number theory
prime numbers
superior algebra
superior algebra unsolved

Suppose that p_1, p_2,.....p_r are distinct primes that divide n. Show that the formula for φ(n) below is correct.

φ(n) = n(1 - (1/p_1))(1 - (1/p_2)*...*(1-(1/p_r)).

This is what I did:

If p is a prime number and k is greater than or equal to one, then we know that it can be written as such:
φ(p^k) = p^k - p^(k-1)

Factoring this formula gives you:
φ(p^k) = p^k(1-p^-1)

Since p^-1 is the same thing as 1/p, you can rewrite the formula as φ(p^k) = p^k(1-(1/p))

So more generally,

φ(n) = n(1 - (1/p_1))(1 - (1/p_2)*...*(1-(1/p_r))

for the "so more generally"-conclusion you should probably mention the multiplicative property of $ \varphi$ on co-prime numbers


Let be $ a,b,c>0$ such that $ ab\plus{}bc\plus{}ca\equal{}1$ . Show that :

$ \frac{a}{a\plus{}\sqrt{b^2\plus{}1}}\plus{}\frac{b}{b\plus{}\sqrt{c^2\plus{}1}}\plus{}\frac{c}{c\plus{}\sqrt{a^2\plus{}1}}\le \frac{(a\plus{}b\plus{}c)^2}{3}$

[quote="caubetoanhoc94"][quote="Jianxing113725"]$ \frac {a}{a \plus{} \sqrt {b^2 \plus{} 1}} \plus{} \frac {b}{b \plus{} \sqrt {c^2 \plus{} 1}} \plus{} \frac {c}{c \plus{} \sqrt {a^2 \plus{} 1}}\leq1$[/quote]

Why ?????????? :([/quote]

Because  $ a^2\plus{}1\equal{}a^2\plus{}bc\plus{}ab\plus{}ca\ge 2a\sqrt {bc}\plus{}ab\plus{}ca\equal{}a(\sqrt b\plus{}\sqrt c)^2$
hence $ \sum\frac {a}{a \plus{} \sqrt {b^2 \plus{} 1}} \le \sum\frac{a}{a\plus{}\sqrt a(\sqrt b\plus{}\sqrt c)}\equal{}1$

[quote="great math"]I think you made a mistake,ddlam because it seems easy but the fact that the inequality you pointed out is in wrong place.  :([/quote]

You 're right, I'm sorry  :oops:


A carpenter is hanging molding that is $ \frac{1}{8}^{\prime \prime}$ thick around the ceiling of a rectangular room. If the ceiling measures 18ft 7in by 14ft 4in and molding is sold in 8-foot lengths, what is the minimum number of lengths the carpenter will need?

first we find the perimeter of the ceiling in inches.
$ 2{[(18)(12) \plus{} 7] \plus{} 2[(14)(12) \plus{} 4]} \equal{} 2(223 \plus{} 172) \equal{} 2(395) \equal{} 790$. since he hangs a molding that is $ \frac{1}{8}$, (you sort of have to visualize this) he can hang $ 2(223)$ but then you have to take off $ 4(\frac{1}{8})$ from the other dimensions. so we take off $ \frac{1}{2}$ from $ 790$ to get $ \frac{1579}{2}$. this is in inches, so convert it to feet.
$ (\frac{1579}{2}inches)(\frac{1 foot}{12 inches}) \equal{} \frac{1579}{24} feet$. since the molding is sold in $ 8$ foot lengths, we divide $ \frac{1579}{24}$ by $ 8$ to get $ \frac{1579}{192}$
after doing some estimation, we find that this approximately $ 8.2$, but we can only buy whole lengths, so we need  $ 9$ lengths


I would use jPicedit. But for the best results this topic should go to the [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX forum[/url]

[quote="mastermathman"]I am looking for help.  Does anyone know how to convert GSP files to Latex?[/quote]You can't directly. Either save to a format LaTeX can read, such as eps, pdf, png, gif, jpg or take a screen shot of the picture, copy into Paint and save in one of these formats. Then include the picture into your tex document - see [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Pictures]LaTeX:Pictures[/url].


What is the difference between using words like thus, so, hence ect.

This would be better placed in the Linguistics forum.  :wink:

I don't think there is a difference. They're all synonyms.

Mostly, you would use "hence" and "thus" when you've used "so" multiple times in your solution (I am assuming you're asking this for when writing a solution). Then after you've used "hence" and "thus" a few times, you can switch back to "so" and so on.[/i]


ratio
geometry
geometric transformation
reflection
trigonometry
parallelogram
similar triangles

I did this with a random triangle with nice dimensions and I got 2/3.  I don't know if that is even right, but I have no idea how to do this for any given triangle.

Does anyone have a good solution for this.  This is a really weird problem.

There are a lot of similar triangles in this configuration. I got an idea of reflect $ C$ in the midpoint $ M$ of segment $ AP$ to $ C'$ and tried to prove that triangles $ C'PC$ and $ RAQ$ are similar. BTW, I still failed to solve this problem  :blush:   :|   :huh:

Ugly problem :wacko:

$ \frac {AB}{PB} \equal{} \frac {BQ}{BC} \equal{} \frac {\sin 60^\circ}{\sin 45^\circ} \equal{} \sqrt {\frac {3}{2}}$ and
$ \angle PBC \equal{} \angle PBA \plus{} \angle ABC \equal{} \angle ABC \plus{} \angle CBQ \equal{} \angle ABQ$ $ \Longrightarrow$
$ \triangle ABQ \sim \triangle PBC$ by s.a.s. with similarity coefficient $ \sqrt {\frac {3}{2}}.$
[quote="April"]... I got an idea of reflect $ C$ in the midpoint $ M$ of segment $ AP$ ...[/quote]
Correct. If $ D$ is reflection of $ C$ in $ M,$ then $ ADPC$ is parallelogram, because its diagonals cut in half at $ M$ $ \Longrightarrow$ $ AD \equal{} CP$ and $ \frac {AD}{AQ} \equal{} \frac {CP}{AQ} \equal{} \sqrt {\frac {3}{2}}.$
$ \angle QAD \equal{} \angle QAB \plus{} \angle BAP \plus{} \angle PAD \equal{} \angle CPB \plus{} \angle BAP \plus{} \angle APC \equal{}$ $ \angle BAP \plus{} \angle APB \equal{} 105^\circ.$

$ \angle CAD \equal{} \angle CAQ \plus{} \angle QAD \equal{} \angle CAQ \plus{} \angle RAC \equal{} \angle RAQ.$
$ \frac {QR}{DC} \equal{} \frac {QR}{2MC} \equal{} \frac {\sqrt {6}}{2} \equal{} \sqrt {\frac {3}{2}} \equal{} \frac {AQ}{AD}$
$ \Longrightarrow$ $ \triangle AQR \sim \triangle ADC$ by S.s.a. $ \Longrightarrow$ $ \frac {AR}{AC} \equal{} \sqrt {\frac {3}{2}}.$

This is pretty ridiculous to do synthetically without an actual diagram or looking at specific trivial cases. Once the answer is known, however, it is quite clear what's happening.

[quote=MellowMelon]Let $ ABP, BCQ, CAR$ be three non-overlapping triangles erected outside of acute triangle $ ABC$. Let $ M$ be the midpoint of segment $ AP$. Given that $ \angle PAB \equal{} \angle CQB \equal{} 45^\circ$, $ \angle ABP \equal{} \angle QBC \equal{} 75^\circ$, $ \angle RAC \equal{} 105^\circ$, and $ RQ^2 \equal{} 6CM^2$, compute $ AC^2/AR^2$.

[i]Zuming Feng.[/i][/quote]

[b]Answer.[/b] $\tfrac{2}{3}$.

Construct parallelogram $CADP$.

[b]Claim.[/b] $\triangle AQR \sim \triangle ADC$.

(Proof) Observe that $\triangle BPA \sim \triangle BCQ$ hence $\triangle BAQ \sim \triangle BPC$. Consequently, $$\tfrac{AQ}{AD}=\tfrac{AQ}{CP}=\tfrac{BP}{BA}=\sqrt{\tfrac{3}{2}}=\tfrac{QR}{DC}.$$ Since $\angle RAC=105^{\circ}$ and $\angle QAD=\angle CPA+\angle QAP=180^{\circ}-\angle (CP, AQ)=180^{\circ}-\angle ABP=105^{\circ}$. Now we can use SSA similarity (since $105^{\circ}>90^{\circ}$) so $\triangle AQR \sim \triangle ADC$ as desired. $\blacksquare$

Clearly, $\tfrac{AC^2}{AR^2}=\tfrac{2}{3}$ as claimed.

[quote=anantmudgal09]This is pretty ridiculous to do synthetically without an actual diagram or looking at specific trivial cases. Once the answer is known, however, it is quite clear what's happening.[/quote]

Well, the angles are nice enough that you [i]can[/i] (with some effort) construct basically everything aside from $R$ fairly accurately (for $R$ you would need to multiply something by $\sqrt6$, which seems like a bit more guesswork). While I agree this problem is fairly ridiculous, I think for me the parallelogram construction was somewhat motivated:

1. I think that given the conditions on $ABP$ and $CBQ$, it's fairly clear that spiral similarities are going to be a big player here, and thus that's what we should be focusing on.
2. I think given a sufficiently nice diagram it's not too difficult to note that the point $N$, which is where the spiral similarity at $B$ taking $P$ to $A$ sends $M$ off to, is going to be rather important - $MANB$ is cyclic, $\angle{NAP}$ is also equal to $105^\circ$, and with a decent enough diagram it looks like $N$ is on $RQ$.
3. Another calculation now reveals that $RQ = 2NQ$. Given this, and the previous collinearity description, it seems natural to guess that $R$ is the reflection of $Q$ across $N$, which turns out to be true!

Note that all the above was not done through any particular special case, although I may have gotten very lucky.

Still quite difficult for a TST 1 though, and [i]very[/i] amusing.

Let $C'$ denote $C$ reflected over $M$ so $CPC'A$ is a parallelogram. The claim is that $CPBC'\sim QBAR$. To see $\triangle CPB\sim\triangle QBA$, apply the Law of Sines. Moreover, the ratio $QR/CC'=AB/BP$ by Law of Sines and the given, and
\[\angle BPC'=60^\circ+\angle APC'=60^\circ+\angle PAC=105^\circ+\angle BAC=\angle BAR,\]
so the claim follows. Then $AR=\frac{QR}{CC'}\cdot C'P=\frac{\sqrt{3}}{\sqrt{2}}AC$, which finishes: $AC^2/AR^2=2/3$.

felt very hard for a tst 4

The answer is $2/3$, or  $$\frac{AC}{AR}=\frac{2}{\sqrt{6}}.$$ Let $C'$ be the reflection of $C$ across $M$. It suffices to show that $$\triangle ACC'\sim\triangle ARQ.$$ Note that there is a spiral similarity at $B$ sening $PA$ to $CQ$, so $$\angle BPC=\angle BAQ.$$ Thus, $$\angle C'AQ=45+\angle C'AP+\angle BAQ=45+\angle APC+\angle BPC=45+60=105.$$

It then suffices to show that $$\frac{AQ}{RQ}=\frac{AC'}{CC'}$$ $$\frac{RQ}{CC'}=\frac{AQ}{AC'}$$ $$\frac{AQ}{CP}=\frac{\sqrt{6}}{2}$$

However, this just follows from the spiral similarity at $B$ sending $PC$ to $AQ$, since the scale factor of the spiral similarity is $$\frac{BA}{BP}=\frac{\sin 60}{\sin 45}=\frac{\sqrt{6}}{2}.$$

remark: the motivation I had for reflecting C was quite complicated. First, I felt that it was very surprising that the quantity is invariant in $\triangle ABC$. Of course, one thing you can do in this scenario is to try special cases. The really nice special case is when $\angle A=180-105-45=30$ and $\angle B=180-75-75=30$, since it makes a lot of things collinear, and even has an added bonus of $\triangle ABC$ itself being nice. This case is actually not hard to solve explicitly, and once you notice that $$\frac{AR}{AC}=\frac{\sqrt{6}}{2}$$ you realize that the problem has already given a $\sqrt{6}$ ratio, so in a sense $CM$ "needs to be twice as long" for a nice ratio to exist. The reflection is then easily motivated.


The point $ D$ lies in the triangle $ ABC$ such that the circles inscribed in triangles $ ABD, BCD$, and $ CAD$ pairwise touch each other. On lines $ BC, CA, AB, AD, BD, CD$, denote the touching points by $ A_1, B_1, C_1, A_2, B_2, C_2$, respectively. Let lines $ B_1C_2$ and $ B_2C_1$ meet at $ E$, and let lines $ A_1C_2$ and $ A_2C_1$ meet at $ F$. Show that the lines $ AF, BE$, and $ C_1$D are concurrent.

[hide="Solution"]
We let $ (I)$ be the incircle of $ \triangle ABC$, $ X$ be the intersection between $ BD$ and $ C_1A_2$, and $ Y$ be that between $ C_1A_2$ and $ AD$.  Allow $ XY$ to meet $ AB$ at $ Z$.  Now, we have that $ AC_1 = AA_2 = AB_1$ from the tangencies.  Similarly, $ CB_1 = CA_1$ and $ BA_1 = BC_1$.  This means that $ A_1$, $ B_1$, and $ C_1$ are the points of tangency between $ (I)$ and $ BC$, $ AC$, and $ AB$ respectively.  Then, we see that $ \frac {AA_2}{A_2D}\cdot \frac {DB_2}{BB_2} = \frac {AB_1}{B_1C}\cdot \frac {CA_1}{BA_1}$, and so by Menelaus's Theorem, $ B_1A_1$ and $ A_2B_2$ intersect $ AB$ at the same ratio.  Therefore, they concur at $ Z$.  Now, by Menalaus's Theorem, we have that $ \frac {DX}{XB}\cdot \frac {BC_1}{AC_1}\cdot \frac {AA_2}{A_2D} = 1$ and $ \frac {DY}{AY}\cdot \frac {AC_1}{C_1B}\cdot \frac {BB_2}{DB_2} = 1$.  Divide these, so $ \frac {DX}{XB}\cdot \frac {AY}{DY}\cdot \frac {BC_1}{AC_1} = 1$, so since $ \frac {ZA}{ZB} = \frac {AC_1}{BC_1}$, we have that $ XY$ passes through $ Z$ by Menalaus's Theorem.  This means that the  intersections  between $ C_1F$ and $ AD$, $ C_1E$ and $ DB$, and $ EF$ and $ AB$ are collinear, meaning that $ \triangle C_1EF$ and $ \triangle DAB$ are perspective.  We conclude that $ C_1D$, $ AE$, and $ BF$ are concurrent. [/hide]


How to make a neat index in TeXnicCenter for an already done document?

MakeIndex is the usual way of creating but you need to add instructions over what to index. Have a look at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=makeindex]Generating an index in (La)TeX[/url]  for this and other ideas. If it's a long document that you have to edit, you might prefer to stick to the easier table of contents :)


Stie cineva ceva mai mult decat e in titlu?
[hide="question"]Subiecte ceva? :P [/hide]

iar vii ma la ONM ? n-ai si tu bac, chestii ? admitere ?

da, stiu eu :D e la C.N. Zinca Golescu, atata stiu in plus. ce ai mai vrea sa stii? :P

in 9 e concursu sau doar in 10? si in 14 se pleaca? un program ar fi bine venit...

cred ca concursul va fi pe 10, iar barajele pe 12 si 13. pe 9 cred ca e sosirea loturilor.

nasol, daca imediat a doua zi e concursul, unii vin de departe.

btw, daca faceti rost de program..

cat timp avem la nationala? 3 sau 4 ore?

La actualul programul ziua z devine ziua z+1....adica sosirea marti....proba miercuri...etc.In sfarsit au aflat si ei ca luni este prima zi dupa Paste :D...oricum tot aia e...eu vruiam o zi 'libera' intre ziua sosirii si ziua probei:(( ( asa a fost in fiecare an )

Cred ca 4 ore,....dar eu nu particip so i might be wrong  :blush:

[quote="mychrom"]foarte ciudat programul. stie cineva raspunsul la intrebarea pusa de mine mai sus? :P[/quote]Pentru nationala 3 ore si jumatate. Pentru baraje, 4 ore probabil.

Prima nationala in 10 ani la care nu particip :(

[quote]Prima nationala in 10 ani la care nu particip :([/quote]

adica vrei sa spui ca nu vei fi prezent la corectat? nu vei fi nici la contestatii? nu vor mai iesi saraci copii plangand? :P

Foarte bine daca ii face pe copii sa iasa plangand, lectie de viata... decat sa primeasca ca prin magie 5-6 puncte...

[quote="pureblack2003"]sigur?
ce naspa ! e printre putinele orase pe care nu am apucat sa le vad  :( de ce a trebuit sa fie Pitesti anu asta?  :mad: auzi tu , 8 insi in camera. sa zica saracu dzeta cand s-o intoarce ca tot p-acolo a stat si el ...[/quote]

tu zici ca a fost naspa, altii ziceau ca a fost bine, cred ca depinde de liceul unde ati picat cazati;)

Si..........ce-ati facut?????????? :D   Cum vi s-au parut subiectele? :?:

[quote="omicroncz"]rezultatele de la baraje se vor afisa pe site-ul om ag? daca le are cineva va rog sa le postati :)[/quote]

primii 6 la seniori:
Adrian Zahariuc- 45 de puncte
Tariuc -bucuresti
Ilie Livia -bucuresti
mai este cineva din bucuresti si cineva din sibiu
Deneanu Andrei-bucuresti 28 puncte

din pacate, aici s-a intamplat ceva foarte trist, dupa cum spunea si domnul Radu Gologan:

Madalina Persu(chiar colega de clasa cu Andrei Deneanu) a obtinut  27.5 puncte si a fost a 7-a....

Am mai retinut doar cateva nume din restul pana la 25:
Vlad Emanuel - cls a XII-a
Prodescu Corneliu
Matican bogdan(ultimu care a prins lotul largit)
si stiu ca au mai prins si cativa de a noua lotul largit

Tot la seniori vorbind, Iurie Boreico a obtinut 46 de puncte si mai stiu de Zamorzaiev care a fost al treilea din Rep. Moldova si a obtinut sub 40 de puncte, dar nu stiu exact cate.

[quote="dragos001"]Mihai tu ce ai facut la baraj k am vazut ca te-ai calificat.[/quote]

tu esti cumva calinescu?:D pei am facut realmente prost, pentru ca pur si simplu nu am lucrat probleme de baraj. tu ce ai facut la baraj?

Da sunt calinescu, la baraj am fost al treilea sub linie....

Hmmm, cine merge la JBMO? sau nu e hotarat inca  :maybe:

[quote="Zamfirmihai"]Madalina Persu(chiar colega de clasa cu Andrei Deneanu) a obtinut  27.5 puncte si a fost a 7-a....[/quote]Am fost si eu in situatia asta, la 0.5 puncte de Balcanida. Dar e ok :) Se poate reveni :D

nu vreau eu sa fiu ala care strica tot. dar cum o fi senzatia cand afli ca nu te duci pt doar 0.5 pt , in conditiile in care ti-ai sacrificat toata viata sa lucrezi la mate? brrr. spix ce-a facut la baraj ? sau perfect_radio ? Gherman Marius si Popa Tiberiu ?

zamfirmihaie, eu am fost la colegiul tehnic dimitrie dima . scris cu litere mici. adica eu vin ca olimpic si ei ma baga acolo, si mai aud si chestii de genu : eh, acolo ai "nimerit" . adica asa ne tratatam unii pe altii ? la "nimereala"? mai ziceam eu ceva acuma ... in fine, am zis eu ceva la femeile alea p-acolo care cica se ocupau de loc  :D am venit marti si miercuri am plecat aks . eram prea frumos ca sa mai stau 5 minute in plus in vagauna aia

[quote="pureblack2003"]nu vreau eu sa fiu ala care strica tot. dar cum o fi senzatia cand afli ca nu te duci pt doar 0.5 pt , in conditiile in care ti-ai sacrificat toata viata sa lucrezi la mate? brrr. spix ce-a facut la baraj ? sau perfect_radio ? Gherman Marius si Popa Tiberiu ?[/quote]Madalina mai are multi ani inainte. Eu eram clasa a XII-a. Plus ca s-a facut o greseala la corectura atunci si de fapt trebuia sa am mai mult dar ... asta e. Te pui pe picioare si te califici la Internationala :D

dumitrescu mihail sau cum il cheama pe baiatu de-a luat anu trecut aur la balcaniada a fost si el in primii 6? deci zahariuc, turea, ilie livia , deneanu si mai cine ?
a fost interesanta premierea? sau tot asa nu asculta nimeni si ii enumera pe toti, fara sa lase pauza intre nume

La aur si argint ii zicea mai rar, dar la bronz ii enumera :)

Eu unul nu m-am calificat pentru lotul largit, oricum nu era in obiectivul meu asa ceva. Felicitari la toata lumea.
La OBM supriza despre care se vorbea am impresia ca e Papara.

bravo spix . am vazut ca mai aveai putin si lesinai. good effort ! m-ai recunoscut, nu ? sau nu zambeai aiurea ca gayu  :D

poate anexeaza cineva clasamentele de la baraj ca sunt curios


Hello all,
I have difficulties understanding the subject of graph automorphism, and I was hoping one of you could help me. For example, I don't understand why the graph defined by $ V\equal{}\{u,v\}$ and $ E \equal{} \{uu,uv\}$ isn't asymmetric. Since $ deg(u) \equal{} 3$ and $ deg(v) \equal{} 1$ and isomorphism conserve degree, the only automorphism should be the identity. Where do I go wrong? :(

Thanks in advance,
Perrin.

Sounds like you're right about all of this. Maybe the graph in question was supposed to be directed? (but I don't think that would fix anything...)

Wikipedia, in particular, claims all asymmetric graphs have at least 6 vertices.

That's my problem; I don't understand why there aren't any asymmetric graphs with less than 6 vertices - unless they mean simple graphs. But they don't say simple anywhere  :(

Regardless, thank you for your reply - it's encouraging to see that my question's at least being read.
Perrin.

Many sources assume that all graphs are simple, using the term multigraph when they want to allow for non-simple graphs.

[quote="JBL"]The word "graph" unmodified most commonly means "loopless with no multiple edges."  In particular, it's under this interpretation that the Wikipedia claim would be true.

Edit: too slow![/quote]

I see! Now it's all clear. And I was so frustrated I couldn't understand what I'm missing. Thanks everyone! :)


limit
linear algebra
matrix
abstract algebra

[b]1.[/b]Sa se calculeze $ \lim_{n \to \infty}{(45+\sqrt{2007})^n}$ unde $ a\in\mathbb{R}$ si $ {a}$ este partea fractionara a lui "a".
[i]Tuduce Florian, Zalau[/i]

[b]2.[/b] Sa se rezolve ecuatia: $ X^n={\begin{pmatrix}2 & 3 \\ 4 & 6\end{pmatrix}}$; $ n\in\mathbb{N}^*$;$ X\in{M_2(\mathbb{R})}$
[i]Tuduce Florian, Zalau[/i]

[b]3.[/b] In multimea $ M_2(\mathbb{Q})$ se considera submultimea $ G={\begin{pmatrix}a&b\\b&a\end{pmatrix}\mid{a^2-b^2=1},a,b\in\mathbb{Q}}$.
a)Sa se verifice ca $ I_2=\begin{pmatrix}1&0\\0&1\end{pmatrix}\in{G}$;
b) sa se arate ca daca $ A,B\in{G}$ atunci $ A\cdot{B}\in{G}$;
c) Sa se arate ca, daca $ X\in{G}$,$ X=\begin{pmatrix}a&b\\b&a\end{pmatrix}$ atunci $ X$ este matrice inversabila si ca
$ X^{-1}=\begin{pmatrix}a&-b\\-b&a\end{pmatrix}\in{G}$;
d) Sa se gaseasca o matrice $ A\in{G},A=\begin{pmatrix}a&b\\b&a\end{pmatrix}$ cu $ b\ne{0}$;
e) sa se arate ca multimea $ G$ este infinita.

[quote="promath"][b]1.[/b]Sa se calculeze $ \lim_{n \to \infty}{(45 + \sqrt {2007})^n}$ unde $ a\in\mathbb{R}$ si $ {a}$ este partea fractionara a lui "a".
[i]Tuduce Florian, Zalau[/i]
[/quote]

La punctul 1) unde apare $ a$ in problema?

$ SOLU{\c{T}}IA$ $ punctelor$ $ {a}$)-$ {c}$) $ fiind$ $ f$. $ simpl{\u{a}}$ $ o$ $ las$ $ {^i}n$ $ seama$ $ elevilor!$.

$ SOLU{\c{T}}IA$ $ punctelor$ $ {a}$)-$ {c}$) $ fiind$ $ f.$ $ simpl{\u{a}}$ $ o$ $ las$ $ ^in$ $ seama$ $ elevilor!$. $ S\u{a}$ $ vin\u{a}$ $ ei$ $ cu$ $ {solu{\c{t}}ii!}$
$ A{\c{s}}$ $ face$ $ {^i}ns{\u{a}}$ $ doar$ $ 2$ $ observa{\c t}ii$, $ care$ $ pot$ $ utile$ $ la$ $ aceste$ $ puncte$:
1). $ \begin{ecuation*} \begin{vmatrix} a & b \\
b & a \\
\end{vmatrix} \end{ecuation*}$ $ = a^2 - b^2 = 1$;     $ 2)$.

$ SOLU{\c{T}}IA$ $ punctelor$ $ {a}$)-$ {c}$) $ fiind$ $ f$. $ simpl{\u{a}}$ $ o$ $ las$ $ pe$ $ seama$ $ elevilor!$. $ S\u{a}$ $ vin\u{a}$ $ ei$ $ cu$ $ {solu{\c{t}}ii!}$
$ A{\c{s}}$ $ face$ $ ins{\u{a}}$ $ doar$ $ 2$ $ observa{\c t}ii$, $ care$ $ pot$ $ fi$ $ utile$ $ la$ $ aceste$ $ puncte$:
1). $ \begin{ecuation*} \begin{vmatrix} a & b \\
b & a \\
\end{vmatrix} \end{ecuation*}$ $ = a^2 - b^2 = 1$;          2). $ {det}{(AB)} = [{det}(A)]$. $ [{det}(A)]$
$ Iat{\u{a}}$ $ acum$ $ solu{\c{t}}a$ $ ultimrlor$ $ 2$ $ puncte$:
$ a^2 - b^2 = 1$ $ \Longleftrightarrow$ $ (a - b)(a + b) = 1$ $ \Longleftrightarrow$ $ a + b = \frac {m}{n}$ $ si$ $ a - b = \frac {n}{m}$ $ \Longleftrightarrow$
$ \Longleftrightarrow$ $ a = \frac {m^2 + n^2}{2mn}$ $ si$ $ b = \frac {m^2 - n^2}{2mn}$     $ unde$:  $ m$, $ n\in$ $ \mathbb{Z^*}$.

$ A{\c{s}}$ $ mai$ $ face$ $ o$ $ ultim\u{a}$ $ {observa\c{t}ie}$: $ (G; .)$ $ este$ $ un$ $ grup$ $ abelian!$


Find all pairs of real numbers $(x, y)$ satisfying the following system of
equations
$2-x^{3}=y, 2-y^{3}=x$.

The function $f(t)=2-t^{3}$ is decreasing on whole $\mathbb{R}$, hence $f(x)=y\land f(y)=x\implies f(x)=f^{-1}(x)$ can have solution iff $f(x)=x$, giving $x^{3}+x-2=0\iff (x-1)(x^{2}+x+2)=0$. The only real solution is $x=1\implies y=1$

If $x=y$ then we must have $x=y=1$, so assume $x\ne y$. We have $(x-y)(x^2+xy+y^2) = (x-y)$ and thus $x^2+xy+y^2=1$. Likewise, we have $4=x^3+y^3+x+y = (x+y)(x^2-xy+y^2+1) = 2(x+y)(x^2+y^2)$. Setting $a=x+y$, we then have $x^2+y^2=2/a$. Moreover,
\[
2=2x^2+2xy+2y^2 = (x+y)^2 +x^2+y^2 = a^2 + \frac2a \implies a^3-2a+2=0.
\]
It is not hard to see that if $a^3-2a+2=0$ ($a\in\mathbb{R}$) then $a<0$. This, however, is a contradiction: $x^2+y^2=2/a<0$. Hence, no solutions unless $x=y$.


algebra
polynomial
superior algebra
superior algebra unsolved

It has been pointed out to me that in any case, if $F$ is the fixfield that

$F^{\times}= F^{\times}^2\cup(-F^{\times}^2)$  where the union is disjoint
Again, this is assuming if we use that

Any ideas?

[quote="fredbel6"]Let $\alpha$ be an automorphism of $\mathbb{C}$ that is involutive

show that $i$ is mapped onto $-i$ ( well it suffices to prove that it is not preserved)

is is true that $\sqrt{2}$ is always preserved ?[/quote]

hint ... show that Fix($\alpha$) is really closed so cannot contain $i$.

[quote="fredbel6"]What precisely do you mean with really closed?[/quote]

:blush: I mean of course "real closed" (see the wikipedia entry for it)
http://en.wikipedia.org/wiki/Real_closed_field

[quote="fredbel6"]Let $\alpha$ be an automorphism of $\mathbb{C}$ that is involutive

show that $i$ is mapped onto $-i$ ( well it suffices to prove that it is not preserved)

is is true that $\sqrt{2}$ is always preserved ?[/quote]

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=sqrt&t=63823]this other topic[/url] for the second question. As for the first one, we could do this:

Let $K$ be the fixed field of $\alpha$. First notice that $[\mathbb C: K]=2$, because every $u\in\mathbb C$ is the root of a polynomial $f(x)=(x-u)(x-\alpha(u))\in K[x]$ of degree $2$. Now assume that $i\in K$, take $u\in\mathbb C\setminus K$ such that $u^{2}\in K$ (it's easy to see that such a $u$ exists), and $v\in\mathbb C$ such that $v^{2}=u$. The minimal polynomial of $v$ over $K$ must have as roots precisely two of the numbers $v,iv,-v,-iv$, but the product of any two of these is outside $K$, and we have a contradiction.


function
arithmetic sequence
algebra solved
algebra

Let $f$ be a bijective function from $N = \{0,1,2,3, \cdots \}$ to itself.
Prove that we can find three integers $a<b<c$ such that $f(a)+f(c) = 2f(b)$.

Pierre.

Nice easy problem...

Assume that we cannot find three integers a, b, c with a < b < c and f(a) + f(c) = 2 f(b).

Since the function f is bijective, it has an inverse function $f^{-1}$, i. e. a function $f^{-1}$ such that $f^{-1}\left( f\left( x\right) \right) =x$ for every $x\in \mathbb{N}$. Just like the function f, this inverse function $f^{-1}$ must also be bijective.

Let $b_0=f\left( 0\right)$, and let $b_1$ be a natural number with $b_1 > b_0$. We construct an infinite arithmetic progression $b_0 < b_1 < b_2 < b_3 < ...$ of natural numbers. Now let $a_i=f^{-1}\left( b_i\right)$ be the images of the elements of this arithmetic progression under the function $f^{-1}$. Then, $a_0=f^{-1}\left( b_0\right) =f^{-1}\left( f\left( 0\right) \right) =0$.

Now, if we had $a_i<a_{2i}$ for some i > 1, then we would have $0<a_i<a_{2i}$ on the one hand (in fact, $a_i>0$, since $a_i<0$ is impossible, and $a_i=0$ would lead to $a_i = a_0$ and thus $b_i = b_0$, what is impossible, too). On the other hand, we would have $b_0+b_{2i}=2b_i$ (since $b_i$ form an arithmetic progression), or, in other words, $f\left( 0\right) +f\left( a_{2i}\right) =2f\left( a_{i}\right)$. This contradicts our assumption that there are no three integers a, b, c with a < b < c and f(a) + f(c) = 2 f(b).

Hence, we must have $a_i \geq a_{2i}$ for every i > 1. We cannot have $a_i = a_{2i}$, since this would yield $b_i = b_{2i}$ and i = 2i. Thus, we must have $a_i > a_{2i}$ for every i > 1. This entails

  $a_{1}>a_{2}>a_{4}>a_{8}>a_{16}>...$.

But since all $a_i$ are numbers $\in \mathbb{N}$, this is not possible by infinite descent. Hence our assumption must be wrong, and the existence of three integers a, b, c with a < b < c and f(a) + f(c) = 2 f(b) is proven.

  Darij

More general , if is a surjective function f : N--> N ,  Then for all a$\in$N , the set $S_a$ = {(b,c) ; (b,c)$\in N^2$ ,  a < b < c , f(a) + f(c) = 2*f(b) } is infinite

I define the sequences b(n),c(n) :
b(0)= min{t ;  t$\in$N and 0 $\le$  t and f(a) < f(t)}  , c(0)= min{x ; x$\in$N and f(x) = 2*f(b0) - f(a)}
b(1) = min{t ;  t$\in$N and b(0) < t and f(b(0)) < f(t)} ,  c(1) = min{x ; x$\in$N and f(x) = 2*f(b(1)) - f(a)}
...........
b(n+1) = min{t ;  t$\in$N and b(n) < t and f(b(n)) < f(t)} , c(n+1) = min{x ; x$\in$N and f(x) = 2*f(b(n+1)) - f(a)}
..........
The sequences  b(n),c(n) are well defined because of f is surjective .
It's easy to see that the sequences b(n) and f(b(n)) are strictly increasing : b(0) < b(1) < b(2) < ... and f(a) < f(b(0)) < f(b(1)) < f(b(2)) < ...
We have also : f(c(n)) = 2*f(b(n)) - f(a) > f(b(n)) $\ge$ f(b(k)) for all k =< n
So because of the definition of b(0),b(1),... , we obtain b(n) < c(n) for all n .
Manifestly there are  an infinite set of m such that a < b(m) < c(m) , b(m) < b(m+1)  and f(a) + f(c(m)) = f(b(m))  (q.e.d)

 :cool: The Concours gnral is the oldest math contest  :cool:

Yes, it goes back to 1747. :)

Some informations on the competition (not only in mathematics) are available here (in french) :
http://www.education.gouv.fr/sec/palmar1.htm

Pierre.


Can someone make a brief explanation of what a debugger is and how it works? Which of the debuggers for Pascal are available on IOI and which would you recommend me to use?
Thank you.

Debugger allows you to execute your program step by step, checking the values of variables and following the execution path of your program at all times. I'm almost sure you've used a debugger before. I don't think you have a debugger guaranteed at IOI though - I used to rely on debugger quite a lot but then I switched from Borland Pascal to FreePascal which had a really #!$@ (= bad) debugger. My friend advised me to print data to screen if I don't have a working debugger and I've used this sort of debugging ever since. It takes some to get used to it but then it works fine.


function
search
induction
integration
calculus
real analysis
real analysis unsolved

Prove that if the function $ f : \mathbb{R}^2 \rightarrow [0,1]$ is continuous and its average on every circle of radius $ 1$ equals the function value at the center of the circle, then $ f$ is constant.

[i]V. Totik[/i]

An idea using fourier transform of distributions is shown in the link above. The other approach, discussed there, to show that $f$ is harmonic, maybe wiil be more difficult than the problem itself. Moreover, I have a doubt about it being true; I mean, given only "the radius $1$ mean value property" is rather not enough to say the function is harmonic!?
A discrete version of this problem was given at Bulgarian TST for the Balkan MO (2003). "In every knot of $\mathbb{Z}^{2}$ lattice is written a real number between $0$ and $1$ with the property every number is the average of its four neighbours. Show that all numbers are equal".
You can see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=5327]here[/url] for many comments and various ideas including using the Krein-Milman theorem, random walks, etc.
I'd like to show here how the idea of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2832968#p2832968]the official solution of the discrete version[/url] (which btw is pretty elementary) can be adapted in this continuous case.

Suppose on the contrary $f$ is not constant. Then there exist $x_1,x_2\in \mathbb{R}^{2}\,,\|x_2-x_1\|=1\,,\, f(x_2) > f(x_1)$. Let us denote $\Delta f(x) := f(x+h) - f(x)$, where $h=x_2-x_1$. We have:
(i) $\Delta f(x)$ also satisfies "radius $1$ mean value property".
(ii) $\Delta f(x)$ is also bounded (between -1 and 1).
Since $\Delta f(x_1) > 0$, and because of (ii), we conclude $\infty > M := \sup_{x\in \mathbb{R}} \Delta f(x) > 0$.

Let us fix $\varepsilon >0$, and choose $x_0$ s.t. $M-\varepsilon < \Delta f(x_0) \leq M$. Imagine for a moment that $\Delta f(x)$ is smooth enough, say Lipschitz. Thus, having $\Delta f(x_0)$ is close to $M$ and "radius $1$ mean value property" will imply $\Delta f(x_0+h)$ is close to $M$ too, i.e. $M-K \varepsilon < \Delta f(x_0+h) \leq M$, where $K$ is a constant, not depending on $\varepsilon$ or $x_0$.
Thus, by induction, we obtain:
\[ \Delta f(x_0+j h) > M-K^{j}\varepsilon \,,\, j=0,1,\ldots \]
Therefore:
\[ f(x_0+j h) - f(x_0)= \sum_{i=0}^{j-1}\Delta f(x_0+i h) > j M - \varepsilon \sum_{i=1}^{j} K^{i} \]
Now, look at the above inequality. The LHS is bounded, while we can make the RHS as big as we want, taking $\varepsilon$ small enough. Thus, we get a contradiction.

Well, going back, the only obstacle to carry out this idea is that we can't be sure $\Delta f$ is a Lipschitz function. To get around it, we could average $f$. So, the proof goes like that:
Suppose $f(x_2)> f(x_1)$, where $\|x_2-x_1\|=1$. Let $\overline{f}(x)= \frac{1}{|B_r|}\int_{B_r} f(x+u)\,d u$, where $B_r$ is the ball with radius $r$, centered at the origin. We can take $r$ small enough, so that $\overline{f}(x_2) > \overline{f}(x_1)$. Of course $\overline{f}$ is bounded. By changing order of integration, one can check that $\overline{f}$ also satisfies the "radius $1$ mean value property". And the most important thing is that $\overline{f}$ is a Lipschitz function, which follows from the boundness of $f$. It means $\Delta \overline{f} (x)$ is Lipschitz  too, and the rest goes as above.


function
algebra
domain
limit
polynomial
calculus
calculus computations

prove that:
$ f: R \equal{} \equal{} > R$
$ x \equal{} > x^{3} \minus{} 3x$
is  surjectif.

The function $ x^3 \minus{}3x$ is continuous on its whole domain and $ \lim_{x\rightarrow \minus{} \infty}(x^3 \minus{}3x) \equal{} \minus{} \infty$ and as $ \lim_{x\rightarrow \infty}(x^3 \minus{}3x) \equal{} \infty$. Also, $ (x^3\minus{}3x)'\equal{}3x^2\minus{}3>0$ for $ |x|>1$, then the function $ x^3\minus{}3x$ is strictly increasing on $ (\minus{}\infty,\minus{}1)$ and $ (1,\infty)$ it follows that the function $ x^3\minus{}3x$ is surjective on its whole domain.

all polynomial with odd degree is surject.
Suppose $ a_n > 0$
Because
$ \lim_{x\to \minus{} \infty} \equal{} \minus{} \infty$
$ \lim_{x\to\ \plus{} \infty} \equal{} \plus{} \infty$


inequalities
function
logarithms
inequalities proposed

Sorry to spoil the fun (you probably want a nicer solution ;)), but anyway:

Denote $f(x,y)=\left(\frac 1 2+\frac 1 x\right)^y$. This function is convex for $x,y\geq 0$, so by Jensen's inequality we have:

$f(x,y)+f(y,z)+f(z,x)\geq 3f\left(\frac{x+y+z} 3,\frac{y+z+x}3\right)=3f(2,2)=3$

Correct? :)

[quote="billzhao"]How do you know that the multivariable function $f(x,y)$ is convex?[/quote]
Sorry, my mistake. I calculated $\displaystyle \frac{\partial^2 f}{\partial x^2}=\frac{2^{2-y}y(x+y+1)\left(\frac{x+2}x\right)^y}{x^2(x+2)^2}>0$ and $\displaystyle \frac{\partial^2 f}{\partial y^2}=2\left(\frac{x+2}{2x}\right)^y\left(\ln\frac{x+2}x-\ln\frac{x+2}x\cdot\ln 2+\ln 2\right)\geq 0$ and $\displaystyle \frac{\partial^2 f}{\partial x\partial y}=\frac{\partial^2 f}{\partial y\partial x}=-\frac{2^{1-y}\left(\frac{x+2}x\right)^y\left(y\ln\frac{x+2}{2x}+1\right)}{x(x+2)}\leq 0$ and for a moment I thought that the Hessian is always positive, but that doesn't have to be true.

[quote="siuhochung"]what is positive definite? I've just read about the Hessian test but I don't undersatnd what it is..[/quote]
Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=195370#p195370]here[/url].

I don't think it is true, in general, that Bojan's function is convex. For example, consider the points $(\frac{2}{3}, 1)$ and $(2, 49)$, whose midpoint is $(\frac{4}{3}, 25)$. Convexity would require $2 + 1 \geq 2(\frac{5}{4})^{25}$, which is clearly not true.

Yes, you are right, I said I made a miscalculation and my solution is wrong.

Hang on a second - don't we have $\frac{1}{2}+\frac{1}{x} \geq \frac{\sqrt{2}}{\sqrt{x}}$ and hence the LHS is at least $3(\frac{2\sqrt{2}}{\sqrt{xyz}})^{\frac{x+y+z}{3}}=\frac{24}{xyz} \geq 3$ ?

[quote="Fiachra"]Hang on a second ..... and hence the LHS is at least $3(\frac{2\sqrt{2}}{\sqrt{xyz}})^{\frac{x+y+z}{3}}=\frac{24}{xyz} \geq 3$ ?[/quote]

Why we have this? I don't think so. :?

I suppose proving this will help:

xyyzzx <= (xyz)(x+y+z)/3.  :)

WLOG suppose that z = min {x,y,z} .

First case : z >=1 .

Then we have : (1/2 +1/x)^y = (1+1/x-1/2)^y >= 1+ (1/x-1/2)y ( by Bernouly inequal )
Similiarly we will have 3 inequalties and add them up we get :
:Sigma: (1/2 +1/x )^y >= 3 + :Sigma: y/x - :Sigma: x/2 >=3

Second case : z <=1

If y<=2 then x>=3 ,so (1/2 +1/z) ^x >=(3/2):^3: >3 .

If x<=2 then we have : (1/2 +1/x) ^y >=1 , (1/2+ 1/z)^x >= 3/2 , and it's easy to show that : (1/2 +1/y)^z > 1/2
Add them up we will get the inequality .

So we can assume that : x,y>=2 .Because x+y<6 so 4>x,y>=2
Then (1/2 +1/z)^x >= (3/2):^2:
        (1/2 +1/y)^z >=1/2 +1/y
        (1/2 +1/x)^y>=(1/2 +1/4) ^4 >1/4
Add all them up we get what we want .

[quote="namanhams"]

.....

.... , (1/2+ 1/z)^x >= 3/2 , .....

.....

[/quote]
I don't think this is true.  :?  We know that $x\leq2$ but it may be less than 1. In which case, $\left(\frac{1}{2}+\frac{1}{z}\right)^x<\frac{3}{2}$.

Oh ,sorry for my mistake .Now i'll correct my proof :

Case x <=2 : if x <=1 then y >=4 ,so (1/2+1/x)^y >= (3/2)^4 >3 and  :)
If x >=1 then follow what i have done

We can suppose that $x$ is the smallest of the three numbers so :
$LHS = [(\frac{1}{2}+\frac{1}{x})^{\frac{y}{x}}]^x+[(\frac{1}{2}+\frac{1}{y})^{\frac{z}{x}}]^x+(\frac{1}{2}+\frac{1}{z})^x$. We know that $a^x$ is convex so we will use the Jensen inequality and we will get that :
$LHS\geq[\frac{(\frac{1}{2}+\frac{1}{y})^\frac{y}{x}+(\frac{1}{2}+\frac{1}{y})^\frac{y}{x}+(\frac{1}{2}+\frac{1}{z})}{3}]^x$.$(*)$ Of course $\frac{y}{x}$ and $\frac{z}{x}$ are greater or equal than $1$ so $(*)$  is greater than :
$(\frac{\frac{3}{2}+\sum\frac{1}{x}}{3})^x$$(**)$ and we have to prove that this inequality is greater than $1$ but we have that $(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq9$ or $(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})\geq\frac{3}{2}$ So $(**)$ is greater or equal than $(\frac{3}{2}+\frac{3}{2})^x$ which is clearly greater or equal than $1$.

hehehe you have a mistake :

y/x >=1 doesn't mean that : (1/2+1/y)^y/x  >=1/2 +1/y .If y>2 then this inequality is absolutely wrong

[quote="sjk"]Let $x,y,z$ be positive real numbers such that $x+y+z=6$. Prove that
\[
\left(\frac{1}{2}+\frac{1}{x}\right)^y+\left(\frac{1}{2}+\frac{1}{y}\right)^z+\left(\frac{1}{2}+\frac{1}{z}\right)^x\geq3.
\][/quote]
OK, here is an elementary solution:
$\left(\frac{1}{2}+\frac{1}{x}\right)^y+\left(\frac{1}{2}+\frac{1}{y}\right)^z+\left(\frac{1}{2}+\frac{1}{z}\right)^x\geqslant$
$\geqslant 3\sqrt[3]{\left(\frac 1 2+\frac 1 x\right)^y\cdot\left(\frac 1 2+\frac 1 y\right)^z\cdot\left(\frac 1 2+\frac 1 z\right)^x}\geqslant$
$\geqslant 3\sqrt[3]{\left(2^y\sqrt{\frac 1 {(2x)^y}}\right)\cdot\left(2^z\sqrt{\frac 1 {(2y)^z}}\right)\cdot\left(2^x\sqrt{\frac 1 {(2z)^x}}\right)}=$
$=3\sqrt[3]{\sqrt{\frac{2^{x+y+z}}{x^yy^zz^x}}}=3\sqrt[6]{\frac{\left(\frac{x+y+z}3\right)^{x+y+z}}{x^yy^zz^x}}\mathop\geqslant^{\text{see [1]}} 3\sqrt[6] 1=3$

References:
[1] http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33540


S19

On a $ 3\times 5$ chessboard, a counter can move one square at a time along a row or column, but not along any diagonal. Starting from some squares, it can visit each of the other 14 squares exactly once, without returning to its starting square. Of the 15 squares, how many could be the counter's starting square?

[hide]Color the squares black and white like a chessboard so that the corners are black. We have $ 8$ black squares and $ 7$ white squares, and every move switches color. So we must start on a black square. It's easy to find a path from each black square. So the answer is $ 8$.[/hide]

Nearly every question relating to chessboards can be approached by comparing black and white squares.

Nice one.. I actually brute forced it. Thanks for the tip  :P


inequalities
trigonometry
modular arithmetic
induction
ceiling function
pigeonhole principle
algebra unsolved

Let $n\in\mathbb{N}$ and $0\leq a_1\leq a_2\leq\ldots\leq a_n\leq\pi$ and $b_1,b_2,\ldots ,b_n$ are real numbers for which the following inequality is satisfied :
\[\left|\sum_{i\equal{}1}^{n} b_i\cos(ka_i)\right|<\frac{1}{k}\]
for all $ k\in\mathbb{N}$. Prove that $ b_1\equal{}b_2\equal{}\ldots \equal{}b_n\equal{}0$.

[quote="enndb0x"]see [code]http://imomath.com/index.php?options=oth|Bul|Ne|Ne&ttn=Bulgaria&p=0[/code][/quote]
It is problem of Bulgari MO 1960-2008 not have problem Bul MO,TST 2009 and Bul TST 2008
if you have them,you can post them?

I and a friend of mine have posted lots of Bulgarian problems at imomath. I have the missing problems. When I've time I'll post them.

I think $ b_i \ge 0$ for $ i \equal{} 1,...n$
Otherwise let $ b_1 \equal{} 1,b_2 \equal{} \minus{} 1,a_1 \equal{} a_2$ which gives a cunterexample :?:

A solution based on $ b_i\ge 0$
The idea is using [b]approximation [/b]
Very sorry for written in Chinese,now I restate the lemma(the proof is not difficult to come up with)

[b]Lemma[/b]:$ x_1,x_2,...,x_n \in R$ we can find infinitely many $ k,k\in N$ and $ c_1,c_2,...,c_n \in Z$
so that $ |kx_i-c_i| \le k^{-\frac{1}{n}}$

For those interested in a proof the lemma above, a special case which is easily generalized is outlined [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=499018]here[/url]

Here's a proof of the slightly stronger claim that given unrestricted $b_i$ with the property below, they must sum to $0$ (of course we recover the problem statement when we impose $b_i \ge 0$) Also, it's worth noting that it is critical we work with $\cos$ and not, say, $\sin$ since then Lemma 2 (and the conclusion) fails to hold.

[b]Lemma 1:[/b] Given real numbers $a_1, a_2, \dots, a_n$ and $\delta > 0$, there exists $k \in \mathbb{N}$ such that \[-\delta < ka_i - ka_1 \pmod{2\pi} < \delta\] for $1 \le i \le n$.

[i]Proof[/i]: We prove by induction on $n$. When $n = 1$ the statement is obvious. Then suppose for some $m \ge 1$ and all $\delta > 0$ there exists $k \in \mathbb{N}$ such that \[-\delta < kb_i - kb_1 \pmod{2\pi} < \delta \; \; \; \; \; \; \; (1)\] for all $1 \le i \le m$.

Let $b_{m + 1}$ and $\delta > 0$ be given. Taking $\delta$ in $(1)$ to be $\frac{\delta}{\lceil \frac{2\pi}{\delta} \rceil + 1}$ we find $k \in \mathbb{N}$ for which \[-\frac{\delta}{\lceil \frac{2\pi}{\delta} \rceil + 1} < k(b_i - b_1) < \frac{\delta}{\lceil \frac{2\pi}{\delta} \rceil + 1}\] for all $1 \le i \le m$.

 By Pigeonhole Principle, for some $a \in \mathbb{N}, a < \lceil \frac{2\pi}{\delta} \rceil + 1$ we have \[-\delta < ka(b_{m + 1} - b_1) \pmod{2\pi} < \delta\] where $k$ is the $k$ chosen above. But \[-\delta < ka(b_i - b_1) \pmod{2\pi} < \delta\] for $1 \le i \le m$ since $a < \lceil \frac{2\pi}{\delta} \rceil + 1$, so we are done since $(1)$ is satisfied for all $1 \le i \le m + 1$ as desired. $\blacksquare$

Then by the continuity of $\cos(x)$ and its periodicity modulo $2\pi$, for all $\epsilon > 0$ there exists $k \in \mathbb{N}$ such that \[\cos(ka_1) - \epsilon < \cos(ka_1), \cos(ka_2), \dots, \cos(ka_n) < \cos(ka_1) + \epsilon \; \; \; \; \; \; \; \; \; (2)\] Furthermore, there exist such $k$ arbitrarily large; given any $\epsilon > 0$, there is some $\delta_0 > 0$ for which $(1)$ implies $(2)$. To find a $k$ value larger than $N$, simply take $\delta = \frac{\delta_0}{N}$ in $(1)$; since \[-\frac{\delta_0}{N} < ka_i - ka_1 \pmod{2\pi} < \frac{\delta_0}{N}\] holds for some $k \in \mathbb{N}$, \[-\delta_0 < kNa_i - kNa_1 \pmod{2\pi} < \delta_0\] so $(2)$ holds for $kN$ which is at least $N$, as claimed. We need just one more lemma to finish:

[b]Lemma 2: [/b]For any $\theta$, \[\max(|\cos(\theta)|, |\cos(2\theta)|) \ge \frac{1}{2}\]

[i]Proof:[/i] If $|\cos(\theta)| \ge \frac{1}{2}$, we're done. Otherwise, \[|\cos(2\theta)| = |2\cos^2(\theta) - 1| = |1 - 2\cos^2(\theta)| \ge |1 - 2(\frac{1}{2})^2| = \frac{1}{2}. \blacksquare\]

Suppose $\sum{b_i} \ne 0$. Using $(2)$ and the construction for arbitrarily large $k$, we can find $k \in \mathbb{N}$ arbitrarily large such that \[\cos(ka_1) - \epsilon < \cos(ka_1), \cos(ka_2), \dots, \cos(ka_n) < \cos(ka_1) + \epsilon\] \[\cos(2ka_1) - \epsilon < \cos(2ka_1), \cos(2ka_2), \dots, \cos(2ka_n) < \cos(2ka_1) + \epsilon \] Then we can choose $\epsilon = \frac{|\sum{b_i}|}{3\sum{|b_i|}}$ and (using Lemma 2) find $k$ arbitrarily large for which either \[|(\sum{b_i})|(\frac{1}{2}) - \frac{|\sum{b_i}|}{3\sum{|b_i|}} \sum{|b_i|} < |\sum{b_i\cos(ka_i)}| < |(\sum{b_i})|(\frac{1}{2} + \frac{1}{3})\] or \[|(\sum{b_i})|(\frac{1}{2}) - \frac{|\sum{b_i}|}{3\sum{|b_i|}}\[ \sum{|b_i|} < |\sum{b_i\cos(2ka_i)}| < |(\sum{b_i})|(\frac{1}{2} + \frac{1}{3})\] where the lower bound for each is $\frac{|\sum{b_i}|}{6} > 0$, contradicting the fact that all $k$  satisfy the problem statement.

The above solutions all use too much thinking!

Here is a troll solution. Squaring both sides of the inequality given, we have that
$$\sum b_ib_j \cos (ka_i) \cos (ka_j) < \frac{1}{k^2}, \qquad (1)$$
where the sum is over all ordered pairs $(i, j) \in [n]^2.$ Observe that $\cos(ka_i) \cos(ka_j) = \frac12 (\cos (k(a_i+a_j)) + \cos (k(a_i-a_j)))$ from the Product-to-Sum Identity.

[b] Lemma. [/b] The sequence given by $x_m= \sum_{i=1}^{m} \cos (i\alpha)$ is unbounded whenever $\alpha$ is not a multiple of $2 \pi$.

[b] Proof. [/b] Let $\omega = e^{i \alpha}.$ Then $x_m$ is the real part of the complex number
$$\frac{\omega^{m+1} - \omega}{\omega - 1}.$$
This is bounded by $|\frac{2}{\omega-1}|$ in magnitude, so we're done.

$\blacksquare$

Using the lemma and the Product-to-Sum Identity, we can sum $(1)$ from $1$ to $N$ for a large positive integer $N$, and rewrite it as
$$(stuff) + \sum_{i=1}^{n} \frac{Nb_i^2}{2}+\sum_{i \neq j, a_i = a_j} \frac{Nb_ib_j}{2} + (trash)_N,$$
where $(trash)_N$ is a quantity dependent on $N$ that is bounded. As for $(stuff)$, it denotes the potential increase of $\frac{Nb_1^2}{2}$ if $a_1 = 0$ and of $\frac{Nb_n^2}{2}$ if $a_n = \pi.$ Since it's easy to see that $\frac2N (stuff) + \sum_{i=1}^{n} b_i^2 + \sum_{i \neq j, a_i = a_j} b_i b_j > 0$ whenever not all $b_i$'s are zero, the above sum is unbounded whenever not all $b_i$'s are zero. However, $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{N^2}$ is bounded above by $\frac{\pi^2}{6}$. Hence, we must have that $\sum_{i=1}^{n} b_i^2 + \sum_{i \neq j, a_i = a_j} b_ib_j = 0$ which implies that $b_1 = b_2 = \cdots = b_n = 0,$ as desired.

$\square$


Halo 2.
The game is good, but seriously its not [i]that[/i] good. People treat as the best game ever, but it is nothing to games like half life. Same thing with ff7. Good Rpg, but there tons better than it.

pretty much any FPS. i just dont understand that genre... the gameplay doesnt vary beyond killing ppl (even in capture the flag, etc), which is interesting for about an hr.

real time strategy rules.
oh, and the civilization series, which is turn-based but still great.

i don't like halo or halo 2 that much.  i don't see why it is so good.

Well honestly, I love Halo 2 - and also HalfLife. While I'll agree that no videogames are really "quality entertainment" i like em. Halo and Halo 2 arent much better than other xbox games - but they ARE better than the others, most of em anyway. The combat system is arguably the best, the graphics are top notch, and the 1P Campaign is almost (in my opinion) movie-like. Put it this way also: for whatever reason, most people seem to have a really really really fun time playing it, especially the multiplayer, which Halo 2 has the most advanced of.

You cant really compare it to Half Life 2 - Half Life 2 is also considered one of the best games ever (1), and also, Half Life is really fun also because it is puzzle-like, and the weapons bring special innovation that can be used in cool ways. In terms of the combat system though, and how the game's FPS aspect goes, I think Halo2 is better. Half Life 2 is good too though.

[quote="ToBe"] multiplayer, which Halo 2 has the most advanced of.[/quote]

If I understood correctly, you're saying Halo 2 has the most advanced multiplayer? In terms of player numbers, there are many PC games/FPSs that have higher numbers. In terms of AI, well, Halo has no AI, wheras almost all other games do. In terms of graphics, there are plenty of PC games that are better. In terms of options, Perfect Dark for N64 was much better, and there are also PC games almost on the PD level. So, what can I say? I disagree. :D


Is there a bijection from $\mathbb{R} \times \mathbb{R}$ to $\mathbb{R}$? (equivalently, is there a bijection from a plane to a line?)

i'm really interested in the proof of the existence (or non-existence)  :P

if the answer is $\mathbb{YES}$, a construction would be appreciated (i have one, but i don't think it works; someone i know also has a construction, but that may not work either)

Cantor Bernstein Theorem (whose proof gives you an explicit bijection by the way).

are talking about this : http://www.math.umn.edu/~garrett/m/algebra/cantor_schroeder_bernstein.pdf ?

one can trivially construct a injection from $\mathbb{R} \to \mathbb{R} \times \mathbb{R}$, but the other way around ( $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ ) causes lots of problems (the injective part is of course the main problem with this function - with or without knowing the theorem you mentioned; btw the proof is nice, but somewhat complicated )

[quote="perfect_radio"] ...  but the other way around ( $\mathbb{R} \times \mathbb{R} \to \mathbb{R}$ ) causes lots of problems (the injective part is of course the main problem with this function ...)[/quote]
Hint  :lol:  ${{X \in R+ \Longrightarrow X = \sum_{k\in Z} x_k2^k}\  ,\   Y \in R+ \Longrightarrow X = \sum_{k\in Z} y_k2^k} $ where $x_k,y_k\in\{0,1\}$ and $\{k \ || \ x_k=1\}, \{k \ || \ y_k=1\}$ are infinite sets.

And now, take for exemple  $f(X,Y) = \sum_{k\in Z}( x_k2^{2k}+y_k2^{2k-1})$

 :cool:  Could you continue ? :cool:

[quote="perfect_radio"]if i understood correctly, your function maps $(X,Y)$ to a number $Z$ by taking alternatively a (in your case binary, but it can be in any base, right ?) digit from $X$ and one from $Y$.

it seems too good (read easy) to be true...
 :P  :pilot:[/quote]

But that's thru, $f$ is injective, even it's too easy !

An other hint :  all positive real number can be written with a infinite number of non zero digits, and this infinite set of non zero digits is unique !

 :cool:  Do you follow me ?  :cool:

what i meant with "it's too good to be true" is that your function is also surjective...

and why do we need another hint?

i was actually thinking of considering the numbers with all digits (including those which are $0$ and of course excluding those which end in a period of $\textrm{base}-1$)  :D


trigonometry
function
inequalities
induction
real analysis
real analysis unsolved

For n = 0,1,2,..., and x real, prove that
\[|\sin (nx)| \le n|\sin (x)| \]

Note that this inequality may be false for other values of n. For instance,
\[|\sin ({\textstyle{1 \over 2}}\pi )| >{\textstyle{1 \over 2}}|\sin (\pi )| \]

$|\sin{(n+1)x}|=|\sin{nx}\cos{x}+\cos{nx}\sin{x}|\leq |\sin{nx}|.|\cos{x}|+|\cos{nx}|.| \sin{x}|\leq |\sin{nx}|+|\sin{x}|$. Done by induction on $n$ :D


tx + 12y = -3

the  equation   above  is   the   equation   of  a  line  in  the   xy - plane  and   t   is  a  constant .
if the slope of the  line is -10, what is the value of t ?

$ tx\plus{}12y\equal{}\minus{}3$
$ 12y\equal{}\minus{}tx\minus{}3$
$ y\equal{}\frac{\minus{}t}{12}x\minus{}frac{1}{4}$

Using slope-intercept form,

$ \frac{\minus{}t}{12}\equal{}\minus{}10$
$ t\equal{}\boxed{120}$


Find all non-constant polynomials P(x) with real coefficients such that

P(P(x))=(P(x))^k ,  where k is any positive integer.

The answer is pretty obvious but I'm having trouble proving it rigorously.

Thanks

[hide="i think theres something wrong!"]are you sure the problem is correct! for any positive integer $ k$?

If we assume that $ P(x) \equal{} a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots\plus{}a_{1}x\plus{}a_{0}$

then according to the equation given we must have $ \deg\left\{P(P(x))\right\}\equal{}\deg\left\{P(x)^{k}\right\}$

but $ \deg\left\{P(P(x))\right\}\equal{} 2n$ and $ \deg\left\{P(x)^{k}\right\}\equal{} kn$ so $ kn \equal{} 2n$ and this gives us $ n \equal{} 0$ or $ k \equal{} 2$, but

$ n \equal{} 0$ is impossible cuz you said the Polynomial must be non-constant!

As a consequence this equation has no solution for non-constant Polynomials!

[/hide]

[quote="sinajackson"]are you sure the problem is correct! for any positive integer $ k$? [....]
but $ \boxed{\deg\left\{P(P(x))\right\}\equal{} 2n}$ [...][/hide][/quote]
but $ deg(p(p(x)))\equal{}n^{2}$

solution:
we have $ p(p(x)) \equal{} (p(x))^{k}$ so $ \forall x\in p(\mathbb{R}):\ p(x) \equal{} x^{k}$ then $ \boxed{\forall x\in\mathbb{R}:\ p(x) \equal{} x^{k}}$ because |$ p(\mathbb{R})| \equal{}\plus{}\infty$

[quote="aviateurpilot"][quote="sinajackson"]are you sure the problem is correct! for any positive integer $ k$? [....]
but $ \boxed{\deg\left\{P(P(x))\right\}\equal{} 2n}$ [...][/hide][/quote]
but $ deg(p(p(x))) \equal{} n^{2}$

solution:
we have $ p(p(x)) \equal{} (p(x))^{k}$ so $ \forall x\in p(\mathbb{R}):\ p(x) \equal{} x^{k}$ then $ \boxed{\forall x\in\mathbb{R}:\ p(x) \equal{} x^{k}}$ because |$ p(\mathbb{R})| \equal{}\plus{}\infty$[/quote]

Hmm, that's not exactly clear what you're doing. You're saying that for all $ x\in P(\mathbb{R})$, $ P(x)\equal{}x^{k}$, then you claim that $ P(x)\equal{}x^{k}$ everywhere. This is valid, but you're not explaining it well. The idea is you consider $ P(x)\minus{}x^{k}$. This is a polynomial, but it has infinite zeros, therefore it must be the zero polynomial, and $ P(x)\equal{}x^{k}$.

[quote="Hamster1800"] [....] Hmm, that's not exactly clear what you're doing. You're saying that for all $ x\in P(\mathbb{R})$, $ P(x) \equal{} x^{k}$, then you claim that $ P(x) \equal{} x^{k}$ everywhere.The idea is you consider $ p(x)\minus{}x^{k}$
 [....][/quote]
here i use that if $ p(x)\equal{}f(x)$ ($ f,p\in C[X]$) has infinite solutioin then $ f\equal{}p$
it's the same of $ p(x)\minus{}f(x)\equal{}0$ ($ f,p\in C[X]$) has infinite solutioin then $ f\minus{}p\equal{}0$  :lol:

[quote="Geeimatree"]can u guys explain how you got x\in P(\mathbb{R}). We are supposed to find polynomials that satify the given, initial equation. You're assuming that all P(R) work.

Can you clarify more. especially the part when you assert that there are infinite values work.[/quote]
if ${ p(\mathbb{R}) =\{x\in\mathbb{R}|\ \exists a\in\mathbb{R},p(a) = x}\}$¨
then $ \forall x\in p(\mathbb{R}),\exists a\in\mathbb{R}:\ p(a) = x$ gives $ p(x) = p(p(a)) = p(a)^{k}= x^{k}$
then $ \forall x\in p(\mathbb{R}),\ p(x)-x^{k}= 0$ ,
and we know that $ p(\mathbb{R})$ is infinit so $ p(x)-x^{k}$ has infintly many zero, then $ \forall x\ \mathbb{R}:\ p(x) = x^{k}$

ah, i get it now. even if only a single polynomial worked, $ P(x)$ has an infinite number of values because $ x\supset\mathbb{R}$ thus $ P(x)\equal{}\infty$. Considering that $ P$ is nonconstant.

and then, as you said, $ P(\mathbb{R})\equal{}\mathbb{R}^{k}$ and the polynomial $ P(x)\minus{}x^{k}\equal{}0$ has infinite number of zeros, so $ P(x)\equal{}x^{k}$ $ \forall$ $ \mathbb{R}$ and only $ x^{k}$ satisfies the equation for nonconstants.

thanks


modular arithmetic
quadratics
advanced fields
advanced fields theorems

Show that if p=x^2+5y^2 then p must me congruent to 1 or 9 (mod 20)

We have $\gcd (x,p)=\gcd (y,p)=1$ and $x^{2}\equiv-5y^{2}\pmod{p}$, so $(xy^{-1})^{2}\equiv-5\pmod{p}$. Therefore $-5$ is a quadratic residue of $p$.

Conversely, if $p$ is congruent 1 or 9 mod 20 then such $x,y$ exist

Can some moderator please move it to Number Theory¿


Let´s now consider the homologation of aldehydes:

devise methods for converting an aldehyde to its next higher homolog,

$RCHO \longrightarrow RCH_{2}CHO$,

and for converting an aldehyde to its next lower homolog,

$RCH_{2}CHO \longrightarrow RCHO$.

These are standard ascent and descent of series reactions.
 RCHO can be reduced to RCH2OH which is converted to RCH2I by P/Iodine followed by conversion to RCH2CN  by nucleophilic substitution.RCH2CN can then by hydrolyzed to the corresponding acid RCH2COOH which can converted to its calcium salt.The calcium salt on heating would give RCH2CHO.
 The descent can be accomplished by conversion of RCH2CHO to RCH2COOH followed by treatment with ammonia to yield the acid amide which when subjected to Hoffmann degradation would yield RCH2NH2.This can be treated with nitrous acid(NaNO2 + HCl) to yield RCH2OH which can then be oxidized to give RCHO.
Kindly bear with the symbols as I am not proficient in Latex.
Sathej

[quote="Sathej"]RCHO can be reduced to RCH2OH which is converted to RCH2I by P/Iodine followed by conversion to RCH2CN by nucleophilic substitution.RCH2CN can then by hydrolyzed to the corresponding acid RCH2COOH which can converted to its calcium salt.The calcium salt on heating would give RCH2CHO.[/quote]

Correct, but some remarks:

1) RCH2CN can be directly reduced to RCH2CHO with DIBALH, followed by hydrolysis.

2) Your synthesis uses 6 steps. However there is a simpler two step procedure. Can you find it?

[quote="Sathej"]The descent can be accomplished by conversion of RCH2CHO to RCH2COOH followed by treatment with ammonia to yield the acid amide which when subjected to Hoffmann degradation would yield RCH2NH2.This can be treated with nitrous acid(NaNO2 + HCl) to yield RCH2OH which can then be oxidized to give RCHO.[/quote]

Some remarks:

1) Reaction of a carboxilic acid with ammonia is not the best way of preparing amides. This is because the product of the acid-base reaction that occurs instead is an ammonium carboxylate, and this species have very low reactivity towards subsequent nucleophilic addition. We can still obtain the amide, but for that we would have to evaporate the solvent and then heat the dry salt. A much better and simpler procedure is to convert previously the acid into the acid chloride (for example, with the stinky thionyl chloride) and then react this with an excess of ammonia.

2) Treatment of RCH2NH2 with nitrous acid to obtain RCH2OH is not certainly a plausible method, because after diazotization nitrogen gas is rapidly lost and the carbocation that forms can (and do) react is many ways: reaction with nucleophiles present in solution, besides water; alkene formation; rearrangement.

3) There is a three step procedure for this convertion.

[quote="Carcul"]However there is a simpler two step procedure.[/quote]

[hide="Solution"]Step 1:

$RCHO+(C_{6}H_{5})_{3}P^{+}-^{-}CHOCH_{3}\longrightarrow RCH=CHOCH_{3}$

Step 2:

$RCH=CHOCH_{3}+H_{3}O^{+}\longrightarrow RCH_{2}CHO$[/hide]

[quote="Carcul"]There is a three step procedure for this convertion.[/quote]

[hide="Solution"]Step 1:

$RCH_{2}CHO+SeO_{2}\longrightarrow RCOCHO$.

Step 2 (Cannizzaro Rearrangement):

$RCOCHO+HO^{-}\longrightarrow RCH(OH)CO_{2}^{-}$.

Step 3 (Blaise-Le Sueur Degradation or alternative oxidation):

$RCH(OH)CO_{2}^{-}+H^{+}+190ºC (or LTA) \longrightarrow RCHO$.[/hide]


inequalities
number theory unsolved
number theory

[quote="N.T.TUAN"]Find the greatest positive integer $A$ with the following property: For every permutation of $\{1001,1002,...,2000\}$ , the sum of some ten consecutive terms is great than or equal to $A$.[/quote]
Let $t_{k}=a_{k}+a_{k+1}+...+a_{k+9}$.
We have $\sum_{k=1}^{100}t_{10k-9}=500*3001=1500500$. Therefore exist k, suth that $t_{k}\ge 15005$ for all permutations. These is better inequalities, because exist permutation, when $a_{k}=1000+(k+1)/2$ if k is odd, and $a_{k}=2001-k/2$ if k is even, when for any k $t_{k}=15005$.


topology
parameterization
real analysis
real analysis unsolved

as you know the order of connectedness of two curve is defined by the equality of terms of their Tailor series up to the $ n$th term.
But I don't know under what conditions we say a curve and a surface have connectedness of order $ n$. :!:

The order of connectedness can be defined geometrically: two subsets $ E$ and $ F$ of a metric space $ X$ have order of connectedness (at least) $ n$ at point $ x_0$ if the [url=http://en.wikipedia.org/wiki/Hausdorff_distance]Hausdorff distance[/url] between $ E\cap B(x_0,r)$ and $ F\cap B(x_0,r)$ is $ O(r^n)$ as $ r\to 0$. Here $ n$ can be fractional.

But this definition may be too abstract for you. Say, the surface has equation $ F(\vec x)\equal{}0$ where $ F$ is smooth and $ \nabla F(x_0)\ne 0$. Let $ \gamma$ be a smooth parametrization of the curve with $ \gamma'(t_0)\ne 0$, where $ \gamma(t_0)\equal{}x_0$ The first nonzero term of the Taylor series of $ F(\gamma(t))$ at $ t_0$ is what you want.


geometry
calculus
integration
trigonometry
parameterization
parametric equation
calculus computations

Tell the parametric representation of the circle $ x^2\plus{}y^2\equal{}r^2$ and compute the area with the integral of type 2.
(I really need all the solution please)

Thanks in advance

Parametric equation of a circle (in the counter-clockwise direction) is
$ \textbf{x}(\theta)\equal{}(r\cos{\theta},r\sin{\theta})$

What is an integral of type 2?


inequalities
function
inequalities solved

Is one to oblige to pass by studies of functions to prove these inequalities?


vector
linear algebra
matrix
inequalities
real analysis
real analysis unsolved

1. It's in the wrong forum. [Ok, I'm wrong if you say so]

2. I think that $A$ is a double-stochastic matrix. I remember seeing something related to this some time ago on the forum. But I may be wrong, so forgive me if I said something stupid.

No, you are wrong, it may seem it contains the word matrix, but the reasoning is analytical.  ;)

You say "no" to the second question also? I think I proved that the sum of elements of a row/column must be $1$. But I remember I couldn't prove that the elements must be positive.
 :blush:

[quote="harazi"]Find all real matrices $A$ such that for any vector $x$, $Ax$ majorizes the vector $x$.[/quote]$Ax \ge x$ and $A(-x) \ge (-x)$ (where the inequality is componentwise) imply $Ax = x$ for all $x$. Hence $A$ is the identity matrix.

I still fail to see where the analytical content ist.

[quote="hpe"]$Ax \ge x$ and $A(-x) \ge (-x)$ (where the inequality is componentwise)[/quote]

Uhh... no. Click [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=14975]here[/url] to find out what majorization is.

If I remember correctly, you can take $x$'s with $1$'s or $-1$'s in some places and $0$'s in other places. But I can't finish my proof.


ratio
geometry
circumcircle
geometry unsolved

In triangle $ ABC$ we have $ a \geq b$ and $ a \geq c.$ Prove that the ratio of circumcircle radius to incircle diameter is at least as big as the length of the centroidal axis $ s_a$ to the altitude $ a_a.$ When do we have equality?

I will denote the centroidal axis by $\mu_a$ and the altitude by $h_a.$ We are asked to show that $\frac{R}{2r}\ge \frac{\mu_a}{h_a}.\ (1)$ However, $E=\frac{abc}{4R}=\frac{1}{2}ah_a=rs=\sqrt{s(s-a)(s-b)(s-c)},$ where $2s=a+b+c.$ Also, we know that $\mu_a^2=\frac{2b^2+2c^2-a^2}{4}.$ Therefore, $(1)$ is equivalent to

$(Rh_a)^2\ge 4r^2\mu_a^2\Leftrightarrow \frac{1}{4}b^2c^2\ge \frac{16E^2}{(a+b+c)^2}\mu_a^2 \Leftrightarrow b^2c^2(a+b+c)\ge (a+b-c)(a-b+c)(-a+b+c)(2b^2+2c^2-a^2).\ (2)$

Let $x=b/a,y=c/a;$ by the statement of the problem we have $0\le x,y\le 1$ and $1\le x+y\le 2.$ Thus, $(2)$ is equivalent to $x^2y^2(1+x+y)\ge (1+x-y)(1-x+y)(-1+x+y)(2x^2+2y^2-1),$ which can be rewritten as $2(x-y)^2(x+y-1)^3+5\left[1-(x-y)^2\right](x+y-1)(1-x)(1-y)+(1-x)(1-y)(x^3+y^3+5xy-5x-5y+4)\ge 0.\ (3)$

All we need to show now is that $f(x,y)\equiv x^3+y^3+5xy-5(x+y)+4\ge 0,$ because all other terms in $(3)$ are $\ge 0.$ But for $1\le x+y\le\frac{5}{3}$ this is obvious because $f(x,y)=(x+y-1)^3+(1-x)(1-y)\left[5-3(x+y)\right]\ge 0,$ while for $\frac{5}{3}\le x+y\le 2$ we have $f(x,y)\ge (x+y)^3-8(x+y)+9\ge \frac{8}{27}.$

Therefore $(1)$ holds under the conditions of the problem. Equality occurs when the triangle is an equilateral triangle.


Hey do you guys know any competition, books or websites which have elementary proofs such as the ones posted in the marathon? (by proofs i dont mean like USAMO kinds which take up hours, something easy like whats posted in the marathon)

The Mandelbrot Competetion (hosted locally) is a great example of an introduction to proofs. (well, the team round is... the individual is usually a joke..)

As far as books go, I think that Art and Craft of Problem Solving by Paul Zeitz is a good start, and I have heard from Alex Anderson (altheman) that Engel's book (not sure of the title) is another, more intermediate one.

If you're looking for BASIC proofs, you should probably just buy AoPS volumes 1 and 2 (the set is pretty cheap, or you could see if the library has a copy)

I started with AoPS one, and worked through it SLOWLY!
That's the trick. Work through each chapter, do all the excercizes, and make sure that you understand each concept thouroughly before progressing. If you get stuck, ask someone to explain the basic premise to you.

I think that after you have mastered AoPS 1 and 2, you should probably move on to ACoPS.
That's a large step up, because that book teachers more proofs than AoPS.

If you're really hardcore into geometry (like I was last year) then you can find a copy of Euclid's Elements.

Try to find one on the internet, and read through it.

It's really insightful, and the best part is that it was written, like, 2000 years ago  :lol:

What a genius that guy must have been.

Anyways, those are my 0.02.

okay thanks for the help, I will definitely look through some of these things


There is a triangle garden O with sides 30, 28, 28 ft. A curve A is to go around the garden. Each point of A is to be exactly 8 ft away from the nearest point of O. What is approximate of A in ft

build 3 rectangles with the 3 sides of the triangle as one of the sides and the width of all 3 rectangles to be 8 (all measurements i give are in feet) (so that we have 8x28,8x28,8x30 rectangles), so that the rectangles lie completely outside the triangle. Then, add sectors (that is, a section of a circle like a pizza slice, with center as tip and 2 straight radii down to the arc) to the 3 vertices with the tips of the sectors coinciding with the vertices of the triangle; these sectors have radii 8. These sectors are also "joining" the 3 rectangles. These the border of the resulting figure (consisting of 3 rectangles  and 3 arcs) is the desired curve, which has length $ 28 \plus{} 28 \plus{} 30 \plus{} 16\pi \equal{}86 \plus{} 16\pi$ (the 3 arcs can be joined to form exactly one full circle)


The sum of $ 101$ consecutive natural numbers is $ p^{3}$, where $ p$ is a prime. Find the smallest of these $ 101$ numbers.

Let the smallest be x

$ 101x\plus{}5050\equal{}p^3$

$ 101(x\plus{}50)\equal{}p^3$

Obviously $ p\equal{}101$, so $ x\equal{}101^2\minus{}50$


[color=darkblue][b]Consider three collinear points $ A$, $ B$, $ C$ and three isosceles triangles in the same plane $ \{\begin{array}{ccc}\triangle AZB & , & ZA=ZB\\\ \triangle BXC & , & XB=XC\\\ \triangle CYA & , & YC=YA\end{array}$.

Prove that the perpendicular lines from the points $ A$, $ B$, $ C$ to the lines $ YZ$, $ ZX$, $ XY$ are concurrently.[/b][/color]

It's an interesting problem, but the only solution I got is with analytic geometry.

Does someone have a different proof :?:

I have solution with no so mutch calculations. It is not essential that points are colliner!
Let AI perp. ZY and CI perp. ZX. We want to prove, that IB perp. XY.
The points I' B' are such that II' perp. XY, and BB' perp. XY. We want to show that B' and I' coinside.
Now:
I'Y^2-I'X^2=IY^2-IX^2=IY^2-IZ^2+IZ^2-IX^2=AY^2-AZ^2+CZ^2-CX^2=AY^2-CX^2=BY^2-BX^2=B'Y^2-B'X^2
But equality: I'Y^2-I'X^2=B'Y^2-B'X^2 is possible, only if B'=I' q.e.d.


Consider the $\mathbb{P}$ a polynomail with integer coefficents  and degree $n$.
and we know $\mathbb{P}(0)=0$. Find all integer roots of $\mathbb{P}(\mathbb{P}(x))$

$P(P(x))=0 \iff P(x)=\"root\"$

so all roots of $P(x)$ are roots of $P(P(x))$

suppose that:exists integer $b$ such as $P(P(b))=0$  and $P(b)=a \neq 0$ of course $a$ is integer root of $P(x)$
then: $b(b-a)H(b)=a$ where $H(b)$ is polynomail with integer cooficients.
$b(b-a)*t=a$
$b^2t-bat=a$
so $a=kb$
$b^2t-b^2kt=kb$
$bt-bkt=k$
so $k=lb$
$bt-b^2lt=bl$
$t-blt=l$
so $l|t$ and $t|l$ then $|t|=|l|$
if $t=l$
$t-bt^2=t$ so $bt=0$ and $a=0$ contradiction.

if $t+l=0$
$t+bt^2=-t$
$2+bt=0$
$b \in \{-2,-1,1,2\}$
and $a=2b$

now we can give answer:
1) all integer roots and $-2$ if $P(x)=x(x+4)( (x+2)K(x)+1)$
2) all integer roots and $-1$ if $P(x)=x(x+2)((x+1)K(x)+2)$
3) all integer roots and $1$  if $P(x)=x(x-2)( (x-1)K(x)-2)$
4) all integer roots and $2$  if $P(x)=x(x-4)( (x-2)K(x)-1)$
5) all integer roots when not exist 1,2,3,4

i'm sorry if it's wrong

[quote="_el_doopa"]then: $b(b-a)H(b)=a$ where $H(b)$ is polynomail with integer cooficients.[/quote]

I don't understand why you can say that...

0 is a root, $a$ is a root, so it's $x(x-a)H(x)=P(x)$, he assumed $P(b)=a$, so $b(b-a)H(b)=a$.  I think the solution is right.


[b]Let a, b, c [/b][u]be positive real numbers satisfying [/u]$\ {abc}\leq\ {1}$. Prove that:
${\frac{a}{c}+\frac{b}{a}+\frac{c}{b}\geq\ (a+b+c)}$

Please post a question only once. And I don't know, but, shouldn't inequalities go in Getting Started??

Let ABCD be a quadrilateral,M is a point on CD(M doesn't coincide C,D).Prove that:
MA+MB<max{CA+CB;DA+DB}

Prove that If we have $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$ and x+y+z=xyz then:
$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=2$

I think math is fun, math is great, life is boring if we don't have math.[/quote][/u]

[quote="Ramanujan"][quote="TRAN THAI HUNG"]Let a, b, c be positive real numbers satisfying $abc \leq 1$ . Prove that:
$\frac{a}{c}+\frac{b}{a}+\frac{c}{b} \geq (a+b+c)$ [/quote][/quote]
Well, famous, http://www.mathlinks.ro/Forum/viewtopic.php?t=85330

:ninja: Let a,b,c be a positive number.Prove that:
$\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\geq\ a+b+c$

Math is my life :lol:

it is very easy.
$\frac{a^2+b^2}{a+b}\geq\frac{a+b}{2}$

I think so, because it's just a small problem.What about the second problem?

[quote="scorpius119"]edit: why does my $\LaTeX$ look so weird?[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88819


Aratati ca ecuatia : $ x^{n}+y^{n}+z^{n}=2$ , are o infinitate de solutii rationale $ \vee n \in \{1,2,3,4\}$ .

Îmi cer scuze că pornesc din nou topicul acesta vechi.Totuși,are cineva o soluție la această problemă? Am postat-o si pe forum in engleză,dar nu am primit nici un răspuns pe care să-l pot considera mulțumitor ...

[hide="hint pt cazul n=3"]$ (1\plus{}6n^{3})^{3}\plus{}(1\minus{}6n^{3})^{3}\plus{}(\minus{}6n^{2})^{3}\equal{} 2$ [/hide]

[hide="hint pt cazul n=4"]la cazul asta considera $ a\plus{}b\plus{}c \equal{} 0$ si $ a^{4}\plus{}b^{4}\plus{}c^{4}\equal{} 2(xy\plus{}yz\plus{}xz)^{2}$, deci $ xy\plus{}yz\plus{}xz \equal{}\minus{}1$... luam $ y \equal{} 1\plus{}tx$ si in final obtinem $ x \equal{}\minus{}\frac{2t\plus{}1}{t^{2}\plus{}t\plus{}1}$.[/hide]

ar mai fi o metoda, nu stiu daca merge (poate chiar e o prostie :blush: ): pp ca ecuatia are un numar de solutii finite si luam tripletul $ (a_{1}, b_{1}, c_{1})$ care maximizeaza suma $ a\plus{}b\plus{}c$ in cadrul solutiilor gasite (evident ca in fiecare caz ecuatia are solutie) si de aici sa plecam de la acest triplet si sa ajungem la altul , pt care suma $ a'\plus{}b'\plus{}c'\geq a_{1}\plus{}b_{1}\plus{}c_{1}$ (contradictie). in acest caz e de preferat s avem si un invariant (spre exemplu $ abc$). sper ca nu am zis prostii :(


geometry
incenter
trigonometry
ratio
calculus
parallelogram
cyclic quadrilateral

Given a right triangle with hypotenuse $ BC$,the incircle of the triangle is tangent to the sides $ AB$ and $ BC$ respectively at $ P$ and $ Q$.$ A$ line through the incenter and the midpoint $ F$ of $ AC$ intersects side $ AB$ at $ E$; the line through $ P$ and $ Q$ meets the altitude $ AH$ at $ M$.Prove that $ AM\equal{}AE$

[quote="apollo"]Given a right triangle with hypotenuse $ BC$,the incircle of the triangle is tangent to the sides $ AB$ and $ BC$ respectively at $ P$ and $ Q$.$ A$ line through the incenter and the midpoint $ F$ of $ AC$ intersects side $ AB$ at $ E$; the line through $ P$ and $ Q$ meets the altitude $ AH$ at $ M$.Prove that $ AM \equal{} AE$[/quote]
In my opinion, $ AM\equal{}ME$, not $ AM\equal{}AE$.

[quote="mr.danh"] In my opinion, $ AM \equal{} ME$, not $ AM \equal{} AE$.[/quote]

I think that it 's $ AM\equal{}AE$. I use an ugly method to prove it (algebraic method). Just wonder if there is a beautiful method (pure geometry).

$ AM \equal{} \frac{(p\minus{}a) \cos \frac{B}{2}}{\sin \frac{B}{2}} \equal{} (p\minus{}a) \cot \frac{B}{2}$ and $ AF \equal{} \frac{\frac{b}{2} \cdot r}{\frac{b}{2}\minus{}(p\minus{}a)} \equal{} \frac{b \cdot r}{a\minus{}c} \equal{} (p\minus{}a) \frac{b}{a\minus{}c} \equal{} (p\minus{}a) \frac{\cos{\frac{B}{2}}}{\sin{\frac{B}{2}}} \equal{} (p\minus{}a) \cot \frac{B}{2}$

[b][size=100][color=DarkBlue]LEMMA. – A triangle $ \bigtriangleup ABC$ is given and let $ E,\ F$ be, the tangent points of its incircle $ (I),$ with the side-segments $ AC,\ AB,$ respectively. Prove that the orthogonal projections of $ B,\ C,$ on the line segments $ CI,\ BI$ respectively, where $ I$ is the incenter of $ \bigtriangleup ABC,$ lie on the line segment $ EF.$[/color][/size][/b]

This well known Lemma, is coming from the past and it is easy to prove it.

$ \bullet$ In our problem now, based on the above Lemma, we conclude that $ AK\perp CI$ $ ,(1)$ where $ K\equiv PQ\cap CI.$

So, it is easy to show that the line segment $ AK,$ bisects the angle $ \angle EAM,$ because of $ \angle NAK = \angle ACN = \angle KCH = \angle KAH$

$ \Longrightarrow$ $ \angle EAK = \angle MAK$ $ ,(2)$ where $ N\equiv AB\cap CI.$ $ ($ from the right triangle $ \bigtriangleup ANC$ and from cyclic quadrilateral $ AKHC$ $ ).$

We will prove that $ \angle AEK = \angle AMK.$

$ \bullet$ Because of the cyclic quadrilateral $ APKI,$ we have $ \angle AKP = \angle AIP = 45^{o}$ $ ,(3)$ $ ($ from $ AP = IP = r$ $ )$ and so, because of $ (1),$ we conclude that $ \angle AKP = \angle PKN = 45^{o}$ $ ,(4)$

We will prove now, that $ EK\parallel BI$ and because of $ BI\perp PQ,$ it is enough to prove that $ EK\perp PQ.$

Let $ L$ be, the intersection point of the line segment $ EF,$ from the line segment $ CP.$

It is easy to show that the points $ E,\ I,\ L,\ F,$  is in harmonic conjugation, because of $ \frac {LI}{LF} = \frac {EI}{EF}$ from $ IP\parallel AC$ and $ AF = CF.$

$ \left (\frac {LI}{LF} = \frac {IP}{CF} = \frac {IP}{AF} = \frac {EI}{EF}\right ).$

So, we conclude that the points $ E,\ N,\ P,\ A,$ as the projections of the points $ E,\ I,\ L,\ F,$ from the point $ C,$ on $ AB,$ are also in harmonic conjugation.

$ \bullet$ Hence, the pencil $ K.ENPA,$ is in harmonic conjugation too and then, we conclude that $ EK\perp PQ$ $ ,(5)$ because of $ PQ,$ is the angle bisector of $ \angle AKN.$

From $ (5)$ and because of $ BI\perp PQ$ $ \Longrightarrow$ $ EK\parallel BI$ $ \Longrightarrow$ $ \angle AEK = \angle PBI$ $ ,(6)$

From the cyclic quadrilateral $ BPIQ$ $ \Longrightarrow$ $ \angle PBI = \angle PQI$ $ ,(7)$

From $ AH\parallel IQ$ $ \Longrightarrow$ $ \angle PQI = \angle AMK$ $ ,(8)$

From $ (6),$ $ (7),$ $ (8)$ $ \Longrightarrow$ $ \angle AEK = \angle AMK$ $ ,(9)$

From $ (2),$ $ (9)$ we conclude that $ \angle AKE = \angle AKM$ and so, the triangles $ \bigtriangleup AKE$ and  $ \bigtriangleup AKM,$ are congruent.

Hence, we conclude that $ AE = AM$ and the proof is completed.

Kostas Vittas.

[size=134][color=darkred][b]Nice problem ![/b][/color][/size]

[quote="apollo"] [color=darkred]The incircle $ w = C(I,r)$ of the $ A$ - right triangle $ ABC$ touches $ [AB]$ and $ [BC]$ at $ P$ and $ Q$ respectively.

Denote the midpoint $ F$ of $ [AC]$ and $ \begin{array}{c} E\in AB\cap IF \\
 \\
M\in PQ\cap AH\end{array}$ . Prove that $ AM = AE$ .[/color] [/quote]

[color=darkblue][b][u]Proof I (synthetic).[/u][/b] The nice [b]Vittasko[/b]'s proof with the well-known property which was mentioned in the his reply !

[b][u]Proof II (syntetic).[/u][/b] The $ C$ - exincirle of $ \triangle ABC$ touches $ AB$ in $ E'$ . Denote $ \{P,R\} = PI\cap w$ . From the well-known property have $ R\in E'C$ .

Since $ PR\perp AC$ and $ I$ , $ F$ are midpoints of $ PR$ , $ AC$ respectively obtain $ E'\in IF\cap AB$ . In conclusion, $ E'\equiv E$ , i.e. $ \boxed {\ AE = p - b\ }$ .

We know $ AP = p - a = r$ . Apply the [b]Sinus' theorem[/b] in $ \triangle AMP\ : \ \ \frac {AM}{p - a} = \frac {AM}{AP} = \frac {\sin\widehat {APM}}{\sin\widehat {AMP}} =$

$ \frac {\sin [180 - (90 - \frac B2)]}{\sin[90 - (90 - \frac B2)]} = \cot\frac B2 = \frac {p - b}{r} = \frac {p - b}{p - a}$ $ \implies$ $ \boxed {\ AM = p - b\ }$ .

[b][u]Proof III (metric).[/u][/b] Suppose w.l.o.g. that $ b > c$ . Show easily (or now !) that $ HQ = \frac {(b - c)((p - a)}{a}$ .

Denote $ S\in PQ\cap AC$ . Show easily (or know !) that $ \frac {SA}{SC} = \frac {p - a}{p - c}$ .

$ \blacktriangleright$ Apply the [b]Menelaus' theorem[/b] to the transversal $ \overline {SMQ}$ and $ \triangle\ AHC : \ \ \frac {SA}{SC}\cdot\frac {QC}{QH}\cdot\frac {MH}{MA} = 1\implies\frac {MA}{MH} = \frac {SA}{SC}\cdot\frac {QC}{QH} =$

$ \frac {p - a}{p - c}\cdot\frac {a(p - c)}{(b - c)(p - a)}$ $ \implies\frac {MA}{a} = \frac {MH}{b - c} = \frac {h_a}{2(p - c)}$ $ \implies MA = \frac {ah_a}{2(b - c)}\implies\boxed {\ AM = \frac {bc}{2(p - c)}\ }$ .

$ \blacktriangleright$ $ I\in EF\Longrightarrow \frac {EB}{EA}\cdot b + \frac {FC}{FA}\cdot c = a$ $ \implies$ $ \frac {EB}{EA} = \frac {a - c}{b}$ $ \implies$ $ \frac {EB}{a - c} = \frac {EA}{b} = \frac {c}{2(p - c)}$ $ \implies$ $ \boxed {\ AE = \frac {bc}{2(p - c)}\ }$ .

[b]Remark.[/b] Here are some remarkable relations in a $ A$ - right triangle : $ \{\begin{array}{ccc} \begin{array}{ccc} r = p - a & ; & r_a = p \\
 \\
r_b = p - c & ; & r_c = p - b\end{array} & ; & \begin{array}{c} S = p(p - a) = (p - b)(p - c) \\
 \\
(p - b)\tan\frac B2 = (p - c)\tan\frac C2 = r \\
 \\
r_a = r + r_b + r_c = p\end{array}\end{array}\ .$

Observe that $ \frac {bc}{2(p - c)} = p - b$ .  Thus, $ \boxed {\ AE = AM = p - b\ }$ , i.e. the $ C$ - exincircle of $ \triangle ABC$ touches $ AB$ in $ E$ .

Show easily that $ M\in BF\Longleftrightarrow a = \lambda c$ , where $ \lambda\in (\frac 32,2)$ , $ \lambda^3 = \lambda^2 + \lambda + 1$ .[/color]

Dear Mathlinkers,
Mathangel wrote :  "I use an ugly method to prove it (algebraic method). Just wonder if there is a beautiful method (pure geometry)."
My question : can we research a purely proof without cross ratio, calculus, report, congruent triangle, trigonometric...?
Any ideas?
Sincerly
Jean-Louis

I've a little ugly solution.

As the incenter of $ \triangle ABC$ and $ FI\cap AB\equal{}E$, we have

$ \frac{AB}{AE}\cdot AC\plus{}\frac{AC}{AF}\cdot AB\equal{}AB\plus{}BC\plus{}CA$

So $ AE\equal{}\frac{bc}{a\plus{}b\minus{}c}$, and $ BE\equal{}p\minus{}a$ (Actually, $ E$ is the tangent point of the excircle)

Use the [b][i]Menelaus Theorem[/i][/b], we have:

$ \frac{BQ}{QH}\cdot\frac{HM}{MA}\cdot\frac{AP}{PB}\equal{}1$

So, $ \frac{HM}{MA}\equal{}\frac{QH}{AP}\equal{}\frac{p\minus{}c\minus{}b\cdot\frac{b}{a}}{p\minus{}a}$

And $ AM\minus{}HM\equal{}AH\equal{}\frac{bc}{a}$

So, $ AM\equal{}\frac{bc(b\plus{}c\minus{}a)}{2(b^2\minus{}a^2\plus{}ac)}\equal{}\frac{b(b\plus{}c\minus{}a)}{2(a\minus{}c)}$

$ AM\equal{}AE\equal{}c\minus{}(p\minus{}a)\equal{}p\minus{}b$ iff $ a^2\equal{}b^2\plus{}c^2$

So, $ AM\equal{}AE$

Let C-excircle touches AC at D, we have $ r_c \equal{} AD \equal{} p \minus{} b \equal{} BP$ hence $ I_cBPD$ is parallelogram therfore $ DP\parallel{}BI_c$ in other hand we have $ PQ\parallel{}BI_c$ hence $ P,Q,D$ are collinear so $ \angle MDA \equal{} \frac {\angle B}{2} \equal{} \angle AMD\Longrightarrow AD \equal{} AM$
Let $ PI$ intersect incircle at $ P'$, cause $ I$ and $ F$ are midpoints of $ PP'$ and $ AC$ hence $ CP'$ passes through $ E$ hence $ E$ is tengency point of C-excircle with $ AB$   hence $ AD \equal{} AE \equal{} AM$

Dear Amir and Mathlinkers,
Is there a little typo in your prove ?
Sincerely
Jean-Louis

yeah  BC must be AC in the  one to last line .

Edited.

PS. anymore  :maybe: ?

[quote="apollo"] [color=darkred]The incircle $ w \equal{} C(I,r)$ of the $ A$ - right triangle $ ABC$ touches $ [AB]$ and $ [BC]$ at $ P$ and $ Q$ respectively. Denote

the midpoint $ F$ of $ [AC]$ , $ H\in BC$ for which $ AH\perp BC$  and $ \begin{array}{c} E\in AB\cap IF \\
 \\
M\in PQ\cap AH\end{array}$ . Prove that $ AM \equal{} AE$ .[/color] [/quote]
[color=darkblue][b][u]Proof IV (synthetic).[/u][/b] $ A \equal{} 90^{\circ}\ \Longleftrightarrow\ B \plus{} C\stackrel{(1)}{\ \equal{} \ }A\ \Longleftrightarrow\ (p \minus{} b)(p \minus{} c)\stackrel {(2)}{\ \equal{} \ }p(p \minus{} a)\ \Longleftrightarrow\ r\stackrel {(3)}{\ \equal{} \ }p \minus{} a$ .

$ \blacktriangleright$ From an well-known general property $ BE \stackrel {(*)}{\ \equal{}\ } r$ obtain $ AE \equal{} c \minus{} r\stackrel {(3)}{\ \equal{} \ }c \minus{} (p \minus{} a)$ , i.e. $ \boxed {\ AE \equal{} p \minus{} b\ }$ .

Denote $ T\in PQ\cap AC$ . Thus, $ \frac {TA}{TC} \equal{} \frac {p \minus{} a}{p \minus{} c}\stackrel{(2)}{\ \equal{} \ }\frac {p \minus{} b}{p}$ , i.e. the point $ T$ is the tangent point of $ C$ - excircle with $ AC$ .

Prove easily that (generally) $ m(\widehat {ATM}) \equal{} \frac {A \minus{} C}{2}$ and $ m(\widehat {AMT}) \equal{} \frac B2$ . Therefore, $ A \equal{} 90^{\circ}\stackrel {(1)}{\ \Longleftrightarrow\ }$ $ \widehat {ATM}\equiv\widehat {AMT}$ ,

i.e. $ \triangle AMT$ is $ A$ - isosceles $ \implies$  $ AM \equal{} AT$ $ \implies$ $ \boxed {\ AM \equal{} p \minus{} b\ }$ . In conclusion, $ AE \equal{} AM \equal{} AT \equal{} p \minus{} b$ .

$ (*)$ [b][u]Remark.[/u][/b] I used the following well-known property (you can prove it easily !) :[/color]

[quote][color=darkred]Let $ \triangle ABC$ with the incircle $ w\equal{}C(I,r)$ . Denote $ \left\|\begin{array}{ccc}
M\in [BC] & , & MB\equal{}MC\\\
L\in MI & , & AL\perp BC\end{array}\right\|$ . Then $ AL\equal{}r$ .[/color][/quote]


abstract algebra
group theory
invariant
vector
superior algebra
superior algebra unsolved

Let $ G$ be a finite group and let $ A$ be an abelian subgroup of $ G$. Show that every irreducible representation of $ G$ is of dimension $ \frac{|G|}{|A|}$ at most.

Let $ V$ be the representation space of the representaiton of G. It induces an representation of A. Since A is abelian, every irreducible representation is 1-dimensional. Pick one of these irreducible representation, say $ W$. the sum $ \sum_{g \in G} gW$ is a nonempty G-invariant vector space, which means that $ \sum_{g \in G} gW \equal{} V$ since the representation of G is irreducible. Now $ aW \equal{} W$ for any $ a \in A$, therefore there are at most $ [G: A] \equal{} \frac {|G|}{|A|}$ distinct $ gW$ in the above sum. Thus $ V$ is of at most $ \frac {|G|}{|A|}$ dimension.


calculus
integration
limit
geometry
geometric transformation
reflection
complex analysis

I believe the following is true although I've not proven it:

Let $ f(z)$ have only simple poles $ |p_1| < |p_2| < \cdots < |p_n|$.  Then show:
\[ \text{P.V.}\hspace{-8pt}\mathop\oint\limits_{|z|=p_n} \hspace{-8pt}f(z)dz=\mathop\lim\limits_{\rho\to 0} \mathop\oint\limits_{C_n} f(z)dz=1/2\left\{\mathop\oint\limits_{|z|=a}
    f(z)dz+\mathop\oint\limits_{|z|=b}
    f(z)dz\right\}\]

where $ |p_{n-1}|<a<|p_n|$ and $ |p_n|<b<|p_{n+1}|$ and the contour $ C_n$ appropriately defined.

That is, the principal-valued integral over a closed contour through a simple pole is the average of the integrals over contours excluding the pole and including the pole.

[edit] Ok, I proved it and it's not as profound as I thought.  I changed the expression to reflect the proof.  Still though I think the results are at odds with what one would intuitively think the principal value is.

[quote="shawtoos"]Still though I think the results are at odds with what one would intuitively think the principal value is.[/quote]

why?


[quote="t0rajir0u"]Question:  Primes $p$ also have the property that $p | 2^{p}-2$.  Can we characterize all positive integers $k$ such that $k | 2^{k}-2$?[/quote]
Some call  them  [b]2-PRP  [/b]or (  (weak) [b]probable prime base 2 [/b]) . This, of course contains all primes, some composites (eg 341 - called pseudo-primes ) and all $F_{n}$ , as shown above.

Same as Farenhajt's : (put slightly differently)

[hide]$2^{2^{n}}\equiv-1 \bmod F_{n}$
squaring the both sides
${2^{2^{n+1}}}\equiv 1 \bmod F_{n}$ ,
(  we have $2^{n}\ge n+1$ so   more squaring will still keep it +1.)
$2^{2^{2^{n}}}= 2^{F_{n}-1}\equiv 1 \bmod F_{n}$
QED.[/hide]

BTW, one interesting and Imortant result (Not too difficult to prove - if you have not seen the proof, see if you can do it without looking up) is
$F_{n}$ is[b] prime if and only if
[/b]
$3^{(F_{n}-1)/2}\equiv-1 \bmod F_{n}$


A sequence of integers is obtained by starting with three digits $ d_1$, $ d_2$, and $ d_3$. The fourth term of the sequence is the units digit of $ d_1\plus{}d_2\plus{}d_3$. Each succeeding term is the units digit of the sum of the three previous terms. What numbers belong in the first, second, and third positions respectively to complete this sequence?

Uh, any three terms can create a sequence...

I think he means it so it is x,y,z,x,y,z,.....[/i]

the sequence is __, __, __, 1, 1, 1

I know that from memory

so the 3rd number has to be 9, since 9+1+1=11 with a units digit of 1. So the 2nd has to be 1, and so does the first.


induction
number theory unsolved
number theory

Let be given two integers $a,b$ and a natural number $n$ such that: $n | a^n-b^n$.Prove that:
$n | \frac{a^n-b^n}{a-b}$

[quote="Andreas"]Since $n > 0$ $a^n \equiv b^n \bmod n \Longleftrightarrow a \equiv b \bmod n$.[/quote]

O.O?

I found a mistake in my initial attempted proof. Anyway, I gave it another try, and I realized the problem is not as easy as I first thought it was.

In the following, I assume $n \geq 2$ (the case $n=1$ being trivial).

$a^{n}-b^{n}=(a-b)(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1})$.

Let $p$ be a prime factor of $n$ and $k$ the largest positive integer such that $p^{k}$ divides $n$. It is enough to show that $p^{k}$ divides $Q=\frac{a^{n}-b^{n}}{a-b}$.

We can assume without loss of generality that $p$ divides $a-b$, which implies that $a=b \ mod \ p$ and $Q=nb^{n-1}\ mod \ p=0 \ mod \ p$.

But then $p$ also divides $2a^{n-1}+a^{n-3}b^{2}+\dots+ab^{n-2}+b^{n-1}$,  $2a^{n-1}+a^{n-2}b+a^{n-4}b^{3}+\dots+ab^{n-2}+b^{n-1})$, and by induction $2a^{n-1}+a^{n-2}b+\dots+a^{2}b^{n-3}+b^{n-1}$. From here we deduce that $p$ divides $a^{n-1}-ab^{n-2}=a(a^{n-2}-b^{n-2})$.

Now if $p$ divides $a$, it also divides $b$ by symmetry. It is then clear that $Q$ is divisible by $p^{n-1}$. But $p^{n-1}\geq n$ (trivial), which gives the result.

Otherwise, $p$ divides $a^{n-2}-b^{n-2}$ and $a+b$; thus it also divides $a^{n-3}+b^{n-3}$, $a^{n-4}-b^{n-4}$, and, by induction, $a-b$ if n is odd. Thus (if $n$ is odd) $p$ divides $a$ and $b$, and we can conclude as before.

Now what is happening in the case $n$ even ? We then have $a^{n}-b^{n}= (a^{k_{n}}-b^{k_{n}})[(a^{k_{n}}+b^{k_{n}}) \dots (a^{2^{n-1}k_{n}}+b^{2^{n-1}k_{n}})]$ where $n=2^{n}k_{n}$ with $k_{n}$ odd.

If $p$ is odd, then we can apply the previous result to $a^{k_{n}}-b^{k_{n}}$, since $p$ divides $k_{n}$. This shows that $p^{k}$ divides $Q$.

If $p=2$, it is pretty obvious that $2^{n}$ divides $\frac{a^{n}-b^{n}}{a^{k_{n}}-b^{k_{n}}}$, and hence $Q$. This concludes the problem.

[quote="julien_santini"]We can assume without loss of generality that $p$ divides $a-b$[/quote]

Why?

Otherwise the problem is solved since $p^{k}$ must divide $Q$.
By the way, in my previous post, there is a typo: the last power of $p$ in the post shall be $p^{k}$, not $p^{n}$. I'm going to correct it.


I think this is a little too hard for Getting Started.

And don't post the pdf of the contest, post one problem in one thread.


For those of you that don't have Lockerz....
Lockerz is a site where you accumulate ptzs.
It didn't launch to the public so some features are still limited.
But you can earn points in the following ways:
Inviting people to Lockerz which gives you 2 ptz(invite only site)
Answer a Dailie which gives you 2 ptz(daily poll)
Log in which gives you 2 ptz daily

I recently redeemed a game for my little brother and all games cost 100 ptz so in 25 days u have enough for a game.
Attached are photos for proof.

If you want to join pm me ur email and I will send you an invite, I won't spam.
Visit here for more info http://www.lockerz.com/faq.

This must be a new site... invites only, and no wikipedia page.

@bluecarneal

But one of my friends had ~500 invites for other people.

Yeah, he invited 500 people to lockerz

And when I was looking at this earlier, he was bugging for me to join.

I think you get free stuff, like some other sites. They might get if from ads?

in 2010 they gonna have a shop to start making money

how do you get past http://ptzplace.lockerz.com/?
i am stuck there with no links

it woin't open until the redemption so once the redemption opens then you can go in and pick a prize

Whta is the prize range, like what is the best and worst?

not quite sure what worst is but around the bottom you can get any video game like Modern Warfare 2 but not any things that need like extra stuff like guitar hero
best is like a mac
attached is a photo of some prizes that appeared on redemption day which was on saturday around 4 (i missed itt) :(

You fail to mention how hard it is to actually buy things. When the re-stock on prizes, they only order 1 or 2 of each thing. So to get what you actually want, or to get somehting that is actually high demand, like that TV or Mac or whatever, you need a bot or something similar to constantly click the second that they make it available online. It's incredibly hard to actually get a good prize.

You might show that bakugan DS game as proof that it isnt htat hard to get something, but that isnt something that most people are aiming for. Most people would aim for some other game of console.

Also to get those prizes you need a LOT of those points. and it takes FOREVER to save that much up. Not to mention that even when you get enough points its almost impossible to get anything that has a good value.

your right but i heard rumors that once they launch, theyll have prizes available 24/7 but at higher prices
and im still really angry that i missed the second redemption on 1/31/10

[quote="bluecarneal"]Have you noticed any obscene/profane/violent content on this site?
How many points do you get from playing games?[/quote]

......

[url]http://www.lockerz.com/ptzredemptionplace[/url]

just typed in ptzredemptionplace at the end...... its funny.

ok none of those bad stuff but the site has limited features because it hasn't yet launched.
they don't have games yet, not sure if they will either
check out their blog to see what kind of stuff is on it: http://blog.lockerz.com/
they're about to launch their video section where u can get points for watching videos
i believe they make their on "webisodes" like a cooking one that they gave some pictures to on their blog at http://blog.lockerz.com/?p=1898

So they pay you for watching their videos, rather than having to give up like 5 pts or something?

Are these supposed to be "educational" videos, or just random vids?

they're not random videos and they give you ptz to redeem for prizes

How many points do you have, and what are you currently saving for?

currently i have 370 savin for a ipod nano which is 425 ptz

just wondering does it cost anything to sign up? If not how do they make money?

it doesn't cost anything to sign up but it's invite only so pm me for an invite
read their faq for how they make money

Hey, sorry if I'm spoiling the fun, but, is that legal?
I didn't understand the FAQ well...

I've noticed that footballer is a really big fan of getting free stuff.  Like, look at his blog..."swagbucks"?

:P

[hide="footballer"]wats legal?[/hide]

dar...

hey hey abacal, i got 40 amazon dollars from swagbucks and a backugon game from lockerz which sold for 25 dollars on ebay so i wouldn't be talkin if i were u

Hm...how long did that take you?  :P

Why not program a bot to get you swagbucks?

[quote=footballer]wats legal?[/quote]

u cant really bot swagbucks cuz u need to do searches "naturally" and you need to enter codes that are valid like the one i posted in wave
and that is in the course of 3 months
but you COULD bot club bing pretty easily

Don't use AoPS for advertising.  [b]footballer[/b] earns more "points" every time he invites a new user - this thread is pure advertising.

not for exactly advertising, this thread can be used for discussions like what to get and when we think the redemption is and etc i dont need the ptz i already have z list


People with green eyes tell the truth, people with blue eyes lie.

There are 5 people in a group so that everybody can see everybody. They say the following:

A: I see three people with green and one person with blue eyes.
B: I see 4 people with blue eyes.
C: I see three people with blue and one person with green eyes.
D: (He says nothing.)
E: I see 4 people with green eyes.
What are the colors of the eyes of these 5 people?

[hide="I believe..."]
Person A: Blue
Person B: Blue
Person C: Green
Person D: Green
Person E: Blue[/hide]

if what hamster1800 says is true
3 blue 2 green

a is lying
b is lying
c is right
d is always right or wrong so he's wildcard
e is wrong

so he's right.

I've gotten 3 greens and 2 blues.

A: B
B: G
C: G
D: B
E: G

That solution is wrong. Look at person C.

Person C says he sees 3 people with blue eyes and 1 person with green eyes but you have 2 other people with green eyes.

Yeah, he's got a point there. I think Hamster's solution works

uhh...you said that green eyes are telling the truth. Yours might work if the problem was wrong.

o sorry, yea hamster ur solution is correct, i posted the wrong problem where green and blue switched "truth" and "lying" values.


Ο εγγεγραμμένος κύκλος $ (I),$ δοσμένου τριγώνου $ \bigtriangleup ABC,$ εφάπτεται της πλευράς $ BC,$ στο σημείο $ A'$ και έστω $ A'',$ το συμμετρικό σημείο του $ A',$ ως προς το μέσον $ M$ του $ BC.$ Αποδείξτε ότι $ IM\parallel AA''.$

Κώστας Βήττας.

Σιλουανέ σε ευχαριστώ για το ''link'' και διαπιστώνω ότι η στραβομάρα μου δεν περιγράφεται. :(

Κώστας Βήττας.


geometry
trapezoid
area of a triangle
Heron\u0027s formula
AMC

i was going through a previous amc 1996 quiz (link below) and i got stuck on  problem 22. i was able to use heron's formula to solve it, but is there a faster way to solve it???

[url]http://www.math.ksu.edu/main/handbook/ProblemSets/AMC8Problems/prb1996amc8.pdf[/url]

thank u all in advance :)

[hide="hint"]let D be the bottom right dot.
The area of the triangle can be calculated as:
[ABC] = [ACD] - [ABD] - [BCD]
[/hide]

[hide="best hint"]The area formula base times height over 2 means that you can shift a point on a line parallel to the side across from it, and the area won't change. Let's just move $ A$ up 3 and right 3. Now the triangle's area should be pretty obvious.

(whether this way is better than the others is up to personal judgment; I just felt like continuing the pattern)[/hide]

[quote="mathemonster"]Most Practical hint: to find [ABC] subtract the area of a 1 x 1 right triangle (.5) and the area of a trapezoid with bases 1,4 and height 2, from a large right triangle of area 6. 6 - 5.5 = 1/2 (B) This is verified by Picks theroem A = I (0) + 3/2 -1 = 1/2 (B)[/quote]

most hints aren't complete solutions?

[hide="big hint"]use [url=http://staff.imsa.edu/math/journal/volume2/articles/Shoelace.pdf]Shoelace Method[/url]
[/hide]


Find the minimum of:

$\displaystyle f(x,y,z)= \frac{x^4 y^2 z^2-2x^2 y z+1}{x^4 y^2 z^2}$

if $x,y,z \geq 0$ and $x+y+z=1$

Seamus, give them a chance.... its only been up for 3 1/2 hours and its somewhere between 2 and 5 o'clock in america... most of them are still in school...

And anyway, if noone else posts a solution, i'll post one of mine...

Ours. i would imagine since we go to the same lectures there would be.... a few likenesses in our solutions! :lol:  :lol:  :lol:

factor it, note it's decreasing, apply weighted AM-GM, done.


Let $E$, $F$, and $G$ be Banach spaces. Show that bilinear mapping $A: E \times F\to G$ is bounded if and only if linear mappings $A_{1,z}: E\to G,A_{1,x}: =A(x,z)$ and $A_{2,w}: F\to G,A_{2,w}: =A(w,y)$ are bounded for every $w\in E, z\in F$.

Try to apply the [url=http://en.wikipedia.org/wiki/Banach-Steinhaus_theorem]Uniform boundedness principle[/url]

Banach-Steinhaus looks good but I just don't see how to apply it. Maybe I'm too tired.


Hi, what is the difficulty level of Olymon. Where does it fit on the
math competion totem pole -  > AIME, < CMO, USAMO, IMO, or something.
Most of the Olymon problems are beyond me, can someone suggest a collection of problems that would lead to Olymon. Any advice you have on taking Olymon is welcome.

Thanks,

Do they still have the International Mathematical Talent search up?  If they do then those are generally easier than Olymon.

read previous solutions

its not that hard, first time i did olymon, only got 1 Q

now can regularly get 5+/7 guaranteed in like 1 day

Hello

Please go to here: http://www.math.toronto.edu/barbeau/writingup.pdf for tips about the olymon.  :)

What is Olymon? Can you tell me briefly what it is about and how to register? What grade levels can enter? Where are they held?

[quote="d.lam.86"]What is Olymon?[/quote]
Olymon, or Mathematical Olympiads Correspondence Program, is for the students who are wishing to be on IMO team.
[quote=""d.lam.86"] Can you tell me briefly what it is about and how to register?[/quote]
Each month, there are about 6 to 8(right?) problems. The problems are posted on internet. you can solve them, and send the solution to Prof. E.J. Barbeau. You don't need to resister. You just send the solutions.
[quote=""d.lam.86"]What grade levels can enter?[/quote]
As far as I know, if you are below university, you can enter it.
[quote=""d.lam.86"]Where are they held?[/quote]
This is for whole Canada.

You can see past and current problems and detailed infomation at [url]http://www.cms.math.ca[/url]. Go to competitions-MOCP.

[quote="billzhao"]It's interesting to think about where the name olymon comes from. I've never actually thought about it until a friend of mine, who doesn't even do math olympiads, saw me doing Olymon, and asked me if Olymon stood for "[b]Oly[/b]mpiad of the [b]Mon[/b]th." Needless to say, I was quite amazed.[/quote]

Really?  That seemed kinda obvious to me (Olymon = Olympiad monthly).  I guess sometimes you just never think of those things.

On a completely related note, my brother didn't know what L and R stood for on SuperNES, N64, & Gamecube controllers, until I told him that L was on the left and R was on the right... he was pretty amazed too.

[quote="d.lam.86"][quote="lightrhee"]
Each month, there are about 6 to 8(right?) problems. The problems are posted on internet. you can solve them, and send the solution to Prof. E.J. Barbeau. You don't need to resister. You just send the solutions.[/quote]

Do they require just the answer or a full detailed solution?[/quote]
You need detailed solution. Each solutions are marked out of 7, like every olympiads. But if you could get a problem, you can just send partial solution.

[quote="TheDreamer"]Just curious, approximately how many people submit OLYMON each month?[/quote]
Ask prof. Ed Barbeau. I think he is the only one who knows the accurate number.

[quote="8th Army"]so you [b]have[/b] to do olymon in order to get into IMO team?[/quote]

No, but it is advantageous as you can get invited to Winter Camp based on Olymon (in addition to a few other things, e.g. CMO, COMC)