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trigonometry
vector
algebra proposed
algebra

Let $f(x) = \frac{1}{x^2 - x + 1}$. We are given  $a,b,c,d \in \mathbb{R}$ st:
$f(a) + f(b) + f(c) + f(d) = 8/3$
$af(a) + bf(b) + cf(c) + df(d) = 4/3$
Evaluate: $(2a-1)(2b-1)(2c-1)(2d-1)$

I feel this is really beautiful! ...and probably posted before... but I couldn't find it using the search $f$... for one, I am not sure what string to put in... enjoy!

Ok...

We see that $\sum \frac{1}{(a-1/2)^2+3/4}=8/3$ and $\sum \frac{a-1/2}{(a-1/2)^2+3/4}=0$.

Let $a'=\frac{(a-1/2)2}{\sqrt{3}}$, then $\sum \frac{1}{a'^2+1}=2$ and $\sum \frac{a'}{a'^2+1}=0$.
Note that, actually, we are to find $a'b'c'd'$.

Suppose $a'=\tan x$ and so on, then these equalities means that $\sum \cos^2 x=2$ and $\sum \sin x\cos x=0$. We can rewrite them as $\sum \cos 2x=0$ and $\sum \sin 2x=0$ :!: .

Square both equalities and sum them up. Then we obtain $4+\sum (\sin 2x\sin 2y+\cos x+\cos y)=0$. It follows that $2+\sum \cos(2x-2y)=0$. Group now some summands: \[\begin{aligned}&(\cos(2x-2y)+\cos(2z-2t))+(\cos(2x-2z)+\cos(2y-2t))=\\&=2\cos(x+z-y-t)\cos(x-y-z+t)+2\cos(x+y-z-t)\cos(x-y-z+t)=\\&=4\cos(x-y-z+t)\cos (x-t)\cos(y-z).\end{aligned}\] Using the fact that $\cos(2x-2t)=2\cos^2(x-t)-1$ and $\cos(2y-2z)=2\cos^2(y-z)-1$, finally we obtain \[\cos^2(x-t)+2\cos(x-y-z+t)\cos (x-t)\cos(y-z)+\cos^2(y-z)=0.\] We know that $|\cos(x-y-z+t)|\leq 1$.

[b]Case 1.[/b] $|\cos(x-y-z+t)|<1$. Then necesarily $\cos(x-t)=0$ and $\cos(y-z)=0$. It immediately follows $\tan x\tan t=-1$ and $\tan y\tan z=-1$, so $a'b'c'd'=1$.

[b]Case 2.[/b] $\cos(x-y-z+t)=1$. Then $\cos(x-t)=-\cos(y-z)$. So $x-y-z+t=2\pi k$, $(x-t)+(y-z)=\pm \pi+2\pi l$. It follows that $x-z=\pm \frac{\pi}{2}+\pi n$, $y-t=\pm \frac{\pi}{2}+\pi m$. So result follows.

[b]Case 3.[/b] Similar to 2.

The answer for the initial problem is $9$.

Hope it is correct.

Right ! :)  But there's a neater way of resolving the ending:

Given,
$\sin 2w + \sin 2x + \sin 2y + \sin 2z = 0$, $\cos 2w + \cos 2x + \cos 2y + \cos 2z = 0$

Let there be vectors of form:
$w' = (sin 2w)i + (cos 2w)j$
$x' = (sin 2x)i + (cos 2x)j$
$y' = (sin 2y)i + (cos 2y)j$
$z' = (sin 2z)i + (cos 2z)j$

$w'$, $x'$, $y'$, $z'$ are unit vectors which sum to 0. The only way for this is WLOG $w'=-x'$, $y'=-z'$, which means that $\tan w \tan x = -1$ and $\tan y \tan z = -1$, leading to the required result  :D


topology
LaTeX
advanced fields
advanced fields unsolved

Let $ X$ and $ Y$ be connected topological spaces. Let $ A$ and $ B$ be proper subsets of $ X$ and $ Y$ respectively.
Prove that $ X \times Y \minus{} (A \times B)$ is connected.

Let $ Z = (X \times Y) \setminus (A \times B)$ and suppose that $ Z = U \cup V$ with $ U,V$ open and $ U \cap V = \emptyset$.

Consider a "vertical" fiber $ F_x = \{x\} \times Y$ for $ x \in X \setminus A$. Then $ F_x \subseteq Z$, and it is obviously homeomorphic to $ Y$, hence connected. The partitioning $ F_x = (F_x \cap U) \cup (F_x \cap V)$ must therefore be trivial: either $ F_x \subseteq U$ or $ F_x \subseteq V$. In the same way, each "horizontal" fiber $ G_y = X \times \{y\}$ for $ y \in Y \setminus B$ is contained in either $ U$ or $ V$.

But now it's trivial to see that either all of those fibers are contained in $ U$, or they're all contained in $ V$, as follows: suppose that $ F_{x_0} \subseteq U$ for some specific $ x_0$ (if this never happens, choose $ V$ instead). Then for any $ y \in Y \setminus B$, the point $ (x_0,y) \in U$, so in particular the fiber $ G_y$ cannot be contained in $ V$, so it must be contained in $ U$. Thus all the horizontal fibers are in $ U$, and using the same argument all the vertical fibers are in $ U$ as well.

But $ Z$ is the union of those fibers, so $ Z \subseteq U$. This shows that $ Z$ is connected.

(PS: I never liked the way $ \text{\LaTeX}$ renders the "setminus" symbol $ A \setminus B$. It's almost (totally?) identical to an inverted quotient symbol $ \Gamma \backslash H$ which means something totally different, i.e. taking a quotient with respect to a group acting on the right. I was taught to draw the set minus symbol much smaller, like one stroke of the $ \times$ symbol. Is there a better symbol for "set minus"?)

[quote="randomgraph"]I never liked the way $ \text{\LaTeX}$ renders the "setminus" symbol $ A \setminus B$. It's almost (totally?) identical to an inverted quotient symbol $ \Gamma \backslash H$ which means something totally different, i.e. taking a quotient with respect to a group acting on the right. I was taught to draw the set minus symbol much smaller, like one stroke of the $ \times$ symbol. Is there a better symbol for "set minus"?[/quote]
About ten years ago, I had the same objection to the $ \text{\LaTeX}$ "setminus" symbol. I tackled it by redefining the symbol using one of the glyphs from the AMSb font. I found at the time that I could get the result I wanted by putting this declaration in the preamble to a file (i.e. before "\begin{document}"). (Of course it would be better to put it in your personal macros file, which you can call whenever you write a document.)

[code]\DeclareSymbolFont{AMSb}{U}{msb}{m}{n}
\DeclareSymbolFontAlphabet{\mathbb}{AMSb}
\begingroup \catcode`"=12
\DeclareMathSymbol{\setminus}{\mathbin}{AMSb}{"72}
\endgroup[/code]

I no longer have the TeX expertise that I used to, and I have no idea whether that code is suitable for modern versions of $ \text{\LaTeX}$.

Interesting - glad to hear I'm not alone. I just found out there's a "smallsetminus" symbol which looks like this: $ A \smallsetminus B$. Still not quite what I would expect but better nonetheless.


Let $ a,b,c$ be positive reals number such that $ a\plus{}b\plus{}c\equal{}1$
Prove that \[ \sum_{cyclic}\frac{1}{a\plus{}bc} \leq \frac{9}{4(ab\plus{}bc\plus{}ca)}\]

[quote="apollo"]Let $ a,b,c$ be positive reals number such that $ a \plus{} b \plus{} c \equal{} 1$
Prove that
\[ \sum_{cyclic}\frac {1}{a \plus{} bc} \leq \frac {9}{4(ab \plus{} bc \plus{} ca)}
\]
[/quote]

We have:
$ \sum_{cyclic}\frac {1}{a \plus{} bc} \equal{} \sum \frac{1}{a(a\plus{}b\plus{}c)\plus{}bc}\equal{}\sum \frac{1}{(a\plus{}b)(a\plus{}c)} \equal{}\frac{2(a\plus{}b\plus{}c)}{(a\plus{}b)(b\plus{}c)(c\plus{}a)}$

Hence, we need to prove

$ 9(a\plus{}b)(b\plus{}c)(c\plus{}a) \ge 8(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$

$ \Leftrightarrow \sum ab(a\plus{}b) \ge 6abc$

Whic is clearly true by AM-GM inequality.

how about the following inequality :
$ a,b,c > 0 , a\plus{}b\plus{}c\equal{}1$
$ \sum \frac{1}{a\plus{}2bc} \leq \frac{9}{4(ab\plus{}bc\plus{}ac)}$
if i didn't mis-calculating somewhere then the above ineq is also true  :)  , and we can not use NguyenDungTn's way  :)

[quote="HTA"]how about the following inequality :
$ a,b,c > 0 , a \plus{} b \plus{} c \equal{} 1$
$ \sum \frac {1}{a \plus{} 2bc} \leq \frac {9}{4(ab \plus{} bc \plus{} ac)}$
if i didn't mis-calculating somewhere then the above ineq is also true  :)  , and we can not use NguyenDungTn's way  :)[/quote]

It must be
$ \sum \frac {1}{a \plus{} 2bc} \leq \frac {9}{5(ab \plus{} bc \plus{} ac)}$

Because
$ \sum \frac {1}{a \plus{} 2bc} < \sum \frac{1}{a\plus{}bc}\le \frac {9}{4(ab \plus{} bc \plus{} ac)}$

[quote="NguyenDungTN"][quote="HTA"]It must be
$ \sum \frac {1}{a \plus{} 2bc} \leq \frac {9}{5(ab \plus{} bc \plus{} ac)}$

[/quote][/quote]
yes , it is :)
[hide="proof"]and here is my proof for $ k \equal{}2$
we use $ p,q,r$ and it is left to prove that $ \frac{3q^2\minus{}2rq}{8r^2\plus{}r\plus{}2q^2\minus{}8rq} \leq \frac{9}{5}$
\leftrightarrow $ 62rq \leq 72r^2\plus{}9r\plus{}3q^2$
we will prove the stronger inequality
$ \frac{19}{9}q^2\plus{}9r\geq 46rq$
let $ q\equal{}\frac{1}{3}(1\minus{}t^2) t \geq 0$ the inequality becomes :
$ \frac{19}{81r}(1\minus{}t^2)^2\plus{}9\geq \frac{46}{3}(1\minus{}t^2)$
we continue setting up a stronger one :
$ \frac{19(1\plus{}t)^2}{(1\plus{}2t)}\plus{}46t^2\geq 19$
which is equivalent to $ t^2(65\plus{}92t)\geq 0$, which is true :)[/hide]
the numbers become complicated when i try to prove for k in general ... :|


LaTeX
inequalities
inequalities proposed

Q: For any numbers a,b,c prove:
a / a' + b / b' + c / c' >= 1 where a' = sqrt ( a^2 + 8*bc), etc.

I guess this is really unconventional to post imo problems in this thread, but yeah i was jsut interested in other solutions to this wonderful problem, since there are undoubtedly several.

My solution:
it easily reduces to
sqrt ( a /a+8) + sqrt (b / b+8) + sqrt (c/c+8) >=1, when abc=1. [set abc=1, and change a-->a^3, b--> b^3, c--> c^3).
now let f(a,b,c) = sqrt ( a /a+8) + sqrt (b / b+8) + sqrt (c/c+8), defined for positive a,b,c.
we prove f(a,b,c) reaches a minimum as a,b,c varies the reals with abc=1 when a=b=c=1; so it is sufficient to prove that f(a,b,c) >= f(sqrt ab, sqrt bc, sqrt ac) (after we get this, suppose we attain a minimum for a system a,b,c with a unequal to b. then we can attain a smaller minimum by replacing a,b,c with sqrt bc, sqrt ac, sqrt ac, contradiction).
so we wish to prove:
sqrt (b/b+8) + sqrt (c/c+8) >= 2 * sqrt (sqrt (bc) / sqrt(bc) +8).  (once we have this, add up the symmetrics to get f(a,b,c) >= f (sqrt ab, sqrt bc, sqrt ac). )
now let x=b/b+8, and y=c/c+8; then this isnt too difficult.
sori it isnt latex-ed, i havnt learnt it yet.

actually, i'm having a bit of trouble proving that 2-variable inequality, my proof was actuali wrong.
any ideas?
the inequality (again) is:
sqrt (a/a+8) + sqrt (b/b+8) >= sqrt ( sqrt (ab) / sqrt(ab) + 8 ) .
hm it seems to be quite tight and not very easy, jensens doesnt work since if u put
f(x) = f(x/x+8), while f is increasing you get by jenson that f(x)+f(y)<= f(x+y/2) which doesn't help. maybe expanding the whole thing, involving squaring several times, may work that that seems a bit tedioius.

I am uploading two short articles I wrote about this inequality a long time ago. The first one contains 5 different solutions and the second one contains a proof a generalization. The first one is in Flemish/Dutch but I am SURE you will be able to figure out what I am actually doing. Maths is universal :D and believe me, I have been looking at maths documents in Vietnamese, Chinese, Polish, Romanian and so on (languages I don't understand at all). (The second one is in English)


topology
real analysis
real analysis unsolved

Hi, could you help me to figure out this question please:

Show that sequences in $ R^{\omega}$ converge in the uniform topology if and only if they converge uniformly.
(In other words a sequence $ (f_k)(n): N\rightarrow R$ converges to
$ f(n): N\rightarrow R$, if and only if for any
$ e>0$ there exist a $ k$ such that if $ k>K$ then for every
$ n$, $ |f_k(n)\minus{}f(n)|<e$.)

Thank you.

Uniform convergence says that for $ k > K(\epsilon)$, $ \|f_k \minus{} f\|_\infty < \epsilon$. That looks like the definition of convergence in the uniform metric $ \|\cdot\|_\infty$.

Thank you, but could you be a little more specific? What is the precise definition of the uniform metric $ \parallel{}.\parallel{}_\infty$ that you mentioned?

I'm sorry for asking this much, but I'm lost and really need help. Thank you.

[color=green]Moderator edit- fixed reference to my mistake[/color]

The uniform norm $ \|x\|_\infty$ is $ \sup_n |x_n|$, and the metric is $ d(x,y) \equal{} \|x \minus{} y\|_\infty$. You kind of need to know this for the notion of the uniform topology to make sense anyway.

[Edit-fixed norm name. Oops.]


I got stuck on a couple of problems and I can't get them.

$ 13\times8^{0.3x}\equal{}663$
$ 8^{0.3x}\equal{}51$
$ \log_{8}{0.3x}\equal{}51$?????

Solve the general equation $ b^{x}\equal{}c$ for $ x$ in terms of $ b$ and $ c$

$ 29^{x}\equal{}0.87$
$ \log_{29}x\equal{}0.87$

[quote="Quevvy"]I got stuck on a couple of problems and I can't get them.

$ 13\times8^{0.3x}= 663$
$ 8^{0.3x}= 51$
$ \log_{8}{0.3x}= 51$?????
[/quote]

$ 8^{0.3x}= 51$ does NOT imply $ \log_{8}{0.3x}= 51$, and you made the same mistake on the third problem. Try $ \log_{8}{51}=0.3x\Rightarrow{}x=\frac{10\log{8}{51}}{3}$, which is the best one can simplify it...

[quote="Quevvy"]Solve the general equation b^{x}=c for x in terms of b and c [/quote]

[hide]By definition of a log, we have $ b^{x}=c\Rightarrow{\log_{b}{c}=x}$[/hide]

[quote="Anaxerzia"][quote="Quevvy"]I got stuck on a couple of problems and I can't get them.

$ 13\times8^{0.3x}= 663$
$ 8^{0.3x}= 51$
$ \log_{8}{0.3x}= 51$?????
[/quote]

$ 8^{0.3x}= 51$ does NOT imply $ \log_{8}{0.3x}= 51$, and you made the same mistake on the third problem. Try $ \log_{8}{51}= 0.3x\Rightarrow{}x =\frac{10\log{8}{51}}{3}$, which is the best one can simplify it...

[quote="Quevvy"]Solve the general equation b^{x}=c for x in terms of b and c [/quote]

[hide]By definition of a log, we have $ b^{x}= c\Rightarrow{\log_{b}{c}= x}$[/hide][/quote]

so $ x =\frac{10\log_{8}{51}}{3}$
$ \frac{10\frac{\log51}{\log8}}{3}\approx 6.303$?

[quote="Quevvy"]so $ x \equal{}\frac{10\log_{8}{51}}{3}$
$ \frac{10\frac{\log51}{\log8}}{3}\approx 6.303$?[/quote]

To input on a calculator, yes one would do that.


calculus
integration
derivative
calculus computations

It can be rewritten as
$ \mathbf{I} \equal{} \int x^{\minus{}n} \left( 1\plus{}x^n \right)^{\minus{}\frac{1}{n}} \, \mathbf{d}x$
and it's fairly easy if you know how to integrate differential binomial. Just substitute
$ t \equal{} \sqrt[n]{\frac{1\plus{}x^n}{x^n}}$
$ \mathbf{I} \equal{} \minus{} \int t^{n\minus{}2} \, \mathbf{d}t \equal{} \frac{1}{1\minus{}n} t^{n \minus{} 1} \plus{} C \equal{} \boxed{\frac{x^{1\minus{}n} \left( 1 \plus{} x^n \right)^{1 \minus{} \frac{1}{n}}}{1 \minus{} n} \plus{} C}$

For $ n\equal{}1$ should be different formula  :blush:

For $ n \equal{} 1$ just separate the fraction in a subtraction of two primitive fractions:

$ \frac {1}{x} \minus{} \frac {1}{x \plus{} 1}$ and you'll get two elementary (Log) integrals (Soln: $ ln (\frac{x}{x \plus{} 1}) \plus{} C$).

What's the sign of $ \frac{x}{x\plus{}1}$?

Forgot the absolute value sign, sorry...  :blush:

[quote="anandghegde"]The value of $ \int \frac {1}{x^n.(1 + x^n)^{\frac {1}{n}}} dx$ where n is a natural number is........

Ans- $ \frac {1}{1 - n}(1 + \frac {1}{x^n})^{1 - \frac {1}{n}}$+c[/quote]

Another way of looking at things:
The given integral be called $ J$.
then ${ J= \int \frac {1}{x^{n+1}}.(1 + \frac{1}{x}^n)^{\frac {1}{n}}} dx$

Now $ \frac{1}{x^{n}} + 1$ be $ t$. you have its derivative off...
ofc differing by sm constant.
That completes the problem.


inequalities
number theory proposed
number theory

Given $ n$ positive interger. Prove that $ \frac {S(8n)}{S(n)}\ge \frac {1}{8}$


AMC USA(J)MO USAMO search function geometry college

I'd like to do a few timed practice USAMOs before the actual test, unfortunately I've spoiled most of them by doing one or two of the questions, so I'll need to use some other test as a substitute.

What national math olympiads are most similar to the USAMO?
What else could I use as a realistic practice?

You might try the Team Selection Test that they give at MOSP.  You can find those off the AMC web site.  http://www.unl.edu/amc :)

TST's are probably (at least in my opinion) quite a bit more difficult than USAMOs. :-) Er...APMO questions are quite nice imo. And other national olympiads are also good starters for training.

APMO stands for Asian Pacific Math Olympiad, and I belive it is a regional olympiad among countries on the Pacific and beyond. Each country selects up to about 30 students to write the exam sometime in March, the top 10 of which comprise that country's team score. Also, it seems that In the United States, invitations go primarily to past MOPers.

All of the tests from 1989 to 2005 can be found [url=http://www.cms.math.ca/Competitions/APMO/]here[/url].

Regional math olympiad. It stands for Asian Pacific Mathematical Olympiad. imo, they are all too hard.

Oh...if you can get your hands on some old MOP test, you can get a pwning out of those. *Just thought of that* :D

The [url=http://www.mathlinks.ro/Forum/index.php?f=217]Advanced Olympiad Section[/url] contains thousands of USAMO-like problems! Right here on THIS site :)

Furthermore, the [url=http://www.mathlinks.ro/Forum/index.php?f=125]National Olympiads[/url] directory (in the Olympiad Forums) contains the most recent olympiad questions given thanks to our users from around the world.

Also use search function and search for NMO :)

Kalva has many olympiad problem from various countries posted as well.

[quote="xxreddevilzxx"]you could try the Candian MO, British MO, or Irish MO, althought all of those are quite easier than USAMO.[/quote]

I know that the irish mo is way easier than the USAMO.

500 Mathematical Challenges is a good book.  "Which Way Did the Bicycle Go?" is also a fun and good book, although its problems tend to be more of a warm-up type.  I don't know where to get the latter though--AMC mailed it to me.  Also, many past USAMOs (especially older ones) are much easier than the current ones.

Check out BAMO (Bay Area Mathematical Olympiad) at: http://mathcircle.berkeley.edu/

BAMO is easier than USAMO. However, it's good exercise for preparing USAMO. You can also find a lot of interesting problems from past monthly contests of Berkeley Math Circle on the same website.

Speaking of BAMO, it's surprising that five of top eight 2005 BAMO winners didn't qualify for USAMO.
http://mathcircle.berkeley.edu/BMC6/pages/BAMO/2005/BAMO_2005_awardees.html

Also surprisingly, among the six California 8th grade USAMO qualifiers last year, only two of them (they are all 9th grader now) made it again this year.

[quote="Fountainhead"]I have a question- I think I qualified for the USAMO for the first time this year, and I have very little experience in olympiad-style problem solving. As a result, I'd like to practice with a few national olympiads that are easier than the USAMO. I know that the Irish, British, and Canadian olympiads have been suggested. Any other suggestions?[/quote]Have you checked out the links on this very forum that I have provided above?!  :?

Yeah, thanks Valentin. Those were very helpful.


Find the smallest positive whole number that does not contain the digit 9, but it is divisible by 999.

[hide]just keep adding 999 to 999 until you get such a number. 111888?[/hide]

Theres a better solution that dosent invove repeated, tedious calculations.

I used brute force with a calculator
[hide]111888[/hide]

Yeah, 111888 is the answer. Could you show what the better method is?

I think there is a better way because there are problems like this that you can't really use calculators to find such a number.

999+999= 1998
+999 = 2997
+999 = 3996

Do you see a pattern?
...
+999 = 99900
+999 = 100899
+999 = 101898
...
+999 = 110889
+999 = 111888

It isn't explained very well, but if you think about what's going on, you'll get the idea.

another method:
with 9:18
with 99:1188
so with 999:111888


Find $ a>0$ such that $ a^{cos2x} \geq 2cos^2 x$ $ \forall x \in R$

[quote="vuthanhtu_hd"]Find $ a > 0$ such that $ a^{cos2x} \geq 2cos^2 x$ $ \forall x \in R$[/quote]
Set $ cos2x\equal{}X$,where $ X$ belongs to $ [\minus{}1;1]$.The inequality becomes
\[ a^X\ge X\plus{}1
\]
\[ \leftrightarrow a\ge \sqrt[X] {X\plus{}1}
\]
Denote
\[ f(X)\equal{}\sqrt[X] {X\plus{}1}
\]so
\[ a\ge max f(X)
\]
Now it is easy (for you) to see the $ max f(X)$ in $ [\minus{}1;1]$ is $ 2$
So $ a\ge 2$

[quote="Allnames"][quote="vuthanhtu_hd"]Find $ a > 0$ such that $ a^{cos2x} \geq 2cos^2 x$ $ \forall x \in R$[/quote]
Set $ cos2x \equal{} X$,where $ X$ belongs to $ [ \minus{} 1;1]$.The inequality becomes
\[ a^X\ge X \plus{} 1
\]

\[ \leftrightarrow a\ge \sqrt [X] {X \plus{} 1}
\]
Denote
\[ f(X) \equal{} \sqrt [X] {X \plus{} 1}
\]
so
\[ a\ge max f(X)
\]
Now it is easy (for you) to see the $ max f(X)$ in $ [ \minus{} 1;1]$ is $ 2$
So $ a\ge 2$[/quote]

Sorry, but you're mistaken! If $ a \equal{} 2$ and $ x \equal{} 30$ (that means $ cos x \equal{} \frac {1}{2}$, the ilequality becomes $ \sqrt {2} \ge \frac{3}{2}$ (!)

Here is my solution:
Set $ t \equal{} cos 2x$, where t belongs to $ [ \minus{} 1;1]$, the ilequality becomes $ a^t \ge t \plus{} 1$

Now we consider two cases
- Case 1: t belongs to $ [0;1]$. We have
\[ a^t \ge t \plus{} 1
\]

\[ \leftrightarrow (a^t)^\frac {1}{t} \ge (1 \plus{} t)^\frac {1}{t}
\]

\[ \leftrightarrow a \ge (1 \plus{} t)^\frac {1}{t}
\]

\[ \leftrightarrow a \ge e
\]
- Case 2: t belongs to $ [ \minus{} 1;0]$
Similary, we have $ a \le e$

So $ a \equal{} e$.
Now we must prove $ e^t \ge t \plus{} 1$
Denote $ f(t) \equal{} e^t \minus{} t \minus{} 1$
Notice that $ f'(t) \equal{} e^t \minus{} 1 \equal{} 0 \leftrightarrow t \equal{} 0$, and then you can easily solve the rest of the problem

[quote="ll931110"][quote="Allnames"][quote="vuthanhtu_hd"]Find $ a > 0$ such that $ a^{cos2x} \geq 2cos^2 x$ $ \forall x \in R$[/quote]
Set $ cos2x \equal{} X$,where $ X$ belongs to $ [ \minus{} 1;1]$.The inequality becomes
\[ a^X\ge X \plus{} 1
\]

\[ \leftrightarrow a\ge \sqrt [X] {X \plus{} 1}
\]
Denote
\[ f(X) \equal{} \sqrt [X] {X \plus{} 1}
\]
so
\[ a\ge max f(X)
\]
Now it is easy (for you) to see the $ max f(X)$ in $ [ \minus{} 1;1]$ is $ 2$
So $ a\ge 2$[/quote]

Sorry, but you're mistaken! If $ a \equal{} 2$ and $ x \equal{} 30$ (that means $ cos x \equal{} \frac {1}{2}$, the ilequality becomes $ \sqrt {2} \ge 2$ (!)

Here is my solution:
Set $ t \equal{} cos 2x$, where t belongs to $ [ \minus{} 1;1]$, the ilequality becomes $ a^t \ge t \plus{} 1$

Now we consider two cases
- Case 1: t belongs to $ [0;1]$. We have
\[ a^t \ge t \plus{} 1
\]

\[ \leftrightarrow (a^t)^\frac {1}{t} \ge (1 \plus{} t)^\frac {1}{t}
\]

\[ \leftrightarrow a \ge (1 \plus{} t)^\frac {1}{t}
\]

\[ \leftrightarrow a \ge e
\]
- Case 2: t belongs to $ [ \minus{} 1;0]$
Similary, we have $ a \le e$

So $ a \equal{} e$.
Now we must prove $ e^t \ge t \plus{} 1$
Denote $ f(t) \equal{} e^t \minus{} t \minus{} 1$
Notice that $ f'(t) \equal{} e^t \minus{} 1 \equal{} 0 \leftrightarrow t \equal{} 0$, and then you can easily solve the rest of the problem[/quote]

If $ a \equal{} 2$ then $ \sqrt {2} \geq \frac {3}{2}$ !!?
$ a\equal{}e$ is true .

similarly ,we have

Problem. Find $ a>0$ such that $ a^x \geq 1\plus{}x\plus{}\frac{x^2}{2!}\plus{}\frac{x^3}{3!}$

We have  $ a\equal{}e$

[quote="Allnames"]\[ a^X\ge X \plus{} 1
\]

\[ \leftrightarrow a\ge \sqrt [X] {X \plus{} 1}
\]
[/quote]
What if $ X < 0$ ?


geometry
circumcircle
incenter
geometry unsolved

Consider an acute, scalene triangle $ ABC$. Let $ H, I, O$ be respectively its orthocenter, incenter and circumcenter. Prove that there is no vertex or there are exactly two vertices of triangle $ ABC$ lying on the circle passing through $ H, I, O$.

It's well known that $ \angle BIC \equal{} 90 \plus{} \frac {\angle A}{2}$, $ \angle BHC \equal{} 180 \minus{} \angle A$ and $ \angle BOC \equal{} 2\angle BAC$. Because of $ BOC \equal{} BHC \equal{} BIC\Leftrightarrow \angle A \equal{} 60$ we deduce that the circumcircle of $ OIH$ pass through $ B,C$ iff $ \angle A \equal{} 60$, and if $ \angle A\not \equal{} 60$ then circumcircle of $ OIH$ dont pass through any vertex ^^
bye  :D


I heard that you can qualify for AIME by taking USAMTS. So if I qualified for AIME through USAMTS,  what would my AMC 12 score be in order to calculate the index score? I heard it's 100 but I wanted to verify the information.

I am pretty sure this has already been asked.... I just couldn't find similar posts.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=133816[/url] the original announcement

Is is just me, or does scoring 68 points through 3 rounds of USAMTS seem much harder than scoring 100 on the AMC-12?  I realize that it's kind of apples and oranges, since the USAMTS is not a timed test and the AMC-12 is, but I'm wondering what other people's perceptions are.

[quote="diophantient"][quote="ehehheehee"]I think it is a lot harder because there is so much less room for error (if you completely miss 2, you are out) and the problems are tougher.  But I think that if they made it relatively the same, there might be some people saying it is unfair.

But USAMTS is free and you get free problems, stuff and a T-shirt.

If I took the AMC 12 and qualified through USAMTS, which one do they take?  I think I remember looking at my index and seeing they took the 100.  It didn't matter as a freshman (or a sophmore), but a 100 would seriously hurt as a junior or senior.  They should take the AMC 12 score, but I just want to make sure.[/quote]
I think they double the score after three rounds, but maybe not :maybe:[/quote]

That'd be awesome.

If you qualify both ways, they take the higher of the two scores.

[quote="ehehheehee"]
But USAMTS is free and you get free problems, stuff and a T-shirt.[/quote]

You get a T-shirt?  Really?  Everyone or just for scoring at a certain level?

USAMTS is an awesome program, AIME qualification aside.  No arguments there.

The USAMTS [url=http://www.usamts.org/TipsFAQ/U_FAQ.php#aime]FAQ[/url] says they take the AMC-12 score if that results in a higher index.

Another thing - on one of my forms it gives a field of "School CEEB Code" for qualifying for the AIME. I googled it and apparently its a listing of high schools. However, if one went to middle school and wanted to attempt AIME qualification through USAMTS, what would they put? I'm sure home-schooled children must qualify through USAMTS also... (I go to a high school for math, would that count?)

Also, for grade do we put the grade for 06-07 or 07-08? I only ask this because it tells past students to update to 07-08, but new students are told to put 06-07...

[quote="Anaxerzia"]Also, for grade do we put the grade for 06-07 or 07-08? I only ask this because it tells past students to update to 07-08, but new students are told to put 06-07...[/quote]
probably a mistake/forgot to update
put your 07-08 grade

[quote="mathmom"]Homeschooled kids are supposed to write in "homeschool" in the CEEB field.  I'm not 100% sure what a middle school student would normally do.  The CEEB code tells AMC where to mail your AIME contest form, so what you write depends on where you plan to take it.  You might find out if your local HS will be offering the AMC contests and would let you write the AIME there if you qualify.  As an alternative, a number of colleges now offer the AMC-10 and/or AMC-12 to students whose schools don't offer them, and then AIME for their qualifiers.  They would probably be happy to let you write the AIME there if you qualified via the USAMTS.  In that case, you are supposed to just type in the name of the college in the CEEB field.
 [url]http://www.unl.edu/amc/b-registration/b1-archive/2006-2007/CU2007/2007-CU-list.shtml[/url]
was the list of colleges participating from last year.[/quote]

Thanks a lot -Now I have a way to register for AMC too  :D  :D


I'm taking the AIME ALT. on coming tuesday. I am wondering
what geometric theorems (other than the ones found on high
school geometry text book) and number theorems i should
know.
And probability...

If you've read thru AoPS I and most of two, I think u'll have mostly everything you need.

I've only done I and parts of II...


Use the theory of symmetric polynomials.

[hide="Really Big Hint:"] http://mathworld.wolfram.com/Newton-GirardFormulas.html

Though I don't know how $\Pi_{4}$ is defined in this situation; but we easily find that $\Pi_{3}= \frac{1}{6}$ and so we will probably be able to brute force $(x+y+z)^{4}$ and factor.[/hide]

how? all u do is square the second equation subtract

you know xyz is 0, (x^2)y^2+(y^2)z^2+(x^2)z^2 is1/4, so 2^2-2(1/4)
you have to play around with the equations.., like cube first one, subtract the third from it, then square the first one, subtract the second from it..

omg.. i forgot (x+y+z)^3.. if i did that right, then xyz=1/6, not 0, so 2xyz=1/3, but that will make the answer 25/6....probably still wrong..

[quote="usaha"]i dont want to double post, but i keep on gettin 25/6, so i just want to know if thats right[/quote]  Oh!  You are right.  I keep on making mistakes :furious:  It should be a $+2(x^{3}+y^{3}+z^{3})$ instead of a $+2(x^{2}+y^{2}+z^{2})$, so you have the right answer.

I'm stupid so I used Newton's Sums  :blush:


Fie $ABC$ un triunghi in care se noteaza cu $M$ mijlocul lui $[BC]$, si cu $D$ picioru bisectoarei din $A$, si fie $E,F$ respectiv pe $(AM), (AD)$ astfel incat $B \in EF$. Sa se arate ca $DE \| AB \Longleftrightarrow BE \perp AD$

Fie $T$ mijlocul segmentului$[AB]$. Atunci avem:
$\indent DE\parallel AB \Longleftrightarrow MT\parallel AC \Longleftrightarrow FT\parallel AC \Longleftrightarrow$ triunghiul $ATF$ este isoscel $\Longleftrightarrow FT=AT=BT \Longleftrightarrow AF \perp FB$


Is $ \gamma$ (Euler constant) irrational?

yes, it is not known if gamma is irrational (forget transcendental). rumor has it the famous mathematician hardy (ramanujan's co-conspirator  :D ) once said he was willing to give up his chair at oxford to anyone who proved gamma was irrational!  :P  even hilbert asserted many times the irrationality of gamma seemed completely "unapproachable" to him.

ok, hardy and hilbert, move aside please, 'cause misan is here! :rotfl:


Since the original tournament ended, Izzy allowed me to make a new one. :) Thanks Izzy!

This 10 Round Tournament is open to all grade levels. Each Round will have 10 questions, each one worth 10 points. So the highest possible score is 100 for each round, and the highest for the whole Tournament is 1000. See if you can reach 1000!

This Tournament is similar the Wickedestjr's tournament in that people who move on to the next round are a few short. (In other words, for each round, there will be a few people who don't make it. The number depends on how many sign up.)

I will give you all until Wednesday, July 23, to sign up. Once everyone who wishes to do so signs up, I will give more information.

So please, sign up!

More info...

In this 10-Round-Tournament, there are 10 Rounds(no kidding). In the first round, the top 20 people will move on(13 people will be eliminated.) In the second, the top 15 will move on(5 people eliminated). In the third, the top 13 people will move on(2 people eliminated). In the fourth, the top 12 people will move on(1 person eliminated). In the fifth, the top  10 people will move on(2 people eliminated). In the sixth, the top 5 will move on(5 people eliminated), in the seventh, top 4, in the eighth, top 3, in the ninth, top 2, and in the tenth,(kind of obviously) [b]the[/b] top person will be declared a winner.

Tomorrow, at about 4 or 5 am forum time, I will post Round 1. Please pm me the answers. You may use a calculator, except where indicated. You may not use any other aids, such as books, the internet, or the magic genie/god ( :D )

Feel free to ask any questions(concerning the tournament, obviously)

well popularity always starts "from unknown" , so...
and i would join, but i have a schedule conflict, so sorry, but i can't join

I will join but how long does this go for?

[quote="AIME15"]I will join.

7 days for over 20 people to join?  if its somebody popular like izzy, then definitely...but you? (no offense, but i dont think 20 peeps will join) Oh well, easier for me.  And if, say, only 15 people join, start at the round where less than 15 people advance to the next round.

Finding 100 problems might be a challenge too.[/quote]
It might not be exactly 10 rounds, but it all depends on how many people join.

YAY!!! i'v been wait 4 this i'll join!!!!

I shall join, but if I forget about it, could you please PM me?

yeah, can you please pm me when it starts? thanks

I'll join.  Can you please pm me too?  I have a tendency to forget about things I sign up for and then fail..  :P

Can I have extra time becuz I'm going to pennslyvania tonight and i'm not coming back till sunday?

due to techinal difficulties i will have to post the problems tomorrow and extend the deadline to monday.

When will you post the problems? Tonight?

Okay so I'm having more technical difficulties and i can only post 2 problems at a time?!?!

1. (EASY!) Simplify $ \frac {6}{9}$
2. I have 11 shirts. How many different ways can I wear them in one school week.

What do you mean by "how many ways can I wear 11 different shirts"?

3. $ \displaystyle\sum_{i\equal{}1}^{5}$
4. What is 1/2 of 1/5 of 1/10 of a billion?

You made a mistake on number 3 -- it's not proper syntax. That is the equivalent of $ 5 \plus{}$.

how come there are ony 4 questions posted...? and I never got any reminders that it was starting. I am confused.

hi ppls

another 10 round tourney?

Can we still join?

guys i thought i posted the rest of the problems!!!!! (oh geez...)
i think i'll need to cancel this and move this to another topic.

Well, I've submitted the 4 answers to the questions so far, and I've been waiting 3 days for the rest. What exactly is the problem?

computer things.
for now, this is canceled.
i apologize.


Se pot scrie numerele de la 1 la 2005(in orice ordine) intr-un sir astfel incat suma oricaror 5 termeni consecutivi ai sirului sa fie multimplu de 3?
      $\ Marius\ Perianu,\ Slatina$

[hide]fie $a_{1},a_{2},\ldots,a_{2005}$ o permutare a numerelor de la $1$ la $2005$, cu proprietatea din enunt
atunci
$\sum_{i=1}^{2005}a_{i}=\sum_{k=0}^{400}(a_{5k+1}+a_{5k+2}+a_{5k+3}+a_{5k+4}+a_{5k+5})$. Dar membrul drept este divizibil cu $3$, deci si membrul stang este divizibil cu $3$.
dar membrul stang este $\sum_{i=1}^{2005}a_{i}=\sum_{j=1}^{2005}j=\frac{2005\cdot 2006}{2}=2005\cdot 1003\equiv 1\,(mod\, 3)$, contradictie.[/hide]

Of, ce pacat ca nu mi-am dat si eu seama...chiar ca era simpla.Dar cu toata agitatia din sala de concurs m-am pierdut. :( Daca o faceam pe-asta, luam locul I.Dar cred ca e destul de bine si locul II.

Multumesc. :) Dar se putea si mai bine.Cea de pe locul I a luat cu 1 punct mai mult decat mine.

ei... si eu am ratat un mare concurs (cel mai mare?) anul acesta din cauza unui punct :)

Poti sa-mi spui si mie despre ce concurs e vorba?Si,...care e numele tau?


inequalities
quadratics
function
calculus
symmetry
calculus computations

The following problem was proposed in Inequalities Marathon at Pre Olympiad  forum

Problem 138: Let $ x,y,z > 0$ such that
\[ x \plus{} y \plus{} z \equal{} xyz\]
Prove that
\[ (x \minus{} 1)(y \minus{} 1)(z \minus{} 1) \le 6 \sqrt {3} \minus{} 10\]

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

This was my solution ,and it is wrong but I dunno why ,hope you will find the mistake.

[hide="solution"]
Observe that
\[ (x\minus{}1)(y\minus{}1)(z\minus{}1) \equal{}xyz \plus{}x\plus{}y\plus{}z \minus{}(xy\plus{}yz\plus{}zx) \minus{}1\leq 6\sqrt 3 \minus{}10\]
\[ \iff xyz \plus{}x\plus{}y\plus{}z \minus{}(xy\plus{}yz\plus{}zx) \leq 6\sqrt 3 \minus{}9\]
\[ \iff 2(x\plus{}y\plus{}z) \minus{}(xy\plus{}yz\plus{}zx) \leq 6\sqrt3 \minus{}9\]

Write $ x\plus{}y\plus{}z\equal{}p$ and $ xy\plus{}yz\plus{}zx\equal{}q$ and $ x^2 \plus{}y^2 \plus{}z^2 \equal{}t$ ,and since $ p^2 \equal{}t\plus{}2q$ ,we have :

\[ 2p\minus{}q \equal{}2p \minus{}\frac{p^2 \minus{}t}{2} \equal{}\frac{\minus{}p^2 \plus{}4p \plus{}t}{2}\]

We consider it as a quadratic function of $ p \ , f(p)\equal{}\frac{\minus{}p^2 \plus{}4p \plus{}t}{2}$ .The maximum attains for $ f'(p)\equal{}\minus{}4p \plus{}4 \equal{}0 \iff p\equal{}1$

But it is obvious that $ p\equal{}1$  is impossible ,because it does not satisfy the given condition $ x\plus{}y\plus{}z\equal{}xyz$[/hide]

[quote="enndb0x"]We consider it as a quadratic function of $ p \ , f(p) \equal{} \frac { \minus{} p^2 \plus{} 4p \plus{} t}{2}$ .The maximum attains for $ f'(p) \equal{} \minus{} 4p \plus{} 4 \equal{} 0 \iff p \equal{} 1$[/quote]  $ t$ is not a constant.

Since you posted it in the calculus section, we might as well look at the Lagrange multipliers solution. (General hint: when there are many circumstances, notably including symmetry across multiple variables, in which Lagrange multipliers is far nicer than one-variable calculus techniques.)

Maximize $ f(x)\equal{}(x\minus{}1)(y\minus{}1)(z\minus{}1)$ subject to $ g(x,y,z)\equal{}xyz\minus{}x\minus{}y\minus{}z\equal{}0.$

We set $ \nabla f\equal{}\lambda\nabla g.$

$ (y\minus{}1)(z\minus{}1)\equal{}\lambda(yz\minus{}1)$
$ (x\minus{}1)(z\minus{}1)\equal{}\lambda(xz\minus{}1)$
$ (x\minus{}1)(y\minus{}1)\equal{}\lambda(xy\minus{}1)$

Solve the first and second equations together:

$ \frac{(y\minus{}1)(z\minus{}1)}{yz\minus{}1}\equal{}\frac{(x\minus{}1)(z\minus{}1)}{xz\minus{}1}$

Either $ z\equal{}1$ or $ (y\minus{}1)(xz\minus{}1)\equal{}(x\minus{}1)(yz\minus{}1)$

$ xyz\minus{}xz\minus{}y\plus{}1\minus{}xyz\minus{}yz\minus{}x\plus{}1$

$ xz\minus{}x\equal{}yz\minus{}y,$ so dividing by $ z\minus{}1$ again, we have $ x\equal{}y.$

Similar work gives us that either $ x\equal{}1$ or $ y\equal{}z$.

If none of $ x,y,$ or $ z$ are $ 1,$ then we have $ x\equal{}y\equal{}z.$ Put that into $ g(x,y,z)\equal{}0$ to get that $ x^3\minus{}3x\equal{}0,$ so $ x\equal{}0$ or $ x\equal{}\pm\sqrt{3}.$

We have the following points that are either critical point or points that we haven't ruled out yet:

$ x\equal{}y\equal{}z\equal{}\minus{}\sqrt{3}$ in which case $ f(x,y,z)<0.$

$ x\equal{}y\equal{}z\equal{}0,$ in which case $ f(x,y,z)\equal{}\minus{}1.$

An unknown number of points for which at least one of $ x,y,$ or $ z$ equals $ 1,$ in which case $ f(x,y,z)\equal{}0.$

$ x\equal{}y\equal{}z\equal{}\sqrt{3},$ in which case $ f(x,y,z)\equal{}6\sqrt{3}\minus{}10.$

We're still not done. The constraint set isn't compact (isn't bounded), so we still have to do something else to assure that we've found the global maximum of $ f.$


The french IMO team 2005 will be formed by :

- Laurent Deproit (34 pts)
- Bruno Le Floch (39 pts, Toumaf on ML)
- Igor Kortchemski (28 pts, bronze IMO 2004, Igor on ML)
- Irène Marcovici (28 pts, yes, a girl!)
- Ilia Smilga (37 pts)
- Kevin Webster (30 pts)

Bruno and Ilia are sufficientely young to try to qualify for the IMO 2006.

Pierre.

We already know the [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=248565#p248565]French IMO team.[/url]  :P Better go and look-up the points for each of the contestants.  :D

Yes, but it deserves a special topic  :P
I've edited it to complete informations.

Pierre.

[quote="pbornsztein"]Yes, but it deserves a special topic  :P [/quote]Indeed it does :) Congrats to all involved, including qualifying (and non-qualifying) students, and training teachers ;)


analytic geometry
rotation
vector
linear algebra
matrix
trigonometry
geometry

so i  have a square with coordinates 4,4 8,4 8,8 4,8 how would i go about rotating it say by 30 degrees?

The best way I know to do it is with a linear algebra technique. First, which point to you want to rotate the square by? If it's (4,4), translate your coordinates so that (4,4) is the origin. Then the square's vertices are (0,0), (4,0), (4,4), and (0,4). Now turn each point into a column vector, for instance $(4,0) = \left[\begin{array}{c}4 \\ 0 \\ \end{array}\right]$. Define the matrix $A = \left[\begin{array}{cc}\cos{\alpha}& \sin{\alpha}\\-\sin{\alpha}& \cos{\alpha}\end{array}\right]$ where $\alpha$ is the degrees/radians in your rotation. Then multiply $A$ by each column vector (in that order $A v$, not $v A$) and you get four new column vectors that constitute the four points of the rotated square.

I am sure someone who actually knows something about geometry can give you a better answer.

To do it without linear algebra, remember that if a point is a distance of $r$ from the origin and at an angle $\theta$, the x and y coordinates are $r\cos\theta$ and $r\sin\theta$.

So if we have a general point $(x, y)$ at an angle of $\theta$ (which you could find by $\arctan \Big(\frac{y}{x}\Big)$) being rotated by an angle of $\phi$, the new coordinates are:

$\Big(\sqrt{x^{2}+y^{2}}\cos (\theta+\phi), \sqrt{x^{2}+y^{2}}\sin (\theta+\phi)\Big)$

You could simplify that using the sum of angles formulas, remembering that $\cos\theta = \frac{x}{\sqrt{x^{2}+y^{2}}}$ and $\sin\theta = \frac{y}{\sqrt{x^{2}+y^{2}}}$

sorry renatzu i dont quite get your post, can you make it any plainer. How can a point have an angle  of [img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/c/b/0/cb005d76f9f2e394a770c2562c2e150a413b3216.gif[/img].

[img]http://i3.photobucket.com/albums/y98/lamefif/mathscopy.gif[/img]

To get a point's angle, draw a ray from the origin to the positive axis and another from the origin to the point in question. The angle between those two rays is the standard definition of a point's angle. It's used in polar coordinates.

The simplified answer in that post is $(x \cos{\phi}+y \sin{\phi}, y \cos{\phi}-x \sin{\phi})$ where $x$ and $y$ are the coordinates of the original point and $\phi$ is the angle of rotation.

consider the positive x-axis to have an angle of 0 (degrees) and the positive y-axis to have an angle of 90 (degrees).  continue onwards so that the negative x-axis has an angle of 180 and the negative y-axis has an angle of 270.  then given a point, say $(2,3)$, we say that its angle $\theta$ is given by $\theta = \arctan{\frac{3}{2}}$ (or more generally, $\arctan{\frac{y}{x}}$ where arctan is the inverse tangent (in degrees).  then the angle of a point is the angle between a ray from the origin passing through that point and the ray from the origin along the positive x-axis.  in addition, we define the radius of a point to be the distance between the point and the origin.  The radius $r$ of a point $x,y$ is given by $r = \sqrt{x^{2}+y^{2}}$.  Then this allows us to define a polar coordinate system where every point in the plane is given by its radius and its angle with the positive x-axis.  Note that if one allows negative radii and one does not restrict the angle to the interval $[0,360)$ then a point can have multiple (actually, infinitely many) polar representations.

what he's trying to say is to use polar coordinates. For a polar coordinate $(r,\theta)$, $r$ represents the distance of the point from the origin, and $\theta$ represents the angle bounded by the line formed by connecting the point to the origin and the positive $x$-axis.

So in your case, you could first assume that the center of your square was the origin. Then by finding the polar coordinates of the other vertices in how renatzu and Teki-Teki said, convert them to rectangular coordinates then translate them to make your center $(6,6)$.


The professor tells Peter the product of two positive integers and Sam their sum. At first, nobody of them knows the number of the other.

One of them says: [i]You can't possibly guess my number[/i].
Then the other says: [i]You are wrong, the number is 136[/i].

Which number did the professor tell them respectively? Give reasons for your claim.

I think something is missing in the statement.
The solutions of the problem are numbers which their sum can be deduced from their product, or their product from their sum.Obviously, the first way is impossible except for 2 and 3, the second imposes that the number can be written as a product in 1 way, so it must a prime number multiplied by 1, hence there can be a solution for the problem if 135 was prime: false. Otherwise it will be ok if we had 138 instead of 136, but that would make the problem trivial...Am i missing something?

Hi,
maskman, I think you have forgotten the statement, that the guessing person is not only using his number to deduce the number of the other, he also uses the knowledge, that the other knows that he [i]can't[/i] deduce the other number by only his own number.

I Hope that my solution is correct

Obviously, if the product is 136, Sam cannot deduce the product by the knowledge, that Peter knows that he can't deduce the product with his sum, so the sum is 136.
A possible solution would be e.g. if Peter had 135.
When Sam said that Peter couldn't deduce the sum, he was right because the only possible case when Peter was able to was if the sum was 1+prime, which is wrong in this case (135 is not prime).
But now Peter knows that Sam knows that Peter cannot deduce the sum by the mere knowledge of the product.
Let Peter's number be $P$. If we consider the set of all $\{x+\frac{P}{x}\mid x|P\}$, then every number in this set except for one must be 1+prime.
This is true for $P=135$, because in this case, Peter would know that Sam has one of the numbers 136, 48, 24, 32. Everyone of these numbers except for 136 is 1+prime. But if Sam had 48, 24 or 32 then Sam could not know that Peter cannot deduce the sum because Sam cannot tell for sure that Peter has no prime number. Hence Peter knows that Sam must have 136.
Now we still have to prove that 135 is the only solution for $P$. But that's actually trivial because if there was another solution $P'$, then the set defined above would contain at least two numbers which do not have the form 1+prime (136 and $P'+1$)
Hence Sam's number must be 136 and Peter's must be 135.

WOW! it's one of the trickiest problems i have ever found and made me feel really stupid!! i would even say that the proposer deserves a special prize :) i also wonder if many students solved it during the competition....


Given $ a,b,c>0$ and $ a\plus{}b\plus{}c\equal{}1$.Prove that:
$ \frac{a}{2b^2\plus{}c}\plus{}\frac{b}{2c^2\plus{}a}\plus{}\frac{c}{2a^2\plus{}b} \ge \frac{9}{5}$
I hope you will like it,my friend :)

[quote="quykhtn-qa1"]Given $ a,b,c > 0$ and $ a \plus{} b \plus{} c \equal{} 1$.Prove that:
$ \frac {a}{2b^2 \plus{} c} \plus{} \frac {b}{2c^2 \plus{} a} \plus{} \frac {c}{2a^2 \plus{} b} \ge \frac {9}{5}$
I hope you will like it,my friend :)[/quote]
Secrets, try it,my friend :)

Yes,I will solve it,and I hope in the evening,I found solution for this peoblem  :wink:

cauchy-swvar we have :
$ \frac {a}{2b^2 \plus{} c} \plus{} \frac {b}{2c^2 \plus{} a} \plus{} \frac {c}{2a^2 \plus{} b} \ge \frac{1}{2(ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} ab \plus{} bc \plus{} ca}$
we uses this :
$ \ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc\le \frac {4}{27}(a \plus{} b \plus{} c)^3; (a \plus{} b \plus{} c)^3 \plus{} 9abc\ge \ 4(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ac)$

[quote="kihe_freety5"]cauchy-swvar we have :
$ \frac {a}{2b^2 \plus{} c} \plus{} \frac {b}{2c^2 \plus{} a} \plus{} \frac {c}{2a^2 \plus{} b} \ge \frac {1}{2(ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} ab \plus{} bc \plus{} ca}$
we uses this :
$ \ a^2b \plus{} b^2c \plus{} c^2a \plus{} abc\le \frac {4}{27}(a \plus{} b \plus{} c)^3; (a \plus{} b \plus{} c)^3 \plus{} 9abc\ge \ 4(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ac)$[/quote]
Nice solution :)

[quote="secrets"]$ \frac {a}{2b^2 \plus{} c} \plus{} \frac {b}{2c^2 \plus{} a} \plus{} \frac {c}{2a^2 \plus{} b} \ge \frac {1}{2(ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} ab \plus{} bc \plus{} ca}$
and we will to prove that:
$ 2(ab^2 \plus{} bc^2 \plus{} ca^2) \plus{} ab \plus{} bc \plus{} ca \le \frac {5}{9}$
We have old ineq:$ ab^2 \plus{} bc^2 \plus{} ca^2 \le \frac {4}{27} \minus{} abc$
And finally we will to show that:
$ ab \plus{} bc \plus{} ca \minus{} 2abc \le \frac {7}{27}$
But this ineq is old.We can prove easily by schur ineq.This proof is completed.Equality occurs if and only $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$
@quykhtn-qa1[hide]hic,chiều nay em ở lớp nghĩ ra,ai ngờ có người giải trước[/hide][/quote]
Thanks secrets :lol:
@ secrets [hide]: em hoc truong nao,vay? :)  :) [/hide]

[hide]Em học trường Lương Thế Vinh ạ.Anh lớp mấy rùi ạ,tên đầy đủ của anh là gì[/hide]


geometry Princeton college Harvard research

Hello everyone,

I am a junior in college. I started off as an engineering major, but ended up switching to math after my sophomore year, because I felt more interested in mathematical proofs than in "numerical math", which is what one usually finds in engineering.

I've always been pretty good at math and now I am wondering whether going for a PhD in Math would be a good idea. My main concern is the uncertainty that I might not be able to actually produce genuinely new results. As a participant in an advanced-level math program this semester (where they essentially teach us graduate-level classes), I have seen how involved advanced mathematical concepts can really be. At first, I felt astonished by the amount of time I had to spend just understanding the definitions involved in the homework problems before even starting to try solving the problem.

At this point (just over a month since the start of the program), I have certainly learned new things and am more comfortable doing the assignments than I was at the start, but there is still a ton of ideas that I find difficult to comprehend. Anyway, it puts me down to see how much time I spend on solved problems (though they come up in these advanced courses) and this leads me to wonder if I'll be able to cope with unsolved problems, which is what research is all about.

I actually did research this summer, but now looking back on it I don't think it was extremely complicated. And, at the same time, my research advisor was very pleased with my progress and congratulated me on being able to think like a mathematician.

I would appreciate any comments, suggestions, words of wisdom anyone can offer.

Thank you!

Some unsolved problems are very hard. Some aren't; they're only unsolved because nobody has really looked at them in depth yet. The PhD path includes enough time to get substantial background in whatever area you choose to focus in, and thus reach some unsolved problems there.

From what you've said, you're capable of going for a PhD. It's not guaranteed to work, but there are plenty of reasonable fallback positions if it doesn't.

[quote="kallo.marinov"]
I actually did research this summer, but now looking back on it I don't think it was extremely complicated. And, at the same time, my research advisor was very pleased with my progress and congratulated me on being able to think like a mathematician.
[/quote]

Research is an art form, and just as in many forms of art, the true artist is rarely satisfied with his or her own work.  I know many people who have proven great new results and even won awards for them, but do not consider their proofs to be very difficult, enlightening, or useful because they are simply too close to their own work.

When you get entirely absorbed in a research problem and revise your own proof twenty times, it becomes so natural to you that there is a tendency to forget how much you have actually done.  So, take a step back and look at your work from the point of view of someone unfamiliar with your results.  You may see where your advisor is coming from.

About the concept of research, it's not necessarily even about trying the "famous" or "semi-famous" unsolved problems necessarily (but of course we do need plenty of good mathematicians for this task, so if that's what you want to do, all the better!). Often, research can be about trying a problem or investigating an area that no-one else has simply even tried before, trying something new, and hoping that this area gives rise to interesting ideas, theory, and new problems. So often you end up doing something completely new, which may be easy and often not that difficult, rather than problems which many mathematicians before you have failed to solve. So keep that in mind, I didn't have this impression that this was the case until I started trying some problems my supervisor gave me, and this really makes research much less intimidating. Many of my lecturers have had most of their research doing things like this, as opposed to solving conjectures/famous problems. Most importantly, you have the freedom to choose the difficulty and the area of what you want to do, so it's important to specialize in what you're absolute best at, rather than being good at everything universally like in Olympiad problems (where it's actually a disadvantage if you're good at Area #1 but very bad at Area #2).

[quote="kallo.marinov"]Thank you for your feedback.

What, in your opinion, might make someone give up on a PhD program they have started?

Also, can you give me some examples of the fallback positions that you mentioned?[/quote]

Many people that enroll in graduate programs often start out as master's students, especially if they are unsure of their math background or have backgrounds from other disciplines. As a first year you will be taking the same required courses. If you have had some of these as an undergrad you can sometimes convince the dgs to bypass these, but in any case you will have to demonstrate at some point that you really have a good grip on the material, which is accomplished by giving prelim exams.

Many students though find that they struggle through first semester classes and get poor grades..this is often the case of those who aren't used to preparing for and taking hard exams in those areas, especially if they are competing against students that [i]are[/i]; some of their classmates have perhaps taken their foundational classes in undergrad, so this makes the playing field uneven for those taking them for the first time. Right away, it seems, doors of opportunity are closed for those who are not well prepared. For those few that have worked hundreds and hundreds of problems, the doors of opportunity will be opened: they have convince their profs that they are competent practitioners of their art and are therefore worthy in further study and investment.

Note  that there is a gradient in levels of grad programs; it's just not that theres princeton and harvard and nothing else. Depending on where you go, the focus on the grad students is to prepare researchers or teachers, or both. This is reflected in how the funding works..what is the proportion to RA's to TA's? Research heavy schools tend to be less forgiving and expect more of you coming in; the other schools expect less, but all have the expectation that you will work very hard, even during the summers.

One's success in grad school is really determined by the first steps: doing well on the math gre subject exam, be somewhat prepared to pass the phd prelimarly exams. Short of that, the student can gradutate with a masters. A fallback position after the masters, there are a few possible pathways. Many students apply for a phd at a less rigorous program, some apply to pedagogy programs or teachers colleges or teach. Just remember that whatever you do, it's competitive.  I wont say its not cutthroat, but people are competing for jobs and opportunities...so its just like any other job or activity where people step on each other to get ahead.

If you are truely unsure about how you will do, why not take a year or two and work, so you can have more time to think about it rather than be forced into doing something. That way, when you begin you can at least have more confidence in yourself!


trigonometry
ratio
Pythagorean Theorem
geometry

25. In triangle JSR, <R is a right angle and the altitude from R meets JS at Z. The lengths of JSR are integers, SZ=1331, and cos(S)=$\frac{p}{q}$ where p and q are relatively prime. Find p+q.

[hide]After playing a bit, we find that $\triangle JRS\sim\triangle RSZ\sim\triangle JZR.$
By the similarity ratios,
$SZ\cdot JS=RS^{2}$
$11^{3}JS=RS^{2}$
Since $JS$ and $RS$ are integers, $JS$ must be of the form $11a^{2}$ for some positive integer $a$ since RHS has to be a perfect square.
Then it follows that
$RS=11^{2}a$
Then by pythagorean theorem,
$RS^{2}+JR^{2}=JS^{2}$
$\iff JR^{2}=11^{2}a^{2}(a^{2}-11^{2})$
Since $JR$ is also an integer, $a^{2}-11^{2}$ must be a perfect square.
Let $a^{2}-11^{2}=b^{2}$ for some positive integer $b.$
Then the numbers $b,11,a$ are a pythagorean triplet.
So by the pythagorean triplet formula thingy,
$a=k(r^{2}+s^{2})$ and $11=2rsk$ or $11=k(r^{2}-s^{2})$
for some integers $r,s,k.$
Since $r,s,k$ are positive integers, $11=2rsk$ is not possible, so
$11=k(r+s)(r-s)$
Two of the 3 factors of the RHS must be 1, but $r+s$ and $r-s$ cannot both be 1, since then $s=0,$ a contradiction.
So
$r+s=11,r-s=1,k=1$
Solving, $r=6,s=5$
Putting this in for $a,$
$a=61.$
Now $\cos\angle S=\frac{RS}{JS}=\frac{11^{2}a}{11a^{2}}=\frac{11}{a}=\frac{11}{61}$
So $p=11,q=61,p+q=\boxed{72}$

[/hide]

That's cool... but where can I learn about this "pythagorean triplet formula thingy" you speak of?

Well, it's a well known fact that all pythagorean triplets can be represented in the form
$c=k(r^{2}+s^{2})$
$b=k(2rs)$
$a=k(r^{2}-s^{2})$
(where $a^{2}+b^{2}=c^{2}$)
[url]http://en.wikipedia.org/wiki/Pythagorean_Triple[/url]


Hi folks,

Mark Qi has taken it upon himself to create a Walton Math Team web site and has created a Problem of the Week contest. The prizes are outstanding. (The pride and honor of winning! :-) I believe that the problems will start somewhat easy and get harder. Feel free to take a look. [url]http://sites.google.com/site/waltonmathteam/potw-1[/url]

If only Mark took the time to UPDATE the PoTW, I've been waiting! Mark, if you read this, I'm still waiting for your solution for our potw, stop slacking!

Jesse Zhao

According to the website, this week is due on Friday Sept. 4th. That means solutions wouldn't be up until after that.

Ya, you got a little extra time to do the first POTW because you knew about it before it was officially announced at Walton. The next POTW will be up on 9/4 or 9/5, depending on Mark's mood. By the way, I'm not completely sure that Mark checks this forum, but I do. ;) By the way, I'm Allen.


Have a look here :
http://mathcentral.uregina.ca/MP/previous/apr06sol.html
The given solution can be easily adapted to the general case of even degree.

Pierre.

@pbornzstein: the links seems to be down.

@orl: is there a "blank" before the $ x^{2n}$? I'm guessing not but it's not quite clear.

Try this one
http://mathcentral.uregina.ca/mp/previous2005/apr06sol.php

Pierre.

[quote="orl"]$ A$ and $ B$ play the following game with a polynomial of degree at least 4:
\[ x^{2n} \plus{} \_x^{2n \minus{} 1} \plus{} \_x^{2n \minus{} 2} \plus{} \ldots \plus{} \_x \plus{} 1 \equal{} 0
\]
$ A$ and $ B$ take turns to fill in one of the blanks with a real number until all the blanks are filled up.  If the resulting polynomial has no real roots, $ A$ wins.  Otherwise, $ B$ wins.  If $ A$ begins, which player has a winning strategy?[/quote]
sorry but I don't understand your symbol $ \_x^{2n \minus{} 1}$.I'm a stupid boy :( !

The notation means $ \_x^{2n \minus{} 1}$ that $ a \cdot x^{2n \minus{} 1}$ where $ a \in \mathbb{R},$ i.e. the place holder underscore denotes a real multiplicative coefficient of the term $ x^{2n \minus{} 1}.$

Maybe somebody wants to write the solution in the ML/AoPS forum?!  :)

[hide="Solution"]

Player B, who tries to force a real root, needs only to ensure that $ P(t) \leq 0$ for some $ t \in \mathbb{R}$. This is because the leading coefficient $ x^{2n}$ forces the polynomial to tend to infinity when $ x \to \infty$, and in particular it takes on positive values; if takes a negative value anywhere it must also vanish at some point, being continuous.

We will show that B indeed has the winning strategy, which is somewhat surprising given that he makes neither the first nor the last move. Since it's difficult to analyze the behavior of the roots as the polynomial is being filled, we'll take the lazy route and only consider B's last move which occurs when there are only two unfilled coefficients - say, the coefficients of $ a^k$ and $ a^l$. So at this point in the game we have

$ P(x) \equal{} Q(x) \plus{} a_k x^k \plus{} a_l x^l$

where $ P(x)$ is the polynomial that the players are constructing, $ Q(x)$ is the part of it which is already filled in and is therefore known at this stage, and $ a_k$ and $ a_l$ are the last two remaining coefficients. Player B gets to choose which one he wants to determine and which value to give it, and subsequently player A will have the last word and determine the last coefficient.

First, suppose that one of $ k,l$ is even and the other is odd. Say $ k$ is odd and $ l$ is even. Notice that

$ P(1) \equal{} Q(1) \plus{} a_k \plus{} a_l$
$ P( \minus{} 1) \equal{} Q( \minus{} 1) \minus{} a_k \plus{} a_l$

so

$ P(1) \plus{} P( \minus{} 1) \equal{} Q(1) \plus{} Q( \minus{} 1) \plus{} 2a_l$

therefore, by choosing an appropriate value for $ a_l$, player B can force $ P(1) \plus{} P( \minus{} 1)$ to be 0 no matter what player A will then do with $ a_k$. So he does that, and then necessarily either $ P(1) \leq 0$ or $ P( \minus{} 1) \leq 0$ which, as we saw, is sufficient for him to be the winner.

Next, suppose that both $ k$ and $ l$ are odd. Once again we're trying to find a [b]positive[/b] linear combination of values of $ P(x)$ such that one of the coefficients $ a_k$, $ a_l$ has no influence. We can't do this with $ P(1)$ and $ P( \minus{} 1)$ so we try

$ P(2) \equal{} Q(2) \plus{} 2^k a_k \plus{} 2^l a_l$
$ P( \minus{} 1) \equal{} Q( \minus{} 1) \minus{} a_k \plus{} a_l$

so now $ P(2) \plus{} 2^kP( \minus{} 1)$ is independent of $ a_k$, indeed

$ P(2) \plus{} 2^kP( \minus{} 1) \equal{} Q(2) \plus{} 2^kQ( \minus{} 1) \plus{} (2^l \plus{} 2^k)a_l$

so player B can choose $ a_l$ appropriately and force $ P(2) \plus{} 2^kP( \minus{} 1)$ to be zero. As before, player A can no longer change that fact, and as before
it guarantees that at least one of $ P(2)$ or $ P( \minus{} 1)$ is non-positive, which implies a real root.

The last case is when both $ k$ and $ l$ are even. Here it's not clear how to find a "good" linear combination as above. However, all's not lost: All player B needs to do is to make sure during the game that at least one of the last two remaining coefficients is odd. This is possible since at the beginning the number of odd coefficients exceeds the number of even coefficients by 1: we have $ n$ odd coefficients and $ n \minus{} 1$ even ones. Whatever coefficient $ A$ picks on the first move, B can ensure that this surplus is maintained (if A picks even, B can pick whatever, and if A picks odd, B picks even), and he can obviously continue to do so on subsequent moves. So B ensures that at least one of the last two remaining coefficients is of an odd power of $ x$ and we're done. Note that it doesn't matter which values B actually chooses to place for the coefficients until his very last move.
[/hide]

I did not want to revive this old thread

But I am really in doubt over my solution which seems too easy and much simpler than the previous one.

[hide="Solution"]
$B$ has a winning strategy.

Since $B$ has the last turn,
Let the sum of the coefficients of all the filled spaces be $z$ (including that of $x^{2n}$ and $1$),
then $B$ simply fills $-z$ as his last number.

And now since sum of the coefficients of the polynomial is $0$,
Pretty obviously $P(1)=0$
which is a real root of this equation and thus we are done.
[/hide]

Unfortunately there are only $2n-1$ blanks, so it's $A$ that has the last move.

I knew I had missed something !

Sorry :wallbash:  :wallbash_red:

$\textbf{\textcolor{red}{Claim:}}$ $B$ always has a winning strategy.\\

$\textbf{\textcolor{blue}{Pf:-}}$ Suppose $A$ starts to fill out a even index , so $B$ must choose a odd index or vice versa. Now suppose $P(x)=Q(x)+a_{j}x^{j}+a_{i}x^{i}$ , where $Q(x)$ has all of it's coefficients determined and $i$ is odd.\\

$\mathrm{Case \hspace{0.1cm} 1:}$ $j$ is even\\

We have $P(1)=Q(1)+a_{j}+a_{i}$ and $P(-1)=Q(-1)+a_{j}-a_{i}$ , so if $B$ chooses $a_{i}=\frac{Q(1)+Q(-1)}{-2}$ then $P(1)+P(-1)=0$ , if both $P(1), P(-1)$ both are $0$ then $B$ wins , otherwise $P(1) , P(-1)$ have opposite signs and since $P(x)$ is continuous by IVT  there would exist some $z \in \mathbb{R}$ s.t. $P(z)=0$ , so again $B$ wins.\\

$\mathrm{Case \hspace{0.1cm} 2:}$ $j$ is odd.\\

so , $2^{j} P(-1)=2^{j}Q(-1)-2^{j}a_{j}-2^{j}a_{i}$ and $P(2)=Q(2)+2^{j}a_{j}+2^{i}a_{i} \implies 2^{j}P(-1)+P(2)=2^{j}Q(-1)+Q(2)+a_{i}(2^{i}-2^{j})$ so if $B$ chooses $a_{i}=\frac{2^{j}Q(-1)+Q(2)}{2^{j}-2^{i}}$ then $2^{j}P(-1)+P(2)=0$,  hence again similar as $\mathrm{ Case \hspace{0.1cm} 1}$ we get $B$ has a winning strategy.\\

Thus, either way, $B$ wins and in conclusion, $B$ always has a winning strategy. $\blacksquare$


geometry
perimeter
conics
hyperbola
geometry proposed

Let $ABCD$ be a circumscribed quadrilateral . Find $M$ inside qudrilateral $ABCD$ such that : $P(MAB)=P(MBC)=P(MCD)=P(MDA)$ , where $P(X)$ is the perimeter of $X$ .

The condition that the quadrilateral is circumscribed is necessary. Indeed:

$P(MAB) = P(MDA) \Rightarrow$
$MA+MB+AB = MD+MA+AD \Rightarrow$
$MB+AB = MD+AD \Rightarrow$
$MD-MB = AB-AD \ \ (1)$

$P(MBC) = P(MCD) \Rightarrow$
$MB+MC+BC = MC+MD+CD \Rightarrow$
$MB+BC = MD+CD \Rightarrow$
$MD-MB=BC-CD$

so, $AB-AD=BC-CD \Rightarrow$
$AB+CD = AD+BC \Rightarrow  ABCD$ is circumscribed, or else there is not a solution.

$P(MAB) = P(MBC) \Rightarrow$
$MA+MB+AB = MB+MC+BC \Rightarrow$
$MA+AB = MC+BC \Rightarrow$
$MC-MA = AB-BC \ \ (2)$

we don't need to check the other pairs of the triangles, because we'll get results which are allready known.
What we need is $(1)$ and $(2):$

$MD-MB = AB-AD \ \ (1)$
$MC-MA = AB-BC \ \ (2)$

Assume that $AB$ is the longest side. Also, set
$t=AB-AD$
$u=AB-BC$
$(t,u>0)$

$(1)$ is an hyperbola with focuses $B,D$ and difference $t$
$(2)$ is an hyperbola with focuses $A,C$ and difference $u$

We usually draw an hyperbola with two branches; one with possitive difference of radiuses and another with negative difference.
You can easy find that the points $A,C$ belong to the "negative branch" of hyperbola $(1)$
(similarly for $B,D$ and hyperbola $(2)$)

We draw the "positive branches" of the two hyperbolas and the intersection of them is the point $M$.

We should also check weither the  point $M$ is in the interior of $ABCD$ but I don't know how  :?:


quadratics
modular arithmetic
number theory proposed
number theory

Prove that there are infinitely many triples of positive integers (m, n, p) satisfying 4mn - m - n = p 2  - 1, but none satisfying
4mn - m - n = p 2

This, I think, has been solved before. Anyway, the first equation is equivalent to $(4m-1)(4n-1)=4p^2-3$, and we can take $3|p$ and $m=1$. The rest follows. As for the second one, its equivalent to $(4m-1)(4n-1)=4p^2+1$, and since $-1$ is a quadratic non-residue $\pmod q$ for any prime $q$ of the form $4k+3$, it means that $x^2+1$ cannot have any prime factors of the form $4k+3$, and we apply this for $x=2p$.


geometry
circumcircle
perpendicular bisector
projective geometry
geometry proposed

Let S be a fixed circle containing a fixed point A. (l) is a fixed line that doesn't cut S and P is an abitrary point in it. Through P, we draw an  abitrary line (m) intersecting S at 2 distinct points C,K. B is a second intersection of PA and S. Let M, N be intersections between CA and (l), BM and (S). Prove that the circumcenter of OKN always lies on a fixed line. (O is the center of S) :D

It has not been proved for so long because it is not true.  The line of the
  centers moves with the motion of point P.

   Pestich.

Ok, I 'll check my note if the problem I posted is corrected. Anyway, thanks pestich.

[quote="pestich"]It has not been proved for so long because it is not true.  The line of the
  centers moves with the motion of point P.
[/quote]

I doubt that's true, man. It looks like no matter how the positions of $P$ and $(m)$ change, $KN$ passes through a fixed point $X$ on $(\ell)$. I'll prove this later. Now for the initial problem (I'll use what I said above without proof):

Take a point $Y\in OX$ s.t. $OX\cdot OY=OK\cdot ON$. Clearly, the circle $(OKN)$ always passes through the two fixed points $O,X$, so its center moves on the perpendicular bisector of $OX$.

[quote="grobber"]
It looks like no matter how the positions of $P$ and $(m)$ change, $KN$ passes through a fixed point $X$ on $(\ell)$. I'll prove this later.[/quote]

This is, indeed, true: Apply Pascal's theorem to the degenerate hexagon $AACKNB$. We find that if $X$ is the intersection between $KN$ and the tangent to the circle through $A$, then $X,\ M=AC\cap BN,\ P=AB\cap CK$ are collinear, meaning that $KN$ passes through the intersection between $\ell$ and the tangent through $A$ to the circle, which is a fixed point. So you see, the position of $P$ doesn't matter.

I am very sorry, for I'm at fault. I misunderstood the text. Very nice
  problem and solution.   There are 2 points P resulting in the same position
  of chord KN.

   Pestich.

I 'm just worried about if my problem was mistyped. But it 's right. My solution is the same as Grobber's. :D


How many languages do you speak?
Please mention them!
As for me, I speak [size=100]हिन्दी [/size],[size=175]বাংলা [/size] (Bengali) and of course English.

k me 4 :D  :D
I speak English,Hindi,Tamil and guess  what................Bengali too!  :D thats becuz ive lived in Bengal for the last 16 years.........

:P  Why are you so bothered about the languages we speak?
I speak English , Hindi , Telugu , Marathi and French.

the most important language i speak.......

$ MATHS$
[url=http://en.wikipedia.org/wiki/List_of_official_languages_by_state]Some others are...[/url]

I speak fabulous Benglish :P ,and I was trying to learn some japani [hide]but without much sucess...coz u totally need to change not only ur accent but evn ur voice :rotfl:  :rotfl: [/hide]


Four men, in different colored overcoats, sat on two facing bench seats in a train. Two sat next to the window and two next to the aisle. The Englishman sat on Mr.Bell's left. Alto womre a tan-colored coat. The man in olive was on the German's right. Mr.Cardinal was the only cigar smoker. Mr.Dunn was across from the American. The Russian was in khaki, and the Englishman started out the window on his left. Who was the man in the rust- colored coat?

Umm...[code]..
..[/code]
Is how the seats were right? so then we know the englishman was not facing a window? or can there be more than one?


[quote="arqady"][quote="Gibbenergy"]My inequality:
Let $a,b,c$ are positive. Prove that:

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{2abc}{(a+b)(b+c)(c+a)}\geq \frac{7}{4}$
 :)[/quote]
I think, it's nicer:
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{4abc}{(a+b)(b+c)(c+a)}\geq 2$ :P[/quote]

  You win me again Arqady! In the general case:
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{kabc}{(a+b)(b+c)(c+a)}\geq \frac{3}{2}+\frac{k}{8}$
  The value of $k$ that the inequality still hold is when k satisfy the inequality:
$(36+k)b^{2}+(3k-144)b+108 <0$ when $1\leq b \leq \frac{3}{2}$ and $k \leq 8$*
The condition $k\leq 8$ in my proof is to make sure that the inequality always at minimize when $a=b$. I am not sure that $k\leq 8$ is the best estimate. Anyway, in order for the quardaric eq have roots, we have the $\delta \geq 0$ leads to $k\leq 72-48\sqrt2 = 4.117$Choose $k=4$ is the nicest.:)

[quote="Gibbenergy"]My inequality:
Let $a,b,c$ are positive. Prove that:

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{2abc}{(a+b)(b+c)(c+a)}\geq \frac{7}{4}$
 :)[/quote]

after expanding it`s schur

[quote="arqady"][quote="Gibbenergy"]My inequality:
Let $a,b,c$ are positive. Prove that:

$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{2abc}{(a+b)(b+c)(c+a)}\geq \frac{7}{4}$
 :)[/quote]
I think, it's nicer:
$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}+\frac{4abc}{(a+b)(b+c)(c+a)}\geq 2$ :P[/quote]

Yes, Arqady, but it is the same with the one I have posted yesterday and created by me six months ago.

http://www.mathlinks.ro/Forum/viewtopic.php?t=115437

[quote="Cezar Lupu"]

Yes, Arqady, but it is the same with the one I have posted yesterday and created by me six months ago.

http://www.mathlinks.ro/Forum/viewtopic.php?t=115437[/quote]
OK, Cezar. I saw Gibbenergy's problem, only.
Now I see, that it is your problem too and you have thought up it earlier.


LaTeX
calculus
derivative
calculus computations

Im in big trouble if I can't answer this problem within the next hr. Please help if u can.

I need help factoring this equation

-5.978(1+5.71e^-.054x) 1+5.71e^.054x

Firstly, this is [b]not[/b] a math. hotline for homework or test problems.
Secondly, either learn $\LaTeX$ or put all patentheses where they belong. What you wrote is
$-5.978(1+5.71e^{-.054}x) \cdot 1+5.71e^{.054}x$, which is the same as $-5.978-5.978\cdot5.71e^{-.054}x+5.71e^{.054}x$. There seems to be no meaningful way to factor this [b]expression[/b] (needless to say, it is not an [b]equation[/b]).
Thirdly, this seems to be related to your other post where you wanted to find the second derivative of something similar. I still do not understand why you need to factor anything to find derivatives instead of just using the standard differentiation rules and what you are going to do with your inflection points. Perhaps, it would help if you stated your initial problem [b]exactly[/b] as you got it.


The inequality is equivalent to
$ \frac{ab}{a\plus{}b}\plus{}\frac{cd}{c\plus{}d}\le \frac{(a\plus{}c)(b\plus{}d)}{a\plus{}b\plus{}c\plus{}d}$
$ \Leftrightarrow (a\plus{}b\plus{}c\plus{}d)(abc\plus{}bcd\plus{}cda\plus{}dab)\le (a\plus{}b)(a\plus{}c)(b\plus{}d)(c\plus{}d)$
$ \Leftrightarrow (ad)^2\plus{}(bc)^2\ge 2abcd$.

Does anyone have proof for

$ (a\plus{}b\plus{}c\plus{}d)(abc\plus{}bcd\plus{}cda\plus{}dab)\le (a\plus{}b)(a\plus{}c)(b\plus{}d)(c\plus{}d)$

without expanding?

[quote="Nguyenhuyen_AG"][quote="April"]Prove that for positive real numbers $ a$, $ b$, $ c$, $ d$, we have \[ \frac{1}{\frac{1}{a}\plus{}\frac{1}{b}}\plus{}\frac{1}{\frac{1}{c}\plus{}\frac{1}{d}}\le\frac{1}{\frac{1}{a\plus{}c}\plus{}\frac{1}{b\plus{}d}}\][/quote]
The inequality is equivalent to
\[ \frac{ab}{a\plus{}b}\plus{}\frac{cd}{c\plus{}d}\le \frac{(a\plus{}c)(b\plus{}d)}{a\plus{}b\plus{}c\plus{}d}\]

\[\left (\frac{a+b}{4}-\frac{ab}{a+b}  \right )+\left (\frac{c+d}{4}-\frac{cd}{c+d}  \right )\ge \frac{a+b+c+d}{4}-\frac{(a+c)(b+d)}{a+b+c+d}\]

\[\frac{(a-b)^2}{a+b}+\frac{(c-d)^2}{c+d}\ge\frac{(a+c-b-d)^2}{a+b+c+d}\]

Which is obviously true by Cauchy-Schwarz.[/quote]

This is an insanely beautiful solution!


hello:d
i'm looking for some olympiade tutorials for beginners, for example: theorems, excercices,..in all levels and difficulties. In english or french.
if you know a site pleaz post the link

thanx

Check out my website: [url]http://www.megus.tk[/url] [I'll invite everyone  :D ] - I intend to build ultimate source for anyone who looks for some materials about olympiads on the Internet. Site is not large now, but is still in development and I'll accept every help.  :)

We will soon restore the downloads section on AoPS - MathLinks. Stay tunned. Untill then you are free to check out the forums here, and also AoPS resources.

[quote="demji"]what would be in the downloads sections?  Where will it be announced?[/quote]there is already a lot of stuff in the download forums. We'll have all that but better presented - remember these are the IMO forums, so all the downloads will be relevant to this level.


[quote="thanh22"]How many permutation {$a_{1},a_{2}, ... ,a_{n}$} of {1,2,...n} which satisfies
$\abs{a_i-a_j} <1$[/quote]
i think you missed something thanh22,and i am agree with zhaoli.
and i think there is a condition for $i,j$. :lol:

For all possible values of $i$, or what?

Hey, friends! I have a nice problem, too.
How many permutations {$a_{1},a_{2},...,a_{n}$) of {1,2,...,n} satisfy
$a_{i}-a_{i+1} > 1$ or $a_{i}-a_{i+1} < -1$ for all $1 \leq i \leq n-1$


Solve in $ R^3 $ the system
\[
\left \{\begin{array}{ll}
x\sqrt{1+y^2}+y\sqrt{1+x^2} = 1\\
y\sqrt{1+x^2}+z\sqrt{1+y^2} = 2\\
z\sqrt{1+x^2}+x\sqrt{1+z^2} = 3
\end{array} \right.
\]

Nice problem! Put $ x=sh(a), y=sh(b), z=sh(c)$ and you will see how easy it becomes. Here sh(x) is the hyberbolic sinus.


I sometimes notice that someone will have posted something and there below it will be an identical post. Why? Is this normal?

Well when the connection's slow (or when the server's unusually busy) we sometimes get impatient and click "Submit" on our posts more than once, thus the information is sent twice and this happens. This also happens when the "Page can not be found" or other errors occur and you click "Back" and get back to your post then try again by clicking "Submit" again.

If this happens just simply delete that post, but some people don't do it (or doesn't notice or forget or realizes after someone replies) thus these posts stay remaining, until a moderator or administrator (usually moderator) come by and clean it up.


this should be in moems or mathcounts, at the most.

[hide="wtf."]so $l+w=21$ and $l=2w$. so $3w=21$ so the width is $7$ and the length is $14$[/hide]

[quote="Interval"]The perimeter of a rectangle is 42 meters.  Find the length and width if the length is TWICE the width.[/quote]

[hide]$\\ 2l+2w=42\\ 2(2w)+2w=42\\ 4w+2w=42\\ 6w=42 w=7\\ \therefore l=14$[/hide]

@purpleplume: People may take Alg at High school, but I still stick with WATR that it should be in MOEMs or MC.

does purple plumes solution look weirdly similar to anirudhs? or is it just my aging eyes... :rotfl:


A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $ 2000$. What is the length of each side of the octagon?

$ \textbf{(A)}\ \frac{1}{3}(2000) \qquad
\textbf{(B)}\ 2000(\sqrt2\minus{}1) \qquad
\textbf{(C)}\ 2000(2\minus{}\sqrt2)$
$ \textbf{(D)}\ 1000 \qquad
\textbf{(E)}\ 1000\sqrt2$

Did you mean to put 2000 - 2x below the square?


limit
floor function
calculus
calculus computations

calculate:
$\lim_{n \rightarrow \infty} n^2 \left( \sqrt{\dfrac{n+1}{n+2}} - \sqrt{\dfrac{n+2}{n+3} \right)}$

and

$\lim_{n \rightarrow \infty} f(\sqrt{n^2+n+1})$, where $f(x)=x- \lfloor x \rfloor$

i'm interested in how you do it

for the first one wouldnt you simplify and then use l'hopitals until it came out right?

because $n<\sqrt{n^2+n+1}<n+1$
so  $\lim_{n \rightarrow \infty} f(\sqrt{n^2+n+1}) =\lim_{n \rightarrow \infty} \sqrt{n^2+n+1}) -n=\lim_{n \rightarrow \infty} \frac{n+1}{\sqrt{n^2+n+1})+n}=\frac{1}{2}$


trigonometry
geometry
geometric transformation
rotation
complex numbers
geometry proposed

Erect outwardly isosceles triangles with vertices $P$ ,$Q$ , $R$
and vertex angles $120$  on the sides $AC$, $BC$, $AB$ of a triangle .Prove that the triangle $PQR$ is regular .

Denote PB=PC=x, QA=QC=y, AR=BR=z. Then $|\angle RBP|=\frac{\pi}6+\beta$, so
$|PR|^{2}=x^{2}+z^{2}-2xz\cos(\frac{\pi}6+\beta)= x^{2}+z^{2}-xz\cos\beta+\sqrt3xz\sin\beta=$
$=\frac{x^{2}}2+\frac{z^{2}}2+\frac{x^{2}+z^{2}-2xz\cos\beta}2+2\sqrt3S =\frac{x^{2}}2+\frac{y^{2}}2+\frac{z^{2}}2+2\sqrt3S$.
Since it's symetrical, the triangle PQR is equilateral.

Just a little remark: if we erect the triangles inwards, the result is the same, we just have to swap several + and - signs in the reasoning.

I think that the problem is wrong, but I'm not too sure. I used complex numbers and got that the affixes of $a,b,c$ are linearly dependent if $PQR$ is gonna be regular. That contradicts the fact that $ABC$ can be [i]any[/i] triangle.

But if the problem is true (and it is :D), then by finding the mistake I made, I would provide a solution.

I have a [b][u]synthetic[/u][/b] solution:
Let $X$ be a point in exterior of $ABC$ s.t. $\triangle ABX$ is equilateral.
$APR \sim ACX$ and $BQR \sim BCX$, so
$RP=\frac{CX\cdot AR}{AX}=\frac{CX\cdot BR}{BX}=RQ$.

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=theorem+napoleon&t=18628
this is very famous :wink:

I found my mistake.  Yay :D

Let $a,b,c$ be the complex numbers associated to $A,B,C$.

Then, we have that $p-c = \frac{\epsilon}{\sqrt 3}(b-c)$, where $\epsilon = \exp \left( \frac{i \pi}6 \right) = \frac{\sqrt 3}2+\frac12 i$. Thus, $p = \frac{b \epsilon+c \overline{\epsilon}}{\sqrt 3}$.

Similarly, $q = \frac{c \epsilon+a \overline{\epsilon}}{\sqrt 3}$ and $r = \frac{a \epsilon+b \overline{\epsilon}}{\sqrt 3}$.

We have that $q-p = \frac{(c-b) \epsilon+(a-c) \overline{\epsilon}}{\sqrt 3}$ and that $r-q = \frac{(a-c) \epsilon+(b-a) \overline{\epsilon}}{\sqrt 3}$.

[b]$\triangle PQR$ is equilateral iff $\epsilon^{4}(q-p) = (r-q)$[/b] iff $(c-b) \epsilon^{5}+(a-c) \epsilon^{3}= (a-c) \epsilon+(b-a) \overline{\epsilon}$ iff $\epsilon^{4}(q-p) = (r-q)$ iff $(c-a) \epsilon^{5}+(a-b) \epsilon^{5}+(a-c) \epsilon^{3}= (a-c) \epsilon+(b-a) \overline{\epsilon}$ iff $(c-a) \left( \epsilon^{5}+\epsilon-\epsilon^{3}\right) = (b-a) \left( \epsilon^{5}+\overline{\epsilon}\right)$ iff $0 = 0$, which is obviously true.

[quote="j555"]I have a [b][u]synthetic[/u][/b] solution:
Let $X$ be a point in exterior of $ABC$ s.t. $\triangle ABC$ is equilateral.
$APR \sim ACX$ and $BQR \sim BCX$, so
$RP=\frac{CX\cdot AR}{AX}=\frac{CX\cdot BR}{BX}=RQ$.[/quote]
I don't understand this ! :roll:
please clarify...

I have proved that $RP=RQ$, and analogically $RP=PQ$, so $\triangle PQR$ is equilateral. What exactly don't you understand?

the deffinition of the point X
you say :
let's take X s.t. ABC is equilateral
this is meaningless

My solution is like perfect radio's one but I found a solution in Englel's book  with tranformations but i don't understand it .So please if anyone want to explain it .
Here it is $P_{120}\circ\ Q_{120}\circ\ R_{120}=I$  since it is translation with fixed point $A$ (WHY??)
Hence we have $P_{120}\circ\ Q_{120}=R_{-120}$
Now construct a regular triangle with base $PQ$ and vertex $R'$
Then $P_{120}\circ\ Q_{120}=p\circ\ q\circ\ q\circ r=p\circ\ r=R'_{-120}$  (WHY??)

and the final is trivial .

Let's suppose that $P$ is opposite to $A$, $Q$ to $B$ and $R$ to $C$.

Also, let $T$ be the $120^\circ$ rotation around some point $O$ and $p = \overrightarrow{OP}$ etc.

The $120^\circ$ rotation about $Q$ (the one that sends $A$ into $C$) is given by $x \mapsto T(x-q)+q$.
The $120^\circ$ rotation about $P$ is given by $x \mapsto T(x-p)+p$.
The $120^\circ$ rotation about $R$ is given by $x \mapsto T(x-r)+r$.

We have that
$\mathcal R_{R}\circ \mathcal R_{P}\circ \mathcal R_{Q}(x) = \mathcal R_{R}\circ \mathcal R_{P}\circ \left( T(x-q)+q \right)$
$= \mathcal R_{R}\circ \left( T^{2}(x-q)+T(q-p)+p \right)$
$= T^{3}(x-q)+T^{2}(q-p)+T(p-r)+r$
$= (x-q)+T^{2}(q-p)+T(p-r)+r$
(since $T^{3}$ is the $360^\circ$ rotation about $O$, i.e. it is the identity)
$= x+T^{2}(q-p)+T(p-r)+(r-q)$.

But $\mathcal R_{R}\circ \mathcal R_{P}\circ \mathcal R_{Q}$ has a fixed point (namely $A$), so $a+T^{2}(q-p)+T(p-r)+(r-q) = a$, i.e. \[T^{2}(q-p)+T(p-r)+(r-q) = 0 .\] Thus, $\mathcal R_{R}\circ \mathcal R_{P}\circ \mathcal R_{Q}$ is exactly the transformation $x \mapsto x$ (the identity).

Now, if we take $Q^\prime$ to be a point such that $\triangle P R Q^\prime$ is equilateral (there are two points actually, but we choose the one whose angle $R Q^\prime P$ has the same orientation as $\angle AQC$), then $\mathcal R_{R}\circ \mathcal R_{P}\left( Q^\prime \right) = Q^\prime$, so $\mathcal R_{Q}^{-1}\left( Q^\prime \right) = Q^\prime$, i.e. $Q^\prime$ is a fixed point of $\mathcal R_{Q}^{-1}$. But $Q$ is the only fixed point of $\mathcal R_{Q}^{-1}$, so $Q \equiv Q^\prime$  :lol:

[quote="silouan"]Thank you very much . It is so nice..... :)
Another question is : All the problems which can be solved with tranformaions can be also solved by complex numbers ??[/quote]
That is an interesting question !I think it is true because all geometric  tranformations(translate,rotate,symmetry...) can be rewritten by complex number we can translate all solution with tranformations to solution of complex(only I think that) but
There is another question :[b]All problems of plane geometry can be solved by tranformations ?[/b]


integration
rotation
calculus
geometry
3D geometry
calculus computations

suppose i want to compute the volume by revolving $ y\equal{}x$ for $ 0\le x\le1$ about the line $ y\equal{}2.$

is it $ \pi\int_0^1(2\minus{}x)^2\,dx$?

The volume from revolving that [i]line[/i] is zero. Please clarify what region you intended to rotate: below or above the line $ y\equal{}x$, and where's the other edge?

ahh, forget this, i got it, i just invented that problem, nevermind.

It's OK, but you could get it (with no integration) as the volume of a truncated cone : $ {{7\pi}\over 3}$.

I also thought that Null meant the area above $ y\equal{}x$ and below $ y\equal{}2$ which when rotated about $ y\equal{}2$ gives a volume of$ \frac{7\pi}{3}$ as Immanuel Bonfils has pointed out.


This was Q1 STEP III 2008

$ \displaystyle \{\begin{aligned} a + b & = 1 \\
ax + by & = \frac {1}{3} \\
ax^2 + by^2 & = \frac {1}{5} \\
ax^3 + by^3 & = \frac {1}{7} \end{aligned} $

Solve for $ a, b, x, y$

Very similar is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=459558#459558]AIME 1990 #15[/url].  Searching for "characteristic equation" should get you results that explain the idea.


When I checked my profile today, I did not see that my entry form or permissions were logged. I sent them in on the day that the solutions for Round 1 were due. Should I send them again?

No.  Please wait until after the Round 1 scores have been posted.  If we still do not list the forms as having been received, then please resubmit them during Round 2.


Ring Theory
superior algebra
superior algebra unsolved

Could you please help me with this:

Let R be a principal ideal ring. Show that if

\[ I_1  \subseteq I_2  \subseteq ...I_k  \subseteq I_{k + 1}  \subseteq ...\]

is an ascending sequence of ideals in R, there is an $ n_0$ such that $ I_n=I{}_n{}_0$ for every

 \[ n \geqslant n_0\]

Consider the union of all the ideals involved in the ascending sequence.

1. Is it an ideal?
2. If it is, then it is principal. What does this imply?

Thanks a lot for the guidelines. I think I can do the question now.

[quote="Soarer"]Consider the union of all the ideals involved in the ascending sequence.

1. Is it an ideal?[/quote]

Yes, it is:

1. 0 is in $ I_1$ , hence in the union (call it U), as well.
2. Suppose a and b are in U. Then there are $ I_k , I_m$ such that a is in $ I_k$ and b is in $ I_m$. Then either
\[ I_k \subseteq I_m\]
or
\[ I_m \subseteq I_k\]
(Say the latter). So both a and b are in $ I_k$ , and so is their difference.
3. Let a be in U. Then a is in some $ I_k$ . Then for any r in R, we have ar is in $ I_k$ , and hence in U.

[quote="Soarer"]
2. If it is, then it is principal. What does this imply?[/quote]

A principal ideal is an ideal generated by one element (say, a). Now, 1 is in R, a in U, so ax1=a is in U. Hence a is in some $ I_k \subseteq I_n$ (hence, for all n>k, a is in $ I_n$ ) So finally we have $ aR \subseteq I_k \subseteq U \equal{} aR$ , which concludes our proof.

Thank you again!


Prove that equation $ x^3\plus{}3y^3\equal{}5$ has infinitely rational solutions

[quote="I love math very much"]Prove that equation $ x^3 \plus{} 3y^3 \equal{} 5$ has infinitely rational solutions[/quote]
To mod: Pls lock this topic because that problem is on Mathematical and Youth last month (The Maths magazine in Vietnam)


algebra polynomial algorithm Polynomials

On Christmas Eve, 1983, Dean Jixon, the famous seer who had made startling predictions of the events of the preceding year that the volcanic and seismic activities of $1980$ and $1981$ were connected with mathematics. The diminishing of this geological activity depended upon the existence of an elementary proof of the irreducibility of the polynomial \[P(x)=x^{1981}+x^{1980}+12x^{2}+24x+1983.\] Is there such a proof?

[quote="Peter"]Is there such a proof?[/quote]What is meant by this? How is 'elementary proof' defined? There are simple algorithms to find this out in a finite amount of time...  :maybe:

Maple did the algorithm for me in 2 minutes: yes it is irreducible (if that even matters).

I suspect the answer is "yes" then. I've tried prodding this question with Eisenstein's criterion without much luck.

My memory is vague... didn't we have a theorem saying $ x^n \plus{} x^{n \minus{} 1} \plus{} 1$ is irreducible over any field, given the right circumstances? This could solve the problem if applicable here (using the idea that a factorisation over $ \mathbb{Z}$ implies a factorisation over the finite field $ GF(2)$)...


linear algebra
matrix
linear algebra unsolved

Let $ A\equal{}\begin{pmatrix} \minus{}3&1&\minus{}1\\ \minus{}7&5&\minus{}1\\ \minus{}6&6&\minus{}2\end{pmatrix}$. Find the eigenvalues of $ A^4$.

From the first and second columns we see that -2 is an eigenvalue of A. Then, from $ tr(A) \equal{} 0$ and $ det(A) \equal{} 16$ we get the other two eigenvalues of A: 4 and -2. Therefore, the eigenvalues of $ A^4$ are $ (\minus{}2)^4 \equal{} 16$, $ (4)^4 \equal{} 256$, and $ (\minus{}2)^4 \equal{} 16$.

[quote="Carcul"]From the first and second columns we see that -2 is an eigenvalue of A.[/quote]
can you make it clearer?

Notice that the first two entries of the first two rows add up to -2, while the same entries in the third row sum to 0: this means that (1,1,0)' is an eigenvector of A with associated eigenvalue -2.


induction
number theory proposed
number theory

Let $ a$ and $ b$ be distinct positive rational numbers such that there exist infinitely many positive integers $ n$ for which $ a^n \minus{} b^n$ is a positive integer. Prove that $ a$ and $ b$ are both positive integers.

The first we have lemma :
If $ p|x \minus{} y$ then
$ v_p(x^n \minus{} y^n) \equal{} v_{p}(n) \plus{} v_p(x \minus{} y)$
 Prove this lemma use induction on n. It is an  well know lemma.
Now consider your problem .
Suppose that $ a \equal{} \frac {x}{m},b \equal{} \frac {y}{m}$
Consider $ A \equal{} \{n\in N |a^n \minus{} b^n\in Z\}$
So for all $ n\in A$ then
$ m^n|x^n \minus{} y^n$
Suppose $ a,b$ is not integer so $ m \not | x,y$
We only consider when $ \gcd(x,y) \equal{} 1$
If $ gcd(x,y) \equal{} d$
$ (\frac {m}{\gcd(m,d)})^n|(\frac {x}{d})^n \plus{} (\frac {y}{d})^n$
It is the same above problem .
So we only need to consider $ \gcd(x,y) \equal{} 1$
Call $ p$ is a prime divisor of $ m$ then from Fermat's theorem
$ x^{p \minus{} 1}\equiv y^{p \minus{} 1}\equiv 1 (\mod p)$
So $ p^n|x^{(p \minus{} 1)n} \minus{} y^{(p \minus{} 1)n}$
Apply the lemma we have
 $ v_p(x^{(p \minus{} 1)n} \minus{} y^{(p \minus{} 1)n}) \equal{} v_{p}(n) \plus{} v_{p}(x^{p \minus{} 1} \minus{} y^{p \minus{} 1})$
So for all $ n\in A$  we have
$ n\leq v_p(n) \plus{} v_p(x^{p \minus{} 1} \minus{} y^{p \minus{} 1})$
It is contradiction when n is large enough .
So $ a,b\in Z$


Show that the function f(x)=(x-a)^2+(x-b)^2+x takes on the value (a+b)/2 for some value of x.

This problem hasn`t got much to do with calculus . Rewrite $f(x)=2x^2-x(2a+2b-1)+a^2+b^2$ you want $f(x)=\frac{a+b}{2}$so simply write  $2x^2-x(2a+2b-1)+a^2+b^2-\frac{a+b}{2}=0$ it remains to prove that $(2a+2b-1)^2-8(a^2+b^2-\frac{a+b}{2})\geq 0$ after a few computation we get $8(a+b+ab)+1\geq 4(a^2+b^2)$ for $a=b=-1$ it is not true so..... you may have a typo.

yeah.there might be something here. it was supposed to use intermediate value theorem.I'll find out.


abstract algebra
algebra
function
domain
superior algebra
superior algebra unsolved

Let R be a domain I, J are ideals in R. Prove that I and J are isomorphic as R-modules iff aI=bJ for some non zero a, b in R.

if $aI = bJ$, for every $x \in I$ there's exactly one $f(x) \in J$ s.t. $a x = b f(x)$. it can be easily checked, that $f$ yields an isomorphism $I \to J$ of $R$-modules. the converse seemed to be a very interesting result ... until I found this too-simple proof ;)

let $f : I \to J$ be an isomorphism of $R$-modules. if $I=0$, the assertion is trivial, so let's pick $0 \neq x \in I$. then for all $y \in I$, we have $x f(y) = f(x y) = f(x) y$, so that $x J = x f(I) = f(x) I$ ...