MIT
college
induction
geometry
geometric transformation

Hallo!

jetzt wo Meldeschluss offiziell vorbei ist (1. Sept.), können wir ja über die Aufgaben der 2. Runde diskutieren! Wie fandet ihr sie und wie habt ihr sie gelöst?  :lol:  :lol:
Ich persönlich fand die 4. Aufgabe am interessantesten und die 2. irgendwie komisch!
naja freu mich auf eure meinung

alex

Also bei der 1 hab ich die Idee ziemlich schnell gehabt. Es war zwar dann noch ein keiner Fehler drin, aber den konnte ich beheben.
Bei der 2 hats eine Zeit gedauert bis ich auf das Muster gekommen bin. Das Muster konnte ich dann eigentlich auch recht schnell beweisen... aber zu dem Zeitpunkt (also als ichs bewiesen hatte) hab ich die Lösung noch nicht so richtig verstanden.  :D

Die 3 war jetzt nicht unbedingt schwer rauszukriegen aber dafür auch nicht so toll aufzuschreiben.

Die 4 hab ich ewig nicht rausbekommen. Aber irgendwann hats doch geklappt  :lol:

Die erste war noch einigermaßen freundlich. Ich habs mit induktion über n gelöst....und du?

Die zweite fand ich scheiße.  :|  Da kommt man zwar schnell auf irgendwas aber es war schlimm alles aufzuschreiben....mega wirr...

Hast du bei der dritten aufgabe den Tetraeder in einen Würfel einbeschrieben, so dass 4 Eckpunkte übereinstimmen?
Das hab ich jedenfalls gemacht und bei mir kam dann raus, dass die restlichen 4 eckpunkte genau nicht zur menge L(L(T)) dazugehören. Außer denen gehören alle Punkte im Raum dazu...
Oder hast du es einfach analytisch ausgerechnet?

Naja zur vierten: Da hab ich einfach alle Kanten mit Pfeilen beschriftet, so dass der Pfeil von der kleineren Zahl zur größeren zeigt. Dann kann man nach schauen ob es im uhrzeigersinn oder gegen den uhrzeigersinn um das dreieck zeigt. Mit dem kommt man dann auf ne ungleichung wo genau das ergebnis raus kommt.

alex

ja bei der 1 hab ich auch Induktion verwendet.  Ich definiere rekursiv eine Folge von Zahlen und beweise eben, dass die Folge die geforderten Eigenschaften hat. Hast du eigentlich gezeigt, dass n eine natürliche ZAhl sein muss ?  :D

Bei der 2 hab ich eine Lösung gesehen die doch recht schön ist. Naja meine is eher so eine haudrauf Lösung  :wink:  P.s. Bei dir kommt nicht zufällig was mt 2er Potenzen vor?

Zur 3. Ja das mit dem Würfel kann gut stimmen (also ich hab auch 4 Punkte die nicht in L(L(T))) liegen. Hab aber die Punkte anders angegeben (mit Vektoren)
Und eigentlich rechne ich auch in der ganzen Lösung kaum was... das ist meistens immer logisch argumentieren.

Zur 4. Dann hast du wohl die schöne/kurze Lösung gefunden? Meine is schon etwas länger ... so 5 Seiten.

Ja bei der 2. kommt bei mir auch was mit ner 2er potenz vor. Ich habe r ung g als produkt der größten gemeinsamen 2er potenz aufgeschrieben.

r = 2^x * R  und g = 2^x*G
Je nach dem ob R und G oder nur einer der beiden ungerade ist, kann man bestimmen wer gewinnt....hast du des auch so ähnlich?

Ja bei der zweiten hab ich auch mit vektoren gerechnet und da kam bei mir zufällig raus, dass der abstand zweier punkte wurzel 2 war. Das hat mich inspiriert...

Ja die 4. ist bei mir 2 seiten lang...das war aber ein ziemlicher glücksfall...falls es richtig ist...;-)
naja wie hast du die letzte gelöst?

Woher kommst du eigentlich?

alex

Ich hab die gleichen Ergebnisse wie du, alex. die erste aufgabe hab ich aber eher straight forward als mit induktion gemacht. (es sei k=  ...100000010001011 (jeweils an 2^i-ter stelle eine 1, aus q(k)=n => q(k^2)=n^2)
Die zwei war eigentlich total einfach, wenn man das muster mal erkannt hat. Es müssen gleichviele 2er Faktoren in g und r sein, damit Bernd gewinnen kann.
Bei der 3 hab ich ziehmlich lang gebraucht, bis ich das mit dem würfel geschnallt hab. ich hab dann in meinem beweis erstmal drei seiten lang erörtert, warum man einen tetraeder stets mit genau einem würfel umschreiben kann.. der rest war mit analytischer geometrie dann schnell gemacht.
die vierte hab ich auf 1 1/4 seiten ;-) zuerst hatte ich eine sehr umständliche beweisidee (wären wohl so 10-15 seiten geworden, auch wenn ich nach wie vor überzeugt bin, dass es geklappt hätte). kommt man auf die idee, dass jedes uhrige dreieck genau zwei seiten hat bei denen die vektoren im uzs zeigen, geht es richtig kurz! (ich hab es anderst formuliert, läuft aber aufs gleiche hinaus)

[quote="aaton"]
naja wie hast du die letzte gelöst?
[/quote]
Ich versuch mal die Datei anzuhängen

[quote="aaton"]
Woher kommst du eigentlich?
[/quote]
Aus der Nähe von Regensburg (genauer wäre Cham, aber das sagt normalerweise keinem was  :) )
Ab Oktober wohne ich dann in München. Fang mein BWL Studium an....  :wink:  Nein natürlich Mathe. (an der TU)

darijs kommentar: BWL und mathe an TU... großer unterschied? ;-)

naja, ich hab die aufgaben nen tag zu spät abgegeben (bzw ich hab sie schon am samstag ind en briefkasten geworfen, aber vergessen, dass samstag ist und es deswegen weder ne nachmittags-noch ne spätleerung gibt... mal schaun)

aufgabe eins ging schnell und hübsch - habs auch über ne rekursiv definierte formel (und nein, keine zweierpotenzen) gemacht.

aufgabe zwei hatten wir zumindest so ähnlich (ich würde sogar behaupten, dass es genau die gleiche war, bin mir da aber net 100% ig sicher) schon mal im pluskurs vor zwei jahren - dementsprechend ging sie sehr schnell.

aufgabe drei ging wohl schnell zu lösen (und mittlerweile hab ich auch ein paar nette wege zum aufschreiben gesehen) - allerdings hab ich se dann analytisch durchgerechnet und ne 15-seiten-lösung _kann_ nicht hübsch sein

und bei aufgabe vier hatte ich am anfang ne hässliche hässliche fallunterscheidung, aber ne woche vor abgabeschluss (nachdem ich auf keinen fall diese fallunterschiedung aufschreiben wollte) ne schöne lösung mit positiven und negativen kanten ;-)

edit. hallo michael ;-)

i don't know german . :(
so is urBundeswettbewerb Mathematik over(seeing the round 2 at top an immediate guess is  yes) so can u post ur questions

[quote="kathi;)"]darijs kommentar: BWL und mathe an TU... großer unterschied? ;-)
[/quote]
 :rotfl:  War klar!

[quote="kathi;)"]
 hallo michael ;-)
[/quote]

Hi Kathi... bist du beim Probestudium?

@pardesi: Yes the constest is over. At least the entry deadline was the 1. September.

Edit: Maybe I did misunderstand you. Did you give us permission to talk about the test  :wink:  or do you want an translation of every task?

actually i wanted the questions(If any ) of the test

Here you go. (Hope you unterstand everything. My english is not the best  :) )

thanks sqrt4  :clap2: .it was really helphul.by the way are these the first round questions?

[quote="sqrt4"]

[quote="kathi;)"]
 hallo michael ;-)
[/quote]

Hi Kathi... bist du beim Probestudium?

[/quote]

nur so halb, ich war am dienstag da und werd morgen nomal hinfahren, gestern und heut war ich zu hause. du bist nicht da, nehm ich an. kommst du morgen?^^

The 1st round question are already on ML, just look for them...
edit: im sorry, i think they aren't. something we have to do...

und noch ein edit: hallo, alex ^^

I've got nothing to do  :wink:  so here are the problems of the 1st round.

P.s. Nein ich komm morgen nicht.

Weiß jemand noch wann die lösungen letztes mal auf der website veröffentlicht wurden?

@kathi: bist du ne organistatorin im q.e.d.??? und warst du in pleinfeld dabei?

[quote="aaton"]Weiß jemand noch wann die lösungen letztes mal auf der website veröffentlicht wurden? [/quote]
hm... hab ich mich auch schon gefragt, weiß ich aber nimmer^^

[quote="aaton"]
@kathi: bist du ne organistatorin im q.e.d.??? und warst du in pleinfeld dabei?[/quote]

ja, genau das bin ich. ;-)  und du heißt nicht michael, sondern alexander und hast auch keinen bruder, der michael heißt ;-)

:rotfl:

Im übrigen ist nur Aufgabe 3 hässlich, der Rest ist eigentlich recht gut machbar.
1.: Einfach eine Zahl angeben und nachweisen.
2.: Bernd gewinnt $ \iff$ $ v_{2}(r)\equal{}v_{2}(s)$, wobei $ v_{2}(n)$ die sogenannte $ 2$-adische Bewertung ist, also die Zahl der Faktoren $ 2$ in der Primzerlegung von $ n$. Beweis per Induktion.
3.: Laut Darij auch schön machbar, er darf sich gerne dazu äußern...
4.: Einfach die Kanten gerichtet betrchten, abzählen. Am besten gleich sowas wie Uhrigkeit eines $ n$-Ecks definieren und das ganze auf einer Kugel betrachten ;)

[quote="ZetaX"]
3.: Laut Darij auch schön machbar, er darf sich gerne dazu äußern...
[/quote]

darij erzählte was von baryzentrischen koordinaten

fast ;-)

[The message is too small. Please make the message longer before submitting.]

Die Lösungen kommen für gewöhnlich nach etwa 3 Wochen ins Netz.

Zur 3: Sei O ein fester Punkt im Raum (der "Ursprung"). Sind A, B, C und D vier Punkte im Raum, die nicht alle in einer Ebene liegen, dann gibt es für jeden Punkt P im Raum ein eindeutiges Quadrupel (x, y, z, w) von reellen Zahlen, sodaß $ x\plus{}y\plus{}z\plus{}w \equal{} 1$ und $ x\cdot\overrightarrow{OA}\plus{}y\cdot\overrightarrow{OB}\plus{}z\cdot\overrightarrow{OC}\plus{}w\cdot\overrightarrow{OD}\equal{}\overrightarrow{OP}$ ist. Dieses Quadrupel (x, y, z, w) heißt die [i]baryzentrischen Koordinaten[/i] des Punktes P bezüglich des Tetraeders ABCD.

(Wie man dieses Quadrupel findet? Indem man $ x \equal{}\frac{\left[PBCD\right]}{\left[ABCD\right]}$, $ y \equal{}\frac{\left[PCDA\right]}{\left[ABCD\right]}$, $ z \equal{}\frac{\left[PDAB\right]}{\left[ABCD\right]}$, $ w \equal{}\frac{\left[PABC\right]}{\left[ABCD\right]}$ setzt, wobei die Abkürzung $ \left[X_{1}X_{2}X_{3}X_{4}\right]$ für das gerichtete Volumen eines Tetraeders $ X_{1}X_{2}X_{3}X_{4}$ steht. Daß das die Lösung ist trau ich dir zu selber nachzuweisen, Kathi ;) )

Nun wird $ x\cdot\overrightarrow{OA}\plus{}y\cdot\overrightarrow{OB}\plus{}z\cdot\overrightarrow{OC}\plus{}w\cdot\overrightarrow{OD}\equal{}\overrightarrow{OP}$  zu
$ \left(x\plus{}y\right)\cdot\frac{x\cdot\overrightarrow{OA}\plus{}y\cdot\overrightarrow{OB}}{x\plus{}y}\plus{}\left(z\plus{}w\right)\cdot\frac{z\cdot\overrightarrow{OC}\plus{}w\cdot\overrightarrow{OD}}{z\plus{}w}\equal{}\overrightarrow{OP}$.
Sei nun U der Punkt mit $ \frac{x\cdot\overrightarrow{OA}\plus{}y\cdot\overrightarrow{OB}}{x\plus{}y}\equal{}\overrightarrow{OU}$. Dann liegt U auf der Geraden AB. Sei V der Punkt mit $ \left(z\plus{}w\right)\cdot\frac{z\cdot\overrightarrow{OC}\plus{}w\cdot\overrightarrow{OD}}{z\plus{}w}\equal{}\overrightarrow{OV}$. Dann liegt V auf der Geraden CD. Obige Gleichung wird nun zu
$ \left(x\plus{}y\right)\cdot\overrightarrow{OU}\plus{}\left(z\plus{}w\right)\cdot\overrightarrow{OV}\equal{}\overrightarrow{OP}$.
Wegen $ x\plus{}y\plus{}z\plus{}w \equal{} 1$, also $ \left(x\plus{}y\right)\plus{}\left(z\plus{}w\right) \equal{} 1$, wird dies zu
$ \frac{\left(x\plus{}y\right)\cdot\overrightarrow{OU}\plus{}\left(z\plus{}w\right)\cdot\overrightarrow{OV}}{\left(x\plus{}y\right)\plus{}\left(z\plus{}w\right)}\equal{}\overrightarrow{OP}$.
Daraus folgt daß P auf UV liegt... Wirklich? Das würde heißen, daß bei der Aufgabe 3 alle Punkte im Raum in L(L(T)) liegen, und dem ist nicht so. Wo ist der Haken? Nunja, wir haben durch $ x\plus{}y$ und durch $ z\plus{}w$ geteilt, und es kann natürlich sein daß eine dieser Zahlen gleich $ 0$ ist.

Das Argument schlägt also fehl wenn $ x\plus{}y \equal{} 0$ oder $ z\plus{}w \equal{} 0$ ist. Analog konstruieren wir ein Argument, das schiefgeht wenn $ y\plus{}z \equal{} 0$ oder $ w\plus{}x \equal{} 0$ ist, und eins, das schiefgeht wenn $ x\plus{}z \equal{} 0$ und $ y\plus{}w \equal{} 0$.

Somit muß ein Punkt P, für den alle drei Argumente schiefgehen, (und wenn ein Punkt P nicht in L(L(T)) liegt, dann müssen für ihn alle drei Argumente schiefgehen) folgendes gleichzeitig erfüllen:

$ \left(x\plus{}y \equal{} 0\text{ oder }z\plus{}w \equal{} 0\right)\text{ und }\left(y\plus{}z \equal{} 0\text{ oder }w\plus{}x \equal{} 0\right)\text{ und }\left(x\plus{}z \equal{} 0\text{ oder }y\plus{}w \equal{} 0\right)$.

Die Fälle, für die dies gilt, lassen sich an der Hand abzählen: das sind erstmal P = A, P = B, P = C und P = D, aber diese liegen trivialerweise in L(L(T)), und dann vier Punkte mit baryzentrischen Koordinaten $ \left(\minus{}\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$ und zyklisch. Das sind die einzigen vier Punkte, die nicht in L(L(T)) liegen...

Übrigens kann das Tetraeder ABCD beliebig sein - so lange nicht A, B, C, D in einer Ebene liegen. Aber ich kann mir eh nicht vorstellen, wo man bei einer Lösung verwenden kann, daß das Tetraeder ABCD regulär ist...

  Darij

[quote="darij grinberg"]
(Wie man dieses Quadrupel findet? Indem man $ x \equal{}\frac{\left[PBCD\right]}{\left[ABCD\right]}$, $ y \equal{}\frac{\left[PCDA\right]}{\left[ABCD\right]}$, $ z \equal{}\frac{\left[PDAB\right]}{\left[ABCD\right]}$, $ w \equal{}\frac{\left[PABC\right]}{\left[ABCD\right]}$ setzt, wobei die Abkürzung $ \left[X_{1}X_{2}X_{3}X_{4}\right]$ für das gerichtete Volumen eines Tetraeders $ X_{1}X_{2}X_{3}X_{4}$ steht. Daß das die Lösung ist trau ich dir zu selber nachzuweisen, Kathi ;) )
[/quote]

danke ;-)

[quote="darij grinberg"]
Übrigens kann das Tetraeder ABCD beliebig sein - so lange nicht A, B, C, D in einer Ebene liegen. Aber ich kann mir eh nicht vorstellen, wo man bei einer Lösung verwenden kann, daß das Tetraeder ABCD regulär ist...

 [/quote]

naja, wenn du es tatsächlich analytisch durchrechnest *unbeteiligt in die luft schau* braucht man es zum einen, um anfangskoordinaten reinzulegen (also hier braucht man es wirklich) und dann, wenn ne eeeeewig lange lösung hast, kannst du sie dadurch abkürzen, indem du verwendest, dass das tetraeder regulär ist und man es deswegen nur für ein windschiefes geradenpaar z.b. durchrechnen muss und für die andern beiden paare das ganze jeweils gleich zum tetraeder liegen muss. wenn es nicht regulär wäre, könnte man das nicht verwenden (hier vereinfacht / verkürzt es die rechnung)

edit [consecutive posts are not allowed in lesser than ten minutes] [quote="aaton"]Naja ich denke in der aufgabe wurde ein regelmäßiges tetraeder genommen, damit man das ganze auch elementar lösen kann. Das regelmäßiges Tetraeder kann man in einen würfel einbeschreiben, so dass die diagonalen gleichzeitig die kanten des Tetreaders sind und 4 eckpunkte des Würfels mit den 4 eckpunkten des Tetraeders übereinstimmen. Die übrigen 4 punkte sind dann genau die Punkte die nicht zu L(L(T)) gehören. Oder ist des falsch?[/quote]

nein, ist richtig

[quote="kathi;)"][quote="darij grinberg"]
Übrigens kann das Tetraeder ABCD beliebig sein - so lange nicht A, B, C, D in einer Ebene liegen. Aber ich kann mir eh nicht vorstellen, wo man bei einer Lösung verwenden kann, daß das Tetraeder ABCD regulär ist...

 [/quote]

naja, wenn du es tatsächlich analytisch durchrechnest *unbeteiligt in die luft schau* braucht man es zum einen, um anfangskoordinaten reinzulegen (also hier braucht man es wirklich) und dann, wenn ne eeeeewig lange lösung hast, kannst du sie dadurch abkürzen, indem du verwendest, dass das tetraeder regulär ist und man es deswegen nur für ein windschiefes geradenpaar z.b. durchrechnen muss und für die andern beiden paare das ganze jeweils gleich zum tetraeder liegen muss. wenn es nicht regulär wäre, könnte man das nicht verwenden (hier vereinfacht / verkürzt es die rechnung)[/quote]

OMG. Ein [i]regelmäßiges[/i] Tetraeder in ein [i]rechtwinkliges[/i] Koordinatensystem einzubeschreiben ist in der Tat ein Ansatz, der mir nicht im Alptraum eingefallen wäre. Aber das Analogieargument geht genauso gut für jedes andere Tetraeder, da muss keine echte Symmetrie vorliegen.

[quote="aaton"]Naja ich denke in der aufgabe wurde ein regelmäßiges tetraeder genommen, damit man das ganze auch elementar lösen kann. Das regelmäßiges Tetraeder kann man in einen würfel einbeschreiben, so dass die seitenflächendiagonalen gleichzeitig die kanten des Tetreaders sind und 4 eckpunkte des Würfels mit den 4 eckpunkten des Tetraeders übereinstimmen. Die übrigen 4 punkte sind dann genau die Punkte die nicht zu L(L(T)) gehören. Oder ist des falsch?[/quote]

Du kannst genauso gut um jedes beliebige (nicht-entartete) Tetraeder eindeutig ein Parallelepiped umbeschreiben. Und ja, die Punkte mit den baryzentrischen Koordinaten $ \left(\minus{}\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$ und zyklisch sind die vier von A, B, C, D verschiedenen Ecken dieses Parallelepipeds.

  darij

[quote="darij grinberg"][quote="kathi;)"][quote="darij grinberg"]
Übrigens kann das Tetraeder ABCD beliebig sein - so lange nicht A, B, C, D in einer Ebene liegen. Aber ich kann mir eh nicht vorstellen, wo man bei einer Lösung verwenden kann, daß das Tetraeder ABCD regulär ist...

 [/quote]

naja, wenn du es tatsächlich analytisch durchrechnest *unbeteiligt in die luft schau* braucht man es zum einen, um anfangskoordinaten reinzulegen (also hier braucht man es wirklich) und dann, wenn ne eeeeewig lange lösung hast, kannst du sie dadurch abkürzen, indem du verwendest, dass das tetraeder regulär ist und man es deswegen nur für ein windschiefes geradenpaar z.b. durchrechnen muss und für die andern beiden paare das ganze jeweils gleich zum tetraeder liegen muss. wenn es nicht regulär wäre, könnte man das nicht verwenden (hier vereinfacht / verkürzt es die rechnung)[/quote]

OMG. Ein [i]regelmäßiges[/i] Tetraeder in ein [i]rechtwinkliges[/i] Koordinatensystem einzubeschreiben ist in der Tat ein Ansatz, der mir nicht im Alptraum eingefallen wäre. Aber das Analogieargument geht genauso gut für jedes andere Tetraeder, da muss keine echte Symmetrie vorliegen.

 darij[/quote]

puh, naja, okay,.... mir _ist_ es eingefallen... aber ich hab sogar überlegt, ein nicht-rechtwinkliges kosy zu nehmen, nur hab mich dann doch dagegen entschieden ;-)

Die Lösungen sind jetzt endlich online!!!  :)  :)
http://www.bundeswettbewerb-mathematik.de


modular arithmetic
number theory proposed
number theory

Two nonnegative integers $a$ and $b$ are [i]tuanis[/i] if the decimal expression of $a+b$ contains only $0$ and $1$ as digits. Let $A$ and $B$ be two infinite sets of non negative integers such that $B$ is the set of all the [i]tuanis[/i] numbers to elements of the set $A$ and $A$ the set of all the [i]tuanis[/i] numbers to elements of the set $B$. Show that in at least one of the sets $A$ and $B$ there is an infinite number of pairs $(x,y)$ such that $x-y=1$.

I believe the problem was originally stated so that the translated version may be misinterpreted.  The original problem stated that $ A$ and $ B$ are such that $ B$ is the set of numbers, such that each of them is tuani to all numbers in $ A$, and $ A$ is the set of numbers such that each of them is tuani to all numbers in $ B$...

daniel73's is right. This was not the original problem.

Here is my solution to the original problem:

Suppose that $ A$ contains two numbers $ a$ and $ a'$ such that $ a\not\equiv a' \pmod {10}$. Being $ b$ an element of $ B$ we have wlog that $ b\plus{}a\equiv 0 \pmod{10}$ and $ b\plus{}a'\equiv 1 \pmod{10}$ because these numbers only contain $ 0$ and $ 1$ in their decimal expression. Consequently $ a'\equiv a\plus{}1 \pmod{10}$. If there is an element $ b'$ of $ B$ such that $ b\not\equiv b' \pmod{10}$ then $ b'\plus{}a\equiv 1 \pmod{10}$ (because $ b'\plus{}a\equiv 0 \pmod {10}\Rightarrow b'\equiv b\pmod{10}$) and $ b'\plus{}a'\equiv 0 \pmod{10}$ and so $ a\equiv a'\plus{}1\pmod{10}$. Contradiction!

We then conclude that if not all numbers in $ A$ have the same residue modulo $ 10$ then all elements in $ B$ have the same residue modulo $ 10$. Suppose now wlog that all numbers in $ B$ have the same residue modulo $ 10$. Choosing one number $ a$ in $ A$ we know that $ a$ is tuanis with all numbers in $ B$ and so, since these all have the same residue modulo $ 10$ we obtain that $ a\plus{}b$ always has $ 0$ as unit digit or always has $ 1$ as unit digit $ \forall \; b\in B$. If $ a\plus{}b\equiv 0 \pmod{10}\; \forall \; b\in B$ then $ a\plus{}1\in A$ because $ a\plus{}b\plus{}1$ only contains $ 0$ and $ 1$ in its decimal expression since $ a\plus{}b$ only contains $ 0$ and $ 1$ in its decimal expression and $ a\plus{}b$ has $ 0$ as unit digit $ \forall \; b\in B$. Analogously if $ a\plus{}b\equiv 1 \pmod{10}\; \forall \; b\in B$ then $ a\minus{}1\in A$.

So we proved that if $ a\in A$ then $ a\minus{}1\in A \vee a\plus{}1 \in A$. As $ |A|\equal{}\infty$ we conclude that $ A$ contains an infinite of pairs $ (x,y)$ such that $ x\minus{}y\equal{}1$.


[quote="mmrproblem"]Find the centroid of the perimeter of any convex quadrilateral.Is it intersection of diagonals?[/quote]
No, to affirmate take any triangle and mark any point on perimeter as  4th vertex.

P.S. But what you mean by centroid? This is to check only whether I've anderstood you correctly.

[quote="mmrproblem"]Find the centroid of the perimeter of any convex quadrilateral.Is it intersection of diagonals?[/quote]

Uh...take the four midpoints, and draw segments connecting opposite midpoints...and take the intersection...

[quote="mmrproblem"]Find the centroid of the perimeter of any convex quadrilateral.Is it intersection of diagonals?[/quote]

I think the answer is that it is not necessarily the intersection of the diagonals. Check out this:

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

There is a way to find a centroid of a quadrilateral, but it takes longer to find it than a triangle.


[quote="gemath"]Given triangle $ABC$ prove that
$sin\frac{A}{2}+sin\frac{B}{2}+sin\frac{C}{2}\le 1+\frac{1}{6}(\frac{h_{a}}{l_{a}}+\frac{h_{b}}{l_{b}}+\frac{h_{c}}{l_{c}})\le \frac{3}{2}$[/quote]
the right part is very easy:$\frac{h_{a}}{l_{a}}\leq 1$

I think it's the length of the angle-bisector

ah yes Hawk Tiger  :) It is length of the angle-bisector. I write the right part for us to see the strong of left part! I seem to see a solution of bodan, don't I ? :roll:

hope you are interested in it !!  :)  :lol:


How many distinct Sudoku boards (that are completed) are possible?

From Wikipedia:
"The number of valid Sudoku solution grids for the standard 9×9 grid was calculated by Bertram Felgenhauer in 2005 to be 6,670,903,752,021,072,936,960. This number is equal to 9! × 7^22 × 2^7 × 27,704,267,971, the last factor of which is prime. The result was derived through logic and brute force computation. The derivation of this result was considerably simplified by analysis provided by Frazer Jarvis and the figure has been confirmed independently by Ed Russell. Russell and Jarvis also showed that when symmetries were taken into account, there were 5,472,730,538 solutions."

Oh ok, thanks for the link! I did not think to look in Wikipedia  :oops: Heh, I thought there might have been a relatively simpler/elegant way to approach the problem that a student might use.


linear algebra
matrix
vector
superior algebra
superior algebra unsolved

Let $ A\in M(n,\mathbb{R})$. Prove: $ A$ is a left zero divisor $ \Leftrightarrow$ $ A$ is a right zero divisor $ \Leftrightarrow$ $ A\neq 0$ and $ \det(A)\equal{}0$

Can someone give me a hint?
Assume $ A$ is a left zero divisor. Then $ A\neq 0$ and $ \exists B\neq 0$ such that $ AB\equal{}0$. How to show that it follows that $ \exists C\neq 0$ such that $ CA\equal{}0$? I'm a bit new to this.

Thanks.

Have you heard about the (classical) adjoint $ \text{adj }A$ of a matrix $ A\in\text{M}_n\left(k\right)$ (for any field $ k$) ? Do you know the identity $ A\cdot\text{adj }A\equal{}\text{adj }A\cdot A\equal{}\det A\cdot I_n$ ?

  darij

I have heard of the adjoint, but never worked with it, and also never saw that identity. I'll go and read about it, and will then try the proof. Thanks.
Is there maybe another way?

[quote="BlueVelvet"]Is there maybe another way?[/quote]

Another argument: Show that

$ A$ is a left zero divisor $ \Longleftrightarrow$ there exists a vector $ v\neq 0$ with $ Av\equal{}0$ $ \Longleftrightarrow$ $ \det A\equal{}0$ $ \Longleftrightarrow$ $ \det A^T\equal{}0$ $ \Longleftrightarrow$ there exists a vector $ w\neq 0$ with $ A^Tw\equal{}0$ $ \Longleftrightarrow$ $ A^T$ is a left zero divisor $ \Longleftrightarrow$ $ A$ is a right zero divisor.

  darij

Show that $ AB \equal{} 0$ implies that $ A$ cannot have full rank and use the equivalence of row and column rank to deduce that there exists $ C$ with $ CA \equal{} 0$ and that the dimension of the image of $ A$ is strictly less than $ n$.

Obviously I need to fresh up /deepen my linear algebra knowledge (this is just a question from ring theory), although darij grinberg's second approach seems doable. Thanks guys.


The side of an equilateral triangle is $ s$. A circle is inscribed in the triangle and a square is inscribed in the circle. The area of the square is:

$ \textbf{(A)} \frac{s^2}{24} \qquad \textbf{(B)} \frac{s^2}{6} \qquad \textbf{(C)} \frac{s^2\sqrt{2}}{6} \qquad \textbf{(D)} \frac{s^2\sqrt{3}}{6} \qquad \textbf{(E)} \frac{s^2}{3}$

[hide="Click for solution"]
Let the length of the radius be $ r$, the length of the altitude be $ a$, and a side of the square be $ x$.

\[ r \equal{} \frac{1}{3} \cdot a \equal{} \frac{1}{3} \cdot \frac{1}{2} s\sqrt{3} \equal{} \frac{s\sqrt{3}}{6} \]

The area of the square is $ x^2$, which equals $ 2r^2 \implies 2\times \left(\frac{s\sqrt{3}}{6}\right)^2 \implies \boxed{\textbf{(B)} \frac{s^2}{6}}$.
[/hide]


Prove that exist $ a$ and $ b$ are prime such that
$ a\minus{}b\equal{}3^{2007}$

that means we have to check that $ 3^{2007}\plus{}2$ is prime :)

[quote="Albanian Eagle"]that means we have to check that $ 3^{2007} \plus{} 2$ is prime :)[/quote]
Can you post proof

It is from Diriclet theorem .
A arithmetic sequence $ an\plus{}b$ where $ \gcd(a,b)\equal{}1$ have infinite prime number.

[quote="TTsphn"]It is from Diriclet theorem .
A arithmetic sequence $ an \plus{} b$ where $ \gcd(a,b) \equal{} 1$ have infinite prime number.[/quote]
?? So?
if the difference of 2 primes is odd then one of them is 2, thus...

and $ 3^{2007} \plus{} 2$ is not prime according to Mathematica


calculus derivative integration vector algebra function domain

$ u: R^n\minus{}>R$ is $ C^1$. How to calculate:

$ \frac{\partial}{\partial r} \int_{B(x,r)}  u(y) dy$,

where $ B(x,r)$ is open r-ball around x. Note $ u$ may not be radial.

Write the integral as $ \int_{B(x,r)} \frac{\partial}{\partial r}u(y) dy$. The gradient $ \nabla u$ is a vector field, and you want to find its integral over a closed surface. Naturally, you should use the divergence theorem. The divergence of $ \nabla u$ is $ \Delta u$.

[quote="mlok"]Write the integral as $ \int_{B(x,r)} \frac {\partial}{\partial r}u(y) dy$. The gradient $ \nabla u$ is a vector field, and you want to find its integral over a closed surface. Naturally, you should use the divergence theorem. The divergence of $ \nabla u$ is $ \Delta u$.[/quote]

I am confused:
1.I don't think one can just move the derivative inside the integral since the integration domain also depends on $ r$.
2.$ \frac {\partial}{\partial r}$ is $ grad$?
3. $ B(x,r)$ is open ball, not sphere , so not a surface.

Actually I need to prove (from Evans book "PDE" section 2.4 (16)): if $ u$ is $ C^3$, then

$ \frac {\partial}{\partial r}(\frac {1}{r^n} \int_{B(x,r)} \Delta u(y) dy) \equal{} \frac {1}{r^n} \int_{\partial B(x,r)} \Delta u dS \minus{} \frac {n}{r^{n \plus{} 1}} \int_{B(x,r)} \Delta u(y) dy$

where $ dS$ means integration on sphere $ \partial B(x,r)$.

so the key is to calculate: $ \frac {\partial}{\partial r} \int_{B(x,r)} \Delta u(y) dy$

please help.

I did not read your question carefully.  :blush: To answer (2): $ \frac{\partial u}{\partial r}$ is not the same as $ \nabla u$, but it's the normal component of $ \nabla u$, which is what counts for surface integrals.

But yes, yours is not a surface integral. Rewrite it using polar coordinates:
$ \int_{B(x,r)}u(y)dy\equal{}\int_0^r \rho^{n\minus{}1} d\rho \int_{S^{n\minus{}1}}u(\rho \theta)\,d\theta$ where the integral over $ S^{n\minus{}1}$ is taken with respect to the surface measure. Differentiating with respect to the upper limit of integration, you get $ r^{n\minus{}1} \int_{S^{n\minus{}1}}u(r \theta)\,d\theta$ which is the integral of $ u$ over the boundary of $ B(x,r)$ with respect to the surface measure.

The iterated integral above could also be written as $ \int_0^r \left(\int_{S^{n\minus{}1}}u(\rho \theta)\,d\theta\right)\rho^{n\minus{}1} d\rho$.

[quote="mlok"]I did not read your question carefully.  :blush: To answer (2): $ \frac {\partial u}{\partial r}$ is not the same as $ \nabla u$, but it's the normal component of $ \nabla u$, which is what counts for surface integrals.

But yes, yours is not a surface integral. Rewrite it using polar coordinates:
$ \int_{B(x,r)}u(y)dy \equal{} \int_0^r \rho^{n \minus{} 1} d\rho \int_{S^{n \minus{} 1}}u(\rho \theta)\,d\theta$ where the integral over $ S^{n \minus{} 1}$ is taken with respect to the surface measure. Differentiating with respect to the upper limit of integration, you get $ r^{n \minus{} 1} \int_{S^{n \minus{} 1}}u(r \theta)\,d\theta$ which is the integral of $ u$ over the boundary of $ B(x,r)$ with respect to the surface measure.

The iterated integral above could also be written as $ \int_0^r \left(\int_{S^{n \minus{} 1}}u(\rho \theta)\,d\theta\right)\rho^{n \minus{} 1} d\rho$.[/quote]

thanks.


For my Honors Precalculus Homework, I have to solve the equation e^x+x=5.   I don't know what to do after I get x=ln(5-x).

hello, here we have $ x\equal{}\minus{}\text{LambertW}(e^5)\plus{}5$ as the solution.
Sonnhard.

Sorry, but I don't really get what you just did.  If someone could please post their work with the solution, that would be helpful.

hello, you can also use your calculator to solve this equation by a numerical way.
On get $ x \approx 1.306558641$.
Sonnhard.


number theory
greatest common divisor
algorithm
Euclidean algorithm

1.  Show that n :^2:  + n - 1 and n :^2:  + 2n have no common factor

2.  a/b and c/d are positive fractions in their lowest terms such that a/b + c/d = 1. Show that b = d.

3.  Prove that (n :^3:  - n)(58n+4 + 34n+2) is a multiple of 3804 for all positive integers n.

[quote="zscool"]3.  Prove that (n :^3:  - n)(58n+4 + 34n+2) is a multiple of 3804 for all positive integers n.[/quote]

Wow.  That's pretty crazy.  Did you come up with that on your own, or did you see it somewhere?

a/b+c/d=1.
1-(a/b)=c/d

its a bit obvious from there

I don't see how that's any more obvious than it was originally.  Post the full solution if you're going to post the solution.

[hide]5^(8n+4) + 3^(4n+2)=625^(2n+1)+9^(2n+1)=(625+9)*(K)=634(K), K=(625^2n-625^(2n-1)9......) So the whole thing=634K(n)(n-1)(n+1). since 3804/634=6. Now we have to prove Kn(n-1)(n+1) is divisible by 6, it must be because one of n and n+1 must divisible by 2, and one of n,n-1,n+1 must divisible by 3.  [/hide]

Solution to #3.

I'll take the easy one I guess.

zscool wrote:1.  Show that n :^2:  + n - 1 and n :^2:  + 2n have no common factor

[hide]gcd(n^2+n-1,n^2+2n)

= gcd(n^2+n-1,n+1)

= gcd(-1,n+1)

Just Euclidean algorithm a few times.

[/hide]

a/b+c/d=1.
1-(a/b)=c/d
b/b-a/b=c/d
c/d=(b-a)/b

is that enough?

Bleh, Ill take 2 too i guess.

Quote:2. a/b and c/d are positive fractions in their lowest terms such that a/b + c/d = 1. Show that b = d.

[hide]a/b + c/d = 1

ad + bc = bd

ad = b(d - c)    (1)

bc = d(b - a)    (2)

Since gcd(a,b)=1, we have b|d from (1) and d|b from (2), so b=d.

Hmm, or is it the other way around? d|b from (1) and b|d from (2)? Im not sure...[/hide]

you had it right originally, (1) says b|d and (2) says d|b, good job.

Scrambled... equating two fractions does not mean they have the same numerator and denominator unless they are in least terms (or you can justify that they are equal), i.e. 1/2 = 2/4 but 2 != 4

[quote="JBL"]zscool wrote:
3. Prove that (n  - n)(5^(8n+4) + 3^(4n+2)) is a multiple of 3804 for all positive integers n.

Wow. That's pretty crazy. Did you come up with that on your own, or did you see it somewhere?[/quote]

All three problems and the other one i posted come from Mexican Olympiads.

Here's a solution to the last problem:

[hide]The first part obviously indicates that the number is a multiple of 6. For the second part, notice that 5^2n + 3^n is always divisible by 634 for n >= 6. (We can verify this easily through mods). Thus the whole thing is always a multiple of 3804.[/hide]

Scratch that sol'n. I failed to notice something...

Okay, here's the actual solution:

[hide] Again, the first part indicates a multiple of 6. We must show that the last part is a multiple of 634. Notice that 634 = 5^4 + 3^2, and that (5^(8n+4) + 3^(4n+2)) = (5^4 + 3^2)(5^(8n)+3^(4n)) - 9*625(5^(8(n-1) + 4) + 3^(4(n-1) +2))). The first part is obviously divisible by 634...and the last part...

Hahaha! The magic of induction...since it clearly works for n = 1, it must work for all n.

Thus the whole thing is always a multiple of 3804.[/hide]

1-(a/b)=c/d
b/b-a/b=c/d
c/d=(b-a)/b

if c/d is in simplest, so must b-a/b.
let x be a factor of d
let n be any number less than x
If c=n(mod x)
then a=(x-n)(mod x)
x-n sure of a heck isnt divisible by x.
so a and c are both simplified in terms of d.
a and c are both simplified in terms of b.
two conclusions:
a) both b and d are prime
b) both b and d are the same
it cant be a because they need a common denominator to add up to 1(I HOPE THATS TOO OBVIOUS TO PROVE)
so they must be equal.

Scrambled...

You said c/d = (b-a)/b. Since (a,b) = 1, (b-a,b) = 1, so both c/d and (b-a)/b are simplified. Thus d = b.


algebra
polynomial
calculus
integration
real analysis
real analysis unsolved

Let $ x_1,...,x_n$ be different algebraic integers with the property that for each $i$ there is a conjugate of $x_i$ that is not among $x_1,...,x_n$. Then the set of $n$ uplets $ (p(x_1),...,p(x_n))$ when p describes the set of polynomials with integer coefficients is dense in $R^n$.

I think the condition that for each $i$ there is a $j$ s.t. $x_i,x_j$ are not conjugates of one another is either too weak or extra :?. What if, say, $n=3$, $x_1,x_2$ are conjugates of one another, and $x_3$ is not a conjugate of $x_1,x_2$?. $(P(x_1),P(x_2))$ must still be dense in $\mathbb R^2$ as $P$ goes through the integral polynomials, even though $x_1,x_2$ are conjugate, so maybe we can do without the condition, or maybe we must say that for all $i\ne j$, $x_i,x_j$ are not conjugate.

By conjugate I mean that their minimal polynomials (integral monic polynomials) are the same. You meant the same, right? :?

I propose first to prove it with this extra condition and after that we will generalize it if you have a proof. IS it so simple even with this condition? My solution is not simple.  :?

I haven't even thought about solving it :), but that condition looked really fishy, that's all.

Here's what I was afraid of: $x_1=\sqrt 3,x_2=-\sqrt 3$, and $x_3$ is some algebraic integer of your choice, different from the previous two. The pairs $(p(x_1),p(x_2))$ look like $(a\sqrt 3+b,-a\sqrt 3+b),\ a,b\in\mathbb Z$. Certainly, these pairs are not dense in $\mathbb R^2$. For one thing, the sum of the components is an integer, so they can'e be too close to pairs of reals with components which add up to $\frac 12$, for example.

What am I missing here? :?

Oh, sorry, I wrongly written the statement. I will edit it.

Excuse me disturbing your Highnesses.

I think, we need an additional condition: $x_i\not\in\mathbb Z$.

But in my opinion, it is the simpliest induction. Indeed, suppose $\{p_i\}_{i=1}^\infty$ give a dense set. Let $f_k\in\mathbb{Z}[x]$ be primitive polynomial for $x_k$, $F=f_1...f_{n-1}$. Then consider the set of points \[(p_{i,a,b}(x_1),...,p_{i,a,b}(x_n)),\] where $p_{i,a,b}(x)=F(x)(ax+b)+p_i(x)$, $a,b\in\mathbb Z$.
I think you got why it works.

No disturbance, your Majesty. I simply don't understand why your Greatness are loosing your time with such simple questions, when open questions are still a lot. :blush:


topology
function
advanced fields
advanced fields unsolved

Consider the subspace $\mathcal{C}([0,1],\mathbb{R})$ of $\mathbb{R}^{[0,1]}$ in the topology of pointwise convergence and take the sequence  $(f_{n})$ of continuous functions given by $f_{n}(x)=x^{n}$. Then the limit of $(f_{n})$ is not in $\mathcal{C}([0,1],\mathbb{R})$. Nonetheless every space has a completion. How does $f$ look like (i.e. the graph) in the completion of our space?

edit:
it's not a metric space :blush:  useful replies are still welcome.

Since I don't know how to "complete" a non-metrizable space, I'd rather change the topology. Each element of $\mathcal{C}([0,1],\mathbb{R})$ can be identified with a compact subset of $\mathbb R^{2}$, namely its graph. When equipped with the [url=http://en.wikipedia.org/wiki/Hausdorff_distance]Hausdorff distance[/url], the set of all compact subsets of $\mathbb R^{2}$ is a complete metric space. The limit of your sequence in this space is the polygonal curve $(0,0)-(1,0)-(1,1)$.

Let's compare pointwise convergence with the Hausdorff convergence. Does one imply the other?


AMC
USA(J)MO
USAMO
MATHCOUNTS
ARML
AMC 10
Harvard

Congrats to all Illinoisians who made USAMO (although I unfortunately didn't make it as an 8th grader with 129/6=189)! Also a big congrats to IMSA who had the most qualifiers!!!! :D

[u]12th Graders[/u]
Yize Leng, Illinois Math and Science Academy (Aurora)
Mendel Friedman, Ida Crown Jewish Academy (Chicago)
Michael Tarczon, Glenbrook South High School (Glenview)
Kiaran Dave, Naperville Central High School (Naperville)

[u]11th Graders[/u]
Seohyun Kim, Illinois Math and Science Academy (Aurora)
Alina Kononov, Illinois Math and Science Academy (Aurora)
Vladislav Kontsevoi, Illinois Math and Science Academy (Aurora)
Brent Terry-Penak, St. Ignatius College Prep (Chicago)
Matthew Waechter, Walter Payton College Prep (Chicago)
Katherine Rudolph, Naperville Central High School (Naperville)
Robin Jia, Oak Park and River Forest High School (Oak Park)
Arjun Puranik, Fremd High School (Palatine)

[u]10th Graders[/u]
Paul Yuan, Illinois Math and Science Academy (Aurora)
Charles Du, University of Chicago Lab High School (Chicago)
Kevin Hu, Naperville Central High School (Naperville)
Ben Mildenhall, Oak Park and River Forest High School (Oak Park)

[u]9th Graders[/u]
Jae Hwang Park, St. Viator High School (Arlington Heights)

Some surprises:

No Ari Tenzer (he's currently a senior at Glenbrook North HS in Northbrook who made USAMO last year as a junior) :(
Only one ninth grader :o
No middle schoolers :(
There are much fewer qualifiers from IL this year than from last year :(
No famously-known Mathcounts alumni/nats team qualifiers from IL besides Arjun Puranik

Hmm...I wonder how many MOPpers this year would be from IL...

[quote="gauss1181"]Hmm...I wonder how many MOPpers this year would be from IL...[/quote]
Not me, that's for sure.

Congratulations to all who made it!

[quote="stickfigure"]no ravi? =0[/quote]
yo stickfigure were you a mathcounts state participant this year? Or are you in high school? And did you make USAMO?

I don't really know about Ravi, but I guess he and his schoolmates do the AMC's and AIME's (Maybe they received a Sliffe Award for this). But I'm not sure if they didn't do it, or if Ravi just didn't score high enough to make USAMO.
Eh, no offense to him or anything but I just kinda feel less stupid and a little more relieved/a.k.a. less phailure type when I'm seeing no middle school qualifiers from IL.

I'm in high school and nope I didn't make USAMO... 135/6  :(

I was at ARML tryouts and I remember seeing ravi completely destroy everybody (except arjun and a few others)... so I immediately thought he'd be a lock for USAMO.

Maybe at the next ARML practice I'll ask him what happened... or actually no, that would seem mean... "hey little seventh grader, I noticed you didn't make USAMO! What happened???"

[quote="stickfigure"]

Maybe at the next ARML practice I'll ask him what happened... or actually no, that would seem mean... "hey little seventh grader, I noticed you didn't make USAMO! What happened???"[/quote]

 :rotfl:  :rotfl:  :rotfl:

Hmph, big mystery. Ravi made National MC last year, and this year: he beasted ARML tryouts, but.....TWO downfalls: no USAMO, no Nats MC. Many are pretty shocked that he missed his chance to go to Disney World... :o

I'm not sure if you can consider missing USAMO when you're a 7th grader to be a "downfall"...

oh and back to the original list... I also noticed that nobody from North qualified. I remember when that school used to be good. :roll:

Yea..I guess Naperville North had way more qualifiers than IMSA in 2008...or was that Naperville Central?

uhh nope, IMSA had the highest # of qualifiers in Illinois in 2008.

and I'm pretty happy for Charles Du. I was friends with him back when we were little and I met him again at some math summer program a little while back. As a freshman, he didn't even make the ARML D team. As a soph now he made ARML A/B and got  150 AMC10 and somewhere between 10 - 12 on the AIME.

Improvement ftw.

hmm yep, but I don't think he was all that bad as a freshie anyway. I think he just tripped up a bit on the arml tryouts that year.

also gauss.... i think its kinda weird how i was on aops talking to you and then i went on fb to comment on alina's status and... you were there too
small world  :wink:

Whoa do you think you guys will be able to pull off >= 1 MOPper this year?

Whoa, Alex, how come you didn't make USAMO this year? Weren't you like on the IMO team a couple of times?

Winston, do you honestly think he's good enough? I don't think a 42 on the IMO reverse-qualifies him for the USAMO =\

wikipedia says that its MOSP and that people still call it MOP cause apparently MOP sounds better... but Kevin said...

[quote]It doesn't matter what you think sounds better. It matters what is RIGHT.[/quote]

EDIT: and sorry math154, i interpreted your post to be saying "hey look, wikipedia titled their article MOP, so its MOP". oops  :oops:

It's fine--it's probably what I would've assumed as well.

Wikipedia is also sometimes wrong.

By the way, if you read that uncyclopedia page, we actually do not play the deceptively-titled card game "Yakuza", we play something else (I think it was the Merry Merchants of Venice or something). But the description is otherwise pretty accurate.

[quote="Aryth"]
By the way, if you read that uncyclopedia page... But the description is otherwise pretty accurate.[/quote]

So i see you were hinting me towards this line on the uncyclopedia page
"Actual participants in MOP are easily able to detect outsiders by their odd mispronouncing of MOP as 'MOSP'. "

Damn. So I guess you win.  :(  :(

It might have been "Peace-Loving Citizens of Sicily."

You're right, probability1.01. I wonder how you know? Were you at MOP?  :huh:

[hide="Disclaimer"]

Sarcasm.  :roll:

[/hide]

[quote="whitesoxfan"]The 9th grader was on the nats team last year, I believe.
Josh, Ravi, and me all phailed on the AIME.[/quote]

Something inside me hurt a lot when I read that last sentence.

</grammar_nazi>

LOL hi Fermi and Winston! @Fermi: way to talk about me behind my back. :mad:  :P
Btw, MOP is way better. I hope to make CGMO this year... if anything, it will give me an excuse not to go with my family on a "vacation" to the middle of nowhere.  :P  Idk if I can though. :(

Same here, except that I have to wait till next year :(

So how'd everyone do?

I got a 5 for my solution to #4...at least it's positive.

How many points were you from blue MOP cutoff?

5. not even close...
i should have gotten #2.  :wallbash_red:  :wallbash_red:  :wallbash_red:

does Illinois have any moppers this year?  :(


algebra
function
domain
inequalities
complex analysis
complex analysis unsolved

find a biholomprphic mapping between the outside of the lemniscate {x+iy:(x^2+y^2)^2=2(x^2-y^2)} to the open unit disc with (0,0) removed

thnx for the help, i have no clue how to deal with the lemniscates :/

please please somebody help! it's urgent!  :huh:
thank you so much in advance!

A small hint: the domain in question can also be described by the inequality $|z^2-1|>1$ ;).

Almost. This one would cover the unit disk twice (it is the same at $z$ and at $-z$).


After about 6 months of planning, the first release of The ImAoPSible Quiz is finally here!

As I said before, it only works on Windows platforms (because the game requires DirectX).  It should be compatible with Vista.  Or if you have a Mac with a program like BootCamp, that might work too.

[b]Size:[/b]
5.82 MB zipped
6.83 MB unzipped

[b]Download it![/b]
[url=http://www.yoyogames.com/games/show/30572]Link 1[/url]
[url=http://www.savefile.com/files/1467842]Link 2[/url]

[color=red][b]WARNING:[/b][/color]
Some of the questions contain humor that may be considered inappropriate for very young children. I would say that the level of "inappropriateness" is at about the same level as the original Impossible Quizzes.

[b]Please do not spoil the answers for others.[/b]

[quote="perfect628"]Yay you showed the question I thought of too...[/quote]
Well... I didn't put your name down in the credits, because I originally thought of it.  I will use one of your questions later in the quiz, though (that is, after question 100).  And you will receive credit for that one.

Oh I didn't really care about that--I was just speculating.

AoPSome!  :D

Hurray, my problem is in there! Unfortunately, it's #98, so I don't even know which one.

Darn, it doesn't like wine. Oh well, were any of my questions used?

Interesting.  I posted this game on another forum, specifically dedicated to the creation of computer games.  Nobody could get past question 26.

And Nerd_of_the_Ages, I did modify the question you gave me, but the overall concept is the same.

By the way, did you find any skips yet?

[quote="lingomaniac88"]Interesting.  I posted this game on another forum, specifically dedicated to the creation of computer games.  Nobody could get past question 26.

By the way, did you find any skips yet?[/quote]

Lol, at the #26 one. I thought that one was one of the easier ones.
Yeah, I got the one in problem <censored>, and the balloon one, but you have to use that one, right?
EDIT: Nvm, you can just pop the balloons in order

I'm currently on about 35.
EDIT: I'm currently on about 65

A lot of the answers for the multiple choice I don't get, I just got them by guessing.

EDIT: Btw, the 3rd mouse game is kinda too hard, IMO. I'm using a mouse pad, I guess that's why.

I can't play it on a mac. :(

I should be able to play online, but the link says only moderators can.  Is there any way to change that?

i was lucky with the 26th one  :P
but i'm getting stuck at #59  :o  :(
no-one else seems to have a problem with that?

Wow.  Klebian's zipping through this.  Questions 99 and 100 are insane, just to let you know.  Although 99 isn't as hard as it looks, if you know what to do...

pooey.. i'm not allowed to play on my computer...

oh well...
i'll play when i get my laptop...

I'm on the 5th mouse game. It's pretty good so far.

let me guess - you used gamemaker...

Which version, pro or not?

(i use v7 pro)

Major lolz at questions 9, 11, 31, 34, 51. (And #1 - I was a bit slow on that)
I'm stuck on 53 because my laptop sucks.

I'm at about Lvl 50~60 on the one with the big green background... no spoilers. I got two skips, used one on the 3rd mouse game and the other on something else. No spoilers.

I HATE the big green background.

[quote="1=2"]I tried it and all I can see are little boxes.

really.[/quote]

Huh?  PM me a screenshot.

This is pretty fun... I'm around level 60 right now.

[quote="lingomaniac88"]PM me a screenshot.[/quote]

I have absolutely no clue how to make a screenshot.

Statistics:
Number of times I have said that I don't know how to make a screen shot: 17
Number of letters that have been replaced by boxes: 47

Basically all the letters were replaced by little boxes their size.

[quote="1=2"][quote="lingomaniac88"]PM me a screenshot.[/quote]

I have absolutely no clue how to make a screenshot.[/quote]

Hit Print-Screen on your keyboard, open up a Paint-style program, and paste.

I've managed to download it, but it says that my graphics card has problems when I try to run it. I've tried this on multiple computers with the requirements asked, but I still can't get it to work.

For graphics, I think you need a DirectX 8 (or later) compatible graphics card with at least 32 MB of memory.  Do you have that?

Erm... How do you tell? I'm not that great with computers...  :maybe:

Um, for me, I guess not.

Our computer is pretty old (2001'ish?) and it has a

Nvidia RIVA TNT2 Model 64.

@hm29168: (this is for my Win XP, you might have another computer)
1. press Winkey + Pause/Break
2. go to "Hardware" tab
3. click "Device Manager"
4. open the category "Display Adapters"
5. read what it says

Thanks. Fixed.  :lol:

This is incredibly fun. I got past #26  :D .

Has anyone gotten this to run under wine?


[quote="Physical Science Textbook, Prentice Hall"]By the time you finish reading this sentence, more than 2 million tons of hydrogen will have exploded in the center of the sun.[/quote]

The sun and other stars use nuclear fusion to produce energy. The explosions in any star are fusion reactions. In the sun, two types of hydrogen, deuterium and tritium, fuse to form helium. This reaction releases enormous amounts of energy.

This energy occurs endlessly on the sun. On earth, it triggers the enormous explosion triggered by a hydrogen bomb.

Can fusion be controlled, though?

Fusion is powered by hydrogen, which is basically one of the only unlimited resources that we have on earth. Hydrogen could be obtained be separating water by running an electric current.

Now comes the big problem. Can we control it?

The energy resulting from a fusion reaction is massive, but not all of it is useful.

According to my book, huge amounts of pressure and heat must be put on the structures of hydrogen to force them together, because the two elements have a powerful repelling force. In one experiment, the temperature achieved to fuse the two atoms was over 70 million degrees, but the resulting energy couldn't boil a cup of water!!!!!

Do you believe that, someday, hydrogen could become a primary source of electricity.

[quote="hwenterprise"]Do you believe that, someday, hydrogen could become a primary source of electricity.[/quote]

Yes, I think so. Moreover, here in Argentina a plant is being installed only for that (to transform $H_2$ and to use it as fuel).

Jose, that's not at all the same use of hydrogen for energy as what he's talking about!

The problem isn't that the energy is too powerful to control.  The problem is to make it react in a small enough way to be able to collect the energy in a way that is useful.  We have been able to build fusion reactors for years.  Reactors that are safe, but they don't produce as much power as they consume.

We can prepare the fuel, reach the desired temperature, pressure and containment time.  We just can't do all of them at the same time.  The reaction temperatures occur at temperatures where the electrons are thermodynamically boiled off, in a plasma state of matter (beyond gas).  I find it interesting that the first date when a commercial fusion reactor was promised the US Congress was 1957.  The atomic researchers have severely underdelivered on that goal.

However, this seems to be the appropriate long term energy source.  It will just destroy the oil companies and the oil economies: this is a major political problem that makes it very difficult to make it happen.  . . . If we could find some way for Exxon and BP to make money off this, it might have a better chance of happening.

[quote="sponge008"]I think it's a good possibility; in fact, if it did work, clean energy would become no problem: fusion plants could produce methane and power power grids, with the methane powering cars, laptops, etc. An experimental fusion plant is being built in France right now.[/quote]

The experimental reactor is called [url=http://en.wikipedia.org/wiki/ITER]ITER[/url], and it's supposedly going to be capable of getting an energy gain.  They're also building a laser-based facility in California called [url=http://en.wikipedia.org/wiki/National_Ignition_Facility]NIF[/url], which is also supposed to be capable of gain.

I think commercial fusion is probably about 30 years away.  It has been 30 years away for as long as most people can remember, but maybe this time it actually *will* be 30 years away.

I hope it doesn't take longer than that.

[quote]but maybe this time it actually *will* be 30 years away.[/quote]

....you never know.....most of man's developments have been at times when he least expected it....

Relevant article in the NY Times today: [url=http://www.nytimes.com/2006/03/08/science/08fusion.html]University to Investigate Fusion Study[/url].  (Note that the article may only be freely available for a little while, like a week or so, and may require a free registration to view.)

For hydrogen to fuse into helium, I believe you have to achieve a temperature of 10,000,000 K.  Antimatter is produced as a by-product. When these positrons (anti-electrons) come in contact with electrons they annihilate creating energy based on E=mc^2. So even extremely small amounts of this could power whole cities for days.

Today, anti-matter is created in particle accelerators on Earth. I believe the largest one is in Switzerland.

Yes, CERN (although I believe an even larger one is under construction)

Question about the antimatter: If high-energy forms antimatter (with it's corresponding matter), then by annihilation wouldn't it only return the energy innitially used to form the matter/antimatter? (e => mc^2 => e)  :? (or am I smoking something?)

Fusion energy doesn't com from antimatter-matter reactions  but from mass defect in the nuclei (even when they occure: the energy for producing this antimatter comes from this).

And noone produces antimatter to win energy from it (it's sometimes a side product of nuclear reactions). But antimatter would be a very good way to store energy in a compact way, e.g. as fuel, but the problems to handle are very big...


number theory
relatively prime
number theory proposed

For which non-negative integers $ n$ and $ m$ does the following equation have an infinite number of solutions in positive integers $ x$, $ y$ and $ z$, with $ gcd(x,y,z) \equal{} 1$?

$ x^ny^m \plus{} y^nz^m \equal{} z^nx^m$

You may assume Fermat's Last Theorem.

(It's easier than you think)

If you have solutions for specific cases of $ n$ and $ m$ please post.

Whenever I put up original problems I dont get any takers. Is it because they are too hard or too boring?
Let me know what you think and I will try to do more suitable ones.

solution follows...

[hide]Firstly take the case $ n = 0$. By FLT this has infinite solutions for $ m = 1$ and $ m = 2$.
In the cases where $ n = m$ FLT also rules out solutions for $ n \ge 3$. We'll come back to the others later.
So now assume that $ 0 \lt n \lt m$.

For this one we use a lemma. Suppose there are integers $ a + b + c = 0$ and the highest powers of a prime $ p$ that divide them are

$ p^u\mid a$, $ p^v\mid b$ and $ p^w\mid c$, then the two smallest out of $ u$, $ v$ and $ w$ must be equal. Proof is by divisibility.

For this problem we nee $ x$, $ y$ and $ z$ to be relatively prime, so if a prime $ p$ divides $ x$ and $ y$ it cannot divide $ z$.

Suppose the highest power of $ p$ that divides $ x$ is $ p^k$ and the highest power that divides $ y$ is $ p^l$.

Now apply the lemma with $ a = x^ny^m$, $ b = y^nz^m$, $ c = - z^nx^m$, so $ u = nk + ml$, $ v = nl$ and $ w = mk$
the two smallest must be equal. Considering all cases we are led to $ nl = mk$.

Taking this result over all primes that divide $ x$, $ y$ and $ z$ we conclude that we can write $ x = d^ie^j$, $ y = e^if^j$ and $ z = f^id^j$
with $ in = jm$. We then have $ e^g + f^g = d^g$ with $ g = jn + im - in$, but with $ 0 \lt n \lt m$ we always have $ g > 2$ so there are no solutions by FLT.

The only remaining cases are $ n = m = 1$ and $ n = m = 2$

$ xy + zy = xz$ has solutions parameterised by $ y = rs$, $ x = r(r + s)$ $ z = s(r + s)$

$ (xy)^2 + (zy)^2 = (xz)^2$ has solutions parametrised by $ y = rs$, $ x = rt$, $ z = st$ for any pythagorean triple $ r^2 + s^2 = t^2$

In conclusion there are an infinite number of solutions iff $ (n,m) = (0,1), (0,2), (1,1)$ or $ (2,2)$ and no solutions in all other cases.[/hide]


Nicole has $ \$40$, and she wants to buy a new calculator. The regular price of the calculator she wants is $ \$36$, and the store is having a $ 20\%$ discount sale today. The state sales tax in her state is $ 5\%$ of the purchase price, and the discount is taken before tax is added. How many dollars will she have left if she buys the calculator? Express your answer to the nearest cent.

Wait, you said 36*0.8=28.8*1.05.
Actually 36*0.8=28.8 and 28.8*1.05=30.24
You were right about subtracting the 30.24 and the answer was correct.


calculus integration geometry logarithms ratio inequalities function

2 Points $ P\left(a,\ \frac{1}{a}\right),\ Q\left(2a,\ \frac{1}{2a}\right)\ (a > 0)$ are on the curve $ C: y \equal{}\frac{1}{x}$. Let $ l,\ m$ be the tangent lines at $ P,\ Q$ respectively. Find the area of the figure surrounded by $ l,\ m$ and $ C$.

[hide="brute force"]
$ l$ and $ m$ have equations
$ y\minus{}a \equal{}\minus{}\frac{1}{a^{2}}(x\minus{}\frac{1}{a})$ and $ y\minus{}2a \equal{}\minus{}\frac{1}{4a^{2}}(x\minus{}\frac{1}{2a})$ respectively

solving for the intersection of $ l$ and $ m$, we have

$ a\minus{}\frac{1}{a^{2}}(x\minus{}\frac{1}{a}) \equal{} 2a\minus{}\frac{1}{4a^{2}}(x\minus{}\frac{1}{2a})$

$ \minus{}4(x\minus{}\frac{1}{a}) \equal{} 4a^{3}\minus{}(x\minus{}\frac{1}{2a})$

$ \minus{}4x\plus{}\frac{4}{a}\equal{} 4a^{3}\minus{}x\plus{}\frac{1}{2a}$

$ \frac{4}{a}\minus{}4a^{3}\minus{}\frac{1}{2a}\equal{} 3x$

$ \frac{7\minus{}8a^{4}}{6a}\equal{} x$

let that intersection point be called $ x_{i}$
now the area is simply

$ \int_{a}^{2a}C dx\minus{}\int_{a}^{x_{i}}l dx\minus{}\int_{x_{i}}^{2a}m dx$

$ \int_{a}^{2a}\frac{1}{x}dx\minus{}\left[\int_{a}^{x_{i}}a\minus{}\frac{1}{a^{2}}(x\minus{}\frac{1}{a}) dx\plus{}\int_{x_{i}}^{2a}2a\minus{}\frac{1}{4a^{2}}(x\minus{}\frac{1}{2a}) dx\right]$

$ \ln{2}\minus{}\left[ ax\minus{}\frac{1}{2a^{2}}(x\minus{}\frac{1}{a})^{2}\Big|_{a}^{x_{i}}\plus{}2ax\minus{}\frac{1}{8a^{2}}(x\minus{}\frac{1}{2a})^{2}\Big|_{x_{i}}^{2a}\right]$

$ \ln{2}\minus{}\left[\frac{7\minus{}8a^{4}}{6}\minus{}a^{2}\minus{}\frac{1}{2a^{2}}\left((\frac{1\minus{}8a^{4}}{6a})^{2}\minus{}(\frac{a^{2}\minus{}a}{a})^{2}\right)\plus{}4a^{2}\minus{}\frac{7\minus{}8a^{4}}{3}\minus{}\frac{1}{8a^{2}}\left((\frac{4a^{2}\minus{}1}{2a})^{2}\minus{}(\frac{2\minus{}4a^{4}}{a})^{2}\right)\right]$

then a motherload of algebra. There is obviously a better way to do this.
[/hide]

[size=134][b]The user Kunny is as a rule to present us nice and interesting problems ![/b][/size]

 [quote="kunny"][color=darkred][b]Two points $ A(a,\ \frac{1}{a})$ and $ B(b,\ \frac{1}{b})$, where $ 0 < a < b$ are on the curve $ \mathcal H\ :\ y =\frac{1}{x}$.
Find the area of the surface surrounded by the curve $ \mathcal H$ and the its tangent lines in the points $ A$ and $ B$.[/b][/color][/quote]
[color=darkblue][b]Answer.[/b] Prove easily that the required area is $ S_{\mathcal H}(a,b) =\ln\frac{b}{a}-\frac{2(b-a)}{b+a}\ .$
[hide="Proof."] Denote the intersection $ T\in AA\cap BB$. Thus, $ \{\begin{array}{ccc}AA & : & x+a^{2}y-2a = 0\\ \ BB & : & x+b^{2}y-2b = 0\end{array}\|$ $ \implies$ $ T(\frac{2ab}{a+b}\ ,\ \frac{2}{a+b})\ .$

Denote the projections $ A_{1}$, $ B_{1}$, $ T_{1}$ on the axis $ Ox$ of the points $ A$, $ B$, $ T$ respectively.  Then the required area is

$ S_{\mathcal H}(a,b) =[\Gamma_{f}]-[ATT_{1}A_{1}]-[BTT_{1}B_{1}]$, where $ [\Gamma_{f}] =\int_{a}^{b}\frac{1}{x}\ dx =\ln\frac{b}{a}$- the subgraph-area of  $ f(x) =\frac{1}{x}$

and $ \|\begin{array}{ccccccc}[ATT_{1}A_{1}] & = &\frac{1}{2}\cdot A_{1}T_{1}\cdot (AA_{1}+TT_{1}) & = &\frac{1}{2}\cdot(\frac{2ab}{a+b}-a)\cdot(\frac{1}{a}+\frac{2}{a+b}) & = &\frac{(b-a)(3a+b)}{2(a+b)^{2}}\ .\\ \ [BTT_{1}B_{1}] & = &\frac{1}{2}\cdot B_{1}T_{1}\cdot (BB_{1}+TT_{1}) & = &\frac{1}{2}\cdot(b-\frac{2ab}{a+b})\cdot(\frac{1}{b}+\frac{2}{a+b}) & = &\frac{(b-a)(a+3b)}{2(a+b)^{2}}\ .\end{array}$

In conclusion, $ S_{\mathcal H}(a,b) =\ln\frac{b}{a}-\frac{4(b-a)(a+b)}{2(a+b)^{2}}$ $ \implies$ $ \boxed{\ S_{\mathcal H}(a,b) =\ln\frac{b}{a}-\frac{2(b-a)}{(a+b)^{}}\ }$.[/hide] [b]Remark.[/b] Observe that the area depends only on the ratio $ t =\frac{b}{a}$ and in this case $ S_{\mathcal H}(t) =\ln t-\frac{2(t-1)}{t+1}> 0\ !$
A nice consequence is the well-known inequality $ t\ge 1\implies\ln t\ge\frac{2(t-1)}{t+1}$.
Generally, from the [b]Lagrange's theorem[/b] exists $ a < c < b$ so that $ \frac{\ln b-\ln a}{b-a}=\frac{1}{c}$. Prove easily that $ \sqrt{ab}< c <\frac{a+b}{2}$.
Therefore, $ \frac{2}{b+a}<\frac{\ln\frac{b}{a}}{b-a}<\frac{1}{\sqrt{ab}}$, i.e. $ \boxed{\ t > 1\implies\frac{2(t-1)}{t+1}<\ln t <\frac{t-1}{\sqrt t}\ }$.[/color]

[size=134][b]Thank you, Kunny ![/b][/size]

Thank you very much, Virgil Nicula. :)

The problem that you proposed has been posed in Japanese university entrance exam as follows.

Nara Women University/Science 1982

[color=blue]Let $ S$ be the area of the region surrounded by two tangent lines at distinct two points $ A\ \left(a,\ \frac{1}{a}\right),\ B\left(b,\ \frac{1}{b}\right)\ (0 < a < b)$ on $ y \equal{}\frac{1}{x}\ (x > 0)$ and the curve. Show that $ S$ can be described by the function of $ k \equal{}\frac{b}{a}$.[/color]


conics
analytic geometry
geometry
circumcircle
parabola
quadratics
ratio

Is given a triangle ABC , P a point . AP,BP,CP cuts BC,CA,AB at A',B'C'. Demostrate that the centroids of PA'B,PA'C,PB'C,PB'A,PC'A,PC'B lie on a same conic if and oly if P lies on any of the median-lines of the triangle ABC. Also construct with ruler and compass the points P on the median AM such that this conic is a parabola.

I've tried to solve this problem by different ways , even by coordinate bashing  but this is too hard for me , I think that this is the best place to post this topic. Thanks in advance

Dear Mathlinkers,
perhaps see

http://forumgeom.fau.edu/
Myakishev A., Woo P. Y., On the Circumcenters of Cevasix Configuration, Forum Geometricorum vol. 3 (2003) 57-63.
Nguyen M. H., Another Proof of van Lamoen's Theorem and its converse, Forum Geometricorum vol. 5 (2005) 127-132.

Ayme J.-L., http://perso.orange.fr/jl.ayme  vol. 2  Le cercle de van Lamoen

Sincerely
Jean-Louis

Thanks .It is a related problem because involves circumcenters  but not centroids ,  it perhaps helps me to visualize an sketch  :)

Let us use barycentric coordinates with respect to $ \triangle ABC.$

$ P \ (u: v: w) \ , \ A' \ (0: v: w) \ , \ B' \ (u: 0: w) \ , \ C' \ ( u: v: 0)$

 $ A_c \ (3u^2 \plus{} 4uv \plus{} 2wu \plus{} wv \plus{} v^2: v(2u \plus{} 2v \plus{} w): (u \plus{} v)w)$

$ A_b \ (3u^2 \plus{} 4uvw \plus{} w^2 \plus{} 2uv \plus{} vw: v(u \plus{} w): w(2u \plus{} v \plus{} 2w))$

$ B_c \ (u(2u \plus{} 2v \plus{} w): u^2 \plus{} 4uv \plus{} wv \plus{} 3v^2 \plus{} 2vw: (u \plus{} v)w)$

$ B_a \ (u(v \plus{} w): 3v^2 \plus{} 2uv \plus{} 4wv \plus{} w^2 \plus{} uw: w(u \plus{} 2v \plus{} 2w))$

$ C_a \ (u(v \plus{} w): u(u \plus{} 2v \plus{} 2w): v^2 \plus{} uv \plus{} 4wv \plus{} 3w^2 \plus{} 2uw)$

$ C_b \ (u(2u \plus{} v \plus{} 2w): v(u \plus{} w): u^2 \plus{} uv \plus{} 4uw \plus{} 3w^2 \plus{} 2vw)$

These 6 points lie on a conic $ \mathcal{C}\equiv Ax^2 \plus{} By^2 \plus{} Cz^2 \plus{} 2Exy \plus{} 2Fyz \plus{} 2Gxz \equal{} 0$ $\iff$ the determinant of the 6x6 linear system formed by the coefficients of $\mathcal{C}$  equals zero. Alternatively, we also can use Pascal theorem to establish the collinearity of the intersection of the opposite sidelines of the hexagon $ A_cA_b...C_b.$ The first determinat $\Delta$ is given by

$ \Delta \equal{} 81uvw(v \minus{} w)(w \minus{} u)(u \minus{} v)(u \plus{} v)^2(v \plus{} w)^2(u \plus{} w)^2(u \plus{} v \plus{} w)^{12}$

If $ P \ (u: v: w)$ does not lie either on the sidelines of  $ \triangle ABC,$ the sidelines of the antimedial triangle  or the line at infinite, then the necessary condition for these six points to lie on a same conic is that

$ v \minus{} w \equal{} 0 \ , \ w \minus{} u \equal{} 0 \ , \ u \minus{} v \equal{} 0$

In other words, if $ P$ lies on any of the three median lines  of $ \triangle ABC.$

Equation of the conic $\mathcal{C}$ when $ P \in AM$ can be written as

$ v(9u^3 \plus{} 58u^2v \plus{} 122uv^2 \plus{} 72u^3)x^2 \plus{} u(18u^3 \plus{} 87u^2v \plus{} 118uv^2 \plus{} 38v^3)(y^2 \plus{} z^2) \minus{} (9u^4 \plus{} 66u^3v \plus{} 164uv^3 \plus{} 36v^4 \plus{} 175u^2v^2)x(y \plus{} z) \minus{} uyz(18u^3 \plus{} 75u^2v \plus{} 98uv^2 \plus{} 34v^3) \equal{} 0$

Its intersection with the line at infinite $ x \plus{} y \plus{} z \equal{} 0$ is

$ (2u(2u \plus{} v): \minus{} u(2u \plus{} v) \minus{} u\delta: \minus{} u(2u \plus{} v) \plus{} v\delta)$

Where $ \delta \equal{} \sqrt { \minus{} u(2u \plus{} v)(u^2 \plus{} 5uv \plus{} 3v^2)}$

Thus, this conic is a parabola $\Longleftrightarrow$ $ \sqrt { \minus{} u(2u \plus{} v)(u^2 \plus{} 5uv \plus{} 3v^2)} \equal{} 0$

This yields three cases:

1) If $ u \equal{} 0,$  $ P \equiv M,$ $ \mathcal{C}$ is a degenerate parabola, which is the product of the parallel lines $ x(2x \minus{} y \minus{} z) \equal{} 0,$ i.e. the sideline $ BC$ and the parallel to $ BC$ passing through centroid $ G.$

2) If $ 2u \plus{} v \equal{} 0,$ the  parabola corresponding to $ ( \minus{} 1: 2: 2)$  degenerates into the product of the lines $ 10x \plus{} y \minus{} z \equal{} 0$ and $ 13x \plus{} 4y \plus{} 4z \equal{} 0,$ respectively.

3) If $ u^2 \plus{} 5uv \plus{} 3v^2 \equal{} 0,$ the roots of this quadratic equation give the wanted points

$ X \ (5 \plus{} \sqrt {13}: \minus{} 2: \minus{} 2) \ , \ Y \ (5 \minus{} \sqrt {13}: \minus{} 2: \minus{} 2)$

Further, they divide the median $AM$ in the ratios

$ \frac{\overline{XA}} {\overline{XM}} \equal{} \minus{} \frac{4}{5 \plus{} \sqrt {13}} \ \ , \ \ \frac{\overline{YA}}{\overline{YM}} \equal{} \minus{} \frac{4}{5 \minus{} \sqrt {13}}$

Would not the calculation be much simpler for midpoints $ K_b, K_c, L_c, L_a, M_a, M_b$ of the segments $ BA', CA', CB', AB', AC', BC'$ ? Hexagon of these midpoints is centrally similar to the hexagon of the centroids with center $ P$ and coefficient $ \frac {_3}{^2}.$ Barycentric of the midpoints can be written without calculation

$ \overrightarrow{K_b} \equal{} (w \plus{} u)(u \plus{} v)\ [0, w \plus{} 2v, w],\ \overrightarrow{K_c} \equal{} (w \plus{} u)(u \plus{} v)\ [0, v, v \plus{} 2w]$

$ \overrightarrow{L_c} \equal{} (u \plus{} v)(v \plus{} w)\ [u, 0, u \plus{} 2w],\ \overrightarrow{L_a} \equal{} (u \plus{} v)(v \plus{} w)\ [w \plus{} 2u, 0, w]$

$ \overrightarrow{M_a} \equal{} (v \plus{} w)(w \plus{} u)\ [v \plus{} 2u, v, 0],\ \overrightarrow{M_b} \equal{} (v \plus{} w)(w \plus{} u)\ [u, u \plus{} 2v, 0]$

When the segment midpoints are on a conic, the intersections $ A \equiv L_cL_a \cap M_bM_a,$ $ D \equiv K_bM_b \cap K_cL_c$, $ E \equiv K_bL_a \cap K_cM_a$ are collinear, which is equivalent to $ (\overrightarrow D \times \overrightarrow E)_x \equal{} 0.$ I could not factor the last expression without error; $ uvw(u \plus{} v)(v \plus{} w)(w \plus{} u)(v \minus{} w)$ factored out OK, but the remaining factor did not make sense.

Thanks to [b]Luis González[/b] for big hint. By Carnot theorem, midpoints $ M_a, M_b, K_c, K_b, L_c, L_a$ of $ AC', BC', BA', CA', CB', AB'$ are on a conic $ \Longleftrightarrow$

$ \overline{AM_a} \cdot \overline{AM_b} \cdot \overline{BK_b} \cdot \overline{BK_c} \cdot \overline{CL_c} \cdot \overline{CL_a} \equal{} \overline{AL_a} \cdot \overline{AL_c} \cdot \overline{CK_c} \cdot \overline{CK_b} \cdot \overline{BM_b} \cdot \overline{BM_a}$

Continuing with directed line segments or barycentric coordinates $ (u, v, w)$ of $ P$ is almost the same. Assuming $ P$ is not on triangle sidelines, i.e., none of $ (u, v, w)$ is zero, substitute

$ \frac {\overline{AM_a} \cdot \overline{AM_b}}{\overline{BM_a} \cdot \overline{BM_b}} \equal{} \frac {v(u \plus{} 2v)}{u(v \plus{} 2u)},$[color=white] . [/color]$ \frac {\overline{BK_b} \cdot \overline{BK_c}}{\overline{CK_b} \cdot \overline{CK_c}} \equal{} \frac {w(v \plus{} 2w)}{v(w \plus{} 2v)},$[color=white] . [/color]$ \frac {\overline{CL_c} \cdot \overline{CL_a}}{\overline{AL_c} \cdot \overline{AL_a}} \equal{} \frac {u(w \plus{} 2u)}{w(u \plus{} 2w)}.$

$ \Longrightarrow\ \frac {v(u \plus{} 2v)}{u(v \plus{} 2u)} \cdot \frac {w(v \plus{} 2w)}{v(w \plus{} 2v)} \cdot \frac {u(w \plus{} 2u)}{w(u \plus{} 2w)} \equal{} 1,$

$ (u \plus{} 2v) (v \plus{} 2w)(w \plus{} 2u) \equal{} (v \plus{} 2u)(w \plus{} 2v)(u \plus{} 2w).$

Whatever's left, after multiplying this through and canceling equal terms, factors as

$ u^2w \plus{} uv^2 \plus{} vw^2 \equal{} v^2w \plus{} uw^2 \plus{} u^2v$

$ (u \minus{} v)(v \minus{} w)(w \minus{} u) \equal{} 0,$

which means at least 2 barycentric coordinates are equal $ \Longleftrightarrow$ $ P$ is on one of the median lines.
__________________________________________________

To find $ P,$ for which the conic is a parabola, parallel project $ \triangle ABC$ to an isosceles right triangle, which takes midpoints to midpoints and parabolas to parabolas. That is, draw isosceles right triangle $ \triangle A^*BC$ on the base $ BC$; $ AA^*$ is direction of the rays. Then rename the triangle back to $ \triangle ABC.$ Put origin at $ A$ and y-axis along the A-median $ AA'$ and WLOG, let $ BC \equal{} 2.$ Let $ P$ be the desired point and let $ CP$ cut $ AB$ at $ C'$ with coordinates $ [c, c].$ The parabola has symmetry axis $ AA'$ and equation $ y \equal{} kx^2 \plus{} q.$ It goes through midpoints $ K_b \equal{} [\frac {_1}{^2},\ 1],$ $ M_a \equal{} [\frac {_c}{^2},\ \frac {_c}{^2}],$ $ M_b \equal{} [\frac {_{c \plus{} 1}}{^2},\ \frac {_{c \plus{} 1}}{^2}]$ of $ BA', AC', BC'.$ Plugging these coordinates to the parabola equation and eliminating $ k, q$ yields $ \frac {c \minus{} 2}{c \minus{} 1} \equal{} \frac {c^2 \minus{} 1}{c^2 \plus{} 2c}$ or $ c^2 \minus{} 3c \minus{} 1 \equal{} 0,$ with roots $ c \equal{} \frac {3 \pm \sqrt {13}}{2}.$ $ CC'$ cuts the A-median at $ P \equal{} [0, p],$ $ p \equal{} 1 \plus{} \frac {c \minus{} 1}{c \plus{} 1} \equal{} \frac {1 \pm \sqrt {13}}{3}$ and $ \frac {\overline{AP}}{\overline{AA'}} \equal{} p.$ From $ \frac {\overline{AC'}}{\overline{BC'}} \equal{} \minus{} \frac {c}{1 \minus{} c},$ barycentic coordinates of $ P$ are $ (u, v, w) \equal{} (1 \minus{} c, c, c).$

I like the problem very much, because for any reasonable point inside any reasonable triangle, the 6 centroids are so close to one conic and yet, only points on medians will do. One theorem and two methods are used to put the points $ P$ generating conics through the centroids on one of the medians. One more method is then used to find points $ P$ generating parabolas:

1. central similarity
2. Carnot theorem (i.e., one of Carnot theorems, for there are at least three different Carnot theorems)
3. barycentric coordinates
4. parallel projection

Barycentric coordinates could be replaced by directed segments, but barycentric coordinates are actually simpler.


[quote="orl"]Prove that $ a < b$ implies that $ a^3 \minus{} 3a \leq b^3 \minus{} 3b \plus{} 4.$ When does equality occur?[/quote]
See here:
http://www.mathlinks.ro/viewtopic.php?t=215124

Easy problem.
[hide="My Solution"]Let $ b \equal{} a \plus{} x$
$ a^3 \minus{} 3a\le b^3 \minus{} 3b \plus{} 4$
$ \iff 3x\le x(3a^2 \plus{} 3ax \plus{} x^2) \plus{} 4$
$ \iff 3x\le x\left(3(a \plus{} \frac {x}{2})^2 \plus{} \frac {x^2}{4}\right) \plus{} 4$
So it is sufficient to prove that:
$ 3x\le x(\frac {x^2}{4}) \plus{} 4$
$ \iff 12x\le x^3 \plus{} 16$
$ \iff 3\sqrt [3]{x^3\cdot 2^3\cdot 2^3}\le x^3 \plus{} 2^3 \plus{} 2^3$
Which is true by AM-GM.
Equality occurs iff $ (a,x) \equal{} ( \minus{} 1,2)\implies \boxed{(a,b) \equal{} ( \minus{} 1,1)}$[/hide]
Edit: Didn't see rachid's post.


A pizza parlor offers six toppings. What is the greatest number
of four-topping pizzas that can be made such that no two pizzas
have the same topping combination?

Not sure if this is right but...

$ \frac{6\cdot5\cdot4\cdot3}{2!}\equal{}180$

I think it is just $ \binom 6 4\equal{}\binom 6 2\equal{}\frac{6 \cdot 5}{2}\equal{}\boxed{15}$.

Thanks PowerOfPi...can you explain this a little more?

You choose 4 out of the 6 toppings to be on the pizza, without regard to order. You were close, because there are $ 6 \cdot 5 \cdot 4 \cdot 3$ ways to choose them with regard to order. But this over counts $ 4!$ times, because that is how many arrangements are possible with a combination of 4 toppings.

[hide=Solution]This is 6 choose 4. $\binom{6}{4} = \binom{6}{2}$ and $\binom{6}{2}=\frac{6!}{2! \times 4!} = \frac{6 \times 5}{2}=3 \times 5=15.$[/hide]


Three straight lines intersect at O, and $ \overline{OD}$ bisects $ \angle COE$. The ratio of $ \angle COB$ to $ \angle BOF$ is 7:2. What is the number of degrees in $ m\angle COD$?
[asy]pair A,B,C,D,E,F,O;
C = dir(60); E = dir(-30); D = dir(15);
A = -D; B = -E; F = -C;
draw(A--D, Arrows); draw(B--E, Arrows);
draw(F--C, Arrows);
label("A",A,W);label("B",B,W);label("C",C,E);
label("D",D,E);label("E",E,E);label("F",F,W);
label("O",O,S);[/asy]

Let $ x \equal{} m\angle COD$. Then $ m\angle COB \equal{} 180^\circ \minus{} 2x$, and $ m\angle BOF \equal{} 2x$, so

$ \frac {180^\circ \minus{} 2x}{2x} \equal{} \frac {90^\circ}x \minus{} 1 \equal{} \frac72\implies\frac {90^\circ}x \equal{} \frac92 \equal{} \frac {90}{20}\implies x \equal{} \boxed{20^\circ}$.

$ m\angle COB \plus{} m\angle BOF \equal{} 180$

Vertical angles: $ m\angle BOF \equal{} m\angle COE$.

Let's assign variables:
$ m\angle COB \equal{} 7x$
$ m\angle BOF \equal{} 2x$

Plugging in:
$ 9x \equal{} 180$
$ x \equal{} 20$

Then,
$ m\angle BOF \equal{} m\angle COE \equal{} 2m\angle COD \equal{} 2x \equal{} 2(20) \equal{} 40$.
$ m\angle COD \equal{} \fbox{20}$.


logarithms
real analysis
real analysis unsolved

Hi. How do you show that $ \beta : \equal{}\prod_{i\equal{}1}^{\infty}(1\minus{}2^{\minus{}i})$ is a positive real number?
Thanks

Because :
$ \sum_{i \equal{} 1}^n2^{ \minus{} i} \equal{} 1 < \plus{} \infty$

Any why does that imply the result? Did you multiply the infinite product out?
Could you clarify a bit please.

For $ x\in\left[0,\frac12\right],$ there is a constant $ c>0$ such that $ \ln(1\minus{}x)\ge\minus{}cx.$

Hence, $ \ln\left(\prod_{i\equal{}1}^{\infty}(1\minus{}2^{\minus{}i})\right)\ge
\minus{}c\sum_{i\equal{}1}^{\infty}2^{\minus{}i}.$

Since that sum is finite (greater than $ \minus{}\infty,$) the product is greater than zero.

The general principle is this: for $ a_i$ of eventually constant sign, $ \prod_{i\equal{}1}^{\infty}(1\plus{}a_i)$ converges iff $ \sum_{i\equal{}1}^{\infty}a_i$ converges.

Also, if $ \sum_{i\equal{}1}^{\infty}|a_i|$ converges, that implies that $ \prod_{i\equal{}1}^{\infty}(1\plus{}a_i)$ converges. And it's part of the conventions of infinite products that a product that tends to zero without any of its factors being zero is referred to as divergent.


$ a,b,m,n$ are positive numbers such that $ \frac {m}{n} \equal{} \frac {b}{a}$ . If we have $ m  \le  b$ or $ \ n  \le  a$ prove that for any real number $ t$ at least one of these two inequalities is always fullfilled.

$ t^2 \ge (t \minus{} m)^2 \plus{} n^2$ or/and $ (t \minus{} b)^2 \plus{} a^2 \ge (t \minus{} m)^2 \plus{} n^2$

This inequality is actually very classical. ( In the sense that it can be solved by contradiction).

Let
\[ \frac{m}{n}\equal{}\frac{a}{b}\equal{}p\]

Now, we might assume that both inequalities are false. After a while, we expect that there exists a bunch of contradictive statements. Indeed, it turns out that

\[ 2tp > (a\plus{}n)(p^2\plus{}1)\]

and

\[ 2tp < n(p^2\plus{}1)\]

Hence, the conclusion follows immediately.


Prove that there is a continuous, strictly decreasing function $ g: \mathbb{R}\to\mathbb{R}$ such that $ g(g(x))\equal{}2x\plus{}2\;\;\forall x\in\mathbb{R}$, but that there is no such function satisfying $ g(g(x))\equal{}x\plus{}1\;\;\forall x\in\mathbb{R}$.

$ \minus{}\sqrt{2}x \minus{}2(\sqrt{2}\plus{}1)$ will do for the first part, and for the second, note that $ g(x\plus{}1) \equal{} g(x)\plus{}1$ by iteration, so $ g$ can't be decreasing.


analytic geometry
graphing lines
slope
rotation
trigonometry

Say an inclined surface is set up at a $60^\circ$ degree angle to the horizontal is held at its center by a frictionless axle, and a ball is put on the raised edge of the ramp and allowed to roll. Between the ball and the ramp, assume there is no friction. The ramp is rotated around its center only by the weight of the ball, and the motion of the ball is then affected by the surface's incline.

If the mass of the ball is $m$, the length of the surface $L$, and the acceleration due to gravity a constant (I think it's $9.8 \ m/s$..), what is the maximal distance that the ball will roll across the surface? When will the ball fall off of the other end of the surface?

If there were no friction between the ball and the ramp, it would slide and not roll.

Now, to solve this problem, you need to know the mass of the surface.

Let's call it $M$.

Then, you find that the inertia of the ramp $I = \frac{1}{12}ML^{2}$.

So, $I \alpha = \tau = xN \cos \theta$, where $x$ is the distance from the center of the ramp.

Thus, $\tau = mg x \cos^{2}\theta$.

And of course, $ma = mg \sin \theta$.

Then, it gets really messy, but if you approximate $\sin \theta \approx \theta$ and $\cos \theta \approx 1$, you get a linear system which can be solved by the eigenvalue method.


inequalities quadratics algebra polynomial trigonometry geometry parallelogram

[hide="Solution"]
Assume that the conclusion is not true, then both $\sum_{i=1}^na_i^2 <1$ and $\sum_{i=1}^nb_i^2 <1$.
Consider the quadratic polynomial $p(x) = (x-1)^2 - \sum_{i=1}^n(a_ix-b_i)^2 = (1-\sum_{i=1}^na_i^2) x^2 - 2 (1-\sum_{i=1}^na_ib_i) x + (1-\sum_{i=1}^nb_i^2)$.
By the given condition, the discriminant is less than $0$ and the leading coefficient is positive, hence $p(x) > 0$ for any real $x$. But $p(1) = - \sum_{i=1}^n(a_i-b_i)^2 \le 0$, we have a contradiction. Q.E.D.

[/hide]

[quote="al.M.V."][quote="fuzzylogic"][hide="Solution"]
Assume that the conclusion is not true, then both $\sum_{i=1}^na_i^2 <1$ and $\sum_{i=1}^nb_i^2 <1$.
[/hide][/quote]

eh, if conclusion is assumed false, then either the first or the second of what you said is true, not necessarily both.[/quote]
If only one is true, then the L.H.S would be negative, whereas the R.H.S is positive, hence either neither or both have to be true.

Another solution:
Suppose instead that $ a_{1}^2 \plus{} ... \plus{} a_{n}^2 \leq 1$ and $ b_{1}^2 \plus{} ... \plus{} b_{n}^2 \leq 1$. We shall prove that $ (a_{1}^2 \plus{} ... \plus{} a_{n}^2\minus{} 1)(b_{1}^2 \plus{} ... \plus{} b_{n}^2 \minus{} 1) \leq (a_{1}b_{1} \plus{} ... \plus{} a_{n}b_{n} \minus{} 1)^2$.
Noting that Cauchy-Schwarz is valid for any, (not necessarily positive) reals, we have:
$ (a_{1}b_{1} \plus{} ... \plus{} a_{n}b_{n})^2 \leq (a_{1}^2 \plus{} ... \plus{} a_{n}^2)(b_{1}^2 \plus{} ... \plus{} b_{n}^2) \leq 1$, so $ a_{1}b_{1} \plus{} ... \plus{} a_{n}b_{n} \leq |a_{1}b_{1} \plus{} ... \plus{} a_{n}b_{n}| \leq 1$.
Let $ X \equal{} a_{1}^2 \plus{}  ... \plus{} a_{n}^2, Y \equal{} b_{1}^2 \plus{} ... \plus{} b_{n}^2$, so $ 0 \leq X,Y \leq 1$.
Since $ 0 \leq X\plus{}Y \minus{} 2\sqrt {XY}$, we have $ XY \minus{} (X\plus{}Y) \leq XY \minus{} 2 \sqrt {XY}$, so $ (X\minus{}1)(Y\minus{}1) \leq (\sqrt{XY} \minus{} 1)^2$, (1)
But Cauchy-Schwarz gives us $ a_{1}b_{1} \plus{} ... \plus{} a_{n}b_{n} \leq \sqrt {XY}$, so since $ f(x) \equal{} (x\minus{}1)^2$ is decreasing for $ x \leq 1$, $ ( \sqrt {XY} \minus{} 1)^2 \leq (a_{1}b_{1} \plus{} ... \plus{} a_{n}b_{n} \minus{} 1)^2$ (2).
Combining (1) and (2) gives us our desired inequality.

Let $A = a_1^2 + a_2^2 + \cdots + a_n^2$, $B = b_1^2 + b_2^2 + \cdots + b_n^2$, and $C = a_1b_1 + a_2 b_2 + \cdots + a_n b_n$. We are given that $(A-1)(B-1) > (C-1)^2$, which can be rearranged to $AB - C^2 > A+B - 2C$. Rewrite this inequality as $(\sqrt{AB} - C)(\sqrt{AB}+C) > (\sqrt{A} - \sqrt{B})^2 + 2(\sqrt{AB} - C)$, so $(\sqrt{AB}-C)(\sqrt{AB}+C-2) > (\sqrt{A} - \sqrt{B})^2 \geq 0$. By the Cauchy-Schwarz inequality, $\sqrt{AB} \geq C$ so we need $\sqrt{AB}+C-2 \geq 0$. But $C \leq \sqrt{AB}$, so $2 \sqrt{AB} \geq 2 \implies AB \geq 1$. Since $(A-1)(B-1) > 0$, either both $A<1,B<1$ or both $A>1,B>1$. In the former case, we have $AB < 1$, a contradiction, so we must have $A,B>1$, as desired.

[hide="easy"]This is the contrapositive of Aczel's Inequality for A=B=1. [/hide]

I'm sorry, but what is [b]Aczel's Inequality[/b]?

Aczel's Inequality states that if $A^2>a_1^2+a_2^2+\cdots+a_n^2$ and $B_2>b_1^2+b_2^2+\cdots+b_n^2$, then $\left(A^2-a_1^2-a_2^2-\cdots-a_n^2\right)\left(B^2-b_1^2-b_2^2-\cdots-b_n^2\right)\le$ $\left(AB-a_1b_1-a_2b_2-\cdots-a_nb_n\right)$.

See the page about it on AoPSWiki.

[hide="Different Solution"] This is a geometric interpretation of the inequality.
Let $\vec{a}=(a_1,a_2,\dots,a_n)$, $\vec{b}=(b_1,b_2,\dots,b_n)$, and $\theta$ be the angle between the two. Note that $\vec{a}\not=\vec{b}$ because that would achieve equality.  Now, partially expanding our inequality, we get
\[\sum_{i=1}^n a_i^2\cdot \sum_{i=1}^n b_i^2-\sum_{i=1}^n a_i^2-\sum_{i=1}^n b_i^2+1>(\sum_{i=1}^n a_ib_i)^2-2\sum_{i=1}^n a_ib_i+1\]

\[\Longrightarrow \sum_{i=1}^n a_i^2\cdot \sum_{i=1}^n b_i^2-(\sum_{i=1}^n a_ib_i)^2>\sum_{i=1}^n a_i^2+\sum_{i=1}^n b_i^2-2\sum_{i=1}^n a_ib_i\]

\[\Longrightarrow |\vec{a}|^2|\vec{b}|^2-|\vec{a}|^2|\vec{b}|^2\cos{\theta}^2>|\vec{a}-\vec{b}|^2\]

\[\Longrightarrow |\vec{a}||\vec{b}|\sin{\theta}>|\vec{a}-\vec{b}|\]

Justifying some of the above steps, the LHS of line $3$ follows from the fact that $\sum_{i=1}^n a_ib_i=\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos{\theta}$.
The RHS is a result of
\[\sum_{i=1}^n a_i^2+\sum_{i=1}^n b_i^2-2\sum_{i=1}^n a_ib_i=\sum_{i=1}^n (a_i-b_i)^2=|\vec{a}-\vec{b}|^2\]
Taking the square root at the end is justified by the fact that all the quantities involved, magnitudes and $\sin{\theta}$, are positive.
Note that the LHS now denotes the area of a parallelogram with sides $\vec{a}$, $\vec{b}$ and the LHS one of the diagonals of this parallelogram.
Considering the sum of the areas of the two triangles with sides $|\vec{a}|$, $|\vec{b}|$, and $|\vec{a}-\vec{b}|$, we have that $|\vec{a}-\vec{b}|h>|\vec{a}-\vec{b}|$.

As noted above, $|\vec{a}-\vec{b}|\not= 0$, so $h>1$. This is one of the legs of a right triangle with hypotenuse $|\vec{a}|$, and similarly for $|\vec{b}|$, so $|\vec{a}|$, $|\vec{b}|>1$, as desired. [/hide]

Let $\mathbf{a}=\begin{bmatrix} a_1 \\ \vdots \\ a_n \end{bmatrix}$ and $\mathbf{b}=\begin{bmatrix} b_1 \\ \vdots \\ b_n \end{bmatrix}$. Then
\begin{align*}
	||\mathbf{a}-\mathbf{b}||^2||\mathbf{a}||^2&=||\mathbf{a}||^2||\mathbf{b}||^2+||\mathbf{a}-\mathbf{b}||^2||\mathbf{a}||^2-||\mathbf{a}||^2||\mathbf{b}||^2\\
	&=||\mathbf{a}||^2||\mathbf{b}||^2+\left(||\mathbf{a}-\mathbf{b}||^2-||\mathbf{b}||^2\right)||\mathbf{a}||^2\\
	&=||\mathbf{a}||^2||\mathbf{b}||^2+\left\langle\mathbf{a}-2\mathbf{b},\mathbf{a}\right\rangle\cdot\left\langle\mathbf{a},\mathbf{a}\right\rangle\\
	&=||\mathbf{a}||^2||\mathbf{b}||^2+\left\langle\mathbf{a}-\mathbf{b},\mathbf{a}\right\rangle^2-\left\langle\mathbf{b},\mathbf{a}\right\rangle^2\\
	&\geq||\mathbf{a}||^2||\mathbf{b}||^2-\left\langle\mathbf{a},\mathbf{b}\right\rangle^2\\
	&=\left(||\mathbf{a}||^2-1\right)\left(||\mathbf{b}||^2-1\right)-\left\langle\mathbf{a},\mathbf{b}\right\rangle^2+||\mathbf{a}||^2+||\mathbf{b}||^2-1\\
	&>\left(\left\langle\mathbf{a},\mathbf{b}\right\rangle-1\right)^2-\left\langle\mathbf{a},\mathbf{b}\right\rangle^2+||\mathbf{a}||^2+||\mathbf{b}||^2-1\\
	&=||\mathbf{a}||^2-2\left\langle\mathbf{a},\mathbf{b}\right\rangle+||\mathbf{b}||^2\\
	&=||\mathbf{a}-\mathbf{b}||^2
\end{align*}
so $||\mathbf{a}||>1$. Similarly $||\mathbf{b}||>1$. Thus $a_1^2+\ldots+a_n^2>1$ and $b_1^2+\ldots+b_n^2>1$.

http://www.artofproblemsolving.com/Wiki/index.php/Aczel's_Inequality.  kill by Aczel inequality

Let $a=\sum_{i=1}^{n} a_i^2 , b=\sum_{i=1}^{n} b_i^2 , c=\sum_{i=1}^{n} a_ib_i$. Assume for contradiction that $a<1$ and $b<1$.
Given inequality can be written as $c^2 - 2c + (a + b - ab) < 0$.
The function $f(x)=x^2 - 2x + (a + b - ab)$ has $2$ positive roots one of which is less than $1$ and that root is $1-\sqrt{(a-1)(b-1)}$.
So $c > 1 -  \sqrt{(a-1)(b-1)}$.
By Cauchy we know that  $ab\geq c^2 > (a-1)(b-1) + 1 - 2\sqrt{(a-1)(b-1)} \implies a + b + 2\sqrt{(a-1(b-1)} > 2$.
But it can easily be proven by computing derivatives that when $0<a<1$ and $0<b<1$,  $a + b + 2\sqrt{(a-1)(b-1)}$ takes its maximum value $2$ when $a=b$. Hence contradiction.

Apparently, essentially the same as an early problem from [url=https://imomath.com/othercomp/Mon/MonMO00.pdf]2000 Mongolian Olympiad, Teacher-secondary level.[/url]

[url=https://artofproblemsolving.com/community/c6h2537349_inequality_in_vectors]See here.[/url]

there's a nice proof of aczel's inequality by constructing yet another quadratic function


geometry
circumcircle
induction
combinatorics unsolved
combinatorics

Show that in any convex polygon there are 3 consecutive vertices, such that all the vertices of the polygon are contained in the circumcircle of these 3 points.

Solution. Let us prove this by induction.

Induction hypothesis: For every convex polygon with $ k$ vertices (points) $ v_1,v_2,\dots,v_i,v_{i\plus{}1},\dots,v_k,v_1$, there exist three consecutive vertices so that circumcircle formed by the these three vertices either has every other point of the polygon inside the circumcircle or exactly on the circumcircle.

In either of the cases we shall say that the circumcircle [i]includes[/i] a vertex. The claim is trivial for $ k\equal{}3$. We shall place the $ k\plus{}1$ th point (new vertex) so that none of the circumcircles of any three consecutive points of the $ k$ polygon includes it and all the $ k\plus{}1$ points form a convex polygon. Let the convex polygon with $ k\plus{}1$ vertices be $ v_1,v_2,\dots,v_i,v_{k\plus{}1},v_{i\plus{}1},\dots,v_k,v_1$. It is clear that the circumcircle including all the points of the $ k\plus{}1$ polygon is the circumcircle of atleast one of the three triangles $ (v_{i\minus{}1},v_i,v_{k\plus{}1}),(v_i,v_{k\plus{}1},v_{i\plus{}1}),(v_{k\plus{}1},v_{i\plus{}1},v_{i\plus{}2})$. Suppose the circumcircles of two of the triangles do not include atleast one point (may be specific to the triangle). We shall arrive at a contradiction by assuming that the circumcircle of the  leftover triangle does not include a point. Suppose it does not include a point (vertex of  $ k\plus{}1$ polygon) say $ v_j$. But the circumcircle of the triangle includes the vertex $ v_j$ as stated by induction hypothesis (it may exclude some other vertex). By similar reasoning we can assert that the circumcircle includes every point of the $ k\plus{}1$ polygon.

Note: Strictly speaking, we should have proved a lemma that any convex polygon with $ k$ vertices can be obtained from a $ k\minus{}1$ convex polygon by placing the new vertex ouside a edge appropriately. I do not know how to prove this from basic principles (Looks quite harder the problem itself). The proof for the converse of the lemma is trivial.


It seems like multiplying the numerator and denominator by $2R$ gets what you want.

[quote="ch1n353ch3s54a1l"][quote="Hikaru79"]Isn't that only true when $2R=1$?[/quote]

That's what i was thinking... and at the same time i didnt' know what R represented.  :dry:[/quote]

$R$ would be the circumradius of the triangle (i.e. the radius of the circumscribed circle). And yeah do what towersfreak said. It's true for any $R$...

[quote="ch1n353ch3s54a1l"]AOPS Vol. 2 Pg. 35

It states that $\sin A = \frac{a}{2R}$ and $\sin B = \frac{b}{2R}$, and then it says $\frac{\frac{a}{2R}-\frac{b}{2R}}{\frac{a}{2R}+\frac{b}{2R}}=\frac{a-b}{a+b}$, but I'm not sure why..[/quote]

Er, okay, simple example:

$\frac{ \frac{3}{2} - \frac{1}{2} }{ \frac{5}{2} - \frac{4}{2} } = \frac{3 - 1}{5 - 4} = 2$.  :)

So, specifically,

$\frac{ \frac{a}{2R} - \frac{b}{2R} }{ \frac{a}{2R} + \frac{b}{2R} } = \frac{ \frac{a}{2R} - \frac{b}{2R} }{ \frac{a}{2R} + \frac{b}{2R} } \times \frac{2R}{2R} = \frac{ \frac{2Ra}{2R} - \frac{2Rb}{2R}} { \frac{2Ra}{2R} + \frac{2Rb}{2R} } = \frac{a - b}{a + b}$.

People who post in Intermediate Topics should know their fraction Basics :D

And the values for $\sin A, \sin B$ are easily derivable by drawing the circumcircle around $\triangle ABC$ and then realizing that the three triangles created by the circumradi are isosceles, meaning their angle bisectors are also altitudes.


let$x_i , y_i\in \mathbb{R}^+\ ,\ i=\{1,2,\dots,n\}$ and $\sum x_i\geq \sum x_iy_i$

Prove that:$\sum x_i \leq \sum \frac{x_i}{y_i}$

[i] Sum must be  cyclic  and not symmetric.[/i]. We ' ll use the so called Engel form of $BCS$

Write

   $\sum \frac{x_i}{y_i} = \sum \frac{x^2_i}{x_i y_i}$

              $\geq \frac{(\sum x_i )^2}{\sum x_iy_i}$    $\geq \frac{(\sum x_i )^2}{\sum x_i} =$      ${\sum x_i}$

 [u]Babis[/u]


What is the minimum value of the function $f(x) = \frac{1}{x(x-1)}+\frac{1}{(1-x)(x-2)}+\frac{1}{(2-x)(x-3)}$ when $1\le x \le 2$

First the domain of $x$ should be defined $1<x<2$.We have $f(x)=-\left(\frac{1}{x}+\frac{1}{x-3}\right)$,
so $f(x) $ is an increasing function for $1<x<2$, yielding $f(1)<f(x)<f(2)$.
Therefore the minimum value doesn't exist.

first of all, that intervall must change to:
$1<x<2$ because $x\not = 1,2$
ans since $f$ is an increasing function, there is no min. ;)

Hmm I must of written the problem wrong. I'm pretty sure that there is a minimum. Oh well I'll post the edited problem once it is corrected.


linear algebra
matrix
algebra
polynomial
complex numbers
linear algebra unsolved

A real nxn matrix and A^3=A+E.prove detA>0.

I think your matrix should have real entries.
$(\det(A))^{3}=\det(A+I_{n})$ so $\det(A)\cdot \det(A+I_{n})\geq 0$.
$\det(A)\cdot \det(A+I_{n}) \cdot \det(A-I_{n})=1\geq 0$ so $\det(A-I_{n})\geq 0$.
On the other hand:$\det(A)=\det(A-I_{n})\cdot \det(A^{2}+A+I_{n})$
from where $\det(A)=\det(A-I_{n})\cdot |\det((A+\frac{1}{2}I_{n})+i\frac{\sqrt{3}}{2}I_{n}|^{2}\geq 0$.
There is a generalisation for this problem, $A^{2k+1}=A+I_{n}$ then $\det(A)\geq 0$.

spix explain the last line of the proof with the imaginary number.

spix do you know a proof of the generalization?

Southstar is right.
In fact Spix has proven that $det(A-I)>0$. Assume that $det(A)\leq{0}$. Then there exists $u\in[0,1[$ s.t. $det(A-uI)=0$. Necessarily $u^{3}=u+1\geq{1}$ and $<1$ a contradiction.

@ §outh§tar: http://www.mathlinks.ro/Forum/viewtopic.php?t=147726
   you can see there how to prove that if $AB=BA$ then $\det(A^{2}+B^{2})\geq 0$ (I just wanted to put $A^{2}+A+I_{n}$ in this form to avoid more complex stuff like the eigenvalues being the roots of $x^{2}+x+1$ which are complex and conjugated..bla bla)
@ redkimchi:
 The generalisation is very similar in proof.
Firstly $A$ is invertible, then:
$\det(A),\det(A+I_{n})$ have the same sign.
$A^{2}=A(A-I_{n})(A^{2k}+....+A+I_{n})$  gives us that $\det(A),\det(A-I_{n})$ have the same sign ($x^{2k}+...+x+1$ has no real roots meaning that the det is positive)($*$),
Afterwards because $A(A^{2}-I_{n})(A^{2(k-1)}+A^{2(k-2)}+...+A^{2}+I_{n})=I_{n}$ , $\det(A)\det(A-I_{n})\det(A^{2(k-1)}+A^{2(k-2)}+...+A^{2}+I_{n}),\det(A+I_{n})$ have the same sign.($x^{2(k-1_{}}+...+x^{2}+1$ has no real roots meaning that the det is positive)($**$)
($*$)$+$($**$)$\to$ $\det(A+I_{n})\geq 0$ which solves the problem.

Hi Spix ; you prove that $det(A^{2}+B^{2})\geq{0}$ using complex conjugates (the proof is wellknown) that is implicitly (but not explicitly) forbidden.

For the original problem:
You can also consider the eigenvalue of $A$.
From:$A^{3}=A+I$,we get the eigenvalue of $A$ must be the root of $x^{3}-x-1=0$,but $x^{3}-x-1=0$ has a positive real root $\lambda_{1}$,and two congugate complex roots $\lambda_{2},\lambda_{3}$.
Also $\lambda_{2},\lambda_{3}$ has the same multiple as the eigenvalues of $A$.
Thus:
$det(A)=(\lambda_{2}\lambda_{3})^{k}(\lambda_{1})^{n-2k}>0$
This solution is also valid for the generalization which $spix$ provided.

[quote="spix"]Wow.
[quote="matin1985"]a^3=a+e
it must be done without using roots of x^3=x+1[/quote][/quote]
 :blush: I haven't seen that

ok if we consider the roots of x^3=x+1...i have several questions

1)A is root of its characteristic polynomial +if Av=av then A^3v=a^3v and that's why a is a root of x^3=x+1(i think this is clear for me) but what isn't...
2)it'known that if all characteristic roots of A are in F(some main number field) then A~Jordan matrix(eigenvalues on the main diagonal and the jordan matrix is triangular). my question is if a is an eigenvalue of A and it's a comlex number(like in this problem there are  conjugate complex numbers) is it still true that this complex root is on the diagonal so many times as it's a root of the characteristic polynomial.


Paintball gun #1 can shoot 60 paintballs per minute.  Paintball gun #2 can shoot 3 paintballs per second.  Paintball gun #3 can shoot 14400 paintballs per hour.  Which gun can shoot the most paintballs in an hour.  WHy?

[hide="click if colts rock"]
gun #2 because in gun #1 it shoots 60 paintballs per minute so times it by 60 (3600)to get how many paintballs per hour. In the second one it shoots 3 paintballs per second. So you would times that by sixty and by 60 again which equals 10800.
The 3rd one tells you already how many paintballs per hour so:
 Gun1: 3600 per hour
 Gun2: 10800 per hour
 Gun3: 1440 per hour
Therefore gun 2 does.
I am probably wrong :blush:[/hide]

[hide="TO COLTS18 ONLY"]you are incorrect, you misread the problem again.  Paintball gun #3 shoots 14400 paintballs an hour, not 1440.[/hide]

[hide="click if colts rock"]
sorry gun 3 becuse there is a extra 0 ( I wrote 1440) So:
Gun1:3600
Gun2:10800
Gun3:14400
Gun 3 is my final answer[/hide]


let a,b,c be positive numbers.prove that


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

Prove that this statement:

If R is an integral domain, then Ring of formal power series is also an integral domain.

If R is a field, then Ring of formal Laurent series is also a field.

1) is right, use the function $ v_X(\sum_i a_i X^i) \equal{} min \{i : a_i \neq 0\}$.
2) is also right. first observe that, for every ring $ R$, the unit group $ R[[X]]^*$ consists of those formal power series whose constant term is a unit in $ R$. now every nonzero element of $ R((X))$ is associated (via some power of $ X$) to some element of $ R[[X]]^*$.

The same argument works for any ring with a discrete valuation.


Let $x, y, z$ be nonnegative real numbers satisfying condition $xy + yz + zx=1$. Find the minimal value of
       $P = \sqrt{(1+x^2)}\sqrt{(1+y^2)}\sqrt{(1+z^2)}+4xyz$

I think inequalities <=> x+y+z+3xyz>=2
This inequalities is quite easy. :D  :?

I think min=2 happen when $x=0,y=z=1$
so we need to prove
$\sqrt{P} \geq 2$
$\leftrightarrow x+y+z+3xyz \geq 2$

It should have been called: A new extremely beautiful problem. Now, let's give a complete solution. It's clear that $\sqrt{(1+x^2)(1+y^2)(1+z^2)}=(x+y)(y+z)(z+x)=x+y+z-xyz$. So, as Hikaru said, we must prove that $ x+y+z+3xyz\geq 2$. Let $ f(x,y,z)=x+y+z+3xyz$. Then if we take y be the maximum amonmg the three numbers, we can easily see that ${ f(x,y,z)\geq f(x+\frac{zx}{y},y,0} $ since it reduces to $ 3y^2\geq 1$ which is obvious. Thus, we only need to show that $ f(x+\frac{xz}{y},y,0)\geq 2$. Use the fact that $ xy+yz+zx=1$ and this will suddenly reduce to $ (y-1)^2\geq 0$.


(For middle schoolers.)

Write the binary numeral for the decimal numeral 245.

my method was [hide] 245 is 256 minus 11, so 100000000 - 1011[/hide]


That's probably an easy problem, but I've got some confusion about what the rank of an operator is, so...
Let [tex]\mathcal{A},\mathcal{B} \in \mathcal{L}( V, V )[/tex], prove that : [tex]\mbox{rank } \mathcal{A}= \mbox{rank } \mathcal{B}+ \dim ( \mbox{Im } \mathcal{A} \cap \mbox{Ker } \mathcal{B})[/tex].
(The book I'm reading didn't define the rank of an operator, so I guess [tex]\mbox{rank } \mathcal{A}= \dim \mbox{Im } \mathcal{A}[/tex], is it true ?)

You have a correct definition of rank. There are several equivalent forms.

The statement you are trying to prove is nonsense, unless there is some relation between $A$ and $B$ you're not telling us about.

In fact I added both sides [tex]\dim \tmop{Ker} \mathcal{B}- \dim \tmop{Ker} \mathcal{A}[/tex] to obtain that [tex]0 = \dim ( \tmop{Im} \mathcal{A} \cap \tmop{Ker} \mathcal{B}) + \dim \tmop{Ker} \mathcal{B}- \dim \tmop{Ker} \mathcal{A}[/tex], which lead to the conclusion that [tex]\tmop{Ker} \mathcal{B}= \{ 0 \} \Rightarrow \tmop{Ker} \mathcal{A}= \{ 0 \}[/tex], for any [tex]\mathcal{B}[/tex] and [tex]\mathcal{A}[/tex], which is absurd. The contradiction, did let me think that I had a wrong definition of operator rank... But aparently, the mistake is in the book (which is kind of odd since it's the second time, I see a ``bad'' problem)...

P.S: I believe there is also a mistake in the Magic Square topic I posted, the case n=2 seems not to work (and this is why I posted it :))


I'm not sure about this...
[hide="solution"]
From the first restriction:
$ x_1: x_2: ...: x_{1005} \equal{} x_1 \plus{} 1: x_2 \plus{} 3: ...: x_{1005} \plus{} 2009$

Therefore each term has been scaled by a factor $ k$

$ \sum x_i \equal{} 2010$ and $ \sum x_i \plus{} 2i \minus{} 1 \equal{} 2010 \plus{} \frac {1005}{2}(2 \plus{} 2(1005 \minus{} 1)) \equal{} 1005^2$

$ \therefore$ our factor $ k \equal{} \frac {1005^2 \plus{} 2010}{2010} \equal{} \frac {1005 \plus{} 2}{2} \equal{} \frac {1007}{2}$

$ \Rightarrow \frac {1007}{2}x_{21} \equal{} x_{21} \plus{} 41 \Longrightarrow x_{21} \equal{} \frac {82}{1005}$...[/hide]

EDIT: Fixed the little typo, answer is still the same though  :) !

um it'd be $ x_n/(x_n \plus{} 2n \minus{} 1)$, not $ 2n \plus{} 1$

Just fix this and I think you have a good solution... just find $ x_{21}$ not $ x_{22}$

EDIT: actually I think that $ 1005^2$ is correct anyway

[quote="GlenthemanN"]okay this is my first go at an 'intermediate' problem but here goes

[hide]

let $ a_n \equal{} a_1, a_2, \ldots, a_{1005} \equal{} 1, 2, 3, \ldots, 2009$

notice we can rewrite the first equation as

$ \frac {x_1}{x_1 \plus{} a_1} \equal{} \frac {x_2}{x_2 \plus{} a_2} \equal{} \ldots \equal{} \frac {x_{1005}}{x_{1005} \plus{} a_{1005}}$

then notice that $ x_n$ must always be some ratio of $ a_n$, therefore

$ x_n \equal{} ka_n$

it follows then that

$ k(a_1 \plus{} a_2 \plus{} a_3 \plus{} \ldots \plus{} a_{1005}) \equal{} 2010$

notice that $ \displaystyle\sum_{n \equal{} 1}^{1005}{a_n} \equal{} 1005^2$

thus:
$ k \equal{} \frac {2}{1005}$

and it follows that $ x_{21} \equal{} ka_{21} \equal{} \frac {82}{1005}$

[/hide]

can someone give me any pointers on explaining myself better, I had a tough time with this one and I also dont know how to show a sequence of numbers right, I always thought in the {} but latex wont allow that :O[/quote]

[hide="This may be"]

$ \left\{a_n\right\}^{1005}_{n\equal{}1}: \equal{} 2n\minus{}1$[/hide]