$x,y$ are two angles in a triangle such that $x+y<90^{o}$.
Prove or disapprove the followings:
(1),$\tan{x}\cdot \tan{y}<1$
(2),$\sin{x}+\sin{y} < \sqrt{2}$
(3),$\cos{x}+\cos{y}>1$

(4),$\frac{1}{2} \tan{(x+y)} < \tan{\frac{x+y}{2}}$

$\tan x$ and $\sin x$ are positive and increasing in the interval $x \in (0, 90^\circ)$, while $\cos x$ is positive and decreasing in this interval. WLOG, assume that $x \le y$. Then $\tan x \le \tan y,\ \sin x \le \sin y,\ \cos x \ge \cos y$.

[color=blue][b](1)[/b][/color] [color=white].[/color] $\tan(90^\circ - y) = \frac{1}{\tan y} > 0$

$\tan x \cdot \tan y = \frac{\tan x}{\tan (90^\circ - y)} \le \frac{\tan y}{\tan (90^\circ - y)} = 1$

Assume that  $\tan x \cdot \tan y = 1$. Then

$\frac{1}{\tan (90^\circ - y)} = \tan y = \frac{1}{\tan x},\ \ \tan (90^\circ - y) = \tan x,\ \ 90^\circ - y = x,\ \ x + y = 90^\circ,$

which contradicts the condition $x + y < 90^\circ$. Therefore, the equality is not possible and $\tan x \cdot \tan y < 1$.

[color=blue][b](2)[/b][/color] Since $0 < x + y < 90^\circ$ by the problem condition and $x \le y$ by our WLOG assumption, we have $0 < \frac{y + x}{2} < 45^\circ$  and $0 \le \frac{y - x}{2} < 45^\circ$. It follows that

$0 < \sin \frac{y + x}{2} < \frac{\sqrt 2}{2} < 1,\ \ 0 < \frac{\sqrt 2}{2} < \cos \frac{y - x}{2} \le 1$

$\sin x + \sin y = 2 \sin \frac{y + x}{2} \cos \frac{y - x}{2} <  2 \cdot \frac{\sqrt 2}{2} \cdot 1 = \sqrt 2$

[color=blue][b](3)[/b][/color] For the same reasons as in (2),

$0 < \frac{\sqrt 2}{2} < \cos \frac{y + x}{2} < 1,\ \ 0 < \frac{\sqrt 2}{2} < \cos \frac{y - x}{2} \le 1$

$\cos x + \cos y = 2 \cos \frac{y + x}{2} \cos \frac{y - x}{2} > 2 \cdot \frac{\sqrt 2}{2} \cdot \frac{\sqrt 2}{2} = 1$

[color=blue][b](4)[/b][/color] See the problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34665]tangent x[/url].


pigeonhole principle
induction
combinatorics unsolved
combinatorics

On a planet there are $3\times2005!$ aliens and $2005$ languages. Each pair of aliens communicates with each other in exactly one language. Show that there are $3$ aliens who communicate with each other in one common language.

A kind of Ramsey theroy question...
Here is my silly idea which I use today:Consider an alien $O_1$,there exist an language $L_1$ which $O_1$ communicates to at least  $3\times2004!$ aliens (let those aliens as Group A)with $L_1$(Easy to prove by Pigeonhole Principle),if any two aliens in Group A communicates to each other with$L_1$,then that two aliens and $O_1$ communicate with each other in one common language$L_1$.

Else among Group A,they can use the other $2004$ languages to communicates to each other only.Consider alien$O_2$ in Group A,there exist an language $L_2$ which $O_2$ communicates to at least  $3\times2003!$ aliens in Group A with $L_2$ (let those aliens as Group B),if any two aliens in Group B communicates to each other with$L_2$,then that two aliens and $O_2$ communicate with each other in one common language$L_2$.

Repeat this logic,until the case"among $3$ aliens,they can only use $1$ language to communicates to each other",which is trivial...

Everyone applied induction on this problem. I wonder if there is any other ways to tackle it. :)

hy

i wonder what were the results of the selection test?

how many people solved 3 or more problems?


In a holiday resort at least 2 people out of any three do not know each other, but there are always at least two people who know each other out of any 7 people. Paul is staying in the resort. Prove that he knows 6 people the most.

[hide]Because 2 out of every 3 people do not know each other, there can not be 3 people that all know the other 2. Since the second part is a group of 7 people, let's picture it as a 6 pointed star, with Paul in the middle. If two people are connected, they know each otehr. Paul is connected to all 6. none of the other 6 know any of the other 5 in the other 6. If we add one more person that paul knows, then the group of the 7 other people does not have any 2 people that know each other, or if that person is connected to another person, then the first restriction is broken. Thus, Paul can only know 6 people. This is assuming that if A knows B then B knows A[/hide]

[quote] Because 2 out of every 3 people do not know each other, there can not be 3 people that all know the other 2. Since the second part is a group of 7 people, let's picture it as a 6 pointed star, with Paul in the middle. If two people are connected, they know each otehr. Paul is connected to all 6. none of the other 6 know any of the other 5 in the other 6. If we add one more person that paul knows, then the group of the 7 other people does not have any 2 people that know each other, or if that person is connected to another person, then the first restriction is broken. Thus, Paul can only know 6 people. This is assuming that if A knows B then B knows A [/quote]

Got beat to it again.


Let $P$ be a point on the circumcircle of a triangle $A_{1}A_{2}A_{3}$, and let $H$ be the orthocenter of the triangle. The feet $B_{1},B_{2},B_{3}$ of the perpendiculars from $P$ to $A_{2}A_{3},A_{3}A_{1},A_{1}A_{2}$ lie on a line. Prove that this line bisects the segment $PH$.

its a very famous as well as useful book on geometry. you should collect it.

[quote="N.T.TUAN"][quote="cvix"]You can find a proof in a book Coxeter, Greitzer "Geometry revisited" in the chapter "More on Simson lines" (or something like that).  :)[/quote]

Post a solution, please!  :blush: Because I haven't got that book![/quote]

It is easy to prove that $QR\parallel P'H,$ where $QP' = QP$ and so, the line $QR$ $($ simson's line of $P$ $),$ bisects the segment $PH.$

kostas Vittas.


1. If m is a positive integer, which of the following could be true?
I. m^2 is a prime number
II. radical m is a prime number
III. m^2=radical m
A) I only B) II only C) III only D) II and III only E) I,II, and III

My answer was E, because I supposed m was 1, so it satisfied all three conditions. However, the answer turned out to be D). then why I. was not being satisfied?

2. Another question, I forgot the definition of even or odd numbers. Is 1345 an odd number because '5' the ending is an odd number? How about 1342? An even number because of '2' the ending? also, even and odd numbers must be integers ,right?

Thanks for help.

one important thing about prime numbers is 1 IS NEITHER PRIME NOR COMPOSITE!
heh

ok the definition is not really well defined, because there are many ways to define them.
my definition is that even numbers are integer multiples of 2 and odd numbers are all other integers
0 is even

ofcourse because of their divisibility by 2, you ould define it based on its last digit

Statement II (when rephrased as “the $ n^{th}$ root of $ m$ is a prime number”) is true for all powers of primes (e.g., $ \sqrt{4}$, $ \sqrt[3]{27}$, $ \sqrt[5]{2625}$), and statement III is true for $ m \equal{} 1$. Statement I is [i]never[/i] true, as $ m^{2}$ has at least three distinct divisors: $ m^{0}$, $ m$, and $ m^{2}$, for $ m > 1$, and $ 1$ is not a prime.

[quote="metafor"]Statement II (when rephrased as “the $ n^{th}$ root of $ m$ is a prime number”) is true for all powers of primes (e.g., $ \sqrt {4}$, $ \sqrt [3]{27}$, $ \sqrt [5]{2625}$), and statement III is true for $ m \equal{} 1$. Statement I is [i]never[/i] true, as $ m^{2}$ has at least three distinct divisors: $ m^{0}$, $ m$, and $ m^{2}$, for $ m > 1$, and $ 1$ is not a prime.[/quote]

but 1 is not a prime.

[quote="iniquitus"]Yes but III does not require m to be prime  :)[/quote]

oh,I see. Thanks.

A prime number is a number that only has two factors: 1 and itself. Therefore, 1 is not a prime, because the latter is the same as the former.

Also, an even number is an integer that can be divided evenly by 2. i.e. It ends in 0, 2, 4, 6, or 8. An odd number is an integer that cannot be divided evenly by 2. It will end in 1, 3, 5, 7, or 9.

*Correction: where it says $ \sqrt[5]{2625}$, it should say $ \textcolor[rgb]{0, 128, 0}{\sqrt[5]{3125}}$. Sorry about that.

Problem 1:

(i) $ m^2$ cannot be prime because $ m^2$ always has $ 1$, $ m$, and $ m^2$ as its divisors. (Not Possible)

(ii) An easy example is $ \sqrt{4}\equal{}2$ and $ \sqrt{9}\equal{}3$. (Possible)

(iii) The only positive integer solution to $ m^2\equal{}m^{1/2}$ is $ 1$. (Possible)

Therefore, the correct answer is D.

(i) is clearly never true. IT will have the $ 3$ divisors $ 1,m,m^2$.

Therefore, two of the divors would have to be the same.

Case 1: $ m=1$. Clearly, $ m^2=1$, and $ 1$ isn't prime.

Case 2: $ m^2=1$. Again, $ m=\pm 1$, which in either case, when squared is $ 1$,

which isn't prime.

Case 3: $ m^2=m$.

If $ m^2=m$, then $ m^2-m=0 \implies m(m-1)=0$.

This means that $ m=0,1$. Neither of these are prime when squared.

Therefore, (i) is never true.

(ii) can clearly be true, the smallest such $ m$ that works is $ 4$.

(iii). If $ m^2=\sqrt{m}$, $ m^4=m \implies m(m^3-1)=0 \implies m=0,1$

By the way, $ m$ could be a root of unity, but that isn't an integer.

In (iii), $ 1$ is fine, because its positive.

Therefore, (ii) and (iii) work, and the answer is D.

(i) doesn't work when $ m=1$ because $ 1$ isn't prime.

A prime number is defined to have $ 2$ and only $ 2$ factors. However,

$ 1$ only has $ 1$ factor, which is $ 1$.

An EVEN integer is defined as being a multiple of $ 2$.

There is a very easy test to find whether a number is odd or even.

We see that any $ n$-digit number can be expressed as

$ a_1 \cdot 10^{n-1} + a_2 \cdot 10^{n-2}+\hdots+a_n \cdot 10^0$,

where $ a_1, a_2, a_3, \hdots, a_n$ are the digits of the number.

For example, $ 1345=1 \cdot 10^3+3 \cdot 10^2 +4 \cdot 10^1 + 5 \cdot 10^0=
1000+300+40+5$.

Clearly, $ 10^n=2^n \cdot 5^n$, so all the terms, with the exception of

the final term, will be divisible by $ 2$, so they can be ignored.

We focus on the last term, $ a_n \cdot 10^0$.

Clearly, $ 10^0=1$, so $ a_n \cdot 10^0=a_n$.

Since $ a_n$ is the final digit, or the units digit, which is the rightmost digit,

it must be divisible by $ 2$ for the number to be.

Therefore, $ a_n=0,2,4,6,8$. If $ a_n$ is not one of these, the number cannot be even.

And yes, when we define EVEN and ODD, we focus on integers only.


Let $a,b,c$  positive numbers .

Find the minimum of $(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})(a+1)(b+1)(c+1)$

[quote="zhaobin"]$\frac{3^4}4$??[/quote]

Post solutions and not answers  :D

Very nice Zhaobin :D . It was my solution also .A nice  small and easy problem I think.


How would you solve the following problem:

Let P(n) and S(n) denote the product and the sum, respectively, of the digits of the integer n. For example, P(23) = 6 and S(23) = 5. Suppose N is a two-digit number such that N = P(N) + S(N). What is the units digit of N?

I encountered this sort of problem before once, and did not fully understand how to solve it.

Thanks.
 - Nazulovesmath

Since we are dealing with digits, you let $ N\equal{}10a\plus{}b$, where $ a$ is the tens digit and $ b$ is the units digit. Now you have $ 10a\plus{}b\equal{}a\plus{}b\plus{}ab$, which simplifies to:
$ 10a\equal{}a\plus{}ab$
$ 9a\equal{}ab$
$ b\equal{}9$

So the units digit is $ \boxed{9}$.

oooh. Now it makes sense!  :idea:
Thanks!!!

For problems like this, usually I construct an algebraic equation and assign variables to the digits.


Gauss
trigonometry
quadratics
algebra
polynomial
modular arithmetic

Evaluate $ \cos \frac{2\pi}{13} \plus{} \cos \frac{6 \pi}{13} \plus{} \cos \frac{8 \pi}{13}$.

A Numeric value or a simplified format? Because most of the questions are to be represented in a certain form under the Evaluate heading.
And To sum the , we need the term cos(4pi/13) right , because we need them to be in an A.P.

We can still sum the 6pi and 8pi term but i dont know how to handle the 2pi term.

[quote="t0rajir0u"][quote="Zed_darkx"]we need them to be in an A.P. [/quote]

We actually don't.  This summation is part of a class of extremely well-known summations (and has a very unambiguously simple value), but has been disguised.

[hide="Hint"] The quadratic residues $ \bmod 13$ are $ 1, 3, 4, 9, 10, 12$. [/hide][/quote]

Well, I got it down to the fact that the sum is evaluating

[hide]$ \frac{x^{12}\plus{}x^{10}\plus{}x^{9}\plus{}x^{4}\plus{}x^{3}\plus{}x^{1}}{2}$, where $ x\equal{}\cos{\frac{2\pi}{13}}\plus{}i\sin{\frac{2\pi}{13}}$[/hide]
But, I don't know how to use your hint here.

This is out of my level. I havent started on anything like that. Once i get the theory maybe i can solve it.

[quote="Dr Sonnhard Graubner"]hello, you can evaluate the term with Mathematica, use the command
FunctionExpand[].
Sonnhard.[/quote]
Mathematica?

Herr Sonnhard was referring to [url=http://www.wolfram.com/products/mathematica/index.html]Wolfram Mathematica[/url].

Allthough it is easier to use FullSimplify:

[code]FullSimplify[Cos[2 Pi/13] + Cos[6 Pi/13] + Cos[8 Pi/13]][/code]

hello, yes, indeed, your result is
$ \frac{1}{4}(\minus{}1\plus{}\sqrt{13})$.
Sonnhard.

[quote="MellowMelon"][hide="Further Hint"]Square it, and now treat it as a problem in combinatorics.[/hide][/quote]

could you elaborate on that?

What do you mean by "treat it as a problem in combinatorics"?

Can some elaborate what the hell everyone is doing in this thread???
It seems to go over the top of my head.

[hide="Further explanation"]You're evaluating this polynomial at a 13th root of unity. So $ x^{13} \equal{} 1$. When you square this sum, you're basically taking all possible pairwise sums of quadratic residues mod $ 13$ (with a few $ 2$s and $ 1$s thrown around). So now you ask yourself, how many solutions are there to
\[ a^2 \plus{} b^2 \equiv c \pmod{13}
\]
given $ c$?

I recommend separating this into the $ c$ is a QR and QNR cases.[/hide]

[hide="For masochists"]Now that you know how to evaluate Gauss sums, maybe you'll be able to solve this problem. Or not.

For an odd prime $ p$, let
$ n_1 \equal{} 1 \plus{} p \plus{} p^2 \plus{} \ldots \plus{} p^{p \minus{} 1}$
$ n_2 \equal{} 1 \minus{} p \plus{} p^2 \minus{} \ldots \plus{} p^{p \minus{} 1}$
Determine all $ p$ such that both of these numbers are prime.[/hide]

[quote="MellowMelon"]I recommend separating this into the $ c$ is a QR and QNR cases.
[/quote]
Could you please elaborate on the "QR" and "QNR" cases? Thanks.
Also, I don't really see how would finding the solutions to the congruence relation solve the problem...
(BTW see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=292136]here[/url] for a related problem)


This is essentially a partition problem:  how many ways can a number be partitioned into three parts, each part between 1 and 6 inclusive?  I've tried to do this before and I don't think I found anything interesting.  That is, I don't think you can get a general formula, except as the coefficient of a generating function, for which I suppose you could write some sort of multiple-sigma-sum expression.  No promises, though.

you know, x+x^2+x^3+x^4+x^5+x^6 can be expressed as x(x^6-1)/(1-x) and i think (if i remember correctly) its some nCm but im too lazy to figure it out now.  And Rep123max? get a calculator that can expand.


function
algebra
polynomial
induction
limit
geometric series

can anyone show me how to get the correct answer of the sum

$ \sum_{i \equal{} 1}^{\infty}\frac{n}{16^{n}}$?
I know the trick is to do something with writing out the first few terms, let the sum be called $ S$. Divide each of the terms by a number (like 2 or 3), and make S - S/3 or S-S/2, which eventually leaves a sequence that can be summed using the proper formula. But I don't want to guess so is there a systematic way of knowing what to divide S by? How can I proceed with this problem? Thanks.

[quote="bythecliff"]is there a systematic way of knowing what to divide S by?[/quote]

Well, look at the summation.  Ideally, you'd want the divided sum to have a similar form as the original sum, so you probably want to divide by [hide="goldfish"] $ 16$. [/hide]

[quote="1=2"]Well, when the denominator is m, you could divide by [hide="tigerlillies."]1/n[/hide][/quote]

I'm not really sure what your variables mean here.  In particular, the denominator is not a number but a function of $ n$, and the original summation itself is written with a confusing mix of variables.  Say, rather, that the original summation takes the form

$ \sum_{i\equal{}1}^{\infty}\frac{i}{16^{i}}$

Then the "denominator" is $ 16^{i}$.

I meant when 16 was replaced by any number m, you could multiply by m.

Except that you wrote $ \frac{1}{n}$ under your hide tag instead, which was somewhat confusing considering the poster's use of $ n$.

[hide="Shazam"]
Let $ S\equal{}\frac{1}{16}\plus{}\frac{2}{256}\plus{}\frac{3}{4096}\plus{}...$
Then $ 16S\minus{}1\equal{}\frac{2}{16}\plus{}\frac{3}{256}\plus{}...$

What is $ (16S\minus{}1)\minus{}S$?

I bet you know how to evaluate the sum of an infinite geometric series, so you can probably take it from here.
[/hide]

There's actually a much more interesting way to look at the problem.

[hide="Specifically..."] Observe that

$ (1\plus{}x)^{2}\equal{} 1\plus{}2x\plus{}x^{2}$
$ (1\plus{}x\plus{}x^{2})^{2}\equal{} 1\plus{}2x\plus{}3x^{2}\plus{}2x^{3}\plus{}x^{4}$
$ (1\plus{}x\plus{}x^{2}\plus{}x^{3})^{2}\equal{} 1\plus{}2x\plus{}3x^{2}\plus{}4x^{3}\plus{}3x^{4}\plus{}2x^{5}\plus{}x^{6}$

What would happen if we were to keep adding terms?  Well, this:

$ (1\plus{}x\plus{}x^{2}\plus{}x^{3}\plus{}... )^{2}\equal{} 1\plus{}2x\plus{}3x^{2}\plus{}4x^{3}\plus{}5x^{4}\plus{}6x^{5}\plus{}...$

Now plug in $ x \equal{}\frac{1}{16}$.  :) [/hide]

Of course we can establish that identity pretty quickly with the shifting and subtracting stuff, but that isn't so rigirous [duh]...we have to justify that the 'tail' of the polynomial goes to zero...which isn't totally obvious because the number of terms in the tail goes to infinity also [just not nearly as fast as the x^n for x=1/16]...

I presume that we would do whatever [some sort of induction] to establish the identity. Then we have to prove that we can ignore the later half of the terms...

[quote="bythecliff"]$ \sum_{i \equal{} 1}^{\infty}\frac{n}{16^{n}}$? [/quote]
It seems to me that this question is adding the same thing over and over again.
So, [hide="Butterflies"]
$ \sum_{i \equal{} 1}^{\infty}\frac{n}{16^{n}}\equal{}\frac{n}{16^{n}}\plus{}\frac{n}{16^{n}}\plus{}\frac{n}{16^{n}}\plus{}\frac{n}{16^{n}}\plus{}...$
Which is equivalent to $ \infty\cdot\frac{n}{16^{n}}$ which is just $ \infty$.[/hide]

I already noted that the original summation is not written correctly.  It should be

$ \sum_{i\equal{}1}^{\infty}\frac{i}{16^{i}}$

[hide="Bumblebees"]
Lets find the summation t0rajir0u has.

Now, $ \sum_{n \equal{} 0}^{\infty}x^{n}\equal{}\frac{1}{1\minus{}x}$ for $ |x| < 1$.

Differentiate term-by-term,
$ \sum_{n \equal{} 1}^{\infty}nx^{n\minus{}1}\equal{}\frac{1}{(1\minus{}x)^{2}}$ for $ |x| < 1$.

Multiply by $ x$ to get,
$ \sum_{n \equal{} 1}^{\infty}nx^{n}\equal{}\frac{x}{(1\minus{}x)^{2}}$ for $ |x| < 1$.

Now let $ x \equal{}\frac{1}{16}$ so,
$ \sum_{n \equal{} 1}^{\infty}\frac{n}{16^{n}}\equal{} ....$
[/hide]

$ \sum_{i \equal{} 1}^{\infty}\frac{i}{16^{i}}$

Given this form it may be easier to break up the summation.

$ \sum_{i \equal{} 1}^{\infty}\frac{i}{16^{i}}\rightarrow\frac{\sum_{i \equal{} 1}^{\infty}i}{\sum_{i \equal{} 1}^{\infty}16^{i}}$

However, I still get $ \frac{\infty}{\infty}$ which is, depending on your defintion, one of $ 1$ (from $ \frac{x}{x}\equal{} 1,x\text{ not equal to }0$), $ \infty$ (from the fact that it [i]is[/i] $ \infty$), or $ undefined$ (as a calculator will tell you). As it would be to obvious for it to $ \infty$ and $ undefined$ is impossible, the answer must be $ \boxed{1}$.

Code for "≠", please. And help with those sums in fractions.  :wink:

[quote="Kyron"]
Given this form it may be easier to break up the summation.
$ \sum_{i \equal{} 1}^{\infty}\frac{i}{16^{i}}\rightarrow\frac{\sum_{i \equal{} 1}^{\infty}i}{\sum_{i \equal{} 1}^{\infty}16^{i}}$
[/quote]

For positive sequences $ a_{j},\; b_{j}$:
$ \sum\frac{a_{j}}{b_{j}}\geq\frac{\sum a_{j}}{\sum b_{j}}$
You cannot break up the summation like that without proving that it must work in that case.

On the same token, $ \sum (a_{j}b_{j})\leq\left(\sum a_{j}\right)\left(\sum b_{j}\right)$
And again, you cannot break up the summation without proving that it must work.

Example: $ \frac{1}{2}\plus{}\frac{2}{4}\equal{} 1$ but $ \frac{1\plus{}2}{2\plus{}4}\equal{}\frac{1}{2}$

[quote="Kyron"]As it would be to obvious for it to $ \infty$ and $ undefined$ is impossible, the answer must be $ \boxed{1}$[/quote]

This is not true; indeterminate forms do not necessarily converge to one. For example, in $ \frac{n}{n^{2}}$ as $ n\to\infty$, both the numerator and denominator go to $ \infty$, but the overall quotient behaves like $ \frac{1}{n}$ and goes to $ 0$. Similarly, $ \frac{n^{2}}{n}$ goes to $ \infty$ and $ \frac{2n}{n}$ goes to $ 2$.

The code for not equal to is given by placing "\not" before the equals sign. "\not" places a diagonal slash through whatever is next. You can even do things like $ 2\not | 5$, $ 2/5\not\in\mathbb{N}$, and even $ \not 0$

@The Zuton Force:
EVERYTHING I'VE BEEN TOLD IS A LIE!

That method (of simplifying sums) was taught to me by my teacher. So, it doesn't work. Does the constant multiple rule work, though? (i.e. $ \sum_{i \equal{} 1}^{\infty}6\left(\frac{2}{3}\right)^{i}\equal{} 6\cdot\sum_{i \equal{} 1}^{\infty}\left(\frac{2}{3}\right)^{i}$)

[quote="The Zuton Force"]This is not true; indeterminate forms do not necessarily converge to one.[/quote]
Yeah... :blush: I forgot about the fact that the $ \infty$ on the bottom was greater than the $ \infty$ on the top. So my answer was wrong [i]even if[/i] my method was right... Guess that explains why my calculator says $ \frac{\infty}{\infty}\equal{} undef$.  :wallbash:

Oh wow, what a bad teacher!  :o

The constant rule works because it is basically the distributive property:
$ (ax\plus{}ay\plus{}az\plus{}...) \equal{} a(x\plus{}y\plus{}z\plus{}...)$

@Kyron, yes the Constant Multiple rules works.

[b]Theorem:[/b] Let $ \{ s_{n}\}$ be a sequence so that $ \sum_{n \equal{} 1}^{\infty}s_{n}$ converges and equal $ s$.
Then $ \sum_{n \equal{} 1}^{\infty}ks_{n}$ converges and is $ ks$.

[b]Proof:[/b] Let $ u_{n}\equal{}\sum_{i \equal{} 1}^{n}s_{i}$, i.e. the sequence of partial sums. And let $ v_{n}\equal{}\sum_{i \equal{} 1}^{n}ks_{i}\equal{} k\sum_{i \equal{} 1}^{n}\equal{} ku_{n}$. Now, $ \lim u_{n}\equal{} s$. Hence by the constant multiple rule for [i]sequences[/i] we have $ \lim ku_{n}\equal{} k\lim u_{n}\equal{} ks$. Q.E.D.

The converse that $ \sum_{n\equal{}1}^{\infty}ks_{n}$ converges implies $ \sum_{n\equal{}1}^{\infty}s_{n}$ [b]almost[/b] works. Can you prove as an excersie when it works and what is an exception.

Also, there exists the special code \neq to give you $ \neq$ if you don't want to do "\not ="

Your answer is correct, Hamster 1800. :)

$ \sum_{i \equal{} 1}^{n}\frac{i}{16^{i}}\equal{}\sum_{i \equal{} 1}^{n}\left\{\frac{a(i\minus{}1)\plus{}b}{16^{i\minus{}1}}\minus{}\frac{ai\plus{}b}{16^{i}}\right\}\longrightarrow b\ (n\longrightarrow\infty)$, where

$ a \equal{}\frac{1}{15},\ b \equal{}\frac{16}{225}$.


geometry
3D geometry
algebra unsolved
algebra

Prove there are infinitely many positive integers which cannot be represented as a sum of three perfect cubes.

Hmm I'm totally stumped on this one, I'm thinking of trying contradiction but I don't know where to start.

the AoPS Vol. 1 has a proof somewhere in chapter 5 or 6

Hmm I'm still totally lost, where do I start with this question?

Show that a perfect cube can only be congruent to $ 0, 1, \minus{}1 \bmod 9$.  What does that tell you about the possible values of a sum of three perfect cubes $ \bmod 9$?

Hmm how do you show a perfect cube can only be congruent to 0 1 or -1 mod 9?

$ x \equiv 0 (mod9) \Longrightarrow x^3 \equiv 0(mod 9)$

$ x \equiv 1(mod 9) \Longrightarrow x^3 \equiv 1 (mod9)$

$ x \equiv 2 (mod9) \Longrightarrow x^3 \equiv \minus{} 1 (mod 9)$

$ x \equiv 3 (mod9) \Longrightarrow x^3 \equiv 0 (mod 9)$

...........*till x=8.

dimitris

Oh yeah I see, but how does that help the original question?

I already told you what the next step is.


For $ n$ belonging to the set of natural numbers,let $ f(n) \equal{} 1^{n} \plus{} 2^{n \minus{} 1} \plus{} 3^{n \minus{} 2} \plus{} .... \plus{} n^{1}$.Determine the minimum value for $ \frac {f(n \plus{} 1)}{f(n)}$

Can n be 0? If n can be 0, then the desired value is clearly 1. If not, the desired number is 3, and thiw can be proved by induction.

No it cant be $ 0$,clearly said that $ n$ is a natural number.

Minimum value is $ \frac {8}{3}$ when $ n \equal{} 2$. When $ n \equal{} 1$ our fraction is equal to $ 3 > \frac {8}{3}$. Let now $ n\geq3$.

$ \frac {1^{n \plus{} 1} \plus{} 2^{n} \plus{} 3^{n \minus{} 1} \plus{} \dots \plus{} n^2 \plus{} n \plus{} 1}{1^n \plus{} 2^{n \minus{} 1} \plus{} 3^{n \minus{} 2} \plus{} \dots \plus{} n} \geq \frac {1^{n \plus{} 1} \plus{} 2^{n} \plus{} 3S_n \plus{} n \plus{} 1}{1^n \plus{} 2^{n \minus{} 1} \plus{} S_n}$

where $ S_n \equal{} 3^{n \minus{} 2} \plus{} \dots \plus{} n$

$ \frac {1^{n \plus{} 1} \plus{} 2^{n} \plus{} 3S_n \plus{} n \plus{} 1}{1^n \plus{} 2^{n \minus{} 1} \plus{} S_n} > \frac {8}{3}$

$ 3\cdot 2^n \plus{} 9S_n \plus{} 3n \plus{} 6 > 8 \plus{} 8 \cdot 2^{n \minus{} 1} \plus{} 8S_n$

$ S_n > 2^{n} \plus{} 2 \minus{} 3n$

$ 3^{n \minus{} 2} \plus{} 4^{n \minus{} 3} \plus{} \dots \plus{} n > 2^n \plus{} 2 \minus{} 3n$

Now if $ n \equal{} 3,4,5$ then inequality is true. If $ n \geq 6$ then $ 4^{n \minus{} 3}\geq 2^n$ and $ 3^{n \minus{} 2} > 2 \minus{} 3n$.


algebra
polynomial
calculus
integration
algebra unsolved

Given $p(x)=a + bx + 3cx^2 + dx^3 + 5ex^4$ for $p(x)\in \mathbb{R}$

if $a+c+e=0$ prove that there is a real root for $p(x)=0$ in $[-1,1]$

The integral between -1 and 1 is 0, so there's must be at least one x in [-1,1 ]where p(x)=0

for example p(x)=(x2-1)^2, p(x) has not roots in (-1,1). Roots +-1.

I am not attentive. I think condition is equality summ even degree's coefficients to 0. In this case integral can be positive and root can be +-1. If Integral equal 0, exist root in (-1,1).

do you mean that

$\int^1_{-1}p(x) dx=0$ so there is root in between ?


integration
function
real analysis
real analysis unsolved

Let $ A \subset \mathbb{R}^n$ measurable set. Prove that $ \mathcal{L}_2(A)\subset \mathcal{L}_1(A) \Leftrightarrow m(A)< \infty$

[quote="csabafarkas"]Let $ A \subset \mathbb{R}^n$ measurable set. Prove that $ \mathcal{L}_2(A)\subset \mathcal{L}_1(A) \Leftrightarrow m(A) < \infty$[/quote]
If $ m(A) < \infty$, then by Hoelder
\[ \|f\|_1 = \int_A |f(x)|dx \le (\int_Af^2(x)dx )^{1/2} (\int_A 1 dx)^{1/2} = (m(A))^{1/2}\|f\|_2 \, .
\]
Suppose now that  $ m(A) = \infty$. It suffices to find a function $ f \in \mathcal{L}_2$  that is not in $ \mathcal{L}_1$.

Define numbers $ R_1,\ R_2,\dots$ by the requirement that $ m(A\cap B_{R_n}) = n$, where $ B_R$ is the ball about $ 0$ with radius $ R$. Set $ A_1 = A \cap B_{R_1}, \, A_k = A \cap (B_{R_k} - B_{R_{k-1}})$ for $ k > 1$. Thus $ m(A_k) = 1$ for all $ k$. Now define $ f: A \to \mathbb{R}$  by  $ f(x) = \frac{1}{k}$ if $ x \in A_k$. Then
\[ \|f\|_2^2 = \sum_k \frac{1}{k^2} m(A_k) = \frac{\pi^2}{6}, \quad \|f\|_1 = \sum_k \frac{1}{k} m(A_k) = \infty\, .\]


if Train A left Station A and headed due east at 60 mph, and Train B left from Station B heading due west at 20 mph, and the 2 stations are 500 miles apart, how long and how far will both trains have traveled untill they collide? (assuming they are both on the same track).

train A =375 Miles
train B=125 Miles
Thugh I calculated it by trial and error,I knew a method to use a bit of simultaneous equations ,if some one can show it here.

i'm too slow, but from now on i'll type how i got it.

can anyone figure out how long it took for them to collide?


[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=86162[/url] is a topic I started about the recent nat geo results among young people in the US and abroad.

You can test yourself here :

[url]http://geosurvey.nationalgeographic.com/geosurvey/templates/question_1.html[/url] to take a 20 question test on the nat geo site.

[url]http://www.sheppardsoftware.com/geography.htm[/url] is a cool site where you can quiz yourself

Now do you know more sites?  I didn't wanna go off topic in Round Table so I started a topic here.

I am looking for geo games in general, but especially games that test you in knowing which countries border which and provinces of China or provinces in general.

[quote="bubka"]There was a game like that in Encarta 2004.
(but surely, you [i]are[/i] serious after all!)[/quote]

 :|  Your messages start getting weirder and weirder.
What do you mean?

why nothing could be simpler. Why else would geography of all things bother you? :D

Let us have a vote. Who else thinks fred is serious?

I'm 10, and I could locate ALL of the countries on the first quiz!
P.S. I got 70% of the test correct.


inequalities
logarithms
integration
inequalities open

Prove that for all positive integers $ n $  we have  $ \displaystyle  \sum\limits_{k=1}^{n}\frac{\sqrt{2^k}}{1+2^k}<\displaystyle  \frac{\pi}{\ln{4}}-\frac{1}{4}+\frac{1}{2}{10}^{-11}\; . $

Trivially, $\sum_{k=1}^n \frac{\sqrt{2^k}}{1+2^k} < \sum_{k=1}^\infty \frac{\sqrt{2^k}}{1+2^k}$. Hence, to prove the statement it is enough to show that $\sum_{k=1}^\infty \frac{\sqrt{2^k}}{1+2^k} < \frac{\pi}{\ln 4} - \frac{1}{4} + 5\cdot 10^{-12}$.

Estimate \[\sum_{k=1}^\infty \frac{\sqrt{2^k}}{1+2^k} < \sum_{k=1}^{80} \frac{\sqrt{2^k}}{1+2^k} + \int\limits_{80}^{\infty} \frac{d x}{2\cosh\left(\frac{\ln 2}{2}x\right)}\] and compute the r.h.s. with nessesary precision to prove that it is less than $\frac{\pi}{\ln 4} - \frac{1}{4} + 5\cdot 10^{-12}$.

For example, in Maple
[code]>evalf(add(2^(k/2)/(1+2^k),k=1..80)+int(1/(2*cosh(ln(2)/2*x)),x=80..infinity)-(Pi/ln(4)-1/4+5*10^(-12)),20);
-0.6840211e-12[/code]
p.s. btw, what is "Alex collection"? Where can I see other items from it?  :?:


A man is moving with a speed u(vector)=ui^  and finds that rain is falling

vertically down along the direction j^(unit vector).if  he doubles his speed

he finds that the rain is falling at an angle $\theta$ with the vertical.

so find the original direction and magnitude of the rainfall

also please tell how to type the vector arrow on the top of a letter

Type vectors with \vec: e.g. \vec{xy} gives $\vec{xy}$.

So, the relative speed of the rain in the first case is $\vec v' = \vec v-\vec u$, and in the second case $\vec v'' = \vec v-2\vec u$. It's easy to see from the vector diagrams (which I don't have the time to draw now) that the angle the rain velocity makes with the vertical is $\theta$, and that the speed of the rain is $v = \frac u{\sin\theta}$.

sorry the question is slightly modified

the question is

find the magnitude and direction of rain with respect to ground

the options are like this

a)ui^-utan$\theta$j^     b)ui^-(u/tan$\theta$)j^

c)ui^                            d)uj^

It's b), of course; that's what you get when you break up the vector $v$ into components...

but how doyou get - sign and the direction

can you solve the problem
if possible any diagram
and what we should do when the speed is doubled
is it not taken into account

Ok, here are the diagrams of vector subtraction you need. The second triangle is isosceles.


calculus
integration
trigonometry
calculus computations

Alright, so I need to show the following:

cos(x) = (1 - tan^2(x/2)) / (1 + tan^2(x/2))

and

sin(x) = (2tan(x/2)) / (1 + tan^2(x/2))

I have no idea how to work through these at all, so if somebody could help me, it would be great!

You can use the following formulae.

\[ \cos 2 \theta=2\cos ^ 2 \theta -1 \]
\[ \sin 2 \theta=2\sin \theta \cos \theta  \]

Then use $1+\tan ^ 2 \theta=\frac{1}{\cos ^ 2 \theta}$

I think he is using trig substitution, where you let $t=\tan\frac{\theta}2$ so that $\sin\theta=\frac{2t}{1+t^2}$ and $\cos\theta=\frac{1-t^2}{1+t^2}$. I am also learning this in my calc class. :)

Thanks for the help, but I'm still not sure where to go with this... some elaboration would be great... thanks!

You do not need integrals for this. Not even calculus is necessary.

You want to show:
$\cos x=\frac{1-\tan^2x}{1+\tan^2\frac{x}{2}}$
$\sin x=\frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}$

Use what kunny posted, replacing his $\theta$ with $\frac{x}{2}$.
Something else that might be helpful: $\tan\theta=\frac{\sin\theta}{\cos\theta}$.

[quote="Bictor717"]
Something else that might be helpful: $\tan\theta=\frac{\sin\theta}{\cos\theta}$.[/quote]

Yes, this is vital, Bictor717. :)


[quote]If $ \sum _{j \equal{} 0}^n x_j \equal{} \frac {1}{2}$ and $ x_j > 0 \; \forall j$, show that $ \prod \frac {1 \minus{} x_j}{1 \plus{} x_j} \geq \frac {1}{3}$[/quote]

using weirstrass the inequality can't be proved as for the denominator the sign of inequality is flipped

[quote="apratimdefermat"]x1+x2+x3+...............+xn=1/2, all xi are +ve reals

PT ((1-x1)/(1+x1))*((1-x2)/(1+x2))*.........*((1-xn)/(1+xn))>=1/3

xn means x subscript n :|[/quote]
[hide="Solution"]
[b]Lemma:[/b] $ \frac{1\minus{}x\minus{}y}{1\plus{}x\plus{}y}\le \frac{1\minus{}x}{1\plus{}x}\cdot \frac{1\minus{}y}{1\plus{}y}$.
[i]Proof:[/i] Upon expansion, this becomes \[ (1\minus{}x\minus{}y)(1\plus{}xy\plus{}x\plus{}y)\le (1\minus{}x\minus{}y\plus{}xy)(1\plus{}x\plus{}y)\] \[ 1\minus{}xy\minus{}x^2y\minus{}x^2\minus{}xy^2\minus{}y^2\le 1\minus{}x^2\minus{}y^2\minus{}xy\plus{}x^2y\plus{}y^2x\] \[ xy(x\plus{}y)\ge \minus{}xy(x\plus{}y)\] which is true.

Now for the problem.  We proceed with induction.  If $ n\equal{}1$, then it becomes $ \frac{1\minus{}\frac{1}{2}}{1\plus{}\frac{1}{2}}\equal{}\frac{1}{3}$, so we are okay.  Say that it works for $ n$ variables.  We will show that it works for $ n\plus{}1$ variables.  By the Lemma, we have that \[ \prod_{i\equal{}1}^{n\plus{}1} {\frac{1\minus{}x_i}{1\plus{}x_i}}\ge \frac{1\minus{}x_n\minus{}x_{n\plus{}1}}{1\plus{}x_n\plus{}x_{n\plus{}1}}\cdot \prod_{j\equal{}1}^{n\minus{}1} {\frac{1\minus{}x_j}{1\plus{}x_j}}\] Yet, we have that $ x_n\plus{}x_{n\plus{}1}\plus{}\sum_{i\equal{}1}^{n\minus{}1} x_i\equal{}\sum_{j\equal{}1}^{n\plus{}1} {x_j}\equal{}\frac{1}{2}$.  Then, by the inductive hypothesis, \[ \prod_{i\equal{}1}^{n\plus{}1} {\frac{1\minus{}x_i}{1\plus{}x_i}}\ge \frac{1\minus{}x_n\minus{}x_{n\plus{}1}}{1\plus{}x_n\plus{}x_{n\plus{}1}}\cdot \prod_{j\equal{}1}^{n\minus{}1} {\frac{1\minus{}x_j}{1\plus{}x_j}}\ge \frac{1}{3}\] as desired. [/hide]


on page 170 of student "introduction to number theory" and page 76 teacher "introduction to number theory" there is a base number subtraction problem I can't figure out

9.3.1
(f)        13a9 (base 12)- 48b (base 12)

                                                         13a9    (both numbers base 12)
                                                         -48b

or

                                                          (1) (3) (10) (9)
                                                           -   (4) (8) (11)

if you take one from the tens digit of the top number and then add 12 to the nine and then subtract you get;

                                                          (1) (3) (9) (21)
                                                           -   (4) (8) (11)
                                                        =                (10)    or a

now nine minus eight is 1* so you have

                                                         (1) (3) (9) (21)
                                                          -   (4) (8) (11)
                                                         =         (1)   a

now you carry twelve from the 1 and get 15-4 as your final problem, this equals 11 or b

the answer should be b1a

but the book says      b2a

*this is the part I think I messed up on

srry, the stupid computer messed up the format, the digits are in the wrong place

i don't really get ur second post? :huh:  :huh:  :o


geometry
limit
integration
trigonometry
function
calculus
real analysis

Evaluate the following limit.

\[\lim_{n\to\infty}\int^{n\pi}_0 e^{-x}|\sin nx| dx \]

Answer is $\frac{2}{\pi}$. Moreover, I can prove that for arbitrary function $f\in L_1(a,b)$ we have
\[\lim_{n\to \infty}\int_a^bf(x)|\sin(nx)|dx=\frac{2}{\pi}\int_a^bf(x)dx.\]
It follows from two facts:
[b]1)[/b] $\int_{t_0}^{t_1}|\sin(nx)|dx\to \frac{2}{\pi}(t_1-t_0)$ for all $t_0,t_1$.
[b]2)[/b] set of staircase functions is dense in $L_1$.

to Myth: does you result go when the integration bounds are not constant numers?
(the upper bound in our case is $n\pi$ )?

To eugene.
Yes, because $\int_{u}^\infty f(x)|\sin(nx)|dx\to 0$ as $u\to \infty$ regardless of $n$.

To kunny.
I know that it is right. :D


Find all partitions of natural numbers larger than 2005 into two infinite subsets in such a way that if $x,y,z$ lie in the same subset then $x^3+y^2+z$ belongs to this set too.

I found one partition :  :lol:
Take the set $A=\{2006,2008,2010,...\}$ the set of even integers $>2005$
and the set $B=\{2007,2009,2011,...\}$ the set of odd integers $>2005$
the two sets $A$ and $B$ satisfy the given condition

Try to prove!
It is moderate easy problem from the preparation of my students for ARO's final round.


Hi,

I just wanted to say hello. My name is David Jimenez, I am a grad student in GATech, and I have been interested in math contest since I realized that they existed (unfortunately, a little late in my life, therefore I didn't do much on them as a student). I am not american, but I live in Atlanta, therefore, I supposed it would be a good place to socialize with people having more or less the same interest on this topic.

Best regards,

Really? You are a grad at GATech? Interesting. I was there Saturday for MATHCounts.

Are you going to be here next saturday for the High School Math Competition?

Let me know. Send me a private message if you want and we can meet sometime!

Best regards,

Same :) It's really cool how young yet old AoPs is.


probability
combinatorics unsolved
combinatorics

Hi! :)

Well, two people throw alternatively one coin n times. It is a normal coin and the result does not depend on the previous one.
What is the probability that the two players get the same number of heads?

Have fun :p

Well, maybe this post should be moved into the "intermediate" section or so... :)

I  try this question.
Let $A_i :$ event of throwing a head i times out of n by first person
And $B_i :$ event of throwing a head i times out of n by second person

$P(A_0B_0+ A_1B_1+\cdot\cdot\cdot+ A_nB_n)= P(A_0B_0) + P(A_1B_1) + \cdot\cdot\cdot + P(A_nB_n)$
$=(\binom{n}{0}(\frac{1}{2})^n)^2 + (\binom{n}{1}(\frac{1}{2})^n)^2 + \cdot\cdot\cdot + (\binom{n}{n}(\frac{1}{2})^n)^2$
$=(\binom{n}{0}^2 + \binom{n}{1}^2 + \cdot\cdot\cdot + \binom{n}{n}^2) (\frac{1}{2})^n$
$=\binom{2n}{n}(\frac{1}{2})^n$

According to what I've done, you're right! ;)

But, since I haven't the solution for this problem, this has to be checked :) (although I think we're both right hehe)


rrusczyk wrote:Prove that the length of median AM in triangle ABC is greater than (AB+AC-BC)/2.

[hide]Let a = BC, b = AC, c = AB, d = AM.  Then from Stewart's Theorem,

a*d^2 + a*(a/2)^2 = c^2*a/2 + b^2*a/2

d^2 = b^2/2 + c^2/2 - a^2/4

We seek to prove that this quantity is greater than (b+c-a)/2.

Since any square of a real number is nonnegative, (b-c)^2  0.  Also, from the triangle inequality, b + c > a, so b + c - a > 0, so 2a(b + c - a) > 0.  Adding these two inequalities yields

(b-c)^2 + 2a(b+c-a) > 0

b^2 - 2bc + c^2 + 2ab + 2ac - 2a^2 > 0

2b^2 + 2c^2 - a^2 > a^2 + b^2 + c^2 + 2bc - 2ac - 2ab = (b + c - a)^2

 > b + c - a

d > [/hide]

anipoh wrote:Another way:
[hide]Extend AM to M' such that MM' = AM.  Connect CM'.
Apply triangle inequality in triangle ACM and M'CM we get the desired result.[/hide]

Wow, very nice.


Consider the sequence of positive integers which stisfies $ a_{n}\equal{}{a_{n\minus{}1}}^{2}\plus{}{a_{n\minus{}2}}^{2}\plus{}{a_{n\minus{}3}}^{2}$ for all $ n\geq 3$

Prove that if $ a_{k}\equal{} 1997$ then $ k\leq 3$.

I think there's a mistake in the problem :huh:
[hide]What if the first three terms are 2, 12, 43?[/hide]

[quote="boxedexe"]The problem asks to prove that [i]if[/i] $ a_{k}\equal{} 1997$ is [i]possible[/i] for some integer $ k$, then we must have $ k\leq 3$.[/quote]
$ 2^{2}\plus{}12^{2}\plus{}43^{2}\equal{} 1997$
But I just realized that the first term is $ a_{0}$, not $ a_{1}$

[hide]
If $ a_{k}$ were to equal 1997 for some $ k\geq 4$, then letting $ x\equal{}a_{k\minus{}2},y\equal{}a_{k\minus{}3},z\equal{}a_{k\minus{}4}$, we would need
\[ 1997\equal{}x^{2}\plus{}y^{2}\plus{}(x^{2}\plus{}y^{2}\plus{}z^{2})^{2}\]
for positive integers $ x,y,z$. In particular, this requires
\[ \left( x^{2}\plus{}y^{2}\plus{}z^{2}\right)^{2}<1997<\left( x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}\frac{1}{2}\right)^{2}\]
But $ \left(44\plus{}\frac{1}{2}\right)^{2}<1997<45^{2}$ which requires $ 44<x^{2}\plus{}y^{2}\plus{}z^{2}<45$, a contradiction.
[/hide]


Let $ABCD$ be a square. Points $E$ and $F$ are chosen on sides $AB$ and $CD$, respectively, such that $AE = CF$.
Let $AD$ intersect $CE$ and $BF$ at $M$ and $N$, respectively. Suppose that $P$ is the intersection of $BM$ and $CN$, find the locus of $P$ when $E$ and $F$ move on the side $AB$ and $CD$, respectively.

$\frac{AM}{BC}=\frac{AE}{EB} \Longrightarrow AM = \frac{AE}{BE}\cdot BC$.

$\frac{DN}{BC}=\frac{DF}{FC} \Longrightarrow DN = \frac{DF}{FC} \cdot BC$.

Let $\angle{AMB}=x$, $\angle{CND}=y$, then we have:
\[ \tan{x}=\frac{AB}{AM}=\frac{BE}{AE}\quad\mbox{and}\quad\tan{y}=\frac{CD}{DN}=\frac{CF}{DF} \]
Therefore we have $\tan{x} \cdot \tan{y}=1$. Thus:
\[ \sin{x}\sin{y}=\cos{x}\cos{y} \]
\[ \cos{x}\cos{y}-\sin{x}\sin{y}=0 \]
\[ \cos{(x+y)}=0 \]
Thus $x+y=90^o$, which gives $\angle{MPN}=90^o$. Therefore the locus of $P$ when $E$ and $F$ moves along $AB$ and $CD$ is a semicircle with diameter $BC$.


1) Two motorists met at 10 a.m at point X. After this meeting, one of them proceeded in the east direction while the other proceeded in the north direction. Exactly at noon, they were 60 Kms apart. Find their Speeds, if the difference of their speeds is 6 Kmph.

2) A bus left point X for point Y. Two hours later a car left point X for Y and arrived at Y at the same time as the bus. if the car and the bus left simultaneously from opposite ends X & Y they would meet in 1.33 hours after the start. How much time did it take the bus to travel from X to Y?

3) A race track is in the form of a right angle triangle. The longer of the legs of the track is 2 Km more than the shorter of the legs (both these legs being on a highway). The start and end points are also connected to each other through a side road. The escort vehicle for the race took the side road and rode with a speed of 30 Kmph and then covered the two intervals along the highway during the same time with a speed of 42 Kmph. Find the length of the race track.

4) Two Indian tourists in the US cycled towards each other, one from a point A and the other from point B. the first tourist left point A 6 hrs later than the second left point B, and it turned out on their meeting that he had traveled 12 Kms less than the second tourist. After meeting, they kept cycling with the same speed, and the first tourist arrived at B 8 hrs later and the second arrived at A 9 hrs later. Find their speeds.

[hide="1"]Let the speed of the slower motorist be $s$. We know $d=st$ (distance = speed x time) and the Pythagorean formula for a right-angled triangle.
Then $(2s)^{2}+(2(s+6))^{2}=60^{2}$
and we can solve for $s$.[/hide]
Don't have time for the rest now... just set up equations and it will work.

[hide="2"]
Assume the distance between X and Y is $1$. The bus need $x$ hours to travel from X to Y, then the car need $x-2$ hours to travel from X to Y. The speed of the bus is $\frac{1}{x}$, the speed of the car is $\frac{1}{x-2}$

$\frac{1}{\frac{1}{x}+\frac{1}{x-2}}=1.33$

OK, you mean $1.3333....$? 'Cause if you solve with 1.33, the answer will be really nasty.[/hide]

[hide="3"]
Assume the two legs are $x$ and $x+2$, then the hypotenuse is $\sqrt{2x^{2}+4x+4}$.

$\frac{x+x+2}{42}=\frac{\sqrt{2x^{2}+4x+4}}{30}$
solve for it, you get $x=-8(reject)$ or $x=\boxed{6}$[/hide]

[hide="4"]Let $C$ be the meeting point, $t$ the time that took $A$-cyclist to come from $A$ to $C$, and their speeds $v_{A}, v_{B}$. Then

\begin{eqnarray}AC &=& v_{A}t\\ BC &=& v_{B}(t+6)\\ AC &=& BC-12\\{BC\over v_{A}}&=& 8\\{AC\over v_{B}}&=& 9\end{eqnarray}

Plugging $(1)$ to $(5)$, $(2)$ to $(4)$ and multiplying $(4)$ by $(5)$ we get

$t(t+6)=72\implies t=6$

Then $(1)$ becomes $AC=6v_{A}$, $(2)$ becomes $BC=12v_{B}$, so $(3)$ gives $6v_{A}=12v_{B}-12\iff v_{A}=2v_{B}-2$. Putting that into $(4)$ we get

$\frac{12v_{B}}{2v_{B}-2}=8\iff 12v_{B}=16v_{b}-16\iff v_{B}=4$, and then $v_{A}=2v_{B}-2=6$[/hide]

[quote="fizsle"]Still not clear about 2.... yes i can form the equations but it kinda gets too messy...any simpler method???[/quote]

Assuming Allan's equations are correct, you end up with a quadratic.  That's not too bad, right?

[quote="fizsle"]Still not clear about 2.... yes i can form the equations but it kinda gets too messy...any simpler method???[/quote]

umm... just solve it, x=4


when is the date for the fifth/final CML? and are we allowed to discuss it yet (my school is on spring break during the normal test date, so we took it earlier)?

[quote="indianamath"]huh? what are you talking about?[/quote]

continental mathleague is a set of 5 6-question tests that are around mathcounts level or maybe easier. i did terrible due to careless mistakes. :mad:

I did that math league in 3rd grade lol.

It was at an old school that I don't remember much about since I moved to a different school.

ya, they have them for all grade levels starting in like 1st or 2nd grade, i believe...

i didn't do too well, yes i got below 26.

I got 28 total, which was 2nd place in 7th grade at my school...I make careless mistakes on these kinds of contests too, but I'm a LOT better than I was last year, probably because of Mr. Fischer engraving the words "don't make careless mistakes" into our brains when we were preparing for chapter and state.

i got 25, because of alot of careless mistakes, well i guess i shouldn't give excuses. what did you get on the last one? everyone said it was really hard, but it was one of only 2 that i got a 6 on.  :blush:

I got 30!!! Last year was soo frustrating, because i missed number 5 on the last one! so i ended up with a 29/30. only one person in my class got #5, but she listed literally 150 entries on the back of her sheet... i looked at the paper when we got it back and i had made a careless mistake and switched the coefficients in the equation or some really retarded thing like that... i was so sad :(

[quote="mathcrazed"]I got perfects on everything -_-...[/quote]

haha me too.


analytic geometry
geometry
integration
calculus

A semicircle with radius 1 and diameter AB lies in the plane. A line parallel to AB intersects the semicircle at points P and Q, dividing the semicircle into two regions with equal area. Find the length of PQ.

Are sure it has a nice solution¿
I'm almost sure that the length you are looking for is a transcendental (even over $\pi$)

Put the coordinate origin $O$ at the circle center and the x-axis along the ray $OB$. The coordinates of the points $A, B$ are $A = (-1, 0)$, $B = (1, 0)$. Let the coordinates of the points $P, Q$ be $P = (-p, q)$, $Q(p, q)$. The circle equation is $x^2 + y^2 = 1$. Since the point $Q$ is on the circle $(O)$, the 1st equation for $p, q$ is

$p^2 + q^2 = 1$

Since areas of the circle cuts below and above the line $PQ$ are equal, the 2nd equation for $p, q$ is

$\int_0^p \sqrt{1 - x^2} \text dx - pq = pq + \int_p^1 \sqrt{1 - x^2} \text dx$

$2 \int_0^p \sqrt{1 - x^2} \text dx - \int_0^1 \sqrt{1 - x^2} \text dx = 2pq$

$\left[x \sqrt{1 - x^2} + \arcsin x \right]_0^p - \frac \pi 4 = 2pq$

$p \sqrt{1 - p^2} + \arcsin p - \frac \pi 4 = 2pq$

Substituting $q = \sqrt{1 - p^2}$ from the 1st equation to the 2nd equation,

$p \sqrt{1 - p^2} + \arcsin p - \frac \pi 4 = 2p \sqrt{1 - p^2}$

$\arcsin p - p \sqrt{1 - p^2} - \frac \pi 4 = 0$

The root of this equation can be found by numerical methods:

$p \doteq 0.914771,\ \ \ PQ = 2p \doteq 1.829542$

[quote="ZetaX"]Well, you need not to use integrals  ;)[/quote]
I know, but it seems to me that the problem calls for it. And Pre-Olympiad high school students either know simple integrals already or they will learn them soon. I will post another geometry problem, which I think really requires the use of integrals. I hope it is not moved to Calculus Tutorials.

Yetti


[quote="RHSLilSweetie07"]The directions were to perform the indicated operations and simplify.

z^1/3*(z^2 - 1)

Is the answer z^7/3 - z^1/3?[/quote]

[hide]$z^\frac{1}{3}*(z^2 - 1)$

$z^\frac{7}{3}-z^\frac{1}{3}$

$\sqrt[3]{z}(z^7-1)$
[/hide]

[quote="boxedexe"]It should be:

$z^{\frac{1}{3}}(z^2-1)\\
=z^{\frac{7}{3}}-z^{\frac{1}{3}}\\
=\sqrt[3]{z}^7-\sqrt[3]{z}\\
=\sqrt[3]{z}(z^2-1)$

Masoud Zargar[/quote]

Oops  :blush:

But wait. That's just getting back where we started...  :?

Lol, I suppose you're correct that we're back to the same place.  I believe the question is already in the simplified form.  The expanded form would be $\sqrt[3]{z^7}-\sqrt[3]{z}$.


MATHCOUNTS
function
algebra
polynomial
number theory
prime factorization

I was wondering if self-creating computer programs on graphing calculators is allowed for the target and team rounds. I know any calculator is okay, but some people I know have complex programs to help them with the competition. Thanks!

Folks at my school- YOU KNOW what I'm talking about

[quote="xpmath"]Yeah, you're allowed to use programs on your calculator...

That being said, since I don't know how to program at all, even on a calculator, I would appreciate if someone gave me a shoelace and a prime factorization program :P . I tried to get it online, but I apparently don't have the software for the N-spire, or something.[/quote]

Use the TI-84 keypad, unless you have TI-[i]n[/i]spire CAS...

I am using the 84 keypad. I don't understand the other one :P

Using the TI-84 keypad, click the program button. then go online and get some basic knowledge. Even if one knows some basics, the calculator is slightly different, though not much. I got most of my programs from somebody else.

The TI-Nspire has prime factorization built in.

Just type "factor(n)" and it'll return the prime factorization.

Oh snap? But is that on the non 84 keypad, because I can't seem to find it on the 84?

Eh is the other keypad even allowed? Considering you have those documents already on there, wouldn't it be somewhat cheating?

Its perfectly fine. The function is built in, and they said the calculator is allowed. Plus that's not the only calculator that can do that. The Ti89 series can also do that.

you do a lot more than prime factorization with nspire...like simultaneous equation solving, prime tester...stuff like that all built in.

Go on ticalc.org and get the following: Omnicalc (includes subfactorial/derangements, base converter I think, factoring command, mod/remainder function, radical simplifier), Noshell (allows you to run archived programs, saving memory and making it more secure), CabaMap (up to 24000ish digits of integer calculations), PlySmlt2 (solves polynomials up to degree 10 and linear equations up to 10 variables). I think they come with instructions...also, delete all the useless applications on your calculator if you haven't so that it runs faster in general. As for Shoelace and other junk, I can send it to you when I feel like it...
[b]
gauss1181:[/b] What's all this fairness for? Programs can be useful; of course they're not the main focus, though.

Of course, all the problems can be solved without programs. It's just that using programs makes it a lot faster and can be useful to double check your answer. I don't know about you, but I feel even a small bit of time is a lot at the higher levels.

If other people don't have calculators, it's their problem. There are lots of things in Mathcounts that are unfair; being in a harder state, going to a school with better people and a good coach, lots of resources, etc.

[quote="gauss1181"]What's the big deal about using calculator programs for MC? They're not really that necessary to be able to solve the problems. Remember that MC problems are solvable without a CALCULATOR PROGRAM....

Also, I don't really think that it's that necessary to program your calculator...Think of the people around you, that will give them an unfair disadvantage if they don't have programmable calculators. :wink:[/quote]

It's faster.


Express the sum of $ 0.\overline{31}$ and $ 0.\overline{8}$ as a common fraction.


The Chinese MO link seemed to be disappeared on the kalve homepage and the files are gone.

Anybody knows what had happened?

nope I have no idea  :? , but this is not "About MathLinks", so I'm moving it to Babel Tower ;)

I asked John Scholes about this a while ago, and he replied:

[quote="J Scholes"]The Australian Maths Trust claims that it owns the copyright to all Chinese olympiad problems in English. They are threatening to sue unless they are removed from the website. I am currenly discussing it with them.[/quote]

It seems that Salinko's hands are reaching far into the Chinese lands ;) :D as far as I figure it out, Australian Math Trust doesn't have anything in common with the Chinese MO ...

I really hate this kind of people, who see in problems and olympiads just a source of profit. Shame on them! They shouldn't be allowed to have math on their name. I'm really astonished by this lack of respect to all those who love elementary mahematics.

Apparently Dr. Andy Liu (who is a Canadian by the way) wrote a[url=http://svc056.bne149v.server-web.com/shop/proddetail.cfm?CFID=1394841&CFTOKEN=21488378&ItemID=35&CategoryID=4&SubCatID=4]book[/url]containing the CMO problems, and the book was publised under the Australian Math Trust. So I guess that's why they claim the copyright.

As well, the Tournament of the Towns and (surprise surprise) Australian Olympiads have been removed from Kalva as well, likely for the same reasons.

damn Austrialians ....  I really don't care much about the Austrilian Olympiad, which sux, but I saw that their book had copyright not on all the Chinese Olympiads.

In plus, who's this Andy Liu guy? How did he get to write a Chinese Olympiads book? I mean did he had copyrights from the Chinese National Olympiad Committee?

Hi Valentin Vornicu, all,

Here is just a little information about Professor Andy Liu ----

Prof. Andy Liu is a board member for the Alberta High School Mathematics Compitition
[url]http://www.math.ualberta.ca/~ahsmc/[/url]

He is a professor at the Universtiy of Alberta.  Here is a link to a picture of him and his reseach Interests
[url]http://www.math.ualberta.ca/Liu_A.html[/url]

Prof. Andy Liu won the 2003 Adrien Pouliot Award for Mathematics Education from
the Canadian Mathematical Society:
[url]http://www.cms.math.ca/MediaReleases/2003/two-prizes.html[/url]
(There is a brief bio. of Prof. Liu on this web page.)

Prof. Andy Liu has been greatly involved with the Canadian IMO teams and won a David Hilbert International Award for the promotion of Mathematics Competitions.  He is also greately involved with other mathematical educational outreach programs in Alberta and in the rest of Canada.

In conclusion, he's done some good work.  I have yet to meet Prof. Liu.  About the book he wrote that contains problems from the Chinese Olympiads, I have no idea whether he got permission from the Chinese National Olympiad Committee.  I don't have access to a copy of his book.  If anyone does, they might want to check the publishing information to see if such permission was given, and report back.

I agree with harazi in that mathematical problems shouldn't be copyrighted.  This just seems silly to me and counter to culture of sharing ideas and problems that is part of the mathematical culture we so love and hold dear.   If one wrote a paper with a problem and solution in it, would one claim copyright to the problem saying that it could not be asked elsewhere in print?  "No Prof. Wiles, you cannot use my problem in your paper that proves my famous question.  If you do, I will sue you and your publisher!",  exclaims P. Fermat ... Very silly indeed.  :P

Yours truly,

mathfreak

at most one could get a copyright on (his) solutions, but not on the questions itselves. It's like putting a copyright on IMO question.

all the people which wrote books on IMO questions should now sue Scholes for putting the imo questions online.

I think money is good, therefore I agree with charging the problems, especially high quality problems. If I can create high quality math problems, I would be happy to receive money for the problems I create too. (but I'm too weak to create problem)

However there are two points to consider when charging for problem:
1) the problems should be available at a cheap price so that all students who like math can enjoy it
2) easy method of payments, I can't apply for a Paypayl account (it's banned in my country) and I do not earn enough money to apply for a credit card account, it's inconvenient for me to pay for goods
If the author can sell the problems at a very cheap price and can arrange for an easy method of payment, I will be happy to buy from them, especially the high quality Russian olympiad problems and solutions.

Considering that many good mathematicians are not rich, it would be nice if they can have an extra source of income by creating high quality problems for students to enjoy. Just make them be available cheaply with an easy method of payment.

[quote="Anonymous"]I think money is good, therefore I agree with charging the problems, especially high quality problems. If I can create high quality math problems, I would be happy to receive money for the problems I create too.[/quote]
[quote="Anonymous"]but I'm too weak to create problem[/quote]

And I'm glad these 2 properties come together :?

[quote="3X.lich"]Attention: I found someone in korea hosting those problems from kalva. if enough people show interests, i would like to put up the link to it for a week. after that, i'll have to remove it for good since the BASTARDS copyrighters may not like it.

BTW, that trust, isn't that one of the organization's primary objective is to promote mathematics education?

later, ill check back here soon.[/quote] obviuosly not ...

[quote="Anonymous"]Considering that many good mathematicians are not rich, it would be nice if they can have an extra source of income by creating high quality problems for students to enjoy. Just make them be available cheaply with an easy method of payment.[/quote]

Many good students are not rich either. You are depriving those students of their rights to solve nice problems.


If 1<a_1<a_2<.....a_k<b_1<b_2<....<b_m are integer such that

 sum a_i  \geq  sum b_i

then

 prod a_i>prod b_i

hello, do you mean

If $1<a_1<a_2<.....a_k<b_1<b_2<....<b_m$ are integers such that

$ \sum_{i=1}^k a_i \geq \sum_{i=1}^{m} b_i$

then

 $\prod_{i=1}^{k} a_i>\prod_{i=1}^{m} b_i$
Sonnhard.


[quote="Superficial"]$ ABC$ is an acute-angled triangle. $ A_2, B_2, C_2$ are the midpoints of the altitudes $ AA_1, BB_1, CC_1$ respectively. Find $ \angle B_2A_1C_2 \plus{} \angle C_2B_1A_2 \plus{} \angle A_2C_1B_2$.[/quote]
[hide="Solution"]
Let the midpoints of $ AC$, $ BC$, and $ AB$ be $ E$, $ D$, and $ F$ respectively.  Furthermore, let $ H$ be the orthocenter of the triangle.  Notice that $ \angle HA_1C \plus{} \angle HB_1C \equal{} 90 \plus{} 90 \equal{} 180$, so $ HB_1CA_1$ is cyclic.  It follows that
\[ BH\cdot BB_1 \equal{} BA_1\cdot BC\implies \frac {BB_1}{2}\cdot BH \equal{} \frac {BC}{2}\cdot BA_1
\]
and so $ BB_2\cdot BH \equal{} BA_1\cdot BD$, so $ HB_2A_1D$ is cyclic.  Also, $ FB_2\parallel AC$ since $ FB_2$ is a midline of $ \triangle ABB_1$.  Similarly, $ B_2D\parallel AC$ and $ FD\parallel AC$, so $ F$, $ B_2$, and $ D$ are collinear.  Similarly are $ F$, $ A_2$, and $ E$ and $ E$, $ C_2$, and $ D$.  From the parallels, since $ BB_1\perp AC$, we have that $ \angle DB_2H\equal{}90$ and similarly, $ \angle HC_2D\equal{}90$.  Thus, $ \angle DB_2H \plus{} \angle DC_2H \equal{} 90 \plus{} 90 \equal{} 180$, so $ HB_2DC_2$ is cyclic.  Hence, $ HB_2A_1DC_2$ is cyclic.  It follows that\[ \angle B_2A_1C_2 \equal{} \angle FDE \equal{} 180 \minus{} \angle FDB \minus{} \angle EDC \equal{} 180 \minus{} \angle ABC \minus{} \angle ACB \equal{} \angle BAC\]Similarly, $ \angle B_2C_1A_2 \equal{} \angle ACB$ and $ \angle C_2B_1A_2 \equal{} \angle ABC$, so $ \angle B_2A_1C_2 \plus{} \angle A_2C_1B_2 \plus{} \angle C_2B_1A_2 \equal{} \angle ABC \plus{} \angle ACB \plus{} \angle BAC \equal{} 180$. [/hide]


Factor the polynomial x^8 +98x^4 +1 into two factors with integer coefficients.

[quote="mdk"] integer (not necessarily
real) coefficients.[/quote]
How can they be integer and not real?


function
induction
algebra proposed
algebra

find all continous functions $f: R\to R\\. f\left(\frac{x+y}{x-y}\right)=\frac{f\left(x\right)+f\left(y\right)}{f\left(x\right)-f\left(y\right)}$ . $\forall{x\not=y}\in\mathbb{R}$ :wink:

(*) $f\left(\frac{x+y}{x-y}\right)=\frac{f\left(x\right)+f\left(y\right)}{f\left(x\right)-f\left(y\right)}$ for all $x\not=y$.

(1) $f$ is injective:
$f(x)\not =f(y)$ for all $x\not =y$ by (*).

(2) $f(0)=0,f(1)=1$:
$x\not =0=y$ in (*) gives $f(1)={f(x)+f(0)\over f(x)-f(0)}=1+{2f(0)\over f(x)-f(0)}$ and $(f(1)-1)(f(x)-f(0))=2f(0)$ constant.
For $f(1)\not =1$ is then $f(x)=f(0)+{2f(0)\over f(1)-1}$ constant, which contradicts (1). So $f(1)=1$ and $f(0)=0$.

(3) $f(-x)=-f(x)$ for all $x\in\mathbb{R}$:
(2) gives $x=0$ and $x\not =0, y=-x$  in (*) gives $0=f(0)={f(x)+f(-x)\over f(x)-f(-x)}$ and $f(-x)=-f(x)$.

(4) $f(xy)=f(x)f(y)$ for all $x,y\in\mathbb{R}$:
(2) gives both $x=1$ or $y=0$, and for $x\not =1,y\not =0$ we have $xy\not =y$, and (*) and (2) give
${f(xy)+f(y)\over f(xy)-f(y)}=f\left({xy+y \over xy-y}\right)=f\left({x+1 \over x-1}\right)={f(x)+f(1)\over f(x)-f(1)}={f(x)+1\over f(x)-1}$ and $f(xy)=f(x)f(y)$.

(5) $f(x)>0$ for all $x>0$ and $f$ is strictly increasing:
(1) and (2) give $f(\sqrt{x})\not =f(0)=0$, and then (4) gives $f(x)=f(\sqrt{x})^{2}>0$ for all $x>0$.
Now for all $x>y\geq 0$ we have $f(x)+f(y)>0$ and also $\frac{x+y}{x-y}>0$ and $f\left(\frac{x+y}{x-y}\right)>0$,
and then (*) gives $f(x)>f(y)$. And then (3) gives the rest.

(6) $f(2)=2$:
Set $f(2)=c$.
(*),(2) and (4) give $f(3)=f\left(\frac{2+1}{2-1}\right)={c+1\over c-1}$, $f(4)=f(2)^{2}=c^{2}$,
$f(5)=f\left(\frac{3+2}{3-2}\right)={{c+1\over c-1}+c\over{c+1\over c-1}-c}={c^{2}+1\over 1+2c-c^{2}}$ and also
$f(5)=f(3)f\left(\frac{4+1}{4-1}\right)={c+1\over c-1}{c^{2}+1\over c^{2}-1}={c^{2}+1\over 1-2c+c^{2}}$.
Therefore $2c-c^{2}=-2c+c^{2}$ and $c=2$. ($c\not =0$ by (1) and (2))

(7) $f(x)=x$ for all $x\in\mathbb{R}$:
(4) and (1) give $f(n+1)=f(n-1)f\left(\frac{n+1}{n-1}\right)=f(n-1){f(n)+1\over f(n)-1}$ for all $n\in\mathbb{N},n>1$.
Then by induction, using (2) and (6), we get $f(n)=n$ for all $n\in\mathbb{N}_{0}$. Then (3) gives $f(n)=n$ for all $n\in\mathbb{Z}$, and using (4) we get $f({m\over n})={1\over n}f(n)f({m\over n})={1\over n}f(m)={m\over n}$ for all $m,n\in\mathbb{Z},n\not =0$. And finally (5) gives $f(x)=x$ for all $x\in\mathbb{R}$.


[quote="myyellowducky82"]When I entered AoPS's URL onto a computer, a screen appeared that said "Oops! This link appears to be broken!" This happened to me on 3 different computers today, but AoPS eventually worked. Some people in the Introduction to Geometry class stated the same thing happened to them! Is AoPS not working or something?[/quote]

it died today.

This happened yesterday too.
Please define "died...."

[quote="vallon22"]i'm gonna guess: "went offline due to technical problems"[/quote]

exactly

So it wasn't my browser or computer's fault? (just to make sure)

The "broken link" screen was very likely caused by issues on AoPS's side, but, to make sure that your browser/computer is functioning, you can always go to a variety of different sites. Chances are, three different sites won't simultaneously experiencing technical problems.  :)

The other sites worked :)
AoPS is working--YAY! Thanks everyone!!

I go on AoPS every day and that's never happened.

Sometimes you will notice that the site doesn't load for a couple of minutes, that is usually due to extremely high traffic (on FTW, Alcumus, Forums, etc.) so the server is trying to automatically balance out this. If you notice such a moment, just come back within a few minutes, and everything should be back to normal.

[quote="myyellowducky82"][quote="Valentin Vornicu"]Sometimes you will notice that the site doesn't load for a couple of minutes, that is usually due to extremely high traffic (on FTW, Alcumus, Forums, etc.)  [/quote]
So even if there's a lot of people on FTW, it will affect the Forum, too?[/quote]A multitude of factors affect the server load and speed. A lot of people using FTW, Alcumus and browsing the Forums at the same time, and also some other backend processes that we have, sometimes generate slowdowns on the site (and, for some users, renders the site unusable.) We are working into avoiding such issues in the future but it's hard considering our base pool of users is increasing by the day :)

What is the average number of requests per second? What is the average duration of a given request? What is the current size of the database for the forum entries? Is there a user request listing by continent, country etc.?


Suppose $ a$ is directly proportional to $ b$, but inversely proportional to $ c$. If $ a \equal{}2$ when $ b \equal{} 5$ and $ c \equal{}9$, then what is $ c$ when $ b\equal{}3$?

since a is directly proportional to b, $ b\equal{}\frac{5a}{2}$

since a is inversely proportional to c, $ ac\equal{}18$

$ b\equal{}3\equal{}\frac{5a}{2} \implies 6\equal{}5a \implies a\equal{}\frac{6}{5}$

$ ac\equal{}18 \implies \frac{6c}{5}\equal{}18 \implies c\equal{}15$

answer : 15


Dear everyone.

please think of the following.

for a triangle ABC (BC=a, CA=b, AB=c)
prove that $ \sqrt{a\plus{}b\minus{}c}\plus{}\sqrt{b\plus{}c\minus{}a}\plus{}\sqrt{c\plus{}a\minus{}b}\leq\sqrt{a}\plus{}\sqrt{b}\plus{}\sqrt{c}$.

thanks.

You can use $ \boxed{\boxed{\sqrt{\frac{x\plus{}y}{2}}\geq \frac{\sqrt{x}\plus{}\sqrt{y}}{2}}}$.

hello, setting $ a \equal{} y \plus{} z,b \equal{} x \plus{} z,c \equal{} x \plus{} y$ we have to prove that
$ \sqrt {2z} \plus{} \sqrt {2y} \plus{} \sqrt {2x}\le \sqrt {y \plus{} z} \plus{} \sqrt {x \plus{} z} \plus{} \sqrt {x \plus{} y}$
but it is
$ \frac {\sqrt {2y} \plus{} \sqrt {2x}}{2}\le \sqrt {\frac {2x \plus{} 2y}{2}} \equal{} \sqrt {x \plus{} y}$
$ \frac {\sqrt {2z} \plus{} \sqrt {2y}}{2}\le \sqrt {\frac {2z \plus{} 2y}{2}} \equal{} \sqrt {z \plus{} y}$
$ \frac {\sqrt {2x} \plus{} \sqrt {2z}}{2}\le \sqrt {\frac {2x \plus{} 2z}{2}} \equal{} \sqrt {x \plus{} z}$
adding these three inequalities we get the desired.
Sonnhard.

[quote="kunny"]You can use $ \boxed{\boxed{\sqrt {\frac {x \plus{} y}{2}}\geq \frac {\sqrt {x} \plus{} \sqrt {y}}{2}}}$.[/quote]

Let $ x\equal{}c\plus{}a\minus{}b,\ y\equal{}a\plus{}b\minus{}c,\ z\equal{}a\plus{}b\minus{}c$, we have $ x\plus{}y\equal{}2a,\ y\plus{}z\equal{}2b,\ z\plus{}x\equal{}2c$, so using the above inequality gives the result.


Polynomial w(x) has no real roots. Prove that polynomials
$v_{\alpha}(x)= w(x)+ \alpha w'(x)+ \alpha^2w''(x)+...+\alph^nw^{(n)}(x)$
(where $n$ is a degree of $w(x)$) have no real roots for all real $\alpha$.

We may WLOG assume $w(x)>0$ for all $x$ and $\alpha\neq 0$. Let's find minimum point of $v_\alpha(x)$. We have $0=(v_\alpha(x))'_x=w'(x)+\alpha w''(x)+...+\alpha^{n-1}w^{(n)}(x)=\frac{v_\alpha(x)-w(x)}{\alpha}$. So we obtain, that for any extremum point $x$ following identity holds: $v_\alpha(x)=w(x)$, but $w(x)>0$ for all $x$. Therefore $v_\alpha(x)>0$ for all extremum points, so it is necessary $v_\alpha(x)>0$ for all $x$ (since $v_\alpha(x)$ has the same leading coefficient as $w(x)$).

It can't be, since $w(x)$ is even degree polynomial (otherwise $w$ has roots) and $\deg v_\alpha=\deg w$. Each even degree polynomial has extremum points.

Hmm... I wrote answer to Risk reply, but it disappeared...  :huh:

[quote="Big Bolzano"]But if $v_{\alpha}(x)$ has no extremum points? It can still has roots.[/quote]
See my reply above yours.

OK. Now it's all clear to me. Thanks again ;]


Hello,

Find all increasing functions from N such as :
$ f(m.n)\equal{}f(n).f(n)$ and $ f(2)\equal{}2$


Let $ S$ denote the set of integers of the form $ x^2\plus{}1,x\in\mathbb{N}$. Prove that for any element $ k$ of $ S$, then $ k^2$ is not in $ S$ except for $ k\equal{}1$.

[quote="Temperal"]Let $ S$ denote the set of integers of the form $ x^2 \plus{} 1,x\in\mathbb{N}$. Prove that for any element $ k$ of $ S$, then $ k^2$ is not in $ S$ except for $ k \equal{} 1$.[/quote]

We claim that the difference between two squares is twice the lower plus one.

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

Now consider an element of $ S$

$ M \equal{} x^2 \plus{} 1$

If $ M^2 \in S \Rightarrow M^2 \minus{} 1 \equal{} x^2$ for some $ x$

Since the difference between two squares is $ 2n \plus{} 1$, This will only be the case when $ x \equal{} 0$


Solve the following system of equations:
$x-\frac{4}{x}=2y-\frac{2}{y}$
$2x=y^{3}+3$

from the first one we get that:
$x-\frac{4}{x}=2y-\frac{2}{y}\Rightarrow x-2y=\frac{4y-2x}{xy}$
so we have two cases:
$\boxed{\text{Case}1: x=2y}$
in this case the second equation becomes:
$4y=y^{3}+3\Rightarrow y^{3}-4y+3=0\Rightarrow (y-1)(y^{2}+y-3)=0$ so we get that $y=1\Rightarrow x=2$ or $y=\frac{-1 \pm \sqrt{13}}{2}\Rightarrow x=-1 \pm \sqrt{13}$
$\boxed{\text{Case}2: xy=-2}$in this case the second equation becomes:
$y^{4}+3y+4=0$...

[quote="BaBaK Ghalebi"]$4y=y^{3}+3\Rightarrow y^{3}-4y+3=0\Rightarrow (y-1)(y^{2}+y-3)=0$ so we get that $y=1\Rightarrow x=2$ or $y=\frac{-1 \pm 2\sqrt{3}}{2}\Rightarrow x=-1 \pm 2\sqrt{3}$...[/quote]

Little modification : $y=\frac{-1 \pm \sqrt{13}}{2}$

[quote="BaBaK Ghalebi"]$y^{4}+3y+4=0$...[/quote]

Little complement : And this equation $f(y)=y^{4}+3y+4=0$ has no real root since $f'(y)$ is strictly increasing with a single root $\sqrt[3]{-\frac{3}{4}}$ and it is easy to show that $f(\sqrt[3]{-\frac{3}{4}})=\frac{9}{4}(\frac{16}{9}-\sqrt[3]{\frac{3}{4}})>0$

[quote="pco"]Little modification : $y=\frac{-1 \pm 2\sqrt{13}}{2}$[/quote]
thanx pco I have edited my post,and I think you should edit yours too...  :wink:

[quote="BaBaK Ghalebi"] thanx pco I have edited my post,and I think you should edit yours too...  :wink:[/quote]

Hum ... I edited too  :blush:

Thanks too


algebra
function
domain
algebra unsolved

Slove this equation: $ 3^x\plus{}2^x\equal{}3x\plus{}2$.

[quote="thanhnam2902"]Slove this equation: $ 3^x \plus{} 2^x \equal{} 3x \plus{} 2$.[/quote]

What is the domain for which we should solve this equation? Integers? Real numbers?
x=0 is one root. And there is a root between x=0 and x=1.


guys i'm presently in my first yr at an institute(within the jee system),in a very good branch ,which is on the verge of becoming an iit,but some how,i'm not impressed by this institute,so i thought of preparing for sat,is it possible?

i can write it only on oct 9,but is 2 months time sufficient?and in the hostel environment it is difficult to study too.How tough is it to get electrical/cs @ Stanford/MIT/Princeton etc?

And what material would you guys recommend?

thanks in advance

The deadline to apply a fresher has passed. You can only apply as a transfer student now, which mind you is VERY VERY VERY VERY difficult, unless you're really really special. And SATs contribute less than 10% in your US applications. Its severely over hyped.

ok,then I have a few questions

1.What exams do I have to write as a transfer student besides SAT reasoning and TOEFL ,?
2.I'm not sure if my institute has a credit transfer system,if it doesn't,will it be a problem?
3.Can all the preparation be done in 2 months ? pl suggest some books/coaching material.
4.I've written a couple of open source software with over 100,000 downloads each   :) ,will that be of any help?
   but i've not participated in any olympiads etc :(
5.What kind of a GPA(on the iit scale of 10) is required for a top college?

Or is it just too late to apply for it??

[quote="deathsaber"]

4.I've written a couple of open source software with over 100,000 downloads each   :) ,will that be of any help?
[/quote]

Cool  :) . Could you tell something more or post a link??

To be honest, I don't know much about transfers :(
But I can surely tell you that if your english is decent, and assuming you're the usual Indian braino, you won't need more than a week for SAT preparation. And for TOEFL? 2 hours maybe? :P

You have to first shortlist the schools for transfer applications..go to their sites ,see their requirements and then give exams. Most of the schools do not take transfer students, some do but only after 2nd yr of your present UG studies..and in any case most of those will require SATsubject test only and not SAT...and that should be very easy for you!

ALL THE BEST TO YOU!!


do you look at our previous contest prefromance?

if so, is it important?

im not very sure on this.
but i can tell you that my contest scores were pretty crappy and i got in. :P
so im sure you'll be fine. just make sure you project the image that you're interested in math and such.

Notice that you don't need to send in or report scores on any contests...


Read the following 5 statements carefully:
(i) Statement (ii) is true.
(ii) At most one of these 5 statements is true.
(iii) All 5 of these statements are true.
(iv)
(v)
The last two statements are printed in invisible ink. Which of the statements are true?

[hide]2 because if i if is true, then ii is true and it contradicts, so i and ii can't be true, making iii not true, therefore the only ones that can be true are iv and v. [/hide]

This is what I did:

[hide] Suppose statement (i) is true. Then statement (ii) would have to be true, but statement (ii) says that at most one statement is true. But in this case, there are two true statements, so both (i) and (ii) are false. Thus, (iii) is also false. (ii) said that at most one statement is true, but since (ii) is false, then there must be more than one statement that is true. Thus, the remaining two statements in invisible ink, (iv) and (v), are both true.

Answer: (iv) and (v)[/hide]

I suppose that everyone has figured out the answer.

These questions are not too challenging for me because they are similar to the ones in a book I like to read.


For two polynomial A(x) and B(x) with integer coefficients (degree of B is less than degree of A), their leading coefficients are a and b, respectively, then prove that when A is divided by B, the coefficients of quotient Q are all integers if and only if b|a.

BTW: I was confused with following notations for polynomials: p(x), P(x), and P(X). Are they different?

Unfortunately, that statement is not true.  For example, try to divide $ x\plus{}1$ into $ x^2\plus{}x\plus{}1$.  It leaves a remainder of $ 1$:

$ x^2\plus{}x\plus{}1\equal{}x(x\plus{}1)\plus{}1$

Also, to answer your question, all three of the notations you gave can be correct.  It doesn't matter what you name the polynomial, and different names are used to distinguish between different polynomials (i.e. $ a(x), C(x), f(x)$), though some letters are used much more often.  Also, the polynomial should be in the variable specified.  For example, $ F(B)\equal{}B^2$ is in $ B$ while $ f(x)\equal{}x$ is in $ x$.

I apologize if I didn't state the question clearly. I mean that the coefficients of quotient polynomial are all integers, not A will be divisible by B. Remainder doesn't matter in this question. As in your example, the quotient x satisfies this statement.

Counterexample:

$ 2x^2 \plus{} 1 \equal{} (2x \plus{} 1) \left( x \minus{} \frac{1}{2} \right) \plus{} \frac{3}{2}$

In general, the quotient is an integer polynomial when $ b \equal{} 1$.

Induct on the degree of $ A(x)$.  Equivalently, the quotient can be explicitly constructed by repeatedly subtracting multiples of the divisor.  For example,

$ 2x^2 \plus{} 1 \boxed{\minus{} 2x}(x \plus{} 1) \equal{} \minus{}2x \plus{} 1$
$ \minus{}2x \plus{} 1 \boxed{\plus{} 2}(x \plus{} 1) \equal{} 3$

hence

$ 2x^2 \plus{} 1 \equal{} (\boxed{2x \minus{} 2})(x \plus{} 1) \plus{} 3$.

Generalize this approach.  :)


function
limit
topology
advanced fields
advanced fields unsolved

Let $ X$ be a seperable metric space, $ f: X\to \mathbb{R}$ a function such that $ \lim_{x\to a}f(x)$ exists for all $ a\in X$. Show that $ S\equal{}\{x\in X | f \text{ is not continous}\}$ is countable.

Maybe I don't get something trivial but what do you mean by this question. Doesn't the condition on existence of $ \lim_{x\rightarrow a} f(x)$ imply that $ f$ is continuous at $ a$ basically by definition?

$ \lim_{x\to a}f(x)$ might exist but be different from $ f(a).$

Clarifying example:

On $ \mathbb{R},$ let $ f(x) \equal{} \begin{cases}\frac1q ,& x \equal{} \frac {p}{q}\text{ in lowest terms} \\
0, & x\text{ is irrational}\end{cases}.$

Then $ \lim_{x\to a}f(x) \equal{} 0$ for all $ a,$ but $ f$ is discontinuous at each rational number. (Which is a countable set, consistent with the proposed theorem.)


$ \mathbb{Q}$ is rational set that is sub topology space of $ \mathbb{R}$.
$ x,y\in \mathbb{Q},x<y$.
There is a set subset $ E$ of $ \mathbb{Q}$. $ x,y \in E$.
$ r$ is a irrational number , $ x<r<y$.
Let $ A\equal{}(\minus{}\infty,r)\cap E, B\equal{}(r,\infty)\cap E$.
why $ A,B$ are both open set in $ \mathbb{Q}$?

Thanks

[quote="water2011"]$ \mathbb{Q}$ is rational set that is sub topology space of $ \mathbb{R}$.
$ x,y\in \mathbb{Q},x < y$.
There is a subset $ E$ of $ \mathbb{Q}$. $ x,y \in E$.
$ r$ is a irrational number , $ x < r < y$.
Let $ A \equal{} ( \minus{} \infty,r)\cap E, B \equal{} (r,\infty)\cap E$.
why $ A,B$ are both open set in $ \mathbb{Q}$?

Thanks[/quote]

[quote="mavropnevma"]This is the very definition of the induced topology: both $ A$ and $ B$ are open subsets of $ \mathbb{R}$, so their intersections with $ \mathbb{Q}$ are open subsets of $ \mathbb{Q}$.[/quote]

Thanks
but we don't know if $ E$ is open .

Just by what you wrote, that is not true. Take for example $ E \equal{} \{x,y\}$, then $ A\equal{}\{x\}$ and $ B\equal{}\{y\}$ and neither is open in the induced topology of $ \mathbb{Q}$. As defined, $ A$ and $ B$ are only guaranteed to be open in the induced topology of $ E$.


Any power of $ 5$ that has two or more digits is or ends with $ 25$. From this, you could could tell that the tens digit is $ \fbox{2}$.


The difference of two positive numbers is 9 and their geometric mean is 5.  Find the arithmic mean of the numbers.

can't figure this out =[

[hide]$x-y=9$
$\sqrt{xy}=5$

The first equation gives

$x^{2}-2xy+y^{2}=81$

and the second

$xy=25$

Now

$x^{2}+2xy+y^{2}=(x^{2}-2xy+y^{2})+4xy=81+4\cdot 25=181$

which, as the numbers are positive, yields

$x+y=\sqrt{181}\iff{{x+y}\over 2}={1\over 2}\sqrt{181}$[/hide]


Determine the greatest value of $ k \in N^*$such that there exist  $ n \in N^*$  such that all numbers $ n,n^2,n^3,....,n^k$ can express in the form $ x^2\plus{}y^2\plus{}1$ where $ x;y$ are positive numbers.

[quote="hoangclub"]Determine the greatest value of $ k \in N^*$such that there exist  $ n \in N^*$  such that all numbers $ n,n^2,n^3,....,n^k$ can express in the form $ x^2 \plus{} y^2 \plus{} 1$ where $ x;y$ are positive numbers.[/quote]
The problem equivalent $ n^a \minus{} 1$ can be represent as sum of two square .
[u]Lemma 1 [/u] Number n can represent as sum of two square if and only if exponent of all prime divisors have form $ 4k \plus{} 1$ is an even number .
[u] Lemma 2 [/u] If $ m,n$ can represent as sum of two square then $ mn$ also .
Consider your problem .

[u]Case 1[/u] $ n\equiv 1 (\mod 2)$ then
$ n^2 \minus{} 1\equiv 3 (\mod 4)$ so it has a prime divisor p such that
i)$ p\equiv 3 (\mod 4)$
ii) $ v_p(n^2 \minus{} 1)\equiv 1 (\mod 2)$
So $ n^2 \minus{} 1$ can not represent as sum of two square.
 [u]Case 2[/u] $ n\equiv 1 (\mod 2)$
Chose $ n \equal{} 99$ then we have
$ 99 \minus{} 1 \equal{} 7^2 \plus{} 7^2$
$ 99^2 \minus{} 1 \equal{} (99 \minus{} 1)(99 \plus{} 1) \equal{} 42^2 \plus{} 56^2$
$ 99^3 \minus{} 1 \equal{} 98.9001$
Because $ 9001$ is a prime form $ 4k \plus{} 1$ so it can represent as sum of two square.
$ 99^4 \minus{} 1 \equal{} (99^2 \minus{} 1)(99^2 \plus{} 1)$
$ 99^5 \minus{} 1 \equal{} 98.97039801$
Because $ 97039801$ is a prime has form $ 4k \plus{} 1$ so it can represent as sum of two square .
Now we will prove that $ \max{k} \equal{} 5$
Suppose $ k \equal{} 6$ satisfy condition .
We prove that $ 3|n$
If $ 3\not |n$ then
$ 3|n^2 \minus{} 1$
So $ v_3(n^6 \minus{} 1) \equal{} v_3(n^2 \minus{} 1) \plus{} 1$
So one of the number $ v_3(n^2 \minus{} 1),v_3(n^2 \minus{} 1) \plus{} 1$ is an odd number .
It mean that one of two number $ n^2 \minus{} 1,n^6 \minus{} 1$can not represent as sum of two square .
So $ 3|n$
But we have $ n^6 \minus{} 1 \equal{} (n^2 \minus{} 1)(n^4 \plus{} n^2 \plus{} 1)$
Because $ 3|n$ so $ \gcd(n^2 \minus{} 1,n^4 \plus{} n^2 \plus{} 1) \equal{} 1$ .
But $ n^4 \plus{} n^2 \plus{} 1\equiv 3 (\mod 4)$ so it has a prime p such that $ v_p(n^6 \minus{} 1)$ is an odd number.
If follow that $ n^6 \minus{} 1$ can not represent as sum of two square .
[b]Result [/b] $ k \equal{} 5$ is the max number satisfy condition.


whenever i get on, it shows that anything that someone has written in the past two days is a new post, even if i have read it. even if i have posted it myself it says "unread". it's really annoying.

Erase your cookies. This is not expected behaviour, unless you have read the post as a guest (ie you were not logged in at the time of reading) :)

Now the unreads are stored in the database, so there should be no problem with them (I have no problem and I browse the site using 4 different browsers :) ).

okay, I'm not having that problem anymore. thanks!


Given that $ \log_{10} 2 \equal{} 0.301$, how many digits are there in $ 5^{44}$?

My proposed question:

Given a number $ a^b$, how can we find the number of digits in $ a^b$ given that we can evaluate one logarithm?

I'm not writing out the entire solution but evaluating a number in log base 10 will give you the number of digits.

$ \log{5^{44}}\equal{}44\log{5}\equal{}44(\log{10}\minus{}\log{2})\equal{}44(1\minus{}.301)\equal{}30.756$

So there are 31 digits.

the first excerpt of Vol 2 explains this VERY well :) (not this exact thing but reading it may help a lot)

This problem is one of the practice problems in Volume 2 so the solutions guide goes through it pretty nicely.

Hmm. The number of digits is the FLOOR of the logarithm (plus 1)... maybe the characteristic means the floor?

[quote="ThetaPi"][quote="solveitall"]In general, the characteristic + 1 of the logarithm gives you the number of digits[/quote]

What's a characteristic of an logarithm?[/quote]

The logarithmic value of a number consists of its characteristic and its mantissa. The integral part of the logarithm of a number, after expressing the decimal part as non negative is called the characteristic, The decimal part, the "mantissa"

For example say log 2.2 will have the characteristic of "0", because while calculating the log of 2.2 the first step you would do is write it as 0.XXX

Hope its clear

Very cool. So it is the floor of the log.

I've never heard those terms before :)

Oh, I remember reading about the characteristic/mantissa before. It's common on "ancient" logarithm tables before the calculator/computer was invented.


[b]A particle $ P$ of mass $ m$ describes a horizontal circle of radius 24cm inside a smooth hemishperical bowl of internal radius 25cm, which is held with its rim horizontal. A fine weightlessness thread tied to the particle passes through a small smooth hole at the bottom of the bowl and supports another particle of mass $ m$ which hands at rest.[/b]

[img]http://i17.tinypic.com/62om63s.jpg[/img]

Show that the speed of the particle on the bowl is a little less that $ 4m/s$

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

[color=red]The sketch given shows a light inelastic string $ OABC$, with one end attached to a fixed point $ O$ and particles of mass $ 3m,2m,m$ are attached to the string at the points $ A,B,C$ respectibley. THe string $ OABC$ lies in a vertical plane which rotates bout the vertical through O with constant angular velocity $ \sqrt{\frac{3g}{a}}$. The particles $ A,B,C$ move in horizontal circles of radii $ a,2a,3a$ respectively. The inclinations to the horizontal of OA, AB , BC are $ \alpha,\beta,\gamma$ respectively[/color]

[img]http://i8.tinypic.com/4qf70xi.jpg[/img]

i) Show that $ \tan\gamma \equal{}\frac{1}{9}$

ii) Find the values of $ tan\beta$ and $ tan\alpha$ in the exact form

iii) what are the tensions in the strings?

First Problem Solution:

Lets make the problem a little more general, and consider a particle describing a horizontal circle of $ r$ inside a bowl of internal radius $ R$ where $ r < R$.

The forces acting on the hanging particle are those due to its weight, $ W = mg$, and the tension in the string, $ T$. Since this particle is at rest, these forces must be balanced. Thus: $ T = W = mg$.

The forces acting on the spinning particle are it's weight, the tension in the string, and the normal reaction from the bowl. This normal reaction is perpendicular to the surface, so acts along the radius of the bowl. Lets say the normal reaction makes an angle of $ \theta$ with the horizontal. This must be the same angle that the radius joining the particle and the centre of the bowl makes with the horizontal direction of the string. The particle, the centre of the circle about which the particle spins, and the centre of the bowl forms a right angled triangle. With respect to $ \theta$, the hypotenuse is $ R$, the adjacent is $ r$ and the opposite, by Pythagoras' theorem must be $ \sqrt{R^{2}-r^{2}}$. From this we can deduce that:

$ \cos{\theta}=\frac{r}{R}$ and $ \sin{\theta}=\frac{\sqrt{R^{2}-r^{2}}}{R}$

The spinning particle is in vertical equilibrium, so the vertical component of normal reaction, $ N\sin{\theta}$ balances the particles weight, $ W = mg$.

$ \Rightarrow N\frac{\sqrt{R^{2}-r^{2}}}{R}= mg$

$ \Rightarrow N =\frac{Rmg}{\sqrt{R^{2}-r^{2}}}$

The horizontal component of normal reaction, $ N\cos{\theta}$, and the tension in the string, $ T = mg$, (the tension is the same at all points) provides the centripetal force, $ F_{c}$

$ \Rightarrow F_{c}=\frac{Rmg}{\sqrt{R^{2}-r^{2}}}\cdot\frac{r}{R}+mg$

$ \Rightarrow\frac{mv^{2}}{r}= mg\left(\frac{r}{\sqrt{R^{2}-r^{2}}}+1\right)$

$ \Rightarrow\frac{v^{2}}{r}= g\left(\frac{1}{\frac{1}{r}\sqrt{R^{2}-r^{2}}}+1\right)$

$ \Rightarrow v^{2}= gr\left(\frac{1}{\sqrt{\left(\frac{R}{r}\right)^{2}-1}}+1\right)$

$ \Rightarrow v =\sqrt{ gr\left(\frac{1}{\sqrt{\left(\frac{R}{r}\right)^{2}-1}}+1\right)}$

In this specific case $ g = 9.8\,ms^{-2}$, $ r = 0.24\,m$, $ R = 0.25\,m$

$ \Rightarrow v =\sqrt{9.8\times 0.24\left({\frac{1}{\sqrt{\left(\frac{0.25}{0.24}\right)^{2}-1}}+1\right)}}$

$ v =\sqrt{9.8\times 0.24\times 4.4286}$

$ v =\sqrt{10.416}$

$ v = 3.2\,ms^{-1}$

This answer I would say is a little greater than $ 3\,ms^{-1}$ rather than a little less than $ 4\,ms^{-1}$

Second problem solution:

Let the tensions in the portions $ OA$, $ AB$ and $ BC$ be $ T_{A}$, $ T_{B}$ and $ T_{C}$ respectively.

The centripetal forces acting on the particles (given by $ Mr{\omega}^{2}$) are:

$ F_{A}\equal{} 3m\cdot a\cdot\frac{3g}{a}\equal{} 9mg$

$ F_{B}\equal{} 2m\cdot 2a\cdot\frac{3g}{a}\equal{} 12mg$

$ F_{C}\equal{} m\cdot 3a\cdot\frac{3g}{a}\equal{} 9mg$

All particles are in vertical equilibrium.

Forces acting on partcle $ C$ are its weight $ mg$ and tension $ T_{C}$.

Resolving vertically gives: $ T_{C}\sin{\gamma}\equal{} mg$.

Resolving horizontally gives: $ T_{C}\cos{\gamma}\equal{} F_{A}\equal{} 9mg$

$ \Rightarrow\tan{\gamma}\equal{}\frac{\sin{\gamma}}{\cos{\gamma}}\equal{}\frac{T_{C}\sin{\gamma}}{T_{C}\cos{\gamma}}\equal{}\frac{mg}{9mg}\equal{}\frac{1}{9}$

(Part (i) has now been answered).

Forces acting on partcle $ B$ are its weight $ 2mg$ and tensions $ T_{C}$ and $ T_{B}$

Resolving vertically gives:

$ T_{B}\sin{\beta}\minus{}T_{C}\sin{\gamma}\equal{} 2mg$

$ \Rightarrow T_{B}\sin{\beta}\minus{}mg \equal{} 2mg$

$ \Rightarrow T_{B}\sin{\beta}\equal{} 3mg$

Resolving horizontally gives:

$ T_{B}\cos{\beta}\minus{}T_{C}\cos{\gamma}\equal{} F_{B}\equal{} 12mg$

$ \Rightarrow T_{B}\cos{\beta}\minus{}9mg \equal{} 12mg$

$ \Rightarrow T_{B}\cos{\beta}\equal{} 21mg$

$ \Rightarrow\tan{\beta}\equal{}\frac{3mg}{21mg}\equal{}\frac{1}{7}$

Forces acting on particle $ A$ are its weight $ 3mg$ and tensions $ T_{B}$ and $ T_{A}$

Resolving vertically gives:

$ T_{A}\sin{\alpha}\minus{}T_{B}\sin{\beta}\equal{} 3mg$

$ \Rightarrow T_{A}\sin{\alpha}\minus{}3mg \equal{} 3mg$

$ \Rightarrow T_{A}\sin{\alpha}\equal{} 6mg$

Resolving horizontally gives:

$ T_{A}\cos{\alpha}\minus{}T_{B}\cos{\beta}\equal{} F_{C}\equal{} 9mg$

$ \Rightarrow T_{A}\cos{\alpha}\minus{}21mg \equal{} 9mg$

$ \Rightarrow T_{A}\cos{\alpha}\equal{} 30mg$

$ \Rightarrow\tan{\alpha}\equal{}\frac{6mg}{30mg}\equal{}\frac{1}{5}$

(Part (ii) has now been answered)

$ \sec^{2}{x}\equal{} 1\plus{}\tan^{2}{x}$

$ \Rightarrow\cos^{2}{x}\equal{}\frac{1}{1\plus{}\tan^{2}{x}}$

Using this formula:

$ \cos^{2}{\gamma}\equal{}\frac{1}{1\plus{}\frac{1}{81}}\equal{}\frac{81}{82}\Rightarrow\cos{\gamma}\equal{}\frac{9}{\sqrt{82}}$

$ \cos^{2}{\beta}\equal{}\frac{1}{1\plus{}\frac{1}{49}}\equal{}\frac{49}{50}\Rightarrow\cos{\beta}\equal{}\frac{7}{5\sqrt{2}}$

$ \cos^{2}{\alpha}\equal{}\frac{1}{1\plus{}\frac{1}{25}}\equal{}\frac{25}{26}\Rightarrow\cos{\alpha}\equal{}\frac{5}{\sqrt{26}}$

From above: $ T_{C}\cos{\gamma}\equal{} 9mg$

$ \Rightarrow T_{C}\cdot\frac{9}{\sqrt{82}}\equal{} 9mg$

$ \Rightarrow T_{C}\equal{}\sqrt{82}\,mg$

From above: $ T_{B}\cos{\beta}\equal{} 21mg$

$ \Rightarrow T_{B}\cdot\frac{7}{5\sqrt{2}}\equal{} 21mg$

$ \Rightarrow T_{B}\equal{} 15\sqrt{2}\,mg$

From above: $ T_{A}\cos{\alpha}\equal{} 30mg$

$ \Rightarrow T_{A}\cdot\frac{5}{\sqrt{26}}\equal{} 30mg$

$ \Rightarrow T_{A}\equal{} 6\sqrt{26}\,mg$

This completes the solution.

thanks for both these nice soln.

Understood everything clearly. appreciate it.

You're welcome. I enjoyed the questions, they were pretty challenging.

Yes, I remeber the experiment only too well. My physics teacher - who has no common sense, and doesn't know much about physics - called it the "whirling bung" experiment. He got 8 adolescent boys - the 9th, me - to whirl bungs around their heads in a cramped physics room. They had to note the time taken for 10 revolutions, at which the hanging bung remained stationary. I gave the challenge to my classmates to try and get the string attached to the whirling bung to be perfectly horizontal (obviously impossible without infinite energy input). The results were very interesting...and involved my teacher gaining a black eye...

[quote="The Zuton Force"]These are nice. The setup with the spinning object and counterweight is often a prescribed experiment in introductory physics classes for measuring centripetal force.

FYI bos1234, you may want to post these in the Introductory Physics forum under "Arts & Sciences"[/quote]

do people visit that part of the forum :maybe:  :huh:

I visit that part all the time...and I am usually the one who will answer...


what is the maximum of the product X1*X2......*Xn

where X1+X2.......+Xn=2002

( 1<=n<=2002)

posted before $maximum=2^{2}3^{666}$ if $x_{i}$ is natural numbers.

can ypu please direct me to where it was posted?

or explain me how to solve it

[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=product+maximum&t=114263]there[/url] you go !


MATHCOUNTS
geometry
perimeter
similar triangles

Prove/Disprove all similar triangles with the same perimeter are also congruent.

Kind a hard but it reenforces the idea of similarity.
Is similarity going to be on MATHCOUNTS?

I like this question.  Yes, similarity is covered in mathcounts.

[hide]Let us take triangle ABC and DEF (similar triangles). their sides are in proportion to one another. Let's say that AB= DE*m, BC=EF*m and CA=FD*m. Since CA+AC+BC=m(EF+FD+DE) we conclude that m equals 1, therfore they are congruent.[/hide]

Yup, that's basically it. You could also do [hide](AB/DE)=(BC/EF)=(CA/FD)  1[/hide]

I'll try to get more similarity problems.


I can't solve this problem!

http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_10A_Problems/Problem_9

Can you guys help me out? Thanks.

Split the two larger right triangles up into smaller triangles.

That's not the real 2004 AMC 10 A problem. The real one asks for the differences in the areas of $ \bigtriangleup ADE$ and $ \bigtriangleup BDC$

In that case:

[hide="Solution"]Let $ x, y, z$ be the areas of $ \bigtriangleup ADE, \bigtriangleup BDC, \bigtriangleup ABD$, respectively.

Then:
$ x \plus{} z \equal{} \frac {8 \cdot 4}{2} \equal{} 16$
$ y \plus{} z \equal{} \frac {6 \cdot 4}{2} \equal{} 12$

The requested difference in areas is $ x \minus{} y$:
$ x \minus{} y \equal{} \left(x \plus{} z \right) \minus{} \left(y \plus{} z \right) \equal{} 16 \minus{} 12 \equal{} 4 \Rightarrow \boxed{B}$[/hide]

[quote="explosiv"]Whew, I though I was doing something completely wrong when I was solving that, none of the answer choices made sense!  If the question was as posted in the wiki, wouldn't it be:

[hide="Solution"]
[img]http://www.artofproblemsolving.com/Wiki/images/0/0e/AMC10_2004A_9.gif[/img]

I used something that was in AoPS Vol 1.

The difference between $ \triangle ABC$ and $ \triangle BDC$ is $ \triangle ADB$, so we need the area of this triangle.

We know that $ EA$ and $ CB$ are parallel because they both form right angles with $ AB$. If we draw a perpendicular from $ D$ to $ AB$ to a point $ F$, it will also be parallel to $ EA$ and $ CB$.

So:

$ 1/EA \plus{} 1/CB \equal{} 1/ DF$

Solving for $ DF$ we get $ DF \equal{} 24/7$

Since it is a perpendicular to $ AB$, and the area is $ (BH)/2$ we get:

$ A \equal{} (4*(24/7))/2$
$ A \equal{} 48/7$

[/hide][/quote]

Yes, you're right.

[hide]But where did you learn that $ 1/EA \plus{} 1/CB \equal{} 1/ DF$? Sorry for my stupidity...[/hide]

[quote="explosiv"][quote]Yes, you're right.

Click to reveal hidden content
But where did you learn that ? Sorry for my stupidity...[/quote]

It can be found in AoPS Volume 1, Chapter 11: Triangles, a.k.a. Geometry as an exercise.  I had just looked over that part when I saw this problem  :P .[/quote]

OK. Thanks. So it just has to do with parallel lines?


function
induction
LaTeX
algebra unsolved
algebra

I have a function from Naturals to Integers, defined recursively:

f(1)=6
f(2)=4
f(n+2)=n(n+1)f(n+1) + 4f(n) - f(n+1)f(n)

Proof that: 2^n | f(n) for all n being Natural.  ( The | means divides
from left to right: 2^n divides f(n) )

I dont master (obviously) induction and have no idea how to apply induction here, because lets say my induction hypothesis is 2^n | f(n), what is my thesis?
Maybe: 2^(n+2) | f(n+2) but where is (n+1)? Its like like this i am jumping over one term?
Please help..

Hy N.T.TUAN thanks for your answer.
Sorry i cant use latex (at least i dont see how).

Let me ask a few things please:

[quote="N.T.TUAN"]
We have $f(k+2)=k(k+1)f(k+1)+4f(k)-f(k+1)f(k)$ and $2^{k+2}|k(k+1)f(k+1)$( because  $2|k(k+1)$), $2^{k+2}|4f(k)$, $2^{k+2}|f(k)f(k+1)$. Done :D[/quote]

When you have $2^{k+2}|k(k+1)f(k+1)$ you take down one $2$ and because $2|k(k+1)$ all the rest follows? But why happens that to the rest in here $2^{k+2}|k(k+1)f(k+1)$?

Following this question, the same, how do you know that $2^{k+2}|4f(k)$ ?

Do you understand my question? Thanks a lot!


vector
superior algebra
superior algebra solved

The axiom of choice is equivalent to the statement: "Every vector space has a vector space basis"
What is the justification of this statement (other than Hamel Basis) if there be any???

The axiom of choice is equivalent to Zorn's Lemma. The later implies that every vector space has a basis, but I don't know how to prove the converse.

Right Amulfiger thats something I have heard,but I dont know the proof.Is the proof very difficult.I have no idea.If its elementary can anybody post the proof please!Any way can you say how Zorn's lemma implies that every vector space has a basis?

Zorn's lemma: If $X$ is a partially ordered set such that every chain in $X$ has an upper bound, then $X$ has a maximal element.

A basis for a vector space is a maximal independent set.

This is straightforward. Take an independent set. Either it spans the whole vector space or it doesn't. If it doesn't span everything, then there is an element outside its span. Appending that element to the original independent set results in a larger independent set. If it does span everything, then it is a basis and any larger set is dependent.

So take the set of all independent sets in our vector space, and partially order it by inclusion. If we have a chain of independent sets, then the union of all of the sets in the chain is still independent.

Proof of that last statement: Suppose the union of an independent chain is dependent. Then there is a nontrivial linear combination that yields zero. However, "linear combination" is defined so that all but finitely many of the coefficients are zero. Hence we are only concered with some particular finite subset of the union, and there must be a member of the chain that includes this set. Since that member of the chain is independent, the linear combination must be trivial.

So the set of independent sets of a vector space satisfies the hypotheses of Zorn's lemma and therefore must contain a maximal element. A maximal independent set is a basis (a Hamel basis).

- all of this was standard textbook material. I haven't tried to answer the more interesting direction yet.