Εάν $ a,b,c,d\geq 0$ και $ a\plus{}b\plus{}c\plus{}d\equal{}4$ αποδείξτε ότι $ \frac{1}{5\minus{}abc}\plus{}\frac{1}{5\minus{}bcd}\plus{}\frac{1}{5\minus{}cda}\plus{}\frac{1}{5\minus{}dab}\leq 1$.

Θα κανω μια προσπαθια με πολλαπλασιαστες Lagrange για να δω ποσο καλα τους εχω κατανοησει, το πιο πιθανο ειναι να κανω καποια λαθη (διοτι πρωτη φορα χρησημοποιω αυτη τη μεθοδο σε καποια δυσκολη ασκηση) οποτε θα ηθελα να με διορθωσετε αν κανω πουθενα κανενα λαθος

Λοιπον

Θετουμε $ f(a,b,c,d) = \sum_{cyc}{\frac {1}{5 - abc}}$, και $ g(a,b,c,d) = a + b + c + d - 4$

Αρχικα ψαχνουμε για ακροτατο στο $ (0, 4)$
Για $ a,b,c,d \in (0, 4)$, εχουμε $ f(a,b,c,d) = \sum{\frac {a}{5a - k}}$, με $ k = abcd$. Αν για καποια $ a,b,c,d \in (0, 4)$ τα οποια ικανοποιουν τη συνθηκη $ a + b + c + d = 4$, η f παρουσιαζει ακροτατο, τοτε για αυτα τα $ a,b,c,d \in (0, 4)$ η g μηδενιζεται, επομενος απο το θεωρημα πολλαπλασιαστων Lagrange, υπαρχει συναρτηση $ F(x) = f(x) + mg(x)$ με ολες τις μερικες της παραγωγους ισες με 0 σε αυτα τα $ a,b,c,d$

Λυνοντας το συστημα $ \nabla F(a,b,c,d) = 0, a + b + c + d = 4$, δηλαδη το $ {\frac { - k}{(5a - k)^2} = \frac { - k}{(5b - k)^2} = \frac { - k}{(5c - k)^2} = \frac { - k}{(5d - k)^2}, a + b + c + d = 4}$ (Λαμβανοντας υπ οψην οτι $ k < 5x$ για x = a,b,c,d, οπως ευκολα προκειπτει με AM-GM και παραγωγηση), παιρνουμε $ a = b = c = d = 1$, αρα η f παρουσιαζει τοπικο ακροτατο (μεγιστο η ελαχιστο) για $ a = b = c = d = 1$ το $ f(1,1,1,1) = 1$

Εξεταζουμε τωρα για ακροτατα οταν ενας απο τους αριθμους $ a,b,c,d$ παρει την τιμη 0 η 4, και στις 2 αυτες περιπτωσεις ενας απο τους αριθμους $ a,b,c,d$ γινεται 0 (εστω χωρις βλαβη της γενικοτητας ο d), και ο τυπος της f γινεται $ f(a,b,c) = \frac {3}{5} + \frac {1}{5 - abc}$. Αν παλι ενας απο τους $ a,b,c$ κανει 0, τοτε η συναρτηση παιρνει την τιμη $ \frac {3}{5} + \frac {1}{5} = \frac{4}{5}$, η οποια ειναι πιθανη θεση ακροτατου της f αφου παρουσιαζεται για τις ακρεες τιμες των μεταβλητων της στο κλειστο διαστημα στο οποιο αυτη οριζεται. Αν κανεις απου τους $ a,b,c$ δεν κανει 0, τοτε παλι ψαχνουμε για ακροτατο της $ f(a,b,c) = \frac {3}{5} + \frac {1}{5 - abc}$ στο (0, 4) υπο τη συνθηκη $ a + b + c = 4$, ετσι εφαρμοζουμε ξανα (οπως παραπανω) τη μεθοδο πολλαπλασιαστων Lagrange και βρησκουμε οτι η f παρουσιαζει τοπικο ακροτατο για $ a = b = c = \frac {4}{3}$, το $ f(\frac {4}{3}, \frac {4}{3}, \frac {4}{3}) = \frac {3}{5} + \frac {1}{5 - \frac {64}{27}}$

Η f ομως, ως συνεχης συναρτηση ορισμενη σε κλειστο διαστημα, απο το Θεωρημα Μεγιστης και Ελαχιστης τιμης, θα παρουσιαζει ενα Ολικο Μεγιστο, και αφου οπως εξηγησαμε τα πιθανα ολικα ακροτατα τις συναρτησης f ειναι οι αριθμοι: $ x = \frac{4}{5}, y = 1, z = \frac{3}{5} + \frac {1}{5 - \frac {64}{27}}$, ο μεγαλυτερος απο τους 3 αυτους αριθμους θα ειναι το ολικο μεγιστο της συναρτησης f. Προφανος ο μεγαλυτερος απο αυτους τους 3 αριθμους ειναι ο αριθμος $ y = 1$, επομενος ο αριθμος 1 ειναι το ολικο μεγιστο της συναρτησης f, και επομενος ισχυει $ f(a,b,c,d) \geq 1 \Leftrightarrow \sum_{cyc}{\frac {1}{5 - abc}} \geq 1$ για καθε $ a,b,c,d \in [0, 4]$ με $ a + b + c + d = 4$ και η αποδειξη ολοκληρωθηκε. Η Ισοτητα προφανος επιτυγχανεται για $ (a,b,c,d) = (1,1,1,1)$

[quote="NickNafplio"]
Λυνοντας το συστημα $ \nabla F(a,b,c,d) = 0, a + b + c + d = 4$, δηλαδη το $ {\frac { - k}{(5a - k)^2} = \frac { - k}{(5b - k)^2} = \frac { - k}{(5c - k)^2} = \frac { - k}{(5d - k)^2}, a + b + c + d = 4}$
[/quote]

Συγχώρεσέ με αν μου ξεφεύγει κάτι, αλλά έχω την εντύπωση ότι (λανθασμένα) έχεις θεωρήσει ότι $ \frac{\partial k}{\partial a} = \cdots = \frac{\partial k}{\partial d} = 0$, δηλαδή ότι το $ k$ είναι κάποιου είδους σταθερά (αλλά φυσικά δεν είναι).

[quote="cmad"][quote="NickNafplio"]
Λυνοντας το συστημα $ \nabla F(a,b,c,d) = 0, a + b + c + d = 4$, δηλαδη το $ {\frac { - k}{(5a - k)^2} = \frac { - k}{(5b - k)^2} = \frac { - k}{(5c - k)^2} = \frac { - k}{(5d - k)^2}, a + b + c + d = 4}$
[/quote]

Συγχώρεσέ με αν μου ξεφεύγει κάτι, αλλά έχω την εντύπωση ότι (λανθασμένα) έχεις θεωρήσει ότι $ \frac {\partial k}{\partial a} = \cdots = \frac {\partial k}{\partial d} = 0$, δηλαδή ότι το $ k$ είναι κάποιου είδους σταθερά (αλλά φυσικά δεν είναι).[/quote]

Ναι, πραγματι δεν ειναι σταθερα το k, και το περιμενα οτι καποιος θα μου πει οτι δεν επιτρεπεται να το χρησημοποιησω σαν σταθερα, αλλα πιστεψα οτι αυτο που εκανα ειναι σωστο επηδη προσφατα ειχα δει μια ασκηση σε καποιο βιβλιο οπου ο συγγραφεας για την αποδειξη μιας αλλης ανισοτητας 3ων μεταβλητων (a,b,c) χρησημοποιουσε ανισοτητα jensen και στην παραγωγηση εθετε λ = abc με (λ)' = 0

[quote="meinstein"]Γενικά η μέθοδος των πολαπλασιαστών Lagrange (αν και ειμαι μεγαλος fan της  :P ) δεν δουλευει συνήθως τόσο ευκολα σε τέτοιες ασκήσεις εκτός αν απλοποιούνται.... Η λύση με βασικές ανισότητες θα είναι σίγουρα πολύ πιο κομψή και σύντομη... Πάντως μην ξεχνάμε ότι την άσκηση δν την κατασκεύασε ο οποιοσδήποτε αλλά ο μαίτρ του είδους Vasc.... :wink:[/quote]

Βασικα μολις το ξανακοιταξα καλα, καταλαβα οτι αυτο που εκανα στη 1η φορα που χρησημοποιησα πολλαπλασιαστες Λαγκραντζ στη συγκεκριμενη ασκηση, οχι απλα δεν επιτρεπεται αλλα ειναι και σοβαρο λαθος που δεν επιτρεπεται να κανει καποιος που εχει κατανοησει καλα την Αναλυση (ακομα και μαθητης Γ Λυκειου). Απλα παρασιρθηκα απο μια ασκηση στο βιβλιο του κ. Στεργιου οπου στη λυση χρησημοποιουσε Jensen και εθετε το γινομενο των μεταβλητων λ. Στην ασκηση με την Jensen ομως δεν υπηρχε προβλημα διοτι παραγωγιζε ως προς χ (το οποιο ειναι ανεξαρτητο απο το λ) για να δειξει οτι η συναρτηση ειναι κυρτη και μετα εθετε οπου χ τα α,β,γ. Εδω περα ψαχνουμε για ακροτατο εχοντας ως μεταβλητες τα α,β,γ,δ, οποτε καθος αυτα μεταβαλονται, μεταβαλεται και το κ, που σημαινει οτι δεν μπορουμε να θεωρησουμε (κ)'=0.

Η Ασκηση μπορει να λυθει και με πολλαπλασιαστες Λαγκραντζ, αλλα ο ποιο απλος τροπος ειναι να θεσουμε x=abc κλπ, οποτε η συνθηκη γινεται $ \frac{1}{x} \plus{} \frac{1}{y} \plus{} \frac{1}{z} \plus{} \frac{1}{w} \equal{} \frac{4}{\sqrt[3]{xyzw}}$ και βγαινει παλι ενα αρκετα πολυπλοκο συστημα το οποιο θελει επισεις αρκετες πραξεις, επομενος σιγουρα θα υπαρχει καποιος πιο απλος τροπος αποδειξης της συγκεκριμενης ανισοτητας.

Συγνωμη αν ταλεπωρησα κανεναν με το λαθος μου.

[quote="NickNafplio"][quote="meinstein"]Γενικά η μέθοδος των πολαπλασιαστών Lagrange (αν και ειμαι μεγαλος fan της  :P ) δεν δουλευει συνήθως τόσο ευκολα σε τέτοιες ασκήσεις εκτός αν απλοποιούνται.... Η λύση με βασικές ανισότητες θα είναι σίγουρα πολύ πιο κομψή και σύντομη... Πάντως μην ξεχνάμε ότι την άσκηση δν την κατασκεύασε ο οποιοσδήποτε αλλά ο μαίτρ του είδους Vasc.... :wink:[/quote]

Βασικα μολις το ξανακοιταξα καλα, καταλαβα οτι αυτο που εκανα στη 1η φορα που χρησημοποιησα πολλαπλασιαστες Λαγκραντζ στη συγκεκριμενη ασκηση, οχι απλα δεν επιτρεπεται αλλα ειναι και σοβαρο λαθος που δεν επιτρεπεται να κανει καποιος που εχει κατανοησει καλα την Αναλυση (ακομα και μαθητης Γ Λυκειου). Απλα παρασιρθηκα απο μια ασκηση στο βιβλιο του κ. Στεργιου οπου στη λυση χρησημοποιουσε Jensen και εθετε το γινομενο των μεταβλητων λ. Στην ασκηση με την Jensen ομως δεν υπηρχε προβλημα διοτι παραγωγιζε ως προς χ (το οποιο ειναι ανεξαρτητο απο το λ) για να δειξει οτι η συναρτηση ειναι κυρτη και μετα εθετε οπου χ τα α,β,γ. Εδω περα ψαχνουμε για ακροτατο εχοντας ως μεταβλητες τα α,β,γ,δ, οποτε καθος αυτα μεταβαλονται, μεταβαλεται και το κ, που σημαινει οτι δεν μπορουμε να θεωρησουμε (κ)'=0.

Θα ηθελα αν μπορουσε να εξεφραζε γνωμη ο μαιτρ Σιλουανος

Η Ασκηση μπορει να λυθει και με πολλαπλασιαστες Λαγκραντζ, αλλα ο ποιο απλος τροπος ειναι να θεσουμε x=abc κλπ, οποτε η συνθηκη γινεται $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \plus{} \frac {1}{w} \equal{} \frac {4}{\sqrt [3]{xyzw}}$ και βγαινει παλι ενα αρκετα πολυπλοκο συστημα το οποιο θελει επισεις αρκετες πραξεις, επομενος σιγουρα θα υπαρχει καποιος πιο απλος τροπος αποδειξης της συγκεκριμενης ανισοτητας.

Συγνωμη αν ταλεπωρησα κανεναν με το λαθος μου.[/quote]

Γεια σας παιδιά  :)  Λοιπόν η άσκηση αυτή είναι κάπως δύσκολη . Ο Vasc αν δεν κάνω λάθος την λύνει με την AC-method , δηλαδή ένα είδος mixing variable αλλά όχι σε απλή μορφή ... Το AC method είναι στοιχειώδες αλλά όχι ευρέως γνωστό . Θα την κοιτάξω και λίγο ακόμα σήμερα και αν δεν βρω κάτι και θέλετε λύση θα ποστάρω την προαναφερθείσα . :wink:

Εψαξα για το AC method, και βρηκα σε ενα ποστ του Vasc αυτο:

[b]Arithmetic Compensation Theorem (AC-Theorem).[/b]Let $ s > 0$, and let $ F(x_1,x_2,...,x_n)$ be
a symmetrical continuous function on the compact set in $ R^n$,
$ S \equal{} ((x_1,x_2,...,x_n): x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} s, x_1 \geq 0,..., x_n \geq 0)$.
If $ F(x_1,x_2,...,x_n) \leq \max{(F(\frac {x_1 \plus{} x_2}{2},\frac {x_1 \plus{} x_2}{2}, x_3,...,x_n),F(0,x_1 \plus{} x_2,x_3,...,x_n))}$
for all $ (x_1,x_2,...,x_n)\in S$ with   $ x_1 > x_2 > 0$, then
$ F(x_1,x_2,...,x_n) \leq \max_{1\leq k\leq n}{F(\frac {s}{k},..., \frac {s}{k},0,...,0)}$  for all $ (x_1,x_2,...,x_n)\in S$

Δεν καταλαβα ομως τι ακριβως σημαινει το $ \max_{1\leq k\leq n}{F(\frac {s}{k},..., \frac {s}{k},0,...,0)}$, δεν καταλαβα δηλαδη αν ενωει το μεγιστο στοιχειο του συνολου των αριθμων $ F(m_1, m_2, ... , m_n)$, οπου $ m_i \in$ {$ 0, \frac {s}{k}$}, για $ i \equal{} 1,2,3...,n$, και $ k \equal{} 1,2,...n$, η αν ενωει κατι διαφορετικο...

[quote="NickNafplio"]Βασικα μολις βρηκα αυτο το αρχειο στο google στο οποιο ο Vasc εξηγει τη μεθοδο αυτη σε διαφορες ασκησεις τετοιου ειδους με διαφορα παραδειγματα (περιεχει και τη αποδειξη της παραπανω ανισοτητας)

Το αρχειο ειναι αυτο: http://reflections.awesomemath.org/2006_3/2006_3_arithmeticcompensation.pdf

Επηδη ομως δυσκολευομαι να κατανοησω τις 2 αποδειξεις του θεωρηματος αυτου που υπαρχουν στο συγκεκριμενο αρχειο (ισος να μη καταλαβαινω καλα τι γραφουν επηδη ειναι στα Αγγλικα, η ισος να γινεται χρηση καποιου πανεπιστιμιακου θεωρηματος το οποιο θεωρειται γνωστο απλα εγω δεν το γνωριζω), θα ηθελα αν γνωριζει καποιος και εχει χρονο να τις εξηγησει λιγο εδω στο φορουμ.[/quote]

Όντως χρησιμοποιεί ένα (γνωστό) πανεπιστημιακό θεώρημα. Θα εφαρμόσω την μέθοδο της απόδεικσης του AC Theorem για να αποδείξω την ανισότητα του Αριθμητικού-Γεωμετρικού Μέσου. (Αυτή η απόδειξη είναι γνωστή, νομίζω την είχα δει στο Hardy-Littlewood-Polya αλλά δεν το έχω τώρα μπροστά μου.)  Δεν θα χρησιμοποιήσω κανένα πανεπιστημιακό θεώρημα αλλά η απόδειξή μου δεν θα είναι πλήρης. Καλείσαι να βρεις τι έχω παραλήψει να δικαιολογήσω. (Αυτή η παράληψη δεν είναι τόσο φανερή αλλά είναι σημαντικη και ειναι εκεί που θα  χρειαστεί το πανεπιστημιακό θεώρημα.)

Από την ομοιογένεια του προβλήματος, αρκεί να δείξω ότι αν $ a_1,\ldots,a_n \geq 0$ με $ a_1 \plus{} a_2 \plus{} \cdots \plus{} a_n \equal{} 1$ τότε $ a_1a_2 \cdots a_n \leq \left( \frac{1}{n} \right)^n$.

Παίρνω $ a_1,\ldots,a_n$ που να ικανοποιούν τις πιο πάνω συνθήκες και να μεγιστοποιούν το γινόμενο $ a_1a_2 \cdots a_n$. Ισχυρίζομαι ότι $ a_1 \equal{} a_2 \equal{} \cdots \equal{} a_n$. Αν δεν ισχύει αυτό τότε (απο συμμετρία) μπορώ να υποθέσω ότι $ a_1 \neq a_2$. Αλλά, ορίζοντας $ a_1' \equal{} a_2' \equal{} \frac{a_1 \plus{} a_2}{2}$, τα $ a_1',a_2',a_3,\ldots,a_n$ ικανοποιούν τις συνθήκες και το γινόμενο τους είναι $ \left(\frac{a_1 \plus{} a_2}{2} \right) \left(\frac{a_1 \plus{} a_2}{2} \right) a_3 \cdots a_n > a_1a_2 \cdots a_n$, άτοπο. Άρα $ a_1 \equal{} a_2 \equal{} \cdots \equal{} a_n \equal{} \frac{1}{n}$ και η ανισότητα ισχύει.


1) What is the pH of a solution that contains $0.818$ M acetic acid $(K_a = 1.76 \times 10^{-5})$ and $0.172$ M sodium acetate?

Could someone explain how to approach problems like these? In my textbook, I haven't been able to find a section that deals with calculating the pH of a solution that contains more than one substance. From what I read, I tried this, and got an answer that was not among the choices:

[hide]
For this problem, I set up the equation:

$K_a = \displaystyle \frac{[H^+][C{}_2H{}_3O{}_2^-]}{HC{}_2H{}_3O{}_2} = 1.76 \times 10^{-5}$

Since $[H^+]$ and $[C{}_2H{}_3O{}_2^-]$ should be equal, calling the concentration $x$:

$\displaystyle \frac{x^2}{.818} = 1.76 \times 10^{-5}$

$x=\sqrt{1.76 \times .818 \times 10^{-5}}$

pH = $-\log{\sqrt{1.76 \times .818 \times 10^{-5}}} = 2.42$
[/hide]

2) This was a problem that our teacher gave us to stump the class:

Calculate the final concentration of $Na^+_{(aq)}$, $C{}_2O{}_4^{-2}_{(aq)}$, $Ba^+2_{(aq)}$, and $Cl^-_{(aq)}$ in a solution prepared by adding $100$ mL of $0.20$ M $Na{}_2C{}_2O{}_4$ to $150$ mL of $0.25$ M $BaCl_2$. Neglect hydrolysis of ions.

for question 1, try to ICE it first before you set up your $K_a$...it's going to be a common ion effect problem...

In the first case you have buffer solution, you may solve it using Henderson-Hasslenalch equation (which is nothing else but rearranged dissociation contanst definition). No need for ICE table, check lectures:

[url]http://www.chembuddy.com/?left=pH-calculation&right=pH-buffers-henderson-hasselbalch[/url]

and example question (different, but in a way similar):

[url]http://www.chembuddy.com/?left=pH-calculation-questions&right=pH-buffer-q1[/url]

Second question is trivial - once you check what is solubility of barium oxalate  :)

yea the first problem should be like:

[hide]

$K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} = \frac{(x)(x+0.818)}{0.172}$

solving gives $x = 3.7 x 10^{-6} M$ so a pH of $5.43$

[/hide]

But thats not the rule..
if for examble one acid was HCl then calculate only the [H+] produced by HCl (unless c is very small)

Here's how to do this problem.

Ka = ([H+][C2H3O2-])/[HC2H3O2]

Substituting the values [C2H3O2-] = 0.172+x and [HC2H3O2] = 0.818-x, and [H+] = x, we can write the following equation,

[x(0.172+x)]/(0.818-x) = 1.76*10^(-5)
As the value of x is miniscule compared to the given initial concentrations of acetic acid and acetate, it can be eliminated in calculating the final concentration of these products. Therefore,

0.172x/0.818 = 1.76*10^(-5)
Solving for x, we get
x = 8.37*10^(-5) M
Now, the pH can be calculated in the following fashion,
pH = -log(x) = [b]4.08[/b]

This problem can also be solved using the Henderson-Hasselbach equation, which is in the following form:
pH = pKa + log([A-]/[HA])

Your second question is a typical AP Chemistry solubility question. Here's how to crack it:

[NaC2O4]initial = 0.20 M, therefore [Na+]initial = 0.40 M
Now, since 150 ml is added to 100 ml of Na+, the new concentration is
[Na+] = (0.40)(100)/(100+150) = [b]0.16 M[/b], which does not change in the reaction as Na+ is a spectator ion.
A similar method can be used to determine [Cl-]
[BaCl2]initial = 0.25 M, therefore, [Cl-]initial = 0.50 M
[Cl-]final = (0.50)(150)/(100+150) = [b]0.30 M[/b]
Now, the important part is determining the final concentrations of the ions that largely precipitate. For this, we must know the equation and the solubility product (Ksp):
BaC2O4 ---> Ba(2+) + C2O4(2-)      Ksp = 2.3*10^(-8)
First, for the purposes of convenience in approximation, we assume that the ions completely precipitate in a stoichiometric reaction:
[Ba(2+)]initial = 0.15 M  [C2O4(2-)]initial = 0.16 M
[Ba(2+)]final = 0.15 M-0.15 M = 0 M
[C2O4(2-)]final = 0.16 M-0.15 M = 0.01 M
Now, we assume that x moles/liter of BaC2O4 dissociates. The equation we get for this dissociation is:
[Ba(2+)]*[C2O4(2-)] = Ksp = 2.3*10^(-8)
Substituting [Ba(2+)] = x and [C2O4(2-)] = 0.01+x, the following equation can be established:
x(0.01+x) = 2.3*10^(-8)
Solving for x (through some approximation), we get
x = [Ba(2+)] = [b]2.3*10^(-6) M[/b]
0.01+x = [C2O4(2-)] = [b]0.01 M[/b]


Find $m \in \mathbb{R}$, if there is only one pair $(x,y)\in \mathbb{R}$ x $\mathbb{R}$, such that $\ y \ge  x^2 + m$ and $\ x \ge y^2 + m$.

It's the same as finding when the parabolas $y = x^2+m$ and $x = y^2+m$ are tangent.

So by substitution we have $y = (y^2+m)^2+m \Rightarrow 0 = y^4+2my^2-y+(m^2+m)$. We want it to only have one solution, so it should be a square. But I'm not sure how to pick an $m$ such that you can factor that into a square...

When the parabolas are tangent, they are both tangent to $y=x$. Take the derivative and set it to 1, so $x=\frac12=y$. Put back in the original equation to get $m=\frac14$.


[quote="shfdfzhjj"][quote="antiparticle"]Prove that the product of $ 9$ consecutive integers can't be a perfect power.

Is the result is generally true for any block of $ n > 1$ consecutive integers? If the general result is true, can you please post the general theorem and the proof. A link to the proof would also suffice.

Thanks![/quote]

Erdos has already sloved. :)[/quote]

 :oops:

 :?: Can you please please post the name, statement and proof of the theorem then?

Thanks!

Check some of the erdos work at:
[url]http://www2.renyi.hu/~p_erdos/Erdos4ht.html[/url]

in particular that result at: [url]http://www2.renyi.hu/~p_erdos/1975-46.pdf[/url]


My teacher  teaching my class C language,once  chellenged us to code a program to calc 10000!
who can help me ! I think that problem is belong to a method of handling large numbers!

for such a small number, use n! = (n-1)! * n is good enough.
It will be faster if you factor the number to it's prime components and then multiply, because then you can do exponentiation.
if you need to handle large numbers, use bigint libraries, like GMP.

mpz_t is a good structure: do indeed go look into GMP.

I have a feeling that one ought to use linked lists in order to compute the values of extremely large numbers in 'C', but I am not so sure. I think I found a solution to a similar problem before (some time back) but I doubt whether this method will work here.

long int, or long float can't hold (store) the value of 10000!.

10000! is not that many digits binary-wise. Allocating enough bits to store the number (approximately 115kBits, so allocate 128 kBits or 16KB to store it), you can perform multiplication directly on the bits with the method shown here:

http://www.cs.princeton.edu/introcs/54programming/

(Ctrl+F "efficient multiplication".)

http://www.cs.sunysb.edu/~skiena/392/programs/bignum.c
Now that you can do arithmetic on bignum-ers implementing factorial is simple.

the simplest and one of the fastest ways of doing 10000! is to simply store the number in base 10^15 using unsigned long longs

[size=150][code]#include <stdio.h>
const unsigned long long base=1000000000000000LLU;
unsigned long long array[3000];
void mult(int t)
{
    unsigned long long carry=0;
    for(int i=0;i<3000;i++)
    {
        carry+=array[i]*t;
        array[i]=carry%base;
        carry/=base;
    }
}
int main()
{
    array[0]=1;
    for(int i=1;i<=10000;i++)
        mult(i);
    int i=2999;
    while(i && array[i]==0)
        i--;
    printf("%llu",array[i--]);
    for(;i>=0;i--)
        printf("%015llu",array[i]);
}[/code][/size]


Two circles touch one another externally at the point A. A straight line through A meets one of the circles at T and the other at S. Prove that the tangent to the first circle at T is parallel to the tangent to the second circle at S.

Just draw the tangent line going through the tangent point, you know arc $ AT$ is equal in measure to arc $ AS$ by vertical angles: then it follows that $ m\angle{AO_{1}T}=m\angle{AO_{2}S}$. (where $ T$ and $ S$ are on the circles with centers $ O_{1}$ and $ O_{2}$, respectively).  From there, it follows that $ O_{1}T\parallel O_{2}S$, and so the perpendicular to one is parallel to the perpendicular to the other.

Rotate $ AT$ at $ A$ through $ 180^{\circ}$ and then stretch it by factor $ \frac{AS}{AT}$. This maps the tangent $ TT_{1}$ to $ SS_{1}$ and $ \angle ATT_{1}$ to $ \angle ASS_{1}$. Hence $ \angle ATT_{1}=\angle ASS_{1}$. Thus $ TT_{1}\parallel SS_{1}$.


Find all  prime numbers  $p$ such  that  $1+p+p^2+p^3+p^4$ is a perfect  square.

[quote="Jutaro"]Excuse me, Mr. Pluricomplex, but how did you conceive the idea of substituting $a=k+p^2$?? Is it a kind of general method or something?? Thanks in advance :P[/quote]
If you have a polynom of degree $2n$, which must be perfect square, try to estimate it by squares of polynoms of degree $n$.

hey try to solve the equation
$y^2=x^4+x^3+x^2+x+1$ in $\mathbb{Z}$
(it's not difficult, and the problem here is just a special case! :) )


The number  1999  is multiplied by a  1060-digit number that has all its digit equal to  1. What is the sum of all the digits of the number obtained?

[hide]1999=2000-1

(2000-1)11111...=22222...000-111...

The last four digits is 0889.  Then we subtract all the rest of the 1's from the answer as there is no other carrying involved.

There are 1060 1's, we have already used 4 of them to subtract 1111 from 2000.  So we have 1056 1's left.  We only have 1059 2's as one of them was one of the last for digits.

So if I didn't mess up, it should be 1059*2-1056+8+8+9[/hide]

That is what I did, and it took me a while, but I figured it out.


1. If a certain car, going with speed v1, rounds a level curve with a radius R1, it is just on the verge of skidding. If its speed is now doubled, the radius of the tightest curve on the same road that it can round without skidding is:
A) 2R1 B) 4R1 C) R1/2 D) R1/4 E) R1

2. An automobile moves on a level horizontal road in a circle of radius 30 m. The coefficient of friction between tires and road is 0.50. The maximum speed with which this car can round this curve is:
A) 3.0 m/s B) 4.9 m/s C) 9.8 m/s D) 12 m/s E) 13 m/s

3. The driver of a 1000-kg car tries to turn through a circle of radius 100 m on an unbanked curve at a speed of 10 m/s. The actual frictional force between the tires and slippery road is 900 N. The car will:
A) slide into the inside of the curve D) make the turn only if it went faster
B) make the turn E) slide off to the outside of the curve
C) slow down due to the frictional force

4. A giant wheel, 40 m in diameter, is fitted with a cage and platform on which a man can stand. The wheel rotates at such a speed that when the cage is at X (as shown) the force exerted by the man on the platform is equal to his weight. The speed of the man (in m/s) is:
A) 14 B) 20 C) 28 D) 80 E) 120

5. A coin is placed on a horizontal phonograph turntable. Let N be the normal force exerted by the turntable on the coin, f be the frictional force exerted by the turntable on the coin, and fs, max be the maximum force of the static friction. The speed of the turntable is increased in small steps. If the coin does not slide, then
A) N increases, f increases, and fs, max stays the same
B) N increases, f increases, and fs, max increases
C) f increases and both N and fs, max stay the same
D) N, f, and fs, max all stay the same
E) N, f, and fs, max all increase

6. The iron ball shown is being swung in a vertical circle at the end of a 0.7-m string. How slowly, in ft/s, can the ball go through its top position without having the string go slack?
A) 1.3 m/s B) 2.6 m/s C) 3.9 m/s D) 6.9 m/s E) 9.8 m/s

7. An 800-N passenger in a car, presses against the car door with a 200 N force when the car makes a left turn at 13 m/s. The (faulty) door will pop open under a force of 800 N. Of the following, the least speed for which the man is thrown out of the car is:
A) 14 m/s B) 19 m/s C) 20 m/s D) 26 m/s E) 54 m/s

Please don't post the same thing in more than one forum. http://www.artofproblemsolving.com/Forum/viewtopic.php?t=306049
Also, I think most of these problems would go in introductory physics.


Wow... This problem is confusing me like none other---any help would be greatly appreciated.

A gas vessel is attached to an open-end manometer containing a nonvolatile liquid of density $0.791 g/mL$. The difference in heights of the liquid in the two sides of the manometer is $43.4 cm$ when the atmospheric pressure is $755 mmHg$. Given that the density of mercury is $13.6 g/mL$, what is the pressure of the enclosed gas in atm?

I know that $P_{gas} = P_{atm} + P_{height}$, but I don't even know if that would help...

You are using the same book as me!

lol

This is a cool problem from a mathematical perspective.  Crazy formula substitution:

$P = \frac{F}{A} = \frac{m*a}{A} = \frac{d*V*a}{A} = \frac{d*A*h*a}{A} = d*h*a$

Can you do something with this?

Yeah, thanks a ton :).

I'm totally the only one in the class who got this problem lol.


Let $f_{n}(x)=\frac{1}{1+\frac{x^{2}}{n}}e^{-\frac{x^{2}}{n}}$, $f(x)=1$, $h_{n}(x)=f_{n}(x)e^{-x^{2}}$ and $h(x)=e^{-x^{2}}$.
It's obvious that $f_{n}\rightarrow f$ and $h_{n}\rightarrow h$ pointwise.
It's also clear that $f_{n}$ doesn't converge uniformly to $f$ since for each $0<\epsilon<1$ and all $n\in\mathbb{N}$ we can find $x\in\mathbb{R}$ such that $x^{2}>n(\frac{1}{\sqrt{1-\epsilon}}-1)$ and then
$\lvert f_{n}(x)-f(x)\rvert=1-\frac{1}{1+\frac{x^{2}}{n}}e^{-\frac{x^{2}}{n}}>1-\frac{1-\frac{x^{2}}{n}}{1+\frac{x^{2}}{n}}>\epsilon$.

My problem is to find out if $h_{n}$ converges uniformly to $h$ and to show this.

Yes, you are right. The correct estimate should be
$\lvert f_{n}(x)-f(x)\rvert=1-\frac{1}{1+\frac{x^{2}}{n}}e^{-\frac{x^{2}}{n}}\geq 1-\frac{1}{(1+\frac{x^{2}}{n})^{2}}>\epsilon$ when $x^{2}>n(\frac{1}{\sqrt{1-\epsilon}}-1)$
This means that $f_{n}$ doesn't converge uniformly to $f=1$ but $e^{-x^{2}}f_{n}$ converges uniformly to $e^{-x^{2}}.$


Let a and b be positive integers
Find all  a, b, such that number is integer  $ \frac {a^{2} \plus{} b^{2}}{ab \plus{} 1}$ also .


Just wondering whether there was a faster way to do it other than listing them all out and counting.

Yes.

The sum of the factors of $ p_1^{k_1}p_2^{k_2}\cdots$ is $ (1\plus{}p_1\plus{}p_1^2\plus{}\cdots\plus{}p_1^{k_1})(1\plus{}p_2\plus{}p_2^2\plus{}\cdots\plus{}p_2^{k_2})\cdots$.

In this case, $ 24\equal{}2^3\cdot3^1$, so the sum is $ (1\plus{}2\plus{}2^2\plus{}2^3)(1\plus{}3)\equal{}15\cdot4\equal{}\boxed{60}$.

I'm not sure if it's just me, but your method seems as hard to execute as simply adding.

When you practice these kinds of problems you become more efficient. I find it easier to do it math154's way, especially under time constraints.


[quote="arqady"]The first is a very known Vasc's inequality from 1989.
The third inequality ( prove that $ \sum_{cyc}\frac {1}{a^3 \plus{} a}\leq\frac {1}{abc}$ for $ a \plus{} b \plus{} c \equal{} 2,$
where $ a,$ $ b$ and $ c$ are positive numbers ) kills easy enough by expanding and homogenization, but it's very ugly for your beautiful inequality.[/quote]

My friend have a nice solution for the third inequality,can you tell me what's the Vasc's method for the first and how about the second.

[quote="shfdfzhjj"]can you tell me what's the Vasc's method for the first and how about the second.[/quote]
Vasc proved two inequalities:
1) $ 2(1\minus{}a\plus{}a^2)(1\minus{}b\plus{}b^2)\geq1\plus{}a^2b^2$ and
2) $ 3(1\plus{}a^2b^2)(1\minus{}c\plus{}c^2)\geq2(1\plus{}abc\plus{}a^2b^2c^2).$ :wink:


search
number theory unsolved
number theory

Determine all positive integers $x,\, y$ such that $y\, |\, x^{2}+1$ and $x^{2}\,|\, y^{3}+1$.

[quote="harazi"]It is very interesting: this problem has been posted at least 4 times on the forum, it generated discussion every single time and there is always no solution. I really start believing that it is not exactly the statement of the problem given in the contest. It would be nice to know whether this problem really admits an explicit solution or we are just playing around by posting it everywhere and trying to solve it.[/quote]
@harazi:As I remmeber that.Orignially ,some one post the problem as $y|x^{2}+1.x|y^{3}+1$and someone replied that you had a proof of it,(But I haven't found the proof on ML.....)

But,This time ,The problem April gived us is the exactly one one the contest(It is not hard.And I have use google to search  Mediterranean Mathematical Competition 2002,and every document shows that the problem is which April gives.)

These problem equavalent to find integer points in surface $y^{3}+1-(yz-1)x^{2}=0$. I think, there are infinetele many solutions.

Actually this problem is not difficult!
Not if you have seen this [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=6413#6413]IMO problem[/url] before  :wink:

well there is a solution on greek forum, maybe Silouan can translate it http://www.mathlinks.ro/Forum/viewtopic.php?t=137635&start=20

I think I found Albanian Eagle's solution.

Clearly $x^2y|(x^2+1)(y^3+1)$ or $x^2y|(x^2+y^3+1)$.
Thus, there is some $a\in \mathbb Z$ such that $ax^2y=x^2+y^3+1$.
Thus, $x^2=\frac{y^3+1}{ay-1}$.
Using the IMO problem, the pairs $(a,y)$ that yield perfect squares are $(1,2),(3,1),(5,2),(3,5)$.
These yield the $(x,y)$ pairs of $(3,2),(1,1),(1,2),(3,5)$ respectively.


Find all $ f: \mathbb{R}\rightarrow\mathbb{R}$ such that
$ f(x\plus{}f(xy))\equal{}f(x)\plus{}xf(y)$
is true for all $ x,y\in\mathbb{R}$.

Would this be a good approach: the function should have the following properties:

$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$

$ f(ax)\equal{}a f(x)$

$ f(1)\equal{}1$

and the only function satisfying this is:

$ f(x)\equal{}x$

this being the solution to the problem.

[quote="milin"]Would this be a good approach: the function should have the following properties:

$ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$

.[/quote]

 could you explain ?

Well, I could try.. By adding these three properties, the left side of the equation can be transformed into the right side, so I think these are the properties the function we're looking at must have. Because, if it doesn't, the equation shouldn't hold. The first one is used to transform

$ f(x \plus{} f(xy)) \rightarrow f(x) \plus{} f(f(xy))$

The second one :

$ f(f(xy)) \rightarrow y x f(1)^2$

This gives us:

$ x f(1) \plus{} x y f(1)^2 \equal{} x f(1) \plus{} x y f(1)$

and therefore:

$ f(1) \equal{} 0$

or

$ f(1) \equal{} 1$

I ignored the trivial solution $ f(x) \equal{} 0$, and the solution is $ f(x) \equal{} x$.

hmm...
I couldn't find $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ but solved.
I found that $ f(x)$ can represent all the real numbers.
I will just show how I solved very briefly....(Sorry for that I am not good at using LATEX :blush: )

Step1)$ f(0) \equal{} 0$
Step2)$ f(1) \equal{} 0 or 1$
Step3)if $ f(1) \equal{} 0$, $ f(x) \equal{} 0$ for all $ x$
Step4)if $ f(1) \equal{} 1$,$ f(x) \equal{} 0$ is equivalent to $ x \equal{} 0$ and $ f( \minus{} 1) \equal{} \minus{} 1$ and $ f(f(x) \plus{} 1) \equal{} f(x) \plus{} 1$
Step5)$ f(x) \equal{} \minus{} f( \minus{} x)$
Step6)$ f(\dfrac{x}{f(x)}) \equal{} \dfrac{f(x)}{x}$
Step7)$ f(\dfrac{x^2}{f(x)}) \equal{} x$

combining Step7 and $ f(f(x) \plus{} 1) \equal{} f(x) \plus{} 1$, we can conclude $ f(x) \equal{} x$ for all $ x$ or $ f(x) \equal{} 0$ for all $ x$

Well....I found my solution was wrong :wallbash:


A primitive Pythagorean triple is a set of three numbers which satisfies the equation a:^2:+b:^2:=c:^2: and is not multiples of other Pythagorean triples. The formula for finding a primitive Pythagorean triple is plugging in an integer above 1 for m and an integer less than m but more than 0 for n into m:^2:+n:^2:, 2mn, and m:^2:-n:^2:. Show why this formula works.

Example: Plugging in 2 for m and 1 for n (the least numbers possible) gives us 5, 4, and 3, which works because 3:^2:+4:^2:=9+16=25=5:^2:.

If m>n, or n>m the proof  follows. It is easy enough to know that (n^2-m^2)^2 + (2mn)^2= n^4 -(2n^2m^2) +m^2 +(4n^2m^2) = n^4 +(2n^2m^2) +m^4 = (n^2 + m^2)^2. However, the entire proof justifying every triple involves trigonometry and is complicated. Maybe too complex for MATHCOUNTS...  just maybe(there are exceptions).  :roll:

[quote]If home is where the heart is and my heart is in this universe, what are you doing in my house?[/quote]

Actually, the proof that that formula represents all triples doesn't require any  trigonometry as I remember it -- it is rather complex, however.


function
logarithms
calculus
calculus computations

Let a, b be the constants and f(x) be a continuous function  and differentiable $ \forall x$ satisfy
$ \{  \begin{array}{l} af(x) = f'(x) \\ f(0) = b \end{array}  $
Find $ f(x)$

Write $ y \equal{} f(x)$ and $ y' \equal{} f(x)' \equal{} \frac{dy}{dx}$ so that $ af(x) \equal{} f'(x)$ can be written as $ \frac{dy}{dx} \equal{} ay$, which is separable and can be solved to give $ f(x) \equal{} be^{ax}$.


For $ a,b,c$ positive reals prove that

 $ \sum_{cyc}\frac {ab\plus{}ac\plus{}4bc}{b^2\plus{}c^2} \geq 4$

P.S. Sorry, but I don't remember if it was already posted

[quote="manlio"]For $ a,b,c$ positive reals prove that

 $ \sum_{cyc}\frac {ab \plus{} ac \plus{} 4bc}{b^2 \plus{} c^2} \geq 4$

P.S. Sorry, but I don't remember if it was already posted[/quote]
$ \sum_{cyc}\frac {ab \plus{} ac \plus{} 4bc}{b^2 \plus{} c^2} \geq 4\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}ab(a\minus{}b)^4\plus{}\sum_{sym}\left(2a^4bc\plus{}6a^3b^2c\minus{}\frac{4a^2b^2c^2}{3}\right)\geq0.$


integration
trigonometry
calculus
calculus computations

$ Find \int\sqrt {1 \minus{} cos x}\ dx$

consider that

$ \sin\left(\frac{x}{2}\right)=\pm\sqrt{\frac{1-\cos(x)}{2}}$

hello, note that $ \cos(x)\equal{}1\minus{}2\sin^2(\frac{x}{2})$.
Sonnhard.

One thing to be careful of: sign.

If you accept uncritically the identity that Mathstud28 wrote and use it to integrate this from $ 0$ to $ 4\pi,$ you would get zero - and that makes no sense.

What we should say is this:

$ \sqrt{1\minus{}\cos x}\equal{}\sqrt{2}\left|\sin\left(\frac{x}2\right)\right|.$

Hence we can't write the antiderivative $ \int\sqrt{1\minus{}\cos x}\,dx$ with a single formula, but will have to break it up into a piecewise-defined function.

$ \int_0^{4\pi}\sqrt{1\minus{}\cos x}\,dx\equal{}\int_0^{2\pi}\sqrt{2}\sin\left(\frac{x}2\right)\,dx \plus{}\int_{2\pi}^{4\pi}\minus{}\sqrt{2}\sin\left(\frac{x}2\right)\,dx$

$ \equal{}4\sqrt{2}\plus{}4\sqrt{2}\equal{}8\sqrt{2}.$

$ \int \sqrt {1 \minus{} \cos x}\ dx \equal{} \int \frac {\sqrt {\sin ^ 2 x}}{\sqrt {1 \plus{} \cos x}}\ dx$

$ \equal{} \pm \int \frac {\sin x}{\sqrt {1 \plus{} \cos x}}\ dx$

$ \equal{} \pm 2\sqrt {1 \plus{} \cos x} \plus{} C$.

or remark $ 1 \minus{} \cos x \equal{} \left(\cos \frac {x}{2} \minus{} \sin \frac {x}{2}\right)^2$.

hello, i have $ \minus{}\frac{\sqrt{2}\sin(x)}{\sqrt{1\minus{}\cos(\frac{1}{2}x)^2}}$.
Sonnhard.

$ \int \sqrt{1\minus{}\cos x}dx$
$ sin^2x \equal{} \frac{1 \minus{} cos2x}{2}$
$ 2\ sin^2 \frac{x}{2} \equal{} 1 \minus{} cos\ x$

$ \int \sqrt{1\minus{}\cos x}dx \equal{} \int \sqrt{2\ sin^2 \frac{x}{2}} dx$
$ \equal{}\sqrt2 \int sin^2 \frac{x}{2})^ \frac {1}{2} \ dx \equal{} \sqrt2 \int sin\frac{x}{2} \ dx$
$ \equal{} \sqrt2 \ . \ 2(\minus{}cos \frac{x}{2}) \plus{} c$
$ \equal{} \minus{}2 \sqrt2 \ cos \frac{x}{2} \plus{} c$

Check my warning about sign. It's not quite true to say that $ \sqrt{\sin^2\left(\frac{x}2\right)}\equal{}\sin\left(\frac{x}2\right).$ Instead, you should say that $ \sqrt{\sin^2\left(\frac{x}2\right)}\equal{} \left|\sin\left(\frac{x}2\right)\right|.$

And that means that the integral must be done in pieces.

[hide]That's the answer my teacher gave us!!![/hide]


number theory
greatest common divisor
least common multiple

Consider the sequence: $x_1=19,x_2=95,x_{n+2}=\text{lcm} (x_{n+1},x_n)+x_n$, for $n>1$, where $\text{lcm} (a,b)$ means the least common multiple of $a$ and $b$. Find the greatest common divisor of $x_{1995}$ and $x_{1996}$.

[hide="Solution"](using $ (a,b)$ for GCD and $ [a,b]$ for LCM)

If $ a$ and $ b$ are two natural numbers and $ c$ = $ [a,b]\plus{}a$, then clearly $ (a,b) \equal{} (b,c)$ by the Euclidean algorithm. By this fact, the GCD of any two adjacent terms in the sequence is equal to $ (x_{1},x_{2})$, which is $ 19$.[/hide]

Something's wrong in that hint. $ x_{3}\equal{} 114$ and $ x_{4}\equal{} 665$.

I think when you substituted the $ x_{n\plus{}1}$ in the LCM, you forgot to add $ x_{n\minus{}1}$ inside of it. That changes a lot.


The Inconvenient Pacho Bike Shop sells bicycles for 100 dollars. If customers are only allowed to pay with 7 dollar and 17 dollar bills, what is the least amount that a customer must overpay to buy 2 bicycles if no change is given?

[quote="ProtestanT"][quote="i_like_pie"][hide]The least possible $ n$ for which the Diophantine $ 17a \plus{} 7b \equal{} n$ has solutions is $ n \equal{} 201 \equal{} 11\cdot17 \plus{} 2\cdot7$. Answer: $ \boxed{\$1}$.[/hide][/quote]
How did you figure that out? (the least possible $ n$ part)[/quote]

Plug and Chug..(Trial and error)?
I mean you can kind of know what to expect...

You know that atleast one of the number $ (200,201,...206$ will work,  because you can express atleast one just by 7s. It's not too difficult to decide which doesn't by some modular arithmetic.

Come to think of it, doesn't 14 7 dollar bills and 6 17 dollar bills make 200 dollars?
 :maybe:
btw, I can't find the answer key to the problems I posted, so I don't have the absolute correct answers.

It seems you are correct. I forgot to check using the "Chicken McNugget Theorem": $ 17\cdot7\minus{}17\minus{}7\equal{}95$, so all integers $ >95$ can be expressed as $ 17a\plus{}7b$ for some $ a$ and $ b$.

[quote="i_like_pie"]It seems you are correct. I forgot to check using the "Chicken McNugget Theorem": $ 17\cdot7 \minus{} 17 \minus{} 7 \equal{} 95$, so all integers $ > 95$ can be expressed as $ 17a \plus{} 7b$ for some $ a$ and $ b$.[/quote]
ah, yes, I didn't even think of using Chicken McNugget.  :blush:

Guys
Are we missing out on somthin here coz Im gettin answer as $0

17a + 7b = 200
ie. 17*6 + 7*14 = 102+98 = 200

Cheerz


A ball is projected from the ground at an angle 45 degrees with the horizontal at a velocity of 20*sqrt(2) m/s. A fixed inclined plane also inclined at 45 degrees with the horizontal is situated at 20 m from the point of projection of the ball.
FIND the maximum height attained by the ball and its displacement at the time of third collision (All collisions are elastic).

Max height = 20 m
Displacement after third collision with respect to initial position = 0

[hide]
Let us fix our origin at the point of projection, with the ground level as our x-axis and the normal to it at the origin, our y-axis.

Since the projectile will follow a 2-dimensional trajectory throughout, it is sufficient if we consider the incline to be a straight line ( a long inclined plank perhaps). Its equation will be $ y\equal{}x\minus{}20$. The particle will follow a parabolic path and its position vector after a time $ t$ will be given by (can be easily proved) $ \vec s \equal{} (u\cos \theta t) \vec i \plus{} (u \sin \theta t\minus{} \frac 12 gt^2)\vec j$, where $ \vec i$ and $ \vec j$ are unit vectors along the x and  y axis respectively, $ \theta$ is the angle of projection and $ u$ the intial velocity. Hece its equation (after putting the values) is $ y\equal{}x\minus{}\frac {x^2}{80}$

The max height is 20m ($ \equal{} \frac {u^2 \sin ^2 \theta}{2g}$) and you can check that the 2 curves meet at $ (40m,20m)$.

The projectile will strike the incline at an angle of $ 45^0$ with respect to the normal drawn to the incline at the point of contact. Since the collision is perfectly elastic it will rebound at the same angle with same spped. It will move straight up into the air and return back to hit the incline again at the same point and once again, follow a parabolic path to collide with the ground at the point of projection.[/hide]


inequalities
function
algebra
polynomial
induction
inequalities unsolved

[hide="Hint"]$ f(x)\equal{}(1\minus{}x)(1\minus{}x_1)(1\minus{}x_2)\cdot ...\cdot (1\minus{}x_{n\minus{}1})\plus{}x_1\plus{}x_2\plus{}...\plus{}x_{n\minus{}1}\plus{}x\minus{}1$ is a linear function.[/hide]

your solution is really good, but I am not familier with linear function.

A linear function is a one degree polynomial, e.g. of the form $ ax\plus{}b$. As the function is linear in each of the variables it's extremal values occur at the endpoints of the interval $ [0,1]$. Since $ f(1)\ge f(0)$, the minimum occurs when all the $ x_i$ are $ 0$, and then the minimum is $ 0$.

I think we can use Cauchy induction for this.
We can assume $ x_1 \plus{} x_2 \plus{} ...x_n \le 1$ or the right side is negative.
The given is obviously true for $ n \equal{} 2$ as
$ (1 \minus{} x)(1 \minus{} y) \ge 1 \minus{} x \minus{} y \iff xy \ge 0$.
Now, suppose the inequality is true for $ n \equal{} 2^k$. We have
$ (1 \minus{} x_1)(1 \minus{} x_2)...(1 \minus{} x_n) \ge 1 \minus{} (x_1 \plus{} x_2... \plus{} x_n)$
$ (1 \minus{} y_1)(1 \minus{} y_2)...(1 \minus{} y_n) \ge 1 \minus{} (y_1 \plus{} y_2... \plus{} y_n)$
Multiplying them,
$ (1 \minus{} x_1)(1 \minus{} x_2)...(1 \minus{} x_n)(1 \minus{} y_1)(1 \minus{} y_2)...(1 \minus{} y_n) \ge [1 \minus{} (x_1 \plus{} x_2... \plus{} x_n)][1 \minus{} (y_1 \plus{} y_2... \plus{} y_n)] \ge 1 \minus{} (x_1 \plus{} x_2... \plus{} x_n \plus{} y_1 \plus{} y_2... \plus{} y_n)$.
So the given is also true for $ n \equal{} 2^{k \plus{} 1}$, which completes the induction.
But for any number m not equal to a power ot two, we can just take the inequality for a power of two greater than m and set all but m terms equal to 0, to get the inequality for $ n \equal{} m$.

Of course, the linear function way is more elegant and faster.

Use simply induction. For n=1 we have the equality. Now suppose that
$ (1\minus{}x_1)(1\minus{}x_2)\cdots (1\minus{}x_n)\ge 1\minus{}(x_1\plus{}x_2\plus{}\cdots \plus{}x_n).$
Then
$ (1\minus{}x_1)(1\minus{}x_2)\cdots (1\minus{}x_n)(1\minus{}x_{n\plus{}1})\ge (1\minus{}(x_1\plus{}x_2\plus{}\cdots \plus{}x_n))(1\minus{}x_{n\plus{}1})\ge 1\minus{}(x_1\plus{}x_2\plus{}\cdots \plus{}x_{n\plus{}1}).$

Having already had Cauchy induction in my mind, I used it right away and made it more complicated that it had to be.  :lol:


Because most game developers don't bother making a mac version since almost everyone has a PC and Mac's have horrible compatibility issues.
Just like Vista.
Oh and because computers suck.
You should play games the old fashioned way.
throw stones at other stones.

It is because... Steve Jobs wants you to pay more and more and more!
Although things might get "better and better and better", your money will get "worse and worse and worse".
:P

[youtube]Id_kGL3M5Cg[/youtube]

LMAO. Sorry this thread reminded me of this video.

It's because Windows has been around longer and is "cheaper". Macs are less expensive in the long run, but some people don't see that. When Windows has an upgrade(Vista), you might need a whole new computer just to run it. For Macs, that is not the case, even though it's more expensive from the start. Also, when you buy a Mac, everything is there, so the cost is raised a hair. For Windows, you have to buy a few more boxes.

And my Mac has not crashed yet.

The Mac Leopard, every version was 127 dollars. They expected people to buy the Ultimate version.

My dad has leopard, I have it, and mom has it.

And more games are coming.

Still, I don't buy extra software, or Vista.
I'll probably start using macs soon if Windows doesn't start fixing the major compatibility issues they have with Vista.

Look kiddo,

A mac pro can't game - doesn't have a good processor. Intel Xeon is for servers, not computers :roll:

As for the iMac, go eat yourself because no upgrades possible without calling a guy to fix it. Uses laptop parts - bah.

Try getting a mac for 700 dolalrs that matches:

(I got these all for $ \$$700):

[list]
[*]Dual 19" Monitors
[*]e-GeForce 8800 GTX 768 MB
[*]4 GB DDR2 Corsair Dominator Memory
[*]Intel Core 2 Quad Processor @ 2.4 GHz
[*]Xion Gaming Case Professional with LightsFX
[*]eVGA 780i Tri-SLI motherboard, X38 chipset[/list]

Try getting that on a mac.

Nope, this computer will run Windows stuff for years to come. I want to see Leopard run on a crapmac from 5 years back. I was able to run Vista pretty smoothly on my sister's 5 year old desktop. We eventually reverted back to XP because Vista was worse than XP at gaming, but still...

Ever wonder how I got this cheap? Buying parts from friends, http://www.microcenter.com, ebay.

Can't do it with a mac. Why? Apple sells 'em. Overclocking? Nope, no one ever mentions it because its just not possible. BIOS on a mac? Too stripped to do anything.

[quote="#H34N1"]Look kiddo,

A mac pro can't game - doesn't have a good processor. Intel Xeon is for servers, not computers :roll: [/quote]

*goes and plays Worms 3D*

WHOA NICE!
INSPIRON FOR $ \$400$!
CRAZY
Wait, unless something is wrong and the desktop's price did drop by a lot.

People should not buy cheap computers. Its called building one.

@1=2: Good juck playing the game from 1900, while I play Crysis - the hot game of the decade.

Besides, even though Macs have a nice pimped out GUI, even though they might have some PC emulator, there will be SO MUCH spent on the graphics and emulator you can't play any game. Also, most PCs are faster than macs. Now suck up your drool ( I don't want your favorite rig wet) and click at least one of these links

[url]http://www.vigorgaming.com/product/rigs-colossus.html[/url]
[url]http://www.vigorgaming.com/product/rigs-forcerecon.html[/url]
[url]http://www.vigorgaming.com/product/rigs-deltaforce.html[/url]
[url]http://www.vigorgaming.com/product/rigs-hornet.html[/url]
[url]http://www.vigorgaming.com/product/rigs-stealth.html[/url]

You will never see a Mac tower with a clear case. Besides, Jobs is too lazy to stick a gaming-grade quad-core processor inside a computer. ( I want the computer on the first link. Fastest gaming rig on market.) However, even with the extremely powerful computer, you will still have frame rate problems. Oh, and have I mentioned- Mac keyboards and mice are crud. I use a Fata1ty Gaming Keyboard and a pretty good mouse.

BEAST DUDE

SPECIAL PROMOTION 1  	Free Shipping
SPECIAL PROMOTION 2 	Free Logitech G11 USB Gaming Keyboard
PROCESSOR 1 	Intel® Core™2 Extreme Quad-Core Processor QX9775 at 3.20GHz, 1600MHz FSB, 2X6M Cache
PROCESSOR 2 	Intel® Core™2 Extreme Quad-Core Processor QX9775 at 3.20GHz, 1600MHz FSB, 2X6M Cache
RIG 	Vigor Cooler Master CM Stacker 830 Screwless Aluminum Chassis
RIG COLOR 	Black
RIG STYLE 	Metal Mesh Side Window for Improved Ventilation
RIG COOLING 	Vigor Twin-Peaks X 2 Powered by Cooler Master Cooling System
RIG POWER 	Enermax Galaxy nVidia-SLI/CrossFire Certified 1000Watt Power Supply
RIG DIMENSIONS 	9.20" Width X 25.20" Depth X 21.10" Height.
MB CORE LOGIC 	Intel D5400XS Intel 5400 FB-DIMM (DDR2) w/1600MHz FSB, Nvidia Quad SLI and Quad CrossFireX Ready
MEMORY 	4GB Kingston KVR800D2D8F5K2/2G X 2 CL5 DDR2 800MHz Memory (4 X 1024MB)
OPERATING HARD DRIVE 1 	WD Raptor 74GB WD RAPTOR S-ATA 10000RPM 16MB Cache Hard Drive
OPERATING HARD DRIVE 2 	WD Raptor 74GB WD RAPTOR S-ATA 10000RPM 16MB Cache Hard Drive
RAID SETTING 	RAID 0 Performance - Striped Drives (2 or More Identical Hard Drives Only)
DATA HARD DRIVE 1 	1TB (1000GB) S-ATA II 7200 RPM Hard Drive
DATA HARD DRIVE 2 	1TB (1000GB) S-ATA II 7200 RPM Hard Drive
RAID SETTING 	RAID 0 Performance - Striped Drives (2 or More Identical Hard Drives Only)
VIDEO CARD CROSSFIRE/SLI SLOT 1 	nVidia GeForce 8800 Ultra 768MB Xtreme Edition @650MHz, DVI and TV Out
VIDEO CARD CROSSFIRE/SLI SLOT 2 	nVidia GeForce 8800 Ultra 768MB Xtreme Edition @650MHz, DVI and TV Out
VIDEO CARD CROSSFIRE/SLI SLOT 3 	nVidia GeForce 8800 Ultra 768MB Xtreme Edition @650MHz, DVI and TV Out
OPTICAL DRIVE 1 	ASUS Internal SATA 2014BLT 20X LIGHTSCRIBE DVD+/- RW + CDRW Rewritable Drive
OPTICAL DRIVE 2 	SONY Internal BWU100A Blu-ray Disc Drive
RECORDING SOFTWARE 	Bundled DVD/CDRW Software
SOUND 	Digital High Definition 3D 7.1 8-Channel Sound
NETWORK PORT 	Onboard Gigabit (10/100/1000Mbps) PCI Network Card
WIRELESS NETWORK PORT 	None.
I/O PORTS 	1 Parallel, 1 Game/Midi, 1 IEEE 1394 Firewire and 6 to 10 USB ports (Varies by Motherboards)
MONITOR 	None.
SPEAKERS 	None.
KEYBOARD 	Standard Black Internet Multimedia Keyboard
MOUSE 	Standard Black Optical Wheel Mouse
FLOPPY DRIVE 	1.44MB 3.5" Floppy Drive
OPERATING SYSTEM 	Microsoft Windows Vista Ultimate 32-bit w/Original DVD
SERVICE 	Standard 3 Year Limited Parts and Life-Time Labor Warranty
STORAGE MEDIA 	None.
SOFTWARE BUNDLE 	None.
ADDITIONAL SOFTWARE 	None.
Price: 	$9250.00

Well, I'm not paying 10K when its gonna be cheaper next year =)

Besides, my Q6600 is fine.

If you really want to play the latest games on a Mac- just buy a copy of a Microsoft OS, and upgrade a few things like the graphics card if necessary. The newer systems have built-in support for dual booting.
If you're not a fanatic and are willing to stick to a shorter list of games and play them longer, any machine is decent for gaming. Popular games do get ported, and compatibility is no big deal due to the shorter list of Mac systems to test. The only problem I have with my five-year-old games is the laptop keyboard.

For all of the "awesome" features of that machine- I wouldn't want one. Why? No wireless network access, and no portability. The first is an absolute dealbreaker (I have no access to wired Internet), and the second is a serious annoyance.
I also prefer the Mac OS, but that's not really relevant to what I need.

You cannot get a Mac that is stripped down and specialized to one purpose like that. You can get a high-end Mac and upgrade it to run pretty much anything well, while also being an excellent computer for all other purposes.

...because Apple couldn't, never did, and most likely will never be able to (so long as Bill Gates's legacy is carried on :) ) encourage developer support the way Microsoft did.  One of the things that led to Microsoft's success is one of the reasons that there aren't the same games for Macs.  Also, like #H34N1 said, they try to give you a computer with all the latest hardware, without worrying about what the hardware was really made for.  They should categorize their products; ex. gaming, office, home, media...but they don't.

If you haven't noticed, a Q6600 STOCK beats a Xeon Woodcrest Quad 3.2 GHz on any gaming benchmark. 3dmark06 also beats. it. The only thing where the Xeon wins is in Premiere's MPEG output benchmark.

Workstation graphics cards are exactly the same hardwarewise as normal video cards, just with a software tweak that quadruples the price :roll:
But yea macs have serious issues. The OS is a bit annoying but I guess it has some good things, but it really pisses me off how they have those sleek cases which are good for looks, and horrible for cooling. The mice they give you have just one clicker for simplicity, but in a world where applications simply cannot be use without right click :mad: The video that comes with them is usually some low end card that is custom built into the motherboard just for them.
BF2142 owns Crysis, even if I haven't played Crysis :P

Dude im in Middle School.
No one I know has even actually seen what the hell is in a computer before.
They just smack their computers and call some repair company when something happens.

Meh, it would make me feel kinda guilty seeing as I got this horrible labtop only last summer.
horrible and expensive, a deadly combo.
It cost around 10000 RMB, which is around... $ \$1300$

Many people actually replace their CPU every 3-4 months, but this is impractical for a lot of other people. However, replacing a whole computer that often is expensive.

On the other hand, the reasoning behind getting a bad $1300 laptop is beyond me...you can get good computers for around$1000.

It's bad in American standards.
In China its like an engineering marvel.
High tech.

Well then the question is: Why does a chinese computer cost 1300 in the first place?

Because new technology in China costs a lot.
Seeing as it's new to them.

I'm surprised by that, given that the counterfeiters there could not copy a functional Intel Core 2XT or a AMD PHENOM-64. Besides, the CORE 2EXTREME only costs about USD 2100 and an AMD PHENOM-64 costs a little under 1 grand.  Besides, SV and the LA basin ( in CA) are the most technologically advanced places IN THE WORLD [hide]if there are no aliens that are better than us.[/hide]

Yeah, but you said WORLD, and we probably would've discovered aliens by now in THE WORLD (or maybe not because we're too concentrated on the Middle East). Anyways, NEVER buy an extreme core, just go and buy a quad-core from intel. Same specs, just that the extreme core is a TON more expensive


analytic geometry
geometry
conics
hyperbola
trigonometry
quadratics
ellipse

where  can   we   post  problems of coordinate geometry?

[quote="acidpop"]hey! same problem here....... but all the same i'm posting a problem right here......

[u]problem[/u]
a line drawn thru the point P(-1,2) meets the hyperbola xy=c^2 at the points A and B.(points A and B lie on the same side of point P) A point Q is chosen on this line such that PA, PQ and PB are in arithematic progression. then the equation of locus of point Q is.......?

i got the answer as 2x=y(1+2x).......dunno if tis' right though...please help!! :([/quote]

sorry  i am of high  school standard and i dont know hyperbola concept but i have a
problem in coordinate geometry and dont know where to post

Let the equation be $y-2 = m(x+ 1)$
 solving this with the hyperbola, we get,
 $\frac{c^2}{x} - 2 = m(x+ 1)$
 or $c^2 - 2x = mx^2 + mx$
 and $mx^2 + (m+ 2)x - c^2 = 0$
 the roots of this equation are the $x-coordinates$ of
 the points of intersection of the line and the hyperbola.
 Our required point $(h, k)$ is the mid-point of these points.
 thus $h = \frac{x_1 + x_2}{2}$
 and $k = \frac{y_1 + y_2}{2}$
 from theory of equations,
 we have $x_1 + x_2 = -(m+2)/m$
 or $h = -(m+2)/2m$
 and $y_1 + y_2 = c^2 (\frac{1}{x_1} + \frac{1}{x_2})$
 which is basically =  $c^2 ( \frac{x_1 + x_2}{x_1x_2})$
 $= c^2 (\frac{-(m+2)/m}{-c^2/m})$
 so $k$ = $(y_1 + y_2)/2 = (m +2)/2$
 or $m = 2k -2$
 substituting this in the first equation, we get,
 $h = -2k/ 2(2k -2)$
 or,
 $2kh -2h + k =0$
 or basically,
 $2h = 2kh + k$
 i.e. $2x = 2xy + y$
 $2x = y(2x + 1)$
 and your answer is correct.
 does anybody have a shorter method?
 As for the co-geo questions try the Geometry Forum in High School or olympiad

hey thanks madatmath!!!!!
so i got it right heh??!! i'm happy!!!!!
 :D  :D

how about keeping this as a co-geo thread?
 somebody else can go ahead and post another question?
 no problem, piece of cake!  :lol:

[quote="madatmath"]
 $2x = y(2x + 1)$
 and your answer is correct.
 does anybody have a shorter method?
 As for the co-geo questions try the Geometry Forum in High School or olympiad[/quote]
Let any point on the line through P is given by $( - 1 + r\cos \theta ,\,2 + r\sin \theta )$.  If it lies on the given hyperbola, we have $( - 1 + r\cos \theta )(2 + r\sin \theta ) = c^2$, which is a quadratic equation in 'r', the distance of the intersection point to P.
Clearly PA and PB are the roots of this equation.  Hence $PA + PB = - \frac{{2\cos \theta - \sin \theta }}{{\cos \theta \sin \theta }}$.  Given PA, PQ and PB are in A.P.
So, $PQ = - \frac{{2\cos \theta - \sin \theta }}{{2\cos \theta \sin \theta }}$
If $Q(h,\,k)$ is point whose locus is to be found, we have $(h,\,k)=( - 1 + PQ\cos \theta ,\,2 + PQ\sin \theta )$.
i.e., $h = - 1 - \frac{{2\cos \theta - \sin \theta }}{{2\cos \theta \sin \theta }}\cos \theta = - \frac{1}{2} - \cot \theta$ and $k = 2 - \frac{{2\cos \theta - \sin \theta }}{{2\cos \theta \sin \theta }}\sin \theta = - \frac{1}{2}\tan \theta$.

Eliminating $\theta$, we get the answer  $2x = y(2x + 1)$.

[quote="madatmath"]nice , very nice.
 it comes out neatly here,
 but parametric forms often create a
 mess in most questions.[/quote]

ok another question

the locus of a point such that the sum of its distances from 2 fixed points(c,0);(-c,0) is 2a units is

a)$x^2$/$a^2$+$y^2$/{$a^2$-$c^2$}=1
b)$x^2$/$a^2$+$y^2$/{$a^2$+$c^2$}=1
c)$x^2$/$a^2$+$y^2$/$c^2$=1
d)$x^2$/$c^2$+$y^2$/$a^2$=1

[quote="madatmath"]It's an ellipse.

 we have $ae = c$
 $b^2 = a^2 - (ae)^2$
  $= a^2 - c^2$

 the reqd. eqn is

 $\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$

 Option $A$[/quote]

how did you know that its an ellipse
is it not a circlewith (0,0) as centre and
what is e here

The definition of an ellipse is the locus of a point
 such that the sum of it's distances from two fixed points is
 constant.  this constant is twice the major axis length, $a$
 $e$ is the eccentricity of the ellipse. [url=http://mathworld.wolfram.com/search/]Eccenticity.[/url]
 check this link.


calculus
integration
trigonometry
function
calculus computations

Calculate

\[\int^{\frac{\pi}{3}}_0 \frac{1}{1+\sin x} dx\]

When I teach the "techniques of integration" part of a calculus class, I like to say that there are only three families of methods, the the order of importance (the order in which you should consider them) goes like this:

1. Algebraic manipulation.
2. Substitution. This includes "natural" substitution, and more forced versions such as (for example) trigonometric substitutions.
3. Integration by parts.

Furthermore, every trigonometric function should be seen as a square root of something. From that point of view, the function in this problem has a square root in the denominator. That brings to the fore a well-known algebraic technique:

$\frac1{1+\sin x}\cdot \frac{1-\sin x}{1-\sin x}=\frac{1-\sin x}{\cos^2x} = \sec^2x-\sec x\tan x.$


geometry
circumcircle
parallelogram
3D geometry
tetrahedron
geometric transformation
homothety

Let $K$ and $M$ be points on the side $AB$ of a triangle $\triangle{ABC}$, and let $L$ and $N$ be points on the side $AC$. The point $K$ is between $M$ and $B$, and the point $L$ is between $N$ and $C$. If $\frac{BK}{KM}=\frac{CL}{LN}$, then prove that the orthocentres of the triangles $\triangle{ABC}$, $\triangle{AKL}$ and $\triangle{AMN}$ lie on one line.

let circumcircle of $\triangle AMN,\triangle AKL$ intersect at $A'\not \equiv A$ it's easy to see that a spiral similarity with center $A'$ ,$MK$ will be $NL$ and caues $\frac{MK}{NL}=\frac{KB}{LC}$ then $KB$ will be $NC$.
hence circumcircle of $\triangle ABC$ will pass through $A'$. it's clear that midpoints of $MN,KL,BC$ are collinear.let $O_1,O_2,O$ be circumcenters of triangle $AMN,AKL,ABC$ respectively.we have in any triangle that $\frac{HG}{OG}=2$ hence it's easy to see (not obvious) that orthocenter of $AMN,ALK,ABC$ are collinear.

[quote="Amir.S"]we have in any triangle that $\frac{HG}{OG}=2$ hence it's easy to see (not obvious) that orthocenter of $AMN,ALK,ABC$ are collinear.[/quote]

Why is it easy to see? I can't see!

We can solve this problem, based on the follows simple idea.

[b][size=109]LEMMA.[/size][/b] – Two lines $(\ell),$ $(\ell'),$ are given and we define three arbitrary points $A,$ $B,$ $C,$ on $(\ell)$ $($ $B,$ between $A,$ $C,$ $)$ and also three points $A',$ $B',$ $C',$ on $(\ell'),$ such that $AB / BC = A'B' / B'C'$ $($ $B',$ between $A',$ $C'$ $).$

Through the points $A,$ $B,$ $C,$ we draw three arbitrary lines $(a),$ $(b),$ $(c),$ parallel each other and also through the points $A',$ $B',$ $C',$ we draw three other ones $(a'),$ $(b'),$ $(c').$ If the directions of these two sets of parallel lines, intersect each other, then the points $K\equiv (a)\cap (a'),$ $L\equiv (b)\cap (b'),$ $M\equiv (c)\cap (c'),$ are collinear.

We can easy to prove this lemma, by [b]Thales theorem[/b].

In your configuration now, we denote as $B',$ $K',$ $M',$ the orthogonal projections of $B,$ $K,$ $M$ respectively, on the sideline $AC$ and also as $C',$ $L',$ $N',$ the ones, of $C,$ $L,$ $N,$ on the sideline $AB.$ So, because of $BB'\parallel KK'\parallel MM'$ and $CC'\parallel LL'\parallel NN',$ we have that  the points $H\equiv BB'\cap CC',$ $H'\equiv KK'\cap LL'$ and $H''\equiv MM'\cap NN',$ as the orthocenters of the triangles $\bigtriangleup ABC,$ $\bigtriangleup AKL,$ $\bigtriangleup AMN,$ based on the above lemma, are collinear and the proof is completed.

Kostas vittas.

PS. This proof is dedicated to [b]Silouanos Brazitikos.[/b]

[quote="Amir.S"] that orthocenter of $AMN,ALK,ABC$ are collinear.[/quote]
I think you haven't been able to claim that yet.

[quote="radio"]Okay, vittasko, I agree that your proof is very nice.

However, I still want to see how to finish the solution in Amir.S's way.[/quote]

I think that the [b]Amir.S[/b]’s solution is correct. Although it is a difficult way to solve this problem, however it is based on the below powerful theorem, very useful to all of us in the future.

[b][size=109]THEOREM.[/size][/b] – A quadrilateral $ABCD$ is given and let $E,$ $F,$ two points on $AB,$ $CD$ respectively $($ $E,$ between $A,$ $B$ and $F,$ between $C,$ $D$ $)$, such that $AE / EB = DF / FC = p.$

We define three points $K,$ $L,$ $M,$ on the segments $AD,$ $EF,$ $BC$ respectively, such that $AK / KD = EL / LF = BM / MC = q.$ Prove that the points $K,$ $L,$ $M$ are collinear, such that $KL / LM = p.$

[b][size=109]REMARK.[/size][/b] – This theorem is true, when the given quadrilateral $ABCD$ is convex, or non-convex and also when it's vertices $A,$ $B,$ $C,$ $D,$ are not in the same plane.

Kostas Vittas.

PS. I have been used the above theorem, for the solution of a problem presented by [b]Tiks[/b], in this forum. You can see at http://www.Mathlinks.ro/Forum/viewtopic.php?t=107395

As I have been promised, I will post here next time, the two elementary proofs of this theorem, I have in mind.

Extremely nice solution Mr Vittas .Thank you for the dedication .

[quote="vittasko"]

[b][size=109]THEOREM.[/size][/b] – A quadrilateral $ABCD$ is given and let $E,$ $F,$ two points on $AB,$ $CD$ respectively $($ $E,$ between $A,$ $B$ and $F,$ between $C,$ $D$ $)$, such that $AE / EB = DF / FC = p.$[/quote]

Okay. Please prove it.

[b][size=109]THEOREM.[/size][/b] – A quadrilateral $ABCD$ is given and let $E,$ $F$ be, two points on $AB,$ $CD$ respectively $($ $E,$ between $A,$ $B$ and $F,$ between $C,$ $D$ $)$, such that $AE / EB = DF / FC = p.$
We define three points $K,$ $L,$ $M,$ on the segments $AD,$ $EF,$ $BC$ respectively, such that $AK / KD = EL / LF = BM / MC = q.$ Prove that the points $K,$ $L,$ $M$ are collinear, such that $KL / LM = p.$

[b][size=109]PROOF 1.[/size][/b] – $($ In my drawing $AB = 5.5,$ $BC = 8.0,$ $AC = 6.3,$ $CD = 2.8,$ $AD = 4.5$ $).$

Through vertices $A,$ $C$ of $ABCD,$ we draw two lines parallel to $CD,$ $AD$ respectively and we denote as $A',$ their intersection point. Through $E,$ we draw two lines parallel to $BA',$ $AA',$ which intersect the segment lines $AA',$ $BA',$ at points $F',$ $E',$ respectively.

Because of $EF'\parallel BA'$ $\Longrightarrow$ $\frac{AE}{EB}= \frac{AF'}{F'A'}$ $\Longrightarrow$ $\frac{AF'}{F'A'}= \frac{DF}{FC}$ $($ because of $\frac{AE}{EB}= \frac{DF}{FC}$ $)$ $\Longrightarrow$ $AD\parallel F'F\parallel A'C$ $,(1)$

Through the point $K,$ we draw a line parallel to $AA'\parallel CD,$ which intersects the segment line $A'C,$ at a point so be it $K'$ and let $L'$ be, the intersection point of $E'C,$ $MK'.$

$\bullet$ From the parallelograms $EF'A'E',$ $F'FCA'$ $\Longrightarrow$ $EE\parallel = F'A'\parallel FC$ and $E'C\parallel = EF$

From the parallelogram $AA'CD$ $\Longrightarrow$ $\frac{AK}{KD}= \frac{A'K'}{K'C}$ $\Longrightarrow$ $\frac{A'K'}{K'C}= \frac{BM}{MC}$ $,(2)$ $($ because of $\frac{AK}{KD}= \frac{EL}{LF}= \frac{BM}{MC}$ $)$

From $(2)$ $\Longrightarrow$ $MK'\parallel BA'$ $\Longrightarrow$ $\frac{E'L'}{L'C}= \frac{BM}{MC}= \frac{EL}{LF}$ $\Longrightarrow$ $LL'\parallel = EE'$ $,(3)$

From $(3)$ and because of $KK'\parallel = AA'$ $,(4)$  and $EE'\parallel AA'$ $\Longrightarrow$ $LL'\parallel KK'$ $,(5)$

From $MK'\parallel BA'$ and $EE'\parallel AA',$ we have also $\frac{ML'}{MK'}= \frac{BE'}{BA'}= \frac{EE'}{AA'}$ $,(6)$

From $(3),$ $(4),$ $(5),$ $(6)$ $\Longrightarrow$ $\frac{ML'}{MK'}= \frac{LL'}{KK'}$ $,(7)$

From $(7),$ we conclude that the points $K,$ $L,$ $M,$ are collinear, such that $\frac{KL}{LM}= \frac{K'L'}{L'M}= \frac{A'E'}{E'B}= \frac{AE}{EB}= p$ and the proof is completed.

Kostas Vittas.

[b][size=109]PROOF 2.[/size][/b] – $($ In my drawing $AB = 3.5,$ $BC = 5.2,$ $AC = 8.2,$ $CD = 9.0,$ $AD = 6.7$ $).$

Through $K,$ we draw two lines $(\ell),$ $(\ell'),$ parallel to $AB,$ $DC,$ respectively.Through the points $E,$ $B,$ we draw two lines parallel to $AD,$ which intersect the line $(\ell),$ at points $E',$ $B',$ respectively. Through the points $F,$ $C,$ we draw two lines also parallel to $AD,$ which intersect the line $(\ell'),$ at points $F',$ $C',$ respectively.

$\bullet$ From the parallelograms $AKB'B,$ $DKC'C$ $\Longrightarrow$ $BB'\parallel = EE'\parallel = AK$ $,(1)$ and $CC'\parallel = FF'\parallel = DK$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $BB'\parallel CC'$ $,(3)$ and $EE'\parallel FF'$ $,(4)$ and let $L\equiv EF\cap E'F'$ be and $M\equiv BC\cap B'C'$

We will prove that $\frac{EL}{LF}= \frac{BM}{MC}= \frac{AK}{KD}= q$ and that the points $K,$ $L,$ $M,$ are collinear, such that $\frac{KL}{LM}= p$

From $(3),$ $(4)$ $\Longrightarrow$ $\frac{BM}{MC}= \frac{BB'}{CC'}= \frac{EE'}{FF'}= \frac{AK}{KD}= q$ $,(5)$

Similarly we have that $\frac{B'M}{MC'}= \frac{BM}{MC}= \frac{E'L}{LF'}= \frac{EL}{LF}= q$ $,(6)$

From $(1),$ $(2),$ we have also that $\frac{KE'}{E'B'}= \frac{AE}{EB}= \frac{DF}{FC}= \frac{KF'}{F'C'}= p$ $\Longrightarrow$ $E'F'\parallel B'C'$ $,(7)$

From $(6,)$ $(7)$ $\Longrightarrow$ the points $K,$ $L,$ $M,$ are collinear such that $\frac{KL}{LM}= p$ and the proof is completed.

Kostas Vittas.

[quote="barasawala"]Well, say the circumcenters of $AMN,AKL,ABC$ are $O_{1},O_{2},O_{3}$. To use your lemma, I think we need to prove $\frac{O_{1}O_{2}}{O_{2}O_{3}}=\frac{MK}{KB}$, which I can't.[/quote]

$($ In my drawing $AB = 11.8,$ $BC = 6.0,$ $AC = 12.3,$ $BK = 1.2,$ $KM = 1.7,$ $CL = 1.3,$ where $K,$ between $B,$  $M$ and $M,$ in the extension of $AB.$ Also $N,$ between $A,$ $C$ and $L,$ between $N,$ $C$ $).$

If $O_{1},$ $O_{2},$ $O_{3},$ are the circumcenters of the triangles $\bigtriangleup ABC,$ $\bigtriangleup AKL,$ $\bigtriangleup AMN$ respectively, we consider their orthogonal projections $O'_{1},$ $O'_{2},$ $O'_{3},$ on the sideline $AB,$ of $ABC.$ So, we have that $\frac{O_{1}O_{2}}{O_{2}O_{3}}= \frac{O'_{1}O'_{2}}{O'_{2}O'_{3}}$ and it is enough to prove that $\frac{O'_{1}O'_{2}}{O'_{2}O'_{3}}= \frac{BK}{KM}$

But this result is true because of $O'_{1}O'_{2}= \frac{BK}{2}$ and $O'_{2}O'_{3}= \frac{KM}{2}$ $($ easy to prove because of the points $O'_{1},$ $O'_{2},$ $O'_{3},$ are the midpoints of the segments $AB,$ $AK,$ $AM,$ respectively $).$

Kostas Vittas.

Dear all my friends.

For reading easiness, I post here the figures of the configurations of all as I wrote about the shobber's problem.

Kostas Vittas.

To Dear vittasko :Thank you very much for writing the rest of my solution.

I was very busy I couldn't type long posts.

[quote="shobber"][color=darkred]Let $ K$ and $ M$ be points on the side $ AB$ of a triangle $ \triangle{ABC}$, and let $ L$ and $ N$ be points on the side $ AC$. The point $ K$ is between $ M$ and $ B$, and the point $ L$ is between $ N$ and $ C$. If $ \frac {BK}{KM} = \frac {CL}{LN}$, then prove that the orthocentres of the triangles $ \triangle{ABC}$, $ \triangle{AKL}$ and $ \triangle{AMN}$ lie on one line.[/color] [/quote]
[color=darkblue][b]Proof.[/b] Let $ ABCD$ be a tetrahedron and the points $ \|\begin{array}{c}
\{M,K\}\subset (AB)\\\
\{N,L\}\subset (AC)\\\
\{U,V\}\subset (AD)\end{array}$

so the the planes $ (MUN)$, $ (KVL)$, $ (BDC)$ are parallely.

The orthocenters of the triangles $ MUN$, $ KVL$, $ BDC$ are collinearly.

For $ D\rightarrow A$ obtain $ U\rightarrow A$ and $ V\rightarrow A$, i.e. in this limit case

the orthocenters of the triangles $ MAN$, $ KAL$, $ BAC$ are collinearly.[/color]

[quote="Virgil Nicula"][quote="shobber"][color=darkred]Let $ K$ and $ M$ be points on the side $ AB$ of a triangle $ \triangle{ABC}$, and let $ L$ and $ N$ be points on the side $ AC$. The point $ K$ is between $ M$ and $ B$, and the point $ L$ is between $ N$ and $ C$. If $ \frac {BK}{KM} = \frac {CL}{LN}$, then prove that the orthocentres of the triangles $ \triangle{ABC}$, $ \triangle{AKL}$ and $ \triangle{AMN}$ lie on one line.[/color] [/quote]
[color=darkblue][b]Proof.[/b] Let $ ABCD$ be a tetrahedron and the points $ \|\begin{array}{c} \{M,K\}\subset (AB) \\
\ \{N,L\}\subset (AC) \\
\ \{U,V\}\subset (AD)\end{array}$

so the the planes $ (MUN)$, $ (KVL)$, $ (BDC)$ are parallely.

The orthocenters of the triangles $ MUN$, $ KVL$, $ BDC$ are collinearly.

For $ D\rightarrow A$ obtain $ U\rightarrow A$ and $ V\rightarrow A$, i.e. in this limit case

the orthocenters of the triangles $ MAN$, $ KAL$, $ BAC$ are collinearly.[/color][/quote]
I think it is true $ ($ to be the planes $ [MUN],$ $ [KVL],$ $ [BDC]$ parallel each other $ ),$ only when $ MN\parallel KL\parallel BC$ and then, the line connecting the orthocenters of the triangles $ \bigtriangleup MUN,$ $ \bigtriangleup KVL,$ $ \bigtriangleup BDC,$ passes through the point $ A.$

But in this particular case, we don’t need the clever artifice of tetrahedron $ ABCD.$

Also, the problem doesn’t state that the pairs of points $ K,$ $ M$ and $ L,$ $ N,$ must be in the same place, with respect to the line segment $ BC.$

Kostas Vittas.

[i]Lemma: Let $ \mathcal{A}\equal{}A_1A_2...A_n$ and $ \mathcal{B}\equal{}B_1B_2...B_n$ be directly similar polygons. Construct $ A_iB_iC_i$ that are directly similar. Then $ \mathcal{A}$, $ \mathcal{B}$, and $ \mathcal{C}$ all have the same spiral center (and they are directly similar).[/i]

[color=darkred]Problem: Let $ ABC$ be a triangle, let $ B_1$, $ C_1$ on $ AB$ and $ AC$. Construct $ B_2$ and $ C_2$ on segments $ AB_1$ and $ AC_1$ such that $ \frac{BB_1}{B_1B_2}\equal{}\frac{CC_1}{C_1C_2}$. Prove that the orthocenters of $ ABC$, $ AB_1C_1$, and $ AB_2C_2$ are collinear. [/color]

[color=blue]Proof: Let $ S$ be the point that such that $ BB_2$ maps to $ CC_2$ under a direct similarity. Then $ BB_1B_2$ and $ CC_1C_2$ are directly similar degenerate triangles constructed atop these similar polygons, so $ BB_1B_2$; $ CC_1C_2$ are directly similar wrt $ S$.

Let $ O$, $ O_1$, $ O_2$ be the circumcenters of $ ABC$, $ AB_1C_1$, $ AB_2C_2$. Clearly, $ OBC$, $ OB_1C_1$, and $ OB_2C_2$ are directly similar since $ OB\equal{}OC$, $ O_1B_1\equal{}O_1C_1$, $ O_2B_2\equal{}O_2C_2$ and $ \angle BOC\equal{}\angle B_1O_1C_1\equal{}\angle B_2O_2C_2\equal{}2\angle BAC$. Therefore, $ OO_1O_2$ is directly similar to $ BB_1B_2$, so they are are both degenerate.

Let $ P$, $ P_1$, and $ P_2$ be the midpoints of $ BC$, $ B_1C_1$, and $ B_2C_2$. Then $ BPC$, $ B_1P_1C_1$, and $ B_2P_2C_2$ are directly similar degenerate triangles, so $ BB_1B_2$ and $ PP_1P_2$ are directly similar, so they are both degenerate.

Let $ G$, $ G_1$, and $ G_2$ be the centroids of $ ABC$, $ A_1B_1C_1$. A homothety center $ A$ with ratio $ \frac{2}{3}$ maps $ PP_1P_2$ to $ GG_1G_2$ so $ GG_1G_2$ is directly similar to $ PP_1P_2$.

Let $ T$ be the spiral center that sends $ GG_1G_2$ to $ OO_1O_2$ (note $ T\ne S$ necessarily). Now by the euler line, $ HOG$, $ H_1O_1G_1$, and $ H_2O_2G_2$ are directly similar triangles, so $ OO_1O_2$ and $ HH_1H_2$ are directly similar, and the result follows.[/color]

Comment: This is a great example of the power of spiral similarity! With the proper knowledge of that topic, this problem is straightforward. Surely, this problem is subject to generalization. That is, we can take any triangle centers of $ ABC$, $ AB_1C_1$, $ AB_2C_2$ that are directly similar to $ BB_1B_2$ and follow the same logic.

Here is my solution
Set the ration $ \frac {MK}{KB} = \frac {NL}{LC} = k$
Call orthorcenters of triangles $ \Delta AMN, \Delta AKL,\Delta ABC$ respect are $ H_{1}, H_{2},H_{3}$.
And $ O_{1}, O_{2},O_{3}$ are circumcenters of triangles $ \Delta ABC, \Delta AMN, \Delta AKL$ respect.
By CANADA MO 2003, easy to prove that $ O_{1}, O_{2},O_{3}$  are collinear and $ \frac {O_{1}O_{2}}{O_{2}O_{3}} = k$
$ \Rightarrow \vec{O_{1}O_{2}} = k\vec{O_{2}}{O_{3}}$
So, we use indentions flowing:
$ \vec{O_{1}H_{1}} = \vec{O_{1}A} + \vec{O_{1}B} + \vec{O_{1}C}$
$ \vec{O_{2}H_{2}} = \vec{O_{2}A} + \vec{O_{2}M} + \vec{O_{2}N}$
$ \vec{O_{3}H_{3}} = \vec{O_{3}A} + \vec{O_{3}K} + \vec{O_{3}L}$
 Hence, we get
$ \vec{H_{1}H_{2}} = k\vex{H_{2}}{H_{3}}$
I think the Lemma of Kostas Vittas can prove by using vector(very short)

an other approach  :
let $S$ the Miquel points of $ABC$ wrt $M-N$ since $\frac{BK}{KM}=\frac{CL}{LN}\iff \frac{\mathcal{P}_{(ABC)}(K)}{\mathcal{P}_{(AMN)}(K}= \frac{\mathcal{P}_{(ABC)}(L)}{\mathcal{P}_{(AMN)}(L)}$ then $SAKL$ is cyclic but  the Steiner line of $S$  is the same wrt $AMN,ABC,AKL$ hence their orthocenters are collinear .

Relabel the points such that the condition reads $\frac{BB_1}{BB_2} = \frac{CC_1}{CC_2}$. The idea is that the Miquel point $M = (AB_2C_2) \cap (ABC)$ is also concyclic with $AC_1B$. So by spiral similarity, it suffices to show that the orthocenters are corresponding parts of the similarity. This immediately follows by checking the $A$-antipodes and midpoints are both corresponding parts too.

After getting that 3 circumcenters are collinear, then just reflect this line about angle bisector of $\angle {A}$, we get desired conclusion!


"Failed to parse (Can't write to or create math output directory): \frac{y} {20} = \frac{1} {k}"

http://www.artofproblemsolving.com/Wiki/index.php/User:Temperal/Introductory_Proportion

I've gotten this several times. Is it a bug?

Might be the math directory has been corrupted? Anyway I'm currently upgrading the software, so by Monday the Wiki will be using the Forum latex parser (and asy parser for that matter).

It’s a very common problem with LaTeX on the AoPSWiki. On average when I add a new AIME problem/solution, I get about 3 ~ 8 of these [color=red]Can't write to or create math output directory errors[/color]. The best way to get rid of them is to just randomly toss in some symbols to change the content in the math tags (white spaces don’t change anything); my favorite to add in is \displaystyle.

If adding \displaystyle to the beginning of the equation doesn’t work, I just add another one until eventually the math will be parse-able.

Hopefully the upgrade will remove this issue  :)

[quote="Valentin Vornicu"]... normal forum $ \text{\LaTeX}$ and Asymptote rendering will be available on Wiki.[/quote]
All hail the mighty site admin! :D

This will make creating images and typing up solutions much easier.

[quote="Valentin Vornicu"][quote="archimedes1"][quote="Valentin Vornicu"]Anyway I'm currently upgrading the software, so by Monday the Wiki will be using the Forum latex parser (and asy parser for that matter).[/quote]
 :thumbup:
I hate those "can't write to math directory" messages.  I think another way of getting rid of them is to put random curly brackets around the LaTeX.[/quote]That is a bad idea. I am currently upgrading the software (probably will be released tomorrow) and normal forum $ \text{\LaTeX}$ and Asymptote rendering will be available on Wiki.[/quote]
Oh, I see what you mean....  So that people won't have to edit out the extra brackets when the new parser comes, right?

[quote="archimedes1"]Oh, I see what you mean....  So that people won't have to edit out the extra brackets when the new parser comes, right?[/quote]Yes.


Substitute each of the capital letters by a different digit from 0 to 9 to satisfy this cryptarithmetic equation, where [b]L[/b] is nonzero.

[b]LOCOMOTION[/b] = ([b]LENTL[/b])*([b]LENTL[/b])

The solution to the above crypt-arithmetic puzzle which adheres to the rules as stated in the puzzle is given below:

[b] L O C O M O T I O N [/b] =  ([b] L E N T L [/b]) * ([b] L E N T L [/b])

[b] 9 4 3 4 0 4 2 6 4 1 [/b] =  [b] 9 7 1 2 9 [/b] * [b] 9 7 1 2 9 [/b]

where '*' denotes multiplication.

[b] L = 9, E=7, N=1, T=2, C=3, M=0, I=6 [/b]


How did the Alabama MATHCOUNTS competition go?

Any AoPSers on the state team?

eh... i would say it went okay... I didn't get on the top ten, but everyone else from Pizitz did... Only one person, Wayne, made it to Nationals, and one person got 5th, which was really sad because she went all the way from 9th to 5th in countdown... and the other people that's going to nationals, I think one or two are from Randalph, and one's from ... eh.. I think maybe chanllenger or something. Pizitz got 1st in team, o well, at least I got a team trophy...  :D  :(

It was a little disorganized, since they had a new coordinator this year. I didn't make nats, oh well, I think the high school I'm going to does AMC, so I'll still have a chance to do math competitively. My team did get 4th though, after Pizitz, Randolph, and Challenger, who each had at least one person on the state team. I was surprised that Altamont wasn't there at State. I think me and Lucy are the only people on AoPS who were at the Alabama MC State competition this year?

i think so, cuz I think me and PTheodora are the only two middle schooler's in this Alabama forum... o well....didn't know somebody from Chanllenger made it, we always see them in math tournaments. I knew the people from Altamont , they were there alright, I had to face a guy from Altamont during chapter countdown.... think he was a bit mad, but anyways, me and Kathy sat by them during lunch because the guy i faced during countdown was kathy's classmate from fifth grade and because we just happen to sat down by them.... they had like people on their team who was on the team last year, but they didn't make top four... am I saying too much?

I've discovered oneil3691 is also from Alabama. He mentions it on page 6 of the Are You Going To Nationals thread.

[url=http://www.artofproblemsolving.com/Forum/topic-29037-75.html]www.artofproblemsolving.com/Forum/topic-29037-75.html[/url]

And he's going to nats. I envy him. :P

Did you know they threw out the countdown round?

omgosh, he's that dude... wait, i think so... lucky, they screwed up the scores last time too...  lucky him. wow, that wouldn't be fair for kathy... and the other fifth place girl...


[/b]HI! I want to participate in IBO 2009. i have participated in ibo 2008 but was not so successful.So i want to improve my skills. from which resources could i prepare for IBO(from which books or web sites,etc.)  :oops:  :)[/b]

By no means am I nearly as talented in biology as just about everyone here. But from what i hear, campbell and reece is THE ibo textbook.

i also heard that campbell and reece i IBO textbook but it is not so deatiled as IBO needs. Campbell is general biology book so,can be used for review. i want to know  which area books(e.g. plant anatomy and physiology books) are used by many successful countries.[b] :?:  :?: [/b]

Biology of Plants by Raven is excellent for botany.

The Molecular Biology of the Cell by Alberts is good for molecular bio, but WAY too detailed for IBO.

Generally, it's not a bad idea to read, reread, and reread Campbell again.

From this year's IBO, I'd also say it's a good idea to review your chemistry and statistics, if you've taken them, and if not, get textbooks and study stoichiometry for chem and t-tests and chi-square tests for stats.

Also, you'll need to prepare for the practical - though I can't say practice dissections anymore, there will likely be something standard like stem and leaf sections and spectrometry.

Oh, and did I mention that you should read Campbell? (just joking, but yes, you really should read it if you haven't)

Would you recommend Lodish for Molecular Biology?  Also, is it necessary to read all of Lodish and Raven; in other words, how much material from the two is tested on the IBO and the USABO?

can i ask a question? Do you (USA biology team) prepare for IBO with those books that listed.  :?:  :|

Let me make this clear: Lodish and Alberts are NOT necessary. While tiny sections may come up occasionally on the USABO and even IBO, they are not worth reading for the sake of biology olympiads. Instead, read Campbell again.

Raven's Biology of Plants, though, can be useful (particularly the second half, starting with the Bryophytes). It goes into topics covered lightly in Campbell, such as the evolution of plants and plant structure.

Team USA's primary resource was Campbell - we pretty much tried to go through all of it again during the IBO. For the IBO, we also tried to prep for the practical, but that won't help so much for USABO (until finals).

If I haven't emphasized it enough, Campbell, Campbell, and Campbell again. Not just reading, not just memorization, but [i]understanding[/i].

which edition of campbell is the best? i have campbell 8th edition  :?:


[quote="sonny"][hide]In 45 seconds it travels 3/4 of a minute. 1 whole evolution is a distance of $8\pi$ so 3/4s would be $6\pi=18.84......in$ so approximately $19$[/hide][/quote]

it's revolution and thanks i mixed up area and circumference forumlas  :ewpu:

[quote="7h3.D3m0n.117"]The second hand of a clock is 4 inches long. To the nearest inch, how far does the tip of the second hand travel in 45 seconds?

What did you get?[/quote]
[hide]
$8\pi\frac{3}{4}$
$6\pi$ is about $19$.
[/hide]


limit
number theory unsolved
number theory

[quote="erdugan"]Does there exist a permutation of natural number like a_1,a_2,a_3,...... such that for any natural number n , n| a_1+a_2+......+a_n[/quote]

Sure, such a permutation exists. Here is a method to build one :

Let $ a_1\equal{}1$
Consider now $ a_k$ known up to $ a_n$ (and so we have that $ k|(\sum_{i\equal{}1}^ka_i)$ $ \forall k\in[1,n])$
Let $ p_n$ be the greatest integer such that $ i\in\{a_1,a_2,...,a_n\}$ $ \forall i\in[1,p_n]$

Let $ a$ any integer such that $ a(n\plus{}1)(n\plus{}2)\plus{}(p_n\plus{}1)(n\plus{}1)\minus{}\sum_{i\equal{}1}^na_i$ is positive, different from $ p_n\plus{}1$ and $ \notin\{a_1,a_2,...,a_n\}$

Let $ a_{n\plus{}1}\equal{}a(n\plus{}1)(n\plus{}2)\plus{}(p_n\plus{}1)(n\plus{}1)\minus{}\sum_{i\equal{}1}^na_i$ and $ a_{n\plus{}2}\equal{}p_n\plus{}1$

Then :
$ \sum_{i\equal{}1}^{n\plus{}1}a_i\equal{}\sum_{i\equal{}1}^na_i\plus{}a_{n\plus{}1}$ $ \equal{}\sum_{i\equal{}1}^na_i\plus{}a(n\plus{}1)(n\plus{}2)\plus{}(p_n\plus{}1)(n\plus{}1)\minus{}\sum_{i\equal{}1}^na_i$ $ \equal{}(n\plus{}1)(a(n\plus{}2)\plus{}p_n\plus{}1)$ and is divisible by $ n\plus{}1$

$ \sum_{i\equal{}1}^{n\plus{}2}a_i\equal{}(n\plus{}1)(a(n\plus{}2)\plus{}p_n\plus{}1)\plus{}p_n\plus{}1$ $ \equal{}(n\plus{}2)(a(n\plus{}1)\plus{}p_n\plus{}1)$ and is divisible by $ n\plus{}2$

And now $ p_{n\plus{}2}>p_n$

So, we have a sequence $ a_n$ which matches the requirement and such that :
- no two elements are identical
- every integer is in the sequence (since $ \lim_{n\to\plus{}\infty}p_n\equal{}\plus{}\infty$)

So this sequence is a permutation of all natural numbers.


MATHCOUNTS has confirmed to me that the last State competition has been completed.  They expect to post the State competition on their Web site within a couple of days.

MATHCOUNTS has given permission to discuss the State competition.

Thank you for your patience, as we awaited this moment!

2006 State Competitions have now been posted, and can be [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=510&z=47]downloaded here[/url].


By generalized Titu's lemma

LHS$\ge \frac{(\sum{a_i})^2}{2(\sum{a_i})}=\frac{\sum{a_i}}{2}=\frac{1}{2}$

Too easy for a TST, isn't it.....

Well,my school exams are near,so I am satisfied with easy ones....

Bye...

Sayantan....

[quote=dwip_neel]\[ \frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+\dots+\frac{a_{n-1}^2}{a_{n-1}+a_n}+\frac{a_n^2}{a_n+a_1}\ge \frac{(a_1 + a_2 + a_3 + a_4 + a_5 + \cdots + a_n)^2}{2(a_1 + a_2 + a_3 + a_4 + a_5 + \cdots + a_n)} = \frac{1}{2}\]

I have used [i][b]Cauchy-Schwartz Lemma[/b][/i] here.[/quote]

It's too easy as a TST problem.

[quote=Xixas]Let $a_1, a_2, \dots, a_n$ be positive real numbers, whose sum is $1$. Prove that \[ \frac{a_1^2}{a_1+a_2}+\frac{a_2^2}{a_2+a_3}+\dots+\frac{a_{n-1}^2}{a_{n-1}+a_n}+\frac{a_n^2}{a_n+a_1}\ge \frac{1}{2} \][/quote]

More generally, if $a_1+\cdots +a_n=b_1+\cdots +b_n=1$ and $a_1,\ldots ,a_n,b_1,\ldots ,b_n>0$, then by the Fractionary form of Cauchy's Inequality we have$$\frac{a_1^2}{a_1+b_1}+\cdots +\frac{a_n^2}{a_n+b_n}\geq \frac{(a_1+\cdots +a_n)^2}{(a_1+\cdots +a_n)+(b_1+\cdots +b_n)}=\frac{1}{2}.$$We have equality iff $a_i=b_i$ for $i=1,\ldots ,n$, in the original problem this is $a_1=\cdots =a_n=\frac{1}{n}$.


What value of $ n$ will make the nine-digit number $ 200,412,0n3$ divisible by 11?

opposite digits must add up and subtract to a multiple of 11

2+0+1+0+3=0+4+2+n

so n=0


geometry
geometric transformation
reflection
analytic geometry
calculus
probability
derivative

I see some variable with a slight short line a after it like: $ r', q'$ but what this mean?

Sometimes in truth tables and sets I have seen this denote the complement of something. Other times it's been just another variable.

[quote="Complex_Ninja"]Sometimes when you translate points. K translated to K'[/quote]
Can you explain more about translating.

like "point A (1,2) and point B (4,6) are reflected over the line x=y. denote their respective reflections A' and B'. find the area of quadrilateral ABB'A'".

even though that's trivial, and not a good example, things like that

Translating a figure is kind of like sliding it over, usually on coordinate planes. For example, if I have square ABCD with coordinates A(0,0), B(2,0), C(2,-2), and D(0,-2), and I translate it two units to the right, I would end up with square A'B'C'D'(pronounced: A prime B prime C prime D prime) with coordinates A'(2,0), B'(4,0), C'(4,-2), D'(2,-2).

In general, it's just another variable, typically used to denote another variable that has been transformed somehow.

In calculus, it denotes the derivative. In probability, the complement.

And of course, in books with bad formatting, it can just merely be a comma.
ex. (I found this in a book once and was extremely confused):

$ \frac{x}{y},$

Thanks guys I know what is translating but I thought its another thing when its have  that symbol.  :)

$ x'$ is pronoucned "x prime". It can mean several things. Some people use it to denote conjugates of complex numbers, and it is used as a shorthand in logic.

In geometry, these can simply be other points, lines, etc., although they are often used to denote changing an existing object. For example, reflecting a point X over something creates another point, that is typically called X'. This can also be used for lines or even figures.

Taking a more advanced approch, $ x'$ is a shorthand for the derivative of x.

$ x'$ is sometimes used in probability to denote the complement, which is just $ 1\minus{}x$.

--BOGTRO--


calculus
derivative
function
trigonometry
analytic geometry
calculus computations

I began reading stuff on partial derivatives and I am a little confused on how to transform from one set of second partials to another set of second partials. Hmm, this may not be clear, let me begin with a problem:

Given $f: \mathbb{R}^2\to\mathbb{R}$ a continuous mapping with continuous second partials. So $f=f(x,y)$. Now let $x=s+t$ and $y=s-t$, and then define $g(s,t)=f(s+t,s-t)$. How can I show that
\[ \frac{\partial^2f}{\partial x^2}-\frac{\partial^2f}{\partial y^2}=\frac{\partial^2g}{\partial t\partial s} \]?

$x(s,t)=s+t$; $y(s,t)=s-t$.

$g(s,t)=f(x(s,t),y(s,t))$.

By the chain rule,

$\frac{\partial g}{\partial s}(s,t)=\frac{\partial f}{\partial x}(x(s,t),y(s,t))\frac{\partial x}{\partial s}(s,t) + \frac{\partial f}{\partial y}(x(s,t),y(s,t))\frac{\partial y}{\partial s}(s,t)$

$=\frac{\partial f}{\partial x}(x(s,t),y(s,t)) + \frac{\partial f}{\partial y}(x(s,t),y(s,t))$.

So, using the chain rule again,

$\frac{\partial^2 g}{\partial t \partial s}(s,t)=\frac{\partial^2 f}{\partial x^2}(x(s,t),y(s,t))\frac{\partial x}{\partial t}(s,t) + \frac{\partial^2 f}{\partial x \partial y}(x(s,t),y(s,t))\frac{\partial y}{\partial t}(s,t) + \frac{\partial^2 f}{\partial x \partial y}(x(s,t),y(s,t))\frac{\partial x}{\partial t}(s,t) + \frac{\partial^2 f}{\partial y^2}(x(s,t),y(s,t))\frac{\partial y}{\partial t}(s,t)$

$=\frac{\partial^2 f}{\partial x^2}(x(s,t),y(s,t)) + \frac{\partial^2 f}{\partial x \partial y}(x(s,t),y(s,t))(-1) + \frac{\partial^2 f}{\partial x \partial y}(x(s,t),y(s,t))+ \frac{\partial^2 f}{\partial y^2}(x(s,t),y(s,t))(-1)$

$=\frac{\partial^2 f}{\partial x^2}(s+t,s-t) - \frac{\partial^2 f}{\partial y^2}(s+t,s-t)$.

Thanks for your reply Aunt Sally. With your notation of handling the variables, I can see which variables were used in the chain rule. But I'm still not sure how can an independent variable like $x$ can turn itself into a function, although it is only a substitution $x=s+t$.

Here is something I worked out today, please correct me if I did something wrong.

Change of rectangular partials to polar partials by making use of $g(r,\theta)=f(r\cos\theta,r\sin\theta)$

In other words, we have $x(r,\theta)=r\cos\theta$ and $y(r,\theta)=r\sin\theta$.

Then,

$\frac{\partial g}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}=\frac{\partial f}{\partial x}\cos\theta+\frac{\partial f}{\partial y}\sin\theta$

$\frac{\partial g}{\partial\theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial\theta}=\frac{\partial f}{\partial x}(-r\sin\theta)+\frac{\partial f}{\partial y}r\cos\theta$

Now the second partials should be

$\frac{\partial^2g}{\partial r^2}=\frac{\partial^2f}{\partial x^2}\cos^2\theta+\frac{\partial^2f}{\partial y\partial x}\cos\theta\sin\theta+\frac{\partial^2f}{\partial x\partial y}\cos\theta\sin\theta+\frac{\partial^2f}{\partial y^2}\sin^2\theta$

$\frac{\partial^2g}{\partial\theta^2}=\frac{\partial^2f}{\partial x^2}r^2\sin^2\theta-\frac{\partial^2f}{\partial y\partial x}r^2\cos\theta\sin\theta-\frac{\partial^2f}{\partial x\partial y}r^2\cos\theta\sin\theta+\frac{\partial^2f}{\partial y^2}r^2\cos^2\theta-\frac{\partial f}{\partial x}r\cos\theta-\frac{\partial f}{\partial y}r\sin\theta$

With this, we can verify easily that $\frac{\partial^2g}{\partial r^2}+\frac1{r^2}\frac{\partial^2g}{\partial\theta^2}+\frac1r\frac{\partial g}{\partial r}=\frac{\partial^2f}{\partial x^2}+\frac{\partial^2f}{\partial y^2}$, with $f=f(x(r,\theta),y(r,\theta))$.

Looks good.

In case you haven't seen it already, you might think this thread about expressing the gradient in spherical coordinates is interesting:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=88887

By the way, you probably know this already, but these computations you're doing are useful in the study of differential equations.  The first problem helps find a general solution to the wave equation, and the second problem you posted is useful when solving the Laplace equation using polar coordinates.

[quote]But I'm still not sure how can an independent variable like  can turn itself into a function[/quote]

Let me slightly rephrase the argument, hoping to clarify this issue.

Suppose that we are given a function $f: \mathbb{R}^2 \to \mathbb{R}$ that is $C^2$ (in other words all of its second-order partial derivatives exist and are continous).

Define functions $x: \mathbb{R}^2 \to \mathbb{R}$ and $y: \mathbb{R}^2 \to \mathbb{R}$ by $x(s,t)=s+t$ and $y(s,t)=s-t$ for all $s, t \in \mathbb{R}$.

Define a function $g: \mathbb{R}^2 \to \mathbb{R}$ by $g(s,t)=f(x(s,t),y(s,t))=f(s+t,s-t)$ for all $s,t \in \mathbb{R}$.

By the chain rule,

$\frac{\partial g}{\partial s}(s,t)=\frac{\partial f}{\partial x}(x(s,t),y(s,t))\frac{\partial x}{\partial s}(s,t) + \frac{\partial f}{\partial y}(x(s,t),y(s,t))\frac{\partial y}{\partial s}(s,t)$
$=\frac{\partial f}{\partial x}(x(s,t),y(s,t)) + \frac{\partial f}{\partial y}(x(s,t),y(s,t))$.

So, using the chain rule again,

$\frac{\partial^2 g}{\partial t \partial s}(s,t)=\frac{\partial^2 f}{\partial x^2}(x(s,t),y(s,t))\frac{\partial x}{\partial t}(s,t) + \frac{\partial^2 f}{\partial x \partial y}(x(s,t),y(s,t))\frac{\partial y}{\partial t}(s,t) + \frac{\partial^2 f}{\partial x \partial y}(x(s,t),y(s,t))\frac{\partial x}{\partial t}(s,t) + \frac{\partial^2 f}{\partial y^2}(x(s,t),y(s,t))\frac{\partial y}{\partial t}(s,t)$

$=\frac{\partial^2 f}{\partial x^2}(x(s,t),y(s,t)) + \frac{\partial^2 f}{\partial x \partial y}(x(s,t),y(s,t))(-1) + \frac{\partial^2 f}{\partial x \partial y}(x(s,t),y(s,t))+ \frac{\partial^2 f}{\partial y^2}(x(s,t),y(s,t))(-1)$
$=\frac{\partial^2 f}{\partial x^2}(s+t,s-t) - \frac{\partial^2 f}{\partial y^2}(s+t,s-t)$.

There.  Now, I never mentioned any "independent variable" that turned into a function.  I never used the letter $x$ to represent anything other than a particular function that I defined in order to help with my argument.  I didn't have to use the letters $x$ and $y$, I could've used, I don't know, $w$ and $z$ instead, and still used the same argument to reach the same conclusion.  Maybe that helps clarify the issue; hopefully at least I'm not confusing you more!

Another thing that might be worth considering is that, instead of using the notation $\frac{\partial f}{\partial x}$, we could use the notation $D_1f$ instead.  And instead of $\frac{\partial f}{\partial y}$, we could call it $D_2f$ instead.  We could use similar notation to express second-order partial derivatives.  This new notation is probably not as useful as a mnemonic device to help you remember your formulas, but sometimes I think that it is logically more clear.

That's an interesting link and I didn't know he posted a similar topic on it a few days ago :blush:

I was able to derive the 3-dim Laplacian in spherical coordinates by applying twice change of variables through cylindrical coordinates. Thanks Aunt Sally, I think I get the idea on how to manipulate with the variables and making sure which variables I'm taking partials with respect to other variables :)

If anyone else is reading this and learning partial derivatives with me, here is a cool problem:

Let $f(|\mathbf{x}|)=g(r)$ and $r=|\mathbf{x}|=\left(\sum_{i=1}^{n}x_i^2\right)^\frac12$ for $n\ge3$.

Show that $\nabla^2f=\sum_{i=1}^{n}\frac{\partial^2f}{\partial x_i^2}=\frac{n-1}rg'(r)+g''(r)$.

PS: I'll post how I derived 3-dim Laplacian in spherical coordinates as well as the solution to above at a later post :)

[quote="10000th User"]$\frac{\partial g}{\partial\theta}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial\theta}=\frac{\partial f}{\partial x}(-r\sin\theta)+\frac{\partial f}{\partial y}r\cos\theta$

Now the second partials should be

$\frac{\partial^2g}{\partial r^2}=\frac{\partial^2f}{\partial x^2}\cos^2\theta+\frac{\partial^2f}{\partial y\partial x}\cos\theta\sin\theta+\frac{\partial^2f}{\partial x\partial y}\cos\theta\sin\theta+\frac{\partial^2f}{\partial y^2}\sin^2\theta$

$\frac{\partial^2g}{\partial\theta^2}=\frac{\partial^2f}{\partial x^2}r^2\sin^2\theta-\frac{\partial^2f}{\partial y\partial x}r^2\cos\theta\sin\theta-\frac{\partial^2f}{\partial x\partial y}r^2\cos\theta\sin\theta+\frac{\partial^2f}{\partial y^2}r^2\cos^2\theta-\frac{\partial f}{\partial x}r\cos\theta-\frac{\partial f}{\partial y}r\sin\theta$

[/quote]

Why/how did you get the extra term $-\frac{\partial f}{\partial x}r\cos\theta-\frac{\partial f}{\partial y}r\sin\theta$ for $\frac{\partial^2g}{\partial\theta^2}$ and not for  $\frac{\partial^2g}{\partial r^2}$?

[quote="10000th User"][quote="swapnillium"]Why/how did you get the extra term $-\frac{\partial f}{\partial x}r\cos\theta-\frac{\partial f}{\partial y}r\sin\theta$ for $\frac{\partial^2g}{\partial\theta^2}$ and not for  $\frac{\partial^2g}{\partial r^2}$?[/quote]Uh... product rule? It's derivative with respect to $\theta$, so you apply product rule...  :ninja:[/quote]

[quote="10000th User"]Now the second partials should be

$\frac{\partial^2g}{\partial r^2}=\frac{\partial^2f}{\partial x^2}\cos^2\theta+\frac{\partial^2f}{\partial y\partial x}\cos\theta\sin\theta+\frac{\partial^2f}{\partial x\partial y}\cos\theta\sin\theta+\frac{\partial^2f}{\partial y^2}\sin^2\theta$

$\frac{\partial^2g}{\partial\theta^2}=\frac{\partial^2f}{\partial x^2}r^2\sin^2\theta-\frac{\partial^2f}{\partial y\partial x}r^2\cos\theta\sin\theta-\frac{\partial^2f}{\partial x\partial y}r^2\cos\theta\sin\theta+\frac{\partial^2f}{\partial y^2}r^2\cos^2\theta-\frac{\partial f}{\partial x}r\cos\theta-\frac{\partial f}{\partial y}r\sin\theta$[/quote]
Ooops ... never mind. I was thinking something else. I just noticed the pattern that

${\Big(\frac{\partial g}{\partial r}\Big)}^2={\Big(\frac{\partial f}{\partial x}\cos\theta+\frac{\partial f}{\partial y}\sin\theta\Big)}^2 = \frac{\partial^2f}{\partial x^2}\cos^2\theta+\frac{\partial^2f}{\partial y\partial x}\cos\theta\sin\theta+\frac{\partial^2f}{\partial x\partial y}\cos\theta\sin\theta+\frac{\partial^2f}{\partial y^2}\sin^2\theta$
had the form $(a+b)^2 = a^2 + 2ab + b^2$

The partial dertivative with respect to theta almost worked out but it had those extra terms. I thought you were using some kind of special "trick" other than the product rule.

my bad... :blush:


function
limit
calculus
calculus computations

\[ \begin{gathered} f\left( x \right) \equal{} \left\{ {\begin{array}{*{20}c} {\frac {1} {{\left| x \right|}}{\text{ }}\left| x \right| > 1} \\
{ax^2 \plus{} b{\text{ }}\left| x \right| \leqslant 1} \\
\end{array} } \right. \hfill \\
{\text{Found a value of a}}{\text{,b to }}f'\left( 1 \right){\text{ exist}}{\text{.}} \hfill \\
\end{gathered}\]
I would know if is a good idea to separate this function without abs. If i commit a wrong, help me please.

thanks

ps: i found 2 values of a (a=1/2, a=-1/2) thats right?thx

As always, the general idea is to look at limits of $ \frac{f(x)\minus{}f(1)}{x\minus{}1}$ from the left and right when $ x\rightarrow 1$. If and only if they both exist and are the same, $ f$ is differentiable at $ 1$. Absolute values don't really play a role here as we're only looking for differentiability at $ 1$.

For $ f$ to be differentiable at $ 1$, $ f$ needs to be continuous at $ 1$, meaning $ a\plus{}b\equal{}\frac{1}{|1|}\equal{}1$ (we could have taken care of that while calculating limits, but it's easier to do it at the beginning).

$ \lim_{x\rightarrow 1\minus{}}{\frac{f(x)\minus{}f(1)}{x\minus{}1}}\equal{}\lim_{x\rightarrow 1\minus{}}{\frac{ax^2\minus{}a}{x\minus{}1}}\equal{}2a$. On the other hand, $ \lim_{x\rightarrow 1\plus{}}{\frac{f(x)\minus{}f(1)}{x\minus{}1}}\equal{}\lim_{x\rightarrow 1\plus{}}{\frac{\frac{1}{x}\minus{}a\minus{}b}{x\minus{}1}}$. Since $ a\plus{}b\equal{}1$, we have $ \lim_{x\rightarrow 1\plus{}}{\frac{f(x)\minus{}f(1)}{x\minus{}1}}\equal{}\lim_{x\rightarrow 1\plus{}}{\frac{\frac{1}{x}\minus{}1}{x\minus{}1}}\equal{}\lim_{x\rightarrow 1\plus{}}{\frac{1\minus{}x}{x\cdot(x\minus{}1)}}\equal{}\lim_{x\rightarrow 1\plus{}}{\frac{\minus{}1}{x}}\equal{}\minus{}1$. So, $ 2a\equal{}\minus{}1$ would mean $ a\equal{}\frac{\minus{}1}{2}$, $ b\equal{}\frac{3}{2}$. This is, of course, the only solution.