Find all composite numbers $n$ having the property that each proper divisor $d$ of $n$ has $n-20 \le d \le n-12$.

[quote="nicetry007"]Suppose $ n = a \cdot b$.

We have

$ n-20 \; \leq \; a \;\leq \; n-12$

$ n-20 \; \leq \; b \;\leq \; n-12$.

multiplying the two inequalities, we get

$ (n-20)^{2}\; \leq \; n \;\leq \; (n-12)^{2}$

The first inequality $ (n-20)^{2}\; \leq \; n$ gives us $ 16 \; \leq \; n \; \leq \;25$. The second inequality $ n \; \leq \;(n-12)^{2}$ gives us $ n\; \leq \;9$ or $ n \; \geq \;16$.

Since we need to satisfy both the inequalities simultaneously, we get $ 16 \; \leq \; n \; \leq \;25$.

Suppose $ n$ is even. Then we have $ n-20 \; \leq \; 2 \;\leq \; n-12$, which gives $ 14 \; \leq \; n \; \leq \;22$ and $ n-20 \; \leq \; \frac{n}{2}\;\leq \; n-12$, which gives $ 24 \; \leq \; n \; \leq \;40$. Since there is no $ n$ that satisfies both the inequalities, $ n$ cannot be even.

This leaves us with $ n = 21$ and $ n=25$.  It is easy to verify that both of them satisfy the required conditions.[/quote]

n is one such composite number
let $ \ d_{1} $ is a proper divisor of n.
$\ d_{2}= \frac{n}{d_{1}} $

so $ \  n-20 \leq d_{1} \leq n-12 $ and $ \  n-20 \leq d_{2} \leq n-12 $
so $ \  (n-20)^{2} \leq d_{1}d_{2} \leq (n-12)^{2} $ [ if $ \ n \geq 20 $]
$ \  (n-20)^{2} \leq  n  \leq (n-12)^{2} $ [ if $ \ n \geq 20 $]

for $ \ n \geq 20 $ from $ \  (n-20)^{2} \leq  n   $ we get $ \ n \leq 25 $
for $ \ n < 20 $ from $ \ d_{1} \leq n-12 $ we see $ \ n \geq 14 $
so $ \  14 \leq n \leq 25 $

so the only possibilities are 14,15,16,18,20,21,22,24,25.

from here easy to see 21 and 25 are only solutions.


number theory
prime numbers
number theory unsolved

Find all positive integers u,v (u > 1) and all prime numbers p,q,r which satisfy the equation p^u = q^v + r^u. The numbers, u,v,p,q,r need not necessarily be different.

Are you sure u>0 and not u>v? Because if not the problem is too challanging.
If p=2, u=v=1,then this problem comes to finding all q,r such that q+2=r (all pairs of twin primes - kind of a hard problem for most of us).

I think Vali is right. But anyway I will check the problem statement again. BTW the problem is from the [i]Mikls Schweitzer competition in 1955[/i].


Yesterday my Alcumus report said that I had passed 5 topics and mastered 9; my Alcumus score was in the 60s. But today when I checked, there were only three subjects on the report page, all of them past due. My Alcumus score had dropped to 50, and all my badges had disappeared. Can somebody please look into this and tell me what happened?

Hello. The developers of Alcumus are looking into the issue. It looks like there might be some database corruption.

Hmm...the same thing happened to me, except i kept my score. But there was practically nothing on my report/badge page until i did a few more problems. Also, most of my videos disappeared except for the ones i passed(all the easy stuff). I've only been a member for a week or so, so i have no clue what is going on. I guess i probably need to re-earn the videos and badge things or something. O well. I don't mind. I like doing math.

The back end server occasionally crashes, which it did today. When this happens, we need to restart it. During the first several hours, it's likely your score won't be accurate as the server has to crunch through all the data to recalculate everything.

does that mean my rank as first in the leaderboard (alcumus score) won't stay there when it finishes 'crunching'? :( sad face.

I checked my Alcumus score and it was 50, 10th out of everybody. I did one problem and it went to 9th...

I still have my points, but everything else is gone.

I checked my alcumus score and it was 50. Then i got 200 more points and now i'm first :)

Alcumus isn't working well right now. We hope the guy who works on the Alcumus server can look at it some time this week.

[quote="PowerOfPi"][my friend] and I have the same ranking (Alcumus score) for some reason, though our Alcumus scores differ by more than one. Why?  :huuh:[/quote]

Yup the exact same thing happened to me.  I'm 26 now too o_O

I also now have no videos available and my whole report has been erased (including the badges).

I am... 9!

But all my badges disappeared!! :( And only 3 topics appear!

The same thing happened to me, i'm now 26th but nothing else disappeared, and also, is you want to access your videos now, I believe there's a complete list at http://www.artofproblemsolving.com/Alcumus/Videos/VideoList.php/.

We definitely get the point already. As levans said, the Alcumus people are already looking into what's going on, so we seriously do not need to continue posting confirmations of the case and we really should just wait.


Let $ABC$ be a triangle. Circle $\omega$ passes through $A$ and $B$ and meets sides $AC$ and $BC$ at $D$ and $E,$ respectively. Let $F$ be the midpoint of segment $AD.$ Suppose that there is a point $G$ on side $AB$ such that $FG\perp AC.$ Prove that $\widehat{EGF}= \widehat{ABC}$ if and only if $\frac{AF}{FC}= \frac{BG}{GA}$.

[quote="April"][color=darkred]Let $ABC$ be a triangle. Circle $\omega$ passes through $A$ and $B$ and meets sides $AC$ and $BC$ at $D$ and $E,$ respectively.
Let $F$ be the midpoint of segment $AD.$ Suppose that there is a point $G$ on side $AB$ such that $FG\perp AC.$
Prove that $\widehat{EGF}= \widehat{ABC}$ if and only if $\frac{AF}{FC}= \frac{BG}{GA}.$[/color][/quote]
Another very nice MOSP problem, and here is my solution  :)

[color=indigo][b]Proof.[/b] Denote: $H\in BC: \,FH\parallel AB,\,K=DI\cap AB$

$\bullet\,(\Longrightarrow)$ Assume that: $\widehat{EGF}=\widehat{ABC}$
We have: $\widehat{EGF}=\widehat{ABC}=\widehat{EHF}=\widehat{EDF}$

So $D,\,E,\,F,\,G,\,H$ are concylic.
$\implies\widehat{DKA}=\widehat{DHF}=\widehat{DGF}=\widehat{FGA}$
$\implies DH\parallel FG\implies DFGH$ is a rectange
$\implies GH\parallel AC\implies FH\parallel AB\implies\frac{AF}{FC}=\frac{BH}{HC}=\frac{BG}{GA}$

$\bullet\,(\Longleftarrow)$ Assume that: $\frac{AF}{FC}=\frac{BG}{GA}$
We have: $\frac{AF}{FC}=\frac{BH}{HC}=\frac{BG}{GA}\implies GH\parallel AC$
$\implies AFHG$ is a parallelogram $\implies DFGH$ is a rectange $(1)$

On the other hand, $\widehat{DEH}=\widehat{DAB}=\widehat{DFH}\implies D,\,E,\,F,\,H$ are concylic $(2)$

From $(1)$ and $(2)$ we have: $D,\,E,\,F,\,G,\,H$ are concylic $\implies \widehat{EGF}=\widehat{EHF}=\widehat{ABC}$[/color]


calculus
integration
trigonometry
calculus computations

can someone give me a hint to calculate these:
$ \int ((x\minus{}s)^2\plus{}y^2)^{\minus{}3/2}ds$, $ \int((x\minus{}s)^2\plus{}y^2)^{\minus{}1/2}ds$

You probably want a trigonometric substitution; something like $ s \equal{} y \tan \theta \plus{} x$.

Hey,
[quote="Clausius"]
can someone give me a hint to calculate these:
$ \int ((x \minus{} s)^2 \plus{} y^2)^{ \minus{} 3/2}ds$, $ \int((x \minus{} s)^2 \plus{} y^2)^{ \minus{} 1/2}ds$
[/quote]
Are: $ {x}$ and $ {y}$; supposed to be constants?

Thanks,

-APSStudent

Hey,
[quote="t0rajir0u"]
The integral is with respect to $ s$, so what you're really asking is whether $ x$ and $ y$ depend on $ s$.  In the absence of other information, we should assume the answer is "no."
[/quote]
Hmm, in that case do we assume that,
$ {y} \equal{} {f(x)}$
$ {x} \equal{} {f(y)}$
Or do we literally treat: $ {x}$ and $ {y}$; as constants?

Thanks,

-APSStudent

Hey,
[quote="t0rajir0u"]
Again, the integral is with respect to $ s$.  What does that mean?
[/quote]
Well since the integral is with respect to $ {s}$ we are only interested in how $ {s}$ varies along the integration. In other words $ {s}$ is the variable of integration. As you have also pointed out that not being given any more information about: $ {x}$ and $ {y}$; we can assume they do not depend on $ {s}$. I would guess the point your driving at (wish you would be more clear :roll:) is that since the integral is with respect to $ {s}$ all other variables are treated as constants, irregardless of their dependability on one another.

Thanks,

-APSStudent

Hey,
[quote="t0rajir0u"]
In the absence of other information, yes.  If $ x$ or $ y$ were dependent on $ s$ or each other we would expect the expressions $ x(y)$ or $ y(x)$ or $ x(s), y(s)$, but these expressions do not appear.  And "irregardless" is not a word.  People who use that term conflate "regardless" and "irrespective."
[/quote]
Thanks for the confirmation.
While you are correct that irregardless is a combination of irrespective and regardless, it is in fact a word (used in "nonstandard speech or casual writing"*).
*Link - [url]http://dictionary.reference.com/browse/irregardless[/url]

Thanks,

-APSStudent

[quote]Usage Note: Irregardless is a word that many mistakenly believe to be correct usage in formal style, when in fact it is used chiefly in nonstandard speech or casual writing. Coined in the United States in the early 20th century, it has met with a blizzard of condemnation for being an improper yoking of irrespective and regardless and for the logical absurdity of combining the negative ir- prefix and -less suffix in a single term. Although one might reasonably argue that it is no different from words with redundant affixes like debone and unravel,[b] it has been considered a blunder for decades and will probably continue to be so.[/b][/quote]

From the link.
T0ra is correct.


I got a private message from a member to explain how to draw geometric figures. Since I used attachments of Sketchpad figures converted to GIF format (the smallest size possible), I have never used Geogebra before. So I tried it for the first time, made 4 figures and posted them. Please, bear with me.

I cannot use certain point names: trying $ X$, $ X_0$ gives message "invalid geogebra input", but cheating with $ X_.$ (X underscore period) works. The same for segment names: Geogebra does not accept $ x$, $ y$, but $ x_0$, $ y_0$ or cheating with $ x_.$, $ y_.$ works.

Trying to color circles, lines, segments works about 30% of the time, but mostly, geogebra does not accept the selected color and picks some other, completely different one, but not at random - always the same. I tried to use the simplest possible colors, such as pure blue, pure magenta, pure green, medium gray, ... But geogebra has a mind of its own for other colors as well.

[color=red]EDIT: Everything is fine, when coloring the circles, lines, segments AFTER the drawing is completed. Beats me.[/color]

Occasionally, geoalgebra does not construct intersection of a segment with a circle, when the other intersection is already constructed. This happened several times, but it is not reproducible. In all cases, after constructing a new segment between the same points, geoalgebra constructed both intersections of the circle with the new segment correctly. Beats me.

Are these geogebra glitches or am I to blame ?

I'm not sure but if they are Geogebra problems we really can't do much about it unfortunately :(

Thank you for the reply. I will try more drawings.

[quote="yetti"]
I cannot use certain point names: trying $ X$, $ X_0$ gives message "invalid geogebra input", but cheating with $ X_.$ (X underscore period) works. The same for segment names: Geogebra does not accept $ x$, $ y$, but $ x_0$, $ y_0$ or cheating with $ x_.$, $ y_.$ works
[/quote]
Some of it is easy to explain: $ x$ and $ y$ are reserved for function variables, so to have them as segment names as well would utterly confuse Geogebra. I'm not sure what the problem with $ X$ is: it works perfectly well on my home computer but not on the site :? On the other hand, $ X_0$ does work on the site. Anyway, there is always a way to cheat by creating a text box positioned where you want instead of a label (sometimes it even looks nicer).

[quote]
Occasionally, geoalgebra does not construct intersection of a segment with a circle, when the other intersection is already constructed. This happened several times, but it is not reproducible. In all cases, after constructing a new segment between the same points, geoalgebra constructed both intersections of the circle with the new segment correctly. Beats me.
[/quote]
Seems like a glitch but may also be due to the mouse location. Make sure that you locate it close enough to the intersection point when clicking.

[quote]
Are these geogebra glitches or am I to blame ?
[/quote]
Don't forget that the site uses the pre-release version. It is assumed to allow more but it has more bugs as well ;)


[quote="1234567890"]I'm enrolled in two courses this term. What courses will be offered next term, and how long will that term last? Is it possible that the next term could have several short, concentrated courses that finish in 6 weeks? Also, do you recommend finishing all the introductory courses before plunging into specific areas in intermediate?[/quote]

If by "next term" you mean this summer, we will not set the schedule for several months.

We are however thinking about running concentrated courses as you suggest and we did so during our first summer.

I think whether or not you take all of the Introductory courses depends on where you are in your own studies.  I would suggest reevaluated which courses are best for you after you have completed the two classes in which you are enrolled.  You will be your own best judge of whether or not other Intro classes are appealing or whether you are in need of Intermediate classes or both.


algebra
function
domain
algebra proposed

Let $n\in{\bf N}^*$. Find \[ \displaystyle\min\limits_{x\in {\bf R}}\sum\limits_{i=1}^n i | x-i|\;. \]

On each of the domains $(-\infty,1]$, $[1,2]$, . . . $[(n-1),n]$, $[n,\infty)$ the expression is a linear function of $x$, which means that on each of those sections the minimum value of the function occurs at one of the endpoints.  So the overall minimum is either $\in\{1,2,\ldots, n\}$ or occurs as $x\to\pm\infty$.  As $x\to\pm\infty$, the function becomes arbitrarily large, so the overall minimum occurs for some $x\in\{1,2,\ldots, n\}$.

Note that the gradient of the function in the section $(m,m+1)$ is $(m+1)^2-\frac{n(n+1)}{2}$, so the function is increasing in the sections $(m, m+1)$ for which $m>\sqrt{\frac{n(n+1)}{2}}-1$ and decreasing in the earlier ones. Hence the minimum of the function is its value when $x$ is the integer part of this expression.

The gradient of the function in the section $(m,m+1)$ is $m(m+1)-\frac{n(n+1)}{2}$ !

Agrippina, your solution is wrong!

Yes! The solution is \[ \left[\sqrt{\frac{n(n+1)}{2}+\frac{1}{4}}-\frac{1}{2}\right] \].


I woz vondering, hu r the ppl that write on Mathlinks?
I min, for example, Lagrangia, are u teacher, student or smb else.
I min, it wud be nice if we wud nou who is each of us and now for example wich grade we are studying.
I'm in the 10th now.
Hope everybody understood what I wrote.

you are probably speaking about the Romanian members of the MathLinks community, aren't you ? :)

Heya there!

You know Dust .. you can always send private messenges and ask... :D

cheers!

I was speaking about all the members.
Anyway, I mean some are really good problem solvers, so I was wondering why they are not in the Romanian IMO Team.
That's it.
BTW, writing in this forum made my English writing very bad ( I had some problems with the writing style :) :D you=u and so on ) :)
:D

they are not in the Romanian IMO Team because ....... well as weird as it looks .... :D :D .... they are not all Romanian!

Arne for example is in the Belgian IMO Team, etc. :)

as for me,i am in 12th grade,and at the crossroads of my career and the life ahead :D

no i dont know imam12 :D
wot made u think so,valentin (the fact that both of us are indians :D

yeah ... in romania I know almost all the pretendats to the IMO Team since say 1996 to today (personally).

it seemed normal for this to happen in india too ... but then again you guys are 10 - 100 times more numerous (as population) then Romania - so that might explain it :D

i am 18 years old and i am just about to start at university (trinity college, cambridge, england :D ) this year

So we started introducing ourselves or what? :D In that case I'm (almost) 18, I'm abt to begin 12'th grade (in Romania that's the last highschool year) AND, more important ( :D ) I'm abt to get my driver's licence..

I almost forgot: Erica, isn't Trinity College where Newton studied? That is cool! (I mean studying at the same place he studied!)

Hey , I'm 18 , from the National College "Vasile Alecsandri" from Galati . I already got my driver's license ( :D ) and now I will be also in the 12th grade. I've been a moderator on this forum for about an year and I am a participant at romanian olympiads from 7th grade (sure was young back then:D) . That's about it , or else it's gonna get boring :)

Okay..

I am 16 years old, I'm 11th grade, from National College "Dinicu Golescu" in Campulung-Muscel, Arges. I have been to the National Olympics in Informatics in the 9th grade. I guess that maths and computers are what I like most, I also love foreign languages especially English and German. I have also been in the college's voleyball team with which I won 2nd place in Romania! :D and after winning it I "retired", as I had some eyes problems :( ( now I am okay :D ). I have become a moderator on this forum this year in August (that's a lucky month for me.. in that month I was born) hehe...

guess that's all for now..

cheers! :D :D

haha, i've worked so hard over the past months that I just might surprise myself

[quote="Valentin Vornicu"][quote="Syntax Error"]haha, my first reply. im in 8th grade at the moment. but its not like i spect to get to imo. unless i move to some 3rd world country from the us[/quote]
don't worry I know some US guys who went since 9th grade - Reid, Tiankai, Caroll - they all scored gold in 9th :D actually they all scored gold in 2001 :)[/quote]

Neal never worries.  I, on the other hand, am worrying a lot.

I'm 16, in 11th grade. I'm generally considered an exceptionally good math student, but in reality I'm not nearly good enough to go to IMO or anything like that. I got 4/15 on the AIME taking it for the first time in 9th grade, which was better than anybody else in my school...

[of course, there is a reason for that. My public school was in the same school district as Thomas Jefferson, in Fairfax County, VA. So all the really smart people in my area went there...]

Heh... and on the 3rd world country thing: yes, that may get you in. I was supposed to go to the qualifying competition for the Nicaraguan team when I lived there in 7th grade... but I was busy over the summer (I would have been beaten anyway--heck, I didn't know how to prove anything--but it is funny to be able to tell people that I came that close...)

On the other hand, it may not work. I wouldn't suggest going to, say, India or someplace like that.


What do you guys think the difficulty of this years AMC 8 was?

~lyra

As a note, since this is not a competition which is considered difficult by mathematical means, please place your answers comparing the competition to previous years.

~lyra

I thought it was easier than previous years.

In reply to Jorian, please read the note above, THIS POLE IS NOT TO RATE THE DIFFICULTY OF THE AMC 8 ITSELF.  It is to rate the 2006 AMC 8's difficulty compared to previous years.

~lyra

I can't answer the poll from personal experience, because I am a coach and not a participant, but it looks like my team did exceptionally well this year, with some new team members and some returning team members, so at the very least it seems that this year's AMC 8 was not unusually hard. (I have not gotten any better as a coach in the last year; let's put it that way.)

When the official statistics are posted

[url=http://www.unl.edu/amc/e-exams/e4-amc08/archive8.shtml]AMC 8 Archive[/url]

it will be interesting to see what the national and state statistics look like. Different minds resonate to different problems, so it's always a little tricky to tell what makes a test easy or difficult.

I just know that this year's AMC 8 was about the same.

After all, it's just a test and depends on how many mistakes you make.

IT was the 1st time I took it, and the only time, but I didn't think it was hard as in terms of difficulty. It was more of how many mistakes you make.

[quote="jhredsox"]there's never been a problem that was hard

most people just make 1 or more mistakes :P
like me

-jorian[/quote]

What are you talking about? Read the statistics, you will find that that statement is very, very, false.

[quote="lyra"][quote="Aryth"]Dang. I bet I got that one problem wrong...

____ got 13 questions correct and 8 questions wrong and he didn't answer 4. What was his AMC 8 score?

 :roll:[/quote]

___ score would be a 13, no matter whether you mark and incorrect answer or you leave a problem blank, only your correct answers are counted.  This is only true in the AMC 8, the AMC 10 and 12 are scored differently, please refer to the AMC webpage for more information: http://www.unl.edu/amc/

~lyra[/quote]
Dude, that was a question on the AMC 8

I know it was a question on the AMC 8, I didn't want the question asker to get confused, and the question asker might want to take a look at the amc website anyway.

~lyra

i did badly due to problems with taking tests when i know they are timed  :what?:  :read:  :wacko: are all pictures of me and tests. man how do you not get like that?
but yeah it was easier than before.

it's okay

When some thing is important, there is a 1 in 50 chance I might freak out. Everyone does. well if they don't then.. i guess they are abnormal  :P


Does there exist a positive integer number n such that :  $ n^6\plus{}26^n\equal{}21^{2009}$

Rewrite it as $ n^{6} \minus{}1\plus{}26^{n}\equal{}21^{2009} \minus{}1$

Now $ n^{6} \minus{}1\plus{}26^{n}$ gives $ 5^n$ $ mod 7$

While $ 21^{2009} \minus{}1$ gives $ \minus{}1$ $ mod 7$.

Contradiction.

What if $ n \equal{} 3$?

$ 5^3\equiv \minus{} 1\pmod{7}$, I think you have a fault in your proof.

Edit: Oh ok, I see

[quote="tonypr"]What if $ n \equal{} 3$?

$ 5^3\equiv \minus{} 1\pmod{7}$, I think you have a fault in your proof.[/quote]
$ 5^n \equiv \minus{}1\pmod{7}$ if $ n\equal{}6k\plus{}3$ but
$ n$ cannot be divisible by 3 as 26 is not divisible by 3 but 21 ..

Hence contradiction...

His proof is correct but incomplete

$ n^6 \minus{}1$ gives $ 0$ $ mod 7$, and $ 26^n$ gives $ 5^n$ $ mod 7$

$ n^6\minus{}1\equiv0\pmod7$ if and only if $ n$ is not a multiple of 7. However, it is clear that $ n$ cannot be a multiple of 7, since otherwise $ 26^n$ would have to be a multiple of 7, so the proof works. (I just wanted to clarify this.)

The problem is from ongoing competition of vietnamese this month' mathematic&youth magazines. The deadline is on the next year' March. So, tnhh is definitely a cheater
I will give you the link later cause I can't access to that page


ii) $ f(x) \equiv 4x\plus{}2 (5)$

notice that $ 4 \cdot 4 \equiv 1 (5)$ then multiple each side by 4
$ 4 f(x) \equiv 4 \cdot 4x \plus{} 8 (5)$
$ 4 f(x) \equiv x \plus{} 8 (5)$
$ 4 f(x)\minus{}8 \equiv x (5)$
$ 4 f(x)\plus{}2 \equiv x (5)$
then $ f^{\minus{}1}(x)\equiv 4 x\plus{}2(5)$

similarly for i) to get
$ f^{\minus{}1}(x)\equiv 3 x\minus{}1(8)$

In general, modular addition ($ f(x)\equal{}x\plus{}k (mod b)$) is always bijective (a.k.a. 1 to 1 and onto), modular multiplication ($ f(x)\equal{}kx (mod b)$ is bijective iff $ k$ is relatively prime to $ b$, and a composition of bijective functions is bijective.


AMC USA(J)MO USAMO ratio geometry HCSSiM AIME

While taking a break from USAMO crud, I decided to update this.  However, I do not know how to properly use the code to retain the indents.  Therefore, I'm just posting the U.S. states (Canadian provinces excluded) with more than 10 qualifiers in 2006 and the overall top 5 rankings from 2001.

CA 57
MA 32
NJ 25
NY 23
TX 22
VA 21
NH 18
MI 17
IL 16
FL 15
WA 15
PA 13
NC 13
MD 12
IN 11

And the rankings are...
1st place: California 214 qualifiers
2nd place: Massachusetts 129 qualifiers
3rd place: Illinois 112 qualifiers
4th place: New York 111 qualifiers
5th place: Virginia 106 qualifiers

Apparently, every single member of last year's cumulative Top Five retained its spot.

If anyone knows how to fix the indents, here's the chart.
[code]
2001        2002       2003        2004        2005     2006
CA 32       CA 30      CA 33       CA 34       CA 28   CA 57
MA 30       NY 27      MA 18       VA 18       IL 17    MA 32
IL 24       MA 24      NY 17       NJ 17        MA 16    NJ 25
NY 19       IL 23      VA 17       IL 16        NJ 14      NY 23
VA 16       VA 21      IL 16       NY 13       TX 14      TX 22
TX 14       NJ 18      NJ 16       NH 11       VA 13     VA 21
NJ 10       TX 18      NH 10                      FL 13      NH 18
               MD 13                                      NY 12     MI 17
              PA 12                                        PA 10     IL 16
                                                                             FL 15
                                                                             WA 15
                                                                             PA 13
                                                                             NC 13
                                                                             MD 12
                                                                             IN 11 [/code]

Um, ok... I don't think that indented right. You might want to attach an excel file.

Most of those states are conspicuously high-population states. It would be interesting to relate the number of USAMO qualifiers from each state to the high-school-age population of each state. Which state does best on a per-capita basis?

Perhaps Virginia should be broken into Northern Virginia and Southern Virginia for this ratio of qualifiers to population  ;) .

[quote="Mildorf"]By USAMO qualifiers in 2001-2005 / population in millions, that top 5 permutes to

1. Massachusetts 15
2. Virginia 11
3. Illinois 7.6
4. New York 4.6
5. California 4.4[/quote]

Wyoming has 4 if you did that way  :D

Yeah, like here in central Illinois we live in cornfield.  So it's not as easy for us to get involved with math when we're too busy looking for the nearest evidence of cities, never mind math communities/good math schools.

[quote="yif man12"][quote="Mildorf"]By USAMO qualifiers in 2001-2005 / population in millions, that top 5 permutes to

1. Massachusetts 15
2. Virginia 11
3. Illinois 7.6
4. New York 4.6
5. California 4.4[/quote]

But Massachusetts also has one of the highest population densities as well. Just another factor that might contribute to the results.[/quote]

How does population density have to do with anything?

Well, population density is correlated to good schools (I think), which is correlated to performance on contests.

Phew, CA is still in the top five after population is taken into account. I was sure dividing by the population would dump us somewhere near last place, since that's where we are in education according to my local newspaper...
What if you don't count qualifiers that only qualify by being top-in-state? If there's any way to keep track of them.

NO, no, no.  When that other person did those rankings by population, he only looked at the top 5 and saw that Calif was at the bottom of the top 5 states with the most qualifiers.  It might possibly be that other states (like NH, for example) have more qualifiers per million than Calif, but were not taken into account because they were not in the top 5 in sheer number of qualifiers.

Mathfanatic, thanks for fixing the indents.  How'd u do that? :?

[quote="bubala"]Mathfanatic, thanks for fixing the indents.  How'd u do that? :?[/quote]

First I used the [ code ]  (no spaces between the brackets) tag to turn the auto-formatting off (any more than one space is usually removed).  Then they still weren't lined up, so I eyeballed (read: trial and error) and corrected until it worked.

NH:
2001: 1 (Exeter: 1)
2002: 7 (Exeter: 6)
2003: 10 (Exeter: 9)
2004: 11 (Exeter: 11)
2005: 9 (Exeter: 8)

Total: 38
Population (2003): 1,287,687
#/Million = 29.51

(If you wanted to be really accurate, we could use the population at each year as opposed to only the 2003 population.)

But NH is somewhat artificial, since Exeter is a complete magnet school: it attracts students from all over the world, not just the DC/VA area (TJ) or, say Illinois (IMSA).  Only two of this year's nine (Joe Ahn and the guy from Manchester West) are from (er, their parents live in) New Hampshire; of the others, three are from the east coast (DE, MA, and NJ), and the rest are farther away (IL, HI, PR, and Korea).

NJ:
2001: 10
2002: 18
2003: 16
2004: 17
2005: 14
Total: 75
Population (2003): 8,638,396

#/Million = 8.68.

Note also that very small "states" (e.g. DC) have an advantage here.  Each year, DC has had one qualifier, but its population is half a million, so its #/million is nearly 10, which puts it pretty high.

Unfortunately for the team I coach, the California qualifiers are not distributed evenly with the population. Southern California accounts for the majority of the population of the state, but only 6 of the 27 or 28 USAMO qualifiers. The Los Angeles metropolitan area, with at least 12 million people, has 4. A large majority of the California qualifiers come from the San Francsico Bay area, and nearly all from the southern portion of the Bay Area: San Jose, Palo Alto, Fremont, etc.

Thank you for the quick and dirty answer on per-capita USAMO qualifiers. The relevant "population," really, is the high-school-age population, and not all states have the same age distribution, so some refinement of the results is possible. Another possible refinement of the results is to take out of the ranking states that only have USAMO qualifiers by the state representation rule in a given year.

Yes mathfanatic, YOUR SCHOOL CHEATED.  ARREST EXETER!!!

That's interesting about Calif.  Why do you think all the qualifiers are clustered in the Southern San Francisco Bay Area?

Because they are very active... Bay Area Math Meet has motivated alot of students to challenge themselves... Last time i tried to move to Los Angeles (when Iwas still in middle school) Orange County because they had good schools (around Irvine), but they are so college-oriented, they only have IB/AP classes, but not those good math teams or math competitions...

[quote="Mildorf"][quote="mathfanatic"]
NH:
2001: 1 (Exeter: 1)
2002: 7 (Exeter: 6)
2003: 10 (Exeter: 9)
2004: 11 (Exeter: 11)
2005: 8 (Exeter: 9)
[/quote]

There seems to be something awry in the 2005 entries.  :D[/quote]

Hehe.  My bad.

[quote]NO, no, no. When that other person did those rankings by population, he only looked at the top 5 and saw that Calif was at the bottom of the top 5 states with the most qualifiers. It might possibly be that other states (like NH, for example) have more qualifiers per million than Calif, but were not taken into account because they were not in the top 5 in sheer number of qualifiers. [/quote]
Oh, darn.

[quote]Note also that very small "states" (e.g. DC) have an advantage here. Each year, DC has had one qualifier, but its population is half a million, so its #/million is nearly 10, which puts it pretty high.[/quote]
How many of the qualifiers qualify through highest-in-state (or in this case, district)? I still think it'd be more accurate if people who don't qualify through index or floor value aren't counted.

My friend was the DC qualifier this year, and I know he made it by having the floor AIME score (and he would've made it as an upperclassman too because of his index being above the cutoff).

It's Jason Kleban--it says on the AMC site.

I see. But that isn't true every year, is it?

[quote="probability1.01"]Could he possibly be Daniel? (I forget his last name, but I think it starts with M :blush: )[/quote]

Daniel Poore?  No M, but I think he's from DC.

Most of the MA qualifiers come from  the eastern part of the state.  But this year Western MA ARML has 3 qualifiers!

[quote="bubala"]
That's interesting about Calif.  Why do you think all the qualifiers are clustered in the Southern San Francisco Bay Area?[/quote]

Culture.  This wasn't the case 10-15 years ago.  Starting in the mid-90s a few things happened:

1) A group of parents and teachers like Paul Zeitz, Tom Rike, Steve Headley, Josh Zucker, Ted Alper, and others chose to create a culture of mathematical excellence there.  I was at Stanford at the time and went to the first couple meetings, then moved out of the area.  They've obviously been very successful.

2) The dot com boom brought in a bunch of talent and parents who value math/science education.  This is reflected to some degree in the interest of their kids.

3) The Miller Middle School built one of the best math programs in the country.  Many of the top students from the Bay Area (like Tiankai Liu) are Miller graduates, and the culture they built there has to some degree inspired students at other schools as well.

We're trying to replicate this in Southern CA with our efforts at SDMC, and Kent Merryfield is doing a ton to do the same in the LA area.  Give us a few years!

A quick addendum to this culture note: From the time I was a sophomore (1987) to the time Mathew graduated high school (mid-90s), Alabama regularly had 1-3 of the 24 MOPers.  Alabama.  Local culture is very important - parents and teachers who would like their students to excel should work hard to build community programs for kids outside just their families and their schools.  It will certainly pay off in the long run.

I had my own experience with Miller Middle School; the California Mathcounts team trained there my year.

In my years (1997-2000), the center of activity in the Bay Area was further north than it is now; there were top students from San Francisco, Berkeley, San Mateo. They aren't there now, and the San Jose/Silicon Valley cluster is much larger now. The majority of the California qualifiers now come from a rather small area.

[quote="bubala"]What is it about Massachusetts then that lends itself to such a smart "math culture"--does anyone know?[/quote]My guess is that they've got all those elite colleges.

Exeter is hardly "away" from great colleges. It's situated in New Hampshire, under the (indirect) influence of Harvard, MIT, Princeton, Yale, UPenn, and Columbia to name a few. It also happens to draw talented students from across the world. Although TJ does only take Virginia students (as far as I know), its program is so well-established that tradition just carries on. TJ is also not exactly a long distance from Princeton, Duke, etc.

Dartmouth is also in New Hampshire, too.

There's a lot of good secondary schools situated in southern and mid New England.

[quote="bubala"]What does that have to do with having great high schools?  For example, Exeter is an amazing HS situated far away from any really great college, and I believe TJ is too, though I'm not entirely sure about that.  Please explain for me.[/quote]

Good universities bring good professors, good professors bring good kids, good kids bring good scores.

>In my years (1997-2000), the center of activity in the Bay Area was further north than it is now;
> there were top students from San Francisco, Berkeley, San Mateo. They aren't there now, and the
> San Jose/Silicon Valley cluster is much larger now. The majority of the California qualifiers now
> come from a rather small area.

I have news for you: it's still shifting! Palo Alto, Cupertino, San Jose are still major players, but Fremont /Milpitas is a much larger factor than previously. When sending out recruiting letters for ARML, it used to be Gunn High School (Palo Alto) and Monta Vista (Cupertino) that required the extra-large envelopes -- they still have substantially more than average, but it is Mission San Jose HS (in Fremont) that gets the biggest share, now.

San Francisco has never been particularly strong since I've been paying attention (c. 1994), though most years there have been a few strong students at  Lowell and (SF) University High School. One year, the top AHSME score in San Francisco was 136 --  but the second highest score in the entire city was below 110...

I don't pretend to know what it means -- there are potentially talented students everywhere, I'm sure!

Ted Alper
SFBA ARML coach, etc.


Prove that there is no positive integer $n$ such that, for $k=1, 2, \cdots, 9,$ the leftmost digit of $(n+k)!$ equals $k$.

For $ m\in\mathbb{N}$, be $ f(m)\equal{}m\cdot 10^{r}$ with $ r$ the unique integer for which $ 1\le f(m)<10$ (i.e. the first digit of $ m$).

Then for $ 2\le i\le9$ we have $ i<f((N\plus{}i)!)<i\plus{}1$, such that \[ 1 < f(N\plus{}i) \equal{} f\left(\frac{(N\plus{}i)!}{(N\plus{}i\minus{}1)!}\right) <\frac{i\plus{}1}{i\minus{}1}\le 3.\] So $ f(N\plus{}i)<f(N\plus{}i\minus{}1)$ and thus $ 1 < f(N\plus{}2) < f(N\plus{}3) < ... < f(N\plus{}9) <\frac{5}4$. Since obviously $ f(ab)\le f(a)f(b)$, is \[ f((N\plus{}4)!)\le f((N\plus{}1)!)f(N\plus{}2)f(N\plus{}3)f(N\plus{}4)< 2\left(\frac{5}4\right)^{3}< 4,\] contradiction.

[quote="Peter"]...... Since obviously $ f(ab)\le f(a)f(b)$........ [/quote]
Excuse me. I don't undertand this one.   $ f(7 \times 13) > f(7)f(13)$ :maybe:

No, $ f(7.13)\equal{}9\cdot8\equal{}7\cdot1.3\equal{}f(7)\cdot f(13)$? I don't understand the objection. Either we have LHS=RHS or we have LHS=RHS/10, so we have the stated inequality?


function
geometry
algorithm
ratio
number theory proposed
number theory

Let  $n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_j^{\alpha_j}$ be the decomposition in prime factors  of a positive integer $n$ .

 If ${\mathbf q}\in {\mathbb N}^*$ define $F: {\mathbb N}^* \to {\mathbb N}^*$ by $\begin{array}{|c|} \hline F(n)=F(n,{\mathbf q})=1+\sum\limits_{\nu=1}^{j}\alpha_{\nu}^{{\mathbf q}}p_{\nu} \\ \hline \end{array}\; .$
Consider the [b] "iterates " [/b] of $F$ that is
$F^2=F\circ F ,\; F^k (x)= (\underbrace{F\circ F \circ \cdots \circ F}_{k-\mbox{{\tiny times}}})(x) \; ,\; k\in {\mathbb N}^*$
[b] i) [/b] [color=blue] [b]  If  ${\mathbf q}=1$  prove (or verify by means of a computer)  that  for each integer $n\; ,\; n>6 ,$
 there exists  ${\mathbf k_0}={\mathbf k_0(n)}\in {\mathbb N}^*$   such that  $F^{\mathbf k_0}(n)={\mathbf 7} \; .$  [/b] [/color]
[b] ii)[/b] [color=blue][b] What about case  ${\mathbf q\ge 2}$  ?  [/b] [/color]

What are the application of the $F$ function you defined ? In what areas of mathematics is it used ?
It looks very interesting :)

[quote="{x}"]........ In what areas of mathematics is it used ?......It looks very interesting :)[/quote]
Here ${\mathbb N}=\{0,1,...\}=$the set of non-negative integers and ${\mathbb N}^*=\{1,2,3,...\}=$ the set of positive integers.

[b]I am not interested to apply Mathematics ![/b] (only if it's possible to  receiev a lot of money).

By the way, [color=red][b] if you prove that there exists an integer ${\mathbf q} \ge 3$  such that for all positive integers $n$ there exists $k_0=k_0(n)$
  with $F^{k_0}(n)=1$ ,    I give you     10 dollars (USA)     . [/b][/color]    .

 However if you want to receive a greater amount of money, please solve (look much simpler problem, but...)
 following  [b]problem proposed in 1932 by Lothar Collatz[/b]  [size=75][b]
(known also as :  Hasse-algorithm,  Problem 3x+1 , Problem from Syracuse , Ulam problem,  Kakutani problem,  Thwaites conjecture )[/b][/size] :
 [color=blue] [b] Let  $T: {\mathbb N} \to {\mathbb N} ,$ where   $\begin{array}{|c|}\hline T(n)=\left\{\begin{array}{lcl} \frac{n}{2} &, & n \; \; \mbox{even}\\ &&\\ \frac{{\bf 3n+1}}{2} &, & n \; \mbox{odd}\end{array}\right.\\ \hline \end{array}\; ,\; \forall n\in {\mathbb N}$  .
 Prove that for all $n\in {\mathbb N}^*$ there exists $k_0$ such that   $T^{k_0}(n)=1$   .
[/b] [/color]

See :
 http://www.newton.dep.anl.gov/askasci/math99/math99009.htm

 http://www.post1.com/~step-by-step

and regarding some  prizes, see :      http://www.math-atlas.org/93_back/prizes.que

There is an award=  10,000 dollars           .
[b]
Regarding this "simple" problem  Paul Erdos has said:    [color=red] "Mathematics is not yet ready for such problems !" [/color]
[Attributed by Paul Halmos , The American Mathematical Monthly, Nov. 1992 ]
[/b]

Thank you for your answer. When asking about the application I meant the mathematical applications not the physical ones :)
Anyway the $10$ dollars prize you propose looks temptating ... (At least more temptating than that of the Collatz-Conjecture, when comparing the prize/difficulty ratio of both problems).

[quote="toumaf"]In fact your $\$10$ prize is impossible to get, because $F(n,q)\neq 1$ for all $q$ and $n$.[/quote]
Of course, thanks for remark !/Proposer

If (m,n)=1 (m and n coprime), then $F(nm,q)=F(n,q)+F(m,q)-1$.
It is easy to show, that $F(n,q) \ge n$ have finite solutions if n is not prime: $n=n(1,q),n(2,q), \dots n(s(q),q)$ for all q. Therefore iterates have one is possible periods if $F(p+1,q) \ge p$ have finite solutions for primes p.

[quote="mathmanman"]Does he get \$10 for this remark? ;)[/quote]
[b]Nice and indeed "good"  question ! [/b]
However , it's not the desired answer of my posted question ! But...if he want, I have the possibility to send him
some books ! [b] I ask him to send me his private e-mail ! [/b]/Alex


there is one?
what about geometric sequence?
(I have no answers).
tnx  :)

That arithmetic sequence doesn't work, even if you allow leading zeroes.  What's the twelfth term?  Anyhow, it is more interesting if you do not allow leading zeroes.

[quote="rem"][quote="JBL"]For the first, it is not hard to see that the answer is no -- the trick is to wait for a point where the size of the terms is "sufficiently large" in comparison with the common difference.[/quote]
You do not want that. ie:
$8,9,10,11$ and we get a palindrome[/quote]
What shall this mean¿

[quote="rem"]Well, for arithmetic sequence:
100, 200, 300, 400...
For geometric sequence:
1,10,100,1000...
Be more specific, please[/quote]
you are wrong if u accept this $010(=10)$
if u don't, what about other arithmetic sequences, can one determine if it containg a palindrome?  :?

srulikbd, please type in English -- I would personally appreciate it.

rem is responding with a mis-understanding of my sketched solution proposal.  So here it is in fuller detail:  To deal with arithmetic sequences, assume for contradiction that there does exist such a sequence.  Then, choose a term of that sequence with many more than twice as many digits as the common difference.  This term will be a palindrome, by assumption.  The next term, however, will not, contradiction.  Is it clear?

Geometric sequences are not so obvious to me ... perhaps do something based on the last digit and the first digit?  I'm not really sure.

[quote="JBL"]rem is responding with a mis-understanding of my sketched solution proposal.  So here it is in fuller detail:  To deal with arithmetic sequences, assume for contradiction that there does exist such a sequence.  Then, choose a term of that sequence with many more than twice as many digits as the common difference.  This term will be a palindrome, by assumption.  The next term, however, will not, contradiction.  Is it clear?
[/quote]
I think you misunderstood the problem. It is not a sequence that consists of only palindromes, but a sequence that has at least one.

Huh ???? I am not able to make much sense out of it.

Assume not all the numbers are palindrome ... Or ..

As long as  there exist a number  $n$ which is not palindrome,  you have
$n,n,n,n \dots$

which by the way is both Arithmetic, and geometric series.  :roll:


geometry
perpendicular bisector
geometry proposed

Let $A_1, B_1$ be variable points on sides $CA, CB$ of $\triangle ABC$ such that
(1) $A_1B_1$ is parallel to $AB$
(2) The perpendicular bisector of $AB$ intersects segment $A_1B_1$ at a point $F$.
Suppose that $A_1B$ and $B_1A$ meet at $E$.  Prove that as $A_1, B_1$ vary, $EF$ passes through a fixed point.

Suppose the points are on the complex plane, with the midpoint of AB=0 and A=a(real number), B=-a, C=x+yi, and $A_1=C+\lambda (a-C), B_1=C+\lambda (-a-C)$.
Since
$\frac{A_1E}{EB}=\lambda $
we have
$E=\frac{A_1-\lambda a}{1+\lambda}=\frac{1-\lambda}{1+\lambda}(x+yi)$
and
$F=iIm(A_1)=(1-\lambda)iIm(C)=(1-\lambda)yi$
Let K=-x+yi which is a fixed point,
by calculating we get $\frac{E-F}{K-F}=(\lambda-1)(\lambda+1)$ is a real number, which completes the proof.□


[quote="Vasc"]Find the smallest value of $ k_1$ and $ k_2$  such that the inequalities below holds for any nonnegative real numbers $ a,b,c$, no two of which are zero:

$ (a) \ \ \ \frac {a(b \plus{} c)}{a^2 \plus{} 2bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} 2ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} 2ab} \minus{} 2 \plus{} k_1(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \minus{} 1) \ge 0$;

$ (b) \ \ \ \frac {a(b \plus{} c)}{2a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{2b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{2c^2 \plus{} ab} \minus{} 2 \plus{} k_2(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \minus{} 1) \ge 0$.[/quote]
Vornicu Schur works very well here. :) My answer is:
a) $ k_1\equal{}\frac{1}{\sqrt{2}}.$
b) $ k_2\equal{}\frac{32}{27}.$
:)

[quote="can_hang2007"][quote="Vasc"]Find the smallest value of $ k_1$ and $ k_2$  such that the inequalities below holds for any nonnegative real numbers $ a,b,c$, no two of which are zero:

$ (a) \ \ \ \frac {a(b \plus{} c)}{a^2 \plus{} 2bc} \plus{} \frac {b(c \plus{} a)}{b^2 \plus{} 2ca} \plus{} \frac {c(a \plus{} b)}{c^2 \plus{} 2ab} \minus{} 2 \plus{} k_1(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \minus{} 1) \ge 0$;

$ (b) \ \ \ \frac {a(b \plus{} c)}{2a^2 \plus{} bc} \plus{} \frac {b(c \plus{} a)}{2b^2 \plus{} ca} \plus{} \frac {c(a \plus{} b)}{2c^2 \plus{} ab} \minus{} 2 \plus{} k_2(\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca} \minus{} 1) \ge 0$.[/quote]
Vornicu Schur works very well here. :) My answer is:
a) $ k_1 \equal{} \frac {1}{\sqrt {2}}.$
b) $ k_2 \equal{} \frac {32}{27}.$
:)[/quote]
Indeed, this is the true answer. :lol:
How Vornicu Schur works?

I will just write the hint, the detail proofs will be left to the readers :)
a) Since $ \frac {a(b + c)}{a^2 + 2bc} - \frac {a(b + c)}{ab + bc + ca} = - \frac {a(b + c)(a - b)(a - c)}{(ab + bc + ca)(a^2 + 2bc)},$ and
$ \frac {a^2 + b^2 + c^2}{ab + bc + ca} - 1 = \frac {(a - b)(a - c) + (b - c)(b - a) + (c - a)(c - b)}{ab + bc + ca},$
we can rewrite our inequality as
$ \sum (a - b)(a - c)( \frac {1}{\sqrt {2}} - \frac {a(b + c)}{a^2 + 2bc}) \ge 0.$
To use the Vornicu Schur method, the terms $ x,y,z$ of $ \sum x(a - b)(a - c)$ must be nonnegative, but unfortunately, the inequality $ \frac {1}{\sqrt {2}} - \frac {a(b + c)}{a^2 + 2bc} \ge 0,$ does not always hold. Therefore, we can't use directly the Vornicu Schur technique. However, with note at the trivial identity
$ (b - c)^2(a - b)(a - c) + (c - a)^2(b - c)(b - a) + (a - b)^2(c - a)(c - b) = 0,$
we can rewrite our inequality in the form
$ \sum (a - b)(a - c) ( \frac {1}{\sqrt {2}} - \frac {a(b + c)}{a^2 + 2bc} + \frac {(b - c)^2}{ab + bc + ca} ) \ge 0,$
It is easy to prove that
$ \frac {1}{\sqrt {2}} - \frac {a(b + c)}{a^2 + 2bc} + \frac {(b - c)^2}{ab + bc + ca} ) \ge \frac {1}{\sqrt {2}} - \frac {a(b + c)}{a^2 + \frac {(b + c)^2}{2}} \ge 0.$
Therefore, we can use the Vornicu Schur technique for $ x(a - b)(a - c) + y(b - c)(b - a) + z(c - a)(c - b) \ge 0,$ where
$ x = \frac {1}{\sqrt {2}} - \frac {a(b + c)}{a^2 + 2bc} + \frac {(b - c)^2}{ab + bc + ca} \ge 0.$
Now, assume that $ a \ge b \ge c,$ we can prove that $ z \ge y$ (the readers can try this) and with the theorem of Vornicu Schur, we can get the result.
Equality holds for $ a = b = c$ and again for $ a = \sqrt {2}b = \sqrt {2}c$ (with its cyclic permutations). :)

b) By the same idea with a), we can rewrite our inequality as
$ \sum (a - b)(a - c) ( \frac {32}{27}\cdot \frac {a + b + c}{ab + bc + ca} + \frac {(a + b + c)(b - c)^2}{2(ab + bc + ca)} - \frac {b + c}{2a^2 + bc}) \ge 0.$
This inequality has the form $ x(a - b)(a - c) + y(b - c)(b - a) + z(c - a)(c - b) \ge 0,$ with
$ x = \frac {32}{27}\cdot \frac {a + b + c}{ab + bc + ca} + \frac {(a + b + c)(b - c)^2}{2(ab + bc + ca)} - \frac {b + c}{2a^2 + bc} \ge 0.$
Now, assume that $ a \ge b \ge c,$ we can prove that $ ax \ge by$ (the readers can try this, too) and with the theorem of Vornicu Schur, we can get the result.
Equality holds for $ a = b = c$ and again for $ a = b = 4c$ (with its cyclic permutations). :)

[quote="Vasc"]Very nice solution, Can_hang.  :lol:
Does this method also work for the inequalities below?

Let $ a,b,c$ be the side-lengths of a triangle. Prove that

$ (a) \ \ \ \ \sum \frac {22a^2 \plus{} b^2 \plus{} c^2}{a^2 \plus{} bc} \ge 36$;

$ (b) \ \ \ \ \sum \frac {a(b \plus{} c)}{a^2 \plus{} 2bc} \minus{} 2 \ge \frac 4{5}(\frac {ab \plus{} bc \plus{} ca}{a^2 \plus{} b^2 \plus{} c^2} \minus{} 1)$.[/quote]
For a), my way still works, Vasc. We can rewrite the inequality in the form
$ \sum \left( 4a\plus{}14b\plus{}14c\minus{} \frac{(b\plus{}c)(22a^2\plus{}b^2\plus{}c^2)}{a^2\plus{}bc}\plus{}\frac{14(b\minus{}c)^2}{a\plus{}b\plus{}c}\right) (a\minus{}b)(a\minus{}c) \ge 0,$
then $ x,y,z \ge 0$ and if we assume that $ a \ge b \ge c,$ we can prove (very easily) that $ z \ge y.$

For b), I haven't checked yet, but I think (with 99% sure  :lol: ) that my way still works.  :lol: I will come back with b) after my teaching this night. :)

For b), we can prove it by rewriting the inequality in the form
$ \sum (a\minus{}b)(a\minus{}c)\left( \frac{4}{5} \cdot \frac{ab\plus{}bc\plus{}ca}{a^2\plus{}b^2\plus{}c^2}\minus{}\frac{a(b\plus{}c)}{a^2\plus{}2bc}\plus{}\frac{2(b\minus{}c)^2}{a^2\plus{}b^2\plus{}c^2}\right) \ge 0.$
Now, assume that $ a \ge b \ge c,$ we can prove that $ bz \ge cy$, and since $ x,y,z \ge 0,$ we can get the result. :)


inequalities
function
real analysis
real analysis unsolved

Prove that the series Σ{ （-1）Λ[sqrt(n)]}/n converges.      []stands for the Gauss's function.
Please tell me your opinion.Thank you!

We can follow this line of reasoning:
the series  $  \sum \frac {-1^{[\sqrt {n}]}} {n^a} $
 converges absolutely if $ \  a > 1 \  $ and diverges if  $ \  a \leq 0 $. We will show that if $ \frac {1} {2} < a  \leq 1 $ the series converges conditionally. Observe that the first three terms of the series are negative, the next five terms are positive, etc. Now grouping the terms of the same sign we get the alternating series $ \displaystyle \sum_ {n=1}^{\infty} (-1)^n C_n \  where \ C_n = \sum_{k=n^2}^{(n+1)^2-1} \frac {1} {k^a} $.
Now if  $ C_n - C_{n + 1} > 0 $ the series for the theorem of Leibniz converges.

\[
C_n - C_{n + 1} = \sum_{k=n^2}^{(n+1)^2-1} \frac {1} {k^a} - \sum_{k=(n+1)^2}^{(n+2)^2-1} \frac {1} {k^a}= \sum_{k=n^2}^{(n+1)^2-1} \frac {1} {k^a} - \sum_{k'=n^2}^{(n+1)^2+1} \frac {1} {(k'+2n+1)^a} = \sum_{k=n^2}^{(n+1)^2-1} \left(\frac {1} {k^a}- \frac {1} {(k+2n+1)^a}\right)  - \frac {1} {((n+2)^2-2)^a} - \frac {1} {((n+2)^2-1)^a} = \sum_{k=0}^{2n}  \left(\frac {1} {(n^2+k^)^a}- \frac {1} {(n+1){}^2k{}^a}\right)  - \frac {1} {((n+2)^2-2)^a} - \frac {1} {((n+2)^2-1)^a} > (2n + 1) \left (\frac {1} {(n^2+2n)^a} - \frac {1} {((n+1)^2+2n)^a}\right) - \frac {1} {((n+2)^2-2)^a} - \frac {1} {((n+2)^2-1)^a}
\]
where the last inequality  follow from the montonicity of the function:
\[
g(x) = \frac {1} {(n^2+x)^a} - \frac {1} {((n+1)^2+x)^a}
\]
on the interval[0,2n]. For sufficiently large n:
\[
C_n - C_{n + 1} > 0
\]

According to the book,the series converges when a>1/2 and diverges when a<1/2.
 So there may be some mistakes in Nené's statements.
 Please tell me your opinion.Thank you!

O.K. ffrogg.
The series, if $ 0 < a \leq \frac {1} {2} $, dosn't converge. Since $ C_n > (2n+1) \frac {1} {(n^2+2n)^a} $, the necessary condition for the convergence of series isn't satisfied.


In his workbook with squares Bernard would like to draw a triangle with
coordinates (0,0),(1,n) and (n,1). Which is the smallest n that he must choose such that there are 1000 or more points with integer coordinates
inside the triangle.

It is taken from Italian selection 2002.

Good bye every body

[hide]With a few sketches, it will be seen that for a given $n$,

 there are $1 + 2 + 3 + ... + (n-1) = \frac{n(n-1)}{2}$ interior lattice points.

Then: $\frac{n(n-1)}{2} > 1000\;\;\Rightarrow\;\;n> 45.224$

Therefore, the least $n$ is $46.$[/hide]

We can use Pick's Theorem to find the number of lattice points inside the triangle. It's easy to find the area of the triangle. Call it $S$. The triangle has $n+1$ points on the sides and say it has $P$ points strictly inside. Then Pick's Theorem says that $S=P+\frac{n+1}2-1$, so we can find $P$ in terms of $n$.


A number has a remainder of 1 when divided by 4, a remainder of 2
when divided by 5, and a remainder of 3 when divided by 6. What is the
smallest number that has the above properties?

$ x\equiv1\equiv \minus{} 3\pmod4,x\equiv2\equiv \minus{} 3\pmod5,x\equiv3\equiv \minus{} 3\pmod6$
$ \implies x\equiv \minus{} 3\equiv\boxed{57}\pmod{60}$.


geometry
geometric transformation
homothety
conics
ellipse

it was about midshapes, does anybody have/know where to get the problems? Did the competitors do a good job (if you saw them?) I would have loved to be in the masters round but needless to say wasn't good enough in other areas of the competition.

I didn't get to see Daniel Li present because I had to go to the Air and Space Museum. Neal was first, he was pretty accurate with his problems and was organized, but he didn't have very good presentation skills. Andrew Ardito did not do very good. Daesun spelled school wrong (schol) but his presentation was pretty good compared to Andrews.

If I had to rank the presentations:

1) Daniel Li - very few mistakes, nice presentation, made subtle observations & impressed judges
2) Daesun Yim - nice, seemed like a clear winner until Daniel came along; made 4 or 5 mistakes
3) Neal Wu - organized, but wasn't well presented.  Nice ideas, but a bunch of costly mistakes
4) Andrew Ardito - bad handwriting, presentation skills, and several mistakes killed Andrew.

Not only that, but after his turn andrew ardito started talking to Neal, and then Treething after he went. He was rude in talking during other's presentation's.

[quote="nat mc"]So, who was the official winner, Daniel Li?[/quote]
yes

More Comments:
-Daniel Li didn't have to write his school because his coach said he was familiar with the projector.
-He was also lefthanded; did that change anything?
-Treething went after Ardito, which should have made him look better.
-Neal solved the problems, but had bad presentation skills.
-Daniel Li studied homothecy in an AoPS class, which gave him an unfair advantage.

Everyone did excellent job with different styles, and each of them has some room to improve to his presentation skills.

[quote="Ignite168"][quote="Boy Soprano II"]"Midshapes?"  What are they?

Yes; I remember Neal wasn't outstanding last year either.  He mumbled and his handwriting slanted on the transparency, which then slanted in the same direction on the projector.  Still, you have to give him credit for winning nationals as a seventh-grader.[/quote]

Take a point and a figure and draw segments from all vertices on the figure to the point. Then, take the midpoint of all those segments and connect them, you have your midshape. There was 7 questions (~4-6 parts to 2-6), Daniel Li said he slowed down at 1 minute left as he didn't get 7b but while he was slow at the beginning, he and Neal answered the most, (got to number 6) Daesun almost got to 6, and Andrew skipped like 20 problems so he was out of it.
I think Daniel did the best job at explaning and presenting, Neal gave the best job at giving straight answers, and Daesun the best at coming up with a different way of doing problems (the 3 other people had a lot of similar answers)

And RC-7th, Daniel actually used the word homothecy in his presentation, so maybe that did give him an unfair advantage :P[/quote]

Yeah, Daniel was good at the 1 minute mark. Everyone else went like:

"So number 6a is like this with an ellipse twice the size, 6b is like this, 6c is like this, 7a is like thi-"
TIME!

*pokes everyone towards RC-7th's other posts XD*

No, he's someone else who got a 41 at nationals that's from Virginia.

In masters, the participents are given a complex problem that they have to find a solution for. They have 30 min. to prepare, 11 to present, and 4 to answer questions.
EDIT: Top 2 written and overall.

Bleh, I meant Daniel Li told us he slowed down at 1 minute left on purpose because he didn't want to present 7b as he didn't get the answer or something. And Daniel Li was trying to hide his AoPS name ;)

I have to say that this year's masters round was incredibly boring.
I only saw Neal, but even though he didn't have great presentation skills (mumbling), he was MUCH better than the professor.
When the professor was presenting, everyone was like, "move your friggin' elbow we can't see the screen!"
It was pretty humorous, actually.

I fell asleep when the professor was going so I was confused when Neal went.


1.A triangle can be cut into two isosceles triangles. One of the angles of the original triangle is 36 . Determine all possible values of the largest angle of the original triangle.
2.A lucky number is a positive integer which is 19 times the sum of digits (in best ten). Determine allthe lucky numbers.
3.Prove that 1596^n + 1000^n -270^n - 320^n is divisible by 2006 for all positive odd integer n.

 :mad:

Can anybody post any hints for number 3?

[hide="Solution to number 3"]$ 2006 \equal{} 2 \times 17 \times 59$, thus it suffices to prove divisibility by 2, 17, and 59.

[b]Divisible by 2:[/b] The expression is clearly divisible by $ 2$ termwise.

[b]Divisible by 17:[/b]
$ 1596 \equiv 15 \pmod{17}$
$ 1000 \equiv 14 \pmod{17}$
$ 270 \equiv 15 \pmod{17}$
$ 320 \equiv 14 \pmod{17}$
$ \therefore (1596^n \plus{} 1000^n) \minus{} (270^n \plus{} 320^n) \equiv (15^n \plus{} 14^n) \minus{} (15^n \plus{} 14^n) \equiv 0 \pmod{17}$

[b]Divisible by 59:[/b]
$ 1596 \equiv 3 \pmod{59}$
$ 1000 \equiv 56 \pmod{59}$
$ 270 \equiv 34 \pmod{59}$
$ 320 \equiv 25 \pmod{59}$
$ \therefore (1596^n \plus{} 1000^n) \minus{} (270^n \plus{} 320^n) \equiv (3^n \plus{} 56^n) \minus{} (34^n \plus{} 25^n) \pmod{59}$
For odd $ n$, we have $ (a \plus{} b)|(a^n \plus{} b^n)$, so $ 59|3^n \plus{} 56^n$ and $ 59|34^n \plus{} 25^n$
Therefore, for odd $ n$, $ (1596^n \plus{} 1000^n) \minus{} (270^n \plus{} 320^n) \equiv 0 \pmod{59}$[/hide]

[quote="watcharapon"]
2.A lucky number is a positive integer which is 19 times the sum of digits [b](in best ten)[/b]. Determine allthe lucky numbers.
[/quote]

What does "in best ten" mean?

[quote="watcharapon"]
2.A lucky number is a positive integer which is 19 times the sum of digits [b](in best ten)[/b].[/quote][quote="RisingStar"]What does "in best ten" mean?[/quote]

I think he means [b]"in base ten"[/b].  :)

For #2:

First note that there are no lucky numbers greater than 999. Assume there is some n-digit lucky number, $ x$, where n > 3. Let $ y$ represent $ 999 \ldots 999$ where there are $ n$ 9's in total.  Note that $ x \leq y$. We note that 19 times the sum of the digits of $ y$ would be $ 19 \times (9n) \equal{} 171n$. Note that $ 10^{n \minus{} 1} \leq x$, since $ x$ has $ n$ digits. It is quite easy to see that $ 171n < 10^{n \minus{} 1}$ when $ n > 3$; this can be rigorously shown with induction. Hence, $ 171n < x$, i.e., $ y < x$. But since $ x \leq y$, we have a contradiction.

Clearly, there are no one-digit lucky numbers. We will now show there are no 2-digit lucky numbers. Assume there exists a 2-digit lucky number. Call it $ x$. We know that $ x \equal{} 10a \plus{} b$, with $ 1 \leq a \leq 9$ and $ 0 \leq b \leq 9$, for some integers $ a$ and $ b$. Hence, $ 19a \plus{} 19b \equal{} 10a \plus{} b$. hence, $ 9a \plus{} 18b \equal{} 0$. Since $ a > 0$, this equation has no solution.

We conclude that all lucky numbers are 3-digit numbers. Let $ x$ be any 3-digit integer. We see that $ x$ is in the form $ 100a \plus{} 10b \plus{} c$, with $ 1 \leq a \leq 9$, $ 0 \leq b \leq 9$, and $ 0 \leq c \leq 9$. We know $ x$ is a lucky number iff $ 100a \plus{} 10b \plus{} c \equal{} 19a \plus{} 19b \plus{} 19c$. Hence, $ 81a \equal{} 9b \plus{} 18c$, so $ 9a \equal{} b \plus{} 2c$. First note that $ b \plus{} 2c \leq 9 \plus{} 18$, so $ b \plus{} 2c \leq 27$. Hence, $ x$ is a lucky number only if $ c \leq 3$.

We now use casework.
[list]
  [*] If $ a \equal{} 1$, then we have that the solutions are $ (1, 1, 4), (1, 3, 3), (1, 5, 2), (1, 7, 1)$, and $ (1, 9, 0)$.
  [*] If $ a \equal{} 2$, then we have that the solutions are $ (2, 0, 9), (2, 2, 8), (2, 4, 7), (2, 6, 6)$, and $ (2, 8, 5)$.
  [*] If $ a \equal{} 3$, then we have that the single solution is $ (3, 9, 9)$. (If $ c < 9$, then since $ 27 \minus{} 2c \equal{} b$, and $ 27 \minus{} 2c > 9$, and $ b \leq 9$, we have a contradiction.)[/list]

Hence, the lucky numbers are $ \boxed{114, 133, 152, 171, 190, 209, 228, 247, 266, 285, and 399}$.

[hide="Not sure..."]Is there only one possible value for the largest angle in the first problem? ($ 72^\circ$)?[/hide]

[hide="#1?"]
I believe there are 5, $ 72^{\circ}, 90^{\circ}, 108^{\circ}, 126^{\circ}, 132^{\circ}$.

Let our triangle be $ \triangle ABC$ with $ \angle A \equal{} 36^{\circ}$. To cut the triangle into two isosceles triangles, a line must be drawn from one of the vertices (either $ B/C$ or $ A$). In the former case, let the dividing cevian extend to $ D$ (it doesn't matter if from $ B$ or $ C$, so without loss of generality let it be from $ B$). Then $ \triangle ABD$ is isosceles, and either $ \angle BDA \equal{} \angle A \equal{} 36^{\circ}$, $ \angle DBA \equal{} \angle A \equal{} 36^{\circ}$, or $ \angle BDA \equal{} \angle DBA \equal{} \frac {180 \minus{} 36}{2} \equal{} 72^{\circ}$. Angle-chasing etc gives the respective cases $ \angle BDC \equal{} 144^{\circ}, 72^{\circ}, 108^{\circ}$. Then using the condition that $ \triangle BDC$ is isosceles, if $ \angle BDC$ is obtuse clearly $ \angle C \equal{} \angle CBD$, which after further angle chasing gives two answers of $ \boxed{126, 108}$. If $ \angle BDC \equal{} 72^{\circ}$, then casework yields the cases $ \boxed{72, 90, 108}$.

In the latter case, we have a cevian $ AD$. Then let $ x \equal{} \angle B, y \equal{} \angle DAB$, angle-chase, and equate angles to fit the isosceles conditions. We end up with $ y \equal{} 12^{\circ}, 36^{\circ}, 48^{\circ}$, of which only the first is non-degenerate, and which yields $ \boxed{132}$.

[/hide]


algebra
function
domain
trigonometry
logarithms
algebra unsolved

Solve the equation: $(1+cos x)^{log_{cos x}sinx}=(1+sin x)^{log_{sin x}cos x}$

By the domain of the logs, $\sin x,\cos x\in(0,1)$, hence $0<x<{\pi\over 2}$. Rewrite the equation thus:

$(1+\cos x)^{\ln^{2}\sin x}=(1+\sin x)^{\ln^{2}\cos x}$

With $f(x)=(1+\cos x)^{\ln^{2}\sin x}$, the equation becomes $f(x)=f\left({\pi\over 2}-x\right)$

$f'(x)=f(x)\cdot\ln\sin x\cdot\left(2\cot x\ln(1+\cos x)-\tan{x\over 2}\ln\sin x\right)$

For $x\in\left(0,{\pi\over 2}\right)$ we have $f(x)>0,\cot x>0,\ln(1+\cos x)>0,\tan{x\over 2}>0,\ln\sin x<0$, therefore $f'(x)<0$ on the whole interval. With $f$ strictly decreasing, we get $f(x)=f\left({\pi\over 2}-x\right)\iff x={\pi\over 2}-x\iff x={\pi\over 4}$


linear algebra
matrix
quadratics
function
calculus
derivative
Asymptote

9th Vietnamese Mathematics Olympiad for College Students 2001

A1.  A is the 3 x 3 matrix a11 = a22 = -a33 = a13 = 1, a23 = 2, other elements zero. Find all 3 x 3 matrices B such that AB + BA = 0.
A2.  A, B are real square matrices such that A2001 = 0 and AB = A + B. Show that det B = 0.
A3.  a, b, c, d are reals such that the quadratic ax2 + (b + c)x + d + e = 0 has a root x ≥ 1. Show that the quartic ay4 + by3 + cy2 + dy + e = 0 also has a root y ≥ 1.
A4.  a1, a2, ... , an are points in Rn (so each ai has n coordinates). A is the matrix whose i, j element is ai·aj. Show that det A is non-negative, and the eigenvalues of A are all non-negative.
A5.  Every element of an n x n matrix is an even integer. Show that none of its eigenvalues are odd integers.
A6.  A is an n x n matrix. The main diagonal elements (aii) are all a + b, where a and b are reals, the elements in the diagonal immediately above the diagonal (the elements ai i+1) are all ab, and the elements in the diagonal immediately below the main diagonal are all 1. The other elements are all 0. Find det A.
B1.  The function f(x) has f ''(x) > 0 for all x > 0 and the graph is asymptotic to y = ax + b as x→∞. Show that f(x) - ax - b has negative derivative for all x > 0, and that f(x) > ax + b for all x > 0.
B2.  p, q are reals such that p > 0, q < 0, p + q < 1. The non-negative real sequence a1, a2, a3, ... satisfies an+2 ≤ p an+1 + q an for all n. Show that the sequence converges and find its limit.
B3.  Let g(x) = (1 + x) (1 + x2) ... (1 + x2001), and f(x) = x2000/g(x). Show that ∫x1 f(t) dt = f(x) for some x ∈ (0, 1).
B4.  A real-valued function f on the reals has a second derivative everywhere and f(x) + f ''(x) ≥ 0 for all x. Show that f(x) + f(x + π) ≥ 0 for all x.
B5.  f(x) is defined for x ≥ 1 and f(1) is non-zero. It satisfies f(x + 1) = 2001 f(x)2 + f(x) for all x. Find lim ( f(1)/f(2) + f(2)/f(3) + ... + f(n)/f(n+1) ).
B6.  f(x) is a differentiable function on [a, b] such that f(x)2 + f '(x)2 > 0 for all x in [a, b]. Show that the function has only finitely many distinct roots.

For B6:

Suppose that f has infinitely many zeroes. Then there must be a sequence of zeroes which converges to a number $t \in [a, b]$. If $f(t) \ne 0$, it is clear that f is not differentiable at t, so f(t) = 0. But then suppose for the sake of contradiction that $f'(t) = c \ne 0$. Then $\exists \delta > 0$ s.t. $\forall x$ with $|x-t| < \delta, \left|\frac{f(x)-f(t)}{x-t}-c \right| < |c|$. However, since we have a sequence of zeroes that converges to t, we can always choose a suitable $x \ne t$ so that $|x-t| < \delta$ while f(x) = 0. In such a case, we're left with $\left|\frac{f(x)-f(t)}{x-t}-c \right| = |c| < |c|$, which is a contradiction. Thus f can not have infinitely many zeroes.

A1:

A = I + [matrix with all 0's, except right column is 1, 2, -2]. Errr... after that it should be pretty easy.

A2:

Errr (A-I)(B-I) = I. Also det(A) = 0. Thus -1 is an eigenvalue of A-I. Then 1/(-1) = -1 must be an eigenvalue of B-I. Thus det(B) = 0.

A3:

I think you need some restrictions on a, b, c, d, e.

A4:

I think the problem statement for this one must be wrong too... since A is the product of the matrix having a_i's as the rows with its transpose. There should be some condition on the a_i.

A5:

Suppose one of the eigenvalues is an odd integer. We just need to prove that a matrix having all entries even except all diagonal entries odd does not have determinant zero. This is readily obvious when you reduce mod 2: the resulting matrix is identity, and so the determinant is 1 mod 2.

A6:

Using the recursive definition of determinant, we get that

$D_{n}= (a+b)D_{n-1}-(ab)D_{n-2}$.

This basically means that $D_{n}$ is a linear combination of $a^{n}$ and $b^{n}$ by recursion.

We can solve for their coefficients by noting that $D_{1}= a+b, D_{2}= a^{2}+ab+b^{2}$. This gets us:

$D_{n}= \frac{a^{n+1}-b^{n+1}}{a-b}$.

B1:

For the first part, it just suffices to show that a is an upper bound for f'(x). If for some t, f'(t) > a, since f' is increasing, we can take x > t so that f(x) - ax - b is arbitrarily large (just integrate to check), contradicting the asymptote condition. Also, look at what happens when $f(t)+\epsilon = at+b$ for some $t$ with $\epsilon > 0$. We already established that a is an upper bound for f', so integrating f', we basically have $f(x) \le ax+b-\epsilon$ for all x > t, which is not very asymptotic. If ever f(t) = at + b, it's a separate case but still easy to check.

B2:

The limit is 0. Let $S_{n}= \sum_{i=1}^{n}a_{i}$. Then we have

$S_{n}-a_{1}-a_{2}\le pS_{n-1}-pa_{1}+qS_{n-2}$
$= pa_{n-1}-pa_{1}+(p+q)S_{n-2}$

$S_{n}< (1-p)a_{1}+a_{2}+pa_{n-1}+(p+q)S_{n-2}$

Since p+q < 1, if we could just show that $a_{n}$ is bounded, then we would have something of the form

$S_{n}< k+rS_{n-2}$

from which we can easily show $S_{n}$ converges.

At this point, there must be some condition missing, because we can easily find a linear recursion which diverges. I'll continue a bit further: Suppose that at some point $a_{k+1}\le a_{k}$. Then $a_{k+2}\le pa_{k+1}+qa_{k}\le (p+q)a_{k+1}< a_{k+1}$, and so on. Thus the only case we need consider is if the sequence $a_{n}$ is strictly increasing.

B3:

Assuming that we want $\int_{x}^{1}f(t) dt = f(x)$, this is pretty easy by intermediate value theorem.

For A4, I didn't notice that it said "non-negative," and not "positive."  :blush:

In that case, since A is the product of a matrix B and its transpose, its determinant is $|B|^{2}\ge 0$. Hmmm I'll have to do the eigenvalues part later...

For B4, WLOG f(0) = 0. Consider the set of all numbers t such that $f(x) \le t \sin x$ for all $x \in [0, \pi]$. Let c be the infimum; we will prove that $f(x) \le c \sin x$ for all $x \in [0, \pi]$. Actually that's trivial.

So now, $f(x)$ and $c \sin x$ should have a common tangent at some point in $[0, \pi]$, since if not, we can make c smaller. This is not immediately obvious, but consider the function $g(x) = \frac{c \sin x-f(x)}{\sin x}$, with $g(0) = c-f'(0)$ and $g(\pi) =-c-f'(\pi)$. It can be proven that g is continuous, and it's always non-negative, and it's all on a closed interval, so it has a non-negative minimum value m achieved on that interval. If m = 0, then it's pretty clear that it must be a tangency point, and if m > 0, we can replace c with c-m.

So let f(x) be tangent to c sin x at x = a. Now suppose $f(\pi) < 0 = c \sin \pi$. Then by mean value theorem, there is some point b between a and $\pi$ where $f'(b) < c \cos b$. But then by MVT again, there is some point d between a and b with $f''(d) <-c \sin d$. This is contradictory, because then

$f''(d)+f(d) \le f''(d)+c \sin d <-c \sin d+c \sin d = 0$.

Thus $f(\pi) \ge 0 \implies f(0)+f(\pi) \ge 0$.

In A4, suppose x is an eigenvector. We have

$BB^{T}x = \lambda x$
$x^{T}BB^{T}x = \lambda xx^{T}$

Since $x^{T}BB^{T}x = (B^{T}x)^{T}\cdot (B^{T}x) \ge 0$, we have $\lambda \ge 0$.

For A3, let p be the quadratic and let q be the quartic, with WLOG a > 0. We only need to consider if both roots of the quadratic are greater than one, since otherwise p(1) = q(1) < 0, and clearly q tends to infinity.

Now note that $q(x) = p(x^{2})+(x-1)(bx^{2}+d)$. We know that $p(x^{2})$ has four roots, two above 1 and two below -1, so that there exists c > 1 with p(c) = p(-c) < 0. In addition, $(c-1)(bc^{2}+d)$ has opposite sign with $(-c-1)(bc^{2}+d)$, so one of q(c) and q(-c) is negative. Then q must have a real root as again q tends to infinity.

Wait a minute... I'm still not understanding something on question B6. What is the statement asking? Are those 2's squares (they can't be?) and also I'm a little confused on probabilty's wording at the beginning of his soluiton...


linear algebra
matrix
limit
real analysis
real analysis unsolved

let $B$ be an antisymmetric matrice. suppose that the sequence $(B^{n})$ has a limit $C$. what can we say about the matrix $C$ ?

is it the right subforum ?
Anyway, if $B$ is antisymetric, $iB$ is hermitian. Hence there exists a unitary transformtaion $U$ such that $B=UDB^{*}$ where $D=diag(id_{1}, id_{2}, .., i d_{n})$ where $d_{i}\in \mathbb{R}$. The limit assumption shows that $(id_{i})^{n}\to 0$. Hence $C=0$.

Sorry if it's not the right subforum, But we were given this exercice in class while studying normes. is there an analytic proof ?  :maybe:


geometry
3D geometry
tetrahedron
octahedron
pyramid
prism

$1) \text {What is the quick way (formula) to find the volume of a regular tetrahedron given the measure of one side,}\ s \ ?$

$2) \text {What is the quick way (formula) to find the volume of a regular octahedron given the measure of one side,}\ s \ ?$

$3) \text {What is the quick way (formula) to find the area of a regular octagon given the measure of one side,}\ s \ ?$

 :ninja:

$\displaystyle \text {What about the voulme of regular pyramid? What's the formula? Thanks alot in advance!!}$ :w00t:

What kind of pyramid, square or triangular or other base?

Note: If a pyramid is equal-sided, its base can have no more than five sides.

Sorry square base. But do you know a fast way for a triangular base?

[hide]The volume of a pyramid is $1\over3$ of a rectanglular prism, so if the square's side length ws $s$ and the height was $h$, the formula would be $\frac{s^2h}{3}$ and same idea for triangular base.[/hide]


algebra
polynomial
inequalities
AMC
USA(J)MO
USAMO
algebra proposed

If $ P(x)$ is a polynomial with integer coefficients and $ a$, $ b$, $ c$, three distinct integers, then show that it is impossible to have $ P(a)\equal{}b$, $ P(b)\equal{}c$, $ P(c)\equal{}a$.

[quote="rachid"][quote="Leon"]If $ P(x)$ is a polynomial with integer coefficients and $ a$, $ b$, $ c$, three distinct integers, then show that it is impossible to have $ P(a) \equal{} b$, $ P(b) \equal{} c$, $ P(c) \equal{} a$.[/quote]

It's rather an easy exercise !!

Lets assume for the sake of contradiction that such a polynomial exists.We know that $ P(a) \minus{} P(b) \equal{} b \minus{} c,P(b) \minus{} P(c) \equal{} c \minus{} a$ and $ P(c) \minus{} P(a) \equal{} a \minus{} b$.Since for any integers $ x$ and $ y$ : $ x \minus{} y | P(x) \minus{} P(y)$,It follows that $ a \minus{} b|b \minus{} c,b \minus{} c|c \minus{} a$ and $ c \minus{} a|a \minus{} b$ which leads to $ a \minus{} b \equal{} b \minus{} c \equal{} c \minus{} a$ or $ a \equal{} b \equal{} c$,Contradiction ! Hence,it's impossible to have $ P(a) \equal{} b,P(b) \equal{} c$ and $ P(c) \equal{} a$ for distinct integers $ a,b$ and $ c$.[/quote]
It doesn't lead to $ a\minus{}b\equal{}b\minus{}c\equal{}c\minus{}a$. It leads to: $ |a\minus{}b| \equal{} |b\minus{}c| \equal{} |c\minus{}a|$. $ |a\minus{}b| \equal{} |b\minus{}c|$ gives $ a\minus{}b\equal{}b\minus{}c$ or $ a\minus{}b\equal{}c\minus{}b$, the last one being absurd because $ a,b,c$ are distinct. $ a\minus{}b\equal{}b\minus{}c \iff a\plus{}b\plus{}c\equal{}3b$. Doing this for $ |b\minus{}c| \equal{} |c\minus{}a|$ yields $ 3b\equal{}3c \iff b\equal{}c$, which is a contradiction.

Consider that $ P(b)\equiv P(a)(mod. b\minus{}a)$. We will assume that the polynomial exists, then we have that $ c \equiv b (mod. b\minus{}a) \implies b\minus{}a|c\minus{}b$. We also have that $ c\minus{}b|a\minus{}c$; $ a\minus{}c|b\minus{}a$. Then we can see that $ |b\minus{}a|\le|c\minus{}b|\le|a\minus{}c|\le|b\minus{}a|$, so all the inequalities are equalities. w.l.o.g. assume that $ a<b<c$, now we have that $ |b\minus{}a|\equal{}b\minus{}a\equal{}|a\minus{}c|\equal{}c\minus{}a\equal{}|c\minus{}b|\equal{}c\minus{}b \implies a\equal{}b\equal{}c$, but they are distinct, contradiction!

Shalom!

And the first problem of the Chilean Iberoamerican Olympiad Team Selection 2008. :roll:

[quote=rachid][quote="Leon"]If $ P(x)$ is a polynomial with integer coefficients and $ a$, $ b$, $ c$, three distinct integers, then show that it is impossible to have $ P(a) \equal{} b$, $ P(b) \equal{} c$, $ P(c) \equal{} a$.[/quote]

It's rather an easy exercise !!

Lets assume for the sake of contradiction that such a polynomial exists.We know that $ P(a) \minus{} P(b) \equal{} b \minus{} c,P(b) \minus{} P(c) \equal{} c \minus{} a$ and $ P(c) \minus{} P(a) \equal{} a \minus{} b$.Since for any integers $ x$ and $ y$ : $ x \minus{} y | P(x) \minus{} P(y)$,It follows that $ a \minus{} b|b \minus{} c,b \minus{} c|c \minus{} a$ and $ c \minus{} a|a \minus{} b$ which leads to $ a \minus{} b \equal{} b \minus{} c \equal{} c \minus{} a$ or $ a \equal{} b \equal{} c$,Contradiction ! Hence,it's impossible to have $ P(a) \equal{} b,P(b) \equal{} c$ and $ P(c) \equal{} a$ for distinct integers $ a,b$ and $ c$.[/quote]

Why do x-y divides f(x) -f(y)?

[quote=soryn]What is the solution if P(a)=a,P(b)=b,P(c)=c?[/quote]

Not terribly interesting $P(X)=x+(x-a)(x-b)(x-c)Q(X)$ ,$Q\in \mathbb{Z}[X]$

This is the solution, from my book  :-D :

[b]My solution:[/b]

Assuming $\exists$ a polynomial $P(x)$ with integer coefficients such that $P(a)=b, P(b)=c, P(c)=a$ for $a,b,c \in \mathbb{Z}$ and $a, b, c$ are distinct.

We know that for any polynomial $P(x)$ with integer coefficients,
$a-b \vert P(a)-P(b)$. Therefore, from the above equations, we get:

$(a-b)\vert b-c \Rightarrow |a-b|\le|b-c|$

$(b-c)\vert c-a \Rightarrow |b-c|\le|c-a|$

$(c-a)\vert a-b \Rightarrow |c-a|\le|a-b|$

Combining the inequalities, we get $|a-b|\le|b-c|\le|c-a|\le|a-b| \Rightarrow |a-b|=|b-c|=|c-a|$.

Thus $(a-b)^2=(b-c)^2 \Rightarrow (a-c)(a+c-2b)=0 \Rightarrow a=c$ or $a+c=2b$.
$a, b, c$ are distinct $\Rightarrow a+c=2b$.
Similarly with the other pairs, $b+c=2a$, $a+b=2c$ Solving the system of equations, we get $a=b=c$, which is not possible as they are distinct integers.

Thus we have a contradiction $(\Rightarrow\Leftarrow)$. [b]Q.E.D.[/b]

more general problem: if integer polynomial $P$ and integer $n$ have that $P(P(...P(n)...))=n$ ($P$ iterated at least once) then $P(P(n))=n$

It would imply that $a - b \mid b - c, b - c \mid c - a, c - a \mid a - b$, so $a - b = b - c = c - a$, but that means $a=b=c$, absurd.


I was wondering if I could see if anyone from my state scored a perfect on the AMC8. Is there a way to find this out? I checked the website, but it only goes back to 1999.

Could I talk to a person from the MAA AMC?

Thanks.

Which state would that be?  In almost every state, every year a few people usually get perfects on the AMC 8.
Every year, the MAA sends out an AMC/AIME/USAMO scores handbook, so you could check that out.

[quote="Jason27603"]In almost every state, every year a few people usually get perfects on the AMC 8.[/quote]
My state isn't one of those states. Over the past 10 years (since 1999), the closest to a perfect was a 24, two years ago (and I know the person).

I would prefer not to say, but I doubt anyone here is from that state.

Wasn't the AMC 8 only available from 1999?  [sarcasm]You must live in California if nobody from your state has ever gotten a perfect.[/sarcasm]

Oh, good point. The AJHSME was available 1985-1998 and the AMC 8 1999-2009+. So is the AJHSME data available anywhere?


function
logarithms
algebra
domain
algebra proposed

Let $f: \mathbb R_{+}\to \mathbb R_{+}$ with$f(x+y)\geq f(x)+f(y),\forall x,y\geq 0$.Prove that $f(x)>f(1)\cdot log_{2}x,\forall x>0$

From hypothesis we have $f$ is increasing and $f(nx)\geq nf(x)\forall n\geq 1\forall x\geq 0$. If $x>1$ then $2^{n}< x\leq 2^{n+1}(n\geq 0)$ and $f(x)>f(2^{n})\geq 2^{n}f(1)\geq (n+1)f(1)\geq f(1)\log_{2}x$. If $x<1$, similary!

But if $x=1$, I can't! :D

$f(x)\geq f([x])+f(\{x\})\geq f(\underbrace{1+\cdots+1}_{[x]\text{ times}})\geq \underbrace{f(1)+\cdots+f(1)}_{[x]\text{ times}}=[x]f(1)$
This leaves to show $[x]>\log_{2}x$ for all $x>0$.
Set $f(x)=2^{x-1}-x$ for $x\in\mathbb{R}$.
Then $f'(x)=2^{x-1}\ln 2-1\leq \ln 2-1=\ln{2\over e}<0$ for $x\leq 1$
and $f'(x)\geq 2\ln 2-1=\ln{4\over e}>0$ for $x\geq 1$.
So $f$ is decreasing on $(-\infty,1]$ and increasing on $[2,\infty)$ with $f(1)=f(2)=0$.
So $f(x)\geq 0$ for all $x\in\mathbb{R}\setminus (1,2)$ and $[x]>x-1\geq \log_{2}x$ for all $x\in\mathbb{R}_{+}\setminus (1,2)$.
But for $x\in (1,2)$ we have $[x]=1=\log_{2}2>\log_{2}x$.
This proves the result.

This problem is wrong :D Function $f\equiv 0$ satisfying.

[quote="Svejk"]Let $f: \mathbb R_{+}\to \mathbb R_{+}$ [/quote]

so  $f(x)$ $\equiv$ $0$ clearly goes against the given condition so ia suppose the problem is right.  :P

[quote="N.T.TUAN"]so if, we have $0>0$   :huh:[/quote]

i couldn't get u . i simply said that range and domain of tthe function was positive reals. :)

[quote="N.T.TUAN"][quote="Svejk"]Let $f: \mathbb R_{+}\to \mathbb R_{+}$ with

$f(x+y)\geq f(x)+f(y),\forall x,y\geq 0$.Prove that
 $f(x)>f(1)\cdot log_{2}x,\forall x>0$[/quote]

$\mathbb{R}_{+}= [0,\infty)$  :)[/quote]
well then thanks because it was just a misconception of symbols  :)


integration
trigonometry
function
real analysis
real analysis unsolved

Calculate ${I=\int_{0}^{\frac{\pi}{2}}(\sqrt{\sin x}}+\sqrt{\cos x})dx$

Please see your identical topic here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=144930


calculus
integration
number theory open
number theory

Is it necessary that for any given integers a and b, the equation
(ap)^2+(bq)^2=(ab)^2
has at most four sets of solution of integral p and q???

The number of solutions equal the number of representation of $ c^2$ as the sum of two squares, where $ c\equal{}\gcd(a,b)$.
Indeed, if $ u^2 \plus{} v^2 \equal{} c^2$ then for $ p\equal{}\frac{bu}{c}$ and $ q\equal{}\frac{av}{c}$ we have $ (ap)^2 \plus{} (bq)^2 \equal{} (ab)^2$.

The number of representations of $ c^2$ as the sum of two squares is not bounded. See http://mathworld.wolfram.com/SumofSquaresFunction.html for details.


inequalities
function
calculus
derivative
inequalities solved

Let Ai (i=1,2,...n) positive real numbers. Let p,q,r,s positive real numbers such that pr=qs. Show that

 ( \sum 1/Ai^r)^p( \sum Ai^s)^q \geq n^(p+q)

Actually it is Hlder inequality:

(\sum xia)1/a(\sum yib)1/b \geq \sum xiyi
where 1/a+1/b=1

In our case a=(p+q)/p and b=(p+q)/q.

I remember seeing a proof which uses some functions and their derivatives and stuff, but I can't remember the proof. Any elementary proofs of this one?

Yeah, Holder. I can find the soln in my room (somewhere  :D ). I'll post it when I find it.

Lemma: for x,y  \geq 0 , p,q>1 with 1/p +1/q = 1
we have x^p/p + y^q/q  \geq xy

For Hlder a = (\sum_{i=1 to n}|a_i|^p )^1/p

b= (\sum_{i=1 to n}|b_i|^q)^1/q

x=a_i/a , y = b_i/b

(a_i/a)^p/p + (b_i/b)^q/q  \geq a_i/a.b_i/b

take the sum i=1 to n

the LHS will be 1/p +1/q =1
the RHS  will be (\sum_{i=1 to n}a_ib_i)/ab

Anyone with a different proof ?

The soln I was looking for is this:

We prove (with the help of derivatives) x^a-ax <= 1-a for a in (0,1). Then replace x with A/B and multiply the ineq with B: A^a*B^b<=A*a+B*b where a.b.A.B are positive and a+b=1. We replace a and b with 1/p and 1/q, then we replace A and B with ai^p/( \sum {1<=i<=n}ai^p) and bi^p/( \sum {1<=i<=n}bi^p) for each i, and then we add up the inequalities we got. The result is Holder's ineq.


[quote="dlumenau"]I am trying to work out a problem, which would help me find a suitable level of initial investment to return a profit of 30%.

Simply, I have put forward a hypothetical situation where if you were to invest £10,000 in a start up company, who returned an amazing 30% in its first year, you would have made a profit of £3,000. Easy enough.

My problem comes when you're trying to make a certain profit based on an assumption that the company will return the same level of profit in its second year.

For example, I want to make £10,000 exactly in the second year. How do I work out the initial investment required?

So far I had the following:

[b]p = profit
i = initial investment

p = (i * 1.3) - i[/b]

Therefore:
[b]
£10,000 = (i * 1.3) - i[/b]

Does that make sense?!

I have no idea!

Thank you

Dan :blush:  :huh:[/quote]

This problem does not belong in this forum. [color=green][It's been moved - moderator.][/color]
As for the solution,
[hide]Suppose that you put in $ \$ x$ into the investment the next year. Then, you can expect a $ 30\%$ profit, or a $ 0.3$ profit. In other words, you will get back $ 0.3x$ profit. Since you want your profit to be $ \$10,000$, we can set up this equation:
$ 0.3x = 10000$
$ x = \frac{10000}{0.3}= 33333.33$

So, you should invest $ \$33333.33$ next year to get the desired profit.[/hide]
Hope that helps. :)


Let $f:[a,b]\to\mathbb R$, $a<b$ primitivable.
If $F$ is a primitive of $f$ then there is at least one $c\in(a,b)$ s.t.
$f(c)\le\frac{\pi}{b-a}\left(1+F^2(c)\right)$.

Let $G(x)=\arctan [F(x)]$. What we must show is that there is some $c\in(a,b)$ s.t. $(b-a)G'(c)\le \pi$. This is easy, since there is some $c\in(a,b)$ s.t. $(b-a)G'(c)=G(b)-G(a)$, and the image of $\arctan$ over $\mathbb R$ is $(-\frac \pi 2,\frac \pi 2)$.


function
combinatorics unsolved
combinatorics

Prove that the number of partitions of $ n$ with a unique less part (the small number just appear one time) and the big part is not more than two times the less part, is equal to the number of partitions of $ n$ with the big part odd and the less part is more than the half of the big part.

Some notes on English: "less" is a word of comparison.  It's used in phrases like "A is less than B."  The word you want is "least."  A synonym is "smallest."  Antonyms are "largest" or "biggest."

The generating function for the first (using $ k$ for the smallest element) is
\begin{align*} \sum_{k \geq 0} \frac {x^k}{(1 - x^{k + 1})(1 - x^{k + 2}) \cdots (1 - x^{2k})} & = \sum_{k \geq 0} \frac {1 + x^k}{(1 - x^{k + 1})(1 - x^{k + 2}) \cdots (1 - x^{2k})} - \sum_{k \geq 0} \frac {1}{(1 - x^{k + 1})(1 - x^{k + 2}) \cdots (1 - x^{2k})} \\
& = \sum_{k \geq 0} \frac {1}{(1 - x^k)(1 - x^{k + 1})(1 - x^{k + 2}) \cdots (1 - x^{2k - 1})} \\
& \hspace{2in} - \sum_{k \geq 0} \frac {1}{(1 - x^{k + 1})(1 - x^{k + 2}) \cdots (1 - x^{2k})} \\
& = \sum_{k \geq 0} \frac {1}{(1 - x^k)(1 - x^{k + 1})(1 - x^{k + 2}) \cdots (1 - x^{2k - 1})} \\
& \hspace{2in} - \sum_{k \geq 0} \frac {1 - x^{2k + 1}}{(1 - x^{k + 1})(1 - x^{k + 2}) \cdots (1 - x^{2k + 1})} \\
& = 1 + \sum_{k \geq 0} \frac {x^{2k + 1}}{(1 - x^{k + 1})(1 - x^{k + 2}) \cdots (1 - x^{2k + 1})}, \end{align*}
and this latter expression is the generating function for the partitions of the second type (using $ 2k + 1$ for the largest element).

Bijective proof, anyone?

Here's my bijective proof. You would understand through the figure below. So-called Proof without words!


Does anyone know of a good website to learn vedic math? I am horrible at mental math and am hoping this will help me.

Vedic math has nothing to do with learning mental math well. Vedic math (the mathematics supposedly mentioned in the vedas, which are some of the scriptures of Hindu religion) is now being used as a brand name in India to promote certain mental math tricks that are equally well known (= equally ignored by the general public, but known to specialists) in the West. It's doubtful whether most veda passages mentioned by promoters of vedic math are even about mathematics at all.

If you want to find good books about mental math, put "mental math," "mental arithmetic," "speed math" and some similar search terms into Google and into Amazon.com and see what you find. I can say that in general books about this topic published by Dover, the reprint publishing company, are better value for the dollar than most newer books on this subject.

Hope this helps!

hmm i dunno, but i have a book called vedic algebra, and it reallly helps for mental math (at least for me).  i use it at competitions in my state in which they have tests designed solely for mental math. but if you cant find any good websites on it, then heed tokenadults advice and just google "mental math" or something

hey pkothari, are you talking about number sense?

yeah, for tmsca and uil stuff. it REALLY helps on estimation and all four types of arithmetic (add, subract, multiply, divide).  the only downfall is that because it was translated, some things are somewhat hard to understand.. but if you try hard enough, it's pretty simple.

try this site jacob:

[url=http://hinduism.about.com/od/vedicmaths/]http://hinduism.about.com/od/vedicmaths/[/url]


integration
calculus
calculus computations

this is an easy one, of-course.  :D

show

$ \sum_{n\equal{}1}^\infty \; \frac{1}{\dbinom{n\plus{}m\plus{}1}{m\plus{}1}}\;\equal{}\;\frac{1}{m}$

$ \sum_{n\equal{}1}^\infty \; \frac{1}{\dbinom{n\plus{}m\plus{}1}{m\plus{}1}} \equal{} \frac{m\plus{}1}{m} \sum_{n\equal{}1}^\infty \; \left[\frac{1}{\dbinom{m\plus{}n}{n}} \minus{} \frac{1}{\dbinom{m\plus{}n\plus{}1}{n\plus{}1}}  \right]$.  :lol:

Generally, for $ a, b, c > 0$, we have $ \sum_{n\equal{}0}^{\infty} \beta(an\plus{}b, c) \equal{} \int_{0}^{1} \frac{x^{b\minus{}1} (1\minus{}x)^{c\minus{}1}}{1\minus{}x^{a}} \, dx$.


limit
function
analytic geometry
calculus
integration
calculus computations

Let us denote by Vn the volume of the region in the first octant that lies beneath the surface

(x/a)^(1/n) + (y/b)^(1/n) + (z/c)^(1/n) = 1

where a, b, c and n are constants.

a.) Find V2.

b.) Show that lim Vn = 0 as n goes to infinity.

One way to look at b):

Let $ E_n$ be the region we are talking about.

Choose any $ (x,y,z)$ such that $ 0 < x\le a,0 < y\le b,0 < z\le c.$ Then

$ \lim_{n\to\infty}\left(\left(\frac {x}a\right)^{\frac1n} \plus{} \left(\frac {y}b\right)^{\frac1n} \plus{} \left(\frac {z}c\right)^{\frac1n}\right) \equal{} 3,$

which means that for sufficiently large $ n,$ this quantity is $ > 1$ and hence this point does not lie in $ E_n.$ We also have that this quantity is an increasing function of $ n,$ which shows that $ \{E_n\}$ is a decreasing sequence of sets: $ E_1\supset E_2\supset E_3\supset\cdots.$

That shows us that $ \bigcap_{n \equal{} 1}^{\infty}E_n$ is a subset of the union of the coordinate planes, which is a set of volume zero.

Given the appropriate theorem about the interchanging of measures and limits of sets (or of interchanging limits and integrals) we conclude that since the volume of the limit (intersection) of the $ E_n$ is zero, we must have that $ \lim_{n\to\infty}\text{Vol}\,(E_n) \equal{} 0.$

Hey,
[quote="Kent Merryfield"]
Which means that for sufficiently large $ n,$ this quantity is $ > 1$ and hence this point does not lie in $ E_n.$ We also have that this quantity is an increasing function of $ n,$ which shows that $ \{E_n\}$ is a decreasing sequence of sets: $ E_1\supset E_n\supset E_3\supset\cdots.$
[/quote]
I did not realize it was mathematically possible to have a sequence of sets, when I think of sequences I usually think of numbers with a pattern (ie. $ {s} \equal{} {2, 4, 8, 16,\dots}$). Since sets can be elements of a sequence does that imply that those sets (or rather their elements) must be restricted to some order to satisfy the overall order of the sequence?

[quote="Kent Merryfield"]
That shows us that $ \bigcap_{n \equal{} 1}^{\infty}E_n$ is a subset of the union of the coordinate planes, which is a set of volume zero.
[/quote]
Is the following the definition of the operator "$ {\bigcap_{i \equal{} 1}^{k}}{{a}_{i}}$"?,
$ {{\bigcap_{i \equal{} 1}^{k}}{{a}_{i}}} \equal{} {{{a}_{1}}\supset{{a}_{2}}\supset{{a}_{3}}\supset\dots\supset{{a}_{k \minus{} 2}}\supset{{a}_{k \minus{} 1}}\supset{{a}_{k}}}$

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715][b]APSStudent[/b][/url]

Don't look for restrictions that might not be there. A sequence of sets is just a list: $ E_1, E_2, E_3,\dots.$ The subscripts are just labels. The sets don't have to be related in any way.

In this particular instance, I did define a [i]decreasing[/i] sequence of sets - that's a statement about set inclusion. But that makes this a very special case. Most sequences of sets aren't increasing, or decreasing, or disjoint, or any such special condition.

The symbol $ \bigcap_{i\equal{}1}^kE_i$: that's [i]intersection[/i]. And the things in that symbol need to be sets.

$ \bigcap_{i\equal{}1}^kE_i\equal{}E_1\cap E_2 \cap E_3\cap \cdots \cap E_k.$

Intersection.

Hey,
[quote="Kent Merryfield"]
Don't look for restrictions that might not be there. A sequence of sets is just a list: $ E_1, E_2, E_3,\dots.$ The subscripts are just labels. The sets don't have to be related in any way.
[/quote]
Ok, thanks for clearing that up for me.

[quote="Kent Merryfield"]
The symbol $ \bigcap_{i \equal{} 1}^kE_i$: that's [i]intersection[/i]. And the things in that symbol need to be sets.

$ \bigcap_{i \equal{} 1}^kE_i \equal{} E_1\cap E_2 \cap E_3\cap \cdots \cap E_k.$

Intersection.
[/quote]
Ok, that makes sense. It seem pretty clear now, that the operator "$ {\bigcap_{i \equal{} 1}^{k}{a}_{i}}$" is simply the continuous intersection of the indexed sets.
In other words,
$ {\bigcap_{i \equal{} 1}^{k}{a}_{i}} \equal{} {{a_1}\cap{a_2}\cap{a_3}\cap\dots\cap{a_{k\minus{}2}}\cap{a_{k\minus{}1}}\cap{a_k}}$

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url]


Find all integer solns of the eqn:
a^2+b^2+c^2=a^2*b^2

Too much of an old problem isn't it? If we suppose that (x,y,z) <> 0, then we obtain a conbtradiction by a Fermat (infinite) descent (we check the parity of all variables included, to conclude that they must all be even, hence the contradiction).

Chjeers,
Thanos

Ok, let's write full solution for this old problem before to move it...

Suppose that (a,b,c) =/= (0,0,0) is a solution.
Let d = gcd(a,b,c). Then (a/d,b/d,c/d) is a solution too, so we may suppose that gcd(a,b,c) = 1. It follows that gcd(a,b) = 1 since any divisor of a and b is a divisor of c.
- If a and b are odd, then the equation leads to 1 + 1 + c 2 = 1 mod[4], that is c 2 = 3 mod[4], which is impossible.
- If a is even and b is odd, then the equation leads to 1 + c 2 = 0 mod[4], which gives the same contradiction.
Thus, the unique solution is (0,0,0).

Pierre.


[i]Find the smalleast  prime that satisfies the condition :

  The number $ A \ \equal{} \ \left[ (3 \ \plus{} \ \sqrt {p} )^{2n} \right] \ \plus{} \ 1$ is divisible by $ 2^{n \plus{} 1}$ for all positive integer $ n$[/i]

It is easy to shek, that $ p\equal{}2$ don't work.
Let $ p$ is odd prime and $ B_n\equal{}(3\plus{}\sqrt p )^{2n},C_n\equal{}(3\minus{}\sqrt p)^{2n},D_n\equal{}B_n\plus{}C_n$.
Then $ D_n$ is natural and $ D_n\equal{}2^{n\plus{}1}d_n$, were $ d_n\equal{}\frac 12 [(\frac{p\plus{}3}{2} \plus{}\sqrt p)^n\plus{}(\frac{p\plus{}3}{2}\minus{}\sqrt p )^n]\in N$.
Obviosly $ A_n\equal{}D_n\minus{}[C_n]$, therefore $ 2^{n\plus{}1}|A_n$ if $ |3\minus{}\sqrt 3|<1$. For example $ p\equal{}5$ work.

What is you don't undestand?
$ (3\plus{}\sqrt p)^2\equal{}2[\frac{p\plus{}3}{2}\plus{}\sqrt p]$.
$ D_n\equal{}(3\plus{}\sqrt p)^{2n}\plus{}(3\minus{}\sqrt p)^{2n}\equal{}2^{n\plus{}1}\frac 12 [(\frac{p\plus{}3}{2}\plus{}\sqrt p)^n\plus{}(\frac{p\plus{}3}{2}\minus{}\sqrt p)^n]\equal{}2^{n\plus{}1}d_n, d_n\in N.$
$ A_n\equal{}D_n\minus{}[C_n]$. Therefore must be $ 2^{n\plus{}1}|[C_n]\equal{}[|3\minus{}\sqrt p|^{2n}]\to 4<p<16$ or $ p\equal{}5,7,11,13.$

[quote="hsiljak"][quote="mai quoc thang"]Rust , i think you are the worst Mathlinker and you participate in Mathlinks just to send the rubbish :huh:

  But the problem seems to be very nice

Can anybody post a solution? :oops:[/quote]

Rust's solutions are in most cases correct and elegant, they're just condensed so you need a pen and paper with you to fill in the steps he omits. And please use milder words.[/quote]

In my opinion, he regularly posts incorrent solutions. He misses some important points of the statement or makes a mistake, but that is apparently due to his english.  So [b]Rust[/b], please reread your solutions before posting them, and try to be more precise about explaining your claims.

and [b]mai quoc thang[/b] don't be so brutal.

I'm sorry to continue this off-topic discussion, but I just have to add this: honestly, I don't read most of Rust's solutions, since they require time and patience. Nevertheless, some of the ideas and techniques he shown in the past are very interesting and good to know.

My words "Rust's solutions are in most cases correct and elegant" should have been "in some cases correct and elegant"... I was a bit on the subjective side there, little bit annoyed with the comment of being the worst Mathlinker...

Once again, sorry for the off-topic.

[quote="mai quoc thang"]Rust , i think you are the worst Mathlinker and you participate in Mathlinks just to send the rubbish :huh:

  But the problem seems to be very nice

Can anybody post a solution? :oops:[/quote]
It is not good for you to use that kind of language to talk to other people, [b]Mai Quoc Thang[/b], :mad:.


geometry
circumcircle
incenter
geometry unsolved

Dear Mathlinkers,
let ABC be a triangle, M be an arbitrary point on the side BC, 0 the circumcircle of ABC, O the center of 0,
M' the meet point of the perpendicular to AB passing through O with the arc ACB,
I' the incenter of the triangle ABM,
1 the circle tangent to MA, MB and 0, (mixtilinear incircle or Thébault' s circle)
U the point of contact of 1 with 0.
Prove that U, I' and M' are collinear.
Sincerely
Jean-Louis

Dear Jean-Louis,
In a book, the author guides how to construct Thebault's circle as this problem (without explaination), and I don't know any construction except the above construction.
And this problem can be easily solved with the note two cyclic quadrilaterals.  :)

reference:
http://www.mathlinks.ro/viewtopic.php?t=215841

N'=⊙O∩UI’
From the link I mentioned above we know:
1.N is the midpoint of arc BC
2.U,B,E,I’ are concyclic
=>  ∠BUI’=∠BUN+∠NUN’=∠NAB+∠NUN’=(∠A+∠B)/2
=>  N’=M’

Dear Plane Geometry and Mathlinkers,
thank for your proof.
My proof use the result of Protassov that you can find on my website :
http://perso.orange.fr/jl.ayme   vol. 2  Un remarquable résultat de victor Protassov.
Sincerely
Jean-Louis


four real numbers are there: $w$, $x$, $y$ and $z$ such that

$w+x+y+z = w*x*y*z = \frac{711}{100}$

If no rounding off and approximations are not allowed, can you find these four variables?

[quote="Treething"]No, you cannot. Two equations, four variables, infinite solutions.

However, if w, x, y, and z must be expressed in the form $\frac{c}{100}$, c being an integer, this problem becomes a well-known one.[/quote]

can you explain a little more into this?

Yeah, the one with money has only been posted a good 100 times here...

[quote="t0rajir0u"]If you don't put any other conditions on $w, x, y, z$, there are an infinite number of solutions, since we can arbitrarily set two of the numbers and we will get a system of two equations in two variables.

The problem to which Treething alludes is phrased in terms of money, if I recall correctly, and it's been solved several times.[/quote]

can you post the link of the problem that you were mentioning?