Trigonometric inequality
trigonometry
circumcircle
Triangle
geometry
IMO
IMO 1974

Let $ABC$ be a triangle. Prove that there exists a point $D$ on the side $AB$ of the triangle $ABC$, such that $CD$ is the geometric mean of $AD$ and $DB$, iff the triangle $ABC$ satisfies the inequality $\sin A\sin B\le\sin^2\frac{C}{2}$.

[hide="Comment"][i]Alternative formulation, from IMO ShortList 1974, Finland 2:[/i] We consider a triangle $ABC$. Prove that: $\sin(A) \sin(B) \leq \sin^2 \left( \frac{C}{2} \right)$ is a necessary and sufficient condition for the existence of a point $D$ on the segment $AB$ so that $CD$ is the geometrical mean of $AD$ and $BD$.[/hide]

Let a point $D$ on the side $AB$.
Let $CF$ the altitude of the triangle $\triangle ABC$, and $C'$ the symmetric point of $C$ through $F$.
We bring a parallel line $L$ from $C'$ to $AB$. This line intersects the ray $CD$ at the point $E$, and we know that $DE=DC$.

The distance $d(L,AB)$ between the parallel lines $L$ and $AB$ is $CF$.

Let $w = (O,R)$ the circumscribed circle of $\triangle ABC$, and $MM'$ the perpendicular diameter to $AB$, such that $M,C$ are on difererent sides of the line $AB$.

In fact, the problem asks when the line $L$ intersects the circumcircle. Indeed:

Suppose that $DC$ is the geometric mean of $DA,DB$.

$DA \cdot DB = DC^{2}\Rightarrow DA \cdot DB = DC \cdot DE$

Then, from the power of $D$ we can see that $E$ is also a point of the circle $w$.
Or else, the line $L$ intersects $w \Leftrightarrow$

$d(L,AB)\leq d(M,AB) \Leftrightarrow$

$CF \leq MN,$ where $MN$ is the altitude of the isosceles $\triangle MAB$.

$\Leftrightarrow \frac{1}{2}CF \cdot AB \leq \frac{1}{2}MN \cdot AB \Leftrightarrow$ $(ABC) \leq (MAB) \Leftrightarrow$ $\frac{AB \cdot BC \cdot AC}{4R}\leq \frac{AB \cdot MA^{2}}{4R}\Leftrightarrow$

$BC \cdot AC \leq MA^{2}$

We use the formulas:

$BC = 2R \cdot \sin A$
$AC = 2R \cdot \sin B$

and $\angle CMA = \frac{C}{2}\Rightarrow MA = 2R \cdot \sin\frac{C}{2}$

So we have
$(2R \cdot \sin A)(2R \cdot \sin B) \leq (2R \cdot \sin\frac{C}{2})^{2}\Leftrightarrow$
$\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}$

For $(\Leftarrow)$

Suppose that $\sin A \cdot \sin B \leq \sin^{2}\frac{C}{2}$

Then we can go inversely and we find that $d(L,AB)\leq d(M,AB) \Leftrightarrow$
the line $L$ intersects the circle $w$ (without loss of generality; if $d(L,AB)=d(M,AB)$ then $L$ is tangent to $w$ at $M$)

So, if $E \in L \cap w$ then for the point $D = CE \cap AB$ we have $DC=DE$ and $AD \cdot AB = CD \cdot DE \Rightarrow$

$AD \cdot AB = CD^{2}$


Let $ {a},{b},{c}$ are positive reals, such that $ {a}\plus{}{b}\plus{}{c}\equal{}1$. Find the maximum value of $ \frac{a^{2}}{b\plus{}1}\plus{}\frac{b^{2}}{c\plus{}1}\plus{}\frac{c^{2}}{a\plus{}1}$

[quote="aleksandar"]Let $ {a},{b},{c}$ are positive reals, such that $ {a} \plus{} {b} \plus{} {c} \equal{} 1$. Find the maximum value of $ \frac {a^{2}}{b \plus{} 1} \plus{} \frac {b^{2}}{c \plus{} 1} \plus{} \frac {c^{2}}{a \plus{} 1}$[/quote]
Denote that $ f(a,b,c)\equal{}\frac {a^{2}}{b \plus{} 1} \plus{} \frac {b^{2}}{c \plus{} 1} \plus{} \frac {c^{2}}{a \plus{} 1}$
Firstly, we will prove that: $ f(a,b,c) \le \max\{f(0,b,c), f(1,b,c)\}$
Actually, We have:
If $ \frac{1\plus{}a}{b\plus{}1}\ge \frac{c^2}{2(a\plus{}1)}$ then:
$ f(1,b,c)\minus{}f(a,b,c) \equal{} (1\minus{}a)\left(\frac{1\plus{}a}{b\plus{}1}\minus{}\frac{c^2}{2(a\plus{}1)}\right) \ge 0$
If $ \frac{1\plus{}a}{b\plus{}1}\le \frac{c^2}{2(a\plus{}1)} \Rightarrow  \frac{c^2}{a\plus{}1}\ge \frac{2(1\plus{}a)}{b\plus{}1} > \frac{a}{b\plus{}1}$ then:
$ f(0,b,c)\minus{}f(a,b,c) \equal{} a\left(\frac{c^2}{a\plus{}1}\minus{}\frac{a}{b\plus{}1}\right) \ge 0$
By the similar way, we get: $ f(a,b,c) \le \max\{f(0,1,1), f(0,1,0), f(0,0,0)....\}\equal{}1$
So, $ \max f(a,b,c)\equal{}1 \Longleftrightarrow a\equal{}1;b\equal{}c\equal{}0 \vee ....$
Done ! :D

btw, someone will say '' This isn't an inequalitie problem''. But this problem was firstly given on some MO as:
Let $ {a},{b},{c}$ are positive reals, such that $ {a}\plus{}{b}\plus{}{c}\equal{}1$. Prove that: $ \frac{a^{2}}{b\plus{}1}\plus{}\frac{b^{2}}{c\plus{}1}\plus{}\frac{c^{2}}{a\plus{}1}\le\ 1$
Thanks again  :lol:

[quote="aleksandar"]btw, someone will say '' This isn't an inequalitie problem''. But this problem was firstly given on some MO as:
Let $ {a},{b},{c}$ are positive reals, such that $ {a} \plus{} {b} \plus{} {c} \equal{} 1$. Prove that: $ \frac {a^{2}}{b \plus{} 1} \plus{} \frac {b^{2}}{c \plus{} 1} \plus{} \frac {c^{2}}{a \plus{} 1}\le\ 1$
Thanks again  :lol:[/quote]

My proof is the following:

$ \frac {a^{2}}{b \plus{} 1} \equal{} \frac {a^{2}}{2b \plus{} a \plus{} c}\leq \frac {a^2}{b \plus{} a \plus{} c}$

Similarly we have:

$ \frac {b^2}{c \plus{} 1}\leq\frac {b^2}{a \plus{} b \plus{} c}$ and

$ \frac {c^2}{a \plus{} 1}\leq\frac {c^2}{a \plus{} b \plus{} c}$

Now, we have to prove that:

$ \frac{a^2}{a \plus{} b \plus{} c} \plus{} \frac {b^2}{a \plus{} b \plus{} c} \plus{} \frac {c^2}{a \plus{} b \plus{} c}\leq 1$

$ \Leftrightarrow a^2 \plus{} b^2 \plus{} c^2\leq a \plus{} b \plus{} c$

which is obvious. (because $ a,b,c \leq 1$ from $ a \plus{} b \plus{} c \equal{} 1$)

 :)

[quote="matrix41"][quote="shaaam"][quote="aleksandar"]btw, someone will say '' This isn't an inequalitie problem''. But this problem was firstly given on some MO as:
Let $ {a},{b},{c}$ are positive reals, such that $ {a} \plus{} {b} \plus{} {c} \equal{} 1$. Prove that: $ \frac {a^{2}}{b \plus{} 1} \plus{} \frac {b^{2}}{c \plus{} 1} \plus{} \frac {c^{2}}{a \plus{} 1}\le\ 1$
Thanks again  :lol:[/quote]

My proof is the following:

$ \frac {a^{2}}{b \plus{} 1} \equal{} \frac {a^{2}}{2b \plus{} a \plus{} c}\leq \frac {a^2}{b \plus{} a \plus{} c}$

[/quote]

how come????? $ a,b,c$ are not [b]nonnegative[/b] numbers right????[/quote]
$ {a},{b},{c}$ are [b]positive reals[/b]

$ a, b, c$ are positive reals , not nonnegative. So you can't say that $ a,b,c \le\ 1$. In fact,  $ a\plus{}b\plus{}c\equal{}1$ implifies  $ 0<a,b,c<1$
That makes this problem difficult.  :(

I think this expression has not maximum value if a;b;c be possible reals number :D

Friends, my above proof was correct. If $ a,b,c$ positive reals, then the sign would be strictly smaller [u]not[/u] smaller and equal to.
That is all. Anyways, here is another solution , I have just found:

Assume that $ a\geq b\geq c$, we have:

$ \frac {a^2}{b \plus{} 1}\leq \frac {a^2}{c \plus{} 1}$, and $ \frac {c^2}{a \plus{} 1}\leq\frac {c^2}{c \plus{} 1}$

then we have to prove that:

$ \frac {a^2}{c \plus{} 1} \plus{} \frac {b^2}{c \plus{} 1} \plus{} \frac {c^2}{c \plus{} 1}\leq 1$

$ \Leftrightarrow \frac {a^2 \plus{} b^2 \plus{} c^2}{c \plus{} 1}\leq 1$

$ \Leftrightarrow a^2 \plus{} b^2 \plus{} c^2 \minus{} c\leq 1$

we can prove it as follows: $ a^2 \plus{} b^2 \plus{} c^2 \minus{} c\leq a \plus{} b \plus{} c \minus{} c \equal{} a \plus{} b\leq 1$

The last inequality is obvious from the condtion $ a \plus{} b \plus{} c \equal{} 1$

Since inequality is cyclic, we have to also assume the case $ a\geq c\geq b$. But proof is almost the same:

$ \frac {b^2}{c \plus{} 1}\leq \frac {b^2}{b \plus{} 1}$ and $ \frac {c^2}{a \plus{} 1}\leq \frac {c^2}{b \plus{} 1}$

we have to prove that $ \frac {a^2}{b \plus{} 1} \plus{} \frac {b^2}{b \plus{} 1} \plus{} \frac {c^2}{b \plus{} 1}\leq 1$

which again turns out proving $ a^2 \plus{} b^2 \plus{} c^2 \minus{} b\leq 1$ , but it is true since $ a \plus{} c\leq 1$

[quote="shaaam"]Friends, my above proof was correct. If $ a,b,c$ positive reals, then the sign would be strictly smaller [u]not[/u] smaller and equal to.
[/quote]
Yes, your above proof was correct with this problem
[quote="aleksandar"]btw, someone will say '' This isn't an inequalitie problem''. But this problem was firstly given on some MO as:
Let $ {a},{b},{c}$ are positive reals, such that $ {a} \plus{} {b} \plus{} {c} \equal{} 1$. Prove that: $ \frac {a^{2}}{b \plus{} 1} \plus{} \frac {b^{2}}{c \plus{} 1} \plus{} \frac {c^{2}}{a \plus{} 1}\le\ 1$
Thanks again  :lol:[/quote]
But not with the problem of the topic :)
[quote="aleksandar"]Let $ {a},{b},{c}$ are [b]positive reals,[/b] such that $ {a} \plus{} {b} \plus{} {c} \equal{} 1$. Find the maximum value of $ \frac {a^{2}}{b \plus{} 1} \plus{} \frac {b^{2}}{c \plus{} 1} \plus{} \frac {c^{2}}{a \plus{} 1}$[/quote]
Anyway, I think this problem is wrong :lol:

Maybe we should leave this problem....
There is a lot of more interesting inequalities, which aren't incorrect.
Lets go to the next level  :)


hallo,
solve in natural numbers: $6 \cdot 2^{x}-y^{2}= 3$.

It's clear that $3|y$, so let $y = 3A$ and now we have $2^{x+1}-3A^{2}= 1$.
We find out that $-3A^{2}\equiv A^{2}\equiv 1 (mod 4) \Rightarrow A$ odd.
Then, $3A^{2}+1 \equiv 4 (mod 8) \Rightarrow x+1 < 3$
So, $x = 1$ and $y = 3$ is the unique solution of the equation.

EDIT : Sorry, thanks for the correction, Moon..

U have made a mistake!
x=1 and y=3 bcz A=1

MOON


geometry
calculus
integration
calculus computations

the radius of a circle is increasing at a constant rate of 0.2 meters per second.  What is the rate of increase in the area of the circle at the instant when the circumference of the circle is 20pi meters?

a particle moves long the x-axis so that any time t>0, its acceleration is given by a(t) = ln(1+2^t).  If the velocity of  the particle is 2 at time t=1, then the velocity of the particle at time t=2 is?

[hide="hints"]
1. $ \frac{dA}{dt}\equal{}\frac{dA}{dr}\cdot\frac{dr}{dt}.$
(A=area, r=radius) by the chain rule

2. velocity is the integral of acceleration.[/hide]


Let $A=\{z\in\mathbb C|z^{18}=1\}, B=\{\omega\in\mathbb C|\omega^{48}=1\}, X=\{z\omega|z\in A, \omega\in B\}$. Find $|X|$.

[hide]$144 ?$

It's the same as counting $8k+3l\ mod\ 144$ with $0 \leq k \leq 17$ and $0 \leq l \leq 47$.[/hide]


calculus
combinatorics proposed
combinatorics

Prove $\sum_{i=m}^{r}(-1)^{i-m }\left(\begin{array}{c}i-1&m-1\end{array}\right)\left(\begin{array}{c}r&i\end{array}\right)=1$
Although this is not too hard to prove it ,I want to compare different methods(from combinatorics,calculus,...)
So [b]any[/b] comment would be welcome.(Feel free to use advanced formulae.)

I did not try, or even read carefully, but at first sight I think this identity can be proved in the same way as I did in :
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=leader+boys&t=20206
or
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=leader+boys&t=20814
or
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=leader+boys&t=16482

Pierre.


Hi everyone...I'm having issues with test case 8 (it's stopped at over 2 seconds...).  :?  I heard that there's a DP solution to it, and I was wondering if someone could explain the concept behind it to me.  Thanks a ton!

I don't know what you consider a DP solution and what not, but the idea of trying all the possible places for the middle of the palindrome should work. At least I think that's what I did and it worked fine. You just start out with one or two characters and take on pairs of adjacent chars if they are equal. You can stop as soon as you hit two different ones, so you won't have to check all the way to the end of the array. $O(N^2)$ but $\Omega (N)$.

Thanks.  :)  I figured out what my issue was.  Was going through the sequence by start and end points, not by center (so it would check through all larger shells of  a substring that you wouldn't know wasn't a palindrome potentially until you got to the center two characters and whatnot).  Not sure what possessed me to do it that way at first :? , but at least it works now.  Oh well!  Thanks again!


[quote="Arvind_sn"]I came across the power of a point theorem, and I'm not sure I totally understand it. Can someone please explain it to me? Thanks. :)[/quote]

[url=http://en.wikipedia.org/wiki/Circle_power]This will maybe help[/url]


2n players are participating in a tennis tournament. Find the number P_n of pairings for the first round.

Do you mean the number of different pairings?

It is just like $ \frac{\binom{2n}{2}*\binom{2(n-1)}{2}*...*\binom{2}{2}}{n!}=\frac{(2n)!}{2^{n}\cdot n!}$

since you choose groups of 2 people for each pair. You must then divide by $ n!$ since you counted like AB, CD and CD,AB.


Source: MMPC 1995 Round 1 (#18)

A marksman fired at a target which had five areas worth scores of 16, 17, 23, 24, and 39 points, respectively.  His total was exactly 100.  How many times did he hit the area worth 16 points?

(A) 0
(B) 1
(C) 2
(D) 3
(E) 4

a, b, c, d, and e are the numbers of times that the target regions were hit.

0 :le: a < 7
0 :le: b < 6
0 :le: c < 5
0 :le: d < 5
0 :le: e < 3

16a + 17b + 23c + 24d + 39e = 100

Modding out with 16, it becomes...

b + 7c + 8d + 7e = 4

Modding out with 7...

b + d = 4
(b, d) = (0, 4), (1, 3), (2, 2), (3, 1), (4, 0)

The only one of these solution pairs that will work is (4, 0).  That would result in having a equaling 2.  The marksman hit the area worth 16 points twice so the answer is C.


quadratics
probability
ratio
geometry
real analysis
limit
analytic geometry

What is the probability that the quadratic equation $ x^2 \plus{} 2bx \plus{} c \equal{} 0$ has real roots?

If $ b, c$ aren't restricted to an interval, there isn't a meaningful way to talk about a probability measure with respect to all of $ \mathbb{R}^2$.  What you're measuring is the ratio of the area of the region such that $ b^2 \minus{} c \ge 0$ to the entire plane, but [i]both regions are infinite in area.[/i]

[i]You want the probability that the minimum value of the quadratic is non-positive. Plug in $ x \equal{} \minus{} b$ into the equation to get the condition $ c \leq b^2$[/i]

[b]The real question:[/b] is there a meaningful way computing the probability of the above?

If you construct a square frame centered around the origin, and consider the probability within that region, and take the limit as you expand this square frame (taking into consideration more and more of the plane), you get that the probability goes towards $ 1$.

On the other hand, you can say that for any value of $ b$, there exists a $ c_*$ such that there are roots for all $ c \leq c_*$ there are real roots and for all $ c > c_*$ there are no real roots. So for any $ b$, the set of satisfactory values for $ c$ is half of the real line. So then one could argue that the probability is $ 1/2$.

One could then respond ot that argument by saying that the graph of $ c \equal{} b^2$ takes up less of the plane than $ c \equal{} |b|$, and section of the plane "under" the latter graph "appears" to cover $ 3/4$ of the plane, so the probability should be at least $ 3/4$.

Catch the drift?

---------------------------------------

[b]Edit:[/b] Beaten. But now I have a question for you, [b]t0rajir0u[/b] ... Suppose someone asks, what is the probability that a random point in the plane satisfies $ \minus{}3 \leq x \leq 3$. Would it still be meaningless?

[quote="TZF"]Suppose someone asks, what is the probability that a random point in the plane satisfies $ \minus{} 3 \leq x \leq 3$. Would it still be meaningless?[/quote]

Yes.  Regardless of how intuitively the answer to this question should be zero, there isn't a meaningful way to compare two infinite Lebesgue measures.  (The related question of whether a random point in a plane lies on a given line is completely meaningful - the probability is said to be zero because the Lebesgue measure of a line with respect to $ \mathbb{R}^2$ is zero.)

I should amend my previous statement:  what I wanted to say is that the Lebesgue measure constitutes the closest thing we have to a [i]uniform[/i] probability measure on $ \mathbb{R}^2$.  We [i]can[/i] talk about probability measures on $ \mathbb{R}^2$ such that the measure of the whole plane is $ 1$ - but these measures cannot be uniform.

On the other hand, what if we wanted to calculate the probability that $ x > 0$ (still the random point example)? Would it still be meaningless?

Intuitively, it is 1/2; experimentally, it is 1/2; but they are still infinite Lebesgue measures, and thus mathematically the probability is meaningless?

Sometimes, such things get defined anyway:

Let $ R$ be the "space" everything happens in (mostly $ R \equal{} \mathbb R^n$ or $ R \equal{} \mathbb Z^n$).
Let $ S_1, S_2, ..., \subset R$ be a fixed (!) sequence $ S$ converging to $ R$ as sets. In most cases, $ S_1 \subset S_2 \subset ... \subset R$. Assume further that we are given some measure $ p$ on $ R$ such that $ 0 < p(S_i) < \infty$ for all $ i$ (especially, $ p$ is defined on all $ S_i$).
Let $ E \subset R$ be the "event" under consideration. Then one can define a probability measure $ P_S(E) \equal{} \lim_{i \to \infty} \frac {p(S_i \cap E)}{p(S_i)}$. Note that it depends on that sequence $ S$, not only on $ p$.

Here it would be meaningful to take $ S_i \equal{} [ \minus{} i,i]^2$.

[quote="Yongyi781"]experimentally, it is 1/2[/quote]

"Experimentally," one generates a "random" point on the plane by picking a random bounded, discrete $ x$-coordinate and a random bounded, discrete $ y$-coordinate.  This corresponds, in the limit (as we pass from the discrete to the continuous) to using the Lebesgue measure to determine the probability measure on a square of [i]finite[/i] size.  But there is no meaningful way to generate a "random" point on the entire plane without specifying a probability measure that cannot be uniform.

It would be good to refer to the [url=http://en.wikipedia.org/wiki/Probability_space]definition of probability[/url].  On a subset $ A$ of the plane of finite Lebesgue measure, it is convenient to specify a probability measure of a subset of $ A$ by taking its Lebesgue measure and dividing by the measure of $ A$.  This is a definition that satisfies all the axioms of a probability measure.

Once we start trying to assign a probability measure to a subset $ A$ of the plane with infinite Lebesgue measure, there is no obvious choice of measure because we can no longer divide by the measure of $ A$.


Let the integer sequence ${a_n}$ defined by:
  $a_1=2$ and $a_n=3a_{n-1}+2n-3$
Find $a_n$.

Define $b_n=a_n+n\implies a_n=b_n-n$. Then $b_1=a_1+1=3.$

Substituting $a_n=b_n-n$ into the given equation we get

$b_n-n=3(b_{n-1}-n+1)+2n-3\implies b_n=3b_{n-1}$

Hence $b_n=b_13^{n-1}=3^n\implies a_n=3^n-n$

[quote="Farenhajt"]Define $b_n=a_n+n$[/quote]
How can you guess this, Farenhajt? Thanks.

I'm looking for a way to eliminate that nasty $n$ from the equation, so I substitute $a_n=b_n+\alpha n+\beta$ and then make a condition that the $n$-terms cancel out. Hence

$b_n+\alpha n+\beta=3(b_{n-1}+\alpha(n-1)+\beta)+2n-3 \implies$
$\implies b_n=3b_{n-1}+(2\alpha+2)n-3\alpha+2\beta-3$

From there $\alpha=-1$, and the easiest would be if $-3\alpha+2\beta-3$ is also zero, giving $\beta=0$


Let $ n$ be a natural odd number.

prove the following identity for Jacobi symbol:

$ (\frac {2}{n}) \equal{} (\frac { \minus{} 1}{n}) (\frac {2}{n \minus{} 2})$

Thanks. The "fraction" symbol means jacobi.

$ (\frac{2}{n}) \equal{} (\minus{}1)^{\frac{n^2\minus{}1}{8}}$
$ (\frac{\minus{}1}{n})\equal{} (\minus{}1)^{\frac{n\minus{}1}{2}}$
$ (\frac{2}{n\minus{}2}) \equal{} (\minus{}1)^{\frac{(n\minus{}2)^2\minus{}1}{8}}$

And $ \frac{n^2\minus{}1}{8} \equal{} \frac{n\minus{}1}{2}\plus{} \frac{(n\minus{}2)^2\minus{}1}{8}$.

[The properties I used are well known properties of the Jacobi Symbol, and follows easily from the corresponding properties of Legendre Symbol]


Hmm, I mailed in my personal reccommendation and quiz solutions and they arrived a long time ago. However, the online application page says that MathCamp has all of my materials except for my quiz solutions and the reccommendation. Is this because there wasn't enough time to update the online applications or did I do something wrong?

Yes, most likely it just hasn't been updated yet.

[quote="MithsApprentice"]Hmm...Can I submit the application and then send in my solutions/recommendations later?[/quote]

Yes, if you want to send in your solutions or recommendations by postal mail, you can select those options in your application and then submit the application.  Note that we will not look at your application until it is complete.


I tried to include a rather large graphic on a page, centered, and it skippeda page and put the graphic on the next page. How could I prevent this?

Do you mean there's a blank page or that it puts the graphic on a new page? Can you give a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157945]minimal example[/url] where this happens?

The first.

[code]
\begin{center}
\includegraphics{graphic.png}
\end{center}
[/code]

That's not a minimal example since there is no problem with that code and no blank page is produced. Please post the complete minimal document with the preamble (otherwise how can anyone know if a package is causing a problem) and a picture so that when it is compiled it produces the problem.

Oh... then:

[code]
\documentclass{article}
\usepackage[pdftex]{graphicx}
\begin{document}
\begin{center}
\includegraphics{graphic.png}
\end{center}
\end{document}[/code]

It's still hard to see what is happening because you haven't given the size of graphic.png nor the default page size.
Firstly, does graphic.png completely fit on a page inside the margins? If not, scale using \includegraphics[scale=0.5]{graphic.png} for example.

If it does fit on the page then try using a float
[code]\begin{figure}
\begin{center}
\includegraphics{graphic.png}
\end{center}
\end{figure}[/code]

It now works, thank you very much!

However, it's aligned to the right? How could I change the alignment?

[quote="stevem"]It's still hard to see what is happening because you haven't given the size of graphic.png nor the default page size. [/quote]There's so little information to go on but the image is probably too big for the page. Either use landscape or scale the image.


limit
analytic geometry
function
real analysis
real analysis unsolved

[quote="hpoincare"]Consider $f$ such that

$f(x,y)=\begin{cases}\frac{x^3y(xy+1)}{x-y^2}\quad&\text{if }x\neq y^2\\0\quad&\text{if }y=x^2\end{cases}$

Find, if exists, $\lim_{(x,y)\to (0,0)}f(x,y)$

I saw that along the curves of the form $y=a_nx^n+\dotsb+a_1x$ and $x=b_mx^m+\dotsb+b_1y$
the limits is zero...

Then I changed to polar coordinates but I couldn't prove anything...

To simplify the function I think we can restrict ourselves to $g$ such that

$g(x,y)=\begin{cases}\frac{x^3y}{x-y^2}\quad&\text{if }x\neq y^2\\0\quad&\text{if }y=x^2\end{cases}$

because $f(x,y)=g(x,y)(xy+1)$ and the factor $(xy+1)$ tends to $1$.

Please help, Lucas[/quote]
I guess you meant to write
$g(x,y)=\begin{cases}\frac{x^3y}{x-y^2}\quad&\text{if }x\neq y^2\\0\quad&\text{if }x=y^2\end{cases}$
Consider $g(x,y)$ for $x = y^2 + y^8$, and send $y$ to zero.


integration
real analysis
real analysis unsolved

If $f \in R[a,b]$,and $f(x)>0,x \in [a,b]$,prove:
\[ \int_a^bf(x)dx >0 \]

What  is $R[a,b]$? If $R$ means "Riemann integrable", see the discussion at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=37648[/url].

[quote="fedja"]What  is $R[a,b]$? If $R$ means "Riemann integrable" .[/quote]
Yes,It is.Thanks for the link


Does anyone play streaming music feeds from the internet while seated at a computer? When I'm on the broadband connection in my office, I'm usually listening to http://www.kusc.org . The evening host, Jim Svejda, is particularly good at finding music that you might not have heard.

I sometimes listen to The Piano Society Radio Show. As of now its Bach, Chopin starts soon... :D

[url]http://www.pianosociety.com/index.php?id=363[/url]

sep 30 seems...so long ago...

anyways, anyone know of sites like singingfish.com that have *free* music? though admittedly, there are only a few full length recordings on that site...


inequalities trigonometry calculus integration algebra function domain

Let $r$ be positive real numbers.In $xyz$ space find the volume of the solid body forming by the all points satisfying the following inequalities.

$x^2+y^2\leqq r^2$
$y^2+z^2\geqq r^2$
$z^2+x^2\leqq r^2$

kunny: do you really intend for that middle inequality to be $y^2+z^2\ge r^2$? Or should all three inequality signs point in the same direction?

If I haven't made an idiotic mistake somewhere, the answer is $16r^3\left(\frac1{\sqrt2}-\frac23\right)$. Anyway, it is just a rather boring exercise in trigonometric substitutions... Do you have a nice geometric solution?

Yes.You are right, fedja. :)

By the way, how many times did you do trigonometric substitution?

[quote="kunny"]
By the way, how many times did you do trigonometric substitution?[/quote]
I did it for all 3 variables: $x=r\sin\alpha$, $y=r\cos\beta$, $z=r\cos\gamma$.
I also restricted the integration to the domain $x,y,z>0$ (finding $1/8$ of the volume first).


[quote="Dimitris X"]Find all the triples of the real numbers $ (x,y,z)$ where $ x,y,z > 3$ and satisfy:

$ \frac {(x \plus{} 2)^2}{y \plus{} z \minus{} 2} \plus{} \frac {(y \plus{} 4)^2}{z \plus{} x \minus{} 4} \plus{} \frac {(z \plus{} 6)^2}{x \plus{} y \minus{} 6} \equal{} 36$.[/quote]
From AM-GM's inequality,we have:

$ \frac {(x \plus{} 2)^2}{y \plus{} z \minus{} 2} \plus{} y\plus{}z\minus{}2 \geq 2x\plus{}4$

$ \frac {(y \plus{} 4)^2}{z \plus{} x \minus{} 4} \plus{} z\plus{}x\minus{}4 \geq 2y\plus{}8$

$ \frac {(z \plus{} 6)^2}{x \plus{} y \minus{} 6} \plus{} x\plus{}y\minus{}6 \geq 2z\plus{}12$

The sum of these inequalities yields :

$ \frac {(x \plus{} 2)^2}{y \plus{} z \minus{} 2} \plus{} \frac {(y \plus{} 4)^2}{z \plus{} x \minus{} 4} \plus{} \frac {(z \plus{} 6)^2}{x \plus{} y \minus{} 6} \geq 36$

With equality if and only if:

$ \frac {(x \plus{} 2)^2}{y \plus{} z \minus{} 2} \equal{} y\plus{}z\minus{}2$,$ \frac {(y \plus{} 4)^2}{z \plus{} x \minus{} 4} \equal{} z\plus{}x\minus{}4$,and $ \frac {(z \plus{} 6)^2}{x \plus{} y \minus{} 6} \equal{} x\plus{}y\minus{}6$

Or,equivalently

$ x\plus{}4\equal{}y\plus{}z,y\plus{}8\equal{}z\plus{}x$ and $ z\plus{}12\equal{}x\plus{}y$

Hence $ (x\plus{}4)\plus{}(y\plus{}8)\plus{}(z\plus{}12) \equal{} 2(x\plus{}y\plus{}z)$,or $ x\plus{}y\plus{}z\equal{}24$

Wich immediately gives $ x\equal{}10,y\equal{}8$ and $ z\equal{}6$.


Alice and Bob play a game involving a circle whose circumference is divided
by 12 equally-spaced points. The points are numbered clockwise, from 1 to
12. Both start on point 12. Alice moves clockwise and Bob,
counterclockwise. In a turn of the game, Alice moves 5 points clockwise and
Bob moves 9 points counterclockwise. The game ends when they stop on the
same point. How many turns will this take?

Alright. Each turn they go 14 spaces altogether. The circle is divided into 12. We want the intersection. So we find LCM(12,14) which is 84.

so they will meet after 84/14=6 moves


probability MATHCOUNTS geometry perimeter ratio

The problems all have a theme, it is to try and see a pattern :D You may use a calculator for all of these:

1. If the digits 1 through 26 are each randomly assigned to a different letter of the alphabet, what is the probability that MATHCOUNTS will equal 2006? Note: MATHCOUNTS means M*A*T*H*C*O*U*N*T*S

2. A school has one hundred lockers numbered from one to one hundred and one hundred students. The first, third, and every other student after that will only open lockers. The second, fourth, and every other student will only close lockers. The first student runs up and opens every locker. The second student runs up and closes every second locker open. The third student runs up and opens every third locker closed. This continues until all the students have gone. What is the sum of the numbers on all lockers that are open?

3. Bob was given a quiz with 100 problems and an equal number of addition, subtraction, multiplication, and division problems. Correct addition and subtraction problems were worth 4 points while correct multiplication and division problems were worth 8 points. Each addition or subtraction problem left blank was worth 1 point while each multiplication or division problem left blank was worth 2 points. Wrong answers were worth 0 points. Bob only got half of the adding and dividing problems he answered right while he got all the subtracting and multiplying problems he answered right. If Bob left exactly 17 answers blank and answered an even number of addition and division problems, what is the positive difference between the greatest amount of points he could’ve scored and the least amount of points he could’ve scored?

4. The geometric mean x of two numbers a and b is calculated by finding x where $\frac{a}{x} = \frac{x}{b}$. What is the sum of all distinct integers that can be arranged in a pair with another positive integer to have a geometric mean of 2006?

5. The perimeter of a regular hexagon is numerically the same as the area of a regular dodecagon where the apothem is $\sqrt{3}$ times the side length which is 2. What is the ratio of the area of a regular hexagon to the perimeter of a regular dodecagon? Express your answer in simplest radical form.

6. Abby has a 24 hour digital clock. When it is 11:59PM, it will display 23:59. When it is 6:00AM, it will display 06:00. If she looks at the clock at a random time of the day, what is the probability the clock will show at least three consecutive increasing or decreasing digits? Express your answer as a common fraction.

They're arranged in terms of difficulty IMO.

Edit: Added numbers, #1 used to be #6, #2 used to be #1, #3 used to be #2, #4 used to be #4, #5 used to be #3, and #6 used to be #5

[quote="slugbug321"][hide]1)385, add all square number one thru one hundred.
2)528
3) I'm not too sure...not that great at geometry :)
4) 1/480
5)2006=2*17*59.  That means that there are three pairs, and then a and b could be 2006.  So there are four, but you have two 2006's, so I would say 8.
6) this seems really trick-questiony to me...i would have to say the chance is 0, because 2006= 2*17*59, and 59>26.[/hide][/quote]
I changed the questions around a bit but from what I see..
[hide]1 is wrong, it's not the classic problem ;), 2 I think is wrong, 4 is wrong, 5 I think you misread the problem, it wants the sum of the numbers and 6 is correct![/hide]

[quote="Iron Fist"]Problem 5 is ninja-tastic.

[hide=" :ninja: "]
Every factor of $2006^2$ can be used. To add up the factors of $2006^2$, you first factor into $(2^2)(17^2)(59^2)$. To find the sum, you do this: $(2^0 + 2^1 + 2^2)(17^0 + 17^1 + 17^2)(59^0 + 59^1 + 59^2)$, which equals $\boxed{7609609}$.
[/hide][/quote]

Correct
[hide="Hint for 1, 2, and 4"]there's a trick that makes the problem easier for all of these[/hide]

[quote="ehehheehee"][hide][hide][hide]0
385
17
7609609
24*radical3
7/180[/hide][/hide][/hide][/quote]

1 and  4 are correct, everything else is wrong.

Number two is [u]not[/u] sum of squares... read carefully...

[quote="Ignite168"]4. The geometric mean x of two numbers a and b is calculated by finding x where $\frac{a}{x} = \frac{x}{b}$. What is the sum of all distinct integers that can be arranged in a pair with another positive integer to have a geometric mean of 2006?
[/quote] [hide="Solution for 4"] If we have $\frac{a}{x}=\frac{x}{b}$, we can cross-multiply to get $x^2=ab$. We can substitute $2006$ for $x$. So we get $2006^2=ab$. From here, we're basically trying to find the sum of the factors of $2006^2=2^2\times17^2\times59^2$. The sum of the factors is $(2^0+2^1+2^2)(17^0+17^1+17^2)(59^0+59^1+59^2)=7\times307\times3541=\boxed{7609609}$. :D [/hide]

[hide="2"] I just tried it for the first 9 lockers and found that all the odd numbered lockers were open, so the answer is $50^2=\boxed{2500}$. Is that right?  :? [/hide]

EDIT: oops. I thought Iron Fist's was for the new number 5.


[quote="dt20022006"]Let $x,y,z > 0$
Prove
  $2(x+y+z)^{3}\geq 9(x^{2}+y^{2}+z^{2}+xy+zx+yz ) \sqrt[3]{xyz}$[/quote]
The following inequality is also true.
Let $x$, $y$ and $z$ are non-negative numbers. Prove that:
\[2(x+y+z)^{3}\geq \left(11(x^{2}+y^{2}+z^{2})+7(xy+zx+yz) \right) \sqrt[3]{xyz}\]


Let $a$ be a non-zero integer, and $n \geq 3$ another integer. Prove that the following polynomial is irreducible in the ring of integer polynomials (i.e. it cannot be written as a product of two non-constant integer polynomials): \[f(x) = x^{n}+ax^{n-1}+ax^{n-2}+\dots+ax-1.\] cheers! :D

[i]Edited by Myth[/i]

This problem is a direct application of the following useful criteria.

[b]Theorem[/b] (Braurer A., [i]On alg. equations with all but one root in the iterior of unit circle[/i], Math. Nachrichten 4 1950/51, 250-257)
Let $a_{1}\geq a_{2}\geq \dots \geq a_{n}$ be natural numbers, $n\geq 2$. Then the polynomial \[p(x)=x^{n}-a_{1}x^{n-1}-a_{2}x^{n-2}-\dots a_{n}\] is irreducible.

This criteria is close to [i]Perron' criteria[/i]. The idea of the proof in the title of the arcticle.

(It is possible I wrote the proof on the forum before)

Searching "irreducible" (by Myth) gives us [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=irreducible&t=34107]this thread[/url].


induction
LaTeX
combinatorics unsolved
combinatorics

[quote="ducpbc"]Let $ A_1,A_2,...,B_1,B_2...$ be sets such that $ A_1 \equal{} \phi,B_1 \equal{} \{0\}$ and:
$ A_{n \plus{} 1} \equal{} \{x \plus{} 1| x \in B_n\},B_{n \plus{} 1} \equal{} A_n \cup B_n \minus{} A_n \cap B_n ,\forall n \in N$

Determine all n such that $ B_n \equal{} \{0\}$[/quote]
What is $ A_{n \plus{} 1}$ if $ B_n \equal{} \phi$ :(  :maybe:

[quote="JBL"]First, $ \{x \plus{} 1 \mid x \in \emptyset\} \equal{} \emptyset$, so I don't see what problem you have with that.  Second, $ 0 \in B_1$ and $ 0 \not \in A_1$, so $ 0 \in B_2$ and $ 0 \not \in A_2$, and by induction $ 0 \in B_n$ for all $ n$.   :roll:

There is a special latex command, \emptyset, that is superior to using \phi for this purpose.[/quote]

No,the conclusion is $ B_n \equal{} \{0\}$,not be $ 0 \in B_n$

p/s: Result is $ n\equal{}2^k$.Who can solve?

ith_power asked about the possibility that $ B_n \equal{} \emptyset$.  I showed that $ 0 \in B_n$ for all $ n$, so this may never occur.  I wasn't trying to answer the original problem  :wink:


Let $a \in \mathbb{C}$ be a "parameter".

Find all possible Jordan forms $J$ over $\mathbb{C}$ of the following matrix, as $a$ varies: \[ \begin{pmatrix} 2 && a^2 - 1 && 0 && -1 \\ 0 && 2 && a^2 - 3a + 2 && 0 \\ 0 && 0 && 2 && 1 \\ 0 && 0 && 0 && a \end{pmatrix}. \]


Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that

(a) the numbers are all different,
(b) they sum to 13, and
(c) they are in increasing order, left to right

First, Casey looks at the number on the leftmost card and says, "I don't have enough information to determine the other two numbers." Then Tracy looks at the number on the rightmost card and says, "I don't have enough information to determine the other two numbers." Finally, Stacy looks at the number on the middle card and says, "I don't have enough information to determine the other two numbers." Assume that each perosn knows that the other two reason perfectly and hears their comments. What number is on the middle card?

$ \textbf{(A)}\ 2\qquad
\textbf{(B)}\ 3\qquad
\textbf{(C)}\ 4\qquad
\textbf{(D)}\ 5$
$ \textbf{(E)}\ \text{There is not enough information to determine the number.}$

[hide="Solution"]
The closest the numbers can be is $3, 4, 6$.  If Casey saw a three, he would have enough information to determine the other two numbers, so the first number is a two or a one.

Tracy has determined that the first number is a two or a one.  The other two numbers sum to $11, 12$.  If Tracy sees a six, the middle has to be five.  If Tracy sees a seven, the middle can be five or four.  If Tracy sees an eight, the middle can be four or three.  If Tracy sees a nine, the middle must be a three and the first must be a two.  Hence, the last number is a seven or an eight.

The middle number could be four, five, or three.  If it was five, the other two numbers must be one and seven.  If it was three, the other two numbers must be two and eight.  Therefore, it is four, and our answer is $\boxed{C}$
[/hide]


geometry
3D geometry
sphere
geometry open

Write sphere equаtion through point $ A(0,\minus{}3,1)$and through $ x^2\plus{}y^2\equal{}16 , z\equal{}0$
 :D

You problem is easy!
We can slove it by some way. I will give you one of them.
$ \bullet$ Let equation of sphere is $ (S): x^2 \plus{} y^2 \plus{} z^2 \plus{} 2Ax \plus{} 2By \plus{} 2Cz \plus{} D \equal{} 0$ with $ A^2 \plus{} B^2 \plus{} C^2 \minus{} D > 0$.

$ \bullet$ Because $ (S)$ through circle $ \begin{cases}x^2 \plus{} y^2 \equal{} 16 \\
z \equal{} 0\end{cases}$ $ \Longrightarrow (S)$ through $ B(0;4;0), C(0; \minus{} 4;0), D(4;0;0)$.

$ \bullet$ So we have $ \begin {cases} \minus{} 6B \plus{} 2C \plus{} D \plus{} 10 \equal{} 0 \\
8B \plus{} D \plus{} 16 \equal{} 0 \\
\minus{} 8B \plus{} D \plus{} 16 \equal{} 0 \\
8A \plus{} D \plus{} 16 \equal{} 0 \end{cases}$
$ \Longleftrightarrow$ $ \begin{cases} A \equal{} 0 \\
B \equal{} 0 \\
C \equal{} 3 \\
D \equal{} \minus{} 16\end{cases}$.

$ \Longrightarrow$ $ \boxed {(S): x^2 \plus{} y^2 \plus{} z^2 \plus{} 6z \minus{} 16 \equal{} 0}$

Hi! the_wonder. You should post your problem at [b]Geometry Unslove Problem[/b].

I`m sorry.
It is really easy ,but I`m only beginer. :blush:
I just couldn`t take it any more.


To the powers that be on this site...a couple of random questions just out of my own curiosity.

1.  According to your database, you've got around 5400 members.  However, a large number have no posts.  Are these people just lurkers, or do they primarily register for your online classes and then just don't care about posting anything?

2.  What are your estimates of how many public or private high school teachers are registered here?  Do you guys pitch your product(s) - both the forum as well as your classes - to public and private high school teachers?  I know the atmosphere in many of the forums would be augmented by having professional educators contributing.  While they wouldn't have time to rack up a ton of posts on a message board, I'm sure many would be interested in having dialogues on different math subjects with both students and outsiders on occasion.

Just some random thoughts I'm throwing out there.  Certainly, I think this site is great and does a good job of doing what it sets out to do - facilitate the learning of math.  I'm just trying to learn more about AoPS.

[quote="AHSIMC"]To the powers that be on this site...a couple of random questions just out of my own curiosity.

1.  According to your database, you've got around 5400 members.  However, a large number have no posts.  Are these people just lurkers, or do they primarily register for your online classes and then just don't care about posting anything?[/quote]

There are also USAMTS participants and Math Jam attendees among that number.  I would guess our lurking percentage is about standard for a message board, maybe a little higher.

[quote]2.  What are your estimates of how many public or private high school teachers are registered here?  Do you guys pitch your product(s) - both the forum as well as your classes - to public and private high school teachers?  I know the atmosphere in many of the forums would be augmented by having professional educators contributing.  While they wouldn't have time to rack up a ton of posts on a message board, I'm sure many would be interested in having dialogues on different math subjects with both students and outsiders on occasion.[/quote]

We do advertise to educators in addition to parents and students.  Suggestions are welcome here, both in terms of avenues and in terms of how to reach teachers.  I'd be happy to hear from the teachers we do have on the site about how best to reach educators.

Just another random thought along these lines.

Do you have to be registered to enroll in classes? Just curious about this, because I've never really known.

i think both issues (lurkers and lack of teachers) are directly connected to the type of student to which you cater your site.

because your site is catered toward the hardcore math student, you ward off some "fence-sitters" who likely may have registered a screen name here, but never post - or perhaps never register, but could have.

this is certainly related to the number of public and private school teachers on here.  i would suspect the best way to attract teachers is through a pull strategy of bringing so many kids on here that the teachers are forced to come and participate, or else risk being "disconnected" in the classroom.  i know you are reaching this point with vestavia, and indeed, if the teachers there weren't on the verge of retiring, they likely would be in the situation of participating on AoPS (in some fashion) or missing out on a cog in the math curriculum those students are getting.

if you truly intend to cater to the already elite students, then you'll have to "push" AoPS on teachers through marketing, which it sounds like you're already doing to some extent.  a teachers-only forum would certainly help, for teachers to discuss either math or any other topic they wanted to, being with other teachers only.  in fact, it would be a great psychological marketing tool to require teachers to register with AoPS and PROVE that they're a teacher somehow, in order to be designated with a teacher credential on this site and have access to the "teachers only" forum.  then, you'd have teachers coming to you, and word of mouth spreading, instead of having to go after them so much.

its an interesting topic.  thanks for responding so quickly.  you guys are doing a great job as it is.  just wanted to pick your brain a little bit about the overall direction of the site.  good little discussion we've got going here.

We have kicked around the idea of having a Educators & Parents Forum on the board - I'd be curious to hear from educators and parents on the board if they would use it (or if any would be interested in moderating it).

I doubt we'd make it a 'teachers only' forum, however.  For many people on the board, the line between student and teacher is very blurred - many of our students are teachers of younger students (and their peers) and might have a lot to learn & contribute to a discussion among traditional educators.

There are many teachers registred and posting on the forum, they just do not reveal that they are teachers. Also, many non-US teachers are active in the Advanced Forums, although it would be great if more teachers would be involved in stimulating other forums too, at the proper level.

Indeed there are over 5400 users registred, and the areas that the site addresses is extremely large, starting from scratch, the lowest level of math, and going up to college questions, so basically each type of student has its own little place to hang out on the site. What I am trying to say is that the site is not oriented for the hardcore student (you can check out http://www.mathlinks.ro which is the same site, only with a different view, specially for the "hardcore" student :) ) but to all students that are connected somehow with mathematics (even if they just want to make their homework).

If you will spend a little more time here you will notice some of the users are very active in some forums, and very little if at all active on all others ...


Solve the system of equations

$ x^2 \minus{} z^2 \equal{} 1$

$ 2xy \plus{} (a \plus{} b)z^2 \equal{} \minus{} 2c$

$ y^2 \minus{} abz^2 \equal{} c^2 \plus{} d^2$,

where $ a,b,c,d$ are real numbers and $ a\neq b$.

We get

$ (a\minus{}b)^2x^4\minus{}[(a\plus{}b)(a\plus{}b\minus{}4c)\plus{}2c^2\plus{}2d^2\minus{}4ab]2x^2\plus{}(a\plus{}b\minus{}2c)^2\equal{}0$

You'd be mad to solve that biquadratic.....


To make it easier for everyone, here are all of the mock tests from the past. They are in no order, and Sprints, Targets, Teams are everywhere. However, I guess you could sort by date, because the authors probably wrote them on the same day or consecutive days. Anyway:

Fun fun fun!!!

[Copies on to Word]

.........On December 9, 1803, the twelfth amendment was proposed, and it was ratified on September 24, 1804. Basically, it eliminated Article II, Section 1, Clause 3, and changed the Presidential election system. It (Article II, Section 1, Clause 3) read, “The Electors shall meet in their respective States, and vote by Ballot for two Persons, of whom one at least shall not be an Inhabitant of the same State with [u]themselves[/u]. And they shall make a List of all the Persons voted for, and of the Number of Votes for each; which List they shall sign and certify, and transmit sealed to the Seat of the Government of the United States, directed to the President of the Senate. The President of the Senate shall, in the Presence of the State and House of Representatives, open all the Certificates, and the Votes shall then be counted. The Person having the greatest Number of Votes shall be the President, if such Number be a Majority of the whole Number of Electors appointed; and if there be more than one who have such Majority; and have an equal Number of Votes, then the House of Representatives shall immediately [u]chuse[/u] by Ballot one of them for President; and if no Person have a Majority, then from the five highest on the List the said House shall in like Manner [u]chuse[/u] the President. But in [u]chusing[/u] the President, the Votes shall be taken by States, the Representation from each State having one [u]Votes[/u]; A quorum for this Purpose shall consist of a Member of Members from two thirds of the States, and a Majority of all the States shall be necessary to a Choice. In every Case, after the Choice of the President, the Person having the greatest Number of Votes of the Electors shall be the Vice President. [u]Buit[/u] if there should remain two or more who have equal Votes, the Senate shall [u]chuse[/u] from them by Ballot the Vice President.” This clause is no longer in effect. All numbers in this problem from now onward are in base 7, unless otherwise noted. This amendment, the 12th (which is in base 10), was passed because the second Jefferson-Adams election resulted in no candidate winning the Presidency, since Burr and Jefferson were not officially running mates (running mates did not exist). What is  ? Express your answer in decimal form.

There. 7.

[quote="Yongyi781"]I especially like Carl Lian's sprint #25!

25. On December 9, 1803, the twelfth amendment was proposed, and it was ratified on September 24, 1804. Basically, it eliminated Article II, Section 1, Clause 3, and changed the Presidential election system. It (Article II, Section 1, Clause 3) read, “The Electors shall meet in their respective States, and vote by Ballot for two Persons, of whom one at least shall not be an Inhabitant of the same State with themselves. And they shall make a List of all the Persons voted for, and of the Number of Votes for each; which List they shall sign and certify, and transmit sealed to the Seat of the Government of the United States, directed to the President of the Senate. The President of the Senate shall, in the Presence of the State and House of Representatives, open all the Certificates, and the Votes shall then be counted. The Person having the greatest Number of Votes shall be the President, if such Number be a Majority of the whole Number of Electors appointed; and if there be more than one who have such Majority; and have an equal Number of Votes, then the House of Representatives shall immediately chuse by Ballot one of them for President; and if no Person have a Majority, then from the five highest on the List the said House shall in like Manner chuse the President. But in chusing the President, the Votes shall be taken by States, the Representation from each State having one Votes; A quorum for this Purpose shall consist of a Member of Members from two thirds of the States, and a Majority of all the States shall be necessary to a Choice. In every Case, after the Choice of the President, the Person having the greatest Number of Votes of the Electors shall be the Vice President. Buit if there should remain two or more who have equal Votes, the Senate shall chuse from them by Ballot the Vice President.” This clause is no longer in effect. All numbers in this problem from now onward are in base 7, unless otherwise noted. This amendment, the 12th (which is in base 10), was passed because the second Jefferson-Adams election resulted in no candidate winning the Presidency, since Burr and Jefferson were not officially running mates (running mates did not exist). What is $ 2^{10}$? Express your answer in decimal form.[/quote] is the answer to this problem 128? 2^10 base 7=2^7 base 10=128?

[quote="math154"]What is $ 2^{10}$? [b]Express your answer in decimal form.[/b] Um...yeah[/quote]

are u saying i'm right or do u think the answer is 1024 or what?

it says that all numbers in the problem are in base 7 because the 12th amendment was removed or something.....so 2^10 means 2^7 base 10?

Really??? Well, were those the "typoes" Yongyi was referring to then?

Chuse

v. t.

See Choose. [Obs.]

OK, I guess it was used...

@mathking123, it's 1024.

EDIT: @mathking123, only the last two lines matter. Disregard everything else

Expressing your answer in decimal form doesn't mean the 10 is in base 10... it's in base 7.

What would be the meaning of decimal form here...?

[quote="CatalystOfNostalgia"]Actually, those are not typos at all. Those spellings were evidently used in the late 1700's and early 1800's, and were spelled that way in the Constitutional Amendment, at least according to my history textbook last year.[/quote]

Another old usage was the spelling "defence", which was one of the many old spellings found in the Declaration of Independence on at least one occasion.

lol mathcrazed is always right; is he agreeing or disagreeing with my answer of 128?

OK, I think this question is sort of ambiguous, but there are only 4 (lol ONLY) possibilities--that I can think of.

1024 in base 7 (2662), or 1024 in base 10. Or maybe the 10 is first converted to base 7 (=13). So then it would be 2^13--in base 7/base 10. Or something else, cause I'm just so stupid like that.

It says the numbers are in base seven and express the answer in base 10(decimal form). The problem was what is 2^10. This means that the 10 is actually a 7 in base 10. So the answer is 2^7=128.

Oh...lol I suck at the easiest problems :blush:

edit: or should I say, UNDERSTANDING the easiest problems

Well, the problem kind of contradicts itself:

[b]All numbers in this problem from now onward are in base 7, unless otherwise noted.[/b]

All numbers from the word "now" onwards are in base 7 so therefore the 7 in "base 7" is in base 7 but there is no 7 in base 7.

As I said--ambiguous. I hate ambiguous problems. Like MCTM (lol not that they're bad problems...). :mad:  :mad:  :mad:

... This thread should be stickied just for having all the mock tests in it =P.  Nice work math154.  Now go add mine and tinytim's xD

I could've been doing more useful things...oh well. xD sure I'll...go add the rest of the test...

edit: argh I can't edit it anymore.


So I was out of math for a couple of weeks because of robotics, but I'm wondering... when's the ARML tryouts this year? Or did I already pass it?

This post might be helpful: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1643138#1643138

Next tryout is 11/14 @ Berkeley.

Darn this appears to be the same time as the Santa Clara Math Contest

yes.  yes it is.  and i'm going to request that you all come to the Berkeley contest instead, not just because it's an ARML tryout but because i scheduled the Berkeley contest first.  i tried to get information from Santa Clara so we could make it possible for people to attend both, but they sabotaged me by scheduling their contest on the same date AFTER they knew i had already scheduled one.  Not cool.  That said, i know Santa Clara is WAAAAAY closer for most of you, so i won't hold it against anyone if they decide to go there instead of driving to Berkeley.  :)  Just keep the schedule in mind for the rest of the year; you'd be surprised how quickly the rest of the dates sneak up on you.  You'd also be surprised how many people last year tried unsuccessfully to get sympathy from me when they realized in February or later that there weren't enough tryout dates left to post the three required scores..

I'm not gonna be able to come to the november 14 tryouts because I'm gonna be away. What's the next time, and how can I register for that?


function
calculus
inequalities
search
inequalities proposed

[quote="hendrata01"]If $a,b,c \geq 0$ such that no two have zero sum, prove that:
$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq \frac{3}{\sqrt{2}}$

I proved it in one line using the convexity of function $f(x) = \sqrt{\frac{x}{1-x}}$. But how do you show that this function is convex without calculus?
Or is there another solution to the original inequality that doesn't involve calculus?[/quote]

How do you show with calculus that a non-convex function is convex?...

  darij

Yes, Jensen fails to prove that inequality. Heck, no matter what you try, you will fail to prove that inequality. (What happens when $a=b=1,\ c=0$?)

The inequality
\[\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq 2\]
is true.
[hide="one way of doing this one"]
\[\sqrt{\frac{a}{b+c}}=\frac{a}{\sqrt{a(b+c)}}\geq \frac{2a}{a+b+c}\]
by AM-GM, and using similar inequalities for the other terms and adding up proves the new one.
[/hide]

Actually, (a/b+c)^k+(b/a+c)^k+(c/a+b)^k>=min(2,3/2^k)is true.
search 'fraction inequality'

[quote=hendrata01]If $a,b,c \geq 0$ such that no two have zero sum, prove that:
$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq \frac{3}{\sqrt{2}}$

I proved it in one line using the convexity of function $f(x) = \sqrt{\frac{x}{1-x}}$. But how do you show that this function is convex without calculus?
Or is there another solution to the original inequality that doesn't involve calculus?[/quote]
[hide=See this]Let $\ $ $a,b,c$ $\ $ nonnegative real number, no two of which are zero . If $k > 0$ , then
$ \sqrt{\frac{a}{kb+c}}+\sqrt{\frac{b}{kc+a}}+\sqrt{\frac{c}{ka+b}}\geq$    $\min\left\lbrace \dfrac{3}{\sqrt{k+1}},\dfrac{{2}}{\sqrt[4]{k}}\right\rbrace$    (Vasile Cirtoaje).[/hide]

[quote=hendrata01]If $a,b,c \geq 0$ such that no two have zero sum, prove that:
$$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq \frac{3}{\sqrt{2}}$$[hide=*]

I proved it in one line using the convexity of function $f(x) = \sqrt{\frac{x}{1-x}}$. But how do you show that this function is convex without calculus?
Or is there another solution to the original inequality that doesn't involve calculus?[/hide][/quote]
Let three non-negative reals $    a, b, c   $ satisfy : $   ab+bc+ca>0    $ . [url=https://artofproblemsolving.com/community/c6h2806790p24748054]Prove that :[/url]
$$\sqrt{\frac{a}{b+c}+\frac{1}{8}\left(\frac{b-c}{b+c}\right)^2}+\sqrt{\frac{b}{c+a}+\frac{1}{8}\left(\frac{c-a}{c+a}\right)^2}+\sqrt{\frac{c}{a+b}+\frac{1}{8}\left(\frac{a-b}{a+b}\right)^2} \ge \frac{3}{\sqrt{2}}$$

Find the minimum value of the positive number $k$ such that the inequality
$$\sqrt{\frac{a}{b+c}+k\left(\frac{b-c}{b+c}\right)^2}+\sqrt{\frac{b}{c+a}+k\left(\frac{c-a}{c+a}\right)^2}+\sqrt{\frac{c}{a+b}+k\left(\frac{a-b}{a+b}\right)^2} \ge \frac{3}{\sqrt{2}}$$holds for all non-negative  real numbers $a, b, c$ such that $ ab+bc+ca>0  . $


Prove that the equation

$x^2y^2=z^2\left(z^2-x^2-y^2\right)$

has no solutions in positive integers.

It's Immediate solve by lemma:
There aren't $a,b,c,d$ st:
$a^2+b^2=c^2$
$a^2-b^2=d^2$
(Which this one is equal:
$a^4-b^4=e^4$ hasn't Positive Integer root!

$x^2y^2=z^2(z^2-x^2-y^2)$
$z^4-z^2(x^2+y^2)-x^2y^2=0$
If you take as a second degree polynomial you get that the discriminat should be integer if you want integer solutions so :
$(x^2+y^2)^2+(2xy)^2=p^2$ and
now let $a=x^2+y^2, b=2xy$ it is easy to see that $a^2-b^2, a^2+b^2$ are both squares so use that lemma by $hungkhtn$ and get that there are no solutions. (To prove the lemma use the fact that if one has $a^2+b^2=c^2$ than $a=m^2-n^2, b=2mn,c=m^2+n^2$) Hope this will help you $Pilot$!


analytic geometry
geometry
trigonometry
probability
AMC
AIME
summer program

So does anyone here take the MAML (Massachusetts Association of Mathematics League) Exam? The first round takes place in a couple of weeks.

I am just wondering because since it will be my first time taking it, I don't really know how the scoring works. There isn't much information on the official website (http://maml.org/). More specifically, I am wondering how the winners of the second round (and thus the overall winners) are chosen. Are the overall winners chosen based on their combined scores on both the first and second rounds, or are they just chosen based on their second round score (since only the top 100 or so students qualify for the second round)?

I think you're talking about the olympiads. Quite frankly I haven't made it far enough to know the answers to your question :), but I can surely answer some. Here's most of the important stuff you need to know:
First level examination is on Tuesday, October 28th at 2:15 pm EST (obviously :lol:)
Second level examination is on Thursday, March 25 of 2004.

Registration fee vary from school to school depending on where you live, and most importantly how much budget the school/math department has :), but it's usually around a dollar

The first level examination is a multiple-choice (I think it's four but they might've changed that). It's 90 minutes long and it will be on:
-elementary and intermediate algebra
-plane, solid, and coordinate geometry
-plane trigonometry
-number theory
-probability
and
-logic

Calculators are allowed on both levels, however the following is not permitted:
-handheld minicomputers such as Palm pilots
-pocket organizers
-lap-top computers
-TI-92
and
-calculator with an ability to output the data in print/has the ability to act as a printer

The test will have 30 problems, 4 points are awarded to each correct answers, no points are added nor subtracted for unanswered question, and 1 point is taken away for each incorrect answer. Thus the maximum points possible is 150 points.

The second level examination will be completely different from the first, and the only information available at this point is that
a.) It's three-hour long test from 9:00 am to 12:00 noon
b.) It will be conducted at "examination centers" in various sections of the state

I hope this is enough...

I'm taking it this year, and am pretty positive I'm gonna make Level II. My goal is...top freshman, which'll probably be somewhere around 20th place or so.

[quote="Fierytycoon"]Thanks for the information. Did you take the MAML last year Tare?[/quote]
Yup, did so horrible I can't even remember my score *the "You are an idiot!" song plays in the background*...just for my self-defence I was in 7th back then, so I guess it's not my fault (desperate save) :lol:

I don't get why Olympiad is so much harder than AMC, I mean it's only for 9th and 10th graders so why make it so hard? Most high schoolers don't even learn trigonometry until later on in 10th, and that's supposing you're in honors. All those mumbo jumbo makes me woozy :lol:

[quote]I don't get why Olympiad is so much harder than AMC, I mean it's only for 9th and 10th graders so why make it so hard? [/quote]

No, any highschooler can take the MAML.

I took MAML each of the last 2 years, so I know a bit about the competition.

Overall winners are based solely on second round scores.  Last year I was 17th on the first round (with a score of 125), and slipped out of the top 20 on the second round. However, second round scores were not always reported, so I don't know how much I missed by.

By the way, I think last year's MAML round 1 was much easier than AMC 12.

[quote="Fierytycoon"]Thanks for the information. And yeah, it doesn't look too bad, the cutoff was 94 last year, so its about the difficult of qualifying for the AIME (perhaps a tad bit harder) I guess.[/quote]

Last year, about 3.3% of students who took the first round qualified for the second (for MAML).

I don't know the corresponding statistic for AMC-12, but I think it's usually about 5%.  So I guess MAML may be a bit harder than AMC-12, but it's hard to compare the two numbers since Massachusetts may not be representative of the nation.

im taking it too.  where did you get this info, tare?  First level is at 2:15 PM?  My school, and I know others, take it during the day.  Also, in your list of banned calculators, you did not include ti-89's, but i believe that it's banned as well.

Fiery: where did you get the cutoff scores?

Correction: TI-89s are also banned. I misread the information; it said that TI-92s are banned, not TI-92 and above. Please disregard my misinformation and excuse it.

The document I have is the courtesy of many people. First our schools math department head, who typed up most of the material that the math depatment heads get (the registration form, how to deal with problems, etc.) and send it to all the math teachers. Then they printed it out. The end :lol:

And yes, 2:15 PM EST to 3:45 PM EST.

Well, there's some stuff at http://maml.org/wwwroot/contestinfo/contestinfo.htm

im not sure the cutoff is listed there, but are you sure that's right?  I know someone said he got 16 right and 1 wrong, i think, and missed it by 1 point or 1 question or something like that.  I didnt work out what that score would be, but im pretty sure it's more than a 96.

Wait a minute, this info is last years! No wonder my info is different...the rules might be still the same but I think there's a lot more different stuff, like dates for example.

The cutoff is on a page that is linked from the url that I gave in my last post. It is on http://maml.org/wwwroot/contestinfo/Level%201%20Press%20Release%202002_rev.htm

[quote="mathnovice"]How do we contact Marty Badoian -- his e-mail address (mbadoian@hotmail.com) doesn't exist.[/quote]I could be way off on this, but I seem to remember it being badoianm@hotmail.com. Try at your own risk.

[quote="TheSouthpaw"]I'm surprised no one got a perfect..

If the cutoff is 94...thats....19 right and 1 wrong. Which isn't so bad, but if you take a look at the exams, in my opinion, they are much hader than AMC 12. Has anyone seen MAML 1999? The last couple questions were near impossible for me.[/quote] Yeah, it was pretty easy.  One kid at my school got a 145. The only question he got wrong was a number theory question, and he had done 2 years of PROMYS.  Incidentally, that was also the only question I got wrong (though I skipped 5 of them) and I also went to PROMYS for a summer.


function
calculus
integration
probability and stats

Let $ X_1,X_2, \cdots X_n$ be independent random variables such that $ X_i$ has $ N(\mu _i, \sigma _i ^{2})$ distribution, where $ \mu _i \in \mathbb{R}, \sigma _i ^{2} > 0, 1 \le i \le n$.

Put $ Y_k \equal{} X_1 \plus{} X_2 \plus{} \cdots X_k$ , $ k \equal{} 1,2, \cdots ,n$

(i) Find the distribution of $ (Y_1,Y_2 \cdots ,Y_n)$
(ii) Find $ Cov (Y_i, Y_j)$

For $ i < j$, $ \text{Cov}(Y_i,Y_j) \equal{} \sum_{k\leq i} \sigma^2_k$.  Just expand everything and note that $ EX_i^2 \equal{} \sigma^2_i \plus{} \mu_i^2$.  If $ i\not \equal{} j$, then $ EX_iX_j \equal{} EX_iEX_j$ by independence, so everything else cancels.

e: maybe simpler: $ \text{Cov}(Y_i,Y_j) \equal{} \text{Cov}(Y_i,Y_i \plus{} (Y_j \minus{} Y_i)) \equal{} \text{Cov}(Y_i,Y_i) \plus{} \text{Cov}(Y_i,Y_j \minus{} Y_i) \equal{} \text{Cov}(Y_i,Y_i) \equal{} \text{Var}(Y_i)$

[quote="Extremal"]it's enough  to find distribution for $ Y_j$ for all $ j$.
[hide="Hint"]
at first try to  find the characterstic function for $ Y_j \equal{} \sum^{j}_{k} X_{k}$
[/hide][/quote]

$ Y_j$ has $ N(\sum^{j}_{i\equal{}1} \mu _{i} , \sum^{j}_{i\equal{}1} \sigma  _{i}^2)$ distribution.

But the $ Y_{j}$s are not independent which is making life difficult.

I get this n-variable integration where I know the integrand and the limits of integration. But I don't know how to evaluate it, basically when the limits are themselves functions of the variables and thus interchanging order of integration is becoming a strict no-no. :mad:


Find all functions $f \colon \mathbb{R} \to \mathbb{R}$ such that for all real numbers $x$ and $y$
\[f(f(x)+y)=2x+f(f(y)-x).\]

put $k=f\left( 0\right) $
By setting $x=0$ we get $f\left( k+y\right) =f\left( f\left( y\right) \right) $ for all $y$.
By setting $y=0$ we get $f\left( f\left( x\right) \right) =2x+f\left( k-x\right) $ for all $x$.
so : for all $x$ $f\left( k+x\right) =2x+f\left( k-x\right) $ with $k=f\left( 0\right) $

Can any one continue ... ?  :lol:

quite simple, and routine.
note that the range of f(x) is non empty, so take 2x from range of f.
and suppose 2x=f(z). then define y=z-f(x). it is then evident that f(f(y)-x)=0.
hence there is some x0 with f(x0)=0.
we shall prove that f(0)=0.
to prove note that replacing x with f(x0) in the original equation we obtain that f(s+y)=ff(y))
further replacing 0 for x we get f(s+y)=ff(y))+2s. and therefore s=0. hence f(0)=0.
now we know that f(f(y))=f(y) and therefore f(x)=x as long as x is in range of f(x).
Now let y=0, and x->-x then -x=-2x+f(x) and so f(x)=x.
-a

Of course after getting f(0)=0 everything is easy. But unfortunately I don't understand how did you get it.

i find it pretty clear. tell me specificly which part you have problems with, so that i can elaborate on it.

Like how ali did at first, we can prove that $f$ is surjective. Suppose $f(a)=0$. Put $x=a$. We get
$f(y)=2a+f(f(y)-a))$
=>
$f(y)-a=a+f(f(y)-a)$
=>
$x=a+f(x)$ for all $x$.


Βρείτε την καλύτερη σταθερά $ k$  έτσι ώστε η ακόλουθη ανισότητα να ισχύει για όλους τους θετικούς πραγματικους $ a,b,c$ :

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3\geq k(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \minus{} \frac {3}{2})$

Η άσκηση είχε προταθεί στο ML contest και έχω βρει μια αρκετά ικανοποιητική λύση .

silouan signwmi alla otan les [color=red]kaliteri [/color]stathera ti ennoeis me ayto? :blush:
an ennoeis na vrw ena "k" pragmatiko gia to opoio isxyei
tote gia $ k \equal{} \minus{} 1$
exoume
 $ a/b \plus{} b/c \plus{} c/a \plus{} a/(b \plus{} c) \plus{} b/(c \plus{} a) \plus{} c/(a \plus{} b) > \equal{} 3/2 \plus{} 3$
to opoio isxyei apo am_gm kai tn anisotita nesbiit(k meta prosthesi kata meli)..
P.S an ennoeis ayto.. :|

Μάλλον εννοεί τη μεγαλύτερη δυνατή σταθερά $ k$. Αυτήν που σφίγγει πιο πολύ την ανισότητα :lol:

PS: By the way, Athinaie το κλάσμα στη LaTeX είναι \frac{αριθμητής}{παρανομαστής} και  το μεγαλύτερο και μικρότερο ή ίσο \geq και \leq αντίστοιχα

orfikos to gnwrizw alla eimai aytin tn stgmi se ena PC p paei me karvuna :rotfl:  kai...katalavaineis i plagia payla dn iparxei...
otan bw apo ton kanoniko m tha to ftiaksw me edit....
apla to ekana viastika gia na dw ti ennooume legontas kaliteri timi..
wste dld opote dw na vrethei i kaliteri timi enooume tn megaliteri dynati?
eyxaristw :)

[quote="Athinaios"]wste dld opote dw na vrethei i kaliteri timi enooume tn megaliteri dynati?
eyxaristw :)[/quote]

Ίσως δεν ήμουνα και εγώ σαφής αλλά όταν λέμε καλύτερη εννούμε αυτή που κάνει την ανισότητα πιο σφιχτή δηλαδή το δεύτερο μέλος όσο το δυνατόν πιο κοντά στο πρώτο . Στη συγκεκριμμένη περίπτωση ψάχνεις πράγματι για το μεγαλύτερο κ .

Θα παρουσιάσω μια απόπειρα λύσης, αν και μη ολοκληρωμένη :( Θέλω βέβαια να μου πείτε αν μπορεί να ολοκληρωθεί με αυτόν τον τρόπο ή αυτό είναι αδύνατο.

Θεωρώ τη συνάρτηση $ G(a,b,c)\geq \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3$ (γνωστό το κόλπο :wink: )

$ G(a,b,c)\geq G(a \plus{} c,b \plus{} c,a \plus{} b)$. Για να το αποδείξω αυτό χρησιμοποιώ απλά ότι

$ G(a,b,c) \equal{} \frac {(a \minus{} b)^2}{ab} \plus{} \frac {(a \minus{} c)(b \minus{} c)}{ac}$ κι έτσι θεωρώντας χωρίς βλάβη της γενικότητας ότι $ c \equal{} min(a,b,c)$ αποδεινκύνεται η παραπάνω ανισότητα (με συγχωρείτε που προσπερνάω αρκετούς αλγεβρικούς μετασχηματισμούς αλλά αυτήν την περίοδο ο χρόνος μου, όπως και όλων των μαθητών είναι περιορισμένος...) .

Τώρα προσπαθώ να συνεχίσω βασιζόμενος στην παραπάνω σχέση.

Ενα πολύ καλό στοιχείο είναι ότι χρησιμοποιώντας την, προκύπτει:
$ G(a,b,c) \plus{} G(a,c,b)\geq G(a \plus{} c,b \plus{} c,a \plus{} b) \plus{} G(a \plus{} c,a \plus{} b,b \plus{} c) \equal{} 2(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \minus{} \frac {3}{2})$

Από εδώ κατ' αρχάς μπορούμε να αποδείξουμε τι γίνεται για την περίπτωση που $ a\geq c\geq b$ γιατί σε αυτήν την περίπτωση εύκολα αποδεικνύεται ότι ισχύει $ G(a,b,c)\geq G(a,c,b)$ κι άρα
$ G(a,b,c)\geq \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \minus{} \frac {3}{2}$ που σημαίνει ότι εδώ κ=1 (παρατηρήστε ότι η μόνη περίπτωση που μας μένει τώρα είναι όταν $ a\geq b\geq c$ γιατί το ότι το $ a \equal{} max$ μπορούμε να το υποθέσουμε χωρίς βλάβη της γενικότητας. Πάντως αυτήν την περίπτωση δεν μπορώ να την αποδείξω.

Τέλος, θυμάμαι επίσης ένα πρόβλημα που είχα δει που έβρισκε το καλύτερο $ k$ τέτοιο ώστε:
$ G(a,b,c)\geq G(a,c,b)$ με $ a,b,c$ πλευρές τριγώνου (δηλαδή καλύπτουμε και αυτήν την περίπτωση...)

Αν και δεν είμαι 100% σίγουρος με πρόχειρους υπολογισμούς παρατήρησα ότι αν σταθεροποιήσουμε τις ποσότητες $ b$, $ c$ και η ποσότητα $ a$ τείνει στο άπειρο, μου φαίνεται ότι το $ k$ τείνει στο 2. Αν κάποιος με βοηθήσει να αποδείξω αυτό παρά να το εικάζω τότε βελτιώνω ακόμα περισσότερο το $ k$ διότι έχω αποδείξει την ανισότητα για $ k \equal{} 2$ με brute force. Παραθέτω πολύ συνοπτικά τη λύση:

Θα αποδείξουμε την ανισότητα:

$ \frac {1}{2}\sum\frac {a}{b}\geq\sum\frac {a}{b \plus{} c}$

Μετά από άπειρες πράξεις και απλοποιήσεις (αν έχετε υπομονή μπορείτε να τις ελέγξετε) καταλήγουμε στο να αποδείξουμε την εξής ανισότητα:

$ a^4bc \plus{} ab^4c \plus{} abc^4 \plus{} a^3b^3 \plus{} b^3c^3 \plus{} c^3a^3 \plus{} a^4c^2 \plus{} b^4a^2 \plus{} c^4b^2\geq2a^3b^2c \plus{} 2a^2bc^3 \plus{} 2ab^3c^2 \plus{} 3a^2b^2c^2$

Παρατηρούμε τώρα ότι από SCHUR:

$ a^4bc \plus{} ab^4c \plus{} abc^4 \equal{} abc(a^3 \plus{} b^3 \plus{} c^3)>\equal{}abc[ab(a \plus{} b) \plus{} bc(b \plus{} c) \plus{} ca(c \plus{} a) \minus{} 3abc] \equal{} a^3b^2c \plus{} a^3bc^2 \plus{} a^2b^3c \plus{} ab^3c^2 \plus{} ab^2c^3 \plus{} a^2bc^3 \minus{} 3a^2b^2c^2$

και με την ίδια λογική:

$ (ab)^3 \plus{} (bc)^3 \plus{} (ca)^3>\equal{}a^3b^2c \plus{} a^3bc^2 \plus{} a^2b^3c \plus{} ab^3c^2 \plus{} ab^2c^3 \plus{} a^2bc^3 \minus{} 3a^2b^2c^2$

Είναι επομένως λοιπόν αρκετό να αποδείξουμε ότι:

$ a^4c^2 \plus{} b^4a^2 \plus{} c^4b^2 \plus{} 2a^3bc^2 \plus{} 2a^2b^3c \plus{} 2ab^2c^3\geq9a^2b^2c^2$

το οποίο είναι προφανές αφού

$ a^4c^2 \plus{} b^4c^2 \plus{} c^4a^2\geq3a^2b^2c^2$ από AM-GM

και

$ 2a^3bc^2 \plus{} 2a^2b^3c \plus{} 2ab^2c^3 \equal{} 2abc(a^2c \plus{} b^2a \plus{} c^2b)\geq6a^2b^2c^2$

από ανισότητα αναδιάταξης άρα QED. :)

Σιλουανέ μπορείς να μου πεις αν είναι όντως το 2 που εικάζω η καλύτερη σταθερά?

Τώρα όσοι γνωρίζουν ανάλυση (εγώ δεν ξέρω πολύ καλά) :(  μπορούν να βρουν το όριο της συνάρτησης

$ f(a)\equal{}\frac{\sum\frac{a}{b}\minus{}3}{\sum\frac{a}{b\plus{}c}\minus{}\frac{3}{2}}$

καθώς το $ a$ τείνει στο άπειρο και τα $ b$,$ c$ είναι σταθερές? :?:

Ο Δημήτρης είναι πολύ κοντά απλά το μπλέκει κάπου και δεν διευκολύνει έτσι μετά . Η ιδέα πάντως είναι σε αυτό το πνεύμα .
Ένα hint που θα δώσω είναι νδο πρώτα ότι $ k\leq 1$ και μετά νδο η ανισότητα ισχύει για $ k\equal{}1$
Κώστα θα σε δυσαρεστήσω αλλά έχεις κάνει λάθος στο expand  :( . (το k βγαίνει 1)

Δηλαδή από ότι βρήκα εγώ η

$ \frac{1}{2}\sum\frac{a}{b}\geq\frac{a}{b\plus{}c}$ είναι ισοδύναμη με

$ a^2b^4\plus{}b^2c^4\plus{}c^2a^4\plus{}a^3b^3\plus{}b^3c^3\plus{}c^3a^3\geq abc(a^3\plus{}b^3\plus{}c^3)\plus{}3a^2b^2c^2$

Μάλλον έχεις δίκιο. (Πάντως δεν το ξαναελέγχω με τίποτα :rotfl: ). Κρίμα... Και θα ήταν μια πολύ ωραία ανισότητα... Πάντως η άσκηση είναι πάρα πολύ ωραία. :)

εγω λυση για κ=1 εχω βρει...το κ<=1...με παιδευει :P

Για κ=1 έχω βρει λύση κι εγώ. (Μη ρωτήσετε ποια λύση... :rotfl: )

Πιστεύω ότι τώρα δεν πρέπει να ψάξουμε το κ<=1 αλλά να βρούμε παράδειγμα αριθμών ώστε για κ>1 να μην ισχύει η ανισότητα...

Ή μάλλον καλύτερα να μελετήσουμε τη συνάρτηση που ανέφερα σε προηγούμενο ποστ. :wink:

Για το ότι $ k\leq 1$ βρήκα λύση θέτοντας $ a \equal{} c$ και $ x \equal{} \frac {a}{b}$ και μου βγήκε $ k\leq x \plus{} 1$ που σημαίνει ότι το $ k\leq 1$ αφού το $ x$ είναι όσο μικρό θέλουμε...(είναι το ίδιο σαν να λέμε έστω $ k\equal{}1\plus{}y$ και να παίρνουμε $ x<y$)

Πιο αναλυτικά, απ' το προηγούμενο ποστ μου ανέφερα ότι $ G(a,b,c) \equal{} \frac {(a \minus{} b)^2}{ab} \plus{} \frac {(a \minus{} c)(b \minus{} c)}{ac}$

Θέτοντας λοιπόν $ a \equal{} c$ ,έχουμε

$ \frac {(a \minus{} b)^2}{ab}\geq k(\frac {2a}{a \plus{} b} \plus{} \frac {b}{2a} \minus{} \frac {3}{2})$ και με $ x \equal{} \frac {a}{b}$,

$ \frac {(x \minus{} 1)^2}{x}\geq k(\frac {2x}{x \plus{} 1} \plus{} \frac {1}{2x} \minus{} \frac {3}{2})\iff$

$ 2(x \plus{} 1)(x \minus{} 1)^2\geq 2k(4x^2 \plus{} x \plus{} 1 \minus{} 3x(x \plus{} 1)) \equal{} 2k(x \minus{} 1)^2\iff$

$ k\leq x \plus{} 1$

QED

Τώρα αν μπορείτε βάλτε και τη λύση σας για $ k \equal{} 1$

για k=1
exoume na apodeiksoume oti
$ \sum\frac{a}{b}\geq\sum\frac{a}{b+c} + \frac{3}{2}$
$ <=>$
$ \sum{a(\frac{1}{b} - \frac{1}{b+c})\geq\frac{3}{2}}$
$ <=>$
$ \sum\frac{ac}{b(b+c)}\geq\frac{3}{2}$
$ <=>$
$ \sum\frac{ac}{b(b+c)}=\sum\frac{a^2c^2}{ab^2c+abc^2}\geq\frac{3}{2}$
αν thesoume x=ac,y=ab,w=bc προκυπτει μια εφαρμογη της andreescu
πιο συγεκριμενα εχουμε
 $ \sum\frac{x^2}{yw+xw}\geq\frac{(x+y+w)^2}{2(xw+yw+xy)}\geq\frac{3}{2}$

Μπράβο και στους δύο σας  :)
Αυτή είναι η λύση μου

Δημήτρη πολύ έξυπνο αυτό για $ 1\geqq$.

Athinaios πραγματικά πολύ ωραία λύση στην ανισότητα αν σου πω ότι εγώ κατέληξα σε muirhead... (γι αυτό υιοθέτησα και το ανάλογο signature) :rotfl:

Το πρόβλημα είναι πάρα πολύ ωραίο πάντως και δικαιούται ωραίες λύσεις. :)

χαχα  :P  ! α και Κωστα οπως και ολοι στο forum  θα προτιμουσα αν το θυμαστε να με φωναζετε Ηλια. :)
Υ.Γ πραγματικα ωραια σκεψη του silouan kai δημητρη για κ<=1!!..παντως εγω θα ηθελα να το ψαξω λιγο με τιμες..(ή σχεση ισοτητα για κ=1..)

Βασικά μπορείς να πεις και ότι αν $ a \equal{} c$ και $ b\rightarrow \plus{} \infty$ τότε ισχύει η ισότητα για $ k \equal{} 1$ άρα αυτή είναι η καλύτερη σταθερά .


geometry
3D geometry
sphere
trigonometry
perimeter
trig identities
Law of Cosines

I'm separating challenges by several threads because of the sources.

In case it looks like spamming, sorry :blush: I hope it's okay.

 :)  :)  :)

1. If the earth is assumed to be a sphere with radius 4000 miles, what is the circumference of a circle of latitude at 60 degrees north?

This question, #2, I actually need a neat way to solve this.

2. In how many different ways can the integer 10 be written as the sum of four positive integers in which the order of terms irrelvant [4+2+2+2 is regarded same as 2+4+2+2]

It's easy but I'm not sure on "exact" number of ways oreasy ways to do this.

 :(

3. A length of a rope can be coiled into circular loops each of diameter D inches. If the same rope is coiled into 12 circular loops, what would their common diameter be?

4. Imagine a rope long enough to circle the Earth (assume that Earth is perfect sphere) along the Equator at an altitude of 1 meter. How much of the rope would have to cut off in order to ensure that the remaining length could circle the Equator while just touching the Earth's surface?

5. What is the sum of the three triangle U, G, and K if:

Triangle U has length 1 and 4 with included angle of 90.

Triangle G has length 1 and 2 wiht included angle of 90.

Triangle K has sides $\sqrt 17, \sqrt 15, 2$. Do this without using calcultor.

6. Three teams, the red team, the blue team, and the gold team, each consisting of two runners, compete in a two-mile relay race. Each runner runs one mile of the relay. The red team's two runners run the mile at 9 mph and 11 mph respectively. The blue team's runner run the mile at 8 and 13 mph respectively. Both gold team runners run the mile at 10 mph. List the team in order they finish - the first, second, third, in that order.

7. Of the nine digits 0,1,2,3,4,5,6,7,8 which two cannot be the units digit of a base-9 representation of a prime number?

Enjoy! :D

[quote="Silverfalcon"]...2. In how many different ways can the integer 10 be written as the sum of four positive integers in which the order of terms irrelvant [4+2+2+2 is regarded same as 2+4+2+2]

It's easy but I'm not sure on "exact" number of ways oreasy ways to do this.

 :(

...

5. What is the sum of the three triangle U, G, and K if:

...
[/quote]
I'll leave #2 to someone else to respond. I'll help you if no one else can do it.
For #5, what do you mean by 'the sum of the three triangles'? Do you mean the sum of the areas of the three triangles?

Yes.

[U]+[G]+[K].

Sorry for poor description. :blush:

Just wondering, is there any easy way to figure out the area of the triangle with sides $\sqrt {15}, \sqrt {17}, 2$ without calculator?

With calculator, this is pretty easy problem but I don't like it without calculator. Any easy ideas?

 :?

A base nine number, minus its units digit, is always divisible by three.  Therefore, base nine numbers with 0 and 6 (adding a number divisible by three to a number already divisible by three) cannot be prime.  Three doesn't work because 3 base nine is prime.

Hmm..

I don't quite get what you said about the base 9. How do you know it's 6??

Sorry for poor comprehension.

S'cool
In elaboration:
Any base nine number's base expansion will yield some form of:
(1st digit)*9^n +(2nd digit)9^(n-1)+(3rd digit)*9^(n-2)....(last digit)*9^0.
example:
$23422_9=2(9^4)+3(9^3)+4(9^2)+2(9^1)+2(9^0)$
The first n-1 terms (excluding the units digit term) of the base expansion of a n-digit base nine number all have factors of nine, and thus both they, and their sum are divisible by three.  Therefore, if the units digit of a base nine number is 0, 3, or 6, (this digit being added to a number already divisible by three) the base nine number will be divisible by three.  However, 3 can be the units digit of a base nine prime number in one number, $3_9$.  Therefore, 0 (not prime by definition), and 6, (divisible by 2 and 3) both can never be the units digit of a prime base nine number.

*Sorry for the initially insufficient explanation.

Sorry, the triangle one should've had sides $\sqrt {17}, \sqrt {5}, 2$.

 :lol: Since the first triangle has area of $1 \cdot 4 \cdot \frac {1}{2} \Rightarrow 2$, and the second triangle has area of $1 \cdot 2 \cdot \frac {1}{2} \Rightarrow 1$, if I can find the area of the last one, I'm done.  :D

Let's call the triangle ABC where AB = c, BC = a, AC = b. (weird naming but). Let c = 2, a = $\sqrt {17}$, b = $\sqrt 5$, then we have:

$c^2 = a^2+b^2-2ab \cos C$

$4 = 5 + 17 - 2 \sqrt {85} \cos C\\
-18 = -2 \sqrt {85} \cos C\\
9 = \sqrt {85} \cos C\\
81 = 85 \cos^2 C\\
\frac {81}{85} = \cos^2 C$

Using Trig Identity, we get:

$\sin^2 C = 1 - \cos^2 C = 1 - \frac {81}{85} \implies \sin C = \frac {2}{\sqrt {85}}$.

So, the area is:

$\frac {1}{2} \cdot \sqrt 5 \cdot \sqrt {17} \cdot \frac {2}{\sqrt {85}} \Rightarrow 1$.

So, the sum of the areas of three triangles would be 1+1+2 = $\fbox 4$.

I hope I did right. :blush:

[quote="xxreddevilzxx"][hide="1"]
lets say O is the center of the Earth, P is the center of the circle we are looking at, and D a point on Earth which is also on the circumference of the circle P. Since OP is perpendicular to PD, and angle POD is 60 degrees, we have a special 30-60-90 triangle. OD = 4000, so PD = $\frac{4000 \sqrt{3}}{2} = 2000 \sqrt{3}$
[/hide][/quote]

Thanks for the answer but my question is since your answer only gives to the radius, wouldn't the answer, the perimeter, is:
[hide]
$4000 \pi \sqrt 3$?

$2000 \sqrt 3 * 2 * \pi$[/hide]

[hide="6"] Using time = distance / speed,
The Red team takes $\frac 19 + \frac{1}{11} = \frac{20}{99}$ hours,
The Blue team takes $\frac 18 + \frac{1}{13} = \frac{21}{104}$ hours, and
The Gold team takes $\frac {1}{10} + \frac{1}{10} = \frac{1}{5}$ hours.
Comparing the times, Gold-1st, Blue-2nd, Red-3rd.[/hide]

I don't know if I did this right, my answer doesn't seem logical but the math seems to work out.

[hide="My answer for 4"]

Let's say that the Earth's radius is $r$.

At every point around the equator, the rope would be one meter above the level, hence its radius would be $r+1$.

So far we have
Earth radius = $r$
the rope's radius = $r+1$

Using $ C=2 \pi r$, we can find

The circumference of the Earth = $2 \pi r$

The circumference of the rope around the Earth is

$C = 2 \pi (r+1)$
$C = 2\pi r + 2\pi$

Subtract the rope's circumference by the Earth's circumference to find the "remaining" length

$2\pi r + 2\pi - 2 \pi r
= 2\pi units$

[/hide]


probability
geometry
MATHCOUNTS
AMC
AIME
number theory
AMC 10

Hey yall, I need some advice.

dynamo729 has $ \$150$ to spend on AoPS books. He would really like to make CD at Nats to make up for a disappointing performance this year, even if he is from a state not known for doing well at Nats. He sucks at Number Theory, Counting, Probability, and Geometry, and hes covered for the rest. What AoPS book should he buy? He has been through Volume 1: The Basics. Also list some other books that are helpful please!

Thanks.

You should probably buy the Introduction books for those topics that u said u need more improvement on.

Volume 1: The Basics is very useful.
Get Volume 2: And Beyond.

Where did you get Volume 1? I borrowed it from a math teacher but I can't find it and since school's over I had to return it, although Volumes 1 and 2 were very useful and I wasn't done reading them, yet. :(

[quote="dynamo729"]He sucks at Number Theory, Counting, Probability, and Geometry, and hes covered for the rest. What AoPS book should he buy? He has been through Volume 1: The Basics.[/quote]

Umm.... I think number theory, counting, probability, and geometry are covered enough for Mathcounts level in Vol. 1. If you can't figure out a problem from mc that's in one of those categories, just look back in Vol. 1. It should have it. (Or maybe it's a counting problem and you just have to go through a gazillion cases, no matter how reluctant you may be to do so.) IMO, Mathcounts questions are not that hard. You just have to go fast. And not make mistakes.

So my opinion is just to do more problems. More and MORE and MORE and MORE and MORE and MORE!!!!!! and MORE!

My teacher had a very old copy of Volume 1 that she let me borrow when I made Nationals this year.

I would suggest getting Volume 2: And Beyond, The Greatest MATHCOUNTS Problems of All Time Handbook, and as many past MATHCOUNTS tests as you can get your hands on.

Do not get Volume 2 if you're just practicing for Mathcounts. Mostly everything in there is above Mathcounts level. Having said that, I think you should try to start making the transition from Mathcounts to other contests like the AMC series. If you want to prepare for that, you should definitely get Volume 2.

For Mathcounts though, I really recommend the Intro to C+P and Intro to Geometry, considering that those are the problems most people have trouble with and come up a lot in Mathcounts (more than Number Theory in my opinion).

yeah.... mathcounts is the only thing i do other than regular math team tournaments.

Oh, I was assuming that he would someday continue onto AMC's.

My current school does nothing but MC and regular tournaments, but the high school I will go to does just about everything so I am going to do other things but not next year as an 8th grader.

Even as an 8th grader, I think it would be good if you started studying and taking AMC's early. You can go to your future high school and ask them to let you take the official test, too. I regret that I didn't follow this advice  :( .

It's a good idea to start practicing for the AMC early. I took the AMC 10 B this year as an eigth grader and I got a score of 118(.5?), one question off from AIME :( . Hopefully I'll make AIME next year. Yeah, just ask your future high school or ask one of your high school friends who are taking the AMC (assuming you do have high school friends).

So, i have the same prob. i got like   350 dollars in my piggy bank (literally)
so, you might think i have enough money to buy every single book on AoPs
but i also want to build a robot, the better the robot the more it costs. my parents bought me intro algebra, and my school library bought the other intros(i made them , told them its 4th grade level :D) but now that school's closed i had to return them  :censored: . Most you are saying that i should buy AoPs 1 and build a better robot, but what about the whole intro series, won't that be in more detail?

thanking you,
Poincare

uh dude...robot??? wtf

Anyway, do at least Vol.1 (if you can, do all the Intros). After that, you do problems, contests, whatever--just keep practicing.

i wanna $ build$ a robot (your "wtf" was just answered) they cost a lot

No sorry if you were offended in any way I meant to say that nobody recommended you build a better robot here...not that you shouldn't. Anyway, just get Vol.1 at least (the rest ARE still very helpful; if you only pick one, I would highly recommend Counting and Probability the most or Geometry--they seem to be the subjects most people aren't that good at).

i wasn't offended :D

but thanks for the advice


Solve the inequality $\log_{\frac12}\left(\log_8\frac{x^2-2x}{x-3}\right)<0$.

Its equivalent to $log_8 \frac{x(x-2)}{x-3} > 1$, or $(x^2 - 2x) > 8x - 24$.  Then $x^2 - 10x + 24 >0$, becoming $(x-6)(x-4) > 0$, where it is clear.


abstract algebra
induction
Ring Theory
topology
superior algebra
superior algebra unsolved

Let $ f: A\to B$ be a ring homomorphism and suppose that $ Spec(B)$ is a Noetherian space. Prove that $ f^*: Spec(B)\to Spec(A)$ is a closed mapping if and only if $ f$ has the going up property (i.e. the conclusion of the "going-up" theorem holds for $ f(A)\subset B$).

"$ \Rightarrow$": This implication does not need the Noetherian assumption.
Let $ f^*$ be a closed mapping. Let $ \mathfrak{p}$ be a prime ideal in $ B$. The induced map $ g: A/(A\cap \mathfrak{p})\rightarrow B/\mathfrak{p}$ is injective and $ g^*: V_B(\mathfrak{p})\rightarrow V_A(\mathfrak{p}\cap A)$ is closed. The image of $ g^*$ is closed and contains the generic point $ \mathfrak{p}\cap A$ of $ V_A(\mathfrak{p})$, hence the image of $ g^*$ is $ V_A(\mathfrak{p})$. In other words $ g^*$ is surjective. From this the going-up property follows immediately.

"$ \Leftarrow$": Let $ f^*$ satisfy the going-up property. Assume that there is a closed subset $ V$ of $ Spec(B)$ such that the image under $ f^*$ is not closed (*).

By Noetherian induction, we can assume that $ V$ is minimal, i.e. for every closed set $ V'$ properly contained in $ V$ the image of $ V'$ under $ f^*$ is closed.

Then $ V$ is irreducible: Otherwise $ V$ is the union of two closed subsets $ V_1, V_2$ properly contained in $ V$. Then $ f^*(V_i)$ is closed, hence $ f^*(V)\equal{}f^*(V_1)\cup f^*(V_2)$ is closed, too. Contradiction.

As irreducible set we have $ V\equal{}V(\mathfrak{p})$ with a prime ideal $ \mathfrak{p}$ of $ B$. As used in the course of the proof of the reverse implication, the going up property can be translated to $ f^*(V(\mathfrak{p}))\equal{}V(\mathfrak{p}\cap A)$. But this shows that $ V$ has closed image under $ f^*$. Altogether a contradiction to assumption (*).


What happened to the AoPS chat proposed in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=98568]this[/url] thread? What about [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=272155]here[/url]?

Are these chat rooms dead?

Is AoPS chat a good idea (especially since FTW is discouraged from being used for chatting)?

I would totally second that. But it would cost a bunch of money just like FTW.

Yeah, chat rooms are so pricey, only very wealthy people can afford them  :wink:

People [i]have [/i] actually expressed an interest in an AoPS chat room so that people wouldn't go on FTW just to chat, so a new chat room might not die like the others. :maybe:

There already is one on euIRC now  :wink:
Channel #mathlinks.

@yunhua98:
[quote="ThinkFlow"]([b]especially since FTW is discouraged from being used for chatting[/b])?[/quote]


inequalities
function
inequalities proposed

I managed to prove the following:

Let $ a_1,a_2,\ldots,a_n$ be nonnegative real numbers such that $ a_1\plus{}\cdots\plus{}a_n\equal{}n$. Then
\[ (a)\quad \frac1{n^2\plus{}(2n^2\minus{}3n\plus{}1)a_1^3}\plus{}\cdots\plus{}\frac1{n^2\plus{}(2n^2\minus{}3n\plus{}1)a_n^3}\ge\frac{n}{3n^2\minus{}3n\plus{}1}\] and
\[ (b)\quad \frac1{n^2\minus{}2n\plus{}2\plus{}(n\minus{}1)a_1^3}\plus{}\cdots\plus{}\frac1{n^2\minus{}2n\plus{}2\plus{}(n\minus{}1)a_n^3}\ge\frac{n}{n^2\minus{}n\plus{}1}.\]
Thank you very much for your inequalities!  :)

[quote="bimbo"]I managed to prove the following:

Let $ a_1,a_2,\ldots,a_n$ be nonnegative real numbers such that $ a_1 \plus{} \cdots \plus{} a_n \equal{} n$. Then
\[ (a)\quad \frac1{n^2 \plus{} (2n^2 \minus{} 3n \plus{} 1)a_1^3} \plus{} \cdots \plus{} \frac1{n^2 \plus{} (2n^2 \minus{} 3n \plus{} 1)a_n^3}\ge\frac {n}{3n^2 \minus{} 3n \plus{} 1}
\]
and
\[ (b)\quad \frac1{n^2 \minus{} 2n \plus{} 2 \plus{} (n \minus{} 1)a_1^3} \plus{} \cdots \plus{} \frac1{n^2 \minus{} 2n \plus{} 2 \plus{} (n \minus{} 1)a_n^3}\ge\frac {n}{n^2 \minus{} n \plus{} 1}.
\][/quote]
Indeed, Bimbo. These inequalities are true. :lol:

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that

$ (a) \ \ \ \ \ \ \ \ \frac 1{9\plus{}10a^3}\plus{}\frac 1{9\plus{}10b^3}\plus{}\frac 1{9\plus{}10c^3} \ge \frac 3{19}$;
[/quote]
Let $f(x)=\frac 1{9\plus{}10x^3}$. Hence, $f''(x)=\frac{1200x\left(x^3-\frac{9}{20}\right)}{(10x^3+9)^3}$.
We see that $f$ is a convex function on $[1,3]$.
Hence, by RCF Theorem it remains to prove our inequality for $b=a$ and $c=3-2a$, which gives
$(a-1)^2(3-2a)(18+15a+20a^2-20a^3)\geq0$, which is true for $0\leq a\leq1.5$. Done!

The second inequality we can prove by LCF Theorem.

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$. Prove that

$ (a) \ \ \ \ \ \ \ \ \frac 1{9\plus{}10a^3}\plus{}\frac 1{9\plus{}10b^3}\plus{}\frac 1{9\plus{}10c^3} \ge \frac 3{19}$;

$ (b) \ \ \ \ \ \ \ \  \  \ \frac 1{5\plus{}2a^3}\plus{}\frac 1{5\plus{}2b^3}\plus{}\frac 1{5\plus{}2c^3} \le \frac 3{7}$.[/quote]

for a)
Let $c \le b \le a$  then $a+b \ge 2$

$\frac 1 {9+10a^3}  +\frac 1 {9+10b^3} \ge \frac {16} {72+10(a+b)^3} \Leftrightarrow$

$(a-b)^2[5(a+b)^4++5ab(a^2+b^2)+30a^2b^2-27(a+b)] \ge 0$

which is true since $5(a+b)^4 \ge 40(a+b)$

hence it is sufficies to prove

$\frac {16} {72+10(3-c)^3} +\frac 1 {9+10c^3} \ge \frac 3 {19} \Leftrightarrow$

$c(c-1)^2(5c^3-35c^2+60c+36) \ge 0$

which is true since $c \le 1$

equality holds for $a=b=c=1$  and $c=0,a=b=\frac 3 2 $

for b)

let $a \ge b \ge c$ then $b+c \le 2$

$\frac 1 {5+2b^3} +\frac 1 {5+2c^3} \le \frac {16} {40+2(b+c)^3} \Leftrightarrow$

$(b-c)^2[(b+c)^4+bc(b+c)^2+4b^2c^2-15(b+c)]\le 0$

which is true since $(b+c)^4 \le 8(b+c)$  and

$2b^2c^2 \le \frac {(b+c)^4} 8 \le b+c$

hence it is sufficies to prove

$\frac {16} {40+2(3-a)^3} +\frac 1 {5+2a^3} \le \frac 3 7 \Leftrightarrow$

$(a-4)(a-2)^2 (a-1)^2 (a+1) \le 0$

which is true since $1 \le a \le 3$

equality holds for $a=b=c=1$  and $a=2,b=c=\frac 1 2$


Wouldn't it be neat with an age field in the profile?

[color=white]Yeah, but does it really matter?  I mean, we're all in this forum and for those of us who've been here a while, we know each other pretty well.  Especially the people who post a lot...like ComplexZeta or Syntax Error.  For all I know, the people I think I know well could be totally different...oh well.[/color]

well, most of us are middle-high-(just graduated high) schoolers anyways, I would think.

Fierytycoon

Some people here are also teachers, so I'd imagine there's a bigger age difference than you might expect. I'm 18.

And the "age field" doesn't inhibit privacy, one does not have to enter any age at all!

[color=white]I wasn't really worried about privacy or anything.  Sorry if I sounded a bit...judgemental on my first post here.  I was in a hurry and scribbled in what I thought [i]really[/i] quickly, so excuse that one.

I know there are teachers here, like the Great Rusczyk and Matthew Crawford.  But I had no idea you were 18, ComplexZeta...I'm barely 12.  Sheesh.  Somehow this always happens...I'm hanging out with some people online and they all turn out to be around 5-6 years older...not like it's a problem or anything.  Anyway, I'm starting to babble, so I'll just leave it at that.[/color]

While we're at it (telling our ages), I'll just say that I'm 14 years old.

-interesting_move

im guessing some of you already know. if you REALLY want to know just go through all of my posts. and youll get an idea.


Let $x_1,x_2,...,x_n$ be positive real numbers . We know that $x_1.x_2...x_n$=1, prove that :
$\frac{1}{n-1+x_1}+\frac{1}{n-1+x_2}+...+\frac{1}{n-1+x_n}\leq1$

http://www.mathlinks.ro/Forum/viewtopic.php?t=19230 .

  darij


Les Cash puts 1200 dollars in a bank account that pays 3% simple interest. Noah Snet puts 1000 dollars in a bank account that pays 5% compund interest compunded monthly. They both leave their money in their accounts for 3 years. What is the difference at the end of the 3 years?

146.53
150.00
162.86
308.00
312.86

Thanks a lot.

Solving this problem is a matter of plugging the numbers in formulas.  I'll tell you that for the continuous interest, you're going to need $ e$.  The other formula is just $ 1200(1.03)^{3}$.  Now you figure out the rest.


Let $x_1,x_2,\cdots ,x_n \in\mathbb Z (n\geq 3)$ satisfying the condition:
\[(-1)^{x_{i+1}}\cdot x_i\cdot x_{i+2}>0,\forall i=1,2,\cdots,n(x_{n+1}=x_1,x_{n+2}=x_2)\]
Prove that:
$\sum_{i=1}^n (|x_i|-x_i)$ is divisible by $4$
Maybe it easy for you!(the problem which was posted by me was easy ;) )

Hmm.. It took a little while to see this, but could it be so simple? :?

We want to show that the number of odd and negative numbers in the sequence is even. This is equivalent to showing that the number of $i$'s for which $x_{i+1}<0,x_ix_{i+2}<0$ (indices are modulo $n$) is even (this is derived from the given relationship).

We go around the circle in a clockwise direction, through $x_1,x_2,\ldots$. When we go from a positive zone to a negative one with at least $2$ terms, we first find $+--$, and, at the end of the negative zone, we find $--+$, so the desired triples $x_i,x_{i+1},x_{i+2}$ come in pairs, meaning that their number is even.

Is this right?


$ a,b,c$ are sides of a triangle then prove
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} 2 abc \le \sum_{sym} a^2b$

$ LHS\minus{}RHS \equal{} \minus{}(a\plus{}b\minus{}c)(b\plus{}c\minus{}a)(c\plus{}a\minus{}b) \le 0$.


function calculus derivative vector geometry trigonometry algebra

If we have a function of two variables and a point on the surface, I understand why the direction of steepest ascent will run in the same direction of the gradient vector. But how do we prove that this direction is perpendicular to the level set of this point? I feel like this should be true for some stupid reason, but I can't seem to find it...

We're assuming, of course, that we have a differentiable function, which means that the function is well-approximated by a tangent plane. The fact that there is a tangent plane - which is given by a linear equation - can be expressed in a number of ways.

One of the useful ways of expressing that notion of a tangent plane is through the relationship of the directional derivative and the gradient.

If $ \vec{u}$ is a unit vector, then the directional derivative of $ f$ in the direction of $ \vec{u}$ is given by $ D_{\vec{u}}f\equal{}\nabla f\cdot\vec{u}.$

Now: when is that directional derivative a large as possible and when is it zero? We can use the geometry of the dot product to say that
\[ D_{\vec{u}}f\equal{}\nabla f\cdot\vec{u}\equal{}\|\nabla f\|\,\|\vec{u}\|\cos\theta\equal{}\|\nabla f\|\cos\theta\]
where $ \theta$ is the angle between $ \nabla f$ and $ \vec{u}.$

We can make this directional derivative as large as possible by making $ \theta\equal{}0$ -- that is, having $ \vec{u}$ point in the same direction as $ \nabla f.$ And we can make this directional derivative zero by making $ \theta$ a right angle, which is to say making $ \vec{u}$ perpendicular to $ \nabla f.$

Now, what's a level curve (or level surface if we're doing this in three dimensions)? A curve or surface along which $ f$ is constant. So if we travel along such a curve, $ f$ does not change, and if we point $ \vec{u}$ in a direction tangent to the level curve (or surface), then $ D_{\vec{u}}f$ must equal zero. And that means the level curve - or at least the tangent direction to the level curve, which is the only thing we can mean - must be perpendicular to the gradient.

This line of thinking also goes into the explanation of why Lagrange multipliers work.


number of ways of distribution
x +y+z+u=31
x=>y=>z=>u

I edited in "8 + 8 + 8 + 7" to bring the total up to 221.

I don't know where your claimed missing other
four ways would be located (in my list). / / / EDIT

Thank you for showing my list and count are wrong.

Hi Arrange your tan;

You are also missing:

23,4,2,2
22,4,4,1
20,6,4,1
19,4,4,4

bringing the count up to 225.


Hello there,

I just started learning about logarithms on my own today and I have 1 question. As always, I would appreciate your help!

1. Convert the following to exponential form:

log x = -2  --> x = 10^(-2)

Question: As per this example, am I allowed to write only log x = -2 to mean log[size=75]10[/size]X = -2 ?

Thank you very much!

Some people use $ \log$ as the natural logarithms $ \ln \equiv \log_{e}$, though generally $ \log \equiv \log_{10}$.

It is assumed that if no base is given, the base of a logarithm is 10 (has to do with scientific notation) you could also use the natural base e (in calculus, I believe that the derivative of the natural exponential function is 1, which is why the number e is so useful) . Hope that helps.

Your responses certainly helped. Thank you very much AstroPhys and Prime factorization!


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose $G$ is a torsion group and $H$ is a subgroup of finite index $m$ such that all elements of $H$ has order at least $m$. Then $H$ is normal.

Order at least $m$¿ What does this mean¿ Should it be at most instead¿  :maybe:

Sorry, all nontrivial element of $H$ has order at least $m$.

Hi everyone.

Let h be an element of H. ord(h) can't be divisible by a prime number $<m$ because there are no elements of order smaller m in H.
Now H contains a normal subgroup N whose factorgroup H/N is isomorphic to a subgroup of $S_{m-1}$. There is no element in $S_{m-1}$ whose order is divisible by a prime number $\geq m$ so every element of H is in the kernel of the projection $H\to H/N$ which means H=N and especially that H is normal.