In the game For The Win, aside from the universal rating, one can also see their daily and monthly rating to try to track their progress there. I was wondering if this might be extended to post ratings.

If one could see in their profile, "Daily Post Rating" and "Monthly Post Rating," in addition to the Overall Post Rating, they could know if they are making progress on the quality of their posts, and have a better understanding of which posts are accepted. Additionally, the Daily Post Rating would narrow the hassle of searching to see "Which one of my posts got rated one!?" to just a couple of posts, since a user could instantly see from their Daily Post Rating when a post is rated a high or low extreme.

Also the number of overall/monthly/daily ratings would be nice.

That would be nice, knowing the posts that have been considered spam, and somewhat getting the idea so you can improve your quality.

Don't be so desperate for ratings, accept the reality - you always say other people that only bad posts are rated, and you yourself just have to worry so much about rating.

If I was an admin, I would immediately ban everyone from seeing any rating (even their own). Itisn't needed, except in those desperate spam cases, when one PM to the admin does the thing.

And, why, I ask why are there so many topics concerning ratings? Make one, do the suggestions in there.

@admins: I was serious about banning post ratings. If it's bad, tell it now.

Oh, I never thought of it that way... I'll start thinking on how much of an effect it has on other people,
check this out; [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=297247]click click[/url]
Bugi, you've already seen it.. I'm adding on an example on what your post explained.

[quote="Dojo"]I like the idea, but I think not enough posts are rated to have a daily rating.[/quote]

The whole purpose is to know precisely when a post gets rated so one can learn to adjust the quality of posts in regards to how that exact post was rated.

Personally I believe that post rating viewing should be only viewable to admins for analysis purposes.

Correct, that's what I recommended.
Why do people need their post rating? Everyone will answer - to see if their posts are going better. But the same people know that (almost) only bad posts are rated, so rating goes down, or remains constant. And then goes the, why why why don't I have 6 rating etc.
People know when they are spamming. And smaller rating is just another way to remind them, not necessary.

[quote="Bugi"]People know when they are spamming.[/quote]
Not necessarily. I think post ratings are okay to have because, if people use it correctly, it can cause a state of competition and make people want to make better posts.

Hm... People have it 6 years (OK, maybe less I don't know), and there's no competition in good posts, just in bad.

And why are we posting here, let's wait for an admin response.

Actually, it's there for the last two years (since May 1st, 2007). In my personal case, it has been stimulating in the positive manner, since I really felt the need to improve my post quality seeing that my posts were badly rated. Even today, when I see other people's rating higher than mine, it makes me want to post useful posts.  :)

You know where the people with ratings above 4.5 are?  Not spending their time in this forum complaining about the post rating system!

See also:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1579827#1579827
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1565092#1565092
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1281922#1281922
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1269799#1269799

I remember in your websites above JBL, Valentin Vorciu mentioned a "maximum of 3 ratings per hour".  Yet, my post rating certainly isn't more than 4, and I tested this by rating(and then cancelling rating) of 10 people and I was allowed to do so.  Is this a bug or did they change the rules?

I also think r15s11z55y89w21 has a good idea with the "Daily Post Rating".  Not only do you save the hassle of searching, you don't necessarily for sure know what your post was rated if only 1 person rates it when you look at it because you have to wait for a few people before one shows up.

My current rating is now steadily rising.  A month ago it was 1.7.  Now it's 2.4.  I wish administrators rated posts more though to differentiate the gap of unfairly rated posts though.  It also highlights good posts to give them more of a weighting. :)


ratio
number theory proposed
number theory

[quote="Johan Gunardi"]Two questions:
1. Did you mean "geometric" instead of "arithmetic"?
2. Is $ M$ is the set of all such progressions?[/quote]

1. No. The problem from the Olympiad says arithmetic progression.
2. Yes, $ M$ contains all such progression.

[quote="Johan Gunardi"]What is the definition of "ratio" in arithmetic progressions?[/quote]

arithmetic progressions is of the following type: $ a_n \equal{} a_1 \plus{} n\cdot d\;\;\;\; \forall n\in Z$

Here $ d$ is the ratio

[quote="Ahiles"][color=darkblue]Let $ M$ be a set of aritmetic progressions with integer terms and ratio bigger than $ 1$.

[b]a)[/b] Prove that the set of the integers $ \mathbb{Z}$ can be written as union of the finite number of the progessions from $ M$ with different ratios.

[b]b)[/b] Prove that the set of the integers $ \mathbb{Z}$ can not be written as union of the finite number of the progessions from $ M$ with ratios integer numbers, any two of them coprime.[/color][/quote]
a) Example $ M_1 \equal{} \{2k\}$ ,$ M_2 \equal{} \{3k\}$ and $ M_3 \equal{} \{4k \plus{} 1\},M_4\equal{}\{6k\plus{}1\},M_5\equal{}\{12k\plus{}11\}$ .
b) Suppose exist a partition satisfy condition . By the condition of problem , exist a sequence which  has infinite terms . Suppose this sequence is $ \{an \plus{} b\}$ where $ a$ is a positive integer . Take a number large enough , it has a prime divisor $ p$ , it is not hard to show that $ a(n \plus{} p) \plus{} b$ is also a term of this sequence and $ p$ is its divisor ,therefore $ \gcd(an \plus{} b,a(n \plus{} p) \plus{} b) > 1$ ,which gives contradiction . That ends part b .

Sorry for my stupidity. :blush:

[quote="TTsphn"]
a) Have many ways to construct solution for this problem ,example $ M_1 \equal{} \{2k\}$ ,$ M_2 \equal{} \{4k \plus{} 1\}$ and $ M_3 \equal{} \{4k \plus{} 3\}$ .[/quote]I think it does not satisfy the given conditions because $ M_2$ and $ M_3$ have the same ratios.

[quote="Ahiles"][color=darkblue]Let $ M$ be a set of aritmetic progressions with integer terms and ratio bigger than $ 1$.

[b]a)[/b] Prove that the set of the integers $ \mathbb{Z}$ can be written as union of the finite number of the progessions from $ M$ with different ratios.

[b]b)[/b] Prove that the set of the integers $ \mathbb{Z}$ can not be written as union of the finite number of the progessions from $ M$ with ratios integer numbers, any two of them coprime.[/color][/quote]

for part (b) "any two of them coprime", I think that "them" refers to the ratios of the progressions instead of the terms in a progression.

my proof

assume there exists such a union of finite number of the progressions from $ M$ with ratios integer numbers, any two of them coprime. Call them $ p_1, p_2,...,p_m$

for every $ i(1 \le i \le m)$, terms of progression $ p_i$ must have the form of $ {a_{i}n\plus{}b_i}$, $ a_i, b_i$ are integers and $ a_i$ is its ratio.

now consider an integer x such that $ x \equiv b_{i}\plus{}1 (mod a_i)$, if such a $ x$ exists, then it does not belong to any of the $ m$ progressions.

But since any two of $ a_i$ are coprime, according to Chinese Residue Theorem such an $ x$ exists. This contradicts that the union of the $ m$ progressions covers $ \mathbb{Z}$

Q.E.D.

Yes ,in this case ,chinese remainder theorem works .  :)


Let $ |U|\equal{}n$. Find the number of ordered triples $ (X,Y,Z)$ of sets such that $ X,Y,Z$ are subsets of $ U$, and

$ X\cup (Y\cap Z')\equal{}X'\cup Y'$ and none of $ X,Y,Z$ are null sets

Assuming $ X^\prime$ means the complement of $ X$ with respect to $ U$...

[hide]The left side contains all elements of $ X$, and the right side contains all elements of $ X^\prime$. Both sides must be equal to $ U$.

We require that any member of $ X^\prime$ be a member of $ Y$ by the LHS and that any member of $ X$ be a member of $ Y^\prime$ by the RHS. This means that $ Y \equal{} X^\prime$, and now we know $ X \neq U$. It remains to find all $ Z$ such that $ X^\prime$ is a subset of $ Z^\prime$. Equivalently, $ Z$ is a (nonempty) subset of $ X$. So if $ |X| \equal{} k$, $ Z$ has $ 2^k \minus{} 1$ choices.

Pick any nonempty $ X$ with $ |X| \equal{} k$. There are $ {n \choose k}$ such $ X$. $ Y$ is uniquely determined. $ Z$ has $ 2^k \minus{} 1$ choices. Using applications of binomial theorem, we can show the resulting sum over all $ k$ from $ 1$ to $ n\minus{}1$ is $ 3^n \minus{} 2^{n\plus{}1} \plus{} 1$.[/hide]


Define $ \{ CDJ \}$ be the area of $ CDJ$.  Then, we can find by SAS that
\[ \{ DEH \} \equal{} \{ HFJ \}
\]
so
\[ \{ DEH \} \plus{} \{ DHFC \} \equal{} \{ HFJ \} \plus{} \{ DHFC \} \Rightarrow \{ CDJ \} \equal{} \boxed{36}
\]


$A,B \in M(n,R)$ wich commute two by two

$D=4A^{6}+12A^{5}B-3A^{4}B^{2}-26A^{3}B^{3}-3A^{2}B^{4}+ 12AB^{5}+4B^{6}$

 Find a necessary and sufficiant condition such that $det(D)=0$

Notice that we can write $D=(A-B)^2(2A^2+2B^2+5AB)^2$, since A e B are commutative matrices.


Let $ a$, $ b$, $ x$ and $ y$ be positive reals. Then

\[ \frac{1}{(x\plus{}a)^2}\plus{}\frac{1}{(y\plus{}b)^2}\ge\frac{1}{xy\plus{}ab}\]

is that always true?

:)

[quote="ElChapin"]Let $ a$, $ b$, $ x$ and $ y$ be positive reals. Then
\[ \frac {1}{(x \plus{} a)^2} \plus{} \frac {1}{(y \plus{} b)^2}\ge\frac {1}{xy \plus{} ab}
\]
is that always true?

:)[/quote]

No, try x=b=2, a=y=1.

But it is true.
a,b,x,y>0

$ \frac{1}{(a\plus{}x)^2}\plus{}\frac{1}{(b\plus{}y)^2}\ge\frac{ax\plus{}by}{(ay\plus{}bx)(ab\plus{}xy)}$

[quote="NextPeace"]But it is true.
a,b,x,y>0

$ \frac {1}{(a \plus{} x)^2} \plus{} \frac {1}{(b \plus{} y)^2}\ge\frac {ax \plus{} by}{(ay \plus{} bx)(ab \plus{} xy)}$[/quote]
By Cauchy Schwarz Inequality, we have
\[ (ay\plus{}bx)(ab\plus{}xy) \ge (a\sqrt{by} \plus{}x\sqrt{by})^2 \equal{}by(a\plus{}x)^2\]
\[ \Rightarrow \frac{1}{(a\plus{}x)^2} \ge \frac{by}{(ay\plus{}bx)(ab\plus{}xy)}\]
Similarly, we have
\[ \frac{1}{(b\plus{}y)^2} \ge \frac{ax}{(ay\plus{}bx)(ab\plus{}xy)}\]
Adding up these inequalities, we get the desired Inequality. :)


Pick any three points on a circle and form a triangle.  Find the probability that the triangle is an obtuse triangle.

An idea:
We set O(1;0) and C((0;0),1)
Let be A(cos $\theta$; sin $\theta$)
For theta fixed, we need to find the set of points where a point B will form an obtuse triangle with O and A, if we call $\phi(\theta)$ the sum of the distance of arcs (I'm not sure of my english traduction of "arcs" in french) where any point B form an obtuse triangle with O and A, the probability of having an obtuse triangle when picking three points on a circle will be $P=\int_{0}^{2\pi} \phi(\theta) d \theta$

[i]Moderator's Note: Use \pi for a lower-case pi: $\pi$ vs. $\Pi$.[/i]


The lines y=-5x+8 and y=ax+3x+7 are perpendicular.  Find a.

need a little help...
please and thx =]

The condition for which two lines intersect orthogonally is that the product of the slope is $-1.$

By the way, this should not be in the Intermediate forum.


inequalities
function
inequalities unsolved

let a,b,c be positive real numbers.Prove that:
$ \sum {ab\sqrt a }  \le \frac{{\sqrt {a \plus{} b \plus{} c} (ab \plus{} bc \plus{} ca)}}{{\sqrt 3 }}$ :maybe:

[quote="BG Yoda"]This is just something like Jensen  :maybe:[/quote]
can you proof here?

[quote="Toan_VN_LC"]let a,b,c be positive real numbers.Prove that:
$ \sum {ab\sqrt a } \le \frac {{\sqrt {a \plus{} b \plus{} c} (ab \plus{} bc \plus{} ca)}}{{\sqrt 3 }}$ :maybe:[/quote]
It's wrong. Try $ a\rightarrow0^\plus{}$ and $ b\equal{}c\equal{}1.$

The right inequality with the same $ LHS$ is
$ \sum\frac{ab}{ab\plus{}bc\plus{}ca}\sqrt{a}\le \sqrt{\sum \frac{ab}{ab\plus{}bc\plus{}ca} a}$  :|

[quote="BG Yoda"]The right inequality with the same $ LHS$ is
$ \sum\frac {ab}{ab \plus{} bc \plus{} ca}\sqrt {a}\le \sqrt {\sum \frac {ab}{ab \plus{} bc \plus{} ca} a}$  :|[/quote]
the function $ f(x)\equal{}\sqrt{x}$ is concave then by Jensen
$ \sum \frac{ab}{ab\plus{}bc\plus{}ca} \sqrt{a} \leq \sqrt{\sum \frac{a^2b}{ab\plus{}bc\plus{}ca}}$
What about this; if $ a\plus{}b\plus{}c\equal{}3$ and $ a,b,c\leq 0$
$ \sum \frac{ba\sqrt{a}}{ab\plus{}bc\plus{}ca} \leq \frac{4}{3}$

[quote="BG Yoda"]The right inequality with the same $ LHS$ is
$ \sum\frac {ab}{ab \plus{} bc \plus{} ca}\sqrt {a}\le \sqrt {\sum \frac {ab}{ab \plus{} bc \plus{} ca} a}$  :|[/quote]
$ \sum \frac{ab}{ab\plus{}bc\plus{}ca}a \ne a\plus{}b\plus{}c,$ ;)


function
limit
inequalities
algebra unsolved
algebra

prove that there doesn't exist functionf : R+->R+ satisgying f(x+y)>=f(x)+yf(f(x))

Clearly, $ f$ is strictly monotone increasing. Setting $ y\equal{}1$ then yields $ f(x)\leq x\plus{}1$, otherwise we would produce the contradiction $ f(x\plus{}1)\geq f(x)\plus{}f(f(x))\geq f(x)\plus{}f(x\plus{}1)>f(x\plus{}1)$.

If $ f(f(x))>1$ holds for some $ x$, we get the contradiction $ 1\geq \lim_{y\rightarrow\infty} \frac{f(x\plus{}y)}{x\plus{}y\plus{}1}\geq \lim_{y\rightarrow\infty} \frac{f(x)\plus{}yf(f(x))}{x\plus{}y\plus{}1}\equal{}f(f(x))>1$. Therefore we must have $ f(f(x))\leq 1$ for all $ x$.

From the original inequality we get $ f(y)\equal{}f(1\plus{}y\minus{}1)>(y\minus{}1)f(f(1))\Longrightarrow\lim_{y\rightarrow\infty}f(y)\equal{}\infty$. Exploiting the monotonicity of $ f$ we must also have $ \lim_{y\rightarrow\infty}f(f(y))\equal{}\infty$, a contradiction.


For integers $ k$ and $ n$ such that $ 1\leq k < n$, let
\[ \binom{n}{k} \equal{} \frac{n!}{k!(n\minus{}k)!}.\]
Then $ \left( \frac{n\minus{}2k\minus{}1}{k\plus{}1} \right) \binom{n}{k}$ is an integer

(A) for all $ k$ and $ n$
(B) for all even values of $ k$ and $ n$, but not for all $ k$ and $ n$
(C) for all odd values of $ k$ and $ n$, but not for all $ k$ and $ n$
(D) if $ k\equal{}1$ or $ n\minus{}1$, but not for all odd values of $ k$ and $ n$
(E) if $ n$ is divisible by $ k$, but not for all even values of $ k$ and $ n$

Thanks in advance!

[quote="JBL"]And what have you tried?

(Unrelatedly, the restriction on $ k$ is very odd -- there is no ambiguity about $ 0! \equal{} 1$, and the combinatorial and algebraic interpretations of the binomial coefficient agree when we extend to $ k \equal{} 0$ and $ k \equal{} n$.)[/quote]

I tried plugging in a few values of $ k$ and $ n$ but didn't get very clear patterns.  Should I be focusing on any specific cases?  Thanks.

[quote="JBL"]I personally would go for simple algebraic manipulations.  One possible thing to use is that adding an integer to your expression won't change whether or not your expression is integral.[/quote]

I rewrote the combinatorics in terms of factorials, but what is the integer approach you're referring to?  Thanks!

Try adding $ 2 \binom{n}{k}$.  (This is the same as rewriting $ \frac{n \minus{} 2k \minus{} 1}{k \plus{} 1} \equal{} \frac{n \plus{} 1}{k \plus{} 1} \minus{} 2$.)


inequalities algebra function domain polynomial

Please don't look until you've tried it.

[hide]

There is only one sensible way to do this, and that is to make all terms in the inequality of degree $2$.

Thus we change it to

$\sum a^2+2\sqrt{3abc}\leq (a+b+c)^2$

which is equivalent to

$\sum ab\geq \sqrt{3abc(a+b+c)}$

which upon squaring is just

$\sum a^2b^2 +2\sum a^2bc\geq 3\sum a^2bc$

so we just need to show

$\sum a^2b^2\geq \sum a^2bc$

which is obvious by bunching or weighted AM-GM

(e.g $a^2b^2+a^2c^2\geq 2a^2bc$ and sum cyclically)

Hope this helps
[/hide]

[quote="mathfanatic"]Prove that, if $a+b+c=1$ and $a,b,c>0$
\[a^2 + b^2 + c^2 + \sqrt{12abc} \le 1.\][/quote]
Hi! This is my solution:
Let` to prove , by Schwarz, [tex]\frac{1}{3}<=a^2+b^2+c^2<=1[/tex] for [tex]a+b+c=1[/tex]. It`s easy![hide]By Schwarz , see that [tex]\frac{(a+b+c)^2}{3}<=a^2+b^2+c^2<=\sqrt{3(a^4+b^4+c^4)}[/tex][/hide]. Now we need to prove that [tex]\sqrt{12.abc}<=\frac{2}{3}[/tex]. It´s easy too. Let´s go! We have [tex]12abc<=\frac{4}{9}[/tex]=>[tex]abc<=\frac{1}{27}[/tex].How [tex]a+b+c=1[/tex], we have [tex](1-a-b)(1-a-c)(1-b-c)<=\frac{1}{27}[/tex] . It´s proved by MA-MG inequalities: [tex](1-a-b)(1-a-c)(1-b-c)<=(\frac{1-a-b+1-a-c+1-b-c}{3})^3=\frac{1}{27}[/tex]. The end.

A different solution:

[hide]The desired is equivalent to $\sqrt{3abc(a+b+c)} \le ab+bc+ca$. Square both sides and cancel. We have $abc(a+b+c) \le a^2b^2 + b^2c^2 + c^2a^2$. This follows from AM-GM.[/hide]

A full solution, like Mildorf's idea:
[hide]
Lemma: $x+y+z \ge \sqrt{3(xy+yz+zx)}$
Proof: Square both sides and cancel terms. We end up with
$x^2+y^2+z^2 \ge xy+yz+zx$
which follows from the trivial inequality.

$a+b+c=1$
$2(a+b+c)=2$
$(2a-1)+(2b-1)+(2c-1)+1=0$

Express $2a-1$ as the difference of the two squares $a^2-(1-a)^2$.

$a^2-(1-a)^2+b^2-(1-b)^2+a^2-(1-b)^2+1=0$.

Since $a+b+c=1$, we have $1-a=b+c$.

$a^2+b^2+c^2+1=(a+b)^2+(b+c)^2+(c+a)^2$
$1 = a^2+b^2+c^2+2(ab+bc+ca)$

Using the lemma with $x=ab, y=bc, z=ca$, we have $2(ab+bc+ca) \ge 2\sqrt{3abc(a+b+c)} = \sqrt{12abc}$.

So $1 \ge a^2+b^2+c^2+\sqrt{12abc}$.[/hide]

[quote="MysticTerminator"]And...I just wanted to butcher an inequality with Lagrange multipliers, so eh.[/quote]

quite entertaining!  :P

[quote="MysticTerminator"]$G(a,b,c) = a + b + c = 1, F(a,b,c) = a^2 + b^2 + c^2 + \sqrt{12abc}$,  $<2a + \sqrt{\frac{3bc}{a}}, 2b + \sqrt{\frac{3ac}{b}}, 2c + \sqrt{\frac{3ab}{c}}>$ $= \nabla F = \lambda \cdot \nabla G = \lambda$ $<1, 1, 1> \Rightarrow 4a^2 + 4\sqrt{3abc} + \frac{3bc}{a} = 4b^2 + 4\sqrt{3abc}$ $+ \frac{3ac}{b} = 4c^2 + 4\sqrt{3abc} + \frac{ab}{c}$ $\Rightarrow 4a^2 - 4b^2 + \frac{3b^2c - 3a^2c}{ab} = 0 \Rightarrow (a^2-b^2) \cdot (4 - \frac{3c}{ab}) = 0 \Rightarrow a = b$ or $4ab = 3c$. Similarly, $b = c$ or $4bc = 3a$ and $a = c$ or $4ac = 3b$. Solutions to this are $(a,b,c) = (\frac{1}{3}, \frac{1}{3}, \frac{1}{3}), (0, 0, 1), (0, 1, 0), (1, 0, 0)$. All give $1$ as the max. And...I just wanted to butcher an inequality with Lagrange multipliers, so eh.[/quote]

The domain is not closed though (because we only have $a,b,c>0$, not $a,b,c\geq 0$) so you have boundary conditions around where $a,b,c=0$ to worry about.  Otherwise your solution is (of course) fine.

Also the general statement of Lagrange is $\nabla F=\lambda \nabla G$

Does anyone besides me find it odd that \del does not produce the same thing as \nabla?

Here's a tip.  This inequality should take everyone about two seconds to see the solution.

Because we have $a+b+c=1$, we can multiply terms by $(a+b+c)^n$ until everything is of the same power.  On the $LHS$, we have terms of power $2$, so make EVERYTHING of power $2$

$1$ becomes $(a+b+c)^2$

$\sqrt {12abc}$ becomes $\sqrt {12abc(a+b+c)}$

Now it's homogeneous, and therefore it's almost certainly solvable by Schur + Muirhead (or weighted AM-GM) and these tools are easy to apply

Kedlaya calls his section on such inequalities the idiot's guide to symmetric polynomial inequalities or something like that, and indeed, using the idiot's method is generally the best way.


Bob has two bags. Bag A contains 8 green marbles and 3 red marbles, Bag B contains 5 green marbles and 7 red marbles. If bob chooses a bag at random, picks out a marble, and see it's red, what is the probability it came from bag B?

There's a 1/2 chance he chooses Bag B, and a 7/12 chance it was red, thus the answer is 1/2 times 7/12 is $ \boxed{\frac{7}{24}}$. Please tell me if I messed up.

The answer they gave was $ \boxed{\frac {77}{113}}$ which makes no sense because $ 113$ is prime and I don't understand how they get that as denominator unless they simplified from a larger number, but I can't imagine having more choices.

There is a 1/2 * 3/11 = 3/22 chance he picked a red ball from Bag A.
There is a 1/2 * 7/12 = 7/24 chance he picked a red ball from Bag B.

If we know he picked a red ball, then the odds that it came from B are

$ \frac {\frac {7}{24}}{\frac7{24} \plus{} \frac3{22}} \equal{} \frac {\frac7{24}}{\frac {113}{264}} \equal{} \frac {7 * 264}{24 * 113} \equal{} \boxed{\frac {77}{113}}$


What is the pH of 10^-7 M HCl? (taking the self ionization of water in consideration)

pH (the power of hydrogen) is defined as -log([H+]). Therefore, the concentration of 10^-7 M HCl =-log(10^-7)=7

yes, that is correct.
but notice $HCl$ is a strong acid, and the $K_a=1$, so if you had something like $10^{-7} M, H_3PO_4$ then it wasn't the same method. :)

wouldn't it have the common ion effect? with the H+ ions from water? and be a little acidic?

no, never say that water has a PH of $7$ which means it nuetral, if water was acitic, then you wouldn't be here :)
($k_w=10^{-14}=[H^+][OH^-]$)

[quote="inteluser"]hm... the $\text{K}_\text{a}$ of $\text{HCl}$ isn't $1 {M}$ at standard conditions... it's $10000 {M}$.

How would one solve this exactly, anyway?[/quote]
yes, i made  a mistake about the $k_a$, the $k_a$ for strong acids are very large numbers. it means that in every 10000 molecules of HCL only one molecule would not disociate and remain as HCl...
because $HCl$ dissociates as following:
$HCl \longrightarrow H^+ + Cl^-$ the $[H^+]$, is the same as the concentration of the original solution therefore the $Ph=-\log{10^-7}=7$

[quote="hmoonster"]I meant the solution being acidic. But isn't pure water neutral with a pH of 7, and when you add in HCl it would become acidic, and the pH of the solution be somewhat lower than 7?[/quote]
yes, but look at this, your acid is barely an acid, its $10^{-7} M$!!! :huuh:
so in one liter of this acid, you only have, $10^-7$ moles of acid, how much do you expect this to effect the PH? :)

[quote="hmoonster"]But the concentration in pure water already [H+] = 10^-7. If by adding HCl 10^-7M of H+ was added more, wouldn't the pH be lower than 7?[/quote]
no, because in water, $[h^+]=[oh^-]$ :huuh:

Ah, I found out how to do it. You can just use the standard equilibrium chart:
Equilibrium [H+] = 1.0 x 10^-7 + x (x is the additional H+ from the self-ionization of H2O)
And the equilibrium [OH-] = x, so plugging that into the [H+][OH-] = 1.0 x 10^-14, you get x = 6.18 x 10^-8. The final pH is 6.79.

Wasn't so hard after all... :|

Thanks to everyone for the input.

argghh :ewpu:
no, i just did the problem i was gonno post the solution, that i saw you've done it...
yes, it was a very easy problem, and i was unbelievalbly stupid before. :wallbash_red:

[quote="inteluser"]hm... the $\text{K}_\text{a}$ of $\text{HCl}$ isn't $1 {M}$ at standard conditions... it's $10000 {M}$.[/quote]

You may safely assume that hydrochloric acid is 100% dissociated in almost all cases.

Best,
Borek

Smarkotan oz gluthozmaz


logarithms
trigonometry
function
calculus
derivative
trig identities
calculus computations

i didnt solve this problem. :( I need help. :(
Thanks and best regards! :wink:
[img]http://img147.imageshack.us/img147/9340/nnzf6ng1.jpg[/img]

differentiate the first term of f, keeping in mind the product rule. then do the same for the second term.. and then just plug in 3pi/8

[hide="mathking123, your approach would lead to this:"]
$ f(x) \equal{} \ln (\sin^2 x) \minus{} \ln (\tan x)$
In using the chain rule, recall that $ \frac {d}{dx}\left(\sin^2 x\right) \equal{} 2\sin 2x$.
$ f'(x) \equal{} \frac {\sin 2x}{\sin^2 x} \minus{} \frac {\sec^2 x}{\tan x}$.

Note that $ \frac {3\pi}{8}$ is not an easy angle to take trig functions of; we would have to use the half-angle identities for $ \sin$, $ \cos$, and $ \tan$ to get the answer this way.
[/hide]

Instead, don't be fooled into taking that derivative so quickly; slow down, and do some algebra:

[hide="A much better approach:"]
What do you do when you see $ \ln a \minus{} \ln b$?

$ f(x) \equal{} \ln (\sin^2 x) \minus{} \ln (\tan x)$
$ f(x) \equal{} \ln \left( \frac {\sin^2 x}{\tan x} \right)$
$ f(x) \equal{} \ln \left( \frac {\sin^2 x}{\frac {\sin x}{\cos x}} \right)$
$ f(x) \equal{} \ln \left( \sin x \cos x \right)$

I don't know about you, but when I got to this point, a smile came across my face.  :lol:
Now take the derivative, using the product rule on $ \sin x \cos x$:

$ f'(x) \equal{} \frac { \minus{} \sin^2 x \plus{} \cos^2 x}{\sin x \cos x}$

Don't plug in for $ x$ just yet: remember that we don't want to handle that $ \frac {3\pi}{8}$ on its own. We use the trig identities $ \cos 2x \equal{} \cos^2 x \minus{} \sin^2 x$ and $ \sin 2x \equal{} 2\sin x \cos x$ to help us:

$ f'(x) \equal{} \frac {\cos 2x}{\frac {1}{2}\sin 2x} \equal{} 2\cot 2x$

[i]Now[/i] plug in for $ x$:

$ f'\left(\frac {3\pi}{8}\right) \equal{} 2\cot \left(\frac {3\pi\cdot 2}{8}\right) \equal{} 2\cot \left(\frac {3\pi}{4}\right) \equal{} 2( \minus{} 1) \equal{} \boxed{ \minus{} 2}$.

(We used the fact that $ \cot \left(\frac {3\pi}{4}\right) \equal{} \minus{} 1$, which is something you can do in your head, unlike the half-angle identity.)

The thing to remember here: [b]do as much algebra as you can[/b] before you dive into the calculus.

[/hide]

[quote="t0rajir0u"][quote="jvenezuela716"]$ f(x) \equal{} \ln \left( \sin x \cos x \right)$
...
The thing to remember here: [b]do as much algebra as you can[/b] before you dive into the calculus.[/quote]

So why didn't you write $ f(x) \equal{} \ln \sin 2x \minus{} \ln 2$?  ;)[/quote]

 :blush: Um... uh... I was... just... testing you... to see if you'd catch me...  :maybe: You passed!

Let $ s_{\omega}$, $ c_{\omega}$, and $ t_{\omega}$ denote the sine, cosine, and tangent functions. Simplifying the expression we have
\begin{eqnarray*}y
&=&2\ln{s_{\omega}}-\ln{t_{\omega}}\\
&=&\ln{s_{\omega}}+\ln{\frac{s_{\omega}}{t_{\omega}}}\\
&=&\ln{s_{\omega}}+\ln{c_{\omega}}\\
\Leftrightarrow && y^{\prime}=\frac{1}{t_{\omega}}-t_{\omega}\\
\end{eqnarray*}


calculus
integration
logarithms
trigonometry
calculus computations

hello, rewrite your integral in the form
$ \int\frac{dx}{(1\minus{}x)\sqrt{9(1\minus{}x)^2\minus{}8x\plus{}7}}$ and set $ t\equal{}1\minus{}x$.
Sonnhard.

Thanks Sonnhard. I had:

$ 1\minus{}x \equal{} \frac{1}{u}$ so $ \ln(1\minus{}x) \equal{} \minus{}\ln u$ and $ \frac{dx}{1\minus{}x}\equal{}\frac{du}{u}$.

Then $ I\equal{}\int \frac {1}{u \sqrt{\frac{8}{u} \minus{}u^{2} \plus{}\frac{9}{u^{2}}}}\ du$

$ \equal{}\int \frac{1}{\sqrt{25\minus{}(u\minus{}4)^{2}}}\ du$

$ \equal{}\sin^{\minus{}1} \frac{u\minus{}4}{5}$

$ \equal{}\sin^{\minus{}1} \frac{4x\minus{}3}{5(1\minus{}x)} \plus{} c.$

hello, i have got
$ \minus{}\arctan\left(\frac{16\minus{}17x}{\sqrt{\minus{}17(\minus{}1\plus{}x)^2\minus{}34x\plus{}33}}\right)\plus{}C$
Sonnhard.


[b]Prove:[/b] For any positive integer $ n$, $ \prod_{d|n}\equal{} n^{\frac{\tau(n)}{2}}$.

I think you mean $ \prod_{d | n}d$.

[hide="Hint"] If $ d$ is a divisor, so is $ \frac{n}{d}$. [/hide]

The book (104 problems in number theory) says $ \prod_{d|n}\equal{} n^{\frac{\tau(n)}{2}}$.

It's a misprint.  Without an argument the $ \prod$ sign has nothing to multiply.  It is as meaningless, as, say, $ \sum_{k\equal{}1}^{n}$.

Aaah. I see. No wonder I couldn't get an answer :!:  :P


ARML
AMC
AIME
USA(J)MO
USAMO
geometry
geometric transformation
\/closed

Canadian contests happen throughout the year. Such as Can. Open., Pascal, Cayley, Fermat, Fryer, Galois, Hypathia, Euclid, CMO, and I'm even going to mention all the regional contests.

Since there seems to be quite a number of Canadian students on this site already, why not start on forum for us to discuss the canadian contests. Because now, it's hard to find a place to do that. Since there's a forum dedicated to AMC, let's make similar one for the Canadian Contests. This way, we can also attract more Canadian students to come to this site.

Futhermore, I do not believe the there is a place on the web that is dedicated to discussion of Canadian math contests.

Please consider this suggestion seriously.

Discussions of other contests and programs are always fair game for the Other Contests and Programs topic, which many people here could well make more use of.

Occam's razor suggests having fewer rather than more topics, as long as the topics are distinct and clear.

I don't feel that the Other Contests forum do justice to the Canadians. If you look in there, 95% of it is about ARML, Mandrebot, and whatnot. We just get buried under all that.

if this is a petition, I will sign it.

there are many canadian contests to talk about...

pretty much the same amount as the ones in USA

I would sign the petition for sure.  Just so many Canadian contests to talk about through out the year.  And although some of them are quite easy, it's still fun to see what different methods people used to solve the last couple of questions.
 *Besides, there are a lot smart people in Canada, not just in US.*

Why is 'Other programs & contests' not sufficient for discussion of Canadian contests (which I would like to hear more about)?

A note to the Canadians: The 'American Mathematics Competitions' Forum is intended for the AMC exams - AMC->8,10,12, AIME, USAMO, not for all contests in America.   Hence, that forum is not meant to mean 'only talk about contests that happen in America'.  The other contests in America that occur are also included in 'Other Programs & Contests'.  We have a special Forum just for the AMC for a variety of reasons, two of which are its importance/prominence and our sponsorship of the AMC.

Perhaps the 'Other' Forum might be tagged 'other US programs' and we have another Forum called 'international programs' or some such - any thoughts on that?  I'm reluctant to start building Forums per country, since the board would quickly get unwieldy.

Perhaps I could just specifically suggest Canadian contests in the subheading of the 'Other Contests' forum?  And maybe one of our Canadians would like to moderate it and introduce us to their programs?

Or maybe Mr.Rusczyk, you could make a sub-forum inside Other Contests...

That might make some things more clear...

Just an opinion..

Subforums are the work of the devil. ;)

Quite honestly I don't see why we can't just talk about Canadian contests in the OTHER CONTESTS section. We even get more exposure this way. If we make a new forum, American readers are less likely to go there. We have what, 5, or 6 regular Canadian posters? I don't think THAT kind of drastic measures are necessary.

[quote="kool_dudy"]maybe you're right....

since people would browse canadian stuffs if it was mixed with the american stuffs...[/quote]

And less Canadians are gonna browse the Canadian topics if it were buried under the American stuff.

The whole point of making a forum is so that we, Canadians, can have our own place, to discuss our own contests. Exposure is unneccesary. Besides, Americans are not going to be too interested in our contests, because, well, they don't participate in them.

This works backwards too. If we just leave our stuff in the "Others" section, then less Canadians are going to bother checking theforum, because they would assume that it's full of American thing that don't matter to them anyway.

[quote="Osiris"]We have what, 5, or 6 regular Canadian posters?[/quote]
And you wonder why that is?  :wink:

[quote="blahblahblah"]I agree . . . , competitor in AMC/CMC competitions since 2002.[/quote]

May I ask which AMC competitions you have been in, and how old you are now? I'd love to know more about the comparison between AMC and CMC competitions.

[quote="rrusczyk"]I'm reluctant to start building Forums per country, since the board would quickly get unwieldy.[/quote]

It's already clear that some users of this site miss forums that are below a certain horizontal position on the forum index page. I will be watching the "views" number and "replies" number for various threads to gauge relative interest in different contests around the world.

The Canadian programs surely would be interesting to more readers of this site if more of those readers knew that those contests were open to people from around the world (as the majority of AMC-sponsored contests are). And anyone anywhere in the world who has a friend interested in math competitions is always welcome to invite that friend to post on this site.

Maybe someone can make a seperate forum for Canadian forums....

Hmm...  The thing is, a Canadian forum could attract more Canadians, but then Americans or people from all over the world would care less about Canadian contests because they wouldn't go into the separate Canadian forum or post anything in it.

A random thought...

But I almost never post in American forums anyway. So. ;)

I really don't care, one way or another - if there is to be a Canadian forum, I'd welcome it; but if there is not to be, I wouldn't mind either. I'd just post what I find in the OTHER CONTESTS section.

True...I don't think anything has gone wrong here yet...

Quote "If it's not broken, don't fix it"

Another quote: "If it ain't broke, you're not trying" :D

[quote="rrusczyk"]Start talking the Canadian contests up more!  I don't think a lot of US people realize that some of them have 'international' divisions - I didn't realize this until I looked at a few last night.[/quote]

I didn't know about this.  Which contests are American students eligible for?  (I know that for the COMC, you have to be either a Canadian citizen or resident, or you have to attend a school in Canada.)

[Edit]: Looks like non-Canadians are eligible for the Cayley, Pascal, and Fermat contests, but they can't win any awards, invitations to the training camp, etc.

Alright, I'll drop this petition for now. But if the demography changes, the issue may be brought back again.

Though, I would appreciate if Mr. Rusczyk would alter the description of "Other Contests & Programs" to reflect the Canadian contests, so that we (and future Canadian members) don't post the messages all over the place.

Thank you.

[quote="billzhao"]I would appreciate if Mr. Rusczyk would alter the description of "Other Contests & Programs" to reflect the Canadian contests.[/quote]

That sounds like a good idea, especially in view of the openness of some of those contests to international participation.

Yes, I would agree about relabling to include some reference to Canadian or international contests.


[quote="vardanch"]Given 81 positive integers all of whose prime factors are in the set ${2,3,5}$, prove that there are $4$ numbers whose product is the fourth power of an integer.[/quote]
It seems similarly a pro of IMO 1985 :)
And  it isn't difficult to have a good solution


The roots of the equation x^3+k*x^2-54*x+216 =0 can be arranged to form a geometric progression. Find the value of k. ( Try two way solve it.)

LaTeX please.

The polynomial is $ x^3+kx^2-54x+216$.

[hide]Let the roots be $ \frac{a}{r}$, $ a$, and $ ar$.  Using Vieta's, $ a^3=-216\Rightarrow a=-6$.  Thus $ -6$ is a root.  Setting the polynomial to zero and plugging in $ x=-6$:

\begin{eqnarray*}
(-6)^3+k(-6)^2-54(-6)+216 &=& 0 \\
-216+36k+324+216 &=& 0 \\
36k &=& -324 \\
k &=& \boxed{-9}
\end{eqnarray*}[/hide]

[quote="lingomaniac88"]LaTeX please.

The polynomial is $ x^3 + kx^2 - 54x + 216$.

[hide]Let the roots be $ \frac {a}{r}$, $ a$, and $ ar$.  Using Vieta's, $ a^3 = - 216\Rightarrow a = - 6$.  Thus $ - 6$ is a root.  Setting the polynomial to zero and plugging in $ x = - 6$:

\begin{eqnarray*} ( - 6)^3 + k( - 6)^2 - 54( - 6) + 216 & = & 0 \\
- 216 + 36k + 324 + 216 & = & 0 \\
36k & = & - 324 \\
k & = & \boxed{ - 9} \end{eqnarray*}[/hide][/quote]
Anyone has different solution than that?
And I am new here. Would anyone tell me how to use LaTeX.


geometry
circumcircle
projective geometry
trigonometry
geometry proposed

Given triangle $ ABC$ and its circumcircle $ (O)$.Let $ P$ and $ Q$ be two arbitrary points; $ A_1,B_1,C_1$ be the intersections of $ AP,BP,CP$ and $ (O); A_2,B_2,C_2$ be the intersections of $ AQ,BQ,CQ$ and $ (O). A_1A_2, B_1B_2,C_1C_2$ intersect and makes triangle $ A_3B_3C_3$. Prove that $ AA_3, BB_3,CC_3$ are concurrent.

According to http://www.mathlinks.ro/viewtopic.php?t=274286 (See the nice proof of my friend tdl) the corresponding sidelines of triangles A3B3C3 and ABC cut in three collinear points, based on desargues theorem we are done

Apply the trigonometry form of Ceva's theorem in $ \triangle A_3B_1C_2$ we have:

$ \frac {\sin{\angle AA_3B_3}}{\sin{\angle AA_3C_3}} \equal{} \frac {\sin{\angle AB_1C_2}}{\sin{\angle AB_1A_3}} \cdot \frac {\sin{\angle AC_2A_3}}{\sin{\angle AC_2B_1}} \equal{} \frac {AC_2}{AB_1} \cdot \frac {\sin{\angle AC_2A_3}}{\sin{\angle AB_1A_3}} \equal{}$
$ \equal{} \frac {AC_2}{AB_1} \cdot \frac {\sin{\angle ACC_1}}{\sin{\angle AC_2B_2}} \equal{} \frac {AC_2}{AB_1} \cdot \frac {AC_1}{AB_2} \equal{} \frac {AC_1}{AB_1} \cdot \frac {AC_2}{AB_2}$

Make the same arguments for $ B_3, C_3$ then multiply those equality together we have:

$ \prod{\frac {\sin{\angle AA_3B_3}}{\sin{\angle AA_3C_3}}} \equal{} \prod{\frac {AC_1}{AB_1}}\cdot \prod{\frac {AC_2}{AB_2}} \equal{} 1$

Which again by the the trigonometry form of Ceva's theorem, means that $ AA_3,BB_3,CC_3$ concurs, QED.

[b]Comment.[/b] I think these problems of this kinds can be solved with the Ceva's theorem. But the nices solutions are still the projective ones  :).


trigonometry
search
algebra proposed
algebra

[quote="mathematica"]Find all real triples such as $(a,b,c)$ so that $a^2-b^2=bc$ , $b^2-c^2=ac$ and $a^2-c^2=ab$.[/quote]
I think
 $(2tcos\frac{2\pi}{7},t,t+2tcos\frac{4\pi}{7}),(2tcos\frac{4\pi}{7},t,t+2tcos\frac{6\pi}{7})$ and $(2tcos\frac{6\pi}{7},t,t+2tcos\frac{2\pi}{7}),$
where $t\in\mathbb R.$

[quote="justdudxit"]How did you come up with that?[/quote]
If any 2 from $\{a,b,c\}$are equal or equal 0 then all numbers are equal 0.
Let $a,b$ and $c$  various and are not equal $0$.
Then $a^2-b^2=bc\Leftrightarrow c=\frac{a^2-b^2}{b}.$
Hence, $b^2-\frac{(a^2-b^2)^2}{b^2}=a\cdot\frac{a^2-b^2}{b}\Leftrightarrow t^3+t^2-2t-1=0$ and
$a^2-\frac{(a^2-b^2)^2}{b^2}=ab\Leftrightarrow t^3+t^2-2t-1=0,$ where $t=\frac{a}{b}.$
$2\cos{\frac{2\pi}{7}},2\cos{\frac{4\pi}{7}}$ and $2\cos{\frac{6\pi}{7}}$  are roots of the equation and ... done.

[quote="mathematica"]Could you tell me how you found the roots of $f(t)=t^{3}+t^{2}-2t-1$ ?[/quote]
And I did not search for them. I already knew, that they are roots (it easily to check up by means of viet's theorem).
For example, it is very difficult to find nice roots of the equation:
$x^{3}+x^{2}-6x-7=0.$ It is desirable to know them a priori. (The hint: this equation is connected with a regular polygon the having 19 sides.)

I think that this problem can also be solved geometricaly:

(Double Angle Theorem) In triangle ABC, the sufficient and necessary condition for $\angle{A}=2\angle{B}$ is $a^{2}=b^{2}+bc$.

[quote="arqady"]For example, it is very difficult to find nice roots of the equation:
$x^{3}+x^{2}-6x-7=0.$ It is desirable to know them a priori. (The hint: this equation is connected with a regular polygon the having 19 sides.)[/quote]Arqady, may you tell me more about those 'nice' roots :roll: ? I've tried hard but without success so far.

Can this equation be solved in the same way :?:

$8x^{3}-4x^{2}-4x+1=0$

[quote="arqady"]The last hint: $\sum_{k=1}^{9}2\cos{\frac{2k\pi}{19}}=-1.$ :wink:[/quote]Sorry but I still don't get it. I was [i]so[/i] far... but I didn't carry on, as this would yield a lenghty 9th degree polynomial. Now if you tell me this is reducible, it might be worth a try. But the "7" as absolute term of $x^{3}+x^{2}-6x-7$ makes me hesitate anyway. :maybe:  :huh:
I'd say, for this one, more remains to 'check up'  :?:

[quote="spanferkel"]
I'd say, for this one, more remains to 'check up'  :?:[/quote]
I think, it is possible to try if to wish to receive good roots :wink:

[quote="spanferkel"]Have you done it?  :D[/quote]
Yes! :lol:
It gives also a nice root of the equation:
$\sqrt{4+\sqrt{4+\sqrt{4-x}}}=x.$ :wink:

Can it be done without "checking up"  :D  :?:


floor function
search
number theory proposed
number theory

Let $p > 2$ be a prime and let $a,b,c,d$ be integers not divisible by $p$, such that
\[
\{ra/p\} + \{rb/p\} + \{rc/p\} + \{rd/p\} = 2
\]
for any integer $r$ not divisible by $p$.  Prove that at least two of the numbers $a+b$, $a+c$, $a+d$, $b+c$, $b+d$, $c+d$ are divisible by $p$.  (Note: $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$.)

I did a quick search and didn't find a solution to this problem.  Anyone have a solution different from Kalva's?

(1)
first, we reduce the problem to a=1, 0<b<=c<=d<p

we call a=1,0<b<=c<=d<p a primitive form
suppose a,b,c,d is not a primitive form, then let ka==1(mod p)
define a',b',c',d' as the smallest positive integers such that a',b',c',d'==ka,kb,kc,kd(mod p)
(without loss of generality we may assume a'<=b'<=c'<=d')
therefore we have reduce a,b,c,d to primitive form a'=1, 0<b'<=c'<=d'<p

if we can prove that a'+d'=p,b'+c'=p, then correspondingly we have a+d==b+c==0(mod p)

if we can prove that {r'a'/p}+{r'b'/p}+{r'c'/p}+{r'd'/p}$\neq$ 2 for some r' not divisible by p, then there is an r not divisible by p corresponding to r' such that
{ra/p}+{rb/p}+{rc/p}+{rd/p}$\neq$2

(2)
these are trivial possibilities, easy to proof
b=1 (a+d=p,b+c=p)
d=p-1 (a+d=p,b+c=p)
b+c=p (a+d=p,b+c=p)
b,c<p/2 (S(r)$\neq$ 2 for some r)
b,c>p/2 (S(r)$\neq$ 2 for some r)
p=3,5,7

(3)
this is not trivial, therefore I shall proof it
$p \geq 11, a=1, 1<b< \frac{p}{2}<c \leq d<p-1$

we can easily deduce that b+c>p

define S(r)={ra/p}+{rb/p}+{rc/p}+{rd/p}
suppose that S(r)=2 for r=1,2,...,p-1, we shall deduce a contradiction

[b]Case I[/b] $[p/b]\neq[p/(p-c)]$

we can prove that [p/b]<[p/(p-c)],[p/(p-d)]<p-1

since a=1, therefore {(r+1)a/p}-{ra/p}=a/p for r=1,2,...,p-1

we have {(r+1)b/p}-{rb/p}=b/p for r=1,2,...,[p/b]-1
and {(r+1)b/p}-{rb/p}=b/p-1 for r=[p/b]

we have {(r+1)c/p}-{rc/p}=c/p-1 for r=1,2,...,[p/(p-c)]-1
and {(r+1)c/p}-{rc/p}=c/p for r=[p/(p-c)]

we have {(r+1)d/p}-{rd/p}=d/p-1 for r=1,2,...,[p/(p-d)]-1
and {(r+1)d/p}-{rd/p}=d/p for r=[p/(p-d)]

from the above, we can deduce that $S([p/b])\neq 2$, a contradiction

[b]Case II[/b] $[p/b]=[p/(p-c)]$

by lemma 2, {(n+1)d/p}-{nd/p}=d/p or d/p-1

if {(n+1)d/p}-{nd/p}=d/p-1 (n<p-2 is the integer defined in lemma 1)
then by lemma 1, we have
{(n+1)b/p}-{nb/p}=b/p-1
{(n+1)c/p}-{nc/p}=c/p-1
since a=1, we have
{(n+1)a/p}-{na/p}=a/p
therefore S(n+1)-S(n)=S(1)-3, this is a contradiction with S(r)=2 for r=1,2,...p-1

if {(n+1)d/p}-{nd/p}=d/p (n<p-2 is the integer defined in lemma 1)
then by lemma 3, we have
{(n+2)d/p}-{(n+1)d/p}=d/p-1
by lemma 1, we have
{(n+2)b/p}-{(n+1)b/p}=b/p
{(n+2)c/p}-{(n+1)c/p}=c/p
since a=1, we have
{(n+2)a/p}-{(n+1)a/p}=a/p
therefore S(n+2)-S(n+1)=S(1)-1, this is a contradiction with S(r)=2 for r=1,2,...,p-1

[quote]
[b]Lemma 1[/b]
if $p \geq 11$ is a prime number and $b,c$ are positive integers that satisfy the following conditions:
(I) $1<b<\frac{p}{2}<c<p-1$
(II) $b+c>p$
(III) $[\frac{p}{b}]=[\frac{p}{p-c}]$
then there exist a positive integer $n<p-2$ such that
$\{\frac{(n+1)b}{p}\}-\{\frac{nb}{p}\}=\frac{b}{p}-1$
$\{\frac{(n+1)c}{p}\}-\{\frac{nc}{p}\}=\frac{c}{p}-1$
$\{\frac{(n+2)b}{p}\}-\{\frac{(n+1)b}{p}\}=\frac{b}{p}$
$\{\frac{(n+2)c}{p}\}-\{\frac{(n+1)c}{p}\}=\frac{c}{p}$
[/quote]
Hint:
To prove Lemma 1, note that
{(r+1)b/p}-{rb/p}=b/p-1 if and only if r=[tp/b] for some integer t
{(r+1)c/p}-{rc/p}=c/p if and only if r=[tp/(p-c)] for some integer t
Let t be the smallest positive integer such that [tp/b]<[tp/(p-c)], we can take n=[tp/b]

[quote]
[b]Lemma 2[/b]
if $d,n$ and $n+1$ are positive integers not divisible by $p$, then
$\{\frac{(n+1)d}{p}\}-\{\frac{nd}{p}\}=\frac{d}{p}$ or $\frac{d}{p}-1$
[/quote]

[quote]
[b]Lemma 3[/b]
if $\frac{p}{2}<d<p$ and $1 \leq n<n+2 \leq p-1$ and $\{\frac{(n+1)d}{p}\}-\{\frac{nd}{p}\}=\frac{d}{p}$,
then $\{\frac{(n+2)d}{p}\}-\{\frac{(n+1)d}{p}\}=\frac{d}{p}-1$
[/quote]

Hmm... It s quite difficult to see the essential idea of your proof.

The idea is very simple.

I proved that
[quote]if p does not divides a+b,a+c,a+d,b+c,b+d,c+d, then there exist a positive integer r not divisible by p such that {ra/p}+{rb/p}+{rc/p}+{rd/p}$\neq$2
[/quote]

Proof:

[b]Defination[/b]
For prime number p>2 and integers a',b',c',d', define (a',b',c',d') as [i]primitive form[/i] iff
(1) a'=1, 0<b'<=c'<=d'<p
(2) {a'/p}+{b'/p}+{c'/p}+{d'/p}=2

[b]Defination[/b]
For prime p>2 and integers a,b,c,d not divisible by p such that {a/p}+{b/p}+{c/p}+{d/p}=2,
We say that (a',b',c',d') [i]is a primitive form[/i] of (a,b,c,d) if
there exist integers a'',b'',c'',d'',k such that
(1) a'',b'',c'',d'' is a rearrangement of a,b,c,d
(2) ka'',kb'',kc'',kd''==a',b',c',d'(mod p)
(3) (a',b',c',d') is a primitive form

[b](1)[/b]
Let p>2 be a prime; Let a,b,c,d be integers not divisible by p; Let {a/p}+{b/p}+{c/p}+{d/p}=2; Let (a',b',c',d') be a primitive form of (a,b,c,d).
If p does not divides a+b,a+c,a+d,b+c,b+d,c+d,
Then b',c'<p/2 or b',c'>p/2 or 1<b'<p/2<c'<=d'<p-1

[b](2)[/b]
If (a',b',c',d') is a primitive form; b',c'<p/2 or b',c'>p/2 or 1<b'<p/2<c'<=d'<p-1;
Then there exist an r' in {1,2,...,p-1} such that {r'a'/p}+{r'b'/p}+{r'c'/p}+{r'd'/p}$\neq$2

[b](3)[/b]
If (a',b',c',d') is a primitive form of (a,b,c,d) and there exist integer r' not divisible by p such that {r'a'/p}+{r'b'/p}+{r'c'/p}+{r'd'/p}$\neq$2,
Then there exist integer r not divisible by p such that {ra/p}+{rb/p}+{rc/p}+{rd/p}$\neq$2

From [b](1)[/b],[b](2)[/b],[b](3)[/b], we have
If p>2 is prime; a,b,c,d are integers not divisible by p;{a/p}+{b/p}+{c/p}+{d/p}=2;p does not divides a+b,a+c,a+d,b+c,b+d,c+d
Then there exist integer r not divisible by p such that
{ra/p}+{rb/p}+{rc/p}+{rd/p}$\neq$2 [b]QED[/b]


trigonometry
inequalities
inequalities proposed

In a triangle prove that: \[\sum a^{2}\geq \frac{(a+b+c)^{3}}{9\sqrt{3}\cdot R}\]

Proof.
Since $\sum{a^{2}}>\frac{(a+b+c)^{2}}{3}$(CAUCHY)
The question is to prove$a+b+c\leq 3\sqrt3R(*)$
Let$a=2R\sin{A},b=2R\sin{B},a=2R\sin{C}$
Since$\sin{A}+\sin{B}+\sin{C}\leq 3\sin{\frac{A+B+C}{3}}=\frac{3\sqrt3}{2}$
Multiply both sides by 2R we get the$(*)$ :)

After putting in $\left\{x+y=a\right\}_{cyc}$ it is a stronger result than both of the weak inequalities: \[8\left(\sum x\right)^{3}\ge 27 \left(\prod x+y \right) \\ \frac{1}{\sqrt{3}}\sum\left( x^{2}+xy \right)\ge 2\sqrt{\sum x^{2}yz}\iff \left(\sum x^{2}+xy\right)^{2}\ge 12 \sum x^{2}yz \]
which are easy with Am-Gm, and Muirhead respectively.
Indeed the inequality is equivalent to: \[\left[\frac{\left(\sum x^{2}+xy\right)^{2}}{12\sum x^{2}yz}\right] \ge \left[\frac{8\left(\sum x\right)^{3}}{27\prod{x+y}}\right]^{2}\]


Deci: am invatat sa bat text in Latex. Cum procedez ca sa inserez aici (intr-un mesaj nou) text matematic?

Multumesc

Pentru a scrie in $LaTeX$ folositi semnele de dolar (SHift+4)

(semn dolar)a^2+b^2+c^2(semn dolar)

ar trebui sa apara $a^{2}+b^{2}+c^{2}$.
Pentru eventualele nedumeriri va invit sa cititi  subforumul LaTeX si implicit intructiunile pe care le doriti.
Sper ca am fost de folos  :P


function
calculus
derivative
real analysis
real analysis unsolved

Let $ f: [a,b] \to \mathbb{R}$ to be a differential function with the property that there exist a number $ x_0 \in (a,b)$ such that . Prove that there exist a number $ k\in (a,b)$ such that

To make life easier define
$ g(x) \equal{} f((b \minus{} a)x \plus{} a) \minus{} f(a)$, (differentiable on $ [0,1]$).
$ g'(x) \equal{} (b \minus{} a)f'((b \minus{} a)x \plus{} a)$.
So, we have to prove that there is a $ k \in (0,1)$ so that $ g'(k) \equal{} g(k)$.
If this is not the case, then (because of the Darboux continuity of $ g'$) we know that $ g'(x) > g(x)$ for $ x\in (0,1)$. (Or $ g'(x) < g(x)$ for all $ x$.)
$ g(0) \equal{} 0$ and $ g'(0) > g(0) \equal{} 0$, so $ g$ has positive $ y$ - values.
There is a $ x_0$ for which $ g'(x_0) \equal{} 0$.
Then $ g(x_0) < g'(x_0) \equal{} 0$.
Then $ g$ must have a zero (say $ x_1$).
Let $ x_1$ be the smallest zero.
Then $ g'(x_1) > g(x_1) \equal{} 0$.
This cannot be true, because for the smallest zero the derivative is negative or zero.
Conclusion: There is an $ k$ with $ g'(k) \equal{} g(k)$

What happened to original problem statement

No idea. The problem was:

Let $ f: [a,b] \to \mathbb{R}$ to be a differential function with the property that there exist a number $ x_0 \in (a,b)$ such that $ f'(x_0)\equal{}0$. Prove that there exist a number $ k\in (a,b)$ such that $ f'(k) \equal{} \frac {f(k) \minus{} f(a)}{b \minus{} a}$


Let $ n$ be a positive integer.Prove that the number of $ k$ for which $ C_n^k$ is odd is a power of $ 2$

See [url]http://mathworld.wolfram.com/Stolarsky-HarborthConstant.html[/url]
If $ b$ is the number of ones in the binary representation of $ n$, then the number of odd $ C_n^k$ is $ 2^b$ (Glaisher).

Also see [url]http://www.cecm.sfu.ca/organics/papers/granville/Binomial/intro.html[/url] for extensions showing that the number of $ C_n^k$ congruent with odd $ j$ modulo $ 4$ is zero or a power of $ 2$. Similarly for modulo $ 8$; conjectures the same for modulo $ 2^s$.

My own question is if the number of $ C_n^k$ exactly divisible by $ 2^s$ is also a power of $ 2$, for any value of $ s$ (for $ s\equal{}0$ we get the question of this topic). It may very well be that it's an easy consequence, or maybe untrue - but I lack the time to delve into it.


probability
logarithms
function
probability and stats

I am reading a lemma and I do not understand one thing.
I will write the relevant parts. If requested, I can write more details to clarify the problem.

Suppose we have P, Q two probability measures on the product space I.
P and Q have identical marginals to the sets $ a\in A$.

In the proof of lemma, there's this part:
we have exploited that $ \sum_{i}p(i) \log \theta_{a}(i_a)$ depends on P through its $ a$-marginal only and therefore is equal to $ \sum_{i} q(i) \log \theta_a(i_a)$
Note: $ \theta(i_a)$ are some functions that depend on the variables $ i_a$ that belong to $ i$ and to some fixed set $ a \in A$.

My question is why those sums are equal.

-In my mind, the marginals of  distribution p, say, are the functions $ p(i_a)$. But I don't see how they are relevant here, the sums are over $ i \in I$, not $ i_a \in A$.

Thank you

ok, somebody helped me to see what was about
The idea is to write the sum over i as a sum over $ i_a$ and its complement, then the logarithm term that depends on $ i_a$ is taken out of the sum that is a marginal.
No further help is needed but I cannot delete the post.


geometry parallelogram rectangle trapezoid

Okay, so there's this puzzle where there are 13 convex polygons using all tangrams. I have found out 7 of them. The parallelogram, a square, the rectangle, the trapezoid, the regular trapezoid, the pentagon that looks like has looks almost like a square, and the hexagon. I can't seem to figure out the others. Is there an easier way? And if you really want to, could you post the others? Heh, thanks...  :D

How would we know what to post when you have the problem?  Seriously, try to draw them out and put it on.

Uh, I hope this might help... Or do you want the other tangram things?

No, I want to see the ones that you can't get.

The ones circled in red are the ones I don't have


[quote="ashrafmod"]x,y,z,t  the real positivs numbers  such that x+y+z+t=1.
Prove that
sqrt(x^2y^2+z^2t^2)+sqrt(x^2t^2+y^2z^2)>=(sin45)/4.[/quote]

$\sqrt{x^{2}y^{2}+z^{2}t^{2}}+\sqrt{x^{2}t^{2}+y^{2}z^{2}}\geq \frac{ \sin 45}{4}$

counterexample:$x=z=10^{-6}, y=t= 0.5-10^{-6}$.


[quote="sri340"]cindy and dave left the dock to canoe downstream. fifteen minutes later tammy left by motor boat with the supplies . since the motor boat travelled twice as fast as the canoe,it caught up with the canoe 3km from the dock. what was the speed of the motorboat ?[/quote]
[hide]
Let $ x$ = the rate of the motor boat . . . (#)

Let $ \frac x2$ = the rate of the canoe . . . (#)

The traveled distance for each is $ 3$ kilometers.

Time = distance/rate

The canoe's rate is $ \frac{3}{(\frac x2)} \equal{} \frac6x$.

The motor boat's rate is $ \frac3x$.

The difference in their times of travel is $ 15$ minutes.

$ 15$ minutes = $ \frac14$ hour.

$ \frac6x \minus{} \frac 3x \equal{} \frac14$.

$ \frac3x \equal{} \frac14$.

$ x \equal{} 12$.

The motor boat's rate is $ 12$ kilometers per hour.  (Answer)

(#) Or, the rate of the canoe could be $ x$, and the rate of the motor
boat could be $ 2x$.  I chose my set-up so a solver would not forget
to multiply the $ x$ result by 2.
[/hide]

cindy and dave left the dock to canoe downstream. fifteen minutes later tammy left by motor boat with the supplies . since the motor boat travelled twice as fast as the canoe,it caught up with the canoe 3km from the dock. what was the speed of the motorboat ?
3=v(m)x(15/60)v(c)/3v(c)
v(m)=36

With reference to my earlier post, the motor boat
moving at 12 km/hr times (1/4)hour  travels 3 km.

The canoe moving at 6 km/hr for (1/2) hour (15 minutes more)
travels 3 km.

motor boat's speed = 12 km/hr.

[quote="sri340"]cindy and dave left the dock to canoe downstream. fifteen minutes later tammy left by motor boat with the supplies . since the motor boat travelled twice as fast as the canoe,it caught up with the canoe 3km from the dock. what was the speed of the motorboat ?[/quote]

Speed of motorboat is twice that of canoe.
So in same time, distance travelled by motorboat = $ 2$(distance travelled by canoe)
As motorboat catches the canoe 3 km from the dock that means in same time canoe has travelled a distance of 1.5 km.
That means earlier 1.5 km has been travelled by canoe in 15 minutes as motorboat started after 15 minutes.
Thus speed of canoe = $ \frac{1.5 km}{15 minutes}$ = $ (\frac{3}{2})(\frac{4}{1}) kmph$ = $ 6 kmph$.
And speed of motorboat = $ (2)(6 kmph) \equal{} 12 kmph$


geometry
rotation
geometric transformation
3D geometry
sphere
analytic geometry
vector

let $A$ and $B$ be two path connected subsets of $R^2$. Prove that there exists
a line that cuts $A$ into $2$ parts of same area and $B$ into $2$ parts of same area.

Consider an arbitrary line L. Let P(A) be the the point such that the normal to L at P(A) divides the area of A into two equal parts. Similarly define P(B). Rotate line L continuously, and P(A) and P(B) will move along L continuously. When the line has done a half-rotation back to its original position, P(A) and P(B) are where they were originally, but the line is oriented oppositely.  Therefore, P(A) and P(B) crossed each other at some time as the line was being rotated. The normal to the line at P(A) = P(B) at this point is the desired line.

[quote="schulmannerism"]Let P(A) be [b]the[/b] point such that..[/quote]

This needs a litle refinement, since there may be more than one such point. The theorem holds in $n$ dimensions, with $n$ bounded measurable sets of positive measure (it has nothing to do with connectedness :?; the sets may consist of as many connected components as we want), and it's called the Ham Sandwich Theorem, I think :).

I think it's fairly easy to give a proof very simialr to the one here with the aid of the Borsuk-Ulam theorem (any continuous map from the sphere $\mathbb S^n$ to $\mathbb R^n$ maps some pair of antipodes to the same point).

"normal" means perpendicular. (I should have just said perpendicular) Therefore P(A) is unique.

I know what normal means :).

What I meant was that there may be more than one line with given direction which bisects a set, even with the path connectedness condition. Think about the case when the set consists of two discs of equal radius connected by a segment. There are infinitely many lines perpendicular to that segment which divide this figure into two parts with equal areas.

and if $A$ and $B$ are open set I thnik it works

Stupid me  :blush:

The simplest thing I can think of is to consider the [i]leftmost[/i] such point on the oriented line that we will rotate.

I believe the following works in three dimensions (with the aid of the Borsuk Ulam theorem, like I said above; if it's ok, this line of thought can easily be generalized to higher dimensions):

Given a unit sphere, to each point $P$ on it we associate the line $OP$, with a coordinate system given by the oriented distance from $O$ (the center of the sphere), the positive direction being $\overrightarrow{OP}$. For each such vector $\overrightarrow{OP}$, we take the normal plane which cuts $OP$ in the highest point and bisects the first region, and also the plane which cuts the line in the lowest point and bisects the region. Now we consider the unique plane which is equidistant to these two. This one cuts the line $OP$ in some point $P_1$. In the same way, we obtain points $P_2,P_3$ corrsponding to the second and third region respectively.

To summarize, to each point $P$ we have associated the points $P_1,P_2,P_3$. Instead of these triples of points, we consider the pairs $\overline{P_1P_2},\overline{P_1P_3}$ (these are oriented distances according to our orientation on the line $OP$). It means that we have a correspondence between the points on the sphere and some pairs of reals. It can be argued that these correspondences are continuous (it may not be such an easy task, but intuitively, it should be Ok).

The Borsuk-Ulam theorem then states that there is a pair of antipodes mapped to the same pair of reals. However, if $P,P'$ are antipodes, then $\overline{P_1P_2}=-\overline{P'_1P'_2}$, so $\overline{P_1P_2}=\overline{P'_1P'_2}$ implies that both of them are $0$, or, in other words, $P_1=P_2$. In the same way we get $P_1=P_3$, and we're done.

I hope it's Ok.


2n points are given in a plane.There isn't any restriction about the positions of the points.Prove that, one can find a line in this plane such that:
i) the line must pass through at least 2 points.
ii) the number of points on 2 sides of plane which are seperated by the line are equal.

I don't think this is true :

Consider a triangle $ ABC$; and 3 points $ A'$, $ B'$ and $ C'$ such that:
$ B\in [A'C]$
$ C\in [AB']$
$ A\in [BC']$

The condition is not verified with the 6 points $ A, B, C, A', B', C'$.


quadratics
number theory
number theory unsolved

Hi! I'm Juan from Argentina. I was reading a book of a spanish mathematician when he introduced quadratic fields theory with the equation $x^{2}+xy-3y^{2}=15$. But that's easy with elementary number theory. So I wrote him and he told me "can you solve this

$x^{2}+xy-3y^{2}=35,$

because if not I'll change it?

Well. I can't. But can you solve that by elementary number theory?

Thanks.
Juan.

If I'm not mistaken that's also very easy. Multiply both sides of the equation by $4$ and see that:
$4x^{2}+4xy-12y^{2}= 4x^{2}+4xy+y^{2}-13y^{2}= (2x+y)^{2}-13y^{2}= 140$

So if there would be a solution to our equation, there would be a solution to the equation $a^{2}-13b^{2}=140$. But $a^{2}-13b^{2}\equiv a^{2}+b^{2}\mod{7}$ and it is well known (or it can be checked by hand) that this implies $a \equiv b \equiv 0 \mod{7}$. But then, $49|a^{2}-13b^{2}$ and $49 \not | 140$, contradiction.

One can check the mod $7$ without transformating the equation, but I think that it is easier in this transformated form.

Thank you! That's very good! Easier than I thought.

Thanks a lot. Good bye!


Find all triples of positive integers (a,b,c) such that:
$(1+1/a)(1+1/b)(1+1/c) = 2$

There are no integers satisfying the equation, because of AM-GM.

I would like to see you apply AM-GM.
http://www.mathlinks.ro/Forum/viewtopic.php?t=42687

sorry it doesn;t work..i made a stupid mistake.. my bad.

do $a\geq b\geq c$ and then u have

$(1+1/c)^{3}\geq 2$ and then u find the largest value of $c$ that works and u find that the solution is any permutaion of

$\{3,4,5\}$

yeah guessed i missed some. but i assume that those solutions can be attained in the same mannor.

maokid, it seems from the link that you actually missed three other distinct solutions  :wink:


search
combinatorics unsolved
combinatorics

In a chess tournament with $ n$ participants, every players must play against the rest two match per person (just like the England Premier League). In every match, the winner get $ 1$ points, the loser get $ 0$ points and if the match ends with a draw then both players get $ 0.5$ point. At the end of the tournament, we notice that the difference between the points in the first round and the second round of every player is not less than $ n$. Prove that those differences are all $ n$.


induction
group theory
superior algebra
superior algebra theorems

This is just a general question... I learned this in Analysis so I thought it might fit in this section.

To prove by induction a statement P(n), the general way is to show it is true for P(1), assume P(k) is true for n=k, then prove that P(k+1) is true for n=k+1.

Is there another way to prove by induction in that you show P(a) is true for a<n, then show it is true for P(n)?  I saw a proof like this for a few theorems given in group theory and ring theory and was wondering if this was the method they used.

Yes.  This is known as strong induction.  Instead of assuming that $ P(k)$ is true, you assume that $ P(1), P(2), ... P(k)$ are true.

Wouldn't it be assuming P(1), P(2),...P(k-1) is true?

Not if you want to prove $ P(k\plus{}1)$.


Darn, everyone where I live is cheering for Red Sox, I naturally thought everyone on AoPS would cheer for Red Sox too... lol. :lol: (It's actually because they come from Boston, which is in Massachusetts, which is where I ... well... you know :D)

Probably not so for someone who lives in New York or Arizona or Kansas or ... Alaska!!!! (Or Hawaii.)

The Red Sox have won 2 of the last 4 World Series.  Time to spread it around a little more.*  I was rooting for the Cubs, but it doesn't look like that's going to happen.

*I know, 2 World Series don't really make up for an 86-year drought.  But I also don't want to see one team win 3 out of 5, either.

Hmmm... so far... the series look somewhat alike...

Red Sox lead Angels, 2-0
Rays lead White Sox, 2-0
Phillies lead Brewers, 2-0
Dodgers lead Cubs, 2-0


In ABC, points P and Q are on AB and AC respectively such that PQ is tangent to the incircle of ABC. Prove that $ p\equal{}AB\plus{}AC\minus{}BC$, where p is the perimeter of $ APQ$.

It's a nice problem  :D and it is not so hard. The result can be easily deduced from the property of the tangents from a point to a circle: these two tangents are equal. After that, it follows some calculation.


[quote="math92"]Given:  $\frac{1}{3}=\frac{1}{a}+\frac{1}{b}$ where $a$ and $b$ are different whole numbers, what are the values of $a$ and $b$?[/quote]
I think there is a general solution to this kind of problems...
Who knows it?

[hide="Answer"]$\frac{1}{3}=\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}\Rightarrow ab=3(a+b)\Rightarrow \boxed{a=4,b=12\text{ or }a=12,b=4}$[/hide]

Here is a more general way to approach it.

Note that the left side is 1/3.  Since 1/a and 1/b are unequal, there is a greater of the two.  Is it possible that the greater of the two fractions is as small as 1/6?  Well in that case, the smaller one can be at most 1/7, and then their sum is too small (just short of 1/3).  [b]Thus the larger of the two fractions has to be either 1/5 or 1/4.[/b]

Case I:  1/5

Then the other fractions is equal to 1/3 - 1/5 = 2/15, which is impossible.  So this case fails.  Thus it has to be case II.

Case II: 1/4

Then the other fraction is equal to 1/3 - 1/4 = 1/12!  woot woot this works, [b]so the only answer is 1/4 and 1/12.[/b] and we have proved it without any guessing games.

Err...

[hide]
$\frac{1}{a}+\frac{1}{b}=\frac{1}{3}$
$\frac{a+b}{ab}=\frac{1}{3}$
$3a+3b=ab$
$0=ab-3b-3b$
Add 9 to both sides
$9=ab-3b-3b9$
$9=(a-3)(b-3)$
Verify that this is indeed true. This "trick" is known among AoPSers as "Simon's Favorite Factoring Trick".
[/hide]

yep thats the easiset way.
If $\frac1x+\frac1y=\frac{a}{b}$, than $(ax-b)(ay-b)=b^2$

lol oh yeah; most of the variants on this problem that i've seen say positive integers :)

[quote="WindSlicer"]Don't forget about negatives...[/quote]

Yes, WindSlicer is right.

The answer is $(a,b)=(-6,2),(2,-6),(4,12),(12,4)$.

thats your definition, whole numbers when i learned it was {1,2,3,4,...}
edit: here look at this
[url=http://mathworld.wolfram.com/WholeNumber.html]Whole Number[/url]

Weird, I learned Whole number was a number with no fractional value (i.e. $-4\frac{1}{2}$ is obviously not "whole"). Which makes logical sense, at least to me. Hmmm, well I'm thinking back and I remember hearing "whole number" when we were talking about fractions of a pie chart in elementary school, and you can't have a negative fraction of a pie chart, so I guess that's right. But I've never seen the term "whole number" used in a contest before, so... either way, it's good to notice that there is a negative answer to this, wether it is extranneous or not.

[quote="kunny"][quote="Palytoxin"]the problem says "whole numbers" from math92, so no negatives[/quote]

Whole number=Integer ;)[/quote]

Actually, whole numbers = intergers without zero
I think at least

I say that we all discard the use of the phrase 'whole numbers' and stick with either integers or natural numbers.

[quote="Barnacle"]oh yeah i forgot about that method as its much faster and foolproof; why is it called simon's favorite factoring trick haha its pretty well-known as just a general tool of manipulation.. is simon some famous mathematician?

still, on these kind of problems to be able to approach them logically as opposed to algebraically is powerful; i dont like the bruteness of algebra as it really trades insight for sort of forcing the system to yield the answer.[/quote]

  In my opinion algebraic manipulation is a form of logical reasoning, however; I can understand certain situations arise ( similar to this one) where such tools as formulas, and "factoring" tricks allow us to derive at such solutions quite quickly, without to much thought into logical reasoning.  Anyways, my initial solution to the problem was posted above by chess64, however; since we seem to be posting a variety of different answers how about another one?

 [hide] We know that $\frac{1}{3}$ = $\frac{1}{a}$ + $\frac{1}{b}$

  We can re-write this as  $\frac{1}{3}$ - $\frac{1}{a}$= $\frac{1}{b}$

  We can simplify $\frac{1}{3}$ - $\frac{1}{a}$ into,

  $\frac{a}{3a}$ - $\frac{3}{3a}$.  So our new equation is:

  $\frac{a}{3a}$ - $\frac{3}{3a}$ = $\frac{1}{b}$.

  Now, let's assume that

  $b$ = $3a$ ( a reasonable assumption ), the only thing we would need to

  satisfy the answer was for $a - 3 = 1$. Meaning $a = 4$, and $b=3a=12$

  Here we used algebraic manipultion to set up the problem in approachable ground, then we made a logical assumption which aided us to the solution

  [/hide]

That's a nice method, but you can't be sure that you have not ignored another solution.

I think you should stop. THat was 7788 words.

Good, he's gone.

[hide]12=a 4=b[/hide]

from 9yo using mom's account :)


[quote="Null"]Given $ xyz \equal{} 8$ then prove that $ \frac1{\sqrt {x^2 \plus{} 1}} \plus{} \frac1{\sqrt {y^2 \plus{} 1}} \plus{} \frac1{\sqrt {z^2 \plus{} 1}} > 1.$

I'm not sure if this holds for any $ x,y,z\in\mathbb R.$ Can someone clarify this? I'd like a hint to solve this.[/quote]

maybe you can let $ x\equal{}\frac{1}{a},y\equal{}\frac{1}{b},z\equal{}\frac{1}{c},abc\equal{}8$,then by Cauchy's inequlity and AM-GM

[quote="Null"]Given $ xyz \equal{} 8$ then prove that $ \frac1{\sqrt {x^2 \plus{} 1}} \plus{} \frac1{\sqrt {y^2 \plus{} 1}} \plus{} \frac1{\sqrt {z^2 \plus{} 1}} > 1.$
[/quote]
We have $ \sqrt{x^2\plus{}1}<\vert x\vert\plus{}1$ (no equality because $ x\not\equal{}0$), thus
\[ \sum_{cyc}\frac{1}{\sqrt{x^2\plus{}1}}>\sum_{cyc}\frac{1}{\vert x\vert\plus{}1}\mbox{.}\]
Setting $ \vert x\vert\equal{}2\frac{a}{b}$, $ \vert y\vert\equal{}2\frac{b}{c}$ and $ \vert z\vert\equal{}2\frac{c}{a}$, with positive real numbers $ a$,$ b$, $ c$ and applying Cauchy-Schwartz, we get
\[ \sum_{cyc}\frac{b}{b\plus{}2a}\equal{}\sum_{cyc}\frac{b^2}{b^2\plus{}2ab}\geqslant\frac{(\sum a)^2}{\sum (a^2\plus{}2ab)}\equal{}1\mbox{,}\]
which proves the desired for any $ x$, $ y$, $ z \in \mathbb{R}$ with $ xyz\equal{}8$.
:)

[quote="Deng Qunyi"][quote="Null"]Given $ xyz \equal{} 8$ then prove that $ \frac1{\sqrt {x^2 \plus{} 1}} \plus{} \frac1{\sqrt {y^2 \plus{} 1}} \plus{} \frac1{\sqrt {z^2 \plus{} 1}} > 1.$

I'm not sure if this holds for any $ x,y,z\in\mathbb R.$ Can someone clarify this? I'd like a hint to solve this.[/quote]

maybe you can let $ x \equal{} \frac {1}{a},y \equal{} \frac {1}{b},z \equal{} \frac {1}{c},abc \equal{} 8$,then by Cauchy's inequlity and AM-GM[/quote]
My problem is how to apply Cauchy, can you tell me how?

At first, if we want to do it like Deng Qunyi suggested, we have afterwards $ abc\equal{}\frac{1}{8}$.
Secondly, I don't think his substitution will be that useful...  :wink:

[quote="Null"]Martin N.
Didn't get your solution well, don't get why you took that substitution, can you give me more details?[/quote]

Ok. Well...my substitution somehow suggests itself...normally, if we have $ xyz\equal{}1$, we set $ x\equal{}\frac{a}{b}$ and so on with positive reals $ a,b,c$. Here, the product $ xyz$ is $ 8\equal{}2^3$ which is why I add the factor $ 2$... :P


integration
search
logarithms
calculus
calculus computations

Consider the differential equation that describes the free fall of a body:

$ \dfrac{dv}{dt}\equal{}g\minus{}cv^r$

if r>1, this differential equation cannot be solved exactly.

...says my extended essay advisor

He does not why it cannot be solved exactly (the book just says so therefore just admit the truth)

I really want to know and explain why it cannot be solved exactly.

This is not true. For $ r\equal{}2$ you have the solution:

$ v(t) \equal{} \sqrt{\frac{g}{c}} \tanh\left(\sqrt{g \, c}\,(t\plus{}C)\right)$

where $ C \in \mathbb{R}$.

What does it mean that it can't be solved exactly anyway? It does have solutions. If you mean these solutions cannot be expressed by "elementary means" this requires huge formal framework to prove

$ \int \frac{dv}{v^r\minus{}\frac{g}{c}} \equal{} \minus{}ct\plus{}C$ does not have a general solution is what he means.

for some specific r, you might be able to do it though (such as 2 above)?

This has been discussed a while ago. Try to search for the thread, a discussion on it was done.

hello, it gives also for $ r\equal{}3$ a solution, see here:
$ t\plus{}1/3\,\ln  \left( v \left( t \right) \minus{}\sqrt [3]{{\frac {g}{c}}}
 \right) {c}^{\minus{}1} \left( {\frac {g}{c}} \right) ^{\minus{}2/3}\minus{}1/6\,\ln
 \left(  \left( v \left( t \right)  \right) ^{2}\plus{}v \left( t \right)
\sqrt [3]{{\frac {g}{c}}}\plus{} \left( {\frac {g}{c}} \right) ^{2/3}
 \right) {c}^{\minus{}1} \left( {\frac {g}{c}} \right) ^{\minus{}2/3}\minus{}1/3\,\sqrt {3}
\arctan \left(  \left( 2/3\,v \left( t \right) {\frac {1}{\sqrt [3]{{
\frac {g}{c}}}}}\plus{}1/3 \right) \sqrt {3} \right) {c}^{\minus{}1} \left( {\frac
{g}{c}} \right) ^{\minus{}2/3}\plus{}{\it \_C1}\equal{}0$

Sonnhard.


inequalities
trigonometry
quadratics
algebra
inequalities unsolved

[quote="idioteque"]If $x, y, z$ are positive reals such that $xyz=x+y+z+2$, prove that

\[ 5(x+y+z)+18\geq 8(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}).  \][/quote]
My ugly proof:
$xyz=x+y+z+2\Leftrightarrow\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=1.$
Let $a=\frac{1}{1+x},$ $b=\frac{1}{1+y}$ and $c=\frac{1}{1+z}.$
Hence, $5(x+y+z)+18\geq 8(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})\Leftrightarrow$
$\Leftrightarrow5\cdot\sum_{cyc}\frac{1-a}{a}+18\geq8\cdot\sum_{cyc}\sqrt{\frac{(1-a)(1-b)}{ab}}\Leftrightarrow$
$\Leftrightarrow5(ab+ac+bc)+3abc\geq8\cdot\sum_{cyc}a\sqrt{bc(a+b)(a+c)},$ where $a+b+c=1.$
But $\sum_{cyc}a\sqrt{bc(a+b)(a+c)}\leq\sqrt{\sum_{cyc}abc(a+b)(a+c)}=\sqrt{abc+abc(ab+ac+bc)}.$
Hence, remain to prove that
$5(ab+ac+bc)+3abc\geq8\cdot\sqrt{abc+abc(ab+ac+bc)}.$
Let $ab+ac+bc=v,$ $abc=w.$ Then $v\geq9w$ and $v^2\geq3w.$
Id est, remain to prove that
 $5v+3w\geq8\sqrt{\frac{v^2}{3}+vw}\Leftrightarrow(v-9w)(11v-3w)\geq0,$ which obviously true. :)

If $x, y, z$ are positive reals such that $xyz=x+y+z+2.$ Prove that
$7(x+y+z)+30\geq 12(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}).$
It is stronger. ;)

I think that we are to prove for $pqr=\frac{qr}{p}+\frac{rp}{q}+\frac{pq}{r}+2,\ 5(pqr-2)+18\geq 8(p+q+r).$

but your substitution does not really reduce the problem, isn't it?

Thanks for replying.

I also managed to find an ugly solution, but this one is really
ugly..

Rewrite the inequality as

\[ 5(x+y+z+2) \geq 8(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}-1) \]
and use the conditon to obtain equivalent forms
\[ \frac{5}{8} \geq \frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}-1}{xyz} \]
\[ \frac{5}{4} \geq \frac{2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})}{xyz}-\frac{2}{xyz} \]

Substituting $x_1=\sqrt{yz}$, $y_1=\sqrt{zx}$, $z_1=\sqrt{xy}$, from
condition (by dividing both sides with $xyz$) we get that
\[ \frac{1}{x_1^2}+\frac{1}{y_1^2}+\frac{1}{z_1^2}+\frac{2}{x_1y_1z_1}=1 \]
Since $x_1y_1z_1=xyz$, we can express $\frac{2}{xyz}$ from
condition. Put that in last equivalent form to get new equivalent
forms :D :

\[ \frac{5}{4}\geq \frac{2}{x_1y_1}+\frac{2}{y_1z_1}+\frac{2}{z_1x_1}-1+\frac{1}{x_1^2}+\frac{1}{y_1^2}+\frac{1}{z_1^2} \]
\[ \frac{3}{2} \geq \frac{1}{x_1}+\frac{1}{y_1}+\frac{1}{z_1} \]

Substitute $a=\frac{1}{x_1}$, $b=\frac{1}{y_1}$, $c=\frac{1}{z_1}$.
Now we got that we have to prove that $\frac{3}{2}\geq a+b+c$ with
condition $a^2+b^2+c^2+2abc=1$. Notice that $0 < a, b, c < 1$. Now
put $a=\cos{\alpha}$, $b=\cos{\beta}$, where $0 < \alpha, \beta < \frac{\pi}{2}$. Observe condition as quadratic equation for $c$ and
see that one solution is negative, so there's actually only one
solution. By trigonometric manipulations you can easily prove (I use
one strange equation and Viete's formulas to prove identitites like
this one, it's:
$\\4R^2x^3+4R(R+r)x^2+(s^2+r^2-4R^2)x+(2R+r)^2-s^2=0$ - with
standard notations, cosines of angles in triangle are solutions.)
that one solution is $c=\cos{\gamma}$ where $\gamma$ is
$\pi-\alpha-\beta$. Now we get that we have to prove well-known
inequality $\cos{\alpha}+\cos{\beta}+\cos{\gamma} \leq \frac{3}{2}$
(use Jensen's inequality on $\cos{x}$ where $x\in [0, \frac{\pi}{2}]$).

[quote="arqady"][quote="idioteque"]If $x, y, z$ are positive reals such that $xyz=x+y+z+2$, prove that \[5(x+y+z)+18\geq 8(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}).\] [/quote]
My ugly proof:
$xyz=x+y+z+2\Leftrightarrow\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}=1.$
Let $a=\frac{1}{1+x},$ $b=\frac{1}{1+y}$ and $c=\frac{1}{1+z}.$
Hence, $5(x+y+z)+18\geq 8(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})\Leftrightarrow$
$\Leftrightarrow5\cdot\sum_{cyc}\frac{1-a}{a}+18\geq8\cdot\sum_{cyc}\sqrt{\frac{(1-a)(1-b)}{ab}}\Leftrightarrow$
$\Leftrightarrow5(ab+ac+bc)+3abc\geq8\cdot\sum_{cyc}a\sqrt{bc(a+b)(a+c)},$ where $a+b+c=1.$
But $\sum_{cyc}a\sqrt{bc(a+b)(a+c)}\leq\sqrt{\sum_{cyc}abc(a+b)(a+c)}=\sqrt{abc+abc(ab+ac+bc)}.$
Hence, remain to prove that
$5(ab+ac+bc)+3abc\geq8\cdot\sqrt{abc+abc(ab+ac+bc)}.$
Let $ab+ac+bc=v,$ $abc=w.$ Then $v\geq9w$ and $v^{2}\geq3w.$
Id est, remain to prove that
 $5v+3w\geq8\sqrt{\frac{v^{2}}{3}+vw}\Leftrightarrow(v-9w)(11v-3w)\geq0,$ which obviously true. :)[/quote]

I don't know what arkady did. But here is a simpler solution:
After letting $a=\frac{1}{1+x},b=\frac{1}{1+y},c=\frac{1}{1+z}$, we have to prove: \[5\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+18 \ge 8\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{bc}},\] This is equivalent to: \[5\left(\frac{a+b}{b}+\frac{a+c}{c}+\frac{b+c}{c}+\frac{b+a}{a}+\frac{c+a}{a}+\frac{c+b}{b}\right) \ge 8\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{bc}}+12.\] By AM-GM: \[5\left(\frac{a+b}{b}+\frac{a+c}{c}+\frac{b+c}{c}+\frac{b+a}{a}+\frac{c+a}{a}+\frac{c+b}{b}\right) \ge 10\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{bc}},\] And the rest is easily shown from AG: \[\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{bc}}\ge 6.\]

What is AG ?

is it AM-GM ? or something else?

[quote="shyong"]What is AG ?
Is it AM-GM ? or something else?[/quote]
[b]AG[/b] is thing, like [b]A[/b]M-[b]G[/b]M.

[quote="Lovasz"]

 we have to prove: \[5\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+18 \ge 8\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{bc}},\] This is equivalent to: \[5\left(\frac{a+b}{b}+\frac{a+c}{c}+\frac{b+c}{c}+\frac{b+a}{a}+\frac{c+a}{a}+\frac{c+b}{b}\right) \ge 8\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{bc}}+12.\] [/quote]

Nice, Lovasz! :)

[quote="Lovasz"]
I don't know what arkady did. But here is a simpler solution:
After letting $a=\frac{1}{1+x},b=\frac{1}{1+y},c=\frac{1}{1+z}$, we have to prove: \[5\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+18 \ge 8\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{bc}},\] This is equivalent to: \[5\left(\frac{a+b}{b}+\frac{a+c}{c}+\frac{b+c}{c}+\frac{b+a}{a}+\frac{c+a}{a}+\frac{c+b}{b}\right) \ge 8\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{bc}}+12.\] By AM-GM: \[5\left(\frac{a+b}{b}+\frac{a+c}{c}+\frac{b+c}{c}+\frac{b+a}{a}+\frac{c+a}{a}+\frac{c+b}{b}\right) \ge 10\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{bc}},\] And the rest is easily shown from AG: \[\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{bc}}\ge 6.\] [/quote]

 :10:

[quote="arqady"]The following inequality is nicer.
For positives $x$, $y$ and $z$ such that $xyz=x+y+z+2$ prove that:
\[x+y+z+6\geq2\left(\sqrt{xy}+\sqrt{xz}+\sqrt{yz}\right)\][/quote]
$\Longleftrightarrow \sum\limits_{cyc}\frac{b+c}{a}+6\ge 2\sum\limits_{cyc}\sqrt{\frac{(a+b)(c+a}{bc}}\Longleftrightarrow$ $2\sum\limits_{cyc}\frac{b+c}{a}+6\ge \left(\sum\limits_{cyc}\sqrt{\frac{b+c}{a}}\right)^2\Longleftrightarrow$ $(b+c+c+a+a+b)\left(\frac 1a+\frac 1b+\frac 1c \right)\ge \left(\sum\limits_{cyc}\sqrt{\frac{b+c}{a}}\right)^2$
which is Cauchy-Schwarz.


geometry
rectangle
3D geometry
topology
real analysis
real analysis unsolved

this is routine stuff, i got a sudden doubt. show that any open set in $\mathbb{R}^{2}$ is a countable union of open rectangles. can you write it as a countable union of disjoint open rectangles? what about closed rectangles?

No, absolutely not. This theorem is true in one dimension because every open connected subset of the real line is an open interval. All you need to do to disprove this for $R^{2}$ is to exhibit an open connected set which is not a rectangle. If it could now be written a disjoint union of open rectangles then we would need more than one rectangle, and it would in particular not be connected.

i understand your point. what is your argument, for the first statement, namely that every open set is just a countable union of open rectangles? i think we can write it as a disjoint union of countably manyCLOSED rectangles.

For the first statement I don't want to write down a complete argument, but you just use that the points with rational components are dense in the open set.

For closed rectangles I think the same counterargument applies because I think that a disjoint countable union of connected closed sets is not connected, but I don't know and I don't have time to think it through, so someone else will have to confirm it.

The interesting arguments, useful in the kind of analysis I know (I come from the Zygmund school of Fourier analysis), involve not disjoint rectangles but nonoverlapping rectangles. Two closed rectangles are called nonoverlapping if their interiors are disjoint.

Theorem (Whitney decomposition). Suppose $\Omega$ is a bounded open set in $\mathbb{R}^{n}.$ Show that there are positive constants $c$ and $C$ such that it is possible to find a sequence of nonoverlapping closed cubes $Q$ that cover $\Omega$ and such that $c\cdot d(Q,\partial\Omega)\le \text{diam}(Q)\le C\cdot d(Q,\partial\Omega).$


Let $ x_{1},x_{2},...,x_{n}$ be real numbers in [0,1]. Prove that the given inequality holds:
$ x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{n} \minus{} (x_{1}x_{2} \plus{} x_{2}x_{3} \plus{} ... \plus{} x_{n \minus{} 1}x_{n} \plus{} x_{n}x_{1}) \leq n/2$

[quote="erbed"]Let $ x_{1},x_{2},...,x_{n}$ be real numbers in [0,1]. Prove that the given inequality holds:
$ x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{n} \minus{} (x_{1}x_{2} \plus{} x_{2}x_{3} \plus{} ... \plus{} x_{n \minus{} 1}x_{n} \plus{} x_{n}x_{1}) \geq n/2$[/quote]

that is wrong. If $ x_{1}\equal{}x_{2}\equal{}...,x_{n}\equal{}1$ => $ x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{n} \minus{} (x_{1}x_{2} \plus{} x_{2}x_{3} \plus{} ... \plus{} x_{n \minus{} 1}x_{n} \plus{} x_{n}x_{1}) \geq n/2$ => It is nonsensical

[quote="bokinhvan3"][quote="erbed"]Let $ x_{1},x_{2},...,x_{n}$ be real numbers in [0,1]. Prove that the given inequality holds:
$ x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{n} \minus{} (x_{1}x_{2} \plus{} x_{2}x_{3} \plus{} ... \plus{} x_{n \minus{} 1}x_{n} \plus{} x_{n}x_{1}) \geq n/2$[/quote]

that is wrong. If $ x_{1} \equal{} x_{2} \equal{} ...,x_{n} \equal{} 1$ => $ x_{1} \plus{} x_{2} \plus{} ... \plus{} x_{n} \minus{} (x_{1}x_{2} \plus{} x_{2}x_{3} \plus{} ... \plus{} x_{n \minus{} 1}x_{n} \plus{} x_{n}x_{1}) \geq n/2$ => It is nonsensical[/quote]
You're right. I changed it.It must be less then or equal, not greater then or equal.


Can someone tell me why $ Tor(Z, Z_2) \cong Z_2$?

It's not. It's zero since $ \mathbb{Z}$ is torsionfree.


I'm just wondering, how many people who are going to the USNCO study camp post on this forum?  (I, through some stroke of amazing fortune, managed to get in, still not sure how that worked.)  how far is everyone in the organic chem book they sent?  im somewhere meandering through the middle of chapter 5.   again, just wondering how many study campers use AoPS, which is traditionally a math site after all...

that actually wasn't me, that was one of my friends...i think?  what kinda name is pwei anyway?


Hi
I was wondering what the difference is between the problems in the handbooks and the the problems in the "Warm Ups and Workouts." Aren't they the same?

More specifically, if you click on this link [url]https://mathcounts.org/Document.Doc?id=471[/url]and scroll to the last page,
what is the difference between the WARM-UPS AND WORKOUTS  and the School Handbook with Solutions Volume II?

im guessing the school handbook would be harder


[hide="my answer"]100/12=8 r 4 so august-sep-oct-nov-dec

December. There's always these type of questions in Math Oympiads(The MOEMS ones)[/hide]

[quote="aznmonkey1992"]what month is it 100 months from now? (august)[/quote]
[hide]
100/12 = 8 r 4

4 months from august is...sept..oct...nov... DECEMBER
[/hide]

[quote="aznmonkey1992"]what month is it 100 months from now? (august)[/quote]

[hide="dont click this"]
[hide="why didnt you listen"]
[hide="you wont listen will you"]

[color=white]December. 12*8=96+4mths=Dec. Please move this to the moems forum, because it's os easy.[/color]
[/hide][/hide][/hide]

First of all, this is a perfect classroom problem. If you think this is easy, don't do it.  ;)

Also, what part of hide your answer don't you understand?  :mad:

I have a question is classroom or moems supposed to be easier I thought classroom was?

yeah, me too. I think classroom is easier than MoEMS.

Read this:

http://artofproblemsolving.com/Forum/viewtopic.php?t=47710

[quote="MCrawford"]MOEMS problems are more basic than the problems in [b]nearly[/b] every other forum on this site.
[/quote]

Uh huh, stop this discussion  ;)

[hide]
i have seen this problem so many times. :D
100/12=8 remainder 4
august plus 4 months is DECEMBER[/hide]


Using pythagorean theorom, other sides are 5, so $ 8\plus{}16\plus{}5\plus{}5\equal{}34$


I heard of those two inequalities.

But what are they? :?

I posted a thread on Cauchy and its proof in the Pre-Olympiad forum not too long ago.

As for Jensen's, I can tell you what it is, but unless you know calculus and understand the geometrical interpretation of certain things in calculus, there's not much point to me telling you.

If you want to get started in inequalities, start with rearrangement and AM-GM, imho.

Concisely, given two monotonically increasing sequences $a$ and $b$, $\displaystyle \sum{a_ib_i} \ge \sum{a_ib_{\delta(i)}}$ where $\delta$ is a permutation of the $i$'s.

Strangely enough, rearrangement isn't covered in either AoPS V2 or Zeitz... I found out about it [url=http://mathdb.org/notes_download/e_algebra_el.htm]here[/url] (go to Inequalities (II).)

Its covered well in Kiran Kedlaya's a>b.

Rearrangement is my new favorite inequality, by far.  I just learned about it last week.  Engel loves it.

Yes, Engel even does Rearrangement for more than two sequencs, which I haven't seen anywhere else, but does help on some problems.


calculus
integration
derivative
inequalities
real analysis
real analysis unsolved

$ f: [a,b]\rightarrow \mathbb{R}$ be continouosly derivative and $ f(a)\equal{}f(b)\equal{}0$ and $ \int_{a}^b|f(x)|^2dx\equal{}1$
prove :$ \int_a^b|f'(x)|^2dx.\int_a^bx^2|f(x)|^2dx>\frac 14$

Maybe you could use Cauchy-Schwarz inequality.

\[ \sqrt {\left(\int_a^b|f'(x)|^2dx\right)\left(\int_a^bx^2|f(x)|^2dx\right)}\geq\left|\int_a^bf'(x)xf(x)dx\right| \equal{} \left|\frac {b|f(b)|^2}{2} \minus{} \frac {a|f(a)|^2}{2} \minus{} \frac {1}{2}\int_a^b|f(x)|^2dx\right| \equal{} \left| \minus{} \frac {1}{2}\right|,
\]
square to get RHS of proposed inequality $ \geq\frac {1}{4}$, with equality iff $ f'(x) \equal{} kxf(x)$ for all $ x\in(a,b)$ and some real-valued constant $ k$.  The diff. equation has solution $ f(x) \equal{} Ae^{\frac{kx^2}{2}}$, which can never be 0 at $ x \equal{} a,b$ unless $ A \equal{} 0$, in which case $ \int_a^b|f(x)|^2dx \equal{} 0$, absurd, so equality cannot be reached, and inequality is strict...  :lol:

Can someone send proof of Caushy-Schwartz? also known as Holder-Inequality.

As a matter of fact, Cauchy-Schwartz-Bunyakovsky (in this case, I ought to mention Bunyakovsky, because this integral form is due to him) is a special case of Hoelder's inequality, not its equivalent.

You can see the proof of Hoelder's general form (not just the integral form!) on wikipedia:

http://en.wikipedia.org/wiki/H%C3%B6lder's_inequality

just copy-paste the link, don't click it  :wink:

A fixed version of the [url=http://en.wikipedia.org/wiki/H%C3%B6lder%27s_inequality]link[/url]. When linking to something with special characters like ' , use alternate text.


[hide="i think the answer is..."]9[/hide]

[quote="MCrawford"]What is the greatest positive integer $n$ such that $n^2$ is less than 95?[/quote]

[hide]$9^2=81$
so it must be 9 or greater

$10^2=100$

so 9 is the answer[/hide]

[hide="da answer!!!"]9 because 95 is in between 102
and 92[/hide]

9 is the answer
 Heres a continuation
Can you find the greatest 6 digit number which is a perfect square of smallest 6 digit number which is a perfect square, Give me a method to find it for any n digit number

[hide="the answer"]$9^2$=81 and $10^2$=100, so the answer is 9[/hide]

Sorry friends ,It should be find the greatest or smallest 6 digit number which is a perfect square

i think it's a program for a Ti calculator.

[quote="math92"][quote="nat mc"]Call [x] Greatest integer function: I'll write it in calculator basic

$
:\text{Prompt n}$
$:\text{If }\sqrt{10^{n-1}}=[\sqrt{10^{n-1}}]$
$:\text{Then}$
$:10^{n-1}\rightarrow{l}$
$:\text{Else}$
$:\{[\sqrt{10^{n-1}}]+1\}^2\rightarrow{l}$
$:\text{End}$
$:\text{If }\sqrt{10^n-1}=[\sqrt{10^n-1}]$
$:\text{Then}$
$:10^n-1\rightarrow{g}$
$:\text{Else}$
$:\{[\sqrt{10^n-1}]-1\}^2\rightarrow{g}$
$:\text{End}$
$:\text{Disp \"Least is\"}$
$:\text{Disp }l$
$:\text{Disp \"Greatest is\"}$
$:\text{Disp }g$

Hope you understand[/quote]
i don't really understand[/quote]
well, in regular words:
n is the number of digits, we are trying to find the least and greatest square numbers

If $10^{n-1}$ is a perfect square, it is the least n digit square.
If not, then $\{[\sqrt{10^{n-1}}]+1\}^2$ is the least n digit square.

If $10^n-1$ is a perfect square, it is the greatest n digit square
If not, then $\{[\sqrt{10^n-1}]-1\}^2$ is the greatest n digit square


trigonometry
calculus
calculus computations

h(x)=(tan((x^2+1)/(3x-2)))^3
Find h'(x). Also, can somebody please help me with why when i plug it into the ti-89 it gives me a direct solution and not a solution with variables?


LaTeX
quadratics
algebra
linear equation
superior algebra
superior algebra unsolved

It's my first post in this section and I'm a beginner in superior Algebra. This may be an elementary exercise but i don't get it...

On set of real numbers we denote the law of composition "$\circ$" by $x\circ y= xy+5x+ay+b$. Find $a,b\in\mathbb{R}$ such the law has unity.

hi i m sorry i do'nt know latex,so i can just give an idea about how to do,i hope its correct  :oops:
So if such a unity exist,then assume its say A,then putting x=y=A u will get a quadratic equation in A,which implies the discriminant is zero as the unity is unique,which will give u a relation between a and b,now again putting x=A,and keeping y variable,u will get a linear equation in y,since the equation holds true for all y,the coefficients r 0,so this will give u relations between A,a,b.At this stage u will be able to get values for A,a,b,the ones i got were A=-4,a=4,b=16,make sure to check if the condition holds true when u put y=A,keeping x vary.
regards
Rohith

[quote="rohith"][...]now again putting x=A,and keeping y variable,u will get a linear equation in y,since the equation holds true for all y,the coefficients r 0,so this will give u relations between A,a,b.[...][/quote]

In this case I got the linar equation $(A+a)y+5A+b$... and do you say that now $A+a=0$ and $5A+b=0$ or what? I'm a little confused   :huh:

We really want $x\circ A=A\circ x=x$ for all $x$; this is
$Ax+5x+aA+b=Ax+5A+ax+b=x$.
Equating the first two, $(5-a)x=(5-a)A$, so $5-a=0$ and $a=5$. Now $Ax+5x+5A+b=x$, so $(A+4)x=-(5A+b)$ for all $x$. This gives $A+4=5A+b=0$, so $A=-4,b=20$.

Our law is $x\circ y=xy+5x+5y+20=(x+5)(y+5)-5$, and the identity $A$ is $-4$

hi,
absolutely,thats what i meant
rohith

Thanks! Well I used only $x\circ A=x$  :oops: silly mistake


Find all positive integers $ n$ such that there exists 100 integers $ x_1,x_2,\cdots,x_{100}$ satisfying $ \sum_{k \equal{} 1}^{100}n^{x_k} \equal{} 1997$.

only 4 solutions:$ n \equal{} 2,8,272,1898$
 :lol:

Nice problem! But who can give a full solution?

Please solve this problem! I'm really interested in it.

[quote="feliz"]Look the equation mod $ n \minus{} 1$. We get $ 100 \equiv 1997$ or $ n \minus{} 1|1897$.[/quote] I really like your solution :clap2:  :clap2:  :wink:

Sorry .

I used wrong RSH.

By the way , my friend showed general type.

Let $ A(S,n) \equal{} \sum_{k \equal{} 1}^{d}a_{k}$ where $ S \equal{} \sum_{k \equal{} 1}^{d}a_{k}n^{k\minus{}1}$.

$ \sum_{k \equal{} 1}^{m}n^{x_k} \equal{} S$ is possible

if and only if $ m \ge A(S,n)$ and $ m \minus{} A(S,n)$ is divisible by $ n \minus{} 1$.

I generalized a note of feliz and made a Mathematica program :)

[code]m = Input["m="]; Print[{m}]; t = 0; For[n = 2, n ≤ m, n++, If[Mod[100, n - 1] == Mod[m, n - 1], {Print[n], t++}]]; Print[t "-tuple"] [/code]

This problem case is $ m\equal{}1997$ :)


If anybody is familiar with context-free grammars and languages, does anyone have an idea on how to do these problems: (Note: 0^i1^j means a string with i 0's and j 1's such as (0^3)(1^5)=00011111. Thanks for your help.)

Give a CFG(context-free grammar) for L = {0^i1^j | j < i OR j > 2i}.

Give a CFG for L = {0^i1^j | i <= j AND j <= 2i}.

Give a CFG for L = {0^i1^j | i < 2j}.

Give a CFG for L = {x#y | x^R(reversal) is a substring of y}. Note that x and y
are elements of {0, 1}* and that # is another symbol, not equal to 0 or 1.

Give a CFG for L = {x#y | y is not equal to x^R(reversal)}. (The comments for the previous
question still hold.)

Show that the following grammar G
G = S -> a | Sa | bSS | SSb | SbS
is ambiguous.

Prove that every string derived by the grammar
S −> 0S1 | 01 | SS
does not begin with 011.

Anybody have a clue? Here are a couple of examples of context-free grammars. e means the empty string

CFG for {w|w contains at least 3 1s}.
S -> A1A1A1A
A -> e | 0A | 1A

CFG for {w|w starts and ends with the same symbol}.
S -> 0A0 | 1A1
A -> e | 0A | 1A

CFG for {w|len(w) is odd}.
A -> 0B | 1B
B -> e | 0A | 1A

CFG for {w|len(w) is odd and its middle symbol is 0}.
S -> 0 | 1S0 | 0S1 | 1S1 | 0S0

CFG for {w|w contains more 1s than 0s}.
S -> 1 | S1 | 1S | 0S1 | 1S0 | SS

CFG for {w|w = w^R(reversal)}
S -> e | 0 | 1 | 0S0 | 1S1

Hm, that's quite a lot of problems. Here's the solution for the first one:

S -> A | X
A -> 0B
B -> 0B1 | 0B | e
X -> Y1
Y -> Y1 | 0Y11 | e

I hope it's correct. At the beginning, if you choose to substitute S for A, you get a word with j < i, if you choose B you get j > 2i.

And this should work for the second one:

S -> 0S1 | 00S1 | e

For the third problem, it is actually a subproblem of the first one so the solution should be obvious...

Number 4 looked somewhat harder but it actually isn't, try this:

S -> AX
A -> 0A0 | 1A1 | #X
X -> 0X | 1X | e

X creates an arbitraty string, A creates two reversed strings and puts #X between them.

I found a solution to #4 but it's little complex... first, you have

S -> A | B | C

where A creates all x#y such that x is longer than y and B creates all x#y with y longer than x. Finally, C works something like this:

C -> 0C0 | 1C1 | 0D1 | 1D0
D -> 0D0 | 1D1 | 0D1 | 1D0 | #

The first time you create two asymetrical symbols C changes to D and the word can be terminated.

To prove #5, you only need to find a word with two different leftmost derivations in G. For example, try the word baaab. #6 is also not very hard, you only need to realize that every S will start with 0.

Thanks for all the replies, they have been very helpful. I also wanted to ask if anyone knows about PDA's(push-down automata) as well, if they would know the solutions to these two problems:

1. Consider the following language:
a*b*c* − {(a^i)(b^i)(c^i) | i >= 0}

• Give a CFG that generates this set
• Give a PDA that recognizes this set

2.Give a PDA for the following language:
{x ∈ {0, 1}* | count(0, x) <= 2 * count(1, x)}

Also, does anybody know what Chomsky Normal Form is? I also need to do the 2 problems:

Give a grammar in Chomsky Normal Form for the language
{a^(i)b^(2i)c^(j) | i, j >=1 }

Give a PDA that recognizes the language of the following grammar:
A -> aA | aE | bAA
E -> e | aEb | EE
where e is the empty string


quadratics
inequalities
calculus
calculus computations

We can do this a number of ways: sketch the graph, square both sides and solve the quadratic inequalites, consider $ x<6$ and $ x>6$ separately, ...

The solution is $ x<6$ or $ x>13$.

[quote="AndrewTom"]The solution is $ x < 6$ or $ x > 13$.[/quote]
I think $ x < \frac {23}{5}, 13 < x$.

hello, the second answer ist right, the solution is given by
$ \minus{}\infty<x<\frac{23}{5} \cup 13<x<\plus{}\infty$
Sonnhard.

Yes, of course, if $ x<6$ and $ x< \frac{23}{5}  \equal{} 4\frac{3}{5}$ then $ x<\frac{23}{5}$. My thanks to both of you.


Let a triangle $ABC$ and an arbitary point $D$ on the side $BC$ .Find a point $M$ at the line $AD$ such that the circumcircles of the triangle $ABM$ and $ACM$ are equal .

Of course that will be when the point A, B, C and M are on the same circle, that is the circumcircle of $\triangle ABC$. The point M must then be the intersection point of this circumcircle and the line AD.

That is obviously correct if "equal" means "coinciding" - but perhaps "equal" means just "congruent", i.e. "equal in size"?

I meant with equal radius .The circles should be 2 not only one .
Thanks

Let $M_{B}$ and $M_{C}$ be the feet of the perpendiculars from $M$ to $AB$ and $AC$ respectively. Then

$R_{ABM}=\frac{AB\cdot BM\cdot AM}{4\cdot\frac{AB\cdot MM_{B}}{2}}$ and $R_{ACM}=\frac{AC\cdot CM\cdot AM}{4\cdot\frac{AC\cdot MM_{C}}{2}}$.

Equating the radii we get

$\frac{BM}{MM_{B}}=\frac{CM}{MM_{C}}\implies \angle MBM_{B}=\angle MCM_{C}$ (equal secants/cosines)

Now basically we have two cases:

1. $\overline{AM_{B}B}\land\overline{AM_{C}C}$ or $\overline{ABM_{B}}\land\overline{ACM_{C}}$

Then the segment $AM$ is seen under the same angles from points $B$ and $C$. Let $C'=\mathcal{S}_{AD}(C)$. Then $AM$ is seen under the same angle from $C'$, but $B$ and $C'$ are on the same side of $AD$. Hence the desired point $M$ is the intersection of the circle $ABC'$ and the line $AD$. Similarly, the same $M$ can be found as the intersection of the circle $AB'C$ and the line $AD$ where $B'=\mathcal{S}_{AD}(B)$ (the point is the same because of the symmetry).

2. $\overline{AM_{B}B}\land\overline{ACM_{C}}$ or $\overline{ABM_{B}}\land\overline{AM_{C}C}$

Then $\angle MBA+\angle MCA=\pi$, hence in that case we have the already described construction - $M$ is the intersection of the circle $ABC$ and the line $AD$.


Ididnt solve this problem.
Thanks and best regards!

Your problem easy too. Now, I'm not free, I will post it.

Not important  :) When you are free you will post :)
Thank you for your help  :)

[hide="hint"]use the sin and cosine laws to find all the sides and angles, also the pythagorean theory.[/hide]

Firstly i am sorry for my poor English  :blush:
I used laws cos and sin but i made wrong.. :(
Thanks and best regards!

i solved this way but i want to learn how can we will solve by law cos  :(
Thanks and best regards!


probability
vector
inequalities
MATHCOUNTS
triangle inequality

This is a WarmUp Problem:
Four line segments are chosen at random from a collection of six line segments having lengths 2,3,5,8,13, and 21 units.  What is the probability that the 4 segments chosen could form a quadrilateral?  Express answer as common fraction.

How do you solve this problem?

What I would do is think of the line segments as vectors and use vector addition.  It's similar to the triangle inequality, but I'm not sure if there's such thing as the "quadrilateral inequality".

vectors don't go in Mathcounts

if this was triangles, the answer would be $ 0$.  However, with this problem, all I know for now is that the denominator is 15, 5, or 3.

I think its just 2/5..

You have to list them-theres only 15 choices so it wont take that long. (i dont know if you listed them correctly)

You have to use the quadrilateral inequality..same bas the traingle one except 4 sides. (and only have to compare the three smallest to the largest)

[hide="MY ANSWER"]2 3 5 8
2 5 8 13
2 8 13 21
3 5 8 13
3 8 13 21
5 8 15 21

So there are 6 possibilities that work.

Denominator: ${ {6\choose2} = 15}$

Therefore, I got $ \boxed{\frac {2}{5}}$.[/hide]

[quote="Ihatepie"]2 3 5 8 13 21

[hide]

something cool I noticed is 2+3=5
3+5=8
5+8=13
etc. etc. but anyways, this is how I would do it.

first I figure out ways that it couldn't be a quadrilateral
So you get 9 ways by listing

2 3 5 13
2 3 8 13
2 3 5 21
2 3 8 21
2 5 8 21
3 5 8 21
2 3 13 21
2 5 13 21
3 5 13 21

then you figure out total possible ways

p!/(p-n)!n!

720/48 or 15

so you get 9/15 or

3/5

[/hide][/quote]

yes so that means your answer would be 1-3/5=2/5 (you forgot to finish it off)

oops  :blush:

/me hates when he gets everything else right and makes a silly mistake.


limit
integration
real analysis
real analysis unsolved

Let $1<p<\infty$ and $q=p/(p-1)$. A sequence $(f_{i})$ converges weakly in $L^{p}$ towards $f\in L^{p}$ if
\[\lim_{i\to\infty}\int f_{i}g\,\mu=\int fg\,\mu \]
for all $g\in L^{p}$. How to prove that if $(f_{i})$ converges in $L^{p}$, it converges weakly in $L^{p}$.