Game Mod:

SHOPPING:

refer to this anytime needed, copy and paste or quote if you will.

SHIPS:

Class 1

Average fighter - 1000 dollars

Good fighter - 2000 dollars

Great fighter! - 3000 dollars

Class 2

Average cruiser - 10,000 dollars

Good cruiser - 20,000 dollars

Great Cruiser! - 35,000 dollars

Class 3

Average Battle cruiser - 100,000 dollars

Good battle cruiser - 210,000 dollars

Great Battle cruiser - 400,000 dollars

SHIELD SHOP:

shield - 1000 dollars

good shield - 10,000 dollars

great shield! - 100,000 dollars

WEAPON SHOP: note: all weapons never run out of ammo

laser - 500 dollars

Mass Driver - 10,000 dollars

Magnetic accelerator - 15,000 dollars

Energy Pulse - 30,000 dollars

Missile - 10,000 dollars

Protius Missiles - 50,000 dollars

Antimatter pulse - 500,000 dollars (OMG!)

UPDATE: all events are now 1000 dollars each

This is almost like EPIC SPACE WAR I; the only thing is that in this one, you can fight each other or aliens, AND BE WARNED, ALIENS HAVE SHIPS TOO!

Good luck, and don’t spam this thread!

And this time, there's no sudden ends to the game. Phew.  :D

[quote="spaceguy524"]

@tcjy: PM me when it is my turn.[/quote]

same for me.

1. spaceguy524
2. Chompy
3. tennis123
4. EggyLv.999
5. cyberspace
6. Bluecarneal
7.
8.
9.
10.
11.
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[quote="spaceguy524"]@tcjy: PM me when it is my turn.[/quote]

same here

1. spaceguy524
2. Chompy
3. tennis123
4. EggyLv.999
5. cyberspace
6. Bluecarneal
7. Zheng
8.
9.
10.
11.
12.
13.

1. spaceguy524
2. Chompy
3. tennis123
4. EggyLv.999
5. cyberspace
6. Bluecarneal
7. Zheng
8. AIME15
9.
10.
11.
12.
13.

Perhaps you could PM anyone when it's their turn instead of random people requesting it :P

1. spaceguy524
2. Chompy
3. tennis123
4. EggyLv.999
5. cyberspace
6. Bluecarneal
7. Zheng
8. AIME15
9. louiethechewy123
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The veteran fighter, louiethechewy123 is back from a mission Epic Space War I !!!

If not enough people join can we play with less people?

Well, 13 is a small number. :P

Well, that's compared to Mafia games anyway...

It won't take that long for the game to fill up.

If no one signs up in a week will this game be dead?
Please sign up, people!!!

Whoops, forgot about that.  :oops:
[b]Unattack[/b]
[b]Attack abcak[/b]

Could I please join? Though dont p.m. me please, and if you could just skip my turn starting Friday for a week plz.? :maybe:

Why don't you want to be PM'd?  It's the only way the mod can inform you that it's your turn.

Because, I dont always check them, and I check my posts on a normal daily basis.Basically, I dont really use p.m.s. :lol: So may I join or not?

Ask spaceguy.

I'm no mod, just the creator lolz  :rotfl:  :rotfl:  :rotfl:

Yes you may join Butler.

westiepaw-92
abcak-92
zheng-92
Eggy-90
cyberspace-92
tennis-78

abcak, choose somebody.

[quote="abcak"]I counter.

You jerk.[/quote]

Huh?

me=:?:

Bump.

Don't want this to die, its a good game

can i replace someone?

pm me when it's my turn

If there's another advertisement here, Then this should get locked.

Well, I guess it's dead. So better create another. SO LONG TESWII!!!


modular arithmetic
number theory unsolved
number theory

Show that if $ p$ is an odd prime,then the congruence $ x^4 \equiv \minus{}1 (modp)$ has a solution iff $ p \equiv 1(mod 8)$

[quote="314 Einstien"]This is to prove that x^4 +1 / ( 8k+1) .
Consider two cases i) x is even x =2m
Then we get that T.P.  16*m^4 +1 / 8k+1
We know that  16*m^4 +1 is congruent to 1(mod 8) also 8k+1  is congruent to 1(mod 8) .
Hence 16*m^4 +1 / 8k+1 .
Similarly consider x =2m+1. and we get similar congruent conditions.
However i am not sure about the following statement if a is congruent to m(mod c) and b is also congruent to m(mod c) . Then either a|b or b|a .
Please do check that result alone.[/quote]

Casually has come across this post. The very crazy text. But the problem is quite reasonable and is interesting for beginning. For solution use the primitive root modulo $p$.

[quote="Ovchinnikov Denis"]$x^8\equiv 1 (mod p), x^4 \neq 1(mod p)$ so order of $x$ modulo $p$ is 8 so $8 | p-1$ =>
$QED$[/quote]

It's correct. But we need to prove inverse statement: if $p \equiv 1 \pmod{8}$, then the congruence $x^4 \equiv -1 \pmod{p}$ is solvable. The simplest way to prove it -- use primitive root modulo $p$.


abstract algebra
superior algebra
superior algebra unsolved

[quote="flowers"]Let R be a group ring AG, where A is a ring, and G is a finite group. Show that R is self-injective iff A is.[/quote]

AG is a free A-module, so a flat A-module, so the restriction of an injective AG-module is an injective A-module.  In particular, if AG is AG-injective, then AG is A-injective, and then so is its A-direct summand, A.

AG is a free AG-module, so a projective AG-module, so if A is A-injective, then Hom_A(AG,A) is AG-injective, but viewing AG as Hom_AG(AG,AG) you get that AG is isomorphic to Hom_A(AG,A) (for instance by Nakayama reciprocity, or just element pushing).

The "injective producing lemma" is very useful for this. It is 3.5 p. 62 in Lam's Lectures on Modules and Rings.  If P is an (R,S)-bimodule that is R-flat and M is an injective S-module, then Hom_S(P,M) is an injective R-module.  This allows for some basic change of rings.

This is theorem 4 of:

Connell, Ian G. "On the group ring."
Canad. J. Math. 15 (1963) 650-685.
http://www.ams.org/mathscinet-getitem?mr=153705
http://www.cms.math.ca/cjm/v15/p650

This is also a part of the result of a paper of Renault, though the main point there was the necessity that G be finite:

Renault, Guy. "Sur les anneaux de groupes."
C. R. Acad. Sci. Paris Sér. A-B 273 (1971) A84-A87.
http://www.ams.org/mathscinet-getitem?mr=288189


geometry
ratio
inequalities
geometric inequality
areas
IMO
IMO 1966

Let $ ABC$ be a triangle, and let $ P$, $ Q$, $ R$ be three points in the interiors of the sides $ BC$, $ CA$, $ AB$ of this triangle. Prove that the area of at least one of the three triangles $ AQR$, $ BRP$, $ CPQ$ is less than or equal to one quarter of the area of triangle $ ABC$.

[i]Alternative formulation:[/i] Let $ ABC$ be a triangle, and let $ P$, $ Q$, $ R$ be three points on the segments $ BC$, $ CA$, $ AB$, respectively. Prove that

$ \min\left\{\left|AQR\right|,\left|BRP\right|,\left|CPQ\right|\right\}\leq\frac14\cdot\left|ABC\right|$,

where the abbreviation $ \left|P_1P_2P_3\right|$ denotes the (non-directed) area of an arbitrary triangle $ P_1P_2P_3$.

I do not think the problem is very difficult :

Wlog, we may assume that [ABC]  = 1.

Let A', B' and C' be the midpoints of the sides BC, CA and AB respectively.
If R is on the segment AC' and Q is on the segment AB', then clearly [AQR] \leq [AB'C'] = 1/4, so we are done in that case.
We have similar conclusion in the case where two of the points P,Q,R lies on the half-sides from the common vertex.

Thus, we may assume that R is on the segment BC', Q is on AB' and P is on CA' (the case R on AC', Q on CB' and P on BA' is similar).
Let's consider R and P to be fixed. From ouir choices of the points, the line parallel to PR through B' meets the side AB in a point which belongs to the segment AC' (look about the respective slope). Thus, for any point in the segment AB', the distance from this point to the line PR is at least the distance from B' to the line PR.
It follows that [PQR] \geq [B'PR].
With a similar reasoning, we prove that :
[PQR] \geq [B'PR] \geq [B'C'P] \geq [A'B'C'] = 1/4.

Since [AQR]+[BRP]+[CPQ]+[RPQ] = [ABC] = 1, we deduce that
[AQR]+[BRP]+[CRQ] \leq  3/4, from which the result follows easily.

Pierre.

Pierre gave too complicated solution ;)
Let $P$, $Q$, $R$ divide sides in ratios $a:1-a$, $b:1-b$ and $c:1-c$ respectively, then if we suppose contrary it will mean that $a(1-b)>1/4$, $b(1-c)>1/4$, $c(1-a)>1/4$. Multiplying them we obtain $abc(1-a)(1-b)(1-c)>1/64$, though $x(1-x)\leq 1/4$ for all $x\in[0,1]$. Contradiction.  :D

[quote="Myth"]Pierre gave too complicated solution ;)[/quote]

'Too complicated' is my third name. :D
('Too long' is my second)

[Moderator edit: Another solution to this problem is to conclude it from the stronger problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=5179 .]

Pierre.

We note that the equality is attained if P,Q,R are the middle points of the sides. Now we define $\overline{AR}=a$, $\overline{BR}=b$, $\overline{BP}=c$, $\overline{CP}=d$, $\overline{CQ}=e$, $\overline{AQ}=f$. Now we suppose that $(\triangle ARQ)>\frac{(\triangle ABC)}{4}$, $(\triangle BRP)>\frac{(\triangle ABC)}{4}$, $(\triangle CPQ)>\frac{(\triangle ABC)}{4}$

Now we have

$\frac{af\cdot sen\angle BAC}{2}>\frac{\frac{(a+b)(e+f)\cdot sen\angle BAC}{2}}{4}$

$\frac{bc\cdot sen\angle ABC}{2}>\frac{\frac{(a+b)(c+d)\cdot sen\angle ABC}{2}}{4}$

$\frac{de\cdot sen\angle BCA}{2}>\frac{\frac{(c+d)(e+f)\cdot sen\angle BCA}{2}}{4}$

Simplifying and adding the inequalities we have

$abcdef>\frac{(a+b)^2(c+d)^2(e+f)^2}{64}$

that is a contradiction, because for AM-GM over abd, afd, bcd, cdf, abe, aef, cbe, cfe we have

$\frac{abd+afd+bcd+cdf+abe+aef+cbe+cfe}{8}\ge \sqrt[8]{a^4b^4c^4d^4e^4f^4}$

$\frac{(a+b)^2(c+d)^2(e+f)^2}{64}\ge abcdef$

we finished :))


Find the number of two-digit positive integers whose digits total 7.

[quote="AIME15"]None of the digits can be larger than 7, and the units cannot be 0, so the answer is 7[/quote]
you mean that the ten's digit can't be 0


How many positive integer multiples of 1001 can be expressed in the form 10^j - 10^i, where i and j are integers and 0 :le: i < j :le: 99?


Does anyone know what scores on the verbal(60 questions) and reading comprehesion(40) tests for the SSAT will give a 90+ percentile?

Sorry, i know this doesnt belong here, but i dont know where it should go.

What is the SSAT?  Do you mean SAT?  By definition, percentile depends on how others do, so it varies from year to year.  I think somewhere around a 670 gets you close.

The SSATs are the standardized test for private and some other secondary high schools. It's 3 sections just like the SAT, and scores for each section are from 250 to 350 points.

@OP, the scores range every year, it just depends how hard the test is. Also you should think in terms of raw score, since like the SAT there's a penalty for wrong answers.


Erica flips a fair two-sided coin. NAte then flips the coin and wins if it matches Erica's flip. If it doesn't match, then Noah flips the coin and wins if it matches Nate's flip. If Noah doesn't win, then Erica flips and wins if it matchs Noah's flip. The game continues until there is a winner. What is the probability that Nate wins the game?

[hide]I think that the probability of Nate winning can be expressed by $\displaystyle \frac{\frac12}{\frac78}=\boxed{\frac47}$[/hide]

[hide]
Not quite sure about this but,
The probability that he wins on the first "set" of flips is
$\frac{1}{2}*\frac{1}{2} = \frac{1}{4}$
On the second
$\frac{1}{2}^5 = \frac{1}{32}$
On the third
$\frac{1}{2}^8 = \frac{1}{256}$

So it looks like an infinite sum of geometric series with first term $\frac{1}{4}$ and ratio $\frac{1}{8}$, so
$\frac{\frac{1}{4}}{\frac{7}{8}} = \frac{2}{7}$[/hide]

I think that the first term of the infinite series is $\displaystyle \frac12$ though. Because it doesnt matter what Erica flips... so the chance for Nate to win on his first flip is $\displaystyle 1*\frac12$... so I think that the sum of the infinite series would be $\displaystyle \frac{\frac12}{\frac78}=\frac47$


hey can anyone help me?

I tried making a new document on TeXnic (Hello world! because im just starting LaTeX) and when I try to compile it, nothing happens. I'm wondering if it is because the bar at the top is blank, while it should say "LaTeX => PDF" (I think) but I can't type anything in there and the dropdown doesn't do anything.

any suggestions?
thx

You probably didn't tell TexnicCenter where to find the latex files when you installed it. [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Help#I.27ve_upgraded_MiKTeX_and_now_TeXnicCenter_won.27t_work]This[/url] will take you through it (you do the same as if you had upgraded MiKTeX).


What is the best way to prepare for Mathcounts?
Work on the 2010 handbook problems or work on Past mathcounts school and chapter problems?
I like to do both but I am short on time.
Our Mathcounts school test will be next week, hopefully I can make to the chapter (my school is a tough school, lots of competitions). The chapter is next month.

If I have to make choice on focusing on 2010 handbook problems or past School/Chapter problems, which one should I choose?

Search the forums for Mathcounts preparation. This question has been asked countless times before.

Hmm... the optimal strategy would be to use the handbook to learn how to solve new kinds of problems, and then focus on speed and accuracy by doing old tests, but since you're low on time, I'd say it's best to practice with old tests and review what you can. I hope this helps!

Yes, I agree with mathluver that doing old tests would be best in your situation.
Good luck! :)


function
Euler
inequalities
number theory proposed
number theory

Hey guys,
i would like to give u guys this problem .
well i got this result when i was doing some research on eulers phi function.
its:
phi(1)+phi(2)+phi(3)+phi(4)+.......+phi(k)> (1+2+3+4+......+k)/2
for all k>4
guys its a very nice result as the difference becomes very narrow and then suddenly widens and then again narrows and so on........
try proving this!!!!
Karthik

For $ k = 5,6,7,8,9,10$ the inequality can be verified instantly. Assume that $ k\geq 11$.

Note that $ \varphi(1) + \varphi(2) + \dots + \varphi(k)$ counts the number of pairs of integers $ (m,n)$ such that $ 1\leq m\leq n\leq k$ and $ \gcd(m,n) = 1$.
Therefore, $ 2(\varphi(1) + \varphi(2) + \dots + \varphi(k)) - 1$ counts the number of pairs $ (m,n)$ such that $ 1\leq m,\, n\leq k$ and $ \gcd(m,n) = 1$.

The total number of pairs $ (m,n)$ with $ 1\leq m, n\leq k$ is $ k^2$. Therefore, the number of pairs with $ \gcd(m,n) = 1$ equal $ k^2$ minus the number of pairs with $ \gcd(m,n) > 1$. It is clear that for prime $ p$ the number of pairs with $ p|\gcd(m,n)$ does not exceed $ (k/p)^2$. Hence,
\[ 2(\varphi(1) + \varphi(2) + \dots + \varphi(k)) \geq k^2 - \sum_{p\,\mbox{is prime}} \frac {k^2}{p^2} = k^2(1 - P(2))
\]
where $ P(\cdot)$ is the [url=http://mathworld.wolfram.com/PrimeZetaFunction.html]prime zeta function[/url]. Since $ P(2)\approx 0.452247 < 0.453$, we have
\[ 2(\varphi(1) + \varphi(2) + \dots + \varphi(k)) \geq k^2(1 - P(2)) > 0.547 k^2 > 1 + 2 + \dots + k,\quad(\mbox{for}\, k\geq 11)
\]
implying that
\[ \varphi(1) + \varphi(2) + \dots + \varphi(k) > \frac {1 + 2 + \dots + k}{2}.
\]


geometry
incenter
circumcircle
Euler
rhombus
exterior angle
angle bisector

A, B and C are the feet of the interior bisectors of a triangle EFG.
Show that the triangle ABC is a right triangle iff one of the angles of EFG is 120°.

$Lemma:$ For a triangle ABC ($BC>AC$) let the interior angle bisectors AD and CF and the exterior angle bisector BE, then D, E and F are collinear. $Prove$ By Menelaus, occur iff and only if $\frac{BD.CE.AF}{DC.EA.FB}=1$, by angle bisector theorem we have $\frac{AB.BC.AC}{AC.AB.BC}=1$, which is true.
Suppose <GEF=120, the angle bisectors are $EA, FB$ and $GC$, and $I$, the incenter. In the triangle $EAF$, $EG$ is an exterior angle bisector and $FI$ is an interior angle bisector, so by the lemma $AC$ is an interior angle bisector,so <EAC=<CAF, similarly, we have: <EAB=<BAG, and we are done.

Let I be the incenter of arbitrary $\triangle EFG$ and $I_{E},\ I_{F},\ I_{G}$ excenters opposite to E, F, G. The circumcircle (O) of $\triangle EFG$ is the common 9-point circle of $\triangle II_{E}I_{F},\ \triangle II_{E}I_{G}$ with orthocenters $I_{G},\ I_{F},$ hence their circumcircles (P), (Q) are congruent. $PI_{G},\ QI_{F}$ are the Euler lines of these 2 triangles cutting each other at their common midpoint O, hence $PQI_{G}I_{F}$ is a parallelogram. Let $A \in FG,\ B \in GE,\ C \in EF$ be feet of internal angle bisectors of $\triangle EFG.$ Quadrilaterals $IFI_{E}G,\ IEI_{F}G,\ IEI_{G}F$ with diagonal intersections A, B, C are all cyclic, having right angles at the common vertices E, F, G, hence $AF \cdot AG = AI \cdot AI_{E},$ $BG \cdot BE = BI \cdot BI_{F},$ $CF \cdot CE = CI \cdot CI_{G}.$ This means AB is radical axis of (O), (P) and AC is radical axis of (O), (Q), so that $AB \perp OP,\ AC \perp OQ.$ The center line PQ of the congruent circles (P), (Q) is the perpendicular bisector of their common chord $II_{E},$ cutting it at its midpoint M. (O) also cuts $II_{E}$ at its midpoint M, being the 9-point circle of $\triangle I_{E}I_{F}I_{G}$ with ortocenter I. Since $II_{E}\equiv EA$ bisects $\angle E,$ M is the midpoint of the arc FG of (O) opposite to E. From the circumcircle (P) of $\triangle II_{E}I_{F}$ and from the cyclic quadrilateral $IEI_{F}G,$

$\angle IPQ \equiv \angle IPM = \frac{\angle IPI_{E}}{2}= \angle II_{F}I_{E}\equiv \angle II_{F}G=\angle IEG = \frac{\angle E}{2}$

and similarly  $\angle IQP = \frac{\angle E}{2}.$ Since $\angle MFG = \angle MGF = \frac{\angle E}{2}$ as well, the isosceles $\triangle FMG \sim \triangle PIQ$ are similar, having equal base angles.

$\angle GEF = 120^\circ\ \Longleftrightarrow\ \angle FMG = 60^\circ\ \Longleftrightarrow\ \triangle FMG$ is equilateral $\Longleftrightarrow\ \triangle PIQ$ is equilateral $\Longleftrightarrow\ PQ = PI = PI_{F}\ \Longleftrightarrow\ PQI_{G}I_{F}$ is a rhombus $\Longleftrightarrow\ OP \perp OQ\ \Longleftrightarrow\ AB \perp AC.$

Could you tell me what $(BIKB')=-1$ and $A'(BIKB')=-1$ means?

[quote="April"][quote="OHO"]Could you tell me what this $(BIKB')=-1$ and $A'(BIKB')=-1$ means?[/quote]
It's harmonious division and harmonically. Do you know it?[/quote]No. :oops:

OK. I have another solution:
$\star\angle B'A'C'=90^\circ$
Let $\Delta$ be the line through $\ A$ and parallel $\ BC. A_{1}B_{1}\cap\Delta=B_{2}, A_{1}C_{1}\cap\Delta=C_{2}.$
We have:
$\frac{AB_{2}}{CA_{1}}=\frac{AB_{1}}{CB_{1}}=\frac{AB}{CB}\frac{AC_{2}}{BA_{1}}=\frac{AC_{1}}{BC_{1}}=\frac{AC}{BC}\implies\frac{AB_{2}}{CA_{1}}.\frac{BA_{1}}{AC_{2}}=\frac{AB}{BC}.\frac{AC}{BC}\implies\frac{AB_{2}}{AC_{2}}.\frac{BA_{1}}{CA_{1}}=\frac{AB}{AC}\implies\frac{AB_{2}}{AC_{2}}.\frac{AB}{AC}=\frac{AB}{AC}\implies\frac{AB_{2}}{AC_{2}}=1$
$\triangle A_{1}B_{2}C_{2}$ have $\angle B_{2}A_{1}C_{2}=90^\circ, AB_{2}=AC_{2}$
$\implies AA_{1}=AB_{2}=AC_{2}\implies\angle AA_{1}B_{2}=\angle AB_{2}A_{1}=\angle CA_{1}B_{2}\implies B_{1}$ is the excenter of triangle $\ ABA_{1}\implies\angle xAC=\angle CAA_{1}=\angle A_{1}AB\implies\angle BAC=120^\circ.$

Very nice. Thank you for this solution.
In the first line, $\frac{AB}{BC}\cdot\frac{AC}{BC}$ should be $\frac{AB}{BC}\cdot\frac{BC}{AC}$

In your first solution, I understand that $BK\cdot IB'=BB'\cdot KI$. Am I right? and could you prove it for me please?

Yes, you are right. You can prove it by applying Menelau's theorem for triangle $\ BIC$ with the line $\overline{A'KC'}$ and $\overline{PB'C'}(P= B'C'\cap BC)$
Note that: $\frac{PB}{PC}=\frac{A'B}{A'C}$.
And you can find some problems and properties of a harmonious division at http://www.mathlinks.ro/Forum/viewtopic.php?highlight=division+harmonious&t=46146

this is quickly baryable. I will post my solution sometime soon.

[quote="April"][quote="Bob75"]$\ A', B'$ and $\ C'$ are the feet of the interior bisectors of a triangle $\ ABC$.
Prove that: $\angle BAC=120^\circ\iff\angle B'A'C'=90^\circ$.[/quote]
Assume that $\ A$ lies on ray $\ Bx$.
$\star\angle BAC=120^\circ$
$\Longrightarrow\angle xAC=\angle A'AC=60^\circ$
$\Longrightarrow A'B'$ is the external bisector of $\angle AA'B$
Similarly $\ A'C'$ is the external bisector of $\angle AA'C$
$\Longrightarrow\angle B'A'C'=90^\circ$
$\star\angle B'A'C'=90^\circ$
Let $I=AA'\cap BB'\cap CC', K=BB'\cap C'A'$
We have:
$\ (BIKB')=-1$
$\Longrightarrow A'(BIKB')=-1$
And $\ B'A'\perp A'K$
$\Longrightarrow\angle AA'B'=\angle CA'B'$
$\Longrightarrow B'$ is the center of $\ B$-excircle of triangle $\ ABC$.
$\Longrightarrow\angle A'AC=\angle xAC=\angle A'AB$
$\Longrightarrow\angle BAC=120^\circ$. :D[/quote]

Nice application of Harmonics! However, is there any angle chasing solution (no harmonic divisions, length bashing, etc.)

Thanks :-)