analytic geometry
graphing lines
slope
LaTeX
trigonometry

Turn
$\displaystyle (x^2+y^2)-9(x^2-y^2)=0$

into a polar form.

[hide]well here is my lame attempt, I have almost no experience with polar, but i think I will need to learn it for a topic next year in math team, but anyways...

when after solving for y we get:
y = +/-sqrt(8/10)x

which isnt that two lines with slopes of +/-sqrt(8/10) so the polar form would just be:

tan(theta) = +/-sqrt(8/10)
or
theta = tan^-1(+/-sqrt(8/10))

haha I need Latex

so did I do that completely wrong or was there anything I did right?[/hide]

Do you mean $= 1$?  If you don't, then the solution is just 2 lines... $\theta = \pm \arctan \pm \frac 2 {\sqrt 5}$

If you do, then
[hide]
Plug in $x = r \cos \theta$ and $y = r \sin \theta$ to get
$r^2 = \frac 1 {5 \sin^2 \theta - 4 \cos^2 \theta}$
[/hide]


[quote="spanferkel"][quote="Sung-yoon Kim"]
Or $\sum \frac{a^{3}}{a^{2}+ab+b^{2}}=\sum \frac{a^{3}+b^{3}}{2(a^{2}+ab+b^{2})}$ :)[/quote]I think that would be too easy.  :( You're making it symmetric, but it is not!  :wink:
to Zhaobin: nice CS application, if only you correct the typo $a^{2}+ab+bc$ by $a^{2}+ab+b^{2}$ :)[/quote]

This is symmetric. $\sum \frac{a^{3}}{a^{2}+ab+b^{2}}-\sum\frac{b^{3}}{a^{2}+ab+b^{2}}= \sum\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}}$
$=\sum a-b =0$.

:oops: you are right (of course  :) ). I'd mixed it up with $a^{2}-ab+b^{2}$. But how do you carry on to get a different proof via $\sum \frac{a^{3}+b^{3}}{2(a^{2}+ab+b^{2})}$ ? This is of course $\sum \frac{(a^{3}+b^{3})(a-b)}{2(a^{3}-b^{3})}$, but then what?

[quote="pvthuan"]Let $a,b,c$ be nonegative numbers, prove that
2) $\frac{a^{4}}{a^{2}+ab+b^{2}}+\frac{b^{4}}{b^{2}+bc+c^{2}}+\frac{c^{4}}{c^{2}+ca+a^{2}}\geq\frac{a^{3}+b^{3}+c^{3}}{a+b+c}$,
[/quote]

$\frac{a^{3}+b^{3}+c^{3}}{a+b+c}=\frac{3abc}{a+b+c}+\sum a^{2}-\sum ab$. So the inequality is equivalent to $\sum\frac{a^{4}}{a^{2}+ab+b^{2}}-\sum a^{2}+\sum ab \geq \frac{3abc}{a+b+c}$ or $\sum \frac{ab^{3}}{a^{2}+ab+b^{2}}\geq \frac{3abc}{a+b+c}$. Applying CS, we obtain $\sum \frac{ab^{3}}{a^{2}+ab+b^{2}}\sum \frac{a^{2}+ab+b^{2}}{ab}\geq (\sum a)^{2}$, so $(LHS) \geq \frac{(\sum a)^{2}}{3+\sum_{sym}\frac{a}{b}}$.

$\frac{(\sum a)^{2}}{3+\sum_{sym}\frac{a}{b}}\geq (RHS)$
$\iff \frac{(\sum a)^{2}}{3+\sum_{sym}\frac{a}{b}}\geq \frac{3abc}{a+b+c}$
$\iff (\sum a)^{3}\geq 3\sum_{sym}a^{2}b+9abc \iff \sum a^{3}\geq 3abc$ which is obciously true.

Kim's solution is so simple and elegant. I will use yours in my book. I also use another inequality proposed by Kim in my book.

Thanks, zhaobin :) and thanks pvthuan for using my inequality!

[quote="pvthuan"]Let $ a,b,c$ be nonegative numbers, prove that
2) $ \frac {a^{4}}{a^{2} \plus{} ab \plus{} b^{2}} \plus{} \frac {b^{4}}{b^{2} \plus{} bc \plus{} c^{2}} \plus{} \frac {c^{4}}{c^{2} \plus{} ca \plus{} a^{2}}\geq\frac {a^{3} \plus{} b^{3} \plus{} c^{3}}{a \plus{} b \plus{} c}$,
[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-59-p-t-viet.html

[quote="The Nguyen"][quote="pvthuan"]Let $ a,b,c$ be nonegative numbers, prove that
3) $ \frac {a^{5}}{a^{2} \plus{} ab \plus{} b^{2}} \plus{} \frac {b^{5}}{b^{2} \plus{} bc \plus{} c^{2}} \plus{} \frac {c^{5}}{c^{2} \plus{} ca \plus{} a^{2}}\geq\frac {a^{4} \plus{} b^{4} \plus{} c^{4}}{a \plus{} b \plus{} c}$.[/quote]
Check this![/quote]
It is false. Try $ a\equal{}13^{12^{1988^{22^{12^{1990}}}}},$ $ b\equal{}1$ and $ c\equal{}0.$


[quote="math92"]Simplify

$\frac{\frac{3}{y}-\frac{6}{y^2}}{\frac{9}{y}}$[/quote]

After finding common denominators and canceling the $y^2$ term (for more detailed explanation, look in Drunken_Math's post), we get
[hide]$\frac{y-2}{3y}$[/hide]

Simplify $\displaystyle \frac{\frac{3}{y}-\frac{6}{y^2}}{\frac{9}{y}}$
Thank you 236factorial for the fraction.
[hide]

$\displaystyle \frac{\frac{3}{y}-\frac{6}{y^2}}{\frac{9}{y}}$
Reduce to the common denominator for $\displaystyle \frac{3}{y} = \displaystyle \frac{3y}{y^2}$

$\frac{\frac{3y}{y^2}-\frac{6}{y^2}}{\frac{9}{y}} * \frac{y}{9} =  \frac{y(3y - 6)}{\frac{9}{y^2}}$

Multiply by y and divide by 3 and you get: $\frac{y - 2}{3y}$[/hide]


But with an added twist!

Now, when Player X offers Player Y a gun, Player Y can keep or switch, Player X then can keep or switch, then Player Y can keep or switch.

Mwahahahahahaha... and the Puppet Master will be your new host...

[quote="budi713"]There is no point in that
player x knows what is in the gun and can still give any gun he chooses to player y

i will join

has RR2 ended?[/quote]

RR2 is currently between me and Brut3Forc3, however, CCM has not been on for a while...

oh and budi- but then player Y can switch/keep

[quote="ernie"]

oh and budi- but then player Y can switch/keep[/quote]
Like he said, there's no point in that, because Player X could just have offered what he wanted to offer in the first place in the old way.

oh lol

nvm, hey, this could happen

X gives A (4-2)
Y switches to B (2-4)
X keeps A (2-4)
Y switches to A (4-2)

However, there is more strategy involved, because X can trick Y into wanting the gun with more bullets, and Y can do the same.

/in.

Also, make there you can put any number greater than or eqaul to  6 of bullets... like

you could put 6-6, you could put 3-3, you could put 5-2...but you can't put somehtign like 0-0 because that isn't at least 6 bullets...

Except tehre might be last round problems...

I'm going to be random this time. RANDOMRANDoMRANDOM

pythag: nope. :P

PM me ur first round bullet choices

only mewto55555 has

[b]People Who Are In[/b]

mewto55555
budi713
pacman2812
ProtestanT
pythag011

bump

only 6 ppl =(

mewto55555- ready
pacman2812- ready
budi713- awaiting
ProtestanT- awaiting
pythag011- awaiting
chenhsi- awaiting

mewto55555- ready to get pwned
pacman2812- ready to get pwned
budi713- ready to get pwned
Brut3Forc3- ready to get pwned
ProtestanT- ready to get pwned
pythag011- awaiting bullet choices
chenhsi- awaiting bullet choices
?????- awaiting sign-up

mewto55555- ready to get pwned
pacman2812- ready to get pwned
budi713- ready to get pwned
Brut3Forc3- ready to get pwned
ProtestanT- ready to get pwned
cf249- ready to get pwned
pythag011- awaiting bullet choices
chenhsi- awaiting bullet choices

bump...

i need one more person to give me bullet choices

This game is formally postponed until further due notice.


function
number theory
greatest common divisor
complex numbers
number theory solved

Let A be the set of positive integer represented by
the form a^2+2*b^2, where a,b are integers with b>0.
Show that if p is a prime number so that p^2 belongs
to A,  then p belongs to A.
**
i think that if P^2=a^2+2*b^2 then
p=c^2+2*d^2  where  c=sqrt((p+a)/2)  and d=sqrt(p-a)
but i can't prove that c,d are integer.

Try Ha Duy Hung's site and check out the number th problems section for sth that might help you (I also thought about this one). Ha Duy Hung is a member here, you can find the address of his site in the memberlist.

Here?s what I meant: Ha Duy Hung put a problem there:

If  n/A^2+kB^2, k is in the set {1, 2, -2} and (A, kB)=1 then there exist C, D s.t. n=C^2+kD^2.

If we prove this for a prime p (not necessarily a natural n) then we are in the case k=2 (I?m talking about the problem you?re trying to solve) and we quickly prove what we have to. The problem on Ha Duy Hung?s site can be proved in the cse n= a prime p by using Thue?s lemma:
If p is a prime and (a, p)=1 then there exist x, y < p^(1/2) s.t. p/ax+y or p/ax-y. Try to prove this, it?s not hard at all. If I?m not wrong, combining all of this gives a proof of the problem from the Romanian selection ?97.

The case p = 2 is easy, so assume p > 2. Note that if p2 = a2 +2b2, then 2b2 = (p - a)(p + a). In particular, a is odd, and since a cannot be divisible by p, gcd(p - a,p + a) = gcd (p - a, 2p) = 2. By changing the sign of a, we may assume p - a is not divisible by 4, and so |p + a| = m2, |p - a| = 2n2 (for certain integers m, n). Since |a| < |p|, both p + a and p - a are actually positive so 2p = m2 + 2n2 and p = n2 + 2(m/2)2.

the solution I tought about uses the fact that $\mathbb{Z}[i\sqrt 2]$ is an [url=http://planetmath.org/encyclopedia/EuclideanRing.html]euclidian ring[/url]. It is well known that if a prime $p$ in $\mathbb{Z}$ cannot be written as $a^2+2b^2$ (which is the norm function for $a+bi\sqrt  2$) then it is also prime (and irreducible for that matter) in $\mathbb{Z}[i\sqrt 2]$.

So, suppose our prime $p$ cannot be written in that form. Then it is prime in the ring $\mathbb{Z}[i\sqrt 2]$ thus $p \mid a+bi\sqrt 2$ or $p \mid a-bi\sqrt 2$ => in either cases we obtain (separating the real part from the imaginary part) that $p \mid  a$ and $p \mid b$ => contradiction. Thus $p$ can be written as $x^2+2y^2$ :)


Determine all rhombuses $ ABCD$ with the given length $ 2a$ of ist sides by giving the angle $ \alpha \equal{} \angle BAD$, such that there exists a circle which cuts each side of the rhombus in a chord of length $ a$.

please specify question clearly. it is a little ambiguous

Let $ ABCD$ be a rhombus, such that $ AB\equal{}BC\equal{}CD\equal{}DA\equal{}2a$. Find all angles $ \alpha \equal{} \angle DAB$ such that there exists a circle in the plane of the rhombus that cuts each side of the rhombus in a chord of length $ a$.


$D$ is a point on the circum circle of triangle $ABC$ in which:

$AD$ is prependicular to $BC$, $AD=BC$ and $\angle BAC=69$

what is :$\angle ABC-\angle ACB$

Please correct me if I am wrong.

We are given that $\angle BAC=69^{\circ}$

Since $AD=BC \implies \angle CAB = \angle ABD = 69^{\circ}$ because equal chords subtend equal angles at the circumference.

Let $\angle CAB=x$

$\therefore$ in $\triangle ABC$ $69^{\circ} + x + \angle CBA = 180 \implies \angle CBA = (111-x)$

Since $\angle DBA=\angle CBA + \angle CBD = (111-x) + \angle CBD = 69^{\circ} \implies \angle CBD=(x-42)$

Let the intesecting point of AD and BC be P

Since we know $AD\perp endicularBC \implies \angle DPB = 90^{\circ}$

$\implies \angle CBD=(90-x)$

$\therefore \angle CBD=(90-x)=(x-42) \implies x=66^{\circ}$

$\angle ABC = (111-x) = (111-66)^{\circ} = 45^{\circ}$

$\angle ACB=x=66^{\circ} \implies t \angle ABC-\angle ACB = (45-66)^{\circ} = -21^{\circ}$

$\angle{ABC}=\angle{ABD}-\angle{DBC}=69^o -\angle{DBC}$.

$\angle{ACB}=90^o-\angle{CAD}$.

Since $\angle{DBC}=\angle{BAC}$, so
\[ \left| \angle{ABC}-\angle{ACB} \right| = \left| 69^o - 90^o \right|=21^o \].

$AD=BC\Longleftrightarrow 2R\sin A=2R\sin \widehat{ABD}\Longleftrightarrow \sin A=\sin \widehat {ABD}\ \ (\bigstar ).$
$\blacktriangleright\ 1^{th}$ case: $[AD]\cap [BC]\ne \emptyset\Longrightarrow m(\widehat {ABD})=90^{\circ}+|B-C|\ .$
$(\bigstar )\Longrightarrow \sin A=\cos |B-C|\Longrightarrow |B-C|+A=90^{\circ}\ .$
$\blacktriangleright\ 2^{th}$ case: $[AD]\cap [BC] =\emptyset\Longrightarrow m(\widehat {ABD})=|B-C|-90^{\circ}\ .$
$(\bigstar )\Longrightarrow\sin A=-\cos |B-C|\Longrightarrow |B-C|-A=90^{\circ}\ .$
From here it is easily ! Therefore,
$\boxed {\ |B-C|= \{ \begin {array}{cc} 90^{\circ}-A & \ if\ \ [AD]\cap [BC]\ne \emptyset \\ 90^{\circ}+A & if\ \ [AD]\cap [BC] = \emptyset\\ \end{array} \ }$

thanks to all of the solvers:

$VIRGIL NICULA$

$SHOBBER$

$SEN$

the problem was one of the geometry problems of the entrance exam of $atomic enerji$ high school...

it was easy. wasnt it? ;)

[quote="Virgil Nicula"]Want you become atomist ? I think you don't want make the atomic bomb !
This problem was not easily for this exam because it has two cases.[/quote]

 :lol: ...no bomb... no war.... no blood... ;)

yahh...

ure right ...

its hard for them...

im highschool student i teach math to the students in other grades.or to the middle school students...

but i dont name my self a teacher...

oh levi ,,,

it seems that ure post is deleted...

we can talk to each other with private messages... ;)


induction
search
combinatorics unsolved
combinatorics

Let $n\geq 2$ be a positive integer. There are $2n$ points in the plane, and $n^{2}+1$ segments joining them. Show that the segments form at least $n$ triangles.

Nice problem  :)

First, we prove that there exists at least $1$ triangle. We proceed both proofs with induction.
For $n = 2$, there exists a point which is connected to $\geq \frac{5 \cdot 2}{4}$, which is $\geq 3$. The result follows easily.
Assume it's true for $n = k$. For $n = k+1$, we have $2k+2$ points and $k^{2}+1+(2k+1)$ segments. Assume the contrary.
Take $2$ points $A,B$ which is connected, and consider all segments that contains at least one of them. There are at most $1+2k$ segments (including $AB$), otherwise there are at least $2k+1$ segments (excluding $AB$) and by $PHP$, there exists another point which are connected to both $A$ and $B$. But now, there are $2k$ points (excluding $A,B$) and $k^{2}+1$ segments which aren't connected to either $A$ or $B$ and by induction hypothesis, there is a triangle. Contradiction and we finished the first part.

Now, let's go back to the original problem. Let the points be $P_{1},P_{2},\cdots,P_{n}$.
By the same reasoning, it's true for $n = 2$. Assume it's true for $n \leq k$. We'll prove for $n = k+1$.
From above, we can take $1$ triangle, WLOG it's $P_{1}P_{2}P_{3}$.
Let $S = \{P_{4},\cdots,P_{2k+2}\}$ and for $i = 1,2,3$, let $S_{i}$ denote the elements in $S$ that are connected to $P_{i}$. If $P_{k}$ is in $S_{i}\cap S_{j}$, then $P_{i}P_{j}P_{k}$ is a triangle.
Thus, the segments form at least $t_{k+1}= |S_{1}\cap S_{2}|+|S_{2}\cap S_{3}|+|S_{3}\cap S_{1}|$ triangles.

By the Inclusion-Exclusion Principle we have
$|S_{1}\cup S_{2}\cup S_{3}| = |S_{1}|+|S_{2}|+|S_{3}|-t_{k+1}+|S_{1}\cap S_{2}\cap S_{3}|$.
Since $|S_{1}\cap S_{2}\cap S_{3}| \geq 0$ and $|S_{1}\cup S_{2}\cup S_{3}| \leq 2k+2-3 = 2k-1$, we have $t_{k+1}\geq |S_{1}|+|S_{2}|+|S_{3}|-(2k-1)$.

If $|S_{1}|+|S_{2}|+|S_{3}| \geq 3k-1$, then $t_{k+1}\geq k$, that is, there are at least $k$ triangle besides triangle $P_{1}P_{2}P_{3}$, and we are done.
We assume that $|S_{1}|+|S_{2}|+|S_{3}| \leq 3k-1$, or $(|S_{1}|+|S_{2}|)+(|S_{2}|+|S_{3}|)+(|S_{1}|+|S_{3}|) \leq 6k-2$.
Hence, by the $PHP$, at least one of $|S_{1}|+|S_{2}|, |S_{2}|+|S_{3}|, |S_{1}|+|S_{3}|$ is $\leq 2k-1$, WLOG it's $|S_{1}|+|S_{2}|$. If we remove $P_{1}$ and $P_{2}$, then we lose at most $|S_{1}|+|S_{2}|+1 = 2k$ segments. Therefore, there are at least $(k+1)^{2}-2k = k^{2}+1$ segments constructed among the $2k$ points $P_{3},\cdots, P_{2k+2}$ and by induction hypothesis, the segments form at least $k$ other triangles besides $P_{1}P_{2}P_{3}$ and our induction is complete.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=458264700&t=6323

Pierre.


algebra
polynomial
superior algebra
superior algebra theorems

I post it in olympiad section but no one answers. I need an answer.
can we proove that a certain polynomial is reducible or not? (for any polynomial,does exsit a general way?)

Er... yes. There are some theorems for it...

Here's one I find useful in particular: [url]http://en.wikipedia.org/wiki/Eisenstein%27s_criterion[/url]

See also here: [url]http://mathworld.wolfram.com/AbelsIrreducibilityTheorem.html[/url] and here:[url]http://mathworld.wolfram.com/IrreduciblePolynomial.html[/url].

A book for polynomials like Barbeau's and and Prasolov's... :wink:


probability
induction
probability and stats

(i) Let $ A$ and $ B$ be events in $ (\Omega,\mathbb{P})$, such that $ 1 > \mathbb{P}(B) > 0$. Show that
$ \mathbb{P}(A) \equal{} \mathbb{P}(A|B)\mathbb{P}(B) \plus{} \mathbb{P}(A|B^c)\mathbb{P}(B^c)$.

(ii) Suppose that a standard dice is cast $ n$ times over. Let $ A_n$ be the event that there are an even number of sixes in $ n$ throws, and let $ p_n \equal{} \mathbb{P}(A_n)$. By considering the events $ A_n$ and $ A_{n \minus{} 1}$ in (i), show that
$ p_n \equal{} (5/6)(p_{n \minus{} 1}) \plus{} (1/6)(1 \minus{} p_{n \minus{} 1})$.

(iii) Show by induction that
$ p_n \equal{} (1/2)(1 \plus{} (2/3)^n)$.

My actual problem is in (ii). I cannot understand where the (5/6) comes from. The probability of the event $ A_n|A_{n\minus{}1}$ seems very difficult to be evaluated. I don't know what i am doing wrong. Thanks anyway for any help.

We split it into two cases:

1) Throw number $ n$ is not a 6: This happens with probability $ \frac {5}{6}$. There is an even number of sixes in the $ n$ throws iff there is an even number of sixes in the $ n \minus{} 1$ first throws. This happens with probability $ p_{n \minus{} 1}$. So there is an even number of throws with a non-six as the last with probability $ \frac {5}{6}p_{n \minus{} 1}$

2) Throw number $ n$ is a 6: This happes with probability $ \frac {1}{6}$. There is an even number of sixes in the $ n$ throws iff there is an odd number of sixes in the $ n \minus{} 1$ first throws. This happes with probability $ 1 \minus{} p_{n \minus{} 1}$. So there is an even number of throws with a siz as the last with probability $ \frac {1}{6}(1 \minus{} p_{n \minus{} 1})$

I have don't know what (i) means, but this looks like schoolwork, so it must be placed there for some reason - i.e. it probably means that we can add these to probabilities to get $ p_n \equal{} \frac {5}{6}p_{n \minus{} 1} \plus{} \frac {1}{6}(1 \minus{} p_{n \minus{} 1})$ :P

$ P(A_n\,|\,A_{n \minus{} 1})$ is the probability of having an even number of sixes after $ n$ throws, given that you have an even number after $ n \minus{} 1$ throws.

Just put yourself in that limited world - assume $ A_{n \minus{} 1}.$ So we already have an even number of sixes. How can we still have have an even number of sixes after the next throw? We have to [i]not[/i] throw another six. And the probability of that happening is $ \frac56.$

On the other hand, $ P(A_n\,|\,A_{n \minus{} 1}^c) \equal{} \frac16.$ Why? If you start with an odd number of sixes, you do have to throw another six to get to an even number.

Edit: simultaneous post. Mathias_DK and I are saying essentially the same thing.


Number $ 1,2,...,2009$ are arranged in a row such that each number is either greater than the number to its left and the number  to its right or less that both of  them (the first and the last can be randomly choosed).How many arrangements are there?


Alexio has 100 cards numbered 1-100, inclusive, and places
them in a box. Alexio then chooses a card from the box at
random. What is the probability that the number on the card he
chooses is a multiple of 2, 3 or 5? Express your answer as a
common fraction.

50 multiples of 2
33 multiples of 3
20 multiples of 5

16 multiples of 6
10 multiples of 10
6 multiples of 15

3 multiples of 30

Then we have 50+33+20-16-10-6+3=74, so we have 37/50

Here is an easy way to get an approximate solution.

1/2 the cards are multiples of 2
~1/3 are multiples of 3
1/5 are multiples of 5

~1/6 are multiples of 2 and 3
1/10 are multiples of 2 and 5
~1/15 are multiples of 3 and 5

~1/30 are multiples of 2,3, and 5

So
p = 1/2+1/3+1/5-(1/6+1/10+1/15)+1/30 = 0.7333.. which is not too far from the correct answer of 0.74

The approximation would be even better if there were 1000 cards!

Personally I think this should be in the Venn diagrams alcumus in C&P instead of Probability with casework in C&P because solving with a venn diagram is a million times easier than the more complex solution in alcumus.

Principle of Inclusion/Exclusion states that the total number of multiples of 2, 3 and 5 between 1 and 100 is

floor(100/2) + floor(100/3) + floor(100/5) -floor(100/10) - floor(100/15) - floor(100/6) + floor(100/30)

This method is useful if you have a calculator.


Solve for $ x$ in the following equation.

$ 3x\sqrt {x \minus{} 1} \plus{} 2(\sqrt {x \minus{} 1})^3 \equal{} 0$

Let $ y\equal{}\sqrt{x\minus{}1}$.
So $ 3(y^2\plus{}1)y\plus{}2y^3\equal{}y(5y^2\plus{}3)\equal{}0 \iff y\equal{}0, \pm\sqrt{\frac{3}{5}}i$.
Thus $ x \equal{} y^2\plus{}1 \equal{} 1, \frac{2}{5}$.

Thanks for the solution.

$ \frac{2}{5}$, however, is actually not an answer, because that would result in the square root of a negative number, which is undefined.  :)

[quote="PhireKaLk6781"]Thanks for the solution.

$ \frac {2}{5}$, however, is actually not an answer, because that would result in the square root of a negative number, which is undefined.  :)[/quote]

Really?
$ \begin{align*} 3 \times \frac {2}{5} \times \sqrt {\frac {2}{5} \minus{} 1} \plus{} 2 \left( \sqrt {\frac {2}{5} \minus{} 1} \right)^3 & \equal{} \frac {6}{5} \times \sqrt {\frac {3}{5}}i \plus{} 2 \left( \minus{} \frac {3}{5} \times \sqrt {\frac {3}{5}}i\right) \\
& \equal{} \left(\frac {6}{5} \minus{} \frac {6}{5}\right)\sqrt{\frac {3}{5}}i \\
& \equal{} 0 \end{align*}$

[quote="PhireKaLk6781"]Thanks for the solution.

$ \frac {2}{5}$, however, is actually not an answer, because that would result in the square root of a negative number, which is undefined.  :)[/quote]
Yeah...you're quite right...but for that you would have to change the question to..."[b]Solve in the real no. system only[/b]"


Give $ a,b,c \ge 0$and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$.Prove that:
(a)$ a^3b \plus{} b^3c \plus{} c^3a \plus{} abc(a \plus{} b \plus{} c) \le 2(a \plus{} b \plus{} c)$
(b)$ a^3b^2\plus{}b^3c^2\plus{}c^3a^2\plus{}3abc \le 2(ab\plus{}bc\plus{}ca)$

Nice hint to creat, secret  :)
Use the lenma of Vasc: $ a^2b\plus{}b^2c\plus{}c^2a\minus{}abc \leq\ 2$ with $ a^2\plus{}b^2\plus{}c^3\equal{}3$

[quote="nguoivn"]Nice hint to creat, secret  :)
Use the lenma of Vasc: $ a^2b \plus{} b^2c \plus{} c^2a \minus{} abc \leq\ 2$ with $ a^2 \plus{} b^2 \plus{} c^3 \equal{} 3$[/quote]
multiplying two sides with $ a \plus{} b \plus{} c$,and using AM-GM ineq,we have (a)
multiplying two sides with $ ab \plus{} bc \plus{} ca$,we have (b)
:D
[hide="to nguoivn"]anh QA ơi,làm thế nào chứng minh cái bổ đề kia thế:D,thông cảm,em vốn dốt :P[/hide]

[quote="nguoivn"][quote="caubetoanhoc94"]My solution:
$ a^3b \plus{} b^3c \plus{} c^3a \le \frac {1}{3}(a^2 \plus{} b^2 \plus{} c^2)^2 \equal{} 3$
We need prove that:
$ 2(a \plus{} b \plus{} c) \ge abc(a \plus{} b \plus{} c) \plus{} 3$(*)
Setting:$ 3 \ge p \equal{} a \plus{} b \plus{} c \ge \sqrt {3}$
So:$ abc \le \frac {p^3}{27}$
$ (*) < \equal{} > 2p \minus{} pr \minus{} 3 \ge 0$
We have:
$ 2p \minus{} pr \minus{} 3 \ge 2p \minus{} \frac {p^4}{27} \minus{} 3 \equal{} \frac {(27 \minus{} 9p \minus{} 3p^2 \minus{} p^2)(p \minus{} 3)}{27} \ge 0$
It's true because:$ p \ge \sqrt {3}$
Done!  :P[/quote]
Your solution is trues but it's not work for ineq (b) . Notice that with secret's problem (b), equality occurs when $ a \equal{} b \equal{} c$ or $ (a,b,c) \equal{} (\sqrt {2}, 1, 0)$
@Toan_VN_LC: [hide]Em co the? tham khao? loi` giai trong STBDT nhe'  :) [/hide][/quote]
Yes.Of course. :lol:

[quote][hide]Em co the? tham khao? loi` giai trong STBDT nhe' [/hide][/quote]
Sorry, I've been thinking about it lately but I don't know what STBDT stands for and where is it.
[hide]BDT chac la` ba^t' dang thuc'[/hide]


This didn't quite fit in the AP Scholar thread...but as I just got my AP scores today I was wondering how other people did, especially in chemistry because that was the hardest test of the ones I took.


Find the distance between the points $ (1,1)$ and $ (4,7)$. Express your answer in simplest radical form.

[quote="markoboy"][hide="answer"]phytagorean theorem!

3^2+6^2=c^2
9+36=c^2
45=c^2
c=3sqrt5

so the answer is 3sqrt5 [/hide]

[b][i][u]please rate my post thanks!![/u][/i][/b]

please be fair though [=

 :bruce:  :bruce:  :bruce:  :bruce:[/quote]
Marko, it really would be nice if you stopped repeating what other people have already stated. I've already seen many posts in which you've done that.


function
real analysis
real analysis unsolved

Let  $ {f(x,y), (x,y)\in[0,1]xR}$ be a real function with: $ 1\geq|f(0,y)|$ and ${ 2\geq|{{df(x,y)}\over{dx}}|}$ for every $ {x}$ and $ {y}$. If $ {y_n}$ is a sequence in $ {R}$, check if the sequence of functions $ {g_n}$, where $ {g_n}(x)=f(x,{y_n}), x\in[0,1]$ has a subsequence that converges uniformely.


modular arithmetic
number theory
algebra unsolved
algebra

I've noticed that the last digit of n^5 is the same as the last digit of n and therefore that the last digit of n^9, n^13, ..., n to the power of (4m - 3) is the same as the last digit of n. Why should this be so, and do similar results hold in bases other than 10? Where can I find out more about this?

This is true by [url=http://www.artofproblemsolving.com/Wiki/index.php/Euler%27s_Totient_Theorem]Euler's Totient Theorem[/url]:
Since $ \phi(10)\equal{}4$, $ a^4 \equiv 1 \pmod{10}$. So $ a^{4k\plus{}1} \equiv a \pmod{10}$.

You can find out more about this in Number Theory sections in books.

Thanks. (I thought that was only true if a is relatively prime to (in this case) 10).

I mean 1, 3, 7, 9, 11, 13, 17, 19, 21, ...are coprime to 10 and are all congruent to 1 mod 10; whereas 2, 4, 5, 6, 8, 1012, 14, 15, 16, 18, 20, ... are not and none of them is congruent to 1 mod 10.

Yes, $ a^4 \equiv 1 \pmod{10}$ iff $ \gcd (a,10) \equal{} 1$.

However, if $ \gcd(a,10) > 1$, then A:$ a \equiv 0 \pmod{2}$ or B:$ a \equiv 0 \pmod{5}$ or possibly both.

If both are true, then $ a \equiv 0 \pmod{10}$, and thus $ a^{4k \plus{} 1} \equiv a \equiv 0 \pmod{10}$.

If only A is true then $ a^{4k \plus{} 1} \equiv a \equiv 0 \pmod{2}$, and ETT gives $ a^4 \equiv 1\pmod{5}$, or $ a^{4k \plus{} 1} \equiv a \pmod{5}$. Thus, the Chinese Remainder Theorem yields $ a^{4k \plus{} 1} \equiv a \pmod{10}$ as desired.

If only B is true then $ a^{4k \plus{} 1} \equiv a \equiv 0 \pmod{5}$, and ETT gives $ a^4 \equiv 1\pmod{2}$, or $ a^{4k \plus{} 1} \equiv a \pmod{2}$. Thus, the Chinese Remainder Theorem yields $ a^{4k \plus{} 1} \equiv a \pmod{10}$ as desired.

So $ a^{4k\plus{}1} \equiv a \pmod{10}$.

Similarly, $ a^{\phi(n)k\plus{}1} \equiv a \pmod{n}$ regardless of whether $ \gcd(a,n)\equal{}1$ is true.


:roll: This is perhaps an interesting question, but I don't know enough number theory to judge its difficulty.  For which odd r, is r or 2r the order of 2 in the multiplicative group Z_2r+1?  This is true for the first couple of primes 2r+1, but note that 2^5=1mod31. Then it becomes kind of irregular.  Is there a pattern at all?

Do you mean the group Z_(2^r-1) for odd r?

No, I do not mean Z_(2^r-1). This is a question that came up while studying patterns of current graphs for n-embeddings (graph theory stuff), and so is probably not a standard problem in algebra or number theory.

Oh I see. I was confused because you were using 31. Sorry. This problem looks interesting. I'll see what I can come up with, but probably I won't be able to solve it.

If I'm understanding the question right, then yes, this should be true for all r. Because 2r = -1 (mod 2r+1); so (2r)^2 = 1 (mod 2r+1).

Another question to ask though, is does the multiplicative group of Z_2r+1 always contain r?

[quote="Mr. Galois"]To clarify things, the order of 2 is the MINIMAL integer n such that 2^n=1mod2r+1.

Also, for Z_(2r+1) to be a group under multiplication, 2r+1 has to be a prime; else the divisors don't have inverses.

But I was also wondering, is it o.k. to say that an integer has a certain order with respect to a modulus even if we're not working in a group?[/quote]

Oh, okay. I read your question as. "For which odd r [b]does r or 2r have the order of 2[/b] in..." Your question is certainly more difficult. But even if 2r+1 is not prime you can still talk about the multiplicative group of any ring by considering just it's invertible elements. In the case of the ring Z_m the multiplicative group is usually denoted by Z_m* and is called the reduced residue system modulo m. [i]Challenge: Can you say how many elements are in the reduced residue system modulo m for each m>1? For which m are all of the non-zero elements invertible? Can you show that the multiplicative group of this Z_m is cyclic (this is a difficult problem)?[/i]

As for your original question, the answer is very complicated. The most you can say is that the order of any element divides the order of the group (which in the case that 2r+1 is prime, is 2r). But whether it is r or 2r, or neither, is very erratic. I don't believe there is any general formula for telling definitely. The best you can hope for is a probabalistic approach. Because :phi:(p-1) of the element will have order p-1 (those such elements are called primitive roots) - these will correspond with your ones having order 2r. And there will be a similar number of elements with order (p-1)/2 (I'm not exactly sure how many but it shouldn't be too hard to figure out) - these will correspond with your ones having order r. [i]Can you show why that is the number or primitive roots in Z_p? (look back to the last question I asked in the first paragraph about cyclic residue systems, go ahead and assume it is true for primes m)[/i].

But telling what the order of a specific element is in a reduced residue system is in general very hard. Very few general techniques are known for finding primitive roots. But I did find one theorem that comes close.

"2 is a primitive root of the primes p and p' where p = 4q+1 for some prime q, and p' = 2r+1 for some prime r congruent to 1 (mod 4)." I found this as an exercise in the book [u]A Course in Number Theory[/u], p. 99 by H.E. Rose. This means that 2 has the order 2r when 2r+1 is a prime, and r = 2q where q is a prime; and also when 2r+1 is a prime, and r is a prime congruent to 1 (mod 4). But it doesn't mean that this is a necessary condition for the order to be 2r, nor does it say anything about when the order is r.

To read more about this, look in any good number theory book about Reduced Residue Systems and Primitive Roots. Particularly the books [u]Number Theory[/u] by Andrews and [u]A Course in Number Theory[/u] by H.E. Rose.


modular arithmetic
AMC
AIME
search
number theory
relatively prime
prime factorization

I know how to find last digit, But I have problem with finding last 2 or 3 digits, I hope if you guy could provide some trick.

Example : $3^{2006}$
$3^1=[3]$
$3^2=[9]$
$3^3=[7]$
$3^4=[1]$
so $3^{2004}3^2=[1][9]$ so the last digit is 9
But how about last 2 digits ?

Take it modulo 100 and observe a pattern with the remainders. For example with three it is

1, 3, 9, 27, 81, 43, 29, 87, 61, 83, 49, 47, 41, 23, 69, 07, 21, 63, 89, 67, 1 and it repeats

Thanks, It is really useful information :D

[quote="scorpius119"]If $a$ and $m$ are relatively prime and $\varphi (m)$ is the number of positive integers less than or equal to $m$ and relatively prime with $m$, then $a^{\varphi (m)}\equiv 1\pmod{m}$.

This helps with this problem because finding the last 2 digits means finding the number $\pmod{100}$. You can either find $\varphi (100)$ directly (it is 40), or split 100 into prime-power mods (so $3^2\equiv 1\pmod{4},\ 3^{20}\equiv 1\pmod{25}\Rightarrow 3^{20}\equiv 1\pmod{100}$). Either way reduces $3^{2006}$ to just $3^6$ in terms of the last 2 digits.[/quote]

$\phi(100)=\phi(2^2 \cdot 5^2) = 100 \left( 1 - \frac{1}{2}\right) \left( 1 - \frac{1}{5}\right) = 100 \cdot \frac{1}{2} \cdot \frac{4}{5} = 40$.

Similarly, for any positive integer $n$, we can take the prime factorization of $n$. Say it's $n=p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_m^{e_m}$. Then \[ \phi(n)=n\left( 1-\frac{1}{p_1}\right) \cdot \left( 1-\frac{1}{p_2}\right) \cdot \ldots \cdot \left( 1-\frac{1}{p_m}\right). \]

Hope that helped. :)

I didn't really understan, more explaination would be great :D

Sorry :D

First, take the prime factorization of $n$. For example, if $n=100$, the prime factorization is $2^2 \cdot 5^2$. So for any $n$, the prime factorization is $n=p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_m^{e_m}$, where $p_i$ are all primes and $e_i$ are the exponents, and there are $m$ primes in the factorization. In the case $n=100$, $p_1=2$, $e_1=2$ and $p_2=5$, $e_2=2$.

Okay, once we have the prime factorization, we can find $\phi(n)$, which is the number of positive integers less than $n$ that are relatively prime to it. (This means that they don't have any common factors with each other. Like 2 and 5 are relatively prime, while 2 and 10 aren't.) The formula for $\phi(n)$ is \[ \phi(n) = n \cdot \left( 1-\frac{1}{p_1} \right)\left( 1-\frac{1}{p_2} \right) \cdots \left( 1-\frac{1}{p_m} \right). \] In other words, multiply $n$ by (1-1/(prime factor)) for all of its prime factors.

Do you understand now? If not, please tell me which part you didn't understand. :)

[quote="FMako"][quote="scorpius119"]If $a$ and $m$ are relatively prime and $\varphi (m)$ is the number of positive integers less than or equal to $m$ and relatively prime with $m$, then $a^{\varphi (m)}\equiv 1\pmod{m}$.

This helps with this problem because finding the last 2 digits means finding the number $\pmod{100}$. You can either find $\varphi (100)$ directly (it is 40), or split 100 into prime-power mods (so $3^2\equiv 1\pmod{4},\ 3^{20}\equiv 1\pmod{25}\Rightarrow 3^{20}\equiv 1\pmod{100}$). Either way reduces $3^{2006}$ to just $3^6$ in terms of the last 2 digits.[/quote]

I don't understand why $3^2 \equiv 1\pmod{4},...$ means that I would understand if you meant $3^20 \equiv 1\pmod{4}$.[/quote]

$9 \equiv 1(\mod 4)$ What's the problem?

Yeah, always try to look for an application of the Binomial Theorem in these types of problems, especially if you have 3's because you can get a 9=10-1 from that. See AIME 1983 for another nice example.

#6: What is the remainder when $6^{83} + 8^{83}$ is divided by 49?

(But don't post the solution here, search it. It's been posted sooo many times. ;))


MIT college blogs Princeton Harvard Yale

[url=http://www.mitadmissions.org/topics/misc/miscellaneous/message_from_ben_and_matt.shtml]Message From Ben And Matt[/url]

[url=http://web.mit.edu/newsoffice/2007/admissions-hastings.html]MIT News[/url]

[url=http://www.boston.com/news/globe/city_region/breaking_news/2007/04/mit_dean_of_adm.html]MIT Dean of Admissions Resigns for Falsifying Resume (Boston Globe)[/url]

Wow.  I plan to apply to MIT next year, so I am especially interested in any news about it.  This piece of news is incredibly sad, and I hope it doesn't derail MIT's goals for decreasing anxiety over the admissions process.

Wow, that sucks for MIT.

I find it ironic that she was someone who wanted to reduce the anxiety over applications and to lessen attempts to beef up them, yet she had "beefed up" her own resume...

Does anyone know why she chose to resign over this now, not earlier, and not later?  Did someone recently look at her credentials?

She was exposed by someone who telephoned MIT. The hiring process must have been too lax in those days; when she was hired for a low-level position that didn't even require a college degree, no one checked for lies on her resume. Her background should have been looked up by journalists when she went on her book tour last year (why wasn't anyone curious about her own college experiences?) but it wasn't. Colleges used to make a LOT of flub-ups like this: they probably still do more hiring flub-ups than shareholder-owned businesses, but from now on I hope no one can become an admission officer at a major university without a resume check.

[quote="Zakary"]I find it ironic that she was someone who wanted to reduce the anxiety over applications and to lessen attempts to beef up them, yet she had "beefed up" her own resume...[/quote]

But in the same token, this shows that she was there before and thus knew how bad the anxiety is and how much beefing up your resume/application can eat away at your conscience (I assume she had qualms about it, at least).  In general, how much should we regard the advice of people who tell you to the opposite of what they did?  Often times society just tosses their advice away because they are hypocrites but I don't think that's what we should be doing.

Anyway, I, along with many other MIT students, were totally shocked by this.  It's really bizarre.  I applaud Marilee Jones for what she has done for MIT admissions in the relatively recent past.  On one hand I think it's irrelevant that she did this with respect to her job now.  But on the other hand, I know that the only possibility was for her to resign.

[quote="joml88"]Often times society just tosses their advice away because they are hypocrites but I don't think that's what we should be doing.[/quote]
I would like to elaborate on this point.  I think that in many cases, people know that they have gotten caught in a trap, and, knowing that they can't get out, do their best to keep others from falling into the same trap.  When society bills these people as hypocrites and ignores their advice, this is exactly the worst possible reaction.  See what mistakes other people have made, and learn from them.  If someone has made a mistake in their life, no matter how large, that in no way invalidates the other, good, things that they have done, or their message.

Well, if she cared so much, why didn't she confess herself? No, she waited for someone else to expose her... This is basic integrity we are talking about here, and I don't think an act like that should be acceptable at any job, let alone a job where she has a responsibility to reject people who make fabrications in their applications.

You are right that something like this shouldn't be tolerated.  Certainly there was no way that they could, in good conscience, keep her on under these circumstances.  The main thrust of the previous posts however, was directed more at the fate of the program.  The main point was that her having lied on her resume should not invalidate the work that she and so many others have done to decrease the stresses of applications.  This is still a worthy goal, and should not be dismissed merely because one of its main ptoponents hasn't followed it.  I am not condoning her lying on her resume, but am saying that it should not cause her work to be abandoned.

from a pragmatic standpoint, it shouldn't affect undergraduate admissions in the near future very significantly, as they'll want to maintain some degree of continuity in admissions standards from year to year.  a reversal of jones's policies or any other significant change won't happen soon, and if it does, it'll be done gradually over several years.

one thing i don't understand is how she reduced stress in college apps (i'm not really against her or anything, i'm just wondering at how anyone can reduce stress)? it's one thing to recognize the existence of stress in college app process, but i'm sure that there were still plenty of kids who were quite stressed about their apps to mit.
if we look at sat's, i'm certain that mit places at least as much emphasis on sat's as the ivies.
if we're saying that mit's placing less emphasis on achievements in math and science and more on other ec's, then maybe mit's trying to become an ivy, which should not be its goal.
i really don't see the differences in policies regarding mit's app process as opposed to other schools' apps, but i understand that such differences might truly exist and i would like to know them.

(on a side note, i was rejected by mit, but i never really had high hopes of getting in; i'm really interested, however, in how my application experience at mit should have been different from my experience of applying to other highly competative institutions.)

I believe Marilee is known for saying that students should try to find the college that is the 'best fit' for them.  So instead of stressing about getting into Elite School X, it's perfectly fine to realize that X might not be the place they'll thrive the most.  And thus, the student should feel fine about going to a less prestigious school where they'll fit in better, have a better time, and possibly learn more.

The MIT admissions people are also very open about their procedures.  Take their blogs, for instance.  They don't try to hide anything and they'll answer just about any question thrown at them.  College admissions are at a fundamental level stressful.  So it's impossible to fully purge all stress from the process.  But their are ways to relieve part of the load, which is what Marilee along with MIT admissions sought/are seeking to do.

[quote]I believe Marilee is known for saying that students should try to find the college that is the 'best fit' for them. So instead of stressing about getting into Elite School X, it's perfectly fine to realize that X might not be the place they'll thrive the most. And thus, the student should feel fine about going to a less prestigious school where they'll fit in better, have a better time, and possibly learn more. [/quote]

This brings up an excellent topic: Is there a concept such as a "best fit" college, and, if so, do colleges know for which students they are their "best fit" colleges?

For the former question, I would answer yes, though I would answer no in the latter case.

If someone really loves math, and does well in math, whether it be in his/her high school, math contests, or math research, then he/she should probably not go to a liberal arts college. If someone detests the countryside, and can't bear not to go to movies and to go out every week, then he/she should probably not go to a school in a rural setting.

However, can the colleges judge this? It seems to me like it is up to the student to find the best match. From personal experience, I think that the answer to the second question above is no. My entire college application was about math and physics (and basketball, but that's non-academic); I was rejected/waitlisted by MIT and Caltech and admitted by Princeton, Harvard, and Yale. First of all, I'm obviously not complaining about not getting into MIT and Caltech - I would have chosen Princeton and Harvard over those two schools anyways. Now, forgetting that this is coming from an AoPS person and that he is discussing his own experience, simply consider the implications of those decisions.

Firstly, mathy, techy schools rejected me while schools with a broader education accepted me despite my application showing exclusively math- and physics-related achievements and personal essays.

Secondly, Princeton, Harvard, and Yale accepted me, though those three schools have many notable differences in their academic programs as well as social environments. Moreover, this is not unique. While at the hosting programs, I met numerous students (about 60% of students there) accepted to at least two of these three schools. This would indicate that P, H, and Y are pretty much the same, whereas we all know that they have numerous significant differences. So which one is the "best fit" school? (This is a rhetorical question; I already decided where I'll go.)

[quote="shbmsc90"]
This brings up an excellent topic: Is there a concept such as a "best fit" college, and, if so, do colleges know for which students they are their "best fit" colleges?

For the former question, I would answer yes, though I would answer no in the latter case.[/quote]

I agree on the former question.  For the latter one, I have to disagree a little bit with you.  The college, i.e. their admissions office and admissions procedure, probably know best about whom their college is a good fit for.  You have told me about one instance of someone whom you think is a good fit for the school did not get in.  I'm pretty sure that's not a statistically significant data set in what we're discussing.  It's one instance and I don't even know enough about your application to know whether or not it contradicts the common perception of what students are best fit for MIT/Caltech.

I'll go back and emphasize that a sample size of one is very insufficient to say much.  Unless you were to look at the whole set of people [i]admitted[/i] to MIT and Caltech, and then somehow show that it contradicts the 'best fit' model there's not much you can really say.

Now that I've gotten that off, I'd like to discuss your specific instance.  If you would have chosen Princeton/Harvard over MIT/Caltech then that suggests you might not be best fit for a tech school.  There must (well, more precisely...there [i]probably[/i] was) be something about you that showed in your application that suggests you're more of an Ivy League person.


trigonometry
ratio
trig identities
Law of Cosines

The sides of a triangle have lengths $ p-q$, $ p$ and $ p+q$, where $ p>q>0$. The largest and smallest angles of the triangle are $ \alpha$ and $ \beta$, respectively. Show by means of the cosine rule that

$ 4(1-\cos \alpha)(1-\cos\beta) = \cos \alpha + \cos\beta$.

In the case $ \alpha = 2\beta$, show that $ \cos \beta = \frac{3}{4}$ and hence find the ratio of the lengths of the sides of the triangle.

Basically, by the Law of Cosines, we can make two equations with $ \cos{\alpha}$ and $ \cos{\beta}$. We get the following:

$ (p\plus{}q)^2\equal{}(p\minus{}q)^2\plus{}p^2\minus{}2p(p\minus{}q)\cos{\alpha}$
$ (p\minus{}q)^2\equal{}p^2\plus{}(p\plus{}q)^2\minus{}2p(p\plus{}q)\cos{\beta}$

Simplifying both equations and expressing the cosine terms in terms of p and q gives the following:
$ \cos{\alpha}\equal{}\frac{p\minus{}4q}{2(p\minus{}q)}$
$ \cos{\beta}\equal{}\frac{p\plus{}4q}{2(p\plus{}q)}$
$ \cos{\alpha}\plus{}\cos{\beta}\equal{}\frac{(p\minus{}4q)(p\plus{}q)\plus{}(p\plus{}4q)(p\minus{}q)}{2(p\minus{}q)(p\plus{}q)}\equal{}\frac{p^2\minus{}4q^2}{(p\minus{}q)(p\plus{}q)}$
$ 4(1\minus{}\cos{\alpha})(1\minus{}\cos{\beta})\equal{}4*\frac{p\plus{}2q}{2(p\minus{}q)}*\frac{p\minus{}2q}{2(p\plus{}q)}\equal{}\frac{p^2\minus{}4q^2}{(p\minus{}q)(p\plus{}q)}$.
Thus $ 4(1\minus{}\cos{alpha})(1\minus{}\cos{\beta})\equal{}\cos{\alpha}\plus{}\cos{\beta}$.

Now if $ \alpha\equal{}2\beta$, then $ 4(1\minus{}\cos{2\beta})(1\minus{}\cos{\beta})\equal{}\cos{2\beta}\plus{}\cos{\beta}$. Simplifying with trig identities, we get
$ 8(1\plus{}\cos{\beta})(1\minus{}\cos{\beta})^2\equal{}(2\cos{\beta}\minus{}1)(\cos{\beta}\plus{}1)$
$ (4\cos{\beta}\minus{}3)(2\cos{\beta}\minus{}3)\equal{}0$
note that cosines can't be greater than 1, so we're left with $ 4\cos{\beta}\minus{}3\equal{}0$ or $ \cos{\beta}\equal{}\frac{3}{4}$.


Let$f(x): R\to R$.And s.t:$f(x^2f(y)-ax^2)=xy+af(x)$ where$a=const$

The problem is wrong, I guess.

f(x^2 f(y) - ax^2 ) = xy+af(x) = -xy + af(-x)

2xy = -af(x) + af(-x)

when x,y is independent.


[quote="rlazertech"]1.	How many of the first 100 positive integers are composite and are also the product of three primes, two of which are the same?

[hide]17 numbers[/hide][/quote]
Well if the number is the product of three primes, it has to be composite.
So you basically go up the primes.
2,2,3
2,2,5
2,2,7
2,2,11
2,2,13
2,2,17
2,2,19
2,2,23
3,3,2
3,3,5
3,3,7
3,3,11
5,5,2
5,5,3
7,7,2
Which there are 15......
I guess if you counted 2,2,2, and 3,3,3 there would be 17...


[quote="lotharek"]More details... $k,n\in\mathbb{N}$?[/quote]
yep..
$k,n\in\mathbb{N}$


Let $ a>2$ be an odd number and let $ n$ be a positive integer. Prove that $ a$ divides $ 1^{a^n}\plus{}2^{a^n}\plus{}\cdots\plus{}(a\minus{}1)^{a^n}$

[hide="Solution 1"]
Let $ S=1^{a^n}+2^{a^n}+\cdots+(a-1)^{a^n}$.
We can express $ S$ as,
\[ (1^{a^n}+(a-1)^{a^n})+(2^{a^n}+(a-2)^{a^n})+\cdots+\left(\left(\dfrac{a-1}{2}\right)^{a^n}+\left(\dfrac{a+1}{2}\right)^{a^n}\right)\]
Then use the fact that \[ a+b|a^k+b^k\] for positive integers $ a$ and $ b$ whenever $ k$ is an odd positive integer.
[hide="Click here for proof of this fact"]
We know that,
\[ a+b\equiv 0\pmod{a+b}\]
Thus,
\[ a\equiv -b\pmod{a+b}\implies a^k\equiv (-b)^k\pmod{a+b}\]
Since $ k$ is odd we know that, $ (-b)^k\equiv -b^k\pmod{a+b}$ and therefore,
\[ a^k\equiv -b^k\pmod{a+b}\implies a^k+b^k\equiv 0\pmod{a+b}\]
From where we get that $ a+b|a^k+b^k$.
[/hide]
Thus,
\[ \begin{eqnarray*}
1+(a-1)&|&1^{a^n}+(a-1)^{a^n}\\
2+(a-2)&|&2^{a^n}+(a-2)^{a^n}\\
&\vdots&\\
\left(\dfrac{a-1}{2}\right)+\left(\dfrac{a+1}{2}\right)&|&\left(\dfrac{a-1}{2}\right)^{a^n}+\left(\dfrac{a+1}{2}\right)^{a^n}\]
We can use this fact because $ a^n$ is always an odd integer when $ n$ is a positive integer and $ a$ is odd. Hence $ S$ is divisible by $ a$.
[/hide]
[hide="Solution 2"]
Note that,
\begin{eqnarray*}
1^{a^n}+(a-1)^{a^n}&\equiv&1^{a^n}+(-1)^{a^n}\equiv 1^{a^n}-1^{a^n}\equiv 0\pmod{a}\\
2^{a^n}+(a-2)^{a^n}&\equiv&1^{a^n}+(-2)^{a^n}\equiv 2^{a^n}-2^{a^n}\equiv 0\pmod{a}\\
&\vdots&\\
\left(\dfrac{a-1}{2}\right)^{a^n}+\left(\dfrac{a+1}{2}\right)^{a^n}&\equiv&\left(\dfrac{a-1}{2}\right)^{a^n}+\left(a-\dfrac{a-1}{2}\right)^{a^n}\equiv \left(\dfrac{a-1}{2}\right)^{a^n}-\left(\dfrac{a-1}{2}\right)^{a^n}\equiv 0\pmod{a}
\end{eqnarray*}
Adding all the equations up we get that,
\[ S\equiv 0+0+\cdots+0\equiv 0\pmod{a}\]
[/hide]


[size=117]22種不可以混合吃的食物!!!

 1. 豬肉*菱角--肚子痛

 2. 牛肉*栗子--引起嘔吐

 3. 羊肉*西瓜--傷元氣

 4. 狗肉*綠豆--會中毒

 5. 兔肉*芹菜--脫發

 6. 雞肉*芹菜--傷元氣

 7. 鵝肉*雞蛋--傷元氣

 8. 甲魚*莧菜--會中毒

 9. 鯉魚*甘草--會中毒

 10. 螃蟹*柿子--腹瀉

 11. 白酒*柿子--會胸悶

 12. 紅薯*柿子--會得結石

 13. 糖精（片）*雞蛋--會中毒、重則死亡 （寒！！！！）

 14. 紅塘*皮蛋--會中毒

 15. 洋蔥*蜂蜜--傷眼楮

 16. 豆腐*蜂蜜--耳聾

 17. 蘿卜*木耳--得皮炎

 18. 馬鈴薯*香蕉--面部生斑

 19. 芋頭*香蕉--腹漲

 20. 花生*黃瓜--會傷身

 21 維生素*蝦--砒霜 （太寒了～～～～）

 22 牛奶*桔子*蘿卜--傷身體

藥品常識！根據美國國家藥典（USP）規定，藥品一旦開罐，罐內所附的棉花和乾燥劑，就必須立刻丟棄，否則它們會因吸附水氣，成為藥罐內的一項污染源！

感謝朋友相告這項常識，我以前從未聽聞,而且還以為乾燥劑可以繼續保持藥品乾燥，請大家相互轉告吧![/size]

[quote="shobber"]干燥剂是不是因为: $\text{CaO+H}_2\text{O}=\text{Ca(OH)}_2$?[/quote]
記得高一時有同學將食品乾燥劑投進塑料水瓶並注入水, 再不停地搖...結果可想而知──白色的液體出現少量白色沉澱, 並發出微熱. (不知我的表述是否正確)

不会吧...我记得我们干的那次瓶子几乎烧烂. 莫非我们那回用的干燥剂有问题!?

[quote="shobber"]不会吧...我记得我们干的那次瓶子几乎烧烂. 莫非我们那回用的干燥剂有问题!?[/quote]
我起初以為"實驗"結果會如你所言, 但當時摸那瓶子只是覺得有點熱.

想問一道簡單的化學題(可我的功課卻做錯了, 亦不懂得如何去改正): :help:

一種有機物具有以下性質: (1) 常溫下是無色液態; (2) 其蒸氣對$\text{H}_2$的相對密度為37; (3) 能與$\text{Na}$反應生成$\text{H}_2$; (4) 與$\text{Na}_2\text{CO}_3$不反應.
a)  推斷該物質屬於哪一類有機物;
b)  計算該物之相對分子質量(式量)並寫出其分子式;
c)  寫出該有機物可能有的同分異構體的結構式.

我还没学有机, 只知道$M=74$.

是丁醇.化学式为$C_4H_9OH$,结构简式有三种,直链型或2-甲基或3-甲基型.
ifai,你上几年级???

[quote="dingdongdog"]ifai,你上几年级???[/quote]
現在上高三了, 不過課程進度慢, 現在仍是用高二的課本. 而去年老師教關於有機的部分不少都已經遺忘了.

[quote="dingdongdog"]是丁醇.化学式为$C_4H_9OH$,结构简式有三种,直链型或2-甲基或3-甲基型.[/quote]
另外, 請問如何推斷出此結果?

说起来惭愧，是用不定方程加上醇的性质做的，而结构则是推的。

答案应该没错吧，虽然我的化学学得不怎样…… :oops:

[quote="dingdongdog"]答案应该没错吧，虽然我的化学学得不怎样…… :oops:[/quote]
may be but i'm not sure. my chemistry isn't good and i'm not interested in it since i go to high school. however, i join the macao chemical olympiad this year...

[quote="ifai"] i join the macao chemical olympiad this year...[/quote]
How was it? Did you qualify for the next round?


calculus
integration
function
limit
real analysis
real analysis unsolved

Sigh. This should be easy but there's some details that I cannot sort out.

Let $f$ be a non-negative integrable function. Show that the function $F$ defined by $F(x)=\int_{-\infty}^{x}f$ is continuous (using the Monotone convergence thm).

[b]MCT as in Royden Chap4, Thm 10.[/b] Let $<f_{n}>$ be an increasing sequence of non-negative measurable functions, and let $f=\lim f_{n}$ a.e. Then $\int f = \lim \int f_{n}$.

a separate question, which may or may not have any link to the above question. Correspondingly, for the decreasing version, what other conditions are required?

For $a<x$, write $\int^{a}_{-\infty}fd\mu=\int^{x}_{-\infty}f\chi_{(-\infty,a]}d\mu$.  This holds for every (countable) sequence of numbers $(x_{n})_{n}\uparrow x$, so...

For the decreasing version you want $f_{n}\in L^{1}$ for some $n$.

lemme try to understand.

blah$^{3}$, you're trying to prove continuity by showing that for every sequence $x_{n}\to x$, $F(x_{n})\to F(x)$. so, for any $x_{n}\to x$, there exists a subsequence either $y_{n}\uparrow x$ or $y_{n}\downarrow x$.

for the first case, use MCT. for the latter, use the decreasing version.

then my next question, how to prove the decreasing version. my guess is to "construct an increasing sequence for the decreasing sequence". but....  :huh:

[quote="blahblahblah"]For the decreasing version you want $f_{n}\in L^{1}$ for some $n$.[/quote]
[quote="mao_zai"]then my next question, how to prove the decreasing version. my guess is to "construct an increasing sequence for the decreasing sequence". but....  :huh:[/quote]
You are right, mao_zai. Suppose $\{f_{k}\}$ is a decreasing sequence on $E$ and without loss of generality let us say $f_{1}\in L^{1}(E)$. Suppose $f_{k}\to f$. Then let $g_{k}= f_{1}-f_{k}$, we have $\{g_{k}\}$ an increasing sequence and you can apply Levi's Theorem. Why we need $f_{1}\in L^{1}$ is that we want to cancel it from both side of an equality.

[quote="mao_zai"]

then my next question, how to prove the decreasing version. my guess is to "construct an increasing sequence for the decreasing sequence". but....  :huh:[/quote]
I am a beginner in learn Royden's real anialysis book :D ,too.So please forgive me if I am wrong.
I don't think the propositive is right.
You can try:
$f_{n}(x)=1,n\le x<\infty;f_{n}(x)=0,othercase$
Then $\int f_{n}=\infty$
But $f_{n}(x)$ is convergent to 0 everywhere,so $f=0$

[quote="mao_zai"][quote]Suppose $f_{k}\to f$. Then let $g_{k}=f_{k}-f$, we have $\{g_{k}\}$ an increasing sequence and you can apply Levi's Theorem.[/quote]

Question: isnt $\{g_{k}\}$ a decreasing sequence too?[/quote]
Sorry, it should be $g_{k}= f_{1}-f_{k}$. I've modified the post.
[quote="zhaobin"]I am a beginner in learn Royden's real anialysis book :D ,too.So please forgive me if I am wrong.
I don't think the propositive is right.
You can try:
$f_{n}(x)=1,n\le x<\infty;f_{n}(x)=0,othercase$
Then $\int f_{n}=\infty$
But $f_{n}(x)$ is convergent to 0 everywhere,so $f=0$[/quote]
As mentioned by blahblahblah, it is critical to have $f_{n}\in L^{1}$ here.


Prove that

a)  $ \frac{m_a^2}{a^2}\plus{}\frac{m_b^2}{b^2}\plus{}\frac{m_c^2}{c^2}\geq\frac{9}{4}$

b)  $ \frac{m_a}{a}\plus{}\frac{m_b}{b}\plus{}\frac{m_c}{c}\geq\frac{3\sqrt{3}}{2}$

[quote="Johan Gunardi"]
\[ \sum\frac {m_a^2}{a^2} \\
\equal{} \sum\frac {2b^2 \plus{} 2c^2 \minus{} a^2}{4a^2} \\
\equal{} \sum\frac {b^2}{a^2} \minus{} \frac14 \\
\ge3 \minus{} \frac34 \\
\equal{} \frac94\]
[/quote]

I think it should be
$ \equal{} \sum_{cyc}{\left(\frac {b^2 \plus{} c^2}{2a^2} \minus{} \frac {1}{4}\right)}$
$ \equal{} \minus{} \frac {3}{4} \plus{} \frac {1}{2}\sum_{sym}\frac {b^2}{a^2}$
$ \ge \minus{} \frac {3}{4} \plus{} \frac {1}{2} \cdot 6$
$ \equal{} \frac {9}{4}.$
...?


inequalities geometry circumcircle incenter trigonometry inradius function

Tried posting this in Pre-Olympiad but thought I'd get more feed back here:
For acute triangle ABC, cevians AD, BE, and CF are concurrent at P.
Prove
$\displaystyle 2\left(\frac{1}{AP}+\frac{1}{BP}+\frac{1}{CP}\right)\leq \frac{1}{PD}+\frac{1}{PE}+\frac{1}{PF}$
and determine when equality holds

equality is given when P is the centroid of ABC. but i think that your inequality is wrong. shouldnt the 2 be multiplyin the right hand side and not the left? actually it has to be or the inequality is wrong???

Yes Marko Avila the centroid is one point of equality but there is another

[quote="amd2357"]For acute triangle ABC, cevians AD, BE, and CF are concurrent at P.
Prove
$\displaystyle 2(\frac{1}{AP}+\frac{1}{BP}+\frac{1}{CP})\leq \frac{1}{PD}+\frac{1}{PE}+\frac{1}{PF}$[/quote]

I'm sorry for telling this to you, but alas, it's fact: An inequality as simple as this one must be either well-known or wrong. This particular one is actually wrong, but there are some funny cases.

For instance, for any acute-angled triangle ABC, if P is the orthocenter or the circumcenter of triangle ABC, then we have

$2(\frac{1}{AP}+\frac{1}{BP}+\frac{1}{CP})\leq \frac{1}{PD}+\frac{1}{PE}+\frac{1}{PF}$,

while for any (not necessarily acute-angled) triangle ABC, if P is the incenter of triangle ABC, we have

$2(\frac{1}{AP}+\frac{1}{BP}+\frac{1}{CP})\geq \frac{1}{PD}+\frac{1}{PE}+\frac{1}{PF}$.

These are nice exercises to prove. ;)

[b]EDIT:[/b] Nobody seems to care for these exercises, so here are my solutions.

We have to prove the following assertions:

[color=blue][b]Assertion 1.[/b] If ABC is an acute-angled triangle with orthocenter H, and if D, E, F are the feet of the altitudes AH, BH, CH of triangle ABC, then

$2(\frac{1}{AH}+\frac{1}{BH}+\frac{1}{CH})\leq \frac{1}{HD}+\frac{1}{HE}+\frac{1}{HF}$.

[b]Assertion 2.[/b] If ABC is an acute-angled triangle with circumcenter O, and if D, E, F are the points where the lines AO, BO, CO meet the sides BC, CA, AB of our triangle, then

$2(\frac{1}{AO}+\frac{1}{BO}+\frac{1}{CO})\leq \frac{1}{OD}+\frac{1}{OE}+\frac{1}{OF}$.

[b]Assertion 3.[/b] If ABC is an arbitrary triangle with incenter I, and if D, E, F are the points where the lines AI, BI, CI (these lines are the angle bisectors of the angles CAB, ABC, BCA) meet the sides BC, CA, AB of our triangle, then

$2(\frac{1}{AI}+\frac{1}{BI}+\frac{1}{CI})\geq \frac{1}{ID}+\frac{1}{IE}+\frac{1}{IF}$.[/color]

[hide="Proof of Assertion 1"][i]Proof of Assertion 1.[/i] Since the triangle ABC is acute-angled, its orthocenter H is an interior point of the triangle ABC. Thus, by the Erdös-Mordell inequality, the sum of the distances from the point H to the vertices of triangle ABC is $\geq$ to the double sum of distances from the point H to the sides of triangle ABC. In other words, $AH+BH+CH\geq 2(HD+HE+HF)$.

Since < AEH = 90° and < BDH = 90°, we have < AEH = < BDH. Also, < AHE = < BHD. Thus, the triangles AHE and BHD are similar. Consequently, AH : HE = BH : HD. In other words, $AH\cdot HD=BH\cdot HE$. Similarly, $BH\cdot HE=CH\cdot HF$. Hence, $AH\cdot HD=BH\cdot HE=CH\cdot HF$. Denote $AH\cdot HD=BH\cdot HE=CH\cdot HF=t$. It is clear that this t is positive. Hence,

$AH+BH+CH\geq 2(HD+HE+HF)$
$\Longleftrightarrow$ $\frac{AH+BH+CH}{t}\geq\frac{2(HD+HE+HF)}{t}$
$\Longleftrightarrow$ $\frac{AH}{t}+\frac{BH}{t}+\frac{CH}{t}\geq 2(\frac{HD}{t}+\frac{HE}{t}+\frac{HF}{t})$
$\Longleftrightarrow$ $\frac{AH}{AH\cdot HD}+\frac{BH}{BH\cdot HE}+\frac{CH}{CH\cdot HF}\geq 2(\frac{HD}{AH\cdot HD}+\frac{HE}{BH\cdot HE}+\frac{HF}{CH\cdot HF})$
$\Longleftrightarrow$ $\frac{1}{HD}+\frac{1}{HE}+\frac{1}{HF}\geq 2(\frac{1}{AH}+\frac{1}{BH}+\frac{1}{CH})$.

Thus, Assertion 1 is proven.[/hide]

[hide="Proof of Assertion 2"][i]Proof of Assertion 2.[/i] Since the triangle ABC is acute-angled, its circumcenter O lies inside the triangle ABC. Also, since O is the circumcenter of triangle ABC, we have AO = BO = CO = R, where R is the circumradius of triangle ABC. Thus, we can equivalently transform the required inequality as follows:

$2(\frac{1}{AO}+\frac{1}{BO}+\frac{1}{CO})\leq \frac{1}{OD}+\frac{1}{OE}+\frac{1}{OF}$
$\Longleftrightarrow$ $2(\frac{1}{R}+\frac{1}{R}+\frac{1}{R})\leq \frac{1}{OD}+\frac{1}{OE}+\frac{1}{OF}$
$\Longleftrightarrow$ $\frac{6}{R}\leq \frac{1}{OD}+\frac{1}{OE}+\frac{1}{OF}$
$\Longleftrightarrow$ $6\leq R(\frac{1}{OD}+\frac{1}{OE}+\frac{1}{OF})$
$\Longleftrightarrow$ $6\leq \frac{R}{OD}+\frac{R}{OE}+\frac{R}{OF}$.

But after the van Aubel theorem (see the references in http://www.mathlinks.ro/Forum/viewtopic.php?t=15211 post #6), applied to the triangle ABC with the cevians AD, BE, CF concurring at the point O, we have $\frac{AO}{OD}=\frac{AE}{EC}+\frac{AF}{FB}$. Thus,

$\frac{R}{OD}=\frac{AO}{OD}=\frac{AE}{EC}+\frac{AF}{FB}=\frac{EA}{CE}+\frac{AF}{FB}$.

Analogously,

$\frac{R}{OE}=\frac{FB}{AF}+\frac{BD}{DC}$;
$\frac{R}{OF}=\frac{DC}{BD}+\frac{CE}{EA}$.

Hence, the inequality in question,

$6\leq \frac{R}{OD}+\frac{R}{OE}+\frac{R}{OF}$,

becomes

$6\leq(\frac{EA}{CE}+\frac{AF}{FB})+(\frac{FB}{AF}+\frac{BD}{DC})+(\frac{DC}{BD}+\frac{CE}{EA})$,

what rewrites as

$2+2+2\leq(\frac{BD}{DC}+\frac{DC}{BD})+(\frac{CE}{EA}+\frac{EA}{CE})+(\frac{AF}{FB}+\frac{FB}{AF})$.

But this trivially follows from the well-known inequality $2\leq\frac{x}{y}+\frac{y}{x}$ for any two positive reals x and y. Thus, Assertion 2 is proven.[/hide]

[hide="Proof of Assertion 3"][i]Proof of Assertion 3.[/i] Since $\measuredangle DAB=\frac{A}{2}$ (angle bisector) and < ABD = B, the sum of the angles in triangle ABD yields

$\measuredangle ADB=180^{\circ}-\measuredangle DAB-\measuredangle ABD=180^{\circ}-\frac{A}{2}-B=90^{\circ}+\frac{180^{\circ}-A}{2}-B$.

But 180° - A = B + C by the sum of angles in triangle ABC; hence,

$\measuredangle ADB=90^{\circ}+\frac{B+C}{2}-B=90^{\circ}-\frac{B-C}{2}$,

and hence $\sin\measuredangle ADB=\cos\frac{B-C}{2}$. Also, trivially, $\measuredangle IAC=\frac{A}{2}$ (angle bisector).

The point I is the center of the incircle of triangle ABC. Let this incircle touch the sides BC, CA, AB of triangle ABC at the points X, Y, Z, respectively. Then, of course, $IX\perp BC$, $IY\perp CA$, $IZ\perp AB$, and IX = IY = IZ = r, where r is the inradius of triangle ABC. In the right-angled triangle IYA, we have $AI=\frac{IY}{\sin\measuredangle IAY}=\frac{IY}{\sin\measuredangle IAC}=\frac{r}{\sin\frac{A}{2}}$, so that $\frac{1}{AI}=\frac{\sin\frac{A}{2}}{r}$, and in the right-angled triangle IXD, we have $ID=\frac{IX}{\sin\measuredangle IDX}=\frac{IX}{\sin\measuredangle ADB}=\frac{r}{\cos\frac{B-C}{2}}$, so that $\frac{1}{ID}=\frac{\cos\frac{B-C}{2}}{r}$. Thus, using the cyclic sum notation, we can transform the inequality in question as follows:

$2(\frac{1}{AI}+\frac{1}{BI}+\frac{1}{CI})\geq \frac{1}{ID}+\frac{1}{IE}+\frac{1}{IF}$
$\Longleftrightarrow$ $2\sum\frac{1}{AI}\geq\sum\frac{1}{ID}$
$\Longleftrightarrow$ $2\sum\frac{\sin\frac{A}{2}}{r}\geq\sum\frac{\cos\frac{B-C}{2}}{r}$
$\Longleftrightarrow$ $2\sum\sin\frac{A}{2}\geq\sum\cos\frac{B-C}{2}$.
$\Longleftrightarrow$ $2\sum\sin\frac{A}{2}-\sum\cos\frac{B-C}{2}\geq 0$
$\Longleftrightarrow$ $\sum(2\sin\frac{A}{2}-\cos\frac{B-C}{2})\geq 0$
$\Longleftrightarrow$ $\sum(\sin\frac{A}{2}(2-\frac{\cos\frac{B-C}{2}}{\sin\frac{A}{2}}))\geq 0$

But by the Mollweide formulas, $\frac{\cos\frac{B-C}{2}}{\sin\frac{A}{2}}=\frac{b+c}{a}$, so that this becomes

$\sum(\sin\frac{A}{2}(2-\frac{b+c}{a}))\geq 0$
$\Longleftrightarrow$ $\sum(\sin\frac{A}{2}(\frac{a-b}{a}-\frac{c-a}{a}))\geq 0$
$\Longleftrightarrow$ $\sum(\frac{a-b}{a}\sin\frac{A}{2})-\sum(\frac{c-a}{a}\sin\frac{A}{2})\geq 0$
$\Longleftrightarrow$ $\sum(\frac{a-b}{a}\sin\frac{A}{2})-\sum(\frac{a-b}{b}\sin\frac{B}{2})\geq 0$
$\Longleftrightarrow$ $\sum((a-b)\frac{\sin\frac{A}{2}}{a})-\sum((a-b)\frac{\sin\frac{B}{2}}{b})\geq 0$.

If R is the circumradius of triangle ABC, then a = 2R sin A by the extended law of sines; thus, $\frac{\sin\frac{A}{2}}{a}=\frac{\sin\frac{A}{2}}{2R\sin A}=\frac{\sin\frac{A}{2}}{2R\cdot 2\sin\frac{A}{2}\cos\frac{A}{2}}=\frac{1}{4R\cos\frac{A}{2}}$. Similarly, $\frac{\sin\frac{B}{2}}{b}=\frac{1}{4R\cos\frac{B}{2}}$. Hence,

$\sum((a-b)\frac{\sin\frac{A}{2}}{a})-\sum((a-b)\frac{\sin\frac{B}{2}}{b})\geq 0$
$\Longleftrightarrow$ $\sum((a-b)\frac{1}{4R\cos\frac{A}{2}})-\sum((a-b)\frac{1}{4R\cos\frac{B}{2}})\geq 0$
$\Longleftrightarrow$ $\sum((a-b)\frac{1}{\cos\frac{A}{2}})-\sum((a-b)\frac{1}{\cos\frac{B}{2}}))\geq 0$
$\Longleftrightarrow$ $\sum((a-b)(\frac{1}{\cos\frac{A}{2}}-\frac{1}{\cos\frac{B}{2}}))\geq 0$.

But this inequality is trivial: If, for instance, $a\geq b$, then $A\geq B$ (since the greater angle in a triangle lies opposite to the greater side), so that $\frac{A}{2}\geq\frac{B}{2}$; since the cosine function is decreasing, this yields $\cos\frac{A}{2}\leq\cos\frac{B}{2}$, and thus $\frac{1}{\cos\frac{A}{2}}\geq\frac{1}{\cos\frac{B}{2}}$, so that $\frac{1}{\cos\frac{A}{2}})-\frac{1}{\cos\frac{B}{2}}\geq 0$. On the other hand, $a\geq b$ yields $a-b\geq 0$. Thus, $(a-b)(\frac{1}{\cos\frac{A}{2}}-\frac{1}{\cos\frac{B}{2}})\geq 0$. Hereby, we were working in the case $a\geq b$. But similarly, in the case $a\leq b$, we also get $(a-b)(\frac{1}{\cos\frac{A}{2}}-\frac{1}{\cos\frac{B}{2}})\geq 0$. Thus, the inequality $(a-b)(\frac{1}{\cos\frac{A}{2}}-\frac{1}{\cos\frac{B}{2}})\geq 0$ holds in every case, so that we have

$\sum((a-b)(\frac{1}{\cos\frac{A}{2}}-\frac{1}{\cos\frac{B}{2}}))\geq 0$.

And Assertion 3 is proven.[/hide]

This solves the problem.

  darij


Find a formula for the area of a quadrilateral triangle depending on $ a,b,c,d,\alpha,\beta,\gamma,\delta$! You may use the lefschetz fixed point formula.


How many five-digit zip codes are possible if the first three digits each exceed 5?

The last five digits each has to be either 6, 7, 8, or 9.

Using these four digits 5 times, there are $ 4^5 \equal{} \boxed{1024}$ ways to do this.

The restriction only applies to the first three digits, so it would be $4^3\times10^2=6400$.


Is there a natural integer $N$ that is written :
$abcca$ in base 5;
$bbab$ in base 8 ?

If such a number exists, write it in base 10.

:)

[hide="Explanation of the Above Answer"]
\[ 625a+125b+25c+5c+a=512b+64b+8a+b\implies 618a-452b+30c=0.  \]

It is clear that $a=2$, $b=3$, and $c=4$ so we have $n=625\cdot 2+125\cdot 3+25\cdot 4+5\cdot 4+2=\boxed{1747}$.
[/hide]


Let $ m, n \in \mathbb{Z}$, and let $ k \equal{} (4n \minus{}1)^3 \minus{} 4m^2$. Suppose there is no prime of the form $ p \equal{} 4t \plus{} 1$ dividing $ m$. Show that $ y^2 \equal{} x^3 \plus{} k$ has no integer solutions.

Can someone show a proof by considering $ \mathbb{Z}[\sqrt{\minus{}k}]$?

The proof I know runs as follows:

1. Suppose (x, y) is an integer solution to $ y^2 \equal{} x^3 \plus{} k$

2. Then reduction mod 4 implies that x is odd so that y is even

3. $ x \equiv \minus{}1 \mod 4 \implies y^2 \equiv 2 \mod 4$, which is not possible so that $ x \equiv 1 \mod 4$

4. A prime $ p \equiv \minus{}1 \mod 4$ must divide the quantity: $ x^2 \minus{} xb \plus{} b^2$, where $ b \equal{} 4n \minus{}1$.

Two more lines are deduced from step 4 and one hits the wall - a contradiction is reached.


Could somebody explalin the difference between IMO short list and long list and the actuall test?

Thanks a lot!

You may want to look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=61819]here.[/url]

[quote="José"]Sorry orl, but when I try to go to the page, it say: The topic or post you requested does not exist[/quote]

There is no problem with the link. You only have to make sure that you are logged into either ML or AoPS.

I don't know why the link does not work. I tried it from different IP addresses and different OS's (Windows XP, Linux) though it should not matter. Anyway here it goes:

[quote="orl"][quote="Silverfalcon"]What is a IMO Shortlist? I just wondered.  :D  :P[/quote]

Before the IMO takes place the IMO Problem Selection Committee chooses a subset of the proposed problems of all countries (IMO LongList, ILL). This subset contains 20-30 problems and is called IMO ShortList (ISL). This ISL is used to select the 6 IMO problems.[/quote]

You also might have a look at the [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=110248#110248]Introduction to the IMO ShortList/LongList project.[/url]

[quote="orl"][quote="climskywell"]apparently I can't reach that site[/quote]

Apparently you must click on the link to have a look at this page. :P[/quote]
Apparently you didn't realize that only moderators can access the forum called the "Moderator's Lounge"  :P


calculus
integration
induction
real analysis
real analysis unsolved

prove, using integral calculus [b]alone[/b], that for all $ a_{k}\in\{\;\mathbb R\;|\;\mathbb C\;\}$, :D

$ \sum_{k\equal{}1}^{\infty}\;\;\frac{a_{1}\cdot a_{2}\cdot a_{3}\cdot a_{4}\cdot\;\ldots\;\cdot a_{k\minus{}1}}{(x\plus{}a_{1})\cdot (x\plus{}a_{2})\cdot (x\plus{}a_{3})\cdot (x\plus{}a_{4})\cdot\;\ldots\;\cdot (x\plus{}a_{k})}\;\;\equal{}\;\;\boxed{\frac{1}{x}}$

[quote="misan"]prove, using integral calculus [b]alone[/b], that for all $ a_{k}\in\{\;\mathbb R\;|\;\mathbb C\;\}$, :D

$ \sum_{k \equal{} 1}^{\infty}\;\;\frac{a_{1}\cdot a_{2}\cdot a_{3}\cdot a_{4}\cdot\;\ldots\;\cdot a_{k\minus{}1}}{(x\plus{}a_{1})\cdot (x\plus{}a_{2})\cdot (x\plus{}a_{3})\cdot (x\plus{}a_{4})\cdot\;\ldots\;\cdot (x\plus{}a_{k})}\;\; \equal{}\;\;\boxed{\frac{1}{x}}$[/quote]

Is this true? Take $ a_{k}\equal{}0$ for all possible $ k$, God assume that $ 0\neq\frac{1}{x}$, uh?  :D

OK, sorry i forgot to mention you cannot have $ a_{k}\equal{} 0$ or $ a_{k}\equal{} 0\plus{}\imath\cdot 0$ :D

by the way, the question i asked is something one needs to establish (amongst several other things) when trying to understand the [i]original [/i]proof of the irrationality of $ \zeta{(3)}$ :|

Let $ P_{n}(x) \equal{}\prod_{k \equal{} 1}^{n}\left( x\plus{}a_{k}\right)$.

By induction or whatever we have that
\[ \sum_{k\equal{}1}^{n}\frac{a_{1}\cdot a_{2}\cdot a_{3}\cdot a_{4}\cdot\ldots\cdot a_{k\minus{}1}}{(x\plus{}a_{1})\cdot (x\plus{}a_{2})\cdot (x\plus{}a_{3})\cdot (x\plus{}a_{4})\cdot\;\ldots\;\cdot (x\plus{}a_{k})}\equal{}\frac{\frac{P_{n}(x)\minus{}a_{1}\ldots a_{n}}{x}}{P_{n}(x)}.\]
That equals $ \frac{1}{x}\minus{}\frac{1}{x}\cdot\prod_{k \equal{} 1}^{n}\frac{a_{k}}{x\plus{}a_{k}}$. If the $ a_{k}$'s and $ x$ are $ > 0$, then the product goes to zero iff $ \sum_{k \equal{} 1}^{\infty}\frac{x}{x\plus{}a_{k}}$ diverges. We can find $ a_{k}$'s for which that converges and $ a_{k}$'s for which that diverges, so your result isn't quite true I'm afraid :maybe:

Did I do anything wrong?

[quote="misan"]$ a_{k}\in\{\;\mathbb R\;|\;\mathbb C\;\}$, :D [/quote]
I'm not entirely sure what this notation means; do you mean $ \mathbb{C}\minus{}\mathbb{R}$?

let $ a_{j}\equal{}a_{i}\equal{}a$ for all $ j$ and $ i$
then we have
$ \sum(\frac{a}{x\plus{}a})^{k}\equal{}\frac{a}{x}$
the left side can diverge if
$ |x\plus{}a|<a$
therefore your problem is wrong

I think misan is using the formal language's notation $ \{a\mid b\}$ mean $ a$ or $ b$. So $ x\in\{\mathbb{R}\mid\mathbb{C}\}$ is really $ x\in\mathbb{R}$ or $ x\in\mathbb{C}$.

I find that hard to believe considering $ \mathbb{R}\subset\mathbb{C}$, so the notation is redundant if interpreted that way.  :|

Anyway, perfect_radio and Extremal make good and troublesome points.  Could the OP clarify the problem?


Greetings All Again (thought I posted my message but it didn't show up, so writing it again).
Could someone please help me understand the financial formula depicted on the image below
http://democourses.ifse.ca/media/images/mathsq16c.gif     ?
I don't understand how one leads to another i.e. the first (most left) equation to the result.
Thanks in advance

Let $1/(1+r) = v$
Present Value of the first Payment = $PMT*v$
Present value of the second pmt  =   $PMT*v^2$
$. . . $
$. . . $

$. . . $
Present value of the nth Payment = $PMT*v^n$

Adding all these we get $PV = PMT*(v + v^2 + v^3 + ... + v^n)$

 Therefore $PV = PMT*(v - v^{n+1})/(1-v)$
           or  $PV = PMT*(1 -  v^n)/r$

            or $PV = PMT*(1 - 1/(1+r)^n)/r$

[quote="salubrious"]Thanks for you reply. I do realize it may be a silly question, but could you please  kindly explain how
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/bae1ff856a31067b92ef206f2139c406.gif[/img]
results in either
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/df6210c014ddde6995845ff0b9e1687c.gif[/img]
or
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/f85b12d6c457b4400de692aede532b75.gif[/img]
?[/quote]

You can type latex directly into the forum and it will be rendered. You don't need to copy and paste images. Just put dollar signs around the formula.

$S=v+v^2+v^3+v^4+...+v^n$

$\frac Sv=1+v+v^2+v^3+v^{n-1}$

$\frac Sv-S=1-v^n$

$S\left(\frac 1v-1\right)=1-v^n$

$S=\frac{v-v^{n+1}}{1-v}$

$S=\frac{1-v^n}{\frac{1-v}{v}}$

$S=\frac{1-v^n}{r}$

[quote="theone853"]possibly he knows no latex because he seems to be a new member[/quote]

How'd he get the formulas if he doesn't know LaTeX?

[quote="theone853"]you said it yourself, he copy and pasted images[/quote]

..images of the formulas written in $\LaTeX$ :) He would probably have compiled them on his computer then taken a picture of the formulae.

actually if you look at the properties of the forumlas he posted and the ones above his posts that looked exactly the same, they have the same link...

Thanks, guys. Appreciate your help.
That's true that I didn't know how to use $Latex$ and just copied & pasted the already generated links to the images

(To get the $\LaTeX$ symbol, type \LaTeX between the dollar signs.)

[quote="salubrious"]Hmm... In my last message, I surrounded word "Latex" by $, but nothing happened  :?[/quote]
Apparently, it wasn't visible in Opera which I generaly use; however, in IE it showed up immediately. Weird, eh  :)


Euler
calculus
function
search
number theory
totient function
relatively prime

The number $e$ is almost certainly not named after Euler. Euler himself named the number, and he was too modest to have named it after himself. Instead, it is believed that he named it after the word ``exponential.'' May I suggest to you [url=http://www.eulerarchive.org/]The Euler Archive[/url]?

Yup $e$ was not Eulers number  it was "the exponent number".  :P

[quote="towbomb"]Well, he has his name on several important things:

-$e$, (a.k.a. the euler constant or euler's number, ~2.718)
-The Euler function (a.k.a. the Totient function), the number of integers less than or equal to n that are relatively prime to n.

This is all I can think of off the top of my head.  Two useful web sites:
[url]http://scienceworld.wolfram.com/biography/Euler.html[/url] for a complete math-related bio, or
[url]http://mathworld.wolfram.com/letters/E.html[/url], where there are no fewer than 90 topics that start with his name.
If all else fails, go to [url]http://mathworld.wolfram.com/search/[/url] and search for "euler" for a complete list of topics that have anything remotely to do with Euler.[/quote]

wolfram.com???

Isn't that latin for "tungsten"?

I wonder why...

The sites were good though.

the "euler" function? Isn't that phi of n?

I m pretty sure wolfram is german.  But your right it does mean tungsten and hence W is tungsten's chemical symbol.

It's also the last name of Stephen Wolfram, the creator of Mathematica and sponsor of the Mathworld website.

Tungsten (wolfram W) is from wolframite, which stops tin (Sn) from reacting.

V+F-E=2 and e^(i*x)=cos x+ i*sin x are two more of Euler's.

i'll tell u some things later but why do u need to know about him?

[quote="pakagawa"]Can anyone tell me what Euler did?

Include:

Important theorems
Things
Events
Politics
Calculus?[/quote]
Theorems:
fermats little theorem
Fermats last for case 3
the proof that there is only 1 integer which a^2+2=b^3
the sum of the reciprocols of primes is divergent
found that 2^32+1 wasnt prime thus disproving one of fermat's theorems
proof that god exists(bogus, but Diderot was no great mathematician)
calculated orbit of Uranus
Bridges of Koinsberg

Events
Euler and Diderot in the court of Russia
half blind at 22?
blind at 66?
Replaced by Joseph Louis Lagrange in Germany

Politics: no clue
Calculus: little notable work done there

[quote="pakagawa"]Can anyone tell me what Euler did?

Include:

Important theorems
Things
Events
Politics
Calculus?[/quote]

Hi pakagawa, Everyone,

A nice place to look for information on famous mathematicians is the The MacTutor History of Mathematics archive (  [url]http://www-groups.dcs.st-and.ac.uk/~history/index.html[/url] ).  It is a wonderful resource.  Here's the link to Euler's biography ( [url]http://www-groups.dcs.st-and.ac.uk/~history/Mathematicians/Euler.html[/url] ).  Here's another short account of Euler's life: ( [url]http://www.maths.tcd.ie/pub/HistMath/People/Euler/RouseBall/RB_Euler.html[/url]).   I hope the links help!  :)

Yours truly,

M.E.B. (formally mathfreak)

[quote="Scrambled"]Calculus: little notable work done there[/quote]

What?  That just isn't true.  He was known as analysis incarnate for a reason!


It is not very inventive :)
We have $\cos^{2}x=\frac{\cos 2x+1}{2}$, $\cos^{4}x=\frac{\frac{\cos 4x+1}{2}+2\cos 2x+1}{4}=\frac{3}{8}+\frac{1}{2}\cos 2x+\frac{1}{8}\cos 4x$. Using now formulas for sum of cos we conclude $\frac{A_{n}}{n}\to \frac{3}{8}$.


What would I have gotten at your state competition? I got a 40, fifth in IN.  I think I would have made it in every other state.  List your state and what rank I would have got.  If your state was 40 40 39 38, list me as 1st.

Honestly, will you stop this? This is the second post like this you've set up. We all know that you --should-- have made it to nationals.

nat mc,

It is ok to be disappointed, but use that feeling to work harder for what ever you are going to do next.  You clearly are very bright and had the misfortune of being in one of the hardest states.  But you also had the great honor of working with quite possibly the best coach out their in Mr. Fischer!!  There is no way to tell how you would have done at Nats.  Our top kid had 30 at state, well below anyone from your states scores, but he was 31st at Nationals, ahead of one of your kids and only 4 or 5 points behing a couple of your top 10 kids.

Just be happy you have received an awesome education and can continue on with the AMC10/12, AIME, etc.......

best of luck, it was very nice meeting you and your whole team at Nationals!

Josh Frost

I know exactly how you feel; first of all, making a bunch of dumb mistakes on the nationals last year and only getting 42nd, when I think I was capable of at least tying with my third place teammate, who was 25th. Second, getting a 3 on the AIME this year. -_- At least I don't think I've lost my only chance of qualifying for the USAMO, because I'm expecting a 150 or not far off on the AMC10 next year (134.5 this year, and my first time ever taking an AMC), and maybe an 8 on the AIME to qualify. :) So, look for pluses like that.

[quote="biffanddoc"][quote="Xantos C. Guin"]GA
34, 34, 33, 31

You would have creamed us.
If I were in DE I would be at nats all 3 years.[/quote]No last year the scores were 45,44,40,39[/quote]

Even I would've made nats from DE last year...

[quote="biffanddoc"][quote="Xantos C. Guin"]GA
34, 34, 33, 31

You would have creamed us.
If I were in DE I would be at nats all 3 years.[/quote]No last year the scores were 45,44,40,39[/quote]
nvm

[quote="biffanddoc"][quote="Xantos C. Guin"]GA
34, 34, 33, 31

You would have creamed us.
If I were in DE I would be at nats all 3 years.[/quote]No last year the scores were 45,44,40,39[/quote]

Wow, what position did they get.  IN got third

26th :D the winner guy Danny Gao got 24th and the rest were under 100

[quote="nat mc"]sorry  :blush: mods please delete this topic
(please don't be so negative to me  :? )[/quote]

Sorry, dude, I didn't mean to bark at you. You deserve nationals more than most people who even went, so I can pretty much understand how you feel. No hard feelings! :)

a kid from my school got 4th in cd, but 5th in written in state, and he missed it by one point  :?  :(

anyway, ny's scores were: 44, 39, 36, 35...

-rorri :?

[quote="bigHC"]phil, how did u do on the national test that was mailed to u?[/quote]

I haven't gotten it yet.  Mr Fish is sneding it in the mail, not email

[quote="bigHC"]ok... be sure to tell me how u did[/quote]
I took it today and got a 33 (19/7) and I know what your response will be, yes I wasn't as nervous as you.

[quote="kpavery"]at the coundown round in our state, south CA, this one guy went frm 10th to 4th

alex was the only one from so. cal. to go to nats.[/quote]

The guy went from 10th to 2nd. I remember because I beat him.

he was that short asian guy?

too bad i didn't get to go against you in  the chapter countdown,alex, i was close

In Utah:
30,30...Well, you get the picture....

You would have gotten 5th in IN (obviously :P )

2nd, but you'd probably lose to Ian, dropping you down to 3rd.


For nonnegative real numbers x,y,z, which satisfy a+b+c=1. Prove that:
xyz(\frac{y}{x}+\frac{x}{y}+1)(\frac{x}{z}+\frac{z}{x}+1)(\frac{y}{z}+\frac{z}{y}+1)\ge 1.

xyz(y/x+x/y+1)(x/z+z/x+1)(y/z}+z/y+1)>= 1.

[quote="worderdos"]For nonnegative real numbers $x,y,z$, which satisfy $a+b+c=1$. Prove that:
$xyz(\frac{y}{x}+\frac{x}{y}+1)(\frac{x}{z}+\frac{z}{x}+1)(\frac{y}{z}+\frac{z}{y}+1)\ge 1$.

${xyz(y/x+x/y+1)(x/z+z/x+1)(y/z}+z/y+1)\geq  1$.[/quote]


integration
calculus
symmetry
calculus computations

Is this right? I was thinking about p-series, and I came to this conclusions.

$\int_{a}^{\infty}\frac{1}{x^{p}}\ dx, p\ge 1$ and $a \ge 0$ converges?

Then as corollary, for $a < 0$, and $-\infty$ as lower and upper bounds of integration respectively, would converge from symmetry for odd powers of p.

Also,

$\int_{0}^{b}x^{n}\ dx, n< 1$ is divergent?

Are these conclusions right?

Not quite.
$\int_{a}^{\infty}\frac{1}{x^{p}}\ dx, p> 1$ and $a > 0$ converges.
$\int_{0}^{b}x^{n}\ dx$, $n\le-1$ is divergent.

Oh I see, for the second one I was thinking of integrating in terms of differentiating! Okay, my bad.


AMC
USA(J)MO
USAMO
AIME
AMC 12
AMC 10
AIME II

[quote="machack"]1. Do they have different indexes for AIME 1 and 2?

2. Are they still gonna count AMC 10 and 12 the same when it comes to calculating the index for underclassmen?[/quote]

not usually for number 1 unless its way different.
I believe they have only used different indexes once in the history of AIME I/AIME  II (go check wikipedia, i might not be right).  But like last year, I thought (and so did many people) that AIME II was easier than AIME I.  That might have also been reflected in the scores/porportions but since it wasnt that big of a difference, I dont think the AMC would make a who new index.

2. yes...its on the AMC website and its been there for the last year, so you cant really say its totally unfair.

[quote="machack"]
So if i get 109 amc12 - 8 aime then i would lose to some guy who gets 120 amc10 - 7 aime   <------ this is totally unfair
[/quote]

Not if the floor score was 8.

Also: I am in 8th grade and took the AMC12, even though I knew I would definitely do better on the AMC10, knowing that they will be counted the same. I find that the first most important thing about the AMCs isn't just qualifying for USAMO (even though many other people would think differently).

Sorry, i don't mean to sound pushy, but about when are the floor scores and indices announced?

is the floor score going to be the same for both aime i and ii?

On the subject of USAMO qualification, in 2005, the index cutoffs were 233 and 220.5 for the AIMEs, and the floor was 9. Does anyone know if anyone scored in the 220.5+ range with an AIME of 8 on the AIME II and they decided to use the floor from AIME I or if there was actually nobody in that range with an AIME of 8?

[quote="Kent Merryfield"]See DPatrick's second post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=197924]here.[/url]

The answer to nearly all of the questions posed in this topic is, "None of us knows. We'll know the answer when everyone knows."[/quote]

i mean that is it a rule that they have to be the same.

[quote="machack"]it is unfair.......
My school only took Amc B tests and i took 12 because i didn't realize that it would matter
So if i get 109 amc12 - 8 aime then i would lose to some guy who gets 120 amc10 - 7 aime   <------ this is totally unfair

Also its like those who took amc10 and got the aime floor are almost [b]guaranteed[/b] to advance, while people who took the harder amc12 are [b]not[/b][/quote]

ok..thats why i said it isnt totally unfair...meaning i still consider it unfair, but had you visited the [b]AMC website[/b] (you should always read the rules, infact i had posted a topic about this a couple of weeks before the A tests asking and the AMCdirector referred me to the rules) or read any of the posts on this forum from last years USAMO cutoff, then you wouldve known and wouldve taken the 10 instead of the 12.  and you, yourself, even posted this link to ask about the question
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=135871 from last year

But im sure an 8 will get in if the floor is set an 8, so for you, you want to hope for an 8 as the floor instead of a 7.  189 might make it in with a 7 if the AIME II was way harder than the AIME I.

(besides how did you get a 109 on the AMC 12?)


logarithms
function
algebra
polynomial
calculus
calculus computations

At how many points do $ y=x^{12}$ and $ y=2^{x}$ intersect? And if you didn't have a graphics calculator how would you deduce this?
I thought it would be two simply because I thought that $ x^{12}$ looks simialr to $ x^{2}$ and $ 2^{x}$ is probably similar to $ e^{x}$ and concluded two intersections but it was wrong. The answer is three. How do you find this out?
thanks in advance

Since $ 2^{x}-x^{12}$ is strictly increasing for $ x<0$ (and changes its sign), there is exactly one intersection over the negative half-line. For positive $ x$ we can write $ \log_{2}x=\frac{x}{12}$ and observe that $ \log_{2}x$ is a strictly concave function. The graph of such a function meets any line at no more than two points. The final piece of the puzzle: $ \log_{2}2=1>\frac{2}{12}$.

mlok should also have mentioned that for large enough $ x,$ $ \log_{2}x<\frac{x}{12}.$ For instance, let $ x=128.$ Clearly, $ 7<\frac{128}{12}.$

Or what is essentially the same thing, for large enough $ x,$ $ 2^{x}>x^{12}.$

This is basic knowledge about rates of growth: any exponential ultimately grows faster than any polynomial (fixed power of $ x,$) and any power of $ x$ ultimately grows faster than a logarithm. My guess is that that's what sit.think.solve wasn't seeing - that the graph of $ y=2^{x}$ does ultimately catch back up to and pass the graph of $ y=x^{12}$ for large enough $ x.$

Yes, I should have said something about growth at positive infinity and probably at zero too.

It was a mistake of the original poster to appeal to "similarity" of graphs. Such an approach has its uses, but not when you count intersections, because even a slight perturbation of one of two functions can change the number of intersections.

In fact, $ x^{2}$ and $ e^{x}$ intersect only once.


geometry
geometric transformation
reflection
linear algebra
linear algebra unsolved

Let T: $ {R^2}\minus{}>{R^2}$ a reflection through $ y\equal{}3x$. Find $ T(x,y)$.

The reflection fixes that line, and reverses a perpendicular line. That gives you two eigenvectors and the eigenvalues that go with them- do you see how to finish?

$ T(1,3)\longrightarrow (1,3)$, $ T(3,\ \minus{} 1)\longrightarrow ( \minus{} 3,\ 1)$.


Solve in $N$ the following equation:

         $1+x+x^2+...+x^n=y^n$ ;)

Of course! What about $1+y+...+y^p=x^n$, Cezar? ;)  :P

Yes,Allek you are right. ;)  Harazi in your problem $p$ is a prime number? :?

No, it is arbitrary. Anyway, I will give you a hint: don't do this at home, it's really dangerous. As far as I know, it has been very recently proved by Maurice Mignotte and Yann Bugeaud that it has a finite number of solutions. I posted it just to see the difference between a trivial and a very hard problem. ;)

Do you know how Maurice Mignote did solve the equation you posted? ;)

 I bet it has something to do with some Galois theory,isn't it? ;)

Oh, great one, don't tell me now you are going to enter the ''easy'' field of transcendental number theory. You can bet on Galois theory, but I may tell you it's not that.  :P  ;)


inequalities
geometry
circumcircle
inradius
trigonometry
geometry proposed

Let $ ABC$ be an acute triangle with circumradius $ R$ . Consider the projections $ D\in BC$ , $ E\in CA$ , $ F\in AB$ of the orthocentre

$ H$ on the sides of $ \triangle ABC$ and denote the inradius $ \rho$ of the triangle $ DEF$ . Prove that $ \boxed {\frac{\rho}{R}\ \le\ \cos A(1\minus{}\cos A)}\ \le\ \frac 14$ .

We have $ \rho \equal{} 2R\cos A\cos B\cos C > 0$, because $ ABC$ is an acute triangle.

$ \Longrightarrow \frac {\rho}{R}\le \cos A(1 \minus{} \cos A)\ \Longleftrightarrow\ 2\cos A\cos B\cos C\le \cos A(1 \minus{} \cos A)$

$ \Longleftrightarrow 2\cos B\cos C\le 1 \minus{} \cos A\ \Longleftrightarrow\ \cos(B \plus{} C) \plus{} \cos(B \minus{} C)\le 1 \minus{} \cos A$

$ \Longleftrightarrow \cos(B \minus{} C)\le 1$, which is obvious. Equality holds for $ \angle B \equal{} \angle C$.

Now, $ \cos A(1 \minus{} \cos A)\le \frac 14\ \Longleftrightarrow\ \cos A \minus{} \cos^2 A \minus{} \frac 14\le 0$

$ \Longleftrightarrow\ 4\cos^2 A \minus{} 4\cos A \plus{} 1\ge 0\ \Longleftrightarrow\ (2\cos A \minus{} 1)^2\ge 0$, which is true.


A secretary types 3 letters and addresses 3 envelopes (to 3 different locations).  Before putting each letter into the appropriate envelope, she drops all the letters and envelopes on the floor.  If she picks them up and inserts each letter randomly into an envelope, what is the probability that each letter will be inserted into its correct envelope?

Could someone try to explain this?  I am a little bit stumped about the best way to find the sample space.

[hide="Answer"]$ \frac{1}{3}$[/hide]

[hide="Solution"]There's obviously only $ 1$ way to put them in correctly. There are $ 3!\equal{}6$ ways to put the letters in. That makes $ 1/6$ probability.

I think your answer is incorrect.[/hide]

[quote="lowtopology"]A secretary types 3 letters and addresses 3 envelopes (to 3 different locations).  Before putting each letter into the appropriate envelope, she drops all the letters and envelopes on the floor.  If she picks them up and inserts each letter randomly into an envelope, what is the probability that each letter will be inserted into its correct envelope?

Could someone try to explain this?  I am a little bit stumped about the best way to find the sample space.

[hide="Answer"]$ \frac {1}{3}$[/hide][/quote]

Your answer could be correct only if two of the letters are the exact same. Otherwise, with 3 letters and 3 locations, there are $ 3!\equal{}6$ ways to arrange them randomly.

Another way to look at it is that given an arbitrary letter is picked up, the probability that the right envelope is picked up is 1/3. For the second pair, it is 1/2. THe third one is guaranteed. Thus, the probability is 1/6.

Thanks for the insights.  I got this problem from a book we were using in math club.  The book said that the answer is $ \frac{1}{3}$ and we thought that was a typo, but weren't entirely sure.  The abundance methods for getting $ \frac{1}{6}$ make me more confident of this.  (Of course, probability is rather difficult to pin down axiomatically, but we have numerous examples of these methods holding.)


um I don't think this is MATHCOUNTS, but since this forum is my favorite forum, I'll post it here.

Using 2 2s, (like 2+2), and any defined functions that you didn't make up and any symbols, make 5. Oh yeah, and $\ 0$ counts as a number too. (.02, .000002, etc).

NOTE: there's probably more than 1 solution, isn't that hard. I got it in 25 minutes without any clues. and I know that lots of ppl here can surpass me in math. :D

There is already a topic on this

http://artofproblemsolving.com/Forum/viewtopic.php?t=48329

 :D Ignore the spam.

ceiling function(ln(22pi))

._.

EDIT: oh i get it

[hide="answer"]
2/.2
[/hide]

WOW the other topic is like this pile of flaming cursing etc  everyone's all ganging up on flamesofdestiny...lol...well if u look up flamesofdestiny's approved solution, that was the one I was looking for.  :)

Oh, and I forgot to put this, but Treething's answer comes out to being 10, so it is wrong.

sry for doubleposting, and this might be considered spam, but I think flamesofdestiny got this problem at the same place I did...does anyone know his name?

:rotfl:  :rotfl: just look up middle schoolers with iqs over 150 who have won lots of national science competitions and won national math field day (you have to read other topic to get) :rotfl:

wait...that might be David...not sure
mainly because his sister goes to Exeter, and Tianka Liu (xcuse the spelling) is? or was at exeter...and he did win national math competitions... and David won some national stuff to...think he was exaggerating a lot though lol


geometry
geometric transformation
reflection
LaTeX
geometry unsolved

$ D$ is in $ \triangle ABC$ where $ A < \frac {\pi}{4}$. $ \angle BDC \equal{} 4\angle BAC$ and $ BD \equal{} CD$. $ E$ is the reflection of $ C$ in $ AB$ and $ F$ is the reflection of $ B$ in $ AC$. Prove that $ AD$ is perpendicular to $ EF$.

[hide]DE^2=BE^2+BD^2-2BDBEcosDBE
DF^2=CF^2+CD^2-2CFCDcosDCF

so DE^2-DF^2=2BCBD(cosDBE-cosDCF)=(2*2RsinA*RsinA/sin2A)*(cos(270-2C)-cos(270-2B))=2R^2*tgA*(sin2B-sin2C)=2R^s*tgA*2*sin(B+C)*cos(B-C)
=4R^2*tgA*cos(B+C)*sin(B-C)=4R^2*sinA*sin(B-C)=4R^2*sin(B+C)*sin(B-C)=AE^2-AF^2

so AD perpendicular to EF[/hide]

OK, I think someone should post a more beautiful solution than this, also, can anyone tell me what do I need to do to use Latex( I even failed download it) . :oops:

[quote="Little Gauss"]This is my another solution.

Define $ G$ such that $ ABC \sim DBG$. ( $ G$ and $ C$ is on same side wrt $ BD$)
Then it is easy to prove that $ DGC\sim AEF$.
Because $ ABD\sim CBG$, $ \measuredangle (AD, CG) \equal{} \measuredangle (BA,BC) \equal{} \measuredangle ABC$.
$ \measuredangle (CG, FE) \equal{} \measuredangle (CD, FA) \equal{} \measuredangle (CD,CA) \plus{} \measuredangle (CA,FA) \equal{} \angle DCA \minus{} \angle CAF \equal{} \angle DCA \minus{} \angle BAC$
Therfore
$ \measuredangle (AD, FE) \equal{} \angle B \plus{} \angle DCA \minus{} \angle A \equal{} \angle B \plus{} \angle C \minus{} (90 \minus{} 2\angle A) \minus{} \angle A \equal{} 90$.[/quote]

Well, also a nice one! Especially constructing $ BDG$ and $ CDG$ similar to $ BAC$ and $ FAE$ respectively.

[quote="greentreeroad"]Anyone help me with the detail of download Latex? I'm poor in english so i confused.[/quote]

Have you ever entered

http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

and

http://www.artofproblemsolving.com/Forum/index.php?f=123

for help?


What is the number of square centimeters in the area of triangle DEF if angle D is a
right angle, DE = 36 cm, and EF = 39 cm?

[hide]Since the leg and the hypotenuse are 36 and 39, the other side must be 15. the area is then 0.5(15)(18)=270. [/hide]


function
floor function
real analysis
real analysis unsolved

Develop a period function f(x)=[x+1/2] into Fourier series.

my teacher told me that the function isn't periodical but i can develop it into a difference of two periodical function.I'll be very happy if you can gve me an idea how to express that function into a difference of two periodical function.

He give me an idea that i should make [x+1/2]=[2x]...

The difference of two periodic functions is still periodic, if the original periods are commensurable, and even with incommensurable periods, you still wouldn't get behavior like $ \lfloor x\plus{}1/2\rfloor.$

It makes no sense to ask about the Fourier series of the integer part of a linear expression. It would make sense to ask about the Fourier series of the fractional part of a linear expression, which would be a "sawtooth wave."

Yes,i undersand what do you say and appreciate your help.This problem was from my Analysis exam(wich was today). I didn't solve that problem and when i speaked with my teacher ...he told me that $ f(x) \equal{} [x \plus{} \frac {1}{2}]$ isnt' periodic but somehow we can express it to be periodical.Now i failed my examen because of this problem  :(

if it was a fractional linear expression it was pretty simple.


Find all functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that
\[ f(x^3\plus{}y^3)\equal{}xf(x^2)\plus{}yf(y^2)\]
for all $ x,y\in\mathbb{R}$.

[quote="wangsacl"]Find all functions $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that
\[ f(x^3 \plus{} y^3) \equal{} xf(x^2) \plus{} yf(y^2)\]
for all $ x,y\in\mathbb{R}$[/quote]

Let $ P(x,y)$ be the assertion $ f(x^3\plus{}y^3)\equal{}xf(x^2)\plus{}yf(y^2)$
Let $ f(1)\equal{}a$

$ P(x,0)$ $ \implies$ $ f(x^3)\equal{}xf(x^2)$ and so $ f(x^3\plus{}y^3)\equal{}f(x^3)\plus{}f(y^3)$ and so $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and so $ f(px)\equal{}pf(x)$ $ \forall p\in\mathbb Q$

$ P(x\plus{}1,0)$ $ \implies$ $ f(x^3\plus{}3x^2\plus{}3x\plus{}1)\equal{}(x\plus{}1)f(x^2\plus{}2x\plus{}1)$ and so new assertion $ Q(x)$ : $ 2f(x^2)\equal{}(2x\minus{}1)f(x)\plus{}ax$

$ Q(x\plus{}1)$ $ \implies$ $ 2f(x^2\plus{}2x\plus{}1)\equal{}(2x\plus{}1)f(x\plus{}1)\plus{}a(x\plus{}1)$ and so $ 2f(x^2)\equal{}(2x\minus{}3)f(x)\plus{}3ax$

Subtracting $ Q(x)$ from $ Q(x\plus{}1)$, we get $ 0\equal{}\minus{}2f(x)\plus{}2ax$ and so $ f(x)\equal{}ax$ and it's immediate to check back that this indeed is a solution.

Hence the unique solution $ \boxed{f(x)\equal{}ax}$

[quote="wangsacl"][quote="pco"]

$ P(x \plus{} 1,0)$ $ \implies$ $ f(x^3 \plus{} 3x^2 \plus{} 3x \plus{} 1) \equal{} (x \plus{} 1)f(x^2 \plus{} 2x \plus{} 1)$ and so new assertion $ Q(x)$ : $ 2f(x^2) \equal{} (2x \minus{} 1)f(x) \plus{} ax$
[/quote]
I don't get this part  :blush:[/quote]

$ P(x \plus{} 1,0)$ $ \implies$ $ f(x^3 \plus{} 3x^2 \plus{} 3x \plus{} 1) \equal{} (x \plus{} 1)f(x^2 \plus{} 2x \plus{} 1)$

But we previously got that $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and $ f(px)\equal{}pf(x)$ with $ p\in\mathbb Q$, so :

$ f(x^3 \plus{} 3x^2 \plus{} 3x \plus{} 1)\equal{}f(x^3)\plus{}3f(x^2)\plus{}3f(x)\plus{}f(1)\equal{}xf(x^2)\plus{}3f(x^2)\plus{}3f(x)\plus{}a$
$ (x \plus{} 1)f(x^2 \plus{} 2x \plus{} 1)\equal{}(x\plus{}1)(f(x^2)\plus{}2f(x)\plus{}a)$ $ \equal{}xf(x^2)\plus{}2xf(x)\plus{}ax\plus{}f(x^2)\plus{}2f(x)\plus{}a$

So $ xf(x^2)\plus{}3f(x^2)\plus{}3f(x)\plus{}a\equal{}xf(x^2)\plus{}2xf(x)\plus{}ax\plus{}f(x^2)\plus{}2f(x)\plus{}a$
So $ 3f(x^2)\plus{}3f(x)\equal{}2xf(x)\plus{}ax\plus{}f(x^2)\plus{}2f(x)$
So $ 2f(x^2)\equal{}2xf(x)\plus{}ax\minus{}f(x)$
So $ 2f(x^2)\equal{}(2x\minus{}1)f(x)\plus{}ax$


linear algebra
matrix
function
real analysis
real analysis unsolved

Suppose  that $ f: \mathbb{ R}^n \to \mathbb{ R}^n$    is   a    $ C^1$    map, and
(1) the Jacobian determinant   of   Jacobian matrix $ \neq 0$   for   every         $ x$.
(about  Jacobian matrix,  please   refer  to http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant )

(2) $ |f(x)| \to \infty$     as  $ |x| \to \infty$.

Prove    that  $ f(\mathbb{ R}^n) \equal{} \mathbb{ R}^n$

Hint: You know  $ f(\mathbb{ R}^n)$ is open by the inverse function theorem. To show $ f(\mathbb{ R}^n)$ is closed (which finishes the proof), suppose $ f(x_k) \rightarrow y$, $ y\in \mathbb{ R}^n$. Now $ (x_k)$ has a subsequence that ...

How  about  the  following  method?

$ a \in \mathbb{R}^n$,     $ g(x) \equal{} {|f(x) \minus{} a|}^2 \equal{} (f(x) \minus{} a)*(f(x) \minus{} a)$,  $ g(x_1) \equal{}$ min$ g(x)$，so

$ Jg(x_1) \equal{} 2 Jf(x_1)*(f(x_1) \minus{} a) \equal{} 0$.

so

 $ f(x_1)\equal{} a$.

[quote="yzymat"]Thanks!  Please  write  down   the  proof [color=red] in  detail[/color]!

How  to  use    (2)   $ |f(x)| \to \infty$     as        $ |x| \to \infty$       ?[/quote]

You are getting free advice here, so please do not tell me what to do.

A useful concept here: a [url=http://en.wikipedia.org/wiki/Proper_map]proper map[/url] between topological spaces.

how to use (2)? use the statement's contrapositive.