Find a matrix whose determinant, when set equal to zero, represents the equation of a circle passing through the points $ (\minus{}3,1)$, $ (2,4)$, and $ (5,\minus{}2)$.

--------

If a determinant is a number, how could it represent the equation of a circle?

[quote="Ravi B"]Some of the entries of the matrix will involve $ x$ and $ y$, so the determinant will be a quadratic expression, not a constant.  As a hint, there is a 4x4 matrix that will work for any 3 given points.[/quote]

Oh, I see how the determinant can yield a circle's equation now.  However, how can you tell that it will be a 4x4 matrix, and what numbers should go where?  Is it trial and error?

Thanks!

I've seen this kind of problem before, which is how I knew about the 4x4 matrix.  As a further hint, here's the first row of the matrix:
\[ \begin{bmatrix} x^2 \plus{} y^2 & x & y & 1 \\  ? & ? & ? & ? \\ ? & ? & ? & ? \\ ? & ? & ? & ? \end{bmatrix}\]
Do you see how to complete the matrix so that plugging in each of the given points for $ x$ and $ y$ makes the determinant zero?  Remember that a matrix with two identical rows automatically has determinant zero.

There was a discussion of this a few weeks ago [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1582110#1582110]here[/url].

Ok, I got it now.  Thanks a lot, everyone!


floor function
limit
function
algebra
calculus
calculus computations

Is  $ {Lim}_{n \to \infty} 2.9999...$( 9 repeated n Times)$ \equal{}3$?

Yes. What gave you reason to doubt this?

Then what is the answer to the following problem:

find  $ [lim_{n \to \infty}2.999....]$ where [.] denotes the greatest integer (floor) function.

[quote="Kent Merryfield"]$ \left\lfloor \lim_{n\to\infty}(2 \plus{} .9 \plus{} .09 \plus{} \cdots 9(.1)^n)\right\rfloor \equal{} \lfloor 3\rfloor \equal{} 3.$

$ \lim_{n\to\infty}\lfloor 2 \plus{} .9 \plus{} .09 \plus{} \cdots 9(.1)^n\rfloor \equal{} \lim_{n\to\infty}2 \equal{} 2.$

Being able to say that $ \lim f(x_n) \equal{} f\left(\lim x_n\right)$ lies very near the heart of what it means for a function $ f$ to be continuous. The floor function is not continuous.[/quote]

I understand your argument. but my problem was visualising the floor.
If $ \alpha$ denotes $ \lim_{n\to\infty}(2 \plus{} .9 \plus{} ...)$, isn't it true that $ \alpha \to 3$ but never $ \equal{} 3$ and so  $ \alpha$ is $ < 3$?

So now we have established the issue: you fundamentally do not understand the meaning of "limit."

When we say that $ \lim_{n\to\infty}x_n=L,$ we are not saying that any $ x_n$ must equal $ L.$ We are saying that we can [i]eventually[/i] make $ x_n$ [i]arbitrarily close[/i] to $ L.$

(The words "eventually" and "arbitrarily close" do have precise mathematical definitions.)

Even though $ \frac1n>0$ for all $ \n\in\mathbb{N},$ we still say that $ \lim_{n\to\infty}\frac1n=0.$ We say that because, by making $ n$ large enough, we can makes $ \frac1n$ as close to zero as we please.

You said:

[quote]If $ \alpha$ denotes $ \lim_{n\to\infty}(2 + .9 + ...)$, isn't it true that $ \alpha\to3\dots$[/quote]

Time out! You can't use that language. If we say that $ \alpha$ equals the limit of something (assuming the limit exists), then $ \alpha$ is just a number. One number, fixed and planted. You can't then talk about $ \alpha$ tending to anything.

Mathematics depends on precise language, and what you've given us is not precise language. Even your expressions: $ \lim_{n\to\infty}(2+.9+\cdots)$ or $ \lim_{n\to\infty}2.999\dots$ - did you notice that I didn't quote those directly in my first response? That I felt compelled to create more precise notation just to make clear what we were talking about?

Imprecise language is not a source of profound paradoxes. It's merely imprecise language.

[quote="Kent Merryfield"]So now we have established the issue: you fundamentally do not understand the meaning of "limit."

When we say that $ \lim_{n\to\infty}x_n = L,$ we are not saying that any $ x_n$ must equal $ L.$ We are saying that we can [i]eventually[/i] make $ x_n$ [i]arbitrarily close[/i] to $ L.$

(The words "eventually" and "arbitrarily close" do have precise mathematical definitions.)

Even though $ \frac1n > 0$ for all $ \n\in\mathbb{N},$ we still say that $ \lim_{n\to\infty}\frac1n = 0.$ We say that because, by making $ n$ large enough, we can makes $ \frac1n$ as close to zero as we please.

You said:

[quote]If $ \alpha$ denotes $ \lim_{n\to\infty}(2 + .9 + ...)$, isn't it true that $ \alpha\to3\dots$[/quote]

Time out! You can't use that language. If we say that $ \alpha$ equals the limit of something (assuming the limit exists), then $ \alpha$ is just a number. One number, fixed and planted. You can't then talk about $ \alpha$ tending to anything.

Mathematics depends on precise language, and what you've given us is not precise language. Even your expressions: $ \lim_{n\to\infty}(2 + .9 + \cdots)$ or $ \lim_{n\to\infty}2.999\dots$ - did you notice that I didn't quote those directly in my first response? That I felt compelled to create more precise notation just to make clear what we were talking about?

Imprecise language is not a source of profound paradoxes. It's merely imprecise language.[/quote]

Thanks for clearing my confusion. :oops: But perhaps you could be a little less harsh. :oops:  :blush:

I assure you, Kent is one of the most helpful people on this forum. He's not being harsh, he's just trying to emphasize an important point.


algebra
polynomial
complex analysis
complex analysis theorems

We know that every polynomial $p(z)$ has a root in $C$. Could you give us a proof without using Liouville theorem? (An geometric proof is my requirement)


Let sequence of numbers $ (u_n)$ such that

$ u_0\equal{}0$
$ u_1\equal{}1$
$ u_{n\plus{}1}\equal{}2u_n\minus{}u_{n\minus{}1}\plus{}1$ for all n be natural number, $ n\ge 1$.

Prove that $ A\equal{}4u_n.u_{n\plus{}2}\plus{}1$ be a square number.

Define $ v_n : \equal{} u_n \minus{} u_{n \minus{} 1}$ for $ n\ge 1$. Then the recursive relation implies
\[ v_{n \plus{} 1} \equal{} v_n \plus{} 1,\,\forall n\ge 1.
\]
Combining this with $ v_1 \equal{} 1$ we conclude that $ v_i \equal{} i\,\forall i$. Thus
\[ u_n \equal{} u_0 \plus{} \sum_{i \equal{} 1}^nv_i \equal{} \frac {n(n \plus{} 1)}{2}.
\]
Therefore $ A \equal{} n(n \plus{} 1)(n \plus{} 2)(n \plus{} 3) \plus{} 1 \equal{} (n^2 \plus{} 3n \plus{} 1)^2$. Q.E.D.

Why is this in the Inequalities forum?


an 84 meter lenght of cable is cut so that one peisce is 18 meters  longer than the other. find the lenght of each peice

Call the shorter piece $ \alpha$. We then have the equation:

$ \alpha \plus{} (18 \plus{} \alpha) \equal{} 84$

Remove the parentheses.

$ 18 \plus{} 2 \alpha \equal{} 84$

Isolate the variable.

$ 2 \alpha \equal{} 84 \minus{} 18$

$ 2 \alpha \div 2 \equal{} 66 \div 2$

$ \alpha \equal{} 33$

Thus the smaller piece is 33 meters long, and the longer piece is $ 33 \plus{} 18 \equal{} 51$ meters long.


you could shorten it down
$1^2=(+-1)^2$
$1=-1$
 :P
this is wrong a bit because $a^2=b^2$
doesnt mean $a=b$


If xyz=6 minimize 2z+2sqrt(x^2+y^2).

The original problem was: A ribbon is tied around a box with volume 6 such that it goes along an edge than along the diagonal of the top side than down the opposite edge and along the diagonal of the bottom side.

Does minimize mean finding the lower or upper bound....?  :|
$ 2z\plus{} \sqrt {x^2\plus{}y^2} \geq 2z \plus{} \frac{x\plus{}y}{\sqrt {2}} \geq 3(xyz)^\frac {1}{3}\equal{}3(6)^\frac {1}{3}$


I need to proof this theorem or find e-book with this proof, please help me:

$ t_{1}: (X,B) \longrightarrow(T_{1},T_{1})$ is sufficient statistic for $ \theta$ in experiment $ \varepsilon \equal{} (X,B,P_{\theta}: \theta \in \Theta )$

$ t_{2}: (T_{1},T_{1}) \longrightarrow(T_{2},T_{2})$ is sufficient statistic for $ \theta$ in experiment which is generate by $ T_{1}$

then proof that $ T_{2}(T_{1}(x))$ is sufficient for $ \theta$ in $ \varepsilon$

I can't edit so i posted again. Sorry.  :blush:  Please help.

I need to proof this theorem or find e-book with this proof, please help me:

$ t_{1}: (X,B) \longrightarrow(T_{1},T_{1})$ is sufficient statistic for $ \theta$ in experiment $ \varepsilon \equal{} (X,B,P_{\theta}: \theta \in \Theta )$

$ t_{2}: (T_{1},T_{1}) \longrightarrow(T_{2},T_{2})$ is sufficient statistic for $ \theta$ in experiment which is generate by $ t_{1}$

then proof that $ t_{2}(t_{1}(x))$ is sufficient for $ \theta$ in $ \varepsilon$

The problem is that your terminology is unclear :( What is $ (T_1,T_1)$ for example? And what is "an experiment generated by $ t_1$"?

[quote="fedja"]The problem is that your terminology is unclear :( What is $ (T_1,T_1)$ for example?And what is "an experiment generated by $ t_1$"? [/quote]

This are sets from statistic definition.

I have following statistic definition:

Statistic $ t$ is measurable transformation $ t: (X,B) \longrightarrow(T,F)$, where $ (X,B,P_{\theta}: \theta \in \Theta )$ is experiment.

Then experiment $ (T,F,P_{\theta}: \theta \in \Theta )$ we call induced (generated) by statistic $ t$.

In this theorem we have two the same sets $ (T_{1},T_{1})$ instead $ (T,F)$

Because $ B$ is $ \sigma$-algebra Borel sets and in this theorem $ t_{2}$ is a statistic so i think $ T_{1}$ and $ T_{2}$ are $ \sigma$-algebras


(A. Myakishev, 8--9) In the plane, given two concentric circles with the center $ A$. Let $ B$ be an arbitrary point on some of these circles, and $ C$ on the other one. For every triangle $ ABC$, consider two equal circles mutually tangent at the point $ K$, such that one of these circles is tangent to the line $ AB$ at point $ B$ and the other one is tangent to the line $ AC$ at point $ C$. Determine the locus of points $ K$.

Locus is a circle with center A and radius of sqrt[(R^2 + r^2)/2].

Let $A(0,0)$; choose the x-axis through the given point $B$, then $B(r,0)$.
Point $C(a,b)$ on the circle with radius $R$, so $a^{2}+b^{2}=R^{2} \quad\quad\quad (1)$.

The first circle, touching $AB$ in $B$, has midpoint $M_{1}(b,n)$ and radius $n$.
The second circle, touching $AC$ in $C$, has midpoint $M_{2}$ and also radius $n$.
Equation of the line $CM_{2}\ :\ y-b=-\frac{a}{b}(x-a)$;
equation of the circle, midpoint $C$ and radius $n\ :\ (x-a)^{2}+(y-b)^{2}=n^{2}$.
Solving the system of this two equations, gives us $M_{2}(a+\frac{bn}{R},b-\frac{an}{R})$.

It is possible to calculate $n$ from $M_{}M_{2}=2n$ or $(a+\frac{bn}{R}-r)^{2}+(b-\frac{an}{R}-n)^{2}=4n^{2} \quad\quad\quad (2)$.

Point $K(x,y)$ is the midpoint of $M_{}M_{2}$:
$a+\frac{bn}{R}+r=2x$ and $b-\frac{an}{R}+n=2y \quad\quad\quad (3)$.

$a+\frac{bn}{R}=2x-r$ and $b-\frac{an}{R}=2y-n$.
From (2): $(2x-r-r)^{2}+(2y-n-n)^{2}=4n^{2}$ and $n = \frac{(x-r)^{2}+y^{2}}{2y} \quad\quad\quad (4)$.

Solving $a$ and $b$ from (3): $a = \frac{R(2Rx-2ny+n^{2}-rR)}{n^{2}+R^{2}}$ and $b=\frac{R(2nx+2Ry-nr-nR)}{n^{2}+R^{2}}$.

Using (1), we have $4x^{2}-4rx+4y^{2}-4ny+r^{2}-R^{2}=0$.
Using (4), we have $2x^{2}+2y^{2}-r^{2}-R^{2}=0$.


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For which cardinalities $ \kappa$  do antimetric spaces of cardinality $ \kappa$ exist?
$ (X,\varrho)$ is called an $ \textit{antimetric space}$ if $ X$ is a nonempty set, $ \varrho : X^2 \rightarrow [0,\infty)$ is a symmetric map, $ \varrho(x,y)\equal{}0$ holds iff $ x\equal{}y$, and for any three-element subset $ \{a_1,a_2,a_3 \}$ of $ X$ \[ \varrho(a_{1f},a_{2f})\plus{}\varrho(a_{2f},a_{3f}) < \varrho(a_{1f},a_{3f})\] holds for some permutation $ f$ of $ \{1,2,3 \}$.

[i]V. Totik[/i]

I posted this one [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1438234035&t=74263]here[/url].


inequalities
geometry
circumcircle
trigonometry
function
Euler
inequalities proposed

Let $ ABC$ be a triangle such that: $ A \ge B \ge \frac{\pi}{3} \ge C$.Prove that:
$ \sqrt{\frac{b\plus{}c\minus{}a}{a}}\plus{}\sqrt{\frac{c\plus{}a\minus{}b}{b}}\plus{}\sqrt{\frac{a\plus{}b\minus{}c}{c}} \ge 3$
Furthermore,we have:
$ \sqrt{\frac{b\plus{}c\minus{}a}{a}}\plus{}\sqrt{\frac{c\plus{}a\minus{}b}{b}}\plus{}\sqrt{\frac{a\plus{}b\minus{}c}{c}} \ge \sqrt{9\plus{}\frac{R\minus{}2r}{R}}$
Where $ r,R$ be the radius and circumradius of a triangle $ ABC$
And ,we can make stronger.I will post later :)

[quote="quykhtn-qa1"]Let $ ABC$ be a triangle such that: $ A \ge B \ge \frac {\pi}{3} \ge C$.Prove that:
$ \sqrt {\frac {b \plus{} c \minus{} a}{a}} \plus{} \sqrt {\frac {c \plus{} a \minus{} b}{b}} \plus{} \sqrt {\frac {a \plus{} b \minus{} c}{c}} \ge 3$
Furthermore,we have:
$ \sqrt {\frac {b \plus{} c \minus{} a}{a}} \plus{} \sqrt {\frac {c \plus{} a \minus{} b}{b}} \plus{} \sqrt {\frac {a \plus{} b \minus{} c}{c}} \ge \sqrt {9 \plus{} \frac {R \minus{} 2r}{R}}$
Where $ r,R$ be the radius and circumradius of a triangle $ ABC$
And ,we can make stronger.I will post later :)[/quote]
Dear Virgil Nicula,hope you like this inequality :) .
May be it's easy :(

[color=darkblue][b]This inequality is very nice and difficult. I will also present a stronger inequality than the proposed one. But first, let's prove the following lemmas :[/b][/color]

[quote][color=darkred][u][b]Lemma 1.[/b][/u] Let $ABC$ be a triangle so that its angles satisfy a relation of the type : $A\ge B\ge 60^{\circ}\ge C$. Prove that:

\[\boxed{\ s\ \ge\ \sqrt 3\cdot (R+r)\ }\ .\][/color]
[/quote]

[color=darkblue][u][b]Proof.[/b][/u] Since the angles of $\triangle ABC$ satisfy $A\ge B\ge 60^{\circ}\ge C \implies \left\{\begin{array}{ccc}
\tan\frac A2-\frac {\sqrt 3}3\ \ge\ 0 \\ \\
\tan\frac B2-\frac {\sqrt 3}3\ \ge\ 0 \\ \\
\tan\frac C2-\frac {\sqrt 3}3\ \le\ 0 \end{array}\right|\bigodot\implies $

$\implies\ \prod\left(\tan\frac A2-\frac {\sqrt 3}3\right)\ \le\ 0\iff\prod\tan\frac A2$ $-\frac {\sqrt 3}3\cdot\sum\tan\frac B2\tan\frac C2+\frac 13\cdot\sum\tan\frac A2-\frac{\sqrt 3}9\ \le\ 0\iff$

$\iff\frac rs-\frac {\sqrt 3}3+\frac 13\cdot\frac{4R+r}s$ $-\frac{\sqrt 3}9\ \le\ 0\iff 9r+3(4R+r)-4s\sqrt 3\ \le\ 0 \iff\boxed{\ s\ \ge \sqrt 3\cdot (R+r)\ }\ .$[/color]

[quote][color=darkred][u][b]Lemma 2.[/b][/u] In any triangle $ABC$ the following inequality holds:

\[\boxed{\sqrt{\frac {b+c-a}a}+\sqrt{\frac {c+a-b}b}+\sqrt{\frac {a+b-c}c}\ge \frac {s+3r\cdot (2-\sqrt 3)}{\sqrt {2Rr}}}\ .\][/color]
[/quote]

[color=darkblue][u][b]Proof.[/b][/u] Apply the [u][b]Jensen[/b]'s inequality[/u] for the [u]convex[/u] function $f(x)=\tan\frac x4$ on the interval $\left(0\ ,\ \frac {\pi}4\right)$ :

$3\cdot\tan\left(\frac {\frac A4+\frac B4+\frac C4}3\right)\ \le\ \tan\frac A4+\tan\frac B4+\tan\frac C4\color{white}{.}$ $\iff \boxed{\ 3\cdot (2-\sqrt 3)\ \le\ \tan\frac A4+\tan\frac B4+\tan\frac C4\ }\ \ (\ast)$ .

On the other hand, we have: $\sum\tan\ \frac A4=\sum\frac{1-\cos\frac A2}{\sin\frac A2}=\sum\frac{1-\frac{s-a}{AI}}{\frac{r}{AI}}=\sum\frac{AI-s+a}{r}=\frac{AI+BI+CI-s}{r}$.

Thus, the $(\ast)$ is now equivalent to: $\boxed{AI+BI+CI \ge s+3r\cdot (2-\sqrt 3)}\ (\ast\ast)$. Since $\boxed{AI=\sqrt {2Rr}\cdot\sqrt{\frac {b+c-a}a}}$.

and so on, the inequality $(\ast\ast)$ finally becomes : $\boxed{\ \sqrt{\frac {b+c-a}a}+\sqrt{\frac {c+a-b}b}+\sqrt{\frac {a+b-c}c}\ \ge\ \frac {s+3r\cdot (2-\sqrt 3)}{\sqrt {2Rr}}\ }$[/color]

[quote][color=darkred][u][b]Problem.[/b][/u] Prove that in any triangle $ABC$ that satisfies a relation of type $A\ge B\ge 60^{\circ}\ge C$, the following inequality holds:

\[\boxed{\sqrt{\frac {b+c-a}a}+\sqrt{\frac {c+a-b}b}+\sqrt{\frac {a+b-c}c}\ \ge\ \frac{6r+\sqrt 3\cdot (R-2r)}{\sqrt{2Rr}}}\ \ge\ \sqrt{9+\frac {R-2r}R}\ \ge\ 3\ .\][/color]
[/quote]

[color=darkblue][u][b]Proof.[/b][/u]

$\blacktriangleright$ Apply [u]lemma [b]1[/b][/u] [b]&[/b] [u]lemma [b]2[/b][/u] : $\implies\sum\sqrt{\frac {b+c-a}a}\ \stackrel{(2)}{\ge}\ \frac {s+3r\cdot (2-\sqrt 3)}{\sqrt {2Rr}}\ \stackrel{(1)}{\ge}\ \frac{6r+\sqrt 3\cdot (R-2r)}{\sqrt{2Rr}}$ .

$\blacktriangleright$ The inequality $\frac{6r+\sqrt 3\cdot (R-2r)}{\sqrt{2Rr}}\ge\sqrt{9+\frac{R-2r}R}$ reduces to : $6r+\sqrt 3\cdot (R-2r)\ge\sqrt{2r(10R-2r)}$

$\iff 3(R-2r)^2+12\sqrt 3\cdot r(R-2r)+36r^2\ \ge\ 20Rr-4r^2\color{white}{.}$ $\iff (R-2r)\cdot\left[3R+(12\sqrt 3-26)\cdot r\right]\ge 0$ .

The last inequality is obvious according to [u][b]Euler[/b]'s inequality[/u] i.e. $\boxed{\ R\ \ge\ 2r\ }$ .Therefore, the problem is now solved :lol: .[/color]


[hide="Large hint"] This should remind you of the identity $ \cos^2 \alpha \plus{} \cos^2 \beta \plus{} \cos^2 \gamma \plus{} 2 \cos \alpha \cos \beta \cos \gamma \equal{} 1$, where $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$.  What are $ x, y, z$ in terms of these cosines?  :) [/hide]

[quote="delegat"][b]t0rajir0u
[/b]

Condition is not $ x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} 2xyz \equal{} 1$ :roll:[/quote]

Wat t0rajir0u means is $ x\equal{}\frac {\cos A \cos B}{\cos C}, y\equal{}\frac {\cos B \cos C}{\cos A}, z\equal{}\frac {\cos C \cos A}{\cos B}$

actually one could have $ \sqrt{xy}\equal{}p$ :wink:

[quote="radio"]Let $ x,y,z > 0$ be such that $ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1$.

Prove that there exist $ p,q,r > 0$ such that $ x \equal{} \frac {p}{q \plus{} r},y \equal{} \frac {q}{r \plus{} p},z \equal{} \frac {r}{p \plus{} q}$[/quote]
Let $ x \equal{} \frac {p}{q \plus{} r}$ and $ y \equal{} \frac {q}{r \plus{} p}.$ It's probably because $ xy\neq1.$  :wink:
Then $ z \equal{} \frac {1 \minus{} xy}{x \plus{} y \plus{} 2xy} \equal{} \frac {r}{p \plus{} q}.$ :wink:


calculus
derivative
integration
real analysis
real analysis theorems

hi
does anyone knows the proof (or where I can fin it) of that:

$\frac{\delta f}{\delta x \delta y}=\frac{\delta f}{\delta y \delta x}$

(not really the right symbol for second derivate $\delta$ , what s the command?)

thanks!

ledurt

[quote](not really the right symbol for second derivate \delta , what s the command?) [/quote]
The right symbol is \partial, and you need a square in there:

$\frac{\partial^2f}{\partial x\partial y}=\frac{\partial^2f}{\partial y\partial x}$

As for the theorem: you need a hypothesis that both of these partial derivatives are continuous in an open set. I found it in Rudin's [i]Principles of Mathematical Analysis[/i] as Theorem 9.34. The proof involves taking integrals of these derivatives and interchanging the order of integration.

:blush:
geez...i'm ashame now, i forgot the $^2$.... :blush:

and thanks for the info, i ll look it up!

ledurt

[quote="Kent Merryfield"]$\frac{\partial^2f}{\partial x\partial y}=\frac{\partial^2f}{\partial y\partial x}$

As for the theorem: you need a hypothesis that both of these partial derivatives are continuous in an open set.[/quote]
A weaker condition: both partial derivatives exist and one of them is continuous.

[quote="Kent Merryfield"][quote="liyi"]We do not need the [i]existence[/i] [u][b]continuity[/b][/u] of both, just the [i]existence[/i] [u][b]continuity[/b][/u] of one partial derivative is okay.[/quote]
You meant to say "continuity", not "existence", right?[/quote]
Yes....sorry for the serious typo...

[quote="liyi"][quote="Ledurt"]eee, what do you mean by "weaker condition" ?
ledurt[/quote]
Kent Merryfield says that $\frac{\partial^2 f}{\partial y\partial x}=\frac{\partial^2 f}{\partial x\partial y}$ if both of these partial derivatives are continuous;
and in fact, the condition can be a little weaker: We do not need the existence of both, just the existence of one partial derivative is okay.[/quote]

 Hey, that's Schwarz's criterion.If you want a proof,i'm here.I also know a similar theorem which is due to Young. ;)


inequalities
calculus
function
inequalities proposed

Edit: I will get back to my initial idea. putting a=x+y, cyclic, we get to prove:

Putting x+y+z=1: $3 \ge \sum_{\text{cyc}} \frac{1}{1 - z + 9z^3}$.

Or putting x+y+z=3: $1 \ge \sum_{\text{cyc}} \frac{1}{3 - z + z^3}$.

I can solve this the ugly calculus-way, or the long muirhead-way, but I guess you have a much nicer solution in thoughts?

Ah darn, it's convex from 0 to 1 I confused the intervals again :D

I'm sorry, I'll check for a better proof.

[quote="xtar"]Let  $a,b,c $  be the sides of a triangle, $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle. Prove that:
\[\frac{3}{s^3}\ge \frac{1}{as^2+9(s-a)^3}+\frac{1}{bs^2+9(s-b)^3}+\frac{1}{cs^2+9(s-c)^3}\][/quote]

Normalize $s$ to $1$, so: given $x + y + z = 1$, we want to prove:
$3 \ge \sum_{\text{cyc}} \frac{1}{1 - z + 9z^3}$
First, if all of the variables are between $0$ and $\frac{2}{3}$, we have: $-\frac{(2x+1)(3x-2)(3x-1)^2}{3} \ge 0 \Rightarrow \frac{1}{1-x+9x^3} \le \frac{5}{3} - 2x$, so adding these inequalities yields the desired. Now, let $f(t) = \frac{1}{1 - t + 9t^3}$. Suppose there is some variable, WLOG $x$, such that $x \ge \frac{2}{3}$. Then $f(x) \le \frac{1}{3}$, so if $f(x) + f(y) + f(z) > 3$, for sake of contradiction, then WLOG $f(y) > \frac{4}{3}$, i.e. $1 - y + 9y^3 < \frac{3}{4}$, i.e. $1 - 4y + 36y^3 < 0$, and we can easily see this never happens (define $g(y) = 1 - 4y + 36y^3$, then note $\frac{dg}{dy} = 108y^2 - 4$, so calculate the critical points, etc.)

hope this works

Yes, but that was exactly what I wanted to avoid. ;) I think if he asks this problem, then there'll be a nicer solution for $3 \ge \sum_{\text{cyc}} \frac{1}{1 - z + 9z^3}$.

(Or when putting x+y+z=3, you get $1 \ge \sum_{\text{cyc}} \frac{1}{3 - z + z^3}$, which also seems nice, but opposes us to the same problem as above)

it's pretty nice ... the only part I actually take offense with is resorting to calculus to prove $36t^3 - 4t + 1 > 0$ but calculus really wasn't necessary for that: simply note that if $t < \frac{1}{4}$, we have $1 > 4t$. Then if $t > \frac{1}{3}$, we have $36t^3 > 4t$. Finally, if $\frac{1}{4} \le t \le \frac{1}{3}$, we have $36t^3 > 2t$ and $1 > 2t$, so in all cases, we're done.

Here is my solution for the problem :
    $ x+y+z =3 $ . prove that : $ \frac{1}{x^3-x+3} + \frac{1}{y^3-y+3} + \frac{1}{z^3-z+3} \leq 1 $

  [b]Case 1: [/b] $ x, y , z \leq 2 $
 We have : $ (x-1)^2(2-x)(2x+3) \geq 0 $
  So $ \frac{1}{x^3-x+3} \leq \frac{5}{9} - \frac{2}{9}x  $
  Similarly , we have another 2 ineqs
  Add them all ---> Q.E.D
  [b]Case 2: [/b]  Suppose $ x,y \leq 1 , z \geq 2 $
 We have $ x^3-x+3 \geq 2.5 $
 So $ \frac{1}{x^3-x+3} + \frac{1}{y^3-y+3} \leq 0.8  (1) $
 Due to $ z \geq 2 $ , we have $ \frac{1}{z^3-z+3} \leq \frac{1}{9}   (2) $
   (1)+(2) ---> Q.E.D

[quote="xtar"][quote="Peter VDD"]I think your inequality should have the reversed sign. :) I will prove it:

\[\frac{3}{s^3}\le \frac{1}{as^2+9(s-a)^3}+\frac{1}{bs^2+9(s-b)^3}+\frac{1}{cs^2+9(s-c)^3}\]

Put $a=x+y, b=y+z, c=z+x$. We get: \[\frac{3}{(x+y+z)^3} \le \sum_{cyc}\frac{1}{(x+y)(x+y+z)^2+9z^3}\]

The inequality is homogenous so we can assume $x+y+z=30$. We get: \[\frac{1}{9000} \le \sum_{cyc}\frac{1}{900(x+y)+9z^3} = \sum_{cyc}\frac{1}{27000-900z+9z^3} \text{, thus } {\sum_{cyc}\frac{1}{3000-100x+x^3} \ge \frac{1}{1000}}\]

But the function $f(x) = \frac{1}{3000-100x+x^3}$ is convex, so Jensen gives us:
\[\dfrac{f(x)+f(y)+f(z)}{3} \ge f\left(\dfrac{x+y+z}{3}\right)=f(10)\]

Which is, written out: \[\sum_{cyc}\frac{1}{3000-100x+x^3} \ge 3 \cdot f(10) = \frac{3}{3000}=\frac{1}{1000}\]

Q.e.D.[/quote]

The inequality is true
The function $f(x) = \frac{1}{3000-100x+x^3}$ is not convex

.[/quote]

   I'm very interested in your assumption $ x+y+z =30 $ , Peter  :lol:

Well, I was looking for f to be convex on [0,1]... so that was true for 30 onwards :P but I forgot I then needed convexity on [0,30] instead :blush:


(chances are you're here and not in our own forum).

because of Missouri's awesomeness compared to Kansas...

Oh definitely. Pew pew. *fires lasers across state line*

And in terms of post count, I am now at the end of WWII. Hmm.

*AHH- gets hit by laser pianoforte shot across MO-KS border*

*falls down to the floor and dies*

*comes back to life*

*shoots laser across the state of Missouri at painoforte's house.*

Remember to play the [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=262852]Mathcounts Lottery[/url].

then pianoforte retaliates with an atom bomb...

but the beastliness of tinytim, justinahmann, and math154 hold it back and completely demolish Kansas

YAY!  :D

So... like, when you get back (which I know is like late evening), could someone post the results, like top indivs, name + school, and top school(s).

(aiming for 2000+ posts)

Congratulations to mewto, math154, tinytim, and mewto's sister (lol).

I'm honored to know all four of you in real life  :)

[quote="mathemagician1729"]Kansas State Results

1. Shyam Narayanan (Owned everyone in CD and had a substantially higher written score than anyone else)
2. Kevin Cao (Hope I spelled that write)
3. Michael Zhou
4. Jack Chen

Sadly I was fourth written, but Jack killed me in CD because I blanked out and now I'm fifth.  :(

Now that I failed Mathcounts, does anyone know any high school competitions (other than the AMC) that I could do?[/quote]

Also, our pwnage team (3rd plce winner, mathemagician1729, me, and other indian guy) got 1st.  lakewood got 2nd.  oxford got 3rd.

Thanks for the info. All BV kids too, should make for easy practices.

Mathe: Sorry for that, that kinda sucks. Do MO ARML, GPML and... there's really not that much.

[quote="tinytim"]1st- Mewto55555
2nd- tinytim
3rd- math154
4th- mewto's sister

Ladue- 1st place team[/quote]

Meet u at nats  :wink: .

I do not believe your state has had its competition yet. Cocky much?


geometry
perimeter
inequalities
congruent triangles

How many non-congruent triangles with perimeter 7 have integer side lengths?

$\left(A\right) 1$
$\left(B\right) 2$
$\left(C\right) 3$
$\left(D\right) 4$
$\left(E\right) 5$

[hide="solution"]$(1, 3, 3)$, $(2, 2, 3)$,  there are 2 answers or $\boxed{B}$.[/hide]

[quote="turboturtle"][hide] for the number 7, we can list the possibilities (since there aren't that many).
115
124
132

214
223

331

so... we also find out that
by triangular inequalities, there aren't any triangles for which two sides added together is less or equal to the third side.
so, we have
223, and 331,
giving the answer 2,
or $B$
[/hide][/quote]
[hide]We can reduce the number of possibilites we need to check. The largest length a side length could be is 3, because if a sidelength is 4, than the sum of the other two sides is 3, which contradicts Triangle Inequality.
Testing, 2,3,3 and 3,3,1 are the only solutions. $B$.[/hide]


[hide]you can write the equation: $4x=x+36$. Solve for x: $3x=36$. Divide by 3 on both sides: $\boxed{x=12}$[/hide]

[hide]12[/hide]

[color=darkred][size=75][Hide your answers and show some work!][/size][/color]

Remember to hide your answers and show your work! :)

[hide=":D - Click to reveal hidden answer..."]

Equation:

4x = x + 36 ????

4x - x = 36

3x = 36

x = 12

The number "x" is 12.

[/hide]

[quote="MCrawford"]The product of a number with 4 is equal to the sum of the number and 36.  What is the number?[/quote]
[hide]i think the number is $12$,and i think it is true.[/hide]

[quote="ashegh"]the solutions are enough i think...

but the problem is so nice ;)[/quote]
yes i am agree with you ashegh. :idea:

mecrawford,  do we have any contest for the basics?

Eival Ashegh and Jensen you 2 are so Strong because after a long calculation I reached to Number $12$.Excellent

Kheili khafanit baba :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D  :D
[color=blue]It means[/color]:So Strong

eivaly az khodete amir joon.to khodet strong hasti ke ma ro strong mibini. :lol:

thanks alot dear amir ;)

jensen means:thanks alot dear amir.ure eyes see us strong,but we are not too...

why are you guys posting here?  ;)  go back to the olympiad section where you belong  :P


analytic geometry
calculus
calculus computations

A particle is moving along the curve $ y\equal{}\sqrt{x}$. As the particle passes through the point $ (4.2)$, its x-coordinate increases at a rate of $ 3 \text{cm/s}$. How fast is the distance from the particle to the origin changing at this point?

I got an answer of $ \frac{9}{\sqrt{5}} \text{cm/s}$ but I don't have the solutions manual so I don't know if this is right. Can someone just let me know what they get?

I used $ x^{2}\plus{}y^{2}\equal{}z^{2}$ with $ y\equal{}\sqrt{x}$ so that $ x^{2}\plus{}x\equal{}z^{2}$.

I differentiated $ z\,\equal{}\,\sqrt{x^2\,\plus{}\,x}$ by $ t$ for the solution and arrived at $ \frac{dz}{dt}\,\equal{}\,\frac{27}{4\sqrt{5}}.$


It is known Gaus's summ
\[ \sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{n}(\cos{\frac {2\pi (i^2 \plus{} j^2)}{n}}) \equal{}Re(\sum_{j\equal{}1}^n\exp(\frac{2\pi i  j^2}{n}))^2\equal{} n\sin{\frac {n\pi}{2}}\]

[quote="Rust"]It is known Gaus's summ
\[ \sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{n}(\cos{\frac {2\pi (i^2 \plus{} j^2)}{n}}) \equal{} Re(\sum_{j \equal{} 1}^n\exp(\frac {2\pi i j^2}{n}))^2 \equal{} n\sin{\frac {n\pi}{2}}\]
[/quote]
another nice identity here
$ \sum_{i \equal{} 1}^{n}\sum_{j \equal{} 1}^{n}(cos(\frac {2\pi(i^2 \minus{} j^2)}{n})) \equal{} n \plus{} n\cos{\frac {n\pi}{2}}$


Let f(n) = ((4n-2)/(n+1))*f(n-1), n >1
Supposed f(2) and f(3) are prime numbers. Prove that f(n) is Not prime for any other integer values n>1.

[hide="solution"]
We notice that $ f(3) \equal{} \frac {5}{2}f(2)$ so the only way both $ f(2)$ and $ f(3)$ are prime is if $ f(2) \equal{} 2$ and $ f(3) \equal{} 5$.  We can then write the unique solution to the recurrence equation:
\[ f(n) \equal{} 2^n\frac {\prod _{k \equal{} 1}^n{(2k \minus{} 1)}}{(n \plus{} 1)!}
\]
The numerator is a product of all odds up to $ 2n \minus{} 1$, and therefore certainly contains primes that the denominator does not.

A question one might ask is if $ f(n)\in\mathbb{Z}$ for all $ n$.  It is, though that requires a few more steps.
[/hide]


If $ x\neq1$, $ y\neq1$, $ x\neq y$ and
\[ \frac{yz\minus{}x^{2}}{1\minus{}x}\equal{}\frac{xz\minus{}y^{2}}{1\minus{}y}\]
show that both fractions are equal to $ x\plus{}y\plus{}z$.


if A. B, C are the angles of a triangle, then sin A + sin B + sin C <= 3 :rt3: /2. this is directly proved from jensen, since sin is concave. how do we prove cos A + cos B + cos C <=3/2? cos is not convex or convace on [0, :pi: ]?

One approach is to apply Jensen to the two smallest angles.

It's basically an application of convexity.  Here it is in the special case we were using:  Let f be a concave function.  Then f(x) + f(y) + f(z) :le: 3 f((x + y + z) / 3).

[quote="Ravi B"]Another approach is to use the identity cos A + cos B = 2 cos((A+B)/2) cos((A-B)/2) :le: 2 cos((A+B)/2), and its permutations.  We just need to show that cos((A+B)/2) + cos((A+C)/2) + cos((B+C)/2) :le: 3/2.  But those angles are all acute, so we can apply Jensen after all.[/quote]

Why are these angles acute?

Also, how does one prove cot A + cot B + cot C :ge: :sqrt:3, where A, B, and C are the angles of a triangle?

[quote]Why are these angles acute?[/quote]

Because (a + b + c)/2 = 90 so (a + b)/2 < 90 and thus is acute.

MysticTerminator wrote:how does one prove cot A + cot B + cot C :ge: :sqrt:3, where A, B, and C are the angles of a triangle?

Hint in spoiler:[hide]Note that cotangent is convex between 0 and 90 degrees.  Assume that angles A and B are acute.  By convexity, we can assume that A = B.  We can then express the inequality purely in terms of C.  Use the identity connecting cot C and cot(C/2) to express the inequality as a quadratic in cot(C/2).[/hide]

I don't know if there is a better method.

Another way to prove
[tex]\cos{A} + \cos{B} + \cos{C} \le 3/2[/tex]
is to use
[tex]\displaystyle\cos{A} + \cos{B} + \cos{C} = \frac{r}{R} + 1[/tex]
and
[tex]2r \le R[/tex].

[quote="MysticTerminator"]cot A + cot B + cot C :ge: :sqrt:3, where A, B, and C are the angles of a triangle?[/quote]

An other solution :

Cot A=(b^2 +c^2 - a^2 )/4S by laws of sines + cosines

So  :Sigma: Cot A = :Sigma: a^2 / 2S (S = surface)

:Sigma: a >= ab + bc+ ca = 2S (:Sigma: 1/sinA)

Now 1/sin(x) is convex so (:Sigma: 1/sinA) >= 3/sin(:pi:/3) = 2 :rt3:


inequalities
induction
graph theory
combinatorics unsolved
combinatorics

The plane is divided into regions by $ n \ge 3$ lines, no two of which are parallel, and no three of which are concurrent. Some regions are coloured , in such a way that no two coloured regions share a common segment or half-line  of their borders. Prove that the number of coloured regions is at most $ \frac{n(n\plus{}1)}{3}$

Note that two regions share a common segment or half-line only if they also share a vertex. Furthermore, every pair of lines forms exactly one vertex and there are $ 1 \plus{} \binom{n \plus{} 1}{2}$ regions. Let $ v_R$ be the number of vertices of region $ R$ and denote $ V \equal{} \sum v_R$. Noticing that each intersection adds four vertices to $ V$ and that each line intersects every other line in distinct points, there $ v \equal{} 4\binom{n}{2}$, and this is $ \le$ the number of sides with double counting (it would be $ \equal{}$, but we have the unbounded regions). Finally, we observe that every side is counted exactly twice, so there are at least $ 2\binom{n}{2}$ sides.

Suppose there were more than $ \frac {n(n \plus{} 1)}{3}$ colored regions. Since no side is counted twice here, there are exactly $ \frac {2n(n \plus{} 1)}{3}$ sides counted. But this is $ > 2\binom{n}{2}$; contradiction.

This seems to make the bound seem relatively weak, so I probably screwed up somewhere. It's late though, so I don't have time to check.

EDIT:

The above is flawed in two ways. Firstly, I only proved there are at least $ 2\binom{n}{2}$ sides with double counting, when I need an upper bound rather than zero bound. It turns out there are *exactly* $ 2\binom{n}{2} \plus{} 2n$ sides.

Furthermore, the last inequality I used isn't actually true.  :oops: I have a solution now, but unfortunately once again it's late and I don't have time to type it up. It rests on the lemma that there are always exactly three regions with 2 sides.

It's worth noting that the desired bound is 2/3 the total number of regions, so this may lead to a nicer proof.

Now that it's morning and I'm more sane, here's my solution:

[b]Lemma:[/b] There are always $ 2n$ unbounded regions.
[b]Proof:[/b] Induction. Note that all other regions have the same number of vertices and sides and unbounded regions have one more side than vertex; hence the total side sum count is exactly $ 2\binom{n}{2}+2n=n(n+1)$ as I noted above.

[b]Lemma:[/b] There are always exactly three regions with two sides.
[b]Proof:[/b] They must be unbounded; induction by considering cases of what unbounded regions the $ n+1$th line cuts through.

So suppose we have chosen $ \frac{n(n+1)}{3}$ regions and none share a side; we want to prove that this consumes more than the number of sides itself. To do this, we use a weighted average. Let $ x$ be the number of regions with exactly three sides; if we can show that $ 6+3x+4(\frac{n(n+1)}{3}-x-3)> n(n+1)$, or rearranging, $ 3x<n(n+1)-18$. Assume for the sake of contradiction $ 3x\ge n(n+1)-18$. However, $ 3x$ is simply the number of total edges we can have in summing over all the 3-regions, with $ n(n+1)$ edges total; we conclude that we can have at most $ 18$ edges in the sum of all other regions. Since there are always exactly 3 regions with 2 sides, we can have at most $ 12$ edges for 4+-sided ones, or at most 3 regions with more than 3 sides.

[b]Lemma:[/b] For all $ n\ge 5$, there are at least 4 4+-sided regions.
[b]Proof:[/b] Induction; use casework similar to our second lemma.

We can check $ n=3$ and $ n=4$ by hand.

[quote="fwolth"][b]Lemma:[/b] There are always exactly three regions with two sides.[/quote]  If we have five lines in the form of a regular pentagon, are there not five such regions?

Also, the first lemma has a very simple explanation: imagine zooming out so far that all of the pairwise intersections of the lines appear to collapse to a single point.

Hmmm this is true. Let's try this then.

Since no three lines concur and no two are parallel, there are $ \binom{n \plus{} 1}{2} \plus{} 1$ regions formed, so it suffices to prove that we have fewer than $ 2/3$ of the regions colored. Now imagine the following process: Pick an arbitrary region and take one region bordering it and call them group $ G_1$. Repeat until we can no longer do so. Then we will have a number of groups $ G_k$ and various one-region "pockets" not included in any group but bordered by regions in groups (we cannot have more than two bordering regions in a pocket or we could make a group of them).

Clearly, only one of the two regions in each group can be colored. If we can show that we can select the groups in a way such that there are more groups than pockets, then we're done. To do this, we prove the stronger statement that in any connected bipartite graph where every vertex has at least degree two (lemma: the graph resulting by interpreting regions as vertices is bipartite. proof: no odd cycles, assume there was one, but each line passes through cycle twice, contradiction) we can remove pairs of adjacent vertices until fewer than a third of them are left.

Noticing the degree-two condition means that the subgraph of lesser order in the bipartite representation has at least half the number of vertices as the one with larger order, it suffices to show that for each vertex on the subgraph of smaller order, we can pick one of the vertices it's connected to on the other side so that we don't pick the same vertex for any two vertices. This is possible because if at any given point we come on a vertex so that all of the vertices (let there be $ v$ of them) it's connected to on the opposite side are already taken, then we would have to have at least $ v$ on the original vertex's side to have no choice, all connected to only that vertex's $ v$ connections. If either the graph isn't connected or there are more vertices on the original vertex's subgraph, contradiction. Otherwise, some connection preventing the original vertex is obligated by another similar situation; however, we can repeat the same logic for this with the same two cases. We clearly cannot have the second cause every time or this would be circular reasoning This concludes the proof.

The last part seems a bit imprecise; I'll try to find a better argument.

[quote="fwolth"][b]Lemma:[/b] There are always $ 2n$ unbounded regions.[/quote]This is essentially all you need to solve the problem. If we "zoom out" as JBL said (or equivalently, place everything in a sufficiently large circle) then the unbounded regions form a $2n$-cycle. If $S\subseteq F$ is the set of colored faces, then $S$ can have at most half of this cycle and thus at most $\frac{1}{2}(2n)=n$ unbounded faces (of degree at least two); everything else in $S$ has at least three edges. It immediately follows that
\[ 3|S| - n \le \sum_{f\in S}\deg{f} \le |E| = n^2, \]as desired.

By Theorem 2(3) [url=http://page.math.tu-berlin.de/~felsner/Paper/tri.pdf]here[/url], the bound is tight for infinitely many $n$ (just include the line at infinity to get the $n$ boundary triangles, and note that no two triangles share a side).


[quote="Badgirl"]:lol: chúc các anh trong đội tuyển thi Toán quốc tế năm nay trúng tủ nhiều nhél, để cho các nước phải sợ Việt nam mình học giỏi.... :lol:[/quote]
chu'c tru'ng tu? nhie^u` ?!

dde^? cho ca'c nu*o*c' pha?i so*. vi.t nam mi`nh ho.c gio?i ?!!
hahaha a^u' tri~   :rotfl:  ddu'ng la` suy nghi~ cu?a ngu*o*i` vi.t nam  :)

[quote="Badgirl"]:lol: chúc các anh trong đội tuyển thi Toán quốc tế năm nay trúng tủ nhiều nhél, để cho các nước phải sợ Việt nam mình học giỏi.... :lol:[/quote]
Ban noi vay la co y gi ??Chang le ban cho rang hoc sinh Vn chi gioi nho trung de thoi sao??Mong ban dung post nhugn cau dai loai nhu the nua..Neu ko toi se delete nhugn topic nhu vay..
EDIT: badgirl qua la 1 girl ..bad....

[quote="nntrkien"]Moi nguoi dung co boi chuyen ra ma de cai nhau nua  :mad: chuyen thi qua lau roi, keira khi nao can delete thi hay thong bao, hoac thong bao ra 1 topic khac.[/quote]Xin loi, chi vi toi qua tuc gian khi chung ta bi noi nhu vay!!! :blush:

[quote="Badgirl"]:lol: chúc các anh trong đội tuyển thi Toán quốc tế năm nay trúng tủ nhiều nhél, để cho các nước phải sợ Việt nam mình học giỏi.... :lol:[/quote]

Badgirl : Cu*ng thie^.t la` so bad ... :lol:

*******************************************

[quote="N.T.TUAN"]Tai cac bac nha minh hay ra de trong sach qua ay ma,cac ban h/s khong co loi gi ca?
Nam 2004 vua roi moi biet la bai 2 lay trong AMM :D[/quote]
Ý bạn là bài 2 năm 2004 hay 2005?
Nếu là 2004, năm ấy chỉ có 1 người làm được bài 2 đấy thôi.

[quote="spdf"]Bai trong AMM la chi tim ham voi a=2[/quote]

Phan day kho nhat roi con gi nua?


Hey if anyone watches this forum, we may still have some openings on our team for ARML this year.  The national championship will be held in Las Vegas the weekend after Memorial Day.  For more information, and to apply to be on the team, go to http://mathleague.org/arml.php


Let x,y,z be positive real numbers such that $x+y+z=1$.
For positive integer n, define $S_n = x^n+y^n+z^n$
Furthermore, let $P=S_2 S_{2005}$ and $Q=S_3 S_{2004}$.
(a) Find the smallest possible value of Q.
(b) If $x,y,z$ are pairwise distinct, determine whether P or Q is larger.

Part b
$P=S_2S_{2005}=(x^2+y^2+z^2)(x^{2005}+y^{2005}+z^{2005})=\displaystyle\sum_{cyc} x^{2007}+\displaystyle\sum_{sym}x^{2005}y^2$
${Q=S_3S_{2004}=(x^3+y^3+z^3)(x^{2004}+y^{2004}+z^{2004})})=\displaystyle\sum_{cyc} x^{2007}+\displaystyle\sum_{sym}x^{2004}y^3$
By Muirhead Inequality,
$\displaystyle\sum_{sym}x^{2005}y^2\ge\displaystyle\sum_{sym}x^{2004}y^3$
Therefore, $P \ge Q$
But x, y, z are distinct.
So equality does not hold, i.e. P is strictly larger than Q.


My school would have easily won regional based on individual scores, but we got a 7 on the Team round, pulling us down to second place.  So we practiced team round, and won the State.  However, we still need better strategy.  What we do is take white paper, and section it into 10 squares.  Each person initials 1 of the first 4 problems, then checks another of the first four problems.  Then we move on to 5-8, and then 9 and 10.  But we need a better strategy.  Our individuals were the best, but our strategy sucks, so we do bad on team rounds.  What strategy did your team use, and were you successful on team round?

[quote="Klebian"]Two people worked 1-5, one one to five, other five to one.
2 people worked 6-10, same way.

I, as captain, did 10 to 6, and then when I was done, I'd look over any problems going on with 1-5.

10 on chapter team round, um... high score (I'll probably post when we're allowed) on state team round.[/quote]

That's what my team did. We got 2nd at chapters.

We just have one person do 1 and 2, another do 3, 4, and 5, one person do 6 and 7, and the last person do 8, 9, and 10.  It's worked out for us.  We got a 9 at chapter. (state yet to come).  If anyone has problems, they share it with the group and we help each other.

We just made it up in 5 seconds and it worked.  I'd be surprised if it was foolproof.

two do 1-5, two do 6-10, organized at first, finish what we can, then at 15 minutes left, chaotically scramble at the other problems, doing problems multiple times because we didn't know that someone else had done it already, scream at others, scream "I'll do number...!", "you check/do this problem!", curse, hit each other with pencils, Scramble at the problems that we feel we can do, finish all but a few problems, reorganize ourselves, and scramble at the last few problems in the last few minutes... :P

not very good strategy.

Yet we still got a 9 on team at chapter this year (dumb mistake on one). State score, uh...

WOAH that matches EXACTLY what we did at states.
Except I ate this kid's lunch when he wasnt looking, then punched me then....... well then some guy had to break it up :lol: . I was worried I was gonna be disqualified but.... I guess not :wow: :w00tb:

Really? We would of been kicked out. Consider you lucky. In fact, this guy another guy got in to a fight in the team round. They were debating in then started to fight. They got kicked out.

here's a supposedly foolproof method, though in practice is not 100%.
make a chart with 1-10 on the side and the initials on the top. two people work bottoms-up and two peole work tops-down. when you get ans answer put it in your box. when you don't know the answer skip the box. it both people working one way skip the box, then you need to collaborate and work together. usually gets all of them done and you are know how your teammates are doing and what needs to be done.
another big tip... don't put down your teammates
there was this one problem about a circle and a coordinate plane on a local competition once.
me: I think i got it!
teammate: it won't work. (hasn't even heard what I was going to do)
me: (thinks) oh, your right!
me: I think I got it
same!!! teammate: it won't work
me: true...
me: I thi--
teammate: it won't work.
I learned afterwards that any of the ways I was thinking of would have worked...:(
moral of the story: it's important not only to have a good strategy and a good team math-wise but to have a friendly team!

Here is the method my team uses which got us 6th at chapter and 10th at state.

The captain divides a piece of paper into ten boxes and numbers them 1-10. Then each person does one of the problems from 1-4. If a person is stuck then they cross out their name and write it on the next available problem.  When someone solves a problem they write their answer in the box and write their name on the next available problem.

When someone finishes a problem and there are no problems left they start checking problems. Once a problem is checked the person who checked it puts a check in it's box and the captain writes the answer on his/her sheet.

This method is probably not foolproof.

what we did was:
first we made a table from 1-10. If we get an answer for it, we put the answer and your initials down. Then if somebody checks it and gets the same answer, you put a check mark next to the problem number, saying that the answer is probably right. If you get a different answer (which suprisingly happens many times) you put down your answers and initials.
Do the problems that you are good at. For example, if you see a figure of a triangle and you are good at geometry, then do it. Same thing if the problem mentions dies, do it if you are good at probabilty/combinatorics. if you are ... lets say not very good at english (like one of my team mates), do the short problems.
if you all get different answers and there is little time left, well... good luck. just take the smartest guy's answer :P
also, in the last minute, check to see that your recorder has the right answers down. because at last year's chapter, our team captain wrote the wrong answer down for one number and we lost by [u][b]0.5 points[/b][/u].  :(

ooh but there is.
You just watch our magical tema round skillz and strategyz at nats this year. Then ill laugh when DELAWARE makes the top 1.

:|  :|  I'm not gonna even respond to that statement [hide]or did I just do it?[/hide]

We don't have any preset things. We just do problems that we think we could do. We just say I got 7 and so on and so forth. We got first at chapter too. Don't know about state yet...

Or you can have the smartest person in your team do all the problems while you take a break :D

At chapter we assigned 2 or 3 to each person, and I ended up having to do like 5 because they didn't understand or know how to do the problem.

There's always one person starting at 1 and working down, one starting at 10 and working up, one doing the evens, and one doing the odds. We got a perfect at chapter, but that was really easy to do. I managed to get through all of the problems, though.

The 4X10 and the 1-5;5-1;6-10;10-6 strategies are the ones my middle school team and GA's team to nationals last year used.

The best strategy is a combination of them. Also, when one team member needs help with a problem, ask the "smartest*" teammate for help.  If you are done with your problems, let others check your answer and work on some of the other problems.

Make sure that someone keeps track of the time.

Also, don't do what DE did at last year's nationals.  Write the team's answers in the [size=200][color=red][b]OFFICIAL[/b][/color][/size] booklet.

*An imaginary penguin, random number generator, your gut instinct, or God (if you are religious) may count as the smartest teammate if all other teammates fail.

For chapter and state, I just did them all and the others checked.

I got 10 at chapter, but since our individual, 33,36,37,46(me!) wasn't that good, we only got 4th.

Advice:win!
The rest of the guys on my team started fighting over who got to do number such and such (yeah, pathetic) but we still got fifth and lost by only a couple of hundreths.lol that was kinda scary.

Our team score was only 55.5 at chapter, yet we still won by a margin of like 8 points against the 2nd place team.  State tomorrow!

Note: We got 5th at chapter, with a 52.?, and 2nd at states.

Good luck tomorrow!

We had a 58... and got 4th.

I guess our chapter was relatively strong.

Our rival team, #6, the last to get recognized (we were the last trophy winners :)) had a 56.25.

My team splits into pairs of people that work well together.  Two take 1-5 and two take 6-10.  One goes from 5 up to 1, one goes from 1 down to 5, one does 10-6 (me) and one does 6-10.  Then whenever they get an answer, they give it to me (team captain) and I right it down, along with an initial for that person so that I know who answered what.  Whenever somebody else gets an answer for that problem, if it's right, I record their initial to so that I know that that problem has been checked; if the answers are different, I check who did the problem and then they work it out together.  If someone can't do a problem, they tell me and I record that they can't do it.  Then, when people finish their assigned problems, I assign them to other problems that either need to be done, need somebody to take time to brute force through them, or that need to be checked.  That way everybody is always working and all the problems get done.

:P we did the same thing, only odds, evens, and 1-5 6-10. we ended up with 2 answers and picked which one was the best. it worked out AWESOME!!! :lol:

 :winner_first: thats how we won 1st place

My strategy is kind of like samath's:

By order of strength from worst to best, arrange your team members A, B, C, D.
Person A does 6-10 and after finishing that works on 1-5
Person B does 5-1 and after finishing that works on 10-6
Person C works 1-10
Person D works 10-1

This ensures an even split of individual brainpower throughout the team, and because the team round tends to get harder towards the end, the difficulty is somewhat evenly matched. In chapter and state, I actually finished all of the problems.

At state we got 2nd with an xx.xx. My coach was pissed because we had 3rd (me  :)), 5th, and sixth individuals and the first team didn't have anyone in countdown.

how come no one's mentioned watches yet? we always get someone that has a watch (usually captain) so we know when time is up and we dont get caught in the middle of checking a problem at the end.

[quote="xxm0rg07hxx"]I have a timer app on my calc.  I have two calcs so one is for the timer.[/quote] [hide][hide][hide] :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: :rotfl: .Talk about being a nerd. The worst I do is play on my graphing calculator in class. If my teachers ask what I'm doing I say I'm finding my average in their class.[/hide][/hide][/hide]
EDIT: No offence (sp?) whatsoever is intended with the nerd comment. I think it's good to be a nerd

[quote="Phelpedo"]I'm not sure exactly what you mean by "foolproof." It's not like there's some magic strategy out there that will guarantee a 10.[/quote]

foolproof just basically means that if someone on the team knows how to do it it gets done, and if there are conflicting answers they get adressed, without complete disorder. the 'foolproof'-ness does not apply if nobody on the team knows how to do the problem.

[hide="This is how our chapter team round went"]
Me: "You guys divide the front page up by yourselfs, I'll start working on the back page."
*I look up 5 minutes later than I'm done with the back page*
Teammate A tells me his answer for #1, Teammate B tells me his answer for #3, I look over at what Teammate C is doing and see that he was DRAWING ON HIS CALCULATOR.
Teammate C sees I'm looking at him and holds it up and goes, "Look, it's <Teammate's A's name> (The picture was semi-pornographic)
I kick Teammate C and say, "I'll work on the rest of the front page, you guys check my answers to the backpage."
After I finished the front page, they just told me they didn't know how to do it but one of them did do #5. So I checked their answers and had to change #1 and #5.
Then, I rechecked all the answers while they fooled around for the rest of the time.
[/hide]

It turns out that we got 18/20 on the Team Round but placed 6th at Chapters anyway. I'll let the school's scores speak for themselves why in format of
Sprint Right/2*Target Right
Me - 28/14
Teammate A - 20/6
Teammate B - 12/4
Teammate C - 12/8
Kid from our school 1 - 12/10
Kid from our school 2 - 12/0


what is the sum of all values of $ x$ for which $ |x\minus{}3| \plus{} |3\minus{}x| \minus{}1 \equal{} 3$?

We have $ |x \minus{} 3| \plus{} |3 \minus{} x| \minus{} 1 \equal{} 3\implies2|x \minus{} 3| \equal{} 4\implies|x \minus{} 3| \equal{} 2\implies x \equal{} 3\pm2$
$ \implies x_1 \plus{} x_2 \equal{} \boxed6$.


floor function
combinatorics unsolved
combinatorics

How much permutations of (1,2....,n) we have that they satisfies:
$s_{i}\leq  s_{\lfloor{i/2}\rfloor}$ for all $2\leq i\leq n$
where $s_{i}$ is the value of the $i$ position in the permutation.
Ex:
$n=3$,  $S=2 1 3$  so  $s_{2}=1$
thanks!

See http://www.research.att.com/~njas/sequences/A056971 or [url]http://mathworld.wolfram.com/Heap.html[/url].  A recurrence and proof is given at the first link.


Prove that any monic polynomial of degree $ n$ with real coefficients is the average of two monic polynomials of degree $ n$ with $ n$ real roots.

I'm not sure of the difficulty of this problem so I appologize if it isn't in the right forum. Thanks for the help.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=337849#337849[/url]

Additionally, please use a more descriptive name than "Problem" - there are plenty of "problems" posted each day.


Find all pairs of positive integers (x,y) such that:
$x^2+y^2-5xy+5=0$

[b] Lemma :[/b] The equation $x^2+y^2-pxy+p = 0$ $(p \in Z^+)$ has positive integer roots if and only if $p=5$

 [i]Proof : [/i]
 Let $T$ be the set of all positive integer roots of the equation : $x^2+y^2-pxy+p=0$ (1) .
 Call $(x_0,y_0)$ be the smallest root of (1) . This means $x_0+y_0 \leq x+y$ , $\forall (x,y) \in T$
 Suppose $x_0 \leq y_0$
 If $x_0=y_0$ , then after some computations , we get a contradiction !
 So $x_0 <y_0$
 Consider (1) as a quadric equation of $y$ .
 By Viette 's theorem : $y_1+y_0 =px_0$
                                  $y_1y_0 =x_0^2+p$
  From this we have $y_1 >0$
 So $(x_0,y_1)$ is also a root of (1)
 By the definition of $(x_0,y_0)$ , we infer $y_1 >y_0 > x_0$
 We have :
  $x_0^2+p-px_0 =y_1y_0 -y_1-y_0 = (y_1-1)(y_0-1)-1 \geq x_0^2-1$
 $\iff 1 \geq p(x_0-1)$
 From here , we infer $x_0 =1 \to y_1 =3 , y_0 =2$ and $p=5$   [/i]

All solution are x1=1;x2=2;x_n=5x_(n-1)-x_(n-2)


limit
logarithms
real analysis
real analysis unsolved

We have that $1+\frac 1 2+\frac 1 3+...+\frac 1 n\approx \ln n$ and $\frac{n}{\sqrt[n]{n!}}\approx e$. The limit is e.

Oh  :blush: ! I'm not clear about the log sum. Can you explain a bit? Thank you.

i think u can get the answer here  :D [url]http://www.mathlinks.ro/Forum/post-361063.html[/url]


As you drive down the road at 13 m/s you press on the gas pedal and speed up with a uniform acceleration of 1.33 m/s2 for 0.62 s. If the tires on your car have a radius of 33 cm, what is their angular displacement during this period of acceleration?
The answer should be in rad.

Thanks for helping!

We have $ v_0\equal{}13ms^{\minus{}1},a\equal{}1.33ms^{\minus{}2},t\equal{}0.62s$. Thus $ s\equal{}v_0t\plus{}\frac{at^2}{2}$ from this we can find $ s$. If the angular displacement is $ \theta$ we have that $ s\equal{}r\theta$. Thus we can find $ \theta$. Sorry I'm in a hurry so skipped the calculations.


probability
function
advanced fields
advanced fields unsolved

I don't understand the difference between the Weak Law of Large Numbers and the Strong Law of Large Numbers. Can someone explain the difference? Also, why does the latter imply the former?

Weak Law of Large Numbers:
\[\mu_{L}(B_{N})\rightarrow0\qquad\text{as}\qquad N\rightarrow\infty\]
Strong Law of Large Numbers:
Let $s_{n}(\omega)$ be the number of heads in the sequence $\omega$ of $n$ flips of a fair coin and $N=\left\{\omega; \frac{s_{n}(\omega)}{n}\rightarrow\frac{1}{2}\text{ as }n\rightarrow\infty\right\}$. Then $\mu_{L}(N^{c})=0$.

($\mu_{L}(A)$ is the measure of $A$.)

The Strong Law of Large Numbers says that the sequence of random variables $S_{n}/n$ converges to $1/2$ [b]almost surely[/b]. The Weak Law of Large Numbers says that the sequence of random variables $S_{n}/n$ converges to $1/2$ [b]in probability[/b].

Here is an example that shows that the convergence in probability doesn't imply the convergence almost surely.

Let $I_{[a,b]}: [0,1]\to\mathbb R$ be the indicator function for the segment $[a,b]$:
\[I_{[a,b]}(t) = \begin{cases}1, & \text{ if }t\in[a,b],\\ 0, &\text{ otherwise}. \end{cases}\]
Consider the following sequence of functions
\[I_{[0,1]}, I_{[0,1/2]}, I_{[1/2,1]}, I_{[0,1/3]}, I_{[1/3,2/3]}, I_{[2/3,1]}, I_{[0,1/4]}, \dots \]
Prove that they converge to $0$ in probability (i.e. $\mu(\{t: f_{n}(t) > \varepsilon\}) \to 0$ as $n\to\infty$ for every $\varepsilon > 0$). However, these functions do [b]not[/b] converge to $0$ almost surely (in fact, for every $t_{0}\in [0,1]$, $f_{n}(t_{0})$ doesn't tend to $0$).

In crude terms, the Strong Law tells you that the places where things screw up are the same for all of the functions and get small. On the other hand, the Weak Law tells you that the places where things screw up get small, but it doesn't tell you that they're the same places for each function (or random variable if you like). The bad places could be all over, like in Yury's example.


probability
quadratics
function
symmetry
ratio
conditional probability

The official contest dates run through the 18th; some teams will be taking it this coming weekend. All public discussion is off limits until at least then.

Can we discuss it now!? Only 20 minutes left... Not like any team can start after reading a post right now...

It's probably been long enough. All of the answers should be in the mail by now.

I think our guys thought this one was difficult.

Having Adam on the team would've helped... it took us (Western Washington) a little too long to get a conceptual grip on this one :|  Our solutions were all pretty rushed.

Problem 6a) and the argument for it was the key to nearly everything. Here's a set of short answers to check against:

1. a) $ P(A) \equal{} \frac23,\ P(B) \equal{} \frac13$

1. b) $ P(A) \equal{} \frac34,\ P(B) \equal{} \frac14$

1. c) $ P(A) \equal{} \frac45,\ P(B) \equal{} \frac15$

2. $ P(A) \equal{} \frac {n}{n \plus{} 1},\ P(B) \equal{} \frac {1}{n \plus{} 1}$

3. a) $ P(A) \equal{} \frac23,\ P(B) \equal{} \frac13$

3. b) $ P(A) \equal{} \frac47,\ P(B) \equal{} \frac37$

4. $ P(A) \equal{} \frac {b}{a \plus{} b},\ P(B) \equal{} \frac {a}{a \plus{} b}$

5. a) $ P(A) \equal{} \frac12,\ P(B) \equal{} \frac14,\ P(C) \equal{} \frac14$

5. b) $ P(A) \equal{} \frac15,\ P(B) \equal{} \frac25,\ P(C) \equal{} \frac25$

5. c) $ P(A) \equal{} \frac13,\ P(B) \equal{} \frac16,\ P(C) \equal{} \frac13,\ P(D) \equal{} \frac16$

6. a) $ P(A_k) \equal{} a_k^{ \minus{} 1}\left(\sum_{i \equal{} 1}^na_i^{ \minus{} 1}\right)^{ \minus{} 1}$

6. b) $ P(A) \equal{} \frac14,\ P(B) \equal{} \frac38,\ P(C) \equal{} \frac38$

7. $ P(A) \equal{} \frac3{14},\ P(B) \equal{} \frac {17}{28},\ P(C) \equal{} \frac1{14},\ P(D) \equal{} \frac3{28}$

8. $ \frac1{1007013}$

9. $ P(A) \equal{} \frac5{19},\ P(B) \equal{} \frac5{19},\ P(C) \equal{} \frac6{19},\ P(D) \equal{} \frac3{19}$

Here's the exact statement:

[quote]8. A tree has the same shape as the tree in Figure 6a. If $ a_1\plus{}a_2\plus{}a_3\plus{}\cdots\plus{}a_k\equal{}2007,$ what is the smallest possible value for $ A_1.$[/quote]
You're right to say that that's not totally clear, but to me it doesn't imply that $ k$ is fixed, and hence it does imply that we seek the overall minimum.

For a given $ k,$ the smallest $ A_1$ would be obtained by setting $ a_1$ as large as possible. That is, we let $ a_1\equal{}2008\minus{}k$ and we let the other $ k\minus{}1$ branches each have length $ 1.$ By the result in 6a), we get

$ P(A_1)\equal{}\frac{\frac1{2008\minus{}k}}{\frac1{2008\minus{}k}\plus{}\frac{k\minus{}1}1}\equal{}
\frac1{\minus{}k^2\plus{}2009k\minus{}2007}$

Now, for what $ k$ is this as small as possible? We can minimize the fraction by maximizing the denominator. The quadratic in the denominator would be maximized by letting $ k\equal{}1004.5,$ but since $ k$ must be an integer, the actual extreme value is achieved by letting $ k$ be either $ 1004$ or $ 1005$ and is the value reported above.

Well, yes, once I got the expression for a given $ k$ it occurred to me that they might want me to minimize it, but even if I had decided on that as the correct expression, we ran out of time.  :|

t0rajir0u - did you guys get that expression for #6a)? That's the "master key" to the set, as it covers every problem except #9.

I'll have to look again at the copies I kept of what we sent, but I don't think we had that.

Yeah.  It was pretty clear after we looked at the numbers on #7.  I don't think we defended any of our answers particularly well, though.

An argument for 6a):

We have a tree with $ n$ edges connected to the root, and each of those $ n$ edges starts a straight branch that leads to a single leaf. We have leaves $ A_1,A_2,\dots,A_n,$ and there are $ a_i$ edges between the root and $ a_i.$

Let's focus on the probability that a random walk starting at the root ends at leaf $ A_1.$ Let $ p$ be that probability.

Let $ x_{i,j}$ be the conditional probability of ending at leaf $ A_1$ given that the current location is the $ j$th node on the $ i$th branch. By convention, let $ x_{i,0} \equal{} p$ for all $ i.$ (That is, think of the root as belonging to all of the branches.) And we will use that $ x_{1,a_1} \equal{} 1$ and $ x_{i,a_i} \equal{} 0$ for $ i\ne0.$

Considering the results of one step leads us to the following equations:

For $ 1\le j\le a_i \minus{} 1,\ x_{i,j} \equal{} \frac12\,x_{i,j \minus{} 1} \plus{} \frac12\,x_{i,j \plus{} 1}$

$ p \equal{} \frac1n\sum_{i \equal{} 1}^nx_{i,1}.$

The first equation can be rearranged to this:

 $ x_{i,j \plus{} 1} \minus{} x_{i,j} \equal{} x_{i,j} \minus{} x_{i,j \minus{} 1}.$

Hence the successive differences are constant and, for each $ i,$ $ x_{i,j}$ is a linear function of $ j.$ (It's an arithmetic sequence.) Considering the boundary values, we get the following:

$ x_{1,j} \equal{} p \plus{} \frac {(1 \minus{} p)j}{a_1}$

$ x_{i,j} \equal{} p \minus{} \frac {pj}{a_i}$ for $ i\ne 1.$

That means that $ x_{1,1} \equal{} p \plus{} \frac {1 \minus{} p}{a_1} \equal{} p\left(1 \minus{} \frac1{a_1}\right) \plus{} \frac1{a_1}$

and $ x_{i,1} \equal{} p \minus{} \frac {p}{a_i} \equal{} p\left(1 \minus{} \frac1{a_i}\right)$ for $ i\ne 1.$

The probability equation around the root node gives

$ p \equal{} \frac1n\left(\sum_{i \equal{} 1}^np\left(1 \minus{} \frac1{a_i}\right) \plus{} \frac1{a_1}\right)$

$ p \equal{} p \minus{} \frac {p}{n}\sum_{i \equal{} 1}^n\frac1{a_i} \plus{} \frac1{na_1}$

$ \frac {p}{n}\sum_{i \equal{} 1}^n\frac1{a_i} \equal{} \frac1{na_1}$

$ p\sum_{i \equal{} 1}^n\frac1{a_i} \equal{} \frac1{a_1}$

So, $ p \equal{} \frac {\frac1{a_1}}{\sum_{i \equal{} 1}^n\frac1{a_i}}$

Since there was nothing special about the first branch, we have our general answer:
\[ \boxed{P(A_k) \equal{} \frac {\frac1{a_k}}{\sum_{i \equal{} 1}^n\frac1{a_i}}}
\]

Assuming I understand the problem correctly, does the following argument work?

Let's assign a weight of 0 to the root and all vertices not on the first two branches.  Let's assign a weight of $ \minus{}1$, $ \minus{}2$, ..., $ \minus{}a_1$ to the vertices on the first branch (as we go away from the root).  Let's assign a weight of $ 1$, $ 2$, ..., $ a_2$ to the vertices on the second branch (as we go away from the root).  We can check that each step, the expected weight does not change.  Thus, when we finish the random walk, the expected weight will be 0.  (The tree being finite helps us here.)

But the expected weight at the end will be $ \minus{}a_1 P(A_1) \plus{} a_2 P(A_2)$.  Because that expected weight is 0, we have $ a_1 P(A_1) \equal{} a_2 P(A_2)$.  By symmetry, all the $ a_k P(A_k)$ must equal each other.  In other words, $ P(A_k)$ is inversely proportional to $ a_k$.  Because the probabilities add up to 1, we get
\[ P(A_k) \equal{} \frac{a_k^{\minus{}1}}{\sum_i a_i^{\minus{}1}} \, .
\]

Ravi: you understood the problem correctly. And that's a very nice argument.

Let me try something related with problem #9.

Description of the tree: The root is connected to node $ x$ on the left and node $ y$ on the right. Node $ x$ is further connected to leaves $ A$ and $ B.$ Node $ y$ is further connected to leaf $ C$ and node $ z.$ Node $ z$ is further connected to leaf $ D.$

Let's assign weights to each node so as to keep the balance about each point. Let the root have weight $ 0.$ Assign weight $ 1$ to node $ x$ and $ \minus{}1$ to node $ y.$

Now to keep the balance about node $ x,$ the three outcomes must average to $ 1.$ To do that, assign weight $ \frac32$ to each of leaves $ A$ and $ B.$

We do the same thing about node $ y.$ Leaf $ C$ has weight $ \minus{}\frac32$ and node $ z$ has weight $ \minus{}\frac32.$ And then to keep the balance about $ z,$ we assign weight $ \minus{}2$ to leaf $ D.$

The expected weight at the end will be

$ \frac32(P(A)\plus{}P(B))\minus{}\frac32P(C)\minus{}2P(D)\equal{}0.$

That's one equation for the four probabilities. We need two more (we always have that the sum of the probabilities is $ 1.$) By symmetry, $ P(A)\equal{}P(B),$ so we need just one more.

Try zeroing the whole $ x,A,B,$ branch. Assign weight $ 0$ to node $ y.$ Then assign weight $ 1$ to leaf $ C$ and weight $ \minus{}1$ to node $ z.$ We get balance around $ z$ by assigning weight $ \minus{}2$ to leaf $ D.$ We now have that

$ P(C)\minus{}2P(D)\equal{}0$ or $ P(C)\equal{}2P(D).$

Replace $ P(B)$ by $ P(A)$ and $ P(C)$ by $ 2P(D)$ and return to our first equation:

$ 3P(A)\minus{}5P(D)\equal{}0.$

Thus the ratio $ P(A): P(D)\equal{}5: 3.$ More completely, $ P(A): P(B): P(C): P(D)\equal{}5: 5: 6: 3.$

Since $ 5\plus{}5\plus{}6\plus{}3\equal{}19,$ that gives us probabilities of $ \frac5{19},\frac5{19},\frac6{19},\frac3{19}.$

---

It looks like to make this work, we'll need to do several assignments of weights, selectively zeroing various parts of the tree. But the assignment of weights is not difficult; each assignment is strictly local.

[quote="t0rajir0u"][quote="Kent Merryfield"]8. $ \frac1{1007013}$[/quote]

I gave the answer in terms of $ k$.  Did the problem imply it wanted the lowest probability among all $ k$?  I didn't think it was clear...[/quote]

D'oh!

Well, I hope the grader understands the ambiguity in the problem statement. When I wrote this event (well over a year ago; I had to reacquaint myself with the problems), I must admit I overlooked that interpretation.

See, this is why I should have students proofread these before they're submitted.

...waaait a second, that'd be a bad idea  :P

[quote="Kent Merryfield"]

Since there was nothing special about the first branch, we have our general answer:
\[ \boxed{P(A_k) \equal{} \frac {\frac1{a_k}}{\sum_{i \equal{} 1}^n\frac1{a_i}}}
\]
[/quote]

I like that approach! Mine was to reduce to a previous problem: note that if you want to compare $ P(A_r)$ to $ P(A_s)$ in a $ k$-endpoint graph, you can disregard whatever the walk does before it restricts itself to the $ 2$-endpoint graph from the earlier problems:

\[ P(A_r | A_r \textrm{ or } A_s) \equal{} \frac{P(A_r)}{P(A_r \textrm{ or } A_s)} \quad \textrm{and} \quad P(A_s | A_r \textrm{ or } A_s) \equal{} \frac{P(A_s)}{P(A_r \textrm{ or } A_s)},\]
but $ P(A_r | A_r \textrm{ or } A_s) \equal{} \frac{1/a_r}{1/a_r\plus{}1/a_s}$ and $ P(A_s | A_r \textrm{ or } A_s) \equal{} \frac{1/a_s}{1/a_r\plus{}1/a_s}$, since we can restrict every $ A_r$-or-$ A_s$ walk to the part contained only in those two branches -- any earlier back-and-forths on other branches don't affect whether we finish at $ A_r$ or $ A_s$. Comparing the conditional probabilities, we get $ P(A_r): P(A_s) \equal{} a_s: a_r$.

That turns out to be quite a bit like Ravi's "weights" argument. Notice that he also cut down to two branches.

Tom Kilkelly emailed a solution set to all of the sponsors. I'll post it here.

If allowed, could you post the problems as well?


function
AMC
AIME
number theory
totient function

I was doing one of the Mock AIMEs and the problem 4 was: Find the remainder when $1+7+7^2+7^3+...+7^{2004}$ is divided by 1000.  The solution used Euler's totient theorem that
$n^{\phi(p)}\equiv1(mod p)$
Is this common on the AIME?

I doubt any of the solutions for actual AIME problems would REQUIRE the use of the totient function; it is however a handy tool in problems such as that one.

[quote="WarpedKlown1335"]Well how do you do that WITHOUT the totient function then?[/quote]

Well that's why it was on a mock AIME instead of a real one =P.

Don't take the Mock AIME's too seriously.  If you really want to prepare for the AIME, look at real AIME problems.

Or just learn the toitent function.

[quote="WarpedKlown1335"]Well how do you do that WITHOUT the totient function then?[/quote]

With more patience.

[hide]We want to calculate $\frac{7^{2005}-1}{6}\bmod1000$, meaning we're gonna have to know $7^{2005}-1\bmod 2000$. (We can't simply calculate mod 1000 and then divide by 6 because 6 and 1000 are not relatively prime.) So we start looking for patterns mod 2000:

7^1=7
7^2=49
7^3=343
7^4=2401=401 mod 2000

That 1 is kinda nice, so we keep it:

7^8=401^2=801
7^12=801*401=1201
7^16=1201*401=1601
7^20=1601*401=1 mod 2000.

Yay, that's really good. So $7^{2005}-1\equiv(7^{20})^{100}\cdot7^5-1\equiv7^5-1\equiv806\equiv4806\bmod2000$, so the answer is $\frac{4806}{6}\equiv801\bmod1000$.[/hide]

If memory serves, there WAS a problem on last year's AIME that was most easily solved with the totient function (but not Euler's theorem.)

8.  A regular n-star is the union of n equal line segments P_1P_2, P_2P_3, ... , P_nP_1 in the plane such that the angles at P_i are all equal and the path P_1P_2...P_nP_1 turns counterclockwise through an angle less than 180 degrees at each vertex. There are no regular 3-stars, 4-stars or 6-stars, but there are two non-similar regular 7-stars. How many non-similar regular 1000-stars are there?


calculus
derivative
Euler
LaTeX
calculus computations

Let $f(x)=\sqrt{x+\sqrt x}$.  You need to use the chain rule on this bad boy.

So, $f'(x)=\frac{1}{2} ({x+\sqrt x})^{-\frac{1}{2}}\frac{d}{dx}\left(x+\sqrt x \right).$
And hence,

$f'(x)=\frac{1}{2} ({x+\sqrt x})^{-\frac{1}{2}}(1+\frac{1}{2}{x^{-\frac{1}{2}}} ).$

And I shall leave the simplification to you.

dont u need to chain it again? since its x+x^.5 after the second step

Is this $()^5$ or $()^{.5}$?

If it is the latter, see my post.  If it is the first, see Euler-Fermat's post.

[quote="=s"][quote="Fermat -Euler"][quote="=s"](x+x^.5)^.5
sorry i dont have those math symbols :([/quote]

you want derivative of $(x+x^5)^5$

=$5(1+5x^4)(x+x^5)^4$[/quote]
no is to the square root=^.5[/quote]
sorry, i dont see it

[quote="=s"]sorry i dont have those math symbols[/quote]
If you can post here at all, you do- the math symbols here are $\LaTeX$, and they're built into the forum software. You only need to use appropriate codes between dollar signs.


calculus
integration
geometry
rectangle
function
real analysis
real analysis unsolved

Prove that if R is an aligned rectangle and f is a continuous function on R, then f is integrable on R.

I'll assume this is a Riemann integral question, so we should be discussing Riemann sums, or upper sums and lower sums. I'll also assume that "aligned" means "with sides parallel to the axes." That's a convenience in the proof but hardly a vital condition.

Somewhere in the proof you're going to use that a continuous function on a compact set is uniformly continuous.

1. For $\epsilon>0$ choose $\delta$ as in the definition of uniform continuity.
2. Prove that any two choices of sample points in a partition with norm $<\delta$ result in double Riemann sums that differ by less than $\epsilon\,\mathrm{area}(R)$.
3. Given two partitions $P_{1},P_{2}$ with norm $<\delta$, let $P$ be their common refinement.
4. Observe that any Riemann sum for $P$ is between the upper and lower Darboux sums for $P_{1}$. [The upper/lower Darboux sum is the Riemann sum with sample points being points of maximum/minimum of $f$ within a subrectangle].
5. Conclude from 2 and 4 that the difference of Riemann sums for $P_{1}$ and $P$ is less than $\epsilon\mathrm{area}(R)$. Same conclusion for $P_{2}$ and $P$.
6. From 5, the difference of Riemann sums for $P_{1}$ and $P_{2}$ is less than $2\epsilon\mathrm{area}(R)$.

Since f is continuous almost everywhere on R, f is Riemann integrable on R.

 :P  :D


[b][size=150][color=red]PE[/color][color=white]RU[/color][color=red] 2007[/color][/size][/b]

[b]   PER1. Daniel Soncco
   PER2. Fernando Manrique
   PER3. Jossy Alva
   PER4. Cesar Cuenca
   PER5. Amilcar Velez
   PER6. Ricardo Ramos [/b]

Felecitaciones a todos,

BLog de Procesos Selectivos de PERU: http://selectivos-peru.blogspot.com/

Saludos,

[color=blue]Acabo de enterarme de que Uruguay no asistirá a la I.M.O. este año. Según se dice, ninguno de los posibles participantes alcanza el nivel requerido para esta competencia y, además, Viet Nam queda muy lejos.
Por lo tanto, ni se molestaron en realizar un examen de clasificación.

Triste noticia, considerando que aportaba materiales a algunos participantes para que pudieran mejorar su nivel…[/color]

[size=150][b][color=yellow]VEN[/color][color=blue]EZU[/color][color=red]ELA[/color][/b][/size]

Como nadie ha posteado la delegación venezolana acá va:

 :arrow: [b]Sofía Taylor
 :arrow: Carmela Acevedo
 :arrow: Gilberto Urdaneta[/b]

saludos!  :lol:

[color=blue]¡Las sorpresas que depara la vida! Al final, según una fuente fidedigna, sí habrá un examen de selección para definir la delegación uruguaya que participará en la I.M.O..
Cualquier cambio imprevisto avisaré.

[b]P.D.:[/b] El examen clasificatorio de Paraguay será mañana (26-5-2.007).[/color]

[quote="R.G.A.M."][color=blue]¡Una buena noticia!
¡Es seguro que yo voy a Viet Nam!

Lástima que, por culpa de un joven muy molesto (que casi me mata a golpes), corro el riesgo de no ir como Paraguay 1.

Cuando tenga la lista completa, la publicaré.

Saludos a todos.[/color][/quote]

q fue lo q te paso con ese muchacho rafael?? :rotfl:

bueno hace varias semanas se hizo el selectivo y la delegacion salvadoreña para esta imo es:

ESA-1: Gabriel Alexander Chicas Reyes (ese soy yo  :wink:)
ESA-2: Mario José Castro Hernández
ESA-3: Alfonso Abraham Alvarenga Gamero
ESA-4: Otto Tang Chen

espero q amigos q ya nos conocemos nos encontremos de nuevo en vietnam  :thumbup:

[color=blue]¡AFortunado eres de no conocerlo!
He sido uno de los menos afectados por su molestia, he comprobado.
Chicos que estaban a cinco metros de él se quejaron al finalizar el examen, pues resultó ser DEMASIADO molesto.
¡El criminal se pasaba silbando y gritando porque no quería decirle cuántos problemas ya tenía resueltos!
He comentado el problema, y espero que tomen medidas, sino… (supongo que no necesito aclarar qué sucederá)

¡Felicitaciones al equipo salvadoreño![/color]

Finalmente salio el equipo de colombia:

1) Diego Cifuentes (IMO 2006, OIM 2006)
2) Santiago Cuellar (IMO 2006, OCCM 2005) (Skuge en Mathlinks)
3) José Alejandro Samper (Yo!!!) (IMO 2006, OIM 2005/2006)
4) Miguel Acosta (IMO 2006, OCCM 2004)
5) Jorge Olarte (OCCM 2006)
6) Federico Castillo $^{2}$ (OIM 2005)

[quote="Killua"]Luego de 10 años Chile vuelve a la IMO... esperemos que hagan una buena actuación  :blush:

1) Francisco Campos
2) Cristobal Parraguez
3) Roberto Villaflor
4) Sebastian Pacheco

Mucha suerte!![/quote]

Bien por chile amigo mio,
espero la delegacion pueda cumplir sus objetivos
en esta edicion de la IMO.

Carlosbr
Lima -PERU

[size=150][b][color=red]DEL[/color][color=white]EGA[/color][color=blue]CIÓN[/color] [color=red]PAR[/color][color=white]AGU[/color][color=blue]AYA[/color][/b][/size]

[b][i]Participantes:[/i][/b]
1. Ratti Bittinger, Gabriela María.
2. Arias Michel, Rafael Guillermo.
3. Caló Caligaris, Maurizio.
4. Jacquet Montiel, Jorge Daniel.

[b][i]Suplente:[/i][/b]
1. Hazevich Mulka, Iván Alexander.

[color=blue][b][i]Observación:[/i][/b] Ésta lista se presentó sin tener en cuenta el orden de los puntajes, debido a que el tiempo apremiaba. Según oí hablar, bien podía merecer estar en la posición primera de la lista de participantes (a pesar del desastre…).

[b]P.D.:[/b] Mañana es el examen de clasificación para la Olimpiada Íberoamericana de Matemáticas en mi país.[/color]


All in the title
$ \displaystyle \frac{\tan x}{1\minus{}\cot x} \plus{} \frac{\cot x}{1\minus{}\tan x} \equal{} \sec x \csc x \plus{}1$

My solution is rather long so if anyone could do it in less than 10+ steps that would be great.

[hide="Discussion"] A general strategy for identities of this type is to write everything in terms of sines and cosines.  I'm also going to do this proof "forward," which is unmotivated (run the steps backwards to see the reasoning):

$ \frac{\sin^3 x \minus{} \cos^3 x}{\sin x \cos x (\sin x \minus{} \cos x)} \equal{} \frac{\sin^2 x \plus{} \sin x \cos x \plus{} \cos^2 x}{\sin x \cos x} \implies$
$ \frac{\sin^2 x}{\cos x (\sin x \minus{} \cos x)} \plus{} \frac{\cos^2 x}{\sin x(\cos x \minus{} \sin x)} \equal{} \frac{1 \plus{} \sin x \cos x}{\sin x \cos x} \implies$
$ \frac{\tan x}{1 \minus{} \cot x} \plus{} \frac{\cot x}{1 \minus{} \tan x} \equal{} \sec x \csc x \plus{} 1$

Two steps okay?  :) [/hide]

Awesome . This was a porblem we had to do in class and here's my overly detailed (as in every step) solution. From a comparison between t0rajir0u's abridged solution and mine I think it's quite obvious that some methods are better than others    :(   (sigh...all that time wasted time I spent coming up with this...)

[hide]
\[ \frac {\tan x}{1 - \cot x} + \frac {\cot x}{1 - \tan x} = \frac {(1 - \tan x)\tan x + (1 - \cot x)\cot x}{(1 - \cot x)(1 - \tan x)}
\]

\[ (1 - \cot x)(1 - \tan x) = 1 - \cot x - \tan x + \cot x \tan x = 2 - \cot x - \tan x
\]

\[ \frac {(1 - \tan x)\tan x + (1 - \cot x)\cot x}{2 - \cot x - \tan x} = \frac {\tan x + \cot x - (\tan^2 x + \cot^2 x)}{2 - (\cot x + \tan x)}
\]

\[ \tan x + \cot x = \frac {\sin x }{\cos x} + \frac {\cos x}{\sin x} = \frac {\sin^2 x + \cos^2 x}{\cos x \sin x} = \sec x \csc x
\]

\[ \tan^2 x + \cot^2 x = \frac {\sin^2 x}{\cos^2 x} + \frac {\cos^x}{\sin^2 x} = \frac {1 - \cos^2 x}{\cos^2 x} + \frac {1 - \sin^2 x}{\sin^2 x} = \frac {1}{\cos^2 x} - \frac {\cos^2 x}{\cos^2 x} + \frac {1}{\sin^2 x} - \frac {\sin^2 x}{\sin^2 x} \\
= \sec^2 x + \csc^2 x - 2
\]

\[ \sec^2 x + \csc^2 x = \frac {1}{\cos^2 x} + \frac {1}{\sin^2x} = \frac {\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \sec^2 x \csc^2 x
\]

\[ \tan^2 x + \cot^2 x = \sec^2 x \csc^2 x - 2
\]

\[ \frac {\tan x + \cot x - (\tan^2 x + \cot^2 x)}{2 - (\cot x + \tan x)} = \frac {\sec x \csc x - \sec^2 x \csc^2 x + 2}{2 - \sec x \csc x}
\]

\[ \frac { - 1}{ - 1} \cdot \frac {\sec x \csc x - \sec^2 x \csc^2 x + 2}{2 - \sec x \csc x} = \frac {\sec^2 x \csc^2 x - \sec x \csc x - 2}{2 - \sec x \csc x}
\]

\[ \frac {\sec^2 x \csc^2 x - \sec x \csc x - 2}{2 - \sec x \csc x} = \frac {(\sec x \csc x - 2)(\sec x \csc x + 1)}{\sec x \csc x - 2} = \sec x \csc x + 1
\]
[/hide]

Well, the first reason that solution's longer than it has to is that $ 1 \minus{} \cot x$ and $ 1 \minus{} \tan x$ both have a common "factor" $ \sin x \minus{} \cos x$.  When you write everything in terms of sines and cosines, this becomes much clearer, so you don't have to put as much effort into adding the fractions on the left :)

[quote="emjay285"]All in the title
$ \displaystyle \frac {\tan x}{1 \minus{} \cot x} \plus{} \frac {\cot x}{1 \minus{} \tan x} \equal{} \sec x \csc x \plus{} 1$

My solution is rather long so if anyone could do it in less than 10+ steps that would be great.[/quote]

$ \frac{a}{1\minus{}a^{\minus{}1}}\plus{}\frac{a^{\minus{}1}}{1\minus{}a}\equal{}1\plus{}a\plus{}a^{\minus{}1}$ and $ \tan x\plus{}\cot x\equal{}(\sin^2 x\plus{}\cos^2 x)\cdot (\sec x\csc x)\equal{}\sec x\csc x$. Using $ a\equal{}\tan x$ finishes the problem.


modular arithmetic
combinatorics proposed
combinatorics

Let $ f : \mathbb{N} \to \mathbb{N}$ be a bijection.  Let $ \text{Fix}(f)$ denote the number of fixed points of $ f$ if it is finite and let $ s_n \equal{} \text{Fix}(f^n)$.  Show that there does not exist $ f$ such that $ s_n \equal{} \varphi(n)$.

What if $ f$ is not required to be a bijection?

$ Fix(f^p) \equiv Fix(f) \pmod p$ for any prime $ p$ - any fixed point of $ f^p$ is either a fixed point of $ f$ or a number in a cycle of length $ p$ in $ f$, and then there are $ p \minus{} 1$ more members who also become fixed.
But it trivializes the problem [if we put $ n \equal{} 1,3$ for example], so I might be wrong....

No, that's correct; this wasn't supposed to be hard, I just thought it was cute :)  What I was thinking was that if $ c_k$ denoted the number of cycles of length $ k$ then $ s_n \equal{} \sum_{d | n} dc_d$, but the Mobius inverse of $ \varphi(n)$ isn't non-negative.  But your argument just takes the case $ n \equal{} p$, which makes it even easier than I thought it was.


let a, b, c sides triangle that a+b=3c and $\alpha$ and $\beta$ are the opposite angle to the sides a and b. show that
$tan\left(\frac{\alpha}{2}\right)\cdot tan\left(\frac{\beta}{2}\right)=\frac{1}{2}$

We have $ \tan\frac{\alpha}{2}=\sqrt{\frac{(p-b)(p-c)}{p(p-a)}}$ and $ \tan\frac{\beta}{2}=\sqrt{\frac{(p-a)(p-c)}{p(p-b)}}$, where $p$ is the semiperimeter of triangle.
So $ \tan\frac{\alpha}{2}\tan\frac{\beta}{2}=\frac{p-c}{p}$, but $a+b+c=4c\Rightarrow p=2c$.
Then $ \tan\frac{\alpha}{2}\tan\frac{\beta}{2}=\frac{2c-c}{2c}=\frac{1}{2}$.

$\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})$
$\tan(\frac{\alpha}{2})=\frac{r}{s-a}$ where $s$ is the semiperimeter in fact $s=\frac{a+b+c}{2}=\frac{3c+c}{2}=2c$. $\tan(\frac{\beta}{2})=\frac{r}{s-b}$. It's known $r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$. Moving members from the LHS to the RHS, we only need to prove that: $r^{2}=\frac{(s-a)(s-b)}{2}$ Substituting yields $P.D.$ that $r^{2}=\frac{(s-a)(s-b)(s-c)}{s}$ be $\frac{(s-a)(s-b)}{2}$ so we only need that $\frac{s-c}{s}=\frac{1}{2}$ but $s=2c$, so it's true.

we well-know that $tan\frac{A}{2}=\frac{r}{u-a},tan\frac{B}{2}=\frac{r}{u-b}$.So,
$tan\frac{A}{2}.tan\frac{B}{2}=\frac{r^{2}}{(u-a)(u-b)}=\frac{u.u.(u-c)r^{2}}{u.u.(u-a)(u-b)(u-c)}=\frac{u-c}{u}=\frac12$
Lokman GÖKÇE


Can you solve this problem?
Find integers a,b,c,n,m such that $a^m+ a^n=b^2+c^2$?

A solution: a=5, b=11, c=3, n=1, m=3.

Let a = u 2 + v 2 , choose n and k arbitrarily (with n>0,k >= 0), and let m = n+2k. Then let b + ci = (u+vi)^n*(a^k+i) where i = :sqrt: (-1).

Another example: a=73, b=41,c=61,m=1,n=2


Two hundred students participated in a mathematical contest. They had 6 problems to solve. It is known that each problem was correctly solved by at least 120 participants. Prove that there must be two participants such that every problem was solved by at least one of these two students.

graph theory surely permit to solve this problem easily.though...

If one student has all of them, or 5/6, it's trivial.

If one got 4/6, then the other 199 will have 120 correct solutions among them for each of the other two problems, so there are many pupils with the other 2 correct.

If none of them got 4/6, the maximum score is 3/6, so the average cannot be above 50% -- but the average was 60%+ for each question. Contradiction.

I may be wrong, but I think that the number 200 is still far from optimized :?
Very easy question in any way :)


1)Let $ A,B,C \in GL_n{\mathbb{(C)}}$ are Hermitian matrices ans such that :
 $ A^3B^5 \equal{} B^5A^3$,$ (A^{2n \plus{} 1} \plus{} B^{2n \plus{} 1} \plus{} C^{2n \plus{} 1})(A^{ \minus{} (2n \plus{} 1)} \plus{} B^{ \minus{} (2n \plus{} 1)} \plus{} C^{ \minus{} (2n \plus{} 1)}) \equal{} I_n$,$ A^{2m \plus{} 1} \plus{} B^{2m \plus{} 1} \plus{} C^{2m \plus{} 1} \equal{} 2009I_n$ with $ m,n,k \in \mathbb{N^{*}}$
Caculate $ P \equal{} A^{2k \plus{} 1} \plus{} B^{2k \plus{} 1} \plus{} C^{2k \plus{} 1}$
 [hide="Answer"]$ \sqrt [2m \plus{} 1]{2009^{2k \plus{} 1}}I_n$[/hide]
2) Let $ a,b,c$ are real numbers and $ X,Y,Z \in GL_n{\mathbb{(C)}}$ are Hermitian matrices satisfying the conditions
 $ aX^{2009} \equal{} bY^{2009} \equal{} cZ^{2009}$, $ XY \plus{} YZ \plus{} ZX \equal{} XYZ$
 Calculate $ P \equal{} aX^{2008} \plus{} bY^{2008} \plus{} cZ^{2008}$
[hide="Answer of me"]$ (\sqrt [2009]{a} \plus{} \sqrt [2009]{b} \plus{} \sqrt [2009]{c})^{2009}I_n$[/hide]

1)
$ A^3,B^5$ commute and $ 3,5$ are odd then $ A,B$ commute and we may assume $ A\equal{}diag(\lambda_i),B\equal{}diag(\mu_i)$ where $ \lambda_i,\mu_i$ are non zero reals. Thus $ C^{2m\plus{}1}$ is a diagonal matrix and $ C$ is also diagonal. We have reduced the problem to the case $ n\equal{}1$.
 :mad:
2)
Assume $ a,b,c$ non all zero ; $ X,Y,Z$ invertible implies $ a,b,c\not\equal{}0$. $ 2009$ is odd then $ Y\equal{}\alpha{X},Z\equal{}\beta{X}$ and the second relation can be written $ (\alpha\plus{}\beta\plus{}\alpha\beta)X^2\equal{}\alpha\beta{X^3}$. $ X$ is invertible then $ X$ is a scalar matrix.........
 re :mad:


Prove that among any $ 7$ real numbers there exist two, say $ x$ and $ y$, such that: $ 0 \le \frac{x\minus{}y}{1\plus{}xy} \le \frac{1}{\sqrt{3}}$.

[quote="moldovan"]Prove that among any $ 7$ real numbers there exist two, say $ x$ and $ y$, such that: $ 0 \le \frac {x \minus{} y}{1 \plus{} xy} \le \frac {1}{\sqrt {3}}$.[/quote]

Let $ x\equal{}\tan a, y\equal{}\tan b$, where $ a,b \in (\minus{} \frac{\pi}{2}, \frac{\pi}{2})$,  the question becomes the existance of $ a,b$ such that $ 0 \le tan(a\minus{}b) \le \frac{1}{\sqrt{3}}) \Leftrightarrow$ existance of $ a,b$ such that $ |a\minus{}b| \le \frac{\pi}{6}$ which is true by dividing $ (\minus{} \frac{\pi}{2}, \frac{\pi}{2})$ into six sub-intervals : $ (\minus{} \frac{\pi}{2}, \minus{}\frac{\pi}{3}],[\minus{} \frac{\pi}{3}, \minus{}\frac{\pi}{6}],[\minus{} \frac{\pi}{6}, 0],[0, \frac{\pi}{6}],[ \frac{\pi}{6}, \frac{\pi}{3}],[\frac{\pi}{3}, \frac{\pi}{2})$ and use pigeon-hole principle.

exactly exist a,b, suth that $ 0\le tan(a\minus{}b)\le tan\frac{\pi}{7}$.

Nice solution. It is a good example for the using of the pigeon-hole principle.


A highway patrol plane flies 3 miles above a level, straight road at a constant rate of 120 mph. The pilot sees an oncoming car and with radar determines that at the instant the line-of-sight distance from plane to car is 5 miles the line-of-sight distance is decreasing at the rate of 160 mph. Find the car's speed along the highway.

My trouble is that I don't see a clear way to relate the speeds of the plane and the car.

Let $ L$ be the horizontal distance from the plane to the car, and let $ h$ be the vertical distance. Let $ z$ be the line-of-sight distance. Then we know that
\[ L^2 \plus{} h^2 \equal{} z^2.\]
We are given that at this instance, $ h \equal{} 3$ and $ z \equal{} 5$. Then $ L \equal{} 4$. Differentiating with respect to time, we get
\[ L \dfrac{dL}{dt} \plus{} h \dfrac{dh}{dt} \equal{} z \dfrac{dz}{dt}\]
The rest should be simple calculations.

Sorry, but I don't quite understand how to substitute the variables back into the equation. Isn't $ \frac{dh}{dt}$ the change in height? If so, then isn't that just 0?

from there we have $ 4x\plus{}0h\equal{} 5(\minus{}160),$$ x\equal{}\frac{\minus{}800}{4}\equal{}\minus{}200$
The car is approaching the helicopter at 200mph


inequalities
number theory open
number theory

Find the smallest $n$ such that the following inequality is false:  $\sigma(n) \le H_{n}+\ln(H_{n})e^{H_{n}}$
where $\sigma(n)$ is the sum of the divisors of $n$ and $H_{n}$ is the $1+\frac12+\frac13+\frac14+...+\frac1n$.

[color=red]Edit by Megus: I've moved your topic to a proper subforum  :D [/color]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=35163, so its equivalent to Riemann's hypothesis (and you should not expect a counterexample).


fie $ABC$ un triunghi si $\omega$ cercul sau inscris, cu centrul in $I$. dreapta $l$ este tangenta la $\omega$ si taie $BC,CA,AB$ in $P,Q,R$. fie $L$ punctul de intersectie al dreptelor $BQ$ si $CR$.
sa se arate ca $IL\bot AP$.

$IL \perp AP \Longleftrightarrow AP$ este polara lui $L$ fata de cerc , deci $A \in$ polara lui $L$ si $P \in$ polara lui $L \Longleftrightarrow L \in$ polara lui $A$, $L \in$ polara lui $P$.

acum fie $D,E,F,H$ punctele de tangenta ale cercului cu $BC,CA,RQ,AB$, deci  polara lui $A$ este $EH$, si polara lui $P$ este $DF$, dar $L$ fiind intersectia diagonalelor patrulaterului circumscriptibil $BCQR$, rezulta conform teoremei lui newton , ca $L \in EH \cap DF$.

@pohoatza: Nu vad de ce ai folosit primul $\Longleftrightarrow$...eu zic ca e adevarat doar intr-un sens. In rest solutia e buna :)
Acum s-a dat o solutie cu polare. Problema iese destul de repede si cu complexe ( in caz ca te plictisesti prea rau :wink: ). Exista si o solutie pur sintetica, fara polare? (bine, e evident ca faza cu polare se poate traduce sintetic dar ma gandeam la alta metoda :) )

:maybe: , eh, ma rog , nu asta conta , am scris in graba.

Data viitoare sa nu te mai grabesti. Nu de alta, dar imi este teama sa nu-ti faci o obisnuinta.


[quote=pluristiq]yes, and...?[/quote]
Find all such a polynomial.
Coefficients are real by the way.

Multiply bith sides by
$(P\left(x+1 \right))(P\left(x+2 \right))\dots(P\left(x+99 \right))$


who will win?

you pick, see if youre right!:lol:

[b]guess federer[/b], i don't follow tennis :D

Through a billion days after it is already over,

(1) Rafael Nadal ESP def. (2) Roger Federer SUI, 7-5 3-6 7-6(3) 3-6 6-2.

Yes, I memed that.

Yeah but Shoujit Banerjee can pwn them all =O

[quote="yaofan"]Rafael Nadal is so much more beastlier than Roger Federer[/quote]

Federer has had better days. The outcome was expected.

both me and rafael are left handed.

I already know who won.

Obviously you know a lot about tennis. :wink:

BTW, Federer got married...


function
real analysis
real analysis unsolved

Function $g: R^{n}\rightarrow R$ is concace.

Find the condition for function $f: R\rightarrow R$ under which composite function $f(g((x))$ is also concave.

$f$ is concave and increasing?
$f(g(\lambda x+(1-\lambda)y))\gneq f(\lambda g(x)+(1-\lambda)g(y))\gneq \lambda f(g(x))+(1-\lambda)f(g(y))$

That can be one condition of this problem.

But is it represent all condition?

i.e. $g(x)=x^{\frac{1}{4}}, f(x) = x^{2}$

then, although $f(x)$ is convex, $f(g(x))$ is concave.

I don't know what does your "all condition" mean.
Besides, $f\circ g$ may be not concave for some concave function $g$, e.g. $x^{\frac34}$
If $f\circ g$ is concave for any concave function $g$, then $f$ must be concave and increasing.
1. Take $g(x)=x$, then $f$ is concave.
2. $f$ must be continuous. Should $f$ be not increasing. Try to prove that there is $p\in\mathbb{R}$ such that $f$ is increasing on $(-\infty, p)$ and decreasing on $(p, \infty)$ (may not be strictly but by no means it could be increasing by assumption). I believe you can find a $g$ such that $f\circ g$ is not concave.

Oh, yes.. you're right. I miss understood your comment.

At first, I wanted to find a nice condition that can define $f$ according to

the specific form of $g$. Now, however, I realized that it's impossible to

find that condition. So, as you said, problem must be changed to

"" for all concave function $g$"" and then $f$ must be increasing and

concave function.

Thanks for your comment, and I'm sorry to my strange problem. ^^


analytic geometry
geometry solved
geometry

let $ABCD$ be a quadrilateral such that length of $AB,BC,CD,AD,AC,BD$ are positive integer numbers.prove that one of these length is $3t$. :D

We will denote $BD, AB, AD, DC, CB$ as $m, n, p, r, q$. Putting coordinates so that $B(0,0), D(m,0)$ is easy to find the length $AC$:
\[ n^2-\frac{(m^2+n^2-p^2)^2}{4m^2} +q^2-\frac{(m^2+q^2-r^2)^2}{4m^2} +2\sqrt{n^2-\frac{(m^2+n^2-p^2)^2}{4m^2}}\sqrt{q^2-\frac{(m^2+q^2-r^2)^2}{4m^2}}  \]
Now suppose that none of $m, n, p, r, q$ is divisible by 3. As $1^2\equiv 1 (\mod 3)$ and $2^2\equiv 3(\mod 3)$,
and $\mathbb{Z}/3\mathbb{Z}$ is a field, is inmediate that the sum of the terms outside the square roots is
$\equiv 0(\mod 3)$, and the terms inside the root are also $\equiv 0(\mod 3)$, and being the result and integer,
it is $\equiv 0(\mod 3)$.

QED

nice solution i relly enjoed it.good job.thanks. :D

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=16468 post #14 problem [b]a)[/b].

  darij


Just for people beginning modular stuff, here's a couple of easy proofs to do.  All of these assume that n is a positive integer.

Prove that the sum formulas will yield integers:
n(n+1)/2
(2n+1)(n+1)n/6

And a tad harder:
(n^2)(n^2-1) will always be divisible by 24 for n>2

Demon wrote:First one is really easy, n :^2:  and n are either both odd or both even, thefore, their sum is even.

Or you could just note that [hide]either n is even or n+1 is even[/hide]

[quote="magnara"](n^2)(n^2-1) will always be divisible by 24 for n>2[/quote]

Are you sure?  What about n=6?  (For it to be divisible by 24, you need three factors of 2 and one factor of 3.. if n is even, but not a multiple of 4, you only have two factors of 2 in the product (n^2)(n-1)(n+1) )

You need 4 consecutive integers (which makes it (n^2 + 2n)(n^2 - 1) or some variant) to ensure the product is divisible by 24.

Yeah, let's just change that n :^2: into an n :^3: .  I was trying to remember an old problem and did so very poorly.

Good - though it's not necessary to consider n mod 6. Try breaking 6 up into 2 times 3, then considering n mod 2 and n mod 3 separately.

NAS wrote:I'll try the second part: (2n+1)(n+1)n/6
[hide]
first let a mod 6 :equiv: 0

if n mod 6 :equiv: a mod 6 then n/6 is an integer

if n mod 6  :equiv: a mod 6 +1 then (2(a +1)+1)(a+1+1)(a+1)/6=(2a+3)(a+2)(a+1)/6 (2a+3) mod 6 :equiv: 3 (a+2) mod 6  :equiv: 2 a multiple of 2 and of 3 so you have a multiple 6

if n mod 6  :equiv: a mod 6 +2 then (2(a+2)+1)(a+2+1)(a+2)/6=(2a+5)(a+3)(a+2)/6 (a+3) mod 6 :equiv: 3 and (a+2) mod 6 :equiv: 2 again you have a multiple of 6

if n mod 6 :equiv: a mod 6+3 then (2a+7)(a+4)(a+3)/6 4*3 mod 6 :equiv: 0 so it's divisable

if n mod 6  :equiv: a mod 6 +4 then (2a+9)(a+5)(a+4)/6 (2a+9) mod 6  :equiv: 3 and (a+4) mod 6  :equiv: 2 so it's divisable

if n mod 6  :equiv: a mod 6+5 then (2a+11)(a+6)(a+5)/6 this one's the simplest because (a+6) mod 6 :equiv: 0
[/hide]
This (AoPS) is my first experence with algabraic proofs any critique on my format/method would be appreciated.

As Osiris said, you can work mod 2 and 3. However lets decide you want to use this mod 6 method, thats fine. But half of the thing about mods is that it reduces the algebra you need to do. Here you don't need to use the letter a at all. As you said, a is 0 mod 6, so you can just use the number 0 wherever a is.

For example, taking one of your lines:

[hide]

if n mod 6  :equiv: a mod 6 +1 then (2(a +1)+1)(a+1+1)(a+1)/6=(2a+3)(a+2)(a+1)/6 (2a+3) mod 6 :equiv: 3 (a+2) mod 6  :equiv: 2 a multiple of 2 and of 3 so you have a multiple 6[/hide]

This can be written as

[hide]

If n:equiv:1 mod 6, then (2n+1)(n+1)n :equiv: (3)(2)(1) :equiv: 6 :equiv: 0 mod 6.

[/hide]

So your proof could be drastically simplified. But you've got the main idea.

I was looking back at an old MA :theta:  test and ran across this problem:

Find the smalest positive ineger x that the solution to the following system of congruencies: x :equiv: 2 mod 3 x :equiv: 3 mod 5 x :equiv: 5 mod 7.

I solved it by finding all the numbers  :equiv:  5 mod 7 then looking for 3 mod 5 and 2 mod 3 I got [hide] 68 [/hide] but there must be a quicker way to do this.

btw thanks for the adivce TripleM and Osiris

Basically, you can't get much better than checking the way you did, though you do it with a bit of algebra. I'm not sure if people want to try your question, so I'll show you how I would do a different example:

x = 3 mod 4, x = 1 mod 5, x = 6 mod 11.

[hide]

Start with the highest modulo, 11. We get x = 11a+6.

Then we go to the next modulo 5. 11a+6 = 10a+a+5+1 = a+1 mod 5. So a must be 0 mod 5. Let a = 5b, and we get x = 55b+6.

Then we go  to the last modulo 4. x = 55b+6 = 52b+3b+4+2 = 3b+2 (mod 4). So 3b+2 = 3 (mod 4), so 3b = 1 (mod 4). 3x0=0, 3x1=3, 3x2=1, 3x3=4, so b is 2 mod 4. Finally, we let b = 4c+2, and thus x = 55(4c+2)+6 = 220c + 116.

So x is 116, 336, 556, etc.

[/hide]

There is another way I think but this is what I do.


Very nice.
[hide]
Case 1: At least one of the fractions $ \dfrac{ix_i}{(x_1\plus{}x_2\plus{}...\plus{}x_n)\plus{}(i\minus{}1)x_i}$ is greater than one.

The inequality is evident.

Case 2: $ \dfrac{ix_i}{(x_1\plus{}x_2\plus{}...\plus{}x_n)\plus{}(i\minus{}1)x_i}\le 1\forall i\in\{1,2,...,n\}$

If $ a$, $ b$ and $ x$ are positive numbers we have that $ 1\ge\dfrac{a}{b}\Leftrightarrow\dfrac{x\plus{}a}{x\plus{}b}\ge\dfrac{a}{b}$ with equality if and only if $ a\equal{}b$.

Then
$ \sum_{i\equal{}1}^n{\dfrac{ix_i}{(x_1\plus{}x_2\plus{}...\plus{}x_n)\plus{}(i\minus{}1)x_i}}\ge\sum_{i\equal{}1}^n{\dfrac{x_i}{x_1\plus{}x_2\plus{}...\plus{}x_n}}\equal{}1$

But equality cannot hold in any inequality since it would imply $ x_1\plus{}...\plus{}x_n\minus{}x_i\equal{}0$, which can't be true because all of the $ x_k$ are positive.

Then the inequality is strict, and we have the result we wanted.
[/hide]

:)


[quote="SnowStorm"]How can I help add problems to the archives?

Do I just need to contact a mod and tell them what/where I've posted something?[/quote]

Um. Here is what I recommend you to do:

1. Post problem ONLY (i.e. no other wording like "this is a nice problem.. etc")
2. In source field, put type of contest (AHSME, COMC, etc..), year, problem # in the source field so I can find them through searching (I already used some problems that you posted, SnowStorm).

I don't really recommend by giving links because when I post problems, I ALWAYS use search button (yeah.. I know not a lot of people do this but this is in Valentin's signature and I really follow them.  :) ) so if you posted them, I'll be able to find them.