trigonometry
geometry
geometric transformation
ratio
parallelogram
circumcircle
cyclic quadrilateral

A convex pentagon $ABCDE$ is given so that $B=E=90^{\circ}$ and $\widehat{BAC}\equiv \widehat{EAD}$. Let $O$ be the intersection point of $BD$ and $CE$. Prove that the lines $AO$ and $BE$ are perpendicularly.

The problem proposed also here and vittasko gave a beautiful solution.
http://www.mathlinks.ro/Forum/viewtopic.php?t=78598

[color=darkred][b]An equivalent enunciation.[/b] In outside of the triangle $ABC$ construct the triangles $MBC\ ,\ CRA\ ,\ APB$ so that

$MB=MC\ ;\ PA\perp PB\ ;\ RA\perp RC\ ;\ \widehat{ABP}\equiv\widehat{ACR}\equiv \widehat{BCM}\ .$ Denote : the middlepoint $D$ of the side $[BC]\ ;$

the intersection $I\in CP\cap BR\ .$ Prove that $DP=DR\ ,\ I\in AM$ and $AM\perp PR\ .$[/color]

[color=darkblue][b]Proof[/b] (shortly, synthetically and easy metrically). Denote the middlepoints $E$, $F$ of the sides $AC$, $AB$ respectively and $m(\widehat{BCM})=x\ .$

$\boxed{DP=DR}\blacktriangleright\ ED=FP=\frac{c}{2}\ ,\ ER=FD=$ $\frac{b}{2}\ ,\ m(\widehat{PFD})=m(\widehat{DER})=$ $A+2\cdot (90^{\circ}-x)$ $\Longrightarrow$ $\triangle PFD\equiv\triangle DER\ \mathrm{(s.a.s)}$ $\Longrightarrow$ $DP=DR\ .$

$\boxed{\ \ I\in AM\ \ }\blacktriangleright$ The pairs of rays $\{\ (AP\ ,\ (AR\ \}$, $\{\ (BM\ ,\ (BP\ \}$, $\{\ (CM\ ,\ (CR\ \}$ are isogonally for the angles $\widehat{BAC}$, $\widehat{CBA}$, $\widehat{ACB}$ respectively.

From the well-known [u]Jakob's theorem[/u] obtain immediately that the lines $AM$, $BR$ and $CP$ are concurrently.

$\boxed{AM\perp PR}\blacktriangleright$ Prove easily that : $MB=MC=\frac{a}{2\cdot\cos x}\ ,\ \frac{AP}{c}=\frac{AR}{b}=\sin x\ ,\ \frac{PB}{c}=\frac{RC}{b}=\cos x$ $\mathrm{\ ,\ }$ $\underline{AP^{2}-AR^{2}=\left(c^{2}-b^{2}\right)\cdot\sin^{2}x}\ \ (1)\ .$

$MP^{2}=BM^{2}+BP^{2}-2\cdot BM\cdot BP\cdot \cos (B+2x)\mathrm{\ ,\ }$ $MR^{2}=CM^{2}+CR^{2}-2\cdot CM\cdot CR\cdot\cos (C+2x)\mathrm{\ ,\ }$ $BM=CM$ $\Longrightarrow$

\begin{eqnarray*}MP^{2}-MR^{2}& = & BP^{2}-CR^{2}+2\cdot BM\cdot \left[CR\cdot\cos (C+2x)-BP\cdot\cos (B+2x)\right]\\\\ & = & (c^{2}-b^{2})\cdot\cos^{2}x+2\cdot\frac{a}{2\cdot\cos x}\cdot\left[b\cdot\cos x\cdot\cos (C+2x)-c\cdot cos x\cdot\cos (B+2x)\right]\\\\ & = & (c^{2}-b^{2})\cdot \cos^{2}x+ab\cdot\cos (C+2x)-ac\cdot\cos (B+2x)\\\\ & = & (c^{2}-b^{2})\cdot\cos^{2}x+(ab\cdot\cos C-ac\cdot\cos B)\cdot\cos 2x-(ab\cdot\sin C-ac\cdot\sin B)\cdot\sin 2x\\\\ & = & (c^{2}-b^{2})\cdot\cos^{2}x-(c^{2}-b^{2})\cdot\cos 2x=(c^{2}-b^{2})\cdot\sin^{2}x\Longrightarrow \underline{MP^{2}-MR^{2}=(c^{2}-b^{2})\cdot\sin^{2}x}\ \ (2)\ .\end{eqnarray*}

From the relations $(1)$ and $(2)$ obtain $AM\perp PR\ .$

[b]Remark.[/b] I used the simple relations : $ab\cdot\sin C=ac\cdot\sin B=2S\ \ ;\ \ ab\cdot\cos C-ac\cdot\cos B=b^{2}-c^{2}\ .$[/color]

Dear all my friends.

I would like to present the translation of the solution, as it has already been posted in the link mentioned by Silouan.

http://www.mathlinks.ro/Forum/viewtopic.php?t=78598

[b][size=117]PROPOSITION.[/size][/b] - A convex pentagon $ABCDE$ is given, such that $\angle B = \angle E = 90^{o}$ and $\angle BAC = \angle EAD.$ If $O,$ is the intersection point of diagonals $BD$ and $CE,$ prove that the lines $AO\perp BE.$

[b][size=117]SOLUTION.[/size][/b] – $($ In my drawing, $AC =7.5,$ $AD = 7.8,$ $CD = 6.3,$ $AB = 4.5$ $).$

The right triangles $\bigtriangleup ABC$ $($ $\angle ABC = 90^{o}$ $)$ and $\bigtriangleup AED$ $($ $\angle AED = 90^{o}$ $),$ are similar and so, we have that $\angle ACB = \angle ADE$ $,(1)$

Through vertices $C,$ $D,$ of the triangle $\bigtriangleup ACD,$ we draw two lines isogonal to $CB,$ $DE$ respectively, with respect to the angles $\angle ACD,$ $\angle ADC$ and we denote as $A',$ their intersection point. Because of $(1),$ we have that $\angle CDA' = \angle DCA'$ and so, the triangle $\bigtriangleup A'CD,$ is isosceles $($ $A'C = A'D$ $).$

We have now, the configuration of the triangle $\bigtriangleup ACD$ and based on it’s sidelines $CD,$ $AC,$ $AD,$ erected outwardly, the triangles $\bigtriangleup A'CD,$ $\bigtriangleup BAC,$ $\bigtriangleup EAD$ respectively, whose their sidelines through the vertices of $\bigtriangleup ACD,$ are isogonal with respect to the corresponded anlge $($ The lines $AB,$ $AE,$ are isogonal with respect to the angle $\angle CAD,$ etc $).$

So, by the Jacobi’s theorem, we have that the lines $AA',$ $BD,$ $CE,$ are concurrent and hence, we conclude that the line $AA'$ passes through the point $O,$ as the intersection point of $BD,$ $CE.$

We denote as $F,$ $G,$ the midpoints of the sidelines $AC,$ $AD$ of $\bigtriangleup ACD,$ respectively and as $B',$ $E',$ the intersection points of the midperpendiculars of $AC,$ $AD,$ from the lines $AB,$ $AE,$ respectively.

We can easy to prove that $B'E'\parallel BE$ and so, it is enough to prove that the line $AA',$ is perpendicular to $B'E'.$

We denote as $K,$ $L,$ the intersection points of the segment line $FG,$ from the lines $A'C,$ $A'D,$ respectively. From $CD\parallel KL,$ $\Longrightarrow$ $\angle KLA' = \angle CDA' = \angle DCA' = \angle LKA'$ $,(2)$

So, the triangle $\bigtriangleup A'KL$ is isosceles $($ $A'K = A'L$ $)$ and if $K',$ $L',$ are the midpoints of $A'C,$ $A'D$ respectively, we have that $KK' = LL'$ $,(3)$ $($ because of $A'C = A'D$ $).$

From the isosceles triangle $\angle up B'AC$ $($ $B'A = B'C$ $)$ and the cyclic quadrilateral $BB'CF,$ we have that $\angle CB'F = \angle BB'F = \angle ACB$ $,(4)$

But $\angle ACB = angle DCA'$ $,(5)$ as isogonal lines with respect to the angle $\angle ACD.$

From $(2),$ $(4),$ $(5),$ $\Longrightarrow$ $\angle CB'F = \angle LKA'$ $\Longrightarrow$  $\angle CB'F = \angle CKF$ $,(6)$

From $(6),$ we conclude that the quadrilateral $B'KCF$ is cyclic and so, we have that $\angle B'KC = \angle B'FC = 90^{o}.$

Hence, the line $B'K$ is perpendicular to $A'K$ and similarly we can prove that the line $E'L,$ is perpendicular to $A'L.$

From the triangle $\bigtriangleup A'B'C,$ by the second theorem of medians, we have that $(A'B')^{2}-(CB')^{2}= 2(A'C)\cdot (KK')$ $,(7)$

Similarly from the triangle $\bigtriangleup A'E'D,$ $\Longrightarrow$ $(A'E')^{2}-(DE')^{2}= 2(A'D)\cdot (LL')$ $,(8)$

From $(7),$ $(8),$ $\Longrightarrow$ $(A'B')^{2}-(AE')^{2}= (CB')^{2}-(DE')^{2}$ $,(9)$ $($ because of $A'C = A'D$ and $KK' = LL'$ $).$

From $(9)$ $\Longrightarrow$ $(A'B')^{2}-(AE')^{2}= (AB')^{2}-(AE')^{2}$ $,(10)$ $($ because of $CB' = AB'$ and $DE' = AE'$ $).$

From $(10),$ we conclude that the line $AA',$ is perpendicular to $B'E'$ and the proof is completed.

[b][size=117]NOTE.[/size][/b] - From all the above, it is clear that the proposition is also true and when the given pentagon $ABCDE,$ is a non-convex one.

Kostas Vittas.

PS. – I am waiting for some other ( simpler ) proofs, to be posted here next time. The above proof, which ( I think ) is presented first time in Hellenic publication, is dedicated to all the members of the Hellenic Mathlinks Forum.

This is very nice problem! I have a solution by Ceva theorem as following
Called lines $CP\cap AR=P',BR\cap AP=R'$ and called $AH$ is altitulde of triangle $APR$ we will show $CP,BR,AH$ are concurrent by Ceva theorem!
Indeed let $\angle PAB=\angle RAC= x$ we consider
$\frac{HP}{HR}\cdot\frac{R'A}{R'P}\cdot\frac{P'R}{P'A}$
Using notation as my fig
$\frac{HP}{HR}=\frac{AP\sin A1}{AR\sin A2}=\frac{AP\cos P1}{AR\cos R1}$
$\frac{R'A}{R'P}=\frac{S_{RR'A}}{S_{RR'P}}=\frac{S_{BR'A}}{S_{BR'P}}=\frac{S_{BR'A}-S_{RR'A}}{S_{BR'P}-S_{RR'P}}=\frac{S_{APC}}{S_{BPR}}=\frac{APAC\sin(x+A)}{PRPB\sin(90-R1)}=\frac{APAC\sin(x+A)}{PRPB\cos R1}$
Similarly
$\frac{P'R}{P'A}=\frac{RPRC\cos R2}{ABAR\sin(x+A)}$
And by $\triangle APB\sim \triangle ARC$ we see easily $\frac{HP}{HR}\cdot\frac{R'A}{R'P}\cdot\frac{P'R}{P'A}=1$ easily.
There is other way to look this problem I will post later !

[quote="Huyền Vũ"]The problem is very interesting.This is the solution
Let $F,K$ be the points on $EC,BD$ that  $AF \bot EC$ $AK \bot BD$ $(1)$
From (1) and  $AB \bot BC$,  $AE \bot ED$ we have $A,B,C,F$ are cyclic  and $A,E,D,K$ are cyclic
So $\angle AFB=\angle ACB$ and $\angle ADE=\angle AKE$ $(2)$
Otherwise triangle $AED$ and $ABC$ are similar so $\angle ACB= \angle ADE$ $(3)$
From (2) and (3) $\angle AKE=\angle AFB$
Thus $90+\angle AKE=90+\angle AFB$ or $\angle EFB=\angle EKB$
So $E,F,K,B$ are cyclic
So $\angle EBO=\angle KFO=\angle KAO$ ($F,A,K,O$ are cyclic)
this means $A,B,K,I$ are cyclic
So $\angle AIB=\angle AKB=90$ or $AO,EB$ are perpendicularly $q.e.d$[/quote]

 [color=red]Yes !

 This solution is delightfull and deserves a gold medal !  :lol: [/color]

 [u]Babis[/u]

Thank you dear Huyen Vu, for a very nice solution.

I was sure that this problem has a synthetic proof, simpler and sorter than the complicated one, I have been presented.

Best regards, Kostas vittas.

Interesting and new results for me, dear Huyen Vu.

As a consequence of the result that the point $T\equiv CQ\cap DP,$ lies on the segment line $AO,$ I see that the segment line $AJ,$ passes through the point $S,$ as the intersection point of $BC,$ $DE.$

Also, the segment lines $CD,$ $BE,$ $PQ,$ are concurrent at one point, so be it, $R.$

I have some ideas about the proofs, but I must check more carefully my notes, before a definite answer.

Kostas Vittas.

[quote="Huyền Vũ"]Some more things about this nice problem
let $P$ be the projection of $D$ on $AB$
let $Q$ be the projection of $C$ on $AE$
$CQ$ meets $DP$ at $T$. $DQ$ meets $CP$ at $J$. We have some following result
[b]1)[/b] $A,I, T,O$ are collinear
[b]2)[/b]The line $AO$ are the isogonal of the line $AJ$ with respect  to the line $AB$ and $AE$
I think they are beautiful results about isogonal lines.[/quote]

[b][size=109]PROOF OF THE RESULT 1.[/size][/b]

$($ In my drawing $AC = 9.5,$ $AD = 8.8,$ $CD = 5.5,$ $AB = 7.7$ $)$

We will prove that the point $T\equiv CQ\cap DP,$ lies on the segment line $AO.$

We denote as $K,$ the intersection point of $AC,$ $DE$ and as $L,$ the one of $AD,$ $BC.$

We consider the pencil $D.ACBP,$ which is intersected from the segment line $BL\parallel DP$ and so, we have that $(D.ACBP) = (L,C,B)$ $,(1)$

$($ We symbolize as $(D.ACBP),$ the double ratio of the pencil $D.ACBP$ and as $(L,C,B),$ the simple ratio $\frac{BL}{BC}$ $).$

Similarly the pencil $C.ADEQ,$ is intersected from the segment line $EK\parallel CQ$ and so, we have $(C.ADEQ) = (K,D,E)$ $,(2)$

It is easy to show that the triangles $\bigtriangleup ABL,$ $\bigtriangleup AEK$ are similar, because of they have $\angle BAL = \angle EAK.$

Because of now, $\angle BAC = \angle EAD,$ we conclude that $\frac{BL}{BC}= \frac{EK}{ED}$ $\Longrightarrow$ $(L,C,B) = (K,D,E)$ $,(3)$

From $(1),$ $(2),$ $(3)$ $\Longrightarrow$ $(D.ACBP) = (C.ADEQ)$ $,(4)$

From $(4)$ and because of the pencils $D.ACBP,$ $C.ADEQ,$ have the segment line $CD,$ as their common ray, we conclude that the points $A\equiv DA\cap CA,$ $T\equiv DP\cap CQ$ and $O\equiv DB\cap CE,$ as the intersection points of the other their homologous rays, are collinear.

Hence, the point $T,$ lies on the segment line $AO$ and the proof of the [b][size=109]result 1,[/size][/b] is completed.

[b][size=109]NOTE.[/size][/b] - Because of $T\equiv AO\cap CQ\cap DP,$ by Desarques’s theorem, we conclude that the triangles $\bigtriangleup APQ$ and $\bigtriangleup ODC,$ are perspective and so, we conclude that the points $B\equiv AP\cap OD,$ $E\equiv AQ\cap OC$ and $R\equiv PQ\cap CD,$ are collinear.

Because of now, the collinearity of the points $B,$ $E,$ $R,$ we conclude that the triangles $\bigtriangleup APQ$ and $\bigtriangleup SCD,$ where $S\equiv BC\cap DE,$ by Desarques’s theorem are perspective and hence, the segment lines $AS,$ $PC,$ $QD,$ are concurrent at one point, so be it $J.$

That is, the point $J\equiv PC\cap DQ,$ lies on the segment line $AS.$

[b][size=109]PROOF OF THE RESULT 2.[/size][/b]

We will prove that the segment lines $AO,$ $AJ\equiv AS,$ are conjugates with respect to the angle $\angle BAE.$

From all the discussion about the problem in this topic, it is clear that all the results, have not been correlated with some property of the pentagon and I think is better to continue by considering the correct alternative configuration of the triangle $\bigtriangleup ACD,$ as a reference triangle and the similar right triangles $\bigtriangleup ABC$ and $\bigtriangleup AED$ $($ $\angle ABC = \angle AED = 90^{o}$ and $\angle BAC = \angle EAD$ $),$ erected on it’s sidelines $AC,$ $AD$ respectively, outwardly to it.

It has already been proved $($ see the Huyen Vu’s best solution $),$ that $AO\perp BE.$

We consider the triangle $ACD$ and it is easy to show that the right triangles $\bigtriangleup AQC,$ $\bigtriangleup APD,$ which they are erected on it’s sidelines $AC,$ $AD$ respectively, inwardly to it, are similar because of $\angle QAC = \angle PAD.$

So, by the same way as before, we have the result that $AJ\perp PQ,$ where $J\equiv DQ\cap CP.$

$($ The problem, as I have been mentioned, is also true for the non-convex pentagon $AQCDP$ $).$

$\bullet$ From the cyclic quadrilaterals $APDE,$ $AQCB,$ we have that $\angle AEP = \angle ADP$ $,(1)$ and $\angle ABQ = \angle ACQ$ $,(2)$

But $\angle ADP = 90^{o}-\angle PAD = 90^{o}-\angle QAC = \angle ACQ$ $,(3)$

From $(1),$ $(2),$ $(3)$ $\Longrightarrow$ $\angle AEP = \angle ABQ$ or $\angle QEP = \angle QBP$ and hence, the quadrilateral $BEQP,$ is cyclic.

That is we have that the segment lines $BE,$ $PQ,$ are antiparallel each other with respect to the angle $\angle BAE.$

Hence, because of $AO\perp BE$ and $AS\perp PQ,$ we conclude that the segment lines $AO,$ $AS\equiv AJ,$ are isogonal conjugates with respect to the angle $\angle BAE$ and the proof of the [b][size=109]result 2,[/size][/b] is completed.

Kostas Vittas.

[quote="Huyền Vũ"]Some more results
$BC$ meets $DE$ at $S$. $BQ$ meets $PE$ at $V$
[b]3)[/b]$S,M,T,V$ are collinear where $M$ is the midpoint of $CD$
[b]4)[/b]$AV$ and $CD$ are perpendicular[/quote]

[b][size=109]PROOF OF THE RESULT 3.[/size][/b]

We will prove that the points $V\equiv BQ\cap PE,$ $T\equiv CQ\cap DP$ and $S\equiv BC\cap DE,$ are collinear.

It has already been prove that the segment lines $CD,$ $BE,$ $PQ,$ are concurrent at the point $R.$

So, because of the collinearity of the points $B,$ $E,$ $R,$ by Desarques’s theorem, we conclude that the triangles $\bigtriangleup VPQ$ and $\bigtriangleup SDC,$ are perspective and hence, the segment lines $VS,$ $PD,$ $QC,$ are concurrent at one point.

That is the segment line $VS,$ passes through the point $T,$ as the intersection point of $PD,$ $QC.$

Hence, the points $V,$ $T,$ $S,$ are collinear and the proof of the [b][size=109]result 3,[/size][/b] is completed.

[b][size=109]NOTE.[/size][/b] – It is clear that the segment line $VTS,$ passes through the midpoint of the segment $CD,$ because of the parallelogram $CSDT.$

[b][size=109]PROOF OF THE RESULT 4.[/size][/b]

The circucircles $(O_{1}),$ $(O_{2}),$ of the cyclic quadrilaterals $ABCQ,$ $AEDP$ respectively, are intersected at one point $($ other than $A$ $),$ so be it $A_{1},$ which is easy to prove that lies on the segment line $CD$ $($ $\angle AA_{1}C = \angle ABC = 90^{o}= \angle AED = \angle AA_{1}D$ $).$

If now, we denote as $(O_{3}),$ the circumcircle of the cyclic quadrilateral $BEQP,$ it is easy to show that the segment lines $AA_{1},$ $BQ,$ $EP,$ as the radical axes of $(O_{1}),$ $(O_{2}),$ $(O_{3}),$ taken per two of them, are concurrent at one point.

Hence, the segment line $AA_{1},$ passes through the point $V,$ as the intersection point of $BQ,$ $EP.$

We conclude now, that $AV\perp CD$ and the proof of the [b][size=109]result 4,[/size][/b] is completed.

Kostas Vittas.

You've done well, Kostas Vittas
Some things about [b]result 1[/b]:
We have $\angle QTA=\angle QPA=\angle QEB$ ($A,Q,T,P$ are cyclic and $B,E,Q,P$ are cyclic)
but $QT$ and $EA$ are perpendicular so $AT$ and $EB$ are perpendicular $(1)$
So we can use (1) and AO,BE are perpendicular to have [b]result 1[/b]
Also we can use Kostas Vittas's proof for [b]result 1[/b] and (1) to prove the original problem.This is another proof for original problem

To prove $PQ,BE,CD$ concurrent at $R$ we can also use [b]result 4[/b]


Madali lang yan eh

just look at the binomial expansion of (x+y)^100
 =  then you will see all the 100Ck  for k:1,2,..100

so if you want to get the sum.  set x=1, y =1 so you will make all the x's and y's to become 1 and you will get what you need
which is 2^100


[i]Ξεκινάω τη στήλη της άλγεβρας με μια άσκηση που έφτιαξα στις ανισότητες , επιπέδου Θαλή για τη Γ΄τάξη ή επιπέδου Ευκλείδη για τη Β΄τάξη.Αν δε βρείτε χρόνο να την κοιτάξετε , κάποια μέρα θα στείλω και την υπόδειξη ή τη λύση.Η άσκηση κάνει και για Βαλκανιάδα junior.Αυτό βέβαια αποτελεί και μια υπόδειξη για τους senior ,αλλά δεν πειράζει![/i]

[color=red]ΑΣΚΗΣΗ - 1[/color]

 Αν $a , b , c >0$ και $a+b+c =1$, να αποδείξετε ότι :

   $\frac{a^{2}+a}{2a+b^{2}+c^{2}}+\frac{b^{2}+b}{2b+c^{2}+a^{2}}+\frac{c^{2}+c}{2c+a^{2}+b^{2}}\leq \frac{3}{2}$

  [u] Μπάμπης[/u]

[color=red]ΑΣΚΗΣΗ 1 - ΛΥΣΗ[/color]

Kathws $a+b+c=1$ ara $2a=2-2b-2c, 2b=2-2a-2c, 2c=2-2a-2b$ ki etsi i pros apodeiksi anisotita (an antikatastisoume ta parapanw ston paranomasti) ginetai:

${\frac{a(a+1)}{(b-1)^{2}+(c-1)^{2}}+\frac{b(b+1)}{(c-1)^{2}+(a-1)^{2}}}+\frac{c(c+1)}{(a-1)^{2}+(b-1)^{2}}\leq \frac{3}{2}$

Apo tin anisotita C-S exoume $(b-1)^{2}+(c-1)^{2}\geq \frac{(a+1)^{2}}{2}$ kai douleuontas kiklika pairnoume gia tous ipoloipous $2$ paranomastes tis analoges sxeseis ki etsi i pros apodeiksi anisotita ginetai:

$\frac{2a}{a+1}+\frac{2b}{b+1}+\frac{2c}{c+1}\leq \frac{3}{2}$

kai ektelwntas ton metasximatismo $a+1=x, b+1=y, c+1=z$ (etsi $x+y+z=4$) pairnoume tin

$\frac{2(x-1)}{x}+\frac{2(y-1)}{y}+\frac{2(z-1)}{z}\leq \frac{3}{2}$

kai kanontas tis prakseis $6-2\big(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\big)\leq\frac{3}{2}$
ap'opou arkei na deiksoume oti $2\big(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\big)\geq \frac{9}{2}$

pou prokiptei apo tin anisotita arithmitikou-armonikou kathws

$\frac{x+y+z}{3}\geq \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$

H isotita profanws isxiei otan $x=y=z=\frac{4}{3}$ 'h oti $a=b=c=\frac{1}{3}$

Alexandros

Κι εγω το ιδιο εκανα!Αλλα δυστυχως με προλαβες cretanman,πανω που προσπαθουσα να την γραψω...επειδη ειμαι νεο μελος εκανα 100 ωρες... :oops:

[quote="cretanman"]
$\frac{2a}{a+1}+\frac{2b}{b+1}+\frac{2c}{c+1}\leq \frac{3}{2}$[/quote]

Απλά γράφω μια άλλη λύση για αυτή χωρίς να θέσουμε .
Αφαιρούμε το 3 και από τα δύο μέλη και η προς απόδειξη γίνεται
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq\frac{9}{4}$
Από Andreescu όμως
$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq\frac{9}{a+b+c+3}=\frac{9}{4}$

 :wink:

Λοιπόν να προτείνω κι εγώ άλλη λύση
1/2α+β^2+γ^2=<1/α^2+γ^2+γ^2 αρα η αποδεικτέα γινεται
2(α^2+α+β^2+β+γ^2+γ)=<3(α^2+β^2+γ^2+1)
που ισχύει μετά τις λίγες πράξεις όπως θα διαπιστώσετε  :D

[quote="ambros"]Λοιπόν να προτείνω κι εγώ άλλη λύση
1/2α+β^2+γ^2=<1/α^2+γ^2+γ^2 [/quote]

Νομίζω ότι δεν ισχύει βάλε για παράδειγμα $a=b=c=\frac{1}{3}$

[quote="ambros"]Λοιπόν παιδιά εκατσα κι ελυσα την ανισότητα ΣΩΣΤΑ ΣΙΛΟΥΑΝΕ  γιατί δεν μπορεί συνέχεια να κάνω λάθη λοιπόν
αρκεί να δείξω ότι
α(α+1)/β^2+γ^2+2α=<3α/2
η οποία μετά απο πράξεις
2βγ=<2α^2+(β+γ)^2
 :D  :D  :D  :D  :D[/quote]
Eίσαι σίγουρος αυτή τη φορά?Καταρχήν δε γράφεις με latex και δε βγάζω τίποτα!Επιπλέον δε είμαι σίγουρος για την ορθότητα των πράξεων...
Μπορείς να το κοιτάξεις σε παρακαλώ?

Μπορείς να γράψεις όλη τη λύση σου σε ένα ποστ ?
Θα το κάνω εγώ μετά σε Latex . ok ?

[color=red]ΑΛΓΕΒΡΑ 2 - ΕΚΦΩΝΗΣΗ[/color]

Estw poliwnimo $P_{n}(x)=x^{n}-a_{n-1}x^{n-1}-\cdots-a_{1}x-a_{0}$ me $n\geq 2$, opou $a_{0},a_{1},\ldots,a_{n-1}>0$ kai $a_{0}+a_{1}+\cdots+a_{n-1}=1$. Na prosdiorisete oles tis rizes $\rho$ tou $P_{n}(x)$ me $|\rho|=1$, kathws kai ton vathmo pollaplotitas kathe tetoias rizas.

Alexandros

Stelio poli swsti i lisi sou!!

To kako einai oti i askisi stin ekfwnisi den dieukrinizei ean thelei na lithei sto $\mathbb{R}$ 'h sto $\mathbb{C}$ dioti sto $\mathbb{C}$ diskoleuoun ligo ta pragmata kathws psaxneis oles ekeines tis rizes pou vriskontai panw ston monadiaio kiklo pou ikanopoioun tin dosmeni poliwnimiki eksiswsi. Einai nomizw arketa eukoli an prokeitai gia to $\mathbb{R}$ giati mporeis akoma kai me paragwgous na deikseis oti to $P_{n}'(1)\neq 0$ opote to $1$ einai riza pollaplotitas $1$.

Pantws epeidi den exw tin lisi apo to diagwnismo, tha tin meletisw ligaki stous migadikous giati einai arketa endiaferousa (an fisika prokiptei apotelesma). An thelete koitakste to! (H ekfwnisi pou exw einai daxtilografimeni se H/Y kai einai i episimi kolla pou dwthike tote ston diagwnismo).

Alexandros

Αλέξανδρε, στη λύση μου θεωρώ ότι το πρόβλημα αναφέρεται στο $\mathbb C$ και με $|a|$ εννοώ το μέτρο του μιγαδικού $a$ και όχι απαραίτητα την απόλυτη τιμή. Έτσι καταλήγω ότι η μόνη [b]μιγαδική[/b] ρίζα $r$ του $P_{n}(x)$ με $|r|=1$ είναι η $r=1$.

P.S. Νομίζω ότι είναι προφανές ότι το πρόβλημα αναφέρεται στο $\mathbb C$, αλλιώς θα αρκούσε να εξετάσουμε μόνο τις περιπτώσεις $r=1$ και $r=-1$.

[quote]Η ισότητα στην πρώτη ανισότητα ισχύει όταν $r^{n}(a_{n-1}r^{n-1}+a_{n-2}r^{n-2}+...+a_{0}) \in \mathbb R_{+}$[/quote]

Diladi isxirizesai oti h isotita $|z_{1}-z_{2}|=|z_{1}|-|z_{2}|$ isxiei otan $z_{1}z_{2}\in \mathbb R_{+}$ ? Auto den ennoeis 'h kanw kapoio lathos? Exei noima na milame gia diataksi stous migadikous an den kseroume oti einai pragmatikoi? An p.x. $z_{1}=5i$ kai $z_{2}=3i$ tote isxiei h parapanw isotita.

Mporeis na to analiseis ligo perissotero an ennoeis kati diaforetiko apo auto pou katalava?

Alexandros

Εγώ απλά είπα (ή μάλλον εννοούσα...) ότι $|z_{1}-z_{2}|=|z_{1}|-|z_{2}|$ αν και μόνο αν $z_{1}\cdot w \in \mathbb R_{+}$, όπου $w$ είναι ο συζυγής μιγαδικός του $z_{2}$.

Το πρόβλημα είναι ότι δεν ήξερα πως συμβολίζεται ο συζυγής ενός μιγαδικού στο $Latex$ (...η παύλα δηλαδή πάνω από τον αριθμό) και έγραψα κατευθείαν αυτό που προκύπτει τελικά.  :roll:
Η ουσία πάντως της λύσης δεν επηρεάζεται. Απλά βάλε όπου $r^{n}$ τον συζυγή του. Από τη δεύτερη κυρίως ισότητα προκύπτει ότι $r \in R_{+}$ και συνεπώς ο συζυγής του $r^{n}$ ισούται με το $r^{n}$ και άρα ισχύει και το πρώτο που έγραψα.

Όσον αφορά τώρα τη διάταξη στους μιγαδικούς που λες, προφανώς δεν έχει νόημα. Εγώ όμως χρησιμοποιώ τα μέτρα των μιγαδικών και όχι τους μιγαδικούς οπότε δεν υπάρχει πρόβλημα.

P.S. Καλά έκανες που το διευκρίνισες...  . :)

Ki egw eixa tin aporia auti otan tin eida RAKAR, alla eutixws exei texnasmataki kai vgainei poli omorfi xwris polles prakseis (kai xwris eksiswseis 16ou vathmou...)! Asxolisou... einai arketa sinarpastiki kai aksizei...! Na anaferw oti to pedio orismou paizei idiaitero rolo sti sigekrimeni askisi gia to texnasma pou xreiazetai! Min pw perissotera kai sas tin xalasw!

Alexandros

cretanman

την έλυσα θέτοντας όπου χ = 2συν8ω , με 8ω ε [ -π/2 , π/2] .
Έτσι την έκανες και εσύ ?

Νίκος Ρ

Eisai swstos Niko! :)

Etsi akrivws... An theleis grapse tin pliri lisi gia to pws to skeftikes na to kaneis auto giati exei simasia na to doun didaktika kapoioi oi opoioi den exoun ksanadei to texnasma auto (gia alles paromoies askiseis kathws episis gia na mi fanei teleiws ouranokatevati idea giati den einai...) kathws episis kai poso evgales to $x$. Nomizw oti kai apo ekei kai epeita exei endiaferon i askisi. Tin idea omws tin anakalipses. Etsi akrivws tin exw doulepsei ki egw kai vrika $2$ trigwnometrikes liseis gia to $x$.

Alexandros

Y.G. Xronia polla kai gia xthes, opws episis kai ston Niko Skompri (nickolas)

Πολύ καλή ιδέα Νίκο .Η εξίσωση τότε καταλήγει αν δεν έκανα λάθος στην
$cos\ m+\sqrt{3}sin\ m=2cos\ 8m$  όπου m είναι το θ που αναφέρεις παραπάνω .Τώρα αυτή έχει νομιζω δύο λύσεις (δεν είμαι σίγουρος όμως  :maybe: )

[quote]Xronia polla kai gia xthes[/quote]

Ευχαριστώ Αλέξανδρε .

Ναι εκεί καταλήγει Σιλουανέ .
Στο συνημμένο αρχείο την έχω ολοκληρωμένη ……….

Niko swstotata,

Epitrepse mou na kanw mia diorthwsi stin 9h grammi tis lisis sou oti pairnoume tin $|\cos{\omega}|+\sqrt{3}|\sin{\omega}|=2\cos{\omega}$ (dioti mporei i gwnia $\omega$ na einai arnitiki ($-\pi /16 \leq \omega \leq 0$) kai to  imitono tote einai arnitiko).

Parathetw kai tin diki mou lisi gia to pws skeftika na tin antimetwpisw arxika. Nomizw oti etsi diafainetai kai o logos gia ton opoio kaneis ton metasximatismo pou anaferei o Nikos.

As paroume $A=\sqrt{2+\sqrt{2+x}}, \ \ A\geq 0$

Tote $A^{2}-2=\sqrt{2+x}$ ara $x=(A^{2}-2)^{2}-2$

Tote h arxiki eksiswsi dinei
$\sqrt{2+A}+\sqrt{3}\sqrt{2-A}=2[(A^{2}-2)^{2}-2]$, ap'opou vgazoume (ligo xontroeidws afou veltiwnetai poli auti i anisotita, alla apo edw fainetai    o logos tou metasximatismou)  $-2\leq A\leq 2$

Apo tin teleutaia auti sxesi pairnoume $0\leq \frac{2+A}{4}\leq 1$ kai $0\leq \frac{2-A}{4}\leq 1$

(Kapoios tha mporouse na pei sto simeio auto oti epeidi $-2\leq A\leq 2$ ara $-1\leq \frac{A}{2}\leq 1$ ara na thesoume  $\frac{A}{2}=cos{a}$ 'h $\frac{A}{2}=cos^{2}{a}$. Kati tetoio omws den mas voleuei dioti emeis theloume na "diwksoume apo ti mesi" ta $2+A$ kai $2-A$ pou emfanizontai katw apo tin riza)

To gegonos loipon auto, (prepei sto $\frac{A+2}{4}$ na thesoume mia timi megaliteri 'h isi apo $0$ kai mikroteri 'h isi tou $1$) mas wthei na thesoume
\[\frac{A+2}{4}=\cos^{2}{a}\]
Tote $A+2=4\cos^{2}{a}$ kai $2-A=4\sin^{2}{a}$

kai kanontas eukoles prakseis $x=(A^{2}-2)^{2}-2$ kai xrisimopoiwntas tin tautotita $\cos{2a}=2cos^{2}{a}-1$ pairnoume to polipothito
\[x=2\cos{8a}\]
Omws $x\geq 0$, ara $\cos{8a}\geq 0$ kai telika $-\frac{\pi}{16}\leq a \leq\frac{\pi}{16}$

Ara i arxiki eksiswsi pairnei tin morfi

$|\cos{a}|+\sqrt{3}|\sin{a}|=2\cos{8a}$, me $-\frac{\pi}{16}\leq a \leq\frac{\pi}{16}$

Eksetazoume tin periptwsi opou $0\leq a\leq \frac{\pi}{16}$ (H alli periptwsi, einai sxedon i idia kai pairnoume akrivws tis idies liseis opws kai apo auti).

Sinepws exoume na lisoume tin eksiswsi
\[\cos{a}+\sqrt{3}\sin{a}=2\cos{8a}\]
'h tin eksiswsi
\[\sin{\big(a+\frac{\pi}{6}\big)}=\cos{8a}\]
'h (Xrisimopoiwntas tin tautotita
$\sin{\omega}=\cos{\big(\frac{\pi}{2}-\omega)}$)
\[\cos{8a}=\cos{\big(\frac{\pi}{3}-a\big)}\]
kai pairnoume tin lisi gia to  a:
\[a=\frac{2k\pi}{9}+\frac{\pi}{27}, \ k\in\mathbb{Z}\ \ 'h \ \ a=\frac{2k\pi}{7}-\frac{\pi}{21}, \ k\in\mathbb{Z}\]
$\bullet$ Ean $a=\frac{2k\pi}{9}+\frac{\pi}{27}, \ k\in\mathbb{Z}$ tote
logw tou periorismou $0\leq a\leq \frac{\pi}{16}$ kai to gegonos oti $k\in\mathbb{Z}$, exoume $k=0$ to opoio simainei oti
$a=\frac{\pi}{27}$ kai
\[\boxed{x=2\cos{\frac{8\pi}{27}}}\]
$\bullet$ Ean $a=\frac{2k\pi}{7}-\frac{\pi}{21}, \ k\in\mathbb{Z}$ tote
douleuontas me ton idio akrivws tropo pairnoume kai pali $k=0$ ara $a=-\frac{\pi}{21}$ kai telika
\[\boxed{x=2\cos{\frac{8\pi}{21}}}\]
Elpizw na pirate kati apo tin sigekrimeni askisi kai gia tetoiou eidous metasximatismous genika (pou exoun pesei kata kairous se diagwnismous, akomi kai tis EME (nomizw to 1995 ston Thali tis 3hs Likeiou me poli poli eukoli ekfwnisi kai lisi)).

Filika,
Alexandros

Y.G. Einai arketa endiaferousa akomi kai meta pou ftanoume sto simeio $\cos{8a}=\cos{\big(\frac{\pi}{3}-a\big)}$ oson afora to $x$ pou emeis psaxnoume na vroume (ara tis times tou $k\in\mathbb{Z}$).

[quote]Epitrepse mou na kanw mia diorthwsi stin 9h grammi tis lisis sou [/quote]

Έχεις δίκιο Αλέξανδρε !

Είχα το νου μου στο πρόσημο του συνημιτόνου και μου ξέφυγε το ημίτονο !!

χρονια πολλα και στον Αμβροσιο.επεσαν μαζεμενες η γιορτες στο φορουμ....ελπιζω να μην διαβαζει για να ειναι ξεκουραστος το σαββατο  :D

[color=red]ASKHSH 9
[/color]
Na vrethoun oi akeraioi $x$ tetoioi wste

$(4-x)^{4-x}+(5-x)^{5-x}+10=4^{x}+5^{x}$

Alexandros

akereoi???.....mipos i8eles na pis pragmatiki gia na exi pio ....endiaferon....
gia $x\geq 5$ ine profanos $4^{x}+5^{x}>4^{5}+5^{5}>10+(4-x)^{4-x}+(5-x)^{5-x}$
gia $x=4,3,2,1,0$ dokimazume me to xeri...kai vriskume oti to$x=2$ ine lisi..
gia $x\leq-1$ ,8etume$x=-k$ ,$k\geq 1$ opote ginete
$(k+4)^{k+4}+(k+5)^{k+5}+10=(1/4)^{k}+(1/5)^{k}$ pu ine profanos adinati me $k\geq 1$....

 :D

Me akeraious tin eixa vrei tin askisi opote tin afisa opws itan. Ontws itan eukoli gia na antistathmisei ligo tin proigoumeni.

Swstos RAKAR.

Kai ksekinw kai thn 10h kai teleutaia askisi gia na kleisei to prwto meros tis Algevras
[color=red]
ASKHSH 10[/color]

Estw $p$ thetikos pragmatikos arithmos kai estw $|x_{0}|\leq 2p$. Gia $n\geq 1$ orizoume

$x_{n}=3x_{n-1}-\frac{1}{p^{2}}x_{n-1}^{3}$

Na ekfrasete tin $x_{n}$ ws sinartisi twn $n$ kai $x_{0}$.

Alexandros

[color=red]ΑΣΚΗΣΗ 10 - ΛΥΣΗ[/color]

Έστω η συνάρτηση $f(x)=3x-\frac{x^{3}}{p^{2}}=x\left[3-\left(\frac{x}{p}\right)^{2}\right]$

Ο επόμενος όρος του $x$ στην ακολουθία θα είναι ο αριθμός $y=f(x)$.

Έστω $x$ κάποιος όρος της ακολουθίας τέτοιος ώστε $|x|\leq 2p$. Θα δούμε ότι και ο επόμενος όρος έχει την ίδια ιδιότητα.

Ο x γράφεται $x=(2p)\sin\theta$ για κάποια γωνία $\theta$

$y = f(x) = x\left[3-\left(\frac{x}{p}\right)^{2}\right]\Rightarrow $

$y = p\cdot\left(\frac{x}{p}\right)\left[3-\left(\frac{x}{p}\right)^{2}\right]$

$= (2p)\sin \theta\left[3-(2\sin \theta)^{2}\right]$

$= (2p)\sin \theta (3-4\sin^{2}\theta)$

$= (2p)\cdot (3\sin \theta-4\sin^{3}\theta) \Rightarrow \boxed{y = (2p)\cdot \sin 3 \theta}$, που είναι $\leq 2p$

Έτσι, το $x=(2p)\sin\theta$ αντιστοιχίζεται στο $y=(2p)\sin 3\theta$

Έστω $\theta_{o}$ η αρχική γωνία. Τότε θα είναι $x_{n}=(2p)\sin(3^{n}\theta_{o})$

Λουκάς

Με πρόλαβες Λουκά !!

Τουλάχιστον, μπορούμε να είμαστε σίγουροι ότι η λύση μας είναι σωστή  :) :)   :D

Eksairetika! Mpravo kai stous dio sas!

Sinexizoume sto epomeno topic me tis askiseis 11-20.

Alexandros


Link http://www.mathlinks.ro/LaTeX/AoPS_L_About.php died , please fix it!

Hmm, well, It's working for me... (now?)

I fixed the problem. Should work fine now. This will go under supports.


There are infinite lines on a planar, then we can color it with 2 colors.
Help me prove it.  :blush:

Could you rephrase the question?
What are we coloring?

I would guess it refers to well known problem of coloring parts in black and white in such a way that neighbourhood parts are colored in different colors.


Let  a,b,c are real numbers such ... $ c^2\plus{}2(a\plus{}b)(a\plus{}c)\plus{}1 \le2(a\plus{}b\plus{}c)$
Prove that ___ $ |b| \ge |a|$.

Solution from school . :D

Η δοσμένη παίρνει τη μορφή $ c^2\plus{}c(2a\plus{}2b\minus{}2)\plus{}2a^2\plus{}2ab\minus{}2a\minus{}2b\plus{}1\leq 0$

Η δακρίνουσα αυτής πρέπει να είναι θετική  :wink:  και $ D\equal{}b^2\minus{}a^2$ qed  :)

Ωραίος Σιλουανέ ...  εγώ έχω μια άλλη λύση.
Θα την αφήσω λίγες μέρες και θα την ανεβάσω μετά  :)

Βαριέμαι να ανεβάσω τη λύση μετά από μέρες ...

Η λύση μου είναι :

Μετα από πράξεις καταλήγουμε στην

$ 2a^2\plus{}c^2\plus{}1\plus{}2(ab\plus{}bc\plus{}ca)\minus{}2(a\plus{}b\plus{}c) \le0$

όπου παίρνοντας το ένα α^2 και αλλάζοντάς του μέλος και προσθαφαιρόντας και στα δυο μέλη το β^2 η ανισότητα γίνεται

$ a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2bc\plus{}2ca\minus{}2a\minus{}2b\minus{}2c\plus{}1 \le b^2\minus{}a^2$,  ή

$ (a\plus{}b\plus{}c\minus{}1)^2\le b^2\minus{}a^2$ <=> $ b^2\minus{}a^2 \ge0$  κ.λ.π


$ \odot O_2$ inscribes $ \odot O_1$ at $ P$, $ AB$ is a chord of $ \odot O_1$ and is the tangent of $ \odot O_2$, $ C$ is the point of tangency.

$ PA$ meets $ \odot O_2$ at $ E$, $ PB$ meets $ \odot O_2$ at $ F$. $ PC$ meets $ \odot O_1$ at $ G$ and $ EF$ meets $ PG$ at $ D$. $ AD$ meets $ \odot O_1$ at $ H$.

Prove that: $ H,F,G$ are collinear.

Dear Mathlinkers,
let Tg the tangnet to O1 at G.
1. Tg // AB // EF (by homothetie or Reim's theorem)
2. DF is the Pascal's line of the degenerated hexagone TgPBAHG.
3. Conclusion : H, F and G are collinear.
Sincerely
Jean-Louis


Find all positive integers $ x$ that leave a remainder of $ 1$ when divided by $ 2, 3$ and $ 5$.

Numbers of the form $ x\equal{}30m\plus{}1$ where m is an integer satisfy the conditions of the problem i think.


What is the value of $ x$ in the equation $ 2x \plus{} \frac{1}{2}x \plus{} 2(1\plus{}x) \equal{} 29$?

Expanding gives $ 2x\plus{}\frac{1}{2}x\plus{}2x\plus{}2\equal{}29$ or $ 4\frac{1}{2}x\equal{}27$. Thus $ x\equal{}\boxed{6}$.


A frog fell down a 20 meter well. Each day the frog climbs 3 meters up. At night it slips down 2 meters from sleeping. At this rate, how many days does it take for the frog to climb out of the well?

[hide="Yup..."]
$17$ (full) days to get to $17$ meters, then the afternoon of the 18th day to reach the top.

So, like $17 \frac{1}{2}$ days. [/hide]

[quote="roadnottaken"][hide="more specifically"]At the end of the eighteenth day, he reaches the top (and presumably climbs over the edge so that he doesn't slip back down), so $18$ days.[/hide][/quote]

Well what do you define as the "end" of the day? (This is why I put my answer as I did.)  :)


I can buy $ 3$ tapes for the price of 2 CDs. I paid $ \$28$ for 2 tapes and 1 CD. How many dollars should I pay for 1 CD?

Ok,

Let t=tapes and c=cd

28=2t+c
3t=2c
56=4t+2c=7t

Therefore, t=8.

28=2(8)+c= 16+c. So, c=12 which is the cost of one cd.


function
limit
calculus
derivative
topology
inequalities
real analysis

A function $f: \mathbb{R}\to \mathbb{C}$ is called a Schwarz function if $f$ is infinitely many times differentiable and if
\[\lim_{|x| \to \infty}\left|x\right|^{n}\left|f(x)\right| = 0 \]
for all positive integers $n$. Let $\mathcal{S}$ be the set of all Schwarz functions.

Show that the map $\mathcal{F}$, which maps each function to its Fourier transform, determines a bijection $\mathcal{S}\to \mathcal{S}$.

it's more than a bijection :  it is an homeomorphism (with Swartz's topology)

[quote="harazi"]Arne, this is Schwartz's space, not Schwarz. Also, I think you fogot to add some derivatives to these functions.[/quote]
[quote="Arne"]How do you mean, I forgot some derivatives?
[/quote]
\[\lim_{|x|\to\infty}|x|^{n}|f^{(k)}(x)|=0 \]

To Arne: the usual definition of Schwartz's space is the set of $C^{\infty}$ functions such that $|x|^{k}|f^{(n)}(x)|$ tends to $0$ for all $k$ and $n$. But apparently here you work with another space.

But is the result I stated (and which I'm trying to prove) correct?

First, a note on spelling: this is Laurent Schwartz (20th century, French), who was a different person than the H.A. Schwarz (19th century, German) who was behind the Cauchy-Schwarz inequality and Schwarz's lemma in complex variables.

The defnintion has to be as RIP and harazi stated it: every derivative goes to zero at infinity faster than any power of $x.$ The result would not be true without the correct definition of the space.

The proof takes advantage of this:

If $g(x)=x^{j}\left(\frac{d}{dx}\right)^{j}f(x),$ then $\widehat{g}(\xi)=C\left(\frac{d}{d\xi}\right)^{k}\left(\xi^{j}\widehat{f}(\xi)\right)$

(The constant $C$ is just a collection of powers of $i$ and of $2\pi.$ The exact details depend on the details of your definition of the Fourier transform.)

In order to claim boundedness and continuity of the transform, we need for the function being transformed to be in $L^{1}.$ We can accomplish that in this proof by investing a couple of extra powers into the decay rate at $\infty.$

The essential thought: the more rapidly the original function decays at $\infty,$ the smoother its transform is. And the smoother a function is, the more rapidly its transform decays at $\infty.$ But that pairing is why the definition of the Schwartz space must involve both derivatives and rates of decay on an equal footing.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finitely generated group of $n*n$ complex matrices .then $G$ contains normal subgroup of finite index  which contains no nontrivial element of finite order.

I like this result :) I know a nice proof of this using $p$-adic specialisations  :P

hi,
I was never aware of the existence of such a lemmma  :blush: ,thanks,I saw on net a proof of this lemma which they call elementary(I wonder how they can say this),neways its here :
[url=http://retro.seals.ch/digbib/view;jsessionid=397B3205E597289E3406EB3E5684E91F?did=c1:42731&p=273]retro.seals.ch/digbib/view;jsessionid=397B3205E597289E3406EB3E5684E91F?did=c1:42731&p=273[/url]

:) There is a half-page elementary solution by Cassels, assuming a 5-page elementary $p$-adic specialization theorem (again by Cassels), which is:

Let $K$ be a field finitely generated over $Q$. Let $S \subset K$ be a fintie set not containing $0$. Then there exist infinitely many primes $p$ (indeed, a set of positive density) for which there is an embedding $\gamma: K \to \mathbb{Q}_p$ such that $|\gamma(s)|_p = 1$ for all $s \in S$; that is, an embedding taking all elements of $S$ to $p$-adic units.

It is a very useful result -- it's most striking and famous application is to the Skolem-Mahler-Lech theorem.

--Vesselin

Nice  :) Can you show (or give some references) how is it used in a proof of Skolem-Mahler-Lech theorem ?

And also I'd like to see solution to original problem using this result  :)


vector
LaTeX
trigonometry
trig identities
Law of Cosines

[hide="solution"]$\begin{eqnarray*}\parallel \vv{a}\parallel \cdot\parallel \vv{b}\parallel \cdot\cos(AOB)&=&\vv{a}\cdot\vv{b}\\ \sqrt{(206^{2}+2007^{2})(2006^{2}+207^{2})}\cos(AOB)&=&206\cdot2006+2007\cdot207\\ \cos(AOB)&=&\frac{165737}{\sqrt{814097\cdot813377}}\end{eanarray*}$
Maybe I made a mistake somewhere... or it's just an awfully computationally heavy problem. BTW, how do you do vectors in LaTeX?[/hide]

I'm terrible at latex and this stuff, so bare with me...

[hide] I would just use the law of cosines and distance formula without using dot product of vectors.  Let's see...

Distance between point O and point A = sqrt( 206^2 + 2007^2)

Distance between point O and point B = sqrt (2006^2 + 207^2)

Distance between point A and B = sqrt ( 1800^2 + 1800^2)

Then use the law of cosines to find the cosine of angle AOB

So...  6480000 = (AO)^2 + (BO)^2 - 2(AO)(BO)cos(AOB)

Then solve...
[/hide]

Couldnt you also find it like this

[hide]
$\cos(AOB) = cos(90-\arctan(\frac{206}{2007})-\arctan(\frac{207}{2006})) = 0.20367.....$

but we need completely exact values right, so cant go about it this way?
[/hide]

It is a [b]very[/b] computationally-heavy problem.  The first solution was correct.


Let ABCD be a convex quadrilateral and all sides of it are constant. Find a formula for maximum and minumum value of the sum of  AC+BD (AC and BD are diagonals of ABCD).

I think I found the minimum value of the sum AC +BD .
 $ S = AC.BD.sin(AC,BD) \leq \frac{(AC+BD)^2}{4} $
 $==> AC+BD \geq 2\sqrt{S} $
 "=" holds AC=BD and AC is perpendicular with BD .

In fact your "=" can be unattainable. Right?
Moreover, $S$ is not fixed.

[quote] Let ABCD be a convex quadrilateral and all sides of it are constant [/quote]
  Maybe I don't understand this sentence !!!???

I think you should exactly estimate $AC+BD$ in terms of $AB,BC,CD,DA$.

[quote="Armo"][quote][quote]Let $ABCD$ be a convex quadrilateral and all sides of it are constant
[/quote]
Maybe I don't understand this sentence !!!???[/quote]
I think that he means that the values of the lenghts of those sides are fixed,no more.[/quote]
How do you think what did I write above?

I just wrote your answer in more details [b]Myth[/b], and note that I answered exactly to the question of [b]nttu[/b]. :)


ARML AMC USA(J)MO USAMO geometry Support

Not that I have anything to say, but it seems that Las Vegas should have its own topic here (rather than the year-old topic that t0rajir0u dragged up).

Southern California will bring four teams this year.

NorCal's got Evan O'Dorney. Prepare to be eviscerated. (all a friendly jest :P )

Is Evan O'Dorney coming? Is he actually on the ARML team? (And if he is on an ARML team, wouldn't it be SFBA, not NorCal?) What about the other USAMO winners/HM? Is Paul Christiano on the SFBA ARML team? Is there an Oregon team and is Eric Larson on it? Is Sam Elder on the Colorado ARML team?

Jeff Manning is definitely on the Southern California team.

From what I heard this weekend at the WA practice, there may be an Oregon team. There's a match between WA and OR next weekend as part of the organizing process.
I also heard that Sam Elder isn't coming this year.

Yes, I can confirm that there will be a (somewhat organized) Oregon team this year (since I'm part of it).

[quote="Kent Merryfield"]Is Evan O'Dorney coming? Is he actually on the ARML team?[/quote]
He came to practice and was put on the A team, so I'm assuming he is.

[quote]Washington and Oregon are having a practice competition next week.[/quote]
It's a pretty long drive from Eugene or Corvallis to Seattle, or vice versa. Where are you meeting?

[quote="t0rajir0u"]Portland (approximately halfway).  Should be a good warm-up.  Most of us are driving down the day before and some of us may join the Oregon team's practice that day.[/quote]

I am just about to send out an update to team.  After talking to Oregon coach and Paul, it is confirmed that Saturday's lecture / practice will be canceled due to various conflict (most WA team members can't make to 12noon anyway, I thought you guys who head there on Saturday are just determined to make contributions to US economy).  There will be an orientation session instead 8:30 - 9:15am (not much a lecture) by Paul.  For team members heading to Portland on Saturday: if this is the first ARML and you have not been to any of the practices before, you may want to be there 8:30am.

[quote="minghu"][quote="paladin8"]I'm trying to tag along with the SFBA team as a coach, so maybe I'll return to ARML once again :).[/quote]

Oh, but you should return to WA team, especially that you missed the opportunity to contribute in senior year 2007 :)[/quote]

True. I'm not sure they'll let me do that though :P. In any case, it'll be fun to meet up with the WA team again if I end up going. Are you coaching it this year?

[quote="t0rajir0u"]Washington and Oregon are having a practice competition next week.  Darryl Wu is probably going to be on the A team, by the way  :D[/quote]

Compared to Southern CA and SFBA, Washington has a relatively smaller list of USAMO qualifiers.  Therefore it is logical to put all USAMO qualifiers on A team.  Don't you envy SFBA's deep pool of talents? (over 60 USAMO qualifiers in the area!)

Following up on what minghu said about USAMO qualifiers - here are the number of USAMO qualifiers by ARML region:

WA: 13
OR: 6, including one HM (Eric Larson)
NorCal: 1
SFBA: 63, including two winners (Evan O'Dorney and Paul Christiano)
SoCal: 19, including one HM (Jeff Manning)
San Diego: 8
AZ/NV/UT: 5
CO: 6, including one winner (Sam Elder)
NM: 4

I don't know how many of these people will actually come to ARML. I also don't know how many different teams AZ/NV/UT is. In the case of Southern California, 17 of the 19 will be on the ARML team (11 of them on the A team.)

Word has it Evan has joined SFBA.  I'm not sure about Paul; he went two years ago, but not last year, and I haven't heard if he's going this year.

We have four teams this year, by the way.  The last two (now named S and F) will each be shorthanded by three unless transportation works out in such a way that six more people can get to Vegas.

In response to the question of AZ/NV/UT, I have no idea if there's a Utah team this year.  Tim's created a "Desert Southwest" region composed of Arizona, New Mexico, and Nevada.  It'll send two teams: One from Arizona, and another with students from both New Mexico and Nevada.  (The region was created to allow those states to send as many full teams as possible.)

[quote="Sly Si"]We have four teams this year, by the way.  The last two (now named S and F) will each be shorthanded by three unless transportation works out in such a way that six more people can get to Vegas.[/quote]
We have four teams, named SoCal A, B, C, and Gamma. (Or to do that right, $ \Gamma.$) Unless someone cancels that I don't know about yet, we'll actually have two extra people, 62 students total. We're covering the transportation shortfall with a few extra cars.

I'm on the Oregon team, and unfortunately Eric Larson's not doing it. Being the pessimist/realist that I am, I personally think that Washington is probably going to kill us... :|

Coach (Manny) said Eric is on ARML roster.  Do you mean he won't be there this coming Sunday (May 18) for the intramural?

No, Eric decided not to be on the ARML team.  He was taken off the roster a few weeks ago.

SD is fielding two teams, including its first ever in division A. We will be arriving in the early afternoon, although our plan is to go out for a [i]very[/i] early dinner before the competition begins. (We could call it a late lunch, but most of the kids will have been awake for ten hours by that point.)

Does Silas require Sheepsheading?  :)

Your schedule is similar to ours. You might try looking us up to see who wants to go to dinner together (our guys seem to be partial to the In-N-Out on the edge of campus.)

As for Silas, I'm pretty sure he's said elsewhere that he's not coming. We are bringing a couple of recent ARML veterans from other regions: Yasha from North Carolina and Tim from Wisconsin (both now Caltech students).

Gah, I would strongly support Sheepsheading, but I'm spending the quarter at Oxford and won't be back until two weeks after ARML.  Next year, though, I'm going, even if I have to schedule all my final exams around it.

By the way, last I heard, SFBA's bringing ARML vets Nathan (from Washington) and Seva (Alabama), in addition to possibly paladin8.

Well, WA did indeed kill us [Team Oregon] yesterday at the Intramural competition. If only Larson was on our team... that'd be soooo sweet.
As of right now, we have 4 USAMO qualifiers on the team, but none made it to MOP, including Brynmor Chapman. Oh well.

Us Nevadans are combining with the New Mexicans for a team.  We're probably not going to do very well, but who knows? 5 of us coming from a school for PG kids, two of whom are taking college level math courses.

I don't know many of the names on the MOP list ... does anybody know how many of them are from the SFBA team, SoCal, ... etc.?

I know Washington has 2 people going to MOP (me and Joy, a freshmen).

[quote="samath"]I don't know many of the names on the MOP list ... does anybody know how many of them are from the SFBA team, SoCal, ... etc.? [/quote]
For the California portion of that list, two names (Johnny Li and Zhifan "Ivan" Zhang) will be on the Southern California team. The other 8 names belong to the SFBA region; I don't know how many of those 8 are coming to ARML.

Well, here's my version of an incomplete list of AoPSers:

JAM
litoxe27182818
Ivan Zhang
jli
bluejay
Zellex
Kimby_102
naitixuy
t\w/o
Brut3Forc3
runeknightx
pythag011
zapi2007
Differ
mws21
Tecnogram888
xsk13

[i]And,[/i]

Kent Merryfield
jmerry
Dr. Kevin
zmli
Yasha

[quote="xscapezaer"][quote="paladin8"]From my going to practices, I recognize the following names:

Vishal Arul
John Boyle
Taylor Han
Evan O'Dorney

There may be more, but those people have been at the practices for sure.[/quote]

Add David Zeng and Lynnelle Ye, and your list's complete. As a member of the SFBA team who epic-ly failed USAMO, I know names :P Out of those 6 SFBAers, 4 are AoPSers, as well...

EDIT: I'm not a coach or anything, but SFBA is sending 4 teams, teams S, F, B, and A, in reverse order of skillz.[/quote]
nope david and lynnelle arent coming
yeah those 4 are our moppers

[quote="DiscreetFourierTransform"]Paladin8, are you one of the coaches?
I can probably come meet up with you on sunday if you are, and give you an incomplete list of all of the AoPSers going to ARML.[/quote]

Sort of. I've only shown up to the last two practices since I didn't know about it before, but I hope to be going to Vegas with you guys. And yeah I'll probably be there tomorrow (Sunday).

[quote="Kent Merryfield"]...the In-N-Out on the edge of campus.[/quote]
You know, I think I ate there in 2000- the Palos Verdes team I was on knew about it.

As for who is going to be there: I count 23 teams from 12 organizations. Here's the list:

Colorado: 2 teams
Desert Southwest: 2 teams (this is NM, AZ, and NV)
Northern California: 1 team (this is Davis/Sacramento)
Oregon: 1 team
Southern California: 4 teams
San Diego: 2 teams
San Francisco Bay Area: 4 teams
Utah: 1 team
Washington: 2 teams
Hong Kong: 1 team
Philippines: 1 team
Taiwan: 2 teams

Of these 23 teams, four are international "guest division" teams, six are Division A teams, and the remaining 13 are Division B teams. The six Division A teams are:

Colorado Granite
Southern California A
Southern California B
San Diego A
San Francisco Bay Area A
Washington Gold


A trapezoid with bases of 6 feet and 8 feet and a height of 4 feet is cut in half. How many square feet are in the area of each half?

Area of a trapezoid with bases of $ 6$ feet and $ 8$ feet and a height of $ 4$ feet : $ (6 \plus{} 8) \times 4 \times \frac {1}{2} \equal{} 28$ square feet.
Thus, $ \frac {28}{2} \equal{} \boxed{14}$ square feet.


calculus
integration
function
trigonometry
real analysis
real analysis unsolved

Find the continuous function which satisfies for all real numbers $x$ as follows.

\[\int^x_0 tf(x-t)dt=\int^x_0 f(t)dt+\sin x+\cos x-x-1\]

Linear convolution equation on $[0,\infty)$: it's time to use Laplace transforms. If $F(s)$ is the Laplace transform of $f(x)$ then:

$\frac1{s^2}F(s)=\frac1sF(s)+\frac1{s^2+1}+\frac{s}{s^2+1}-\frac1{s^2} -\frac1s.$

Solving for $F(s)$,

$F(s)=\frac{s+1}{(s-1)(s^2+1)}=\frac1{s-1}-\frac{s}{s^2+1}.$

So $f(x)=e^x-\cos x.$

At least that's the answer for $x\ge0.$ I haven't considered $x<0$ yet.


probability
expected value
combinatorics unsolved
combinatorics

$n, k$ are positive integers. $A_0$ is the set $\{1, 2, ... , n\}$. $A_i$ is a randomly chosen subset of $A_{i-1}$ (with each subset having equal probability). Show that the expected number of elements of $A_k$ is $\dfrac{n}{2^k}$

Should $\dfrac{n}{2k}$ be $\dfrac{n}{2^k}$?

My mistake, I've corrected it, thanks Ravi.

I don't really understand... what if $\frac{n}{2^k}$ is not an integer?

It doesn't matter.  The average (expected value) doesn't have to be an integer, even if the number of elements is always an integer.

this is rather trivial. by linearity of expectation it suffices to show that $P(1\in A_k)=\frac{1}{2^k}$. This is seen easily inductively, as $P(1\in A_k)=P(1\in A_{k}|1\in A_{k-1})\cdot P(1\in A_{k-1})$ and also noting that $P(1\in A_{k}|1\in A_{k-1})=\frac{1}{2}$, since for any set $X$ containing $1$ the number of subsets of $X$ containing or not containing $1$ is the same.

We k-times make a subset of the set A, in each step each number has probability $\frac 1 2$ to be chosen. So after k steps, it's $\frac 1 {2^k}$. Because of that, the number of elemtnts has binomial distribution and the expected value is
$E=\sum_{p=0}^n p{n \choose p} \left( \frac 1 {2^k}\right)^p\left( 1-\frac 1 {2^k}\right)^{n-p}$, the first element of sum is 0, so
$E=\sum_{p=1}^n n{n-1 \choose p-1} \left( \frac 1 {2^k}\right)^p\left( 1-\frac 1 {2^k}\right)^{n-p}$
$E=\frac n {2^k}\sum_{p=0}^{n-1}{n-1 \choose p} \left( \frac 1 {2^k}\right)^p\left( 1-\frac 1 {2^k}\right)^{n-1-p}=\frac n {2^k} \left( \frac 1 {2^k}+\left(1- \frac 1 {2^k}\right)\right)^{n-1}$, QED

Induct on $k$. $k=0$ given.  Assume that $E[|A_k|]=\frac{n}{2^k}$. If $|A_k|=m$ then by linearity of expectation  \[E[|A_{k+1}|]=\sum_{i=0}^{m}i\frac{\dbinom{m}{i}}{2^m}=\frac{1}{2^m}\sum_{i=0}^{m}i\dbinom{m}{i}=\frac{1}{2^m}(m2^{m-1})=\frac{m}{2}\].

from the well-known identity $\sum_{i=0}^{m}i\dbinom{m}{i}=m2^{m-1}$. Let $p_k(m)$ be the probability that $|A_k|=m$. Then

\[E[|A_{k+1}|]=\sum_{m=0}^{n}p_k(m)\Bigg(\frac{m}{2}\Bigg)=\frac{1}{2}\sum_{m=0}^{n}p_k(m)m=\frac{1}{2}E[|A_k|]=\frac{1}{2}\Bigg(\frac{n}{2^k}\Bigg)\]

which is $\frac{n}{2^{k+1}}$ so done.


even though -1 works, the domain restriction still carries over.

plug in $(\sqrt{x})^2$ into your calcuator..., there won't be any negative x values..

its kid of like sayng $\frac{(x-1)^2}{x-1}$  you can simplify it into x-1 but it still has the domain of x can't be 1

Actually the main reason I asked was because my calculator (TI-89) [b]did[/b] give a value.
In the y= editor I typed $y_1=x^2$, $y_2=\sqrt{x}$, and $y_3=y_1(y_2(x))$.  For $x=-1$, $y_3=-1$.
Since the TI-89 is the only calculator I've tested this on that deals with complex numbers, I assumed the reason other calculators gave an error was just because they couldn't deal with $i$.

But thanks for the answer, and I agree completely with your example $\frac{(x-1)^2}{x-1}$, but I still think $(\sqrt{x})^2$ is different because instead of involving division by zero, it involves $i$.

Can anyone show a proof of why this is, or is it just a definition?

Does the answer change if I said the domain is $C$ instead of $R$?

[quote="Jimmy"]If $f(x)=x^2$ and $g(x)=\sqrt{x}$.  Then $f(g(x))=x$, but does the domain restirction of $g(x)$ carry over into $f(g(x))$?  I don't think it should.  Here's an example, if $x=-1$, then $g(x)=i$ and $f(i)=-1$, thus $f(g(x))=x$ holds true.[/quote]The domain restriction does carry over, unless $f$ and $g$ are complex-valued functions (Is there a term for this?), as you have treated them, in which case you would normally write $f(z)=z^2$ and $g(z)=\sqrt{z}$. (Someone correct me if I'm wrong; I don't really know much about functions of a complex variable.)

[quote="Jimmy"]Does the answer change if I said the domain is $Z$ instead of $R$? ( I can't figure out how to type a cursive R for real numbers and the Z that represents complex numers.)[/quote]Actually, $\mathbf{C}$ is the set of all complex numbers. $\mathbf{Z}$ is the set of all integers.

You can use \mathbb{} to get that special font (blackboard bold):

$\mathbb{C, Z, R, N}$

[quote="AntonioMainenti"]Actually, $\mathbf{C}$ is the set of all complex numbers. $\mathbf{Z}$ is the set of all integers.[/quote]
Thanks I fixed this.

[quote="Jimmy"]If $f(x)=x^2$ and $g(x)=\sqrt{x}$.  Then $f(g(x))=x$, but does the domain restirction of $g(x)$ carry over into $f(g(x))$?  I don't think it should.  Here's an example, if $x=-1$, then $g(x)=i$ and $f(i)=-1$, thus $f(g(x))=x$ holds true.[/quote]
Defining the domains and codomains of functions or mappings clearly distinguishes whether there is a restriction or not. It is useful to write something like $f:\mathbb{X}\rightarrow\mathbb{Y}$ or $f:\mathsf{F}^n\rightarrow\mathsf{F}^m$, where you specify domains, they can tell you more information, rather than guessing what is restricted or what is extended.


inequalities
function
limit
algebra unsolved
algebra

hello!
find all the functions suchs that :
i)$ \forall (x,y) \in \mathbb{R} : f(x\plus{}y)\leq f(x)\plus{}f(y)$
ii)$ \lim_{x\rightarrow 0}\frac{f(x)}{x}\equal{}1$

Then $ f(x)=x$.
[hide="Sketch of proof"]Resolve condition (ii) in terms of the $ \varepsilon$-$ \delta$-definition of a limit to see that $ \lim_{x\to 0}f(x)=0$. (i) easily gives $ f(0)\ge 0$. The inequality $ f(0)\le f(h)+f(-h)$ yields for $ h\to 0$ the reverse inequality $ f(0)\le 0$, hence $ f(0)=0$. Then (ii) means that $ f$ is differentiable at $ x=0$ with $ f'(0)=1$.

$ f$ is everywhere differentiable with $ f'\equiv 1$, this implies $ f(x)=x$. To see this fix $ x_0\in \mathbb R$ and use \[ f(x_0)\le f(x_0\pm h)+f(\mp h)\le f(x_0)+f(\pm h)+f(\mp h)\] with $ h>0$ to build up the inequalities of the difference quotient \[ \frac{f(-h)}{-h}\le\frac{f(x_0+h)-f(x_0)}{h}\le\frac{f(h)}{h}\] for $ h>0$. The case $ h<0$ works in exactly the same way. Altogether this implies that $ f$ is differentiable at $ x_0$ with $ f'(x_0)=1$.[/hide]


Situation is as follows: we have three towers and $2<n<61$ disks. We are solving standard problem using well known algorithm based on the legend: "The temple priests of hanoi have to transfer a tower consisting of $64$ fragile disks of gold from one part of the temple to another, one disk at a time. The disks are arranged in order, no two of them the same size, with the largest on the bottom and the smallest on top. Because of their fragility, a larger disk may never be placed on a smaller one, and there is only one intermediate location where disks can be temporarily placed. It is said that before the priests complete their task the temple will crumble into dust and the world will vanish in a clap of thunder".

It is known, that number of moves ($k$) to solve problem with $n$ disk is  $k=2^n-1$.

The question is: if we know the number of moves already made how to find out what is the situation on towers - how to find out which disks are on the first, which are on the second,  and which are on the third tower?

I think you have asked an unsolvable question - although I could be wrong. Basically , if we can solve your question, we know [i]how[/i] to solve the tower of hanoi - a thnig whcih I don't think is possible!!!


algebra
polynomial
invariant
geometry
geometric transformation
homothety
superior algebra

Having $A,B \in M_{2n}(R)$ and $a \in R^*,|a| \ne 1$ ,such that $AB=aBA$, and $detA \ne 0$.
Prove that $det(A+B)=detA+detB$

Suggested by Silviu

I imagine that passing to determinants in the equality $A(A+xB)=(A+axB)A$ says everything.  ;) (I haven't seen many polynomials invariant under a homothety).

From $AB=aBA$ and $det(A) \neq 0$ we have $(a^{2n}-1)det(B)=0$ but

$|a|\neq 1$ so $det(B)=0$

From the post of harazi take the determinant and simplify by $det(A)\neq 0$

$det(A+xB)=det(A+xaB)$ for any x real , the polynomial  invariant by homothety are
constant

$det(A+B)=det(A)$

Finally we have $det(A+B)=det(A)+det(B)$ since $det(B)=0$


advanced fields
advanced fields unsolved

This is very nice problem ! but I don't understand it ...
I want to know how do we get $ cos^{\minus{}1}\frac{1}{7}$ from this proof. Thanks very much !  :)


I  am  using  Miktex +texniccenter +ghostcript.  the packages  were  downloaded   from the  packges management.
the following  code gets  4 warnings(not mistake).

Why?

[code]\documentclass{beamer}

\title{Gauss-Bonnet }

\author{Math}

\date{\today}

\begin{document}

\begin{frame}

\titlepage

\end{frame}

\section*{Outline}

\begin{frame}

\tableofcontents

\end{frame}

\section{AAAA}

\subsection{AAA1}

\begin{frame}

\frametitle{Gauss-Bonnet in 20th Century}

\begin{itemize}

\item<1->

(Gauss-Bonnet)Let $\Sigma$ be a smooth closed oriented surface in $R^3$, then

$$\int_{\Sigma}k\mbox{d}s = 2\pi\chi (\Sigma).$$

\item<2->

Hopf, 1925: For a hypersurface $M^n$ in $R^{n+1}$($n$ even), one has

$$\int_{M}k\mbox{d}v =\frac{1}{2}\mbox{vol}(S^n)\chi (M),$$

where $k$ is the Gauss-Kronecker curvature.

\end{itemize}

\end{frame}

\subsection{AAA1}

\begin{frame}

\frametitle{Gauss-Bonnet in 20th Century}

\begin{itemize}

\item<1->

(Gauss-Bonnet)Let $\Sigma$ be a smooth closed oriented surface in $R^3$, then

$$\int_{\Sigma}k\mbox{d}s = 2\pi\chi (\Sigma).$$

\item<2->

Hopf, 1925: For a hypersurface $M^n$ in $R^{n+1}$($n$ even), one has

\end{itemize}

\end{frame}

\section{AAAA}

\subsection{AAA1}

\begin{frame}

\frametitle{Gauss-Bonnet in 20th Century}

\begin{itemize}

\item

(Gauss-Bonnet)Let $\Sigma$ be a smooth closed oriented surface in $R^3$, then

$$\int_{\Sigma}k\mbox{d}s = 2\pi\chi (\Sigma).$$

\item

Hopf, 1925: For a hypersurface $M^n$ in $R^{n+1}$($n$ even), one has

$$\int_{M}k\mbox{d}v =\frac{1}{2}\mbox{vol}(S^n)\chi (M),$$

where $k$ is the Gauss-Kronecker curvature.

\end{itemize}

\end{frame}

\end{document}

[/code]

You really ought to post the warnings as we can't guess what they are.
Solution is found [url=http://groups.google.co.uk/group/comp.text.tex/browse_thread/thread/a734f9c5b5016fc9/6a0ef006e290f29b?hl=en&lnk=gst&q=beamer+font+warning#6a0ef006e290f29b]here[/url] using a different font or, to use non-standard sizes in the current font [code]\usepackage[T1]{fontenc}
\usepackage{fix-cm}[/code]
You should not be using \$\$...\$\$ as that is obsolete. See [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=dolldoll]Why use \ [ ...\ ] in place of \$\$...\$\$?[/url]

Many thanks!

For  the  beamer, I download   3  packages  beamer , pgf and  xcolor.

the 4  warnings  are:

1. package pgf  warning:  this  package is  obsolete and no longer needed on input line 13;

2. package hyperref  warning: option 'pdfpagelabels' is turned  off
                                 because  \thepage is  undefined

3.  Latex Font warning: Font shape 'OT1/cmss/m/n'  in size <4>  not available

4  Latex Font warning: Size substitutions with differences
(Font)                       up  to 1.0pt have occurred.

[color=red]
If  I  write  the  code   below:[/color]
[code]\documentclass{beamer}
\usepackage{times }
\usefonttheme{professionalfonts}
\title{Example Presentation Created with the Beamer Package}
............
[/code]

[color=red]Then  the  first  two  warnings   still  exist[/color]

If I write the code below:
[code]\documentclass{beamer}
\usepackage[T1]{fontenc}
\usepackage{fix-cm}
\title{Example Presentation Created with the Beamer Package}
............
[/code]

and  remove  all  \$,   the first two warnings still exist

1. See [url=http://www.nabble.com/This-package-is-obsolete...-td17642354.html]here[/url]
2. \documentclass[hyperref={pdfpagelabels=false}]{beamer}


$a,$ $b$ and $c$ are non-negative numbers such that $ab+ac+bc=3.$ Prove that:
$(a^{2}+1)(b^{2}+1)(c^{2}+1)\geq8.$

Nice  :)

Expand to get the equivalent \[a^{2}b^{2}c^{2}+\left(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\right)+\left(a^{2}+b^{2}+c^{2}\right)+1 \geq 8\]Now let $a+b+c = m$ and $abc = n$. Clearly, $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}= (ab+bc+ca)^{2}-2abc(a+b+c) = 9-2mn$ and $a^{2}+b^{2}+c^{2}= (a+b+c)^{2}-2(ab+bc+ca) = m^{2}-6$. Therefore, we have to prove that $n^{2}+9-2mn+m^{2}-6+1 \geq 8$, which is equivalent to $|m-n| \geq 2$.
Now remark that $m^{2}= (a+b+c)^{2}\geq 3(ab+bc+ca) = 9$, so $m \geq 3$, and $3 = (ab+bc+ca) \geq 3\sqrt[3]{a^{2}b^{2}c^{2}}$, so $1 \geq n$. Therefore $m\geq 3 > 1 \geq n$, so we only need to prove that $m-n \geq 2$. And that's obvious now.

Could you post the statement of Holder's inequality and its proof :?:

So did you make up the problem with the solution in mind, or the other way around?

[quote=shalomrav][quote=Understandingmathematics][quote=arqady]OK, Kurt Gödel.
My proof:
by Holder:
 $(1+1+1+1)(a^{2}b^{2}+a^{2}+b^{2}+1)(a^{2}+a^{2}c^{2}+c^{2}+1)(b^{2}+c^{2}+b^{2}c^{2}+1)\geq$
$\geq(ab+ac+bc+1)^{4}=256.$ :)[/quote]

but we have to prove that $(a^{2}+1)(b^{2}+1)(c^{2}+1)\geq8.$....where we proved that in your proof ???
sorry if this is a stupid question but i am not getting..i think i am missing something simple,can you pls give more details....[/quote]

$(a^{2}+1)(b^{2}+1)=
(a^{2}b^{2}+a^{2}+b^{2}+1)$[/quote]

thanks ...


The normal boiling point of methanol(CH3OH) is 64.7 degree celsius, and its molar enthalpy of vaporization is 71.8kJ/mole. Calculate the change in enthalpy when 33.5 grams of liquid methanol is vaporized into a gas at its normal boiling point.

The molar mass of methanol is 32.042 g/mol, and so $ n_{methanol} \equal{} \frac {33.5\,g}{32.042\,g/mol} \equal{} 1.04\,mol$. Therefore,

$ \Delta H \equal{} 1.04\,mol \cdot 71.8\,kJ/mol \equal{} 74.7\,kJ$.


geometry
LaTeX
area of a triangle
geometry proposed

This problem was posted by Levi. Actually, the post contained flawed LaTeX code which forced the compiler into an endless loop, so I had to delete it. Here is the actual problem:

[color=blue]Consider a line $AB$ and three collinear (in this order) points $X,Y,Z\not\in AB$. Then, prove that:

- If the line AB does not intersect the segment XZ, then $XZ\cdot (AYB)=XY\cdot (ABZ)+YZ\cdot (XAB)$.
- If the line AB intersects the segment XY, then $XZ\cdot (AYB)=XY\cdot (ABZ)-YZ\cdot (XAB)$.
- If the line AB intersects the segment YZ, then $XZ\cdot (AYB)=-XY\cdot (ABZ)+YZ\cdot (XAB)$.

[b]Remark.[/b] Hereby, the notation (UVW) denotes the area of a triangle UVW.[/color]

  Darij

Maybe it is better to say:

$\triangle{AYB}=\lambda \triangle{AZB}+(1-\lambda)\triangle{AXB}$ where $\lambda = \frac{XY}{XZ}$.


Is there exists a partition of $ Z_\plus{}$ into three non-empty subsets $ A_1,A_2,A_3$ such that
for all $ x\in A_i$ ,$ y\in A_j$ then $ xy \plus{} x\plus{} y \in A_k$ ( $ i,j,k$ are three pairwise numbers in the set {1,2,3} )

Suppose such a partition exists, and add $ 1$ to each of its elements.  It is then a partition of $ \mathbb{Z}_{>1}$ such that for all $ x \in A_i, y \in A_j$ we have $ xy \in A_k$.  Say that $ 2 \in A_1$ without loss of generality.

Suppose further that some other positive integer $ a \in A_2$.  Then $ 2a \in A_3, 2a^2 \in A_1, 4a \in A_2$ and $ 8a, 2a^3, 4a^2 \in A_3$.  But then $ 4$ is not in $ A_1$ (or else $ 8a \in A_2$, contradiction), $ 4$ is not in $ A_2$ (or else $ 8a \in A_1$, contradiction), and $ 4$ is not in $ A_3$ (or else $ 4a \in A_1$, contradiction).

We conclude that $ A_2$ is empty, also a contradiction.  Hence no such partition exists.


function
geometry
real analysis
real analysis theorems

Prove that:

A real-valued function $f$ on $[a,b]$ is Riemann-integrable if and only if it is bounded and continuous almost everywhere.

[quote="pleurestique"]continuous almost everywhere means what? continuous on a dense subset?[/quote]

obviously! jesus christ!

Almost everywhere means that the stated property does not hold for at most a set of measure zero. I'm not entirely sure if this is equivalent to what you said. (I'm a beginner in this area) But looking strictly at the definitions I would say no. Sorry if I said a dumb thing.

No, they are not equivalent; density is a topological notion of largeness whereas almost everywhere is a measure theoretic notion.  For example, for every $t\in I=[0,1]$, there exists a dense subset of $I$ having (Lebesgue) measure $t$.  We can even find uncountable subsets of $[0,1]$ with Lebesgue measure zero.

As for the proof of the stated theorem, it can be found in almost any real analysis textbook.  Probably on the internet, as well.

i've also heard people (professional mathematicians included) use "almost everywhere" to mean "at all but a finite number of places"... i didn't know the phrase actually had an official definition, heh... or does it?

It does have the official definition 'except on a set of measure zero'.  The definition which only includes finitely many points (I admit, I've never heard this usage) would thus be subsumed in the measure-theoretic definition, since any set consisting of finitely many points has Lebesgue measure zero.


function
inequalities
integration
real analysis
real analysis unsolved

Here's (part of) a cool problem from Stein's book:

A sequence $\{f_{k}\}$ of measurable functions on $\mathbb{R}^{d}$ is Cauchy in measure if for every $\epsilon > 0$,
$m(\{x : |f_{k}(x)-f_{l}(x)| > \epsilon \}) \to 0$ as $k, l \to \infty$.

We say that $\{f_{k}\}$ converges in measure to a (measurable) function $f$ if for every $\epsilon > 0$
$m(\{x : |f_{k}(x)-f(x)| > \epsilon \}) \to 0$ as $k \to \infty$.

If $\{f_{k}\}$ converges to $f$ in measure, does the sequence also converge to $f$ in $L^{1}$?

My additional question (maybe stupid):
- What about $L^{2}$? Do either directions of
"$f_{k}\to f$ in $L^{2}$ $\Leftrightarrow$ $f_{k}\to f$ in measure"
hold?

Convergence in measure is very weak; it does not even imply that the functions are integrable, let alone close in $L^{p}$ norm. For example, $f_{n}(x)=\frac{1}{nx}$ converges to $0$ in measure on the interval $[0,1]$.

Convergence in $L^{p}$ for any $p$ implies convergence in measure by something called Chebyshev's inequality, like \[m(\{x : |f_{k}(x)-f(x)| > \epsilon \}) \le \epsilon^{-1}\int |f_{k}(x)-f(x)|\,dm(x)\]


vector
analytic geometry
algebra
polynomial
conics
Galois Theory
geometry solved

Let $\triangle ABC$ find $M$ in plane which satisfy equation:
$MA\vec{MA}+MB\vec{MB}+MC\vec{MC}=\vec{0}$

It is really hard with me!!!

So you are asking for the points M in the plane of triangle ABC such that the barycentric coordinates of M with respect to triangle ABC are ( MA : MB : MC ).

I have finally done the computation. For a triangle ABC with a = 2, b = 2, c = 3, the required point M belongs to the families of points with barycentric coordinates

$\left(2: 2: 3u\right)$, where u is a root of the polynomial $162x^4+243x^3+162x^2-56$;
$\left(325314v^3-83052v^2-143592v-15376: 2: 17880v\right)$, where v is a root of the polynomial $1749x^4-108x^3-784x^2-168x-16$.

Both of the polynomials are irreducible and have the Galois group $S_4$, so their roots cannot be constructed with compass and ruler. Thus, the points M cannot be constructed with compass and ruler either.

It gets even worse if triangle ABC is non-isosceles; for a triangle ABC with a = 3, b = 4, c = 5, I obtain a polynomial of degree 10. Its Galois group is $S_{10}$, if I haven't done some stupid mistake, so the point shouldn't have a construction using conics either.

By the way, the Galois groups were computed using [url=http://www.math.tu-berlin.de/~kant/kash.html]KASH[/url], a nice program where you can just type
[code]Galois (1337*x^3-23*x+1);[/code]
and you get the Galois group of $1337x^3-23x+1$.

  Darij

Hi Darij, I understand your trying to solve it !! Through your word I can see that this problem is really hard. I think the contruction of M by ruler and compass is less important than show the barycentric of M, In Catersian coordinates, this work is equivalent with solving the system equation as you solved.
We knew the famous problem is Fermat problem, Fermat problem is similar this problem
First we can see that $F$ is point that satifies equation
$\frac{\vec{FA}}{FA}+\frac{\vec{FB}}{FB}+\frac{\vec{FC}}{FC}=\vec{0} (*)$
and then ETC we can see that F has barycentric:
$(a\sec(A-\frac{\pi}{6}),b\sec(B-\frac{\pi}{6}),c\sec(C-\frac{\pi}{6}))$
this much we can say that we can solved (*) and the root is this barycentric. I think that !!!


In a game of tic-tac-toe, where X and O take turns, or moves, marking spaces in a 3x3 grid. The player who succeeds in placing three respective horizontal, vertical, or diagonal row wins the game. If X goes first, how many sequences of mobes are there that end with X winning in 5 total moves of both players?

There are eight total ways to place the X's. Each of these have 6 ways to order the X moves. And for each of these, there are 6 x 5 = 30 ways to move with O's. So there are  8 x 6 x 30 = 1440 total sequences of moves.


Use energy conservation arguments to verify that a cylindrical sound (where the wave fronts are cylindrical shells originating from a line source) wave must be written on the form.

$ \delta P(r,t)\equal{}\frac{\delta P_0}{\sqrt{r}} cos(kr\minus{}wt)$

Solution:

$ P\equal{}I*2\pi rl$

$ I\equal{}K*A^2$

$ A\equal{}\sqrt{\frac{I}{K}}\equal{}\frac{1}{\sqrt{r}} \sqrt{\frac{P}{2\pi lK}}\equal{}\frac{C}{\sqrt{r}}$

$ \psi\equal{}Acos(kr\minus{}wt)\equal{}\frac{C}{\sqrt{r}} cos(kr\minus{}wt)$


A surveyor goes to a house where a woman open the door (S: Surveyor, W: Woman)
[list]S: How many children do you have?
W: Three
S: Age of each?
W: The product is 36 and the sum is equal to the number of my neighbor's house[/list]
The surveyor goes to the check the number of the neighbor's house, and then comes back to the woman and says
[list]S: There is no enough data.
W: Ohh! Sorry, you are right, sorry. The older plays the piano.[/list]
Now the surveyor has enough data to findout the age of the children. What is the age of each child?

[hide]We know from the way the woman phrases her last comment that there is only one oldest child. So we're looking for three numbers which multiply to be 36, as well as sum to a sum that is shared by another triple which also multiplies to 36 but has more than one largest number. 9, 2, 2, satisfies this through 6, 6, 1 and since the question implies that the solution is unique, I won't bother making sure there's no other possibility. The children are 9 years, 2 years, and 2 years old.[/hide]

[quote="joml88"]Ah, yes.  I have seen this problem in the [i]Art and Craft of Problem Solving[/i] and also [i]In Code[/i].  It's a fun problem...I wont give a solution so others who haven't already seen it can try it.[/quote]

Heh, I too have seen this problem in the [i]Art and Craft of Problem Solving![/i]

[quote="joml88"]Ah, yes.  I have seen this problem in the [i]Art and Craft of Problem Solving[/i] and also [i]In Code[/i].  It's a fun problem...I wont give a solution so others who haven't already seen it can try it.[/quote]joml, would you recomend [i]In Code[/i], I mean, I have it here at home, but I have been to lazy to start reading it, and now I have to give it back to the library, then, would you recomend me to check it out again?[quote="nr1337"]Heh, I too have seen this problem in the [i]Art and Craft of Problem Solving![/i][/quote]OK, we all agree it is in the book [i]Art and Craft of Problem Solving[/i]!

By the way LynnelleYe, good solution  ;)

I really enjoyed reading [i]In Code[/i] a lot.  A lot of the book would probably be repetitive for you, however.  She discusses puzzle, like the ones in this thread, that motivated her early in her life.  You probably have seen them all before and could skip that part.  She also explains a lot of the basic math and basic cryptology.  When I read the book a lot of the math was new to me but if you've seen most of it before (as you probably have) then you can just skim through those parts.  The rest of the book is very enjoyable also.  I would recommend it as a read.  I think that a lot of my joy from reading the book was the puzzles and basic math I hadn't seen before.  So this book would probably interest the beginner rather than people with more experience though I still recommend it as a read to anyone.

[quote="Ihatepie"]can also be 6, 3, 2 u have 2 know what his neighbors number is[/quote]

If it's 6,3,2 the neighbor's number is 11 and 6,3,2 is the only solution if that's the case so the first part is enough information and it doesn't fit the puzzle.

As delivered here,

this is a flawed problem with the line

"The older plays the piano."  It is awkward.

The word "older" necessarily means there is a choice between

two quantities.  Instead, the use of the word "oldest" would

have meant the choice to be among (potentially) three or

more.  Even with twins who arrived later on in the family, the

oldest would occupy #1 spot as oldest, and the younger twins

would tie at the #3 spot in the age ranking.

A parent with younger twins and a third child older

than the twins would instead say, "My oldest child plays the

piano."

And if there's not enough logical information to go on from

there for this puzzle, then one can redo the puzzle's set-up.

box of rocks,

not to be rude or anything, but shouldn't it be eldest, as opposed to oldest?


find all positive integer $n$ satisfies $n$ divide $2^n+1$

do you know whether this problem has a solution or not? i mean, there are "too many" integers $n$ satisfying your request.
if you solved IMO 2000/5 then you will know that there are infinitely many integers $n$.. for example take $n=3^m$ and then you have $3^{m+1}| 2^{3^m}+1$...
 or you can take with that $m$ an integer $t$ that is a product of primes that divide $2^{3^m}+1$, $t$ not divisible by 3, and you will have that $nt| 2^n+1$ and therefor $nt|2^{nt}+1$..

I remember this problem appeared on forum before.
One more example: take an arbitrary $x_0$ s.t. $x_0\mid 2^{x_0}+1$ and define $x_{n+1}=2^{x_n}+1$.


Find the condition of $ x$ that $ x^{x^{x^{x^{.^{.^{.}}}}}}$
a)Has no limit.
b)Has only 1 limit.
c)Has 2 limits.

http://www.mathlinks.ro/viewtopic.php?t=77942

Pretty informative topic, and the answer to your question is given.

a)$ x > e^{\frac {1}{e}}$
b)$ x\in\left[e^{\frac {1}{e}},\frac {1}{e^e}\right]$
c)$ x < \frac {1}{e^e}$


Basically, you take the first term, which is B, and use that as your answer.

Or:

(B*B)*(C*C)=B*C=B

Hm?
How about inputting some numbers for B and C?
B=2, and C=1. The expression gives 4 not 2.

[hide]
\[ (B\cdot B)(C\cdot C)\equal{}B^2\cdot C^2\equal{}(B\cdot C)(B\cdot C)\equal{}B^2
\]
[/hide]

the table was this:

   * A  B  C
   A B  A  C
   B A  C  B
   C B  C  A

so it should have been B*B=C and C*C=A  so C*A should have been C..... but the answer says B.

[quote="GameBot"]In the accompanying numeration system, B * C = B. What is the value of (B * B) * (C * C)?[/quote]
C*A is B, according to your chart


Hello:

I have been using the excellent contest resource  that AOPS hosts. ( http://www.artofproblemsolving.com/Forum/resources.php ) The variety of problems present in there is truly amazing.

As I was going through it, however, I noticed that some of the problems were missing. For example, none of problems from the 2002 AMC 12 A and B are present. I have found them online, and I was wondering if there was anyway I could upload them onto the resource. I have some free time to do all the uploading, I think :)

Also, when I browse through the problems, I have difficulty viewing images. For example, #s 12 and 13 on AMC 12 2006 A have images along with them, but I simply see a "External_Image". Is that only a problem with me or are other users also affected by it?

Thanks,

Gautam

whoops. I browsed around and I did find something that answered my first question on the AOPS wiki.

but what about that second question

No, the images don't work for anyone because they were uploaded when AoPS was using a different server.  When they switched to a new server the images weren't transferred over.  It seems that at some point someone's going to have to upload new images into the galleries.  Either that or get the images from the old server back working.  However that would be a job for the admins, they're busy, and it would take at least few hours/days to do that.


linear algebra
matrix
search
vector
algebra
polynomial
induction

What conditions are necessary or sufficient for a real(or any field that is not closed) matrix to be triangularizable?
Anyone knows or has any problems that have a solution involving triangularizable matrices?
I found these here(loup blanc's posts), but only vaguely understand them:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=144008
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=633174368&t=184775
I would also kindly ask not to use any concepts more advanced than fields (vector spaces and God knows what).
Thanks.

triangularizable := similar to a triangular matrix = the characteristic polynomials splits completely = the minimal polynomial splits completely

The necessary and sufficient condition for a matrix to be triangularizable is simply that all of its eigenvalues belong the field in question. We can also write that as the fact that the characteristic polynomial can be completely factored into linear factors over the field in question. (The characteristic polynomial "splits".)

The proof isn't that hard. If the matrix is triagularizable, we can read the eigenvalues off of the main diagonal of the triangular form. For the converse, the proof proceeds by induction on the size of the matrix. Try it yourself. Suppose you have one eigenvector, $ Av\equal{}\lambda v.$ Extend $ v$ to a basis, which is to say, create an invertible matrix $ P$ with $ v$ as its first column. Now compute $ P^{\minus{}1}AP.$ What does it look like?

Okay.
I have a question:
1.loup blanc wrote in his solution from my first link:
[quote]Proof: $ rank(AB \minus{} BA)\leq1$ then A and B are simultaneously triangularizable over $ \mathbb{C}$.[/quote] I do not understand why this condition is necessary. Is there a difference between "A is triangularizable" and "A and B are simultaneously triangularizable"?

And some conclusions. Please correct me if I am wrong.
1.So basically any matrix whose elements belong to a closed field (like $ \mathbb {C}$, for example) is triangularizable .
2.If its eigenvalues belong to the field it's triangularizable;furthermore if the eigenvalues are distinct then it's diagonalizable.

The term "simultaneously triangularizable" means that there is some $ M$ such that $ M^{\minus{}1}AM$ and $ M^{\minus{}1}BM$ are [b]both[/b] upper triangular. It's a much stronger condition.

Your conclusions are correct. For the second, I like this statement: A matrix is diagonalizable if and only if its minimal polynomial splits as a product of distinct linear factors. Since the minimal polynomial is a factor of the characteristic polynomial, this happens whenever the characteristic polynomial splits this way.


calculus
integration
trigonometry
calculus computations

Find the value of $a$ for which $\int_{0}^{\frac{\pi}{2}}|a\sin x-\cos x|\ dx\ (a>0)$ is minimized.

When $a\sin(x)-\cos(x) < 0$, the sign should be reversed. So, letting $c=\tan^{-1}(\frac{1}{a})$, (which is the point of sign change of integrand)

$F(a) = \int_{0}^{\pi/2}|a\sin(x)-\cos(x)|dx$
$= \int_{0}^{c}(a\sin(x)-\cos(x))dx+\int_{c}^{\pi/2}-(a\sin(x)-\cos(x))dx$
Calculating the integral, we find
$F(a) =-2(a\sin(c)-\cos(c))+a+1$

To find the lowest rung, we solve $\frac{dF}{da}= 0$ for $a$. Calculating the derivate, we find
$\frac{1}{2}+\cos(c) = \frac{a\sin(c)-\cos(c)}{1+a^{2}}$
Plugging $a = \frac{1}{tan(c)}$, we get
$\frac{1}{2}+\cos(c) = 0$
Solving this for $a$ (which should be positive), minima occurs at $\frac{1}{\sqrt 3}$.

[color=darkblue]Define $x\ .s.s.\ y\Longleftrightarrow xy>0\ \vee \ x=y=0\ .$([u]same sign[/u]). If $a\le 0$, then $F(a)=\int_{0}^{\pi}{2}(\cos x-a\sin x)\ dx=1-a$ and $\boxed{\ \min_{a\le 0}F(a)=F(0)=1\ }\ .$
If $a>0$, then $a\sin x-\cos x\ .s.s.\ \tan x-\frac{1}{a}=\tan x-\tan\left(\arctan\frac{1}{a}\right)\ .s.s.\ x-\arctan\frac{1}{a}$ and in this case
$F(a)=\int_{0}^\frac{\pi}{2}\left|a\sin x-\cos x\right|\ dx=$ $\int_{0}^{\arctan\frac{1}{a}}(\cos x-a\sin x)\ dx+\int_{\arctan\frac{1}{a}}^\frac{\pi}{2}(a\sin x-\cos x)\ dx=$ $-1-a+2\sqrt{a^{2}+1}\ .$
But $F'(a)=-1+\frac{2a}{\sqrt{a^{2}+1}}\ .s.s.\ 2a-\sqrt{a^{2}+1}\ .s.s.\ 4a^{2}-(a^{2}+1)=$ $3a^{2}-1\ .s.s.\ a-\frac{\sqrt 3}{3}$, i.e. $a_{\mathrm{min}}=\frac{\sqrt3}{3}\ .$
Therefore,  $\boxed{\ \min_{a>0}F(a)=F\left(\frac{\sqrt 3}{3}\right)=\sqrt 3-1\ }\ .$ In conclusion, $\min_{a\in R}F(a)=F\left(\frac{\sqrt 3}{3}\right)=\sqrt 3-1\ .$[/color]

Your answers are correct. :)
Here is my solutuon.

For $t$ such that $a\sin t-\cos t=0\ \left(0<t<\frac{\pi}{2}\right),$

$\int_{0}^{\frac{\pi}{2}}|a\sin x-\cos x|\ dx$

$=\int_{0}^{t}|a\sin x-\cos x|\ dx+\int_{t}^{\frac{\pi}{2}}|a\sin x-\cos x|\ dx$

$\geq-\int_{0}^{t}(a\sin x-\cos x)\ dx+\int_{t}^{\frac{\pi}{2}}(a\sin x-\cos x)\ dx=(2\cos t-1)a+2\sin t-1.$

Thus determinimg the value of $t$ such that this equation takes the constant number independent of the value of $a,$ we obtain $t=\frac{\pi}{3}\Longleftrightarrow a=\frac{1}{\tan t}=\frac{1}{\sqrt{3}}.$

If $a$ takes this value, the given definite integral have the minimum value of $\sqrt{3}-1.$


let $ABCDE$ be a convex pentagon...

prove there exist three diagonals which are the sides of a triangle.

Cute !

Let's take $XY$ the longuest diagonal of the pentagon. There must be 2 vertices between $X$ and $Y$, say $Z$ and $T$. $XT$ and $ZY$ are diagonals and $XY$, $XT$ and $ZY$ work : we just have to prove that $XY \leq XT +ZY$ which is clear.

thanks...

im glad .because u know what $extermal$ is... ;)


trigonometry
geometry
perimeter
trig identities
Law of Sines

A certain triangle has angles $ 180\minus{}3x,2x,x$ and opposite sides $ n\plus{}1,n\plus{}2,n$ respectively. Find its perimeter.

By the law of sines, we obtain
$ \frac{n\plus{}1}{sin3x}\equal{}\frac{n\plus{}2}{sin2x}\equal{}\frac{n}{sinx}\equal{}\frac{1}{sin2x\minus{}sin3x}\equal{}\frac{1}{sin3x\minus{}sinx}$
So
$ sin2x\minus{}sin3x\equal{}sin3x\minus{}sinx$
$ sin2x\plus{}sinx\equal{}2sin3x$
$ sin(3x: 2)cos(x: 2)\equal{}2sin(3x: 2)cos(3x: 2)$
$ sin(3x: 2)(cos(x: 2)\minus{}2cos(3x: 2))\equal{}0$
Thus $ cos(x: 2)\equal{}2cos(3x: 2)\equal{}2(4cos^{3}(x: 2)\minus{}3cos(x: 2))$
$ cos^{2}(x: 2)\equal{}7: 8$
$ sinx\equal{}\frac{\sqrt{7}}{4}$
$ sin2x\equal{}\frac{6\sqrt{7}}{16}$
$ sin3x\equal{}\frac{5\sqrt{7}}{16}$
So we have $ n\equal{}4$ and the perimeter is $ 15$


algebra
polynomial
superior algebra
superior algebra solved

I hope with commute you mean $P(Q(x))=Q(P(x))$, because otherwise my solution is wrong...

Proof:
Let be $P(x)=x^m+a_{m-1} x^{m-1} + ... + a_1 x + a_0$
Now assume that there is an $a_i \neq 0, i<m$.
Then the second biggest exponent occuring in $P(Q(x))$ is $ni$ and the second biggest exponent occuring in $Q(P(x))$ is $(n-1)m+i$.
But this would give $ni=(n-1)m+i \iff (n-1)i=(n-1)m \iff i=m$, contradicting $i<m$.

There's also a proof [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=30555]here[/url], in post number $3$. :)

I have another proof. Suppose that $ P(x) $ commutes with $ Q(x) := x^{n} $.
I will prove that all roots of $ P(x) $ are zero. Suppose that it is not
so, i.e $ P(x) $ has a nonzero root $ a \in \mathbb{C} $. Then because of
$ P(x^{n}) = (P(x))^{n} $, we get that all $ n $ roots of $ z^{n} = a $ are
also roots of $ P(x) = 0 $. Now observe that if $ P(x) $ commutes with
$ Q(x) = x^{n} $, then for every $ k \in \mathbb{N} $, $ P(x) $ commutes with
$ Q_{k}(x) := x^{N} $, where $ N = n^{k} $. Now if we take such a $ k $, that
$ N > deg(P) $, then we will have that $ P(x) $ has at least $ N $ different
roots, so $ P(x) \equiv 0 $.
Therefore if $ P(x) $ is nonzero polynomial then it has the only root zero,
and because it is monic, then $ P(x) = x^{m} $ for some $ m $.

[quote="Leva1980"]We consider only monic polynomials.

 Prove that if for $ n>1 $ a polynomial $ P(x) $ commutes with a polynomial
$ Q(x) = x^{n} $, then it is of a form $ P(x) = x^{m} $ for some natural $ m $.[/quote]
I think you forgot the trivial solution P(x)=0 ;)

But I consider only monic polynomials  :D !


Sorry for the lameness, but can anyone show me why 0! = 1?? My teacher refused to explain she said just take it as a given.

Given:$1!=1$
$1\cdot0!=1$, because $n!=n\cdot(n-1)!$. On top of that, that equation is the identity property of multiplication, which means that $0!=1$ as well.

hmm...curse your awesome clarity. :ninja:

How many ways can you choose zero things? Exactly one way, right? Just
take none of them - anything else you do is wrong. :)

I understand n! = n (n-1)!.
but in that case,
0! = 0*(0-1)! and i thought you couldnt do negative factorials.
[quote]On top of that, that equation is the identity property of multiplication, which means that 0!=1 as well.[/quote]
is it just a given?

0!=1
It's just a definition, no proof involved. It's defined that way, so that the combinations and permutations work out and fit all the rules, like nCr=nC(n-r). If 0! were defined otherwise, then we would always have to state something like "excluding the 0 case", and we would have to have separate definitions for the 0 case.

It simplifies a lot of formulae.  For example, we need $0!=1$ to say that $\binom{n}{n}=\frac {n!}{n!\cdot 0!}=1$

The simplest explinations are:

1.  1! = 1*0! = 0! = 1.
2.  There is only 1 way to choose 0 objects out of n: do nothing!

There's also the calculus way... :D

\begin{eqnarray*}
0!  &=& \Gamma (1) \\
&=& \int_0^\infty t^{1-1} e^{-t}\, dt\\
&=& \int_0^\infty e^{-t} \, dt\\
&=& 1
\end{eqnarray*}


let $ f_n$ be the n-th Fibonacci number.

for $ q > 1$ let $ L_q(z)$ be the Lambert q-series:
\[ L_q(z) \equal{} \sum_{n \equal{} 1}^{\infty}\frac {z^n}{q^n \minus{} 1} \equal{} \sum_{n \equal{} 1}^{\infty}\frac {z}{q^n \minus{} z}
\]
in the disk $ |z| < q$.

prove that there exists 3 numbers $ A,B,q$ of the form $ q_1 \plus{} q_2\sqrt {5}$, $ q_1,q_2\in Q$, such that
\[ \sum\frac {1}{f_n} \equal{} AL_q(B)
\]
(probably $ A \equal{} \minus{} \frac {5 \plus{} \sqrt {5}}{2}$ , $ B \equal{} \frac {1 \plus{} \sqrt {5}}{2}$ , $ q \equal{} \minus{} \frac {3 \plus{} \sqrt {5}}{2}$ would work fine)


[quote="gauravpatil"]does anyone have shortlist 2007 ?please post iff available :D
 :lol:[/quote]
will not be [b]officially[/b] available till July 2008 :P


Let $ f$: $ \mathbb{N^*} \to \mathbb{N^*}$ such that:
$ \forall n \in \mathbb{N^*}$: $ f(n\plus{}f(n))\equal{}f(n)$

1)- Prove that if the set $ f(\mathbb{N^*})$ is finite then f is periodic
2)- Give an example of a such function but not periodic

Some one knows where was it proposed?

$ 1)$
if we take $ (a_i)$ such that $ a_{i \plus{} 1} \equal{} f(a_i) \plus{} a_i$ and $ a_0 \equal{} n$
then $ f(a_{i \plus{} 1}) \equal{} f(a_i)$ then $ \forall i\in\mathbb{N}: \ f(n) \equal{} f(a_i)$
and $ a_{i \plus{} 1} \equal{} f(a_i) \plus{} a_i \equal{} f(n) \plus{} a_i$ gives $ a_i \equal{} if(n) \plus{} n$
then $ \forall i\in\mathbb{N}: \ f(n) \equal{} f(n \plus{} if(n))$
we suppose that $ f(\mathbb{N}^*) \equal{} \{b_1,b_2,...,b_h\}$ finite and we take $ T \equal{} ppcm(b_1,b_2,...,b_h)$
so $ f(n \plus{} T) \equal{} f(n)$ because $ \exists m:$ such that $ T \equal{} mf(a)$

$ 2)$
example:
we take $ f$ such that $ f(n) \equal{} 2^{a \plus{} 1}\iff 2^a|n\ and\ 2^{a \plus{} 1}\not|n$
or $ \forall n,k\in\mathbb{N}: \ f(2^m(2k \plus{} 1)) \equal{} 2^{m \plus{} 1}$

so if $ n \equal{} 2^m(2k \plus{} 1)$
$ f(n \plus{} f(n)) \equal{} f(2^m(2k \plus{} 1) \plus{} 2^{m \plus{} 1}) \equal{} f(2^{m}(2k \plus{} 3)) \equal{} 2^{m\plus{}1} \equal{} f(n)$

Sorry for not typing LATEX, as i'm a new member and i don't have a math career.

The first part is quite easy, for the second part, i got the same sample as aviateurpilot did. I just show here the road to this sample:

In fact, what we need is an unbounded sequence a1<a2<...< an<.... such that

 i. The sets  Si={ ai+ k.f(ai)| k=0,1,.... } are pairwise disjoint

ii.  Union of S1,..,Sn,... covers  N*

Hence, for t belongs to [0,1) we have

  1+t+t^2+.....= sum: ( t^ai. ( 1+t^f(ai)+... )

that yields

 1/(1-t) = sum: (t^ai/ 1- t^f(ai))

Remark that

1/(1-t^2^n)= 1/(1-t^2^(n+1)) + t^2^n/  1- t^2^(n+1)

Therefore

1/(1-t)= sum: (t^2^i/ 1- t^2^(i+1))

So it 's evident to choose   ai= 2^i and f(ai) = 2^(i+1)

We could so that there are infinite number of nonperiodic funtion f such that f(n+f(n)=f(n) holds for every positive integer n


Determine k and solve the equation x^3 - kx^2 + 20x +12 = 0, if one of hte zeros is the triple of another.

[hide="solution?"]Assuming one root is $ b$, one is $ a$ and the other is $ 3a$, the factored form of the polynomial is:
$ (x\minus{}a)(x\minus{}3a)(x\minus{}b) \equal{} 0$
Expanding this new polynomial leads to:
$ x^{3}\minus{}4ax^{2}\plus{}3a^{2}x\minus{}bx^{2}\plus{}4abx\minus{}3a^{2}b \equal{} 0$
After sorting like terms, we have
$ x^{3}\minus{}(4a\plus{}b)x^{2}\plus{}(3a^{2}\plus{}4ab)x\minus{}3a^{2}b \equal{} 0$, compared to the original,
$ x^{3}\minus{}kx^{2}\plus{}20x\plus{}12 \equal{} 0$

So:
$ k \equal{} 4a\plus{}b$
$ 20 \equal{} 3a^{2}\plus{}4ab$  and
$ \minus{}12 \equal{} 3a^{2}b$

At this point, I'm sure there's a better method (I would appreciate it if someone could show me one such method, I'm drawing a blank right now), but using guess and check, I arrived at $ a \equal{}\minus{}2$ and $ b \equal{}\minus{}1$, which leads to $ k \equal{}\minus{}9$.

Replacing variables with their values, the original equation, when factored, is:
$ (x\plus{}2)(x\plus{}6)(x\plus{}1) \equal{} 0$
So, the roots are $ x \equal{}\minus{}2,\minus{}6,\minus{}1$[/hide]

Using the second two equations, you can easily eliminate $ b$, leaving a cubic in $ a$. You may not have found all solutions with your guess and check.

Doing that gave me $ a^{3}\minus{}10a\minus{}8\equal{}0$, but how would I go about solving that?

[quote="Lazarus"]It's a depressed cubic. You could bash it through the unholy mess that is Cardano's formula.[/quote]

He also has already found one root, so you can factor it out and solve the quadratic ;)

So, factoring out one root always leaves me with a quadratic that can be factored into the two other roots. Does this mean these are the only solutions, or am I missing something?

Also, could someone point me towards Cardano's forumla? It's not in the AoPS Wiki.


geometry
parallelogram
geometry unsolved

In a parallelogram $ ABCD$, $ M$ is the midpoint of $ BC$ and $ N$ an arbitrary point on the side $ AD$. Let $ P$ be the intersection of $ MN$ and $ AC$, and $ Q$ the intersection of $ AM$ and $ BN$. Prove that the triangles $ BDQ$ and $ DMP$ have equal areas.

indeed we have to prove PQ||BC,
but it's obvious that  (triangle)ANP~(triangle)PMC and (triangle)ANQ~(triangle)BQM; so ...


abstract algebra
topology
calculus
integration
algebra
function
domain

I've got stuck trying to prove following theorem which I found in lecture notes:

[i]Let $ A, R$ be commutative rings with unity, and $ \lambda: R\to A$ be a homomorphism of such rings. We consider R-algebra $ (A,\lambda)$. The following conditions are equivalent:

(i) $ A$ is faithfully flat R-algebra (i.e. $ A$ is faithfully flat as R-module, with respect to multiplication $ ra \equal{} \lambda(r)a,$ for $ r\in R, \ a\in A$).

(ii) Map $ \text{Spec}(\lambda): \text{Spec}(A)\ni \mathfrak{q} \mapsto \mathfrak{q}^{c}: \equal{} \lambda^{ \minus{} 1}(\mathfrak{q}) \in \text{Spec}(R)$ is surjective.

(iii) $ \text{Spec}(\lambda)(\text{Max}(A)) \equal{} \text{Max}(R),$ where $ \text{Max}(A)$ is set of all maximal ideals in $ A.$[/i]

Here is the definition of faithfully flat module which I use:
[i]Let $ R$ be a commutative ring with unity.
R-module $ N$ is called faithfully flat if it is flat and the following equivalent conditions are true:
(a) $ N \neq \mathfrak{m}N$ for all $ \mathfrak{m}\in \text{Max}(R).$
(b) For every R-module $ M$ if $ M\otimes N \equal{} 0$ then $ M \equal{} 0.$
(c) For every R-modules $ M, M'$ and homomorphism $ f: M'\to M$ if induced homomorphism $ f\otimes 1_{N}: M'\otimes N\to M\otimes N$ is an isomorphism, then $ f$ is an isomorphism.[/i]

Now I can show that $ (i)\implies (ii),$ and it is easy to see that $ (ii)\implies \text{Spec}(\lambda)(\text{Max}(A)) \supset \text{Max}(R),$ and $ \text{Spec}(\lambda)(\text{Max}(A)) \supset \text{Max}(R)\implies (a),$  but I've no idea how to show that $ (ii) \implies \text{Spec}(\lambda)(\text{Max}(A)) \subset \text{Max}(R)$ and that from (iii) follows the flatness of $ A$... :|

I'll appreciate:
-any suggestions how to finish this proof,
-any other approaches to this theorem.
:)

I think you are loosing too much information in the step (i) $ \rightarrow$ (ii), since "faithfully flat = flat + surjective on Spec's". Suppose $ Spec(\lambda)$ is a bijective closed immersion, e.g. $ \lambda: R\to A \equal{} R/nil(R)$, then there is no way back from (ii) to (i) in general. Assuming that $ R$ is a finitely generated $ k$-algebra, then this example above generalizes to that (iii) $ \rightarrow$ (i) fails to be true in general, by Hilbert's Nullstellensatz.

So I think it is best to check the different assertions of the equivalences. :) A very blunt guess to amend the claims is to add to (ii) and (iii) that $ \lambda$ be flat. ;)

Thank You for immediate response:)

I thought of possibility that the flatness of $ A$ should be assumed before conditions $ (i) \minus{} (iii)$ are stated, but I couldn't provide any example.

I'm at very beginning of discovering some basic notions of commutative algebra, so it took me a while, to find out how it works:

If $ A \equal{} R/\text{nil}(R)$ and $ \lambda: R\to R/\text{nil}(R)$ is canonical projection, then for every R-module $ M$ we have 'standard' isomorphism $ M\otimes_{R}A\cong M/\text{nil}(R)M.$
In particular for every ideal $ I\triangleleft R$ we have $ I\otimes_{R}A\cong I/\text{nil}(R)I.$
Now if $ A$ is flat over $ R$ then for every $ I\triangleleft R$ 'standard' epimorphism $ I\otimes_{R}A\to IA$ is isomorphism (in fact it is equivalent to flatness of $ A$, but this time we need only this much easier implication).
Hence, when $ A$ is flat, the homomorphism $ g: I/\text{nil}(R)I\ni a \plus{} \text{nil}(R)I \mapsto a(1 \plus{} \text{nil}(R))\in IA$ as a composition of isomorphisms above is itself isomorphism.
But $ \text{ker}\, g \equal{} (\text{nil}(R)\cap I)/\text{nil}(R)I,$ so $ A$ cannot be flat if $ \text{nil}(R)\cap I\supsetneq \text{nil}(R)I.$
Putting $ k$ field, $ R \equal{} k[X,Y]/(X^{2}Y), \ I \equal{} (Y^{2} \plus{} (X^{2}Y))$ we obtain concrete example:
it's not hard to see that $ \text{nil}(R) \equal{} (XY \plus{} (X^{2}Y)),$ and then $ XY^{2} \plus{} (X^{2}Y) \in (\text{nil}(R)\cap I)\setminus (\text{nil}(R)I)$
(the less trivial (but still) part: if $ XY^{2} \plus{} (X^{2}Y) \in \text{nil}(R)I$ then $ XY^{2} \minus{} fXY^{3} \equal{} gX^{2}Y$ for some $ f,g\in k[X,Y],$ thus $ XY^{2} \in (X^{2}, Y^{3})$ - contradiction).

Since the induced map: $ \text{Spec}(R/\text{nil}R)\to \text{Spec}(R)$ is bijective (even homeomorphic) and preserves maximal ideals, we have $ \text{Spec}(\lambda)(\text{Max}(A)) \equal{} \text{Max}(R),$ and this proves that (ii) and (iii) are not equivalent with (i).

Assuming flatness of $ A$ we have $ (a) \implies (i)$ and hence I've got the proof of the theorem with $ "\supset"$ instead of $ " \equal{} "$ in (iii).
But I'm still confused about proving $ "\subset "$ part of (iii) using (i) and (ii) :oops:  May I ask You for help with this problem? :maybe:

[quote="ifk"]Assuming flatness of $ A$ we have $ (a) \implies (i)$ and hence I've got the proof of the theorem with $ "\supset"$ instead of $ " = "$ in (iii).
But I'm still confused about proving $ "\subset "$ part of (iii) using (i) and (ii) :oops:  May I ask You for help with this problem? :maybe:[/quote]
Here is a counterexample for this inclusion: $ \lambda: \mathbb Z\to A=\mathbb Q\times \mathbb Z$. Since $ \mathbb Q$ is a localization of $ \mathbb Z$ w.r.t. $ \mathbb Z\setminus\{0\}$, it is a flat $ \math Z$-algebra, so is $ A$. Using (b) above $ A$ is a faithfully flat $ \mathbb Z$-algebra. Now the maximal ideal $ \{0\}\times\mathbb Z$ maps to $ \{0\}$ via $ Spec(\lambda)$.

[quote="ifk"]If $ A \equal{} R/\text{nil}(R)$ and $ \lambda: R\to R/\text{nil}(R)$ is canonical projection, then for every R-module $ M$ we have 'standard' isomorphism $ M\otimes_{R}A\cong M/\text{nil}(R)M.$
In particular for every ideal $ I\triangleleft R$ we have $ I\otimes_{R}A\cong I/\text{nil}(R)I.$
Now if $ A$ is flat over $ R$ then for every $ I\triangleleft R$ 'standard' epimorphism $ I\otimes_{R}A\to IA$ is isomorphism (in fact it is equivalent to flatness of $ A$, but this time we need only this much easier implication).[/quote]
From here it is easy to conclude without any tedious calculation with elements: Letting $ I\equal{}nil(R)$ we get $ I/I^2\equal{}0$ which implies $ I\equal{}0$, by the Nakayama lemma (with the additional assumption that $ I$ is finitely generated - which is the case in your example).

Your posts are really helpful, thanks again:)

So, that's why I couldn't prove it... :|
I hope in this case I'm the only one who waste so many time trying to prove false statement. If not, I'm sorry...

Anyway, the flatness of $ \mathbb{Q}\times\mathbb{Z}$ follows also from the fact, that it is torsion-free $ \mathbb{Z}$-module, and $ \mathbb{Z}$ is a PID.
And then faithfully flatness is also an easy application of (a) above.

Remark about NAK is also illuminating:)

[quote="ifk"]Anyway, the flatness of $ \mathbb{Q}\times\mathbb{Z}$ follows also from the fact, that it is torsion-free $ \mathbb{Z}$-module, and $ \mathbb{Z}$ is a PID.[/quote]That's nice. :) At first I had the general thing $ B\to Quot(B)\times B$ for an integral domain $ B$ in mind with the general reasoning as above, then I changed it to $ B \equal{} \mathbb Z$ to keep the counterexample as simple as possible.