algebra
polynomial
Rational Root Theorem

Prove that $ \sqrt[3]3\plus{}\sqrt 2$ is irrational.

hello, set $ \sqrt[3]{3}\plus{}\sqrt{2}\equal{}x$ and $ x$ is a rational number, then
$ x^3\plus{}6x\minus{}3\equal{}\sqrt{2}(3x^2\plus{}2)$
$ \frac{x^3\plus{}6x\minus{}3}{3x^2\plus{}2}\equal{}\sqrt{2}$
this is a contradiction.
Sonnhard.

We can split this expression into 2 parts.

For the $ \sqrt[3]{3}$ part:
It is a solution of the equation $ x^3\equal{}3$, or $ x^3\minus{}3\equal{}0$.
By the Rational Root Theorem, the possible rational roots are $ \pm1$ and $ \pm3$.
Using synthetic division and testing all of those roots for factor divisibility, we find that none of those give a remainder of 0. (too lazy to write out the division process)
Therefore there are no rational roots in $ x^3\minus{}3\equal{}0$, so the real roots must be irrational. Since $ \sqrt[3]{3}$ is a real root of that equation, it's irrational.

For the $ \sqrt{2}$ part: It's a solution of $ x^2\equal{}2$, or $ x^2\minus{}2\equal{}0$. By the RRT, the possible rational roots are $ \pm1$ and $ \pm2$. Using synthetic division, we find none of those values are roots because they don't yield remainders of 0. So there are no rational roots in $ x^2\minus{}2\equal{}0$. Thus the real roots must be irrational. $ \sqrt{2}$ is a real root of the equation, so it's irrational.

The sum of two irrational numbers is irrational. Therefore $ \sqrt[3]{3}\plus{}\sqrt{2}$ is irrational, thus completing the proof.  $ \blacksquare$

[quote="gauss1181"]The sum of two irrational numbers is irrational.[/quote]  Oy vey!

I don't know what oy vey means, but I'll hazard a guess that you have a problem with that statement.

Obviously $ \sqrt {2}$ and $ 2 \minus{} \sqrt {2}$ are both irrational, but their sum is rational.  Hence your idea is flawed.

However, you could say that $ \sqrt [3]{3}$ and $ \sqrt {2}$ are linearly independent over the rationals, and prove it as follows.

Suppose there are some integers so that $ a\sqrt [3]{3} \plus{} b\sqrt {2} \equal{} 0$.  This means that $ a\sqrt [3]{3} \equal{} \minus{} b\sqrt {2}$ and $ 9a^6 \equal{} 8b^6$.  Divide both sides by $ 9b^6$ and let $ \frac {a}{b} \equal{} c$.  Thus the problem is reduced to $ c^6 \equal{} \frac {8}{9}$ if $ c\in\mathbb{Q}$.  However, if $ c$ is reduced, then every factor in the top of $ c$ occurs 6 times in $ c^6$ in the top and is not canceled by anything on the bottom.  So that means that the reduced form of $ c^6$ must have a 6th power in the numerator.  However, 8 isn't.  Hence there are no integers $ a, b$ asked for and $ \sqrt [3]{3}$ and $ \sqrt {2}$ are independent over the rationals.  This means that for any nonzero pair $ a, b$, $ a\sqrt [3]{3} \plus{} b\sqrt {2}$ is not a rational.  For the answer, just choose $ a \equal{} b \equal{} 1$.

I hope this is correct...  :wink:

Sonnhard's answer would be a good answer if he had explained how he got to the cubic equation in $ x$. :|

He isolated $ \sqrt[3]{3}$ and cubed  :wink:

[quote="tenniskidperson3"]However, you could say that $ \sqrt [3]{3}$ and $ \sqrt {2}$ are linearly independent over the rationals, and prove it as follows.

Suppose there are some integers so that $ a\sqrt [3]{3} \plus{} b\sqrt {2} \equal{} 0$. ... Hence there are no integers $ a, b$ asked for and $ \sqrt [3]{3}$ and $ \sqrt {2}$ are independent over the rationals.  This means that for any nonzero pair $ a, b$, $ a\sqrt [3]{3} \plus{} b\sqrt {2}$ is not a rational.  For the answer, just choose $ a \equal{} b \equal{} 1$.[/quote]
I don't follow your logic.  I agree that the only integer solution of $ a \sqrt[3]{3} \plus{} b\sqrt{2} \equal{} 0$ is $ a \equal{} b \equal{} 0$.  How does that imply that $ \sqrt[3]{3} \plus{} \sqrt{2}$ is irrational?  If you wanted to make this approach work, you would have to show that $ \sqrt[3]{3}$, $ \sqrt{2}$, and $ 1$ are independent over the rationals.  For example, $ 2 \minus{} \sqrt{2}$ and $ \sqrt{2}$ are also independent over the rationals (in the sense you have written), but their sum is rational.


modular arithmetic
number theory unsolved
number theory

Given an arbitrary $ n\in \mathbb{N}$ and two single digit integers $ a,b\in \mathbb{N}$

Prove that there exist $ x> n$and $ y$ which is a muliple of 5 , $ x,y\in \mathbb{N},$ such that $ x^2\minus{}y^2\equal{}nab$

This problem is equivalent to proving that $ nab$ could be factorised into the product $ uv$ where $ u \equiv v \pmod{10}$. It turns out that this is not true. For a counter example, take $ a \equal{} 2$, $ b \equal{} 5$, $ n \equal{} 13$. (In this counter example, you cannot even find $ x,y$.)

[quote="mathangel"]This problem is equivalent to proving that $ nab$ could be factorised into the product $ uv$ where $ u \equiv v \pmod{10}$[/quote]

Wolud you please expain it in details.


Ελπίζω να μη βγαίνει πιο εύκολα απ' ότι έχω σκεφτεί γιατί αυτό που έχω σκεφτεί δεν είναι και τόσο εύκολο. :maybe:

Έστω πολυώνυμο

$ x^3\plus{}kx^2\plus{}12x\minus{}8\equal{}0$

Με δεδομένο ότι οι ρίζες του είναι όλες πραγματικοί να βρεθούν οι πραγματικοί αυτοί.

[quote="Protonios"]Ελπίζω να μη βγαίνει πιο εύκολα απ' ότι έχω σκεφτεί γιατί αυτό που έχω σκεφτεί δεν είναι και τόσο εύκολο. :maybe:

Έστω πολυώνυμο

$ x^3 \plus{} kx^2 \plus{} 12x \minus{} 8 \equal{} 0$

Με δεδομένο ότι οι ρίζες του είναι όλες πραγματικοί να βρεθούν οι πραγματικοί αυτοί.[/quote]

Kosta to k ti einai??akeraios, pragmatikos??

Αν απομονώσεις το κ και κάνεις μελέτη της συνάρτησης που προκύπτει στο β μέλος βλέπεις από την γραφική παράσταση ότι για κ>15/2 η οριζόντια ευθεία ψ=κ τέμνει την συνάρτηση σε 3 σημεία άρα έχεις 3 ρίζες απροσδιόριστες αφού και το κ δεν προσδιορίζεται Μήπως υπάρχει κάποια  παράληψη στην εκφώνηση?

Φυσικό είναι να έχει τρεις ρίζες .. αυτό είναι προφανές απο το θεμελιώδες θεώρημα της αλγεβρας.  Αν θέσεις όπου χ το $ y\minus{} \dfrac{k}{3}$τότε θα φτάσεις σε μια εξίσωση της μορφής
 $ y^3\plus{}py\plus{}q\equal{}0$ όπου $ p\equal{}12\minus{}\dfrac{k^2}{3}$ και $ q\equal{}\minus{}8\plus{}\dfrac{2k^3\minus{}108k}{27}$

Αυτή είναι ελλειπές πολυώνυμο 3ου βαθμού και αν προσέξετε είναι όμοιο με τον κύβο

$ (a\plus{}b)^3\equal{}3ab(a^2\plus{}b^2)\plus{}(a^3\plus{}b^3)$ όπου 3αβ=-p και $ a^3\plus{}b^3\equal{}\minus{}q$ άρα ψαχνουμε για τις λυσεις πλέον y=a+b

αν υψώσουμε σε κύβους το 3ab θα έχουμε $ a^3b^3\equal{}\minus{}\dfrac{p^3}{27}$ και αν υποθέσουμε πως a^3 και b^3 ρίζες του τριωνύμου $ (z\minus{}a^3)(z\minus{}b^3)$ τότε έχουμε το τριώνυμο:

$ z^2\plus{}(a^3\plus{}b^3)z\plus{}a^3b^3$ δήλαδη $ z^2\minus{}qz\minus{}\dfrac{p^3}{27}$ Βρείτε την διακρίνουσα του τριωνύμου και θα έχετε 2 ριζες τις

$ a^3\equal{}q/2\plus{}\sqrt{\Delta}$
$ b^3\equal{}q/2\minus{}\sqrt{\Delta}$

αφού γνωρίζουμε τα α και β βρίσκουμε το y και αντιστοιχα το χ. Tώρα αφού θες ολοι να εχει πραγματικές ρίζες πρέπει Δ<0 γιατί διαφορετικά το α εχει 3 λύσεις (1 πραγματικη και 2 μιγαδικες) και το β τα ίδια... Οπότε μόνο το Δ<0 σε βολέυει για να είναι πραγματικές οι ρίζες.... Αυτά, τα έγραψα πολυ περιληπτικά,συγχωρέστε με για τυχόν λάθος σε κάποιο πρόσημο...

Επειδή μίλησα με τον Κώστα και έχω καταλάβει την άσκηση θα προσπαθήσω να δώσω κάποιες απαραίτητες διευκρινίσεις (Κώστα εύχομαι να μη σε πειράζει...  :oops: ).

Κατ' αρχάς το $ k$ είναι οποιοσδήποτε πραγματικός.

Κατά δεύτερον νομίζω ότι πρέπει να γίνει μια διευκρίνηση ότι οι ρίζες πρέπει να είναι θετικές (πιθανόν βέβαια να μη χρείαζεται αυτό αλλά Κώστα αν το διευρύνεις και στους αρνητικούς πές το...εγώ πάντως αδυνατώ :blush: αν και πιστεύω ότι ισχύει...).

Με τα παραπάνω υπόψην είμαι (99%) σίγουρος ότι το πρόβλημα είναι σωστά δοσμένο.

Τέλος θέλω να πω ότι η άσκηση είναι πολύ ωραία και ο Κώστας την έχει βάλει σε μια πολύ όμορφη και έξυπνη μορφή :wink:

Λοιπόν ότι ακολουθεί παρκάτω είναι ότι πιο κομψο έχω βρει (αν κάτι τόσο ugly μπορεί να θεωρηθεί κομψό)

Λοιπόν από Vieta έχουμε $ x_1\plus{}x_2\plus{}x_3\equal{}\minus{}k$ άρα ο $ k<0$  (αφού όπως είπε ο mim οι ρίζες είναι θετικές)
Έστω $ f(x)\equal{}x^3\plus{}kx^2\plus{}12x\minus{}8$
$ f'(x)\equal{}3x^2\plus{}2kx\plus{}12$ .Από το θεώρημα Rolle έχουμε ότι η f' έχει δύο ρίζες άρα η διακρίνουσα αρνητική άρα $ k^2\geq 36$ άρα $ k\leq \minus{}6$ . Τότε οι ρίζες της f' είναι  $ k_1\equal{}\frac{\minus{}k\minus{}\sqrt{k^2\minus{}36}}{3}$ και $ k_2\equal{}\frac{\minus{}k\plus{}\sqrt{k^2\minus{}36}}{3}$

Τότε λόγω της μονοτονίας θα πρέπει να υπάρχει μία ρίζα σε καθένα από τα επιμέρους διαστήματα $ (\minus{}\infty,k_1)$
$ (k_1,k_2)$ kai $ (k_2,\plus{}\infty)$

Παρατηρούμε ότι $ f(k_1)f(k_2)\equal{}\frac{16(k\plus{}6)^2(15\minus{}2k)}{27}\geq 0$ άρα πρεπει

An $ f(k_1)f(k_2)>0$ τότε η f έχει το μία πραγματική ρίζα επομένως πρέπει $ f(k_1)f(k_2)\equal{}$ δηλαδή $ k\equal{}\minus{}6$

Για $ k\equal{}\minus{}6$ $ f(x)\equal{}(x\minus{}2)^3$   :D   qed

Αν και μπορούμε με $ ab \plus{} bc \plus{} ca\geq 3\sqrt [3]{a^2b^2c^2}$ δηλ $ abc\leq 8$ ktl  :blush:

Εαν είναι έτσι οπως τα λεει ο Silouan τότε μια νομίζω πιο εύκολη λύση είνα: Η εξίσωση μετασηματίζεται σε κ=φ(χ) η φ΄(χ)>0 για χ>0 οπότε η ευθεία ψ=κ τέμνει σε ενα μονο σημείο την φ (φ(R+)=R) αρα η εξίσωση δεν μπορεί να εχει δυο η και περισσότερες διακεκριμένες λύσεις και επειδή ειναι πολυώνυμο 3ου βαθμου θα είναι της μορφής (χ-ρ)^3 οπότε για να ταυτίζεται με την χ^3 + κχ^2 +12χ -8 πρέπει εύκολα ρ=2

Αυτό με την ανισότητα είχα σκεφτεί Σιλουανέ. Κάπως διαφορετικά βέβαια αλλά ήταν τσάμπα κόπος απ' ότι μου είπε και ο Δημήτρης. Ωραίος. :wink:

Εδώ θέλω να κάνω την εξής ερώτηση... αν πράγματι οι ρίζες μπορεί να είναι αρνητικές το πολυώνυμο έχει και άλλες πιθανές ρίζες κι αν ναι μπορούμε να τις βρούμε:?:

Για κ>15/2 έχουμε 3 διακεκριμένες ρίζες οι οποίες είναι οι τετμημένες των σημείων τομής της φ(χ) και της ψ=κ και όπως φαίνεται απο την γραφική παράσταση για κάθε διαφορετικό κ υπάρχει και άλλη τριάδα λύσεων 2 αρνητικές και 1 θετικήΗ εύρεσή τους μπορεί να γίνει συναρτησει του κ αλγεβρικά όπως γράφτηκε στα προηγούμενα post's


I have a few questions I am having trouble with can anyone help!!  Assume the diameter of the field of vision in your microscope is 2mm under low power.  If one Bacillus cell is 2um, how many Bacillus cells could fit end to end accross the field?  How many 10um yeast cells could fit across the field?  If anyone knows the answer could you show me all of your work I can't seem to get these.  Thank you!!!!

Remember, milli- is one thousandth. micro- is one millionth. there are 1,000 micrometers in a millimeter. so the diamater of your view field is 2,000 micrometers.  divide that by two for the bacillus and you will be able to see 1,000 bacilli end on end in your viewfield. divide 2,000 by 10 for the yeasts and you will be able to fit 200 yeast cells in your view field. hope that helps.


Find the sum of all palindromes of five digits, that is, find

10001+10101+...+99999

[hide]You can tell that there will be 9*10*10=900 total palindromes, since the first 3 digits will define the last 2 digits of a 5-digit palindrome. Now each digit place can be calculated independently to find the sum.
For each possible value of the units digit (1-9), there will be 100 combinations. Thus, each digit from 1-9 will be added in the units place 100 times, and in total will add up to 45*100=4500. This logic applies similarly for the first digit.
For each possible value of the tens digit (0-9), there will be 90 combinations, so each digit from 0-9 will be added 90 times, giving 45*90=4050 for the tens digit. Similar logic applies for the hundreds and thousands digit.
Adding up all the sums in each digit gives $ 4500\cdot(10000\plus{}1)\plus{}4050(1000\plus{}100\plus{}1)\equal{}49500000$.[/hide]

[hide="Another Way"]

Another way to solve this problem would be to find the average, and then easily find out the sum. To find the average, we can find the average of each digit. Doing this, the average is 55000. There are $ 9 \times 10 \times 10 \times 1 \times 1$ or $ 900$ 5-digit palindromes, so $ 55000 \cdot 900 \equal{} \boxed{49500000}$. [/hide]


Few problems have already been posted, so I decided to post rest of the problems too (as given here: http://www.ichb.ro/concursul-stelele-matematicii-2008.html ).

Problem 1. http://www.mathlinks.ro/viewtopic.php?t=250823
Problem 2. http://www.mathlinks.ro/viewtopic.php?t=250836
Problem 3. http://www.mathlinks.ro/viewtopic.php?t=250837

Problem 4. http://www.mathlinks.ro/viewtopic.php?t=250839
Problem 5. http://www.mathlinks.ro/viewtopic.php?t=245324
Problem 6. http://www.mathlinks.ro/viewtopic.php?t=250841

Why are most of the threads deleted above? I already asked Cezar Lupu to post the Stars of Mathematics 2008 problems but he did not get around it yet. I appended the problems with solutions.

Sorry orl, error in linking. I've edited the original post.


Find the derivative of the equation y = (1 +  :rt5:)/ 2x

[tex]\displaystyle \frac{1+\sqrt{5}}{2x} = \frac{1+\sqrt{5}}{2}*x^{-1}[/tex]

So, wouldn't that just be [tex]\displaystyle -\frac{1+\sqrt{5}}{2x^2}[/tex]?

[quote="nr1337"][tex]
\displaystyle \frac{dy}{dx} = f^{\prime}(\phi \cdot x^{-1})
= x(0) - \frac{1+\sqrt{5}}{2x^2}
= -\frac{1+\sqrt{5}}{2x^2}
= -\frac{\phi}{x^2}
[/tex]

by the the product rule for derivitaves.  (phi = golden ratio)[/quote]

There is no need for the product rule here, prob's solution is correct...

Well... you could use any rule you want, but it's superfluous.

I could take the derivative of f(x)=x^2 using the product or quotient rule... that doesn't mean it should be done, and it would be a waste of time.


analytic geometry
graphing lines
slope
calculus
calculus computations

Suppose that the tangent line at a point P on the curve y = x^3 intersects the curve again at a point Q. Show that the slope of the tangent at Q is four times the slope of the tangent at P.

I know i didn't show any work for this question but i was wondering i could get an idea on how should this be started?

Let the tangent point $P(a,\ a^{3})$ on the curve $y=x^{3},$ the tangent line at $P$ is $y=l(x).$ Solving the equation $x^{3}=l(x),$ we have the another abscissa of the intersection point $Q$ is $Q(b,\ b^{3}),\ b$ is expressed by $a.$

Just wondering if u meant abscissa cause i was trying to look for the definition.

Sorry can you find the equation of tangent line at $P$?


geometry
linear algebra
matrix
analytic geometry
area of a triangle
Heron\u0027s formula

Find area of a triangle formed by 3 lines whose equations are:
$y-x=6$
$x-2y=3$
$x+y=6$

1. Find the intersection of each pair of lines. This will give you the three needed points.
2. Find the distance between those points ($d_{A,B}=\sqrt{(a_x-b_x)^2+(a_y-b_y)^2}$)
3. Apply Heron's formula to get the area ($[ABC]=\sqrt{s(s-a)(s-b)(s-c)}$)

hmm easier way than finding distances... that gets kinda messy

find intersections, then use shoelace method to find area, or that matrix thing where you plug in the $x$ coordinates in the first column, $y$ in the second, and $1$ in the third. the determinant is supposed to be the area too  :D

There's a still faster way for these particular lines :yup:


Let $ X,Y$ be topological spaces, $ f: X\to Y$ and let $ x\in X$. Prove that if the sequence $ \left\langle f(x_n)\right\rangle$ converges to $ f(x)$ for every sequence $ \left\langle x_n\right\rangle$ that converges to $ x$, then $ f$ is continuous at $ x$.

Is it required that $ X$ satisfy the first axiom of countability?

First-order countability is sufficient, but not necessary for the above statement to be true. It is false in general topological spaces.
http://en.wikipedia.org/wiki/Sequential_space


Does anyone know where I can get hold of a copy of 2008 National test without having to pay money?  :maybe:

Hm... you can politely ask one of our coaches who was there this year! :lol:

I was just wondering if someone put them up on a site or something...does anyone know where I can get [b]any[/b] nationals test?

They cannot be released publicly. it's against MATHCOUNTS copyright policies to post tests publicly. If you really want a copy of the Nationals tests, PM someone and ask them if they have any. Tell them to email them to you, since exchanging must be done privately.

If you don't get why MATHCOUNTS doesn't allow people to view Nationals test, go under MATHCOUNTS Resources Here! under the Stickies thread.

All I currently know about is that the Minnesota Mathcounts website carries all the chapter and state tests as of 2005 (I think). Just google "minnesota mathcounts" and there you'd find it.

Additionally, the site Mathcounts Central has some problems similar to the past test problems, just with some modifications made. Google "mathcounts central" and you should find it.

If you really wanted to know the real reason why MATHCOUNTS doesn't allow people anywhere, even on AoPS, to release copies of the previous nationals tests or any other MATHCOUNTS competition, here's the reason why:

MATHCOUNTS is only a non-profit organization that's run by very few people who only volunteer to work at MATHCOUNTS. They do not get paid very much for it. And MATHCOUNTS has to sell the tests on their online store to help make money and support the organization. Now, that might leave you wondering why they are doing that. The true fact is, they aren't simply only making the money, they just need it for the supporting of their organization, and they do that by charging fees for its products rather than the competitions, because the competitions are all big and are covered for by the sponsoring engineering companies. It's obviously not fair to MATHCOUNTS if they were to only give out the tests for free and let people post them publicly on the website, because then otherwise MATHCOUNTS would be losing money and not be able to run their organization in terms of the non-competition components. When someone is asking, "why not free past tests?", that person is simply asking a similar thing in terms of why Ford doesn't give people their cars for free. MATHCOUNTS only allows release of chapter and state tests back to a certain recent year; they cannot just release ALL of the tests from the 21st century.

So basically, the moral of the story is, if you want past tests, ask a friend to send them to you via email.


Find all continous functions $f: \mathbb{C} \to \mathbb{C}$ such that for all $z\in \mathbb{C}$ we have:

\[ f(z)=f(\Re \{z\})+f(i \ \Im \{z\}) \]

Note that f(0)=0
f(x+iy) = f(x) + f(iy)
Define h(x):R->R such that h(0)=0 and g(y):I (pure imaginary) -> C such that g(0)=0. And such that both of them are continious.
then define f(x+iy)= h(x) + g(iy)
it is easy to see that this is a set of solutions.

f(x+iy) = f(x) + f(iy) implies that we are done once we know f(x) on R and on I.
Clearly f has to be continous on both. Hence the class of solutions above is also the only one.

Note: This not similar to cauchy problem. Infact once you remove the continouty condition the problem remains the same:
the set of solutions would still be the same except this time h and g only have to vanish at zero.

-a


aya har noghteye faza ra mitavan ba yek adade haghighi namayesh dad :?:
(aye $R$ va $R^{n}$ham arzand?)

Inja ro negah kon :

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=line+plane&t=63979


Let $a,b,c$ be real numbers such that $ab\not= 0$ and $c>0$. Let $(a_{n})_{n\geq 1}$ be the sequence of real numbers defined by: $a_{1}=a, a_{2}=b$ and
\[a_{n+1}=\frac{a_{n}^{2}+c}{a_{n-1}}\]
for all $n\geq 2$.
Show that all the terms of the sequence are integer numbers if and only if the numbers $a,b$ and $\frac{a^{2}+b^{2}+c}{ab}$ are integers.

Sorry for bumping this thread :)

Define $ \alpha = \frac{a^{2}+b^{2}+c}{ab}$. Notice that $ a_{n+2}a_{n}-a_{n+1}^{2}= c = a_{n+3}a_{n+1}-a_{n+2}^{2}$ for all values of $ n$. Hence $ \frac{a_{n}+a_{n+2}}{a_{n+1}}= \frac{a_{n+1}+a_{n+3}}{a_{n+2}}$ for all values of $ n$, so that $ \frac{a_{n}+a_{n+2}}{a_{n+1}}$ is independent of $ n$. Evaluating for $ n = 1$ gives $ \frac{a+\frac{b^{2}+c}{a}}{b}= \alpha$, so that $ \frac{a_{n}+a_{n+2}}{a_{n+1}}= \alpha$ or $ a_{n+2}= \alpha a_{n+1}-a_{n}$. If $ \alpha$ is an integer, then all terms of the sequence are trivially integers. Suppose that $ \alpha$ is not an integer and that all terms are integers. Write $ \alpha = \frac{p}{q}$ where $ p$ and $ q$ are coprime integers and $ q > 1$. An easy induction shows that $ q^{n}$ divides $ a_{n+1}$ for all values of $ n$, so $ q^{2n}$ divides $ a_{n+2}a_{n}-a_{n+1}^{2}$. Since $ a_{n+2}a_{n}-a_{n+1}^{2}= c$, this is a contradiction (for $ n$ large enough).

[quote=Arne]Sorry for bumping this thread :)
An easy induction shows that $ q^{n}$ divides $ a_{n+1}$ for all values of $ n$[/quote]
Excuse me, but I'm so blind that I can't see how this simple induction goes.
$q^n|a_{n+1},\ q^{n-1}|a_n$ doesn't imply $q^{n+1}|a_{n+2}= \alpha a_{n+1}-a_{n}$


LaTeX
superior algebra
superior algebra unsolved

hello , can you help me
Let A be a ring  prove that


function
calculus
integration
calculus computations

Given f(x) is a convex function which is twice differentiable and $ f'(x) > 0$ in an interval [a,b] prove that

$ \frac {f(b) \minus{} f(a)} {b \minus{} a} \geq \frac {f(x) \minus{} f(a)}{x \minus{} a}$ for $ x \in [a,b]$

Geometrically it is plain to see, but I would like to have a calculus approach. I have a hunch it has to do with Taylor Series

I'm not familiar with the term "convex" describing a function.  Does it mean the same thing as "concave down"?  (I'm familiar with the terms concave up and concave down.)

BTW, the average value of a "nice" function $ g(x)$ on the interval $ [u, v]$ is

$ \frac{\int_u^v g(t) \ dt}{v \minus{} u}$

So, you are asking how to show this:

$ \frac{\int_a^b f'(x) \ dx}{b \minus{} a} \geq \frac{\int_a^x f'(t) \ dt}{x \minus{} a}$

Notice that the interval $ [a, x]$ is just a portion of the entire interval $ [a, b]$.

I hope this reasoning is correct.

Rewrite it in the form $ (x \minus{} a)(f(b) \minus{} f(x)) \ge (f(x) \minus{} f(a))(b \minus{} x)$
$ (x\minus{}a)f(b) \plus{} (b\minus{}x)f(a) \ge f(x) (a\minus{}b)$
$ \frac{x\minus{}a}{a\minus{}b} f(b) \plus{} \frac{b\minus{}x}{a\minus{}b}f(a) \ge f(x)$
Then by definition of convexity.

The term "convex" means the same as "concave up". The region above the graph of a convex function is a convex set.


limit calculus derivative LaTeX induction algebra polynomial

Let $ k, n \in \mathbb{N} $ and $ f(x)= \frac{1}{x^k-1}$. Define $P_n(x) = (x^k-1)^{n+1}\cdot f^{(n)}(x)$.
Find $ \lim_{x \to 1} P_n(x)$.

Note : $ f^{(n)}(x) $ is the $ n^{th} $ derivative of $ f(x) $.

[i]Edited by Myth[/i]

Hi, nttu!

Please, read http://www.mathlinks.ro/viewtopic.php?p=140098 .
And properly name problems next time!

Regards,
Myth

I've found $n! (-1)^n k^n$, am I right?

Pierre.

[quote="pbornsztein"]I've found $n! (-1)^ k^n$, am I right?[/quote]
I can't check so short solution.  :lol:  :lol:  :lol:

Hmmm... bad latex code. I will edit it.
I will write the computations if it is ok, otherwise it is boring.

Pierre.

Why do you say it is boring? Actually, it very simple and your answer is correct.

Ok! I will write it.
We can write $P_{n}=P'_{n-1}(x)(x^k-1)-nkx^{k-1}P_{n-1}(x)$. So $P_n(1)=-nkP_{n-1}(1)$, hence result!  :D

Ok, so it is easy to prove by induction that $f^{(n)}(x) = \frac {Q_n(x) } {(x^k - 1)^{n+1}}$ where $Q_n(x)$ is a polynomial satisfying the relation $Q_{n+1}(x) = Q'_n(x)(x^k - 1) - (n+1)kx^{k-1}Q_n(x).$

Thus $P_n(x) = Q_n(x)$, and $P_{n+1}(1) = -k(n+1)P_n(1)$.
From $P_0(x)=1$, we easily deduce that $P_n(1) = n!(-1)^n k^n$ as claimed.

Pierre.

Aaargh....too late. I wrote it for nothing!  :(

Pierre.


Find all nonnegative integer $ x,y,z,t$ satisfying:
\[ 2^x3^y \minus{} 5^z7^t \equal{} 1
\]

See here http://www.mathlinks.ro/Forum/viewtopic.php?p=155821#155821 .


A cellular telephone company has two plans. Plan A charges $11 a month and$0.21 per minute. Plan B charges $20 a month and$0.10 per minute. After how much time, to the nearest minute, will the cost of plan A be equal to the cost of plan B?

(1) 1 hr 22 min         (3) 81 hr 8 min
(2) 1 hr 36 min         (4) 81 hr 48 min

[hide]$11+0.21x=20+0.10x$

$0.11x=9$

$x\approx1\text{ hour }22\text{ minutes}\Rightarrow\boxed{\text{A}}$[/hide]


The radius of a particular circle inscribed in an equilateral triangle is 2 units.  What is the perimeter of the triangle?  Express your answer in simplest radical form.

Answer: $ 12\sqrt3$

My answer: $ 6\sqrt3$; Why:

Note that the circumcenter is the same as the median, so the length of the altitude is $ 3\cdot2 \equal{} 6$. To calculate [i]half of a side[/i], we have $ \frac6{\sqrt3} \equal{} 2\sqrt3$, so each side is of length $ 4\sqrt3$, and the perimeter is $ 12\sqrt3$.

These posts are pointless; if someone asks for answers he/she will post them.
---EDIT---
Please don't mind this post; I didn't understand the problem. I'm sorry!
I'll ask a mod to delete it.

Uh, in your drawing, you've drawn the circle that the triangle is inscribed in, i.e. the circumcircle.

akhil0422, you can just press that weird envelope thing in the bottom right corner to delete the post. That's only if your post is the most recent in the topic, so too bad.

[geogebra]1f8febb3f082adca6fce197599da501edfe7caed[/geogebra]

Forgive me for my awful Geogebra skills :( .

[hide="solution"] That triangle that has the center of the incircle as a vertex is a 30-60-90 triangle. Since the radius is two, the other leg is $ 2\sqrt {3}$ and the side of the equilateral triangle is thus $ 4\sqrt {3}$. The perimeter is then $ 12\sqrt {3}$.[/hide]


Please, does anyone know anything about this terrible inequality:

If $a,b,c\geq0$, then $(a^2+b^2+c^2)^2\geq3(a^2b+b^2c+c^2a)$.

You probably meant this one: $(a^2+b^2+c^2)^2\geq3(a^3 b+b^3 c+c^3 a)$ (look at the powers on the right hand side; the inequality must be homogeneous). This is very hard indeed; it was discussed before on

http://mathlinks.ro/viewtopic.php?t=6026

(this is actually the discussion [i]Advanced Section \ Inequalities \ Proposed & Own Problem \ ineq3[/i]).

  Darij

[quote="darij grinberg"]http://mathlinks.ro/viewtopic.php?t=6026&sid=5b5def6f246dd8cb31da60cc7a35d8b7

But it doesn't look like the link wants to work,
  Darij[/quote]

The link should work if you remove the last part:

http://mathlinks.ro/viewtopic.php?t=6026


geometry
circumcircle
geometric transformation
reflection
parallelogram
geometry proposed

Contruct equilateral triangle $ BCD, CAE, ABF$ outside a triangle $ ABC$. Call $ M, N, P$ are intersection of $ AD, BE, CF$ with circle $ (DEF)$.
Prove that: $ AM \plus{} BN \plus{} CP \equal{} AD$.

use these :
if a point $ P$ is inside a circle and three distinct chords called $ AB,CD,EF$ passes through it such that $ \angle APC\equal{}\angle CPE\equal{}60^\circ$ then we have :
$ PA\plus{}PD\plus{}PE\equal{}PB\plus{}PC\plus{}PF$
if a point $ P$ is on the circle $ (O)$ and $ ABC$ is an equilateral triangle with circumcenter $ (O)$ then one of the distances $ PB,PA,PC$ is equal to the sum of two others.

[b]An interesting result:[/b]
Denote $ J,L,K$ the reflections of $ M,N,P$ wrt $ A,B,C$, respectively. Let $ T,O$ be Fermat point and circumcenter of triangle $ ABC$. Prove that the center of $ (JLK)$ lies on $ OT$.

$T \equiv AD \cap BE \cap CF$ is the 1st Fermat point of $\triangle ABC,$ i.e. $T$ is the common point of circles $\odot( DBC),$ $\odot( ECA),$ $\odot(FAB).$ It's clear that $T$ is also the 1st Fermat point of $\triangle DEF$ and $\triangle MNP.$ Let $P_1,Q_1,R_1$ be the circumcenters of $\triangle TNP,$ $\triangle TPM,$ $\triangle TMN$ and let $P_2,Q_2,R_2$ be the circumcenters of $\triangle TEF,$ $\triangle TFD,$ $\triangle TDE.$ Equilateral $\triangle P_1Q_1R_1$ and $\triangle P_2Q_2R_2$ are the outer Napoleon triangles of $\triangle MNP$ and $\triangle DEF,$ respectively.

Let $U$ be the second intersection of circumcircles $(R_1),(Q_2)$ and $V \equiv DE \cap TU.$ Since $\angle MDV=\angle MNE=\angle MUV,$ it follows that $V \in \odot(UMD).$ Similarly, $MP \cap UT$  lies on $\odot(UMD)$ $\Longrightarrow$ $V \equiv MP \cap DE$ lies on $UT.$ Analogously, intersections $[T,S] \equiv (R_2) \cap (Q_1),$ $W \equiv MN \cap DF$ and $T$ are collinear. But, by Pascal theorem for the cyclic hexagon $DFPMNE,$ the points $V,W,T$ are collinear $\Longrightarrow$ $U,S,T$ are collinear. Thus, it follows that $R_1Q_2 \perp UTS \perp Q_1R_2.$ Since $Q_1R_1 \perp AT \perp R_2Q_2,$  then $R_1Q_1R_2Q_2$ is a parallelogram $\Longrightarrow$ $R_1Q_1$ and $R_2Q_2$ are congruent and parallel $\Longrightarrow$ $\triangle P_1Q_1R_1$ and $\triangle P_2Q_2R_2$ are congruent. Thus, by Viviani's theorem for $T$ inside  $\triangle P_1Q_1R_1$ and $\triangle P_2Q_2R_2$ we get

$TM+TN+TP=\sqrt{3} \cdot R_1Q_1=\sqrt{3} \cdot R_2Q_2=TD+TE+TF$

$\Longrightarrow \ TA+TB+TC+AM+BN+CP=TD+TE+TF \ \ (\star)$

By Ptolemy's theorem for cyclic quadrilaterals $TBDC, TCEA, TAFB$ we obtain:

$TD=TB+TC \ , \ TE=TC+TA \ , \ TF=TA+TB$

Combining these with $(\star),$ we get  $AM+BN+CP=TA+TB+TC=AD.$


find all integer solutions of
\[ 2x^2+5y^2=11(xy-11) \]

2x^2+5y^2=11(xy-11)
Answer below

[hide]Rearranging you get 2x^2+-11xy+5y^2=-121
Factor to (2x-y)(x-5y)=-121
Since x and y have to be integers, 2x-y and x-5y also have to be integers, so you have to represent -121 as a product of two integers, in which case they are either 11 and -11, which by using 2x-y=11 and x-5y=-11, you get y=11/3 and x=22/9, which aren't integers. The other case is 2x-y is +-1 or +-121, in which the only integer answers come from 2x-y=+-1 and x-5y=-+121 which gives y=+-27 and x=+-14.[/hide]


trigonometry
calculus
calculus computations

Let $ a\in\mathbb{R}$, and $ (x_{n})_{n\in\mathbb{N}_{0}}$ bounded sequence, such that

$ x_{n\plus{}1}(1\minus{}a x_{n}) \equal{} x_{n}\plus{}a$,$ n\in\mathbb{N}_{0}$.

Prove that $ (x_{n})_{n\in\mathbb{N}_{0}}$ is periodic.

[hide="hint"]Let $ a \equal{}\tan\alpha$ and $ x_{n}\equal{}\tan t_{n}$.  Then our equation is $ \tan t_{n\plus{}1}\equal{}\tan(t_{n}\plus{}\alpha)$, so we can take (for example) $ t_{n}\equal{} t\plus{}n\alpha$.[/hide]


calculus
integration
floor function
trigonometry
function
real analysis
advanced fields

compute:
\[ \int_{[0,2]}(\lfloor\sin x\rfloor \plus{} 1_C)d\theta\]

where $ 1_C$ is a char. function of $ C$ - Cantor set;
$ \theta\equal{}\lambda\plus{}\delta_{1/6}$, here $ \lambda$ is the Lebesgue measure, $ \delta$ means Dirac-delta

Looks like a silly problem.  $ \lfloor \sin x\rfloor$ agrees with $ 1_{\{\pi/2\}}$ on $ [0,2]$, so your integrand is just a characteristic function of a (Lebesgue) null set, so you don't need to worry about the integral wrt Lebesgue measure.

Next question:  Is $ 1/6$ in the Cantor set?

It can be shown that the Cantor set $ C \equal{} \{\sum_{k \equal{} 1}^{\infty}\frac {a_k}{3^k}~|~a_k\in\{0,2\}\}$. Therefore $ \frac {1}{6} \equal{} 0.0\overline{1}_3 \notin C$


trigonometry geometry quadratics function calculus AMC AIME

I'm planning to take the next COMC

COMC = Canadian Open Mathematis Challenge

For some students who know it, you would probably also know Pascal, Cayley and Fermat.

I took Pascal this year and I did really bad...getting only a 110.

With this kind of math level, what should I do to prepare for 2004's COMC?

Much help is needed.  Thanks to all.

your best bet is to
1) learn all the basic stuff (trig, geometry, quadratics, functions)
2) do previous contests
3) buy a book

i think CMS publishes books like "problems and how to solve them" (something like that)... those arent bad...  (try not to buy books that contains only problems)
A little more advanced book would be A Mathematical Mosiac by Ravi Vakil
with some hard work, you should obtain at least an 60 on next year's COMC

Also, do you know if this past year's COMC problems are online anywhere?

Yes, I've been there, but I haven't seen the COMC problems.

[quote="bugzpodder"]A Mathematical Mosiac by Ravi Vakil[/quote]

I have a copy of that book. It isn't really contest oriented, it's more like bedside reading.

Suprisingly, COMC doesn't really a lot of advanced knowledge, most of its stuff are pretty basic. Of the past two years, stuff like log or trig never came up. But nevertheless familiarize yourself with them.

The most important advice: do [b]a lot[/b] of problems.

"most of its stuff are pretty basic"
We all know ppl like Yufei is a genius (IMO team in grade ten), but for some of us this stuff is pretty hard.  For example, in this year's COMC one needs to know about characteristic equations (only a few, -- I guess Yufei included -- can derive the result without seeing it before)

and you DEFINATELY needs to familiarize yourself with basic topics if you want to be any good at math.  it doesnt matter if it has a small chance of showing up in contests or not.

"It isn't really contest oriented, it's more like bedside reading."
exactly why I recommended it....  I like books that raises one's interests...  Bill, only a few of us can dive into books like Problem Solving Strategies, Mathematical Olympiad Challenges, IMO/Putnam from 1969 - 1989 (or whenever).  but I am sure that you have all of those books too already too, right?

[quote="bugzpodder"]IMO team in grade ten[/quote]

:lol: Don't rush to judgement yet, I'm still a "potential" team member, and as far as I know, there's about 15 of them (yourself included :)).

On the other hand, I'm still waiting for the CMO results. CMS told me that their press release on the CMO would come out before May, but it's still nowhere in sight (the CMO site still says [i]The 35th CMO took place on Wednesday March 26th, 2003[/i] :evil:) .

Thanks.  Somehow I didn't see that.

[quote="kool_dudy"]You need Adobe Acrobat...

Here's a link to 2003's COMC...

[url]http://contest-cemc.uwaterloo.ca/past_contests/2003/2003_comc_contest_e.pdf[/url][/quote]

Thanks to all your suggestions.

The book that bugz talked about...you know where I can get in besides online stores...

[quote="Osiris"]COMC 2003 #10 was Newton Sums, right? That required specific knowledge. I studied them (and subsequently forgot), and spent an unforgivable amount of time on a).

#9 (or whichever the goat-pasture problem was) was also ridiculous. Wouldn't it be so much easier with Calculus?

Anyhow, it wouldn't hurt to delve a little into specific knowledge. Not too much though.

I only got 59. IT was pure luck that I got invited to CMO; I messed up on time management. I concur with billzhao: do a lot of problems so you know exactly how to proceed when you do them on the contest.[/quote]

How exactly did you get invited?  I know people with 60-62 who did not get invited.

I wrote CMO 2003, maybe that gave me a leg up? I don't know.

I think Olymon get can you invited. But I can't really be sure.

Orisis, on your CMO invitation letter, did it say the reason that you were invite for?

It actually stated that I did well on the COMC, which I myself suspected of being bogus. Maybe it was the Olymon.

Do any of you have any more suggestions?

Thx

I'm going to be doing to COMC hopefully this year (will have finished gr11 math since I did 10 and 11 this year). I printed off a huge stack fo questions (Fermat, AMC, AIME, USAMO, RMO, Balkan, IMO, COMC) and will just work through it as far as I get over the holidays. If I can't get one, I'll ask on the site and be happy to get a nice, quick answer:)

btw, I got 136 on Pascal and 138 on Cayley

You guys should start doing OLYMON problems if you haven't already:

http://cms.math.ca/Competitions/MOCP/

You should send in your solutions so Mr. Barbeau can mark them and give feedback.

[quote="kool_dudy"][url]http://www.cemc.uwaterloo.ca[/url]

Go to Results for err....results

and Past Contest & Solutions for past contests and solutions[/quote]

This site has COMC going back to 2000-2001, for COMC's back until 1996, go to http://www.cms.math.ca/Competitions/ and click on COMC

On an unrelated note, hi all, I'm new here... (I'm actually in the same boat as some of you as I will be taking the COMC next year (in my gr 11 year)) ... hope to have many future exciting discussions (if that's what you do here)


Two medians drawn from vertices A and B of triangle ABC are perpendicular. Prove that side AB is the shortest side of ABC.

[hide="Hint"]If the medians are AA' and BB' and the centroid is T, then AT:TA'=2:1, BT:TB'=2:1.
So, we got 3 right-angled triangles in which we can use Pythagora's theorem. ...[/hide]


trigonometry
inequalities
function
inequalities proposed

1) Let $ x,y,z$ be three real numbers such that $ x,y,z \ge - \frac {_1}{^4}$ and $ x + y + z = 1.$ Find the maximum value of $ \sqrt {4x + 1} + \sqrt {4y + 1} + \sqrt {4z + 1}.$

2) Let $ a,b,c,d,e$ be real numbers such that $ a + b + c +d + e =8$ and $ a^2 + b^2 + c^2 + d^2 + e^2 = 16.$ Find the maximum value of $e.$

3) $ a,b$ are real numbers. Find the maximum and minimum value of $ a \cos \phi + b \sin \phi.$

Solution to 3)

Let $ r > 0$ and $ \theta$ be two reals such that $ a \equal{} r\cos\theta , b \equal{} r\sin\theta$. :roll:
We have,
$ a^2 \plus{} b^2 \equal{} r^2(\sin^2\theta \plus{} cos^2\theta) \equal{} r^2\implies r \equal{} \sqrt {a^2 \plus{} b^2}$. :lol:
Now, $ A \equal{} a\cos\phi \plus{} b\sin\phi \equal{} r\cos\theta\cos\phi \plus{} r\sin\theta\sin\phi \equal{} r\cos(\theta \minus{} \phi)$
We have, $ \minus{} 1\le\cos( \theta \minus{} \phi )\le1\implies \minus{} r\le A\le r$.
$ \implies \minus{} \sqrt {a^2 \plus{} b^2}\le A\le \sqrt {a^2 \plus{} b^2}$. :oops:
[edited to fix a mistake]

Solution to 1)

From cauchy's inequality,
$ A^2 = (\sum\sqrt {4x + 1})^2\le 3(4 + 3) = 21$        $ (\as a + b + c = 1)$
Hence $ A\le \sqrt {21}\implies A_{max} = \sqrt {21}$
[hide]
The condition $ x, y, z\ge - \frac {1}{4}\implies$ no radical of the LHS is negative.[/hide]

hello, we have $ \sqrt{4x\plus{}1}\plus{}\sqrt{4y\plus{}1}\plus{}\sqrt{4z\plus{}1}\le \sqrt{21}$ and the equal sign holds if $ x\equal{}y\equal{}z\equal{}\frac{1}{3}$.
Sonnhard.

for the first problem, we have the function $ f(a)\equal{}\sqrt{a}$ is concave the by Jensen,
$ \sum \sqrt{4a\plus{}1} \leq \sqrt{21}$  with equality if $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$
for the second problem, it's really very nice,
by Cauchy-Schwarz,
$ 4(a^2\plus{}b^2\plus{}c^2\plus{}d^2)\geq (a\plus{}b\plus{}c\plus{}d)^2$
$ \Leftrightarrow 4(16\minus{}e^2)\geq (8\minus{}e)^2$
$ \Leftrightarrow 5e^2\minus{}16e \leq 0$
$ \Leftrightarrow e \leq \frac{16}{5}$


Prove that two congruent regular hexagons can be cut up into (altogether) $6$ parts such that these $6$ parts can be composed to form an equilateral triangle (without gaps or overlaps).

Correct. A nice solution, BTW. Mine was a bit more difficult:

  Darij

[quote="darij grinberg"][b]PS.[/b] The attachments upload mechanism works somewhat strangely, so the order of the pictures is wrong... hence please look first at the second picture and then at the first one  :) [/quote]

Darij, the mechanism doenst work strangely. You worked strangely naming the first one 3c and the second one 3b. ;) The mechanism just sorts them alphabetically. :)

No. I renamed them later when I saw them in wrong order, since I thought this would change it. But it didn't. I think that the order in which the figures appear in the message is opposite to the order they get uploaded.

[Edit: Yes, that's exactly the case. Now I have uploaded them in reverse order and everything is fine  :D ]

  Darij


how was the chemistry .
could u do all?
did  u know of such reactions before

Was decent comparitively
One of my friends i.e. Chemrock got 58 in the first paper  :roll:  and 25 in the second  :(
But he's the highest  :P

I spent a hell lot time in Physics
in first paper about 1 and half hour
second paper about two hours just for physics
net result just a 102 out of possible 162  :rotfl:  in physics
Also some rearrangements seemed completely new at that point but coming home i realised i had done them before it was sheer lack of time mangaement  :o

Don't worry we messed it up with Maths.
I got headstarts to all problems but I was not able to finish the problems of co-ordinate geometry :( .
Chemistry solutions to the second paper some questions look wrong.
Like the acetal and hemiacetal order of hydrolysis question.
We all know that hemiacetal is hydrolysed the least. He is surely worng in that one.

hey so how much are you all getting net and if if that person pardesi is getting 102 in physics itself is he expecting something like 330 in the net like our lappy!!! i cant stand this how many people are there like this person in our town who are getting more than 200???? :ninja:  :huh:
too much geeks here ah???

Nooooo Pardesi is not that sort  :maybe:
He said he had time only for Physics and messed up the other two ( :maybe: ) Read carefully .
And by the way he's not From Chennai
 :D

P.S. 77 is one of my Luckiessssttttt numbers.
P.S. @ pardesi Srini is my friend and a mokka master

hi srinitrs
i hope i had done the other two i used more than 75 percent of time just for physics and then did some maths
the co-geo questions were really long so ,,,
chemistry i couldn't even have a look at some questions  which infact were very easy  :rotfl:  :rotfl:
so not much expectationts

hi da!!
i got alarmed at your 102 mark as i am getting only around 70 in it( which is my best mong the three) :( and in chem first paper my performance was speakable but dont ask abt the second paper i flopped like hell loose and got all four assertion reasonings wrong :rotfl:  and getting in single digits for the second paper chem part maths of course all time worst performance and thus only getting around 125 to 130!! highest here is a person called lakshmikanth i hope rohit would have told abt him he is expecting about 333 and is getting 137 in physics alone(72 in 1st paper and 65 in the second) what a man!!! and of course there's another mastro here namely G.Vignesh the maths genius whos getting around 190 and our chem rakesh stands next getting around 160 i think!! how have people in your state done (ie your friends???) :D

Same here join the club .  :blush:
Even I got all 4 assertion and reasonings wrong in the second paper both in Chemistry as well as in Physics  :o

so how much are you getting da,rohit???and do anyone of you know the exact cut off for NSEP na NSEC??? :lol:

most of us here are in 150-190 mark range.

Well in chem
1st test 33
2nd test 23  :blush:
People say Chem cutoff will be 175-180

I kinda slogged hard for organic so am getting
Paper I - 35
Paper II - 25 (thanks to Match)

Physics is what I should have done well but sadly nup!! Co-Geo was a terror. Luckily there was enough Trig in paper II to compensate.


What were the MOP qualifying scores from 2006-2009?

Um ya, this should probably be moved to the highschool intermediate forum.

But since its already here, I would also like to know that.

There is a [url=http://www.artofproblemsolving.com/Forum/index.php?f=133]AMC Forum[/url] which is where this should be posted.

I don't know where to find that kind of information, sorry.

Another question: How do you qualify for USAMO?

It should be moved to the AMC forum. People there know more about this.

Ummm... While this is being moved...

@powerofpi, You qualify by adding your AMC10/AMC12 score(the higher one) to your AIME score*10. So if I got a 140 on AMC10 and an 8 on AIME, then I would be guaranteed(I think) USAMO qualification because my index would be greater than 215...

Red MOP (freshmen only) cutoff is usually 8-10.

Blue MOP cutoff is usually around 18.

Black MOP cutoff is usually around 28-30.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276193]This[/url] states the scores for USAMO Winner status, HM status, and MOP cutoff levels for 2009.

Qualification for USAMO is based on your AMC and AIME scores.  To BBND: That would qualify for the USAJMO, but not the USAMO.  See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=279691]this[/url] for more about the USAJMO and USAMO cutoff.

Also, it's impossible to get 140 on the AMC 10 :wink:

Ok.... So if I score an 8 on USAMO when I'm in 8th grade, then do I make RED MOP?

Only freshman are allowed in red. All others must make blue or black, except for seniors, who must make black.

See [url=http://en.wikipedia.org/wiki/Mathematical_Olympiad_Program]here[/url].


$ (X_t)_{t\in [0,\infty]}$ is a markov process that means $ \forall\quad 0\leq s<t$, $ \mathbb{P}(X_t\in B|\mathcal{F}_s)\equal{}\mathbb{P}(X_t\in B|X_s)$, where $ \mathcal{F}_s\equal{}\sigma(X_u, 0\leq u\leq s)$.

    Can we get the following statement from this definiton?
    $ \forall B\in \mathcal{G}_t\equal{}\sigma(X_u, u\geq t)$, $ \mathbb{P}(B|\mathcal{F}_s)\equal{}\mathbb{P}(B|X_s)$

if we can, how to prove it?

any hints are welcome.

thanks

it is true if $ B \equal{} \{ \omega : X_{t_1} \in A_1, X_{t_2} \in A_2, \ldots ,X_{t_n} \in A_n \}$ where $ t_k \geq t$. But the sets of this form is a stable under finite intersection and generate $ \sigma(X_u, u \geq t)$. Then the monotone class lemma allows you to conclude..

thanks.

I know this, but I still can't show that is true for such a $ B$

can any one give me some hints?

many thanks


If $x^{3}+2x^{2}+kx-3 = 0$, find the sum of all distinct $k$ such that $k$ is an integer and the cubic equation has exactly one rational root for $x$?

[hide="Solution"]
The only rational roots can be $-3,-1, 1, 3$.
By the remainder theorem, the remainder when $f(x)$ is divided by $x+3$ is $-3k-12$.  We want that to be 0, so one solution is ${k =-4}$.
Try -1: ${k =-2}$
Try 1: ${k = 0}$
Try 3: ${k =-14}$
The sum is $\boxed{-20}$
[/hide]

[quote="miyomiyo"][hide="Solution"]
The only rational roots can be $-3,-1, 1, 3$.
[/hide][/quote]

I'm stuck there. I know where the sum and the products of roots in the polynomial are. why is it those four numbers?

[quote="jeez123"]The theorem that, if a rational number p/q, where p and q have no common factors, is a root of a polynomial equation with integral coefficients, then the coefficient of the term of highest order is divisible by q and the coefficient of the term of lowest order is divisible by p.

Neat!!![/quote]

I prefer to think of it as the factors of $\frac{p}{c}$ where p is the last coefficient and c is the first.


quadratics Vieta algebra polynomial trigonometry inequalities geometry

What are the common formula/relationships used in math problems? For example, the sum of the roots of a quadratic equation is $ \minus{}\frac{b}{a}$

You might want to specify a branch; there are too many common formulas used in math in general to list.
It's better to know the more general theorems from which common formulas are derived though.

For example, if you knew Vieta's formulas, you'd know that the sum of the roots of $ ax^2\plus{}bx\plus{}c$ is $ (\minus{}1)^1\cdot\frac{b}{a}\equal{}\minus{}\frac{b}{a}$. But you'd also know that the product of the roots is $ \frac{c}{a}$, and you'd be able to apply this to $ n$ degree polynomials, etc. This is significantly more useful than just memorizing $ \minus{}b/a$.

then to refine my question:
what are the common theorems? or where can I find them/learn them (including AoPS chapters)?
or should I just read AoPS again?

The branches I think i need these theorems/formulas the most are probably trig, circles (not power of a point), and quadratics.

oh, and @ grn_trtle
I didn't know about Vieta's formulas until you posted  :blush:

what i did to get $ \frac{\minus{}b} {a}$ was:
\[ \frac{\minus{} b \plus{} \sqrt {b^2 \minus{} 4ac}} {2a} \plus{} \frac{\minus{} b \minus{} \sqrt {b^2 \minus{} 4ac}} {2a} \equal{} \frac{\minus{}2b} {2a} \equal{} \frac{\minus{}b} {a}\]

thanks   :D
-aznzhou

About Trig ...
Try to memorize some basic concepts and identities , and from them you can have further identities .
Example  : If you don't know $ \sin^{2}\alpha \plus{} \cos^{2}\alpha \equal{} 1$ then ...kill yourself  :wink:  (just joking,you can still learn)

hahaha  :lol:

well, at least I  know that...

"kill yourself" was a little unnecessary :)

Vieta's is a good thing to know. It states that if $ P(n)=a_nx^n+\cdots+a_1x+a_0$, and if the roots to that polynomial are $ r_1, r_2,...,r_n$, and if we define:

$ S_1=r_1+r_2+\cdots+r_n$
$ S_2=r_1r_2+\cdots+r_nr_1$
$ \vdots$
$ S_n=r_1r_2\cdots r_n$

(If you're familiar with some of the weirder notations, this is saying $ S_k=\sum_{\text{cyc}}\prod_{j=0}^{k}a_j$).

Then $ S_k=(-1)^n\cdot\frac{a_{n-k}}{a_n}$.

Try some examples to illustrate this for yourself. It holds for any polynomial as long as the coefficients are integers (or real, I can't remember...)

As enndb0x said, trig identities are a very good thing to know. Familiarity with those is important to a lot of problems. Knowing the Law of Sines:
\[ \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=2R\]

and the Law of Cosines:
\[ a^2=b^2+c^2-2ab\cosh\]

is a good thing as well. It's also sometimes helpful to understand the relationship between trig and complex numbers; like Euler's formula, which says that $ e^{i\theta}=\cos\theta+i\sin\theta$.

Aside from trig, it's helpful to know inequalities like the famous and frequently abused AM-GM:
\[ \frac{x_0+x_1+\cdots+x_n}{n}\geq\sqrt[n]{x_0x_1\cdots x_n}\]

and some of the more obscure ones like Cauchy.

Some random thing that might come in handy..in no specific list cause it takes to long to remember then..then sort...ect

Shoelace Therm.
[url]http://www.artofproblemsolving.com/Wiki/index.php/Shoelace_Theorem[/url]

Fermat's Little Thrm.
[url]http://www.artofproblemsolving.com/Wiki/index.php/Fermat's_Little_Theorem[/url]

Wilson's Thrm:
[url]http://www.artofproblemsolving.com/Wiki/index.php/Wilson's_Theorem[/url]

Area Formula's
[url]http://en.wikipedia.org/wiki/Area[/url]

Pick's Thrm:
[url]http://en.wikipedia.org/wiki/Pick's_theorem[/url]

Ptomely's Thrm
[url]http://www.cut-the-knot.org/proofs/ptolemy.shtml[/url]

If we're talking number theory, there are some big ones you need to know.

Fermat's Little Theorem says $ a^{p\minus{}1}\equiv 1\pmod p$ if $ p$ is prime and $ a$ is integral.

Euler's Theorem says that $ a^{\varphi(n)}\equiv 1\pmod n$ if $ a$ is coprime to $ n$, where $ \varphi(n)$ is the totient of $ n$.

Carmichael's Theorem says that $ a^{\lambda(n)}\equiv 1\pmod n$ if a is coprime to $ n$, where $ \lambda(n)$ is the Carmichael function.

Wilson's Theorem says that $ (p\minus{}1)!\equiv 1\pmod p$ iff $ p$ is prime.

[quote="grn_trtle"]
\[ \frac {\sin A}{a} \equal{} \frac {\sin B}{b} \equal{} \frac {\sin C}{c} \equal{} 2R
\]
[/quote]

I think you mean $ \frac{a}{\sin A} \equal{} \frac{b}{\sin B}\equal{}\frac{c}{\sin C} \equal{} 2R$.

@lifeisacircle: Yeah, you're right. I always get it right on paper but when I'm TeXing...  :huh:

@r15s11z55y89w21: $ R$ is the circumradius.

There is a lot. You must specify a field for yourself and start studying it from the beginning you can't just go the advanced formula without knowing the basic one  :) .

Thanks everyone, especially grn_trtle :)

[quote="grn_trtle"]and the Law of Cosines:
\[ a^2 = b^2 + c^2 - 2ab\cos C
\][/quote]

Fixed the $ \text{\LaTeX}$. :)


I liked how the first one was in a different languauge...

This already looks like that topic PWNED (sorry its deleted now)

Am I allowed to upload something 8 kb and 325 kb here?

[quote="MysticTerminator"][quote="fredbel6"][quote="bubka"]I sense trouble. Please stop spamming![/quote]
Amazingly, I agree with the former champion of spam : this sounds like spam  :spam:[/quote]

Although we appreciate your efforts to further the development of this thread, we would appreciate not having it degenerate into vulgar spam. Thus, we ask that you discontinue any subsequent posts in this manner.[/quote]
 :surrender: I don't get it :|   What is the fundamental difference between my posts and the previous ones?  Am I the top spammer now? :yoda:

THIS USED TO BE A POLL THAT's WHEN I ORIGINALLY POSTED!
What happened to this thread?

[quote="Iversonfan2007"][quote="MysticTerminator"]please post in my topic![/quote]

whoa what a great topic good job MT for posting such an interesting one on this forum![/quote]

yup, I try

ok. ok. I'm sorry. I looked the topic a bit, and now I see I was wrong. :blush:

I think this is a topic about spammming topics.

-I see in your eyes the same fear that would take the heart of me. A day may come when the courage of spammers fails, when we forsake our friends and break all bonds of fellowship, but it is not this day. An hour of woes and shattered topics, when the age of spammers comes crashing down, but it is not this day. This day we spam!

(Altered quote from Lord of the Rings: Return of the King if you missed it)

LETS MAKE A LIST OF ALL THE SPAMMERS:

I START:

MYSTIC TERMINATOR, ...

4) Treething (well he isn't a spammer because he doesn't post enough, but his posts are generally spam.)

[quote="akalra1"]6) gff posters[/quote]
7)darn that includes me

9) ignite

it's too bad none of you get his internet references


Kelly's average score on four Spanish tests is 85.5. The average of her three
highest scores is 87, and her two lowest scores are the same as each other. What is
the average of her two highest test scores?

[hide="Solution"] The sum of the scores of all her tests is $85.5\times4=342$. The sum of the highest three tests is $87\times3=261$. Therefore, the lowest score must be $81$ and so must the next lowest. The sum of the highest two tests must be $342-81\times2=180$. So the average must be $\boxed{90}$ :D [/hide]

[hide]Sum of 4 tests=(85.5)*4=324.
Sum of three highest scores=87*4=261.
Lowest score=324-261=81.
Sum of two lowest scores=81*2=162.
Sum of two highest scores=324-162=180.
Average of two highest scores=90.[/hide]

that's easy  :lol:
[hide="solution"]w+x+y+z=342
261+w(or x)=342, so w=x=81
subtracting that from the original equation we get
y+z=180
180/2=[color=red]90[/color]  :D [/hide]


counting
distinguishability
Alcumus
complementary counting
cars
needing the answer
Gamebot

Six cars pull up to a red light, one at a time. At the light, there are three lanes, one left-turn lane, one straight-going lane, and one right-turn lane. How many ways can the cars stack up so that all three lanes are occupied?

Note that if the first car turns left and the second goes straight, this is considered different from the first car going straight and the second car turning left. In other words, the cars are distinguishable, but pull up to the intersection in a fixed order.

[hide="this may not be right, but..."]
We can use complementary counting.

Without restrictions, each car has 3 choices, so there are $ 3^6\equal{}729$ ways to arrange them.

Now we count the number of ways we can have only 1 or 2 lanes filled. Let's choose one of the lanes to exclude. We have 3 ways to do this. Then, each car has 2 choices, so we have $ 2^6$ ways to do this. But this overcounts the cases when we have only one lane filled - excluding lane 1 then filling lane 3 is the same as excluding lane 2 then filling lane 3. This happens for each pair of lanes, and there are 3 such pairs. Thus, we must subtract 3, giving us $ 3 \cdot 2^6\minus{}3\equal{}189$.

Finally, we subtract to get $ 729\minus{}189\equal{}\boxed{540}$.
[/hide]

Is that right?

Yep, its right.  Reminds me of a certain AMC 10A problem I couldn't get...

I was thinking differently...
I listed out the different ways that the cars could be arranged amongst the 3 lanes so they added up to 6...

The table I had...
[size=150]Left         Straight        Right
1                  2                  3
1                  1                  4
2                  1                  3
2                  3                  1
3                  2                  1
3                  1                  2
4                  1                  1
1                  4                  1
2                  2                  2
1                  3                  2[/size]

So I got ten ways to arrange the cars...
and knew that there were 6 spots which could be arranged in each spot differently.
[b]I did 6!*10 and got 7200[/b]
I am way off will somebody help me see where I am going wrong? I looked at the way Alcumus solved it and was a bit confused.
Thanks,
Rohan

[quote="professordad"]The thing is, there are not $6!$ ways to arrange each condition.  For example, when we have $1$ left, $2$ straight, and $3$ right, we have $1! \cdot 2! \cdot 3!$ ways to arrange the cars, because all of them are in different lanes.  Only if all the cars are in the same lane can you say there are $6!$ ways.[/quote]

Why does 6! only work if they are in 1 lane?

@ Rohan

$6!$ is what is wrong in your approach.

Notice the question states "the cars are distinguishable, but pull up to the intersection in a fixed order." KEY WORDS: "fixed order"

The thought process behind $6!$ is that the first car has six places to choose where to arrive, the second car has five choices, etc. However, the cars come up in a fixed order.

Therefore, for example, where the order is $1-2-3$. The first car has to be first in line in one of the three lanes. So it only has three choices, not six. With rigorous casework, we can figure out the total amount of options for each of the ten different ways the cars can be arranged.

Rigorous casework usually translates to convenient complementary counting, which is the case in this problem.

I still don't really understand the meaning of "fixed order." How is "fixed order" reflected in the solution?

[quote=wizwilzo][quote=txdra]If you were to use casework by splitting the number of cars into lanes with a minimum of 1 per lane:

4-1-1/1-4-1/1-1-4  would yield 6C4 x 2C1 x 1C1 x 3 = 15 x 2 x 1 x 3 = 90
3-2-1/3-1-2/2-3-1/2-1-3/1-3-2/1-2-3  would give 6C3 x 3C2 x 1C1 x 6 = 20 x 3 x 1 x 6 = 360
2-2-2 adds 6C2 x 4C2 x 2C2 x 1 = 15 x 6 x 1 x 1 = 90

for a total of 540 ways[/quote]

I cant really understand the solution. Perhaps using parenthesis?[/quote]

I agree, please use some latex so we can understand it more clearly.

Uh I'm pretty sure that the user you're talking to is inactive now
Don't revive threads to talk to inactive users ;)

Anyhow, this is just a simple distinguishability question; just sort with 6 distinguishable candies and 3 indistinguishable boxes

[hide]The subject states "complementary counting," so let us try to use complementary counting to solve the problem. There are $3^6=729$ possibilities if the cars are in a fixed order but are distinguishable. If the cars leave $1$ lane unoccupied, there are $2^6$ ways the cars could go into the other $2$ lanes. However, we must subtract $2$ to account for the possibility where the cars leave $2$ lanes unoccupied.Therefore, there are $3(2^6-1-1)$ ways for the cars to leave $1$ lane unoccupied. Also, we must consider the possibility where $2$ lanes are unoccupied, or where $1$ lane is occupied. There are $3$ possibilities. Therefore, there are $\boxed{729-[3(2^6-1-1)+3]=540}$ possibilities.=[/hide]

[hide]Hm, I may have gotten this by luck but I first chose 3 cars out of the six to go into the lanes so the lanes aren't empty, and then I multiplyed 6C3 by 27 (3^3, the other cars can go where they want.) That gave me 540 :/[/hide]

[quote=Possible][quote=V_V]Can someone give a clear, detailed explanation of why we are over counting?[/quote]

I think the above solutions are pretty clear[/quote]

I really don't understand them

A month later and now I have understood!!

I don't understand the Alcumus solution...The problem specifically says that the cars are distinguishable, so if we label them Cars 1 through 6, then the case where Car 1 is behind Car 4 (in any lane) is not counted in the total...

@above

Since the problem says the cars are distinguishable, it matters which cars go into each lane, but it doesn’t matter which order they’re in.

For example, for the 2-3-1 case (2 cars in the left lane, 3 in the middle, and 1 on the right), it matters which 2 cars are in the left lane. Cars 1 and 4 in the left lane is different than Cars 2 and 3 in the left lane. However, it doesn’t matter if Car 1 is behind or in front of Car 4 because the problem tells us that they always pull up in the same order.

I hope this helps!

I've definitely fakesolved this but [hide = cheese?]We must satisfy the condition that there must be at least one car in each lane, so let us pick 3 cars arbitrarily such that they each occupy one distinct lane. The number of ways to pick these 3 cars is $\binom{6}{3} = 20$. After this, we must assign either $L, S$ or $R$ to the $3$ remaining cars, which can be done in $3^3 = 27$ ways. Thus, the answers is $20 \cdot 27 =
\boxed{540}.$[/hide]


inequalities
rearrangement inequality
algebra proposed
algebra

Let $A$ be the set $\{1,2,\ldots,n\}$, $n\geq 2$. Find the least number $n$ for which there exist permutations $\alpha$, $\beta$, $\gamma$, $\delta$ of the set $A$ with the property: \[ \sum_{i=1}^n \alpha(i) \beta (i) = \dfrac {19}{10} \sum^n_{i=1} \gamma(i)\delta(i) . \]
[i]Marcel Chirita[/i]

Cute. For $n = 28$ it works: take $\alpha(i) = \beta(i) = \gamma(i) = i$ and $\delta(i) = 29 - i$, then we get \[ \sum_{i = 1}^{28}\alpha(i)\beta(i) = \sum_{i = 1}^{28}i^2 = 7714 \] and \[ \frac{19}{10} \cdot \sum_{i = 1}^{28}\gamma(i)\delta(i) = \frac{19}{10} \cdot \sum_{i = 1}^{28}i(29 - i) = \frac{19}{10} \cdot 4060 = 7714. \] However, it cannot work for $n \leq 27$. Indeed, by the rearrangement inequality we have \[ \frac{19}{10} \cdot \sum_{i = 1}^n\gamma(i)\delta(i) \geq \frac{19}{10} \cdot \sum_{i = 1}^ni(n + 1 - i) = \frac{19}{10} \cdot \frac{n(n + 1)(n + 2)}{6} \] and \[ \sum_{i = 1}^n\alpha(i)\beta(i) \leq \sum_{i = 1}^ni^2 = \frac{n(n + 1)(2n + 1)}{6}. \] Now, since $n \leq 27$, \[ \frac{19}{10} \cdot (n + 2) > 2n + 1 \] (since this is equivalent to $n < 28$), so the right-hand side will always be "too big".

So, the solution is $n = 28$.


probability calculus integration inequalities

Is there a method for easily computing geometric probability in three dimensions?
For example, solve the problem:

Three mathematicians go to a coffee shop some time in between 1 and 2. Each one stays exactly 10 minutes and then leaves. What is the probability that they are all there at some point?

I know the concept, but is there a way to do it other than trying to get volume (because that's kinda hard to visualize)?

[quote="Singular"]integrals

i dont know too much about it though, but i know its possible[/quote]
[quote="Altheman"]i had a similar problem on my mock aime...you can draw the intersection if you are good with 3d stuff, you just have to set up some inequalities with algebra to make sure that you are graphing it correctly[/quote]
Could someone post the solution to this problem using either of these methods?


hi i am longtime reader but first time posting :blush:

I would like some links to nice sites with good formulas, i saw many problems that i have no idea how to do, so if you can provide some help, it would be greatly appreciated :blush:


linear algebra
matrix
limit
absolute value
linear algebra solved

Let $A \in M_n(R)$ be matrix such that $\exists k \in N :kA^k=A^{k-1}+A^{k-2}+\cdots+A+I_n$.
Prove that $\exists B=\lim_{n \rightarrow \infty}A^n$ and ${B=B^2}$

1 is a root of $f(x)=-kx^k+x^{k-1}+...+x+1$
What can you say about the modulus of the root in C (comparing modulus of the root with 1) ?
Does $f$ has multiple roots in C ?

Try  :D
My solution for this one isn't simple

The limit exists simply because if a sequence satisfies $x_{n+k}=x_{n+k-1}+\ldots+x_{n+1}+x_n,\ \forall n$, then it converges. We can write $x_n$ as a linear combination (with fixed coefficents) of the $n$'th powers of the roots of $kx^k-x^{k-1}-\ldots-x-1=0$, which has all its roots with absolute value $\le 1$ (simple application of the triangle inequality).

We now know that there is a real matrix $B=\lim_{n\to\infty}A^n$. The sequence $A^{2n}-A^n$ tends to $B^2-B$, but it also tends to $O_n$ as $n\to\infty$, so we have $B^2-B=O_n$.

[quote="grobber"]The limit exists simply because if a sequence satisfies $x_{n+k}=x_{n+k-1}+\ldots+x_{n+1}+x_n,\ \forall n$, then it converges.[/quote]
Could you please explain how exactly from this fact it follows that $\displaystyle\lim_{n\to\infty}A^n$exists? What you denote by $x_n$

You apply that fact to each sequence $(x_n(i,j))_n$, where $x_n(i,j)$ is the $(i,j)$ entry in $A^n$.


Solve in $ \mathbb{R}$ the equation: $ \left ( \frac{1}{3} \right ) ^{x\plus{}1}\plus{}\left ( \frac{1}{12} \right ) ^{x}\minus{}3\left ( \frac{\sqrt{3}}{12} \right ) ^{x}\equal{}3$

hello, $ x \equal{} \minus{} 2$ is the only real solution.
Sonnhard.


The five digit number $ 31d26$ is divisible by 3. Find the sum of all possible values of $ d$.

The divisibility rule of 3 states that the sum of the digits must be divisible by 3. So,

$ 3\plus{}1\plus{}d\plus{}2\plus{}6 \equiv 0 \mod3$

$ d\plus{}12 \equiv 0 \mod 3$

$ d \equiv 0 \mod 3$

Therefore, the possibile values of $ d$ are: $ 0,3,6,9$, so the sum is

$ 0\plus{}3\plus{}6\plus{}9 \equal{} \boxed{18}$.

I think this is glitched.  This is the exact same question as in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=289065, and it was asked at the exact same time...


inequalities
search
inequalities proposed

Prove or disprove for $ n\equal{}2k\plus{}1$, $ k\in \mathbb{Z}^\plus{}$.

Let $ S\equal{}\{1,2,\ldots n\}$.

Then

\[ \sum_{i\in S}{a_i^n} \plus{} n(n\minus{}1)\pi_{i\in S}{a_i} \ge \sum_{i,j\in S}{a_i^{k}a_j^{k\plus{}1}}\]

If it's true, then can the RHS be strengthened to $ \sum{a_i^{n\minus{}1}a_j}$?

It doesn't seem your inequality is true... if I understand your notation correctly. what happens when one of the variables is 0?

product is \prod not \pi

Well, 2 edits (since I can't edit anymore):

it should be $ n$, not $ n(n\minus{}1)$, and all $ a_i$ are positive.

if you put $ a_1\equal{}1\equal{}a_2\equal{}...\equal{}a_{n\minus{}1}$ and $ a_n\equal{}0$ then you get $ n\minus{}1\geq (n\minus{}1)(n\minus{}2)$
on the other side I think you should see this maybe...
http://www.mathlinks.ro/viewtopic.php?search_id=1047964584&t=110149

Hmm, maybe I am a bit deluded. But after seeing Schur and

\[ a^4\plus{}b^4\plus{}c^4\plus{}d^4\plus{}2abcd\ge \sum_{sym}a^2b\]

 I was wondering if there existed generalizations.

did you see my link above?
aren't you content with [url=http://www.mathlinks.ro/viewtopic.php?search_id=2015658919&t=14008]this[/url]?

Oh, I didn't read very carefully. :oops: Sorry.


It is given that  $ \begin{array}{l} \Delta CAB \sim \Delta EFD \end{array}$.

If....

$ AC\equal{}x\plus{}y\plus{}z$
$ AB\equal{}z\plus{}6$
$ BC\equal{}x\plus{}8z$
$ EF\equal{}3$
$ DF\equal{}2y\minus{}z$
$ DE\equal{}y\plus{}2$

Find for $ x^2\plus{}y^2\plus{}z^2$

Mathematica-bashing gives $ \minus{} 88/9$, among other (not pretty) solutions.

Edit: Um, fail....

If the numbers are not real...but a length should be real.  Move to appropriate forum if nobody can solve here?

So Here's a [hide="Hint"]$ AC \equal{} EF, AB \equal{} FD,$ and $ BC \equal{} ED$..So what you've got??? [/hide]

And If you want to see the Solution already [hide="Click Here!"]
[hide="Continue...."]
Since $ \begin{array}{l} \Delta CAB \sim \Delta EFD \end{array}$, it follows that $ AC \equal{} EF, AB \equal{} FD,$ and $ BC \equal{} ED$. Thus, You'll end up having a system of linear equations:

$ x \plus{} y \plus{} z \equal{} 3$
$ z \plus{} 6 \equal{} 2y \minus{} z$
$ x \plus{} 8z \equal{} y \plus{} 2$

Solving the system yields $ x \equal{} \minus{} 2, y \equal{} 4,$ and $ z \equal{} 1$

Where $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 21$

[/hide][/hide]

 :lol:

You said that the two triangles were similar, not congruent. :wink:


1,$ sin3x\plus{}(4sin2x\plus{}\sqrt{11})cos3x\equal{}0$

2,$ cos5x\equal{}sinx\plus{}sin3x\plus{}sin5x$

3,$ 4(cosx)^3\plus{}4(cosx)^2\plus{}7\sqrt{2\minus{}(cosx)^2}\equal{}3cosx\plus{}12$

Please, hoangvn.fix, could you give us your own clever solution to these three equations, instead of my ugly numeric calculus one ?

Thanks

hello, the solution of your last equation was allready posted, $ x\equal{}2k\pi$ and $ k\in \mathbb{Z}$.
Sonnhard.


Consider a set $ A$ of positive integers such that the least element of $ A$ equals $ 1001$ and the product of all elements of $ A$ is a perfect square. What is the least possible value of the greatest element of $ A$?

No, the set $ \{1001, 1008, 1053, 1100 \}$ fullfils that condition but probably there is another optimal one.

Another "small" set is $ \{1001, 1008, 1014, 1053, 1056 \}$

mszew's example of $ \{1001, 1008, 1014, 1053, 1056\}$ is optimal.

Since $ 1001 = 7 \cdot 11 \cdot 13$, $ A$ must contain another multiple of 7, another multiple of 11, and another multiple of 13.  We list all multiple of 7, 11, and 13 between 1002 and 1056, and their prime factorizations:
\begin{align*}
1001 &= 7 \cdot 11 \cdot 13, \\
1008 &= 7 \cdot 2^4 \cdot 3^2, \\
1015 &= 7 \cdot 5 \cdot 29, \\
1022 &= 7 \cdot 2 \cdot 73, \\
1029 &= 7^3 \cdot 3, \\
1036 &= 7 \cdot 2^2 \cdot 37, \\
1043 &= 7 \cdot 149, \\
1050 &= 7 \cdot 2 \cdot 3 \cdot 5^2, \\
1057 &= 7 \cdot 151, \\
1012 &= 11 \cdot 2^2 \cdot 23, \\
1023 &= 11 \cdot 3 \cdot 31, \\
1034 &= 11 \cdot 2 \cdot 47, \\
1045 &= 11 \cdot 5 \cdot 19, \\
1056 &= 11 \cdot 2^5 \cdot 3, \\
1014 &= 13^2 \cdot 2 \cdot 3, \\
1027 &= 13 \cdot 79, \\
1040 &= 13 \cdot 2^4 \cdot 5, \\
1053 &= 13 \cdot 3^4.
\end{align*}

We see that no multiple of 11 smaller than 1056 will lead to a better bound on $ A$.  For example, the number 1012 contains the prime factor 23, and no other number in the list above contains another prime factor of 23.  Therefore, the bound of 1056 is optimal.

[quote="nsato"]mszew's example of $ \{1001, 1008, 1014, 1053, 1056\}$ is optimal.

Since $ 1001 = 7 \cdot 11 \cdot 13$, $ A$ must contain another multiple of 7, another multiple of 11, and another multiple of 13.  We list all multiple of 7, 11, and 13 between 1002 and 1056, and their prime factorizations:
\begin{align*} 1001 & = 7 \cdot 11 \cdot 13, \\
1008 & = 7 \cdot 2^4 \cdot 3^2, \\
1015 & = 7 \cdot 5 \cdot 29, \\
1022 & = 7 \cdot 2 \cdot 73, \\
1029 & = 7^3 \cdot 3, \\
1036 & = 7 \cdot 2^2 \cdot 37, \\
1043 & = 7 \cdot 149, \\
1050 & = 7 \cdot 2 \cdot 3 \cdot 5^2, \\
1057 & = 7 \cdot 151, \\
1012 & = 11 \cdot 2^2 \cdot 23, \\
1023 & = 11 \cdot 3 \cdot 31, \\
1034 & = 11 \cdot 2 \cdot 47, \\
1045 & = 11 \cdot 5 \cdot 19, \\
1056 & = 11 \cdot 2^5 \cdot 3, \\
1014 & = 13^2 \cdot 2 \cdot 3, \\
1027 & = 13 \cdot 79, \\
1040 & = 13 \cdot 2^4 \cdot 5, \\
1053 & = 13 \cdot 3^4. \end{align*}
We see that no multiple of 11 smaller than 1056 will lead to a better bound on $ A$.  For example, the number 1012 contains the prime factor 23, and no other number in the list above contains another prime factor of 23.  Therefore, the bound of 1056 is optimal.[/quote]
Who says that every number in the set must be a multiple of either 7, 11 or 13 ?

The number should be 1040.
We need another multiple of 13. $ 1014\equal{}2*3*13*13$ does not help. $ 1027\equal{}13*79$ will ask for another multiple of $ 79$, a number greater than 1040.


think of any 2 numbers. add them to get a 3rd number. now add second and third to get a fourth. continue this procedure untill you get to tenth number. add all ten numbers if you tell me your 7th number i will by multiplying by 11 tell u wat ure sum is. why does this always work?


LaTeX AMC AIME algebra function domain geometry

srry :blush:
i didn't know
last yr they were posted same day

Not yet Alexander. Problems shouldnt be discussed now.

I think everyone should've done by now... so can we discuss the problems now?

[quote="nneonneo"]Just be patient...[/quote]
BTW why can't we discuss?

The problems were discussed last yr the same day starting at 5 pm!

[quote="nneonneo"]I'm not a moderator or an authority; I'm just recommending that you wait a bit. It's only 4:50 here, and it'll be about 3:50 in B.C.[/quote]
Yeah i know, but isn't it written in the morning?

Cynthia, how did u do?

Pretty much, but my solutions are real bad, so I may have marks off for that...

I'm pretty sure Cynthia's just being modest  :)

I'm telling the truth....I've never got perfect for my solution..... :blush:
How did you guys do? I'm sure you did great!

We posted in the other forum.

You did really well too! Nice job.

Bah, if nobody says anything, then we should start discussing. It's almost 6 here.

[quote="cynthia1991"]I didn't figure out #3b, and I made a really stupid mistake on #4a, but hopefully it doesn't affact the whole idea....so I may get partial mark for that.[/quote]

Hey hey!!! Cynthia's on AOPS! Wooooo.... congrats on AIME.... and CMO.......and......Say, how'd u do on APMO?

When you will can post the problems?  :D

the one that i see is 5b, that was posted by lomos_lupin, very nice problem!

I don't have a copy as well as everyone else, so we don't know yet, I think.

I can post the problems in a few days, I am quite busy now.

The problems I posted on the day of the contest I didn't save so I have to type them up again.

My math teacher stole my English copy, so I got the French one :P
I don't read French very well, so here are the questions verbatim.

39e Olympiade mathématique du Canada
le mercredi 28 mars 2007

1. Quel est le nombre maximum de dominos 2 x 1 qui peuvent être placés sur un damier 8 x 9 si six d'entre eux sont placés comme illustré dans le diagramme ci-dessous? (les dominos ne doivent pas se chevaucher) Chaque domino doit être placé horizontalement ou verticalement afin de couvrir deux places adjacentes du damier.
[img]http://img100.imageshack.us/img100/9489/39thcmoquestion1ty9.gif[/img]
(now then, if anyone knows how to do that in LaTeX...)
(I did that in Word and took a screenshot. I really want to know how to do it in LaTeX)

2. Soit deux triangles qui satisfont aux conditions suivantes:
  (a) deux côtés d'un triangle sont égaux (en longueur) à deux côtés du deuxième, et
  (b) les triangles sont semblables, mais pas nécessairement congruents.
Montrer que le rapport des côtés qui sont liés par la similarité est un nombre entre $\frac{1}{2}(\sqrt{5}-1)$ et $\frac{1}{2}(\sqrt{5}+1)$.

3. Supposer que $f$ est une fonction à valeurs réelles qui satisfait
\[f(xy)+f(y-x)\geq f(y+x) \]
pour tous nombres réels $x$ et $y$.
  (a) Donner un polynôme non constant qui satisfait cette condition.
  (b) Montrer que $f(x)\geq 0$ pour tout nombre réel $x$.

4. Si $a$, $b$, sont deux nombres réels tels que $ab\neq1$, on définit l'opération $*$ par
\[a*b = \frac{a+b-2ab}{1-ab}. \]
Commencer par une liste de $n$ $(\geq 2)$ nombres réels dont chaque entrée $x$ est tel que $0<x<1$. Pour chaque choix de deux nombres $a$ et $b$ dans la liste; on enlève les deux nombres et on place le nombre $a*b$ à la fin de la liste réduisant ainsi sa longueur par un. Répéter le processus jusqu'à ce qu'il reste un seul élément dans la liste.
  (a) Démontrer que le dernier élément qui reste est toujours le même indépendamment du choix de la paire des nombres à la chaque étape.
  (b) Supposer maintenant qu'on affaiblit la condition sur les nombres $x$ dans la list $S$ à $0<x\leq 1$. Qu'est-ce qui se passe si $S$ contient exactement un 1?

5. Dans un triangle $ABC$, le cercle inscrit touche les côtés $BC$, $CA$ et $AB$ en $D$, $E$ et $F$, respectivement. Soit $\Gamma$, $\Gamma_{1}$, $\Gamma_{2}$ et $\Gamma_{3}$ les cercles circonscrits aux triangles $ABC$, $AEF$, $BDF$ et $CDE$ respectivement.
Soit $A$ et $P$ les points d'intersection de $\Gamma$ et $\Gamma_{1}$, $B$ et $Q$ les points d'intersection de $\Gamma$ et $\Gamma_{2}$, et $C$ et $R$ les points d'intersection de $\Gamma$ et $\Gamma_{3}$.
  (a) Démontrer que les cercles $\Gamma_{1}$, $\Gamma_{2}$ et $\Gamma_{3}$ s'intersectant en un point commun.
  (b) Démontrer que $PD$, $QE$ et $RF$ sont concourants.

hey, the basic idea is there, it's just in French.

And it's practically my first time using LaTeX. If what I've done looks crappy, please do tell me.

Someone please tell me how to do that funky diagram in LaTeX...

Translated version (by me :P)
39th Canadian Mathematics Olympiad
Wednesday, March 28, 2007

1. What is the maximum number of 2 x 1 dominoes which can be placed on an 8 x 9 board if six are already placed as shown by the diagram below? (The dominoes cannot overlap) Each domino must be placed horizontally or vertically so that they cover two adjacent squares of the board.
[img]http://img100.imageshack.us/img100/9489/39thcmoquestion1ty9.gif[/img]

2. There are two triangles satisfying the following properties:
  (a) Two sides of one triangle are equal (in length) to two sides of the other, and
  (b) the triangles are similar, but not necessarily congruent.
Show that the ratio of similarity between the two triangles is a number between $\frac{1}{2}(\sqrt{5}-1)$ and $\frac{1}{2}(\sqrt{5}+1)$.

3. Suppose $f$ is a real-valued function satisfying
\[f(xy)+f(y-x)\geq f(y+x) \]
for all real numbers $x$ and $y$.
  (a) Find a non-constant polynomial satisfying this condition.
  (b) Show that $f(x)\geq 0$ for all real $x$.

4. If $a$, $b$, are two real numbers such that $ab\neq1$, define the operation $*$ as
\[a*b = \frac{a+b-2ab}{1-ab}. \]
Begin with a list of $n$ $(\geq 2)$ real numbers where each entry $x$ satisfies $0<x<1$. For each choice of $a$ and $b$ in the list; remove the two numbers and replace them by the number $a*b$ at the end of the list, reducing the list size by one. Repeat this process until there is only a single element in the list.
  (a) Prove that the last element remaining in the list is independent of the choice of number pairs at each step.
  (b) Suppose now that the condition on the numbers $x$ in the list $S$ is relaxed to $0<x\leq 1$. What happens if $S$ contains exactly one 1?

5. In a triangle $ABC$, the inscribed circle touches $BC$, $CA$ and $AB$ at $D$, $E$ and $F$, respectively. Let $\Gamma$, $\Gamma_{1}$, $\Gamma_{2}$ and $\Gamma_{3}$ be the circumcircles of triangles $ABC$, $AEF$, $BDF$ and $CDE$ respectively.
Let $A$ and $P$ be the points of intersection of $\Gamma$ and $\Gamma_{1}$, $B$ and $Q$ be the points of intersection of $\Gamma$ and $\Gamma_{2}$, and $C$ and $R$ be the points of intersection of $\Gamma$ and $\Gamma_{3}$.
  (a) Prove that the circles $\Gamma_{1}$, $\Gamma_{2}$ and $\Gamma_{3}$ intersect in a common point.
  (b) Prove that $PD$, $QE$ and $RF$ are concurrent.

nice exam!

interesting.. i'm solving easily the last four problems, but the first i don't find one easy way to solve. my solution have a lot of cases.

my ideas for the first is see a lot of cases. for the second see two cases and do any computations. for 3, a) $x^{2}+4$ holds, b) let $y = \frac{(x-1)}{x}$. for 4a, see that $(a*b)*c = (a*c)*b$, and for 4b, $1*x = x$.
my solution to 5 are in this post:

http://www.mathlinks.ro/Forum/viewtopic.php?t=141156

there's a solution to Question 1) on Page 2 of this thread.

[quote="rem"]OK. So you paint it in chessboard color -- black and white.
then you divide it into two pieces -- first bounded by the bottom left corner square and the second right-most square in top row and the dominoes already placed square (not including those squares). The other piece -- same as first but rotationally symmetric over center. Then, each piece has 12 white, 16 black squares or vice versa. So it can contain max 12 complete dominoes. Also, it can contain at most 2 half-dominoes from other dominoes not completely lying in the piece. So each piece has at most 12*2+2=26 half-dominoes. So 2 pieces contain at most 52 half-dominoes. There are 4 squares not belonging to the pieces and the dominoes already placed, so at most 4 more half-dominoes. Finally, each domino already placed gives 2 half-dominoes, there are 6 of them, so 12 more half-dominoes. So, at most 52+4+12=68 half-dominoes, so at most 34 dominoes.
The example with 34 dominoes can be quite easily found by trying to cover as much area as possible.
So, answer is 34.[/quote]
--posted at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=801150#801150

This year's CMO seems pretty easy. Is there anyone who actually solved all the problems? I think a lot of people are really closed to getting perfect....

maybe some of the grade 12's..(the ones who have lots of CMO experience)

how much time do you have for the CMO?
(sorry for asking such a dumb question;I couldn't find it anywhere :blush: )


How many 15-letter arrangements of 5 A's, 5 B's, and 5 C's have no A's in the first 5 letters, no B's in the next 5 letters, and no C's in the last 5 letters?

\[ \text {(A) } \sum_{k=0}^{5} \binom{5}{k}^3 \qquad \text {(B) } 3^5\cdot2^5 \qquad \text {(C) } 2^{15} \qquad \text {(D) } \frac{15!}{(5!)^3} \qquad \text {(E) } 3^{15}  \]

[hide]Suppose there are $k$ B's and $5-k$ C's in the first 5 letters.  Then there are $k$ C's and $5-k$ A's in the next 5 letters, and $k$ A's and $5-k$ B's in the last 5 letters.

For each value of $k$, there are $\binom{5}{k}$ ways to arrange each group of 5 letters.  Therefore, the number of arrangements is (A), $\sum_{k=0}^{5}\binom{5}{k}^3$.[/hide]


conics
parabola
LaTeX
symmetry
analytic geometry

[hide]Substituting, we get

2b+c=-1
4b+c=-13

Where b=-6. Thus $c=\boxed{11}$. [/hide]

[quote="ryanprov"]How do I do the box around the answer?[/quote]

It's \boxed{} in $\LaTeX$.

[quote="easyas3.14159..."]A parabola with the equation $y=x^2+bx+c$ passes through the points $(2,3)$ and $(4,3)$. What is $c$?[/quote][hide="More geometric answer"]Due to parabola symmetry, the x-coordinate of the vertex is 3, so $\frac{-b}{2}=3 \Rightarrow b=-6$. Plugging in b, x, and y, $3=4-6\times2 + c \Rightarrow 3+12-4 = c \Rightarrow c=11$.[/hide]


So i scored constantly in 139-150 range on the AMC 10, but my highest score in AIME is only 4, can you guys offer some tips in doing well on the AIME please?

Gauss i think all of the movement in your sig detracts from the impact of your posts

[quote="gauss1181"]Do past tests....
Also read the Intermediate books and AoPS Volume 2.[/quote]

Do past tests should be changed to Study past tests.

With such a high AMC score, but a low AIME score, you should consider taking the AIME classes from AoPS.

How'd you get so good at AMC 10?

Same.  But if you can finish an AMC 10 in 45 minutes...

What are you capable of on AMC 12?

Work through old AIME tests, and study the solutions.

blah, don't read the solutions until you have attempted to solve the problem. then read the solution and make sure you understand ever aspect of it.

If you have no clue at all, ask on the forum.

Also you can try reading AMC Contest Problem Book V. It contains past AMC 12's and some AIMEs.

[quote="cdymdcool"]i don't know, i'm just a fast problem solver, i usually finish AMC 10 like in 45 minutes but in AIME i felt it's so long and lost patience, AIME is also way harder than AMC 10.[/quote]Mental endurance is mostly something that comes from practice. Do not just study old tests. Do harder problems, including olympiad-style ones. You'll start to get better at those and you'll be able to focus for longer too.


geometry
parallelogram
analytic geometry
graphing lines
slope

so, if a parallelogram has points at (0,0), (2,5), (10,0), and (m,n), what is the length of the longest diagonal?
Please tell me how to do it.

This is the answer
[hide]13[/hide]

[hide]ok so we know that the other point (m,n) has to be at (12,5) the longest diagonal therefore is just the distance betwee the origin and point (12,5) you could use distance formula or right triangles, using right triangles we have a special right trianle where the hypotenuse is the diagonal of the paralelogram so 12^2 + 5^2 = 13^2 so we have the length of the longest diagonal is 13 units![/hide]

[hide="Sketchy Solution"]Visualize the picture... There are actually two possible parallelograms, but since I'm assuming $ m,n>0$, it will yield the desired answer.  It's easy to see that in counter-clockwise order, the vertices are $ (0,0),(10,0),(m,n),(2,5)$.

Thus, we have that the slope from $ (10,0)$ to $ (m,n)$ should be the same as the opposite side.

So add $ 2$ to $ 10$ and $ 5$ to $ 0$ to get $ (12,5)$.
The longest diagonal is then 13.[/hide]


There is a triangle with side lengths 4, 6 and 8.
A=the length of the median to the side of length 8
B=the length of the angle bisector to the side of length 6.
find B/A.
I need help finding B. I already got A.

If you figured out A, shouldn't B be easy?
My way involves Stewart's, if there isn't another way, this needs to be moved.

this is eric right? where did you find this problem?

this is not eric, its chris jinzi! I fou8nd the problem at the vestavia test


superior algebra
superior algebra unsolved

These are representations of D4, are they equivalent?

S(x)

 13  -34
  5   -13

S(y)

7  -16
3  -7

U(x)

0  -1
1   0

U(y)

0  1
1  0

yes, $ T\equal{}\left(\begin{array}{cc}3 & 5\\1 & 2\end{array}\right)$ will perform the required basis change, i.e. $ TU(\ldots)T^{\minus{}1}\equal{}S(\ldots)$


arithmetic sequence
number theory proposed
number theory

Prove that there exists infinitely many primes $ p$ such that: \[ 13|p^3\plus{}1\]

[quote="Omid Hatami"]Prove that there exists infinitely many primes $ p$ such that:
\[ 13|p^3 + 1
\]
[/quote]
It goes straight from Diriclet's theorem about primes in a arithmetic sequence. But I think I need a simple solution in a mathematical contest . Here is my solution .
Easy to check that $ 13|p^3 + 1 \Leftrightarrow p\equiv \{12,10,4\} (\mod 12)$
Suppose there exist only finite prime have one of the form $ 12 + 13k,10 + 13k,4 + 13k$
Exist an integer $ n_0$ such that for all $ p\geq n_0$ then p hasn't these forms.
Call $ p_1,...,p_t$ is all primes have one of the form $ 13k + 12,13k + 10,13k + 4$
[u]Result 1[/u] 13 is a quadric reside mod p iff $ p(\mod 13)$ is $ {1,12,3,10,4,9\}}$
[u]Result 2[/u] $ 3^x (\mod 13) \in\{1,3,9\}$
Call $ A$ is set of prime in these forms 1$ 13k + 1,13k + 3,13k + 9$
 Now consider the number $ A_k = (13\prod_{i = 1}^t p_i + 2)^2 - 13$
Easy to check that $ A_k\equiv 4 (\mod 13)$
If $ A_k$ has prime only in these form $ 13k + 1,13k + 3,13k + 9$ then exist an natural x such that
$ A\equiv 3^x (\mod 13 )$  it gives $ A_k$ in A ,this  is contradiction .
So have many primes satisfy condition .

Can I get a full solution if I use Dirichlet's Theorem?

[hide]Note that all primes of the form $ p\equal{}13k\minus{}1$ satisfy $ 13|p^3\plus{}1$. But, by Dirichlet's Theorem, there are infinitely many prime in the arithmetic progression $ \{12,25,38,51,64,\ldots\}$, all of whose terms are of the form $ 13k\minus{}1$. Our proof is complete.[/hide]

[quote="Johan Gunardi"]Can I get a full solution if I use Dirichlet's Theorem?

[hide]Note that all primes of the form $ p \equal{} 13k \minus{} 1$ satisfy $ 13|p^3 \plus{} 1$. But, by Dirichlet's Theorem, there are infinitely many prime in the arithmetic progression $ \{12,25,38,51,64,\ldots\}$, all of whose terms are of the form $ 13k \minus{} 1$. Our proof is complete.[/hide][/quote]
My teacher says, that if we know a know theorem(e.g. Dirichlet's), which leads to a solution, we can just write it. But this is a national olympiad so the situation is a little different, because your teacher cannot "defend" your solution.... Anyhow I would write that solution down, and try to come up with a solution like TTsphn's when/if I had finished the rest of the problems.. :)


For triangle ABC and a D inside the triangle, $ \angle$DAC=40, $ \angle$DCA=10, $ \angle$DCB=20, $ \angle$ABC=50. Find $ \angle$DBC.

Thanks.

I think its not possible. Let $ x \equal{} \angle DBC$ so that $ \angle ABD \equal{} 50 \minus{}x$. Then $ \angle BDC \equal{} 160 \minus{}x$ and $ \angle ADC \equal{} 130$. Therefore, $ \angle ADB \equal{} 360 \minus{} \angle ADC \minus{} \angle BDC \equal{} 70 \plus{} x$.  But then $ \angle BAD \equal{} 180 \minus{} \angle ABD \minus{} \angle ADB \equal{} 60$ so $ \angle BAC \equal{} \angle BAD \plus{} \angle DAC \equal{} 60 \plus{} 40 \equal{} 100$ which means the triangle is not acute.

$ E$ is circumcenter of $ \triangle ABC$.
$ F$ on $ AC$ s.t. $ \angle CEF\equal{}40^\circ$.

[hide="Find angle"]$ \angle DBC \equal{} 40^\circ$[/hide]

/Kouichi Nakagawa

Why is A, D, E on one line?

$ \angle BEC \equal{} 2(180^\circ \minus{} \angle BAC) \equal{} 160^\circ$.
So, $ \angle ECB\equal{}10^\circ$ i.e. $ \angle ECA\equal{}40^\circ$.

$ \angle DAC \equal{} 40^\circ$.
And, $ EC\equal{}EA \iff \angle ECA \equal{} \angle EAC \equal{} 40^\circ$.

So, $ \angle DAC \equal{} \angle EAC \equal{} 40^\circ$ i.e. $ A$, $ D$, $ E$ are colinear.

[b][color=red]Correction[/color][/b]
Sorry, I found a typo in my post.

$ \times : \angle DBC \equal{} 40^\circ$
$ \bigcirc : \angle DBC \equal{} 30^\circ$

[b][color=red]Epexegesis (Solution)[/color][/b]
$ F$ is circumcenter of $ \triangle DEF$.
Because, $ \angle CDE \equal{} \angle CAD \plus{} \angle ACD \equal{} 40^\circ \plus{} 10^\circ \equal{} 50^\circ$.
and $ \angle CFE \equal{} 180^\circ \minus{} (\angle FEC \plus{} \angle FCE) \equal{} 180^\circ \minus{} (40^\circ \plus{} 40^\circ) \equal{} 100^\circ$.
Hence, $ 2 \angle CDE \equal{} \angle CFE$.

And, $ \triangle DEF$ is equilateral triangle.
Because, $ FD \equal{} FE$ and $ \angle DFE \equal{} 2 \angle DCE \equal{} 2 \times 30^\circ \equal{} 60^\circ$.

Therefore, $ \triangle AFE \equiv \triangle BDE$.
Because, $ \triangle ABE$ is equilateral triangle, too.
So, $ AE \equal{} BE$, $ FE \equal{} DE$ and $ \angle AEF \equal{} \angle BED$ by $ SAS$.

Thus, $ \angle DBC \equal{} \angle DBE \minus{} \angle CBC \equal{} \angle FAE \minus{} \angle CBC \equal{} 40^\circ \minus{} 10^\circ \equal{} \boxed{30^\circ}$.


A collection S of integers is defined by the following three rules:
(1): 2 is in S
(2): For every x in S, 3x and x+7 are also in S
(3): No integers except those defined by rules (I) and (II) are in S.
What is the smallest integer greater than 2004 which is NOT in S?

[hide="big hint"]
Consider the set in mod 7
[/hide]


geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
geometry unsolved

Let $ ABC$ be a triangle. Its incircle has center $ I$ and touches $ AC$ and $ BC$ at $ D,E$ respectively. Let $ K,L$ be the reflection of $ D,E$ with respect to $ I$ respectively. Suppose that $ A,B,K,L$ are concyclic. Prove that $ AC\equal{}BC$ or $ AC\plus{}BC\equal{}3AB$.

The incircle $ (I)$ touches $ AB$ at $ F.$ $ (M)$ is cirumcircle of the $ \triangle IAB.$ $ (I), (M)$ intersect at $ P, Q$ and $ PQ \perp AIM.$ Inversion in $ (I)$ takes $ P, Q$ into themselves, $ A, B$ into the midpoints $ A', B'$ of $ DF, EF$ and $ (M)$ into the midline $ (A'B' \equiv PQ) \parallel DE \parallel KL$ of the intouch $ \triangle DFE.$ Assuming $ AC \neq BC,$  perpendicular bisectors of $ AB, KL$ are different from each other. Circumcenter of $ ABKL$ is intersection of the perpendicular bisectors of $ AB, KL,$ while circumcenter of $ ABPQ$ is  intersection of the perpendicular bisectors of $ AB, PQ,$  Perpendicular bisector of $ KL, PQ$ is common $ \Longrightarrow$ the circumcenters of $ ABKL, ABPQ$ are identical $ \Longrightarrow$ their circumradii $ MA \equal{} MB$ are equal $ \Longrightarrow$ $ (M)$ is circumcircle of $ ABKL$ (and $ P \equiv K,\ Q \equiv L$ identical). $ (I_a)$ is the A-excircle of the $ \triangle ABC,$ touching $ AC, BC$ at $ D_a, E_a.$ $ s \equal{} \frac {_1}{^2} (BC \plus{} AC \plus{} AB)$ is semiperimeter of the $ \triangle ABC.$ $ II_a$ is a diameter of $ (M).$ Circumcircles of $ CDIE, ABKL$ are circumcircles of congruent isosceles $ \triangle IDE \cong \triangle IKL,$ they are congruent $ \Longrightarrow$ $ AI_a \equal{} 2\ AI,$ $ s \equal{} AD_a \equal{} 2\ AD \equal{} 2 (s \minus{} AB)$ $ \Longrightarrow$ $ AC \plus{} BC \equal{} 3\ AB.$

Nice. I can't see why you need the inversion, though.

I did not. Unnecessary proving well known $ A \in IM$ by $ A'B' \parallel DE$ $ \Longrightarrow$ $ (A'B' \equiv PQ) \perp AI.$


So did anyone take it? I know we cant discuss till friday night

Well, this being oddly the only discussion about I can find, I'm posting here

I only took regional (1st yr school) and thought it was pretty easy, really liked the cars and the last polynomial 1.


determine the maximum value of $k$

s.t $|(1+\sqrt{n})sin(\pi \sqrt{n})|>k$ for all positive integer $n$ which is not square number.

Thia problem is in VIMC 1995  :)
  $ \frac{\pi}{2} $ is the best constant

I posted some time ago a solution on the forum, which I also put in Old and New Inequalities, but it has too many formulas to put it again.

Wow; It looks having complicated solution. Can anyone post solution?

and What is VIMC?


[color=darkblue]Let $ x,y,z\in Z^\plus{}$. Find $ (x;y;z)$ of this system: $ \begin {cases} xy\minus{}x!\plus{}y\equal{}1\\ yz\minus{}y!\plus{}z\equal{}1\\x^2\minus{}2y^2\plus{}2x\minus{}4y\equal{}2 \end {cases}$[/color]

[quote="thanhnam2902"][color=darkblue]Let $ x,y,z\in Z^ +$. Find $ (x;y;z)$ of this system: $ \begin {cases} xy - x! + y = 1(1) \\
yz - y! + z = 1(2) \\
x^2 - 2y^2 + 2x - 4y = 2(3) \end {cases}$[/color][/quote]

from $ (3) \rightarrow x$ be even number

from $ (1) \rightarrow y$ be odd number $ \rightarrow t y!$ be odd number $ \rightarrow y=1$

thus, $ x=2$ and $ z=1$ :wink:  $ (x,y,z)=(2,1,1)$

I don't see how you got $ y!$ is odd directly from these equations, bunhiacovski. Instead, I offer a second solution:
From $ (3)$ we can see $ x>y$, obviously. From $ (1)$ we gather $ y|x!\plus{}1$. Since $ x>y$, $ y|x!$ so $ y|1$. Therefore, $ y\equal{}1$. Then we get from $ (1)$ $ x\equal{}x!$, or, from $ (3)$, $ x^2\plus{}2x\equal{}8$, which lead to conclusion $ x\equal{}2$. After that, from $ (2)$ we gather $ z!\plus{}z\equal{}2$ and, obviously, $ z\equal{}2$. So, the only solution to this is $ (2,1,1)$.

[quote="mornik"]I don't see how you got $ y!$ is odd directly from these equations, bunhiacovski. [/quote]
Equation $ (2)$ is $ (y\plus{}1)z \minus{} y! \equal{}1$. Since $ y$ is odd, $ y\plus{}1$ is even, and thus $ y!$ is odd.


If $ a \plus{} b \plus{} c \equal{} 1$ , $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2$ and $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$

Compute:

$ a^4 \plus{} b^4 \plus{} c^4$

[hide="just a hint"]
From the first two we can find $ ab\plus{}bc\plus{}ca$.
Then from euler identity we can find $ abc$.

Then $ a^4\plus{}b^4\plus{}c^4\equal{}(a\plus{}b\plus{}c)^4\minus{}4(a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca)\plus{}2(ab\plus{}bc\plus{}ca)^2\plus{}4(a\plus{}b\plus{}c)abc$ :) [/hide]

This actually looks like a Newton's sums problem...

[hide="Solution"]
Let s_n = a^n + b^n + c^n. Let $ a$. $ b$ amd $ c$ be the roots of a cubic polynomial:
\[ P(x) \equal{} x^3 \plus{} px^2 \plus{} qx \plus{} r\]
By Vieta's we know that $ p \equal{} \minus{} s_1 \equal{} \minus{} 1$. Through Newton's Sums, we know that:
\[ s_2 \equal{} \minus{} ps_1 \minus{} 2q\]

\[ 2 \equal{} 1 \minus{} 2q\]

\[ q \equal{} \frac { \minus{} 1}{2}\]
Using Newton's Sums again:
\[ s_3 \plus{} ps_2 \plus{} qs_1 \plus{} 3r \equal{} 0\]

\[ 3 \minus{} 2 \minus{} \frac {1}{2} \plus{} 3r \equal{} 0\]

\[ r \equal{} \frac { \minus{} 1}{6}\]
Finally, we want $ s_4$, so we can write the Newton's sums again and substitute in the values that we know:
\[ s_4 \plus{} ps_3 \plus{} qs_2 \plus{} rs_1 \equal{} 0\]
Solving for $ s_4$, we see that $ s_4 \equal{} \frac{25}{6}$, as desired.

[/hide]

[quote="Obel1x"][quote="Dimitris X"][hide="just a hint"]
From the first two we can find $ ab \plus{} bc \plus{} ca$.
Then from euler identity we can find $ abc$.

Then $ a^4 \plus{} b^4 \plus{} c^4 \equal{} (a \plus{} b \plus{} c)^4 \minus{} 4(a \plus{} b \plus{} c)^2(ab \plus{} bc \plus{} ca) \plus{} 2(ab \plus{} bc \plus{} ca)^2 \plus{} 4(a \plus{} b \plus{} c)abc$ :) [/hide][/quote]

Actualy I did obtain:

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ac) \implies ab \plus{} bc \plus{} ac \equal{} \minus{} \frac {1}{2}$

After some computations I got

$ a^4 \plus{} b^4 \plus{} c^4 \equal{} 4 \minus{} \frac {1}{2} \plus{} 4abc \equal{} \frac {7}{2} \plus{} 4abc$

I don't know Euler's identity,that's why i messed up. :(  Can you give any for instance,thank you.[/quote]

Eulers identity is the following dear friend:
$ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)\equal{}\frac{1}{2}(a\plus{}b\plus{}c) \left( (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2 \right)$ :wink:

From $ ab\plus{}bc\plus{}ac \equal{}\minus{}\frac{1}{2}$we obtain: $ a^2b^2\plus{}b^2c^2\plus{}a^2c^2\equal{}\frac{1}{4}\minus{}2abc$ ... $ (1)$

From $ (a^2\plus{}b^2\plus{}c^2)^2\equal{}2^2$ and using ...$ (1)$ we get:

$ a^4\plus{}b^4\plus{}c^4\equal{}\frac{7}{2}\minus{}4abc$

Now from Euler's Identity:

$ a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}bc\minus{}ca) \implies abc\equal{}\frac{1}{6}$

substituting $ abc\equal{}\frac{1}{6}$ in ... $ (2)$ finally we reach the result:

$ a^4\plus{}b^4\plus{}c^4\equal{}\frac{25}{6}$


Find the no. of all 6 digit natural nos. such that the sum of their digits is 10 and each of the digits 0,1,2,3 occurs at least once in them.

The 6- digit number has 0,1,2,3 occurinf at least once.

If the number only consists of 0,1,2,3, then the following are the possible options:
(a)0,1,1,2,3,3
(b)0,1,2,2,2,3

If the number also consists of digits other than 0,1,2,3, then the following is the only option:
(a)0,0,1,2,3,4

The number of 6-digit numbers possible with $ \{0,1,1,2,3,3\}$ without the leading digit being $ 0$ is
$ \binom{5}{1}\frac{5!}{2! 1! 2!}$

The number of 6-digit numbers possible with $ \{0,1,2,2,2,3\}$ without the leading digit being $ 0$ is
$ \binom{5}{1}\frac{5!}{1! 3! 1!}$

The number of 6-digit numbers possible with $ \{0,0,1,2,3,4\}$ without the leading digit being $ 0$ is
$ \binom{5}{2}\frac{4!}{1! 1! 1! 1!}$

Adding up the three expressions, we get the required answer = 150 + 100 + 240 = 490

since the sum of the nos. is only 10
therefore other than rearranging the nos. 0,1,2,3
the only combinations are 4,0  3,1  1,3 and 2,2

I also solved this question. And I think my answer is right. Are you fully confirmed about your answer? My answer is 956. This message is for nicetry007.