Let $ABC$ be a triangle. For a fixed point $D\in (BC)$ construct the position of the point $L\in AD$ for which $\widehat {DLB}\equiv \widehat {DLC}\ .$

[b]Construction.[/b] Find the point $B'$ such that $B$ and $B'$ are symmetric about the line $AD$. The desired point $L$ is the intersection point of the lines $AD$ and $CB'.$

[b]Proof.[/b] Because of the symmetry, we have $\angle DLB=\angle DLB'$ (both $D$ and $L$ are on the symmetrial axis), but as the points $C, B', L$ are collinear, $\angle DLB'=\angle DLC\implies \angle DLB=\angle DLC$.

[b]Discussion.[/b]
1) If $B'\equiv C$, any line passing through $C$ gives a valid solution for $L$, hence in that case there are infinitely many solutions. That will happen if and only if $\triangle ABC$ is an isosceles with $AD$ being its symmetrial axis, and any point of the line $AD$ satisfies the given condition.
2) If $CB'$ is parallel to $AD$, there will be no solution. In that case let $M=BB'\cap AD$. It's obvious that $\triangle BB'C$ and $\triangle BMD$ are both right, therefore $M$ being the midpoint of $BB'$ and $DM\parallel CB'$ imply that $D$ is the midpoint of $BC$. We conclude that if $AD$ is the median of $\triangle ABC$ where $AB\neq AC$, there's no solution.
3) In all the other cases the line $CB'$ is uniquely determined and intersects the line $AD$, hence we'll have a unique solution for $L.$

[b]Note.[/b] Obviously, the same result is obtained if we find $C'$ symmetric to $C$ and then intersect the line $BC'$ with the line $AD.$

The case $(b=c)\ \vee\ (DB=DC)$ is evidently. Indeed, if $b=c$ then there are a $\infty$ solutions (the all points of the line $AD$) and if $b\ne c$ and $DB=DC$ then there is no solution ($\emptyset$). Suppose $(b\ne c)\ \wedge\ (DB\ne DC)$. Construct the harmonical conjugate point $E$ of the point $D$ w.r.t. the points $B$ and $C$. Then the appreciated point $L$ is the projection of the point $E$ on the line $AD\ !$

Virgil, are you saying that my construction is wrong, or you're just offering another solution?

done similarly to Farenhajt , drawing the perpendicular from B to AD and trying to find the circumference with center L using simmetry

[quote="Virgil Nicula"]No, the your solution is clearly and shorterly !
You must construct only a symmetrical point of the point w.r.t. a line.

I presented a another more pretentious solution !
I must construct only a perpendicular from a point on a line
(the construction of the conjugate is made without compasses).

Understood it you ? $\mathrm {Score\ : \ 1\ -\ 1\ .}$ Sorry, I joke ![/quote]

Yes, Virgil, all clear :) Thanks


[quote="santosguzella"]Find all pairs $ (a,b)$ of positive integers such: $ a^{{b}^{2}}=b^{a}$[/quote]

From  $ a^{b^{2}}=b^{a}$, we get  $ a=b^{\frac{a}{b^{2}}}$ and so $ a=b^{\frac{p}{q}}$ for some positive coprime integers $ p$ and $ q$

So $ b^{\frac{p}{q}b^{2}}=b^{b^{\frac{p}{q}}}$ and so $ \frac{p}{q}b^{2}=b^{\frac{p}{q}}$  or $ \frac{p}{q}=b^{\frac{p-2q}{q}}$

Since RHS is a rational number, we need $ b=c^{q}$ for some positive integer $ c$ and so $ \frac{p}{q}=c^{p-2q}$ and so either $ p=1$ or $ q=1$

$ 1).$ Case $ p=1$ : $ q=c^{2q-1}$ which gives immediately :
$ q=1$ and $ c=1$
And if $ q>1$, either $ c=1$ and equality is wrong, either $ c>1$ and $ c^{2q-1}>q$

$ 2).$ Case $ q=1$ : $ p=c^{p-2}$ which gives immediately :
$ c=1$ $ \implies$ $ p=1$
$ c=2$ $ \implies$ $ p=4$
$ c=3$ $ \implies$ $ p=3$
$ c>3$ $ \implies$ $ c^{-1}\neq 1$, $ c^{0}\neq 2$, $ c^{1}\neq 3$ and $ c^{p-2}>p$ $ \forall p>3$

So we just have three different solutions :
$ p=1$, $ q=1$, $ c=1$ $ \implies$ $ a=1$ and $ b=1$ : $ 1^{1^{2}}=1^{1}$
$ p=4$, $ q=1$, $ c=2$ $ \implies$ $ a=16$ and $ b=2$ : $ 16^{2^{2}}=2^{16}$
$ p=3$, $ q=1$, $ c=3$ $ \implies$ $ a=27$ and $ b=3$ : $ 27^{3^{2}}=3^{27}$

Problem 5 of IMO 1997
[url]http://www.mathlinks.ro/resources.php?c=1&cid=16&year=1997[/url]


hi, i'm trying to self-study AP psych for this year.  I known I only have like 6 weeks but besides review book, what is a good text book?  any recommendations?

Thanks in advance

My previous high school used [i]Psychology[/i] by Peter Gray for the AP course.


let $f: R\to R$ such that
     $|f(x+y)-f(x)-f(y)|\leq 1$
 prove that there exist $g: R\to R$ st
      $|g(x)-f(x)|\leq 1$ and $g(x+y)=g(x)+g(y)$

what do you think of  it?
maybe it is very hard ;)


search
inequalities proposed
inequalities

[quote="Vasc"]Let $ a,b,c$ be positive real numbers, and let $ x \equal{} \frac {2a}{b \plus{} c}$, $ y \equal{} \frac {2b}{c \plus{} a}$, $ z \equal{} \frac {2c}{a \plus{} b}$. Find the greatest value of $ p$ such that
$ \frac 1{x^2} \plus{} \frac 1{y^2} \plus{} \frac 1{z^2} \minus{} 3 \ge p(x \plus{} y \plus{} z \minus{} 3)$[/quote]
Let $ a\equal{}b\equal{}1,c\equal{}\sqrt{3}\plus{}1$, we have $ p \le \frac{5}{2}\plus{}\frac{3\sqrt{3}}{2}$, we can prove it by using SOS or LCF-RCF method. :)

[quote="can_hang2007"][quote="Vasc"]Let $ a,b,c$ be positive real numbers, and let $ x \equal{} \frac {2a}{b \plus{} c}$, $ y \equal{} \frac {2b}{c \plus{} a}$, $ z \equal{} \frac {2c}{a \plus{} b}$. Find the greatest value of $ p$ such that
$ \frac 1{x^2} \plus{} \frac 1{y^2} \plus{} \frac 1{z^2} \minus{} 3 \ge p(x \plus{} y \plus{} z \minus{} 3)$[/quote]
Let $ a \equal{} b \equal{} 1,c \equal{} \sqrt {3} \plus{} 1$, we have $ p \le \frac {5}{2} \plus{} \frac {3\sqrt {3}}{2}$, we can prove it by using SOS or LCF-RCF method. :)[/quote]
Why are you know $ a\equal{}b\equal{}1,c\equal{}\sqrt{3}\plus{}1$?
What's LCF - RCF method?
Please, !
Thanks!

[quote="can_hang2007"] Let $ a \equal{} b \equal{} 1,c \equal{} \sqrt {3} \plus{} 1$, we have $ p \le \frac {5}{2} \plus{} \frac {3\sqrt {3}}{2}$, we can prove it by using SOS or LCF-RCF method. :)[/quote]
This is, indeed. :wink:

[quote="vipCD_A1"][quote="can_hang2007"][quote="Vasc"]Let $ a,b,c$ be positive real numbers, and let $ x \equal{} \frac {2a}{b \plus{} c}$, $ y \equal{} \frac {2b}{c \plus{} a}$, $ z \equal{} \frac {2c}{a \plus{} b}$. Find the greatest value of $ p$ such that
$ \frac 1{x^2} \plus{} \frac 1{y^2} \plus{} \frac 1{z^2} \minus{} 3 \ge p(x \plus{} y \plus{} z \minus{} 3)$[/quote]
Let $ a \equal{} b \equal{} 1,c \equal{} \sqrt {3} \plus{} 1$, we have $ p \le \frac {5}{2} \plus{} \frac {3\sqrt {3}}{2}$, we can prove it by using SOS or LCF-RCF method. :)[/quote]
Why are you know $ a \equal{} b \equal{} 1,c \equal{} \sqrt {3} \plus{} 1$?
What's LCF - RCF method?
Please, !
Thanks![/quote]
[color=green]Have look at Vasc'book or Search in the foum.[/color]


[quote="NextPeace"]Let $ x_1,x_2,\cdots,x_n,k$are positive numbers. prove that
$ \frac {1}{(x_1 \plus{} k)^n} \plus{} \frac {1}{(x_2 \plus{} k)^n} \plus{} \cdots \plus{} \frac {1}{(x_n \plus{} k)^n}\ge\frac {n}{2^{n \minus{} 1}(x_1x_2\cdots x_n \plus{} k^n)}$

It is a generalization of $ \frac {1}{(x \plus{} z)^2} \plus{} \frac {1}{(y \plus{} z)^2}\ge\frac {1}{xy \plus{} z^2}$.
But I think this inequality is so weak. Do you have idea which a nice(and strong) generalization of it?
I am really want to know more nice generalization.  :blush:[/quote]
Nice inequality, NextPeace. This inequality is a nice corollary of my old result:
\[ \sum^{n}_{i \equal{} 1} \frac {x^{n \minus{} 1}_{i \equal{} 1}}{\prod_{j\ne i} (x_i \plus{} x_j)} \ge \frac {n}{2^{n\minus{}1}}.
\]
This inequality is my generalization for the PHONGVAN's Inequality:
\[ \frac {a^3}{(a \plus{} b)(a \plus{} c)(a \plus{} d)} \plus{} \frac {b^3}{(b \plus{} c)(b \plus{} d)(b \plus{} a)} \plus{} \frac {c^3}{(c \plus{} d)(c \plus{} a)(c \plus{} b)} \plus{} \frac {d^3}{(d \plus{} a)(d \plus{} b)(d \plus{} c)} \ge \frac {1}{2},
\]
for $ a,b,c,d > 0.$
:)

$ x_{i} \equal{} e^{y_{i}}$라고 하면 곱이 합이되고
$ k \equal{} e^{t}$인 $ t$가 존재하고, 저런식으로 치환하면 젠센으로 안되나?

흠-_-;안해봐서 될지는...

[quote="NextPeace"][quote="can_hang2007"][quote="NextPeace"]Let $ x_1,x_2,\cdots,x_n,k$are positive numbers. prove that
$ \frac {1}{(x_1 \plus{} k)^n} \plus{} \frac {1}{(x_2 \plus{} k)^n} \plus{} \cdots \plus{} \frac {1}{(x_n \plus{} k)^n}\ge\frac {n}{2^{n \minus{} 1}(x_1x_2\cdots x_n \plus{} k^n)}$

It is a generalization of $ \frac {1}{(x \plus{} z)^2} \plus{} \frac {1}{(y \plus{} z)^2}\ge\frac {1}{xy \plus{} z^2}$.
But I think this inequality is so weak. Do you have idea which a nice(and strong) generalization of it?
I am really want to know more nice generalization.  :blush:[/quote]
Nice inequality, NextPeace. This inequality is a nice corollary of my old result:
\[ \sum^{n}_{i \equal{} 1} \frac {x^{n \minus{} 1}_{i \equal{} 1}}{\prod_{j\ne i} (x_i \plus{} x_j)} \ge \frac {n}{2^n}.
\]
This inequality is my generalization for the PHONGVAN's Inequality:
\[ \frac {a^3}{(a \plus{} b)(a \plus{} c)(a \plus{} d)} \plus{} \frac {b^3}{(b \plus{} c)(b \plus{} d)(b \plus{} a)} \plus{} \frac {c^3}{(c \plus{} d)(c \plus{} a)(c \plus{} b)} \plus{} \frac {d^3}{(d \plus{} a)(d \plus{} b)(d \plus{} c)} \ge \frac {1}{2},
\]
for $ a,b,c,d > 0.$
:)[/quote]

Yes, can_hang. My solution is same with you  :)

*oh, sorry, I have a mistake :oops: . My solution is not same with you. Can you post your solution, can_hang?

To. Heebeen, Yang
해보세요 ㅋㅋ[/quote]
My solution is the followiong:
For any $ 1 \le i \le n$, the Holder Inequality yields
\[ (x_i\plus{}k)^n \equal{}\left( \sqrt[n]{x_1x_2\cdots x_n} \prod_{j \ne i} \sqrt[n]{\frac{x_i}{x_j}} \plus{}\sqrt[n]{k^n} \prod_{j \ne i}  \sqrt[n]{1}\right)^n \le (x_1x_2\cdots x_n \plus{}k^n) \prod_{j \ne i} \left( \frac{x_i}{x_j}\plus{}1\right) .\]
Using this inequality, we can reduce our inequality to
\[ \sum^{n}_{i\equal{}1} \frac{1}{\prod_{j \ne i}\left( \frac{x_i}{x_j}\plus{}1\right)} \ge \frac{n}{2^{n\minus{}1}}.\]
Setting $ y_i\equal{}\frac{1}{x_i},$ this inequality becomes
\[ \sum^{n}_{i\equal{}1} \frac{y^{n\minus{}1}_{i}}{\prod_{j \ne i} (y_i\plus{}y_j)} \ge \frac{n}{2^{n\minus{}1}},\]
which is just the inequality I have posted. :)

P/s: I have mistake when typing my inequality. It should be $ \frac{n}{2^{n\minus{}1}}$, I have edited it.

[quote="can_hang2007"][/quote][quote="NextPeace"][quote="can_hang2007"][quote="NextPeace"]Let $ x_1,x_2,\cdots,x_n,k$are positive numbers. prove that
$ \frac {1}{(x_1 + k)^n} + \frac {1}{(x_2 + k)^n} + \cdots + \frac {1}{(x_n + k)^n}\ge\frac {n}{2^{n - 1}(x_1x_2\cdots x_n + k^n)}$

It is a generalization of $ \frac {1}{(x + z)^2} + \frac {1}{(y + z)^2}\ge\frac {1}{xy + z^2}$.
But I think this inequality is so weak. Do you have idea which a nice(and strong) generalization of it?
I am really want to know more nice generalization.  :blush:[/quote]
Nice inequality, NextPeace. This inequality is a nice corollary of my old result:
\[ \sum^{n}_{i = 1} \frac {x^{n - 1}_{i = 1}}{\prod_{j\ne i} (x_i + x_j)} \ge \frac {n}{2^n}.
\]
This inequality is my generalization for the PHONGVAN's Inequality:
\[ \frac {a^3}{(a + b)(a + c)(a + d)} + \frac {b^3}{(b + c)(b + d)(b + a)} + \frac {c^3}{(c + d)(c + a)(c + b)} + \frac {d^3}{(d + a)(d + b)(d + c)} \ge \frac {1}{2},
\]
for $ a,b,c,d > 0.$
:)[/quote]

Yes, can_hang. My solution is same with you  :)

*oh, sorry, I have a mistake :oops: . My solution is not same with you. Can you post your solution, can_hang?

To. Heebeen, Yang
해보세요 ㅋㅋ[/quote][quote="can_hang2007"]
My solution is the followiong:
For any $ 1 \le i \le n$, the Holder Inequality yields
\[ (x_i + k)^n = \left( \sqrt [n]{x_1x_2\cdots x_n} \prod_{j \ne i} \sqrt [n]{\frac {x_i}{x_j}} + \sqrt [n]{k^n} \prod_{j \ne i} \sqrt [n]{1}\right)^n \le (x_1x_2\cdots x_n + k^n) \prod_{j \ne i} \left( \frac {x_i}{x_j} + 1\right) .
\]
Using this inequality, we can reduce our inequality to
\[ \sum^{n}_{i = 1} \frac {1}{\prod_{j \ne i}\left( \frac {x_i}{x_j} + 1\right)} \ge \frac {n}{2^{n - 1}}.
\]
Setting $ y_i = \frac {1}{x_i},$ this inequality becomes
\[ \sum^{n}_{i = 1} \frac {y^{n - 1}_{i}}{\prod_{j \ne i} (y_i + y_j)} \ge \frac {n}{2^{n - 1}},
\]
which is just the inequality I have posted. :)

P/s: I have mistake when typing my inequality. It should be $ \frac {n}{2^{n - 1}}$, I have edited it.[/quote]

Thanks for your posting, can_hang  :lol:
My proof is similar with yours, but I was proof
$ sum^{n}_{i = 1} \frac {1}{\prod_{j \ne i}\left( \frac {x_i}{x_j} + 1\right)} \ge \frac {n}{2^{n - 1}}$ directly(with out transposition).
But your transposition is very nice! Thanks  :)


AMC
AIME
email
AMC 12
AMC 12 B
probability
AIME II

Hey y'all,

I was just wondering if these e-mails should have already been received by whatever math teacher was designated when they signed up for the AMC. The reason why I'm concerned is that last year two students that had qualified for the AIME weren't notified until after the test was administered.

Also, what steps should I (or can I?) take if the e-mails should have already arrived but the math teacher hasn't received them?

Thanks!

Yes, the teacher in charge of the AMC at your school should already know your score and therefore, whether or not you qualified for AIME (you should already know your unofficial score if you copied your answers onto the test booklet). If noone at your school knows, it probably is not even administering the AIME, but it wouldn't hurt to email the AMC director.

Well, it won't hurt, but it won't do any good, because it is much easier, much more sensible,  and more polite to ask your school contest manager.  Your school contest manager has a report (in most cases) on about 30-60 students, while I have about 250,000.

I will delete any such requests, you have much better avenues for finding out your score.

Steve Dunbar
Director, American Mathematics Competitions

Yes, just to clear things up, I didn't intend for any discourtesy  :blush: . I meant that you should email the AMC only if no one at your school knew anything, which isn't very likely. Sorry if I was unclear.

Well, we've been having technological issues at our school recently (overload caused all of the information that teachers had on their computers to be wiped and the internet wasn't working for a while). I'm in constant contact with our school AMC coordinator (because I'm heavily involved in the math team and she's the sponsor for that) so I'm assuming that if she had gotten the e-mail, she'd have reminded me (and I prescored my AMC12B as a 109.5, so yeah..), but I'll ask her tomorrow if she can check. The problem I think lies in the fact that she can only check her e-mail at school and I'm not quite sure if the internet has been fixed, yet.

The postal mail follow-up, with the AIME contests and a computer print-out (identical to the information in the email) should be in the possession of your teacher.  We do not rely solely on email.  Ask, your teacher has the information.

Steve Dunbar
American Mathematics Competitions

Mk, like I expected, she said that she hadn't gotten the information yet (had a math competition yesterday, so I asked her on the way there).

So.. what do I do?

Bump...

What a bummer if I can't take the AIME...

In order to help you, we need some basic information, for instance the school contest manager, your school name, your school CEEB, your city and state, whether your school took the A-date contests or the B-date contests, etc.

Have your teacher call the AMC office,  we have an 800 number published on everything we print.  Check on the results for your school, check on the package tracking if available, check to see who signed for the package when it arrived, and we can work from there.

The AIME Alternate date is still available.

Steve Dunbar
Director, American Mathematics Competitions

Ok, well she hasn't been at school for the last 3 days (according to my current math teacher, the math competition took a lot out of her, and the death of her mother recently isn't helping, I'm sure) and I don't know what to do. I know that you seem to be against students finding out what they got, but is there any way that I could find out if I qualified for the AIME. I mean, the probability that I will be able to take the AIME II is slim b/c I think we'll be on spring break, but I just want to know.

Honestly, there should be a way for students to find out. Not everyone has a coordinator that's up to date (like I said, last year she caused 2 people to not be able to take the AIME, and it looks as if I'll be in the same position this year).

Oh well, it's life, right?

I suggest you talk to:
1.  another math or science teacher in the school, perhaps the department chair
2.  a principal or vice-principal,
3.  a counselor.

I believe that somewhere in your school the package with the results is waiting to be opened.

If you identify your school, your school CEEB number or some other identifier we can check our records to determine when and to whom that package was delivered.  Without that information, I cannot help you determine your scores.

Steve Dunbar
Director, American Mathematics Competitions


logarithms
limit
trigonometry
calculus
derivative
real analysis
real analysis unsolved

\[ \frac{{(1 \plus{} x)^{\frac{1}{x}}  \minus{} e}}{x}
\]

Calculate without derivates

You can use $ (1\plus{}x)^\frac{1}{x}\equal{}e\minus{}\frac{e}{2}x\plus{}\frac{11e}{24}x^2\plus{}...$, so
$ \frac{(1\plus{}x)^\frac{1}{x}\minus{}e}{x}\equal{}\frac{\minus{}\frac{e}{2}x\plus{}...}{x}\to \minus{}\frac{e}{2}$ for $ x\to 0$

But to calculate the first expansion, you actually use derivatives.

thanks
i`m still waiting for a method to do it without derivates

i make it easier a little, if i didn`t missed any sign or number it should look like this :

\[ x \cdot \frac{{\ln (1 \plus{} x) \minus{} x}}{{x^2 }}
\]

If you are allowed to use the facts $ e^x = 1+x+\mathcal{O}\left(x^2\right)$ and $ \log (1 + x) = x-\frac{x^2}{2}+\mathcal{O}\left(x^3\right)$ then,

\begin{eqnarray*} \lim_{x\to 0}\frac {(1 + x)^{\frac {1}{x}} - e}{x} & = & \lim_{x\to 0}\frac {e\left(e^{\frac {\log (1 + x)}{x} - 1} - 1\right)}{x} \\
& = & \lim_{x\to 0}\frac {e}{x}\left(e^{ - \frac {x}{2} + \mathcal{O}\left(x^2\right)} - 1\right) \\
& = & \lim_{x\to 0}\frac {e}{x}\left( - \frac {x}{2} + \mathcal{O}\left(x^2\right)\right) \\
& = & - \frac {e}{2} \end{eqnarray*}

no, i can`t use it :|
you started using derivates near the end

No, I didn't differentiate anything; I used only the assumptions stated at the top.  Well if you aren't allowed to use those two facts, both can be proven from definitions without using any calculus.  But what's the point of reinventing the wheel?

[quote="JoeBlow"]But what's the point of reinventing the wheel?[/quote]
I'm in agreement with JoeBlow on this. In both this topic, and to an even greater extent on night's earlier topic, I've found the request to do it "without using derivates" to be entirely artificial and perhaps pointless.

Put another way: any method whatsoever for finding $ \lim_{x\to0}\frac{\sin x\minus{}x}{x^3}$ has in effect computed the third derivative of the sine at zero; it doesn't matter what methods were used.

that was the teacher request, with derivates is very easy


calculus
integration
logarithms
trigonometry
calculus computations

Prove that
$ \int _{0}^{1/4\,\pi }\!{\frac {\ln  \left( 1\plus{}\tan \left( u \right) \right) }{\cos \left( u \right) \sin \left( u \right) }}{du}\equal{}1/12\,{\pi }^{2}$
and
$ \int _{0}^{1/4\,\pi }\!{\frac { \left( \ln  \left( 1\plus{}\tan \left( u \right)  \right)  \right) ^{2}}{\cos \left( u \right) \sin \left( u \right) }}{du}\equal{}1/4\,\zeta  \left( 3 \right)$

First one:

Since for $ x \in \left[0, \frac{\pi}{4} \right]$ we have $ 0 \leq \tan x \leq 1$, we can write:

$ I \equal{} \int ^\frac{\pi}{4} _ 0 \frac{\ln \left(1 \plus{} \tan x\right)}{\sin x \cos x} \, dx \equal{} \int ^\frac{\pi}{4} _ 0 \frac{1}{\cos ^2 x} \sum ^\infty _ { n \equal{} 1} \frac{(\minus{}1)^{n\plus{}1}}{n}\tan^{n\minus{}1} x \,dx$

Now, due to absolute convergence of the sum, we can write:

$ I \equal{}  \sum ^\infty _ { n \equal{} 1} \frac{(\minus{}1)^{n\plus{}1}}{n}  \int ^\frac{\pi}{4} _ 0 \frac{\tan^{n\minus{}1}x }{\cos ^2 x} \,dx$

The integral can be easily computed since we have:

$ \int ^\frac{\pi}{4} _ 0 \frac{\tan^{n\minus{}1}x }{\cos ^2 x} \,dx \equal{} \int ^\frac{\pi}{4} _ 0 \tan^{n\minus{}1}x \frac{d}{dx}\left(\tan x\right) \,dx \equal{} \frac{\tan ^n x}{n} \Big| ^\frac{\pi}{4} _ 0 \equal{} \frac{1}{n}$

Plugging that in, we have:

$ I \equal{} \sum ^\infty _ { n \equal{} 1} \frac{(\minus{}1)^{n\plus{}1}}{n^2} \equal{} \frac{ \zeta(2)}{2} \equal{} \frac{\pi ^2}{12}$


function
search
algebra unsolved
algebra

Prove or disprove there exists a continuous function which is nowhere differentiable.

Did you try googling that? You'll find lots about Weierstrass' example. You can also search the forum, since I'm sure people have posted before on the subject.

There is a famous example by Weierstrass showing that there are continuous nowhere differentiable functions, or we may use the Baire Category Theorem.  Would you actually like to see an example?

Thanks for letting me to know this is actually very famous. I just searched in mathworld and found [url=http://mathworld.wolfram.com/WeierstrassFunction.html]this[/url].

That function is still *somewhere* differentiable.  There are functions that are nowhere differentiable.

Put $\varphi(x)=|x|$ for $-1\leq x\leq 1$, and make it periodic by setting $\varphi(x+2)=\varphi(x)$.

Now define $f(x)$ as

$f(x)=\sum^{\infty}_{i=0} \left(\frac {3}{4}\right)^i\varphi(4^ix)$

This is continuous but nowhere differentiable.


I was wondering if someone could name a few math competitions that satisfy the following:

[list]
[*] over the Internet
[*] free
[*] does not require to sign up a school[/list]

apart from the iTest.

Thanks in advance!


inequalities calculus algebra function domain

for reals a, b > 0 prove that ((a+1)/(b+1))b+1 :ge: (a/b)b

I'm curious if there are non-calculus methods:

[hide]Everything that follows is in reference to the positive real numbers as our domain for x.

Re-write our equation as bb/(b + 1)b + 1 :ge: ab/(a + 1)b + 1.  This needs to be true for all a not equal to b, so we need to show that f(x) = xb/(x + 1)b + 1 has a global maximum at b.  To do this, we can just examine f'(x).

f'(x) = (b*xb - 1*(x + 1)b + 1 - xb*(b + 1)*(x + 1)b)/(x + 1)2b + 2.  The denominator is always positive, so we need only examine the numerator, b*xb - 1*(x + 1)b + 1 - xb*(b + 1)*(x + 1)b = xb*(x + 1)b*(b(x + 1) - x*(b + 1)) = b*(x + 1)b*(b - x).  This is positive when x  b, and zero when x = b, which gives us a global maximum at x = b, which proves our theorem.[/hide]

((a+1)/(b+1))^(b+1)   :ge: (a/b)^b

[hide] (b+1)log(b)(a+1)-(b+1)log(b)(b+1) :ge: blog(b)a-blog(b)b

 I'm going to simplify the right and left and sides seperately

First the left

(b+1)log(b)(a+1)=(b+1)(log(b)a+log(b)1) = (b+1)(log(b)a)

and -(b+1)log(b)(b+1)=-(b+1)log(b)b-(b+1)log(b)1=-b+1

So we have the left side equal to

(b+1)(log(b)a)-b+1 :ge: blog(b)a-blog(b)b

the right hand side is blog(b)a-b

adding b to both sides equals

(b+1)(log(b)a)+1 :ge: blog(b)a

dividing by log(b)a (which must be positive, therefore no sign change is necessary)

equals

(b+1)+log(a)b  :ge: b

Subtracting (b+1)

log(a)b :ge: -1

Which is obviously true[/hide]

This is what I got, is it wrong?

I believe that this is a viable non-calculus solution.

logb(a+1) = logba+logb1  is not true!  log(a*b) = log(a) + log(b).  log(a + b) isn't anything!

thanks for the sol'n Joel, that question was on a contest i took wednesday, and I thought I could do it by fixing b and varying a, but that did not work out for me.

Well, that's essentially what I did, but I made it look nice first :).  Was this a contest where calculus was expected?

I don't know what that means in terms of my question.

By the way, if you're a freshman at Cornell interested in math, you wouldn't happen to know Konstantin Getmanchuck (not sure of spelling there), would you?  Skinny Russian guy from Brooklyn?

oops, i misread your original post thinking you asked what contest it was.

Anyway, i do not know Konstantin, did he go to Stuy with you?

Here are two more proofs.  Actually, both proofs are basically one proof in different words.  I think they qualify as non-calculus proofs, but you tell me.

First proof: [hide]We will use the weighted AM-GM inequality, which goes like this.  Let x and y be nonnegative numbers.  Let p and q be nonnegative numbers such that p + q = 1.  Then

px + qy :ge: xpyq.

Let's plug in x = a/b, y = 1, p = b/(b+1), and q = 1/(b+1).  Then we have

(a+1)/(b+1) = px + qy :ge: xpyq = (a/b)b/(b+1).

Raising both sides to the power b+1 gives the desired inequality.

[/hide]

Second proof: [hide]We will use the concavity of the log function.  Note that

(a+1)/(b+1) = b/(b+1) (a/b) + 1/(b+1) (1).

By concavity, we have

ln((a+1)/(b+1)) :ge: b/(b+1) ln(a/b) + 1/(b+1) ln(1) = b/(b+1) ln(a/b).

Taking exponentials on both sides gives

(a+1)/(b+1) :ge: (a/b)b/(b+1).

Again, raise both sides to the b+1 power to finish the proof.

[/hide]

Konstantin went to Brooklyn Tech, but I know him from math stuff (NYC math team, CCNY math camp).

I thought about trying to use the shape of the log function but couldn't figure out how.  Nice proofs!

hmm, i actually think i know who you are talking about, there is this kid that I see around who fits the description and sometimes wears a nysml shirt, could be the same person.


Find all points on the y-axis that are 5 units from the
point (-4, 4).

What about if the question asked for all points from the same point given above but this time from the x-axis?

[hide]So you have one point $(-4,4)$. The distance has to be 5 units and on the y axis. In order to do this problem we do the distance formula backwards. Lets call the other point $(0,x)$. Plugging it in to the distance formula we have $5=\sqrt{(-4)^{2}+(4-x)^{2}}$. Solving for x we have two solutions. $x=1,7$. So the two points are $(0,1)$ and $(0,7)$[/hide]

In order to do this for the x axis we have to set the $(x,0)$ and the distance fromula backwards from there.

So, definitely use the distance formula.

Got it.

Thanks.

before you do it for the x-axis, notice that (-4,4) is the same distance from the y-axis as it is from the x-axis.  all you have to do now flip the whole cartesian coordanates around the line y=-x.


LaTeX
inequalities unsolved
inequalities

Let a,b,c be positive numbers. Prove that
a(b+c)/(b^2+bc+c^2)+b(a+c)/(a^2+ac+c^2)+c(a+b)/(a^2+ab+b^2)>=2.
( I am sorry because I don't know how to use latex now ) :blush:

[quote="CHICKEN"]Let a,b,c be positive numbers. Prove that
$ \frac {a(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2} \plus{} \frac {b(a \plus{} c)}{a^2 \plus{} ac \plus{} c^2} \plus{} \frac {c(a \plus{} b)}{a^2 \plus{} ab \plus{} b^2}\geq2.$
[/quote]
By Cauchy-Schwartz we obtain:
$ \sum_{cyc}\frac {a(b \plus{} c)}{b^2 \plus{} bc \plus{} c^2}\sum_{cyc}\frac {a(b^2 \plus{} bc \plus{} c^2)}{b \plus{} c}\geq(a \plus{} b \plus{} c)^2.$
Id est, it remains to prove that $ (a \plus{} b \plus{} c)^2\geq2\sum_{cyc}\frac {a(b^2 \plus{} bc \plus{} c^2)}{b \plus{} c},$ which equivalent to
$ \sum_{sym}(a^4b \minus{} a^3b^2)\geq0.$
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25553


Three of the following test scores are Cyprian's and the other three
are Margaret's: 85, 87, 92, 93, 94, 98. Cyprian's mean score is
90. What is Margaret's mean score?

The scores total $ 85 \plus{} 87 \plus{} 92 \plus{} 93 \plus{} 94 \plus{} 98 \equal{} 549$

Cyprian's total $ 90 \cdot 3 \equal{} 270$

Margaret's total $ 549 \minus{} 270 \equal{} 279$

Her mean score is $ \frac {279}{3} \equal{} 93$

OR

The scores total $ 85 \plus{} 87 \plus{} 92 \plus{} 93 \plus{} 94 \plus{} 98 \equal{} 549$

The average score is $ \frac{549}{6} \equal{} 91.5$

Both took three tests, so Margaret's average score is $ 91.5 \plus{} (91.5 \minus{} 90) \equal{} 93$

we first find the mean of all six scores.
that mean will also be the mean of cyprian's and margaret's scores.
we can use it to find margaret's mean.
mean of 85, 87, 92, 93, 94, 98 is
$ 90\plus{}\frac{\minus{}5\plus{}\minus{}3\plus{}2\plus{}3\plus{}4\plus{}8}{6}\equal{}90\plus{}\frac{9}{6}\equal{}90\plus{}\frac32$
so since it's $ \frac32$ greater than cyprian's mean, it's $ \frac32$ less than margaret's making margaret's mean $ 90\plus{}3$
93.
preview edit: cyprian's total is $ 90\cdot3$, since each person has 3 scores.

Let x be margaret's mean score.
We have:
(90)(3) + 3x = 85+87+92+93+94+98
270 + 3x = 549
3x = 279
x = 93

Mission's accomplished! :clap: Thank you. :clap2: Thank you  :winner_second:

An easier way is the guess and check (quite ironic). Notice that Margaret must have three scores such that the difference from 90 adds to 0. So 85 would be -5 (90-85), 92 would be 2 (92-90). So you see that 85, 92, and 93 have differences that add up to zero. Just average 87, 94, and 98 and get $\boxed{93}$


trigonometry
integration
calculus
calculus computations

For real $a, b$ define the sequence $x_{0}, x_{1}, x_{2}, ...$ by $x_{0}= a, x_{n+1}= x_{n}+b \sin x_{n}$. If $b = 1$, show that the sequence converges to a finite limit for all $a$. If $b > 2$, show that the sequence diverges for some $a$.

$x_{n+1}-x_{n}$ we can mean $x'(n)$. then $\int_{0}^{n}\frac{dx(n)}{sinx(n)}$~$\int_{0}^{n}bdx$ $\Rightarrow$ $ln(|tan\frac{x(n)}{2}|)|_{0}^{n}$~$bn$  $\Rightarrow$ $ln(tan(x_{n}/2))-ln(tan(a/2))$ ~ $nb$, this is idea and then is simpl. I think so.


Sequence ${a_{n}}$ is:
$a_{1}=2$
$a_{n}=2(n+a_{n-1})$, for all $x\geq 2$
Prove that $a_{n}\leq 2^{n+2}$, for all $n$, who $n$ is natural numbers.
 :wink:


inequalities
AMC
USA(J)MO
USAMO
inequalities proposed

if a,b,c>0 prove the following inequality:

       (1+a/b)(1+b/c)(1+c/a)>=2(1+(a+b+c)/2 3:sqrt(abc)). I hope tou do understand what I wrote.

It's not USAMO, it's APMO, but the author is common: Titu Andreescu. Why is it tricky? Direct application of AM-GM?  :?

Of course I can. One solution is based on the fact that the inequality is homogenous and we can take $abc=1$. Then use the fact that $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}\geq x+y+z$ when $xyz=1$, which is trivial. Use this twice and you have the result. Another solution follows from the fact that $\frac{a}{b}+\frac{a}{c}+1\geq\frac{3a}{\sqrt[3]{abc}}$. Add them and use AM-GM once again.

[quote="ehku"]Easy:
[tex]a(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \ge \frac{3a}{\sqrt[3]{abc}}[/tex]

 :lol:[/quote]

That's just GM-HM...


I used the following identity:
3S=4sqrt[(ma+mb)(ma+mc)(mb+mc)(ma+mb+mc)] and I got the following inequality which I can't figure out what to do. Perhaps this is not a good idea!

I got to:
[(ma^2+mb^2+mc^2)/(ma+mb+mc)]^4>=4(ma+mb)(ma+mc)(mb+mc)(ma+mb+mc) which I cannot prove! Perhaps someone could!

tschuess

Hey!Nobody knows to solve this problem.Is someone interested to see the solution?Shall I put the solution?I am waiting your opinion.

   Regards!

I would love to see a solution! This one has bugged be enough! I think we should see a solution! :D

4 Lagrangia
    I am going to put a solution of my problem
    We know that ma^2+mb^2+mc^2=3(a^2+b^2+c^2)/4
    After we use the ineq \sum a^2>= \sum ab,we use the fact that ab>2S
because sinC<1 Idem for bc,ac.
    So we have  \sum ab>6S That means  \sum ma^2>9S/2 (1)
    Then from Cauchy we have  \sum ma^2>=( \sum ma)^2/3 (2)
     Now we will (1)*(2) and we will have what we are loking for.

     I am still waiting other solutions!

well.. has anyone tried this one using the identity I gave? Cause with that identity we get an inequality only in ma,mb and mc which should be easier to solve ( I think :D).. though I can't see how..I  tried denoting ma+mb+mc=x, sum(ma*mb)=y and prod(ma)=z but got nowhere! perhaps someone will find something!
cheers!

[quote]I created this problem and I want to know if it's not too easy.

Let ABC be a triangle with standard notation. Prove that:

\sqrt 3S/2 \leq (ma*ma+mb*mb+mc*mc)/(ma+mb+mc) [quote]
  Here is my solution:
   ma^2+mb^2+mc^2=3*(a^2+b^2+c^2)/4\geq 3*4*3^(1/2)*S/4=3^(3/2)*S
   By using CS  ma+mb+mc\leq  sqrt(3*(ma^2+mb^2+mc^2))
   Then  (ma*ma+mb*mb+mc*mc)/(ma+mb+mc)\geq  sqrt(ma^2+mb^2+mc^2)*3(-1/2)\geq sqrt(S)*3^(1/4)

lagrangia, can u post a proof of your identity?

Lagrangia you used that fact [u] ab>2S[/u] So in that problem that you gave there is no equality sign.


In your formula where do you shut the door ?  ")"


limit
function
calculus
calculus computations

How do I get $\lim_{x\rightarrow 1}\frac{x^{p}-1}{x^{q}-1}$ $=$ $\frac{p}{q}$ without using L'Hopital's rule? ($p$ and $q$ being integers)

easiest way is to use the substitution $y=x-1$, and then apply the binomial theorem...

$\lim_{x\to 1}\frac{x^{p}-1}{x^{q}-1}=\lim_{y\to 0}\frac{(1+y)^{p}-1}{(1+y)^{q}-1}=\lim_{y\to 0}\frac{(1+py+\textbf o(y^{2})+\cdots)-1}{(1+qy+\textbf o(y^{2})+\cdots)-1}$

$=\lim_{y\to 0}\frac{p+\textbf o(y)+\cdots}{q+\textbf o(y)+\cdots}=\frac{p}{q}$

$\lim_{x\rightarrow 1}\frac{x^{p}-1}{x^{q}-1}=\lim_{x\rightarrow 1} \frac{\frac{x^{p}-1}{x-1}}{\frac{x^{q}-1}{x-1}}$

$=\lim_{x\rightarrow 1}\frac{(x^{p})'|_{x=1}}{(x^{q})'|_{x=1}}=\frac{p}{q}$, which can be obtained by the difinition of differential coefficient.

[quote="misan"]
$ \lim_{x\to 1}\frac{x^{p}-1}{x^{q}-1}=\lim_{y\to 0}\frac{(1+y)^{p}-1}{(1+y)^{q}-1}=\lim_{y\to 0}\frac{(1+py+\textbf o(y^{2})+\cdots)-1}{(1+qy+\textbf o(y^{2})+\cdots)-1}$
[/quote]
I agree with your answer and method, but I must tell some correction

I note that
[quote="misan"]
$ \lim_{x\to 1}\frac{x^{p}-1}{x^{q}-1}=\lim_{y\to 0}\frac{(1+y)^{p}-1}{(1+y)^{q}-1}=\lim_{y\to 0}\frac{(1+py+\textbf o(y^{2})+\cdots)-1}{(1+qy+\textbf o(y^{2})+\cdots)-1}$
[/quote]
(... about defination)
here :
$ (1+y)^{p}-1=1+py+\textbf o(y^{2})+\cdots)-1$ this is not correct.
. but this
$ (1+y)^{p}-1=1+py+p(p-1)y^{2}/2+\textbf o(y^{2})+\cdots)-1$
is correct.
for example:
let's consider function
$ f(x)=x^{2}$ in point $ 0$
I can say that
$ f(x)=o(x)$ in point $ 0$ because
$ \lim_{x\to 0}\frac{f(x)}{x}=0$
but in you case
you say
$ f(x)=o(x^{2})$ (this is not true because $ \lim_{x\to 0}\frac{f(x)}{x^{2}}=1$)
but you can say  $ f(x)=O(x^{2})$ because it's mean $ |f(x)|<Cx^{2}$ for the some constant $ C$)

And in our case
$ \lim_{y\to 0}\frac{(y+1)^{p}-1}{(y+1)^{q}-1}=\frac{1+yp+o(y)-1}{1+qy+o(y)-1}=\frac{p+\frac{o(y)}{y}}{q+\frac{o(y)}{y}}=\frac{p}{q}$


inequalities
triangle inequality
inequalities proposed

Let a,b,c denote the sides of a triangle . Prove that the quantity $(a/(b+c)) + (b/(c+a)) + (c/(a+b))$ lies between  2 and  3/2.Does equality hold at either limit?

even if a,b,c are not sides of triangles, it's still true. This has been posted for many times already.

No, for one half of the inequality, \[ a/(b + c) + b/(c + a) + c/(a + b) < 2, \] a, b and c must be sides of a triangle, I think...

oops i messed up with another problem. Sorry.  :blush:

[quote="cezar lupu"]If $a, b, c>0$ then the following inequality holds:

       $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b} <2$ ;)[/quote]

$(a,b,c)=(1,1,\infty)$..

Sorry, Siuhochung is right here :oops: . Actually,

$\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a} <2$ ;)

the solution i gave when i saw by the first time this inequality consisted in a brute-force multiplication and then using the triangle inequality... although there is a very simple one.

by the triangle inequality we have that $\dfrac{a}{b+c}<\dfrac{2a}{a+b+c}$, then taking the cyclic sum gives the result...  :D

yeah I know this was easy! I'll come up with something better next time!


Prove that there are exactly [b]7[/b] planes that are equidistant from 4 non-coplanar points in space.

You might want to refine what you mean by "equidistant from 4 points".

If the definition for [u]a plane equidistant from 4 points[/u] is [u]the 4 perpendicular segments dropped from the given 4 points on the plane are all the same length[/u] (hence "equidistant"), then there are indeed 7 such planes. I'll leave this for someone to have fun with :)

Come on now. This shouldn't be that difficult.
[hide="a hint"]
Any three points lie on the same plane.
Now consider 3 points in space. No matter where the 4th point is chosen, the 4 points are co-planar. {why?}
Therefore any 3 points should not be collinear.
Now with 4 such points, we can form 4 triangles with one point not in the plane on which the triangle lies.
Construct a line perpendicular to the plane and passing through this point. The plane passing through the midpoint of this line segment and parallel to the plane on which the triangle lies is a plane that satisfies our condition.
Thus we have found 4 such planes.

Now find the other three.
[/hide]
10000th User's interpretation is correct.

[hide="Continuation from above"]
The other 3 occur when the plane divides space in half with two points in either half. There's $ \frac{1}{2}\binom{4}{2} \equal{} 3$ ways that can happen.  :)
[/hide]

[hide="Solution"][b]3 points on one side and 1 on the other:[/b]
The three points define a plane. Let this plane be P, and let the point be Q. Let Q' be the projection of Q onto P, and let M be the midpoint of QQ'. There is a unique plane L passing through M that is parallel to P.

Thus, this case results in four possible planes -- one for each choice of P.

[b]2 points on each side:[/b]
Let P and Q be two points, which we will put on one side of the plane, and let R and S be two points on the other side of the plane. Let v be the direction vector of PQ and u be the direction vector of RS. Then, the family of planes with normal vector uxv is parallel to both line PQ and line RS.

Let A and B be points on PQ and RS, respectively, such that AB is the unique such line segment that is perpendicular to both PQ and RS (AB has direction vector uxv). Let M be the midpoint of AB. Then, the plane with direction vector uxv that passes through M is the unique plane equidistant from P Q R and S.

This case results in three possible planes -- one for each way of dividing four points into two groups.[/hide]


function
calculus
derivative
algebra unsolved
algebra

[quote="willydavidjr"][b][size=150]Let f(x)=ax^2 - 4ax + b, (a>0) be defined in 1<=x<=5.
Suppose the average of the maximum value and the minimum value of the function is 14, and the difference between the maximum value and minimum value is 18. Find the values of a (a>0) and b.[/b][/size] ;)

[/b][/quote]
Let $\displaystyle f(x)=ax^2-4ax+b, (a>0)$ be defined in $\displaystyle 1\leq x\leq 5$

Suppose the average of the maximum value and the minimum value of the function is 14, and the difference between the maximum value and minimum value is 18. Find the values of $a$ and $b$.

LaTeXed for easier reading.

Solution:
y=ax^2-4ax+b, a>0
first derivative=2ax-4a=0, x=2
for x>2, f(x) is an increasing function
for x<2, f(x) in a decreasing function

so your max value is f(5)=5a+b
your min value=f(1)=-3a+b
(max value+ min value)/2=14, a+b=14   (i)

max value - min value=18, a=9/4
put x=9/4 into the equation (i), to get the value of b:
b=47/4


geometry
geometric transformation
rotation

A man of mass 80 kg mass is standing on the rim of a circular platform rotating about its axis. The platform with the man on it rotates at 12 r.p.m. How will the system rotate, if the man moves to the platform's centre? What work will the man perform in changing his position? The mass of the platform is 200 kg and the radius is 1.2 m.

Apply angular momentum conservation to find the new angular velocity.
Then find the angular kinetic energies of new system and old system and then find the work done. ??

as the man moves to the axis of rotation the disk starts moving faster and faster..........
this is due to the law of conservation of angular momentum........


You may of heard of this problem before, so if you already know the answer please give others a chance.

A surveyor goes to a special recipient and asks, "what are the ages of your children?"
She replies,"Well, the product of my daughters'  ages is 36."
the surveyor says, "Please tell me more."
She replies, "It wouldn't matter if I told you the sum anyway, it won't do much good."
The surveyor said, "Well, can you tell me the smallest bit more of information?"
She replies, "Oh no! I have to go. I need to pick up my daughter Annie from swimming lessons, and I have to pick up the twins from daycare."

What are the ages of the children?

[hide="answer"]
I have seen this before but worded differently,
Let's say that the age of the twins is $ x$, and the age of the daughter is $ y$.

So that $ x^2\cdot y\equal{}36$, now we find the prime factorization of $ 36$ which is $ 2^2\cdot 3^2$.

Meaning that $ x^2$ has to be either $ 1$,$ 4$,$ 9$,$ 36$ and the $ y$ values are $ 36$,$ 9$,$ 4$,$ 1$ respectively.

Now we cross out the impossible $ y$ values like $ 36$,$ 4$,$ 1$. Because first she wouldn't be picking up a $ 36$ year old and normally $ 4$ year olds and $ 1$ year olds do not take swimming lessons. Then the two twins are each $ 2$ years old and the daughter is $ 9$ years old. I believe this answer fits best with the description of the children.

Might be wrong, might be right. I am putting my solution based on a previous question like this.
[/hide]

[hide] This is how you have to solve these problems : Because she says she has to pick up her twins, meaning that they have the same age. Let the age of 1 twin be A and the age of Annie be B. A x A x B = 36; A^2 x B = 36. So we have to find two numbers that multiply to 36. so lets write the factors of 36 as pairs where 1 pair is a perfect square in the form (A,B) {(36,1) (1,36) (4,9) (9,4)} So the answer has to be one of these. The lady says that her twins are at a daycare and the older daughter is at a swimming lesson. so it cannot be the pair (36,1) because 6 year-olds don't go to day care and a 1 year old doesnt go to swimming lessons alone. It can't be the pair (1,36) because you don't leave 1 year olds at daycare when they are so young. You are left with (4,9) and (9,4).
The swimmer must be older than the twin because swimmers are older than kids atdaycare. Therefore the answer is {2,2,9} [/hide]

Or we note that by the second clue, (2,2,9) and (1,6,6) are the possible solutions, since we must have two solutions to the first clue summing to the same value.

Usually the problem implies that there is an oldest child. However, in this case 6 year olds don't attend daycare (actually, I'm not sure about that). So (2,2,9).

What would the answer be if the twins consisted of one boy and one girl?
The problem states that the product of the daughters' ages is 36. Then the answer would be xy=36

What if the twins were boys!!!
Then, the daughter's age would be 36!!!  :D

To rts2007:
   That's a good point you brought up, but the way the problem is stated, it tells you the product of the daughters' ages not the daughter's age. That tells me that there must be at least two daughters, but the problem does not tell you what constitutes the pair of twins. So it is possible that there could be only one girl.

If there were a boy and a girl, there would be multiple possibilites.

I agree with both bogtro and rts.a major reason.

There is nothing in the statement of the problem that gives you the idea that the solution is limited to only one answer.

Guys, the chlildren are all females.

And the information requires a bit of logic.

BOGTRO quotes "Or we note that by the second clue, (2,2,9) and (1,6,6) are the possible solutions, since we must have two solutions to the first clue summing to the same value."
Isn't it possible that the twin girl could be 4 and the older daughter could be 9?
Do you see anywhere in the problem where it says, without a doubt, that there are three girls?

Without doubt, this is an interesting problem!

-rts2007


prove that if $ [x \plus{} y ] \equal{} [na \plus{} b]$for every $ \in N* ; a,b,x,y \in R$, then $ \equal{} a , y \equal{} b$

[quote="long14893"]prove that if $ [x \plus{} y ] \equal{} [na \plus{} b]$for every $ \in N* ; a,b,x,y \in R$, then $ x\equal{} a , y \equal{} b$[/quote]
no one interested :(?


The circumradius and the inradius of a certain right triangle are 17/2 and 3, respectively. Find the length of the shortest height. Express your answer as a mixed fraction.

$ \text {(A) } 5 \frac{1}{7} \qquad \text {(B) } 7 \frac{1}{17} \qquad \text {(C) } 7 \frac{2}{19} \qquad \text {(D) } 8\frac{1}{14} \qquad \text {(E) } 10\frac{1}{4}$

I am $ 97\frac{1}{13}\%$ sure that the AMC's do not have this problem in their archive.

Well, since its a right triangle, the circumradius  is 289

and also (a+b-17)/2=3

so we have a+b=23
and a^2+b^2=289

Squaring the first equation you get 289+2ab=529

so 2ab=230 and ab=120, area is 60 (thats irrelevent)

since its a right triangle, ab=120=17*h, so h=7 1/17 B


Wich of the following represents the largest # of moles

a) 10g NH3
b) 10g N2
c)10g CU
d)10g Br2
e) All these contain the same number of moles

??????

find the molar mass of each, devide the gram by molar mass to find the MOLE


i was wondering what kind of grades/scores/etc. it takes to get a merit-based scholarship, especially when applying to an out of state college. for instance, i am planning to apply to florida and possibly duke (with clemson being my in-state choice).  would the following get a free-ride?

-gpa: 4.9 weighted, 3.95 unweighted
-sat: 1310 - 720 math, 590 english (only taken once and i will have to take the new one, which will hopefully be easier)
-member of math team and i will join mu alpha theta my senoir year.
-not too many volunteer activities
-a few notable acedemic awards

what grade are you in?
i just took the SAT and i'm in 9th right now, got 1330(760 on math)( i hate verbal)

i believe you can get in to Clemson very easy(i lived in SC for 2 yrs so i'm a big clemson fan!!!)

you can probably get into FL if you get a higher score on the new SAT(which is harder).

i don't think you can't get into duke(my #1 choice) because of the few extracurricular activites(i joined math, academic, beta, and plan to join Key club, and mu alpha beta and business clubs next yr)

your GPA is very very good, my is only 3.7

duke is very competitive(20-25%)

Currently, you almost certainly wouldn't get accepted to Duke, let alone a free ride.  Certainly not with that SAT and the lack of volunteering/extracurriculars.  But you have time to bump that stuff up, so start soon.

Did you take the PSAT your junior year?
If you qualify for national merit scholar you can get (basically) full rides to Arizona State Univ. and Oklahoma Univ.
Florida has a pretty nice NMS package too.

The most sensible way to answer your questions is to call up the admissions departments of the schools in question.  Now that admissions are nearly done, they probably will be at a relative ebb of work, and I'm sure there is an admissions officer there who wouldn't mind talking about this to you.

You're a junior now, right? Having higher SAT scores would help IMMENSELY in making merit scholarships available to you.

Furious, I know plenty of people from South Carolina who have gotten into Duke with SAT scores similar to yours. So don't let anyone discourage you from applying.

Talk to your guidance couselor and keep a look out for outside sources of funding and scholarships. Whether you can get a "free-ride" or not - who knows. Just apply to where you want to go and fill out as many scholarships applications as you can.

[quote="gauss202"]Furious, I know plenty of people from South Carolina who have gotten into Duke with SAT scores similar to yours. So don't let anyone discourage you from applying.[/quote]

Gauss, the last thing I would want to do is be discouraging.  Last year's SAT 1 midrange (25-75%) at Duke was 1350-1510.  Considering that a large number of students in the lower 25% are athletes and whatnot, and also considering that he's said that his extracurriculars are relatively weak, that does not bode well - even given that admissions are marginally less selective when it comes to the Carolinas.

Now, he does have plenty of time to fix the lack of extracurriculars and to up his SAT by 100 or even 150 points, both of which would help him a lot.  In particular, I think that a boost of 100 in the verbal score would be quite welcome.  I'm just saying that realistically, admissions would be a reach right now, and a full ride would be completely out of the question.

All I'm saying is that none of us are admissions officers. We don't really know what a school will decide on any particular student. I know of a student last year from South Carolina who got admitted to Duke with an SAT score very similar to his, and grades that were probably not quite as good. So admissions decisions can be based on a lot of factors that are not transparent to us.

I'm not giving him a gaurauntee that he will get admitted, I'm just saying that its certainly possible. So he should continue to work on his scores and application profile, and then apply when the time comes. It is unlikely that he (or anyone) will get a "full-ride" solely from the University - but if he gets in and he's really interested in going, there are always other financial aid options and outside scholarships that he can apply for.

I know furious personally and I don't think it's unrealistic for him to get accepted to Duke.  I can almost guarantee that he will finish as either 1st or 2nd in his class.  He took AP US History last and got a 5.  And he will most likely get 5's on AP Euro, AP Stats and AP Bio this year.  Next year he will be taking AP Chem, AP Gov,  and AP Calc and will probably get 5's on these too.  So AP wise he is looking very strong.  He can definitely raise his SAT score.  An 800 should be easy for him to attain and if he practices his verbal that will go up also.  Don't know about writing but I think he should do decent on that section.  His extracurricular record isn't spectacular but does have math team for that.  He will have been on the math team all 4 years.  His accomplishments are very nice (compared to average people not AoPSers ;) ).  His biggest weak spot is volunteering.  I don't think has any volunteer work yet.

And to the OP I would say, definitely build a plan to READ, READ, READ, and READ to increase your verbal score, which ought to be possible in the amount of time you still have.

As of right now, tutoring is my [i]only[/i] form of community service.  I tutor for about 2 and a half to 3 hrs a week.  I didn't have to look for opportunities rather they came to me.  I should have about 30 hrs at the end of this year.  This sounds kind of superficial but what is a good number to have for a transcript?  I also have a bunch of hours from helping out at church when I was in middle school.

thanks for all the comments.

-i am a junior
-i do feel my sat will go up on the new one. first of all, i expect to make an 800 math and my verbal will go up since analogies are gone. i was in the 25th percentile for analogies (yuck).
-my psat score was only a 206, so i won't be a national merit scholar.
-i will probably be 2nd in my graduating class, i doubt i can pass  joml88.
-i did student volunteer for a year (it's a class where you help a teacher) and volunteered at a marathon once.
-i don't see me picked up tutoring cuz i am not good at it as i do many things (especially math) different from the way i should be tutoring it. for not tutoring, i had to drop out of the National Honor Society (or as i call it National Tutoring Society as all you did was tutor the athletes if you didn't have anyone else to tutor).

is a full-ride to florida out of the question?

I think you are in full-tuition "merit scholarship" range for some state universities with your current stats, especially now that I have seen your classmate's description of your course load. I'm not sure which private universities would offer "merit" money at your range, but some of those are better at offering "financial need" discounts than the public universities are, so you should apply to both kinds of schools and see what you are offered.

Duke is way to hard to get into with the score of 1310, you can hope and maybe get lucky(one or two lucky ones get in) but don't bet on it,  you have to raise it up 100 pts at least.
but the your APs and GPA and Class rank very help a lot
everybody going to duke are all A students and are the best in their own school, that's why colleges care so much about EXTRA activities, like commuity services(at least 4 contuines yrs at one place), clubs, interests, they want well-rounded students
try to find volunteer work at the hospital, theatre, sports clubs.etc

[quote="tokenadult"]For this thread, on the subject of volunteering, do a lot of you AoPSers look at oppportunities to tutor math as a volunteer activity? The inner city schools in my town seem always to be looking for tutors (or so they say in radio public service announcements), and I wonder if that would be suitable for a teenager to do as a community service activity, or if schools are mostly looking for adult tutors.[/quote]

FWIW, I can say that I had a very small amount of 'real' volunteering under my belt when I applied to universities - probably less than 20 hours.  What I did have was sustained involvement in a variety of activities - I was editor in chief of my school's newspaper, I created my school's math club (an activity I continue to be involved in), I had been involved with student council in the past, I was a (founding) member of a mentorship program for younger students - heck, I was even a pretty competitive table tennis player.  To these causes, I gave hundreds of hours.

I considered tutoring as a volunteer activity, but tutoring was my main source of income during high school - I was tutoring up to 8 hours a week at one point and making about 200 dollars a week, income that I was unwilling to give up for the sake of resume padding.


Nearly 2 weeks left to the great competition and seems like everybody is so busy with preparation except me, that nobody is starting the topic  :D

Who knows, may be we will get the answer to the topic after IMC  :lol:

[b]Edit:[/b] National University of Uzbekistan - me  :oops: (no team leader)

I am. Won't be much of a trip, though :P
Am guessin' i'm gonna go by bike..

There should be a team from Ukraine.

Students:
1. FESHCHENKO Ivan (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, gold medal at SEEMOUS-2008, 1st prize at IMC-2007
2. RADCHENKO Danylo (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, gold medal at SEEMOUS-2008, gold medal at IMO-2007, bronze medal at IMO-2006
3. SLOBODIANIUK Sergii (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, silver medal at Iran ISMO-2008, 1st prize at IMC-2007, 2nd prize at IMC-2006, gold medal at IMO-2005
4. KRAVETS Oleksandr (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, silver medal at Iran ISMO-2008, 1st prize at IMC-2007, 2nd prize at IMC-2006, bronze medal at IMO-2005
5. TYMOSHKEVYCH Larysa (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, bronze medal at Iran ISMO-2008, 2nd prize at IMC-2007, 2nd prize at IMC-2006
6. SHYSHATSKYI Iurii (Kyiv Taras Shevchenko National University) - gold medal at IMO-2008, bronze medal at IMO-2007
7. IURCHENKO Ivan (Kyiv Taras Shevchenko National University) - 1st prize at IMC-2008, gold medal at SEEMOUS-2008, 1st prize at IMC-2007
8. TANTSIURA Maksym (Kyiv Taras Shevchenko National University) - 2nd prize at IMC-2008, silver medal at SEEMOUS-2008
9. LIVINSKYI Ivan (Kyiv Taras Shevchenko National University) - 2nd prize at IMC-2008, 2nd prize at IMC-2007, 2nd prize at IMC-2006
10.KOVALENKO Oleg (Dnipropetrovsk Oles Gonchar National University)
11.APYSHKOV Mykhailo (Donetsk National University)
12.DZHULGAKOV Dmytro (National Technical University "Kharkiv Polytechnic Institute")

Leaders:
1.BONDARENKO Andrii (Kyiv Taras Shevchenko National University)
2.MITIN Dmytro (Kyiv Taras Shevchenko National University)
3.RIABYCH Nataliia (Ministry of Education and Science of Ukraine)

Also there should be a team from Odessa I.I.Mechnikov National University (2 students and 1 leader) and National Technical University of Ukraine "Kyiv Polytechnic Institute" (2 students and 1 leader).

This will most likely be the team from Zagreb, there are ten of us this year (if we, unlike before Vojtech, manage not to lose a couple of guys on illnesses or, believe it or not, burglaries):

Nikola Adzaga, second year
Goran Drazic, third year
Nikola Grubisic, fourth year
Zeljko Kereta, third year
Rudi Mrazovic, fourth year
Marin Misur, second year
Melkior Ornik, first year
Petar Sirkovic, second year
Sasa Stanko, second year
Luka Zunic, second year

Our team leader will be Matija Basic.

Seen a lot of ppl out there. Gonna be tough))
Tomorrow is a first day. Good luck everyone!

P.S.: my room number is 449.


quadratics
algebra
quadratic formula
special factorizations
polynomial
sum of roots

Please help me with this:
The sum of all the integers up to n (including n) is equal to 136. what is n? (you cannot use calculators)

[hide="how far i got"]First i know that $ \frac{n(n\plus{}1)}{2}$ is the sum all integers including n. So, we have $ \frac{n(n\plus{}1)}{2}\equal{}136$ but once i solve i get $ n^2\plus{}n\equal{}272$ which i can't solve because if i use the quadratic formula i can't do the square root part so can someone please help me with a completing the square approach?[/hide]

you got to $ n^2 \plus{} n \equal{} 272 \implies n^2 \plus{} n \minus{} 272 \equal{} 0$

sum of the roots of $ ax^2 \plus{} bx \plus{} c \equal{} 0$ is $ \minus{} \frac {b}{a}$.

a=1, b=1.

answer : -1

EDIT :

oh meh sry i misread the prob

whered -1 come from

First of all, memorize your perfect squares. That makes it really REALLY obvious that $ n \equal{} 16$ (just by estimation).

Secondly, I don't see why you cant do the quadratics.

$ \frac {(n)(n \plus{} 1)}{2} \equal{} 136$

$ (n)(n \plus{} 1) \equal{} 272$

$ n^2 \plus{} n \minus{} 272 \equal{} 0$

$ n \equal{} \frac { \minus{} 1\pm\sqrt {1^2 \plus{} 4\times 272}}{2}$

$ n \equal{} \frac { \minus{} 1\pm\sqrt {1089}}{2}$

$ n \equal{} \frac { \minus{} 1\pm33}{2}$

The answer's obviously positive, so the root is obviously $ 16$.

EDIT: Lol fantasy, sum of first n whole numbers, not sum of roots  :)

EDIT2: Completing the square is too beast and takes too long for a problem like this  :wink:

But Fantasy, since there are 2 solutions you don't want the sum, you want each of them individually.

EDIT: Sorry, tex. I didn't see your post. But I stand by what I said. I think Fantasy just misread it.  :lol:

Yeh, I figured that out... 3 mins after my post  :wink:

EDIT: I'm sure editing a lot... It's ok super  :)

the method asked for was completeing the square

$ n^2\plus{}n\equal{}272$
i hope you already know this, but to complete the square $ x^2\plus{}bx$, add $ \frac{b^2}{4}$
i will not explain it here, but you can ask someone else

so we add $ \frac14$ to both sides

$ n^2\plus{}n\plus{}\frac14\equal{}\frac{1089}{4}$
the left side may be simplified to (n+1/2)^2
$ (n\plus{}\frac12)^2\equal{}\frac{1089}{4}$

now taking the square root of 1089 either requires memorization, a calculator (which we cant use) or a slight bit of work
to take the square root of 1089 (assuming it is a natural number), first note that the square root is between 30 and 40 since it is between $ 30^2$ and $ 40^2$
then, since it ends in a "9", the square root ends in 3 or 7
then test 33 and 37 and you find it is 33

now taking the square root of both sides in the previous equation...
$ n\plus{}\frac12\equal{}\pm \frac{33}{2}$

which yields solutions 16 or -17 (but -17 is extraneous), so 16 is the answer

For more on completing the square, click [url=http://www.artofproblemsolving.com/Forum/weblog.php?ppage=3&sep=%2F&w=1172&start=39]here[/url]. See "Quadratics Part 3".

i know how to complete the square  :D
anyway thanks to every single one who posted here!
mods, you can safely delete this if you think its getting too spammy but i think it isn't

[quote="Poincare"]Please help me with this:
The sum of all the integers up to n (including n) is equal to 136. what is n? (you cannot use calculators)

[hide="how far i got"]First i know that $ \frac {n(n \plus{} 1)}{2}$ is the sum all integers including n. So, we have $ \frac {n(n \plus{} 1)}{2} \equal{} 136$ but once i solve i get $ n^2 \plus{} n \equal{} 272$ which i can't solve because if i use the quadratic formula i can't do the square root part so can someone please help me with a completing the square approach?[/hide][/quote]

I have another solution:

$ 1 \plus{} 2 \plus{} 3 \plus{} .... \plus{} (n \minus{} 1) \plus{} n \equal{} 136$

$ \begin{array}{l}
  \leftrightarrow \frac{{n(n \plus{} 1)}}{2} \equal{} 136 \leftrightarrow n^2  \plus{} n \minus{} 272 \equal{} 0 \\
  \leftrightarrow (n \minus{} 16)(n \plus{} 17) \equal{} 0 \leftrightarrow \left[ \begin{array}{l}
 n \equal{} 16* \\
 n \equal{}  \minus{} 17 \\
 \end{array} \right. \\
 \end{array}$

With $ n\equal{}16$ The sum of all the integers up to n (including n) is equal to 136.

ALTERNATIVE SOLUTION!

We can divide out 2 repeatedly from 134 and find it factors into $ 8\cdot{17}$

That becomes equal to $ \frac{16\cdot{17}}{2}$ which means our answer is $ 16$

what mewto said gave me an idea to find the square root quickly, prime factorize  :|


Let a,b,c,d be relatively prime to m=ad-bc. Prove that the pairs of integers (x,y) for which ax+by is a multiple of m are identical with those for which cx+dy is a multiple of m.

[hide="One line"] $ (x, y) \equal{} (dt\minus{}bs, as\minus{}ct)$ [/hide]


[b]You can see now an interesting Romanian site
http://www.mateforum.ro with the moderator [u]Cezar LUPU[/u] ![/b]

Congratulations on your opening site! :)

kunny

[b]Thank you ![/b] [u]You are the first my quest ![/u]


For $ a>b>c>0$, prove :

$ \frac{\sqrt{a}}{(a\minus{}b)(a\minus{}c)}\plus{}\frac{\sqrt{b}}{(b\minus{}c)(b\minus{}a)}\plus{}\frac{\sqrt{c}}{(c\minus{}a)(c\minus{}b)}$

$ <0<\frac{1}{\sqrt{a}(a\minus{}b)(a\minus{}c)}\plus{}\frac{1}{\sqrt{b}(b\minus{}c)(b\minus{}a)}\plus{}\frac{1}{\sqrt{c}(c\minus{}a)(c\minus{}b)}$

I doubt that we could possibly factorize these expressions in the nicer form. Nevertheless, set $ x \equal{} \sqrt {a}$,            $ y \equal{} \sqrt {b}$,             $ z \equal{} \sqrt {c}$.

From the given condition, it follows $ x > y > z$

By the way, it is advisable, not compulsory to give $ x > y > z$ because after the factorization, we get the following

$ \sum_{cyc} \frac{\sqrt{a}}{(a\minus{}b)(a\minus{}c)} \equal{}\minus{}\frac{1}{(\sqrt{a} \plus{} \sqrt{b})(\sqrt{b}\plus{} \sqrt{c})(\sqrt{c}\plus{}\sqrt{a})}$

the same method for the second inequality.

As we can see, the condition given is not critically necessary even though the fulfilled constrain must be
[i]x,y, z are different from each other.[/i]


inequalities
function
induction
inequalities proposed

I solved it by a simple induction.
The step of induction is trivial so what remains is just prove that:

Let $f$ be a convex function on $[a,b]$.
for $x,y,0 \in [a,b]$

$f(x)+f(y) \le f(x+y)+f(0)$


limit
integration
calculus
Support
function
Euler
real analysis

Find the limit:

$\lim_{n\rightarrow \infty}\frac{\sum_{j=1}^{\infty}\frac{j^{n}}{e^{j}}}{n!}$.

${\frac{\sum_{j=1}^{\infty }\frac{j^{n}}{e^{j}}}{n!}=\frac{\text{Li}_{-n}\left(\frac{1}{e}\right)}{n!}=\frac{\text{Li}_{-n}\left(\frac{1}{e}\right)}{\Gamma (n+1)}}$

MAking approximations I think the result is $1$.

Don't have a proof. :)

My first thought:

$n!=\int_{0}^{\infty}x^{n}e^{-x}\,dx=\sum_{j-1}^{\infty}j^{n}e^{-j}+\int_{0}^{\frac12}x^{n}e^{-x}+\sum_{j=1}^{\infty}\int_{j-\frac12}^{j+\frac12}\left(x^{n}e^{-x}-j^{n}e^{-j}\right)$

The integral from zero to $\frac12$ is trivial, of course. I would conjecture that the sum of integrals is small compared to $n!$, which would support yagaron's conjecture that the overall limit is 1. I don't yet have an argument for that that I'm happy with.

The function $f(t) = t^{n}e^{-t}$ increases on the segment $[1,n]$; and it decreases on $[n,+\infty)$. Therefore,
\[\int_{1}^{n}f(t) dt+f(1) \leq \sum_{j=1}^{n}j^{n}e^{-j}\leq \int_{1}^{n}f(t) dt+f(n). \]
Note that $f(n) = (n/e)^{n}= O(n!/\sqrt{n})$. Hence
\[\sum_{j=1}^{n}j^{n}e^{-j}= \int_{1}^{n}t^{n}e^{-t}dt+o(n!). \]
Similarly,
\[\sum_{j=n+1}^{\infty}j^{n}e^{-j}= \int_{n}^{\infty}t^{n}e^{-t}dt+o(n!). \]
Combining these two estimates we get
\[\sum_{j=1}^{\infty}j^{n}e^{-j}= \int_{1}^{\infty}t^{n}e^{-t}dt+o(n!) = (\Gamma(n+1)-O(1))+o(n!) = n! (1+o(1)). \]

${\frac{1}{\Gamma (n+1)}\sum_{j=1}^{\infty }\frac{j^{n}}{e^{j}}\approx \frac{1}{\Gamma (n+1)}\int_{1}^{\infty }\frac{j^{n}}{e^{j}}\, dj=\frac{\Gamma (n+1,1)}{\Gamma (n+1)}\approx 1}$

As $n$ get larger the approximation turn in a equality.

we can use this, don't we?

Nice.

Yury's solution is elegant and it is  based on , or probably I should say simillar to, the method of studying Euler-like sequences.

Please, take time to read the previous posts in the thread before posting yourself. Do you see any flaws in Yury's proof? If not, you'd better check your own argument before posting a different answer. Also, just posting an unsupported answer after somebody posted a full solution is hardly ever appropriate :spam:

:lol:
OK
$\int_{0}^{x}t^{x}e^{-t}dt$~$\frac{1}{2}\Gamma(1+x)$
and $n!=\Gamma(1+n)$
when $x\to+oo$ and $n\to+oo$

This would correspond to $\sum_{j=1}^{n}\frac{j^{n}e^{-j}}{n!}$, not to the sum $\sum_{j=1}^\infty$.


In a triangle, the sum of the length of 2 sides is 10, and the height of the third side is 3.
Find the maximum value of the radius of the circumcircle of the triangle.

Let $ a\plus{}b\equal{}10, \ h_c\equal{}3$

$ \frac{a}{\sin A}\equal{}\frac{b}{\sin B}\equal{}\frac{a\plus{}b}{\sin A\plus{}\sin B}\equal{}2R$

But $ \sin A\equal{}\frac{h_c}{b}\equal{}\frac{3}{b}, \ \sin B\equal{}\frac{h_c}{a}\equal{}\frac{3}{a}$

Then $ R\equal{}\frac{1}{2}\cdot\frac{a\plus{}b}{\frac{3}{a}\plus{}\frac{3}{b}}$

$ R\equal{}\frac{1}{6}ab\equal{}\frac{1}{6}a(10\minus{}a)\equal{}\minus{}\frac{1}{6}a^2\plus{}\frac{5}{3}a$

$ R_{max}\equal{}\frac{25}{6}$


calculus
integration
trigonometry
real analysis
real analysis unsolved

Evaluate: \[ \int\frac{\cos 2x}{8\left(\sin^6x\plus{}\cos^6x\right)\plus{}5\sin 4x}\]

[hide]As an exercise, try to show that   $ \sin^6x \plus{} \cos^6x \equal{} \frac{5 \plus{} 3 \cos(4x)}{8}$. Therefore,

$ P \equal{} \int\frac{\cos(2x)}{5 \plus{} 3 \cos(4x) \plus{} 5 \sin(4x)}\,dx$.[/hide]


The expression $\frac{7n + 18}{2n + 3}$ takes integer values for certain integer values of $n$. What is the sum of all such values of the expression?

Source: Australian Math Competition, S28, 2000

[hide]$2n + 3 \mid 7n + 18$ $\Longrightarrow$ $2n + 3 \mid 15$.
That gives 8 values of $n$: 6, 1, 0, -1, -9, -4, -3, -2.
So the sum should be $28$. [/hide]

Nope... the question asks for the sum of all the possible values of $\frac{7n + 18}{2n + 3}$ which are integers. And there are more than 4 values as well. :)

[quote="Andreas"]$2n + 3 \mid 7n + 18$ $\Longrightarrow$ $2n + 3 \mid 15$. [/quote]
Umm...how did you get this? :?  :blush:

Err, one possible integer value of the expression occurs when $n = -2$. Then $\frac{7n + 18}{2n + 3} = \frac{-14 + 18}{-4 + 3} = -4$ ,which is an integer. We want to find all the sum of all such expressions which are integers when $n$ is an integer.

I've edited my previous post. Please, look it.  :)

Ooh, didn't see that.  :blush:  It's absolutely correct.  :)


vector
geometry
geometric transformation
topology
group theory
abstract algebra

How to show that the collection (group) of orthogonal matrices in R^n is orientable?

Every Lie group is orientable as you can choose a local orientation in an arbitrary point and then move the orientation to each other point by the map on homology induced by multiplication.

I think I get the general idea of extending a local orientation using the multiplication,
however, it's not quite clear to me how to do it formally.
could you give me a more detailed example?

and I'm not familiar with the term Homology...

no need to know anything about homologies to solve the problem.
let $ G$ be a lie group, and consider a basis $ \{E_1,\dots,E_n\}$ for $ T_eG$ (the tangent space to $ G$ in the identity), and define $ n$ vector fields by $ (X_i)_g \equal{} dL_gE_i$ for $ i\equal{}1,\dots,n$, where $ L_g$ is the left translation in $ G$. since $ \{(X_i)_g\}$ is linearly independent for every $ g\in G$, it follows that the tangent space is trivial (as a vector bundle), in particular $ G$ is orientable.

now, the fact that $ O(n)$ is a lie group should be standard.

[quote="ma_go"]let $ G$ be a lie group, and consider a basis $ \{E_1,\dots,E_n\}$ for $ T_eG$ (the tangent space to $ G$ in the identity), and define $ n$ vector fields by $ (X_i)_g \equal{} dL_gE_i$ for $ i \equal{} 1,\dots,n$, where $ L_g$ is the left translation in $ G$. since $ \{(X_i)_g\}$ is linearly independent for every $ g\in G$, it follows that the tangent space is trivial (as a vector bundle), in particular $ G$ is orientable.[/quote]
in effect, what we're doing is choosing (in a smooth fashion) a basis at each point on the manifold?
 and is that the meaning of "trivial tangent space"?
or is it that $ G$ is diffeomorphic to $ TG$?

[quote="ma_go"]now, the fact that $ O(n)$ is a lie group should be standard.[/quote]
obviously

[quote="Gilad.R"]in effect, what we're doing is choosing (in a smooth fashion) a basis at each point on the manifold?
and is that the meaning of "trivial tangent space"?
or is it that $ G$ is diffeomorphic to $ TG$? [/quote]

in order:
1. no, we're doing more: we're finding linearly independent sections in any point. finding a basis at each point "in a smooth fashion" is orientability: finding such a basis should be the triviality of the principal $ O(n)$-bundle associated to the tangent bundle of the manifold. for a (tough) review on this, you can take a look at either kobayashi-nomizu's classical book (first volume, i don't remember the title) or steenrod's "fiber bundles", for something easier (and of course less complete), try wikipedia.

2. a bundle $ F\to E \stackrel{\pi}{\to} B$ is trivial if it's isomorphic to the trivial bundle $ F\to F\times B \stackrel{\pi_2}{\to} B$, e.g. if there's an homeomorphism $ \phi: E\to F\times B$ between total spaces commuting with projections, that is $ \pi \equal{} \pi_2\circ\phi$. in the simpler case of vector bundles (which is what we're dealing with), being trivial is equivalent to have $ {\rm rk}\, E \equal{} \dim F$ linearly independent (at each point) sections: this is a good exercise to get a handle on definitions.

3. $ G$ can't be diffeomorphic to $ TG$ unless $ \dim G \equal{} 0$, by a simple consideration on dimensions.

3. I meant $ G\times T_{1}G$ is diff. to $ TG$...

anyways, thanks!


geometry
3D geometry
sphere
analytic geometry
symmetry

Given a cube of unit side. Let $A$ and $B$ be two opposite vertex. Determine the radius of the sphere, with center inside the cube, tangent to the three faces of the cube with common point $A$ and tangent to the three sides with common point $B$.

By "sides", do you mean the edges or the faces?

well, you could do a coordinate plane, and intersect the sphere with the lines from B with a random radius, and note the symmetry that the intersections have; if it is tangent to one, it is tangent to all of them, we know that it is in the form $(x-r)^2+(y-r)^2+(z-r)^2=r^2$, if A is the origin, and it is tangent to the positive planes

Not completely correct. A far more concise method would be to take a sphere and both inscribe and circumscribe cubes. We can easily find the sidelength of each cube (the radius of the sphere is half the sidelength of one and half the long diagonal of the other). Now we halve the difference between the sidelengths and add this value to the sidelength of the smaller cube. This is the sidelength of the desired cube.

The inverse operation is used to go from the cube to the sphere.  :)  Hope that wasn't to wordy.

[quote="Altheman"]i don't really understand how that works because when you circumscribe of inscrible a cube in a sphere, you never have the sphere tangent to the edges, just tangent at the verticies or centers of the faces...?[/quote]

This problem is that it is tangent to three of the faces intersecting at one point, and the three edges intersecting at the opposite point.  Nevertheless, seethelight's solutions it not correct...
The sphere would be tangent to three faces of the cube intersecting at one vertex.  However, it would pass through the opposite vertex of the cube, rather than being tangent to the three edges.

Whoops, I meant that the first cube's sidelength was equal to the diameter of the cube, and that the second cube's shorter diagonal was also equal to the diameter. It works then. :blush:


[quote]Let $ a,b,c$ be positive real numbers. Prove that
\[ a \plus{} b \plus{} c \plus{} \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \ge 4\left( {\frac {a}{{1 \plus{} ab}} \plus{} \frac {b}{{1 \plus{} bc}} \plus{} \frac {c}{{1 \plus{} ca}}} \right).\][/quote]

$ a \plus{} \frac {1}{b} \equal{} a\left(1 \plus{} \frac {1}{ab}\right) \ge \frac {4a}{1 \plus{} ab}.$


MATHCOUNTS quadratics algebra polynomial AMC AIME HMMT

This is, in my opinion, one of the best MathCounts problems ever.

If $ x^{2} \minus{} 5x \plus{} 8 \equal{} 1$, find $ x^{4} \minus{} 10x^{3} \plus{} 25x^{2} \minus{} 9$.

i think i got this at the competition, but i forgot if i did. i know that it's based on some clever multiplication/factoring, but i dont remember. can anyone please post a solution?

ugh, how did I not get this at the competition?

[hide]
We can change this equation to $ x^2\minus{}5x\equal{}\minus{}7$, and squaring both sides gives us $ x^3\minus{}10x^3\plus{}25x^2\equal{}49$.  Should be obvious that the thing we're looking for is $ \boxed{40}$
[/hide]

[hide]
We try to complete the square for the first equation.

So we have $ (x - 2.5)^2 = - 0.75$. Yeah, $ x$ is nonreal, deal with it.

Note that the second equation is $ x^2(x - 5)^2 - 9$.

Let $ y = x - 2.5 = - 0.75$. Now our answer is

\begin{align*} & (y - 6.25)^2 - 9 \\
= & ( - 7)^2 - 9 \\
= & \fbox{40} \\
\end{align*}

[/hide]

Beaten. With a simpler, although typoed, solution. :)

[hide="Hint for alternate but essentially the same solution"] Write the desired expression as $ (x^{2}\minus{}5x\plus{}a)(x^{2}\minus{}5x\plus{}b)\plus{}c$. Do you see the motivation for this?[/hide]

[quote="CatalystOfNostalgia"][hide="Hint for alternate but essentially the same solution"] Write the desired expression as $ (x^{2} \minus{} 5x \plus{} a)(x^{2} \minus{} 5x \plus{} b) \plus{} c$. Do you see the motivation for this?[/hide][/quote]

OHH YEAH! That's how I solved it at states. I forgot how exactly I did it.

how is this one of the best mathcounts problems ever =P?  This is in essence similar to a countdown round problem in 2005 with Patricia Li vs. David Benjamin, where she squared some expression to get 49, while David did the problem and got 56. xD...

Note that since the $ x$ term is not in the second expression and that $ 10\equal{}2*5$ while $ 25\equal{}5^2$, you should definitely go for $ (x^2\minus{}5x)^2$.

Or note that if you do $ ((x^2\minus{}5x)\plus{}8)((x^2\minus{}5x)\minus{}8)$, you also cancel out the $ x$ terms.

This gives us $ x^4\minus{}10x^3\plus{}25x^2\minus{}64\equal{}1*(1\minus{}16)\equal{}\minus{}15$

Thus to get the desired expression, we add $ 55$ to come up with $ 40$.

Okay, somehow, I misread a sign, did the exact same thing, but put -40, even though the sign thing shouldn't have made a difference at all.

I suck.

For the friendly moderator who altered the subject line, it was State.

[hide="HUGE HINT"]

square the quadratic  :P

[/hide]

my method is a little similar to mathcrazed's.

i used polynomial long division and got

$ x^4 \minus{} 10x^3 \plus{} 25x^2 \minus{} 9$ = $ (x^2 \minus{} 5x \plus{} 8)(x^2 \minus{} 5x \minus{} 8) \plus{} 55$

= and we know that $ (x^2 \minus{} 5x \plus{} 8) \equal{} 1$ and $ (x^2 \minus{} 5x \minus{} 8)$ is 16 less cuz the constant on the first one is 8 and on the second one is -8 making a diff. of 16 since all the other terms are the same in both.

so its just $ 1\times(\minus{}15)$ + $ 55$ = $ 55\minus{}15$ = $ \fbox{040}$  :D there i made it an AIME problem.

[quote="cognos599"]my method is a little similar to mathcrazed's.

i used polynomial long division and got

$ x^4 \minus{} 10x^3 \plus{} 25x^2 \minus{} 9$ = $ (x^2 \minus{} 5x \plus{} 8)(x^2 \minus{} 5x \minus{} 8) \plus{} 55$

= and we know that $ (x^2 \minus{} 5x \plus{} 8) \equal{} 1$ and $ (x^2 \minus{} 5x \minus{} 8)$ is 16 less cuz the constant on the first one is 8 and on the second one is -8 making a diff. of 16 since all the other terms are the same in both.

so its just $ 1\times( \minus{} 15)$ + $ 55$ = $ 55 \minus{} 15$ = $ \fbox{040}$  :D there i made it an AIME problem.[/quote]
How did you use polynomial long division to factor the 4th degree polynomial? :huh:
EDIT: nevermind. Just being out of my mind, oops :oops:  :mad: I know how to do long division!

Yeh, I thought this was a really slick problem, with, ofcourse, a really slick solution. The complex root may have tricked some people. One of the best.. definitely.

Although you have motivation that the quadratic factors of the quartic start with x^2-5x, a general way of factoring quartics is just writing it as $ (x^{2}\plus{}ax\plus{}b)(x^{2}\plus{}cx\plus{}d)$, equating coefficients, and solving a system of equations. The system of equations can get messy, but if you know you're looking for integers, you can usually reduce it to a small number of cases.

Try this on HMMT Algebra 2006...it totally kills two of the harder problems  :P


Calculate ${\lim_{\theta \to \frac{3 \pi }{4}}\, \frac{\text{CF}}{\text{CD}}}$

this problem is very hard, but very beautiful :)

It's my own. i spent some time making this :)

PS: This problem has 2 solutions. because the limit of the rigth and the left is different.

I put just one limit to simplificate. but actually is 2

i don't see why it is a limit... the problem is solved by putting in that value of theta, making DE and AF parallel, then constructing point C

basically, <CED=60, <FEA=90, < DEA=135,

since CD and CF are chords, it is trivial now to compute the lenghts since we have their arc measures

[quote]PS: This problem has 2 solutions. because the limit of the rigth and the left is different.[/quote]

the limit is because we have a triangle, not parallel lines.

and there is 2 possible solutions calculated from the limit

Yagaron, you know the solution of your own problem and such problems should be posted in the section [url=http://www.mathlinks.ro/Forum/index.php?f=47]Geometry Proposed & Own Problems[/url].

hey dude, i know that.
i clicked on the wrong link and didnt noticed it

a commom mistake


[quote="AIME15"]Yeah.  Even to me, 24 is fail.  And I'm not that smart.

Pythag, what did you get?[/quote]

I thought I failed.. but it turns out that I got 25.

Anyways, some people got the wedge problem wrong because they thought the wedge was a cone...

[quote="pythag011"][quote="numbermaster"]Thanks for voting. Good luck on the AMC10/12/beyond![/quote]

Hmm... I think you should remove the labels on the scores because to most AoPSers, 24 is fail...[/quote]

!!!!!!!! :)

im so angry because i got the most simple questions wrong

so i got a 20-21

[quote="abacadaea"]HOPEFULLY i got a 25, but i cant rememer my answers...:(

What # was the wedge problem?[/quote]

#21

cut a cylinder into half, answer is 1/2 of the volume of cylinder :D

I would vote 25, but I did not officially take it. (I took it only because I was co-proctoring and my coach was taking it too - we finished at about the same time, which was 23 minutes in, 17 minutes remaining!)

I brought new meaning to the word epic-fail. You know #8? The one with the bar graph where you had to find the average sales? Yeah. 24/25. After a perfect score last year.  :( :sob:  :(

Hey, I don't actually know what my score was...maybe you guys can help me figure it out. (Note: I had never taken the AMC before and I was just happy to qualify for the AIME, so my score is probably shameful by your standards  :oops: ) Okay, so I emailed the AMC people with my information and requested my scores, and this is what they sent back:

[i]American Mathematics Competitions
2008 AIME Results

Student Name AMC 12 AMC 10 AIME
INDEX*
(My Name) 97.5 A 4
137.5 [/i]

What does that mean? I took the AMC 12 if that helps with the interpretation at all...
Thanks!

I believe that means that you scored a 97.5 on the AMC 12 and a 4 on the AIME, making your USAMO index 97.5+4*10=137.5 :)

Strangely, that's pretty much the exact same score I got last year...97.5 and 6 on AIME.

Congratulations! :) It's not very often that a 6th grader gets 25.

uh when i was in 6th grade i took the amc 8 unofficially and got a 25 :D

[quote="BOGTRO"]I got a 25!!!!!!!
[/quote]

[sarcasm] that was completely unexpected [/sarcasm]  :rotfl:

24.... fail...
I got the second to last one wrong somehow. The one with the 10 cards and the dice and multiplying for a perfect square.... so embarrassed...

Yea when I was in 6th grade I got like a 15 and I got a 22 last year so...

Ok so...

6th grade: 13 (O_O how'd a get a 13???  :rotfl: )
7th grade: 22
8th grade: not taken, 2009

I got a 16. I have no clue how i got SUCH a bad score.

I got 24 because I numbered BEST OF LUCK like:
1234556789

I can actually see that's the work I did on the packet.

yep thats exactly what I did, hope I don't do it on the amc 10 too.

So when do the perfect scorer people's names come out?

I want to see how badly my middle school phailed this year after all of our smart people left.

But I heard it was reaaaaaaaaallllllllllllyyyyyyy easy, compared to that of last year, and even during last year's, I was like, oh this difficulty is game over.

25. hmm.. this year's was a lot easier though.. maybe because calculators weren't allowed this year..

When does the AMC 8 list the perfect scorers on the website?

there's a 6th grade in my school that is crazy smart
he got a 25 on AMC8

anyway, i thought that this one was a lot easier than last year's
but i don't think its because of the calcs
weren't they first banned in 2006?

No, I think this was the first year they were banned, because people used them last year.

Why on the answers on the site for #12 it gives 3 answers?  :huh:  :huh:  :huh:

I beleive that when the test first came out calculators were not allowed.  I think the test was a lot easier than past tests.  I thought it was like a school competition in Mathcounts.  My kids had 7 perfects including one 6th grader.

success!!!
25/25 :bruce:  :bruce:  :bruce:


\[ \triangle ABC\Longrightarrow \sum \frac{(b+c)^2}{b+c-a}\le \frac{2R}{r}\cdot (a+b+c). \]

Levi, your inequality follows immediately from this one: \[ (b+c)^{2}(a+b-c)(a+c-b)\leq 4a^{2}bc.  \] ( this is GL.18 from my book  :D )
So, from here we have \[ \frac{(b+c)^{2}}{b+c-a}\leq abc\cdot\frac{a}{2pr^{2}}.  \]
Writing the analogous onesand by summing them up, we get the desired inequality. ;)

[quote="edcwconan"][quote="Ji Chen"][quote="Virgil Nicula"]$ \triangle ABC\Longrightarrow$

$\sum \frac{(b+c)^2}{b+c-a}\le \frac{2R}{r}\cdot (a+b+c). $[/quote][quote="Cezar Lupu"]Levi, your inequality follows immediately from this one: \[ (b+c)^{2}(c+a-b)(a+b-c)\leq 4a^{2}bc.  \] [/quote]$4a^{2}bc-(b+c)^{2}(c+a-b)(a+b-c)=(b-c)^2\left[(b+c)^2-a^2\right]\geq 0.$[/quote]

Mr.Chen,I really want to know how can you put every inequalities into sum of squares.I read you book about
Inequalities, that is fairly good.Also, I am a student in Jiangsu Province.
Can you tell me some skills in it?Thank you very much.[/quote]

$4a^{2}bc-(b+c)^{2}(c+a-b)(a+b-c) = 4a^{2}bc - (b+c)^2[a^2-(b-c)^2] = (b-c)^2(b+c)^2 +a^2[(b+c)^{2}-4bc] = (b-c)^2[(b+c)^2-a^2]$


Hello everyone. Just became a member and this was the closest sub-forum i could find that relates to music and mathematics. Anyways, I've been highly interested in the relationship between math and music (particulary piano performance) ever since the Mozart sonata study in 1993 (mozart effect). I didnt really research it until the mid-90's but i heard about it when the study was published. Just had a few questions:

Are there any composers (ie:bach, greek composer yanni) that you feel or think have an effect on mathematical performance? if so, why?

I've been playing the piano on and off since i was 9 (now in my 20's) and since i dont have time to dedicate myself to a huge repertoire, i focus my playing to mozart and bach.

if you play the piano or know alot about this topic, please chime in. thx

-david

I've never really thought about that before. But I have noticed there are some pieces you can just relax and play, and some you really have to think about as you play. I think maybe Bach has as an effect, because when I play inventions, I have to think about which hand has the melody in each measure, and how it will fit in the the big picture.

Thanks for this topic, I will be thinking about it more now.  :read:

I'm not exactly sure what you mean, but I know that different styles and periods of time require a different way of playing.

Bach, for example, is very metronomical
when I play Bach, the notes fall either right on the beat or off the beat
the tempo is very fixed (not sure if that's the best way to explain.)

However, composers like Chopin are different
When I play Chopin, I feel as if I can strech out the beat a bit
it is not as metronomical as Bach's pieces

Is this what you mean by mathematical performance?

Well, of course, scientists have had an expiriment where they
found out that piano players do better on tests
than those who don't.

Some people say that Mozart and Bach make you smarter.
I think maybe that's because they're very organized.
They have a exact rhythm.
However, sometimes you they make you kinda bored.

Some Romantic styles make you really peaceful, which
is why I guess they call it the Romantic period.
I love listening to those pieces.
There's a lot of those pieces too!
(Chopin being my favorite)

The Contemporary music is either really crazy (lots
of flats, sharps, polyrhythms), or similar to the
Romantic period. If you listen to modern pieces,
they make your blood pressure go up.

Hope I made it make more sense!  :D

i actually listen to alot of romantic piano music (mainly chopin) but since i usually only have about half hour to play piano a day. i focus on mozart/bach. i am actually looking into pieces of music that are conducive to learning. i love music and have audiophile headphones/great music collection/excellent mp3 player,etc.  any more suggestions would help. thx everyone

I've also been noticing things when I listen to Mozart/Bach because I really listen to how all the melodies combine into the piece.

Hm. I'd never really thought about that before, but I guess I think better doing my homework after playing a piece where I have to concentrate on many melodies at the same time.

I have very good rhythm, and my teacher says I must be a good math student. I usually laugh, but I really think that might have to do with it. Somehow, math and piano go together.


Let [tex]p[/tex] be an odd prime. Show that the polynomial

[tex]x^{p-1}+\cdots+x+1[/tex]

can not be factored into a product of two polynomials of degree > 0 with integer coefficients.

[quote="koopa"]Make the linear changle x = y + 1. Then use Eisenstein.[/quote]

A specialized theorem like Eisenstein's Criterion should be quoted when used. Such theorems often slip people's memories.

[b]Eisenstein's Criterion[/b]

Suppose we have a polynomial with integer coefficients, [tex]f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0[/tex]. If there exists a prime number [tex]p[/tex] that divides all [tex]a_i [/tex] except [tex]a_n[/tex], and [tex]p^2[/tex] does not divide [tex]a_0[/tex], then [tex]f(x)[/tex] is irreducible.

Erin Schram

This is not a [b]challenge[/b]! It is a pretty standard exercise in algebra. My intention was to provide high school students something for thoughts.
I believe that one of the best ways to learn math is proving theorems before looking at the proofs in the book. So I am restating the exercise in several steps as follows:

1. Prove the [b]Eisenstein's Criterion[/b].

[b]Eisenstein's Criterion[/b] Suppose we have a polynomial with integer coefficients, [tex]f(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_2x^2+a_1x+a_0[/tex]. If there exists a prime number [tex]p[/tex] that divides all [tex]a_i [/tex] except [tex]a_n[/tex], and [tex]p^2[/tex] does not divide [tex]a_0[/tex], then [tex]f(x)[/tex] is irreducible.

2. Let [tex]p[/tex] be an odd prime, show that p divides [tex]\displaystyle {p\choose k}[/tex] for [tex]1\le k \le p-1[/tex].

3. Using [b]Eisenstein's Criterion[/b] (with a variable change) to prove the following:

Let [tex]p[/tex] be an odd prime. Show that the polynomial [tex]x^{p-1}+\cdots+x+1[/tex] is irreducible over [tex]\mathbb{Z}[/tex].

[quote="JBL"]An excellent theorem that many of you can probably prove yourself, by the way.[/quote]

Good point! I posted [b]Gauss Lemma[/b] for primitive polynomials last night. I removed it after I saw the replies in this thread. The proofs of [b]Gauss Lemma[/b] and [b]Eisenstein's Criterion[/b] are pretty elementary and accesible for high school students (although I learned them in college :-)).


Does anyone have experience conducting an orchestra, or at least has been taking conducting lessons for a while? Care to give me some tips before I start? Some specific questions I have:

-How do you practice? About how long do you practice a day?
-What is considered a good piece for starters? Or good composers in general?
-What other musical talents are necessary/helpful for conducting? I don't play the piano, and I'm afraid that it might make conducting hard for me. I do play the violin, though.
-How long does it take on average to be able to conduct an orchestra and perform?
-How much competition is there for young conductors?

Thanks in advance!

I assistant-conduct my school orchestra.

[quote="zephyredx"]-How do you practice? About how long do you practice a day?[/quote]

I usually practice conducting for maybe 10 -15 minutes at a time. I conduct by taking piano reductions of my cello concerti, listening to the orchestral parts, and cuing as neccessary. Sometimes, I conduct music from youtube. It also helps to watch famous conductors, such as Herbert von Karajan. You can imitate them and develop your own style. There are also many good books on conducting that you can buy off of amazon.

[quote="zephyredx"]-What is considered a good piece for starters? Or good composers in general?[/quote]

For me, I started conducting my easy school music. You can find easy music (although somewhat modern and unknown) at http://www.jwpepper.com. An easy, yet more well known composer would be Vivaldi.

[quote="zephyredx"]-What other musical talents are necessary/helpful for conducting? I don't play the piano, and I'm afraid that it might make conducting hard for me. I do play the violin, though.
[/quote]

Playing any instrument usually helps. However, you must have a deep understanding of the songs you plan to conduct. Personally, I play Cello+Piano. You should have basic knowledge of the instruments in your symphony or orchestra though. Conducting is not all about flapping your arms, jumping around, and making speeches in front of audiances. You must have leadership skills to pull your orchestra or symphony through.

[quote="zephyredx"]-How long does it take on average to be able to conduct an orchestra and perform?[/quote]

I cannot answer you question as I am only an assistant conductor. Usually, my school takes around 2~3 months to prepare 3~4 songs.

[quote="zephyredx"]-How much competition is there for young conductors?[/quote]

It really depends on the city. My city has only one youth symphony and that's it. And the assistant conductor there is 18. I'm sure that you can probably find some school or youth symphony if you are in your early teens though.


saludos
lquisiera saber si se tienen las longlist , pues las deart of solving problem  no entra
si alguien las tiene agradeceria que las colgaran
adan

si le das a "resources"  en la esquina derecha, te saldran ahi los longlists y shortlists

Aqui estan los links:

-  [url=http://www.artofproblemsolving.com/Forum/resources.php?cid=17]IMO SHORTLIST[/url]

y tambien:

- [url=http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=18]IMO LongList[/url]

este ultimo aun no cuenta con material ....
pero me parece que si se tiene el longlist de la IMO 1967, no se porque no sta aqui..

Carlos Bravo :)
Lima-PERU


abstract algebra
group theory
number theory
least common multiple
superior algebra
superior algebra solved

[quote="the game"]If an abelian group has subgroups of order m & n then it will have a subgroup of order which is the lcm of m & n.

Please donot use the properties of finite abelian groups then the entire flavor of he problem will be lost.Just use the elementary concepts( i mean the defn & stuff about groups).

P.S. I have encountered many solutions of this problem including a few by myself but all of them are wrong.Can anybody here do it.....[/quote]

Is this the problem where in the later editions, Herstein says that he received more correspondence about this problem than all the others in the book?

Yup thats I called it famous.But theres no harm in trying.

[quote="the game"]If an abelian group has subgroups of order m & n then it will have a subgroup of order which is the lcm of m & n.

Please donot use the properties of finite abelian groups then the entire flavor of he problem will be lost.Just use the elementary concepts( i mean the defn & stuff about groups).

[/quote]

Although you didn't want this (:)), you can consider it just a harmless comment.

Using the structure theorem for finite abelian groups, it's trivial to show that for any divisor $d$ of the order of the group, the group has a subgroup of order $d$ (and the converse also holds, of course). From the hypothesis we find that $m,n$ divide the order of the group, so their least common multiple also divides the order of the group, and that's that.

Here's how to do it without the heavy tools- just some basic stuff and a bit of number theory:
(We treat the abelian group as a $\mathbb{Z}$-module)
Lemma: Suppose $m$ and $n$ are relatively prime. Then $G$ has a subgroup of order $mn$ if and only if it has a subgroup of order $m$ and a subgroup of order $n$.
Proof. If $H$ has order $m$ and $K$ has order $n$, $H+K$ has order $mn.$
If $H$ has order $mn$, $mH$ has order $n$ and $nH$ has order $m$.

This allows us to reduce to prime powers. Once there, choose the larger one at each prime, then recombine to form the desired group of order $\mathrm{lcm}(m,n).$

Wow that is one hell of a solution .Thts something ingenious i wanted.

I don't quite follow: what are $mH$ & $nH$?

In the $\mathbb{Z}$-module view, they're the submodules $\{mx: x\in H\}$ and $\{nx: x\in H\}$. In the group picture, they're the subgroups of $m$th and $n$th powers of elements in $H$ respectively.

How to count order of mH & nH ?
I don't quite follow.


algebra
partial fractions
algebra unsolved

(i) The coefficients in the series

$ S\equal{}\frac{1}{3}x\plus{} \frac{1}{6}x^{2} \plus{}\frac{1}{12}x^{3} \plus{} ... \plus{} a_{r}x^{r}\plus{}...$

satisfy a recurrence relation of the form $ a_{r\plus{}1} \plus{} pa_{r}\equal{}0$. Find the value of $ p$.

By considering $ (1\plus{}px)S$, find an expression for the sum to infinity of $ S$ (assuming that it exists). Find also expressions for the sum of the first $ n\plus{}1$ terms of $ S$.

(ii) The coefficients in the series

$ T\equal{}2\plus{}8x\plus{}18x^{2}\plus{}37x^{3} \plus{} ... \plus{} a_{r}x^{r} \plus{} ...$

satisfy a recurrence relation of the form $ a_{r\plus{}2} \plus{} pa_{r\plus{}1} \plus{} qa_{r} \equal{} 0$. Find an expression for the sum to infinity of $ T$ (assuming thatit exists). By expressing $ T$ in partial fractions, or otherwise, find an expression for the sum of the first $ n\plus{}1$ terms of $ T$.

(i) find $ p$ by solving the equation $ a_{r\plus{}1}\plus{}pa_r\equal{}0$ for two particular consecutive terms. then the fact that $ a_{r\plus{}1}\plus{}pa_r\equal{}0$ implies that $ (1\plus{}px)S\equal{}\frac{1}{3}x$, giving $ S\equal{}\frac{x}{3(1\plus{}px)}$. the sum of the first $ n$ terms of $ S$ is found in a similar way.

(ii) this is done in a similar fashion to (i). here you'll need two equations to solve for $ p$ and $ q$. then consider $ (1\plus{}px\plus{}qx^2)S$


Are they moon-quake fault lines?

[img]http://www.pinktentacle.com/images/moon_kaguya_3.jpg[/img]

Nice inference. :lol:
Actually this "lines" are "rays" made of the dust the was ejected from the impacts that caused the craters. If these rays are present around a crater, it can be assumed that it is a relatively new crater. Of course, there are exceptions, like what material the rays are made of.  Different materials have different albedos.  :wink:

[hide="Image"][img]http://apod.nasa.gov/apod/image/0503/moon8_mandel_big.jpg[/img][/hide]
[url=http://en.wikipedia.org/wiki/Ray_system]Wikipedia - Ray Systems[/url].

EDIT Since the picture was added to the original post.
Those still look like rays from a crater, the material from the impact was just large enough to create more defined streaks like that.
EDITTED EDIT The moon does not have crust plates like the Earth so, they could not be quake lines.

The "ray" theory seems plausible to explain the rays outwards around the craters.

However, it does not seem to explain the lines in the image of the original post.
if the tracks were created by ejected big rocks, then we should be able to see that at the end of each track, there should be a big boulder.

Also, we cannot completely rule out the moon-quake fault line idea. Even though there are no tectonic plates on the moon due to its cold core,
but each meteorite impact (which caused those huge craters) would have caused a big quake on the moon in the distant past.
Since there is no weathering on the moon, the fault lines have survived for billions of years, to be seen today.
A close look in the image of the original post shows that the lines are not straight, some resemble cracks instead of rays.

About the quake idea, then where are the assumedly big craters from those big impacts? The Moon's suface does remain unchanged over time unless there is an impact, like you said, so the craters should be there still also.
Another thing is that this "cracks" [i]in the surface [/i]do not look like they are more than [i]on the surface [/i]of the Moon.  The mare on the Moon are cause by large impacts, where the molten layer beneath flowed onto the surface and hardened.

The rays of a crater could still create this "lines", the material that created it does not have to be large just needs to be moving fast enough, the distance of the picture does not do justice to seeing the "possible boulders" that would be at the end of the trails. :wink: