I suggest something like the following:

Titles containing some words like easy, hard, problem, Number Theory, Algebra,... are only allowed in titles that have some minimum length without that words. Or just don't allow easy and hard everytime as content.
Similar for e.g. "primes" as title, it's not helpful I think.

At least, it gets annoying how many topics called "easy" there are...

Will we also get some introduction for how to post here in the nearer future¿

Indeed it sounds like a good idea. I'll implement it soon (ie a word list such that if a topic title contains only those words the topic cannot be posted).

For now I suggest:
[code]hard, easy, difficult, not, yet, help, please[/code]

Do you have other suggestions?

what about

[code]problem, question, homework, nice, new, old, tough, proof, prove, and, me, a, the, me, you, it, is, this, interesting, try, show, for[/code]

?

  darij

Some more:

never useful:
[code]very, some, true(?), false(?), I, interesting[/code]

no sense alone (also not in pairs¿):
[code]equation/equal, sequence, find, sum, prime [factor], number,...[/code]
(all that, there is a lot more)

Something more general:
Are there exceptions when simply not allowing zero-or-one-word-titles¿ (words having at least lets say 5 characters)
And/Or requiering some minimum length¿ (17 characters¿)

The problem is not to forbid too much.
But when the people that conflict with it get linked to some place where this is explained in a good way, this should work (after some time at the beginning ;) ).

PS: more words will come when I see them

I'd keep the list pretty short, and make sure that the error message is very clear as to why the post wasn't accepted.

Just thought that I would say, an easy way around the eliminating easy limitation would be to post "Easy Problem" or "Easy Question", etc.  I believe that lots of combinations of the suggested words should also be limited, because they mean basically the same thing.

Sample message:

[code]Please be more specific in your topic title.  Thanks.[/code]

Banning the word "easy" completely from titles, this will not be possible.
But what do you think abou the at-least-$\zeta$-words-rule in general¿ ($\zeta \in \{2,3,4,...\}$ constant)

I don't think that the list should be to short: there are words that should never make sense, and when there are more of them, I see no problem in banning all.

The best seems to combine minimum word number with banned (not even allowed) and bad words (are not counted as words, so not for the number of words). Additionally, another condition could be that there are twice that much (or more ;) ) good words than bad words.
E.g. a title like "Easy problem on equations, is it really true¿" has a nice length, but should be forbidden, too.

But the message of b-flat is to short, it should explain in at most theree sentences what's going on, giving a link to some whole topic about that and also show what was not correct (e.g. " bad words: 'blah', 'spam' ").

O.K.  I agree with what you say  :)

How would this be?

[code]Please be more specific in you title.  [words that are blocked go here] should not be part of the title.  Also, please make sure the title contains [good words/characters here].  For more information, visit [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100147[/url].

Thank you.[/code]


vector
superior algebra
superior algebra theorems

Homomorphism: A transformation of one set into another that preserves in the second set the operations between the members of the first set.

Do homomorphisms have to preserve ALL the operations of the first set?

That definition is not really accurate. But if you speak about X-homomorphisms, it should preserve the structure of X (e.g. X=group or vector space over some field).


Heres a cool problem:

There are a group of positive integers, that are not necessarily distinct, and that add up to 40. There can be any amount of positive integers in this group. What is the maximum product of the positive integers that add up to 40?

There are a group of positive integers, that are not necessarily distinct, and that add up to 40. There can be any amount of positive integers in this group. What is the maximum product of the positive integers that add up to 40?

x + y = 40
y= 40-x

F(x) = xy
=x(40-x)
=40x -x^2
derive to F'(x)= 0
2x=40
x=20
Max product = 20 x 20 = 400

First, let's see that it's better to split a number greater than 3 into the sum of two smaller numbers:
in facts, if $ n$ is even, we observe that $ (n/2)^2 > 2(n/2)$ (so the best choice is take its half part) ; if $ n$ is odd, it's better to split it into $ (n \minus{} 1)/2$ and $ (n \plus{} 1)/2$: $ n < (n^2 \minus{} 1)/4$.

So, there are $ a$ 2 and $ b$ 3 in the product. Besides, see that 3+3 has a greater product than 2+2+2, so we can substitute 3 "2" in 2 "3".

If the sum is 40,
the best choice is to write 40 as $ 3 \cdot 12 \plus{} 2\cdot2$ and the product becomes $ 4\cdot 3^{12}$  :)

Is it right?

EDIT: harwi, why are you taking into consideration just two integers? Have I misread the question?

sorry , I didn't read the question well.


Ean $n_{1},n_{2},\ldots,n_{k}$ opou $k\in \mathbb{N}^{*}, k\geq 2$, einai diaforetikoi metaksi tous thetikoi akeraioi oi opoioi den diairountai apo prwtous arithmous megaliterous tou $3$, tote na deixthei oti

$S=\frac{1}{n_{1}}+\frac{1}{n_{2}}+\cdots+\frac{1}{n_{k}}<3$.

Alexandros

[quote="silouan"]
Προφανώς
$S< S_{1}\cdot S_{2}=\frac{1}{1-\frac{1}{2}}\cdot\frac{1}{1-\frac{1}{3}}=3$

το ζητούμενο[/quote]

Apolitws swsto me mia mikri paratirisi. Einai: $S_{1}<\sum_{i=0}^{\infty}\frac{1}{2^{i}}=\frac{1}{1-\frac{1}{2}}$
kai $S_{2}<\sum_{i=0}^{\infty}\frac{1}{3^{i}}=\frac{1}{1-\frac{1}{3}}$

Ara $S\leq S_{1}S_{2}<\frac{1}{1-\frac{1}{2}}\cdot\frac{1}{1-\frac{1}{3}}=3$

Poli wraia Silouan,

Alexandros


:surf: let triangle ABC: $C\leq B \leq A \leq 90$
and $M=cos(\frac{A-B}{2})sin(\frac{A}{2})sin(\frac{B}{2})$
Find min of M :rotfl:

nobody find this problem????
We prove $cos(A-B)\leq cosA+cosB$

first i guess min of M=$\frac{1}{4}$
we need to prove $M\geq\frac{1}4{}$
$\Leftrightarrow cos(A-B)\geq cosA+cosB$
(Sorry!!!I made mistake "$\leq$" in second reply)
we have $2sin(A+B)cos(A-B)=sin2A+sin2B$
$\Leftrightarrow sinCcos(A-B)=sinAcosA+sinBcosB$
$\Leftrightarrow cos(A-B)=\frac{sinA}{sinC}cosA+\frac{sinB}{sinC}cosB$
$\Leftrightarrow cos(A-B)\geq cosA+cosB$
Because $\frac{\pi}{2}\geq A\geq B\geq C$
Besides I have another solution but long use function.


we set ourselves coordinate axes. of the two colors, at least one, say red, must occur infinitely many times on the $ x$ axis. denote by $ S$ the set of these red points. now, let $ X$ be the set of points that are directly above the points in $ S$ on the line $ y\equal{}1$. if more than one point in $ X$ is red, we're done; otherwise, all points in $ X$ except possibly one of them are black. let $ Y$ be the set of points directly below the points in $ S$ on the line $ y\equal{}\minus{}1$. if more than one point in $ Y$ is red, we're done; otherwise, all points in $ Y$ except possibly one of them are black. but with all points but one in $ X$ and all points in $ Y$ but one black, there must exist a black rectangle, and again we're done.

I always find great pleasure in solving coloring problems in $ \mathbb{R}^n$.  I will generalize the problem using $ n$ colors instead of two.

[u]Solution.[/u] Consider a set of $ n^{n\plus{}1}\plus{}1$ parallel lines with $ n\plus{}1$ points that lie atop one another.  Then there exist at least two monochromatic points on each line.  Since there are $ n^{n\plus{}1}$ sequences of colors, and we have $ n^{n\plus{}1}\plus{}1$ parallel lines, two of them must have the same color sequence.  And the conlcusion follows. $ \Box$

right. though... there's really no need for it to be $ \mathbb{R}^n$ for a specific $ n$.

The $ \mathbb{R}^n$ in the first sentence is not related to the solution.  $ n$ in the solution is simply the number of colors used in $ \mathbb{R}^2$.

[quote="sylow_theory"]I seen this problem awhile ago, I find it hard. (I'm gonna write the problem so that even color blind can understand).

[i]Suppose every point in $ \mathbb{R}^2$ is colored either black or white. Prove there exists a rectangle with all four vertices the same color[/i].

This has been bugging me, what if [u]rectangle[/u] is replaced by [u]square[/u]. Is it still true?[/quote]
The original problem is still true if you replace $ \mathbb{R}^2$ with a 4x7 lattice grid. (A 4x6 lattice grid will not suffice.)

As for the part that's bugging you, it's bugging me too...


Let a, b, c are the besides in a triangle. Prove that
$2<\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}-\frac{a^{3}+b^{3}+c^{3}}{abc}\leq 3$

[quote="kcbu"][quote="N.T.TUAN"]2nd inequality is equivalent to $\sum a(a-b)(a-c)\geq 0$, it is true by Schur's theorem.[/quote]
Thanks, but i need a solution for 1st ineq.[/quote]

It is well known (not a known inequality but it is famous and in fact the only one I know of  :) )

That,

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\geq \frac{3}{2}$

Now the Harmonic mean (HM-AM),

$\frac{3}{\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}}\geq \frac{3}{2}$

Thus,

$\frac{1}{\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}}\leq \frac{1}{2}$

Take reciprocals,

$\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\geq 2$

(I hope I did this right, I am brand new to these inequalities stuff).

sylow_theory,

What do you do? I don't understand  :huh:

First maybe equivalent to $(a+b-c)(b+c-a)(c+a-b)>0$, I think so.

[quote="me@home"]How are a,b,c the angles? Meaning radians in terms of pi? This seems really unnatural. Do you mean sides?

Also, why did you write "easy" if you can't solve it...[/quote]
OMG, sorry, a, b, c, are besides not angles.
I solved but i need another solution  :oops:

I am true :D
[quote="N.T.TUAN"]

First maybe equivalent to $(a+b-c)(b+c-a)(c+a-b)>0$, I think so.[/quote]


$ a_i>0$
$ \sum_{i\equal{}1}^{n}{a_i}\equal{}1$
Prove:
$ \sum_{i\equal{}1}^{n}{\frac{a_i}{\sqrt{1\minus{}a_i}}} \ge \sqrt{\frac{n}{n\minus{}1}}$

[quote="alpha-beta"]$ a_i > 0$
$ \sum_{i \equal{} 1}^{n}{a_i} \equal{} 1$
Prove:
$ \sum_{i \equal{} 1}^{n}{\frac {a_i}{\sqrt {1 \minus{} a_i}}} \ge \sqrt {\frac {n}{n \minus{} 1}}$[/quote]
If $ f(x)\equal{}\frac{x}{\sqrt{1\minus{}x}}$ then, $ f''(x)\equal{}\frac{4\minus{}x}{4\sqrt{(1\minus{}x)^5}},$ which is positive for $ 0<x<1.$ :wink:

But arqady,We only prove by cauchy,
$ \sum\frac {a_i}{\sqrt {1 \minus{} a_i}} \equal{} \sum\frac {1}{\sqrt {1 \minus{} a_i}} \minus{} \sum\sqrt{1 \minus{} a_i}$
$ \ge \frac {n^2}{\sum\sqrt {1 \minus{} a_i}} \minus{} t$ with $ t \equal{} \sum\sqrt {1 \minus{} a_i}$
We can find minimum $ \frac {n^2}{t} \minus{} t$ very easy.

If you're going to use Cauchy, the following is much easier :)

$ ( \sum \frac {a_i}{\sqrt {1 \minus{} a_i}} )(\sum \frac {a_i}{\sqrt {1 \minus{} a_i}})(\sum a_i (1 \minus{} a_i )) \geq ( \sum a_i )^3 \equal{} 1$

This, combined with $ \sum a_i \equal{} 1$, yields $ (\sum \frac {a_i}{\sqrt {1 \minus{} a_i}})^2 \geq \frac {1}{1 \minus{} \sum {a_i}^2}$

That the right side is not less than $ {\frac {n}{n \minus{} 1}}$ is trivial.

(Use $ n(\sum a_i^2 ) \geq (\sum a_i )^2 \equal{} 1$ )


inequalities
induction
search
inequalities proposed

Prove that if $a_1,a_2,...,a_n$ are positive real numbers such that $a_1a_2...a_n=1$ then for all positive integer $1\le k \le n$ we have
\[ (n-1)(a_1^k+a_2^k+...+a_n^k) +n\ge (a_1+a_2+...+a_n)(a_1^{k-1}+a_2^{k-1}+...+a_n^{k-1}).  \]
Futhermore, prove that $k=n$ is the best constant for which the above inequality is true.

Homogenise and you will take
http://www.mathlinks.ro/Forum/viewtopic.php?t=73295

Yes, your problem which you post is exactly Suranji Inequality (very famous). But in fact my problem is more general and I think more difficult.

Notice that if we choose $k=2$, we get a very nice and famous one
\[ (n-1)(a_1^2+a_2^2+...+a_n^2)+n\sqrt[n]{a_1^2a_2^2...a_n^2} \ge (a_1+a_2+...+a_n)^2 \]
It involves Turkevici Inequality for $n=4$.

I hope this problem ca by solved by someone in a near day.

[quote="Albanian Eagle"]we should have n>k>n\(n-1) real hunghtn
and this generalisation just follows form suranji itself
(put $a_i ^k = b_i ^n$  ;) )
 :)[/quote]

Oh, surprisingly, I don't realize this relation. Very good, Albanian.  :)

So this problem was solved....

But I'm still bemused.
[b]circumstantiate[/b],please.Thanks.

How to prove Suranji's Inequality? I mean a solution different from Harazi's solution in Old and New Inequality? I'm have a very nice solution (in my book :).

well, some days ago Hawk Tiger showed me a nice way to prove Suranyi (again with induction) and that's the shortest proof I've ever seen for it. I hope he will post it :)

I solved it by Karamara Inequality, no induction. It is in Hungkhtn's English Version book but I will wait by two month before post my solution.

Can anyone post the proof of Suranyi's inequality now, please? :maybe:
Thank you.

search because it is already in the forum. :wink:

Can you please post a link, because I cannot find it...


[quote="irantst"]hi find all f:R--->R that for any real x,y
(x^y)-f(f(y))=y*f(x)[/quote]

Where is this problem coming from ?

$ x^y$ is not defined for any $ x,y\in\mathbb R$, so this is meaningless. In what kind of contest did you encounter it ?

[quote="irantst"]hi find all f:N--->N that for any real x,y
(x^y)-f(f(y))=y*f(x)[/quote]

Do you just try to solve the problems you found in your strange class / contest / book ?

Let $ P(x,y)$ be the assertion $ x^y\minus{}f(f(y))\equal{}yf(x)$

$ P(1,1)$ $ \implies$ $ f(1)\equal{}1\minus{}f(f(1))<1\notin\mathbb N$ and so no solution .


a+b>c
b+c>a
a+c>b
a>0 b>0 c>0

PROVE $ (a\plus{}b\plus{}c)(a^2 \plus{}b^2 \plus{}c^2 )\ge a^3 \plus{}b^3 \plus{}c^3$

hello, after expanding your inequality
$ (a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\minus{}(a^3\plus{}b^3\plus{}c^3) \geq0$ we get $ ab^2\plus{}ac^2\plus{}a^2b\plus{}bc^2\plus{}a^2c\plus{}b^2c\geq0$ which is true because the conditions $ a>0,b>0,c>0$.
Sonnhard.

Sorry! i made a mistake. $ (a\plus{}b\plus{}c)(a^2 \plus{}b^2 \plus{}c^2)\ge2(a^3 \plus{}b^3 \plus{}c^3 )$

hello, afteryour correction is your inequality equivalent to
$ \minus{} a^3 \plus{} ab^2 \plus{} ac^2 \plus{} a^2b \minus{} b^3 \plus{} bc^2 \plus{} ca^2 \plus{} cb^2 \minus{} b^3\geq0$ This is true because we can rewrite in the following form
$ a^2( \minus{} a \plus{} b \plus{} c) \plus{} b^2(a \plus{} c \minus{} b) \plus{} c^2(a \plus{} b \minus{} c)\geq0$ and this is true because of the triangle inequality.
Sonnhard.


trigonometry
inequalities
real analysis
real analysis unsolved

study the convergence of

$u(n)=\sin(1+\sin(2+\sin(3+...+\sin(n)...)))$

This is a nice problem. It is a pity that nobody replied for a month. Here is a
[hide="Hint"] We always have $|\sin x-\sin y|\le |x-y|$ and for most $x,y$, this inequality can be improved quite a lot. Use it to estimate $|u(n)-u(n+1)|$[/hide]

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=121224[/url]
In fact it's easy to prove that $|u_{n+1}-u_{n}|\leqslant C\cdot q^{n}$ with some $q\in(0;1)$

Here's proof of my statement. In [b]Rust[/b]'s notation \[|x_{n+1}-x_{n}|=|x_{n+1}(y_{n,1})-x_{n+1}(0)|=|\frac{\partial x_{n+1}}{\partial x}(\theta y_{n,1})|\cdot|y_{n,1}|\le\sup_{[-1;1]}|\frac{\partial x_{n+m}}{\partial x}|\]
\[\frac{\partial x_{n+m}}{\partial x}=\cos(1+\sin(2+\ldots)\cdot\cos(2+\sin(3+\ldots)\cdot\ldots\]
For $0\le k\le\frac{n-\frac{\pi}2-1/2}{\pi}$ let $m$ be the nearest integer to $\frac{\pi}2+\pi k$. For such $m$
\[|\cos(m+\ldots)|\le\max_{t\in[\frac{\pi}2+\pi k-1,5;\frac{\pi}2+\pi k+1,5]}|\cos t|=0,9974949866\ldots.\]


Thanks you for helping me for a long time.
Given two circles $(O_{1})$, $(O_{2})$ which meet together at $A$ and $B$. Let $PQ$ and $MN$  are two common tangent ($P,M \in (O_{1})$, $Q,N \in (O_{1})$). $MM'$ is diameter of $O_{1}$.
Let $O_{3}$, $O_{4}$ are the circumcircle of $APQ$ and $AMN$ respectively. Prove that
$O_{1}O_{2}$, $O_{3}O_{4}$, $M'N$ are concurrent .

$\angle AMN = \angle MBA, \angle ANM = \angle NBA \Rightarrow \angle MAN+\angle PAQ = 180^{o}\\ \Rightarrow \angle MAN+\angle MBN = 180^{o}\Rightarrow R_{(AMN)}= R_{(APQ)}\text{(Of course }MN = PQ ) \\ \angle O_{3}AP = 90^{o}-\angle PQA = 90^{o}-\angle QNA \\ \angle O_{4}AM = 90^{o}-\angle MNA = 90^{o}-\angle NQA \\ \angle PAM = 360^{o}-\angle MAP-\angle PAQ-\angle QAN = 180^{o}-\angle QAN \\ \Rightarrow O_{3}AO_{4}= 180^{o}\Rightarrow (O_{3}) \text{ touches }(O_{4}) \text{ at }A \\ \text{We have }\angle PAO_{3}-\angle QAO_{3}= (90^{o}-\angle PQA)-(90^{o}-\angle QPA )\\ = \angle QPA-\angle PQA = \frac{\angle AO_{1}P-\angle AO_{2}Q}{2}= \angle O_{2}AQ-\angle O_{1}AP \\ \text{So }\angle O_{1}AO_{3}= \angle O_{2}AO_{3}\Rightarrow \frac{\bar{SO_{1}}}{\bar{SO_{2}}}=-\frac{R_{1}}{R_{2}}\text{ (\{S\}=}O_{3}O_{4}\cap O_{1}O_{2}) \\ \text{Easy to prove that }S \in M'N$

Yes, good solution. Congratulation and thanks for solved my problems. :)

P/s: I wonder that why I can't illustrate this problem by Skectchpad or Cabri.  It seem wrong if I draw in computer.
Can anybody help me?

I've drawn it by Sketch Pad and it's true.
It's a RAR file because MathLinks doesn't allowed to attach GSP file  :D


If a is 50% larger than c and b is 25% larger than c, then a is what percent larger than b?

$ a\equal{}\frac{3c}{2}, b\equal{}\frac{5c}{4}$

$ c\equal{}\frac{4b}{5}$

$ a\equal{}\frac{3}{2}\times \frac{4b}{5} \equal{} \frac{6b}{5}$

Therefore $ 20\%$.


function
real analysis
real analysis unsolved

we considere a function $ f: I\minus{}>R$ ($ I$ is an intervalle)

$ 1$) let $ E$ be the differential equation :$ x'\equal{}f(x)$
Proof that a solution $ \phi$ of $ E$ has to be monotonic.

$ 2$)  let $ F$ be the differential equation : $ x\equal{}f(x')$
proof that a solution $ \psi$ of $ F$ has to be convexe or concave.

[quote="fedja"]No need to bump it up :P. This is fairly standard, so I'm also surprised nobody posted a solution yet. Perhaps, it just doesn't look attractive enough... :?[/quote]

Or maybe it is incomplete. I think it will be possible to find degenerate f, for which there is no solution to either E or F.

[quote="othman"]I don't like this "BAD HABIT" of saying that everything is STANDARD or easy,just because you have read a lot of problems(maybe without even trying to solve them) :)[/quote]

Not everything, far from it. But look for yourself. Take part 1. What does the condition really say, if you drop all antourage? It says that $ x(s)\equal{}x(t)\implies x'(s)\equal{}x'(t)$. This forces us to try to find two equal values of $ x$ with different values of $ x'$ for any differentiable non-monotone function. What is our intuition here? If $ x$ is not monotone, it should cross some level going both up and down. Essentially we are done. The rest is just routine
[hide="technicalities"]
If $ x$ is not monotone, we can find $ a<b<c$ such that $ x(a)>x(b)$ and $ x(b)<x(c)$ (replacing $ x$ by $ \minus{}x$ if necessary). Thus, for every $ x(b)<y<\min(x(a),x(c))$, there exist $ s\in(a,b)$ and $ t\in(b,c)$ such that $ x(s)\equal{}x(t)\equal{}y$. Moreover, we can choose the least such $ s$ and the largest such $ t$ (again, because of our intuition that it is the first crossing that should be down and the last one that should be up). Then we must have $ x'(s)\le 0$ (because $ x$ must be above $ y$ to the left of $ s$) and $ x'(t)\ge 0$. Thus, $ x'(s)\equal{}x'(t)\equal{}0$. That also should hold for [i]every[/i] crossing of $ y$ then, i.e., at every point $ s$ where $ x(b)<x(s)<\min(x(a),x(c))$. Now assume for definiteness that $ x(c)\le x(a)$. Let $ b'$ be the largest point on $ (b,c)$ with $ x(b')\equal{}x(b)$ and let $ c'$ be the least point on $ (b',c)$ with $ x(c')\equal{}x(c)$. Then $ x(b)<x(s)<x(c)$ for all $ s\in(b',c')$ whence $ x'(s)\equal{}0$ there whence $ x(b)\equal{}x(b')\equal{}x(c')\equal{}x(c)$ by Rolle, which is absurd.
[/hide]
I leave part 2 to somebody else.

The point is that there is no crazy twist in the problem, no unexpected idea. It just tests your basic knowledge of elementary analysis (this is what I mean by "standard"). If you cannot solve it "on the fly", your analysis training is just not adequate yet. Anyway, my "standard :P" post was written primarily in response to your "no one :rotfl:" post, the message being that there [b]are[/b] people on the forum (plenty of them, actually), who can solve this in no time but prefer to leave it to less experienced AoPSers. :)

As to just "reading a lot of problems without an attempt to solve them", that would be a bad habit, indeed. But I plead "not guilty" :P.


geometry
geometric transformation
combinatorics unsolved
combinatorics

Three boxes with at least one marble in each are given. In each step we double the number of marbles in one of the boxes, taking the required number of boxes from one of the other two boxes. Is it always possible to have one of the boxes empty after several steps?

In short, the answer is yes.
This is the problem М115 journal "Kvant", issue 11, 1971.
The solution in Russian is given at
http://www.problems.ru/view_problem_details_new.php?id=73650

Thanks for that, but I cannot read Russian. Can you give the outline of the proof, maybe?

Sorry, I'm quite busy atm.
Try [url=http://babelfish.yahoo.com/translate_url?doit=done&tt=url&intl=1&fr=bf-home&trurl=http%3A%2F%2Fwww.problems.ru%2Fview_problem_details_new.php%3Fid%3D73650&lp=ru_en&btnTrUrl=Translate]Babelfish translation[/url] - it generates acceptable text, but breaks some formulas - check them up the original text.


Is it possible to break the infinite lattice into 1x2 dominoes in such a way that each line that goes along the lattice only crosses a finite number of dominoes?

The answer, perhaps surprisingly, is yes.  It's not hard to come up with an actual example.  (I found it by trying to prove it couldn't be done.)


There are 500 bottles.
1 is toxic.
There are 5 tries per rat.
You need the least number of rats to find the toxic bottle in the 500.

Good Luck :!:  :!:
 :ninja:

[quote="jonathanchou711"]Isn't it ummmm.. I dunno. I'm going to guess 100, but prob. not.

But correction: You probably meant that you could be sure taht you can get teh poison bottle. Cause I could get it on my first try. DIE RATS!!  :P[/quote]
Hmmm... I got lower than that

[quote="jonathanchou711"]There's always a trick to these question. Like robbers and gold and one of the bags is filled with like some sort of thing that kills them...  :maybe:  :ninja:[/quote]
Yeah I agree with you, there is always a key. :maybe:  :rotfl:

[quote="joyofpi"]This might take more but I'm not sure... but what if you were to mix two of the bottles, like some of each and feed the mixture to a rat and if it dies then you can test it on several other rats and then so on and so forth. I think that would take more rats... I'm not sure...[/quote]
You are getting close

[quote="mz94"]pour everything into 1 jar. you foudn your poisoned jar. 0 rats. gg[/quote]
Sorry you are incorrect


Ring Theory
algebra
function
domain
superior algebra
superior algebra unsolved

$ R$ is commutative ring with unity and is a principa ideal ring.if $ I$ and $ J$ be prime ideals so that $ I \subset  J$ .proove $ J$ is a maximal ideal.

By replacing $ R$ with $ R/I$ it is the same to show that any nonzero prime ideal $ J$ in a PID is maximal. Let $ J \equal{} (x)$ and suppose $ (y) \supset (x)$; then $ yz \equal{} x$ for some $ z \in R$.

Suppose $ z \equal{} kx$ for some $ k \in R$. Then $ yk \equal{} 1$, implying $ (y) \equal{} R$; this is also false by assumption.

Therefore $ z \not\in (x), y \not\in (x)$, and $ yz \in (x)$. This contradicts that $ (x)$ is a prime ideal; therefore we have shown $ (x)$ is maximal.

Alternately: we can show that in any domain, every maximal ideal contains no nonzero prime ideals. Let $ M$ be a maximal ideal; then we may localize at $ M$ and obtain a domain with a single maximal ideal; therefore let us assume the ring $ R$ is a local PID. Therefore $ R$ is a discrete valuation ring, with maximal ideal $ (\pi)$ for some $ \pi$. Every nonzero element in $ R$ can be written $ u\pi^n$ for some unit $ u$. Clearly $ (u \pi^n)$ is only prime if $ n \equal{} 1$, in which case $ (u \pi) \equal{} (\pi)$. Therefore the only prime in $ R$ is $ M$.[/hide]


logarithms
function
algebra unsolved
algebra

Solve in R : $ c^{\log_b x}\plus{}a\cdot c^{\frac{a\cdot \log_b x\plus{}1}{a}}\plus{}a\cdot\log_b x\cdot c^{\log_b x}\equal{}a$ , a>0 , b>1 and c>=1 .

[quote="trbst"]Solve in R : $ c^{\log_b x} \plus{} a\cdot c^{\frac {a\cdot \log_b x \plus{} 1}{a}} \plus{} a\cdot\log_b x\cdot c^{\log_b x} \equal{} a$ , a>0 , b>1 and c>=1 .[/quote]

So $ c^{\log_b x}(1\plus{} a\cdot c^{\frac 1a} \plus{} a\cdot\log_b x) \equal{} a$

So $ 1\plus{} a\cdot c^{\frac 1a} \plus{} ay \equal{} ac^{\minus{}y}$ where $ y\equal{}\log_b x$

LHS is an increasing function while RHS is a decreasing one. So we have at most one solution to this equation.

And since we have the trivial one $ y\equal{}\minus{}\frac 1a$, we got the unique solution $ \boxed{x\equal{}b^{\minus{}\frac 1a}}$


Joe and Millie each have some baseball cards, and $ \frac34$ of the number of cards Joe has is equal to $ \frac23$ of the number of cards Millie has. What is the least number of cards they could have altogether?

First, we must find $ \text{lcm}(3,4)$.  This is clearly $ 12$.

Now, you have $ 12 \div \frac34 \plus{} 12 \div \frac23$ as our answer.  This is $ 16 \plus{} 18$, or $ \boxed{34}$.

$ \frac {3}{4}\cdot J \equal{} \frac {2}{3}\cdot M$

$ 9J \equal{} 8M$

Therefore J=8 and M=9 when we minimize. 8+9=17

EDIT: Thanks math!

[quote="mewto55555"]$ \frac {3}{4}\cdot J = \frac {2}{3}\cdot M$[/quote]

What you really want is:

[code]$\frac34J=\frac23M$ or $\frac34\cdot J=\frac23\cdot M$[/code]

$ \frac34J=\frac23M$ or $ \frac34\cdot J=\frac23\cdot M$

I know, I'm fanatical about optimization...

But in general, you must separate all commands from variables by a space or a bracket around the variable, etc.

You could just cross multiply to get $\frac{3}{4}x={\frac{2}{3}y}$.  You'd get $\frac{9+8}{12}=\frac{\boxed{17}}{12}$


geometry
3D geometry
geometric transformation
rotation

Carl has no life and glues together four unit cubes, lined up face to face. How many non-congruent solids can he make? Rotations do not form new solids.

[hide="comments"]The answer key sayd 8, but I only count 7...[/hide]

[hide]I get 8.  You probably forgot that the weird 3D one (an isosceles L with an extra cube stuck on top (3rd dimension) of the leg) AND ITS MIRROR IMAGE are two different ones.  You cannot get the mirror image with any number of rotations. (i thin k)

The list:
2D figures:
1. line
2. long L
3. T
4: Z - ish
5. 2x2 square
now for the 3D ones:
6.isosc. L with a cube on top at the vertex
7. '' on top at the leg-end
8.  MIRROR IMAGE of 7.[/hide]

Oh, I was suspecting that. Gah I can't imagine these things, and I didn't have any blocks handy...

Yeah... these spatial things are hard to imagine sometimes.  Spatial problems are not like any other MC problems; you should practice them separately.


Let a,b,c>0 prove that:
 $ \frac{a^3}{b^2\minus{}bc\plus{}c^2}\plus{}\frac{b^3}{c^2\minus{}ca\plus{}a^2}\plus{}\frac{c^3}{a^2\minus{}ab\plus{}b^2}\geq{a\plus{}b\plus{}c}$

it is true by chebysev if a,b,c is in triangle :wink:

[quote="xiaohui"]Let a,b,c>0 prove that:
 $ \frac {a^3}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^3}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {c^3}{a^2 \minus{} ab \plus{} b^2}\geq{a \plus{} b \plus{} c}$[/quote]
$ \sum_{cyc}\frac {a^3}{b^2 \minus{} bc \plus{} c^2}\equal{}\sum_{cyc}\frac {a^4}{ab^2 \minus{} abc \plus{} ac^2}\geq\frac{(a^2\plus{}b^2\plus{}c^2)^2}{\sum(a^2b\plus{}a^2c\minus{}abc)}\geq a\plus{}b\plus{}c,$
where the last inequality is true by Schur.

[quote="arqady"][quote="xiaohui"]Let a,b,c>0 prove that:
 $ \frac {a^3}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^3}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {c^3}{a^2 \minus{} ab \plus{} b^2}\geq{a \plus{} b \plus{} c}$[/quote]
$ \sum_{cyc}\frac {a^3}{b^2 \minus{} bc \plus{} c^2} \equal{} \sum_{cyc}\frac {a^4}{ab^2 \minus{} abc \plus{} ac^2}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(a^2b \plus{} a^2c \minus{} abc)}\geq a \plus{} b \plus{} c,$
where the last inequality is true by Schur.[/quote]
Can U detail ? If use Schur ; follow me ; we have to prove that
$ (a^\plus{}b^2\plus{}c^2)^2 \ge\ (a^3\plus{}b^3\plus{}c^3)(a\plus{}b\plus{}c)$
But This Ineq is wrong :maybe:

[quote="onlylove_math"][quote="arqady"][quote="xiaohui"]Let a,b,c>0 prove that:
 $ \frac {a^3}{b^2 \minus{} bc \plus{} c^2} \plus{} \frac {b^3}{c^2 \minus{} ca \plus{} a^2} \plus{} \frac {c^3}{a^2 \minus{} ab \plus{} b^2}\geq{a \plus{} b \plus{} c}$[/quote]
$ \sum_{cyc}\frac {a^3}{b^2 \minus{} bc \plus{} c^2} \equal{} \sum_{cyc}\frac {a^4}{ab^2 \minus{} abc \plus{} ac^2}\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{\sum(a^2b \plus{} a^2c \minus{} abc)}\geq a \plus{} b \plus{} c,$
where the last inequality is true by Schur.[/quote]
Can U detail ? If use Schur ; follow me ; we have to prove that
$ (a^ \plus{} b^2 \plus{} c^2)^2 \ge\ (a^3 \plus{} b^3 \plus{} c^3)(a \plus{} b \plus{} c)$
But This Ineq is wrong :maybe:[/quote]
$ (a^2\plus{}b^2\plus{}c^2)^2\geq(a\plus{}b\plus{}c)\sum_{cyc}(a^2b\plus{}a^2c\minus{}abc)\Leftrightarrow\sum_{cyc}(x^4\minus{}x^3y\minus{}x^3z\plus{}x^2yz)\geq0,$ which Schur.

I think the problem is wrong . Because With $ a\equal{}1 , b\equal{}10 , c \equal{} 100$ the problem is wrong

[quote="thanhak5"]I think the problem is wrong . Because With $ a \equal{} 1 , b \equal{} 10 , c \equal{} 100$ the problem is wrong[/quote]
Are you sure?

[quote="xiaohui"]Let $ x,y,z>0$, prove that

$ \frac {x^3}{y^2 \minus{} yz \plus{} z^2} \plus{} \frac {y^3}{z^2 \minus{} zx \plus{} x^2} \plus{} \frac {z^3}{x^2 \minus{} xy \plus{} y^2}\geq{x \plus{} y \plus{} z}$.[/quote]
Without lost the general, we may assume $ x\equal{}\min\{x,y,z\}$. Using the arithmetic mean - geometric mean inequality,

$ \sum{\frac{x^3}{y^2\minus{}yz\plus{}z^2}}\plus{}\frac{4}{(y\plus{}z)^2}\sum{x\left(y^2\minus{}yz\plus{}z^2\right)}\geq\frac{4}{y\plus{}z}\sum{x^2}$,

we only need to prove :

$ \frac{4}{y\plus{}z}\sum{x^2}\geq\sum{x}\plus{}\frac{4}{(y\plus{}z)^2}\sum{x\left(y^2\minus{}yz\plus{}z^2\right)}$.

[hide="Proof."]$ 4(y\plus{}z)\sum{x^2}\minus{}(y\plus{}z)^2\sum{x}\minus{}4\sum{x\left(y^2\minus{}yz\plus{}z^2\right)}\equal{}(y\minus{}z)^2(3y\plus{}3z\minus{}5x)\geq 0$.[/hide]


We realize that this expression is symmetrical at $ \frac {1}{2!3!}$. That means we only need to evaluate $ 2\left(\frac {1}{5!} \plus{} \frac {1}{4!} \plus{} \frac {1}{2!3!}\right) \equal{} 2\left(\frac {1}{120} \plus{} \frac {1}{24} \plus{} \frac {1}{12}\right) \equal{} 2\left(\frac {1}{120} \plus{} \frac {5}{120} \plus{} \frac {10}{120}\right) \equal{} 2\left(\frac {16}{120}\right) \equal{} 2\left(\frac {2}{15}\right) \equal{} \boxed{\frac {4}{15}}$

[quote="RoFlLoLcOpT"]We realize that this expression is symmetrical at $ 2\left(\frac {1}{120} \plus{} \frac {1}{24} \plus{} \frac {1}{12}\right) \equal{} 2\left(\frac {1}{120} \plus{} \frac {4}{120} \plus{} \frac {10}{120}\right) \equal{} 2\left(\frac {15}{120}\right) \equal{} 2\left(\frac {1}{8}\right) \equal{} \boxed{\frac {1}{4}}$[/quote]

I never thought of the symmetry, so that's why I had no idea how to do this  :blush:   But the problem here is that $ \frac{1}{24}\equal{}\frac{5}{120}$, and not $ \frac{4}{120}$.  Therefore, the answer is actually $ \boxed{\frac{4}{15}}$

Well, there's another thing you could have seen.
If you look at the terms again, you notice something related to values of 5C1 5C2 5C3..etc, only 5! is missing, so multiply by it and divide by it and it becomes 1/5! * ( 2^5) --> Using summation of nCr = 2^n
which is 32/120 = 4/15 :)

LaTeX edited:

$\frac{5!}{5!}\cdot\left(\frac{1}{5!}+\frac{1}{4!}+\frac{1}{3!2!}+\frac{1}{2!3!}+\frac{1}{4!}+\frac{1}{5!}\right)$

$=\frac{\dbinom{5}{0}+\dbinom{5}{1}+\dbinom{5}{2}+\dbinom{5}{3}+\dbinom{5}{4}+\dbinom{5}{5}}{5!}$

$=\frac{2^5}{5!}$

$=\frac{32}{120}$

$=\boxed{\frac{4}{15}}$

This method tends to be easier when there are a lot of terms in the expression.


a circle passes through the vertex C of a rectangle ABCD and touches AB and AD at M and N respectively.If the distance between C and MN is 5 cm. find the area of the rectangle.

thank you.....

are you sure there is enough information?

i dnt know.......it was given in a school test

[hide="diagram"][asy]pair A=(0,0), B=(0,45+45*sin(65/180*pi)), C=(45+45*cos(65/180*pi),45+45*sin(65/180*pi)), D=(45+45*cos(65/180*pi),0), E=(45-45*cos(65/180*pi),45+45*sin(65/180*pi)), F=(45+45*cos(65/180*pi),45-45*sin(65/180*pi)), M=(0,45), N=(45,0), O, P; draw(circumcircle(M, N, C)); O=circumcenter(M,N,C); draw(A--B--C--D--cycle); draw(O--M--N--O--C--M--F--C--N--D--F--N--E--F); draw(M--E); P = intersectionpoint(M--N,C--(C+3*(M-N)*dir(90))); draw(P--C); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,NW); label("F",F,NE); label("M",M,NW); label("N",N,SW); label("P",P,SW); label("O",O,SW); dot(A^^B^^C^^D^^E^^F^^M^^N^^P^^O);[/asy][/hide]Let a perpendicular down from $ C$ onto $ MN$, call this point $ P$.
Let the center of the circle be $ O$.
Let $ E$ and $ F$ be the intersection of $ BC$ and $ CD$ with circle $ O$, respectively.
Now we will do some angle chasing. Let $ \angle BME = \beta$ and $ \angle DNF = \alpha$. Since $ ON\perp ND$, $ \angle DNF = \angle NMF = \angle NEF = \angle NCF = \alpha$ and similarly $ \angle BME = \angle MNE = \angle MFE = \angle MCE = \beta$.
Since $ \triangle OEN$ is isosceles, $ \angle ONE=\angle OEN=\alpha$. $ ON\perp ND, CD \perp ND \implies ON \parallel CD \implies \angle ONC = \angle NCD=\alpha$. So $ \angle MNC = \angle MNE+\angle ENO+\angle ONC = \beta+\alpha+\alpha$.
We already know that $ \angle ECF = 90$ and $ \angle MCN = \frac{1}{2}\angle MON = 45$, so $ \alpha+\beta = 45$. This means $ \angle PCN = \beta$. $ PC = 5$ by definition so $ NC = \frac{5}{\cos\beta}\implies DC = \frac{5\cos\alpha}{\cos\beta}$.
Similarly we obtain $ \angle PCM = \alpha \implies MC = \frac{5}{\cos\alpha}\implies BC = \frac{5\cos\beta}{\cos\alpha}$.
So $ [ABCD]=BC\cdot DC = 25$

thank you for the solution....but i have a question
is it necessary that EF will pass through the centre O

yes, because $ \angle ECF = 90$, so $ \angle EOF=2\cdot \angle ECF = 180$, which means $ EF$ passes through $ O$.


Determine the least natural number $n$ for which the following result holds:

No matter how the elements of the set $\{1,2, \dots ,n\}$ are coloured red or blue, we can find integers $w,x,y,z$ in the set (not necessarily distinct) of the  same colour such that
$w+x+y=z$

[hide]
We claim that $n = 11$. We will attempt to construct a failing case for $n = 11$ and show that it is impossible. First, we note that $1+1+1 = 3$, $2+2+2 = 6$, and $3+3+3 = 9$ so none of the pairs $(1,3)$, $(2,6)$, or $(3,9)$ can be the same color. Also, $3+3+2 = 6+1+1$ so $(3,2)$ and $(1,6)$ cannot be red-red and blue-blue or vice versa. Consequently, we can obtain (assuming WLOG that $1$ is blue):

[code]1 2 3 4 5 6 7 8 9 10 11
B B R X X R X X B X  X
[/code]

since if $2$ was red and $6$ was blue we would have the red-red and blue-blue case. Now $2+1+1 = 4$, so $4$ is red. And $3+4+4 = 11$, so $11$ is blue.

[code]1 2 3 4 5 6 7 8 9 10 11
B B R R X R X X B X  B
[/code]

But this gives us $9+1+1 = 11$ and they are all blue, so contradiction. Hence we cannot construct a failing coloring with $n = 11$.

For $n = 10$, we have

[code]1 2 3 4 5 6 7 8 9 10
B B R R R R R R B B
[/code]

So $n = 11$ is the smallest.

[/hide]

Paladin:  It's a good proof, but how did you find the smallest n?  Did you just keep trying cases?

Well, I constructed a case for like $n = 5$ or something... then I realized it was too small so I added numbers one-by-one until it didn't work, tried to recolor and if it worked I kept going. So pretty much just experimenting to find the answer; a good way to solve olympiad problems.


function
number theory open
number theory

In the following $x_0$  and $x_1$ are integers and $A,B$ are given rational numbers. By  [ X ]  is denoted the integer part of  $X$ . Suppose that
\[ x_{k+1}= \left[ Ax_k + Bx_{k-1} \right] \; ,\; \; k=1,2,...,n,... .  \]
[color=green] [b]It's possible to find  $x_n$ as a function of variables $n,x_0 ,x_1,A,B$ ? [/b] [/color]

I think flip meant Unsolved instead of Open?

Anyway I know there must be one for every value of A,B although I guess the general way of writing will be extremely complex...

Guess it'll start with:

...in the fibonacci sequence the n-th term is given by: $\displaystyle F(n)=\frac{(\frac{1+\sqrt{5}}2)^n-(\frac{1-\sqrt{5}}2)^n}{\sqrt{5}}$

and writing it as a lineary combination of fibonacci sequences.
Although I have to admit I didn't get it fixed myself yet.

[quote="flip2004"]I have'nt a solution of this problem.Regards,proposer.[/quote]Then please next time post the problem in the open problems section. :)

[quote="Peter VDD"]Valentin, this [b]IS[/b] the open problems section :D:D[/quote]yes, because I have moved the topic. ;)

lol

well, he originally posted it in the open problems section, and you replied: "Is this an open question??"

And the times I replied were in open section as well :P  (never even noticed it's been gone since then) :)


Given an acute triangle $ ABC$. Let $ P$ be the interior point of $ ABC$ such that $ \hat{APB}\equal{}\hat{BPC}\equal{}\hat{CPA}\equal{}120^o$.
Prove that the Euler Line of triangles $ ABP,BCP,CAP$ are concurrent at the centriod of Napoleon Triangle.

[quote="¬[ƒ(Gabriel)³²¹º]¼"]see here:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=6188[/url][/quote]

Thank's Very much :)


geometry
circumcircle
perpendicular bisector
geometry proposed

[b]let ABC be a triangle. and let P be a point inside. let M=max{PA,PB,PC}. Find P that M has its min value! :maybe: [/b]

for an acute triangle , i believe it is the cicuncenter of the triangle :|

i think we need  to think about the perpendiculars bisectors
but i´m not sure of it :maybe:
and , by inside of the triangle , you mean not at it´s sides?

I don´t whether this is right or not...my guess is that it is not totally right,or maybe it is all wrong :blush: .can anyone tell me whether i´m wrong or not?

 for an acute triangle ABC, P is the cicuncenter of ABC .suppose for absurd that it is not the circumcenter O and call this point X.
   Then , WLOG X lies in the triangle OAB . But then CX  is bigger than CO , absurd.
  Now , if the triangle is obtuse, let AB be it´s longest side .
Consider the perpendicular bisector of AB . Let D be the point such that |AD-CD| is as least as possible.If CD>AD=BD,choose one point E at CD such that |EC-EB| is as least as possible.
Since ABC is obtuse, we´ll have that  EA<EC.
So now the maximun value of EC, EB ,EA is as elast as possible.
so P=E .
Is this right? :maybe:

In a triangle ABC, C=90, A=30 and BC=1 cm. let A'B'C' be a triagnle inside of ABC, so that A' is on BC and so for B' and C'. If M=max{A'B',B'C',C'A'}. find the min value of M.

my solution:(sorry my English is not well in geometry)

We want to find the smallest in A'B'C' 's, that they have the same shape (A'B'C' ~ A"B"C")

First draw A'B'C' as an example. Let C1 be the points that (It is a part of a circle) for any point of

them like E we have A'EB'=90. and C2 the points that B'FC'=30 and C3 that C'GA'=60(EFG stands

for the original ABC). We know that there is a point like P that it is on C1,C2 and C3. Because all

EFG's have the same shap and ABC~EFG. the smallest A'B'C' can be seen when we have the biggest

EFG and then we have the longest EF. If O1 is the center of C1 and O2,O3 for C2,C3. We can say:

EF =< 2 * O1 O2: And If we want to have the equality we have to choose E when EF || O1 O2. then

EB'P=90 , and then for others we have the same thing.

Then we know that B'C'= 2R1 * sin(A) ,  AP=2R1 => B'C' = AP sin(A)

And we find it the same for others. So the problem is changed to what I said At first. :blush:  :blush:  :roll:  :maybe:

I hope some one solve it for me! I am really interested in it! :huuh:  :help:

Sorry I don't have pro painting programs!:D

I don't have a good program to draw a good one! sorry!
And the names have problem in the pic! But you can understand what I tried to say!:D


calculus
derivative
limit
function
calculus computations

Suppose f(x) =
x^3 if x < 1
ax^2 + bx + c if x ≥ 1

Find the values of a, b and c such that f''(1) exists.

You don't have to calculate the limits explicitly, by mean value theorem you can just pass limits in your other expressions for the derivatives to get the left and right derivatives at the problematic point.


X,Y and Z are positive real numbers such that  XYZ=1.

Find the least values of

1. I/(X+1) + 1/(Y+1) +1/(Z+1)

2. 1(/X+1)(Y+1) + 1/(Y+1)(Z+1) +1/(Z+1)(X+1)

you mean:
[quote]Given $x,y,z \ | \ x,y,z \in \mathbb{R^+} \ | \ xyz=1$ minimize:
$\mathfrak{1) } \sum_{cyc}\frac{1}{x+1}$
$\mathfrak{2) } \sum_{cyc}\frac{1}{(x+1)(y+1)}$[/quote]
 :P

Well here's how I see it:
We're most certainly gonna use: $\prod_{cyc}{(x+1)}=xyz+\sum_{cyc}{xy}+\sum_{cyc}x+1=2+\sum_{cyc}{xy}+\sum_{cyc}x$

I'll finish up tomorrow, but I'll give [hide="my unproved answers:"]Just set $x=y=z \implies x^3=1 \implies x=y=z=1$
So answer to #1: $\frac32$
#2: $\frac34$[/hide]
Those are just my first guesses...

for the first one , if $x,y\rightarrow \infty$ then $z\rightarrow 0$ so the whole expression will $\rightarrow 1$

for the second one , we do the same way and the expression $\rightarrow 0$

[quote="xxreddevilzxx"][hide="1"]
3-term AM-GM yields
$\frac{S}{3} \ge \sqrt[3]{\frac{1}{xyz+xy+yz+zx+x+y+z+1}}$, $S = LHS$
$\frac{S}{3} \ge \sqrt[3]{\frac{1}{xy+yz+zx+x+y+z+2}}$
$S \ge 3\sqrt[3]{\frac{1}{6+2}}$, since $xy+yz+zx+x+y+z \ge \sqrt[6]{[xyz]^3} = 6$
$S \ge 3 \cdot \frac{1}{2} = \frac{3}{2}$
[/hide][/quote]

if $xy+yz+xz+x+y+z\geq 6$ then $\frac{1}{xy+yz+xz+x+y+z}\leq \frac{1}{6}$
so your solution is not correct . Moreover , see my above post which implies minimum at $1$

 ;)

According to shyong's post, there is [b]no[/b] minimum because $x,y,z$ are positive real numbers. 1 is a limit point but it's not actually achieved by the inequality (least upper bound)

[quote="10000th User"]1 is a limit point but it's not actually achieved by the inequality (least upper bound)[/quote]

Are you sure you understand what limit point means?  There are better phrases you could use there.


The numerator of a fraction is 1 less than the denominator. if the  numerator  and the denominator are both increased by 4 ,the new fraction will be 1/8 more than the original fraction . find the original fraction

[hide] Let $ x$ be the denominator of the first fraction. Then,

$ \frac{x\minus{}1}{x}\plus{}\frac{1}{8}\equal{}\frac{x\plus{}3}{x\plus{}4}$.

Solving, we have

$ \frac{x^2\plus{}3x\minus{}4\minus{}x^2\minus{}3x}{x^2\plus{}4x}\equal{}\minus{}\frac{1}{8}$

$ \implies x^2\plus{}4x\equal{}32 \implies (x\plus{}8)(x\minus{}4)\equal{}0$.

Upon inspection, $ x\equal{}\minus{}8$ does not work as a solution, so the original fraction is $ \boxed{\frac34}$.[/hide]


Two distinct positive integers $ x$ and $ y$ are factors of 36. What is the least product $ xy$ that is not a factor of 36?

both 2 and 4 are factors of 36, but 8 is not.

answer : 8


calculus
derivative
integration
calculus computations

$ f \in C^0 $ , differentiate
\[
\displaystyle \int^{\int^x_0  f(t) \, dt}_0  f(u) \, du
\]

The answer is

\[f\left(\int^x_0 f(t) dt\right) f(x)\].


Hello, I have a question that I have help with.

It states: If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have?

[hide="Full explanation"]If we can factor $n=2^{a}\cdot 3^{b}\cdot 5^{c}\cdots$, then, by combinatorics, $n$ has $d(n)=(a+1)\cdot (b+1)\cdot (c+1)\cdots$ divisors.
Since only $2$ and $3$ are important in this problem, let's say
$d(n)=(a+1)\cdot (b+1)\cdot k$
$d(2n)=(a+2)\cdot (b+1)\cdot k$
$d(3n)=(a+1)\cdot (b+2)\cdot k$
$d(6n)=(a+2)\cdot (b+2)\cdot k$

Let's factor $d(2n)=28$ and $d(3n)=30$ in three different factors with all the different possibilities:
$28=1\cdot 1\cdot 28=1\cdot 2\cdot 14=\boxed{1\cdot 4\cdot 7}=2\cdot 2\cdot 7$
$30=1\cdot 1\cdot 30=1\cdot 2\cdot 15=1\cdot 3\cdot 10=\boxed{1\cdot 5\cdot 6}=2\cdot 3\cdot 5$

I boxed the only way (just a little casework) that we can get
$d(2n)=(a+2)\cdot (b+1)\cdot k=28$
$d(3n)=(a+1)\cdot (b+2)\cdot k=30$
with $a=5$, $b=3$, $k=1$ ($n=2^{5}\cdot 3^{3}=864$).

Therefore, $d(6n)=7\cdot 5\cdot 1=35$.[/hide]

[quote="lordWings"]
Notice that there are two possible values of $n$:
$n=2^{3}\cdot 3^{5}=1944$
$n=2^{5}\cdot 3^{3}=864$[/quote]

Not quite. For the first value, the numbers of divisors of $2n$ and $3n$ are reversed.

You're right. I just edited the mistake so that the reasoning will be clear.


[quote="zhaobin"]I think we can just prove:
$(a+b)(b+c)(c+a) \ge (a+1)(b+1)(c+1)$
am I right?[/quote]

Yes, you're right. The problem was easy, although it took me about a day to think about AM-GM  :blush:

Valentin, can you tell me now whose problem is this?  :?

[quote="Soarer"]I think so, and this is easy, it's simply
$(a+1)(b+1)(c+1) \ge 8$ and $(a+a)(a+b)(a+c) \ge (a+1)^3$[/quote]
your solution are nicer than mine :P

I didn't understand what Soarer said in his post... Not thinking clearly now. can you develop?

ok.my pleasure
$(a+a)(a+b)(a+c) \ge (a+\sqrt[3]{abc})^3=(a+1)^3$ the first one is holder.
write any two similar.then we have
$2a2b2c(a+b)^2(a+c)^2(b+c)^2 \ge (a+1)^3(b+1)^3(c+1)^3 \ge 8 (a+1)^2(b+1)^2(c+1)^2$
which is equal to $(a+b)(b+c)(c+a) \ge (a+1)(b+1)(c+1)$

I didn't think, that after using AM-GM inequality our ineq would true, and I didn't use it.

[quote="Vasc"]Actually, the inequality holds for more general constraint
$ab+bc+ca \geq 3$.[/quote]
yes.with it we can get
$\frac 1 3(a+b+c)(ab+bc+ca) \ge ab+bc+ca$
and $\frac 1 3(a+b+c)(ab+bc+ca) \ge a+b+c$
so
$\sum_{syms}a^2b \ge \frac 2 3(a+b+c)(ab+bc+ca) \ge ab+bc+ca+a+b+c$
which is equal to what I wrote :)

[quote="zhaobin"] so
$\sum_{syms}a^2b \ge \frac 2 3(a+b+c)(ab+bc+ca) \ge ab+bc+ca+a+b+c$
which is equal to what I wrote :)[/quote]But this result doesn't imply the inequality
$(a+b)(b+c)(c+a) \geq (a+1)(b+1)(c+1)$.

OK，it is wrong. :)
but another one:
I think we can wlog $ab+bc+ca=3$ if  $ab+bc+ca=3k^2(k>1)$,then we can multiply $\frac 1 {k^3}$ in both side,let  $x=\frac a {k}$,then we need to prove $(x+y)(y+z)(z+x) \ge (x+\frac 1 k)(y+\frac 1 k)(z+\frac 1 k)$ $xy+yz+zx=3$ which is weaker than the case $ab+bc+ca=3$.
$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc$
so we need to prove:
$(a+b+c)(ab+bc+ca) \ge a+b+c+ab+bc+ca+1+2abc$
equal to
$(a+b+c-1)(ab+bc+ca-1) \ge 2+2abc$
with $ab+bc+ca-1 = 2$so we need to prove
$2(a+b+c) \ge 4+2abc$
which is true(because $a+b+c \ge 3$,$abc \le 1$)
hope it is right this time :)


hello, your equation is equivalent with
$ {\displaystyle (1915+76\cos(4x)+68\cos(8x)-12\cos(12x)+\cos(16x))\csc(x)^{10}\sec(x)^{10}\sin\left(\frac{\pi}{4}-x^\right)^2\sin\left(\frac{\pi}{4}+x\right)^2=0}$
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, your equation is equivalent with
$ {\displaystyle (1915 + 76\cos(4x) + 68\cos(8x) - 12\cos(12x) + \cos(16x))\csc(x)^{10}\sec(x)^{10}\sin\left(\frac {\pi}{4} - x^\right)^2\sin\left(\frac {\pi}{4} + x\right)^2 = 0}$
Sonnhard.[/quote]
Yes! I see, thank you. :P


Natural languages have a huge amount of redundancy and words are often unnecessarily long. These both limit the efficiency in which people may communicate, but help to prevent ambiguity and confusion.

Many have tried to construct languages or coding schemes in which it is faster to communicate. For example
[url]http://en.wikipedia.org/wiki/Speedtalk[/url]
[url]http://en.wikipedia.org/wiki/Arahau[/url]
[url]http://en.wikipedia.org/wiki/Ithkuil[/url]
[url]http://www.iiap.res.in/personnel/srik/Lin.html[/url]

I have read somewhere, though I cannot remember, that these attempts of efficiency are doomed to fail. Is it possible to communicate at a significantly higher rate than natural languages but still retain the same accuracy and low probability of error?

For that matter, do there exist natural languages that differ significantly (i.e. two times faster) in the speed of communication? It seems to me that languages with fewer phonemes (major ones being Japanese, Spanish) have fewer options for word construction, and thus must have longer words, but compensate through a high speed of speech. This is possible because the speaker doesn't have enunciate so much and distinguish each phoneme clearly from the next, since there are fewer of them.

Here are some Language Log posts that are relevant to your question.  BTW, Language Log is a collaborative linguistics blog run by a number of prominent linguists.  It's both wonderfully entertaining and a good jumping off point for research on wide-ranging linguistics questions like yours.

This one, "Speech rate and per-syllable information across languages," looks right on target for what you were asking:
[url]http://languagelog.ldc.upenn.edu/nll/?p=22[/url]

This set of posts, about comparative efficiency of different languages measured in terms of text size, might also be interesting:  [url]http://languagelog.ldc.upenn.edu/nll/?p=93[/url]


logarithms
inequalities
function
calculus
calculus computations

Prove the following inequality for $0<a<b.$

\[\sqrt{ab}<\frac{b-a}{\ln b-\ln a}<\frac{a+b}{2}.\]

[quote="kunny"]Prove the following inequality for $0<a<b.$
\[\sqrt{ab}<\frac{b-a}{\ln b-\ln a}<\frac{a+b}{2}. \]
[/quote]

This is an old problem and if i am not mistaken it is from St.Petersburg MO.One solution is obtained by setting $t=\frac{b}{a}$, and after obtaining inequalities only in terms of $t$ and then analyzing the functions using 1st and 2nd derivatives.


modular arithmetic
induction
algebra
linear equation
number theory proposed
number theory

For each positive integer $ n$, prove that there exists an $ n$-digit number with all its digits odd which is divisible by $ 5^n$.

We prove it by induction. The first few are $ a_1\equal{}5,a_2\equal{}75,a_3\equal{}375$ etc.

Suppose $ a_n$ exists and is an n-digit number divisible by $ 5^n$. Then $ a_n\equal{}5^nk$, k odd.

Consider $ f_n(b)\equal{}10^nb\plus{}5^nk$, which is what we get by sticking the digit $ b$ in front of $ a_n$. We have $ f_n(b)\equal{}5^n(2^n b \plus{}k)$. We want $ 2^nb\plus{}k$ to be divisible by 5, and we have a solution for  $ b\pmod5$ since we have a linear equation and $ 2^n$ is invertible mod 5. So there is a residue class for $ b\pmod5$ such that $ f_n(b)$ is divisible by $ 5^{n\plus{}1}$. Then either its residue in the range 0 to 4 is odd, or its residue in the range 5 to 9 is odd.

This sequence is also unique. We prove it by induction again.

Suppose $ a$ is an odd digit and $ b$ is an $ n$ digit number with all odd digits such that $ 10^na\plus{}b$ (an $ n\plus{}1$ digit number) is divisible by $ 5^{n\plus{}1}$. Either $ v_5(10^na)\equal{}n$ or $ n\plus{}1$ depending on whether $ a\equal{}5$ or not. Either way, we need $ 5^n\mid b$ since if $ v_5(b)\equal{}m<n$ then $ v_5(10^na\plus{}b)\equal{}m<n\plus{}1$.

So for each $ a$, each $ b$ which works is an $ n$ digit number divisible by $ 5^n$. By the inductive hypothesis, for each $ a$, there is at most one $ b$. Then from the existence proof, there is exactly one way to choose $ a$ given $ b$.


[quote="chess64"]Prove that for real numbers $a$, $b$, and $c$,
$a^ab^bc^c\ge(abc)^{\frac{a+b+c}{3}}$.[/quote]
The numbers must positive or $0$.
Wolog $a \le b \le c$
Then $a^ab^bc^c = a^ab^bc^{\frac{2c-a-b}{3}}c^{\frac{a+b+c}{3}}$
$\ge a^ab^bb^{\frac{2c-a-b}{3}}c^{\frac{a+b+c}{3}}$
$\ge a^ab^{\frac{3b+2c-a-b}{3}}c^{\frac{a+b+c}{3}}$
$= a^ab^{\frac{b+c-2a}{3}}(bc)^{\frac{a+b+c}{3}}$
$\ge a^aa^{\frac{b+c-2a}{3}}(bc)^{\frac{a+b+c}{3}}$
$= a^{\frac{a+b+c}{3}}(bc)^{\frac{a+b+c}{3}}$
$=(abc)^{\frac{a+b+c}{3}}$, qed

[hide="Solution"]I'll just assume you meant positive numbers since reals actually don't work.

WLOG let $a \geq b \geq c$.

For any positive numbers $x$ and $y$ such that $x \geq y$, we have

$x^x y^y \geq x^y y^x$
since this is equivalent to
$x^{x-y} \geq y^{x-y}$,
which is obvious.

We then have
$a^a b^b \geq a^b b^a$, $b^b c^c \geq b^c c^b$, and $a^a c^c \geq a^c c^a$.

Multiplying the three inequalities together, we get

$a^{2a} b^{2b} c^{2c} \geq a^{b+c} b^{a+c} c^{a+b}$.

Multiplying the inequality by $a^a b^b c^c$, we get

$a^{3a} b^{3b} c^{3c} \geq (abc)^{a+b+c}$,

which is equivalent to the problem statement.[/hide]

This isn't completely my solution though since I've seen this problem before...  :roll:

http://www.mathlinks.ro/Forum/viewtopic.php?t=46247

Soarer pointed a really good way: Take log and Chebyshev.


If $ a,b,c>0$, then $ \prod(\sqrt{a}\plus{}\sqrt[3]{a}\plus{}\sqrt[6]{a})\leq\prod(b\plus{}2)$

We also have $ \prod_{cyc}(\sqrt{a}\plus{}\sqrt[3]{b}\plus{}\sqrt[6]{c})\leq\prod(a\plus{}2)$:)

[quote="bakerbakura"]We also have $ \prod_{cyc}(\sqrt {a} \plus{} \sqrt [3]{b} \plus{} \sqrt [6]{c})\leq\prod(a \plus{} 2)$:)[/quote]

By Holder inequality we have

$ (a^6\plus{}1\plus{}1)(a^6\plus{}1\plus{}1)(a^6\plus{}1\plus{}1)(1\plus{}b^6\plus{}1)(1\plus{}b^6\plus{}1)(1\plus{}1\plus{}c^6) \ge (a^3\plus{}b^2\plus{}c)^6$

$ (b^6\plus{}1\plus{}1)(b^6\plus{}1\plus{}1)(b^6\plus{}1\plus{}1)(1\plus{}c^6\plus{}1)(1\plus{}c^6\plus{}1)(1\plus{}1\plus{}a^6) \ge (b^3\plus{}c^2\plus{}a)^6$

$ (c^6\plus{}1\plus{}1)(c^6\plus{}1\plus{}1)(c^6\plus{}1\plus{}1)(1\plus{}a^6\plus{}1)(1\plus{}a^6\plus{}1)(1\plus{}1\plus{}b^6) \ge (c^3\plus{}a^2\plus{}b)^6$

Multiplynig the three inequalities we obtain the desired result


ratio
geometry
invariant
analytic geometry
trigonometry
geometry theorems

Could someone post good proofs for these theorems?

Ok... this might help more. Hem Hem.  :lol:

[url=http://planetmath.org/encyclopedia/ProofOfCevasTheorem.html]Proof of Ceva's[/url]

the simplest proof for Ceva's theorem is to
express the product in terms of the
ratios of the areas of the 6 pieces
into which the lines divide the triangle.
This is in a sense the "correct" proof, as the
statement of the theorem is affine (i.e.
invariant under linear transformations,
and so dependent only on ratios of areas).

You can prove menelaus theorem in the same
way.

barycentric coordinates have a lot to do with
this, as well.

i have a very nice solution by using inversive geometry.but it may take time to write it.i will post it as soon as

possible  ;)

there is another version of ceva using trigonometry. Every cevian dives each angle of the $\triangle{ABC}$ in two Angles.
$A$ into $\angle {a_1}$ and $\angle {a_2}$
$B$ into $\angle {b_1}$ and $\angle {b_2}$
$C$ into $\angle {c_1}$ and $\angle {c_2}$

Then we have that the 3 cevians are concurrent if anf only if:
$\frac{sin a_1}{sin a_2} \frac{sin b_1}{sin b_2} \frac{sin c_1}{sin c_2}=1$
 you can prove using the sine law.

Hello
Paul A. Clement, found in The American
Mathematical Monthly, v. 65, num.8, October, 1958 the following Theorem.

 Given D on BC, E on CA, and F on AB  (possibly extended sides) of
triangle ABC the perpendiculars to the sides at D, E, and F are concurrent
iff

       (AF)^2 + (BD)^2 + (CE)^2 = (FB)^2 + (DC)^2 + (EA)^2.

And the proof???

[quote] Given D on BC, E on CA, and F on AB  (possibly extended sides) of
triangle ABC the perpendiculars to the sides at D, E, and F are concurrent
iff $(AF)^2 + (BD)^2 + (CE)^2 = (FB)^2 + (DC)^2 + (EA)^2$.[/quote]
This was known long before 1958.   It follows from the Pythagoras theorem applied to the 6 right triangles with vertex at the point of concurrency.


There are exactly $4n+2$ paths in space, and there are $2n^{2}+6n+1$ intersections of exactly two of them. What is the maximum number of intersections of three paths?

I failed the SMT :( .

Well it was super hard for a stupid guy like me, but you can probably solve it in under a minute :( .

'Paths' seems sort of vague but maybe...

${{{4n+2}\choose 2}= 8n^{2}+6n+1}$ Ways that two paths intersect. Since exactly two intersect at $2n^{2}+6n+1$ points, there's still $6n^{2}$ pairs left. However, each intersection of three, say A,B, and C, includes 3C2 = 3 pairs (AB, BC, AC). So that would be $2n^{2}$ intersections to cover all the pairs.


let us choose arbitrarily n vertices of a regular 2n-gon
and colour them red. the remaining vertices colour blue.
next arrange all red-red distances into a non-decreasing sequence and do the same with blue-blue distances.
prove that the sequences are equal.

Let $ A$ be the set of red vertices, $ B$ be the set of blue vertices.  Label the vertices $ v_1 , v_2 ... v{}_2{}_n$ going clockwise around the polygon.

Let $ E_k = \{ (v_i, v_i_ + _k) | v_i, v_i_ + _k \in A \}$
Let $ F_k = \{ (v_i, v_i_ + _k) | v_i, v_i_ + _k \in B \}$

So $ E_k$ is the set of red-red edges where the vertices are k steps around the polygon from each other and $ F_k$ is the set of blue-blue edges where the vertices are k steps around the polygon from one another.
So the problem is equivalent to showing that:

$ | E_k | = | F_k |$    for any k

Define $ S_k = \{ (v_i, v_i_ + _k) | v_i \in A , v_i_ + _k \in B \}$

Claim that $ | E_k | = n - | S_k |$

This is because for each red vertex, the vertex k steps clockwise round the polygon must be either red or blue, so $ | E_k | + | S_k | = |R| = n$

But then similarly $ | F_k | = n - | S_k |$
(this time we look at the vertex k steps [b]anticlockwise[/b] round the polygon from each blue vertex)

So we are done.  I made a bit of a meal out of this, it's pretty simple really.


in the city of medlers there is $100$ medlers.one day we told each medlers one different news.(none of news are same).
$A$ can tell anything that he know to $B$ with calling him.in each call only one of the medlers can tell the other everything that he knows.
with how many phone calls at least all of them will know all of the news?

i thinks this problem is nice and i think the answer is $198$ or $197$.if anybody has any idea plz send. :alien:


A pulley that has a diameter of 8 inches is belted to a pulley that has a diameter of 12 inches. The 8-inch-diameter pulley is running at1,548 revolutions per minute. If the speeds of the pulleys vary inverselyto their diameters, how many revolutions per minute does the larger pulley make?

since they are inversely proportional,
8*1548 = k = 12 * r
r = 1032


function
algebra
floor function
complex analysis
complex analysis unsolved

Does exist a function $ f$ which is analytic on $ \mathbb{D}$- the unit disc such that : $ f(\frac{1}{n})\equal{}\frac{1}{[\sqrt{n}]}$ ? where [x] is the floor function of $ x$

Just think of what $ f(0)$ and $ f'(0)$ should be in this case.


I'm trying to solve this nice equality :

C(n, 0)^2 + C(n, 1)^2 + C(n, 2)^2 + ... + C(n, n)^2 = C(2n, n)

where :

                        n!
C(n, k) = ___________
                    k! (n - k)!

Can anybody please give me a hint...?

This is actually a very famous formula, and it probably should be in Intermediate, not Advanced.  Anyhow, I don't want to give you too big a hint, but you're going to need to use a counting argument somewhere  --  you know, show that the two sides each count (in different ways) the number of ways to do something, and since the number of ways of doing that thing is fixed, you will show that the two sides are equal.

[hide]Hints :

don't forget that C(n,k) = C(n,n-k)

a) combinatorics way :

count (in two different ways) the number of ways to choose n persons from a group of n men and n women.

b) algebraic way :

look from coefficient of x^n in (1+x)^2n = (1+x)^n*(1+x)^n[/hide]

It's easier to prove (try a conbinatorial way) a more general formula
C(m+n,k)= C(m,0)*C(n,k) + C(m,1)*C(n,k-1)+...+ C(m,k-1)*C(n,1)+C(m,k)*C(n,0).
It's Valdermode formula if i'm not mistaken.
Then for m=n=k and C(n,k)=C(n,n-k). You get result.

The name of the formula is Vandermonde's Equation.  It's not particularly easier or harder to prove the generalization.

[hide]

Consider 2n objects, of which we want to choose n.

Then we can choose them in C(2n,n) ways:

We also can choose them in:

C(n,0)*C(n,n)+C(n,1)*C(n,n-1)+.......+C(n,n)*C(n,0) ways

But since C(n,k) = C(n,n-k), we have

C(n,0)*C(n,0)+C(n,1)*C(n,1)+.......+C(n,n)*C(n,n)

= C(n,0)^2+C(n,1)^2+.....+C(n,n)^2 ways

And we are done.

[/hide]

Thanks guys, but I think that i mislead something here.

Well the idea is that by observing  :

(1 + x)^a * (1 + x)^b = (1 + x)^(a+b)

to prove the equality......

but thanks anyway, now i got a lot of information that i didn't know before.

On the other hand. Why are you telling me that that problem belong to Interm.,  not to Advance, if that kind of problems belong to Discret Mathematic?

[quote="Andy"]Well the idea is that by observing  :

(1 + x)^a * (1 + x)^b = (1 + x)^(a+b)

to prove the equality......[/quote]

Try expanding out each of the three terms and see.

Just because something has a fancy name like "Discrete Mathematics" doesn't make it difficult -- this question is a fairly standard combinatorics result, and the Intermediate forum has lots of combinatorics on it.  I mean, it could go either place -- I just learned it a long time ago, so I think it's Intermediate :)

I trust Joel's judgment, so off it goes to Intermediate.

Isn't this one of the first things stated in AoPS Vol. 2 Combinatorics Chapter


calculus
analytic geometry
graphing lines
slope
ratio

Could someone help me with this problem:

Find an equation for the family of lines tangent to the circle with center at the origin and radius 3.

I cheated and used calculus, so I'll just post my answer. Hopefully someone can find a nice geometrical or algebraic solution.

[hide="general equation"]On the $xy\text{-plane}$ the general equation I think would be \[\sqrt{r^2-x_0^2}\left(y\pm\sqrt{r^2-x_0^2}\right)=\pm{x}_0(x-x_0).\]

$r$ is a positive real number, $3$ in this case. Also, $0\le{x}_0\le{r}$.[/hide]

I have a solution that is in the form [tex]y = mx + b[/tex] which is the slope-intercept form of the equation of a straight line.

[hide]

First, draw the circle given in the problem and then draw a line tangent to the circle to represent the general case of a line tangent to a circle of radius 3. Draw the tangent so it will intercept the y-axis at [tex]b[/tex].

Next draw a line from the origin or center of the circle to the point where the tangent touches the circle. This line will be normal or perpendicular to the tangent. We will call this point [tex](x,y)[/tex].

Now draw a line from [tex](x,y)[/tex] perpendicularly down to the x-axis.

This will form two similar right triangles.

the larger triangle will have its hypotenuse along the y-axis with a length of [tex]b[/tex] and the line perpendicular to the tangent from the origin will form one of the legs of that right triangle. The length of that leg is [tex]3[/tex].

The smaller triangle's hypotenuse is the same line that forms the given leg on the larger triangle, which is [tex]3[/tex]. It's legs are lengths [tex]x[/tex] and [tex]y[/tex].

We can now find the other leg of the larger triangle in terms of [tex]b[/tex].

Using Pythagorean's Theorem, the other leg is [tex]\sqrt {b^2 - 9}[/tex]

We can now set up ratios between the larger triangle and the smaller triangle.

[tex] \frac {b}{3} = \frac {3}{y}[/tex] so [tex]y = \frac {9}{b}[/tex]; also [tex]\frac {b}{\sqrt {b^2 - 9}} = \frac {3}{x}[/tex] so [tex] x = \frac {3 \sqrt {b^2 - 9}}{b}[/tex]

The next step is to find [tex]m[/tex] or the slope in the formula in terms of [tex]b[/tex]

Substituting into the formula:

[tex]\frac {9}{b} = m(\frac {3 \sqrt {b^2 - 9}}{b}) + b[/tex]

[tex]m = \frac {9 - b^2}{3 \sqrt {b^2 - 9}}[/tex]

The general equation would then be [tex]y = \frac {9 - b^2}{3 \sqrt {b^2 - 9}}x + b[/tex]

This is the solution you would have if you were to make it in terms of the y-intercept or [tex]b[/tex].

There may be other ways to represent a general equation.

This is for values [tex] 3 \leq \ b \ < \ \infty[/tex] and [tex] -3 \geq \ b \ > \ - \ \infty[/tex]

[/hide]

[hide]If you could find the x- and y-intercepts, you could say $b|x|+a|y|=ab$, where $a,b$ are the x- and y-intercepts, respectively. But I'm too lazy. :) [/hide]


Please do not discuss the AIME solutions or any "cheating" on this post.

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

Mr. Dunbar, are you saying that we cannot discuss it at all, including "I think I did well" and "That problem was hard"?

I am taking him to mean "Do not discuss it at all."


1) Prove $ \displaystyle{\frac{a^3\plus{}a}{a^4\plus{}3a^2\plus{}1}, a\in \mathbb{N}}$ is fraction in its lowerst terms

2)Find $ a,b$ coprime such that $ \displaystyle{\frac{a\plus{}b}{a^2\minus{}ab\plus{}b^2}\equal{}\frac{7}{13}}$

[u][b]Solution for number 1[/b][/u]:

It suffices to show that $ gcd(a^3\plus{}a,a^4\plus{}3a^2\plus{}1)\equal{}1$.

In the contraty,let's suppose that there is some $ d>1$ such as:

    $ d|a^3\plus{}a$ and $ d|a^4\plus{}3a^2\plus{}1$.

Then $ d$ also divided the difference:   $ d|a^4\plus{}3a^2\plus{}1\minus{}a(a^3\plus{}a)\equal{}2a^2\plus{}1$.

Moreover $ d$ also divides:  $ d|2(a^3\plus{}a)\minus{}a(2a^2\plus{}1)\equal{}a$.

Finally $ d|(2a^2\plus{}1)\minus{}2a^2\equal{}1$,therefore d=1 and our proof is complete.

Let me try the Number 2 question
I assume that they are positive(is it??).

Expression equivalent to  $ 13(a \plus{} b) \plus{} 21ab \equal{} 7(a \plus{} b)^2$. In  $ (\bmod (a \plus{} b))$  we get that  $ 21ab \equiv 0(\bmod (a \plus{} b))$  or  $ 21ab \equal{} k(a \plus{} b)$  or  $ {{21ab} \over {(a \plus{} b)}} \equal{} k$  for  positive  $ k$. Since  $ ab$  is coprime with  $ (a \plus{} b)$  then we got only  $ 21$  divisible by  $ (a\plus{}b)$, which means  $ a\plus{}b\equal{}3$  which has not solutions and  $ a\plus{}b\equal{}7$  for which we get 2 solutions set  $ (a,b) \equal{} \left\{ {(3,4),(4,3)} \right\}$.

Sorry if I mistaken.
I hope this solution is right... :blush:


geometry parallelogram rhombus trigonometry rectangle

The area of a parallelogram is 28 and the length of the largest diagonal is 14. Find the length of the smallest diagonal.

I think this problem has solution only if we consider that parallelogram is a rhombus. I'm wrong?

[quote="4everwise"]Um, there are numerous possible side lengths for the parallelogram. The smallest occurs when the small diagonal is length 4 and we have a rhombus.

Is that what your question was?[/quote]

I think i used the wrong word.

Trying again: If $D$ and $d$ are the diagonals of that parallelogram and $D=14$ find $d$ if the area of the parallelogram is 28.

So, i guess the author of this question supposed that parallelogram is a rhombus, but that is not necessarily one, right?

[hide]The diagonals of the parallelogram bisect each other so it can be split into 4 triangles with side lengths 7, d/2, and some other length. It $\theta$ denotes one of the angles of the intersection of the diagonals, then the other three angles have measures $180^\circ -\theta$, $\theta$, and $180^\circ -\theta$. The total area will be $7d\sin \theta=28\Rightarrow d=4\csc \theta$.

Maybe the problem wanted the minimum value of d (which will occur at $\theta=90^\circ$, which would make the parallelogram a rhombus), which is 4.[/hide]

Thank you scorpius. Thats what i thought. And no, the problem don't want the minimum value of d.

Also, can't you write the area of a parellagram as the product of the diagonals over 2
therefore
$\frac{14*d}{2}=28$
so
$d=\boxed{4}$

That formula is only for a rhombus, I think.

[quote="AMPIStiffler"]Well, wouldn't the formula always work for any parallelagram?
If anyone could clarify, that would be great. :)[/quote]

No that's not true, it's only for a rhombus (that's how the formula is derived). And if you want to see proof, think about a rectangle with sides $3$ and $4$ ;)

Generally, for any quadrilateral, we have $(\text{Area})=\frac{1}{2}d_1d_2\sin{\theta}$, where $d_1$ and $d_2$ are the diagonals and $\theta$ is the angle formed by the two diagonals.

In a rhombus, $\theta=90^\circ$, so $\sin{\theta}$ becomes $1$ and thus the formula becomes $\frac{1}{2}d_1d_2$.

Edit: scorpius119 has already posted the same idea :P

[quote="frt"]Generally, for any quadrilateral, we have $(\text{Area})=\frac{1}{2}d_1d_2\sin{\theta}$, where $d_1$ and $d_2$ are the diagonals and $\theta$ is the angle formed by the two diagonals.

In a rhombus, $\theta=90^\circ$, so $\sin{\theta}$ becomes $1$ and thus the formula becomes $\frac{1}{2}d_1d_2$.[/quote]I was just thinking about using that in a problem! Too bad it didn't turn out as well as I wanted, hehe.  :blush:


We are given a circle $ \Omega  $, a chord $ AB $, the midpoint $ M  $ of $ AB $ and a chord  $ CMD  $ through $ M $. If the tangents to the circle at $ C $and $ D $ interest line $ AB $ at $ E $ and $ F $ , prove that  $AE = BF$.

[i]This is a not a new problem .It is very old and very known, but it is a good introductory problem for Olympiads and especially for pupils 13-15 years old. I post it for our new members and visitors. Enjoy it. [/i]

Babis

Pretty cute.  Note that F, M, O, D are concyclic as are E, M, O, C.  Then $\angle MEO = \angle MCO = \angle MDO = \angle MFO$ so OEF is isosceles.  Then the result follows.

Let me tell you : Congratulations!

I think that this is the best (probably the unique) solution.

    Babis

I didnt understand,please explain,a figure would help.

Please tell me : Do you know the theory of inscriptible quadrilaterals ?If no , do not try any more. If yes , then O is the center of the circle , so , as in the hint(post 2)  , two quadrilaterals are inscriptible. Play with the angles ...

  If you have problem , write again. I' ll do something for you.

Babis

[quote="stergiu"]Please tell me : Do you know the theory of inscriptible quadrilaterals?[/quote]
No. :?
Can you please explain?

Stergiu is talking about inscribed quadrilaterals, more commonly known as cyclic quadrilaterals.  A cyclic quadirlateral is a quadrilateral for which there exists a circle that goes through all four vertices. Cyclic quads have some special properties that other quads do not have.  For instance, ABCD is cyclic iff angle CBD equals angle ACB.  Cyclic quads can often be very useful is solving vertain types of problems, as probability1.01 demonstrates.  Try to draw the diagram yourself and follow the proof.

There is somewhere a topic with the same problem , but in a little different formulation. I think it was << North China - 2005 >>. You can see  there 3 nice proofs, not all new.

 Babis

I suppose $A\in (ME),\ B\in (MF)$ and I note $L\in CE\cap DF$. From the Menelaus's theorem applied

to the transversal $\overline {DMC}$ in the triangle $LEF$ we obtain the relation $\frac {DF}{DL}\cdot \frac {CL}{CE}\cdot \frac {ME}{MF}=1$. But $LC=LD$.

Thus, $\frac {ME}{MF}=\frac {CE}{DF}$. But $EC^2=EA\cdot EB,\ FD^2=FA\cdot FB$. Therefore, $\left(\frac {ME}{MF}\right)^2=\frac {EA\cdot EB}{FB\cdot FA}\ \ (1)$.

I note $EA=y,\ MA=MB=x,\ BF=z$. The relation (1) becomes

$(x+y)^2z(z+2x)=(x+z)^2y(y+2x)\Longrightarrow$

$[2(x^2+yz)+x(y+z)](y-z)=0\Longrightarrow y=z\Longrightarrow AE=BF.$

Levi thanks!
 You taught me  that in this forum I must never say << ... probably the unique solution ! ) >> ;)

Here's a diagram (made in KSEG)

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=274[/img]


Hi, i'm sorry if there is a previous thread similar to this one, but i couldn't find one, so i thought I'd ask. So: taking the SAT, I did very well on math (800) and a 630 on verbal, which placed me in the 80-something percentile. I'm thinking of taking the SAT again because there is such a huge gap, between my scores. However, English is my second language (i only live in the U.S. since 8th grade), so I was wondering if 630 was a good enough score and what you think.

ursina

630 is more than fine.  It also matters what grade you're in and how competetive the schools you're looking at are.  But an 800 and 630 (as an immigrant) will get you in virtually anywhere.

I didn't catch how old you are in your message, and of course that makes a difference, but it is clear that you are above eighth grade. I'm wondering from how you signed your message if you are a girl. If you intend to major in a field such as math or engineering that has traditionally had more male applicants than female applicants, being a girl should be an admissions advantage in itself.

An 800 math with a 630 verbal is indeed already quite a good score for a first-generation immigrant. I think if you indicate what kind of reading you are doing and what else you are doing to make an effort to improve your English, admissions officers will take that into account when evaluating your scores. Certainly you are writing better English than many participants (who possibly are younger than you) who post on this forum, so you are making a good start in your learning of English. I think if you keep up what you are doing, and especially if you work hard at something that increases your English ability in your school course work or your extracurricular activities (for example, debate or speech), you should have little to worry about when applying for college.

Good luck.

I might recommend that you take the TOEFL or equivalent exam.  That's the "out" for people for whom English is a second language.

Ok, to clarify, yes i am a girl and i'm a junior in high school. I'm in Honors English and was even offered to take the AP English language this year (I reclined, deciding to focus on my other three APs). I'm taking the SAT II writing in june. I am also on the model UN team, which involves a lot of speaking.
I'm thinking of studying business or economics.

You would have a particularly good-looking application for a business program with your previous international experience, and there are certainly many strong economics programs that would be glad to get a young woman with the scores you have mentioned. It sounds like you are making good preparations already.


To DPatrick:

In the "are you ready" test for Intermediate Counting and Probability, it mentions that you have a book recommendation for sigma and sets. I wanted to ask you what that is.

Also, I really liked having the textbook to prepare for class in Introduction to Counting and Probability. I'm wondering how I'll be able to prepare for class without a textbook for Intermediate.

I have really enjoyed Intro. Thanks!

The whole "having a book to supplement the class" is actually very new.  AoPS classes have gone on for a long time, and they are extremely enjoyable and worthwhile without a book.  I think it would be a very bad idea not to take a class for that reason.

The bit about "sigma and sets" is just a mild warning that the level of algebra in the Intermediate Course is a bit more sophisticated than that of the Introduction course.

In AoPS Volume 1, you can read Chapter 24 for sigma notation and Chapter 27 for the basics of set theory.  This will also be covered in our forthcoming Introduction to Algebra book, which will be available later this year.

The Intermediate Counting textbook should be available in late 2006 or early 2007, and the online class will likely be offered Spring 2007 using the new book.  I agree that the classes are enhanced with the book.  If you prefer that style of class, you might consider taking Intro Number Theory this summer instead (which should have its book available later this spring) and deferring Intermediate Counting until next spring.  Or you could take the Intermediate Counting class now and follow it up by reading the book in the spring.

I took intermediate c&p last fall, and it's great without a book. I don't think you should wait to take it just because the book isn't out yet.

Though I might suggest trying to familiarize yourself with the topics a little beforehand... it really helps you out alot.  I personally wish I had haha.

Same applies for pretty much any class.  If you know what the topic is beforehand.. its the best thing you can do to help yourself... learn some of the material before class starts and get a head start.

I personally didn't do that well in Intermediate Counting, but I did wind up learning alot more than I thought I had, just by looking at the transcripts any time I was sorta fuzzy in my understanding on a topic.


AMC USA(J)MO USAMO MATHCOUNTS AIME ARML USAMTS

[size=150]Hey everybody!  [/size]Okay, so here's my problem.  I enjoy doing math competitions, but I dont really have time to do all of them.  So I was wondering what your guys's favorite math competitions were and why...  because if I can only do one, it would be nice to do the best   :)   thanks!
~[size=117]becky[/size]

What type of competition do you mean? In-school, mathmeets, AMC sort of stuff, etc.

Personally, I (duh) like the Mandelbrot for in-school, I like Charleston for mathmeets (that doesn't mean much, though, because I only go to two), and I like the USAMO best of all national competitions, even though I only got like a 2 this year.

Well, my favorite competition was MATHCOUNTS, that is before I got to 8th grade and got killed at nationals. Now, I look forward to next year and I am looking to do AMCs, Mandelbrot, and ARML.  8-)

I'd have to say MathCounts, since I won it, but my second favorite is the AIME (Didn't make USAMO this year)

good job winning nationals. great comeback against andrew chien.

Thanks. Were you there? I still can't believe it...

nope. the top 4 CA people were too good... thats y. AND i missed the first sprint question... sad

My favorite's Mathcounts(excluding the countdown round - totally stink at it, but I'm working on it). Then, when you're from Oklahoma, your school doesn't really suggest any competitive math contests to you, but I've been trying to get the school to let me do AMC 10 next year instead of AMC 8. And I was a Nats. You made a great comeback! Congrats!

china boi............ you most likely will be able to. and if you AND I can we dont have anything to lose cause amc 8 is kinda pointless all you get is a certificate thing. and who knows maybe you AND I can take the aime or amc 12 or somat. well nothing to lose eh. lau. lool. nuff said

[quote="Chinaboy"]Neal (or Synatx), would you please make more sense by using right grammar. What can I do if they don't let me take the AMC 10 or 12? Complain (I don't think they'll listen to a kid) or just pass up the whole chance by taking AMC 8? Or is there also an alternative?  :?:  :?:  :?:  :?:[/quote]
agreed :)

I definitely suggest complaining, not to some random teacher but directly to the principal. It takes some guts but you don't have anything to lose!

AIME is the most important one for me right now.

same for me. fiery are you practicing from aime test in the test format or just do individual problems?

[quote="tetrahedr0n"]my favorite contest is imo, but obviously ive never taken it.  :D[/quote]
How do you know you'll like it if you've never taken it?

I have no favorites currently, but if I must take one at this point I'd choose the middle school Math Team since it's fun, entertaining, and gives out tons of food/drink/soda/candy/etc. and it's just so much fun, especially with the hair spray and all :)

[quote]fiery are you practicing from aime test in the test format or just do individual problems?[/quote]

test format of course. If i wanted to do individual problems, I can find tons of those from other sources, but there are only a limited number of past AIME's. Plus, I find it very important to have experience in actually taking the test under test conditions.

The Mandelbrot Competition (http://www.mandelbrot.org) is a really wonderful contest, and unlike many of these that people are naming (particularly IMO and AIME), you don't need to have already done something in order to do it.  It's taken 4 times a year, and you've unfortunately missed the first date already.  It really is a great contest, though.

[quote="Tare"][quote="tetrahedr0n"]my favorite contest is imo, but obviously ive never taken it.  :D[/quote]
How do you know you'll like it if you've never taken it?[/quote]

Probably just the novelty of being good enough to get to it.  That is, everyone at IMO is either 1) really good at math or 2) in a country with approximately six people that are less than 18 years of age.  :wink:   Since the latter case almost never happens, making it to IMO is a sign of math superiority, which is obviously a good thing.  If that made sense at all.
Why are we revitalizing all these old threads?

What a long time since someone posted here!

I personally like ARML and USAMTS, but the AMCs are OK too.  Those are the only contests that I know of.

USAMO, cuz its a big confidence booster and last competition of year b4 ARML

i like amc8 its really easy ^.^ mathcounts is okay too. and kramier or wtv, our class (from miller middle) watched you countdown on tv last year. mucho congRats


please send me [i][b]direct[/b][/i] prove of steiner-lehmus theorem(if two angle bisectors of a triangle are equal, then the triangle
 is isosceles.)

Let [AD] and [BE] the angle bisectors of triangle ABC, then: $ |AD|^2 \equal{} \frac {bc.(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)}{(b \plus{} c)^2}$ and $ |BE|^2 \equal{} \frac {ac.(a \plus{} b \plus{} c)(a \plus{} c \minus{} b)}{(a \plus{} c)^2}.$
If |AD|=|BE|, then:
$ \frac {bc.(a \plus{} b \plus{} c)(b \plus{} c \minus{} a)}{(b \plus{} c)^2} \equal{} \frac {ac.(a \plus{} b \plus{} c)(a \plus{} c \minus{} b)}{(a \plus{} c)^2}\Rightarrow \\
\Rightarrow b.(a \plus{} c)^2.(b \plus{} c \minus{} a) \equal{} a.(b \plus{} c)^2.(a \plus{} c \minus{} b)\Rightarrow \\
\Rightarrow b.(a \plus{} c)^2.(b \plus{} c) \minus{} ab.(a \plus{} c)^2 \minus{} a.(a \plus{} c)(b \plus{} c)^2 \plus{} ab(b \plus{} c)^2 \equal{} 0\Rightarrow \\
\Rightarrow (a \plus{} c)(b \plus{} c).[b(a \plus{} c) \minus{} a(b \plus{} c)]\plus{}ab.[(b\plus{}c)^2\minus{}(a\plus{}c)^2] \equal{} 0\Rightarrow\\
\Rightarrow c(a \plus{} c)(b \plus{} c)(b \minus{} a)\plus{}ab(b\minus{}a)(a\plus{}b\plus{}2c) \equal{} 0 \Rightarrow a \equal{} b.$

thank you!but I dont want Algebraic solution.I want Geometric solution! :(

Dear Mathlinkers
after Coexter's book "Geometry revisited" p. 14-16 no direct proof exist or is known...
What do you think?
Sincerely
Jean-Louis

http://www.mathematik.uni-bielefeld.de/~sillke/PUZZLES/steiner-lehmus

Lots of proofs. Find one for yourself  :D

See and [url=http://forumgeom.fau.edu/FG2005volume5/FG200525.pdf][color=red][b]here[/b][/color][/url] ....


Two number cubes, each with the digits 1-6 on the six faces, are rolled. What is the probability that the product of the numbers on the top faces will be greater than 12? Express your answer as a common fraction.

Since you have to count out 23 sets, wouldn't it be easier to just count the pairs that have a product $ >12?$

it might be harder because you have to check twice as many numbers for greater than 12.
where as if you have a number less than 12, you can find factor pairs easily.

First, find all the possible ways to get 2 numbers, so 6*6=36.
Now, find all the ways 2 numbers on a die can be multiplied to get a number >12.
3 can be multiplied by 2 numbers.
4 can be multiplied by 3 numbers.
5 can be multiplied by 4 numbers.
6 can be multiplied by 4 numbers.
2+3+4+4=13

So  the probability is 13/36.


calculus
integration
trigonometry
trig identities
calculus computations

Hi all, I've recently encountered an integral which I eventually solved.  However, I'm searching for any alternative novel solutions, especially one's that utilize a unique combination of power substitutions.

$I= \int [ \frac{15x-20}{x^2 } ]^{ \frac{1}{6}}$

it turns out that the power 1/6 turns out to be the major inconvenience.  I took out the 15 factor and substituted x=1.5sec^2@ ,this way one is left with soley sin2@ as the fractional power term, not to mention the 1/6 being reduced to 1/3.  Then subsequently u^3=sin2@ (to be general), this get's rid of the annoying 1/6 exponent to merely terms consisting of square root exponents through the usage of trig identities and algebraic simplifications....trig substitutions and so on.

I was quite fortunate to solve it since I'm not too familiar with power substitutions.

hello, do you mean $ \int\sqrt[6]{\frac{15x\minus{}20}{x^6}}\,dx$?
Sonnhard.


There are $n$ cards lying down on a table.  Persons $x$ and $y$ switch turns ($x$ goes first).  When it is either $x$ or $y$'s turn, they must flip either 1, 3, or 4 cards (Flipped cards cannot be flipped again; only laid down cards can).  The winner is one who flips the last card.  How many cards $n$ are recquired for $y$ to win the game no matter what $x$ does?

Masoud Zargar

For $y$ to win, it must be $n\equiv 2\pmod{4}.$

Proof:

Let's look one move by $x$ and the subsequent move by $y.$ If $x$ flips $1$ card, $y$ will flip $3$. If $x$ flips $3$ cards, $y$ will flip $1.$ If $x$ flips $4$ cards, $y$ will flip $4$ if possible (or otherwise win the game - see later). In any case, after every $y$'s move the number of unflipped cards remains congruent modulo $4$ to the initial number of cards. After some $y$'s move the number of unflipped cards will necessarily reach $2$, as $n\equiv 2\pmod{4}$, and then $x$ will be forced to flip $1$ card. In the next move $y$ wins.

Would you algeabraically please explain more?

Tell me what part isn't clear, since there's not much algebra there.

After every $y$'s move the number of unflipped cards is smaller by $4$ or $8$ ($1+3$ or $3+1$ or $4+4$), because this is his strategy. Therefore, since the initial number of cards is congruent to $2$ modulo $4$, eventually there will remain only $2$ unflipped cards, in which situation $y$ wins as described.

[b]ADDITIONAL NOTE[/b]: Actually, after some more analyzing, I see that $y$ can win ALSO if $n\equiv 0\pmod{8}$, in the following manner: If $x$ flips $1$ card in his first move, $y$ will also flip $1$ and face him with the number of cards of the form $4k+2$, which is the winning situation for $y$, as we've seen. If $x$ flips $3$ cards in his first move, $y$ will also flip $3$ and again reach $4k+2$ unflipped cards. If $x$ flips $4$, then $y$ should also flip $4$ and force $x$ either to flip $1$ or $3$ and thus to lose, or if $x$ keeps flipping $4$, $y$ will keep doing that too, and the last four cards will go to $y$ since $n\equiv 0\pmod{8}$.

If $n=4k+1$ or $n=4k+3$, then $x$ can assure his win by flipping $3$ cards in the first case or $1$ card in the second case and facing $y$ with $4k+2$ cards, which we know is a losing situation for him.

If $n=8k+4$, then the game is lost by whoever flips $1$ or $3$ cards first, as already seen in the case $n\equiv 0\pmod{8}$, and if both keep flipping $4$, the last four will go to $x$.

[u]The final conclusion[/u]: $y$ wins if $n\equiv 0,2,6\pmod{8}$ and $x$ wins if $n\equiv 1,3,4,5,7\pmod{8}$

Oh, sorry, it is my stupidity.  :blush: I got it, thanks.

Please see the additional note in my previous post.

Only one more thing, this is similar to a contest question that had multiple choice answers that included both 32 and 34, which one would be correct, since both 32 and 34 satisfy $n\equiv 0,2,6\pmod 8$?

Then the author of the problem must be held responsible :)

First move if $n=32=8\cdot 4$:
a) If $x$ takes $1$, $y$ takes $1$ and reaches $30=4\cdot 7+2$ and hence he wins.
b) If $x$ takes $3$, $y$ takes $3$ and reaches $26=4\cdot 6+2$ and hence he wins.
c) If $x$ takes $4$, $y$ takes $4$ and reaches $24=8\cdot 3$. If in the second move $x$ takes one or three, see cases a) and b). If $x$ keeps taking four, $y$ will be a copycat and finally take the last $4$.

First move if $n=34$:
This is $4\cdot 8+2$ and is already described in the first post.

Therefore BOTH answers are correct.

But actually, I've just discovered I was wrong. For example, if $n=6$, then $x$ wins by taking four.

Let me think some more. I'll patch it up.


algebra
polynomial
quadratics
Rational Root Theorem

Factor completely: $x^3-7x^2-85x+91.$

I did it with synthetic division and got lucky, but is there a better way?

I would do it with synthetic division too. I don't know if there's an easier way.

Okay, try this: I just discovered it.

Expanding $(x-a)(x-b)(x-c)$ and collecting like terms, we have $x^3-x^2(a+b+c)+x(ab+ac+bc)-abc$!

So we can set $abc=-91=7 \cdot 13 \cdot -1$ and $a+b+c=7$, so by a little experimentation, we have $(x-1)(x+7)(x-13)$

Awesome! Now I can factor monic trinomials with degree 3! \/\/o0+!

NIce way, that's just like quadratics! Except a little longer.

$(a, b, c) = (-7, 13, 1)$ because the last term is positive 91.

The order is insignificant because the values are interchangeable.  I just thought your previous post said 7, 13, and -1 as factors.

He stated that $-91=7\cdot 13\cdot -1$, but his factored equation uses $(13, -7, 1)$.

A third degree polynomial factors (over the rationals) if and only if it has a linear factor, and that linear factor gives you a root. This makes the Rational Root theorem your best tool, since it immediately reduces the problem to testing a finite collection of possible linear factors. If none of them work, the cubic has no rational roots and is guaranteed to be ugly. If one works, you divide it out and get a quadratic which is easy to handle.

in this case particularly, just check whether $-1$ or $1$ are roots of the polynomial (it's easy to add up al coëfficients)...
then you're left with a second degree polynomial, which is easy to solve :)

[quote="4everwise"]Chess's idea is really cool! I useed to just guess for the real root because there has to be at least one and use synthetic division, but that's nice!

 :10:

And of course, we can all factor out the leading co-efficient to allow us to find the roots of any trinomials of degree 3.  ;)[/quote]
Try to guess as little as possible.
For the polynomial $P(x)=a_1x^n+a_2x^{n-1}+...+a_nx+a_{n+1}$ a could strategy is to try the factors of $\frac{a_{n+1}}{a_1}$ as zeros

Could you explain this further? I haven't factored in a while, but this seems really familiar.

If you look at chess's post you can see that the last term of the polynomial is the product of the roots, so you try the factors of the x^0 term  as roots


-1(x-1)=((3/x)-1)(x-3)
solve for x.

im having some trouble with this after i multiply it out.

Thats a bit hard to decipher w/o $\LaTeX$, but I'll give it a shot:

[hide]$-1(x-1)=(\frac{3}{x}-1)(x-3)$
Use distributive property on left and FOIL on right
$-x+1=\frac{3x}{x}-x-\frac{9}{x}+3$
Now combine like terms and solve
$1=6-\frac{9}{x}$
$-5=-\frac{9}{x}$
$\boxed{x=\frac{9}{5}}$[/hide]


Nine lines, parallel to the base of a triangle, divide the other sides into 10 equal segments and the area into 10 distinct parts. Find the area of the original triangle, if the area of the largest of these parts is 76.

$S(\triangle ABC) - \left(\frac{9}{10}\right)^2S(\triangle ABC) = \frac{19}{100}\ S(\triangle ABC) = 76$

$S(\triangle ABC) = 400$


Factor

               $ b^8 \plus{}b^6 \plus{} b^4 \plus{}b^2\plus{} 1$ .

[hide="Here is the solution"] $ (b^4\plus{}b^2\plus{}1)^2\minus{}b^2(b^4\minus{}1)^2$

It is trivial to factor it from here...[/hide]

But i found it by guessing . i'm interested in how one could come up with something like that.

hello, i have $ (1\minus{}b\plus{}b^3\minus{}b^3\plus{}b^4)(1\plus{}b\plus{}b^2\plus{}b^3\plus{}b^4)$.
Sonnhard.

Let $ P(b) \equal{} b^8 \plus{} b^6 \plus{} b^4 \plus{} b^2 \plus{} 1$ and let $ \omega \equal{} e^{2\pi i/5}$ be the fifth root of unity; then you could note that $ P(\omega) \equal{} P(\omega^2) \equal{} P(\omega^3) \equal{} P(\omega^4) \equal{} 0$, so $ (b\minus{}\omega)(b\minus{}\omega^2)(b\minus{}\omega^3)(b\minus{}\omega^4) \equal{} b^4\plus{}b^3\plus{}b^2\plus{}b\plus{}1 | b^8 \plus{} b^6 \plus{} b^4 \plus{} b^2 \plus{} 1$. Then long divide.

See (33) and (34) [url=http://mathworld.wolfram.com/CyclotomicPolynomial.html]here[/url].

You could just treat it as a geometric series, sum it, and then divide out the denominator factors in the resulting fraction.  :)

Consider the equation $ b^8 \plus{} b^6 \plus{} b^4 \plus{} b^2 \plus{} 1 \equal{} 0$. As b = 0 is not a solution of this equation, divide both terms by $ b^4$. Then we get

$ b^4 \plus{} b^2 \plus{} 1 \plus{} \frac {1}{b^2} \plus{} \frac {1}{b^4} \equal{} 0$

$ \equal{} > \left(b^2 \plus{} \frac {1}{b^2}\right)^2 \minus{} 2 \plus{} \left(b^2 \plus{} \frac {1}{b^2}\right) \plus{} 1 \equal{} 0$

$ \equal{}> \left(b^2 \plus{} \frac {1}{b^2}\right)^2 \plus{} \left(b^2 \plus{} \frac {1}{b^2}\right) \minus{} 1 \equal{} 0$

and now make $ b^2 \plus{} \frac {1}{b^2} \equal{} x$ to get $ x^2 \plus{} x \minus{} 1 \equal{} 0$ whose solutions are $ x \equal{} \minus{} \phi$ and $ x \equal{} \phi \minus{} 1$. Now for each of these solutions $ x_i$ solve the equation $ b^4 \minus{} x_i b^2 \plus{} 1 \equal{} 0$ by making $ b^2 \equal{} u$.

treat as a geometric series with $ r \equal{} x^2, a_1 \equal{} 1$. Then the sum of the first 5 terms is equal to the desired expression $ \equal{} \frac {a(1 \minus{} r^5)}{1 \minus{} r} \implies \frac {(1 \minus{} x^{10})}{1 \minus{} x^2},$ dividing out gives the desired result.


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

find all matrix A of order n $ (n \geq 2)$ such that det(A+M) =detA +det M for all matrix M

You didn't specify what type of matrices, so I'll assume $ \mathcal{M}_n(\mathbb{C})$, but you can change $ \mathbb{C}$ with a field of characteristic zero easily below.

First, $ A \equal{} 0$ is solution. We then show this is the only solution.
Assume $ A\neq 0$ is solution.
If $ A\in \mathrm{GL}_n(\mathbb{C})$, then taking $ M \equal{} A$ you get $ 2^n \det A \equal{} 2 \det A$ which is impossible.
If $ A\in \mathcal{M}_n(\mathbb{C}) \setminus \mathrm{GL}_n(\mathbb{C})$. Take $ M \equal{} \lambda I_n$, $ A$ is nilpotent because $ 0$ is its only eigenvalue. To conclude look at a Jordan block with $ 1$ is position $ (i,i \plus{} 1)$ and $ 0$ everywhere else, take $ M \equal{} \mathrm{diag}(K, 1,1,\ldots, 1)$ with $ K \equal{} \begin{pmatrix} 1 & 0 \\
1 & 1 \end{pmatrix}$.

Any field of characteristic zero? Try any field.

Consider cases in which $ M$ has exactly $ k$ ($ 0<k<n$) nonzero entries, all equal to $ 1$, and all in distinct rows and columns.
From $ k\equal{}1$, we get that all $ (n\minus{}1)\times (n\minus{}1)$ minors of $ A$ are zero.
From $ k\equal{}2$ and the previous result, we get that all $ (n\minus{}2)\times (n\minus{}2)$ minors of $ A$ are zero.
...
From $ k\equal{}n\minus{}1$ and the previous results, we get that all $ 1\times 1$ minors of $ A$ are zero. These are just the entries of $ A$, and $ A$ is zero.

thank you , and here is my solution.
if $ A\equal{}[a_{ij}]_{n \times n}$ satisfies the condition, it is easy to refer det A =0.
consider $ M\equal{}\begin{pmatrix} 0 & 0 & 0 &... &0\\0 & m & 0 &... &0\\0 & 0 & m &... &0\\...\\0 & 0 & 0 &... &m\end{pmatrix}$
$ 0\equal{}detM\equal{}det(A\plus{}M)\equal{}det \begin{pmatrix}a_{11} & a_{12} & a_{13} &... &a_{1n}\\a_{21} & a_{22}\plus{}m & a_{23} &... &a_{2n}\\a_{31} & a_{32} & a_{33} \plus{}m&... &a_{3n}\\...\\a_{n1} & a_{n2} & a_{n3} &... &a_{nn}\plus{}m\end{pmatrix}$
this is a polymial of m with order n-1. because P(m)=o for all m so the coefficient of $ m^{n\minus{}1}$ is equal to 0 i.e $ a_{11}\equal{}0$. similarly, we can have $ a_{ij} \equal{}0$ for all i,j.
so A is 0 :lol:

... and what if it's a finite field? A polynomial can evaluate to zero everywhere without being the zero polynomial.

For $ M\equal{}A$ then $ \det(A)\equal{}0$
 If $ A \neq 0$ then the exist a column of $ A$ is orther $ 0$ call $ C_j \neq 0$ We have   supplement $ (n\minus{}1)$ column {$ C_1,C_2,...,C_{j\minus{}1},C_{j\plus{}1},...,C_n$} to {$ C_1,C_2,...,C_{j\minus{}1},C_j,C_{j\plus{}1},...,C_n$} is $ n$ column Independent .
 Put $ M\equal{}(C_1 C_2 ... \minus{}C_j ...C_n)$ then $ \det(A\plus{}M)\equal{}0 \neq \det(M)$ Therefore $ A\equal{}0$

[quote="zengzeng"]For $ M \equal{} A$ then $ \det(A) \equal{} 0$[/quote]
Not with evey field...


Triangle ABC  is acute-angled and O is a point in its plane.

U,V,W are the distances from O to the sides of the triangle.

Prove that 4{(U+V+W)^2 } +(AB+BC+CA)^2  >=4{(OA+OB+OC)^2}

[quote="Michael Niland"]Triangle $ABC$  is acute-angled and $O$ is a point in its plane.

$U,V,W$ are the distances from $O$ to the sides of the triangle.

Prove that $4(U+V+W)^2  +(AB+BC+CA)^2  \geq 4(OA+OB+OC)^2$[/quote]

 :)


Starting with any three digit number $ n$ (such as $ n\equal{}625$) we obtain a new number $ f(n)$ which is equal to the sum of the three digits of $ n$, their three products in pairs and the product of all three digits.

(i) Find the value of $ \frac{n}{f(n)}$ when $ n\equal{}625$. (The answer is an integer!)

(ii) Find all three digit numbers such that the ratio $ \frac{n}{f(n)} \equal{}1$.

[quote="AndrewTom"]Starting with any three digit number $ n$ (such as $ n \equal{} 625$) we obtain a new number $ f(n)$ which is equal to the sum of the three digits of $ n$, their three products in pairs and the product of all three digits.

(i) Find the value of $ \frac {n}{f(n)}$ when $ n \equal{} 625$. (The answer is an integer!)

(ii) Find all three digit numbers such that the ratio $ \frac {n}{f(n)} \equal{} 1$.[/quote]

So $ f(\overline{abc})\equal{}(a\plus{}1)(b\plus{}1)(c\plus{}1)\minus{}1$

(i) $ f(625)\equal{}7\cdot 3\cdot 6\minus{}1\equal{}125$ and $ \frac{625}{f(625)}\equal{}5$

(ii) We have to solve $ 100a\plus{}10b\plus{}c\equal{}(a\plus{}1)(b\plus{}1)(c\plus{}1)\minus{}1$
So $ 100a\plus{}10b\equal{}a(b\plus{}1)(c\plus{}1)\plus{}b(c\plus{}1)$

If $ (b\plus{}1)(c\plus{}1)<100$, we need $ 10b<b(c\plus{}1)$ and so $ c>9$, impossible.
So $ (b\plus{}1)(c\plus{}1)\ge 100$ and so $ b\equal{}c\equal{}9$

And it is easy to see that this is indeed a solution. Hence the answer $ \boxed{f(\overline{a99})\equal{}\overline{a99}}$ for any $ a\in\{1,2,3,4,5,6,7,8,9\}$