Show that in any triangle ABC whose angles A,B,C are in (0,  :pi: /2) we have:

( :sqrt: tan(A)+:sqrt: tan(B) + :sqrt: tan(C) )/ ( :sqrt: ctan(A)+:sqrt: ctan(B) + :sqrt: ctan(C) ) >=  :rt3: .

my solution in spoiler

[hide]Let a= :sqrt: tan(A), and similarly define b and c. Because of our angles, a, b, and c are positive. Since tan(A)tan(B)tan(C)=tan(A)+tan(B)+tan(C), we get abc= :sqrt: (a^2+b^2+c^2). The inequality becomes (a+b+c)/(1/a+1/b+1/c) :ge:  :rt3: , i.e. [(a+b+c) abc]/(ab+bc+ca) :ge:  :rt3:. To homogenize, sub for abc:

[(a+b+c) :sqrt: (a^2+b^2+c^2)]/(ab+bc+ca) :ge:  :rt3: .

Then, square, clear denominators, and move everything to the left:

(a+b+c)^2 (a^2+b^2+c^2)-3(ab+bc+ca)^2 :ge: 0.

By factoring, we see the truth of this last statement:

[(a^2+b^2+c^2)-(ab+bc+ca)][(a^2+b^2+c^2)+3(ab+bc+ca)] :ge: 0.

The left factor is non-negative (by rearrangement for example), and the right factor is positive.[/hide]


function
real analysis
real analysis theorems

How should I understand the following theorem? Let $ C$ be the Cantor dust. There is no continuous function from the interval $ [0,1]$ onto $ C$. I can understand if we wanted to map $ C$, but this "reverse" case..

Is this straightfoward what is going on here?

Thanks in advance,

Los

Well $ C$ is a subset of some $ \mathbb{R}^n$, right? Both are metric spaces, so the definition of continuity being used here is the usual one. Equivalently:
[quote]For any surjective function $ f: [0,1]\to C$, there exists an $ x\in [0,1]$ and an $ \epsilon > 0$ such that for each $ \delta > 0$ there exists a $ y\in [0,1]$ such that $ |x \minus{} y| < \delta$ and $ |f(x) \minus{} f(y)| > \epsilon$[/quote]
where the norms are understood to be the norms on $ [0,1]$ and $ C$ respectively.  I imagine this text is trying to prove that Cantor dust is totally disconnected?


A sequence $ (a_n)$ is defined as follows:

$ a_1\equal{}p$ is a prime number with $ 300$ nonzero digits, and for each $ n\geq 1$, $ a_{n\plus{}1}$ is twice the decimal period of $ \frac{1}{a_n}$. Find $ a_{2003}$.

Quite strange, because p is not bounded :maybe:
Can anyone find a upper bound for $ a_2$?

Well, since $ a_{n\plus{}1}$ is twice the decimal period of something, and that decimal period doesn't exceed $ 1$, doesn't that provide an upper bound of $ 2$?

[quote="radio"]that decimal period doesn't exceed $ 1$[/quote]
I don't understand why :( .Could you please explain it :huh:
[quote="radio"]doesn't that provide an upper bound of $ 2$?[/quote]
What is "$ 2$"?

What is a decimal period? Is it $ 6$ or $ 142857$ or anything else For $ 1/7\equal{}0.14285714285714...$

What? The decimal period is how often the decimal repeats, so of course it's greater than or equal to one.

Well, I'm not that good at English, maybe I misunderstand the problem. Anyway I took it from Russian MO. Please continue it.

[hide="Discussion"] The decimal period of $ \frac{1}{p}$ is at most (and divides) $ p\minus{}1$.  This gives $ a_2 | 2(p\minus{}1) \equal{} 4 \cdot \frac{p\minus{}1}{2}$.  The $ 4$ does not affect the decimal period of $ \frac{1}{a_2}$, so continuing this reasoning we have the inequality

$ a_{n\plus{}1} \le \frac{a_n}{2} \implies$
$ a_{2003} \le \frac{a_1}{2^{2002}}$.

It remains to find an upper bound on $ p$, which I have to admit is stumping me. [/hide]

Ah! I think the problem means that p have 300 digits which are all nonzero, but not p has many digits among which 300 are nonzero, right?

Obviously $ p\ne 2,5$. Multiplying by $ 2$ does not change the decimal period of a decimal...Let $ k\equal{}\text{ord}\, _{p}10$, that is $ k$ is the minimal positive integer such that $ p|(10^k\minus{}1)$. It is well known that $ k$ is the decimal period of $ p$. So the answer is $ 2k$...?


floor function
function
calculus
quadratics
algebra

Sketch:
Clearly $ x$ is a rational number, moreover, $ x$ is of the form $ \frac{k\minus{}3}{4}$. Now after considering $ 4$ cases of what residue $ k$ gives modulo $ 4$, we're about to get $ 4$ equations.

This problem seems unanswered yet, so i will write my solution. Corrections are always welcome. Thanks :)
Notice that the problem is equivalent to
\[x = \frac{\lfloor x^3 \rfloor - 3}{4} \longleftrightarrow \lfloor x \rfloor = \lfloor \frac{\lfloor x^3 \rfloor - 3}{4} \rfloor \]
Also, notice that
\[(x+1)(x^2-x-3) = x^3-4x-3 \geq \lfloor x^3 \rfloor -4x-3 = 0\]
\[(x+1)(x^2-x-3) \geq 0\]
This gives
\[\frac{1-\sqrt{13}}{2} \leq x \leq -1 \text{ or } x \geq \frac{1+\sqrt{13}}{2}\]
So,
\[-2 \leq \lfloor x \rfloor \leq -1 \text{ or} \lfloor x \rfloor \geq 2\]

Now, consider the function $f(x) = x^3 - 4x - 3$. We will find on which interval does $f$ increase, so we need to take derivative, then $f'(x) > 0 \longleftrightarrow 3x^2 - 4 > 0 \longleftrightarrow (x\sqrt{3} - 2)(x\sqrt{3} + 2) > 0$, then $-\frac{2}{3}\sqrt{3} \geq x \text{ or } x \geq \frac{2}{3}\sqrt{3}$.
Notice again that $f(3) = 27 - 12 - 3 = 12 > 0$. Since $3 > \frac{2}{3}\sqrt{3}$, and $f$ is increasing on $[\frac{2}{3}\sqrt{3}, +\infty)$, then $f(x) > 0$, so $x < 3$.
Noticing that $f(-2) = -8 + 8 - 3 = -3 < 0$ and using similar reasoning, gives $x > -2$.
So, we have
\[-2 \leq \lfloor x \rfloor \leq -1 \text{ or } 2 \leq \lfloor x \rfloor < 3\]
This means, the possible values for $\lfloor x \rfloor$ are $-2 , -1, 2$

If $\lfloor x \rfloor = -2$, then $-2 \leq x < -1$, so $-8 \leq x^3 < -1$ and $-8 \leq \lfloor x^3 \rfloor < -1$.
So, $\lfloor x^3 \rfloor = -8, -7, -6, ..., -2$.
Plugging each value to $\lfloor x \rfloor = \lfloor \frac{\lfloor x^3 \rfloor - 3}{4} \rfloor$, we see that the only possible values are $-5, -4, -3, -2$.

If $\lfloor x \rfloor = -1$, then $-1 \leq x < 0$, so $-1 \geq x^3 < 0$ and $1 \leq \lfloor x^3 \rfloor < 0$.
Checking, we see that $-1$ is the only possible value for $\lfloor x^3 \rfloor$ in this case.

If $\lfloor x \rfloor = 2$, then $2 \leq x \leq 3$, so $8 \geq x^3 < 27$ and $8 \leq \lfloor x^3 \rfloor < 27$.
Thus, $\lfloor x^3 \rfloor = 8, 9, 10, \cdots, 26$.
Checking again, we see that the possible values are $11,12,13,14$.

So
\[\lfloor x^3 \rfloor = -5, -4, -3, -2, -1, 11, 12, 13, 14\]
Hence, since $x = \frac{\lfloor x^3 \rfloor - 3}{4}$, then we substitute those values there and get :
\[\boxed{x = -2, -\frac{7}{4}, -\frac{3}{2}, -\frac{5}{4}, -1, 2, \frac{9}{4}, \frac{5}{2}, \frac{11}{4}}\]

I don't think there's anything wrong in MrRTI's solution. He just gave a correct proof for reducing the set of [b]possible solutions[/b] to the 9 numbers he stated as "solutions". The mistake is just that not all these numbers really satisfy the equation because the conditions you elaborated are just [b]necessary [/b]but not [b]sufficient [/b]ones. So, you just need to plug in all your 9 values to see that only $1.25$ and $1$ will be solutions.

[quote=efang]Holddddd up if you're saying I spent 30 minutes on this freaking problem just to get it wrong at midnight I will quite possibly fall into depression and eat all of my halloween candy.

First off $x = -2$ doesn't even work. Nor does $x = \frac{-7}{4}, -\frac{3}{2}, 2, \frac{9}{4}, \frac{5}{2},$ and $\frac{11}{4}$ (trust me I calculated each of those individually to check for any mistakes).

Here is my solution:

The restriction of $\lfloor{x^3}\rfloor = 4x+3$ tells us that $4x+3 \le x^3 < 4x+4$. It also tells us that $4x + 3$ must be an integer which means that $4x$ is an integer.

This means that $x = \frac{a}{4}$ for some integer $a$.

First let's investigate $4x +3 \le x^3$

By rearranging the numbers we have $x^3 - 4x - 3 \ge 0$. By RRT we find that $-1$ is a root of $x^3 - 4x - 3$. Factoring we get $(x+1)(x^2-x-3) \ge 0$.

By the quadratic equation the solutions for $x^2 - x - 3 =0$ are $x = \frac{1 \pm \sqrt{13}}{2}$.

Thus this condition is satisfied if $\frac{1-\sqrt{13}}{2} \le x \le -1$ or $\frac{1+\sqrt{13}}{2} \le x$

Now let's investigate $x^3 < 4x + 4$

We have that when $x = 2, 8 < 12$ but when $x = 3, 27 > 16$. Obviously $x^3$ increases faster than $4x + 4$ so $x < 3$. Note that this bound is not strong and in fact $x \le c$ for some $2 < c < 3$

Now we want to find the values of $x$ in the range $\frac{1-\sqrt{13}}{2} \le x \le -1$ such that $x = \frac{a}{4}$ for some integer $a$.

Note that $3.6^2 = 12.96 \approx 13$. Thus $\frac{1-\sqrt{13}}{2} \approx \frac{1-3.6}{2} = -1.3$

Thus the possible values of $x$ in this range are $-1.25, -1$ which we can check to see that they both work.

Now we approximate $\frac{1+\sqrt{13}}{2} \approx \frac{1+3.6}{2} = 2.3$

Thus the possible values of this range are $2.5$ and $2.75$. But wait! We have to check if these values indeed satisfy $x^3 - 4x - 4 < 0$ because we never determined the value $a$ such that $x \le a$. Plugging in $2.5$ we see that the condition is not satisfied. As again, $x^3$ grows faster than $4x+4, x=2.75$ won't work either.

Thus the solutions are simply $\boxed{x = -1.25, -1}$

Hm it's too late right now (even with daylight savings) for me to have the energy to read the above solution so I might do that later and try finding what's wrong.[/quote]

efang's right. Also, you could have noted that the fractional part of x lies in the set S = {$0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}$} since 4x+3 must be an integer. From there you note that $n^3 < 4n+3 + 4s < (n+1)^3$, where x = n+s, n is an integer, and s $\in$ S. Separating the bound into two polynomials and casework bashing gives you x = -1.25 and -1.

I think I have a somehow easier to understand (for people more in the beginners level like me) solution that isn't too long
[hide=why do I always induct inequalities like at JBMO]
*Idk how to type floor function, so absolute value(||) here means floor function

Claim 1: For $x\geq 3$, we have $LHS>RHS$
Proof by Induction: Clearly, for x element of $[3,4)$ (I shud really learn latex better...),
We have $|x^3| \geq 3^3 >4 \cdot 4+3 >4x+3$
Suppose it is true for x element of $[k,k+1)$, where k is a positive integer that is at least 3
Then when x element of $[k+1,k+2)$, we have
$|x^3| \geq (k+1)^3=k^3+3k^2+3k+1$
$>(4k+3)+3k^2+3k+1=3k^2+7k+4$ (Since it is true for x=k)
$\geq 9k+7k+4=16k+4$ (As $k\geq 3$)
$=4(k+2)+(12k-4)>4(k+2)+3$ (Again, since $k\geq 3$)
$>4x+3$
Thus, the Claim is proved

Claim 2:For $x\leq 3$, we have $LHS<RHS$
Proof: Let $k=-x$
It suffices to show $|k^3|>4k-3$ by flipping the signs, where $k\geq 3$
But this follows immediately from Claim 1
Thus, the Claim is proved

Hence, by these 2 Claims, x must be an element of $(-3,3)$
Clearly, the LHS is an integer, so 4x+3, and hence, 4x must be an integer
So x must be of the form $\frac{k}{4}$, where k is an integer
Combining this with the range, we get that there's only 23 possible values
Checking (a lot of them can be eliminated in groups by a bit of bounding),
we see that $x=-\frac{5}{4},-1$ are the only solutions
.

[/hide]
Brb. Ok done

[quote=orl]Find all real numbers $ x$ such that $ \lfloor x^3 \rfloor \equal{} 4x \plus{} 3$.[/quote]
Problem is "find all $n\in\mathbb Z$ such that $n+4>\frac {n^3}{64}\ge n+3$" (and then $x=\frac n4$)

Which is $256>n^3-64n\ge 192$

And it is not very difficult to get that only $n\in\{-5,-4\}$ fits, and so $x\in\{-\frac 54,-1\}$

This is what I got ( I think so it is wrong)
Observe LHS is an integer so rhs should also be. so is 4x so is of always of the form k/4


limit
integration
function
logarithms
calculus
calculus computations

For each $ n$, it's a Riemann sum for $ \int_{1}^{2}\frac{1}x\,dx$.

Use Taylor law with function $ ln(x\plus{}1)$

TTsphn, that's not obvious. I'd need to see more details before I'd believe your method.

(It is the right answer, of course, for the reason jmerry gave.)

Well, is it the right answer? The following comment is directed to xavier9977: Everyone who as answered you has assumed that you meant something different than what you said. The answers jmerry and TTsphn gave were to the following problem:

$ \lim_{n\to\infty}(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n})$

But what you actually said was this:

${ \lim_{n\to\infty}(\frac{1}{n}+1+\frac{1}{n}+2+\cdots+\frac{1}{2}n})$

And that limit would be infinity. Now, I think you probably did mean the first problem, with the limit of $ \ln 2,$ because the second version doesn't make much sense as a problem. In the future:

(1) You could learn to input your expressions in $ \text{\LaTeX}.$ It's really not that hard.

(2) Failing that, if you insist on in-line text expressions, [b]put in the parentheses![/b] In-line text expressions typically need lots of parentheses and layers of parentheses in order to be unambiguous.

Here 's a proof for $ \lim_{n\rightarrow\infty}\left(\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}...\plus{}\frac{1}{2n}\right) \equal{}\ln(2)$ without using advanced knowledge such as Riemann sums :lol:.

Let $ x_{n}\equal{} 1\plus{}\frac{1}{2}\plus{}..\plus{}\frac{1}{n}\minus{}\ln(n)$

We have the identity:
$ \left(\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}...\plus{}\frac{1}{2n}\right) \equal{} x_{2n}\minus{}x_{n}\plus{}\ln(2)$ $ (*)$

It's well-known that $ \lim_{n\rightarrow\infty}x_{n}\equal{}\gamma$ ,where $ \gamma$ is Euller's constant

So from $ (*)$ we have that:

$ \lim_{n\rightarrow\infty}\left(\frac{1}{n\plus{}1}\plus{}\frac{1}{n\plus{}2}\plus{}...\plus{}\frac{1}{2n}\right) \equal{}\lim_{n\rightarrow\infty}(x_{2n}\minus{}x_{n}\plus{}\ln(2)) \equal{}\gamma\minus{}\gamma\plus{}\ln(2)$ ,and the conclusion follows

[quote="Svejk"]... without using advanced knowledge such as Riemann sums
...
It's well-known that $ \lim_{n\rightarrow\infty}x_{n}\equal{}\gamma$ ,where $ \gamma$ is Euller's constant[/quote]

 :roll:

Of course, what JBL is trying to say is that the concept of Riemann sums, a basic foundational tool in integral calculus, is in fact less advanced than the proof that $ x_{n}$ tends to a limit.

[color=darkred]This limit is a well-known aplication of the definition and the main property $ (*)$  of the number $ e$.[/color]

[color=darkblue]Denote $ L_{n}\equal{}\sum_{k \equal{} 1}^{n}\frac{1}{n\plus{}k}$. Since $ \left(1\plus{}\frac{1}{n}\right)^{n}\nearrow e$ and $ \left(1\plus{}\frac{1}{n}\right)^{n\plus{}1}\searrow e$ $ \implies$ $ \left(1\plus{}\frac{1}{n}\right)^{n}< e <\left(1\plus{}\frac{1}{n}\right)^{n\plus{}1}$ $ \implies$ $ \boxed{\ \frac{1}{n\plus{}1}<\ln (n\plus{}1)\minus{}\ln n <\frac{1}{n}\ }\ \ (*)$

$ \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{}$ $ \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{}$ $ \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{}$ $ \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{}$

Particularly, for any $ k\in\overline{1,n}$ obtain  $ \frac{1}{n\plus{}k\plus{}1}<\ln (n\plus{}k\plus{}1)\minus{}\ln (n\plus{}k) <\frac{1}{n\plus{}k}$ $ \implies$ $ \sum_{k \equal{} 1}^{n}\frac{1}{n\plus{}k\plus{}1}<\sum_{k \equal{} 1}^{n}\left[\ln (n\plus{}k\plus{}1)\minus{}\ln (n\plus{}k)\right] <\sum_{k \equal{} 1}^{n}\frac{1}{n\plus{}k}$.

Thus, $ \L_{n}\minus{}\frac{1}{n\plus{}1}\plus{}\frac{1}{2n\plus{}1}<\ln (2n\plus{}1)\minus{}\ln (n\plus{}1) < L_{n}$ $ \implies$  $ \ln\frac{2n\plus{}1}{n\plus{}1}< L_{n}<\ln\frac{2n\plus{}1}{n\plus{}1}\plus{}\frac{n}{(n\plus{}1)(2n\plus{}1)}$ for any $ n\in N^{*}$ $ \implies$ $ L_{n}\rightarrow\ln 2$.[/color]

[quote="TTsphn"]${ \sum_{n = 1}^{\infty}\frac{(-1)^{n-1}}{n}}=\frac{1}{n+1}+..+\frac{1}{2n}$[/quote]
Say what?

In the first place, that's not literally true; the left hand side is exactly $ \ln 2$ and the right hand side is merely an approximation. But how are the sides even connected? (And be aware that rearranging a conditionally convergent series is not a convincing proof of anything.)

He meant $ \sum_{k \equal{} 1}^{2n}\frac{(\minus{}1)^{n\minus{}1}}{n}$ in the left hand side, i.e.
\[ \sum_{k\equal{}1}^{2n}\frac{(\minus{}1)^{n\minus{}1}}{n}\equal{}\sum_{k\equal{}n\plus{}1}^{2n}\frac{1}{k}.\]
This even has a name, but I don't remember it right now.

$ L_n\equal{}\sum_{k\equal{}1}^n\frac {1}{n\plus{}k}\equal{}\frac 1n\cdot\sum_{k\equal{}1}^n\frac {1}{1\plus{}\frac kn}\equal{}\frac 1n\cdot\sum_{k\equal{}1}^nf\left(\frac kn\right)$ , where $ f(x)\equal{}\frac {1}{1\plus{}x}$ .

In conclusion, $ L_n\rightarrow\int_0^1\frac {1}{x\plus{}1}\ \mathrm {dx}\equal{}\ln 2$ .


function
search
Putnam
real analysis
real analysis unsolved

let $ f: \mathbb{R}\rightarrow \mathbb{R}$ be a $ C^{3}$ function (3 times continuously derivable).
Prove that there exist $ a \in \mathbb{R}$ such that : $ f(a)f'(a)f''(a)f^{(3)}(a) \geq 0$

This is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=936377842&t=88773]Putnam 1998 A3.[/url]


Find the area of the figure formed by all solutions to $ \parallel{}x|\minus{}3|\plus{}\parallel{}y|\minus{}4|\le 5$.

[hide="solution"]
We have $ \parallel{}x|\minus{}3| \plus{} \parallel{}y|\minus{}4| \le 5$
Now suppose $ x>0$ then clearly $ x \equal{} |x| \equal{} |\minus{}x|$
So the graph is symetrical around the $ x$ and $ y$ axis

So we can solve $ |x\minus{}3|\plus{}|y\minus{}4| \le 5$ for $ x,y \ge 0$ then multiply by $ 4$.

Now translate $ x\minus{}3\to x'$ and $ y\minus{}4 \to y'$ (which does not affect area). The equation becomes

$ |x'| \plus{} |y'| \le 5$ with $ x'\ge \minus{}3, y' \ge \minus{}4$

This is rombus with two corners cut off.

The desired Area is; $ \frac{1}{2}(2\cdot 5)^2 \minus{} \frac{1}{2}\cdot 4 \cdot 2 \minus{} 1 \equal{} 45$

Now we multiply by four to get $ \boxed{180}$[/hide]

The alternative is doing annoying casework for every quadrant separately and sketching (that way you will also convince yourself more easily that the figure is indeed symmetrical).


[hide]
There are 4 lines going downwards, and we need to choose 2 of these to be sides of our rectangle.  There are 3 lines going across, and 2 of these also need to be chosen, so there are $ \binom{4}{2}\binom{3}{2}\equal{}18$ rectangles
[/hide]

I tried that on a two by two so there would be three lines to choose from on both sides of the rectangle so would it be (3C2)*(3C2), but does that not equal nine? when the answer is 5.

you forgot the one by two rectangles  :D

hmm is there a formula for this like there is for the number of squares?

when x is the length of the shortest side of the grid:

1^2+2^2+3^2......+x^2

he did.

The message is too small. Please make the message longer before submitting.

Am I missing something or wouldn't the answer to this be 20?

You would have the \[ \binom{4}{2} \binom{3}{2}\equal{}18 \] for the rectangles already listed, plus these 2:

[asy]for ( int x = 0; x < 4; ++x )
{
    for ( int y = 0; y < 3; ++y )
    {
        dot((x, y));
    }
}
draw((1,0)--(2,1)--(1,2)--(0,1)--cycle, blue);
draw((2,0)--(3,1)--(2,2)--(1,1)--cycle, red);[/asy]

[quote="neurosurgeon1118"]Well for this problem it didnt take me too long to list everything out but if the grid was larger is there a formula for this?
Problem: How many rectangles are in a two by three grid?
My method: there are 6 1x1's
7 2x1's
2 3x1"s
2 2x2's
1 2x3
adding,6+7+2+2+1 we get [b]18.[/b]
[/b][/quote]
i've experimented with this b4
if its an x by y grid, then the number of rectangles that can be formed is
$ \dfrac{x(x \plus{} 1)}{2}\times\dfrac{y(y \plus{} 1)}{2}$
i've proved it b4, but im too lazy to do it again

[quote="levans"]Am I missing something or wouldn't the answer to this be 20?

You would have the
\[ \binom{4}{2} \binom{3}{2} \equal{} 18
\]
for the rectangles already listed, plus these 2:

[asy]for ( int x = 0; x < 4; ++x )
{
    for ( int y = 0; y < 3; ++y )
    {
        dot((x, y));
    }
}
draw((1,0)--(2,1)--(1,2)--(0,1)--cycle, blue);
draw((2,0)--(3,1)--(2,2)--(1,1)--cycle, red);[/asy][/quote]

I think that the problem meant one of those grids of squares, but good point

Ahh, makes sense then.  So the original problem would have had an image which would have looked like:

[asy]for ( int x = 0; x < 4; ++x )
{
    draw((x,0)--(x,2));
}
    for ( int y = 0; y < 3; ++y )
    {
        draw((0,y)--(3,y));
    }[/asy]


superior algebra
superior algebra unsolved

Let $A \in M_{n}(K)$, where $K$ is a comutative field. If $A^{2} =A$ which are the eingenvalues of $A$?


Hello. Can you please help me???

the question: How many edges does a cylinder have?

my hypothesis: I think 0, because at the top of the cylinder, the sharp part is continuous, not just a point. So it might be 2 too, because some people say that it has 2 sides.

can you help me? Thanx!

[quote="Termination Terminator"]Hello. My son came home today with a probably that both me and him don't really understand.

the question: How many edges does a cylinder have?

 ![/quote]
Such questions, unless one defines what an "edge" is, IMO , are worthless.   Unfortunately such questions keep popiing up...FYI  here  is one  such discussion which might give some pestpective:
[quote]Number of Cylinder Edges

Date: 04/01/2002 at 12:16:20
From: David XXXX
Subject: Number of edges in a solid cylinder

Dear Sir:

My 8-year-old son was asked "how many edges are there on a solid
cylinder?" on a recent math examination.

His answer was "2" and it was marked as incorrect.

He truly believes in his answer and has asked for my assistance in
researching.

Can you assist?
Thank you.

Date: 04/01/2002 at 12:33:11
From: Doctor Peterson
Subject: Re: Number of edges in a solid cylinder

Hi, David.

It depends on how "edge" was defined in his class, which may not agree
with his intuitive definition.

Often, an edge is required to be straight, in which case a cylinder
has no edges. Unfortunately, elementary texts are not always very
careful about definitions, and they can ask questions like this that
are really worthless. The only definition of "edge" that would make
sense in this context would be the one your son is naturally using (a
boundary between smooth surfaces making up an object), which would
allow a cylinder to have two edges. Asking this question with the
other definition only invites confusion, so I wish they wouldn't ask
it. I'd like to hear how they did define the word.

You can read some other discussions of related issues here:

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/

 [/quote]

[quote="mathmom"]I would say that a cylinder has 3 faces (top, bottom, and curved side part) and 2 edges (the circles around the top and bottom faces).[/quote]

He said edges actually  :P .

These kinds of questions are truely worthless. I'm scared schools make children do. It all depends on what you definition of "edge" is. If it has to be straight than 0. If it just borders two faces than 2. I hope that helps.

As Gyan said, it all depends on the definition and/or the context of the discussion.  I think all you should assume Middle Schoolers will encounter in their curriculum is that an edge is a straight line between two vertices on a polygon or a polyhedron.

If the teacher had been discussing Mobius Strips, and said something like "A Mobius Strip is a cylinder with one edge.  How many edges does a standard cylinder have?"  Well.  There's plenty to complain about in this question, but the answer would probably be two.

If the teacher had been discussing regular polygons with an ever-increasing number of sides and said "So, a circle is a polygon with an infinite number of sides.  Since a circle is the top and bottom face of a cylinder, how many edges does a cylinder have?"  There's plenty to complain about in this question, too, but the answer would probably be infinity.

If the teacher had just given the standard definition.  "An edge is a straight line connecting two vertices in a polygon or a polyhedron.  So, how many edges does a cylinder have?"  There's a deliberately deceiving question, but I would say the answer is zero.

If the teacher was talking about elementary graph theory, and had shown how to represent a cylinder by a graph, then an edge is an entirely different thing, and we need to know how the cylinder had been represented on the graph.

Seems like most of us agree the question is difficult to answer with neither definition nor context.


calculus
derivative
function
calculus computations

Well first off, I am new here (and thankful I found this forum...hopefully it will help me survive this year of Calculus :) ).  I have a takehome test that is due tomorrow, and I need to finish this before I study...and I need a bit of help.  Here is the prompt:

Graph some multiples of e^x and their derivatives.  Look at the graphs of e^nx for different values of n and their derivatives.  What do these graphs show about the simplicity of the natural exponential function?

I know the derivative of e^x is e^x, but the n in there tossed me off a bit (the confusing part of this whole derivatives lesson was the dy\dx, etc because the letters tend to throw me off).  Is this just as simple as graphing e^x, etc and comparing, or is there a step that I'm missing because I was told by everyone that they were having problems with it (and if it really is as easy as I think it is I will be really happy :D ).

The Chain Rule states that the derivative of e^nx is equal to the derivative of nx * e^nx. So, for example, if n= 2, then the derivative of e^2x would be equal to 2 (derivative of 2x) * e^2x.

I hope that this helps you understand it.


modular arithmetic
number theory
greatest common divisor
relatively prime
number theory unsolved

Hello, I'm having trouble with what I assume is a rather easy proof involving the primes. The problem, which comes from Burton's "Elementary Number Theory", is as follows:

[quote]If $ p \not\equal{} 5$ is an odd prime, prove that either $ p^2 \minus{} 1$ or $ p^2 \plus{} 1$ is divisible by $ 10$.[/quote]

I first tried expressing an odd prime $ p$ as $ 2k \plus{} 1$ for some integer $ k$ and then substituting that into the above expressions, but I didn't get any results. Could someone give me some advice? Thanks!

[quote="caffeineboy"]Primes except $ 5$ and $ 2$ are $ \equiv1,3,7, 9 \pmod {10}$.[/quote]

I can certainly see how a solution can easily follow from this, but could you direct me to a proof of this fact?

Just a small note: if you want to show that something is divisble by $ n$, it is often less work to show that it is divisble by the prime powers dividing $ n$. Thus you could prove the following:
a) For any odd number $ p$, the numbers $ p^2\minus{}1$ and $ p^2\plus{}1$ are always even.
b) For any $ p$ not divisible by $ 5$, at least [and by the way at most] one of the numbers $ p^2\minus{}1$ and $ p^2\plus{}1$ is divisible by $ 5$.

[quote="caffeineboy"]$ 1,3,7,9$ are the only numbers less than 10 and relatively prime to it. Suppose we have a number $ 10x \plus{} a$ where 10 isn't relatively prime to $ a$. It is divisible by the GCD of $ a$ and 10. The exceptions are the prime divisors of 10, because when we factor out the GCD we merely get 2 or 5 times 1. The GCD here isn't more than 10 and so we can guarantee these to be the only exceptions.[/quote]

That makes sense. Thanks!

And one quick way to show part (b) of ZetaX's post is because the product $ (p^2 \minus{} 1)(p^2 \plus{} 1) \equal{} p^4 \minus{} 1$ is divisible by 5, by Fermat's Little Theorem.

Thank you for the advice, everyone. This is all very helpful.


There are $ 2n$ people at a party. Each person has an even number of friends at the party. Prove that there are two people who have an even number of common friends at the party.

prove the opposite:
[b]there are $ n$ people at a party,each pair of these people have an odd number of common friends,prove that $ n$ is odd...[/b]
(in this case,we will see that each person has an even number of friends...)
now the rest is some double counting...


Consider a polynomial
\[ p(x) \equal{} \sum_{k \equal{} 0}^{d} a_kx^k
\]of degree $ d \equal{} 4n \plus{} 2$ such that the coefficients are real and satisfy $ a_k \equal{} a_{d \minus{} k}$. Show that if all the roots of $ p$ are nonreal, then there exists a $ z\in\mathbb{C}$ such that $ p(z) \equal{} 0$ and $ |z| \equal{} 1$.

[hide="Hmm"]Divide through by $ x^{2n\plus{}1}$, then sub $ u\equal{}x\plus{}\frac{1}{x}$ to get a new polynomial $ q(u)$ with degree $ 2n\plus{}1$. Since the degree is odd, $ q$ must have at least 1 real root. Let this real root be $ r$. We have $ x\plus{}\frac{1}{x}\equal{}r \iff x^2\minus{}rx\plus{}1\equal{}0$. We know that the roots of this quadratic in $ x$, corresponding to two roots of $ f(x)$, must be nonreal. But since they are roots of a quadratic with real coefficients, they must be conjugates. Let $ z$ be a root of this polynomial. By Vieta's, $ z\cdot\bar{z}\equal{}|z^2|\equal{}1 \Rightarrow |z|\equal{}1$, and we are done.[/hide]


trigonometry
Putnam
inequalities unsolved
inequalities

find the minimum value of $ |{sinx\plus{}cosx\plus{}tanx\plus{}cotx\plus{}secx\plus{}cosecx}|$
for x is real number.


calculus
integration
logarithms
function
trigonometry
calculus computations

can you integrate :)

$ \int_0^\infty\;\frac{\ln\,(1\plus{}x^2)}{e^{2\pi\,x}\minus{}1}\;dx$

Let $ I = \int_0^\infty\;\frac {\ln\,(1 + x^2)}{e^{2\pi\,x} - 1}\;dx$. Then tranform our function to series

$ \frac {\ln (1 + x^2)}{e^{2\pi x} - 1} = \frac {e^{ - 2\pi x}\ln (1 + x^2)}{1 - e^{ - 2\pi x}} = e^{ - 2\pi x}\ln (1 + x^2)\sum_{k = 0}^{\infty}{e^{ - 2\pi x k}} =$
$ = \ln (1 + x^2)\sum_{k = 1}^{\infty}{e^{ - 2\pi x k}}$

and get

$ \textbf{(i)}\; I = \sum_{k = 1}^{\infty}{\int_{0}^{\infty}{\ln(1 + x^2)e^{ - 2\pi x k}dx}}$

Now put $ 2\pi x k = y$. This gives $ dx = \frac {1}{2\pi k}dy$ and

$ I = \frac {1}{2\pi}\sum_{k = 1}^{\infty}{\frac {1}{k}\int_{0}^{\infty}{\ln\left(1 + \left(\frac {y}{2\pi k}\right)^2\right)e^{ - y}dy}}$

Now represent $ \ln\left(1 + \left(\frac {y}{2\pi k}\right)^2\right)$ as series

$ \ln\left(1 + \left(\frac {y}{2\pi k}\right)^2\right) = \sum_{m = 1}^{\infty}{\frac {( - 1)^{m + 1}}{m}\left(\frac {y}{2\pi k}\right)^{2m}}$

and

$ \int_{0}^{\infty}{\frac {( - 1)^{m + 1}}{m}\left(\frac {y}{2\pi k}\right)^{2m}e^{ - y}dy} = \frac {( - 1)^{m + 1}}{m(2\pi k)^{2m}}\int_{0}^{\infty}{y^{2m}e^{ - y}dy} =$
$ = ( - 1)^{m + 1}\frac {(2m)!}{m(2\pi k)^{2m}}$

so

${ \int_{0}^{\infty}{\ln\left(1 + \left(\frac {y}{2\pi k}\right)^2\right)e^{ - y}dy}} = \sum_{m = 1}^{\infty}{( - 1)^{m + 1}\frac {(2m)!}{m(2\pi k)^{2m}}}$

finally

$ I = \frac {1}{2\pi}{\sum_{k = 1}^{\infty}{\frac {1}{k}\sum_{m = 1}^{\infty}{( - 1)^{m + 1}\frac {(2m)!}{m(2\pi k)^{2m}}}}}$

I hope we can evaluate this double sum, misan. And some more about it

If we integrate $ \textbf{(i)}$ by parts we get

$ I = \frac {1}{\pi}\sum_{k = 1}^{\infty}{\frac {1}{k}\int_{0}^{\infty}{\frac {x}{1 + x^2}e^{ - 2\pi k x}dx}}$

This is all I can say right now.

To evaluate first from double sum, maybe we should use generating function

$ f(x) = \sum_{m = 1}^{\infty}{( - 1)^{m + 1}\frac {(2m)!}{m}x^{2m}}$

and series will be equal to $ f\left(\frac {1}{2\pi k}\right)$.

We alsow know that $ \ln(1 + x^2) < x^2$, so

$ I < \frac {\zeta (3)}{4\pi ^{3}}$

when you represent $ \ln \left( 1 \plus{} \left( \frac {y}{2\pi k}\right)^{2} \right)$  as a series, I mean $ \ln (1 \plus{} z) \equal{} \sum ( \minus{} 1)^{n}\frac {z^{n}}{n}$ it's true  iff when
$ \minus{} 1 < z \leq 1$ thus you need  $ |\frac {y}{2\pi k}| < 1$ but interval of integration is $ [0,\infty]$  :( , hence there is big $ y$ such that this series representation of
$ \ln$  is not true.

also this series
[quote="Yuriy Solovyov"]...
$ f(x) \equal{} \sum_{m \equal{} 1}^{\infty}{( \minus{} 1)^{m \plus{} 1}\frac {(2m)!}{m}x^{2m}}$
...[/quote]
alsways will diverges, since each terms for fixed $ x$ , $ \frac {(2m)!}{m}x^{2m}$ tends to $ \infty$ as $ m \to \infty$ since $ m! \approx ... m^{m}....$
 :roll:

Yes, I see, thank you.

Try another way.

$ \ln(1 + x^2) = \ln(1 + x i) + \ln(1 - x i)$

$ \textbf{(ii)}\;$ $ \int_{0}^{\infty}{\ln(1 + x^2)e^{ - 2\pi k x}dx} = \int_{0}^{\infty}{\ln(1 + x i)e^{ - 2\pi k x}dx} + \int_{0}^{\infty}{\ln(1 - xi)e^{ - 2\pi k x}dx}$

$ \textbf{(iii)}\;$ $ \int_{0}^{\infty}{\ln(1 + ax)e^{ - 2\pi k x}dx} = \int_{0}^{\infty}{\ln (1 + ax)d( - \frac {1}{2\pi k}e^{ - 2\pi k x})} =$
$ = - \frac {1}{2\pi k}e^{ - 2\pi k x}\ln(1 + ax)|_{0}^{\infty} + \frac {a}{2\pi k}\int_{0}^{\infty}{\frac {1}{1 + ax}e^{ - 2\pi k x}dx}$

$ \textbf{(iv)}\;$ $ \int_{0}^{\infty}{\frac {1}{1 + ax}e^{ - 2\pi k x}dx} = \{t = 1 + ax;\;dx = \frac {1}{a}dt\} = \frac {1}{a}e^{\frac {2\pi k}{a}}\int_{1}^{\infty}{\frac {e^{ - \frac {2\pi k}{a}t}}{t}dt}$

Easy to see that $ - \frac {1}{2\pi k}e^{ - 2\pi k x}\ln(1 + ax)|_{0}^{\infty} = 0$ for all $ a$. Then

$ \textbf{(iii)} = \frac {1}{2\pi k}e^{\frac {2\pi k}{a}}\int_{1}^{\infty}{\frac {e^{ - \frac {2\pi k}{a}t}}{t}dt}$

Now we should put here $ a = i$ and $ a = - i$. Then use expression

$ \int_{1}^{\infty}{\frac {e^{ - ix\cdot t}}{t}dt} = - \textbf{Ci}(x) + i\cdot (\textbf{Si}(x) - \frac {\pi}{2})$

finally we get

$ \int_{0}^{\infty}{\ln(1 + x^2)e^{ - 2\pi k x}dx} = - \frac {1}{\pi k}\textbf{Ci}(2\pi k)$

Now we can back to our series. We get

$ \boxed{I = - \frac {1}{\pi}\sum_{k = 1}^{\infty}{\frac {\textbf{Ci}(2\pi k)}{k}}\approx 0.00876942058}$

P.S. $ \textbf{Ci}(x) = \gamma + \ln x + \int_{0}^{x}{\frac {\cos t - 1}{t}dt}$

We can alsow define

$ I(x) = - \frac {1}{\pi}\sum_{k = 1}^{\infty}{\frac {\textbf{Ci}(2\pi k x)}{k}}$

Then

$ I'(x) = - \frac {1}{\pi x}\sum_{k = 1}^{\infty}{\frac {\cos(2\pi k x)}{k}} = \frac {1}{\pi x}\ln (2\sin(\pi x))$

[quote="Yuriy Solovyov"]

$ \boxed{I \equal{} \minus{} \frac {1}{\pi}\sum_{k \equal{} 1}^{\infty}{\frac {\textbf{Ci}(2\pi k)}{k}}\approx 0.00876942058}$

[/quote]
So called Yuriy's constanta . :)


trigonometry
geometry
inequalities
inequalities proposed

Suppose $a,b,c \in \mathbb R^+$and \[\frac1{a^2+1}+\frac1{b^2+1}+\frac1{c^2+1}=2\]

Prove that $ab+ac+bc\leq \frac32$

[quote="RaMlaF"]If $a=b=c= \sqrt{2}$ then $ab+bc+ca = 6$  :S[/quote]

when $a=b=c=\sqrt{2}$   :arrow:   $\sum\frac{1}{a^2+1}=1\neq 2$  :P

Yes, I guess we all see that Shyong, but there was some confusion since Omid Hatami had a typo in his post.

[quote="nttu"]This problem is not hard !
  Expand the hypothesis , we get : $ a^2b^2+b^2c^2+c^2a^2+2a^2b^2c^2 =1 $
  Put $ ab =2cos C ; ac= 2cos B , bc =2cos A $
 The ineq becomes $ cos A+cos B +cos C \leq \frac{3}{2} $  ;)[/quote]

I guess you mean $ab=\cos C$ instead of $2\cos C$  :)

This problem has been in Iran exam in (1376,problem 4)
I think it worth repeating it cause it has a nice solution with Caushy.
Omid Why did you guys repeat this problem :D ?

Solution .$ab+bc+ca=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}\le \frac{3}{2}$ so its enough to prove

 $(a+b+c)^2 \le (a^2+1+b^2+1+c^2+1)$

Let $x=a^2+1,y=b^2+1,z=c^2+1$

The problem is equal to :
Let $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$ prove

$\sqrt{x+y+z} \ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{y-1}$

which follows from a simple but nice caushy

$\sqrt{x+y+z} *\sqrt{(\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z})=3-(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=1} \ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$

NTTU=I am not sure that you can use that subtitution with $Cos$

you are so cool lomos_lupin(Arman). :lol:  :rotfl:

Thanks but There is nothing to be cool about  :|  but its strange that problem was repeated in the official exam , i am more interested in the solutions of the contestants ,Omid Would you please give one of the nice solutions?

:(  ;)  this is not only problem that was repeated Arman there are many problems.
as we see in Geometry .I think the reason that OMID HATAMI didn't send Graph problems was this.
most of the problem (or maybe all of them) of graph problems was in Bondy's Book and ...

$So\ Easy\ to\ get\ gold\ this\ year.$

Will there be any more rounds for the Iranian Mathematical Olympiad 2005? I know that there used to be three rounds with 6 problems in the past (with the third round very hard). Did that change? If so, that's a pity. I don't like this "algebra exam", "geometry exam", ... etc

[quote="Arne"]Will there be any more rounds for the Iranian Mathematical Olympiad 2005?[/quote]
yes,there is.
Actually in iran there are four rounds.4th round is very hard and cute.

I am glad to hear that. Which round was this one?

[quote="nttu"][quote="lomos_lupin"]

NTTU=I am not sure that you can use that subtitution with $Cos$[/quote]

 Why aren't you sure ?  :whistling:[/quote]

Why,There exist Angles like $A,B,C$ such that $A+B+C=180$ and $ab=CosC,bc=CosA,ca=CosB$?
I think you reached this Existance from $a^2b^2+b^2c^2+c^2a^2+2a^2b^2c^2=1$ but how?

[quote="lomos_lupin"]
I think you reached this Existance from $a^2b^2+b^2c^2+c^2a^2+2a^2b^2c^2=1$ but how?[/quote]

Indeed, take $x = bc$, $y = ca$ and $z = ab$, then $x^2 + y^2 + z^2 + 2xyz = 1$ which implies that $x$, $y$ and $z$ are the cosines of the angles of a triangle.

If $a,b,c\ge 0 $ such that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=2$  prove that \[{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le \frac{3}{2} \le  \frac{1}{1+2a}+\frac{1}{1+2b}+\frac{1}{1+2c} }.\]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2938972#p2938972]here[/url]

The following inequality is also true.
If $a,b,c > 0 $ such that $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$  prove that \[abc(a-1)(b-1)(c-1)\le 8.\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=404216&start=80


anything from the closing minutes before you open the booklet to the end. strategies.

discussion started  :lol:

The closing minutes before I open the booklet: Wait until I'm allowed to open the booklet.

After I open the booklet: Solve the problems.

After the test is over: Compare answers with other people.

After the test is over, are we suppose to get the solutions manual? Or do we just get to check the answer key?

rub lucky penny, eat smarties, sharpen numerous pencils...

This might be the most useless topic ever.

While strategy is important to an extent, its not everything. A little time management goes a long way. Thats about all you can do to better your chances. spend the most time on questions you know how to do.

Solve the problems. Do some math. Get over it.

I dont believe we're supposed to get solutions manual.

oh ok, but do they let us see the answers?

[quote="paladin8"]I hold my booklet up to the light so I can read the questions before time starts; that way I get a head start and can easily get 150.[/quote]

Oh man, that is pretty slick, thanks for sharing. I've always wondered how people get 150...

When i first saw this thread, I thought it was referring to non-superstitious things.

E.G. I have a lunchtime pre-countdown diet.

Nothing for AMC though.

I thought they send a solution manual with each pack of 10 amc exam booklets (I saw one after the test)

[quote="blootzboy"]i alredy wrote AMC 12 A am i still allowed to write the AMC 12 B? or will they like disqualify me to write AIME or somethin cuz they see i wrote both AMCs[/quote]

Yup, it's perfectly fine to do both.

[quote="Pakman2012"]I dont believe we're supposed to get solutions manual.[/quote]

As soon as your proctor has put all the answer sheets in the envelope to send or whatever, they can give you the solutions manuals (they send you one for every test you get).

I'm going to sleep in and skip school. I'll show up right before we start taking the exam. Well, that's what I did on the A anyway.

[quote="blootzboy"]k thanks and do they count the higher score, average score, or lower score?[/quote]
higher

[quote="Pakman2012"]I'm really superstitious. so before every important test i take, I kill a hobo with a hammer. you know, the usual.[/quote]

man not the hobos

[quote="Pakman2012"]I'm really superstitious. so before every important test i take, I kill a hobo with a hammer. you know, the usual.

i also sleep in, and don't worry bout the rest of the day[/quote]

If you kill a hobo, a travelling band of stray cats will come to sit upon your doorstep and you will get a group of the wandering men sitting in front of your house and sniffing conspicuously. Then, dust storms will ruin your crops.


inequalities
Cauchy Inequality
inequalities unsolved

Prove that for a,b,c>0, a+b+c=3 We have
[code] \sqrt{ \frac{ a^{2}b }{a+c}}+ \sqrt{ \frac{ b^{2}c }{b+a}}+ \sqrt{ \frac{ c^{2}a }{c+b}} \leq frac{3 \sqrt{3}}{4} \sqrt{ \frac{(a+b)(b+c)(c+a)}{a+b+c}}[/code]

[quote="ThangTongHopK41"]Thanks you, can you show me the laxtext code of this problem. I don't know typing laxtex[/quote]

Actually, you had it right, but you forgot to include dollar signs at the beginning and end of it.

In fact, all I did was copy your code, and add one dollar sign at the beginning, another at the end, and then insert a backslash that you had missed.

Can you post the proof ò thí problem. It's hard for me

[quote="PTynan89"]$a, b, c > 0$; $a+b+c = 3$.

Prove: $\sqrt{ \frac{ a^{2}b }{a+c}}+\sqrt{ \frac{ b^{2}c }{b+a}}+\sqrt{ \frac{ c^{2}a }{c+b}}\leq \frac{3 \sqrt{3}}{4}\sqrt{ \frac{(a+b)(b+c)(c+a)}{a+b+c}}$[/quote]
This is a problem in Mathnfriend contest, so you are not allowed to discuss it here. :)

Isn't this already homogenized? Why do you need the condition on a,b,c?

Oh yes, I change the start Inequality to a homegeneous one, this inequality, we don't need a+b+c=3

That is not humanly possible :maybe:
Therefore you used mathematica? What kind of software can do that!? it could certainly solve many olympiad inequalities in seconds...


What is the number of ways to put 6 rooks on a 8*8 board such that no two can attack each other?

[hide="almost certainly the wrong answer"]

First rook: $ 64$ ways. Second rook: $ 49$. And so on. Divide by $ 6!$ to account for overcounting with permutations. So $ \frac{8^2 \cdot 7^2 \cdot 6^2 \cdot 5^2 \cdot 4^2 \cdot 3^2 \cdot}{6!} \equal{} \boxed{188160}$.[/hide]

You could count the number of ways by using the rook polynomial. Or consider in how many columns the rooks must be placed in, in how many ways can we select the number of such columns and in how many ways can we position the rooks within the selected columns.

[hide="Explicit formula for the number of k non-attacking rooks on a m x n board."]For a $ m\times n$ board, and $ k\leq\min\{m,n\}$ the number of ways in which $ k$ rooks can be arranged so that they don't attack each other is $ \frac {m!n!}{k!(m \minus{} k)!(n \minus{} k)!}$, under the assumption that all the rooks do not differ from each other (otherwise we multiply by $ k!$). So with $ k \equal{} 6, m \equal{} n \equal{} 8$ the result is 564 480.[/hide]
[hide="A simpler approach."]
Rooks must be placed on six different columns. We can choose 6 out of 8 columns in $ {8 \choose 6}$ ways. We can place the first rook in one of the selected columns in 8 ways, the second rook in 7 ways, ..., and the last rook in 3 ways. So the total number of such arrangements is $ {8 \choose 6}8\cdot7\cdot6\cdot5\cdot4\cdot3 \equal{} 564480$. That is, under the assumption that all the rooks are considered the same. If they differ, just multiply by $ 6!$.
[/hide]

leoxnlin, your solution works fine (and it's not hard to see that it's a specialization of the solutions goranm gave), but you botched the arithmetic.  :)

Oh. wow. Thanks.  :oops:

I used a calculator.

This is kind of sad.


Let $ a, b$ two positive real numbers. Prove that $ a^{a}+b^{b}\geq  a^{b}+b^{a}$ .

[quote="K81o7"][quote="junggi"]$ (a^{a-b}-1)\ge b^{b}(b^{a-b}-1)$
since $ b,a-b\ge0$
[/quote]

Try $ b$ to be some fairly large number, $ a$ to be just slightly larger, and you'll find this isn't true.

Even if you meant to say $ a^{a-b}-1 \ge b^{a-b}-1$, it wouldn't suffice to show that: suppose both expressions are negative? (ie if $ b \le a < 1$)[/quote]
woops typo
i did mean the latter one you mentioned.
and yea good point
hm


Prove that

$\sum\frac{p^{2}a^{2}}{l_{a}^{4}}\geq 12$,

where $a$, $b$, $c$ are the sidelengths of a triangle $ABC$, and where $p=\frac{a+b+c}{2}$ is its semiperimeter, and $l_{a}$, $l_{b}$, $l_{c}$ are its angle bisectors.

Let's use a well-known formula la2 = bc (1- a2/(b+c)2). Applying bc  \leq (b+c)2/4 we get la2 \leq p(p-a)

So it's sufficient to prove that  \sum p 2 a 2 /(p(p-a))2 \geq 12.

\sum (a/(p-a))2 \geq 1/3 (\sum a/(p-a))2 =
1/3 (\sum (p-b+p-c)/(p-a))2 = 1/3 (\sum ((p-b)/(p-a) +(p-a)/(p-b)))2 \geq 1/3 6 2 = 12


analytic geometry
trigonometry
advanced fields
advanced fields theorems

What would a coordinate system be for the n-dimensional ball?

For n=2, we have polar coordinates $(r,\theta) \in [0,\infty) \times [0,2\pi)$.
For n=3, we have spherical coordinates $(r,\theta,\phi) \in [0,\infty) \times [0,2\pi) \times [0,\pi)$.

What would be the general case?

There are plenty of ways to do it, but none are standard.

For example, here are two different "polar" coordinate systems for $\mathbb{R}^{4}$:

$(x,y,z,w)=(r\cos\theta_{1}\cos\theta_{2},r\cos\theta_{1}\sin\theta_{2}, r\sin\theta_{1}\cos\theta_{3},r\sin\theta_{1}\sin\theta_{3})$, with $\theta_{1}$ restricted to $[0,\frac{\pi}2]$

$(x,y,z,w)=(r\cos\theta_{1},r\sin\theta_{1}\cos\theta_{2}, r\sin\theta_{1}\sin\theta_{2}\cos\theta_{3},r\sin\theta_{1}\sin\theta_{2}\sin\theta_{3})$, with $\theta_{1},\theta_{2}$ restricted to $[0,\pi]$.

The number of possibilities grows out of control quickly, and none of them is intrinsically better than the others.


1.  Which of the following is an ellipsis?

A) x^2 + 2y^2 = 5

B) . . .

============================================

2.  Which of the following is a hyperbole?

A) "I'm so hungry I could eat a horse."

B) x^2 - y^2 = 4

============================================

3.  Which letter choice contains ellipses?

A)  x^2 + 2y^2 = 1/3  &
    3x^2 + 2x^2 = 1

B)  . . . & . . .

C)  both A) and B) from above

=============================================

4.  An associate of mine asked me, "Do you like Pat?"
    I answered "No."
What should my associate (logically) go away thinking?

A)  I definitely harbor some type of ill will toward Pat.

B)  I do not necessarily have any animosity toward Pat;
I may be indifferent toward Pat.

=============================================

5.  Why do mathematicians grow their plants in boxes? **

A) They want to wreck tangles growing out of the plant.

B) They want to get square roots.

==============================================
/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/

** Part of this one was written by some other author.

[hide="Problem 1"]B, because the other is an ellipse or something (I have really bad spelling.)[/hide]

[hide="Problem 2"]B, because the other is a hyperbola.[/hide]

[hide="Problem 3"]C, because the plural from is the same.[/hide]

[hide="Problem 4"]B, I'm pretty sure.[/hide]

[hide="Problem 5"]B, I'm pretty sure.[/hide]

One of your lettered responses is incorrect.

Please find out for yourself and correct it,

because that will be more educational for

you, IMO.

The correct answer to the second
question of the five questions is "A",
BUT the other portion of that answer
given by you "because the other is
a hyperbola" IS correct.  Your parts
were/are inconsistent with each other.

Wow, I'm a bit of an idiot. Did you just make all of these up by yourself, except for # 5?

In the second question, I looked up "I'm so hungry I could eat a horse." because I was looking for the exact wording of an example of a hyperbole.

And other than the B) answer and the fifth question that goes with it,
I made up the rest.


articles
geometry
geometric transformation
dilation
puzzles

A [u]New York Times[/u] headline for January 1, 2200 reads as follows:

100-Year-Old Queen Comes to Visit U.S. Today

However, the article later states that the queen was born on November 28, 2175.

How is this possible?

EDIT:  Of course, you may ask yes/no questions.

Yes, I know the answer.
In fact, I made up the problem on my own.
A hint:
[hide]Why would the time be 200 years in the future?[/hide]

[hide="Is"]US Today some company, corporation, or TV Program?[/hide]

Does it involve a time machine? [b]No.[/b]

Is U.S. Today some company, corporation or TV program? [b]No.  U.S. stands for United States, and "today" means the present time (Jan. 1, 2200).[/b]

Ihatepie asks if it involves a time machine.  So I'll ask if it involves time dilation.

Does any altering or abnormal traveling of time happen at all?

Are years still considered the same length of time in the future?

The article had a typographical error, which they had to correct the next morning?

You never know.

Does it involve time dilation?  [b]I'm not sure what you mean by time dilation, but I don't think so.[/b]

Is the queen alive? [b]Yes.[/b]

Is the queen human? [hide][b]If you define "human" as a member of the species [i]Homo sapiens[/i], then yes.  If you define human as someone born on planet Earth, then no. (this should be a hint)[/b][/hide]

Are all the times in A.D. and not B.C.? [b]Yes.[/b]

Are years still considered the same length of time in the future? [b]Yes.[/b]

Does any altering or abnormal traveling of time happen at all? [b]No.[/b]

The article had a typographical error, which they had to correct the next morning?  [b]No.[/b]

Does the queen live on Mercury, where years are about 1/4 as long as on Earth?

Yes. Ravi B is correct.
The queen was born to human colonists on the planet Mercury on Nov. 28, 2175.  When she came to visit the United States (and Earth in general) on Jan. 1, 2200, she was 100 Mercurian years old, but only 24 Earth years old.

Good job, Ravi B!


analytic geometry
combinatorics unsolved
combinatorics

A rational point in tne plane is one with rational coordinates .Divide the rational points into three disjoint sets so that every line contains points from at most two sets ,but every circle with rational center contains points from all three sets

Maybe you mean every disk? :? Since it is possible that circle with rational center doesn't contain any rational points.

I solved this one for WOP, remember.

Pierre.

I remember. And I really hope someone will write a more nice solution here ;)

can you post your solution Myth;i find it too difficult for me


[hide="shortcut if you have a calculator"]Let $ n$ be the number of envelopes there are, then the answer is $ \frac {n!}{e}$ rounded to the nearest integer[/hide]

[hide="non-calc solution"]We use complementary counting to find the number of ways the secretary can put the letters in the envelopes such that AT LEAST one letter is in the correct envelope.

Casework:

i) All four letters are in the correct envelope: There is only $ 1$ way for this to happen.

ii) It is impossible for $ 3$ of the letters to be in the correct envelope but the 4th to be in an incorrect envelope (see why?)

iii) Two of the four letters are in the correct envelope: We must choose 2 of the 4 letters to be placed correctly and with the remaining two letters, we only have $ 1$ way of putting them into incorrect envelopes. So the number of ways for this to happen is $ \binom42\times{1} \equal{} 6$

iv) One letter is put into the correct envelope: We choose one of the 4 letters to be placed in its correct envelope and with the remaining three there are $ 2$ ways for those to all be placed in incorrect envelopes. Thus, there are $ \binom41\times{2} \equal{} 8$ ways for this to happen.

Adding the number of ways for all 4 of these cases to happen gives $ 15$ ways for at least one of the letters to be placed correctly. Thus the answer is $ 4! \minus{} 15 \equal{} \boxed{9}$ [/hide]

[quote="sunehra"]In how many ways can a really mixed up secretary manage to get 4 letters in 4 envelopes, one letter per envelope, so that no letter is in the correct envelope?[/quote]
This is an easy question of derangements.
See:
http://en.wikipedia.org/wiki/Derangements


Hi all. It seems (Former) Yugoslavia forum will need a mod. I've sent Valentin a PM on Wednesday, but he seems not to have enough time (or he simply forgot...) and I've recieved no response, so I decided I'll post here.

Former Yugoslavia subforum got somewhat more active in the previous month or two (since all the people are preparing for various international competitions) and, with that came Pointless Post Almighty. (first post on http://www.mathlinks.ro/viewtopic.php?t=210598 - another guy decided to make use of that topic, so he started a discussion on number theory). On http://www.mathlinks.ro/viewtopic.php?t=208275 (it's in English) we (some of the more active subforum members) have previously voted on the issue and there seems to be a general consensus that hsiljak should become a moderator, even though I'm sure most people would do a good job (it's just that hsiljak is always around and willing to help).

Thanks a lot to anybody willing to assign us a mod.

Isn't hsiljak already the moderator?
Look at the http://www.mathlinks.ro/viewtopic.php?t=208275


The product of two consecutive odd positive integers is 23 more than their sum. What is the product of these two integers?

Call the smaller odd integer $ x$.  The larger one is obviously $ x \plus{} 2$.  Using our given information, we have
\[ (x)(x \plus{} 2) \equal{} x \plus{} x \plus{} 2 \plus{} 23\implies x^2 \plus{} 2x \equal{} 2x \plus{} 2 \plus{} 23\]
Moving everything over to the left side, we get
\[ x^2 \minus{} 25 \equal{} 0\]
Factoring:
\[ (x \plus{} 5)(x \minus{} 5) \equal{} 0\implies x\equal{}\pm5\]
Since $ x$ has to be positive, $ x \equal{} 5$, and $ x \plus{} 2 \equal{} 7$.  So our answer is $ 5\cdot7 \equal{} \boxed{35}$

Alternatively, note that the integers' product would have to be at least 23, so the smallest possible pair is (5,7), which works. Thus, the answer is 35.


function
binomial coefficients
algebra
system of equations
real analysis
real analysis unsolved

How to show that
\[ \sum_{k \equal{} 1}^nk^m \equal{} \frac {n^{m \minus{} 1}}{m \plus{} 1} \plus{} \frac {1}{2}\binom{m}{1}B_2n^{m \minus{} 1} \plus{} \frac {1}{4}\binom{m}{3}B_4n^{m \minus{} 3} \plus{} \frac {1}{6}\binom{m}{5}B_6n^{m \minus{} 5} \plus{} \ldots
\]
where $ B_k$ are the Bernoulli numbers and the sum contains the positive powers of $ n$?

Difference it. You get a system of equations involving the coefficients and binomial coefficients, which the Bernoulli numbers also satisfy.

It's not too hard to guess the form $ \sum_{k\equal{}1}^n k^m\equal{}\frac{n^{m\plus{}1}}{m\plus{}1}\left(1\plus{}\binom{m\plus{}1}{1}a_1n^{\minus{}1}\plus{}\binom{m\plus{}1}{2}a_2n^{\minus{}2}\plus{}\cdots \plus{}\binom{m\plus{}1}{m}a_mn^{\minus{}m}\right)$ by calculating small $ m$ cases, and we can difference this to get a recursion for $ a_i$. From there, you just have to consider the exponential generating function for the $ a_i$, which brings out the Bernoulli numbers.


I have seen this problem somewhere and Im just wondering how to solve it.  Suppose that you have a number, lets say 3^1995, you add the digits, when you get the sum, add again its digits.  Suppose that you keep on adding until one digit is left, what is that digit?  Is there any good technique or shortcut to solve this type of problem?

Well, digit that you end up with will be the remainder when the original number is divided by 9, or if the remainder is 0 you will end up with 9.
9 l 3^n where n>=2 so the digit that you end up with is 9.

Yes. The sum of the digits is always (mod 9) of the original!


I was just wondering, for anyone who went to PROYMS last year, does the reccommendation have to be filled out by our current math teacher, or can it be filled out by any other math teacher we have had. I am a sophmore, and I am applying to PROYMS '05 (according to the director, application forms will be uploaded on Monday), but as far as the reccommendation goes, I feel that my math teacher freshman year is a much more competent teacher (knowledge-wise), and knows me better than my current teacher, who absolutely discourages any other math-related activites besides the boring classroom exercises. I was wondering, if it would be okay to ask my math teacher from last year to write me a reccommendation or not.

Thanks in advance!

I can't speak specifically to PROMYS, but I've never ever heard of a program that required you get a recommendation from your current teacher.  I have heard of a requirement that you get a recommendation from someone who was in fact your teacher (as opposed to someone who knows you through some other capacity), but never that it need be your teacher at the time you apply.

This is what I was thinking as well, because on the first page of last year's application itself, it says "Please have [b]one[/b] of your mathematics teacher write a letter of reccommendation for you on the enclosed form". It does not specifially state that it has to be your [u]current[/u] math teacher, but I was just making sure.

If that's the wording, then any math teacher (past or present) will do, although you probably don't want to go all the way back to 4th grade :)

Hehe, if only I could :D...nah, I think I will stick with freshman year :)

Thanks for the help!


Find all continuos function $f: (0,+\infty)\rightarrow R$ such that :
$\frac{f(x)+f(y)+f(z)}{3}=f(\sqrt{\frac{x^2+y^2+z^2+xy+yz+zx}{6}})\forall x,y,z>0$
 :?


linear algebra
matrix
integration
algebra
polynomial
linear algebra unsolved

Let $A = (a_{ij}) \in M_n(R)$, $a_{ij} = \frac{1}{i + j}$. Prove or disprove $det(A) > 0$
What about $a_{ij} = \frac{1}{x_i + x_j}$, where $x_i, x_j > 0$
Can be found more general cases ?

notice that $\frac{1}{i+j}=\int_{0}^{1} x^{i+j-1}dx$.
So if you call $A_x$ the matrix with coefficients $x^{i+j-1}$, $A=\int_{0}^{1} A_x dx$.

But $YA_xY^t=\frac{1}{x} \sum_{i,j} y_iy_jx^{i+j}=\frac{1}{x} (\sum_i y_ix^i)^2$.
Hence, if $YA_xY^t=0$ then $x$ is a zero of the polynomial $P(X)=y_1X+y_2X^2+..+y_nX^n$.
This is why $YAY^t=\int_{0}^{1} YA_xY^t dx$ cannot be equal to zero unless $Y=0$. So $A$ is a positive definite matrix.

(The same proof works with the $\frac{1}{x_i+x_j}$)

When $\displaystyle a_{ij}=\frac{1}{a_i+b_j}$ we have $\displaystyle D=\frac{V(a_1,a_2,...,a_n)V(b_1,b_2,...,b_n)}{\prod^n_{i,j=1}(a_i+b_j)}$ where $\ V(...)$ is the Vandermonde determinant. D is called Cauchy's determinant.

Yes, of course, all these are very nice isn't it ? :)

Can we say similar things about :
$a_{ij} = \frac{1}{i^2 + i \cdot j + j^2}$
$a_{ij} = \frac{1}{x_i^2 + x_i \cdot x_j + x_j^2}$
$a_{ij} = \frac{x_i - x_j}{x_i^k - x_j^k}, k \in N, k \geq 2$
and more for matrices $B = (a_{ij}^p), p \in N, p \geq 2$ ?

[I hope you don't find this boring, i found them really interesting  ]

The problem is discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=435219#p435219]here[/url]


calculus
integration
trigonometry
logarithms
calculus computations

Evaluate the following definite integrals.

(1) $ \int_0^{\frac{\pi}{2}} x\cos x\ dx$.

(2) $ \int_1^2 \frac{1}{x(x\plus{}1)}\ dx$.

(3) $ \int_0^1 \frac{1}{1\plus{}e^x}\ dx$.

(4) $ \int_0^{\frac{\pi}{4}} \cos 2x \cos 3x \ dx$.

[hide="Number 3"]$ I \equal{} \int_0^1 \dfrac{1}{1 \plus{} e^x}\; dx \equal{} \int_2^{1 \plus{} e} \dfrac{du}{u(u \minus{} 1)} \equal{} \int_2^{1 \plus{} e} \left( \dfrac{1}{u \minus{} 1} \minus{} \dfrac{1}{u} \right) \; du \;$ with ($ u \equal{} 1 \plus{} e^x$)

$ I \equal{} \left[ \log_{e} \left(u \minus{} 1\right) \minus{} \log_{e} \left(u\right) \right]_2^{1 \plus{} e} \equal{} \left(1 \minus{} \log_{e} \left( 1 \plus{} e \right) \right) \plus{} \log_{e} 2 \equal{} 1 \plus{} \log_{e} \dfrac{2}{1 \plus{} e}$[/hide]

[hide="Number 3"]$ \int_0^1 \frac{1}{1\plus{}e^x}\,dx \equal{} \int_0^1\frac{1}{e^x(1 \plus{} e^{\minus{}x})}\,dx \equal{} \minus{} \int_0^1\frac{\minus{}e^{\minus{}x}}{1 \plus{} e^{\minus{}x}}\,dx$

$ \equal{} \minus{} [\ln(1 \plus{} e^{\minus{}x})]_0^1 \equal{} \ln 2 \minus{} \ln(1 \plus{} 1/e) \equal{} \ln\frac{2e}{e\plus{}1}$[/hide]

3[hide]

$ \int_0^1\frac {dx}{1 \plus{} e^x}$

$ \equal{} \int_0^1\frac {1 \plus{} e^x \minus{} e^x}{1 \plus{} e^x}\,dx$

$ \equal{} \int_0^1\,dx \minus{} \int_0^1\frac {e^x}{1 \plus{} e^x}\,dx$

$ \equal{} 1 \minus{} \ln(e^x \plus{} 1)\bigg|_0^1$

$ \equal{} 1 \plus{} \ln(2) \minus{} \ln(e \plus{} 1)$

$ \equal{} 1 \plus{} \ln\left(\frac {2}{e \plus{} 1}\right)$

[/hide]

4[hide]

$ \int\cos(2x)\cos(3x)\,dx$

$ \equal{} \frac {1}{2}\int\bigg[\cos(x) \plus{} \cos(5x)\bigg]$

$ \equal{} \frac {1}{2}\bigg[\sin(x) \plus{} \frac {1}{5}\sin(5x)\bigg]$

$ \Rightarrow\int_0^{\frac {\pi}{4}}\cos(2x)\cos(3x)/,dx$

$ \equal{} \frac {1}{2}\bigg[\sin\left(\frac {\pi}{4}\right) \plus{} \frac {1}{5}\sin\left(\frac {5\pi}{4}\right)\bigg]$

$ \equal{} \frac {\sqrt {2}}{5}$[/hide]


ratio
geometry
geometric transformation
homothety
circumcircle
rotation
parallelogram

Let $A,B,C$ be points on a fixed circle with $B,C$ fixed and $A$ variable. Points $D$
and $E$ are on segments $BA$ and $CA$, respectively, so that $\overline {BD} = m$and $\overline {CE} = n$ where $m$ and $n$ are constants. Points $P$ and $Q$are on $BC$ and $DE$, respectively, so that , $\frac{CP}{PC}=\frac{DQ}{QE}=k$
also a constant. Prove that the length of $PQ$ is a constant.

[quote="Mathx"]... so that $\frac{CP}{PC}=\frac{DQ}{QE}=k$ also a constant...[/quote]
This is a typo, it should be $\frac{BP}{PC}=\frac{DQ}{QE}=k$ is a constant.

Let (B), (C) be circles centered at B, C and with radii m, n. The locus of points R with a constant ratio of distances $\frac{RB}{RC} = \frac m n$ is a circle (Z) centered on the line BC and passing through the internal and external homothety centers G, H of the circles (B), (C). Let the locus circle (Z) meet the circumcircle of the triangle $\triangle ABC$ at a point R, so that $\frac{RB}{RC} = \frac m n = \frac{BD}{CE}$. Since the angles $\angle RCE \equiv \angle RCA = \angle RBA \equiv \angle RBD$ spanning the same arc RA of the circle (O) are equal, the triangles $\triangle RBD \sim \triangle RCE$ are similar by SAS. Therefore, the angles $\angle BRD = \angle CRE$ are equal and consequently, the angles $\angle CRB = \angle ERD$ are equal. Thus we conclude that the triangles $\triangle BCR \sim \triangle DER$ are similar by SAS. As a result, the angles $\angle EAD \equiv \angle CAB = \angle CRB \equiv \angle ERD$ are equal, which means that the quadrilateral DEAR is always cyclic, regardless of the (allowed) position of the point A on the arc BC.  (The other intersection R of the circumcircles (O), (S) of the triangles $\triangle ABC, \triangle ADE$ different from the point A is the rotation center of the spirally similar segments BPC, DQE and we see that it is fixed.) Thus the triangles $\triangle RBD \sim \triangle RCE \sim \triangle RPQ$ formed by the fixed rotation center R and 2 corresponding points B, D or C, E or P, Q of the spirally similar segments BPC, DQE are spirally similar. This implies $\frac{PQ}{PR} = \frac{BD}{BR}$ and $PQ = \frac{BD \cdot PR}{BR}$, where the BD = m is fixed by the problem condition and BR, PR are also fixed, because the points B, P, R are fixed.

Formidably ! As usual, you gave again a very nice  ([i]even syntetical[/i]: must keep silent those who say that you give only metrical solutions) answer.
[i]Yetti[/i], you have a small mistake against the end:

$\triangle RBD\sim \triangle RPQ\Longrightarrow \frac{RB}{RP}=\frac{BD}{PQ}=\frac{RD}{RQ}\Longrightarrow \overline {\underline {\left| QP=\frac{RP}{RB}\cdot BD\right| }}$.

It is excusably by your speed of the writting !

Thanks for pointing it out, I could not resist to fix it. BTW, the above solution could be used as a slightly different proof of the proposition (a) in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56605]Nice (Circle passes through a constant point)[/url].

here is my solution:

   Draw lines parallel to both BD and CE at points P and Q.  We have a fixed

  parallelogram  in the intersection, since BD and CE are fixed and BC and DE

  are  divided in the same ratio.  PQ is a diagonal in a fixed parallelogram.

  Thank you.

   M.T.


Hello everybody!
Please can someone put or send (link, file [pdf, etc]) about the simple properties of the matrices... like addition, multiplication, division etc principles.

Thank you anyway!

davron latipov
Khujand, Tajikistan

ummm, maybe this will help: http://www.cut-the-knot.com

http://en.wikipedia.org/wiki/Matrix_%28mathematics%29

Wikipedia is always a good bet.

Thank you all very much!!

Davron Latipov


symmetry
analytic geometry
calculus
calculus computations

Prove that every cubic curve has rotational symmetry about its point of inflection.

I'm thinking of translating the point of inflection to the origin and then rotating through 180 degrees.

y = f(x) = ax^3 + bx^2 + cx + d.
f'(x) = 3ax^2 + 2bx + c
f''(x) = 6ax + 2b = 0 gives x = -b/3a = X.
f'''(x) = 6a is not equal to zero.
Substituting, we find the y coordinate of the point of inflection, Y.
Then f(x) ---> f(x + X) - Y translates the point of inflection to the origin, giving y = ax^3 + (c - b^2/3a)x which is clearly symmetric about the origin since y(-x0 = - y(x).


geometry
incenter
trigonometry
geometry proposed

Let $ ABC$ be a $ A$-right triangle. Denote the midpoint $ M$ of $ [AC]$ and $ D\in BC$ for which

$ AD\perp BC$ . The $ A$-exincircle and the $ B$-exincircle touch the sideline $ AC$ in $ E$ , $ F$ respectively.

Let $ G$ be the second intersection  between $ DF$ and the circle with the diameter $ [AC]$ .

Prove that the quadrilateral $ DMGE$ is cyclically and $ C$ is incenter of the triangle $ DGE$ .

$ FD.FG\equal{}FA.FC\equal{}(p\minus{}c)(p\minus{}a)\equal{}(b/2\minus{}(p\minus{}a)).((p\minus{}b)\plus{}(p\minus{}a))\equal{}FM.FE$ thus $ DMGE$ is cyclic.
$ \angle GEC\equal{}\angle MDG\equal{}\angle MGD\equal{}\angle CED$
$ \angle DGE\equal{}\angle DME\equal{}2\angle DAM\equal{}2\angle DGC$
$ \rightarrow$ QED

[b][u]Another similar problem[/u].[/b]

Let $ ABC$ be a $ A$-right triangle. Denote the midpoint $ M$ of $ [AB]$ and $ D\in BC$ for which

$ AD\perp BC$ . The incircle and the $ B$-exincircle touch the sideline $ AB$ in $ F$ , $ E$ respectively.

Let $ G$ be the second intersection  between $ DF$ and the circle with the diameter $ [AB]$ .

Prove that the quadrilateral $ DMGE$ is cyclically and $ A$ is incenter of the triangle $ DGE$ .

[b][u]Remark[/u].[/b] Observe that $ (B,F,A,E)$ is a harmonical division because $ \frac {FA}{FB}\equal{}\frac {EA}{EB}\Longleftrightarrow$

$ \frac {p\minus{}a}{p\minus{}b}\equal{}\frac {p\minus{}c}{p}\ \Longleftrightarrow\ p(p\minus{}a)\equal{}(p\minus{}b)(p\minus{}c)\ \Longleftrightarrow\ \tan\frac A2\equal{}1\ \Longleftrightarrow\ A\equal{}90^{\circ}$ .

So the proof is simply, is a particular aplication of the properties of a harmonical division.

For example, $ FM\cdot FE\equal{}FA\cdot FB$ a.s.o. It is analogously and at the previous problem !


prove that the sequence (1+1/n)^{n+1} is decreasing (maybe using bernoulli's inequality)


probability
geometry
3D geometry
octahedron
tetrahedron

The labels $1, 2, ... , 8$ are randomly placed on the faces of an octahedron (one per face). Find the probability that no two adjacent faces (sharing an edge) have adjacent numbers, where $1$ and $8$ are also considered adjacent.

Nevermind, for some reason I thought this topic was in the Intermediate Forum...

Let us consider the equivalent problem of numbering the vertices of a cube.
We put the $1$ at some vertex.  The $2$ can then go anywhere not connected by an edge to the $1$.  Then the $3$ cannot be connected by an edge to the $2$, and so on.  When we are done, we will have completed a cycle by walking only on diagonals of the cube.  The graph of the vertices of a cube connected by its diagonals looks like this:
[img]http://www.geocities.com/quasidave04/graph.JPG[/img]
One can think of this as two tetrahedra with corresponding vertices connected.
So what we would like to do is count how many hamiltonian cycles this graph has.  Then, if we multiply that number by 16 ways to put $1$ through $8$ in each cycle, we will have the total number of numberings that satisfy the requirements of the problem.

Well, take a look at the edges of the graph which I have colored purple (the ones that connect corresponding vertices on opposite tetrahedra).  Clearly, each cycle we find will either go through exactly 2 or 4 of these.
If 2:
We need to choose 2 of the purple edges, and then there are 2 ways to connect them on each tetrahedron.  This gives $c(4,2)\cdot 2\cdot2 = 24$ cycles.
If 4:
Now, as we walk our cycle, we need to alternate between blue and purple edges.  We just need to find how many cycles a single tetrahedron has, and multiply that by 2 to account for whether we start on the left tetrahedron or the right one.  We get 6.

Thus, our probability is
\[\frac{16(24 + 6)}{8!} = \frac{2(30)}{7!}=\frac{4\cdot3\cdot5}{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}=\frac{1}{84}\]


Solve this cryptarithm, where each letter denotes a different decimal digit from 0 to 9. None of O and L can be zero.

(OUTLAY)/(LAYOUT) =  5/8

[hide]O=3
U=8
T=4
L=6
A=1
Y=5

The equation becomes 384615/615384=5/8, which is true.

How I came up with this:
Because 13 (5+8) evenly divides 999999,  simply rewrite 5/8 as $ \frac {\frac {999999}{13}\cdot5}{\frac {999999}{13}\cdot8}$ and simplify.[/hide]

Or like this: LAY=a,OUT=b
$ 8000a\plus{}8b\equal{}5000b\plus{}5a$
$ 7995a\equal{}4992b$
$ 205a\equal{}128b$

This has more than one solution with different digits :wink:


What is the standard way to write a Halmos square right-aligned on the same line as the last line of a proof? For example,

..........
...and so by Lemma 1, the proof follows.                          [black square here]

I like to use: \hfill $\blacksquare$

for more info on hfill look in the [tex]\LaTeX[/tex] guide under spacing.

Why not use the layout provided by AMS Theorem package for this purpose to provide nice layouts of proofs? Include in your document [code]\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsthm}[/code] then use [code]\begin{proof}
and so by Lemma 1, the proof follows.
\end{proof}[/code] and an open square will automatically be placed in the correct place.

You can also use [code]\begin{proof}[Proof of Main Theorem][/code] for example.

Prefer a black square to an open square? Then use  [code]\renewcommand{\qedsymbol}{\blacksquare}[/code] earlier on

Thanks for the suggestion. I usually define my own proof environments though so I can customize certain things. :)

I just use \begin{proof} ... \end{proof} with the amsthm package.


Compute the superiour limit of the sequence $\sqrt[n]{|a^n-b^n|}$ for given $a,b$ complex numbers.

It's $\max(|a|,|b|)$, right?

Assume we have $|b|\ge |a|$ and we have taken a $|b|$ outside the root, so that we are left with $\sqrt[n]{|z^n-1|}$ with $|z|\le 1$. It's clear that if $|z|<1$ this sequence even has a limit, which is $1$, and if $|z|=1$ the sequence $(z^n)_n$ has a subsequence converging to some $t\ne 1$ (there are two cases to consider: $z$ can have as argument either a rational multiple of $\pi$, or an irrational one; either way, it works), which means that our sequence has a subsequence converging to $\sqrt[n]{|t-1|}=1$, so the superior limit is at least $1$. On the other hand, since $|z^n-1|\le 2$, the superior limit is at most $1$, and from ehre we find it to be $1$.

[quote="grobber"]It's $\max(|a|,|b|)$, right? [/quote]

What about $a=b\ ?!$  :D
 :cool:


vector
geometry
perimeter
angle bisector
geometry solved

Let ABC is a triangle such that AB < AC. D lies on the line segment AC that CD = AB. Let E, F are the midpoints of BC, AD, respectively. Calculate the angle EFC follow the angle BAC.

Well, here's one :):

We use vectors: let $x,y$ be the vectors starting from $A$, oriented along $AB,AC$ respectively, and having length $1$. Let $a=|\overrightarrow{BC}|$ (the length of the side $BC$). We have $\overrightarrow{FE}=\frac 12(\overrightarrow{AB}+\overrightarrow{DC})=\frac a2(x+y)$, and this vector is collinear with the angle bisector of $A$, meaning that $EF$ is parallel to the angle bisector, so $\angle EFC=\frac{\angle BAC}2$.

Great! So that's the 10th solution I've known. My solutions are in grade 9. Would you find more?

I guess we could simply show the paralllelism between the bisector of $\angle A$ and $EF$ by computing $CE$ and $CF$ and then using Thales, but I don't to write such an ugly proof :). I'll think about others.

Grobber, you definitely know this problem. Since CD = AB, and the points E and F are the midpoints of the segments BC and AD, it is clear that the segment ED bisects the perimeter of triangle ABC. On the other hand, proving that $\measuredangle EFC = \frac{\measuredangle BAC}{2}$ is equivalent to showing that the line ED is parallel to the angle bisector of the angle CAB. So it remains to show that the perimeter bisector through the midpoint E of the side BC of triangle ABC is parallel to the angle bisector of the angle CAB. And this is a classical problem. See, for instance, (your own!) [url=http://groups.yahoo.com/group/Hyacinthos/message/9629?expand=1]Hyacinthos message #9629[/url].

  darij

Another simple solution:
With usual notations: we can easily compute that the FC = b+c/2 and EC = a/2. Construct R on CA extended s.t. AB=AR. CR=b+c, and CB=a so then it follows EFC is similar to BRC. But ARB is iscoceles and thus <BAC=2<BRA. But <BRA=<EFD since EFC is similar to BRC. so <EFC=<BAC/2.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ a group and $ H_{1}$ and $ H_{2}$ are only non-trivial subgroup of $ G$ with $ |H_{1}|\equal{}2$ and $ |H_{2}|\equal{}n$ , n prime . Find $ |G|$ .

I think these groups are only cyclic group with order 2n

Let $ k_{1}, k_{2}$ be the indices of $ H_{1}, H_{2}$ in $ G$ respectively. Then we have,
$ |G| \equal{} k_{1}|H_{1}| \equal{} k_{2}|H_{2}| \implies 2k_{1} \equal{} nk_{2} \implies 2\left(\frac {k_{1}}{k_{2}}\right) \equal{} n \implies k_{2} \equal{} 2$ or $ k_{1} \equal{} k_{2}$.
We must have $ k_{2} \equal{} 2$ because $ k_{1} \equal{} k_{2} \implies |H_{1}| \equal{} |H_{2}|$, a contradiction.
Hence, $ k_{1} \equal{} n \implies |G| \equal{} 2n$ as promised.

We have $ n | k_1$ and $ 2 | k_2$.....but whay $ k_1\equal{}n$ and $ k_2\equal{}2$ ?

What you said is true except how do you use the fact that n is prime to finish off the problem?
The following equation is true: $ 2\left(\frac {k_{1}}{k_{2}}\right) \equal{} n$.
Since n is prime, it can only be expressed as a product of itself and 1.
n is prime if and only if either of the following holds:
1) $ \left(\frac {k_{1}}{k_{2}}\right) \equal{} 1$
2) $ k_{2} \equal{} 2$.

1) leads to contradiction so 2) must be true and then the conclusion in the previous post follows.


\[ \underline {\overline {\left| \ \left(\forall \right)x\in \left( 0,\frac{\pi}{2}\right),\ \left( \forall \right)k\ge \frac{1}{24}\Longrightarrow 1-\frac{x^2}{2}<\cos x < 1-\frac{x^2}{2}+kx^4.\ \right| }}  \]

Prima parte  se poate rezolva la nivel de clasa a IX - a :

$16(1-\cos x)=32\sin^2\frac x2\le\< 32\left( \frac x2\right)^2\Longrightarrow1-\cos x<\frac{x^2}{2}.$

Oricum si asta si problema zilei nu mi se par asa "mari inegalitati", dupa umila mea parere avand in vedere ca ele rezulta din dezvoltarea in serie Taylor a lui $\cos x.$ ;)

Ai dreptate, deoarece nu s-a precizat nivelul !. Cine stie, poate Vornicu doreste o solutie fara analiza, si atunci ce te faci ?

De acord cu cine ? Cu Caesar sau cu mine ?

Da, daca se doreste o rezolvare mai la nivel de clasa a-10-a desi nu prea vad cum, este bine venita precizarea.
Oricum, ambele inegalitati sunt foarte utile. ;)

Iata o incercare de a aduce inegalitatea initiala la o forma mai frumoasa sau ... poate mai accesibila:

$\cos x<1-\frac{x^2}{2}+\frac{x^4}{16},\ 0<x<\frac{\pi}{2}\Longleftrightarrow 16(1-\cos x)>x^2(8-x^2)\Longleftrightarrow$

$32\sin^2 \frac x2>x^2(8-x^2)\Longleftrightarrow \ (x=2t)\ : \ 2\sin ^2 t>t^2(2-t^2),\ 0<t<\frac{\pi}{4} \Longleftrightarrow$

$1-\cos 2t>2t^2-t^4\Longleftrightarrow \cos 2t<(t^2-1)^2,\ 0<t<\frac{\pi}{4}$, a.s.o.

De aici incolo continua tu, Caesar (sau ... ), la nivelul clasei a IX - a !


If $ a,b,c>0$ and $ a\plus{}b\plus{}c \ge 1$ then:

$ 2(a\plus{}b\plus{}c)\plus{}(\frac {1} {a}\plus{}\frac {1} {b}\plus{}\frac {1} {c})\ge7$

[quote="moldovan"]If $ a,b,c > 0$ and $ a \plus{} b \plus{} c \ge 1$ then:

$ 2(a \plus{} b \plus{} c) \plus{} (\frac {1} {a} \plus{} \frac {1} {b} \plus{} \frac {1} {c})\ge7$[/quote]
$ a\plus{} \frac {1} {a}\ge 2$ so $ 2(a \plus{} b \plus{} c) \plus{} (\frac {1} {a} \plus{} \frac {1} {b} \plus{} \frac {1} {c})\ge a\plus{}b\plus{}c\plus{}6\ge 7$

[quote="moldovan"]If $ a,b,c > 0$ and $ a \plus{} b \plus{} c \ge 1$ then:

$ 2(a \plus{} b \plus{} c) \plus{} (\frac {1} {a} \plus{} \frac {1} {b} \plus{} \frac {1} {c})\ge7$[/quote]
Let $ a \plus{} b \plus{} c \equal{} k$
We want to find the minimum :)
So we have:
$ 2k \plus{} (\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})\frac {(a \plus{} b \plus{} c)}{k} \equal{} 2k \plus{} \frac {3}{k} \plus{} \frac {\frac {a}{b} \plus{} \frac {a}{c} \plus{} \frac {b}{a} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {c}{b}}{k} \ge 2k \plus{} \frac {9}{k}$

So we want to find minimum for $ f(k) \equal{} 2k \plus{} \frac {9}{k}$ (We don't even have to assume $ k \ge 1$)
We know $ f'(k) \equal{} 2 \plus{} \frac { \minus{} 9}{k^2}$. $ f'(k) \equal{} 0$ and $ k > 0$ $ \iff$ $ k \equal{} \sqrt {4.5}$.
So $ 2(a \plus{} b \plus{} c) \plus{} (\frac {1} {a} \plus{} \frac {1} {b} \plus{} \frac {1} {c}) \ge 2\sqrt {4.5} \plus{} \frac {9}{\sqrt {4.5}} \equal{} \sqrt {72} \equal{} 8.48528$

So it's a bit stronger, and without the condition $ a \plus{} b \plus{} c \ge 1$ (Minimum is attained with $ a \equal{} b \equal{} c \equal{} \sqrt {\frac {1}{2}}$)

Edit: LOL it's the most stupid solution ever seen.... By AM-GM : $ 2(a \plus{} b \plus{} c) \plus{} (\frac {1} {a} \plus{} \frac {1} {b} \plus{} \frac {1} {c}) \ge 6\sqrt [3]{abc} \plus{} 3\sqrt [3]{\frac {1}{abc}} \ge 2\sqrt {18} \equal{} \sqrt {72} > 7$...
Edit 2: Didn't see the post by dgreenb801...

$ 2a\plus{} \frac{1}{a} \ge 2\sqrt{2}$ by AM-GM, so the left side is at least $ 6 \sqrt{2}$. You don't even need $ a\plus{}b\plus{}c\ge 1$


The measures of the angles of a particular convex quadrilateral are in the ratio 3:5:7:10. What is the degree measure of the largest angle?

[hide] We have that $ 3x\plus{}5x\plus{}7x\plus{}10x\equal{}360 \implies 25x\equal{}360 \implies x\equal{}\frac{360}{25}$. Thus,
$ 10(\frac{360}{25})\equal{}\boxed{144^\circ}$.[/hide]


There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?

Note - If the answer comes out in factorials, then don't cancel them.

Yep, definitely old AMC10 problems:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=53532700&t=77548

(I searched for "peg* triang* board" to find that thread.)

I have a hard time interpretating this problem (particularly word problems in general; usually take me a while to understand them fully): what does a triangular peg board look like?  Is it just a triangle?
I think problems of this type should always include a picture!   :!:


Let $X$ be a connex set and $f: X\to Y$ be a local constante function.Show that $f$ is constante.

If $f(x_0)=y_0$, say, then $f^{-1}(y_0)$ is closed because $f$ is continuous, open because of the hypothesis of local, and non-empty because it contains $x_0$, so it must be the whole $X$.

I'll post "another" solution:
Let $c\in E$.Define $g: E\to R$ as following:$g(x)=1$ if $f(x)=f(c),g(x)=0$ otherwise.
$g$ is local costante,so it is continuous.Since E it is connex we know that $g(E)$ is a connex set of R(intervale).So $g(x)=1, \forall \in E$ i.e $f$ is constante.


Find all pairs of positive intergers $ (a,b)$ such that $ 5a^b\minus{}b\equal{}2004$

well... use inequality...it's the obvious idea!
Let $ a > 1$ then u can prove that the difference:$ 5a^{b} \minus{} b$ is strictly increasing.u can conclude that $ b < 10$.the rest is easy.


Find the number of DIGITS in $N$ where $N$ is the product of all the positive divisors of 100 , 000 , 000.

[hide]81 factors so

$100,000,000^{40} \cdot  10,000$

Giving a total of $40 \cdot 8 + 5 = 325$ digits.[/hide]

How did you know that there are exactly 81 factors :?:

That's equal to $2^8 \cdot 5^8$ adding one to each exponent and multiplying gives us the number of factors.

yes, in general
if $A=p_1^{\alpha_1}p_2^{\alpha_2}p_3^{\alpha_3}...$
then $N=(\alpha_1+1)(\alpha_2+1)(\alpha_3+1)...$ is the number of divisors.(obviously the number of proper divisors is $N-1$)

I didn't exactly get the  $\displaystyle 100,000,000^{40} \cdot 10,000$ part....How did you get that 40 ?? :oops:

for  every divisor a(a is not 10000) there's one more divisor different from a so that axb=10000000

I still dont understand the stuff properly.Could someone give a detailed explanation ?

100000000 is equal to $2^8 \cdot 5^8$ which clearly has 81 factors.

Since all the factors except 10000 require another, different factor to multiply to 100000000, we have 40 pairs of factors that multiply to 100000000 and 10000. Hence, the product of all the factors is $100000000^{40} \cdot 10000$ which has $8 \cdot 40 + 4 = 324$ zeros and then the last 1 making $325$.


calculus
integration
geometry
similar triangles
calculus computations

[img]http://img443.imageshack.us/img443/4364/aditea5.jpg[/img]

a. slice the solid parallel to the triangular faces. Sketch one slice and calculate its volume in terms of x, the distance of the slice from one end. Then write and evaluate an integral giving the volume of the solid.
b. repeat part a for horizontal slices. Instead of x, use h, the distance of a slice from the top

[quote="EquinoX"]a. slice the solid parallel to the triangular faces. Sketch one slice and calculate its volume in terms of x, the distance of the slice from one end. Then write and evaluate an integral giving the volume of the solid.[/quote]

These slices are all equal-sized triangular pieces having: base=2 cm, height=3cm, and thickness = dx. So the integral is

\[V=\int_{0}^{4 cm}\frac1{2}(2 cm)(3 cm)dx = \left[\frac1{2}(2 cm)(3 cm)x \right]_{x=0}^{4 cm}=12 cm^{3}\]

[quote="EquinoX"]b. repeat part a for horizontal slices. Instead of x, use h, the distance of a slice from the top[/quote]

These slices are paralell to the 2 cm X 4 cm rectangular face of the solid and are themselves rectangular pieces, having length = 4 cm, thickness = dh, and width = ???. Let call the width w, now to find w in terms of h use similar triangles to get... you finish.


For me, I can't quote some people, such as SplashD...

I've noticed that too: for SplashD's post, the posted today at ...., quote, mark as rread, back to top- the stuff at the bottom right is missing...

I had a topic about that recently. It's because he has a hide tag at the start of his sig...
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=108884

You're also using a hide tag in your signature, except it's extremely small. Please don't do this. It will make it harder for people to reply to questions you have, etc. Thanks for your cooperation.

[quote="anirudh"]what do you mean  :maybe: [/quote]

hidden user

[quote="anirudh"]how would you know that :maybe:[/quote]

your screen shot  :P


calculus
derivative
probability
expected value

I would like to discuss the Stock Market here, it appears no-one really visits the econ forum so I am posting here, it is also pretty much finance.

I am trying to begin by understanding how does a company decide how much shares they want to sell and how much they charge for them?

Why is a company interested in how a stock does later on if they dont get profit from it? Why does the stock drop and rise if people can always buy it at a higher price? I seem to be miscomprehending a lot here so please help!! U ppl must think I am crazy by now.

Stocks fal because people want to sell it at a certain price (high enough).  :D

I used to watch hours of stocks, but that was when I was 8 or 9, so I only knew that when the stocks fall, it's bad, and when it rises, it's good.

well, a company knows how much to charge after undergoing an Initial Public Offering, which analyzes the revenue of the corporation and uses that to determine a stock price.

how exactly does it do that, I mean with the revenue?

Also, factorial236, if it wants to sell it at a higher price, the price should always increase. Suppose you buy stock at $10 per share, and sell it at a higher price to make a profit say at $20. the next person sells it at an even higher price to make a profit say 30 etc etc. So doesnt it always keep increasing? Like why does the value fall?

The value falls when nobody wants to buy it at 30 dollars - then it must be offered for lower.. its the basics of the "buy low sell high".

Also, some stocks prices "fall" (although not really) when a company splits its stocks - so instead of having 2 stocks worth 100 total, they have 4 stocsk worth 100 total, at 25 each. Any shareholder's stocks in that company (the quantity) will double, but the value will halve.

What I am trying to understand is that a given value of a stock at a particular time is when some person or company bought that stock. Won't there be someone who will agree to pay  more for a stock?

I mean cant u just make a stock represent some number in a thuoght experiment and always try to increase it? Like I said in my last post, wont everyone benefit in that case?

stock prices also have to do with how the company's doing. an IPO is like a thought-out price. btw, if you sell stocks at a high price, fewer people will buy them, so, you lose money. if you sell stocks at a low price, they're practically worthless.

So then if a stock is really cheap it shuold have a lot of buyers for some time till it gets high?

not necessarily, the company could be in terrible financial shape, in which case it would be very risky to buy stocks.

but why do stock prices depend on their profits? What is the actual reason for this?

Now, I wanted to ask. I am reading a book called "How Wall Street Works" and am trying to make sure I am understanding everything. Are brokers and market makers (traders I guess) almost the same ppl, they generally buy stocks or bonds (at least the traders do) and sell to other individuals and business?

Also, what is the difficulty in being a trader? I am not saying it is not difficult I am just curious as to why it is difficult and why does a trader need to have advanced mathematics skills? Is that all traders or just quants and what is the difference between traders and quants? Sorry for asking questions that are proly to easy.

Don't worry about asking these questions they may seem easy but they can be pretty hard to work out unless you actually get to see the inside of financial institutions.  I'm still trying to work similar stuff out myself.  My understanding is,

Market makers are the banks who actually "create" financial instruments and sell them to their customers, they don't really want to take on alot of risk when they do this so they then have to some how hedge the exposure they have to this instruments by buying and selling simpler financial instruments that are equivalent to the more complex one that they have sold.  This way banks often make their money out of transaction costs rather than taking risk.  This means that the more deals they do the better so a trader at a bank is also often a market maker in that he wants to do deals so when you phone a bank up and ask them for a price on a tradable security they will give you a quote - its very bad form for them not to trade with you because its in their best interest to create liquidity.

A broker is someone who merely links up buyers and sellers and tries to do it in the best way for their client.  So if you want to buy 100 000 shares or 100 000 options you may go to a broker and they will actually do the trades for you, trying to get it at the right price.  Remember that  although a share may be quoted at a price of 100 on an exchange that just means that the last trade went through at 100.  I reality there are a bunch of people want to sell certain amounts for certain prices and guys want to buy so you could have

A wants to sell 20 000 shares at 110
B wants to sell 40 000 shares at 108
C wants to sell 100 000 shares at 105

D wants to buy 20 000 shares at 100
E wants to buy 30 000 shares at 95
F wants to buy 50 000 shares at 90

If C gets really desperate to sell he might lower his price to 100 but he will only sell 20 000 of his share there, to get all of them sold he needs to go down to 90.  Brokers take all of these sorts of things into account.  Also you need to have a license to trade on an exchange so you have to do that through a broker.

Traders don't necessarily need to be good at maths.  A foreign exchange trader who just trades currencies may need a lot more economics knowledge and not so much maths but a derivatives trader may need to know more maths because derivatives can be very complicated instruments.  In the banks that I've seen there are a number of different roles on the trading floor, a quant doesn't trade she builds the models that the bank uses to price the instruments that they are selling (this is when they are creating a new instrument that hasn't got a price already) so a quant works out and implements models this is where the most maths goes on by far.  Then there are guys who struture deals for clients.  These are less mathematical roles than a quant in general but they still require a good understanding of the instruments and are very analytical.  You have to create an instrument that pays out certain cashflows to your customer and you have to build that out of simple instruments.  Then when that is done the trader is the guy that actually buys and sells these components.  That is a simple version of what I understand happens in the South African market things could be pretty different in a more developed market.

When a company issues share that company is then owned by the public.  The board members that run that company are then just employees of the shareholders and the shareholders decide if they keep their jobs or not.  The shareholders want the share price to do well because then they get a good return on investment.  If the share price doesn't do well the shareholders will vote for new management.

In theory the value of a share of a company is just the discounted expected value of the future dividend payments that you will recieve if you hold that share - just like anything the value of it is just the expected value of the cashflows that thing will give you.  The dividends that are payed out is the profit that is left over after reinvestment, so the dividends depend on the profit.  Ofcourse it is very hard to predict future profits and dividends so the share price is set by supply and demand.  So on one hand a share is only really worth what someone else is willing to pay you for it.  This is why people shouldn't do your thought experiment although it does happen sometimes.  If you just keep valuing a share higher and higher than its intrinsic value there might be a stage where every body else chickens out and you are holding the share but know one else wants to buy it from you at a higher price.  Then you are stuck with some thing that you paid a lot for but no one will buy back from you for anywhere near that price and if you hold on to it the value of the dividends you get from it are nowhere near its cost to you.  This happens in bubbles and crashes.  Everyone believes that IT companies will make huge profits in the future so the share prices go up.  Then people start to realise that everyone is getting carried away but they will still buy the share above what they think is fair value because they believe someone else will pay a higher price for it later then suddenly market sentiment changes and everyone comes down to earth and the price drops drastically.


The four digit numeral 3AA1 is divisible by 9.  What does a represent?

[hide] a number is dvisible by 9 if its digits are divisible by 9.  So 4 + 2a must be divisible by 9.

4 + 2a = 9y

Now we can note many things about about why "y=2", ill let you do that for yourself. But our new equation is:

4 + 2a= (9*2)=18

4 + 2a=18

2a=14

a=7

So a=7 and our number is 3771.   [/hide]

[hide]A number divisible by 9 has to have its digits add up to 9, so looking at the number 3AA1, we can see that either A+A = 5 or A+A = 14.
A can't be a fraction, so 2A = 14.
A = 7, and the number is 3771. [/hide]

The definition of a number being divisible by nine is that the sum of the digits is divisible by nine
so 3+1+2a=4+2a is divisible by nine
a is a one digit number
4+2a=9,18,24,32......
2a=9-4,18-4,24-4,32-4
2a=5, 14, 20, 28
a=2.5 , 7, 10, 14
but because a is a one digit number a is 7
so the number is 3771 :lol:


geometry
circumcircle
incenter
inradius
geometric transformation
homothety
ratio

Let $ O$ and $ R$ be respectively the center and radius of the circumcircle of $ \triangle ABC$ and let $ Z$ and $ r$ be respectively the incenter and inradius of $ \triangle ABC$.
Denote by $ K$ the centroid of the triangle formed by the points of tangency of the incircle and the sides.
Prove that $ Z \in OK$ and $ \frac{OZ}{ZK}=\frac{3R}{r}$

I'm looking for a solution that doesn't have complex number.Thanks a lot.

Dear Mathlinkers,
first ideas...
1. the centroid of the intouch triangle is the Weill's point X(354) by ETC.
2. K, O and Z are collinear (Gob's theorem)
Sincerely
Jean-Louis

[hide="Solution"][quote]Let $ O$ and $ R$ be respectively the center and radius of the circumcircle of $ \triangle ABC$ and let $ Z$ and $ r$ be respectively the incenter and inradius of $ \triangle ABC$.
Denote $ A_1,B_1,C_1$ be the points of tangency of the incircle with the sides $ BC,CA,AB$ respectively. Denote $ K$ be the centroid of the triangle $ A_1B_1C_1$. Prove that $ Z \in OK$ and $ \frac {OZ}{ZK} = \frac {3R}{r}$[/quote]
[i]Solution:[/i]
$ AZ,BZ,CZ$ meets the circumcircle at $ A_2,B_2,C_2$ respectively. $ A_1A_2$ meets $ OZ$ at $ P$.
$ ZA_1\parallel OA_2\Rightarrow \frac{PA_1}{PA_2}=\frac{PZ}{PO}=\frac{ZA_1}{OA_2}=\frac{r}{R}$
Similarly, $ A_1A_2,B_1B_2,C_1C_2$ meet at $ P$ on the line $ OZ$ (doesn't lie on segment $ OZ$) satisfying
$ \frac{PA_1}{PA_2}=\frac{PB_1}{PB_2}=\frac{PC_1}{PC_2}=\frac{PZ}{PO}=\frac{r}{R}$
Thus, the triangle $ A_2B_2C_2$ becomes the triangle $ A_1B_1C_1$ after the homothety of center $ P$ and ratio $ k=\frac{r}{R}$. So if we denote H be the orthocenter of the triangle $ A_1B_1C_1$ then $ \frac{PH}{PZ}=\frac{r}{R}$ (since $ Z$ is the orthocenter of triangle $ A_2B_2C_2$).
$ \frac{PH}{PZ}=\frac{PZ}{PO}=\frac{r}{R}$
$ \Rightarrow \frac{PZ-PH}{PZ}=\frac{PO-PZ}{PO}\Rightarrow \frac{HZ}{PZ}=\frac{OZ}{PO}$
$ \Rightarrow \frac{OZ}{HZ}=\frac{PO}{PZ}=\frac{R}{r}$
On the Euler line of triangle $ A_1B_1C_1$, $ ZK=\frac{1}{3}ZH$
Hence $ \frac{OZ}{ZK}=\frac{3R}{r}$
[/hide]

[hide]Construct 3 excircles $ (I_a), (I_b),(I_c)$ of triangle $ ABC$. Denote $ (O',R'), G$ the circumcircle and centroid of triangle $ I_aI_bI_c$
We have $ Z, O$ are the orthocentric and Nine-point circle's center of triangle $ I_aI_bI_c$. then $ ZO$ is the Euler line of triangle $ I_aI_bI_c$
But $ \Delta I_aI_bI_c$ and $ \Delta DEF$ have $ DE//I_aI_b, EF//I_bI_cFD//I_cI_a$ so they are similar $ (1)$ and Euler line of triangle $ DEF$ is parallel to Euler line of triangle $ I_aI_bI_c$. Moreover, they have a same point $ Z$ therefore $ \Delta I_aI_bI_c$ and $ \Delta DEF$ have a common Euler line.
Thus $ K,Z,O$ are collinear.
On the other side, since $ (O)$ is the Nine-point circle of triangle $ I_aI_bI_c$ then $ R' \equal{} 2R$
From $ (1)$ we obtain $ \frac {ZK}{O'G'} \equal{} \frac {DE}{I_aI_b} \equal{} \frac {r}{2R}$
$ \Rightarrow \frac {ZK}{\frac {2}{3}OZ} \equal{} \frac {r}{2R}$
$ \Rightarrow \frac {ZK}{OZ} \equal{} \frac {r}{3R}$
We are done![/hide]

Let $ D,E,F$ respectively be the tangency points of $ (Z)$ with $ BC,CA,AB$. Denote $ M_a,M_b,M_c$ respetively be the midpoints of $ EF,FD,DE$. Consider the inversion through pole $ I$, power $ k \equal{} r^2$. We have $ \mathcal {I}(Z,k): D\mapsto D$, $ E\mapsto E$, $ F\mapsto F$ and $ A\mapsto M_a$, $ B\mapsto M_b$, $ C\mapsto M_c$. Therefore $ \mathcal {I}(Z,k): (ABC)\mapsto (M_aM_bM_c)$. Note that $ (M_aM_cM_c)$ is the $ 9$- point circle wrt $ \triangle DEF$ and let $ O'$ be its center $ \Longrightarrow$ $ O',Z,O$ are collinear. But $ O'$ lies on the $ Euler$ line $ ZK$ of $ \triangle DEF$. Thus, $ O,Z,K$ are collinear. $ \square$

Now since $ (M_aM_bM_c)$ is the image of $ (ABC)$ by the inversion through pole $ Z$, power $ k$, hence $ (M_aM_bM_c)$ is also the image of $ (ABC)$ through the homothety with center $ Z$, ratio $ k_1 \equal{} \frac {k}{\mathcal {P}_{Z/(O)}}$ $ \equal{} \frac {r^2}{2Rr}$ $ \equal{} \frac {r}{2R}$ $ \Longrightarrow$ $ \frac {ZO'}{ZO} \equal{} \frac {r}{2R}$. But in $ \triangle DEF$, we have $ \triangle DEF$ is the image of $ \triangle M_aM_bM_c$ by the homothety with center $ K$, ratio $ k_2 \equal{} \minus{} 2$, hence $ \mathcal {H}(K,k_2): O'\mapsto Z$ $ \Longrightarrow$  $ \frac {KZ}{KO'} \equal{} 2$ $ \Longrightarrow$ $ \frac {KZ}{ZO'} \equal{} \frac {2}{3}$ $ \Longrightarrow \frac {KZ}{ZO} \equal{} \frac {r}{3R}$. $ \square$

[b]REMARK-[/b] The problem would be more interested if instead of "Prove that $ \frac {KZ}{ZO} \equal{} \frac {r}{3R}$", we will ask to prove: "$ OZ\ge 6KZ$".


modular arithmetic
number theory
relatively prime
number theory unsolved

For an odd positive integer $n>1$, let $S$ be the set of integers $x$, $1\leq x\leq n$, such that both $x$ and $x+1$ are relatively prime to $n$
Prove that\[\prod_{x\epsilon S}x\equiv 1(\bmod n)\]

I remember posting this a long time ago (and receiving a solution), but I can't find it now (it realy was a very long time ago :)).

If $(x,n)=(x+1,n)=1$, then $(x^{-1},n)=(x^{-1}+1,n)=1$ as well, where $x^{-1}$ is the inverse of $x$ in the multiplicative group of invertible residues $\pmod n$, so for each $x$ in that product, $x^{-1}$ is also in the product. This means that we can pair the $x$'s with the $x^{-1}$'s, the product of these pairs gives $1$, and there's only one problem: those residues $x$ for which $(x,n)=(x+1,n)=1$ and $n|x^2-1$. If this happens, we have $n|(x-1)(x+1)$, but we know that $(x+1,n)=1$, so $n|x-1$, which means that $x=1\pmod n$, so there are no such residues, except maybe for $1$, which makes no difference to the product.


O is the circumcentre of triangle ABC. A' is the midpoint of BC. A'' is the intersection of AA' with the circumcircle of ABC. The line through A' perpendicular to AO meets the tangent at A'' at a point A'''. Similarly define B''', C'''. Prove A''', B''', C''' are collinear.

Discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=16566 .

  darij