be kasany ke dar doreye tabestany medale tala kasb kardand tbrik migooyam.
asamy darandegan medale ta be sharh zir ast: (be tartib horoofe alefba)
[size=142]
1. Kasra Ahmady Kamalabadi
2. Milad Bakhsizade
3. Mohammadreza Takapouyi
4. Shayan Dashmiz
5. Mahyar Sefidgaran
6. Mohammadhossein Shafinia
7. Saman Farhat
8. Hesam Firoozi
9. Pouria Kabir
10. Mohammad Mansouri
11. Sima Mehri
12. Saeed  Hadikhanlou[/size]

rasty kasy midoone 15% taraz baraye noghre va tala hanooz hast ya bardashte shod?    :)

Salam,

tabrik migam be hameye unai ke emsal tala gereftan  :winner_first: va albate be unai ke noghre ya bronz gereftan ham!

dar zemn fek konam 2 nafaro ja endakhti  :D

13. Nima Ahmadi Pour Anari
14. Arman Fazeli

[quote="A_SAFAKAR"]
6. Mohammadhossein  Shafinia
7. Saman Farhat
[/quote]

Congrats!!!! :showoff:  :thumbup:

akhe mage har ki harkie ke harki khast ye sal zoodtar biyad.mage na agha Nima?
khob ishoon nokhbe hast ghabool vali nemizaran ke nokhbe ha zoodtar beran daneshgah.akhe inja ke majarestan nis ke.motavajjehe manzooram shodin?

To maryam rashin:
Ta oonjai ke man etela daram nima sale dige miad daneshgah.(1 sal zudtar)
ke kare kamelan maghulie.Guya yeki dige az bachehaye sal dovomi ham dare hamin kara mikone.
1sal zudtar vared daneshga shodan gheire marsum hast vali gheire momken na.masalan aghaye Iman Eftekhari ham 1sal zudtar varede daneshga shodan.

manam be hameye bachehayi ke tala gereftan tabrik migam va hamchenin be oonayi ke noghre ya bronze shodan!!! :)

Man in 12 nafare ki shumo onhoro tabrik mekuned man ham onhoro tabrik mekunam va ba onho muvafaqiyat meguyam.  :) GOOD LUCK :!:


:huh: i need help with multiplying negitive numbers with a positive number. the problem is 6,869 x -7. plz help me!!!!!!!!![/u]

A $positive\times positive=positive$.
A $negative\times negative=positive$.
A $positive\times negative=negative$.

So $6,869\times(-7)=-(6,869\times7)$.


Are there any good summer programs for high school or college astronomy?

one of my students did this one and had a great experience:
http://www.summerscience.org/home/index.php


The incircle of triangle ABC  touches BC CA and AB at D,E,F respectively.

P is any point inside triangle ABC.The line PA meets the incircle in two points.Of these points let X be the one closer to A.Similarly Y and Z are points where PB, PC respectively meet the incircle.

Prove DX EY FZ concurrent

http://www.mathlinks.ro/Forum/viewtopic.php?t=249

  darij


Find all surjective function $f : N \Rightarrow N$ satisfying the condition
for $m$  $divides$  $n$ $\Rightarrow$ $f(m)$   $divides$  $f(n)$; m,n is N

*Sigh* Already posted.. (even with an "iff" condition) :
http://www.mathlinks.ro/Forum/viewtopic.php?t=4937

 :roll:


There exist positive integers $A,B$ and $C$, with no common factor greater than $1$, such that \[A \log_{200} 5 + B \log_{200} 2 = C. \] What is $A+B+C$?

$\textbf{(A)}\ 6 \qquad
\textbf{(B)}\ 7\qquad
\textbf{(C)}\ 8 \qquad
\textbf{(D)}\ 9 \qquad
\textbf{(E)}\ 10$

[hide]This is equal to $\log_200{5^A \cdot 2^B}=C$, since we want it that $A,B$ and $C$ don't have any common factors and we know that $200=2^3 \cdot 5^2$, we can say that $C=1$, so $1+2+3=\boxed{6}$[/hide]


calculus
integration
function
inequalities
real analysis
real analysis unsolved

Let $f:[0,\infty)\to [0,\infty)$ a continuous function such that $f(f(x))=x^2$ for all $x\geq0$.Prove that:

    a) $\int_{0}^{1}f(x)dx\leq\frac{1}{2}$
    b) there is $c\in (0,1)$ such that $f(c)=\frac{1}{2}$  ;)

$x^2 < f(x) < x < 1 < y < f(y) < y^2 ,\forall x\in ]0,1[ \   \forall y\in ]1,\infty[ $
 :cool:  Could you say why ?   :cool:

A hint:

               prove that $f(x)\leq x$ for any $x\in [0,1]$ ;)

I really don't think we need hints for this one, Cezar. Don't super-estimate the problem. ;)

i wasn't estimating this problem.In fact,I consider it quite easy. ;)

Additional question: is it true that $\int_{0}^{1}{f(x)dx}\leq\sqrt{2}-1$?

That's exactly why some of us haven't responded. The more interesting question: how much better than $\frac13<\int_0^1 f(x)\, dx<\frac12$ can we do?

well,I haven't thought yet.did you prove it? The only idea that comes into my mind is to use the relation:
  $\int_{0}^{1} (2x+1)f(x) dx=1$ ;)

By Chebyshef's inequality we only get $\int_{0}^{1}f(x)dx\leq\frac{1}{2}$ ;) and by Cauchy we get $\int_{0}^{1} f^2(x)dx>\frac{13}{3}$ if I haven't made a mistake in my calculations. ;)

From $f(f(x))=x^2$ $f$ is injective on $[0,+\infty[$, the function is continous
so it is monotonous.

Take the limit in
$x^2 < f(x) < x < 1 \forall x\in ]0,1[ $
with continuity

$f(0)=0$, $f(1)= 1$,  $f$ is monotonous it is increasing
there exist $c \in ]0,1[$ such that $f(c)=1/2$

In fact for any $r\in [0,1]$ there exist $c_{r}$ such that $f(c_{r})=r$

moubi,that's exactly one of the solution I gave. ;)


How can i write  under  a number  (example 111...1)  the comand  \underbrace  with a  note  (n digits)?
Thank you!

Subscript _ puts text under the brace
[code]\underbrace{111\dots 1}_{n \text{ digits}}[/code]\[ \underbrace{111\dots 1}_{n \text{ digits}} \]


linear algebra
matrix
invariant
linear algebra unsolved

Let $A,B$ be two matrices over an infinite field $F$, such that they are similar(not necessarily over $F$)-i mean there exists the matrix $U$(not neccesary over $F$ ) such that $UAU^{-1}=B$. Prove that there exists $V$ over $F$ such that $VAV^{-1}=B$.

nice question, wish i could do it

is infinity necessary?
i was thinking of jordan forms but that will not help i fear

$UA=BU$ i was thinking of applying all automorphisms of the bigger field $K$ that fix $F$ pointwise to this equation
and then adding them up

problem : we do not know whether there are finitely many
               we do not know whether the sum will be invertible

we do know that the sum of all those $U^theta$ will be a matrix in the field $F$

one can prove this one with the theory of invariant factor. I don't know if there is a very elementary proof (except for the case of $\mathbb{R}$ and $\mathbb{C}$). Anyway, I think someone has asked the question some times ago but no very elementary proof were found  :)

The problem is very difficult if the field is not infinite, but for infinite fields it is not that difficult: just look at the system $AU=BV$ as a linear system in the coefficients of $V$ and apply Rouche-Fontenay theorem.

Let $ A,B \in M_n{\mathbb{(C)}}$    such that the exists the matrix  $ U \in GL_n{\mathbb{(C)}}$ such that $ UAU^{\minus{}1}\equal{}B$. Prove or disprove that there exists $ V \in GL_n{\mathbb{(R)}}$   such that $ VAV^{\minus{}1}\equal{}B$
[b]If It is correct show for me ! Thanks [/b]!


Um ok so in general, how many points of contact are there between L lines and C circles. I don't mean LL and CC points but just when you have a line through a circle.

Is it just 2LC?

It isn't. Let's call x the number of lines and y the number of circles. Where the circles intersect the circles, each one intersects the other 2 times, so it is y(y-1). When lines intersect lines, each one intersects the other once, so it is x(x-1)/2. When lines intersect circles, you get two points per line/circle, so it is what you mentioned before or 2xy. 2xy+y^2-y+x^2/2-x/2. If I were you, I'd not memorize the formula, but figure out how to do it every time.

Edit: oh wait you were just talking about their intersection. So yes, you're right.

I think that is what mathemagician was asking: how many LC points.

EDIT: just saw your "EDIT". Ok that's right


limit
inequalities proposed
inequalities

Let $ a$, $ b$, $ c$ be the three sides of a triangle. Determine all possible values of \[ \frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}\]

[quote="April"]Let $ a$, $ b$, $ c$ be the three sides of a triangle. Determine all possible values of
\[ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}
\]
[/quote]
we have: $ a^2\plus{}b^2\plus{}c^2 \geq ab\plus{}bc\plus{}ca$
and $ |a\minus{}b| <c$ then $ a^2\plus{}b^2\minus{}c^2 <2ab$
Thus, $ a^2\plus{}b^2\plus{}c^2<2(ab\plus{}bc\plus{}ca)$
$ 1\leq F\equal{}\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}<2$

Is the problem famous?

I myself have posted it before, but I can't find my post.

I remember some Vietnamese answered it.

[quote="broniran"][b]Hong Quy[/b], I don't think this is complete solution. In my opinion, we are asked to find the min and the maximum and you haven't found the maximum. You only proved that it is <2.[/quote]

I'm not sure, but we can make $ \frac{\sum a^{2}}{\sum ab}$ arbitrarily close to $ 2$ by letting $ a\equal{}c$, $ b\rightarrow 0^{\plus{}}$. :D

As a matter of fact, $ 2$ can be attained on a degenerate triangle where $ a \equal{} c, b \equal{} 0$ or any permutation.

If we are not to consider a degenerate triangle, we fix $ a \equal{} c$. Thus, $ b$ can be brought arbitrarily close to $ 0$.

Now, $ \lim_{b \to 0^ \plus{} } \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \equal{} \lim_{b \to 0^ \plus{} } \frac {2a^2 \plus{} b^2}{a^2 \plus{} 2ab} \equal{} 2.$

Hence $ \frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}$ can get arbitrarily close to $ 2$.

or see that $ \sum a^2 \equal{} 2(\sum bccosA)$ so just take a triangle with $ \angle B \equal{} \angle C$ tend to zero and $ \angle A$ to get $ \pi$..

I have just found it. :lol:
[url]http://www.mathlinks.ro/viewtopic.php?t=271586[/url]

$ab \plus{} bc \plus{} ca\le a^2 \plus{} b^2 \plus{} c^2=2( bccosA+cacosB+abcosC)<2(ab\plus{}bc\plus{}ca)$.


How many permutations are there of the group of letters A,A,G,E,E,E,M, supposing spelling does not matter?


Suppose that $ a_{1},\ldots,a_{n}$ are nonnegative real numbers such that $ a_{1}\geq\cdots\geq a_{n}.$ Prove the inequality
\[ \prod_{i}\frac{a_{i}+a_{i+1}+a_{i+2}}{3}-\prod_{i}\frac{a_{i}+a_{i+1}}{2}\geq \frac{a_{3}\cdots a_{n}}{24}\sum_{i}(a_{i}-a_{i+2})^{2}\]
where the products and the sum are cyclic.

[color=darkred][i]edit.[/i] Sharpened the right hand side by substituting $ a_{n}^{n-2}$ with $ a_{3}\cdots a_{n}.$[/color]

Notice that the inequality is very sharp even for small values of $ n,$ something that the original Mongolian inequality isn't. Take $ n=3,$ and the Mongolian inequality is ordinary AM--GM, while the generalization becomes
\[ \frac{1}{432}(a+b-2c)(11(a-b)^{2}+5(b-c)^{2}+5(c-a)^{2})\geq 0.\]
Of course, just as the original, it becomes sharper as $ n$ increases.

Another question would be: which is the smallest value of $ \lambda$ we can substitute $ 24$ on the right hand side so that the inequality remains valid? I believe, although had not proved, that $ 24$ is indeed the optimal value.

[quote="bodan"]Suppose that $ a_{1},\ldots,a_{n}$ are nonnegative real numbers such that $ a_{1}\geq\cdots\geq a_{n}.$ Prove the inequality
\[ \prod_{i}\frac{a_{i}+a_{i+1}+a_{i+2}}{3}-\prod_{i}\frac{a_{i}+a_{i+1}}{2}\geq \frac{a_{3}\cdots a_{n}}{24}\sum_{i}(a_{i}-a_{i+2})^{2}\]
where the products and the sum are cyclic.
[/quote]

[color=darkred][i]comment.[/i] The right hand side is not clear when it comes to $ n=4.$ In this case we do [b]not[/b] count $ ac$ and $ bd$ twice, so the sum is
\[ \frac{cd((a-c)^{2}+(b-d)^{2})}{24}. \]
In other cases the sum is exactly what it says it is.[/color]


Let $a>b>c>0$ be positive integers. Prove the existence of a triple $(x,y,z)\ne (0,0,0)$ of integers such that $ax+by+cz=0$ and $\max(|x|,|y|,|z|)<\frac{2}{\sqrt{3}}\sqrt{a}$. Is the constant $\frac{2}{\sqrt{3}}$ optimal?

I think there's typo in the problem since statement is trivial because we can take $(x,y,z)=(0;c;-b)$.

I think it is not correct. For example c big, $b=c^{2},a=c^{4}$. Then if $|x|,|y|,|z|<c$ $|ax+by+cz|>|x|(c^{4}-c^{3}-c^{2})>0$, therefore if $ax+by+cz=0$ x=0, by analogue y=0 and z=0.

I think I will edit this one until the end of the world... Sorry again! Anyway, Rust, I don't get your point: how do you know that $|by|\geq |x| c^{3}$ or something like that? I think you supposed an order between $|x|,|y|,|z|$, which is not asked by the problem.

Ok, but could you please post your solution? Thank you for posting it in a comprehensible way.

I give my solution with $C=\frac{12}{5}$ and I don't know solution with $C<2$.


How do we apply epsilon delta criterion to show f(x) = x. and similar functions.
it is easy for function with exponents like x*2 and x^3 because we can get them in the form of
\[ \mid f(x) \minus{} f(x_0)\mid \leq C\mid x \minus{} x_0\mid\] and then define $ \delta$ to be equal to min$ \{1, \frac{\epsilon}{C}\}$ where  C is obtain by restricting x,
$ x_0 \minus{} 1 < x < x_0 \plus{}1$.

Why is it that for $ \delta>0$,  $ \delta \leq 1$

Similar functions, meaning what?  $ f(x)\equal{}x$ is probably the easiest function you can do; If $ \left|f(x)\minus{}f(a)\right|<\epsilon$, then $ \left|x\minus{}a\right|<\delta$, so choose $ \delta\equal{}\epsilon$.


A work crew of 3 people requires 3 weeks and 2 days to do a job.  How long would it take a work crew of 4 people to do the same job if each person of both crews works at the sam rate as each of the others?  Not: each week contains six work days.

[hide]3 * [(6 *3) + 2]= 60 man-powered days

60 man-powered days /  4 people = 15 days.

Thus the answer is 15 days or 2 weeks and 3 days.

P.S.  "man-powered days" sounds pretty cool..[/hide]

[hide]3 * 20 = 4 * ?
? = 15 days, or 2 weeks, 3 days. [/hide]


[quote="cancer"]In convex quadrilateral $ABCD$, $\angle DAB=\angle ABC, AD=DC$. $E$ is the midpoint of $BC$. $DE$ intersects $AB$ at $F$. Show that $FC^{2}=FA\cdot FB$[/quote]
Let $\angle DAC=\angle DCA=\alpha,\angle CAB=\beta$. Then, we have
$\angle ABC=\alpha+\beta,\angle BCA=\pi-\alpha-2\beta$
By Law of sine
\[\frac{EB}{EF}=\frac{\sin\angle EFB}{\sin\angle EBF}=\frac{\sin\angle DFA}{\sin\angle FAD}\;\;\;\;(\angle BAD=\angle ABC=\pi-\angle CBF) \]

\[=\frac{DA}{DF}=\frac{DC}{DF}\]
Therefore $\frac{EC}{EF}=\frac{EB}{EF}=\frac{DC}{DF}$
On the other hand
\[\frac{\sin\angle CFE}{\sin\angle ECF}=\frac{EC}{EF}\]
and then
\[=\frac{DC}{DF}=\frac{\sin\angle DFC}{\sin\angle DCF}\]
Consequently $\sin\angle BCF=\sin\angle DCF$ then $\angle BCF=\pi-\angle DCB-\angle BCF=\beta\Rightarrow FC$ tangent to circle $(ABC)$ thus $FC^{2}=FA\cdot FB$.

The line paralell to $ BC$ meets $ AB$ at $ M$. $ MC$ meets $ DF$ at $ K$
We have $ \angle DAM=\angle CBA=\angle DMA$ so $ DM=DA=DC$
thus $ \angle DCM=\angle DMC=\angle MCB$ then $ CM$ is the bisector of $ \angle BCD$
otherwise $ KE/KD=CE/DM=BE/DM=EF/DF$ so $ CF$ is the external bisector of $ \angle BCD$
then $ \angle MCF=90$
we have $ \angle ACF=90+\angle ACM=90+\angle ADM/2=\angle DMB=\angle CBF$
So triangle $ FBC$ and $ FCA$ are similar
then $ FC^{2}=FB*FA$ $ q.e.d$


[b]What is Gmail?[/b]

As part of Google's mission to organize the world's information and make it
universally accessible and useful, we're testing an email service called
Gmail.

Gmail is a free, search-based webmail service that includes 1,000
megabytes (1 gigabyte) of storage. The backbone of Gmail is a powerful
Google search engine that quickly recalls any message an account owner
has ever sent or received. That means there's no need to file messages in
order to find them again.

When Gmail displays an email, it automatically shows all the replies to that
email as well, so users can view a message in the context of a
conversation. There are no pop-ups or untargeted banner ads in Gmail,
which places relevant text ads and links to related web pages adjacent to
email messages.

[b]The Contest[/b]
I have a couple Gmail invitiations that are just aching to be given away.
But, of course, demand for invitations exceeds supply. I've decided to give
away (at least) one invitation right here, to the person who can guess,
figure, or divine the correct number. Here are the rules:

The contest begins when this post is finished, and ends at midnight
(exactly) EST, Wednesday, June 30th (The contest and the month of June
end at the same time, to avoid any confusion). EST is GMT-5:00 for
international members. After this time, entries [b]will not[/b] be accepted.

Each entry is to be an email with the subject line "[b]AoPS Gmail
Contest[/b]" (please use this subject line), emailed to
confuted@gmail.com. Only entries to this address will be accepted. Do not
PM me your entries, and do not post your entries in this thread, because
they will not be counted if you do. The body of the email should contain a
real number [i]between 0 and 1000[/i]. Non-integers are, of course,
allowed. For my ease, please don't put anything else in the email; just a
number.

When the contest ends, I will compute my number according to the
following rules:

[tab][b]*[/b]First, I will take the root-mean-square (RMS) of all the
numbers. That means I'll be taking the square root of the sum of the
squares of the numbers divided by the number of entries.
([url=http://mathworld.wolfram.com/Root-Mean-Square.html]details from mathworld[/url])
[tab][tab][b]-[/b]If the RMS>500, I will multiply it by 4/5 to get my number.
[tab][tab][b]-[/b]If the RMS<500, I will multiply it by 5/4 to get my number.
[tab][tab][b]-[/b]If the RMS=500, my number will be the smallest number submitted to the contest.

The prize (one gmail invitation) will go to the person with the number
closest to mine. I will compute this by taking the absolute value of (my
number)-(your number), so it doesn't matter if you "go over" my number.

One entry per member. Only enter if you do not have a gmail account
already, and you are interested in one. It goes on the honor system. Feel
free to use your friends, the internet, your textbooks and your calculator
to help you improve your guess.

The winner will be announced July 1.

[i]edit: I attempted to eliminate the scrolling problem some members were having with this post[/i]

[b]edit #2: theone853 has generously offered one of his invitations to the person taking second place in this contest. So keep the entries coming![/b]

[quote="JBL"]Just for the record, "between" means "exclusive," right?  So, don't anyone go guessing 0 or 1000.[/quote]
Indeed.

[quote="bleumoose"]And what are you going to do if two people send the same number?[/quote]
There's nothing special about that unless they are both equally close, in which case I suppose I'll try to get them both an invite.

[quote="JBL"]Don't worry, that will happen with probability 0, unless uncountably many people start guessing ;)[/quote]

Somehow I don't think that integers will be guessed with probability zero. ;)

[quote="Yasha"]Anyone noticing that the lines on this post don't wrap until way off the screen. I think it has to do with Confuted's original post, parts of which might have been copied from somewhere. It would be really nice if someone would fix this.[/quote]
I made it in Round Table and moved it here, and I copied and pasted the gmail description, but it doesn't seem that those things should affect it. It wraps fine for me...

[quote="zabelman"]How did you come up with the rules for this contest?[/quote]
I wanted to give away an invite in a contest, but I didn't want it to be strictly math because then the younger members here who aren't very experienced wouldn't stand a chance against the people who are answering questions in the advanced forum. So I asked Simon to help me come up with something that could be approached like that - everyone has roughly equal chances, but people can perhaps improve their chances if they think about it a lot. Or they might think about it a lot and not improve their chances at all. Mathematically, it's very tough to predict what my number will be, especially with all the random guesses.

[quote="bleumoose"]Actually, think about it- 1000 numbers is not a lot, and there are quite a few "special" numbers that a lot of people might chose as "lucky" (666 comes to mind... *evil grin*)... so the same "lucky" numbers may be chosen by two people, in which case... you see my point? Or, someone may try and mathematically choose a number, and two people may do the same math to get the number, and therefore would have the same number and...[/quote]
There are an infinite number of choices; guesses need not be integers. Regardless, if you could only guess integers, there would be 998 choices, not 1000.

this page doesn't wrap for me either... strange.

[quote="ComplexZeta"]Actually there would be 999. ;)[/quote]
For purposes of making my figure of 998 correct, 43 is no longer considered a number.  :twisted:

People wishing to enter have 28 minutes left to do so.

Just entered now....with 5 minutes to spare!!!  :D

The AoPS Gmail Contest is over, and I have the results.

There were 19 entries, with an RMS of approximately 461.927.
Multiplying this number by 5/4 to get my number gave 577.409 as the
"number to get." JMI from AoPS had the closest entry with 564.75, so
(s)he wins a gmail invitation. Failure from AoPS didn't fail this
time, because (s)he had the second closest entry at 610.527726 (thanks
for all those decimals, failure). Anyway, failure also wins a gmail
invitation, courtesy of theone853 from AoPS. Congratulations to the
winners, and "maybe next time" to the losers.

(A mod may now de-sticky this thread)

[quote="TripleM"]Can you post the guesses, just so we can see what they were like?[/quote]
150 :pi: , 502, 610.527726, 100, 441, 273, 716.128, 456, 564.75, 117, 43.346432181, 511, 521.0019194, 485, 419.72, 426, 17, 714, 207
If it's of interest, the later entries in that list were the ones that were recieved first, and 150 :pi: was the last entry.

Darn, it looks like I was third.  My number was the 521...

I didn't do much math either.  I just thought about it a little and thought that it would be close to 500.  So I took the power of phi closest to 500.

[quote="yif man12"]good thing i didn't enter. i didn't do any math and was just gonna guess 23. i would've embarrassed myself a ton.[/quote]
That would have changed the RMS, though, making your entry [i]somewhat[/i] less ridiculous ;)

[quote="starrynightskies"]what is the person who got 3rd's email? (confuted answer please)
i have a gmail invites too, and since im bored...
if confuted does not wish to give me the email, i will be very inuslted because he let theone, and i geve gmail to theone in the first place[/quote]
joml88 took third, and Matt Folz (not exactly sure who that is) took 4th.

yea, but whats his/her email? if you dont want to say out loud, pm me

my e mail is joml8@hotmail.com .... if that is what your asking

OH MY GOSH... Aops should be used as a time archive because you can see so many "old parts of history" on this website... like, remember when Gmail was still really new? Exactly, this thread gives us a great example... wow gives me so much nostalgia.

Hard to believe that 15 years ago Gmail had 1gb of storage... sounds really little nowadays but it probably was a LOT back then, much more than 15gb of storage is right now. :)


Determine all continous functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy the relation:
\[2f(x\plus{}y)\plus{}f(2x\minus{}y)\plus{}f(2y\minus{}x)\equal{}9f(x)\plus{}9f(y),\]
for all $x,y \in \mathbb{R}$.

[quote="moldovan"]Determine all continous functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ which satisfy the relation:

$ 2f(x \plus{} y) \plus{} f(2x \minus{} y) \plus{} f(2y \minus{} x) \equal{} 9f(x) \plus{} 9f(y), (\forall) x,y \in \mathbb{R}.$[/quote]

Let $ P(x,y)$ be the assertion $ 2f(x \plus{} y) \plus{} f(2x \minus{} y) \plus{} f(2y \minus{} x) \equal{} 9f(x) \plus{} 9f(y)$

$ P(0,0)$ $ \implies$ $ f(0) \equal{} 0$
$ P(x,x)$ $ \implies$ $ f(2x) \equal{} 8f(x)$
$ P(x,\minus{}x)$ $ \implies$ $ f(\minus{}x)\equal{}\minus{}f(x)$
$ P(x,2x)$ $ \implies$ $ f(3x) \equal{} 27f(x)$

It's very easy to check that $ f(x) \equal{} x^3$ is a solution.

Let then $ g(x)$ any solution. $ h(x) \equal{} g(x) \minus{} g(1)x^3$ is also a solution such that $ h(1) \equal{} h(0) \equal{} 0$
Then $ h(2) \equal{} 8h(1) \equal{} 0$ and $ h(3) \equal{} 27h(1) \equal{} 0$ and $ h(4) \equal{} 8h(2) \equal{} 0$ and so $ h( \minus{} 4) \equal{} h( \minus{} 3) \equal{} h( \minus{} 2) \equal{} h( \minus{} 1)$ $ \equal{} h(0) \equal{} h(1) \equal{} h(2) \equal{} h(3) \equal{} h(4) \equal{} 0$

$ P(\frac x2,1)$ may be written $ 2f(\frac x2 \plus{} 1) \plus{} f(x \minus{} 1) \plus{} f(2 \minus{} \frac x2) \equal{} 9f(\frac x2) \plus{} 9f(1)$
$ \iff$ $ 2f(x \plus{} 2) \plus{} 8f(x \minus{} 1) \plus{} f(4 \minus{} x) \equal{} 9f(x) \plus{} 72f(1)$
$ \iff$ $ f(x) \equal{} \frac 12(9f(x \minus{} 2) \minus{} 8f(x \minus{} 3) \plus{} f(x \minus{} 7) \plus{} 72f(1))$

Applying this to $ h(x)$, we get $ h(n) \equal{} 0$ $ \forall n\in\mathbb Z$ and so $ h(n2^p) \equal{} 0$ $ \forall n,p\in\mathbb Z$ and so $ h(x) \equal{} 0$ $ \forall x\in\mathbb R$ (continuity and considering that $ \{n2^p,n,p\in\mathbb Z\}$ is dense in $ \mathbb R$).

So $ g(x) \equal{} g(1)x^3$ and it's easy to check back that this indeed is a solution.

And the only solutions are $ \boxed{f(x) \equal{} ax^3}$


Asymptote
absolute value
complex analysis

How do I describe the mapping of $f(z)=e^{z}$ on the region $\{z=x+iy: \,\,x\geq 1,\,\, 0\leq y \leq \pi\}$?

Is there a nive geometric aspect?

I can give you a hint.  Geometrically the conformal mapping $e^{z}$ is simple to understand if you write it as $e^{x+iy}$.  Now, for example, if you fix $x$ and vary $y$ say between $0$ and $2 \pi$ you should get a circle right?  Also, if you fix $y$ and vary $x$ say between $0$ and $\infty$ you will get a ray emanating from the origin.  This should be enough information to solve the problem.  Good luck.

thanks a lot!

[b]fedja[/b]: What is that image code called. I mean how does it work?
Please give me some guidance.


2. ii) Find k such that the equations are consistent (that is, they can be solved):

x+2y+z=k
2x+y+4z=6
x-4y+5z=9

This was already solved in intermediate.
[quote="HTA"]hope that i didnt make some idiot mistakes
from $x-4y+5z=9$ we obtain $x=9+4y-5z$ ,
since $2x+y+4z =2(9+4y-5z)+y+4z=6$ , we get $z=\frac{4+3y}{2}$
thus $x+2y+z=9+4y-5z+2y+z=9+6y-2(4+3y)=1$
so we obtain $k=1$[/quote]


algebra
polynomial
irrational number
number theory
prime factorization
relatively prime
Rational Root Theorem

Can anyone name the therome that says that the square root of any natural number will be always a natural or an irrational number?

I don't know if there is one, though I believe I know how to prove it.

If you prime factor a number into $a^{p_a}b^{p_b}c^{p_c}...$, if $p_a,p_b,p_c,...$ are all even, then the when you take the square root, you get $a^{\frac{p_a}{2}}b^{\frac{p_b}{2}}c^{\frac{p_c}{2}}...$.  Since $p_a,p_b,p_c,...$ are all even, ${\frac{p_a}{2},\frac{p_b}{2},\frac{p_c}{2}}...$ are all integers.
If $p_a,p_b,p_c,...$ are not all even, then the when you take the square root, you get a prime factorization with $something^{something.5}$, so it is irrational..

That's not a proof though...

That ends with prooving that $x^{\sqrt{y}}$ is irrational for $\sqrt{y}\not\in \mathbb{N}$, which I don't know how to do.  I believe there is an easier way.  Do you want me to show you my way?

There's a stronger and more general theorem: the rational root theorem.

Here's what the rational root theorem says:

Consider the polynomial $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ in which all of the coefficients $a_k$ are integers and $a_n\ne0.$ If this polynomial has any rational root - any number $x=\frac pq,$ with $p$ and $q$ relatively prime - such that $P(x)=0,$ then $q$ divides $a_n$ and $p$ divides $a_0.$

Corollary to the rational root theorem: consider a [i]monic[/i] polynomial, which is to say a polynomial of the form $P(x)=x^n+a_{n-1}x^{n-1}+a_1x+a_0.$ If $x$ is a rational number such that $P(x)=0,$ then $x$ must be an integer.

Now consider the equation $x^n-m=0$ for a positive integer $m.$ The positive root of this is $\sqrt[n]{m}.$ By the corollary to the rational root theorem, this can only be rational if it is an integer.

If you want to prove this, focus on proving the most general case: the original rational root theorem. If you want to look it up, you should be able to find it in number theory textbooks (and it is surely a theorem of number theory.)

Something based off of what AoPS book teaches:
$x$ is some natural number

Suppose there exists some $\sqrt{x}$ such that it can be written as the quotient of two integers in[b] lowest terms[/b].

We then have $\sqrt{x} = \frac{p}{q} \Rightarrow x = \frac{p^2}{q^2} \Rightarrow p^2 = xq^2$

In order for this to be true and the $\frac{p}{q}$ rational, $p^2$ must be divisible by $x$

let $r$ be the quotient when $p^2$ is divided by 2

$xr^2=q^2$

Now $q$ must also be divisible by $x$

This means that our original assumption that there is some $p/q = \sqrt{x}$ was false, since $p/q$ was not in lowest terms.  This shows us that $\sqrt{x}$ cannot have a  root in the form $p/q$ in lowest terms when $q$ does not equal 1

When $q=1$ our case is changed...
We then have $\sqrt{x} = p \Rightarrow x = p^2$

Which is an entirely possible circumstance for all perfect squares.


geometry
3D geometry
sphere
prism
vector
tetrahedron
linear algebra

Hm, is it possible to have the Shoelace Theorem that applies to volume in 3D? This obviously won't work for spheres or any other circular-related prisms but it might be useful in working with the 3D plane.

The Shoelace Theorem is based off of the properties of the cross product.  Volume isn't related to the cross product; it's related to the [i]triple scalar product[/i], which isn't a binary operation.  And I don't know what you mean by "3D plane."

[quote="t0rajir0u"]And I don't know what you mean by "3D plane."[/quote]

Whoops, I meant 3D space. A three-dimensional plane would be quite interesting, indeed.

Just as the area of a triangle formed by vectors $ \vec{a}$ and $ \vec{b}$ is

$ \frac {1}{2}|\vec{a}\times\vec{b}|$

the volume of a tetrahedron formed by vectors $ \vec{a},$ $ \vec{b}$, and $ \vec{c}$ is

$ \frac {1}{6}\vec{a}\cdot (\vec{b}\times\vec{c})$

Also, if $ \vec{a} \equal{} (a_1,a_2,a_3),\vec{b} \equal{} (b_1,b_2,b_3),\vec{c} \equal{} (c_1,c_2,c_3)$, then $ \vec{a}\cdot (\vec{b}\times\vec{c})$ is the determinant of the matrix

$ a_1~~a_2~~a_3$

$ b_1~~b_2~~b_3$

$ c_1~~c_2~~c_3$

The determinant viewpoint is the one that generalizes; the only way to go higher with binary operations is to introduce the wedge product, and work with spaces of different dimensions.

In order to use these determinants to handle a complicated polytope, you need a dissection into simplices (with orientation). This can always be done without adding vertices, but you may need a lot of them. You also have to pay attention to orientation: $ ABCD$ is the negative of $ ACBD$.

What is Shoelace Theorem? I've wiki it, but there's no article about it.

There's a stub article about it called [[shoestring]] (it needs a lot of work), but there are [url=http://wikipedia.org]numerous[/url] [url=http://mathworld.wolfram.com]other[/url] [url=http://planetmath.org]resources[/url] [url=http://google.com]available[/url] that you could use to find out about it.

But, the article is very understandable (Although I agree with you about the messiness of it). I've checked Shoestring/Shoelace Theorem on two (Hey, I've use wikipedia b4 to find about this btw) of the links you provided but there are no info about it.


geometry
inradius
trigonometry
circumcircle
cyclic quadrilateral
geometry solved

k, just use that inradius * semiperimeter = area of any triangle. Then use Ptolemy's Theorem to get that AD = BD + CD. Now use the sine angle formula to solve each area, and you get a equation at the end that cancels out completely. Kinda messy, but straight-forward.

Actually, the triangle ABC needs not to be equilateral. Generally, we have the following fact:

[color=blue][b]Japanese Theorem.[/b] If ABCD is a cyclic quadrilateral, and $r_a$, $r_b$, $r_c$, $r_d$ are the inradii of triangles DAB, ABC, BCD, CDA, respectively, then $r_a+r_c=r_b+r_d$.[/color]

[i]Proof.[/i] This theorem is pretty well-known, here is a proof for the sake of completeness:

Let R be the circumradius of the cyclic quadrilateral ABCD. Of course, this R is simultaneously the circumradius of each of the four triangles DAB, ABC, BCD, CDA.

Since the quadrilateral ABCD is cyclic, we have < ACB = < ADB. Denote < ACB = < ADB = x. Similarly, denote < BDC = < BAC = y, < CAD = < CBD = z and < DBA = < DCA = w.

It is well-known that in a triangle with angles A, B, C, inradius r and circumradius R, we have $\frac{r}{R}=\cos A+\cos B+\cos C-1$. Applying this to the triangle DAB, which has the inradius $r_a$ and the circumradius R, we get $\frac{r_a}{R}=\cos\measuredangle ADB+\cos\measuredangle DAB+\cos\measuredangle DBA=\cos x+\cos\measuredangle DAB+\cos w$. Similarly, $\frac{r_c}{R}=\cos z+\cos\measuredangle BCD+\cos y$. But since the quadrilateral ABCD is cyclic, < DAB = 180° - < BCD, so that cos < DAB = - cos < BCD and cos < DAB + cos < BCD = 0. Hence,

$\frac{r_a+r_c}{R}=\frac{r_a}{R}+\frac{r_c}{R}=\left(\cos x+\cos\measuredangle DAB+\cos w\right)+\left(\cos z+\cos\measuredangle BCD+\cos y\right)$
$=\left(\cos x+\cos y+\cos z+\cos w\right)+\left(\cos\measuredangle DAB+\cos\measuredangle BCD\right)$
$=\cos x+\cos y+\cos z+\cos w$.

Similarly, $\frac{r_b+r_d}{R}=\cos x+\cos y+\cos z+\cos w$. Hence, $\frac{r_a+r_c}{R}=\frac{r_b+r_d}{R}$, so that $r_a+r_c=r_b+r_d$, and the Japanese Theorem is proven.

Another proof is sketched in §10.2.7 of Paul Yiu's "Notes on Euclidean Geometry", avaliable from http://www.math.fau.edu/yiu/Geometry.html .

[b]PS.[/b] You can upload figures as attachments to your posts or into the MathLinks [url=http://www.mathlinks.ro/Forum/album.php]Photo Album[/url], you don't need external servers for that (moreover, some external servers will remove your pictures as soon as the number of views becomes too large, what is inevitable on MathLinks).

  darij

In my opinion, it is a little bit easyer to prove it using Karno's theorem:[i]the sum of the distanses between circumcentre of triangle and it's sides is equal to the sum of exradius and inradius.[/i]

[quote="darij grinberg"]It is well-known that in a triangle with angles A, B, C, inradius r and circumradius R, we have $\frac{r}{R}=\cos A+\cos B+\cos C-1$[/quote]

I didn't know this identity. Can anyone proof it?


trigonometry
inequalities
algebra unsolved
algebra

Let $ n$ be a natural number and $ a$ be such that $ 0<a<\frac{\pi}{n}$. Prove that

$ \sin a\sin 2a...\sin na<\frac{1}{n^n\sin^n\frac{a}{2}}$

[quote="Jumbler"]Let $ n$ be a natural number and $ a$ be such that $ 0 < a < \frac {\pi}{n}$. Prove that

$ \sin a\sin 2a...\sin na < \frac {1}{n^n\sin^n\frac {a}{2}}$[/quote]

$ \sin a\plus{}\sin 2a\plus{}...\plus{}\sin na\equal{}\frac{1}{2\sin \frac{a}{2}}(\cos \frac{a}{2}\minus{}\cos \frac{(2n\plus{}1)a}{2})$
$ \equal{}\frac{1}{\sin \frac{a}{2}} \sin na \sin (n\plus{}1)a<\frac{1}{\sin \frac{a}{2}}$

So   $ \sin a\sin 2a...\sin na \le (\frac{\sin a\plus{}\sin 2a\plus{}...\plus{}\sin na}{n})^n$ (By AM-GM)
$ <(\frac{1}{n \sin \frac{a}{2}})^n\equal{}\frac {1}{n^n\sin^n\frac {a}{2}}$

It suffices to show that the geometric mean of the LHS < 1/(n*sin(a/2))

By GM-AM, it suffices to show that [ sin(a) + sin(2a) ... + sin(na) ] / n < 1/(n*sin(a/2))
<==> [ sin(a) + sin(2a) ... + sin(na) ]< 1/sin(a/2)
using the simple sin(x) < x bound,

we then only need to check n = 1, n = 2, n = 3. n = 1 is trivially true
for n = 2

[sin (a) + sin (2a)] [sin(a/2)] = 2sin(3a/2)cos(a/2)sin(a/2) = sin(3a/2)sin(a) which is definitely less than 1

for n = 3

[sin (a) + sin (2a) + sin(3a)] [sin(a/2)] = sin(a)sin(3a/2) + sin(3a)sin(a/2) = [cos(a/2)-cos(5a/2) + cos(5a/2) - cos(7a/2)]/2 = [cos(a/2) - cos(7a/2)]/2 which is definitely less than one.


induction
IMO 2009
modular arithmetic
number theory
IMO Shortlist
IMO
Hi

Let $ n$ be a positive integer and let $ a_1,a_2,a_3,\ldots,a_k$ $ ( k\ge 2)$ be distinct integers in the set $ { 1,2,\ldots,n}$ such that $ n$ divides $ a_i(a_{i + 1} - 1)$ for $ i = 1,2,\ldots,k - 1$. Prove that $ n$ does not divide $ a_k(a_1 - 1).$

[i]Proposed by Ross Atkins, Australia [/i]

Let $ n\equal{}pq$ such that $ p|a_1$ and $ q|a_2\minus{}1$. Suppose $ n$ divides $ a_k(a_1\minus{}1)$.
Note $ q|a_2\minus{}1$ implies $ (q,a_2)\equal{}1$ and hence $ q|a_3\minus{}1$. Similarly one has $ q|a_i\minus{}1$ for all $ i$'s, in particular, $ p|a_1$ and $ q|a_1\minus{}1$ force $ (p,q)\equal{}1$. Now $ (p,a_1\minus{}1)\equal{}1$ gives $ p|a_k$, similarly one has $ p|a_i$ for all $ i$'s, that is $ a_i$'s satisfy $ p|a_i$ and $ q|a_i\minus{}1$, but there should be at most one such integer satisfies them within the range of $ {1,2,...,n}$ for $ n\equal{}pq$ and $ (p,q)\equal{}1$. A contradiction!!!

too easy no?

$ a_1\equiv a_1\cdot a_2\equiv a_1\cdot a_2\cdot a_3\equiv \dots \equiv a_1\cdot\dots\cdot a_k\equiv a_1\cdot\dots\cdot a_{k \minus{} 2}\cdot a_k\equiv\dots\equiv a_1\cdot a_{k}\equiv a_k$.

Therefore, $ 0 < |a_1 \minus{} a_k| < n$ is divisible by $ n$. Contradiction.

P.S Please add 's' in the statement, integer[b]s[/b]

We prove inductively that $ n|a_1a_l \minus{} a_1$ for $ l\equal{}2, 3, \ldots, k$. We know that it's true for $ l\equal{}2$. Suppose that it's true for $ l$. Since $ n|a_1a_l\minus{}a_1$ and $ n|a_la_{l\plus{}1} \minus{} a_l$ we know that also $ n|(a_1a_l\minus{}a_1)a_{l\plus{}1} \minus{} a_1a_l\plus{}a_1$ and $ n|a_1a_la_{l\plus{}1} \minus{} a_1a_l$. Substracting gives exactly what we want and the induction proof is complete.
Since $ n|a_1a_k \minus{} a_1$ we can't have $ n|a_k(a_1\minus{}1)\equal{}a_ka_1 \minus{} a_k$ since it would imply $ n|a_1\minus{}a_k$, which is impossible.

It is an easy problem!!! I solved it!!! It's very similar to ychjae's, but I complicated stuff with prime factors of n...

Another wording for the same approach : all equalities are considered modulo $ n$.
We know that $ a_{i\plus{}1}a_i \equal{} a_i$ for all $ i \leq k\minus{}1$, thus clearly $ a_ia_{i\minus{}1} \cdots a_1 \equal{} a_1$ for all $ i \leq k$.
But, leaving $ a_i$, this also gives $ a_i(a_{i\minus{}1} \cdots a_1) \equal{} a_ia_1$ for all $ i \leq k$.

It follows that $ a_ka_1 \equal{} a_1$.
Since $ k \geq 2$ and $ a_k,a_1$ belong to $ \{1, \cdots , n \}$ we cannot have $ a_k \equal{} a_1$, thus $ a_ka_1 \neq a_k$, as desired.

Pierre.

too easy I think, it only took me 15 minutes. my approach is similar to ychjae's. i used chinese remainder theorem.

Same solution as ychjae, but to finish the problem I used chinese remainder theorem as well as johan

I think it is simple, yet very nice problem

Suppose the contrary is true.Then we have $ n^{k} |\prod^{k}_{i \equal{} 1} a_{i}(a_{i}\minus{}1)$ which imply
there exist j such that $ n|a_{j}(a_{j} \minus{} 1)$ Then by considering two cases $ "n|a_{j}"$ and $ "n|a_{j } \minus{} 1"$ we have $ a_{j} \equal{} a_{j \minus{} 1}$ or $ a_{j} \equal{} a_{j \plus{} 1}$.A contradiction.

There is a simple method to prove the last.
$ p|a_{i} \minus{}a_{i\plus{}1}$, $ q|(a_{i}\minus{}1)\minus{}(a_{i\plus{}1}\minus{}1)$, so $ pq|a_{i} \minus{}a_{i\plus{}1}$. A Contradiction!!

I solved it in the exam from this way:
clearly  a1=a1.a2 (mod n)  then a1.a3=a1.a2.a3 (mod n) and from a2.a3=a2 (mod n)  we have a1=a1.a2=a1.a3 (mod n) and we have from the same way a1=a1.a2=a1.a3=...=a1.ak (mod n) so a1=a1.ak (mod n)  (*)
   For the sake of contradiction we may assume that  ak=a1.ak (mod n) then we have a1=ak (mod n) from (*) ,contradiction. :lol:

Suppose $ n|a_k(a_1 \minus{} 1)$.

If $ \gcd (a_1,n) \equal{} 1$, we have $ n|a_2\minus{}1$, then $ \gcd (a_2,n) \equal{} 1$ and so $ n|a_3 \minus{} 1 \rightarrow a_2 \equal{} a_3$, a contradiction.
If $ \gcd (a_1,n) \equal{} n$, we have $ \gcd (a_1\minus{}1,n)\equal{}1$, hence $ n|a_k \rightarrow a_1\equal{}a_k\equal{}n$, contradiction again.

Now, we know that $ 1 < \gcd (a_1,n) < n$. This means that there is a prime $ p$ such that $ p$ divides both $ a_1$ and $ n$, yet the exponent of $ p$ in the prime factorization of $ a_1$ is less than the exponent of $ p$ in the prime factorization of $ n$.

Hence, $ p|a_2\minus{}1$. But then $ p$ does not divide $ a_2$, so $ p|a_3\minus{}1$, and so on, until we get $ p|a_1\minus{}1$, which is absurd because we said $ p|a_1$.

[b]uglysolutions[/b]

why $ p$ divides $ a_2\minus{}1$?

Because $ n|a_1(a_2 \minus{} 1)$ and the exponent of $ p$ in the prime factorization of $ a_1$ is less than the exponent of $ p$ in the prime factorization of $ n$, hence there must be at least one more $ p$ factor in $ a_2 \minus{} 1$.

I mean, if $ p^{\alpha}\parallel{}n$ and $ p^{\beta}\parallel{}a_1$, we have $ p^{\alpha \minus{} \beta}|a_2 \minus{} 1$, since $ \alpha > \beta \rightarrow \alpha \minus{} \beta \geq 1$ we can say $ p|a_2 \minus{} 1$.

[b]Solution: [/b]Let $i$ be some element in $\{1,2,\cdots,k-1\}.$ We are given that $n\mid a_i(a_{i+1}-1),$ which is equivalent to having $a_i(a_{i+1}-1)\equiv0\pmod{n}$, so that $$a_i\equiv a_ia_{i+1}\pmod{n}.$$ As a result, for any $i,j\in\{1,2,\cdots,k-1\}$ such that $i<j$, we obtain that $$x_ix_{i+1}\cdots x_j\equiv x_ix_{i+1}\cdots(x_jx_{j+1})\pmod{n}.$$
Repeating this $k-j$ times gives $$x_ix_{i+1}\cdots x_j\equiv x_ix_{i+1}\cdots x_k\pmod{n},$$ so setting $i=j$ implies that $$x_i\equiv x_ix_{i+1}\cdots x_k\pmod{n}.$$

Now suppose by contradiction that $n$ divides $a_k(a_1-1)=a_1a_k-a_k$, so that $a_k\equiv a_1a_k\pmod{n}.$ But then, using our above results, we see that
\begin{align*}
a_1&\equiv a_1a_2\cdots a_k\pmod{n}\\
&\equiv a_2a_3\cdots a_{k-1}(a_ka_1)\pmod{n}\\
&\equiv a_2a_3\cdots a_{k-1}a_k\pmod{n}\\
&\equiv a_2\pmod{n},
\end{align*}
contradicting the assumption that the $a_i$ are distinct integers in the set $\{1,2,\cdots,n\}.$ Thus our original claim that $n\mid a_k(a_1-1)$ is incorrect and the result follows. $\blacksquare$

Assume otherwise, and take indices $\pmod{k}$. Now take $\pmod{n}$, and note that we have
\[a_1 \equiv a_1a_2 \equiv a_1a_2a_3 \equiv \cdots \equiv a_1a_2a_3\cdots a_{k-1}a_k \equiv \cdots \equiv a_1a_k \equiv a_k \pmod{n},\]
impossible. $\square$

Two solutions: 1. Assume it does. We have $$a_i\equiv a_ia_{i+1}\pmod n,\dots\implies a_1\equiv a_1a_2\equiv\dots a_1a_2\dots a_{k-1}a_k\equiv a_1a_2\dots a_{k-2}a_k\equiv\dots\equiv a_1a_k\equiv a_k\pmod n,$$ contradiction.
2. Again WLOG it does. Let $n=ab$ s.t. a|a_1, b|a_2-1, since gcd(b,a_2)=1, to have n divide a_2(a_3-1) we must have b|a_3-1, ... so b|a_i-1 for all i. Then a|a_1,b|a_1-1 implies gcd(a,b)=1. Hence n|a_i(a_{i+1}-1) for all i from 1 to k (instead of k-1 now, where indices are taken mod k), from b|a_{i+1}-1 we must have a|a_i since gcd(b,a_i)=1 for all i. Since a,b are relatively prime, a_i=1 mod b, a_i=0 mod a, by CRT there's only one solution mod n, yet k is more than one. $\blacksquare$

$a_1*a_2-a_1 \equiv 0 \pmod n$
$a_1*a_2 \equiv a_1 \pmod n$

Similarly,
$a_2*a_3 \equiv a_2 \pmod n$
$a_1*a_2*a_3 \equiv a_1*a_2 \equiv a_1 \pmod n$
Eventually, get that $a_k*a_{k-1}*...*a_1 \equiv a_1 \pmod n$

If $a_k(a_1-1) \equiv 0 \pmod n$, then
$a_1*a_k \equiv a_k \pmod n$
$a_1*a_k*a_{k-1} \equiv a_k*a_{k-1} \equiv a_{k-1} \pmod n$
$a_1*a_k*a_{k-1}*a_{k-2} \equiv a_k*a_{k-1}*a_{k-2} \equiv a_{k-1}*a_{k-2} \equiv a_{k-2} \pmod n$
Use similar logic to get that $a_1*a_k*a_{k-1}*...*a_2 \equiv a_2 \pmod n$
But this product also $\equiv a_1 \pmod n$
So $a_1 \equiv a_2 \pmod n$
But $a_1$ and $a_2$ are between $1$ and $n$, so this is impossible and proof by contradiction

Assume by contradiction $a_k a_1 - a_k \equiv 0~(\text{mod } n)$. We begin with the following:

[b][color=blue]Claim:[/color][/b] For $i = 2,~3,~\ldots~, k$, we have $n \mid a_1 a_i - a_1$.

We prove this using induction. Our base case, $i = 2$, is given.

We wish to prove $i = j+1$, or $n \mid a_1 a_{j+1} - a_1$, given $i = j$ holds:

\begin{align*}
0 &\equiv (a_{j+1} - 1)(a_1 a_j - a_1) \\
&\equiv (a_1 (a_j a_{j+1}) - (a_1 a_j)) - (a_1 a_{j+1} - a_1) \\
&\equiv (a_1 a_j - a_1) - (a_1 a_{j+1} - a_1) \\
&\equiv -(a_1 a_{j+1} - a_1) \quad (\text{mod } n).~{\color{blue} \blacksquare}
\end{align*}

Thus we know $j = k$ holds, and by our initial assumption \[a_1 \equiv a_k a_1 \equiv a_k~(\text{mod } n),\] which is a contradiction, as $a_1$ and $a_k$ cannot differ by a multiple of $n$. $\blacksquare$

Another nice N1.

Notice that the given condition implies the following chain of congruences.
\begin{align*}
    a_ka_{k-1}&\equiv a_{k-1}\\
    a_ka_{k-1}a_{k-2}&\equiv a_{k-2}\\
    & \vdots\\
    a_ka_{k-1}\cdots a_1&\equiv a_1
\end{align*}
Then, assume that $n$ does divide $a_k(a_1-1)$, this gives a similar chain of congruences.
\begin{align*}
    a_1a_k&\equiv a_k\\
    a_1a_ka_{k-1}&\equiv a_{k-1}\\
    &\vdots\\
    a_1a_ka_{k-1}\cdots a_2 &\equiv a_2
\end{align*}
Putting these two together, $$a_1 \equiv a_ka_{k-1}\cdots a_1\equiv a_2 \implies a_1\equiv a_2$$, but this is impossible since $a_1,a_2\in \{1,2,\cdots,n\}$. Thus, it is a contradiction and in-fact, $n$ does not divide $a_k(a_1-1)$ .

We proceed with proof by contradiction. Suppose $a_k(a_1-1)\equiv 0 \pmod{n}$. From the given, we have
\begin{align*}
a_1a_2-a_1 & \equiv 0 \pmod n \\
a_2a_3-a_2 & \equiv 0 \pmod n \\
&\vdots \\
a_{k-1}a_k-a_k & \equiv 0 \pmod n
\end{align*}
We can manipulate the second equation as follows
\begin{align*}
a_2a_3-a_2 & \equiv 0 \pmod n \\
a_1a_2a_3-a_1a_2 & \equiv 0 \pmod n \\
a_1a_3-a_1 & \equiv 0 \pmod n
\end{align*}
Following a similar substitution argument, we have
\[a_1a_i-a_1 \equiv 0 \pmod n\] for all $2\leq i \leq k$. Examine the case where $i=k$. We have
\[a_1a_k-a_1 \equiv 0 \pmod n\]
and
\[a_1a_k-a_k \equiv 0 \pmod n.\]Subtracting gives
\[a_k-a_1 \equiv 0 \pmod n.\] However, since $a_i$ comes from $\{1, 2, \dots, n\}$, this is impossible. Thus we have a contradiction. $\Box$

[hide=Solution]
We have $n|a_1(a_2-1)$ and $n|a_2(a_3-1).$ Since $a_2$ and $a_2-1$ are relatively prime, for every prime factor $p{}$ of $n{}$, either $v_p(a_2)=0$ or $v_p(a_2-1)=0.$ If $v_p(a_2)=0,$ then $v_p(a_3-1)\geq v_p(n),$ and if $v_p(a_2-1)=0,$ then $v_p(a_1)\geq v_p(n).$ Since $v_p(xy)=v_p(x)+v_p(y)$ for positive integers $x$ and $y,$ we have $v_p(a_1(a_3-1))=v_p(a_1)+v_p(a_3-1)\geq v_p(n)$ for all prime factors $p{}$ of $n.$ (This is because at least one of $v_p(a_1)$ and $v_p(a_3-1)$ is $\geq v_p(n).$) This implies that $n|a_1(a_3-1).$
Similarly, for any positive integer $i$ such that $1<i<k$ such that $n|a_1(a_i-1)$ and $n|a_i(a_{i+1}-1),$ we have $n|a_1(a_{i+1}-1).$ By induction, this means that $n|a_1(a_k-1)=a_1a_k-a_1.$
We have $a_1a_k-a_1\equiv0\pmod n,$ so $a_1a_k-a_k\equiv a_1-a_k\pmod n,$ and we want to prove that this cannot be $0\pmod n.$ Since the minimum value for $a_1-a_k$ is $1-n$ and the maximum is $n-1,$ the only way for $a_1-a_k\equiv0\pmod n$ would be if $a_1=a_k,$ contradiction. Thus, $n$ cannot divide $a_1-a_k$ and therefore cannot divide $a_1a_k-a_k=a_k(a_1-1).$ $\Box$
[/hide]

Assume for the sake of contradiction that $a_k \equiv a_ka_1 \pmod n$. We can write out the congruences: \[a_1 \equiv a_1a_2 \pmod n\] \[a_2 \equiv a_2a_3 \pmod n\] \[\cdots\] \[a_{k-1} \equiv a_{k-1}a_k \pmod n\] \[a_k \equiv a_ka_1 \pmod n.\] It is easy to see that \[a_1 \equiv a_1a_2 \equiv a_1a_2a_3 \equiv \cdots \equiv a_1a_2a_3\cdots a_k \pmod n.\] Then, \[a_k \equiv a_k(a_1a_2a_3\cdots a_{k-1}) \Rightarrow a_k \equiv a_1a_2a_3\cdots a_k \equiv a_1 \pmod n.\] But this contradicts the fact that $a_1, a_2, a_3, \cdots, a_k$ are distinct modulo $n$, so we're done. $
\blacksquare$

Let $p$ be a prime dividing $n$, and take all indices modulo $k$.

[color=#00f][b]Claim:[/b][/color] The prime $p$ cannot divide both $a_i$ and $a_{i + 1} - 1$ for any $i$, and further, $p$ must divide either/or.
[i]Proof:[/i]We make the first observation that $p$ cannot divide both $a_i$ and $a_{i + 1} - 1$ for some $i$. If we had such situation, observe that as $p \mid a_{i + 1} - 1 \implies p \mid a_{i + 2} - 1, \dots, \implies p \mid a_k - 1$ and inductively, $p \mid a_1 - 1 \implies p \mid a_i - 1$. Hence $p \mid a_i$ and $p \mid a_i - 1$, which is a clear contradiction. $\square$

Therefore $p$ either divides $a_i$ or $a_{i + 1} - 1$, for any prime $p$. Let $n = pq$, for primes $p, q$. For general $n$, the idea is to take $p$ to an arbitrary power. And for the number of primes dividing $n$, this is a matter of pure induction.

[color=#f00][b]Case 1:[/b][/color] ($p \mid a_1$ and $q \mid a_1$). Then $p, q \mid a_i$ for all $i$, from the above claim. However this is a contradiction, as due to the fact that $\gcd(p, q) = 1$, we must have $pq = n \mid a_i$ for all $i$. This contradicts the fact that $a_i \in \{1, 2, \dots, n\}$. $\square$

[color=#f00][b]Case 2:[/b][/color] ($p \mid a_1 - 1$ and $q \mid a_1 - 1$). Then $n = pq \mid a_i - 1 < n$, which is another contradiction. $\square$

[color=#f00][b]Case 3:[/b][/color] ($p \mid a_1$ and $q \mid a_1 - 1$). Then from the above claim, $n = pq \mid a_i(a_{i + 1} - 1)$. However then we have that \[a_2 \equiv a_2a_3 \equiv a_2a_3a_4 \equiv \cdots \equiv a_1a_2 \cdots a_k \equiv a_1 \pmod n,\] so that $a_1 \equiv a_2 \equiv \cdots a_k \pmod n$, which is another contradiction. $\blacksquare$

[hide = Remark] Upon revisiting the third case, we don't need to use any fancy primes or anything of the sort to solve this problem. Rather, we find naturally that $a_1 \equiv a_1a_2 \equiv \cdots \equiv a_1a_2 \cdots a_k \equiv a_2 \pmod n$, and we are finished, as this gives rise to a contradiction to the number of multiples of $n$ found in $\{1, 2, \dots, n \}$. $\blacksquare$ [/hide]

For the sake of contradiction, assume that $n \mid a_k(a_1-1)$. Notice we have

\[a_2a_3-a_2 \equiv 0 \pmod{n}\]
\[\implies a_1a_2a_3-a_1a_2 \equiv 0 \pmod{n}\]
\[\implies a_1a_3-a_1 \equiv 0 \pmod{n}\]

In a similar fashion, we find that

\[a_1a_i-a_1 \equiv 0 \pmod{n}\]

for all $i$. At this point, we find that

\[a_1a_k - a_1 \equiv a_1a_k-a_k \equiv 0 \pmod{n}\]
\[\implies a_1 \equiv a_k \pmod{n},\]

a contradiction as $a_i$ is in the set $\{1,2,\dots, n\}$. $\square$

Assume for the sake of contradiction that $n \mid a_k (a_1 - 1)$. Thus we have \[n \mid a_i (a_{i + 1} - 1)\] for all $1 \leq i \leq k$ (taking indices modulo $k$), so it follows that \[a_i a_{i + 1} \equiv a_i \pmod{n}\text{.}\] Now we have \[a_1 a_2 \dotsm a_k \equiv a_1 a_2 \dotsm (a_{k - 1} a_k) \equiv a_1 a_2 \dotsm (a_{k - 2} a_{k - 1}) \equiv \dots \equiv a_1 a_2 \equiv a_1 \pmod{n}\text{.}\] Similarly, we have \[a_1 a_2 \dotsm a_k \equiv a_2 a_3 \dotsm (a_k a_1) \equiv a_2 a_3 \dotsm (a_{k - 1} a_k) \equiv \dots \equiv a_2 a_3 \equiv a_2 \pmod{n}\text{.}\] Therefore \[a_1 \equiv a_2 \pmod{n}\text{.}\] Contradiction.

For the sake of contradiction, assume that $n\mid a_k(a_1-1)$. So, $n\mid a_i(a_{i+1}-1)$ for each $i$ if indices are taken $\pmod{k}$.

Let $p$ be a prime such that $p^r\mid n$. Suppose $p^x\mid a_i$ with $r>x>0$. Then, $p\nmid (a_i-1)$, so $p^r\mid a_{i-1}$, which means $p\nmid (a_{i-1}-1)$, which gives $p^r\mid a_{i-2}$ as forced. Continuing like this, we obtain that $p\mid a_i$ for all $i$. Similarly we may proceed if $p^x\mid a_i-1$. Hence, for any prime $p$ with $p^r\mid n$, we have either $a_i\equiv1\pmod{p^r}$ for all $i$ or $a_i\equiv0\pmod{p^r}$ for all $i$. Call a prime $q$ with $q^v\mid n$ as ``cold" if $q^v\mid a_i$ and ``warm" if $q^v\mid(a_i-1)$. Let $d$ be the product of all cold prime powers, and $\frac{n}{d}$ be the product of all warm prime powers. So, we have
\[a_i\equiv0\pmod{d},\ \ \  a_i\equiv1\pmod{\frac{n}{d}} \Longrightarrow a_i\equiv jd\pmod{n}\]
where $n\mid d(jd-1)$. However, it follows that $a_1\equiv a_2\equiv\cdots\equiv a_k\pmod{n}$, which contradicts the fact that $k\geq2$ and the numbers are distinct from set $1,2,\ldots,n$.

it is very easy to found
$a_1 \equiv a_2a_3 \equiv . . . \equiv a_1a_2a_3...a_k \equiv a_k (\text{mod n})$
Henceforth, $a_1 \equiv a_k (\text{mod n})$
Easy contradiction.

We can easily prove by induction that $a_1(a_i-1) \equiv 0 \pmod{p}$ for any $i\in \{1,\ldots,k\}$. Now if $n \mid a_k(a_1-1)$, using the fact that $n\mid a_1(a_k-1)$ we get $n_1 \equiv a_k \pmod{n}$ a clear contradiction.

$$\mathbb{Q.E.D.}$$


abstract algebra
superior algebra
superior algebra unsolved

Let R be a commutative ring with ideals A and B

Show that the following is an exact sequence of R-modules
$ 0\to A\cap B\to R\to R/A\times R/B\to R/(A \plus{} B)\to 0$

where $ i: A\cap B \to R$ denotes inclusion
$ f: R\to R/A\times R/B$
$ f(x) \equal{} (x \plus{} I,x \plus{} J)$
$ g: R/A\times R/B\to R/(A \plus{} B)$
$ g(r \plus{} A,r \plus{} B) \equal{} r \minus{} s \plus{} (A \plus{} B)$

conclude that there is an induced exact sequence of R modules
$ 0\to R/(A\cap B)\to R/A\times R/B\to R/(A \plus{} B)\to 0$

I know I'm suppose to show that the im i = ker f and im f = ker g
so far i think that $ im$ i $ \equal{} A \cap B$ and $ ker f \equal{} A\cap B$
$ im$ f $ \equal{} \{(x \plus{} A,x \plus{} B) | x \in R\}$ and $ ker g \equal{} \{(x \plus{} A,x \plus{} B) | x \in R \}$

where do i go from here?

you just repeated the claim; you can easily write down the proofs if you think over the definitions. for example, $ (r\plus{}A,S\plus{}B)$ lies in the kernel of $ g$ iff $ r\minus{}s \in A\plus{}B$, say $ r\minus{}s\equal{}a\plus{}b$, but then $ r\plus{}A\equal{}b\plus{}s\plus{}A, s\equal{}b\plus{}s\plus{}B$.


Given a, b, c >1 prove that:

$\frac{\sum_{a,b,c}^{\text{cyc}}\left((a^{2}+1)(1-4bc)+7ab\right)-3}{2(a+b+c)}\geqslant \sum_{a,b,c}^{\text{cyc}}a^{2}(b+c)$

Is there some Latex/image that I'm not seeing?

It's ok now, it was my problem- on the computer I was using before, the latex wasn't showing. I can see it now.  :)

Nobody wants to solve it?  :( I thought it was a pretty cool problem.


modular arithmetic
quadratics
Congruences

Suppose there is such $k$ and let $t$ be the order of $2$ mod $2p+1$. Then $t|2p+1-1=2p$ and $t|k$ so in particular $t<2p$. This implies that $t=p$ since, as one can easily verify, $t$ can't be $1$ or $2$. But then we also have:
$2^{p+1} \equiv 2 \mod{2p+1}$
and since $p+1$ is even it follows that $2$ is a quadratic residue mod $2p+1$ which is impossible because $2p+1 \equiv 3 \mod{8}$ and $2$ is a quadratic residue only for primes which are $1$ or $7 \mod{8}$.

[quote="crazyfehmy"][quote="Peter"]Let $p$ be a prime number of the form $4k+1$. Suppose that $2p+1$ is prime. Show that there is no $k \in \mathbb{N}$ with $k<2p$ and $2^k \equiv 1 \; \pmod{2p+1}$.[/quote]
I think that $p$ must be a prime in the form of $4r+1$ instead of $4k+1.$[/quote]
Yes,I think so


algebra
polynomial
quadratics
number theory
relatively prime
quadratic formula

The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r,$ where $m, n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0.$  Find $m+n+r.$

[quote="Andreas"]$2000x^6 + 100x^5 + 10x^3 + x - 2 = (100x^4 + 10x^2 + 1)(10x^2 + x - 2) = 0$
[/quote]
How do we find the factorization of such an expression?
Is there any ultimate way?

[quote="Andreas"]$ 2000x^6 \plus{} 100x^5 \plus{} 10x^3 \plus{} x \minus{} 2 \equal{} (100x^4 \plus{} 10x^2 \plus{} 1)(10x^2 \plus{} x \minus{} 2) \equal{} 0$.
The first factor has $ \Delta < 0$, so solutions come from the second one.
$ x_{1/2} \equal{} \frac { \minus{} 1 \pm \sqrt {161}}{40}$.
$ m \plus{} n \plus{} r \equal{} \minus{} 1 \plus{} 161 \plus{} 40 \equal{} 200$.[/quote]

How can you find it?
The question tell you "exactly two real root", is that your way?

The two real roots must come from either $ (100 x^4\plus{}10 x^2 \plus{}1)$ or from $ (10 x^2\plus{}x\minus{}2)$

The first root has real roots iff $ y^2\plus{}y\plus{}1$ has nonnegative real roots ($ y \equal{} 10 x^2)$. However, that equation in $ y$ has no real roots.

On the other hand, $ 20 x^2\plus{}x\minus{}2$ does have two real roots. This becomes clear when you note that at $ x\equal{}0$ the polynomial is negative, but for very positive or very negative $ x$, the polynomial is positive.

Using the quadratic formula, $ r_{1,2} \equal{} \frac{\minus{}1 \pm \sqrt{161}}{40}$

can we use vieta here instead of just factorizing it?

[quote=songssari][quote=MSTang]Probably not, since there are six roots to the original equation....[/quote]

Then, what technique can we use if we cannot factorize it? Any ideas....?[/quote]

Bump? I don't think this question (and others like it) have been answered. Thanks in advance! :)

[quote=First]Read the last paragraph of the [hide=wiki solution]https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_II_Problems/Problem_13[/hide] but IMO if you have the intuition to notice how you can turn this polynomial into a symmetric polynomial you could easily find the factorization using $a^3+b^3=(a+b)(a^2-ab+b^2)$ and $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$ easily.[/quote]

I don't understand that last paragraph. First of all, what's a symmetric polynomial? And also I don't understand the rest of that solution

[hide=Very Complicated Complex Bash Solution I posted on the Wiki]
It would be really nice if the coefficients were symmetrical. What if we make the substitution, $x = -\frac{i}{\sqrt{10}}y$. The the polynomial becomes

$-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2$

It's symmetric! Dividing by $y^3$ and rearranging, we get

$-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})$

Now, if we let $z = y + \frac{1}{y}$, we can get the equations

$z = y + \frac{1}{y}$

$z^2 - 2 = y^2 + \frac{1}{y^2}$ and $z^3 - 3z = y^3 + \frac{1}{y^3}$

(These come from squaring $z$ and subtracting $2$, then multiplying that result by $z$ and subtracting $z$) Plugging this into our polynomial, expanding, and rearranging, we get

$-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})$

Now, we see that the two $i$ terms must cancel in order to get this polynomial equal to $0$, so what squared equals 3? Plugging in $z = \sqrt{3}$ into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying $z = -\sqrt{3}$, we see that it also works! Great, we use long division on the polynomial by

$(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)$ and we get

$2z -(\frac{i}{\sqrt{10}}) = 0$.

We know that the other two solutions for z wouldn't result in real solutions for $x$ since we have to solve a quadratic with a negative discriminant, then multiply by $-(\frac{i}{\sqrt{10}})$. We get that $z = (\frac{i}{-2\sqrt{10}})$. Solving for $y$ (using $y + \frac{1}{y} = z$) we get that $y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}$, and multiplying this by $-(\frac{i}{\sqrt{10}})$ (because $x = -(\frac{i}{\sqrt{10}})y$) we get that $x = \frac{-1 \pm \sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = \boxed{200}$
[/hide]

p9: Rewrite as $(2000x^6-2)+(100x^5+10x^3+x) = 2((10x^2)^3 - 1^3) + x(100x^4+10x^2+1)$. By Difference of Cubes, $(10x^2)^3 - 1^3 = (10x^2-1)(100x^4+10x^2+1)$, so our expression is equivalent to $(20x^2+x-1)(100x^4+10x^2+1)$. It is clear that $100x^4+10x^2+1$ cannot have real roots since for all reals $x^2 \geq 0$, so we are looking for the roots of $20x^2 + x-2$. The root in the desired form is $\frac{-1 + \sqrt{161}}{40} \implies \boxed{200}$.

[quote=OlympusHero]It makes sense to try factoring, and I get what #4, #15, #16 did, but these solutions manipulate in ways that it would take me a while to think of trying if I even tried it at all. Is there a motivation to doing any of these? My first thought would probably be to bash undetermined coefficients.[/quote]

Yeah, i’m wondering this as well. I don’t think I would’ve thought of difference of cubes as a motivation if i were to take the exam.

Solved from [b]Polynomials in the AIME.[/b]
I love complex numbers in computational they're so easy lol (this is just a sketch i did in my url space)
$$2((10x^2)^3-1)+x(100x^4+10x^2+1)=2((10x^2-1)(100x^4+10x^2+1))+x(100x^4+10x^2+1)=(20x^2+x-2)(1000x^6-1)/(10x^2-1)=0.$$ If $1000x^6-1=0$ then $x^6=1/10^3$ or $x^2=w/10$ where w is a cube root of unity. to be real w=1 but contradiction since that would make 10x^2-1 in the denominator 0. Hence instead it's just solution to first factor which is $(-1+\sqrt{1+160})/40$ so solution is $\boxed{200.} \blacksquare$


Gasiti toate functiile $f: R \to R$ care au primitive si  verifica relatia $f^2(x)=x^2,\forall x\in R$.

E clar ca $f(x)=x$ pe o multime $A$,sa zicem si $f(x)=-x$,in rest.
Totul ar fi simplu daca as arata ca daca A nu este interval,$f$ nu are proprietatea lui Darboux.
Cum pot sa arat asa ceva:daca $A$ nu este interval, $f$ nu are Darboux?

$f$ nu se anuleaza pe $(-\infty,0)$ si nici pe $(0,\infty)$, deci pastreaza semn constant pe fiecare dintre aceste intervale.
Obtii prin urmare $4$ solutii: $id_R$, $-id_R$, $|\,\cdot\,|$ si $-|\,\cdot\,|$.


Γειά σας παιδια.  Συγχαρητήρια για όλα όσα κάνετε και εύχομαι το καλύτερο. !!!!
Θα προσπαθήσω να συμβαλλω και εγώ στην προσπάθεια σας όσο περισσότερο μπορώ...

Καλώς ήρθες  :)


function
topology
real analysis
real analysis theorems

If we want to find the maximum of a continuously differentiable  function $ f(x_1,\ldots,x_n)$
on an closed set $ S$ in $ R^n$.
where $ S \equal{} \{s\in R^n: g(s) \equal{} 0 \}$, and $ g$ is continuously differentiable

Lagrange method
$ \nabla (f \plus{} \lambda g) \equal{} 0$

can be used to find the extreme value at the interior of $ S$

we need to take care of the boundary of $ S$ individually.

my problem is if only one extreme value is found by Lagrange method.

and it can be shown to be a local maximum.

I think the value is also a global maximum.  :?:
is this correct?

(if there are more than one extreme value is found, then we need to consider the boundary for
the global maximum. But if only one extreme value is found,
from the geometric consideration, it seem to be a global extreme value.?)

[quote="eeshift"]my problem is if only one extreme value is found by Lagrange method.

and it can be shown to be a local maximum.

I think the value is also a global maximum.
is this correct?[/quote]
So are you asking if one local maximum implies that it is a global maximum?  In a bounded region, it would be a global maximum if it's the only one.  But not necessarily on the entire surface.  For example, you could take some quintic equation, and find some extreme value in an interval which happens to be a max.  Then it's a global maximum in that interval but on the real line, it doesn't have to be.

Correct me if I misunderstood your question.

To JRav.

I express my problem more clearly. :oops:

if there is only one extreme value, and it's a local maximum.

then i think it's also a global maximum  even $ S$ is not bounded

for example $ f(x_1,x_2) \equal{} \minus{} (x^2_1 \plus{} x^2_2)$
the set $ S \equal{} \{(x_1,x_2): x_1 \equal{} x_2\}$

then by Lagrange method, there is only one extreme value at $ (0,0)$
and it's a global maximum on $ S$

As for your example, the quintic equation on the real line

if there is a local maximum, then there is a local minimum
hence, at least two extreme values.

If there is only one extreme value, then yes, it must be a global.

The problem with your example is that it is just that-an example.  Coming up with infinitely many examples that work does not prove anything-except that you haven't found a counterexample yet.

[quote="JRav"]If there is only one extreme value, then yes, it must be a global.
[/quote]

may you prove this or explain  :maybe:

[quote="JRav"]
The problem with your example is that it is just that-an example.  Coming up with infinitely many examples that work does not prove anything-except that you haven't found a counterexample yet.[/quote]

yes, i know this.  I just want to give an example to express my problem more clearly.  :)

The tricky part of the whole thing, of course, is that we're dealing with multiple dimensions. In one dimension, it's easy.

A simplifying assumption:
- $ S$ is a connected smooth manifold embedded in $ \mathbb{R}^n$. Any points where that breaks down are automatic critical points anyway. Now, the critical points are the points where $ df$ is zero on $ S$, and we don't need to look closely at the embedding any more. Locally, everything looks like $ \mathbb{R}^d$.

This is a problem in the relatively obscure field of [url=http://en.wikipedia.org/wiki/Morse_theory]Morse theory[/url]. One thing we can say: if $ f$ has a strict local maximum at some point $ x$, and $ f(y) > f(x)$, there is some saddle point $ z$. To find such a point, look at the maximum over curves $ \gamma$ between $ x$ and $ y$ of the minimum of $ f$ on $ \gamma$.
Your original statement is almost correct- but we absolutely need connectedness.

[quote="jmerry"]

This is a problem in the relatively obscure field of [url=http://en.wikipedia.org/wiki/Morse_theory]Morse theory[/url]. One thing we can say: if $ f$ has a strict local maximum at some point $ x$, and $ f(y) > f(x)$, there is some saddle point $ z$. To find such a point, look at the maximum over curves $ \gamma$ between $ x$ and $ y$ of the minimum of $ f$ on $ \gamma$.
Your original statement is almost correct- but we absolutely need connectedness.[/quote]

Thanks. Though i don't quite understand.

i am confused about

if $ f$ has a strict local maximum at some point $ x$, and $ f(y) > f(x)$,
 there is some saddle point $ z$

for example
$ f(x_1,x_2) \equal{} x^3_1 \minus{} x_1$
$ S \equal{} (a , 0), a\in \mathbb{R}$

then we have one strict local maximum and one strict local minimum
but no saddle point.

Did I misunderstand ?

I also assumed that $ S$ has at least two dimensions. That saddle point becomes a minimum in one dimension.

thanks again, jmerry.

how about the example

$ f(x_1,x_2,x_3)\equal{}x^3_1\minus{}x_1\minus{}x^2_2\minus{}x^2_3$

$ S\equal{}(a,b,0)$ and $ a,b\in \mathbb{R}$

then one strict local maximum and one strict local minimum
no salddle point

No, there's no local minimum there. Take another look at those critical points.

oh, there are one strict local maximum and one salddle point  :)


Question: Why can't I put percents within $ \text{\LaTeX}$?

$ 100%
$   is 100% in latex

$ \boxed{100%}
$     is 100% in latex boxed

The reason here is that % is a reserved character; it comments out the following code in $ \text{\LaTeX}$. As for most such characters, the way to get it is to preface with a backslash.

So, if I put "100%=1" in $ \text{\LaTeX}$, the parser sees 100 and prints it, then sees % and ignores the rest of the line to print nothing else. For your boxed example, it sees an open { with no closing }, which is unparseable and generates an error.


This might seem VERY easy but I'm convinced that it might not be that easy under exam conditions. For example, the South African scores in 2004 were:

RSA1: 7 1 7 7 7 0 - 29 - Silver
RSA2: 5 1 5 6 7 3 - 27 - Silver
RSA3: 7 0 1 7 7 1 - 23 - Bronze
RSA4: 0 0 7 7 7 1 - 22 - Bronze

Keep in mind that the PAMO contestants are in general very young and unexperienced. South Africa does have quite a strong IMO team (sometimes :D for example, in 2004, not in 2005) but they send 4 promising youngsters (which are not on the IMO team) to PAMO.

And this question in particular - yes, it isn't hard, and South Africa scored only 2/28 on it, but if it was REALLY trivial, they would have scored more marks there. So, I guess it isn't!

When I wrote that PAMO i didn't even know about completing the square  :oops:

By the way, day 2 question 2 was in the SAMO of a year or two before, somehow noone noticed...

In general the PAMO is much easier than the IMO, but as Arne said, South Africa (which has the strongest team in Africa) only sends its second, more inexperienced team, so the contentants are not that strong. Usually South Africa's PAMO team goes to the IMO the following year.


when  sinB=4/5 , what's the degree of the B?

hello, taking a calculator and typing $ \sin^{\minus{}1}(\frac{4}{5})$ and you will get the right answer, the modus must be DEGREE.
Sonnhard.


Let $ a, b , c \geq 0 $ and $ a + b + c=1 $ . Find the maximum value of the expression :
     $ \frac{a^2+1}{b^2+1} + \frac{b^2+1}{c^2+1} + \frac{c^2+1}{a^2+1} $

I am sure "Pre-Olympiad" is not right place for such inequality. Inequalities->Advanced section is more appropriate section.

We will prove that $\max=\frac{7}{2}$ (example: $a=1,\ b=c=0$).

There are two general cases:

[b]1)[/b] $b\leq a$, $b\leq c$ and $a\geq \frac{1}{3}$. We have $\frac{b^2+1}{c^2+1}\leq 1$, so we need to show that
\[\frac{a^2+1}{b^2+1}+\frac{c^2+1}{a^2+1}\leq \frac{5}{2}.\]
It is clear that
\[LHS\leq \frac{a^2+1}{0^2+1}+\frac{(b+c)^2+1}{a^2+1}=a^2+1+\frac{(1-a)^2+1}{a^2+1}.\]
And we obtain
\[a^2+1+\frac{(1-a)^2+1}{a^2+1}\leq \frac{5}{2}.\]
This inequality is equivalent to correct inequality $(a-1)(2a^3+2a^2+3a-1)\leq 0$, since $\frac{1}{3}\leq a\leq 1$.

[b]2)[/b] $b\leq a$, $b\leq c$ and $a\leq\frac{1}{3}$. We have
\[\frac{b^2+1}{c^2+1}+\frac{c^2+1}{a^2+1}\leq\frac{x^2+1}{c^2+1}+\frac{c^2+1}{x^2+1},\]
where $x=\frac{a+b}{2}\leq\frac{1}{3}$, $c=1-2x$.

We will show $RHS\leq\frac{5}{2}\left(1-\frac{8}{5}x^2\right)$. Indeed, this inequality is equivalent to
\[S=x(31x^3+27x-6-38x^2-16x^4+16x^5)\leq 0\]
for $0\leq x\leq \frac{1}{3}$. Actually, we can write
\[\frac{S}{x}=-16x^4(1-x)-3(3x-1)(3x-2)-x^2(11-31x)<0.\]

Now we will prove that
\[\frac{a^2+1}{b^2+1}+\frac{5}{2}\left(1-\frac{8}{5}x^2\right)\leq \frac{7}{2}.\]
Indeed,
\[LHS\leq \frac{(a+b)^2+1}{0^2+1}+\frac{5}{2}\left(1-\frac{8}{5}x^2\right)=4x^2+1+\frac{5}{2}-4x^2=\frac{7}{2}.\]

That's all.

Hi Myth,
 I think in the second case , you can solve similarly with the first one .
   $ \frac{b^2+1}{c^2+1} + \frac{c^2+1}{a^2+1} \leq \frac{(a+b)^2+1}{c^2+1} + c^2+1 $
   After that we'll prove :
   $ \frac{(a+b)^2+1}{c^2+1} + c^2+1 \leq \frac{5}{2} $
  Is it true ?

Hi!
[quote="romano"]$ \frac{(a+b)^2+1}{c^2+1} + c^2+1 \leq \frac{5}{2} $ [/quote]
Anyway, it willn't help us, because third fraction $\frac{a^2+1}{b^2+1}\geq 1$.

Yes , you had a beautiful proof ! I think this problem has shorter solutions because this is a problem for lower secondary schools .

[quote="romano"]this is a problem for lower secondary schools .[/quote]
Unbelievable!

Anybody! Show nice short solution for this problem.

Romano, in which book did you find this nice problem ?

There is a reason that makes me think the inequality in question is very unsharp. In fact, you can homogenize it into the form

$\frac{a^2+\left(a+b+c\right)^2}{b^2+\left(a+b+c\right)^2}+\frac{b^2+\left(a+b+c\right)^2}{c^2+\left(a+b+c\right)^2}+\frac{c^2+\left(a+b+c\right)^2}{a^2+\left(a+b+c\right)^2} \leq \frac72$.

Then, multiply with the denominator and get

$\left( a^{2}+\left( a+b+c\right) ^{2}\right) ^{2}\left( c^{2}+\left(a+b+c\right) ^{2}\right) $
$+\left( b^{2}+\left( a+b+c\right) ^{2}\right) ^{2}\left( a^{2}+\left(a+b+c\right) ^{2}\right) $
$+\left( c^{2}+\left( a+b+c\right) ^{2}\right) ^{2}\left( b^{2}+\left(a+b+c\right) ^{2}\right) $
$\leq \frac{7}{2}\left( a^{2}+\left( a+b+c\right) ^{2}\right) \left(b^{2}+\left( a+b+c\right) ^{2}\right) \left( c^{2}+\left( a+b+c\right)^{2}\right) $.

But this is clear, since

$\left( a^{2}+\left( a+b+c\right) ^{2}\right) ^{2}\left( c^{2}+\left(a+b+c\right) ^{2}\right) $
$+\left( b^{2}+\left( a+b+c\right) ^{2}\right) ^{2}\left( a^{2}+\left(a+b+c\right) ^{2}\right) $
$+\left( c^{2}+\left( a+b+c\right) ^{2}\right) ^{2}\left( b^{2}+\left(a+b+c\right) ^{2}\right) $
$-\frac{7}{2}\left( a^{2}+\left( a+b+c\right) ^{2}\right) \left( b^{2}+\left( a+b+c\right) ^{2}\right) \left( c^{2}+\left( a+b+c\right)^{2}\right) $
$=-\frac{1}{2}\left(6ab^{5}+6a^{5}b+6ac^{5}+6a^{5}c+6bc^{5}+6b^{5}c+38abc^{4}+38ab^{4}c\right. $
$+38a^{4}bc+21a^{2}b^{4}+34a^{3}b^{3}+23a^{4}b^{2}+23a^{2}c^{4}+34a^{3}c^{3}+21a^{4}c^{2}$
$+21b^{2}c^{4}+34b^{3}c^{3}+23b^{4}c^{2}+82ab^{2}c^{3}+82ab^{3}c^{2}+82a^{2}bc^{3}$
$\left. +82a^{2}b^{3}c+82a^{3}bc^{2}+82a^{3}b^{2}c+124a^{2}b^{2}c^{2}\right)$,

what is definitely $\leq 0$. Now, this is a completely idiot proof, but it leaves me with the impression that the inequality is far too weak because of the many clearly nonnegative terms. It can be probably made way stronger. What do you think?

  Darij

[quote="darij grinberg"]There is a reason that makes me think the inequality in question is very unsharp. In fact, you can homogenize it into the form

*stuff*

  Darij[/quote]

As myth said, let a=1, b=c=0, which shows that 7/2 is attainable.  This reminds me of another inequality which I handled using Schur which had 'extra' terms, and the following inequality from the 1999 Canadian Mathematical Olympiad.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15825

http://www.cms.math.ca/Competitions/CMO/solutions/sol_1999.pdf

{this is my first post}
put x=a 2 +1
y=b 2 +1
z=c 2 +1
1 <= x,y,z <= 2 (because a+b+c=1 then 0 <= a,b,c <= 1)
let x=max(x,y,z}
simply to prove that x/y + y/z +z/x  <=  1 + x + 1/x (if x,y,z >= 1)
x  <= 2 so 1 + x + 1/x  <= 7/2

Thank you very much Minhkhoa  :)

Very nice solution. ;)


Umm the forums are showing supoer small fonts and i cant see anything im typing, so i dont know if this is the right place but is anyone else having this problem?

If you're using Internet Explorer, go to View - then to Text Size, and change it to a bigger size. I use Medium and that works for me. That might fix it.

Also, on the Forum page, near the top right, it says "Font Size".  You can choose one of three options.


i use my school library's computer and a mac is logged in as someone else called vanity, but i can't log out. i've tried closing the browser and refreshing, but neither has worked for me or another computer also logged in as vanity. what should i do?

You have to click log out on the message board (it's near the top).  Whoever used your computer last had the 'log in automatically' box clicked, so that computer logged you in as someone else.


algebra
polynomial
geometry
3D geometry
modular arithmetic
LaTeX
blogs

I am going to post some problems that my teacher gave me. Is that okay?

[hide]1. Determine $ (r \plus{} s)(s \plus{} t)(t \plus{} r)$ if r, s, and t are the tree real roots of the polynomial $ x^3 \plus{} 9x^2 \minus{} 9x \minus{} 8$.

2. If the five digit number 5DDDD is divisible by 6, then find the digit D.

3. Ed is thinking of an integer from 1 to 25 inclusive. Eehern asks him if the number is a multiple of 3, if it is a perfect cube, and whether it is a prime. After receiving the replies Eehern can deduce Ed's number. What is that number?

4. Determine the minimum possible value of the sum $ \frac {a}{2b} \plus{} \frac {b}{4c} \plus{} \frac {c}{8a}$ where a, b, and c are postive real numbers.

5. Suppose that in triangle ABC we have AB=15. AC=14, and BC=13. What is the radius of the circle through A and B that is tangent to line AC? (No pic in the beginning.)

6. Avogadro's day is celebrated on October 23, a fact of interest chiefly to chemistry students. On what day of the week will Avogadro's day fall in the year 2023? In the year 1999, January 18th was a Monday.

7. A carnival game is set up so that a ball put into play has an equaly likely chance of landing in any of the 60 slots at the base. The operator of the game allows you to choose a certain number of balls and put them all into play. If every ball lands in a seperate slot and you receive 1 buck for each ball played, other wise there is no payoff. How many balls should you choose to maximize your expected winnings?

8. The speed of light is $ 3*10^8$ m/sec. Written in scientific notation, what is the velocity (in m/sec) of an object moving at 12% of the speed of light?

9. Jonathan begins jogging at a steady 3 m/sec down the middle of lane 1 of a public track. Laura starts even with him in the center lane 2 but moves at 4 m/sec. At the instant they begin, Ellis is located 100 meters down the track in lane 4 and is heading towards them in his lane at 6 m/sec. After how many seconds will the runners lie in a straight line?

10. Find the sum of all primes less than 20 which are factors of $ 19^{99} \plus{} 99^{19}$.[/hide]

hello, proof to 4) we have $ \frac{a}{2b}\plus{}\frac{b}{4c}\plus{}\frac{c}{8a}\geq\frac{3}{4}$ this is equivalent with
$ 2ab^2\plus{}4a^2c\plus{}bc^2\minus{}6abc\geq0$ wich is true by AM-GM.
Sonnhard.

Good job! I never thought of it that way...  :blush:

Problem 1
[hide]
$ (r + s)(s + t)(t +r) = 2rst + r^2s + r^2 t + s^2r + s^2t + t^2r + t^2s = (r + s + t)(rs + rt + st) - rst = \frac{-b}{a}\cdot\frac{c}{a} - \frac{-d}{a} = -9*-9-8 = 72$[/hide]
Problem 2
[hide]
$ 50000 \equiv 2 \pmod{6}$ and $ 1111 \equiv 1 \pmod{6}$, so the digit D is 4.[/hide]
Problem 7
[hide]
If the expected winnings for $ n$ balls is $ W(n)$, then the expected winnings for $ n+1$ balls is $ W(n)(\frac{60 - n }{60})(\frac{n+1}{n})$. Then, the maximum winnings are by choosing the first $ n$ such that $ (\frac{60 - n}{60})(\frac{n+1}{n}) = \frac{-n^2  + 59n + 60}{60n} < 1 \Rightarrow n^2 - 59n - 60 > - 60n \Rightarrow n^2 +n - 60 > 0 \Rightarrow n = 8$. 8 balls will maximize the winnings, with an expected value of $ \$$4.91 as compared to an expected $ \$$4.87 and $ \$$4.79 for 7 and 9 balls respectively.[/hide]
Excuse my crappy $ \text{\LaTeX}$

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=251095]#1[/url]
(I have posted this problem with solution in my blog, but with different method)

[quote="varunrocks"]is the answer to #3, 3!
well if he says
q's
1:yes
2:no
3:yes.
then the number must be 3.[/quote]
If it is a multiple of three, there is no reason to ask whether or not it is a perfect cube or not, as there are no such numbers less than 25.  :maybe:

hmmm......is it just me or are the last few really hard? :D

[hide="Problem 8"]First, multiply:
\[ .12 \cdot (3 \cdot 10^8)  \equal{} .36 \cdot 10^8\]

Now, adjusting to fit the requirements of scientific notation,
\[ .36 \cdot 10^8 \implies \boxed{3.6 \cdot 10^7}\]

[/hide]

There's a very simple solution to #1 which I believe no one has put up just yet.

$ r + s = -9 - t $
$ s + t = -9 - r $
$ r + t = -9 - s $

The product is $ (-9 - t)(-9 - r)(-9 - s) $

If $ r, s, t $ are the roots of $ x^{3} + 9x^{2} - 9x - 8 $, then $ x^{3} + 9x^{2} - 9x - 8 = (x - r)(x - s)(x - t) $

Plug in $ x = -9 $

The answer is $ 73 $.


symmetry
conics
parabola
geometry
power of a point
radical axis
geometry proposed

Let c1 and c2 be twoo circles and let P be on the same plain of teh circles. Trace, if it is possible (find the condition of existence for a real solution) , the line(s) through P than determine equal chords  on c1 and c2.

Nice problem!

At first, we should notice that, if the circles have a common chord, then of course, every point $P$ on the radical axis has got the desired property.
So we have to check the cases in where the two equal chords are distinct.

Notice that, if one circle is in the interior of the other circle, then we can not have equal chords (the smaller circle's chord is included in the other chord)

$\hline$
Let:
$C_{1}= (A,a)$
$C_{2}= (B,b)$

Assume that $a\geq b$. Let $M$ be the midpoint of $AB$ and $d$ the radical axis of $C_{1},C_{2}$

Let $l$ be a line wich determine equal chords on $C_{1}$ and $C_{2}$.

We will see that an $'iff'$ condition is that $MK \perp l$, where $K \in l \cap d$

$\hline$

The line $l$ intersects the circle $(A,a)$ at the points $A_{1},A_{2}$ and the circle $(B,b)$ at the points $B_{1},B_{2}$
The points $A_{1},B_{1}$ are in the interior of the segment $A_{2}B_{2}$

Let $K$ be the midpoint of $A_{2}B_{2}$
Of course, $K$ is also the midpoint of $A_{1},B_{1}$, so we have:

$\begin{array}{r}KA_{1}= KB_{1}\\ KA_{2}= KB_{2}\end{array} \}\Rightarrow  KA_{1}\cdot KA_{2}= KB_{1}\cdot KB_{2}$

So, clearly the point $K$ is on the radical axis of $C_{1}$ and $C_{2}$.

Let $C,D$ be the midpoints of $B_{1}B_{2},A_{1}A_{2}$ respectively

We have:
$\begin{array}{r}AD \perp CD \\ BC \perp CD \end{array} \}\Rightarrow  AD \parallel BC$

Since $M$ is the midpoint of $AB$ and $K$ is the midpoint of $CD$, we get $KM \parallel AD \parallel BC$, hence $KM \perp CD$

$\star$ We said that $KM \perp CD$, but we didn't check the case $K\equiv M$. Since $K$ is on the radical axis, the only case such that $K\equiv M$ is when the circles are congruent.
Indeed, if $a=b$, then we can easy see (by symmetry) that, every line passing through $M$ determines equal chords on $C_{1}$ and $C_{2}$.

So, we have a necessary condition:
if $K \in d \cap l$, then $KM \perp l$

We will show that this condition is also sufficient.

$\hline$

[color=blue][size=150]Construction[/size][/color]

$KM \perp KP \Rightarrow $ We can easy find the point $K$ if we draw the circle $z$ with diameter $PM$. Then $K \in z \cap l$

There is a discussion about the existence of $K$, and another discussion about weither or not the line $l$ intersects the circles

[hide="The first discussion is simple"]Let $O$ be the midpoint of $PM$ (the center of $z$) and $O'$ its projection onto $d$.
$OM<OO' \Rightarrow $ no solution for $K$
$OM=OO' \Rightarrow $ one solution $(K \equiv O')$
$OM>OO' \Rightarrow $ two solutions

Let $Q$ be the parabola with focus $M$ and directrix $d$.

It is clear that, if $P\in Q$ then we have one solution for $K$.
If $P$ belongs to the region which includes the focus $M$ then we have not solution.
In the rest case, we have two solutions.[/hide]

Once we found the point $K$, we draw the line $l$, such that $K\in l$ and $l \perp KM$.

$\hline$

[color=blue][size=150]Proof[/size][/color]

We find the projections $D,C$ of the points $A,B$ (respectively) onto the line $l$.
Since $AD \parallel MK \parallel BC$, the point $K$ is the midpoint of $CD$
We will show that $AD^{2}-BC^{2}= a^{2}-b^{2}$

$K$ is on the radical axis of $(A,a)$ and $(B,b) \Rightarrow  \boxed{2MH \cdot AB = a^{2}-b^{2}}$

We draw the circles $w_{1}= (A,AD), w_{2}= (B,BC)$.

From the construction, the segmetns $KD,KC$ are tangent to the circles $w_{1}$ and $w_{2}$.
Also $KD=KC\Rightarrow K$ is on the radical axis of $w_{1},w_{2}$.

so we have $\boxed{2MH \cdot AB = AD^{2}-BC^{2}}$

as a result, $AD^{2}-BC^{2}= a^{2}-b^{2}\Rightarrow $

$a^{2}-AD^{2}> b^{2}-BC^{2}\ (1)$

[color=blue][size=150]Discussion[/size][/color]

There are three possible cases

$1)\ AD=a$. Then $(1)$ implies $BC=b$. That is, the line $l$ is a common tangent of $C_{1}$ and $C_{2}$

$2)\ AD>a$

$(1) \Rightarrow BC>b$. That is, the line $l$ [u]does not[/u] have common point with the circles $C_{1},C_{2}$.

Obviously, in the cases $1)$ and $2)$ we cannot discuss about chords

$3) AD<a$

$(1) \Rightarrow  BC<b$. The line $l$ intersect the circles $C_{1},C_{2}$.

Let $A_{1},A_{2}\in C_{1}\cap l$ and $B_{1},B_{2}\in C_{2}\cap l$

$A_{1}D = A_{2}D = \sqrt{a^{2}-AD^{2}}$
$B_{1}D = B_{2}D = \sqrt{b^{2}-BC^{2}}$

$(1) \Rightarrow  A_{1}A_{2}= B_{1}B_{2}\Rightarrow $ The chords are equal

$\hline$

$\bullet$ Now, assume that the circles have no common points.
We draw the common tangents of the circles.
Let $E,E'$ the points where the external common tangents intersect $d$.
And $I,I'$ the points where the internal  common tangents intersect $d$.
(Assume that $E,I$ are on one side of $AB$ and $E',I'$ on the other side)

If $K \in \{E,I\}$, then the line $l$ is one of the common tangents, so the length of $A_{1}A_{2}$ is zero (no solution).
If $K \in (IE)$, then we have solution $A_{1}A_{2}= B_{1}B_{2}$
If $K \in (HI)$ or $K$ is between $E$ and infinity, then we have not solution.

We'll have the same results, if we replace $E$ by $E'$ and $I$ by $I'$

$\bullet$ If the circles touch each other then $I \equiv I' \equiv H$. So we have solutions when $K\in (E'E)$.
$\bullet$ If the circles intersect each other, then again it should be $K\in (E'E)$ (because then there are not $I, I'$).

$\hline$

Summary, we have:

$C_{1}= (A,a)$
$C_{2}= (B,b)$

$a\geq b$

For any point $K \in (EE')$ we define the line $l$ which passes through $K$ and is perpendicular to $KM$.
$\star$ If the circles have not common points then we shall choose $K$ such that $K \not \in II'$
Then the line $l$ cuts from the circles $C_{1},C_{2}$ equal chords.

$\star$ If the circles are congruent, then any line passing through $M$ determines equal chords.
$\star$ If the circles intersect each other, then their radical axis determines equal chords.
$\star$ If $C_{2}$ is in the interior of $C_{1}$, then there is not solution.


[hide]
$f(n) = f(n+16) \Rightarrow f(n) = f(n+64)$

$f(n) = f(n+9) \Rightarrow f(n) = f(n-63)$

Combining them, we get $f(n) = f(n+64) = f(n+64-63) = f(n+1)$. By induction, $f(n)$ must be constant.

This holds true for $f(n) = f(n+a)$ and $f(n) = f(n+b)$ with $(a,b) = 1$. [/hide]

If I have only $f(n) = f(n+16)$
Is it enougth to conclude that $f$ is constant?
if not please show some example for$f$

thank you
Tummy

Of course not. Take, for example, $f(n) = 0$ if $n$ is even and $f(n) = 1$ if $n$ is odd.

[quote="RohanSingh"]Let f be a function from the set of Natural numbers to real numbers. If $f(n+9)=f(n)$ and $f(n+16)= f(n)$ for every n then show that f is a constant function.[/quote]

[hide]
The 1st and 2nd equation together imply that $f(n+7) = f(n)$. This combined with the first equation implies $f(n + 2) = f(n)$. And $f(n+2) = n$ and $f(n + 9) = f(n)$ implies that $f(n + 1) = f(n)$(just apply it 4 times). Thus the function is constant. [/hide]


I saw this on a different forum before, want to try it here.

The idea is to post an image that defeats the image above it.  For example, if the last post is of a mouse you could post a cat, or a snake, or a mousetrap, or a biologist...
If you need to upload images http://www.imageshack.us is a good place to do so.

[img]http://edbatista.typepad.com/edbatista/images/2005/05/Lab%20Rat.jpg[/img]

[img]http://lisa.jpl.nasa.gov/gallery/images/stellar-mass-black-hole.jpg[/img]

Black hole pwnage.

[img]http://www.sav.sk/uploads/a0054117906/Einstein%20-%20jazyk%20cb.jpg[/img]
Brains, not brawn.

[img]http://tell.fll.purdue.edu/JapanProj/FLClipart/Adjectives/strong.gif[/img]

Brawn, not brains.

i'm tempted to post a picture of extreme side effects of steroids, but i should know better...

Math defeats all :lol:[img]http://www.jiskha.com/images/subjects/mathematics/mathematics.gif[/img]

I have looked all over the web......cant find one that beats math....hmm......nothing beats math!!!no fair!!

That's not math; it's just a bunch of Greek letters. And everyone knows the Romans completely pwned the Greeks.

[img]http://www.pbs.org/wgbh/pages/frontline/shows/religion/maps/art/roman.gif[/img]

[img]http://www-tc.pbs.org/wgbh/pages/frontline/shows/choice2004/art/p_bushmainp.jpg[/img]

We all know Bush would annhilate the Romans with his military power  :P  :D  ;)

[img]http://money.cnn.com/2003/09/04/pf/saving/pepsi_monkey_game/monkey.03.jpg[/img]

'Nuff said.

[img]http://www.themoderatevoice.com/files/joe-mosquito.jpg[/img]
owned by bloodsucker

Owned by anti aircraft gun. It's just slightly overkill.

[img]http://forum.apan-info.net/spring99/images/spring99/high/fe12.jpg[/img]

http://www.kill-more-people.de/images/terror/Master_Chief_over_looking_9_11_by_mpincopallino.jpg

Yeah. Can't attach it.

[img]http://www.familieswithpurpose.com/images/motherdaughter.jpg[/img]

Best picture I could get...
(Your mom)

[img]http://www.ritilan.com/archives/images/2005/03/08/mr-t-mom.png[/img]

(I knew Chuck would come up, so this one is directed at him as well)

[img]http://ironmaiden.webvis.net/images/Albums/Iron-Maiden/Iron-Maiden.jpg[/img]

Iron Maiden=Awesome Rock Music

Awesome Rock Music > All

On a side note, would you say some people's obsession with pi is...[i]irrational?[/i]

[quote="krustyteklown"]ahh, lets continue with the opinions:
[img]http://www.eminemitalia.it/images/hiphoptimeline/dr_dre.jpg[/img]

Dr. Dre's beats>rock[/quote]
[img]http://www.baseballhalloffame.org/games/images/dale_holman_bat.jpg[/img]

I was going to make some sort of wordplay with the bat beating Dr. Dre in the face. You know, cause he is a rapper, and you referenced his beats. Instead I will just leave this picture of a bat here and leave you to draw your own conclusions.

Baseball, tarnished by steroids. Owned.

[img]http://www.seapin.org/drugs/steroids.jpg[/img]


This is really a nice problem. It does not require any knowledge.

Given 2003 points in the plane, no three of which are collinear and no four of which concylic. Prove that there exists a circle trough three of these points such that exactly 1000 of the given points are inside this circle and the other 1000 are not inside this circle.

[quote]It does not require any knowledge.
[/quote]

...

Yes it does.

 8-)

No, it does not.

(Ok, there is a circle trough any 3 points in the plane - is that knowledge ?)

Well ? I expected an immediate solution :?

Hehe. You'll be surprised by the simplicity of the solution. 8-)

I think I have a solution, but it's not "simple", it uses some plane geometry (angle inside and ouside a circle), I give it nevertheless (I'm sure you have a 2 lines solutions  :evil: ) :

Let A,B,C concyclic.

If M is outside circle (and on the same side as C) angle (A,B,M) < (A,B,C).
If M is inside circle (and on the same side as C) angle (A,B,M) > (A,B,C).

Let A, B two points of the 2003 such as all other points are on one side of line(AB). Order (increasingly for example) all angle (A,Pi,B). Take P the 1001th point. Then there are 1000 points with smaller angle (so outside the circle (A,B,P)) and 1000 points with biger angle (so inside the circle (A,B,P)).

would an [hide]algorithmic proof[/hide] work?

How do you mean ? My solution is essentialy one drawing.

oh. i was thinking along the lines of: choose any old circle, if there are too many points in inside, choose the "next smallest" one, if there are too little, choose the "next biggest" one. I didn't know how to quantify "next smallest" and "next biggest" though. I was thinking something like: choose the point whose distance from the center is closest to the radius, but I didn't know which point it should replace. Anyways.

Consider two fixed (adjacent) points from the convex hull of the given points. Then consider the 2001 circles through these two points and each time one of the 2001 other points.

OK ? Do you see it ?


let $ \lfloor c\rfloor$ the biggest integer which is not bigger than c.

find the range of x when $ \left\lfloor \frac {3}{x} \right\rfloor \plus{} \left\lfloor \frac {4}{x} \right\rfloor \equal{} 5$.

Thanks.

Let $ k\equal{}\frac{1}{x}$,the question becomes $ \lfloor 3k \rfloor \plus{} \lfloor 4k \rfloor\equal{}5$,
obviously $ k$ is positive and $ \lfloor 4k \rfloor \ge \lfloor 3k \rfloor$

So Case 1: $ \begin{cases}\lfloor 3k \rfloor\equal{}2 \\
\lfloor 4k \rfloor\equal{}3 \end{cases}$ $ \Leftrightarrow \begin{cases} \frac{2}{3} \le k <1 \\
\frac{3}{4} \le k <1 \end{cases}$ $ \Leftrightarrow \frac{3}{4} \le k <1$ is the solution.

Case 2: $ \begin{cases} \lfloor 3k \rfloor \le 1 \\
\lfloor 4k \rfloor \ge 4 \end{cases}$ $ \Rightarrow \begin{cases} 3k<2 \\
4k \ge 4 \end{cases}$, impossible.

So $ \frac{3}{4} \le k <1 \Leftrightarrow \frac{4}{3} \ge x >1$ are the only solutions.


MATHCOUNTS
geometry
articles
probability
AMC 10
AMC

I got one question. Whenever  I practice on the old AMC tests or pastAHSME tests, my score varies from good to very bad. And I saw that this occurred because I'm so familiar with this test format even though I know how this works.

How do you people who took this test before do? Like I remember from Mathcounts that to qualify to next contest (like chapter to state, state to national, blahblah), I have to use the time carefully and get most right as possible on the Sprint Round.

I enrolled on Intro to Geometry class which is an awesome class that helped me A LOT on my geometry and I have Art of Problem Solving book one and the Mathematical Circles (Russian Experience). I do math chat every Friday (since last week) and constantly practicing for my AMC by using old AMC tests.

But the question that I get is that whenever I do the practice tests, my score goes up and down so easily. Yesterday, I took one AMC-10 test and I totally flunk on it :( and today, I took the another one and I got 109 points (which isn't too bad but not that good neither). And so, I'm here wondering what should I do.

I read the articles at GoodleyMath0Catch, which stated a lot of good things but most of them I'm doing now and also the articles by Richard Ruscryzk(sorry, I can't spell it now. yawn).

I talked with some people too. All they're saying is "you're going to make it in no time. I didn't even know how the format was."

Simply, I need a practical test taking strategy of AMC other than solving how many and leaving how many.

Any kind of reply would be helpful.

Thanks.

I really appreciated it.

Make sure you understand the solution to each and every problem.  All of them.  If you have to, memorize the solutions.  As humans, we tend to create trends.  Something that worked in one problem will likely show up again.

Yeah - Sometimes that happens to me also. I always try to understand the solutions to problems I don't understand, and if I still don't, I post it here or ask a friend. I think it's very important to look at problems you didn't get right, and think: Do I understand the solution, and if you don't, think, what should I do to understand the method or solution. Like Magnara said.

Yeah, I don't know why. I took another AMC test today and got 112.5, which so far my best score :D Of course, it is usually not a cutoff score so I have to aim higher but still, it seems improving..

All I need to do is solve math problems I have... I just can't squeeze my time to it..

And don't answer questions unless you are at least 75% sure that you are right.  If you just guessed the whole thing, you would get a 30 on average; if you leeave things blank, you can get a 62.5.  Of course, I'm not suggesting that you leave everything blank, but don't answer questions you have no idea of doing.

Ouch 30. You would have to know all the answers to get something that bad... or a 0. You would need to be a genious to get a 0. The probability of that is about $(\frac{4}{5})^{25}$, which although considerably higher than all right, is rediculously low.

Also, it is usually a good idea to, after the test, try your best to solve the ones you didn't answer.  I did that in 7th grade, and in 8th grade got a 136.  And in 8th grade I did all the ones I didn't answer and it has, I think, helped me for future tests.


geometry
circumcircle
geometric transformation
reflection
perpendicular bisector
angle bisector
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14381[/img]

 :(

Let $ h_a$ be the perpendicular to the triangle sideline $ a \equal{} T_aH_a$ at $ H_a.$ The $ \triangle H_aBH_c \sim \triangle ABC$ are similar by SAS $ \Longrightarrow$ $ \angle T_aH_aH_c \equal{} \angle CAB \mod 180^\circ.$ Draw the circumcircle $ (P)$ of the $ \triangle T_aH_aH_c.$ Perpendicular bisector of the segment $ H_cT_a$ cuts this circle at diametriaclly opposite points $ Q_1, Q_2.$ Circles $ (Q_1), (Q_2)$ with radii $ Q_1T_a, Q_2T_a$ can cut the A-altitude line $ h_a$ at up to 4 points $ A_1, A_2, A_3, A_4,$ such that $ \angle T_aA_iH_a \equal{} \frac {\angle T_aH_aH_c }{2}$ or $ \angle T_aA_iH_a \equal{} 90^\circ \minus{} \frac {\angle T_aH_aH_c }{2}.$ $ A_iH_c$ cuts the line $ a \equal{} T_aH_a$ at $ B_i$ and reflections of $ A_iB_i \equal{} A_iH_c$ in $ A_iT_a$ cut the line $ a \equal{} T_aH_a$ at $ C_i,$ yielding up to 4 different triangles $ \triangle A_iB_iC_i,\ i \equal{} 1, 2, 3, 4.$ For some of these triangles, $ T_a$ might be fooot of the external angle bisector of the angle $ \angle C_iA_iB_i$ and these can be eliminated by inspection. But configurations of the given points $ T_a, H_a, H_c$ exist, such that all 4 solutions are acceptable.

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14648[/img]

 :thumbup: