inequalities
Euler
geometry
incenter
circumcircle
inequalities proposed

\[ \underline {\overline {\left|\ IH^2+IO^2\le HO^2\ \right| }} \] [u]A metrical solution.[/u]

$IH^2=4R^2+4Rr+3r^2-s^2,\ IO^2=R^2-2Rr,\ HO^2=9R^2-2s^2+2r^2+8Rr$

$R(4R+r)^2\cdot H\Gamma ^2=4R^2\left[ R(4R+r)^2-2s^2(2R-r)\right]\ge 0\Longrightarrow s^2\le \dfrac {R(4R+r)^2}{2(2R-r)}\ \ \ (1)$

$IH^2+IO^2\le HO^2\Longleftrightarrow IH^2+IO^2-HO^2\le 0\Longleftrightarrow s^2\le 4R^2+6Rr-r^2\ \ \ (2)$

$\dfrac {R(4R+r)^2}{2(2R-r)}\le 4R^2+6Rr-r^2\Longleftrightarrow R(4R+r)^2\le 2(2R-r)(4R^2+6Rr-r^2)\Longleftrightarrow$

$8R^2-17Rr+2r^2\ge 0\Longleftrightarrow (R-2r)(8R-r)\ge 0\ \ \ (3)$  [u]TRUE.[/u]

From the relations $\overline {1,3}$ results the above remarkable inequality. This inequality shows that the point $I$ belongs to the disk with the diameter $HO\ !.$

Question: Is there a syntetical demonstration of this remarkable inequality ?.

[quote="Ph-An = Levi"]\[ \underline {\overline {\left|\ IH^2+IO^2\le HO^2\ \right| }} \][/quote]

This is equivalent to the fact that the triangle HIO is right-angled or obtuse-angled at its vertex I, i. e. to the inequality $\measuredangle HIO\geq 90^{\circ}$. But this is easy to see: According to http://www.mathlinks.ro/Forum/viewtopic.php?t=39447 , we have $\measuredangle GIH>90^{\circ}$; but since the point G lies inside the segment HO (this follows from the properties of the Euler line), we have $\measuredangle HIO>\measuredangle GIH$. Thus, $\measuredangle HIO>90^{\circ}$, and we are done.

By the way, please don't forget about defining the notations you use: You are referring to a triangle ABC with incenter I, orthocenter H, circumcenter O and centroid G.

  darij

[b]Darij: [/b][i]This is equivalent to the fact that the triangle HIO is right-angled or obtuse-angled at its vertex I.[/i]

I wrote "[u]I belongs to the disk with the diameter HO[/u]". Thanks. I will respect your recommendations.


Of course it is wrong. The problem is:

 Given $a, b, c$ three positive real numbers, prove the following inequality

                            \[ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq\frac{3}{2}\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right).  \]

  I have proposed this problem. :D

 [b] Remark. [/b]

      This inequality given in the contest is stronger than the following one given at Iran Olympiad in 2005.

     Given $a, b, c$ three positive real numbers, prove the following inequality

               \[ \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^{2}\geq\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right). \]

yes.but still easy.after computing the ineq is 2sigma(a^2/b^2)+sigma(a/c)>=3sigma(a/b).this is equivalent with 2sigma(a^4c^2)+sigma(a^3b^2c)>=3sigma(a^3bc^2).but 2a^4c^2+ab^3c^2>=3a^3bc^2 from am-gm.writing the other 2 ineq like this one and suming we get the desired ineq

You could also expand and simplify, and the ineq is killed by Muirhead.

[quote="Valentin Vornicu"]Prove that for all positive real numbers $a,b,c$ the following inequality holds \[ \left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 32 \cdot \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b \right) .  \][/quote]

This is very easy!
Just use $x^2+y^2+z^2\geq xy+yz+zx\ (x>0,\ y>0,\ z>0)$ and $xyz=1.$ :D :  ;)

$x=\frac{a}{b}$;$y=\frac{b}{c}$;$z=\frac{c}{a}$  :arrow:  $xyz=1$

The inequality becomes: $2(x+y+z)^2 \geq 3(xy+yz+zx) + 3(x+y+z)$

We know $(x+y+z)^2 \geq 3(xy+yz+zx)$  (1)

Also by AM-GM: $x+y+z \geq 3$  :arrow:  $(x+y+z)^2 \geq 3(x+y+z)$  (2)

By summing (1) and (2) we get $2(x+y+z)^2 \geq 3(xy+yz+zx) + 3(x+y+z)$

which is what we wanted to prove

I used the same notation as M4RIO and got to this:
$2(x^2+y^2+z^2)+xy+xz+yz \geq 3(x+y+z)$ which is true considering that $2x^2+yz \geq 3x$

With substitution x=a/b   y=b/c  z=c/a
2(x+y+z)(x+y+z)>=3(xy+xz+yz)+3(x+y+z)2(x*x+y*y+z*z)+(xy+yz+xz)>=3(x+y+z)
X*x+x*x+yz>=3*∛(x*x*x*x*y*z)=∛(x*x*x )=x   (y*z=b/a  x*x*x*x=a/b*b/a*a/b*a/b*a/b=x*x*x)
2(x*x+y*y+z*z)+(xy+yz+xz)>=3(x+y+z)

[quote="PlatinumFalcon"][quote="Valentin Vornicu"]Prove that for all positive real numbers $a,b,c$ the following inequality holds \[ ( \frac ab + \frac bc + \frac ca )^2 \geq \frac 32 \cdot  ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b ) .  \][/quote]
Note that the left hand side can be rewritten as $\frac{3}{2}\cdot( \frac{a}{b}+\frac{b}{a}+\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b})$. Now, the right can be expressed as $\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}+2( \frac{a}{c}+\frac{b}{a}+\frac{c}{b})$. We move the common terms of $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}$ to the left, giving $\sum_{cyc} \frac{a^2}{b^2}\ge \sum_{cyc} \frac{3}{2}\cdot \frac{a}{b}-\frac{1}{2}\cdot \frac{b}{a}$. We can prove the inequalities by showing that the corresponding terms also satisfy the inequality; i.e. $\frac{a^2}{b^2}\ge \frac{3}{2}\cdot \frac{a}{b}-\frac{1}{2}\cdot \frac{b}{a}$. Setting $\frac{a}{b}=x$, we have $x^2\ge\frac{3}{2}x-\frac{1}{2}\cdot\frac{1}{2}$, which is equivalent to $2x^3-3x^2+x=x(2x^2-3x+1)\ge 0$. Since the discriminant of $F(x)=2x^2-3x+1$, $\Delta=9-8=1$, $F$ will always be positive, so the result follows.[/quote]

The ineq. becomes
$ x^2\ge\frac{3}{2}x-\frac{1}{2x} $
 The generating cubic function should be $f(x)=2x^3-3x^2+1=(x-1)^2(2x+1)\geq 0$

Or AM-GM

$2x^3+1=x^3+x^3+1\geq 3\sqrt[3]{x^3.x^3.1}=3x^2$,

the result follows.

Letting $x = a/b, y = b/c, z = c/a$, we see we have $xyz = 1$ and $x,y,z$ are positive. Our inequality now becomes
\[(x+y+z)^2 \ge \frac{3}{2}(x+y+z+xy+yz+zx)\]
Rearranging yields
\[2x^2 + 2y^2+2z^2 + xy + yz + zx \ge 3(x+y+z)\]
Now we can just AM-GM: we have $x^2 + x^2 + yz \ge 3x$, and cyclically summing, we get the desired inequality.

[quote=Valentin Vornicu]Prove that for all positive real numbers $a,b,c$ the following inequality holds \[\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 32  \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b \right) .  \][/quote]
Let  $a,b,c$ be positive real numbers. Prove that\[{\left(\frac{b}{a}+ \frac{a}{b}\right)\left(\frac{c}{b}+ \frac{b}{c}\right)\left(\frac{a}{c}+ \frac{c}{a} \right)}\ge \frac{4}{3}\left(\frac{a+b} {c}+\frac{b+c}{a}+ \frac{c+a}{b}\right)\]
Prove that for all positive real numbers $a,b,c$ the following inequality holds $$  \left(a+ \frac
 1a\right)\left(b+ \frac 1b \right)\left(c+ \frac 1c \right)
\geq \frac 43 \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b \right) .  $$
[size=50]  Jichen  p/5916256821[/size]

Prove that for all positive real numbers $a,b,c$ the following inequality holds
$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 95  \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b -1\right) $$
$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 94 \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b -2\right) $$
$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac {18}{11} \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b-\frac{1}{2} \right) $$
$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq 2 \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b - \frac {3}{2}\right) $$
$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac{12}{7}  \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b - \frac {3}{4}\right) $$
$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq  \frac {12(11-4\sqrt 6)}{5} \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b-3 \right) $$

[hide=Solution]Denote $\frac{a}{b}=x, \frac{b}{c}=y, \frac{c}{a}=z$. Then we want to show $(x+y+z)^2 \geq \frac{3}{2}\left(\frac{1}{z}+y+\frac{1}{x}+z+\frac{1}{y}+x\right)$, or $(x+y+z)^2 \geq \frac{3}{2}\left(\frac{xy+xz+yz}{xyz}+x+y+z\right)^2$. However $xyz=1$, so this is equivalent to $(x+y+z)^2 \geq \frac{3}{2}(xy+xz+yz+x+y+z)$, or $2(x+y+z)^2 \geq 3(xy+xz+yz+x+y+z)$, equivalent to $2(x^2+y^2+z^2)+4(xy+xz+yz) \geq 3(xy+xz+yz)+3(x+y+z)$, or $2(x^2+y^2+z^2)+(xy+xz+yz) \geq 3(x+y+z)$. Now we can just use the bound $x^2+y^2+z^2 \geq xy+xz+yz$, so it suffices to show $3(xy+xz+yz) \geq 3(x+y+z)$, or $xy+xz+yz \geq x+y+z$, which is obviously true.[/hide]

[quote=sqing]Prove that for all positive real numbers $a,b,c$ the following inequality holds
$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 95  \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b -1\right) $$[/quote]

Prove that for all positive real numbers $a,b,c$ the following inequality holds$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq  \frac{3}{k+1} \left (\frac{a+kb}c + \frac{b+kc}a + \frac{c+ka} b\right) $$
Where $k>0.$
$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq  \frac{a+2b}c + \frac{b+2c}a + \frac{c+2a} b$$

Prove that for all positive real numbers $a,b,c$ the following inequality holds$$\left( \frac ab + \frac bc + \frac ca \right)^3 \geq  \frac{9}{k+1} \left (\frac{a+kb}c + \frac{b+kc}a + \frac{c+ka} b\right) $$
Where $k>0.$
$$\left( \frac ab + \frac bc + \frac ca \right)^3 \geq  \frac{a+8b}c + \frac{b+8c}a + \frac{c+8a} b$$

[quote=sqing]Prove that for all positive real numbers $a,b,c$ the following inequality holds$$\left( \frac ab + \frac bc + \frac ca \right)^2 \geq  \frac{3}{k+1} \left (\frac{a+kb}c + \frac{b+kc}a + \frac{c+ka} b\right) $$
Where $k>0.$[/quote]

[quote=sqing]Prove that for all positive real numbers $a,b,c$ the following inequality holds$$\left( \frac ab + \frac bc + \frac ca \right)^3 \geq  \frac{9}{k+1} \left (\frac{a+kb}c + \frac{b+kc}a + \frac{c+ka} b\right) $$
Where $k>0.$[/quote]

[quote=Valentin Vornicu]Prove that for all positive real numbers $a,b,c$ the following inequality holds \[ \left( \frac ab + \frac bc + \frac ca \right)^2 \geq \frac 32 \cdot \left ( \frac{a+b}c + \frac{b+c}a + \frac{c+a} b \right) .  \][/quote]
[url=https://artofproblemsolving.com/community/c6h3007466p27012468]h[/url]
[url=https://www.awesomemath.org/wp-pdf-files/math-reflections/mr-2023-02/mr_2_2023_problems.pdf]S624[/url] Let $a,b,c>0$. Prove that$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \sqrt{\frac{a+b}{2c}}+ \sqrt{\frac{b+c}{2a}}+ \sqrt{\frac{c+a}{2b}}$$


Geia se olous. Eimai kainoyrgia sto site. Eimai prwtoeths foithtria plhroforikhs. Apofasisa na dhmosieysw ayto to mhnyma giati to oneiro moy einai na symmetasxw se mia olympiada mathimatikwn. To panepisthmio moy symmetexei sto diagwnismo SEEMOUS. Apofasisa loipon na ksekinhsw na meletw systimatika. To provlhma einai oti den kserw ti akrivws prepei na diavasw kai perissotero poia vivliografia wste na eimai kala proetoimasmenh. Exw koutaksei kai ta palia themata kai moy fainontai ola teleiws agnwsta. Eimai poly apogohteymenh giati kaneis de me vohthaei, oyte oi kathightes sto panepisthmio:( . den prepei na yparei mia proetoimasia ek meroys twn kathightwn gia osous theloyn na symmetasxoyn? sas parakalw VOHTHISTE ME!

Καλως ηρθες στο φορουμ :welcomeani:  . Να' σε σιγουρη πως θα μαθεις ή μαλλον θα καταλαβεις πολλα για τα μαθηματικα αφου τα παιδια εδω εχουν μεγαλες ικανοτητες και πολυ πειρα.

Αυτο βεβαια που φοβαμαι ειναι οτι το φορουμ εδω ειδικευεται περισσοτερο στην προετοιμασια για σχολικες ολυμπιαδες παρα για πανεπιστημιακες. Προσωπικα δεν κανω προετοιμασια οποτε συγχωρεσαι με αμα λεω βλακειες...

Οσο για αυτο που λες οτι θα επρεπε να υπαρχει προετοιμασια εκ' μερους των καθηγητων, ετσι που ειναι τα ελληνικα πανεπιστημια μην περιμενεις και πολλα...

Υ.Γ. Επιτελους και ενα κοριτσι στο φορουμ! :P

Ευχαριστώ πολύ για την απάντηση σου. Στο εξωτερικό κάνουν 1-2 φορές μάθημα την εβδομάδα για προετοιμασία. Εγώ παλεύω μόνη μου. Έχω βρει κάποια βιβλία αλλά κυριολεκτικά είναι ακαταλαβίστικα. Θέλω πρώτα να μάθω τα βασικά. Μέχρι πριν ξεκινήσω να τα διαβάζω πίστευα ότι έχω κάποιο ταλέντο γιατί έλυνα αρκετά θέματα διαγωνισμών σχολικού επιπέδου. Τώρα όμως έχω χάσει την αυτοπεποίθηση μου.

Κατι που εμαθα εδω στο φορουμ ειναι οτι δεν υπαρχει ταλεντο σε τιποτα. Ολα ειναι θεμα προσπαθειας, κατανοησης και ποιοτικης δουλειας(ποιοτικης οχι ποσοτικης). Μην απογοητευεσαι, τα παιδια εδω νομιζω θα σε βοηθησουν. Τσεκαρε για αρχη τα stickies των ολυμπιαδων που εχουν βεβαια βιβλια για σχολικες ολυμπιαδες αλλα οπως λες και εσυ καλυτερα να αρχισης απο τα βασικα(θα σου σου συνηστουσα να μην δωσεις μεγαλο βαρος στην ευκλειδια γεωμετρια καθως απο οτι ξερω αυτη δεν εξεταζεται στους πανεπιστημιακους :wink: ).

Οντως ο filosofimenos εχει δικιο,μην απογοητευσαι,ποτε δεν εινα αργα.Εαν ψαξεις λιγο στο forum θα βρεις πολλα topic στα οποια διατυπωνονται παρομοιες αποριες με τις δικιες σου και προτεινονται διαφορα βιβλια.Επισης θα βρεις πολλα θεματα και θα καταλαβεις πανω κατω το πνευμα των διαγωνισμων αφου απο οτι ξερω ο SEEMOUS εχει πανω κατω την ιδια υλη με τις IMO.(Εαν εξαιρεσεις την Ευκλειδια γεωμετρια)

Κατ αρχας καλως ηρθες στο φορουμ :)

Η Υλη για τον Seemous ειναι:
Γραμμικη Αλγεβρα
Αναλυτικη Γεωμετρια
Ανωτερη Αλγεβρα (Ομαδες, Δακτυλιοι, Σωματα)
Απλη Αναλυση - Calculus
Συνδιαστικη
Λιγα πραγματα απο Θεωρια Συνολων και απο Πιθανοτητες
Και νομιζω (χωρις να ειμαι σιγουρος) λιγα πραγματα Πραγματικη και Αριθμιτικη Αναλυση\

Απο βιβλια τωρα, για αλγεβρα θα σου πρωτεινα το "Εισαγωγη στην Αλγεβρα" του Fraileigh (Πανεπιστημιακες Εκδοσεις Κρητης), ενω για Συνδιαστηκη το "Διακριτα Μαθηματικα" του κ. Παπαιωαννου (Εκδοσεις ΕΜΠ), αυτα διαβαζω και εγω αυτες τις μερες, μου τα πρωτειναν καποιοι γνωστοι οι οποιοι ειχαν διακρισεις παλιοτερα στον Seemous και στον IMC

Αν θελεις μπορεις να ρωτησεις και στο subforum ΠΡΟΒΛΗΜΑΤΑ ΓΙΑ ΦΟΙΤΗΤΕΣ, διοτι οπως ειπε και ο Φιλοσοφιμενος, εδω ασχολουνται περισσοτερο με Ολυμπιαδες για μαθητες

ΕΧΩ ΤΗΝ ΕΝΤΎΠΩΣΗ ΟΤΙ ΘΑ ΓΊΝΟΥΝ ΚΑΙ ΦΕΤΟΣ ΜΑΘΗΜΑΤΑ
(ΟΠΩΣ ΚΑΙ ΠΕΡΣΥ).
ΤΑ ΜΑΘΗΜΑΤΑ ΓΊΝΟΝΤΑΙ ΣΑΒΑΤΟ ΚΑΙ ΚΥΡΙΑΚΗ ΣΤΗΝ ΜΑΘΗΜΑΤΙΚΗ ΕΤΑΙΡΕΙΑ
ΑΠΟ ΚΑΘΗΓΗΤΕΣ ΤΟΥ ΜΑΘΗΜΑΤΙΚΟΥ ΤΜΗΜΑΤΟΣ ΚΑΙ ΤΟΥ ΠΟΛΥΤΕΧΝΕΙΟΥ.
ΓΙΑ ΑΚΡΙΒΕΣΤΕΡΗ ΕΝΗΜΕΡΩΣΗ ΘΑ ΕΠΑΝΕΛΘΩ.

vasika egw spoudazw sthn kerkyra plhroforikh kai den exw th dynatothta na symmetasxw sta mathimata. nicknafplio eyxaristw gia tis polytimes plhrofories sou. tha prospathisw na vrw ta vivlia poy mou eipes. an kserei kaneis exw apories se dyo palia themata tou seemous. opoios mporei as episkeftei to akoloytho link

http://www.seemous.eu/assets/files/2008/seemous08_exam.pdf

kai as diavasei ta dyo prwta provlhmata me tis lyseis tous. esw tis ekshs erwthseis. opoios mporei as vohthisei...

ERWTHSH PRWTH:

TI EINAI AYTO TO MIN KAI MAX STH SYNARTHSH? APO POIO GNWSTIKO PEDIO EINAI AYTO TO PROBLHMA? SE GENIKES GRAMMES TO KATALAVAINW ALLA KOLLAW STO MIN KAI MAX

ERWTHSH DEYTERH:

STO DEYTERO PROVLHMA: PROKEITAI GIA ANALYTIKH GEWMETRIA? H LYSH TOY MOY FAINETAI TELEIWS EKSWGHINH...POIOS MPOREI NA ME DIAFWTISEI LIGO?

[quote]
TI EINAI AYTO TO MIN KAI MAX STH SYNARTHSH? APO POIO GNWSTIKO PEDIO EINAI AYTO TO PROBLHMA? SE GENIKES GRAMMES TO KATALAVAINW ALLA KOLLAW STO MIN KAI MAX
[/quote]

Υπάρχει ένα γνωστό θεώρημα στην ανάλυση που λέει ότι κάθε συνεχής συνάρτηση σε κλειστό φραγμένο διάστημα είναι φραγμένη και λαμβάνει τις ακρότατες τιμές της. Με άλλα λόγια, αν $ f$ συνεχής στο $ [a,b]$, τότε υπάρχει $ c \in [a,b]$ ώστε $ f(x) \leq f(c)$ για κάθε $ x \in [a,b]$ και υπάρχει $ d \in [a,b]$ ώστε $ f(x) \geq f(d)$ για κάθε $ x \in [a,b]$. Γράφουμε $ min_{x \in [a,b]}f(x) \equal{} f(d)$ και $ max_{x \in [a,b]}f(x) \equal{} f(c)$.

[quote]
STO DEYTERO PROVLHMA: PROKEITAI GIA ANALYTIKH GEWMETRIA? H LYSH TOY MOY FAINETAI TELEIWS EKSWGHINH...POIOS MPOREI NA ME DIAFWTISEI LIGO?
[/quote]

Η λύση χρησιμοποιεί και ανάλυση και (αναλυτική) γεωμετρία. Το πρόβλημα ρωτάει το εξής: Δίνεται ένα κυρτό πολύγωνο $ P_0$. Ας υποθέσουμε (για καλύτερη κατανόηση του προβλήματος) ότι το $ P_0$ είναι τρίγωνο. (Άρα το $ n$ στη λύση ισούται με 3.) Μετά ορίζουμε τρίγωνο $ P_1$ του οποίου οι κορυφές είναι τα μέσα των πλευρών του $ P_0$, ορίζουμε τρίγωνο $ P_2$ του οποίου οι κορυφές είναι τα μέσα των πλευρών του $ P_1$ κτλ. (Δοκίμασε να σχεδιάσεις τα $ P_1$ και $ P_2$.) Μας ζητάει η άσκηση να δείξουμε ότι υπάρχει μοναδικό σημείο που βρίσκεται μέσα σε όλα τα τρίγωνα.  Η λύση προχωράει ως εξής:
0)  :D  Μαντεύουμε ότι το μοναδικό σημείο είναι το κέντρο βάρους του $ P_0$.
1) Δείχνουμε ότι το κέντρο βάρους του $ P_0$ είναι το κέντρο βάρους του $ P_1$ άρα και του $ P_1$ κτλ. Άρα το κέντρο βάρους του $ P_0$ όντως βρίσκεται μέσα σε όλα τα τρίγωνα
2) Δείχνουμε ότι τα τρίγωνα "μικραίνουν". Πιο συγκεκριμένα, αν $ O$ το κέντρο βάρους του $ P_0$ και $ R_k$ η ακτίνα του μικρότερου κύκλου που περιέχει το $ P_k$, τότε $ R_k \rightarrow 0$. Άρα κάθε σημείο διαφορετικό του $ O$ είναι έξω από το $ P_k$ για $ k$ αρκετά μεγάλο.

Συνέχιζε να κοιτάζεις προβλήματα και ότι απορίες έχεις εμείς εδώ είμαστε. (Αν μπορούμε βέβαια να τις λύσουμε.)

Νομίζω καλύτερα να μεταφερθεί στους φοιτητές η συζήτηση.

Η ΣΥΖΗΤΗΣΗ ΟΠΩΣ ΒΛΕΠΕΤΕ ΜΕΤΑΦΕΡΘΗΚΕ ΣΤΑ ΠΡΟΒΛΗΜΑΤΑ ΓΙΑ ΦΟΙΤΗΤΕΣ. Demetres ΕΥΧΑΡΙΣΤΩ ΜΕ ΒΟΗΘΗΣΕΣ ΑΡΚΕΤΑ. ΤΕΛΙΚΑ ΙΣΩΣ ΝΑ ΜΗΝ ΕΙΝΑΙ ΤΟΣΟ ΠΟΛΥΠΛΟΚΑ ΤΑ ΠΡΟΒΛΗΜΑΤΑ. ΑΛΛΑ ΟΠΩΣ ΕΙΝΑΙ ΛΥΜΕΝΑ ΕΠΙΣΗΜΑ ΕΞΑΚΟΛΟΥΘΩ ΝΑ ΜΗΝ ΤΑ ΚΑΤΑΛΑΒΑΙΝΩ. ΞΕΡΕΙ ΚΑΝΕΙΣ ΚΑΝΕΝΑ ΚΑΛΟ ΒΙΒΛΙΟ ΑΝΑΛΥΤΙΚΗΣ ΓΕΩΜΕΤΡΙΑΣ ΠΟΥ ΝΑ ΕΞΗΓΕΙ ΤΟ ΑΝΤΙΚΕΙΜΕΝΟ ΜΕ ΚΑΤΑΝΟΗΤΟ ΤΡΟΠΟ?

ρε συ NickNafplio ([hide]eisun sthn eme otan kaname grafhmata me ton papaiwannyu?[/hide]) δεν το κοβω να χρειαζονται πολυ ομαδες δακτυλιοι κτλπ εκτοσ απο κτ βασικα οποτε το βιβλίο του Φραλευ παραπάει (ένα εναλλακκτικό στο θεμα ειναι αυτο:[url]http://noether.math.uoa.gr/Undergraduate/programma-spoydn/propty3c7iaka-mathmata/ypo3c7reotika-mathmata-1/alg_nea[/url])

τωρα για αναλυτική:[url]http://server.math.uoc.gr/~pamfilos/Geo2000.pdf[/url]

συνδιαστική: κπ απλα πραγματα [url]http://combinatorics.scientist.gr/[/url]
ή [url]http://eclass.uoa.gr/modules/document/document.php?openDir=/10301218616j9[/url] ή [url]http://discrete2008.files.wordpress.com/2008/10/notes.pdf[/url]

[url]http://eclass.uoa.gr/modules/document/file.php/MATH206/Διάλεξη%20%2002%20Αθανασιάδης/Notes%20Athanasiadis.pdf[/url](Nicknafplio το βιβλίο διακριτων του  Παπαιωαννου ξερεις που μποω να το βρω?)

τωρα γενικα για τον διαγωνισμο
κπ σημειωσεις απο τη μαθηματα προετοιμασιας που γινονταν περσυ στην εμε [url]http://eclass.uoa.gr/modules/document/document.php[/url]

και κπ προβληματα καταλληλα για τον διαγωνισμο (πολυ καλα δεν ειναι ολα κ τοσο δυσκολα αλλα ειναι μεσα στο πνευμα του διαγωνισμου) [url]http://www.math.northwestern.edu/~mlerma/problem_solving/putnam/training_problems-2008.pdf[/url]
το οποιο τον βρηκα σε αυτη την χρησιμη σελιδα [url]
http://www.math.northwestern.edu/~mlerma/problem_solving/putnam/training.html[/url]
[url]http://www.math.northwestern.edu/~mlerma/problem_solving/[/url]

Σημερα οσοι ηταν στο μαθημα στην ΕΜΕ πηραν ενα αλλο βιβλιο του κ. Παπαιωαννου το οποιο περιεχει τα παντα γυρω απο τη Θεωρια Γραφηματων και ειναι επισεις παρα πολυ καλο. Το βιβλιο των διακριτων ομως που αναφερεται γενικα στη Συνδιαστικη υπαρχει στη βιβλιοθηκη του ΕΜΠ (εγω απο εκει το πηρα και το εβγαλα φωτοτυπιες), αρκετοι αλλοι φοιτητες του ΕΜΠ ομως πηγαν στο γραφειο του και του το ζητησαν και τους το εδωσε. Εξω απο το ΕΜΠ δεν γνωριζω αν μπορεις να το βρεις καπου.

Για την Αλγεβρα τωρα, ρωτησα τον κ.Φελλουρη πριν κατι μερες και μου ειπε οτι Δακτυλιοι και Μοναδικες Παραγοντοποιησεις ειναι στην υλη οπως επισεις και καποιες εφαρμογες τους στη Θ. Αριθμων... απο κει και περα δεν γνωριζω

ΝICK NAFPLIO σε ποιό μάθημα γραφτήκατε στην ΕΜΕ? κάνετε ήδη προετοιμασία? γιατί εγώ να είμαι τόσο άτυχη? γιατί δεν μπορώ να βρώ έναν άνθρωπο να με βοηθήσει? ειμαι πολύ απογοητευμένη.

laracroft μπες εδω και κατεβασε το επισυναπτομενο αρχειο:
http://www.hms.gr/modules/news/article.php?storyid=365

Τα μαθηματα προετημασιας για τον Seemous γινονται καθε Σαββατο και Κυριακη 4-7μμ στο κτηριο της ΕΜΕ και παει οποιος θελει καθε φορα χωρις εγγραφη, αν βρεθεις Αθηνα καποιο Σαββατοκυριακο μπορεις να ερθεις κι εσυ.

NickNafplio σε ευχαριστώ πολύ. Δυστυχώς είναι πολύ δύσκολο να έρθω Αθήνα. Αν μπορεί κάποιος που πηγαίνει να ανεβάζει τις σημειώσεις εδώ σε ηλεκτρονική μορφή αύτό θα ήταν τέλειο. αν πάλι όχι θα ήθελα απλά να ξέρω για ποια θέματα μιλάνε οι καθηγητές ώστε να ψάχνω και εγώ την αντίστοιχη θεωρία. ευχαριστώ πολύ

[quote="laracroftheoriginal"]Dimitris T σε ευχαριστώ θερμά για όλα τα links...προσπαθώ να μπω στη διεύθυνση http://eclass.uoa.gr/modules/document/document.php για να πάρω τις περσινές σημειώσεις της ΕΜΕ που ανέφερες αλλά θέλει κάποιο κωδικό τον οποίο δεν έχω αφού ανήκω στο ιόνιο πανεπιστήμιο. πως αλλιώς μπορώ να τις πάρω. μπορείς να τις βάλεις ως attachment? σε εκλιπαρώ...[/quote]
εκει κολλησες... :P  κανε εγγραφη μαθητη (με  ψευτικο αριθμο μητρωου.. εναν αριθμο οποιον θες εσυ δεν το ελεγχουν)
και θα μπορεις να δεις τισ σημειωσεις (δυστυχως δεν εχει σε ολα τα μαθηματα) καθως κ να μπαινεις στις σελιδες των μαθηματων ολων των τμηματων του εκπα :wink:

[quote="Papadopoulos"][quote="Dimitris X"]Επισης θα βρεις πολλα θεματα και θα καταλαβεις πανω κατω το πνευμα των διαγωνισμων αφου απο οτι ξερω ο SEEMOUS [color=red]εχει πανω κατω την ιδια υλη με τις IMO[/color].(Εαν εξαιρεσεις την Ευκλειδια γεωμετρια)[/quote]

Όχι, όχι. Η ύλη του SEEMOUS είναι αρκετά διαφορετική από τις IMO.
Π.χ. η Ανάλυση (και όχι μόνο όρια, παράγωγοι, ολοκληρώματα αλλά και εψιλοντικοί ορισμοί, σειρές, σειρές συναρτήσεων και λοιπά) είναι στον SEEMOUS αλλό όχι στην ΙΜΟ. Επίσης η Γραμμική Άλγεβρα είναι στο SEEMOUS και όχι στην ΙΜΟ, και [color=red]οι διαφορές των δύο διαγωνισμών δεν σταματούν μόνο εδώ[/color].[/quote]

Νομιζω οτι εχω δει θεμα γραμμικης Αλγεβρας στο βιβλιο IMO Compedium, οχι προβλημα IMO αλλα προβλημα σε IMO Shortlist, αλλα δεν ειχε να κανει με διανυσματικους χωρους και γραμμικες απεικονισεις, νομιζω ηταν ενα γραμμικο συστημα με ν αγνωστους και με κατι παραμετρους...

Παντος νομιζω οτι οι διαφορες αναμεσα στους 2 διαγωνισμους ειναι μονο αυτα που ανεφερες + το οτι στον SEEMOUS δεν μπαινουν θεματα Ευκλειδιας/Γραμμικης Γεωμετριας.

[quote="NickNafplio"]

Παντος νομιζω οτι οι διαφορες αναμεσα στους 2 διαγωνισμους ειναι μονο αυτα που ανεφερες + το οτι στον SEEMOUS δεν μπαινουν θεματα Ευκλειδιας/Γραμμικης Γεωμετριας.[/quote]

Όχι μόνο. Υπάρχουν και άλλες διαφορές. Π.χ. στον διαγωνισμό SEEMOUS 2007 υπήρχε ένα θέμα με πολυώνυμα πάνω από το σύνολο [b]Q[/b] των ρητών που η λύση του απαιτούσε να ξέρει κανείς τι θα πει αριθμήσιμο σύνολο και ότι το [b]Q[/b] είναι αριθμήσιμο ενώ το [b]R[/b] δεν είναι. Γενικά, η Θεωρία Cantor είναι εντός ύλης SEEMOUS.
Επίσης η αφηρημένη Άλγεβρα (ομάδες, δακτύλιοι, σώματα, διανυσματικού χώροι πάνω από σώματα, σχέσεις και κλάσεις ισοδυναμίας, χώροι πηλίκα και παρεμφερή) είναι εντός ύλης SEEMOUS.

Οι δύο διαγωνισμοί έχουν βέβαια μεγάλη επικάλυψη ύλης, αλλά ας προσθέσω ότι η έμφαση στα διάφορα κεφάλαια που είναι κοινά στους δύο διαγωνισμούς, είναι συχνά διαφορετική.
Προσοχή, δεν θέλω με αυτά να πω ότι ο SEEMOUS είναι πιο δύσκολος από την ΙΜΟ. Ίσως να συμβαίνει και το αντίθετο. Πάντως αυτό που είναι βέβαιο είναι ότι αν πάει κανείς στον SEEMOUS "μόνο" με τις γνώσεις ΙΜΟ, τότε σε κάποια θέματα
που θα πέσουν, μάλλον δεν έχει ελπίδα να τα λύσει ακόμη και αν είναι σχετικά "εύκολα".

Τάκης

demetres ξανακοίταζα λίγο τα προβλήματα του διαγωνισμού και ειδικά αυτό με τα πολύγωνα. είπες οτι στη λύση χρησιμοποιείται και αναλυτική γεωμετρία και ανάλυση. σε κανένα όμως βιβλίο δεν μπορώ να βρώ την ανάλυση που χρησιμοποιείται στο πρόβλημα. μήπως μπορείς να μου πεις τουλάχιστον ποια θεωρήματα χρησιμοποιούνται. επίσης μπορείς να μου πεις αν η αναλυτική γεωμετρία που χρησιμοποιείται είναι απλή ή πολύ δύσκολη? δηλαδή την περιέχει ένα εισαγωγικό βιβλίο αναλυτικής γεωμετρίας?

[quote="laracroftheoriginal"]demetres ξανακοίταζα λίγο τα προβλήματα του διαγωνισμού και ειδικά αυτό με τα πολύγωνα. είπες οτι στη λύση χρησιμοποιείται και αναλυτική γεωμετρία και ανάλυση. σε κανένα όμως βιβλίο [b]δεν μπορώ να βρώ την ανάλυση που χρησιμοποιείται στο πρόβλημα.[/b] μήπως μπορείς να μου πεις τουλάχιστον ποια θεωρήματα χρησιμοποιούνται. επίσης μπορείς να μου πεις αν η αναλυτική γεωμετρία που χρησιμοποιείται είναι απλή ή πολύ δύσκολη? δηλαδή την περιέχει ένα εισαγωγικό βιβλίο αναλυτικής γεωμετρίας?[/quote]

Η Ανάλυση χρησιμοποιείται στο σημείο όπου ζητά να αποδειχθεί ότι οι ακτίνες τείνουν στο 0.
Αυτό το σημείο είναι μεν απλό, αλλά δεν παύει να είναι Ανάλυση.
Αποδεικνύεται χρησιμοποιώντας το γεγονός ότι η κάθε ακτίνα του περιγεγραμμένου κύκλου είναι το μισό της προηγούμενης. Έτσι προκύπτει η ακολουθία $ \frac{1}{2^n}$, η οποία τείνει στο 0.

Η αλήθεια είναι ότι η ερώτηση γιατί η $ \frac{1}{2^n}$ , και όμοια η $ \frac{1}{n}$, τείνει στο 0, έχει διαισθητικά φανερή απάντηση. Όμως οι Μαθηματικοί δεν αφήνουν τίποτα στη διαίσθηση: Η σωστή και πλήρης απόδειξη ότι οι $ \frac{1}{2^n}$  και $ \frac{1}{n}$ τείνουν στο μηδέν ΧΡΕΙΑΖΕΤΑΙ την Αρχιμήδεια ιδιότητα του R (γνωστότερη ως Αξίωμα Αρχιμήδη ή Αξίωμα Αρχιμήδη-Ευδόξου).
Το θεώρημα αυτό το συναντούν οι Μαθηματικοί, ενώ σε Μηχανικούς [b]συνήθως[/b] δεν αναφέρεται (ως έξω από τους μαθηματικούς στόχους τους).

Τάκης

[quote="laracroftheoriginal"]Τάκη σε ευχαριστώ πολύ για την απάντηση σου. αν πας στο http://www.seemous.eu έχει τα θέματα και τις λύσεις και μου φαίνονται πολύ περίπλοκα. εγώ σπουδάζω πληροφορική και δυστυχώς δεν κάνουμε πολλά μαθηματικά. το θέμα είναι οτι σκεφτόμουν να πάω στο διαγωνισμό το μάρτιο του 2010. βέβαια αυτό που με απασχολεί είναι αν σε ένα χρόνο που μου απομένει θα έχω φτάσει σε αυτό το επίπεδο (δηλαδή να λύνω τέτοια προβλήματα). όταν κοιτάω τα παλιά θέματα και τις αποδείξεις τους μου φαίνονται πραγματικά ακαταλαβίστικα. πιστεύεις οτι αξίζει τον κόπο να κάνω μια προσπάθεια ή είναι χάσιμο χρόνου και καλύτερα να ασχοληθώ μόνο με την πληροφορική; είμαι πολύ μπερδεμένη και δεν μπορώ να βρω κανέναν να με βοηθήσει.[/quote]

Ναι, θα σου συνιστούσα ανεπιφύλακτα να πας στον SEEMOUS ή οποιονδήποτε άλλο διαγωνισμό. H εμπειρία είναι μοναδική, θησαυρός θα έλεγα, και το ταξίδι στο εξωτερικό χρήσιμο.

Στον διαγωνισμό θα συναντήσεις άλλους της ηλικίας σου, έλληνες και ξένους, με κοινά ενδιαφέροντα και υψηλές ικανότητες.

Είναι ΑΠΟΛΥΤΑ φυσιολογικό τα θέματα να σου φαίνονται δύσκολα. ΕΙΝΑΙ δύσκολα.

Αν δεις τον διαγωνισμό Putnam στην Αμερική (που τρέχει πολλά χρόνια και υπάρχουν έγκυρα στατιστικά στοιχεία) θα δεις ότι κάποιοι από αυτούς που πρώτευσαν, αργότερα στη ζωή πήραν βραβείο Νόμπελ ή Fields Medal.
Αυτούς τους γίγαντες δεν θα τους φτάσουμε! Αλλά μας δίνεται η ευκαιρία να
δώσουμε τον καλύτερό μας εαυτό, να μάθουμε πολλά πράγματα, να αντιληφθούμε τι θα πει ευγενής άμιλλα και λοιπά.
Ας προσθέσω ότι μη περιμένεις να λύσεις όλα τα θέματα. Στον Putnam, από τους 3500 διαγωνιζόμενους μόλις 100 λύνουν πάνω από 1/3 των θεμάτων!
Αν όμως πάρεις μετάλλιο (αν δεν κάνω λάθος στον SEEMOUS, οι μισοί παίρνουν κάποιο μετάλλιο) τότε θα σε ακολουθεί όλη σου την ζωή.

Το Μάρτιο του 2010 θα είσαι δευτεροετής φοιτητής; Ο SEEMOUS είναι μόνο μέχρι και δευτεροετείς. Αν είσαι τώρα δευτεροετής, ΤΡΕΧΑ ΦΕΤΟΣ.

Τάκης Π.

του χρόνου θα είμαι δευτεροετής. δεν θέλω να πάω εκεί και οι άλλοι να τα λύνουν και εγώ να κοιτάω τους τοίχους. και εκτός αυτού οι άλλοι στο εξωτερικό έχουν καθηγητές που τους βοηθούν ενώ εγώ όχι. αυτό είναι άδικο. θέλω να πάω καλά προετοιμασμένη. ξέρεις περίπου τι να διαβάσω? γιατί ενώ διαβάζω θεωρία αριθμών, ανώτερη άλγεβρα, άνάλυση και γραμμική άλγεβρα πάλι κινέζικα μου φαίνονται τα θέματα.

Εμεις παντος αν και ειμαστε πολυτεχνικη σχολη, εχουμε ασχοληθει αρκετα και με την Αρχιμηδεια Ιδιοτητα των Πραγματικων Αριθμων και ακομα περισσοτερο με τους εψιλοντικους ορισμους. Το γιατι οι παραπανω ακολουθιες συγκλινουν στο 0 προκειπτει αμεσα απο τον ορισμο του οριου. Μπορουμε να χρησημοποιησουμε και με την Αρχιμηδεια ιδιοτητα στην αποδειξη, αλλα δεν ειναι απαραιτητο

Lara, επηδη ο διαγωνισμος ειναι πολυ κοντα και δυσκολα θα τα προλαβεις ολα, απο Γραμμικη Αλγεβρα θα σου πρωτεινα να διαβασεις περισσοτερο Πινακες, Διανυσματικους χωρους πανω σε Σωματα, και γραμμικες απεικονισεις, απο Αναλυση οποσδηποτε παρα πολυ καλα εψιλοντικους ορισμους οριων και ολοκληρωματα, λιγα πραγματα απο Συνδιαστηκη, και οτι προλαβεις απο ομαδες/δακτυλιους απο τον Fraileigh. Η Θεωρια Αριθμων πεφτει παντα σε σχολικες Ολυμπιαδες, και στον Putnam/IMC, γι αυτο πιστευω οτι θα μπορουσε να μπει θεμα Θεωριας Αριθμων και στον Seemous, επηδη ομως 2 χρονια τωρα δεν εχει μπει, θα σου πρωτεινα να την αφησεις για το τελος...

[quote="laracroftheoriginal"]Nick εσείς έχετε βοήθεια από καθηγητές εκεί; σε ρωτάω γιατί έχει μεγάλη σημασία να έχεις μια υποστήριξη. εγώ δυστυχώς εδώ δεν έχω καμία υποστήριξη. πάντως πιστέυω ότι για να διακριθεί κανείς σε αυτό το διαγωνισμό πρέπει να ασχολείται όλη μέρα με αυτό, πράγμα που εγώ είναι αδύνατον να κάνω γιατί και η πληροφορική θέλει πολύ διάβασμα. δε νιώθω πάντως οτι μπορώ να προετοιμαστώ σωστά και για αυτό σκέφτομαι να μην πάω. θέλω ο,τι κάνω να το κάνω σωστά. το να πάει κάποιος σε αυτό το διαγωνισμό χωρίς ένα δάσκαλο είναι σαν να πηγαίνει ένας αθλητής σε αγώνες χωρίς προπονητή. γίνεται κάτι τέτοιο; σαφώς και όχι. τέλος πάντων κάτι τελευταίο από γεωμετρία έχεις καμία ιδέα τι να διαβάσω?[/quote]

To να ασχοληθεισ με τον διαγωνισμο μονο κερδισμενη μπορεις να βγεις (εχω ακουσει οτι κ η θεωρητικη πληροφορικη θελει βαρβατα μαθηματικα).
Τωρα για τον διαγωνισμο το βιβλίο the art and craft of problem solving [url]http://www.amazon.com/Art-Craft-Problem-Solving/dp/0471135712[/url] ειναι φοβερο και μαλλιστα απευθεινεται σε αυτους που δεν εχουν ασχοληθει ξανα με προβληματα ολυμπιαδων.Δεν χρειαζεται να ξερεις σχεδον καθολου θεωρια για να διαβασεις αυτό το βιβλίο.Σε καλυπτει ακομα κ σε θεωρητικο υποβαθρο για τον διαγωνισμο (εκτοσ απο γραμμικη κ ανωτερη αλγεβρα)...να γανταστεισ μεσα σε 3 chapters σου μαθενει καλη γεωμετρία ξεκινώντας απο το [u]μηδεν [/u]
Το βιβλίο ειναι ακριβο γι αυτο προσπαθησε να το βρεις σε καμια βιβλιοθηκη κ να το φοτωτυπησεις...

για γραμμικη και ανωτερη αλγεβρα πιστευω θα εισαι καλυμενη (για τισ αναγκες του διαγωνισμου πιστευω) αν διαβασεις
απο εδω...[url]http://users.uoa.gr/~pkrikel/[/url] (μην διαβασεις με την σειρα τα κεφαλαια)

[quote="NickNafplio"]Εμεις παντος αν και ειμαστε πολυτεχνικη σχολη, εχουμε ασχοληθει αρκετα και με την Αρχιμηδεια Ιδιοτητα των Πραγματικων Αριθμων και ακομα περισσοτερο με τους εψιλοντικους ορισμους. Το γιατι οι παραπανω ακολουθιες συγκλινουν στο 0 προκειπτει αμεσα απο τον ορισμο του οριου. Μπορουμε να χρησημοποιησουμε και με την Αρχιμηδεια ιδιοτητα στην αποδειξη, αλλα δεν ειναι απαραιτητο

[/quote]

Το ότι μαθαίνεις εψιλοντικούς ορισμούς και Αρχιμήδεια Ιδιότητα στο ΕΜΠ, σε καθιστά έναν από τους τυχερούς Μηχανικούς, σε παγκόσμια κλίμακα. Εύγε στους δασκάλους σου. Χαίρομαι ιδιαίτερα γιατί τα καλά Μαθηματικά χρειάζονται σωστή μελέτη των δομών.

Όμως δεν έχεις δικιο όταν γράφεις

[i]Μπορουμε να χρησημοποιησουμε και με την Αρχιμηδεια ιδιοτητα στην αποδειξη, αλλα δεν ειναι απαραιτητο[/i]

Η απόδειξη ότι η 1/n τείνει στο 0, από τον ορισμό, είναι απλή υπόθεση. Δεν αντιλέγω. Παρατήρησε όμως ότι είναι ΙΣΟΔΥΝΑΜΗ της Αρχιμήδειας.

Στην μία κατεύθυνση, το σημείο στο οποίο γίνεται χρήση της Αρχιμήδειας όταν δείχνουμε ότι η 1/n τείνει στο μηδέν, είναι στο βήμα που λέει ότι:

 [i]για δοθέν ε>0 επιλέγουμε [/i] $ n_0$ [i]φυσικό αριθμό > 1/ε. [/i]

Η ύπαρξη τέτοιου φυσικού είναι διαισθητικά φανερή ΑΛΛΑ ΓΙΑ ΤΑ ΥΨΗΛΑ ΜΑΘΗΜΑΤΙΚΑ ΘΕΛΕΙ ΑΠΟΔΕΙΞΗ. Και αυτή ακριβώς είναι η Αρχιμήδεια ιδιότητα.

Θυμίζω, η απόδειξή της Αρχιμήδειας πάει μέσω αξιώματος πληρότητας και λέει: αν δεν υπήρχε τέτοιος $ n_0$ τότε το σύνολο [b]Ν[/b] των φυσικών αριθμών θα ήταν άνω φραγμένο υποσύνολο του [b]R[/b]. Άρα θα είχε supremum. Αν s = sup[b]N[/b] τότε .... και λοιπά, με άτοπο. (Η απόδειξη οφείλεται στον Γερμανό Μαθηματικό Stolz, πάνω από 100 χρόνια πριν).

Συνήθως στις Πολυτεχνικές Σχολές δεν γίνεται αυτή η θεωρία. Αν την διδάχθηκες, τότε, ας το επαναλάβω, είσαι από τους τυχερούς.

Σε μερικές Πολυτεχνικές Σχολές η παραπάνω ιδιότητα του [b]Ν[/b], ότι δηλαδή το [b]Ν [/b] είναι μη φραγμένο, ούτε καν δηλώνεται. Λαμβάνεται ως δεδομένο.

Παραδείγματος χάριν ένα κρυφό σημείο όπου γίνεται χρήση του ΧΩΡΙΣ ΝΑ ΔΗΛΩΝΕΤΑΙ είναι σε εκφράσεις της μορφής " έστω [α] το ακέραιο μέρος του αριθμού α" .

Όσο και αν σου φαίνεται περίεργο, πίσω από την ύπαρξη τέτοιου ακεραίου βρίσκεται η Αρχιμήδεια. Είναι στο "αθώο" βήμα "έστω ν φυσικός μικρότερος ή ίσος του α, με ν+1 >α "
Το ερώτημα είναι γιατί υπάρχει τέτοιος ν. Με άλλα λόγια, ΓΙΑΤΙ ΥΠΑΡΧΕΙ ν
με ν > α-1; Η απάντηση είναι: Αρχιμήδεια ιδιότητα.

Ελπίζω να μην σε κούρασα. Αν δεν τα γνωρίζεις αυτά, θα ήταν χρήσιμο να ρωτούσες τους Καθηγητές σου στο ΕΜΠ, ιδίως αυτούς που η ειδικότητά τους είναι Ανάλυση. Τέτοια θέματα γι ' αυτούς είναι ρουτίνα.

Τάκης.

Βασικα κ. Παπαδοπουλε, στο μυαλο μου ειχα μια αποδειξη οπου για καθε ε>0 θεωρουμε το ακεραιο μερος του 1/ε αυξημενο κατα 1, αυτος ειναι ενας θετικος ακεραιος μεγαλυτερος του 1/ε, αλλα μαλον η υπαρξη του αριθμου αυτου ειναι συνεπεια της Αρχιμηδειας Ιδιοτητας

Lara (Νικολετα, σωστα?) οταν εχεις μια βοηθεια εχεις περισσοτερες πιθανοτητες να πετυχεις, σε αλλη περιπτωση πρεπει να ασχολησε πολυ καιρο με το αντικειμενο για να μπορεσεις να αντιμετωπισεις με επιτυχια τα δυσκολα θεματα που μπαινουν σε τετοιους διαγωνισμους. Θεωρω οτι αν πεσει Γεωμετρια τοτε θα πεσει καποιο θεμα διανυσματικου λογισμου, δεν νομιζω να μπει τιποτα αλλο γιατι ολα τα υπολοιπα ειναι πολυ θεωρητικα πραγματα, καλο θα ηταν ομως να ξερεις καλα καποια βασικα πραγματα απο Ευκλειδια Γεωμετρια (η να θυμασε καλα αυτα του Λυκειου) οχι επηδη θα πεσει θεμα με τριγωνα/κυκλους κλπ αλλα επηδη πολλες φορες μπαινουν ασκησεις που συνδιαζουν Αναλυση η Συνδιαστικη με λιγο Γεωμετρια (βλεπε Seemous 2007 4ο θεμα, Seemous 2008 2ο θεμα, Putnam 2008 1ο θεμα 2ης μερας)

Nick, εγώ θα πάω (αν τελικά το αποφασίσω), το 2010. έχω ένα χρόνο μπροστά μου. πιστεύεις οτι μου φτάνει για να καλύψω την ύλη? αλλά σκέφτομαι οτι πρέπει να λύσω και πολλές ασκήσεις. εσύ δεν μου απάντησες, στο δικό σου πανεπιστήμιο έχετε βοήθεια όσον αφορά τους διαγωνισμούς; και αυτοί οι εψιλοντικοί ορισμοί τι είναι; πρώτη φορά τους ακούω δεν μας τους έχει αναφέρει κανείς ακόμα. πιστέυω ότι έχω ταλέντο στα μαθηματικά αλλά απλά δεν έχω συνηθίσει ακόμα αυτά του πανεπιστημιακού επιπέδου. το χάσμα ανάμεσα στα μαθηματικά του λυκείου και τα πανεπιστημιακά είναι τεράστιο και για αυτό το λόγο χρειάζονται και καθηγητές με όρεξη. πάντα θα μελετώ μαθηματικά πάντως και ελπίζω κάποια στιγμή στο μέλλον να μπορώ να λύνω με ευκολία τα θέματα αυτών των διαγωνισμών και ακόμη δυσκολότερα...αυτό θα είναι για μένα σαν να έχω κερδίσει ένα μετάλλιο και ούτως ή άλλως αυτό είναι που μετράει και όχι το μετάλλιο.

Αν εισουν κι εσυ σε σχολη μαθηματικων τοτε θα μπορουσα να σου πω με σιγουρια οτι προλαβαινεις ανετα να προετοιμαστεις για τον Seemous 2010, αλλα μιας και εισαι στην Πληροφορικη δεν γνωριζω... Οι καθηγητες μας βοηθανε σε οτι αφορα την υλη του ετους που βρησκομαστε, η υλη του διαγωνισμου περιεχει και αλλα πολλα πραγματα τα οποια εγω τουλαχιστον τα διαβαζω μονος μου, υπαρχουν ομως και τα μαθηματα προετοιμασιας που γινονται στην ΕΜΕ απο τα οποια μαθαινουμε αρκετα πραγματα και μας λυνονται αρκετες αποριες.

αποφάσισα τελικά να διαβάσω λίγο πιο χαλαρά για την ολυμπιάδα γιατί έχω καταφέρει ως τώρα να παραμερίσω το πανεπιστήμιο θεωρώντας το πολύ πιο εύκολο. το καλοκαίρι πάλι θα δω σε τι επίπεδο θα έχω φτάσει και ίσως τότε προετοιμαστώ πιο σκληρά. αλλά και πάλι δύσκολο το κόβω χωρίς καμία βοήθεια. στο εξωτερικό πάντως έχουν ειδικές ώρες που κάνουν προετοιμασία για τέτοιους διαγωνισμούς.

Νικολέτα

[quote="NickNafplio"] Θεωρω οτι αν πεσει Γεωμετρια τοτε θα πεσει καποιο θεμα διανυσματικου λογισμου, δεν νομιζω να μπει τιποτα αλλο γιατι ολα τα υπολοιπα ειναι πολυ θεωρητικα πραγματα, καλο θα ηταν ομως να ξερεις καλα καποια βασικα πραγματα απο Ευκλειδια Γεωμετρια (η να θυμασε καλα αυτα του Λυκειου) οχι επηδη θα πεσει θεμα με τριγωνα/κυκλους κλπ αλλα επηδη πολλες φορες μπαινουν ασκησεις που συνδιαζουν Αναλυση η Συνδιαστικη με λιγο Γεωμετρια (βλεπε Seemous 2007 4ο θεμα, Seemous 2008 2ο θεμα, Putnam 2008 1ο θεμα 2ης μερας)[/quote]

Ακριβώς. Να τονιστεί ότι η [color=blue][b]κλασσική Ευκλείδια Γεωμετρία είναι εκτός ύλης SEEMOUS[/b][/color]. Το μόνο που μπορεί να πέσει είναι ένα θέμα που ίσως, σε κάποιο κομμάτι του, να χρειάζεται Ευκλείδια. Αλλά Ευκλείδια όπως στις Βαλκανιάδες ή στην ΙΜΟ [b][color=blue]δεν θα πέσει[/color][/b].

Τ.Π.


do we take the geometry class daily or weekly?

That's listed on the Online Courses page.

[quote="Magic-girl189"]On which days of the week do I take the class[/quote]

The Introduction to Geometry class that just started is on Fridays.


hmm... I dunno

I'm pretty sure this topic is not intermediate but I am unsure where to send it.  Any ideas?

humm... A french can me translate "bijection" in english?  :blush:

The word 'bijection' is correct iff you want an injective ( $f(x)=f(y) \iff x=y$ ) and surjective ( $ \forall x \exists y : f(y)=x$ ) mapping.
But I don't know what you want...

[quote="bénabar"]Either a bijection of $]a, b[$ to $\mathbb{R}$. What imply the existence of this bijection?[/quote]

Use the $\tan$ function.

A bijection between two sets is a function between them that is both one-to-one (injective) and onto (surjective). The notation ]a,b[ means the set of real numbers x such that a < x < b.

I was badly made understand :( . I know what is a bijection :). My question was the properties which imply this bijection on ]a, b[. For exemble, equipotence.

For the record $]a,b[ = (a,b) = \{ x \in \mathbb{R} \mid a<x<b \}$ :) I remember I was also puzzled by this (French) notation at first ...
Post edit: Oops didn't notice Gauss202's post upthere :)


function
topology
real analysis
real analysis unsolved

1.
Let $ G\subset\mathbb R^n$ be an open set and $ f: G\to\mathbb R$ be a real function. Show that the set $ \{x\in G : f$ is not continuous at $ x\}$ is a $ F_\sigma$ set.

2.
Suppose that $ f$ is a continuous function on $ \mathbb R$. Show that the set $ \{x : f$ is differentiable at $ x\}$ is a $ F_{\sigma\delta}$ set.

How is question 2 related to question 1?
Or can I do something on question 2 similar to question 1?
Thank you.


Let $ ABC$ be a triangle, and $ I$ its incenter. Let the incircle of $ ABC$ touch side $ BC$ at $ D$, and let lines $ BI$ and $ CI$ meet the circle with diameter $ AI$ at points $ P$ and $ Q$, respectively. Given $ BI \equal{} 6, CI \equal{} 5, DI \equal{} 3$, determine the value of $ \left( DP / DQ \right)^2$.

[asy]size(250);defaultpen(fontsize(8));
pair A=(0,0), B=(0,0), C=(3*sqrt(3)+4,0), D=(3*sqrt(3),0), E, F, P, Q, I=(3*sqrt(3),3);
E=4*expi(pi+2*angle(I-C))+C; F=3*sqrt(3)*expi(2*angle(I));
A=extension(F,B,C,E);
Q=intersectionpoint(0.1*(I-C)+I--20*(I-C)+I,circumcircle(A,E,I));
P=intersectionpoint(0.1*(I-B)+I--20*(I-B)+I,circumcircle(A,E,I));
draw(I--A--B--C--A);draw(B--P--F--I--E--Q--C);draw(Q--D--P..I--D);
draw(incircle(A,B,C));
draw(circumcircle(A,I,E));
dot(A^^B^^C^^D^^E^^F^^P^^Q^^I);
label("A",A,(0,1));label("B",B,(-1,-1));label("C",C,(1,-1));label("I",I,(-1,1));label("D",D,(0,-1));label("E",E,(1,0));
label("F",F,(-1,0));label("P",P,(1,0));label("Q",Q,(-1,0));[/asy]
$ DP\equal{}FP$ because $ BP$ is the angle bisector of $ \angle FBD$ and $ FB\equal{}BD$. Similarly $ DQ\equal{}EQ$.
$ \triangle BIA \sim \triangle BFP$ so $ \frac{BI}{IA}\equal{}\frac{BF}{FP}$.
$ \triangle CIA \sim \triangle CEQ$ so $ \frac{CI}{IA}\equal{}\frac{CE}{EQ}$.
$ \left(\frac{DP}{DQ}\right)^2\equal{}\left(\frac{FP}{EQ}\right)^2\equal{}\left(\frac{BF\cdot CI}{BI\cdot CE}\right)^2\equal{}\left(\frac{3\sqrt{3}\cdot 5}{6\cdot 4}\right)^2\equal{}\frac{75}{64}$

hmm... I found a different solution but it's pretty long so here's a sketch:

1. $Q, F, D$ and $P, E, D$ are collinear

2. Connect $AP$ and $AQ$ to get right triangles $\bigtriangleup API$ and $\bigtriangleup AQI$

3. Find $AB, BC$ and $CA$ by letting $AF = AE = s$ and calculating the area of ABC in two different ways: $[ABC] =
3(3\sqrt3 + s+4) = \frac{1}{2}(s+4)(3\sqrt3+4)(\sin(2\angle ICD)) = \frac{1}{2}(s+4)(3\sqrt3+4)(\frac{12}{25})$ (the A =
rs formula and the 1/2 ab sin C formula). From here, bash out $s$

4. Finally, length chase $IQ$ and $IP$ by repeatedly using the Pythagorean theorem.

5. Finally, use the Law of Cosines with the angle $\angle QID$ to get an answer of $\frac{75}{64}$ or $139$

[quote = HMMT Geometry 2008]
Let $ABC$ be a triangle, and $I$ its incenter. Let the incircle of $ABC$ touch side $BC$ at $D$, and let lines $BI$ and $CI$ meet the circle with diameter $AI$ at points $P$ and $Q$ respectively. Given $BI = 6, CI = 5, DI = 3$, determine he value of $\left( \frac{DP}{DQ} \right)^2.$[/quote]

Let $Y$ be the foot of the perpendicular from $I$ to $AB$ and let $X$ be the foot of the perpendicular from $I$ to $AC.$
[b][color=#f00]Claim: [/color][/b]$P,X,$ and $D$ are collinear.
Notice that $AQYIXP$ is circumscribed in a circle with diameter $AI.$ We see that:
\begin{align*}
    \angle XAP &= \angle PIX \\ &= 180 - \angle XID - 60 \\ &= 180 - (180 - 2 \angle ICD) - 60 \\ &= 2 \angle ICD - 60.
\end{align*}
\begin{align*}
    \angle IAX &= 180 - 2 \angle YBI - 2 \angle ICD \\ &= 120 - \angle ICD.
\end{align*}
Therefore,
\begin{align*}
    \angle IXP + \angle IXD &= 180 - \angle IAX - \angle XAP \\ &= 180 - (120 - \angle ICD) - (2 \angle ICD - 60) + \angle ICD \\ &= 180,
\end{align*}
so, as desired, $P,X,$ and $D$ are collinear. $\blacksquare$

[b]Note: [/b]By the same reasoning, we know that $Q,Y,$ and $D$ are collinear.

By power of a point on the cyclic $QYXP$, we see that:
$$DQ \cdot DY = DX \cdot DP \Longleftrightarrow \frac{DP}{DQ} = \frac{DY}{DX}.$$
Since $IC$ bisects $DX$, we know that:
$$DX = 2 DC \sin (\angle DCI ) = \frac{24}{5}.$$
Since $\triangle BYD$ is equilateral
$$DY = 3 \sqrt{3}.$$
It follows that:
$$\left(\frac{DP}{DQ} \right)^2 = \left( \frac{DY}{DX} \right)^2 = \frac{75}{64},$$
giving us an answer of $\boxed{\frac{75}{64}}.$


20 messages is a small sentbox. Maybe enlarge it a bit? The increased server strain will be virtually unnoticeable.

That's weird...I have 101 messages in my sentbox right now...

I have 160.  I believe it is a function on how many posts a user has?

Well, it says 100% full but there are more and more... (102 in inbox, 26 in sentbox both 100% full)


If 2 cats are as cool as 5 alpacas and 3 pandas are as cool as 7 cats, how many pandas are as cool as 70 alpacas?

1)Write out what you know

2c=5a
3p=7c

2)Try putting all of them into one equation

6p=14c=35a

3)double it
12p=28c=70a

[b]You can also substitute it:
2c=5a
3p=7c
You can divided the first equation by two to get:
c=5a/2
Then you can substitute (c=5a/2) into the second equation:
3p=7(5a/2)
Then you simplify:
3p=35a/2
Then you multiply by 2 and get:
6p=35a
Then you multiply by two one more time to get:
[i]70a=12p[/i] [/b]


If a polygon of n sides has (1/2)n(n - 3) diagonals, how many sides will a polygon with 65 diagonals have?

Also, is there a polygon with 80 diagonals?

[hide]Since a polygon has $\frac{n(n-3)}{2}$ diagonals, we simply set this equal to 65 and solve for n using the quadratic formula...

You will get the equation $n(n-3) = 130$, which you can solve using the quadratic formula or by factoring $13(13-3) = 130$ therefore, $n = 13$ sides. (We ignored the negative solution because a polygon cannot have a negative number of sides)

A polygon cannot have 80 sides as when using this same method, you will see that n turns out to be a decimal, which is not allowed when taking the number of sides in a polygon!

- NAWAHD WERDNA[/hide]

I am learning so much on this site.

Thanks.

Just wondering if anyone wanted a proof.

[hide="Proof"]Given a n-sided polygon (convex, not sure if concave had work). The number of lines that can be formed by choosing two random points are $\binom{n}{2}$ since we wish to discard the sides we have the number of diagonals equal to $\binom{n}{2}-n$ which simplifies to to our desired equation.[/hide]

BanishedTraitor,

I thank you for providing the prove to the given question.


calculus
integration
algebra solved
algebra

Solve the equation:
$\sqrt{3x^2-2} + \sqrt[3]{x^2-1}= 3x-2$

[quote="t0rajir0u"][hide="Solution"]
Guess that every term involved is integral (it's often a good strategy when dealing with ugly equations  :P).  We have $3x^2 - 2$ is a square.  Checking several values gives $x = 1, 3$ as our first few.  Plugging in and checking:

$1 + 0 = 1$, so $x = 1$ is our first solution.
$5 + 2 = 7$, so $x = 3$ is our second solution.

Examining the equation's limits to infinity (or simply plugging a few values of $x > 3, x < 1$ will make it clear that there are no more solutions.
[/hide][/quote]

   I don't think that can be called a solution my dear....How funny u are... :rotfl: next  time plz think clearly before post...ok..


Under what conditions

[list]$ \sqrt{a\plus{}\sqrt{a^2\minus{}b}}\plus{}\sqrt{a\minus{}\sqrt{a^2\minus{}b}}\equal{}\sqrt{2a\plus{}2\sqrt{b}}$?[/list]

Squaring both sides, we have

$ a \plus{} \sqrt {a^2 \minus{} b} \plus{} a \minus{} \sqrt {a^2 \minus{} b} \plus{} 2(\sqrt {a^2 \minus{} (a^2 \minus{} b)}) \equal{} 2a \plus{} 2\sqrt b \equal{} 2a \plus{} 2\sqrt b$, so I think this is true for all $ a,b\in\mathbb C$?

I'm not sure though.

I think that the intension of the problem is to include the restrictions generated by the fact that $ \sqrt{x}\ge 0$ and $ x\ge 0$. Things get tricky when you take square roots of complex numbers.

I know; that's what i originally thought, but I've tried several complex numbers, and they've all worked...I'm too lazy to actually plug in $ (a,b) \equal{} (x \plus{} yi,z \plus{} wi)$ or something and simplify.

It gets messy because when you plug in the values to the original equation, the square root function outputs a principal value. I'm not sure if there is a generally accepted definition, but I believe it is if $ z\equal{}r\cdot e^{i\theta}$ where $ r\ge 0$ and $ \theta\in[0,2\pi)$, then the square root function outputs $ r^{\frac{1}{2}}\cdot e^{\frac{i\theta}{2}}$. Therefore, we automatically have that the imaginary part is non-negative by this convention.

Another thing to mention is that when you take the square roots, you are increasing the domain. For example, if we start with $ x\equal{}1$, then square it to get $ x^2\equal{}1$, so $ x\equal{}\pm 1$. But of course you got an additional solution. It is possible that something analogous happened when you squared it.

I know that it's not clear in statement, and you both are thinking about complex numbers. But let's think only about real numbers.

I'm using the formula

[list]$ \sqrt{x\pm \sqrt{y}}\equal{}\sqrt{\frac{x\plus{}z}{2}}\pm\sqrt{\frac{x\minus{}z}{2}},$[/list]
where $ z\equal{}\sqrt{x^2\minus{}y}.$


Let (x_n)_n be a sequence defined for any n \geq 1 x_n={nsqrt(2)}, where {x}=x-[x].
Prove that x_n contains at least one subsequence which is convergent to 0.

and a quick generalisation: if r \in R\Q then proove that the sequence (a_n)_n, a_n={nr} contains at least one subsequence convergent to 0.

Ohhh, this is a very hard problem and especially new. I wonder why does it have two authors, when Kronecker is its author! :D

i mean... i thought about the generalisation and i think... but i have a solution only if r=sqrt(k), k=/=p^2,p\in N.
let q=[r], the integer part of r
then (r-q)^n->0
but (r-q)^n=A_n.r-B_n, A_n,B_n \in Z.
now we have that {(r-q)^n}={A_n.r}->0... so the subsequence we need is a_{A_n}...
is it correct??

This proof is similar to Jordan's proof for the fact that if a isn't a power of 10 then there is an n s.t. an begins with any given sequence of digits (I just mentioned it so that no one thinks it's original :D):

Take the interval [0,1] and make a circle out of it (of length 1). We look at the numbers (mod 1) because we're only interested in their fractional parts. We start from the point denoting 0 and we place a segment of length r, with r irrational. We continue doing this. We obviously can never stop because then r would be rational (I mean to say that there are infinitely many points). Another observation tells us that there are 2 such points as close to each other as we want them to be (because if there were a minimum distance between 2 such points then there would be finitely many on the circle).

Let's assume 0<=a<b<=1. We want to show that there exists a point inside the segment (a,b) (I want to show that {{nr} | n\in N} is dense in (0,1)). There exist 2 points, An and An+k s.t. |AnAn+k| < b-a. Then it's obvious that |An+(t-1)kAn+tk|=|AnAn+k| for all natural t, so we sort of go along the circle in steps smaller than b-a, so we can't possibly "skip over" the interval (a,b). There must exist a t s.t. An+tk\in (a,b).

Hope it's clear..

This shows us that the sequence {nr}n\in N is dense in (0,1). There's a stronger result, called Weyl's thm, saying that that sequence is evenly distributed in (0,1), but I can't prove that.


In a group of $n\geq 4$ persons, every three who know each other have a common signal. Assume that these signals are not repeatad and that there are $m\geq 1$ signals in total. For any set of four persons in which there are three having a common signal, the fourth person has a common signal with at most one of them. Show that there three persons who have a common signal, such that the number of persons having no signal with anyone of them does not exceed $[n+3-\frac{18m}{n}]$.

Re-interpreting in terms of graphs, $2$ people know each other if and only if there is an edge between their vertices in the corresponding graph $G$ and three people have a common signal if and only if they form a $K_3$/triangle in $G$. For a triangle $ABC$ in $G$, let $x(ABC)$ denote the number of vertices which are not contained in a triangle with any of $A,B,C$. Then $n-3-x(ABC)$ vertices form a triangle with at least one of $A,B,C$ (aside from $A,B,C$ themselves).

Notice further that any fourth vertex $D$ can form a triangle with at most one of $A,B,C$ as otherwise the given condition is violated. Thus, among vertices contributing to $x(ABC)$, we can partition them into those adjacent to $A$, those adjacent to $B$, and those adjacent to $C$. Furthermore, any vertex contributing to $x(ABC)$ and adjacent to $A$ belongs to exactly one triangle with $A$, as otherwise the condition is again violated. We can argue similarly for $B$ and $C$.

We now count pairs containing a triangle $ABC$ in $G$ and a vertex in $V(G) \setminus \{A,B,C\}$ incident to a triangle containing one of $A$, $B$, or $C$. As noted above, for a given triangle $ABC$, the number of pairs with the triangle being $ABC$ is $n-3-x(ABC)$. Letting $\mathcal{T}$ be the set of triangles in $G$, the count is then:

\[\sum_{ABC \in \mathcal{T}} (n-3-x(ABC))\]

As mentioned above, for a vertex $v$, the set of triangles containing $v$ are such that the pairs of the other two vertices are disjoint from each other. We can then get the count by looking at individual vertices. Letting the number of triangles incident to $v$ be $T(v)$, we choose the triangle around the vertex for the pair in $T(v)$ ways and we choose the vertex in the pair in $2(T(v)-1)$ ways. Then:

\[\sum_{ABC \in \mathcal{T}} (n-3-x(ABC)) = \sum_{v \in V(G)} (2T(v)^2-2T(v))\]

Notice that $\sum_{v \in V(G)} T(v)$ counts each triangle $3$ times, once for each vertex, so it is $3m$. Then:

\[m(n-3) - \sum_{ABC \in \mathcal{T}} x(ABC)  = \sum_{v \in V(G)} 2T(v)^2 - 6m\]
\[\Longrightarrow m(n+3) - \sum_{ABC \in \mathcal{T}} x(ABC) = 2\sum_{v \in V(G)} T(v)^2\]

Using Cauchy-Schwarz on the vectors $<T(v_1), \ldots, T(v_n)>$ and $<1, \ldots, 1>$, we get $\sum_{v \in V(G)} T(v)^2 \ge \frac{9m^2}{n}$. Rearranging the above:

\[\sum_{ABC \in \mathcal{T}} x(ABC) \le m(n+3) - \frac{18m^2}{n}\]

By the pigeonhole principle, some $ABC \in \mathcal{T}$ must be such that $x(ABC) \le \left\lfloor \frac{m(n+3)-\frac{18m^2}{n}}{m}\right\rfloor = \left\lfloor n+3 - \frac{18m}{n} \right\rfloor$.


Square $S_{1}$ is $1\times 1.$  For $i\ge 1,$ the lengths of the sides of square $S_{i+1}$ are half the lengths of the sides of square $S_{i},$ two adjacent sides of square $S_{i}$ are perpendicular bisectors of two adjacent sides of square $S_{i+1},$ and the other two sides of square $S_{i+1},$ are the perpendicular bisectors of two adjacent sides of square $S_{i+2}.$  The total area enclosed by at least one of $S_{1}, S_{2}, S_{3}, S_{4}, S_{5}$ can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m-n.$

[asy]
size(250);
path p=rotate(45)*polygon(4);
int i;
for(i=0; i<5; i=i+1) {
draw(shift(2-(1/2)^(i-1),0)*scale((1/2)^i)*p);
}
label("$S_1$", (0,-0.75));
label("$S_2$", (1,-0.75));
label("$S_3$", (3/2,-0.75));
label("$\cdots$", (7/4, -3/4));
label("$\cdots$", (2.25, 0));[/asy]

I'd like to think the original problem was with an infinite series of squares. If that's the case, it's:

$1/4+\frac{3/4}{1-\frac{1}{4}}=\frac{5}{4}\implies m-n=1$

@nate
we're looking for m-n

@ Zuton Force
I think the problem was only the first five
http://www.kalva.demon.co.uk/aime/aime95.html

Oh yeah, you're right... the "..." after the first five squares in the diagram got me  :o

In that case its what nate said, and $m-n=255$

[quote="4everwise"]The total area enclosed by at least one of $ S_{1}, S_{2}, S_{3}, S_{4}, S_{5}$ [/quote]
gosh, if i'm taking this text i'd be screwed... i didn't even UNDERSTAND what the problem meant by "at least one of"... shouldn't it be "ALL OF" instead?

This original wording makes me think it means the union of area(s) of at least one of the squares S1 to S5 could be written in the form $ m/n$, and we are supposed to find the total area, but apparently we can find the area of 1 square, or the union of any 2 squares, or any 3 squares, up to 5 squares, (since it says AT LEAST), but then there would be more than one possible answer...

looking at the solution it makes me think the problem meant to find the total area enclosed the union of (ALL OF) $ S_{1},...,S_5$... is this interpretation correct?

Wait, so am I supposed to count only the first five squares?

[quote="4everwise"]The total area enclosed by at least one of $S_{1}, S_{2}, S_{3}, S_{4}, S_{5}$ can be written...[/quote]

does that answer your question?

We can start by finding the total area of each square. That is just $1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \frac{1}{256}$. But we must subtract off the overcounted square areas, which have area $\frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \frac{1}{1024}$. This gives a result of $1+\frac{1}{4} - \frac{1}{1024} = \frac{1024+256-1}{1024} = \frac{1279}{1024}$. Subtracting, we get an answer of $\boxed{255}$.


I've always wondered this:

What do people immediately think of when they hear South Carolina.  Be honest, I don't care if you say domestic violence or high DUI mortality rates.  Just tell the truth.  I mean, SC isn't a remarkable state in any way.  For example, when I hear NY, I think Brooklynn accent and traffic.

Please post your response, as well as the state you are from.

Poverty and Myrtle Beach. Quite contradictory...

State: NC (woot)

Povery is interesting.  I associate SC with a lot of bad things, but not poverty.

[quote="junggi"]When I hear SC, I think of soul caliber(the game).[/quote]

Let me rephrase.  What do you think of when you hear "South Carolina?"

I think of my family or my high school usually.

I don't see poverty as that unusual.  SC education is not great and I would think that this is one of the first things that people think about when they hear SC.  "Uneducated" and "poverty" often times go together.  However, there aren't many places in the US that I would considered in poverty.  Having lived in SC, however, there weren't many instance of poverty.  There was a lot of lower class people, but I consider poverty to be below lower class.

I honestly know less about SC than pretty much any other state, even if it is our neighbor. So I think of North Carolina (aren't I sad).

[quote="mysmartmouth"][quote="Inquisitor"]I, being an ignorant Brit, think of Hurricanes (both the meterological phenomenon and the hockey team - though do they play in the North...?)[/quote]

You say that you are an ignorant Brit.  Yet most Americans (or at least those in SC) could not name the countries in the UK, let alone the political divisions analgous to states in the US).  :maybe:  That shows how bad US (or SC) geography class is.[/quote]
No, it is the entire country. Also, I think of this one friend of mine who has a REALLY thick country accent.

Well I went to hs in Texas and now I am in NJ for university. When I hear SC I think of you people because you're the only South Carolinans (?) I know, and I don't have any other connection with the place besides I went there (Savannah River Site, then Charleston) for the award trip for National Science Bowl my junior year. But that was ages ago.

By the way, I have decent knowledge of British geography, but that's probably because I've studied Roman history a lot and the Romans conquered part of Britain in the first century AD. And because I used to play a computer game that had all of the countries divided into their respective parts. I might be able to remember all of teh counties in Ireland too. :P

Hicks and rednecks.

Afterwards, Myrtle Beach.

SC=University of Southern California

Particularly how they lost to Oregon 33-31 last weekend in college football...

random people on aops who post in the sc forum alot

sc=south california? jk its south carolina
south carolina=football

I think of the color blue for some reason :huh: . Sky blue, like the color of the uniforms of UNC...Hmm...


I thought they were pretty easy..got a 25

But some of the questions I missed when going through it the first time, and I was very lucky to have seen them when double-checking.

This year was really easy; got 24 b/c 8+9 $ \neq$ 24...

That means next year will be supppper hard.

How come next year will be super hard?

I got a 18...  :) . But I rushed through it in about 25 minutes, so I guess I deserve it. They were pretty easy, but I rushed so much (because they were easy) that I got many wrong.

I got a 25 also, but i guessed on number 24 (the dice and tiles problem) and got it right luckily!  :clap2:
I thought the questions were pretty simple, the only problem was that i almost ran out of time.
Do you guys know when and if the scores will be posted online? I want to see what my friends got...

i thought it was extremely easy.

25!
[hide="click only if you got a 25"] YESSSSSS!!!! [/hide]

Next year will be hard because this year was easy.  They probably made it easy because it's the first year w/ the no-calc rule.  Maybe that would've helped me with 8+7...

A typo in the calculator is just as easy to make. :wink:

Don't value the AMC8 that much, as it is not a good indicator.

However, it is a fun thing to brag about, and you can't really brag about 24 :(

24 :mad:

I can't believe that I thought that the word [i]square [/i]meant the word [i]same[/i]! And that I missed this while checking!!!

meh.....$ 23$....sad....but i made 2nd in my school....

Hm, [hide=" "]I don't know why I'm here...[/hide] my brother got 18, which tied for third (two people in my school got 22, two people got 19, and two people tied for 18). I know, epic phail :P

Our school had 3 24's, and some 22 and 21's.  Our class average (8th graders taking geometry except im in 7th) was 16.8

i got a 25 (yay)

i thought that the questions were REALLY easy this year compared to some of them last year
i finished with about 10 minutes to spare and went back to check it
didn't find any errors though =)

25! Yay!

The test was pretty easy, #25 was a joke, #24 was more guess/check, the rest were not particuary hard.

This year was ridiculously easy even for an AMC 8.

Our school hasn't even gotten our results back yet. I think I should have gotten 20 more or less though.

21 best in my school kinda because i was the only one there to take it.

i got a 22 but there were at least 4 25's and a looooot of 24's and 23's in my school :( x_x 2 of them were stupid mistakes, the other one was the last one (i guessed on it because i was lazy  :rotfl: )

24.... :mad: After a 25 last year too. Anyway, math154 is right, because they got rid of calculators, this was a contest of who could not make stupid mistakes.

not really
you don't need calculators because all the probs in the test do not require the use of a calculator (now i sound like my math teacher....)

anyway, it's multiple choice
chances are, if you make a silly mistake, the answer you get will not be a choice

Actually no, they always are choices.

and what I meant was, now that there are no calculators, they significantly "dumbed down" the test.

[quote="vallon22"]not really
you don't need calculators because all the probs in the test do not require the use of a calculator (now i sound like my math teacher....)

anyway, it's multiple choice
[b]chances are, if you make a silly mistake, the answer you get will not be a choice[/b][/quote]

The AMC board doesn't randomly choose answer choices. In particular, on casework counting problems, they will use 5 consecutive choices, which means that any one mis-count will be devastating. :wink:

ok
you've proved me wrong
but calculators aren't exactly going to help u on counting problems  :wink:

i'm not sure what the problems were like before the no calcs rule, since i have not done any of the probs...
was it that much harder?

Yeah. I finished in $ 20$ minutes, and checked all the problems two times, still missing one small thing on a casework problem (because I thought $ 7<7$ on problem $ 24$).

If I had a calculator, I would've finished in $ 25$ minutes (because I would type everything in $ 3$ times), and then have sufficient time to program for all the counting problems. :wink:

Next year they'll probably allow calcs and then like 2 years later, then ban them again... :rotfl:

Wait..so in the years before we were allowed to use calcs?  i got an average of 23 on those test without calc, ad thats what I got on my Real test this year.

Yeah, this is the first year with the no calc rule.

[quote="vallon22"]really?!?!

my school couldn't use a calc last year....
my proctor said it's been that way since 2006.....[/quote]

Then you just got ripped of.  :P

On the front of the test, thought, it clearly says you can use a calculator.  Did anyone rad the instructions?

EDIT: It also says no problems on the test requires the use of a calculator.

*sigh* 23 points.

I know after looking at the test, like AIME I thought 8+9=24... wow

AND I just had to think of 14 wrong.

 :(

[quote="Dojo"]

I know after looking at the test, like AIME I thought 8+9=24... wow
[/quote]

Whoa that is scary.

We are beast at math!  However, we suck at addition.

24...
i am 1st, with 2 20s, an 18, and 2 15s for other people :(
i got #14 wrong

[quote="BOGTRO"]Would anyone have any idea why there are 3 answers listed for #12?

I checked the scoring distribution for NJ:
#       A          B           C           D             E         OMIT       ?
12   12.14     23.27     29.77*    18.81*    14.26*     1.56      0.18

* indicates a correct answer
? indicates unable to read, either because of a double or not completely erased blah blah blah answer.

3 *'s??

I'm SURE I got 25, not to mention my coach said I did, but

SCORE   5th Grade    6th Grade    7th Grade    8th Grade
  25           0                 0               1               4
.................................................................................
(I'm in 6th BTW)[/quote]

The question asks for when it will not rise more than .5 feet, after 5 bounces it is not.  Thus 5, 6, and 7 are possible answers.  If the said [i]first[/i], then 5 would be the answer.


I'm pretty sure there is a quick formula to find the exterior angle of a regular polygon. i just forgot it. can anyone give it to me thanks.

To create an exterior angle, you extend one of the sides of a polygon and find the angle formed.  It will be $180^\circ$ minus an interior angle of the polygon.  An interior angle of a polygon will have $\left(\frac{180(n-2)}{n}\right)^\circ$ (divide the polygon into n-2 triangles to get the total number of degrees, then because all angles are congruent, divide by $n$ for each angle).  Thus, you have:

$180^\circ-\frac{180(n-2)}{n}^\circ\Rightarrow \frac{360}{n}$

[quote="vishalarul"][hide="Proof"]The supplement of an exterior angle is equal to the interior angle, which has measure $180\frac{n-2}{n}$.
Measure of exterior angle $=180-180\frac{n-2}{n}$
$=180(1-\frac{n-2}{n})$
$=180{\frac{2}{n}}$
$=\frac{360}{n}$ [/hide]

P.S. How do you write degrees with Latex?[/quote]
^\circ

Extension: Prove that the sum of all interior angles in an n-gon is $180(n-2)$. :D

[hide="for convex n-gons"]Pick a random vertex. Draw all possible diagonals from that vertex. There are n-2 of those diagonals.Each of these diagonals form a triagle, so n-2 triangles are formed. The angles of the triagles make of the angles of the n-gon and we know that the sum of int. angles of a triangle is 180. So the sum of int. angles of an n-gon is $(n-2)180$[/hide]


geometry
trigonometry
ratio
IMO Shortlist
geometry proposed

Let $P$ be a convex polygon. Prove that there is a convex hexagon which is contained in $P$ and which occupies at least 75% of the area of $P$.

By a theorem of E. Sas (1939), any convex body $K$ in the plane contains a convex $n$-gon $P_n$ of area at least $\frac {n} {2\pi} \sin \dfrac {2\pi} {n}$ of the area of $K$, this bound being precise if and oly if $K$ is an ellipse. For $n=6$, this ratio is $\approx 0.827 > \dfrac {75} {100}$. See [url]http://books.google.ro/books?id=E3-BheG-CoYC&pg=PA14&lpg=PA14&dq=Sas+theorem+convex+Agarwal+ellipse&source=bl&ots=TOcKOnF5Gt&sig=V0Npk5WX1WZKrcuuBsrk4CPitbs&hl=en&sa=X&ei=sCHGT9yyJ8Xh4QTDgLW4BQ&redir_esc=y#v=onepage&q&f=false[/url].

As such, with bound $3/4$, it is a problem from the 2004 IMO Shortlist [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=143293&sid=8c5e0c71a28f1b41f1d4c41708594037#p143293[/url]

Pretty funny that this problem was also p6 on polish MO in 2005.


Given:

(sinX * cosX + cosX)/(sinX) = 2*cotX

Verify each identity. In other words...
Proove that the two statements are equlivalent to one another.

i tested if for some random value of X, and it didn't work

wOOOPs!!!!

silly me.

should be>>

(sinX * [b]cotX[/b] + cosX)/(sinX) = 2*cotX

Ah, now that it's fixed, it's easy.
[hide]
$\frac{\sin x \cot x+\cos x}{\sin x}= \frac{\sin x \frac{\cos x}{\sin x}+\cos x}{\sin x}$

$= \frac{2\cos x}{\sin x}= 2\cot x$.[/hide]

[hide]
$2\cos x=2\cos x$
$\cos x+\cos x=2\cos x$
$\sin x\frac{\cos x}{\sin x}+\cos x=2\cos x$
$\sin x\cot x+\cos x=2\cos x$
$\frac{\sin x\cot x+\cos x}{\sin x}=2\frac{\cos x}{\sin x}$
$\frac{\sin x\cot x+\cos x}{\sin x}=2\cot x$
[/hide]

What I always do (bad habit) is do it like that, and work it down to a tautology.  Then I write out my final proof backwards, so it works out logically.

[quote="tjhance"]are you sure this is correct  :maybe: or am i just making some dumb error?

[hide]
$\frac{\sin x\cos x+\cos x}{\sin x}=2\cot x$
$\frac{\sin x\cos x+\cos x}{\sin x}=2\frac{\cos x}{\sin x}$
$\sin x\cos x+\cos x=2\cos x$
$\sin x\cos x = \cos x$
$\sin x=1$

so that only works if $\sin x=1$...
[/hide][/quote]
Actually since you divided by cos x it would work for where
cos x=0.

[quote="k8reindeer"]In my math class at school, my teacher tells me I must work on only one side of a proof (like this, or when we were doing induction). I understand why: working on both sides assumes equality, it doesn't prove it. I'm wondering why you guys aren't doing that... is it one of those rules that doesn't really hold up outside a math classroom?[/quote]

Well you can't really end up dividing both sides by 0 if you're working on one side at a time or something else dumb like that, but otherwise, there's really no use for working on only one side.


superior algebra
superior algebra solved

hi all,

let A and B be 2 matrices of M(n,R) so that:
A^2 + b^2=A*B  and
A*B - B*A is invertible.
Prove that n is a multiple of 3.

it is from IMC u can find it in official site Hint : let $C=A+\omega B$


How do you prove that there exist a circle determined by four points say A,B,C,D :oops:

It depends on the information you have.
- You can check if the circumcircles determined by two different triangles (e.g. ABC and BCD) coincide (if the four [url=http://mathworld.wolfram.com/PerpendicularBisector.html]perpendicular bisectors[/url] of the quadrilateral ABCD cross each other at the same point).
- If you already know (or can compute in another way) the area of the quadrilateral ABCD, you can check if [url=http://mathworld.wolfram.com/BrahmaguptasFormula.html]Brahmagupta's formula[/url] for cyclical quadrilaterals holds.

If you know the angles formed by the points, you could check to see if opposite angles equal 180 degrees, and if you know the distances between the all the line segments, you could check to see if Ptomley's holds. In AoPS 2, chapter 4 is entitled cyclic quadrilaterals, and it is one of the few chapters that I think that i have come close to mastering, so it isn't too hard. :D

[quote="jgao"]circle determined by four points say A,B,C,D :oops:[/quote]

Circles are determined by three points.

With no other conditions, the problem is still false.  Let $E = D$ and obviously the quadrilateral cannot be cyclic.

[quote="jgao"][quote="mustafa"]The question is really how do you prove that a quadrilateral is cyclic.[/quote]

The actual problems was actually.

Given four points $ABCD$, let $AC$ and $BD$ intersect at $E$. Prove that quadrilateral $ABCD$ is cyclic.[/quote]
Umm, this isn't necessarily true. $AC$ and $BD$ will always intersect, no matter what (well, unless they're parallel, of course :P ), so that doesn't imply that the quadrilateral is cyclic.  :wink:

i think he is asking for what is necessary to prove that the quadrilateral is cyclic

[first note that $ABCD$ must be convex]

$ABCD$ cyclic $\Leftrightarrow AE\cdot EC=BE\cdot ED\Leftrightarrow AED\sim BEC\Leftrightarrow AEB\sim DEC$

Note: I left out the angle relationships have nothing to do with $E$


In my view you should be able to prove that the solns are equivalent. There can be NO paradox in Maths( unlike Twin Paradox in relativity)

[quote="eminem"]this question was submitted by someone else( all i remember that he was an american) quite some time back . I didnt get the solution yet

let
$x^4 + x^2 + 1$ = 0

hence
applying quadratic formula

$x=\sqrt{(-1+\sqrt-3)}/2$ or $\sqrt{(-1-\sqrt-3)}/2$ or $-\sqrt{(-1+\sqrt-3)}/2$ or  $-\sqrt{(-1-\sqrt-3)}/2$

now
$x^4 + x^2 + 1$ = $x^4 + 2x^2 + 1-x^2$
                            = $(x^2+1)^2-x^2$
                            = $(x^2 -x +1)(x^2 +x +1)$
hence $x=(1+\sqrt-3)/2  or  (1-\sqrt-3)/2  or  (-1+\sqrt-3)/2  or  (-1-\sqrt-3)/2$

how come there are so many solutions?there shoud be just 4 solutions[/quote]

Indeed , they are just 4 solutions namely $\frac{1+\sqrt{-3}}{2}$ ,$\frac{1-\sqrt{-3}}{2}$ , $\frac{-1+\sqrt{-3}}{2}$ ,$\frac{-1-\sqrt{-3}}{2}$ :P

[quote="Rushil"]can you prove that the solns. are equivalent[/quote]

[hide="hint"] Try to square those solution I gave(of course you must add a square root sign outside the number) and you will notice something  :D [/hide]

[quote="Rushil"]square which? those in sqrts ?[/quote]

[hide="hint"]
This  :arrow: $\frac{1+\sqrt{-3}}{2}$ .  Just like $a = \sqrt{a^2}$[/hide]

[quote="riddler"]look at this,
$(y-x)^2=(x-y)^2$
that what you're trying to say? :huh:
so then that means
$(y-x)=(x-y)$
which is wrong assumption,
You are trying to prove $2$ different things equal by a wrong way.[/quote]

[hide]
Lets see example . Let $\sqrt{-1+\sqrt{-3}}/2 $ = $\frac{\sqrt x - \sqrt y}{2}$ . Squaring both side and simplify,

$-2-2\sqrt{-3}$ = $x+y-2\sqrt{xy}$ . Comparing coeffienct , $x+y=-2$ , $xy=-3$

solving this we have $x=1,y=-3$ and $x=-3,y=1$  So you will have it become

$\sqrt{-1+\sqrt{-3}}/2 $ = $\frac{1 - \sqrt {-3}}{2}$

OR

$\sqrt{-1+\sqrt{-3}}/2 $ = $- \frac{1 - \sqrt {-3}}{2}$

working all out and you will see that theres only 4 pairs ( 4 will repeat ) .

*Note here that we didnt cancel off "additional roots" because they are imginary roots which since imaginary roots have no positive or negative sign . [/hide]

[quote="riddler"]So you mean to say in every quadratic equation, $4$ similiar roots will appear,
but I understand what you said :D  cool![/quote]

A bit wrong in the statement when you say $every$ quadratic equation . :P  Actually you dont need to work out all 4 because theres 2 of them which differ only but positive negative sign . So working out two with positive sign can give you the equivalent result you want . :P

[quote="Rushil"]There can be NO paradox in Maths( unlike Twin Paradox in relativity)[/quote]

I don't think so. Consider the Russel paradox, or the Banach-Tarsky paradox. There's plenty.

yo  Shyong , i got your point. they look like there are 6 solutions but there are only 4 thats by squaring and then square rooting . the first 2 are the same as the last ones

thanks


induction
combinatorics proposed
combinatorics

The numbers $ 0$, $ 1$, $ \dots$, $ n$ ($ n \ge 2$) are written on a blackboard. In each step we erase an integer which is the arithmetic mean of two different numbers which are still left on the blackboard. We make such steps until no further integer can be erased. Let $ g(n)$ be the smallest possible number of integers left on the blackboard at the end. Find $ g(n)$ for every $ n$.

For each $ n$, $ g\left(n\right)$ equals to $ 2$ ,when $ n$ is a power of $ 2$, and $ 3$ when otherwise.

First, we will do the easier side, that is,
we have the steps of operations that leave $ g\left(n\right)$ numbers on the board.

Suppose that $ n$ is a power of $ 2$. We can just fix $ 0$ and $ n$ so we can erase $ 1, 3, ..., n\minus{}1$.
Then, like induction, we can erase $ 2\cdot 1,...,2\cdot (n\minus{}2)/2$ and so on.
(We erase those of $ 1(mod 2)$ then $ 2(mod 4)$ then ... then $ 2^{m\minus{}1} (mod 2^{m})$)
By this way, we can erase all but $ 0$ and $ n$ which is what we wanted.

When in the case that $ n$ is not a power of $ 2$, let $ n\equal{}2^{m}\plus{}a$ where $ a, m$ is a natural number with $ 0 < a < 2^{m}$.
We will not erase $ 0,2^{m}$ nor $ n$.
We then erase $ n\minus{}1, n\minus{}2, ... 2^{m}\plus{}1$ respectively (or not erase any of them if $ a\equal{}1$)
and erase $ 1, ... , 2^{m}\minus{}1$ in the same way as the first case.
By this way, we can leave $ 0, 2^{m}$ and $ n$ as we claimed.

Now, for the latter side, that is, we will show that $ g\left(n\right)$ can't be less than we claimed.
Notice that $ 0$ and $ n$ can't be erase -- this yields $ g\left(n\right) \geq 2$

The last part to be tackled is when $ n$ is not a power of $ 2$.
Since $ n$ is not a power of $ 2$, we have a prime $ p>2$ that $ p\mid n$
Suppose, for sake of contradiction, that $ g\left(n\right)\equal{}2$.
Then, the left numbers are $ 0$ and $ n$.
If there is a series of steps that can leave only $ 0$ and $ n$, then we can do it backward by writing the integer number that is arithmetic mean of two different numbers on the board (which, initially, are $ 0$ and $ n$)
and repeat this process until we have all the numbers $ 0,1,...,n$
But notice that, by writing in this fashion, all the new number we get must still be divisible by $ p$
In particular, we can't get $ 1$ => contradiction!  :lol:


[quote="tchebytchev"]solve the system:
$x(x+y)^2 = 9$
$x(x^3 - y^3 ) =7$[/quote]
Maybe $y(x+y)^2 = 9$ and $y(x^3 - y^3 ) =7$?
If so it is oral entrance examination in the Moscow University, 1979. ;)


Hi - This is my first post on this forum, and I hope I'm doing so under the proper heading.

I'm looking for an explanation of the card trick known as "Any Way You Count 'Em."  Clearly the explanation is a mathematical one.  It's been years since I've tackled this sort of problem, so I would greatly appreciate it if someone would enlighten me.  The description of the trick follows.  Try it, it's fun!

--------------------------
Shuffle the cards well. Holding them face down, turn over the top card and place it face up on the table. Think of it as a stack. Starting with its face value, deal face up on top of it as many more cards as needed to reach 10. For instance if it's a 3, deal seven cards on top of it; if it's a 5, deal five cards. Face cards count as 10, so no more cards are needed. An ace counts as 1 and needs nine more cards.

Continue making stacks as above, keeping them separate, until the deck is exhausted. If there are not enough cards to complete a final stack, keep that incomplete stack in your hand. Now choose at random any three stacks that contain at least four cards each, and turn these stacks face down.

Gather all the remaining cards in any order and add them to the cards (if any) in your hand. Pick any two of the three face-down stacks on the table, and turn up the top card on each of those two piles. Add their values together. Discard that many cards from those in your hand, and then discard 19 additional cards.

Count the remaining cards in your hand. Now turn up the top card of the third stack. Its value will equal the number of cards in your hand.
---------------
reference : http://www1.superb.net/~cardtric/tricks/intermed/countem.htm

[hide]Let a, b, and c be the values of the top cards of the three stacks left over. When these three stacks were being made, 10-a cards were placed on top of card a, 10-b cards on card b, and 10-c cards on card c. So the stacks have 11-a, 11-b, and 11-c cards, for a total of 33-(a+b+c) cards. So you are holding in your hand 52-(33-(a+b+c))=19+a+b+c cards. Now you discard 19+a+b cards, leaving c cards, and c is the top card of the third stack.
Using this reasoning though, I don't see why the left-over stacks must have at least 4 cards each...I didn't use/assume it at any time in my reasoning, did I?[/hide]

What do we do with jacks queens and kings


The function $ f : N \rightarrow N$ satisfies $ f(1) \equal{} 1, f(2) \equal{} 2$ and $ f(n \plus{} 2) \equal{} f(n \plus{} 2 \minus{} f(n \plus{} 1) ) \plus{} f(n \plus{} 1 \minus{} f(n) )$.
Show that $ 0 \leq f(n \plus{} 1) \minus{} f(n) \leq 1$. Find all $ n$ for which $ f(n) \equal{} 1025$.

$ f(3)\equal{}f(3\minus{}2)\plus{}f(2\minus{}1)\equal{}2$

$ f(4)\equal{}f(2)\plus{}f(1)\equal{}3$

$ f(5)\equal{}f(2)\plus{}f(2)\equal{}4$

$ f(6)\equal{}f(2)\plus{}f(2)\equal{}4$

We will prove inductively that $ f(n)\equal{}4$ for all $ n\ge 5$.

Suppose that $ f(5)\equal{}\ldots\equal{}f(k)\equal{}f(k\plus{}1)\equal{}4$

$ f(k\plus{}2)\equal{}f(k\plus{}2\minus{}4)\plus{}f(k\plus{}1\minus{}4)\equal{}f(k\minus{}2)\plus{}f(k\minus{}3)\equal{}4$

So the statement follows. There is no value $ n$ for which $ f(n)\equal{}1025$

You are right, my induction was wrong, because exactly of those small examples, my f(k-3) can be 3, for k=6, and k+2=8


find the condition that a box of square base of length $ a$ and height $ h$ kept on an incline of inclination $ \theta$ topples before it starts sliding.the frictional coeffecient between the surfaces is $ \mu$.The box's mass is $ m$ .

Is the answer like this?

$ tg\theta=\frac{\mu a}{a+\mu h}$

If not,please post the answer.

[hide="ans"] $ \tan \theta > \frac{a}{h}$[/hide]


I had to do a homework, but i haven't really understood these types of exercises. Can you give any help by solving it?
Prove if this function is an euclidian distance:

$d\left|{f(x)g(x)}\right| = \sup \left\{{\left|{f(x) = g(x)}\right|}\right\}$ in $C_{[0,1]}$

This forum isn't the right place. Try the advanced algebra section in the college playground.


Shouldn't ratings of deleted posts still count towards calculating the post rating, in order to differentiate those who made productive posts from those with spam, just with the spam deleted? Basically, I'm suggesting that when a post is deleted, the data for the ratings of the posts aren't removed.

Yes, I agree. What is the current situation? Do the deleted posts count or not?

Right now, they don't, which is why I'm suggesting this.

[b] With no offense, [/b]Well, Those who have 'lower' post rating (I meant overall rating) doesn't necessarily mean that they spam, it might be the case they were too busy engaged in other activities than 'posting' regularly in AOPS forum. Such an assumption (lower rating,spamming assumption) seems to be highly incoherent, because it is only true to a certain extent.

After viewing the rating of the posts of a user, one should also take care to view the number of posts made by him/her. A large number of posts with lower rating might (in all probability) be 'spamming'.

Yes, and I am trying to close the ambiguity, since spam is most likely to be deleted, and my suggestion would keep their post rating representative of a "spammer"; in the case with few posts, anybody could see that they have had few posts.


function
probability
calculus
derivative
probability and stats

Say for $ i = 1, 2, 3...$

We have

$ X_i = 1$ if $ p = 1/2$
$ X_i = -1$ if $ q = 1/2$

And we have ${ M_k = \sum_{i=1}^{n}\ X_i}$ for $ k = 1, 2 ,3,...$ and $ M_0 = 0$

Find $ E[M_k ^3]$

I know that we cannot do $ E[M_k]*E[M_k]*E[M_k]$   and that taking  ${ ( \sum_{i=1}^{n}\ X_i} ) ^3$ is just insane...

Does anybody know the trick around this?

I was thinking of using a moment generating function, but that too seems messy...

Their has to be a simple way to do this...any ideas?

Moderator, please move this to the Probability & Statistics Forum. [Now obsolete]

Expanding $ \left(\sum_{i} X_i\right)^3$ is not particularly insane: you get one copy of each $ X_i^3$ term, three copies of each $ X_i^2X_j$ term for $ i \neq j$ and six copies of each $ X_iX_jX_k$ term for $ i \neq j \neq k \neq i$.  Calculating the expectation of each of these terms is simple, and since expectation is linear you can distribute the expectation and add up.

The easier "method" is just to note that your expression takes the value $ i$ exactly as frequently as it takes the value $ \minus{} i$.

$ M_n\equal{}(n\minus{}2k)$ with probability $ \frac{1}{2^n}\binom{k}{n}$ hence $ E(M_n^3) \equal{} \sum_{k\equal{}0}^n \frac{1}{2^n}\binom{k}{n} (n\minus{}2k)^3$.

This is a good use of a transform method.

The moment generating function for $ X_i$ is $ E(e^{tX_i})=\cosh t.$

Adding independent random variables causes us to multiply moment generating functions:

$ m{}_M{}_n(t)=\cosh^n(t).$

Finally, the moment generating function does, in fact, generate moments:

$ E(M_n^3)=\left.\left(\frac{d}{dt}\right)^3M_n(t)\right|_{t=0}.$

At this point, we note that the third derivative of a smooth even function must be a smooth odd function and to be zero at zero. Hence $ E(M_n^3)=0.$

We note that by the same argument, we have that $ E(M_n^k)=0$ for $ k$ odd. And if $ k$ is even, we would still use this path to a computation.


find all pairs of natural numbers x,y for which
$ x^2 \equal{} 4y \plus{} 3 \cdot lcm(x,y)$

[hide="Hidden by mod"]Put $ x \equal{} ad, y \equal{} bd$ where $ (a,b) \equal{} 1$. Then $ [x,y] \equal{} abd$, hence

$ a^2d^2 \equal{} 4bd \plus{} 3abd\implies a^2d \equal{} b(3a \plus{} 4)$

Since $ (a,b) \equal{} 1$, this implies $ b|d\implies d \equal{} kb$ for some natural $ k$, so thus

$ a^2k \equal{} 3a \plus{} 4\implies k \equal{} {3a \plus{} 4\over a^2}$

It's easily shown that for $ a\geqslant 5$ the RHS is less than $ 1$, hence $ a\in\{1,2,3,4\}$. Checking shows that only $ a \equal{} 1$ and $ a \equal{} 4$ give integer values for $ k$.

Case 1. $ a \equal{} 1\implies k \equal{} 7\implies (x,y) \equal{} (kab,kb^2) \equal{} (7b,7b^2)$. Since $ [7b,7b^2] \equal{} 7b^2$, it's easily checked that the initial equation is satisfied.

Case 2. $ a \equal{} 4\implies k \equal{} 1\implies (x,y) \equal{} (kab,kb^2) \equal{} (4b,b^2)$. However, the initial condition states $ 1 \equal{} (a,b) \equal{} (4,b)$, hence $ b$ must be odd, and with $ [4b,b^2] \equal{} 4b^2$, the checkup also holds.

Conclusion: $ (x,y) \equal{} (7n,7n^2)$ or $ (x,y) \equal{} (4(2n \minus{} 1),(2n \minus{} 1)^2)$ where $ n\in\mathbb{N}$[/hide]


Does problem difficulty rise as the rounds progress? I'm assuming otherwise, due to the similar correlation in scores between Round 1 and later rounds last year.

[quote="surge"]Does problem difficulty rise as the rounds progress? [/quote]

It will.

[quote="surge"]I'm assuming otherwise, due to the similar correlation in scores between Round 1 and later rounds last year.[/quote] Possibly because many people drop out in the later rounds as they are unable to solve the problems.

[quote="JSteinhardt"][quote="Treething"]oh joy i can't answer anything in the first round[/quote]

That's why you have an entire month.[/quote]
I think he's being sarcastic. :lol:

Do not discuss the first round problems; not even the difficulty of them.


[quote="barcelona"]Prove that if $ a,b,c > 0$ then the polynomial $ ax^{p} \plus{} bx^{q} \plus{} cx^{s} \equal{} (x^2 \plus{} x \plus{} 1)*Q(x)$ ,where $ Q(x)$ is some ppolynomial.
PP:I don't know is that is true,so can you help me?[/quote]

That's obviously wrong in general. You need conditions on $ a,b,c,p,q,s$

For example, $ p\equal{}q\equal{}s$ would imply $ (x^2\plus{}x\plus{}1) | (a\plus{}b\plus{}c)x^p$, which is false
.

Sorry!I forgot that  $ p\ne q\ne s$ and also $ a,b,c$ gives a diferent remainder by modulo 3(ex:$ p\equal{}6,q\equal{}7,s\equal{}8$).

[quote="barcelona"]Sorry!I forgot that  $ p\ne q\ne s$ and also $ a,b,c$ gives a diferent remainder by modulo 3(ex:$ p \equal{} 6,q \equal{} 7,s \equal{} 8$).[/quote]
Try to forget nothing next time ... :(

Still wrong : Choose $ p\equal{}2,q\equal{}1,s\equal{}0$ and $ a\equal{}3,b\equal{}2,c\equal{}1$ : $ 3x^2\plus{}2x\plus{}1$ is obviously not divisible by $ x^2\plus{}x\plus{}1$

[quote="barcelona"]So  $ p\ne q\ne s\ne 0$.
Now I think that's the pproblem.[/quote]

Are you kidding ?

$ x^3\plus{}2x^2\plus{}3x\equal{}(x\plus{}1)(x^2\plus{}x\plus{}1)\plus{}x\minus{}1$ is not divisible by $ x^2\plus{}x\plus{}1$

What is true, but quite elementary, is that if $ p,q,r$ are pairwise different modulo $ 3$, then $ x^2\plus{}x\plus{}1$ divides $ x^p\plus{}x^q\plus{}x^r$


geometry
3D geometry
probability
rectangle
algebra
difference of squares
special factorizations

1) What is the largest square number that is under 50?

2) What is the largest cube number that is under 50?

3) Is there a very quick way to multiply 59 and 61? If so, tell me how you would go about multiplying those numbers.

4) Jim draws a triangle. One side is 13 feet long, another side is 5 feet long, and the third side is 12 feet long. What is the largest angle of the triangle? Express your answer in degrees.

5) I created a candy that is 80% sugar by weight. One piece of my candy is 100 pounds (yum). How many pounds of sugar does one piece of my candy have?

6) Albatross and Frankinfueter each have a fair coin. They both flip it. What is the probability that both Albatross and Frankinfueter's coins end up heads?

7) A rectangle's length is twice its width. The rectangle's area is 18. What is the length of the rectangle?

Please, if you're way above this level, don't post the answers here. If you absolutely must post to increase your post count (j/k :P ), spoiler post.

1) What is the largest square number that is under 50?
[hide] 7^2 = 49[/hide]

2) What is the largest cube number that is under 50?
[hide]3^3 = 27[/hide]

3) Is there a very quick way to multiply 59 and 61? If so, tell me how you would go about multiplying those numbers.
[hide]the closet thing that I could think of is to multiply 59 times 60 then add 59 to it, OR 61 times 61 minus 61[/hide]

4) Jim draws a triangle. One side is 13 feet long, another side is 5 feet long, and the third side is 12 feet long. What is the largest angle of the triangle? Express your answer in degrees.
[hide]13^2 = 12^2 + 5^2 so the triangle is a right triangle, and so it's 90 degrees[/hide]

5) I created a candy that is 80% sugar by weight. One piece of my candy is 100 pounds (yum). How many pounds of sugar does one piece of my candy have?
[hide]4/5 = x/100  when you solve it x = 80lb[/hide]

6) Albatross and Frankinfueter each have a fair coin. They both flip it. What is the probability that both Albatross and Frankinfueter's coins end up heads?
[hide]the probabliy of one of them flipping head is 1/2 and so when you want both of them to happen, you mulitiply and get 1/4[/hide]

7) A rectangle's length is twice its width. The rectangle's area is 18. What is the length of the rectangle?
[hide] 18 = x*1/2x = 1/2x^2 so x^2 would be 36, and x would be 6[/hide]

[quote="LuCy4EvA"]

3) Is there a very quick way to multiply 59 and 61? If so, tell me how you would go about multiplying those numbers. The closet thing that I could think of is to multiply 59 times 60 then add 59 to it, OR 61 times 61 minus 61.[/quote]

That way is correct, but there exists a better way.

Think of the difference of squares.
AKA: $x^2 - y^2 = (x-y)(x+y)$.

Expanding on what white horse king said:
[hide]59=(60-1), 61=(60+1), so you want to find what (60-1)*(60+1) is.  Using the difference of squares, it is 60^2-1^2=3600-1=3599[/hide]

This was one of the COP Championship Questions actually.

[quote="COP_CHAMPIONSHIP"]Given that {abcd}^2 = uvwx, find what {abce}^2, where e = d+1, in terms of uvwx and abcd.[/quote]

Answer

[hide]
{abce^2} = {abcd+1}^2 = {abcd}^2 +2abcd + 1
= {abcd}^2 + 2uvwx + 1
[/hide]
I think that was the answer

See if you only do on the first way, you can't do this problem. But realizing the factoring by 1 or other integere within the FOIL itself is very useful.

thanx a lot! never really thought about that :D


If the following relations hold for the nonnegative numbers x{i},i=1,2,..n,
1) x{1}=x{2}=1/n
2)if i is odd,then x{i}+x{i+1}=x{i+2}+x{i+3}

prove that

Π(from i=1 to n) of (x{i})^2) <= [Σ(from i=1 to n) of (x{i})^(2n+1)]^n

(where x{i} means ''x sub i'').


Can one find a bijection from $\mathbb{Q}$ onto $\mathbb{Q}\setminus{0}$ which is order-preserving?

See these two topics:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=order&t=89433[/url]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Sierpinski&t=48347[/url]

You can use a back-and-forth procedure to find an order isomorphism between $\mathbb Q$ and $\mathbb Q\setminus\{0\}$.


Can somebody confirm the leaks of harry potter.  Does harry die

Which ones are true/false

Can we compare tommorrow?

I personally the book is going to be bad, but that doesnt stop me frmo readaing it....too bored..

Okay get this clear:
[hide="Spoiler"]Harry, Ron and Hermione all survive. Voldemort, Lupin, Moody, Hedwig die. That much has been confirmed til now.[/hide]
And yes of course the leak is true, with a huge probability.

You can easily search and get it online...not saying that you should...well...i didn't at the least to say.

well now the books' out.....anyone want a pdf version?

Yeah. Although my sis already has the book.

Do you have a pdf?

[quote="Asiodea"]wouldn't posting a pdf be illegal?[/quote]
Yes, but apparently some people don't care...


modular arithmetic
real analysis
advanced fields
advanced fields unsolved

Suppose that $X$ is a measurable space with measure $\mu$ and $T$ is a measure-preserving transformation on $X$.

Call a transformation is [i]compressible[/i] if there exists a measurable set $E \subset T^{-1}E$ such that $\mu(T^{-1}E \setminus E) > 0$. Otherwise the transformation is incompressible.

(1) Prove that if $T$ is one-to-one and $T^{-1}$ is measurable, then $T$ incompressible implies $T^{n}$ incompressible for all $n \in \{1, 2, 3, \ldots\}$. (Halmos, Annals, 1947, p. 738)

(2) The same, but without the assumption that $T$ is one-to-one.

[quote="julien_santini"]Why do we need to assume measure-preservation ? This would force $E$ to be of infinite measure in the definition of a compressible operator, wouldn't it ?[/quote]
We want it to be measure-preserving because Halmos wants to use this result in his book on ergodic theory, and ergodic theory is concerned only with measure-presering transformations. I believe what you say about the measure of $E$ is correct. I am afraid I'm only just starting to learn ergodic theory, so I can't give you a lecture explaining the reasons for everything, but I'll look over some stuff and try to explain the point of compressibility later. Right now I'm really stressed and in a hurry, sorry.

I hope this works as a proof for (2).

Notice (I believe Halmos also made this comment in his book "Lectures on ergodic theory" :)) that compressibility is equivalent to [i]dissipativity[/i]: the existence of a set $F$ of positive measure with the property that $T^{-k}F,\ k\ge 0$ are disjoint (in what follows, “disjoint” may have to be replaced by “such that each two have intersection with measure zero”; nothing will be lost if we still write “disjoint” instead of that). The proof will take the following form: I’ll show that if $T^{n}$ is dissipative for some $n\ge 2$, then $T$ is dissipative.

Let $F$ be a set with positive measure such that $T^{-kn}F,\ k\ge 0$ are mutually disjoint. It follows that for each $0\le u<n$ the sets $T^{-(kn+u)}F,\ k\ge 0$ are mutually disjoint as well. Now let $1\le p\le n$ be the largest positive integer with the property that we can find non-negative integers $0=a_{1}<a_{2}<\ldots<a_{p}$ giving distinct residues modulo $n$ such that $\mu\left(\bigcap_{i=1}^{p}T^{-a_{i}}F\right)>0$. Finally, let $G=\bigcap_{i=1}^{p}T^{-a_{i}}F$ for such a maximal system $\{a_{i}\}$.

We have $\mu(G)>0$. If we show that $T^{-k}G,\ k\ge 0$ are disjoint, we’re done. Moreover, because $T$ is measure-preserving, this is equivalent to showing that $\mu\left(G\cap T^{-k}G\right)=0,\ \forall k\ge 1$. We have $T^{-k}G=\bigcap_{i=1}^{p}T^{-(a_{i}+k)}F$. Clearly, one of the following holds: (a) we can find $i,j\in \overline{1,p}$ such that $a_{i}\equiv a_{j}+k\pmod n,\ a_{i}\ne a_{j}+k$, in which case $G,T^{-k}G$ are disjoint because one is a subset of $T^{-a_{i}}G$ and the other is a subset of $T^{-(a_{j}+k)}G$ and these are disjoint, or (b) we can find $j\in\overline{1,p}$ such that $a_{j}+k$ is different modulo $n$ from all $a_{i},\ i=\overline{1,n}$, in which case the two sets have intersection with measure zero by the maximality of  $p$.


Let $ ABC$ be a triangle. $ W_a$ is a circle with center on $ BC$ passing through $ A$ and perpendicular to circumcircle of $ ABC$. $ W_b,W_c$ are defined similarly. Prove that center of $ W_a,W_b,W_c$ are collinear.

$ \mbox{The lines } W_aA, W_bB, W_cC \mbox{ is tangents to circumcircle of triangle }ABC. \mbox{The circles is Appolonian circles of triangle }ABC.$

It has been [url=http://www.mathlinks.ro/viewtopic.php?p=117527#117527]posted[/url] before, under a different disguise.


O1,O2,O3 are three circles，the internal tangent of each two form a hexagon.
prove the 3 diagonals are concurrent

[quote="¬[ƒ(Gabriel)³²¹º]¼"]It was posted [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=140096]here[/url]. I know a nice solution By April of this  :lol:[/quote]
I don't see a solution there,can you post APRIL's nice proof?
thanks in advance!

mybe we can use this:
Monge-d'Alembert Theorem :
The external center of similitude of three circles are collinear

Does someone have some idea?
 Thank you, Mathlinkers!


Suppose than $ v, \mu$ are signed measures, prove that v and $ \mu$ are mutually singular if and only if $ \mid v \mid$ are mutually singular.

Proof so far.

Suppose that v and $ \mu$ are mutually singular, then there exists $ E,F \in \mathbb {M}$ such that E and F are disjoint, $ E \cup F \equal{} X$ such that $ \mu (E) \equal{}0$ and $ v(E) \equal{} 0$.

Now, $ \mid v \mid \equal{} v^\plus{} \plus{} v^\minus{}$, so can I say that $ \mid v \mid (E)$ = 0 and that would conclude the claim?

And would that mean $ v^\plus{}$ (or $ v^\minus{}$ )and $ \mu$ are mutually singular?

Thank you!

The definition of [i]mutually singular[/i] requires two disjoint measurable sets each of which is a null set for one of the measures. Being a null set [b]for a signed measure[/b] is not the same as having measure $ 0$. A set $ E$ is null for $ \nu$ if $ \nu(F) \equal{} 0$ for any measurable $ F\subset E$. [At least in Folland's book].

What you should prove first is that a measurable $ E$ is null for $ \nu$ if and only if $ |\nu^ \plus{} (E)| \equal{} 0$.  This is a statement about one signed measure, not two; and Hahn  decomposition should help.

Once you have the above fact, the problem reduces to a tautology.

So the Hahn decomposition says that if v is a signed measure, we can obtain:

1.  $ X \equal{} P \cup N$

2.  P , N disjoint.

where P is positive set for v and N is negative set.

So now, I'm trying to get E is v-null iff $ \mid v ^\plus{} \mid (E) \equal{} 0$

If E is v-null, then $ v(F) \equal{} 0 \ \ \ \forall F \subset E$.

$ \mid v \mid \equal{} v^\plus{} \plus{} v^\minus{}$, but what is $ \mid v^\plus{} \mid$ ?  Would that be just $ v^\plus{}$?

Plus is a typo - I meant $ |\nu|(E)\equal{}0$.


Suppose $ A$ and $ B$ are non-zero digits and $ \underline{A5}\cdot\underline{B1} \equal{} 1995$. Find $ A \plus{} B$

To do this, we should prime factorize 1995. First, dividing by 5 gives us 399 (2000-5). Then, dividing by 3 gives us 133. Then, dividing by 7 gives us 19 (140-7). So, the prime factorization is $3\cdot5\cdot7\cdot19.$ To end in a 5 with a factor means using the 5. Let's list out the 5's (when multiplying) (that are 2 digits): 15, 35, and 95. Now, let's list out the 1's: 21. Thus, B is 2. Now, just divide $1995 \div 21 = 9$something. Thus A is 9 (something is 5 but it is not really relevant now). $9+2 = \boxed{11}.$


I am a newcomer to this forum and I was studying some of the posts that were submitted by some of your members. I noticed that a good majority of the message of a lot of the posts is not showing up on my screen. Could anyone tell me how to correct this problem and tell me why this is happening?
Your responses would be greatly appreciated. Thanks.

Perhaps you are referring to spoiler text (which is invisible)?  To see it, click and drag your mouse over it.  For example, I am putting a word [hide](congratulations!)[/hide] in this sentence in spoiler font.

People sometimes write solutions and hints to problems in spoiler text so that others can choose not to look, in case they want to solve the problem by themselves.

hi, Ravi B:

That is a really clever idea to do that. I am a member of the sosmath forums and they don't have anything like that so it is new to me. But I can understand why some members prefer that. I think that you can better remember how to solve a problem if you can solve it on your own without assistance from others. Though it's nice to read other posts for a check. Thanks.

Welcome to the forum.  I think you will enjoy it.


Prin picioarele cevienelor perpendiculare pe dreapta lui Euler a unui triunghi se duc paralele la laturile corespunzatoare ale triunghiului ortic. Demonstrati ca cercurile circumscrise celor 3 triunghiuri formate dintr-o paralela si doua laturi ale triunghiului initial sunt concentrice.
         Nu va spun care e centrul :P

Conform acestei [url=http://www.mathlinks.ro/Forum/album.php?mode=attach&id=6167]imagini[/url], centrul trebuie sa fie $H$. Sper ca nu vrei si o demonstratie... :D