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integration
calculus
function
topology
real analysis
real analysis unsolved

a measure $m$ on a metric space $(X,d)$ is said uniformly distributed if the measure of an open ball $B(x,r)$ depends only of $r$. Prove that $2$ borel measures on a metric space $(X,d)$ that are uniformly distributed are proportional.

I found the following version on the web:

A positive Borel regular measure $\mu$ on a metric space $(X,d)$ is said to be uniformly distributed (I’ll cal it “UDM”) if $0<\mu(B(x,r))=\mu(B(y,r))<\infty$ for all $x,y\in X$ and all positive $r$, where $B(x,r)$ is the open ball of radius $r$ centered in $x$.

Prove that a [i]complete[/i] metric space has at most one UDM (up to scalar multiples).

I’ll write down the proof for the even more restrictive case when the metric space is compact. This isn’t as big a deal as it looks (I think :)), because a complete metric space with a UDM is easily seen to have the Heine-Borel property (i.e. every closed bounded subset is compact), so the proof shouldn’t be too hard to extend to the general case, by restricting attention to a bounded ball and working away from the boundary. I’m not sure if this works as smoothly as it sounds, but I’ll look into it later.

Let $(X,d)$ be a compact metric space with two UDM’s $\mu$ and $\nu$, which must be finite. By Radon-Nikodym, we can write $\nu=\mu_{c}+\mu_{s}$, where $\mu_{c},\mu_{s}$ are positive regular Borel measures, $\mu_{c}\ll\mu$ ($\mu_{c}$ is absolutely continuous with respect to $\mu$) and $\mu_{s}\perp\mu$ ($\mu_{s}$ and $\mu$ are singular). Moreover, $\mbox{d}\mu_{c}=f\mbox{d}\mu$ for some $0\le f\in\mathcal L^{1}(\mu)$. Also, let $K$ be the support of $\mu_{s}$. It is a compact nowhere dense subset of $X$ such that $\mu(K)=0$.

For all open balls $B(x,r)$ which do not cut $K$, we have $\nu(B(x,r))=\mu_{c}(B(x,r))$ which, in turn, equals $\int_{B(x,r)}f\mbox{d}\mu$. The important thing to notice here is that the latter integral does not depend on $x$ as long as $d(x,K)\ge r$. I claim that this implies $f$ constant (in this context, this means “constant amost everywhere with respect to $\mu$”, of course; it’s also equivalent to “constant $\mu$-a.e. on $X\setminus K$”, since $\mu(K)=0$). From this, it would follow immediately that $\nu$ is a scalar multiple of $\mu$, the scalar being the a.e. constant value of $f$.

We now have to prove that the properties of $f$ that we’ve discovered imply that it must be constant. Assuming the contrary, we can find compact sets $S_{1},S_{2}\subset X\setminus K$, both having positive measure, such that $f\le a_{1}$ on $S_{1}$ and $f\ge a_{2}$ on $S_{2}$, for some positive reals $a_{1}<a_{2}$. Now consider the following

[i]Lemma[/i]

In the present hypotheses, let $S$ be a compact subset of $X$ with $\mu(S)>0$. Then, for any $\varepsilon>0$, we can find $\delta>0$ such that whenever $r<\delta$, there is an $x\in X$ with $\mu(S\cap B(x,r))>(1-\varepsilon)\mu(B(x,r))$ (that is, small open balls of any radius can be filled arbitrarily well by $S$).

Assuming the lemma for now, we continue the proof. Choose two balls $B(x_{i},r)\subset X\setminus K,\ i=1,2$ of the same small radius such that $\mu(S_{i}\cap B(x_{i},r))>(1-\varepsilon)\mu(B(x_{i},r))$ for a convenient $\varepsilon>0$. If this $\varepsilon$ is small enough, the integral of $f$ over $B(x_{1},r)$ will be smaller than $\frac{a_{1}+a_{2}}2\mu(B(x_{1},r))$, while the integral over $B(x_{2},r)$ will be larger than the same value. This contradicts the fact that the integral of $f$ on any ball of radius $r$ which doesn’t cut $K$ is the same.

[i]Proof of the Lemma[/i]

For an arbitrary set $A\subset X$, denote by $\chi_{A}$ the characteristic function of $A$. Also, in order to simplify the notation a little bit, we’ll denote by $\mu(r)$ the $\mu$-mass of any open ball of radius $r$.

For every $r>0$, notice that the map $f_{r}(x)=\mu(S\cap B(x,r))$ is continuous. Let’s integrate it over $X$ with respect to $\mu$. The integral is $\int_{X\times X}\chi_{S}(y)\chi_{B(x,r)}(y)\mbox{d}\mu(y)\mbox{d}\mu(x)$. Now change $\chi_{B(x,r)}(y)$ into $\chi_{B(y,r)}(x)$ (they are clearly equal), and change the order of integration (the two-variable function involved is measurable and non-negative, so this is allowed). The result is $\mu(r)\mu(S)$ if we also assume (and we do from now on) that $\mu(X)=1$. In other words, the integral of $g_{r}=\frac{f_{r}}{\mu(r)}=\frac{\mu(S\cap B(x,r))}{\mu(r)}$ over $X$ equals $\mu(S)$, for any $r$.

Now choose a compact $T\subset X$ disjoint from $S$, and such that $\mu(X\setminus T)<\frac{\mu(S)}{1-\varepsilon}$. Take $\delta<d(T,S)$. Then, for every $r<\delta,\ g_{r}=0$ on $T$. This means that $\mu(S)$ equals the integral of $g_{r}$ over $X\setminus T$. If $g_{r}$ were to be $\le 1-\varepsilon$ everywhere, then this integral would be at most $(1-\varepsilon)\mu(X\setminus T)<\mu(S)$, a contradiction.


inequalities
LaTeX
inequalities unsolved

-(n^2+n-1)/(n^2+2n+2)(n^2+1)<=0 when n>=1  can you prove this? sorry for non-latex writing

If you mean that $ \frac {n^2 \plus{} n \minus{} 1}{(n^2 \plus{} 2n \plus{} 2)(n^2 \plus{} 1)} \geqslant0$ $ (n\geqslant1)$

We get that $ (n^2 \plus{} 2n \plus{} 2)(n^2 \plus{} 1) \geqslant0$ for all $ (n\geqslant1)$

So $ n^2 \plus{} n \minus{} 1$ must be positive, to solve this ineq, it's up to $ n^2 \plus{} n \minus{} 1\geqslant0$

Choose the positve interval which $ n\geqslant1$

 :lol:  Don't forget to try Latex  :lol:


[quote="mecrazywong"]When will this weird translation be replaced?[/quote]
I think it is pretty good!
What do you think the new name should be?

[quote="liyi"]other topics?
其他议题?[/quote]
maybe simply 其他?

“无心呢喃”怎么样？

  英语怎么说？

这样吧：Math Problems——好题难题经典题，题题绽艳
        Other Topics——家事国事天下事，事事关心
对得不大好，虽然对仗基本工整，但平仄有点问题，读起来不大顺。希望哪位古典文学的巨擘能给出更好的提议。


Λιγα ευκολακια τωρα για ξεκουραση...
   1)Αν a,b,c>0 και a+b+c=1 ν.δ.ο $5(a^{2}+b^{2}+c^{2})\leq 6(a^{3}+b^{3}+c^{3})+1$
   2)a,b=θετικοι ακεραιοι ν.δ.ο αν ο αριθμος (a+1)/b+b/a ειναι ακεραιος τοτε ειναι ο αριθμος 3
   3)N.δ.ο σε καθε τργωνο ισχυει a^2/(b+c-a)+b^2/
(a+c-b)+c^2/(a+b-c)>=3R(3)^(1/2)

 ΡΑΛΛΗΣ  :D

Gia tin (2):

Estw $\frac{a+1}{b}+\frac{b}{a}=k$. Tote $(a+1)a+b^{2}=kab \ \ (1)$ ap'opou
$a|b$ ki etsi $b=b_{1}a$. Tote antikathistwntas stin sxesi $(1)$ kai kanontas prakseis pairnoume $[(k-1)b_{1}-1]a=1$ ap'opou $a=1$ kai $(k-1)b_{1}-1=1$ ara eite $k=3$ kai $b_{1}=1$ (ara $b=1$) eite $k=2$ kai $b_{1}=2$ (ara $b=2$). Vazontas tin deuteri stin $(1)$ den tin epalitheuei ki etsi i monadiki lisi tis einai $a=1,b=1$ kai etsi $k=3$ pou einai kai to zitoumeno.

Alexandros

(Y.G. Exw prosthesei ena poli kalo eisagwgiko vivlio thewrias arithmwn sto site pou anaferw se proigoumeno post, kathws episis kai mia sillogi askisewn thewrias arithmwn oi opoies einai alites. Euxaristw ton stelio (stedes) pou mou ta esteile).

Για την πρώτη τώρα .H anisotita einai isodinami me
$5(a^{2}+b^{2}+c^{2})\leq 18abc+6(a^{2}+b^{2}+c^{2})-6(ab+bc+ca)+1$

η αποία είναι ισοδύναμη με $4(ab+bc+ca)\leq 1+9abc$
ή $(b+c-a)(c+a-b)(a+b-c)\leq abc$
που είναι ισοδύναμη με την ανισότητα Schur.  :D

Aς λυσουμε και την τριτη.

Αρχικα τσεμπισεβ, οποτε $\text{LHS}\geq \frac{1}{3}(a^{2}+b^{2}+c^{2})(\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c})$

Τωρα $(\frac{1}{b+c-a}+\frac{1}{a+c-b}+\frac{1}{a+b-c})\geq \frac{\sqrt{3}}{2r}$

και μετα πραξεις με ΑΜ-ΓΜ.....

Αγαπητέ Ράλλη Γεια σου
Μια ερώτηση έχω από που βρίσκεις όλα αυτά τα προβλήματα ειναι ο ατελείωτος  τροφοδότης προβλημάτων μας  :D αν θέλεις σε παρακαλώ να μου δημοσιεύσεις τις πηγές σου θα σε ευχαριστούσα πολύ
καλή συνέχεια αν κι μάλλον από όσο σε ξέρω είναι αυτονόητη για σενα

[quote="cretanman"][quote="socrates"][quote="cretanman"]Gia tin (2):

Estw $\frac{a+1}{b}+\frac{b}{a}=k$. Tote $(a+1)a+b^{2}=kab \ \ (1)$ ap'opou
$\boxed{a|b}$  .[/quote]

μηπως $a|b^{2}$ ή ειναι $(a,b)=1$?   :maybe:

Δε το πιανω.... σορρυ.... :)[/quote]

Exeis apolito dikio! Einai lathos o isxirismos oti $a|b^{2}\rightarrow a|b$. Parasirthika apo to oti ean p prwtos kai $p|a^{2}$ tote $p|a$. Diorthwnw loipon tin askisi.

Apo tin sxesi $(a+1)a+b^{2}=kab$ exoume oti $b|a(a+1)$ kai epeidi $(a,a+1)=1$ ara $b|a$ 'h $b|a+1$.

(I)  $\textbf{Ean b / a}$ tote $a=rb$ gia kapoio $r\in \mathbb{N}$ opote i parapanw sxesi ginetai $(rb+1)rb+b^{2}=krb^{2}$ ap'opou $r^{2}b+r+b=krb$ diladi $b|r$ kai etsi $r=bl$ gia kapoio $l\in \mathbb{N}$.
 Tote $l(kb-1-b^{2}l)=1$ ki etsi $l=1$ kai $kb-1-b^{2}l=1$ ap'opou $b(k-b)=2$ kai etsi kai stis dio periptwseis ($b=1, k-b=2$ eite $b=2, k-b=1$) pairnoume $k=3$.

(II) $\textbf{Ean b / a+1}$ tote $a+1=rb$ gia kapoio $r\in \mathbb{N}$ opote i parapanw sxesi ginetai $(rb-1)rb+b^{2}=k(rb-1)b$ ap'opou $k-r=b(kr-r^{2}-1)$ ara $b|k-r$ ki etsi $k-r=bl$ 'h $k=bl+r$ gia kapoio $l\in \mathbb{N}$. Tote h teleutaia sxesi ginetai $l=(bl+r)r-r^{2}-1$ ki etsi $l(br-1)=1$ ap'opou $l=1$ kai $br-1=1$ diladi $br=2$. Kai stis dio periptwseis ($b=1, r=2$ eite $b=2, r=1$) exoume $k=bl+r=3$.

Alexandros[/quote]

πολυ ωραια!!!!!!!
 :lol:

Να σημειωσω οτι το προβλημα 3 προερχεται απο ρουμανικό ΤΣΤ,
δημιουργος νομιζω ειναι ο laurentiou Panaitopol και
ειναι μια απο τις αγαπημενες μου καθως ειναι πολυ κλειστη!!!

 :)

ΓΕΙΑ ΣΕ ΟΛΟΥΣ!
Φιλε,ambro,τα προβληματα ειναι απο μια συλλογη προβληματων που εχω απο ρουμανικους διαγωνισμους..
Οσον αφορα την 3η
Θετωντας a=2RsinA,.....
εχουμε sin^2A/(sinB+sinC-sinA)+....>=3riza(3)
Oμως το α μελος ειναι >= του (sinA+sinB+sinC)  απο cauchy-shwratz
Ομως sinA+sinB+sinC>=3ριζα(3) απο ΑΜ-ΓΜ και sinAsinBsinC>=3PIZA(3)/8
Θελω να πω οτι παρεμβαλλονται πολλα,αρα δεν μπορει να θεωρηθει και πολυ σφιχτη....

ΡΑΛΛΗΣ :D

Ανάποδα: $\sin A \sin B \sin C \leq  \frac{3\sqrt{3}}{8}$

Οντως :blush: Βλακεια.....Ισως τελικα ειναι λιγο σφιχτη....

Λοιπον,επειδη θελω να μην αφηνω αλυτες τετοιες ασκησεις, καθησα χθες μεχρι της 2 το βραδυ,και βρηκα μια  αλλη λυση.Εχουμε +λεμε

a^2/(b+c-a)+b^2/(a+c-b)+c^2/(a+b-c)=1/2(a^2/(t-a)+b^2/(t-b)+c^2/(t-c))
me  2t=a+b+c
episis a^2/(t-a)=ta/(t-a)-a,.......opote

a^2/(t-a)+b^2/(t-b)+c^2/(t-c)=t[a/(t-a)+b/(t-b)+c/(t-c)]-2t

Γνωριζουμε ομως οτι
                              $ab+bc+ca=t^{2}+r^{2}+4Rr$    (1)
                              $(t-a)(t-b)(t-c)=tr^{2}$    (2)
                              $abc=4tRr$    (3)

οποτε ευκολα υπολογιζουμε   με την βοηθεια των 1,2,3 τελικα οτι
a^2/(t-a)+b^2/(t-b)+c^2/(t-c)=4t(R-r)/r

Επομενως πρεπει να αποδειξουμε  οτι [2t(R-r)]/r>=3R(3)^(1/2)   (4)

 oμως $(a+b+c)^{3}\geq 27abc$  οποτε απο 3 εχουμε
          $t^{2}\geq 27/2Rr$  και επειδη $R\geq 2r$
           $t\geq 3r(3)^{(}1/2)$   (5)

Eπομενως η σχεση προς αποδειξη ειναι 2t(R-r)>=3Rr(3)^(1/2)  (4)
γινεται λογω της 5 , 6r(3)^(1/2)(R-r)>=3Rr(3)^(1/2)
                           <=>$2R-2r\geq R$
                           <=>$R\geq 2r$
που προφανως ισχυει.Παλι βεβαια παρεμβαλονται αρκετα πραγματα......(αυτη την φορα ομως ,ολες οι ανισοτητες ειναι με την ιδια φορα)

ΡΑΛΛΗΣ :D


algebra
polynomial
superior algebra
superior algebra unsolved

Find the minimal polynomial of 1+i over Q

here is what I did plz correct me
Since [Q(1+i):Q]=[Q(1,i):Q]=4 the min polyn of 1+i must have degree 4
I let u=1+i then u-i = 1 squaring we get u^2-2ui-1=1 ==> u^2+2=2ui squaring again we get u^4+8u^2+4=0

I'm not too sure about what I'm doing here please give me some feedbacks

Thank you

The problem is that [Q(1,i):Q] = 2, since Q(1,i) = Q(i).

[quote="sally"]I let u=1+i then u-i = 1 squaring we get u^2-2ui-1=1[/quote]

It would've been more useful here to write $ u \minus{} 1 \equal{} i$.  Squaring, we then get $ u^2 \minus{} 2u \plus{} 1 \equal{} \minus{}1 \Leftrightarrow u^2 \minus{} 2u \plus{} 2$, which is our actual minimal polynomial.


I may be crazy, but I think this is just Cauchy-Schwarz for both $a_i$ and $b_i$

[quote="Sung-yoon Kim"]Let $a_{1}\leq b_{1}\leq a_{2}\leq b_{2}\leq ... \leq a_{5}\leq b_{5}$ be real numbers. Prove that
$ (\sum_{i=1}^{5}(a_{i}+b_{i}))^{2}\geq 5 (\sum_{1 \leq i < j \leq 5}a_{i}a_{j}+b_{i}b_{j})$.[/quote]

If there are no equalities, then the remained inequality woyld be valid still. i.e., if $a_{1}<b_{1}<a_{2}<b_{2}< \cdots <a_{5}<b_{5}$, then $ (\sum_{i=1}^{5}(a_{i}+b_{i}))^{2}> 5 (\sum_{1 \leq i < j \leq 5}a_{i}a_{j}+b_{i}b_{j})$

And for this strict inequality problem, there is very nice solution. However, the method cannot be applied to the original one. You will see that there is another equality condition except $a_{1}=b_{5}$.

[quote="Sung-yoon Kim"]Let $a_1 \leq b_1 \leq a_2 \leq b_2 \leq ... \leq a_5 \leq b_5$ be real numbers. Prove that
$\displaystyle (\sum_{i=1}^5 (a_i+b_i))^2 \geq 5 (\sum_{1 \leq i < j \leq 5} a_ia_j+b_ib_j)$.[/quote]

Well, anyone has any idea about this inequality?

[quote="Sung-yoon Kim"]Let $a_1 \leq b_1 \leq a_2 \leq b_2 \leq ... \leq a_5 \leq b_5$ be real numbers. Prove that
$\displaystyle (\sum_{i=1}^5 (a_i+b_i))^2 \geq 5 (\sum_{1 \leq i < j \leq 5} a_ia_j+b_ib_j)$.[/quote]
It's a very nice inequality, Kim. I found that it can be proved by Cauchy-Schwarz.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose that an arbitrary group  G has two subgroups  H and K each of finite index in  G. Show that the intersection  H \cap K has finite index in  G. In particular, show that  [G: H\cap K] < [G:H] [G:K].

There's an obvious monomorphism $G/(H \cap K) \to G/H \times G/K$ (sets of left-cosets).

In this case, a monomorphism is a injective homomorphism.

P.S.: The proposition is in fact false. But it would be right, if you wrote $\leq$ instead of $<$. There are counterexamples where $|G: N\cap H|=|G: H\parallel G: N|$, so it must be a $\leq$ and can't be a proper inequality.

For that map to be a homomorphism, we need all sugroups here to be normal; that wasn't assumed here.

Without that assumption, we still have a natural injective map $(a\cdot H\cap K)\mapsto (a\cdot H,a\cdot K)$ between the two quotients. This map isn't necessarily a homomorphism, because both sides aren't necessarily groups.


So I want to get latexrender on my server

I have full access to it

I set up GS, ImageMagick, and Miktex under C:\latexrender\gs, C:\latexrender\ImageMagick, and C:\latexrender\latex

I am on Server 2003.

My site's absolute location would be C:\Site\Site2

I'm on Apache/PHP 5.23

I used stevem's latexrender thingie and in class.latexrender.php, I have:

[code]
    var $_latex_path = "/latexrender/latex/";
    var $_dvips_path = "/latexrender/latex/dvips/";
    var $_convert_path = "/latexrender/ImageMagick/";
[/code]

What are these paths supposed to be?
Any help would be greatly appreciated.

mk we solved the issue in an IM session just now, and now I'm going to hijack this topic :D

I've got latexrender installed, but the gif image quality is horrible. I'm guessing it's ImageMagick's fault, but how do I customize it?

png is fine, but I [i]think[/i] someone told me not to use png for some reason..

hmm.. I thought it was weird, since gifs were just as high-quality as pngs on my OS X server.

[quote="#H34N1"][code]
    var $_latex_path = "/latexrender/latex/";
    var $_dvips_path = "/latexrender/latex/dvips/";
    var $_convert_path = "/latexrender/ImageMagick/";
[/code]

What are these paths supposed to be?
Any help would be greatly appreciated.[/quote]
They should be the (absolute) paths to where latex.exe, dvips.exe and convert.exe respectively can be found. There's also a line
[code]var $_identify_path= ...[/code]which should point to identify.exe
Quoting from the readme.txt:
[code]# For Windows the paths in class.latexrender.php must use \\ or / not just a single \
# For example
#    var $_latex_path = "C:\\texmf\\miktex\\bin\\latex.exe";
# or
#    var $_latex_path = "C:/texmf/miktex/bin/latex.exe";[/code]

Ya I got that now but I need to get it working in a way that gif works

There should be no problem with gifs. Post a copy of a gif and a png to compare but zip them up first to ensure that the forum doesn't change the quality when it shows them.

No. WHen I copy and paste the image into Paint, it displays fine. o.o

Okay I zipped both a png and gif into one

It seems it may be the way that browsers interpret the transparency. PNGs were designed to have a much better transparency. You could try increasing the density
[code]var $_formula_density = 120;[/code]but that may not help. It also seems to work better on a Linux system.
You could also try some options of [url=http://www.imagemagick.org/script/convert.php]ImageMagick's convert program[/url] or ask in an ImageMagick forum.
If png works well you could just stick to that as browsers now support the png format.

How does AoPS work magic?

And also, aren't gifs significantly smaller than pngs?

AoPS uses the same system but (I believe) on Linux. It could be a problem with Windows or the particular version of ImageMagick, I suspect it is a Windows problem.

[quote="#H34N1"]And also, aren't gifs significantly smaller than pngs?[/quote]Not necessarily - see attached created on a Linux system

hehe I tried increasing the density yesterday, and it just made the entire image bigger :lol:

but is it really because of differences across browsers/OS's? I'm on Ubuntu Hardy and I get these:

png:
[img]http://i28.tinypic.com/2wna3pd.png[/img]

gif:
[img]http://i26.tinypic.com/2hrkgt0.gif[/img]

Increased density gif (I think this one is 320):
[img]http://i29.tinypic.com/f4inwo.gif[/img]

the GIF looks unreadable to me (FF3, Windows)

The PNG looks like what I got.

[quote="undefined117"]hehe I tried increasing the density yesterday, and it just made the entire image bigger :lol:
Increased density gif (I think this one is 320):
[img]http://i29.tinypic.com/f4inwo.gif[/img][/quote]That looks to be excellent quality but I expect the density is rather high. What happens with a density of 150? Also try reducing the font. If it's already at 10pt use the extarticle class to get 8pt. Then try compromising between sizes and densities.

Try a PNG at that size. It owns on quality

Ya I'm switching server to Ubuntu 8.04 server edition

hopefully I'll get this thing resolved

or I'll just ask undefined


[quote="Megus"]It started in September as some of you may know and now first serie is over [to those who don't what I'm talking about: in Polish Olympiad there are three stages, first one is divided into three series each containing 4 problems which you solve at home and send to the proper place, next two stages are "normal"  :) ] so I think I can translate it for if you want to. Where should I publish translated problems ?[/quote]

Thanks for publishing all problems from the various rounds on MathLinks.   :)

The best procedure might be to post all the problems in the respective sub-forums and to post links to the problems in this sub-forum or if you like even in this thread.  :)

No problem orl  :D  - Ok I translated these problems and here are links to appropiate topics:

First problem [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18082[/url]
Second problem [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18083[/url]
Third problem [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18084[/url]
Fourth problem [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=18085[/url]

Maybe they are not very challenging but that's just first stage of the Polish National Olympiad 2004  :) . I hope this will be useful for someone. Please post your opinions about the problems and whatever else you want

P.S. I've added these translated problems to my website [url]http://www.megus.tk[/url] in Math Problems section so you can find them all in one place easily.  :cool:


linear algebra
matrix
LaTeX
algebra
polynomial
vector
invariant

Hi,  I have the following matrix:

  ROW 1:   1   3   3

  ROW 2:   3   1   3

  ROW 3:  -3  -3  -5

(Sorry I had to edit my post because I screwed up the Latex)

And I'm asked to find an invertible 3x3 real matrix P such that P^-1AP is in rational form.

Anyways, I computed the char. polynomial, and I arrived at:

P(x) =  (x+2)^2(x-1)

Minimal polynomial to be:

M(x) = (x+2)(x- 1)

Now, I calculated the null spaces associated with:

A+2I and A-I.  Now, for A-I, I had a space generated by the following vector:

(-1, -1, 1)

But for A+2I I get the zero vector and similarly for (A+2I)^2.  I'm at a loss at how to proceed and construct P in this case.  Can somebody help me?

My second question is the following:

If T is diagonalizable linear operator on V, prove that every T invariant subspace has a complementary T invariant subspace as well.

Any help woudl be appreciated!

 [/list]

You should have gotten a two-dimensional subspace for $ A \plus{} 2I$: $ A \plus{} 2I \equal{} \begin{bmatrix}3 & 3 & 3 \\
3 & 3 & 3 \\
\minus{} 3 & \minus{} 3 & \minus{} 3\end{bmatrix}$ (quote for working matrix code)
That's a rank-one matrix; the null space is orthogonal to the row space.

As for the second question- there's no need to assume the first subspace is $ T$-invariant. Choose a basis to make it diagonal, and show that every subspace has a complementary subspace spanned by some subset of the basis.


trigonometry
inequalities
inequalities proposed

For $0\leq x,y\leq 1$, let:
$f(x,y)=xy^{2}\sqrt{1-x^{2}}-x^{2}y\sqrt{1-y^{2}}$
Find the minimum constant $c$ for which the following condition holds:

For any integer $n>1$ and any real numbers $a_{1}, a_{2}, .., a_{n}$ such that $0\leq a_{1}<a_{2}<...<a_{n}\leq 1$ we have:
$f(a_{1}, a_{2})+f(a_{2}, a_{3})+...+f(a_{n-1},a_{n})<c$

[quote=kunny]This problem is Japan Mathematical Olympiad Preliminary 2006, Problem 11.  :lol:  [/quote]
 can you give the links ?

fast calculation
$c=\frac{\pi}{2}$
right? :P  :wink:

put $\sin \alpha_{i}=a_{i}$
then $f(a_{i},a_{i+1})=\sin \alpha_{i}\sin \alpha_{i+1}\sin (\alpha_{i+1}-\alpha_{i})\le \alpha_{i+1}-\alpha_{i}$
thus our required sum is less than $\alpha_{n}-\alpha_{1}<\frac{\pi}{2}$
to see that it can not be improved take $n$ to infinity and $a_{1}$ to $1$ as limit. :)

Can you derive the inequality: \[f(a_{i},a_{i+1})=\sin \alpha_{i}\sin \alpha_{i+1}\sin (\alpha_{i+1}-\alpha_{i})\le \alpha_{i+1}-\alpha_{i}\]

we know that $\sin x\le 1$ and $\sin x <x$ for $x>0$


Two kings are placed on a normal ($8 \times 8$) chessboard, a white one at $a1$, a black one at $h8$. White starts moving (like a 'normal' king at chess) and it's not allowed to enlarge the distabce between the kings (measured by how much moves they are away from each other). Now the goal for every player is to reach any one of the border lines ($a,h,1,8$) he is not starting at (e.g. white has to reach row $8$ or column $h$). Which player has a winning strategy and why¿

I don't get it.. Why can't the white king simply go upwards from $a_1$ to $a_2$, then to $a_3$, and so on, until it reaches $a_8$? :? How could black stop this? I'm sure I'm overlooking something..

Sorry  :blush:  :blush:  :blush:
I've edited it:
The distance between them is not allowed to get bigger.


[tex] \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq 3(a^x+b^x+c^x)^\frac{1}{x}[/tex] Is this what do you mean?

:blush:  May be I can solve it later, but for the moment I can't :oops:

[quote="leibniz"]$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq 3(a^x+b^x+c^x)^\frac{1}{x}$ Is this what do you mean?[/quote]
It should be $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\geq 3\left(\frac{a^x+b^x+c^x}3\right)^\frac{1}{x}$. (Compare with the x=3 case discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=38369]here[/url].)
By trial & error, I find that x<6,536524: for $(a,b,c)=(0.9169,0.8765,1.1127)$ (normed to $a^x+b^x+c^x=3$) this yields a counterexample with $x=6,536524$.

I guess it should be very hard to solve for $x=5$ or $x=6$ :huh:  reminds me a similar problem [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=382825#p382825]"superhard"[/url]

spanferkel, your example is very good.But I think with $x=5;6$ is very difficult

[quote="nthd"]Find all x such that with a,b,c>0 we have:
a^2/b+b^2/c+c^2/a>=3(a^x+b^x+c^x)^(1/x)

I solved with x=4,but I think x<7.[/quote]
See here  Gladkich's proof:
http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=%2Aeneq%2A&t=38369

It was here:
[quote="spanferkel"]
It should be $\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq 3\left(\frac{a^{x}+b^{x}+c^{x}}3\right)^\frac{1}{x}$. (Compare with the x=3 case discussed [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=38369]here[/url].)[/quote]

I have a dicussion: using mixing variable is a good method in this case!(because we have x<7)


solve this equation:
$a^3-b^3-c^3=3abc$
$a^2=2(b+c)$

Make $a=x$, $b=-y$ and $c=-z$. Thus we have:

$x^3+y^3+z^3=3xyz$. And so we´ll have

$(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]=0$.

But we have $x^2+2(y+z)=0$.

First case give us $x+y+z=0$. And so this implies $x=0$ or $x=2$.

Second case give us $x=y=z$. And this also implies $x=0$ or $x=-4$.

And the solutions are $(0,k,-k)$, $(2,k,2-k)$ and $(-4,4,4)$, for any real k.

[quote="gopherhole112"]Wow! that was beautiful. props.[/quote]
yes i am agree with you. :D

let $(a,b,c)$ be the answer of this equation we have:
$a^3-b^3-c^3=3abc \rightarrow a^3-b^3-c^3-3abc=0 \rightarrow (a-b-c)(a^2+(b-c)^2+ab+ac+bc)=0 \rightarrow a=b+c=\frac{a^2}{2} \rightarrow a=2,b=c=1$

[quote="mffmartinelli"]I think that you forgot some solutions, no?!
Excuse me, but can we explain me?
Thanks![/quote]
i think i didn't forget anything. :D

Yes, you forgot per example the case when $a=-4$, $b=4$ and $c=4$.


calculus
derivative
analytic geometry
graphing lines
slope
calculus computations

Find the equation for the tangent line of $ f(x)$ at $ A$, where $ f(A)$ is defined and is differentiable.

I was reading limits and continuity chapter, and I came across some derivative concepts, so I decided to make this problem.

That's not a novel problem, in fact, you're assumed to know how to do it.  They will give you a problem and say "Find the equation of the tangent line at the point whatever".  It's given by
$ y_{1}\minus{}y_{0}\equal{}m(x_{1}\minus{}x_{0})$, where $ m$ is slope obtained by evaluating $ f'(A)$.

ah. O.K.

I did know how to do it...

[hide="how to do it"]We have the point $ (A,f(A))$ and we find that the slope of the tangent line is $ f'(A)$.

$ f(A)\equal{}f'(A)*A\plus{}b$

$ b\equal{}f(A)\minus{}f'(A)*A$

The equation is

$ g(A)\equal{}f'(A)x\plus{}f(A)\minus{}f'(A)*A$[/hide]

[quote="JRav"]What you have looks something like a first-order linear approximation but not exactly like it.[/quote]

what he has there is correct :?:


modular arithmetic
number theory unsolved
number theory

Let $ n\in\mathbb{N}$ and $ n > 1$. Denote $ S_n \equal{} 1^n \plus{} 2^n \plus{} \ldots \plus{} \left(n \minus{} 1\right)^n$. Prove that:

a) $ n^2\mid S_n$ for every odd $ n$;
b) there exist infinitely many n such that $ n^3\mid S_n$.

Solution to a-)
$ (n \minus{} k)^n \equal{} \dbinom{n}{0}( \minus{} k)^n \plus{} \dbinom{n}{1}( \minus{} k)^{n \minus{} 1}n \plus{} n^2(stuff)\equiv \minus{} k^n\mod(n^2)$
$ \Leftrightarrow n^2\mid (n \minus{} k)^n \plus{} k^n$.
So $ n^2\mid \sum_{k \equal{} 1}^{n \minus{} 1} (n \minus{} k)^n \plus{} k^n \equal{} 2\sum_{k \equal{} 1}^{n \minus{} 1} k^n$, but $ n$ is odd, and the result follows.

[quote="SAPOSTO"]Let $ n\in\mathbb{N}$ and $ n > 1$ $ S_n \equal{} 1^n \plus{} \ldots \plus{} (n \minus{} 1)^n$. Prove that:

a) $ n^2|S_n$ for every odd n;
b) there exist infinitely many n such that $ n^3|S_n$.[/quote]

In $ \mathbb{Z}/n^3\mathbb{Z}$ with $ n$ odd we have $ S_n \equal{} \frac {1}{2}\sum_{i \equal{} 1}^{n \minus{} 1}{n^2i^{n \minus{} 1}}$. (First point is trivial).
Take $ n \equal{} pq$ with $ 2 < p < q$ primes such that $ p \minus{} 1 \nmid q \minus{} 1$ and $ q\minus{}p>2$. Then $ S_{pq} \equal{} \sum{i^{pq \minus{} 1}} \equal{} \sum{i^{q \minus{} 1}} \equal{} 0$ in $ \mathbb{Z}/p\mathbb{Z}$. (the same thing with $ q$).

@Patrick, i have edited, thank, i was meaning that if $ p \nmid y$ then the sum of the $ y$- power of $ p$-consecutive integer is a multiple of $ p$  :blush:

[quote="bboypa"]  Then $ S_{pq} \equal{} \sum{i^{pq \minus{} 1}} \equal{} \sum{i^{q \minus{} 1}} \equal{} 0$ in $ \mathbb{Z}/p\mathbb{Z}$. [/quote]

I dont understand why.

For example, with $ p\equal{}3$ and $ q\equal{}5$, $ S_{15} \equal{} \sum_{i\equal{}1}^{14}{i^{3*5 \minus{} 1}} \equal{} \sum_{i\equal{}1}^{14}{i^{5 \minus{} 1}} \equal{} 1\neq 0\pmod 3$

[quote="bboypa"] @Patrick, i have edited, thank, i was meaning that if $ p \nmid y$ then the sum of the $ y$- power of $ p$-consecutive integer is a multiple of $ p$  :blush:[/quote]

I'm sorry but I continue to misunderstand.

Let $ p\equal{}2$ and $ y$ odd. $ 2 \nmid y$ and $ 1^y\plus{}2^y$ is the sum of the $ y$- power of $ 2$-consecutive integer but  is [b]not [/b]a multiple of $ 2$

Let $ p\equal{}3$ and $ y\equal{}8$. $ 3 \nmid 8$ and $ 1^8\plus{}2^8\plus{}3^8$ is the sum of the $ 8^{th}$- power of $ 3$-consecutive integer but  is [b]not [/b]a multiple of $ 3$

I hope I'll not bother you but I really dont understand your demo.

[quote="pco"][quote="bboypa"] @Patrick, i have edited, thank, i was meaning that if $ p \nmid y$ then the sum of the $ y$- power of $ p$-consecutive integer is a multiple of $ p$  :blush:[/quote]

I hope I'll not bother you but I really dont understand your demo.[/quote]

Scuse me for the delay Patrick, i have not seen your replay before  :P
However i am doing a lot of copying mistake in this period (i have miss a (-1)): i correct (you surely know it) in:
"if $ p$ is a odd prime and $ p \minus{} 1 \nmid y$ then the sum of the $ y$- power of $ p$-consecutive integer is a multiple of $ p$."

Cheers!


Let $ x,y \in R, x,y \ge 1, x \plus{} y \equal{} S$. Find min, max:
$ P \equal{} \lfloor x \rfloor ! \lfloor y \rfloor !$

_________
Edited!
I'm sorry!

[quote="NguyenDungTN"]Let $ x,y \in N, x,y \ge 1, x \plus{} y \equal{} S$. Find min, max:
$ P \equal{} \lfloor x \rfloor ! \lfloor y \rfloor !$[/quote]

I think I must be confused, because I don't understand where the floors are coming from. Since, if $ x \in \mathbb{N}$ then $ \lfloor x \rfloor \equal{} x$.

Anyway, ignoring that

Consider $ x \plus{} y \equal{} S,\ x \ge y \ge 2$

Then $ x!y! \le (x \plus{} 1)!(y \minus{} 1)!$, since this has the effect of multiplying the product by a larger number than it is dividing by

Therefore, the maximum occurs for $ x \equal{} S \minus{} 1, y \equal{} 1$

By the same logic

The minimum occurs for

$ P \equal{} \begin{cases} x \equal{} y \equal{} m, & \mbox{if } S \equal{} 2m \\
x \equal{} y \plus{} 1 \equal{} m \plus{} 1, & \mbox{if } S \equal{} 2m \plus{} 1. \end{cases}$

Hence, we get the inequality

$ \begin{cases} (\frac {S}{2}!)^2 \leq x!y! \leq (S \minus{} 1)! & \mbox{if } S \equal{} 2m \\
(\frac {S \minus{} 1}{2}!)(\frac {S \plus{} 1}{2}!) \leq x!y! \leq (S \minus{} 1)! & \mbox{if } S \equal{} 2m \plus{} 1 \end{cases}$


I know this problem has been posted before, but I've searched and not been able to find any satisfactory solution. Can  someone help me out?

Sure : http://www.mathlinks.ro/Forum/viewtopic.php?t=74482 (for example.. :) )

[quote="Phelpedo"]a^b=b^a[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?t=63955
http://www.mathlinks.ro/Forum/viewtopic.php?t=33195
http://www.mathlinks.ro/Forum/viewtopic.php?t=33195
http://www.mathlinks.ro/Forum/viewtopic.php?t=21966
http://www.mathlinks.ro/Forum/viewtopic.php?t=63798
http://www.mathlinks.ro/Forum/viewtopic.php?t=74482
http://www.mathlinks.ro/Forum/viewtopic.php?p=618360

some of them could be merged...

[quote="Phelpedo"]a^b=b^a[/quote]

Because I cannot edit the previous one post, I send an improved bunch:

http://www.mathlinks.ro/Forum/viewtopic.php?t=63955
http://www.mathlinks.ro/Forum/viewtopic.php?t=33195
http://www.mathlinks.ro/Forum/viewtopic.php?t=33195
http://www.mathlinks.ro/Forum/viewtopic.php?t=21966
http://www.mathlinks.ro/Forum/viewtopic.php?t=63798
http://www.mathlinks.ro/Forum/viewtopic.php?t=74482
http://www.mathlinks.ro/Forum/viewtopic.php?p=618360
http://www.mathlinks.ro/Forum/viewtopic.php?t=148787

My old problem about $ a^{b}=b^{a}.$
Let $ a>1,$ $ b>1,$ $ a\neq b$ and $ a^{b}=b^{a}.$ Prove that $ a^{b}>e^{e}.$


prove that sequential compactness is true $ \iff$ "covering" compactness for metric spaces is true.

Compactness $ \Rightarrow$ Sequential Compactness: assume the contrary, that the space $ X$ is compact but not sequentially compact. then there exists a sequence $ (x_n)$ with no convergent subsequence. in particular, every point occurs only a finite number of times in $ (x_n)$, and in fact it makes no difference if we assume that each point occurring in $ (x_n)$ occurs exactly once. moreover, since $ (x_n)$ has no convergent subsequence, it follows that for each $ i\in \mathbb{N}$, there is an open ball $ U_i$ centered at $ x_i$ such that $ x_j\not\in U_i$ if $ j\neq i$. every point in $ (x_n)$ is thus an isolated point and so the set $ \{x_n\}_{n\in \mathbb{N}}$ is closed. its complement, $ U$, is therefore open. the collection $ \{U\}\cup \{U_i\}_{i\in \mathbb{N}}$ of sets forms an open cover of $ X$, but every finite subcollection of these sets excludes infinitely many $ x_n$ - a contradiction.


linear algebra
matrix
combinatorics proposed
combinatorics

Let $ A=\left(a_{i,j}\right)_{1\leq i\leq n,\ 1\leq j\leq m}$ be an $ n\times m$ matrix whose entries come from the set $ \left\{0,1\right\}$, such that if $ a_{i,j}=1$ for some integers $ i$ and $ j$ satisfying $ 1\leq i\leq n$ and $ 1\leq j\leq m$, then $ a_{i^{\prime},j^{\prime}}=1$ for any integers $ i^{\prime}$ and $ j^{\prime}$ satisfying $ 1\leq i^{\prime}\leq i$ and $ 1\leq j^{\prime}\leq j$.

Prove that there exist $ n+m$ rational numbers $ x_{1}$, $ x_{2}$, ..., $ x_{n}$, $ y_{1}$, $ y_{2}$, ..., $ y_{m}$ such that $ x_{1}\geq x_{2}\geq ...\geq x_{n}$ and $ y_{1}\geq y_{2}\geq ...\geq y_{m}$, and such that for any integers $ i$ and $ j$ satisfying $ 1\leq i\leq n$ and $ 1\leq j\leq m$, we have

$ a_{i,j}=\left\{\begin{array}{c}1,\text{ if }x_{i}+y_{j}\geq 0;\\ 0,\text{ if }x_{i}+y_{j}<0\end{array}\right.$.

  Darij

Indeed, trivial: let the corners of the diagram (that is, the pairs $ (a, b)$ such that $ a_{a, b}= 1$ but $ a_{a+1, b}=a_{a, b+1}= 0$) be $ (a_{0}, b_{0}), (a_{1}, b_{1}), \ldots, (a_{k}, b_{k})$ with $ a_{i}< a_{j}$ and $ b_{i}> b_{j}$ whenever $ i < j$.  Choose arbitrary different values (obeying the required ordering) for the $ x_{a_{i}}$, and let $ y_{b_{i}}=-x_{a_{i}}$.  Then if $ a_{i}< a < a_{i+1}$ set $ x_{a}= x_{a_{i+1}}$ and if $ b_{i}< b < b_{i+1}$ set $ y_{b}= y_{b_{i}}$.

I think that takes care of it (maybe up to some typos and bad notation), yes?

I think this is correct, except that $ b_{i}< b < b_{i+1}$ should be $ b_{i}> b > b_{i+1}$.

Nice solution, I had a similar idea but was not bold enough to lead it to its conclusion. :P

My solution works as follows: You can easily see that either the first column of the matrix consists of 1's only, or the last row of the matrix consists of 0's only. Induct over $ n+m$.

  Darij


A right triangle and a square has the same area. Prove that the perimeter of the triangle is greater than the perimeter of the square.

[hide="Hint"]
Apply AM-GM twice.  :wink: [/hide]

[hide="Im not quite seeing the Am-Gm twice idea but here is what I came up with"]
This inequality has no equality case; however, the sharper one $P_{\triangle}\ge \frac{P_{square}}{(2\sqrt{2}-2)}>P_{square}$ does.

Indeed, consider the set of right triangles with area $=\frac{P_{square}}{(2\sqrt{2}-2)}$. It is well known that out of these, the one with largest area is isosceles (easy to prove, just a little algebra/inequalities). That's it!
Let the side of the square $=a$, so $P_{square}=4a$ so all the triangles have area $a^{2}$. The unique isosceles triangle has sides $\sqrt{2}a,\sqrt{2}a,2a$ for $P_{isosceles}=(2+\sqrt{2})a=\frac{P_{square}}{(2\sqrt{2}-2)}$. Thus \[P_{\triangle}\ge P_{isosceles}=\frac{P_{square}}{(2\sqrt{2}-2)}>P_{square}\][/hide]

[hide]Let a and b be legs of right triangle, s be side of square (obviously all non-negative).
We know they have same area, $\frac{ab}{2}=s^{2}\implies s=\frac{\sqrt{2ab}}{2}$.

We want to prove that $a+b+\sqrt{a^{2}+b^{2}}\ge 4s=2\sqrt{2ab}$.

From AM-GM, $a+b\ge\sqrt{2}\sqrt{2ab}$,
From QM-GM, $\sqrt{a^{2}+b^{2}}\ge\sqrt{2ab}$.
$\implies a+b+\sqrt{a^{2}+b^{2}}\ge(1+\sqrt{2})\sqrt{2ab}>2\sqrt{2ab}$.?[/hide]

[hide="AM-GM Solution"]
Let the triangle have legs $a$ and $b$. The problem reduces to proving that
\[a+b+\sqrt{a^{2}+b^{2}}> 4 \sqrt{\frac{1}{2}ab}= \sqrt{8ab}\]
But by AM-GM we have that
\[a+b \geq 2 \sqrt{ab}\]
\[a^{2}+b^{2}\geq 2ab \implies \sqrt{a^{2}+b^{2}}\geq \sqrt{2ab}\]
Thus,
\[a+b+\sqrt{a^{2}+b^{2}}\geq (2+\sqrt{2})\sqrt{ab}> (\sqrt{2}+\sqrt{2})\sqrt{ab}= \sqrt{8ab}. \ \ \text{ QED.}\]
[/hide]

okay I suspected the AmGm included something like that which completely drops the equality case...


Consider the function $ f(x) \equal{} 2x^2 \minus{} 4x \plus{} 9$.  Evaluate $ 2f(3) \plus{} 3f(\minus{}3)$.

$ f(3) \equal{} 2 \times 3^2 \minus{} 4 \times 3 \plus{} 9 \equal{} 18 \minus{} 12 \plus{} 9 \equal{} 15$

$ f(\minus{}3) \equal{} 2 \times (\minus{}3)^2 \minus{} 4 \times (\minus{}3) \plus{} 9 \equal{} 18 \plus{} 12 \plus{} 9 \equal{} 39$

Thus, the answer is $ 2 (15) \plus{} 3 (39)$, or $ \boxed{147}$.

You could also notice that $ 2(f(x)\plus{}f(\minus{}x))\plus{}f(\minus{}x)\equal{}2(4x^2\plus{}18)\plus{}f(\minus{}x)$, which could possibly reduce the time taken by 0.0001 seconds. :wink:


analytic geometry
geometry
HMMT
Duke
college
graphing lines
slope

This shotlist will take sometime for all problems to be updated. I'll just post whenever I'll see the nice ones.

1. A convex quadrilateral is drawn in coordinate plane such that each of its vertices $(r,s)$ satsify the equation $ry = 24$ and $r^2+s^2 = 73$. Find the area of this quadrilateral.

2. Find the largest integer $k$ such that $2^k$ is one of factor of $3^{512} - 1$.

3. In equation $x^3-ax^2+bx-48$, let $l,m,n$ be the three positive roots with $l<m<n$. Find the minimum value of $\frac{1}{l} + \frac{2}{m} + \frac{3}{n}$.

HMMT Guts Round Problems

4. This is from 1986 AIME.

What is the sum of the solutions to the equation:

$\sqrt[4] x = \frac{12}{7 - \sqrt[4] x}$

5. This is from 1987 AIME.

Find $3x^2y^2$ if $x$ and $y$ are integers such that $y^2 + 3x^2y^2 = 30x^2 + 517$.

6. This is from Cayley 2000.

For the system of equations $a^2+a^2b^2+a^2b^4 = 525$ and $a+ab+ab^2 = 35$, the sum of the real $b$ values that satisfy the equation is what?

7. This is from Michian Math Prize Competition in 1999.

Let's denote $z$ as a complex number such that $\left(z + \frac{1}{z}\right)^2 = 3$. Then compute the value of $z^3 + \frac{1}{z^3}$.

8. This is from HMMT Team Round

A ladder is leaning against a house with its lower end 15 feet from the house. When the lower end is pulled 9 feet farther from the house, the upper end slides 13 feet down. How long is the ladder (in feet)?

9. This is from Harold Reiter's Page.

Tweleve poitns are arranged along the edges of 3 x 3 square as shown. How many triangles have all three of their vertices among these poitns?

[code]* * * *
*-----*
*-----*
* * * *[/code]

I made "----" to make them in right shape. Treat "---" as empty spaces. Only *'s are the points.

10. This is from Duke Math Meet Team Round

In how many ways can 1000 be written as sum of consecutive integers?

11. This is from Math League Sample Problem Set.

What is the least positive integer $n > 1$ for which the expression $\sqrt{1 + 2 + 3 + \cdots + n}$ simplifies to an integer?

12. This is from Part II Maryland State Math Competition 1997.

It took 2322 digits to number the pages of a dictionary (all pages are numbered). How many pages are there in the dictionary?

13. This is from [i]Mathematical Circles[/i]

What is the remainder when $3^{1989}$ is divided by 7?

14. Me

How many 6-digit numbers do not have all digits of equal parity?

15. This is from [i]Mathematical Circles[/i]

How many ways are there to place a set of chess pieces on the first row of a chessboard? The set consists of a king, a queen, two identical rooks, two identical knights, and two identical bishops.

16. This is from FMO.

Within the set $\mathbb {R}_0$, the equation $\frac{a}{\frac{b}{c}} = \frac{\frac{a}{b}}{c}$ holds iff for which condition:

A) a = b = c
B) a = b = c = 1
C) c = 1
D) c = 1 or c = -1
E) The equation holds true for all $a,b,c \in \mathbb {R}_0$

17. This is from FMO

Assuming the real numbers $x$ and $y$ satisfy $\frac{x}{y} = \frac{2y}{x+y}$, how many values can $\frac{y}{x}$ possibly attain?

18. This is from FMO.

Fibonacci numbers $a_n$ are defined by the relations

\[a_n = a_{n-1} + a_{n-2}\]
\[a_1 = a_2 = 1\].

The value of $a_{13}$ is:

A) Even
B) Square
C) Divisible by 13
D) Prime
E) Divisible by 3

19. This is from MMPC 1994.

The lines with equations $y = 2x+3, y = 6-x$ and $y=0$ meet at points $A,B$ and $C$. Find the area of $\triangle ABC$.

20. This is from MMPC 1993.

Easy geometry problem to end this shortlist.

Radii OA, OB, and OC divide the circle into three congruent parts. Let D be the point on the arc AB such that arc AD has half the measure of arc DB. Find the measure of $\angle OCD$.

1
[hide]It's easy to see that coordinates that satisfy the equations are $(3,8), (-3,-8), (8,3), (-8, -3)$

Their slopes are perpendicular and finding the length and width of the rectangle we see that the area is $5 \sqrt 2 \cdot 11 \sqrt 2 = 110$[/hide]

State your answer for #2 with more proper explanation.

I don't think your answer is right neither but I'll wait till I see the better explanation.

Yes!

Good job on Number 4. Mind on showing your work? (I personally like this problem which was #1 in real AIME)

[quote="tarquin"]2.
$(3^{512}+1$[/quote]

Close but note that:

$3^{512} + 1$ is not part of the factors in $3^{512} - 1$.

So what's your final answer? :)

Hmm is #3

[hide]$\frac {13}6$[/hide]

I assumed that the roots had to be integers =o

tarquin, you're right for $k=11$.

But you're wrong about for number 3.

I take it that for #6, $b$ must satisfy both equations?

$6.$

[hide]$525=3\cdot 5^2\cdot 7$, $35=5\cdot 7$

$a=5$

$b^2+b+1=7 \Rightarrow b^2+b-6=0 \Rightarrow b=2, b=-3$

$b^4+b^2+1=21 \Rightarrow b^4+b^2-20=0 \Rightarrow (b^2+5)(b^2-4)=0 \Rightarrow b=2, b=5i$

Therefore, the only solution that satisfies both equations is $b=2$.[/hide]

Incorrect.

Anyone also wants try #3? Hint: AM-GM Inequality on three variables.. That actually gives it away.

[quote="Silverfalcon"]7. This is from Michian Math Prize Competition in 1999.

Let's denote $z$ as a complex number such that $\left(z + \frac{1}{z}\right)^2 = 3$. Then compute the value of $z^3 + \frac{1}{z^3}$.[/quote]

[hide]
$z+\frac{1}{z}=\sqrt{3}$
$(z+\frac{1}{z})^3=3\sqrt{3}$
$z^3+3z^2(\frac{1}{z})+3z\frac{1}{z^2}+\frac{1}{z^3}=3\sqrt{3}$
$z^3+3z+\frac{3}{z}+\frac{1}{z^3}=3\sqrt{3}$
$z^3+3(z+\frac{1}{z})+\frac{1}{z^3}=3\sqrt{3}$
$z^3+3\sqrt{3}+\frac{1}{z^3}=3\sqrt{3}$
$z^3+\frac{1}{z^3}=0$?
[/hide]

All first 10 problems are finally up! :)

Solve them because they are pretty good problems.

10
[hide]If $m$ is the last term and $n$ is the first term then

$(m + n)(m - n + 1) = 2000$

If you let $m+n = p$ then

$m-n = \frac {2000 - p}{p}$

Subtracting the two equations we have

$2n = \frac {p^2 - 2000 + p}{p}$

$n = \frac 12 p + \frac 12 - \frac {1000}{p}$

Subbing the value for $n$ into $m - n = \frac {2000 - p}{p}$ yields $m = \frac 12 p - \frac 12 + \frac {1000}{p}$.

The values of $p$ that will make $m$ and $n$ integers are values that force $\frac {1000}{p}$ to have a fractional part of $.5$. The numbers that satisfy this condition are numbers that only have one more factor of $2$ than $1000$ and do not contain any extra $5$'s.

There are exactly $\fbox {4}$ values that work.[/hide]

You're wrong, tarquin.

Okay, the problems that need correct answers are:

3,6,8,9,10

Please keep trying.

I think I forgot to count the negative values...

No.

Here is a hint.

[hide="Look at hint"]

The reason I posted this problem is because there is easy way to solve this kind of problem.

Note that sum of consecutive integers is same as $\frac{pq}{2}$ where $p$ and $q$ are some numbers. So, we have:

$\frac{pq}{2} = 1000 \Rightarrow pq = 2000$.

Now, I hope you can finish from here.[/hide]

All 20 problems are finally up!! :D  :D  :D

16
[hide]$\frac {a}{\frac bc} = \frac {\frac ab}{c}$

$ac = \frac ac$

$c = \pm 1$

Soooo D.[/hide]

Number 10:

[hide]
I can prove this if you want to, but the number of ways to form n using consecutive numbers is the number of odd factors of 2n. In this case, n = 1000, so 2n = 2000 and the number of odd factors is 3
So is the answer 3?
[/hide]

Actually, the number of odd factors is 4... $2000 = 2^4 \cdot 5^3$

And that's exactly the same way I did it in my solution :(

I guess I was kind of slow on this, but here is an alternate solution to number 4:
[hide]Let $y=\sqrt[4]{x}$.  Then we have $y=\frac{12}{7-y}$, so $y^{2}-7y+12=0$, and $y=4,3$.  So $\sqrt[4]{x}=4$ or $\sqrt[4]{x}=3$.  Since the imaginary solutions are always conjugates and will sum to $0$, the answer is simply $4^{4}+3^{4}=256+81=337$.[/hide]

Solution for number 10..

[hide="View"]

From the last post I made on this problem, we had:

${pq} = 2000$ and since $2000 = 2^4 \cdot 5^3$, so the answer is $(4+1)(3+1) = 20$.[/hide]

I'll give some help on some of the problems because shortlists suppose to have answers.

[hide="Answer to number 9"]

First, we have to choose 3 points out of 12 points on the diagram. Note that it's $\binom{12}{3}$. But also realize that we can pick three points in same edge which would make a straight line. The number of ways to do that in ONE edge is $\binom{4}{3}$. We have four sides $\Rightarrow 4 \binom{4}{3}$.

$\therefore \binom{12}{3} - 4 \binom{4}{3} = \fbox {204}$. [/hide]

 :) Come on guys.. This is going to help you for your problem solving so keep trying.

Not sure on these answers, but...
13. [hide]
By Fermat's Little Theorem, 3^6 = 1 mod 7
Therefore, 3^1989 mod 7 = ((3^6)^331 * 3^3) mod 7 = (1^331 * 27) mod 7 = 27 mod 7 = 6 mod 7
So the answer is 6?
[/hide]

15. [hide]
This is equivalent to placing 8 objects, 6 of which are in pairs of 2. This makes the problem's solution:
8!/(2!2!2!0 = 5040?
[/hide]


Given integer n>1, prove that we can find integer k>1 and a1, a2, ......ak >1 such that  a1 + a2 +........ak = n(1/a1 + 1/a2 +....1/ak)

thanks alot.

Are you sure $ a_1, a_2, ... a_k > 1$?  I do not think there is a solution for $ n \equal{} 2$.  For fixed $ k$ the LHS is at least $ 2k$ while the RHS is at most $ k$.

If the condition is relaxed to $ a_1, a_2, ... a_k \ge 1$, then $ a_i$ can be the divisors of $ n$.

take $ a_i$ the divisors of $ n$, if a_i is a divisor then $ n/ai$ is also a divisor of n
 so $ \sum_{i}a_i\equal{}\sum_{i}n/a_i$


algebra
polynomial
Vieta
algebra unsolved

[b][color=blue]
If $\alpha$, $\beta$, and $\gamma$ are the roots of $x^3 - x - 1 = 0$,
compute $\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}$.
[/color]
[size=75](Canada National Olympiad 1996)[/size]
-------------------------------------------------------
[/b]
[b][u]Solution:[/u][/b]

If $\alpha ,\beta$ and $\gamma$ are the roots of $x^3 - x - 1 = 0$,
then $\frac{{1 + \alpha }}{{1 - \alpha }},\frac{{1 + \beta }}{{1 - \beta }}$ and $\frac{{1 + \gamma }}{{1 - \gamma }}$ are the roots of
$\left( {\frac{{u - 1}} {{u + 1}}} \right)^3 - \left( {\frac{{u - 1}} {{u + 1}}} \right) - 1 = 0 \Leftrightarrow$

$\Leftrightarrow u^3 + 7u^2 - u + 1 = 0$

Then we have:
$\frac{{1 + \alpha }} {{1 - \alpha }} + \frac{{1 + \beta }} {{1 - \beta }} + \frac{{1 + \gamma }} {{1 - \gamma }} = - 7$

[quote="metru"][If $\alpha ,\beta$ and $\gamma$ are the roots of $x^3 - x - 1 = 0$,
then $\frac{{1 + \alpha }}{{1 - \alpha }},\frac{{1 + \beta }}{{1 - \beta }}$ and $\frac{{1 + \gamma }}{{1 - \gamma }}$ are the roots of
$\left( {\frac{{u - 1}} {{u + 1}}} \right)^3 - \left( {\frac{{u - 1}} {{u + 1}}} \right) - 1 = 0$[/quote]

Why? :huh:

If $u = \frac{1+\alpha}{1-\alpha}$, then $u-u\alpha=1+\alpha\Longleftrightarrow \alpha\cdot (1+u)=u-1$, so $\alpha = \frac{u-1}{u+1}$. Since $\alpha^3-\alpha-1=0$, we must have \[ \left(\frac{u-1}{u+1}\right)^3-\left(\frac{u-1}{u+1}\right)-1=0. \]

here is my friend's solution .

Letting $S=\sum\frac{1+\alpha}{1-\alpha}$ , then $S+3=\sum\frac{2}{1-\alpha}=2\frac{\sum (1-\beta)(1-\gamma)}{(1-\alpha)(1-\beta)(1-\gamma)}$

Since $f(x)=x^3-x-1=(x-\alpha)(x-\beta)(x-\gamma)$ , $f(1)=(1-\alpha)(1-\beta)(1-\gamma)=-1$

So the denominator of $S+3$ is $-1$ while the numerator (after expanding and using vieta) we find that it is $4$ .So $S+3=-4$ $\implies S=-7$

 :)

[quote="shyong"]here is my friend's solution .

Letting $S=\sum\frac{1+\alpha}{1-\alpha}$ , then $S+3=\sum\frac{2}{1-\alpha}=2\frac{\sum (1-\beta)(1-\gamma)}{(1-\alpha)(1-\beta)(1-\gamma)}$

Since $f(x)=x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma)$ , $f(1)=(1-\alpha)(1-\beta)(1-\gamma)=-1$

So the denominator of $S+3$ is $-1$ while the numerator (after expanding and using vieta) we find that it is $4$ .So $S+3=-4$ $\implies S=-7$

 :)[/quote]

Can some body please explain how
$f(x)=x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma)$

Just because \alpha , \beta , \gamma are roots of f(x) doesn't mean above is true.

[quote]
Can some body please explain how
$f(x)=x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma)$

Just because \alpha , \beta , \gamma are roots of f(x) doesn't mean above is true.[/quote]

This is just euclidean division on polynomials , $P(\alpha) = 0 $ means $x-\alpha$ divides $P(x)$ i.e , there exists a polynomial $Q$ (of degree $2$ ) such that $P(x) = (x-\alpha) Q(x) $ for all $X$ (This is because the ring of polynomials is euclidean just like $\mathbb{Z}$ Do the same thing , since $\beta $ is another root of $P$ We have necessarily $Q(\beta)=0$ (since $\beta \ne \alpha $ ) and so on .

Note that $\sum\limits_{sym}^{} \frac{1+\alpha}{1-\alpha} = \sum\limits_{sym}^{}\frac{1-\alpha + 2\alpha}{1-\alpha} = \sum\limits_{sym}^{}1+\frac{2\alpha}{1-\alpha} = 3+ 2\sum\limits_{sym}^{}\frac{\alpha}{1-\alpha}$.

We can rewrite the equation $x^3-x-1=0$ as $x=x^3-1$. Hence, $3+2\sum\limits_{sym}^{}\frac{\alpha}{1-\alpha} = 3+2\sum\limits_{sym}^{}\frac{\alpha^3-1}{1-\alpha} = 3-2\sum\limits_{sym}^{} \frac{\alpha^3-1}{\alpha-1}$

Since $1$ is not a root of the equation, we can divide and get

$3-2\sum\limits_{sym}^{} \alpha^2+\alpha+1 = 3-2(s^2-2(\alpha\beta + \beta\gamma + \gamma\alpha) + s + 3) = 3-2(0^2-2(-1) + 0 +3) = -7$

@below thanks!

[quote=NewBeginning]

$3-2\sum\limits_{sym}^{} \alpha^2+\alpha+1 = 3-2(s^2-2(\alpha\beta + \beta\gamma + \gamma\alpha) + s + 1)$[/quote]

$\sum_{sym}(\alpha^2+\alpha+1)=s^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)+s+3$

Isn't this simply vieta? :D
$Solution:$
Firstly let $P(x) = x^3 - x- 1$ . Then $\alpha,\beta,\gamma$ are the roots of $P(x)$. For brevity, let $\alpha =a ,
\beta =b, \gamma=c$.(Lazinesss :P )
Then by vieta's theorem, we have that:
$\longrightarrow \sum_{cyc} a = 0$

$\longrightarrow \sum_{cyc} ab = -1$

$\longrightarrow  abc = 1$.

Expanding the given equation we have:

$\dfrac{\sum_{cyc}(1+a)(1-b)(1-c)}{(1-a)(1-b)(1-c)}$

$\iff  \dfrac{7}{-1}$

Hence the answer is  [size=150]$-7$[/size]

Let $P$ be the polynomial $x^3-x-1$. Then $$\sum_{\mathrm{cyc}} \frac{1+\alpha}{1-\alpha} = \frac{\sum_{\mathrm{cyc}} (1+\alpha)(1-\beta)(1-\gamma)}{(1-\alpha)(1-\beta)(1-\gamma)} = \frac{3+3\alpha\beta\gamma - \sum \alpha - \sum \alpha \beta}{P(1)} = \boxed{-7}.$$

Ans:  -7
1+a÷1-a=t
=> 1÷a=t+1÷t-1=>a=t-1÷t+1
Putting the value of 'a' in given Equation, we get The sum of roots as -7

[hide=solution]Let $\frac{1+\alpha}{1-\alpha}=r$, $\frac{1+\beta}{1-\beta}=s$, and $\frac{1+\gamma}{1-\gamma}=t$. Then solving gives $\alpha=\frac{r-1}{r+1}$, $\beta=\frac{s-1}{s+1}$, and $\gamma=\frac{t-1}{t+1}$. We know that $\alpha$, $\beta$, and $\gamma$ are roots of $x^3-x-1=0$, so $r$, $s$, and $t$ are roots of \[\left(\frac{x-1}{x+1}\right)^3-\left(\frac{x-1}{x+1}\right)-1=0.\] Simplifying, we get $-x^3-7x^2+x-1=0$, so $r+s+t=\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}=\boxed{-7}$.[/hide]

@cinnamon_e,
very nice job! I'm highly impressed. Oh...if you don't mind accepting my friend request....
Later

[quote=shobber]If $\alpha$, $\beta$, and $\gamma$ are the roots of $x^3 - x - 1 = 0$, compute $S=\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}$.[/quote]

Let $P(x)=x^3-x-1$, then $\frac{P’(x)}{P(x)}=\sum\frac1{x-\alpha}$, therefore
$$S=-3+ 2\frac{P’(1)}{P(1)} =-7$$

By Vietas Relation, we get $$\alpha+\beta+\gamma=0$$  $$\alpha\beta+\beta\gamma+\gamma\alpha=-1$$ $$\alpha\beta\gamma=1$$
We need to find the value of $\sum_{cyc}{}\frac{1+\alpha}{1-\alpha}.$ Now we have to find an equation whose roots are $\frac{1+\alpha}{1-\alpha},$ $\frac{1+\beta}{1-\beta},$ $\frac{1+\gamma}{1-\gamma}$
Now, $\frac{1+\alpha}{1-\alpha}=t\implies\alpha=\frac{t-1}{t+1}$ Putting the value of $\alpha$ in given $eq^n$ we get $$\left(\frac{t-1}{t+1}\right)^3-\left(\frac{t-1}{t+1}\right)-1=0.$$
After simplify we get our reqd. equation as $$t^3+7t^2-t+1=0.$$ therefore $$\sum_{cyc}{}\frac{1+\alpha}{1-\alpha}=-7$$ Is the correct answer....
VIETAS RELATION :love:

Taken from my notebook:
Clearly,
\[
\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}=3+2\left(\frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta}+\frac{\gamma}{1-\gamma}\right).
\]
We also know that
\begin{align*}
	\alpha & =\alpha^3-1, \\
	\beta & =\beta^3-1, \\
	\gamma & =\gamma^3-1.
\end{align*}
After substituting, our desired result becomes,
\begin{align*}
	3+2\left(\frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta}+\frac{\gamma}{1-\gamma}\right) = & 3+2(-\alpha^2-\alpha-1-\beta^2-\beta-1-\gamma^2-\gamma-1).
\end{align*}
By Vieta's, we have
\begin{align*}
	\alpha^2+\beta^2+\gamma^2 & = 2, \\
	\alpha+\beta+\gamma & =0.
\end{align*}
Hence, our desired result becomes
\[3+2(-\alpha^2-\alpha-1-\beta^2-\beta-1-\gamma^2-\gamma-1)=3+2(-5)=\boxed{-7}\]

[quote=Ianis][quote=Filipjack]I think solutions like [url=https://artofproblemsolving.com/community/c6h77670p445349]#2[/url] and [url=https://artofproblemsolving.com/community/c6h77670p27549629]#16[/url] are actually slightly incomplete. It should be mentioned that the roots are distinct in order for the proof to actually work. Otherwise there is no guarantee that the multiplicities of the roots are preserved.[/quote]

Anyway, that's easy. Right now I can think of two ways of proving it.

[b]Elementary:[/b]
If $f(x)=x^3-x-1$ had a multiple root then it would be a root of $f'(x)=3x^2-1$, but the roots of $3x^2-1$ are not roots of $x^3-x-1$. Hence $\alpha \neq \beta \neq \gamma \neq \alpha$.

[b]Advanced:[/b]
By the Rational Root Theorem we have that the only possible rational roots of $f(x)=x^3-x-1$ are $\pm 1$, but $f(1)=f(-1)=-1$, so $f$ does not have rational roots and therefore it is irreducible in $\mathbb{Q}[x]$. From $\operatorname{char}\mathbb{Q}=0$ we get that $f$ is separable over $\mathbb{Q}$. Hence $\alpha \neq \beta \neq \gamma \neq \alpha$.[/quote]

Well, yeah, it's easy, but I didn't make that remark because that step was hard, but rather to emphasize this common (and subtle) error: If $r_1, \ldots, r_n$ are the roots of some polynomial $p$ and I prove that [center]$r_i$ is a root of $p$ $\implies$ $\varphi(r_i)$ is a root of $q,$ for $i=\overline{1,n},$[/center]

this won't actually prove that the multiplicity is preserved. If $r_1$ is a double root, for instance, the above implications merely say twice that $\varphi(r_1)$ is a root of $q,$ but they don't actually say that $\varphi(r_i)$ is a double root of $q.$

Anyway, we don't really need the roots to be distinct. That was just a quick fix. Using the factorization of $p,$ we can easily see that $p\left(\tfrac{X-b}{a}\right)$ has the roots $ar_1+b,\ldots,ar_n+b$ (with multiplicities preserved) and $X^np \left(\tfrac{1}{X} \right)$ has the roots $\tfrac{1}{r_1},\ldots\tfrac{1}{r_n}$ (again, with multiplicities preserved; here we assume that the roots are non-zero), so we get the result if we use the right compositions.

By [i]Vieta's formula[/i], we get $\left\{ \begin{array}{l}
\alpha  + \beta  + \gamma  = 0\\
\alpha \beta  + \beta \gamma  + \gamma \alpha  =  - 1\\
\alpha \beta \gamma  = 1
\end{array} \right. ,$ then we have$$\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}=\frac{{3 - (\alpha  + \beta  + \gamma ) - (\alpha \beta  + \beta \gamma  + \gamma \alpha ) + 3\alpha \beta \gamma }}{{1 - (\alpha  + \beta  + \gamma ) + (\alpha \beta  + \beta \gamma  + \gamma \alpha ) - \alpha \beta \gamma }}=-7.$$

By Vieta's formula
\begin{align*}
\alpha\beta\gamma = 1 \\
\alpha\beta + \beta\gamma + \alpha\gamma = -1 \\
\alpha + \beta + \gamma = 0
\end{align*}

and expressing $\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}$ with a common denominator we get

$$ \frac{3+3\alpha\beta\gamma-\alpha\gamma-\beta\gamma-\alpha\beta-\alpha-\gamma-\beta}{1-\alpha-\beta-\gamma+\alpha\beta+\alpha\gamma+\gamma\beta-\alpha\beta\gamma} = \frac{3+3+1}{-1} = \boxed{-7} $$

We have

\begin{align*}
&\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma} \\
= \ &\frac{1+\alpha}{1-\alpha} + 1 + \frac{1+\beta}{1-\beta} + 1 + \frac{1+\gamma}{1-\gamma} + 1 - 3 \\
= \ & -2\left(\frac{1}{\alpha-1} + \frac{1}{\beta-1} + \frac{1}{\gamma-1} \right)-3
\end{align*}

The polynomial with roots $\alpha-1$, $\beta-1$, and $\gamma-1$ is $(x+1)^3-(x+1)-1=x^3 + 3 x^2 + 2 x - 1$. Then, the polynomial with roots $\frac{1}{\alpha-1}$, $\frac{1}{\beta-1}$, $\frac{1}{\gamma-1}$ is $-x^3+2x^2+3x+1$, so our answer is

\[-2\left(\frac{1}{\alpha-1} + \frac{1}{\beta-1} + \frac{1}{\gamma-1} \right)-3 = -2(2)-3 = \boxed{-7}.\]

By Vieta's formula, we know that  $u=a+b+c=0$,$v=ab+bc+ca=-1$ and $w=abc=1$.
And since the desired sum can be rewrited as
$\dfrac{3+3w-v-u}{1+v-u-w}$ we find that $S=-7$

If $\alpha, \beta$ & $\gamma$ are the zeroes of $$f(x) =ax^3+bx^2+cx+d,$$ then $\frac{k+\alpha}{k-\alpha}, \frac{k+\beta}{k-\beta}$ and $\frac{k+\gamma}{k-\gamma}$ are the zeroes of $$g(x) = (ak^3+bk^2+ck+d)x^3-(3ak^3+bk^2-ck-3d)x^2+(3ak^3-bk^2-ck+3d)x-(ak^3-bk^2+ck-d).$$
$$\boxed{\therefore \frac{k+\alpha}{k-\alpha}+\frac{k+\beta}{k-\beta}+\frac{k+\gamma}{k-\gamma}=\frac{3ak^3+bk^2-ck-3d}{ak^3+bk^2+ck+d}}$$

Let $r$ be a root of $f(x)=x^3-x-1$. Through addition, we have that
\[\frac{1+r}{1-r}=-1+\frac{2}{1-r},\]
however, since $r$ is a root of $f(x)$, we have that
\[r^3-r-1=0 \iff r^3-r=1 \iff -r^2-r=\frac{1}{1-r} \iff -2r^2-2r=\frac{2}{1-r},\]
meaning that
\[\frac{1+r}{1-r}=-1-2r-2r^2,\]
which gives that
\[\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma} = -3-2(\alpha+\beta+\gamma)-2(\alpha^2+\beta^2+\gamma^2).\]
By Vieta's formulas, we have that $\alpha+\beta+\gamma=0$ and $\alpha^2+\beta^2+\gamma^2=0^2-2(-1)=2$, giving us an answer of $-3-2(0)-2(2),$ or $-7$.

Let a,b,c be the roots, so we have that a+b+c=0, abc=1 and ab+bc+ac=-1
Let H be our expression so H=7/(1-a)(1-b)(1-c)
It is easy to check that  (1-a)(1-b)(1-c)=-1  and finally we get H=-1


Sometimes, I'll type in an answer after 5 seconds and it will say I answered after 20

Shouldn't this be in a different forum? FTW Problems Discussion is for discussion of the math problems, not internet problems. This should be in the questions and support forum, if any.

This should be in another forum, but I played a guy named like joshissunny or something and I would answer in like 10 seconds, and then like 30 seconds later, his answer pops up as if he had answered in 5 seconds. Actually, it seems like your theory is just the opposite.

I think that is just lag. He probably didn't see the question until 35 seconds after you did.


Hey guys, can some one please tell me how to prove this expression in this problem if the centre of mass is  exactly on middle of AB ? :huh:  Thanks!!

Hope this help :D

[img]http://i37.photobucket.com/albums/e71/phucnv87/p113chap15Halliday.png[/img]


inequalities geometry circumcircle Russia

Let O be the circumcenter of a triangle ABC. Points M and N are choosen on the sides AB and BC respectively so that the angle AOC is two times greater than angle MON. Prove that the perimeter of triangle MBN is not less than the lenght of side AC

http://www.mathlinks.ro/Forum/viewtopic.php?t=30486 .
(I removed your other post, please don't cross-post.)

  darij

An extremely nice problem: Basically, rotate triangles $BMO$ and $BNO$ to triangles $AM'O$ and $CN'O$. By the properties of rotation and the given condition, it is easy to see that $MB=M'A$ and $NB=N'C$ and $MO=M'O$, $NO=N'O$, and $\angle{MON}=\angle{M'ON'}$. The latter 3 conditions give $MN=M'N'$, so $MB+BN+NM=M'A+N'C+M'N'=AM'+M'N'+N'C'$. Since any broken line starting at A and ending at C  is longer than $AC$ itself, the problem is solved.

Very quirky but also pretty nice.

The idea is to rotate $AOM$ to $BOM'$ such that $A$ coincides with $B$, and define $N'$ similarly. Then $\triangle OM'N" \cong OMN$, and thus $$BC \leq BM'+M'N'+N'C = OM+MN+NO$$ by triangle inequality.

This is a really cool problem.
[hide = solution]
Let $X, Y$ denote the reflections of $B$ in lines $OM$ and $ON$ respectively. We remark that $X, Y$ both lie on $\odot(ABC)$ and that $XB, YB$ are perpendicular to lines $OM, ON$ respectively. It follows that $\angle XBY = 180^{\circ}-\angle B$, or that $XY = AC$. Here comes the finish:\[BM+MN+BN = XM+MN+NY \geq XY = AC\] and thus the perimeter of $\triangle BMN$ is not less than $AC$. $\square$
[/hide]


Find the least integer $k$, such that the quintic $P(x)=(x-a)(x-b)(x-c)(x-d)(x-e)$ has $k$ non-zero coefficients, if  $a,b,c,d,e$ are distinct integers.


function
combinatorics unsolved
combinatorics

(1)

Provided n positive integers. On a circle 2n distinct points $P_1, P_2, P_3, ..., P_{2N}$ are chosen.
Determine the number $c_n$ including all possibilities that n chords are drawn such that that the points $P_1, ..., P_{2N}$ are the endpoints and none of these chords intersect.

(2)

There are $3n$ points on the circumference. In how many ways one can select three points which form a triangle such there are no intersections among the triangles ?

(3)

In how many ways one can select $s \cdot n (s,n \in \mathbb{N}, s \geq 2)$ points on the circumference of a circle which form a $s-gon$ such there are no intersections among the $s-gons$ ?

For an elementary approach consider the similar problem which makes us of an recurrence relation.

Shortest route in a rectangular grid / way of shortest length in a rectangular lattice

This solution of the problem is based on this introducing problem.

For a non-elementary approach apply power series:

reference:

http://mathworld.wolfram.com/PowerSeries.html
http://mathworld.wolfram.com/LinearRecurrenceEquation.html
http://mathworld.wolfram.com/GeneratingFunction.html

For the first one only:

c_n satisfies famous realation:
c_n = c_0.c_(n-1) + c_1.c_(n-2) + ...+ c_(n-1).c_(0) (*), where c(0) = 1.
To solve this relation, let take generation function of sequence c_n:
f(x) = c_0+c_1.x + c_2.x^2 + ...
Then by (*) we have
              (f(x))^2 = c_0.c_0 + (c_0.c_1+c_1.c_0)x + (c_0.c_2+c_1.c_1+c_2.c_0)x^2 + ... = 1 + c_2x + c_3.x^2 = (f(x)-1)/x
=>  f(x) = (1 - sqrt(1-4x))/(2x).
Expand right hand by power series, we get  c_n = nC2n/(n+1).

Namdung

The number Namdung refers to is called Catalan number. There are many problems being connected with this numbers. Often they can be mapped into each other by means of bijections.

source:

http://mathworld.wolfram.com/CatalanNumber.html

[quote="Namdung"]For the first one only:

c_n satisfies famous realation:
c_n = c_0.c_(n-1) + c_1.c_(n-2) + ...+ c_(n-1).c_(0) (*), where c(0) = 1.
To solve this relation, let take generation function of sequence c_n:
f(x) = c_0+c_1.x + c_2.x^2 + ...
Then by (*) we have
              (f(x))^2 = c_0.c_0 + (c_0.c_1+c_1.c_0)x + (c_0.c_2+c_1.c_1+c_2.c_0)x^2 + ... = 1 + c_2x + c_3.x^2 = (f(x)-1)/x
=>  f(x) = (1 - sqrt(1-4x))/(2x).
Expand right hand by power series, we get  c_n = nC2n/(n+1).[/quote]
The same approach can be aplied in the general case too.
Namdung! Please, write it!

Thank you!


Let $ a$, $ b$, $ c$ be positive reals such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.
Prove that
$ \frac{1}{1\minus{}a^2}\plus{}\frac{1}{1\minus{}b^2}\plus{}\frac{1}{1\minus{}c^2}\plus{}\frac{1}{1\minus{}ab}\plus{}\frac{1}{1\minus{}bc}\plus{}\frac{1}{1\minus{}ca} \geq 9$.

Posted in http://www.mathlinks.ro/Forum/viewtopic.php?t=36873&start=20 .

  darij


algebra
polynomial
number theory
relatively prime
algebra unsolved

Is there a polynomial $ f$ of degree $ 1999$ with integer coefficients, such that $ f(n), f(f(n)), f(f(f(n))),\cdots$ are pairwise relatively prime for any integer $ n$?

We consider polynimial $ f(x) \equal{}x(x\minus{}1)g(x)\plus{}1$ where $ g(x)\in Z[x]$ deg$ g(x)$ =1997
$ f(n) \equal{}n(n\minus{}1)g(n)\plus{}1$ hence $ \gcd (f(n),n)\equal{}1$
$ f(f(n))\equal{}f(n)(f(n)\minus{}1)g(f(n))\plus{}1$ because $ n |(f(n)\minus{}1)$ hence $ \gcd((f(f(n)),n)\equal{}1$ and $ n|(f(f(n))\minus{}1)$
....


Here is the IMO shortlist problems 1989, I got it from internet.

[url=http://imo2009.de/]IMO 2009[/url] will be in Germany. Here is a new version of the above International Mathematial Olympiad (IMO) Longlist 1989 attachment from Braunschweig, Germany, which is a bit more comprehensive. For Windows there is a [url=http://www.stdutility.com/download/stduviewer.exe]STDU viewer.[/url]


Determine the highest power of $1980$ which divides:

$\frac{(1980n)!}{(n!)^{1980}}$

Hint: By combinatoric , it's easy to prove that $1980! \mid \frac{(1980n)!}{(n!)^{1980}}$


geometry
3D geometry
number theory solved
number theory

Let a,b,c be integer such that: a/b + b/c + c/a = 3.
Prove that a.b.c is the cube of an integer.

You are right Harazi :
http://www.mathlinks.ro/Forum/viewtopic.php?t=2592&highlight=cube

I didn't remember at all.... :D  :?

Pierre.


A few facts about my own typing habits:

I almost never use my right pinkie to type letters. Instead, I use it to hold the shift key whenever I need to. I use my ring finger to type apostrophes, and my middle finger to type semicolons. Strangely enough, I use my left ring finger to type the letter "q", but I use my pinkie to type the letter "a".

I almost never use the left shift key. If I need to capitalize a letter typed with my right hand, I stretch my finger out and hold both keys with the right hand. The left hand? Same thing, except that now I'm using both hands.

I use my middle finger to type periods, like the one at the end of this sentence.

What about you? What are your typing habits?

Hmm, I tend to type pretty by the book I think, using the correct fingers for all of the keys
As for shift I always use the one on the right, always using the right pinkie
I've actually never thought about this before, typing habits and the such, thanks for bringing it up.
Odd questions for people:
Are people loud typers?
How fast do people here type?
Does anyone use anything besides QWERTY?
(Sorry for asking more questions than answering, maybe I should've posted those as another topic?)

I don't even really know QWERTY, I found a typing arrangement that was convenient for me. I can still type pretty fast though.
As for the loudness of my typing, it depends what keyboard i'm using. If I use like a mac keyboard, I'm almost silent, whereas my home computer keyboard is vey very very loud.

My typing habits are similar to the OP's.

I shift and use 4th finger for q, `, 1, and tab, with the left hand, and ;, [, ], -, =, comma, period with the right hand. I'll actually type ' with my 5th finger, but most of the time I use that key I end up typing ", for which the 5th finger is anyway occupied with pressing the shift.

I never use left shift, control, or alt, except sometimes when I'm pressing ctrl alt del. I always press spacebar with my right thumb.

I use middle finger whenever I press enter several times in a row.

I'm usually not very loud, since I don't hammer on the keys, but I still do push them until they contact the bottom barrier, so it does make some sound.

I type 80-120 wpm, depending on if I know what I'm typing or if I'm thinking it up as I go along.

I have a pretty weird set up, my left hand stays still and generally hits all keys to the f column (and occasionally t) using three fingers. My other hand also uses 3 fingers but moves a lot around the keyboard to compensate for the rest of the space. (playing guitar makes this sooo much easier) I still get 150 wpm, so I'm happy.

I use my left pinky to hold shift or control, 4th finger to hit tab and alt, and my left thumb to hit space. I use my 4th finger on my right hand to hit backspace, pinky for all the other keys below it. Everything else is pretty normal.

I switched from using my right thumb to hit space to my left because I used to make too many mistakes (and I still do :( ) so extending my fingers to hit backspace would result in the joint of the thumb hitting the key, which did strange things to the skin.

i either type by the rules, but i have this habit of using my right hand to reach most of the keys in the middle, including the y b and n at some points. i also have this weird habit of never following the rules when using shift. idk why. i also sometimes type using a pencil. no joke

I kinda type by the rules, except my pinky finger never seems to do anything except press the really far-out keys like Shift and Enter. However. I often go by instinct and type however I feel, which means that how I type is dictated by the key I'm pressing more than anything else. My left thumb hits the space key and my right thumb somehow avoids all contact with keys. When I slow down my typing seems really awkward, but I still type well without many mistakes (probably around 80-100 wpm when I know what I'm going to say, but that's pretty rare).

i type every thing with my right nostrill

if thath is possible

i also like ot use my right pinky toe

I usually type by the rules, except when I'm feeling lazy. In the latter case, I take whatever is available, e.g. a pencil, a straw, a ruler, and stab the keyboard to get the letters.  :P  I never use the right shift, always the left, and I type around 100 WPM.

Around 120~130 WPM.

Fastest=146 WPM.

Use either shifts to type as fast I can.

Lol i go at about 80 WPM and people think I'm fast.  I only use left shift.

Strangely enough, I don't use my right pinkie for the right shift key all the time, either.

Usually, when I'm typing a Shift-enabled punctuation, I use the right ring finger for Shift and hold down the index finger for the punctuation mark itself.

I don't always use the right pinkie to type Enter, either. Sometimes, it will be with the right pinkie, when I'm typing a line break.

About my speed, I can probably type at about 100 to 120 WPM. 140 WPM, really? That's 700 characters per minute, or about 12 per second.

When typing LaTeX, my typing speed is probably about 40 or 50 WPM.

[quote="sdkudrgn88"]When typing LaTeX, my typing speed is probably about 40 or 50 WPM.[/quote]

that might just be because you have not muh experience using latex. i was pretty slow with latex too to begin with, but now that i've been using it for about a year or two now i type latex just about as fast as normal words

I, uh, can type fast.  I just type by the book.

Wow I feel inferior, I can only type 80-90 WPM :( However if I get a better keyboard I suspect I may be able to get 90 solid.

[quote="MewxRetasu"]o__O

I'm just the opposite of you... I never use the right shift key, I use the left shift key all the time. If I'm capitalizing a letter I usually type with my pinky (q, a, z), I move my whole left hand so that I can press the shift key with my pinky and just type the letter with my ring finger...

And when I enter numbers, I take both hands away from the keyboard, scrutinize it for a second, and then start picking out numbers, one by one, with my right middle finger.

:D[/quote]

I always use my left pinky to use the Control button. No exceptions. To press Ctrl+Alt+Delete, I always use my ring finger and index finger to hold down Ctrl and Alt respectively, and use my index finger to type "Delete".

For arrow keys, I always use my right index finger for left, ring finger for right, and middle finger for up and down. Same thing with WASD, except that it's my left hand instead, and I use index for D.

On the numpad, I always use index finger for left hand column (147), middle for the middle column (258), and ring finger for the right hand column (369).

As for the Escape key, I always use my left middle finger, for some reason.

[quote="sup3rcrash3r"]I type almost normally, but I use my left hand to reach the "y" key. Also, I use shift keys correctly, except when typing "I", where I use left shift half the time and right shift the other half.

I actually don't know my speed but I'm pretty sure it is around 90wpm. Except with numbers because I have to jam the function key down just to type with a messed up number pad (laptop).[/quote]

I also use my left hand to type Y. I sometimes use it to type a capital U, alternating between left and right. I almost always use the numpad for numbers.

The leftmost seven number keys, I type with my left hand, and I use the right shift to access the symbols.

My fastest (as of last year) was about 80 WPM...I basically type how you're supposed to...yeah...

[quote]I always use my left pinky to use the Control button. No exceptions. To press Ctrl+Alt+Delete, I always use my ring finger and index finger to hold down Ctrl and Alt respectively, and use my index finger to type "Delete".

For arrow keys, I always use my right index finger for left, ring finger for right, and middle finger for up and down. Same thing with WASD, except that it's my left hand instead, and I use index for D.

On the numpad, I always use index finger for left hand column (147), middle for the middle column (258), and ring finger for the right hand column (369).

As for the Escape key, I always use my left middle finger, for some reason.[/quote]

For Ctrl+Alt+Delete, I take my hands away from the keyboard, then press Ctrl with my left ring finger, Alt with my left pointer finger, and Delete with my right middle finger.

I do the same thing as you do for arrow keys and for the number pad.

For Escape... *short pause* What do I do? o__O *reaches for Escape key* Oh. Uh... for some reason, I use... my right pointer finger? What? xD

I think my WPM is ~80-90... yay?

I, uh, don't do touch typing. I'm trying to, but for now I type everything with my middle fingers :lol:

[quote="MewxRetasu"]For Ctrl+Alt+Delete, I take my hands away from the keyboard, then press Ctrl with my left ring finger, Alt with my left pointer finger, and Delete with my right middle finger.[/quote]

Wait, scratch that, that's what I do too.

When making short and useless posts (like this one), I type with my right hand alone and keep my left hand warm :D . It's actually not much faster than when I type with both hands, considering my typo rate with both hands. But when making long reports and stuff, I type with both hands.

Strange, I always type with both hands.

Actually, I think my casual typing speed, with error correction, is about 70 or 80 WPM. I don't type that fast.

[quote="nikeballa96"]This is what happens when I type a sentence and don't backspace. I am looking away from the computer, at my kitchen, and typing as fast as I can. I don't care how many typoes I made (although i think I hacve only made two so far), but I will press enter, right when I'm done with this post. I feel really guilty of spamming, and keep forgetting that this is the FUN FACTORUY 9and posts don't count).

MY GUESS OF HOW MANY TYPOES:
six.[/quote]

No, you only made three.

Posts in the Fun Factory don't count? I never knew that.

i only use my pinkies to press the shift keys, but i mainly use my middle finger to type the letters, but i use my ring finger and pointer finger also. i use my right pointer to press the space bar, and my right ring finger to backspace. i almost never use my thumb, though.

I always use the right shift key and never use the left.  Use right hand for Spacebar unless unless I am ending a sentence with a period or with a question mark.  I also move my hand out of place do hit question mark since I always use right shift key.  I type around 90 wpm.

Please do not revive dead threads.

Mods, please lock this.

Hey guys which finger do you use for the following keys: E, I, 3, 8?

Whichever finger is closest/most convenient. Usually the second finger, but I just used my middle finger to type 8.

yeah, about that....how about you show us a picture of you touching the "8" key?

Seriously, stop continuing revived threads.

Locked.


Let a(k+1),...an>0 fix; a1,...ak>0 .Find min:
      P= $\sum$ $\frac{ai}{aj}$ (i $\neq$ j)

you have post this two times by mistake.

I saw this same post in the Algebra unsolved.


integration
calculus
logarithms
calculus computations

ok thanks.

So how would I be able to split this into partial fractions?

$\int \frac{du}{u^{2}-1}\,dx$ where u = $e^{-}x$

$\int \frac{du}{u^{2}-1}\,dx$ where u = $e^{-x}$

[b]Solution[/b]

[hide]So our fraction is $\frac{1}{u^{2}-1}=\frac{1}{(u-1)(u+1)}$, thus we can spit it into two fractions which would look something like:

$\frac{1}{(u-1)(u+1)}=\frac{A}{u+1}+\frac{B}{u-1}$
$1=A(u-1)+B(u+1)$; This must hold true for all values of $u$.

Let $u=1 \Rightarrow 1=2B \iff B=\frac{1}{2}$
Let $u=-1 \Rightarrow 1=-2A \iff A=-\frac{1}{2}$

$\frac{1}{(u-1)(u+1)}=-\frac{1}{2(u+1)}+\frac{1}{2(u-1)}$

So our integral becomes:

$\frac{1}{2}\int \left( \frac{1}{u-1}-\frac{1}{u+1}\right) ~dx$
$=\frac{1}{2}\int \left( \frac{1}{e^{-x}-1}-\frac{1}{e^{-x}+1}\right) ~dx$
$=\frac{1}{2}\int \left( \frac{e^{x}}{1-e^{x}}-\frac{e^{x}}{e^{x}+1}\right) ~dx$
$=-\frac{1}{2}\int \frac{e^{x}}{e^{x}-1}~dx-\frac{1}{2}\int \frac{e^{x}}{e^{x}+1}~dx$

Let $w=e^{x}-1 \Rightarrow dw=e^{x}~dx$
Let $v=e^{x}+1 \Rightarrow dv=e^{x}~dx$

$=-\frac{1}{2}\int \frac{1}{w}~dw-\frac{1}{2}\int \frac{1}{v}~dv$
$=-\frac{1}{2}\left[ \ln |w|+\ln |v| \right]+C$
$=-\frac{1}{2}\left[ \ln |e^{x}-1|+\ln |e^{x}+1| \right]+C$
$=\boxed{-\frac{\ln |e^{2x}-1|}{2}+C}$[/hide]

[quote="skimnc"]
$=-\frac{1}{2}\int \frac{e^{x}}{e^{x}-1}~dx-\frac{1}{2}\int \frac{e^{x}}{e^{x}+1}~dx$

Let $w=e^{x}-1 \Rightarrow dw=e^{x}~dx$
Let $v=e^{x}+1 \Rightarrow dv=e^{x}~dx$[/quote]
You can use $\int\frac{f'(x)}{f(x)}~dx=\ln\left|f(x)\right|+k$


In $\triangle ABC$, $AB = 13, BC = 14$ and $CA = 15$. Also, $M$ is the midpoint of side $AB$ and $H$ is the foot of the altitude from $A$ to $BC$. The length of $HM$ is

$\text{(A)} \ 6 \qquad \text{(B)} \ 6.5 \qquad \text{(C)} \ 7 \qquad \text{(D)} \ 7.5 \qquad \text{(E)} \ 8$

[hide]
this triangle can be split into two right triangles, 9-12-15 and 5-12-13
so $AM = MB = \frac{13}{2}$
let the midpoint of HB be N
then MN is half of 12, which is 6
using pythagoras, HM is $\sqrt{6^2+\frac{5}{2}^2} = \frac{13}{2}$
B
[/hide]

[hide="Answer"]The 13-14-15 triangle can be split into a 5-12-13 and a 9-12-15 connected at the 12. Since $M$ is the midpoint of $AB$, and $\triangle ABH$ can be inscribed in a semicircle, then $AM$, $BM$, and $HM$ are all radii of the semi-circle, and $HM=6.5\Rightarrow \boxed{B}$.[/hide]

I don't understand why "splitting" is necessary. Here is a HINT to the easiest method (which I used and solution packet did as well; this only takes 3 seconds):

[hide="Hint"]
What can you tell about the median drawn to the hypotenuse in the right triangle? [/hide]

[hide="Answer"]
It's half of the hypotenuse. [/hide]

[quote="JesusFreak197"]So, you need to know that the 13-14-15 triangle is composed of two right triangles, which is all I meant when I said to "split" it.[/quote]

Well.. The problem told you that $AH$ forms a right triangle, right? So, we know that $\triangle ABH$ is a right triangle with $AB$ as a hypotenuse.

I know what you mean by "split." I just made an above post because that idea is pretty "slick" and well-known.  :)


Hello,

If you are in teh [b]Puzzles & Brainteaser[/b] forum, then probably you like a lot Problem Solving, even other than the directly related to Math. Then, this is your oportunity.  [url=http://www.otuzoyun.com/pqrst/]PQRST[/url] is a quarterly and online puzzle competition open to everybody. Competition will be in January, April, July and October.  [url=http://www.otuzoyun.com/pqrst/PQRST13.pdf]PQRST 13[/url] has started.Send your answers until April 30th, Saturday at 20:00 (GMT+02). For further information, visit the page itself, that is, again [url]http://www.otuzoyun.com/pqrst/[/url].

JUST IN CASE... DO NOT DISCUSS THE PROBLEMS FROM PQRST 13 HERE IN THE FORUM UNTIL MAY 2ND, just in case there are any problems with the international hour system, if it is may 2nd in your place, no matter where in the world you are, you can be sure that it is passed 20:00 of April 30th (GMT+02).

Good luck in your problem solving!

I changed the post from announcement to normal, because the contest already finished. By the way, did someone in here participate? I solved two of them, but I was too busy, and I forgot to submit them (any way, it would be just over 100 points, and regularly those are the last places :blush: ). It would be nice to see a MathLinker on the first positions of the Contest! :first:

That looked like fun but by the time I saw it I only had a couple hours left.  I'll have to remember to try it next quarter.


find all infinite arithmetic sequences consisted of positive integers ($ a_n$) such that for all  $ n$ we have :
$ a_n\plus{}a_{n\plus{}1}\plus{}...\plus{}a_{n\plus{}9} | a_na_{n\plus{}1}...a_{n\plus{}9}$
try also to find as many as possible integer $ c$ for which such a sequence doesn't exist with:
$ a_n\plus{}...\plus{}a_{n\plus{}c} | a_n...a_{n\plus{}c}$
and try to find an integer $ c$ for which such a sequence exists.and if there doesn't exist an integer $ c$ with such property,prove its non-existence!

arithmetic sequence means arithmetic progression?

i mean $ a_n\equal{}a\plus{}(n\minus{}1)d,a_1\equal{}a$. for all $ n \in Z^{\plus{}}$


calculus
integration
derivative
geometric series
real analysis
real analysis unsolved

Determine the $z \in \mathbb{C}$ with $|z|=1$ such that $\sum_{i=1}^\infty\frac{z^{n}}{n}$ converges.

All $z\ne 1$ --- use summation by parts.

[url=http://en.wikipedia.org/wiki/Summation_by_parts]Wikipedia link[/url]. This is like integration by parts: when you have a product of two sequences $z^{n}$ and $\frac{1}{n}$, you replace one with an "antiderivative" (sequence of partial sums), and the other with "derivative" (sequence of differences). This wins the game, because the partial sums of $z^{n}$ are easy to control (geometric series), and the differences $\frac{1}{n+1}-\frac{1}{n}$ are of order $\frac{1}{n^{2}}$.

This problem has been done before  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=121138]here[/url].


Hello and welcome to Legend of the Black Dragon.  This game is based off a RPG called Legend of the Red Dragon.
           RULES:
    You start out as lvl 1, the place that you will usually be in is the forest, where your battles will take place.  From you battles you will get experience and gold.  There will be many events in the forest that you don't know about  :P .  There is also a town, inside the town is a bank, a weapon shop, a armor shop, a healing place, a Inn, a players arena, the daily news, and training place.  If you die, you must wait 3 days before reincarnation.  In the training place, you have a master.  If you defeat your master, you will level up.  You can also question you're master and he will tell you the # of experience points you still need in order to beat him or he can tell you if you're ready.  The daily news show the daily events and in the players arena, you can fight other players.  You will lose health so you can heal for some money.  Once you get enough cash, you can purchase items at the armor of weapon shop.  You can deposite or withdraw gold from you're bank.  The Inn is the place you'll be if you don't want other players to attack you.  Also inside the Inn, is a pretty girl called Daisy.  You can flirt with her and if you are successful, you get a little experience.  If you are not, you lose experience  :o .  You can buy potions in the inn to increase your attack, defense, or hit point levels.

So for example, you would type:
"Go to forest"  you now have the options to look for something to kill, go to bank, go to healing place, or return to town.
"Look for something to kill"  I will now select a creature that you will battle with.  You will attack first because you're just that cool and you have the options to attack, run, or look at your profile.

BTW, you start out with a stick as your weapon and a coat as your sheild   :rotfl:
Oh yeah, and once you get to your max lvl 7, you will fight the BLACK DRAGON!

This is basically it.  I will allow 7 players to play.  If you have any questions, feel free to ask!  NOTICE THAT THIS WILL GET PRETTY BORING AFTER YOU HAVE DISCOVER AND EXPLORED EVERYTHING.  IF YOU WANT TO QUIT, JUST TELL ME.

OK so I'm co-moding.  We are NOT multies.

ok im in but thisll take forever since this isn't a bot like print

1.footballer
2.andrewjjiang97
3.
4.
5.
6.
7.

Get in quick!

alright

1.footballer
2.andrewjjiang97
3.percy314
4.
5.
6.
7.

I've got many evil plans in mind
muahhahahahha

If no one comes, we will have 6 people.

[hide] IF YOU SAY I LIKE WAFFLES IN UR NEXT POST, ILL GIVE YOU 500 EXTRA COINS![/hide]

OH YAY WAFFLES ARE AWESOME NOW GIVE ME MONEY XD

1.footballer
2.andrewjjiang97
3.percy314
4.Chompy
5.
6.
7.

I is here since I has nothing to do
(Actually I have quite a bit of stuff to do, but I just dont feel like it...)

Somebody spent time on ICC... and didn't use it constructively

BTW, I'm not joining.

There are 2 more slots open.  The deadline for joining is January 2nd.  However, i will allow people to join or replace during game play.  Hurry, hurry, hurry!

[quote="Butler"]Hmmmmm..........I am just saying, remember the Golden Rule........... Anyway, if you really want to quit this post in the replacements thread....[/quote]
ur so cool butler... u know the golden rule.
I'm gonna copy butler for a few moments.
butler gets shapeshifting skillz
andrew gets teleporting skillz.
blue gets invisibility skillz.
meewhee gets mind reading skillz.

butlers thinking how piegraph is stupid.
andrew is thinking of icecream
blue is thinking of the beauty of meewhee.
:P

:P

uses invisibility skillz to kill the black dragon.

I help bluecarneal by teleporting and hitting the dragon and teleporting away when it tries to hit me.

Well Pie doesn't like this game so I'll finish it up for him.

Pie doesn't like Butler either.

OK so you guys meet the legendary Black Dragon!

BD's health: 1
Your guy's health: 100 I think?

Andrew hits the BD for 1 health.  He wins!

Oh, turns out it was a bomb.  you guys are all dead.  but so is the black dragon.  So it is a tie.

wth abcak.

ok the legendary Black Dragon has 1000 health.

You guys health: 20 i think.

Wait for everyone to post.

[quote="piegraph"]bleh...
lets just continue this so my i can brag to everyone how many pages my game got lol.[/quote]

*sighs*

im comparing this to holy cow and setting a goal for clans

AUGH!!!!!!!!!!                            I use my dragon summoner on the Black Dragon, then I drop infinite bombs on it.

ok, im inviting a mod to lock this thank you everyone for your time playing and stuff, I'm glad to be the mod of this game :)


Describe how you would construct a triangle given the lengths of two of its sides and the length of the median to the third side.

Choose a point on a piece of paper, get a compass and draw an arc below whose radius is the length of side 1, next do the same for side 2, and finally, for the median's length.

Then, draw a straight line intersecting all 3 arcs at the bottom, which should contain the 3rd side.

However, draw this line so that it intersects the middle arc at its midpoint.

A bit easier: In the desired triangle extend the median $AD$ over $D$ to $E$ such that $DE=DA$. Then $ABCE$ is a parallelogram, hence you can construct auxilliary triangle $AEC$ whose sides are $2m_{a},b,c$. From there it should be easy.


I mean like say you find an error in one of the books...

Either in here, or the preferred way is to send a quick email to books@artofproblemsolving.com with the typo/error you found.

Umz... In Counting and Probability on page 57 it says icosahedon not icosahedron.

[quote="louiethechewy123"]Shouldn't you post it in support?[/quote]No, that's for support with the website issues.


In a triangle $ ABC$, prove that

$ \cos A \plus{} \cos B \plus{} \cos C \leq \frac{3}{2}$.

[quote="AndrewTom"]In a triangle $ ABC$, prove that

$ \cos A \plus{} \cos B \plus{} \cos C \leq \frac {3}{2}$.[/quote]

Since in triangle $ ABC$ this equality holds:
\[ \cos \alpha \plus{} \cos \beta \plus{} \cos \gamma\equal{}1\plus{}4\sin \frac{\alpha }{2} \cdot \sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}\]
the inequality is equivalent with
\[ 4\sin \frac{\alpha }{2} \cdot \sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2} \leq \frac{3}{2}\minus{}1\equal{}\frac{1}{2}\]
therefore
\[ \sin \frac{\alpha }{2} \cdot \sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2} \leq \frac{1}{8}\]
which is true, hence we're done.

$ \lvert\ \overrightarrow{e_1}\rvert\equal{} \lvert\ \overrightarrow {e_2}\rvert\equal{}\lvert\ \overrightarrow{e_3}\rvert\ \equal{}1$
$ \overrightarrow{e_1}$ collinear $ \overrightarrow{AB}$ ,$ \overrightarrow{e_2}$ colinear $ \overrightarrow{CA}$
It is known $ (\overrightarrow{e_1}\plus{}\overrightarrow{e_2}\plus{}\overrightarrow{e_3})^2 \ge 0$
$ \overrightarrow{e_{1} ^{2}}\plus{}\overrightarrow{e_{1} ^{2}}\plus{}\overrightarrow{e_{1} ^{2}}\plus{}\overrightarrow{2e_{1}e_{2}}\plus{} \overrightarrow{2e_{2}e_{3}}\plus{}\overrightarrow{2e_{1}e_{3}}  \plus{}2\cos(\pi \minus{}B)\plus{}2\cos(\pi \minus{}C)\plus{}2\cos(\pi \minus{}A) \ge  0$
$ \cos A\plus{}\cos B \plus{}\cos C \leq \frac{3}{2}$ :lol:

hello, we easily can prove the following formula
$ \cos(A)\plus{}\cos(B)\plus{}\cos(C)\equal{}1\plus{}\frac{r}{R}$
and since $ 2r\le R$ we get the inequality above.
Sonnhard.


For each of the middle $ 83$ cubes, they have a lateral surface area of $ 4$.

For the $ 2$ cubes at the end, they have a surface area of $ 5$, because only one face is covered up.

Multiplying and adding the total, we get $ \boxed{342}$.