geometry
incenter
conics
ellipse
circumcircle
geometry proposed

Given a triangle $ \Delta ABC$ with incenter $ I$. An arbitrary line passes though $ I$ and cuts $ AB$ and $ AC$ at $ X,Y$.the lines $ CX$ and BY meet at $ Q$. $ M,N,L$ are the projections of $ Q$ on the sides $ BC,CA$ ans $ AB$.Show that :

$ \frac{1}{QM}\equal{}\frac{1}{QN}\plus{}\frac{1}{QL}$

Any idea? Can we prove that the locus of Q is an ellipse??

Angle bisectors $ BI, CI$ cut $ CA, AB$ at $ E, F.$ As $ I$ is its own isogonal conjugate, the isogonal conjugate $ Q^*$ of $ Q$ is on $ EF$ - see [url]http://www.mathlinks.ro/viewtopic.php?t=261474[/url]. Let $ U, V, W$ be feet of perpendiculars from $ Q^*$ to $ BC, CA, AB.$ $ \triangle MNL, \triangle UVW$ have common circumcircle $ (P)$ centered at midpoint $ P$ of $ QQ^*$ $ \Longrightarrow$ powers of $ Q, Q^*$ to $ (P)$ are equal $ \Longrightarrow$ $ \overline{QM} \cdot \overline{Q^*U} \equal{} \overline{QN} \cdot \overline{Q^*V} \equal{} \overline{QL} \cdot \overline{Q^*W}$ Since $ Q^*U \equal{} Q^*V \plus{} Q^*W,$ the conclusion follows. Locus of $ Q$ is isogonal circumhyperbola of the line $ EF.$

Sorry for mistake. I didnt remember it was hyperbola.

\ [quote="Tiger100"] [color=darkred]Given are $ \Delta ABC$ with incenter $ I$ and two points $ X\in (AB)$ , $ Y\in (AC)$ so that $ I\in XY$ . Denote  the projections

$ M$ , $ N$ , $ P$  of the point $ Q\in CX\cap BY$ on the sides $ BC$ , $ CA$ , $ AB$ respectively. Prove that $ \frac {1}{QN}\plus{}\frac {1}{QP}\equal{}\frac {1}{QM}$ .[/color] [/quote]

[color=darkblue][b][u]Proof (with areas)[/u].[/b] Prove easily that or is well-known the implication $ \left\|\begin{array}{c}
X\in (AB)\\\
Y\in (AC)\\\
I\in XY\end{array}\right\|\Longrightarrow$

$ \boxed {\ b\cdot\frac {XB}{XA}\plus{}c\cdot\frac {YC}{YA}\equal{}a\ }$ . Denote $ \delta_d(X)$ - the distance from the point $ X$ to the line $ d$ . Therefore,

$ \left\|\begin{array}{c}
\frac {QM}{QN}\equal{}\frac ba\cdot \frac {a\cdot QM}{b\cdot QN}\equal{}\frac ba\cdot\frac {[BQC]}{[AQC]}\equal{}\frac ba\cdot\frac {\delta_{QC}(B)\cdot QC}{\delta_{QC}(A)\cdot QC}\equal{}\frac ba\cdot\frac {\delta_{QC}(B)}{\delta_{QC}(A)}\equal{}\frac ba\cdot\frac {XB}{XA}\\\\
\frac {QM}{QP}\equal{}\frac ca\cdot \frac {a\cdot QM}{c\cdot QP}\equal{}\frac ca\cdot\frac {[BQC]}{[AQB]}\equal{}\frac ca\cdot\frac {\delta_{QB}(C)\cdot QB}{\delta_{QB}(A)\cdot QB}\equal{}\frac ca\cdot\frac {\delta_{QB}(B)}{\delta_{QB}(A)}\equal{}\frac ca\cdot\frac {YC}{YA}\end{array}\right\|\ \bigoplus$

$ \ \implies\ \ \frac {QM}{QN}\plus{}\frac {QM}{QP}\equal{}1\ \ \implies\ \ \boxed {\ \frac {1}{QN}\plus{}\frac {1}{QP}\equal{}\frac {1}{QM}\ }$ .[/color]

[quote="Tiger100"] [color=darkred]Given are $ \Delta ABC$ with incenter $ I$ and two points $ X\in (AB)$ , $ Y\in (AC)$ so that $ I\in XY$ . Denote  the projections

$ M$ , $ N$ , $ P$  of the point $ Q\in CX\cap BY$ on the sides $ BC$ , $ CA$ , $ AB$ respectively. Prove that $ \frac {1}{QN} \plus{} \frac {1}{QP} \equal{} \frac {1}{QM}$ .[/color] [/quote]

[color=darkblue]Here is an easy extension :[/color]

[quote="Virgil Nicula"] [color=darkred]Given are $ \Delta ABC$ with an interior point $ P$ and two points $ X\in (AB)$ , $ Y\in (AC)$ so that $ P\in XY$ . Denote  $ D\in AP\cap BC$ and the projections

$ M$ , $ N$ , $ P$  of the point $ Q\in CX\cap BY$ on the sides $ BC$ , $ CA$ , $ AB$ respectively. Prove that $ \boxed {\ \frac {CD}{CA}\cdot\frac {1}{QN} \plus{} \frac {BD}{BA}\cdot\frac {1}{QP} \equal{} \frac {PD}{PA}\cdot\frac {1}{QM}\ }$ .

[b][u]Remark[/u].[/b] If $ (\alpha ,\beta ,\gamma)$ is the barycentrical coordinates of the point $ P$ , then the upper relation becomes $ \boxed {\ \frac {\beta}{b}\cdot\frac {1}{QN} \plus{} \frac {\gamma}{c}\cdot\frac {1}{QP} \equal{} \frac {\alpha}{a}\cdot\frac {1}{QM}\ }$ ,

i.e. $ \boxed {\ \frac {\beta}{[CQA]} \plus{} \frac {\gamma}{[AQB]} \equal{} \frac {\alpha}{[BQC]}\ }$ . The proof is similarly with upper proof.[/color][/quote]

[color=darkblue]Here is a particular case when $ P: \equal{} G$ - the centroid of $ \triangle ABC$ :[/color]

[quote][color=darkred]Given are $ \Delta ABC$ with the centroid  $ G$ and two points $ X\in (AB)$ , $ Y\in (AC)$ so that $ G\in XY$ . Denote the projections

$ M$ , $ N$ , $ P$  of the point $ Q\in CX\cap BY$ on the sides $ BC$ , $ CA$ , $ AB$ respectively. Prove that $ \boxed {\ \frac {1}{[CQA]} \plus{} \frac {1}{[AQB]} \equal{} \frac {1}{[BQC]}\ }$ .[/color][/quote]


function
Putnam
algebra unsolved
algebra

Prove that there is a unique function $f : R^+ \rightarrow R^+$ satisfying $f( f(x) ) = 6x - f(x)$ for all $x.$

an old putnam question, if i remember correctly.
I shall prove that f(x)=2x is the only solution.
suppose that f(x0) != 2x0 for some x0>0.
then define xi=f(xi-1) for i>=1.
clearly xi+1 + xi= 6xi-1
thus xi = (f(x0)+3x0)/5 (2^n) + [[2x0 - f(x0) ]/ 5 ](-3)^n
Now note that for large enough n, the first term is negligible (??) and sign of xi will be the same as that of [2x0-f(x0)]*(-1)^n. but this term will change sign consequtively and therefore xn<0 for some n. Contradiction.
thus f(x)=2x.


Anyone can solve this?
HOw to solve this:

If 0 < a < b < c < 1, how can i verify that (c-a)/(1-ca), (b-a)/(1-ab), (c-b)/(1-cb) are the three sides of a triangle?

Letting $\frac{c-a}{1-ac}=x$ , $\frac{c-b}{1-bc}=y$ , $\frac{b-a}{1-ab}=z$

First we show that $\text{max\{x,y,z}\}=x$

or same as $x-y=\frac{(b-a)(1-c^2)}{(1-ac)(1-bc)}>0$ and $x-z=\frac{(c-b)(1-a^2)}{(1-ac)(1-ab)}>0$

so $x>y$ , $x>z$

Then we will show $y+z>x$ which is equivalent to

$y+z-x=\frac{(c-b)(b-a)(c-a)}{(1-ac)(1-ab)(1-bc)}>0$ .

Hence $x,y,z$ are three side of a triangle .


γιατί να μην συναντηθούμε;

νομιζω οτι ειναι υπεροχη ιδεα αυτο, τα παιδια  που θα δωσουν Αρχιμηδη  ειναι τυχερα  γιατι  αυτος ειναι και ενας σκοπος των διαγωνισμων  η επικοινωνια, ποιοι ειναι μεσα ?

Θα ξέρουμε αύριο μάλλον..... :!:
Πάντως καλή ιδέα . Όλη η κοινότητα του <<ελληνικά>> ενωμένη :)

[quote="fotis"]Α ωραία και εγώ θέλω να σας γνωρίσω.....Βέβαια πρέπει να περάσω στον Αρχιμήδη...Αλήθεια ξέρει κανείς πότε βγαίνουν τα αποτελέσματα;[/quote]

ΈΝΑς ΑΠΌ ΤΟ ΔΣ της ΕΜΕ μου είπε ότι θα βγουν Σάββατο , αλλά χθες ένας από την επιτροπή διαγωνισμών μου είπε για το άλλο Σάββατο . ΥΠΟΜΟΝΗ !

  [b]Καλά αποτελέσματα !!![/b]

 [i]Α! Άκουσα - από έμπιστο άτομο της ΕΜΕ -  ότι φέτος ίσως πάρουν μεγαλύτερο ποσοστό μαθητών στον Αρχιμήδη σε σχέση με πέρισυ. Αυτό είναι καλή είδηση. [/i]

[quote="stergiu"]...
 [i]Α! Άκουσα - από έμπιστο άτομο της ΕΜΕ -  ότι φέτος ίσως πάρουν μεγαλύτερο ποσοστό μαθητών στον Αρχιμήδη σε σχέση με πέρισυ. Αυτό είναι καλή είδηση. [/i][/quote]
ειδικα για εμας που θα μπορουσαμε να γραψουμε καλυτερα...
στο "νεο" χημειο θα γινει παλι ο αρχιμιδης?

να ρωτισω κατι...επιτρεπονται παιδια που να μην ειναι απο το artofproblemsolving? επισης ειχα απορια που σκεφτεστε να παμε...

ο διαγωνισμος θα γινει στο νεο χημειο, και φυσικα γιατι να μην ερθουν και αλλοι darksaga,?  :)

prokeite gia sinantisi tis tomis aops-archimidi i ola ta meli tou aops kai alloi?


If $P$ is a monic polynomial of degree $n>1,$ having $n$ negative roots ( counting multiplicities), show that:

                                            \[ P'(0)P(1)\geq 2n^{2} P(0). \]

  Determine the equality case.  ;)

$P(x)$ is expressible as $\prod_{k=1}^{n} (x + a_k)$, where $a_k > 0$.

$P(1) = \prod_{k=1}^{n} (1 + a_k)$
$P(0) = \prod_{k=1}^{n} a_k$
$P'(0) = \sum_{k=1}^{n} \displaystyle\frac{ P(0) }{a_k}$

So, dividing both sides by $P(0)$, we have $\left( \sum_{k=1}^{n} \displaystyle\frac{1}{a_k} \right) \left( \prod_{k=1}^{n} (1 + a_k) \right) \ge 2n^2$.

Aaand I'm stuck.   :P  I'm terrible at inequalities.


For how many integer value of $ m$ does the lines $ 13x\plus{}11y \equal{} 700$ and $ y \equal{} mx\minus{}1$ intersect in a point with integer valued coordinats?

A. None
B. 1
C. 2
D. 3
E. Infinitely many

[hide]
We have a system:
\[ 13x \plus{} 11y \equal{} 700 \\
y \equal{} mx \minus{} 1
\]
Rearranging and multiplying, we have
\[ 11y \equal{} 700 \minus{} 13x \\
11y \equal{} 11mx \minus{} 11
\]
Setting them equal, we have
\[ 700 \minus{} 13x \equal{} 11mx \minus{} 11
\]
Rearranging, we have
\[ 24mx \equal{} 711
\]
Clearly, $ 24$ does not divide $ 711$, and thus there are no integer values of $ m$ that work $ \Rightarrow \boxed{A}$. [/hide]

Note from Silverfalcon: Please hide your answer!

123456789, your answer is wrong. LHS has $ \minus{}13x$ while RHS has $ 11mx$. Their sum does not equal to $ 24mx$.

[hide="Correct solution"]
(C)

Solve both equations for $ y$ to get:

$ \frac{\minus{}13x}{11} \plus{} \frac{700}{11} \equal{} mx \minus{} 1$

Simplifying this and solving for $ m$ gives us:

$ \frac{711}{11x} \minus{} \frac{13}{11} \equal{} m$. Now, key here is to factor out $ \frac{1}{11}$:

$ \frac{1}{11} \left(\frac{711}{x} \minus{} 13\right) \equal{} m$

Note that 13 is 2 in mod 11. So, $ \frac{711}{x}$ must be 2 in mod 11 or otherwise, $ m$ will not take integer value. There are 6 factors of 711 and only 9 gives us the result (79) that is 2 in mod 11. Now, are we done? NO.

Remember that the problem doesn't change for -9 as well. So there are actually 2 values of $ m$ so the correct answer is C. [/hide]


Number Theory Problem
The natural number y was obtained from the number x by rearranging its digits. It is known that

Prove that x is divisible by 50.

Hello first post the q properly in olympiad problem discussion topic.
Don't open separate topics for such a simple problem[it can be easily guessed] :D  :)

sorry to say

but what is known???????? :huh:


Middle-schoolers:  Two questions follow.  You are invited to [b]answer only one[/b] of them.  If you think the first question is easy, please don't answer it.  Give the second question a try.  Please hide your answer.

Q1:  "Using pennies, nickels, and dimes, how many ways are there to make 17 cents change?"

Q2:  "Using pennies, nickels, dimes, quarters, and half-dollars, how many ways are there to make 50 cents change?

I'm guessing KBabe answered question one.

[hide="Here is my answer to question two."]p=pennies
n=nickels
d=dimes
q=quarters
h=half-dollars
Whenever the sum doesn't add to 50, just add the appropriate number of pennies
1h
2q
1q+2d+1n
1q+2d+
1q+1d+3n
1q+1d+2n
1q+1d+n
1q+1d
1q+5n
1q+4n
1q+3n
1q+2n
1q+n
1q
5d
4d+2n
4d+n
4d
3d+4n
3d+3n
3d+2n
3d+n
3d
2d+6n
2d+5n
2d+4n
2d+3n
2d+2n
2d+n
2d
d+8n
d+7n
d+6n
d+5n
d+4n
d+3n
d+2n
d+n
d
10n
9n
8n
7n
6n
5n
4n
3n
2n
1n
50p
So 50.
[/hide]

[hide="to KBabe"]Close.  But, look for another one.[/hide]
[hide="to RC-7th"]Great job!  A very systematic approach.[/hide]

[quote="RC-7th"]rcv, do you have a faster way? I know that I didn't have to list all the possibilities to count them, but it's still a lot of casework.[/quote]

[hide]For a faster method -- or to help check your work -- I would suggest you begin with a systematic listing, and then see if you can find any patterns.

Beginning with your approach, let's combine certain elements of your list into groups of cases:

1h+... <-- 1 case

2q+0d+... <-- 1 case

1q+2d+... <-- 2 cases
1q+1d+... <-- 4 cases
1q+0d+... <-- 6 cases

0q+5d+... <-- 1 case
0q+4d+... <-- 3 cases
0q+3d+... <-- 5 cases
0q+2d+... <-- 7 cases
0q+1d+... <-- 9 cases
0q+0d+... <-- 11 cases

Excluding the half-dollar, we have 3 groups (2q, 1q, and 0q).  Do you see the patterns within each group?

What if you were asked to find all ways to make change for 75 cents, using just quarters, dimes, nickels, and pennies.  I think we could extend this grouping fairly easily, without enumerating all answers:

3q+0d+... <-- 1 case

2q+2d+... <-- 2 cases
2q+1d+... <-- 4 cases
2q+0d+... <-- 6 cases

1q+5d+... <-- 1 case
1q+4d+... <-- 3 cases
1q+3d+... <-- 5 cases
1q+2d+... <-- 7 cases
1q+1d+... <-- 9 cases
1q+0d+... <-- 11 cases

0q+7d+... <-- 2 cases
0q+6d+... <-- 4 cases
0q+5d+... <-- 6 cases
0q+4d+... <-- 8 cases
0q+3d+... <-- 10 cases
0q+2d+... <-- 12 cases
0q+1d+... <-- 14 cases
0q+0d+... <-- 16 cases

Without enumerating the values, I think there would be $1+12+36+72=121$ ways to make change for 75 cents (excluding the half-dollar).
[/hide]

[quote="KBabe"][hide] 5 ways:
1d+1n+2pennies
1n+12p
2n+7p
3n+2p
17p
[/hide][/quote]

one more: 1d+7p

[quote="KBabe"]one more: 1d+7p[/quote]

Good job!

Hi RCV,

In reading your "making change" post I have two questions for you and the students.  The simple one is I think you just had a typo on 75cents, but I thought it would help some with the trick that showed me the typo as at first I was confused.  When counting using 1 quarter you had 1,3,5,7,9,11 sets of cases for a total of 35. I noticed that since this is the sum of odd numbers starting with 1(1+3+5+7+9+11), the answer had to be a square number, not 35.  Since you are adding the first 6 odds, their sum is 6^2 =36.  You must have had that in your work since the total is correct.

My real question is in the meaning of making change.  My kids recently took an Old target round in which a question asked how many ways could you make change for a quarter.  The official answer did not inlcude using just 1 quarter, since 1 quarter is not "change" for 1 quarter, but rather the same(not changed).

While a half dollar is a way to make 50 cents, it is a great example of how reading the question carefully can make a big difference.  I am not really sure how they would word "making change for 75 cents".  To me if you are changing 75cents that implies you already have 75 cents in come configuration, and therefore one of the list of ways to make 75 cents would not count.  Does that make sense to anyone :)

Oh well, enough rambling, I guess I would just say be careful if it is asking you to make 75 cents, or make change for 75 cents!

Umm...You double posted so I'm going to delete one.  :)

[quote="frost13"]. . . I noticed that since this is the sum of odd numbers starting with 1(1+3+5+7+9+11), the answer had to be a square number, not 35.  Since you are adding the first 6 odds, their sum is 6^2 =36.  . . .[/quote]
Good catch.  And a very good observation on summing the odd numbers!

[quote="frost13"]My real question is in the meaning of making change.[/quote]
I understand your point about the wording.  I think my original query was unambiguous -- "how many ways are there to make 50 cents change".  But the wording of my follow-up was sloppy -- "find all ways to make change for 75 cents".

In my experience, modern MATHCOUNTS competition questions are rarely worded ambiguously.  As frost13 points out, "change in the amount of 25 cents" isn't quite the same as "change for a quarter".  Read the questions carefully.


Let $ f: \mathbb R\to\mathbb R$ be continuous. If $ \lim_{x\to\infty}\dfrac{f(x)}{x^3}\equal{}\lim_{x\to\minus{}\infty}\dfrac{f(x)}{x^3}\equal{}0,$ then prove that $ \exists\,x\in\mathbb R\mid x^3\plus{}f(x)\equal{}0.$

Let's denote  $ g(x) \equal{} x^3 \plus{} f(x)$, continous function. For $ x\neq 0$ we can write $ g(x)\equal{}x^3(1\plus{}\frac{f(x)}{x^3})$ , so $ g(\minus{}\infty)\equal{}\minus{}\infty ,\ g(\plus{}\infty)\equal{}\plus{}\infty$, so there is $ x_0$ such that $ g(x_0)\equal{}0$
 :cool:


Given n number $ a_1;a_2: ..: a_n$ in[-1;2] such that their total sum is 0.Let $ u_k$=$ \frac{a_k. \sqrt{4k\minus{}1}}{(4k\minus{}3)(4k\plus{}1)}$ for k=1,2,..n Prove that $ \minus{}\frac{\sqrt{n}}{2} \leq u_1\plus{}u_2\plus{}..\plus{}u_n \leq\frac{\sqrt{n}}{2}$

What is the source of this problem? I think there is a forum more appropriate for it than the AMC forum.


geometry
incenter
circumcircle
geometric transformation
reflection
rectangle
parallelogram

Consider a triangle ABC and I be its incenter.Let D,E,F be the tangency point of incenter to $ BC,AC,BA$.Let M be the point such that:$ \vec{MI}\equal{}\minus{}2\vec{EI}$,N be the points such that $ \vec{NI}\equal{}\minus{}2\vec{FI}$.Prove that $ (MBI),(NCI)$ are tangent to each other.

Let $ K$ be orthocenter of the intouch $ \triangle DEF$ with circumcircle $ (I),$ $ D', E', F' \in (I)$ diametrically opposite points to $ D, E, F,$ and $ I_D, I_E, I_F$ reflections of its circumcenter $ I$ in $ EF, FD, DE.$ $ EFF'E'$ is a rectangle, $ EF \equal{} E'F'$ and $ EF \parallel E'F'.$ $ \triangle KEF \sim \triangle II_EI_F$ are centrally congruent by SAS, $ I_EI_F \equal{} EF$ and $ I_EI_F \parallel EF.$ Then $ I_EI_FF'E'$ is a parallelogram and $ E'I_E \parallel F'I_F.$ Circumcenters $ Q_B, Q_C$ of $ \triangle IMB, \triangle INC$ are intersections of perpendicular bisectors $ p_B, p_C$ of $ IB, IC$ with perpendicular bisectors $ t_E, t_F$ of $ IM, IN,$ i.e., tangents of $ (I)$ at $ E', F'.$ $ B, C$ are poles of $ FD, DE$ WRT $ (I)$ $ \Longrightarrow$ $ I_E, I_F$ are poles of $ p_B, p_C$ WRT $ (I)$ and $ E', F'$ are poles of $ t_E, t_F$ WRT $ (I)$ $ \Longrightarrow$ the circumcenters $ Q_B, Q_C$ are poles of parallels $ E'I_E \parallel F'I_F$ WRT $ (I),$ therefeore $ Q_B, Q_C$ are collinear with center $ I$ of the polar circle.


The points 1,2,... 1000 are paired up at random to form 500 intervals [i,j]. What is the probability that among these intervals is one which intersects all the others?

Very complicated problem.. are you sure this is Intermediate? Lol

I'm with aidan: can we get this moved to one of the Combinatorics forums?

Looking at small cases, it appears that the answer will be [hide="answer?"]$ \frac{2}{3}$[/hide] but I haven't a clue how to prove it in general.


Anybody is having all the identities of the binomial coefficients ?
A resource link ?

Thanx for the link. But getting all the identities under one topic is not a bad idea.

The idea that one could write down all identities involving binomial coefficients is very strange.  :|  Mathematics is not just a list of formulas, and I'm not sure what end is served by producing such lists.   :maybe:

I see two problems with this, one is that the list will go on forever, and the second is that, to someone who doesn't know a particular identity, simply knowing it isn't all that useful.

I sure hope you don't plan on memorizing this list, because the majority of these identities seem pretty useless. You're better off learning the combinatorics behind the identities rather than the identities themselves.

I don't know much about combinatorics or the Bionomial theorem, but what I do know is by the looks of it, that list is neverending.  What is the use of this?

I also know that it can continue for quite sometime though I cannot agree on its endlessness. The intention is not to memorise all of them. Neither it is that they will be presented without proofs. The ones I have given are elementary, and that's the reason proofs are not presented. I am sure, as we proceed, we will come across many new and beautiful identities.
Members must be aware that there is an 'Inequalities Marathon' going for quite sometime, which really does not seem to end. I consider it as a very delightful and worthwhile exercise.
There is also a book titled "Summation of Series" By L. B. W. JOLLEY, which is a collection of more than 1100 formulas for various mathematical series/summation without proof. Many will question as to what purpose it is serving. Only when one looks through the pages of the book will come to know about its worth as numerous ideas start percolating down. It is really an amazing work.
Finally, I dare not call the 'Binomial Theorem Identities' useless. I request those who are not interested, to refrain themselves from participating. Other Mathlinkers who will consider this 'useless' persuit of unknown yet beautiful identities worthwhile may please come forward. And of course with proofs, if you have a nice one.

 :(  :(

$ \binom{n}{k}\equal{}\binom{n\minus{}1}{k\minus{}1}\plus{}\binom{n\minus{}1}{k}$

The above identity governs Pascal's Triangle.

Thanx gurmies. Please number the identities you send serially. Your entry will be as this -

$ 13.\ \quad C_{k}^{n}\equal{}C_{k\minus{}1}^{n\minus{}1}\plus{}C_{k}^{n\minus{}1}$

For [9] vide [url]http://www.mathlinks.ro/viewtopic.php?t=63957[/url]

This is from pfanni

$ 14. \quad\sum_{i \equal{} k}^{n}iC_{k}^{i} \equal{} (n \plus{} 1)C_{k \plus{} 1}^{n \plus{} 1} \minus{} C_{k \plus{} 2}^{n \plus{} 2}$

[quote]Original pfanni's version $ \sum_{i \equal{} k}^{n}i{i\choose k} \equal{} (n \plus{} 1){n \plus{} 1\choose k \plus{} 1} \minus{} {n \plus{} 2\choose k \plus{} 2}$[/quote]
Have I converted correctly ?
Thanx pfanni.

Solution is at [url]http://www.mathlinks.ro/viewtopic.php?search_id=1117761205&t=315391[/url]

These are from makar:

$ 15. \quad \frac {\sum_{r \equal{} p \plus{} 1}^{m}C_{p \plus{} 1}^{r}C_{q}^{m \minus{} r}}{(m \minus{} p \minus{} q)\sum_{r \equal{} p}^{m}C_{p}^{r}C_{q}^{m \minus{} r}} \equal{} \frac {p \plus{} 1}{p \plus{} q \plus{} 2}$
$ 16. \quad \ \sum_{r \equal{} 0}^{n}C_{r}^{n \plus{} r}\left(\frac {1}{2}\right)^{n \plus{} r} \equal{} 1$

Thanx makar.

vide [url]http://www.mathlinks.ro/viewtopic.php?search_id=1117761205&t=302880[/url]
[url]http://www.mathlinks.ro/viewtopic.php?search_id=1117761205&t=302881[/url]

This is from hxy09's friend:

$ 17. \quad  \sum_{k\equal{}0}^{n}C_{k}^{n}(k\plus{}1)^{k\minus{}1}(n\minus{}k\minus{}1)^{n\minus{}k}\equal{}n^{n}$
hxy09's original:
[quote]$ \sum_{k\equal{}0}^{n}\binom{n}{k}(k\plus{}1)^{k\minus{}1}(n\minus{}k\minus{}1)^{n\minus{}k}\equal{}n^{n}$[/quote]
Thanx hxy09.

Solution at [url]http://www.mathlinks.ro/viewtopic.php?search_id=1117761205&t=296996[/url]


What is the condition for a root to be primitive based on roots of unity?
E.g solve$ {x^6}\equal{}1$ what are the roots and the primitive roots

hello, use that $ x^6\minus{}1\equal{}(x\minus{}1)(x\plus{}1)(x^2\plus{}x\plus{}1)(x^2\minus{}x\plus{}1)$
Sonnhard.

Roots are $ e^{i\frac{2k\pi}{6}} \; k\equal{}0,1,2,3,4,5$. Root of unity is primitive if and only if $ (k,n)\equal{}1$. In our case primitive roots are $ e^{i\frac{2\pi}{6}}, e^{i\frac{5k\pi}{6}}$

In fact, the precise definition is that $ \omega$ is a primitive $ n$th root of unity if and only if it is not a $ k$th root of unity for any positive integer $ k < n$. Which, of course, is the same thing as $ \gcd(k,n)\equal{}1$.


Let $ a,b,c > 0$ and $ abc\equal{}1$ prove that $ \frac{a^{3}c}{(b\plus{}c)(c\plus{}a)}\plus{}\frac{b^{3}a}{(c\plus{}a)(a\plus{}b)}\plus{}\frac{c^{3}b}{(a\plus{}b)(b\plus{}c)} \geq \frac34$

After clearing denominators we obtain:
$ 4\sum_{cyc}(a^4c \plus{} a^3cb)\ge 3(\sum_{cyc}(a^2b) \plus{} \sum_{cyc}(a^2c) \plus{} 2abc)$ multiply $ LS$ by $ 1 \equal{} a^{\frac {2}{3}}b^{\frac {2}{3}}c^{\frac {2}{3}}$
$ \Leftrightarrow 4\sum_{cyc}(a^4c) \plus{} 4\sum_{cyc}(a^3cb)\ge 3\sum_{cyc}(a^{\frac {8}{3}}b^{\frac {5}{3}}c^{\frac {2}{3}}) \plus{} 3\sum_{cyc}(a^{\frac {5}{3}}c^{\frac {8}{3}}b^{\frac {2}{3}}) \plus{} 6a^{\frac {5}{3}}b^{\frac {5}{3}}c^{\frac {5}{3}}$
$ \Leftrightarrow \sum_{cyc}(a^4c \plus{} a^3bc \plus{} b^4a \minus{} 3a^{\frac {8}{3}}b^{\frac {5}{3}}c^{\frac {2}{3}}) \plus{} \sum_{cyc}(a^4c \plus{} a^3bc \plus{} c^3ab \minus{} 3a^{\frac {8}{3}}b^{\frac {2}{3}}c^{\frac {5}{3}}) \plus{} (a^4c \plus{} b^4a \plus{} c^4b \minus{} 3a^{\frac {5}{3}}b^{\frac {5}{3}}c^{\frac {5}{3}}) \plus{} (a^3bc \plus{} b^3ca \plus{} c^3ab \minus{} a^{\frac {5}{3}}b^{\frac {5}{3}}c^{\frac {5}{3}})\ge 0$  (*)
Each of the expressions in the brackets is greater than or equal to $ 0$ by AM-GM and therefore inequality (*) is true, hence the original inequality is true.

I think that  I 've solved it problem before, the technique I use is substituting $ a\equal{}\frac{x}{y} ; b\equal{}\frac{y}{z} ; c\equal{}\frac{z}{x}$. From here, using algebraic transformation, we get a homogenerous inequality which is easier to solve (I think so,  :D ) and get rid of the hypothesis $ abc\equal{}1$.


Consider any permutation $\sigma$ of $\{0,1,2,\dots,9\}$ and for each of the 8 triples of consecutive numbers in this permutation, consider the sum of these three numbers. Let $M(\sigma)$ be the largest of these 8 sums. (For example, for the permutation $\sigma = (4, 6, 2, 9, 0, 1, 8, 5, 7, 3)$ we get the 8 sums 12, 17, 11, 10, 9, 14, 20, 15, and $M(\sigma) = 20$.)

(a) Find a permutation $\sigma_1$ such that $M(\sigma_1) = 13$.

(b) Does there exist a permutation $\sigma_2$ such that $M(\sigma_2) = 12$?

Ok, let me post it.

Suppose that $M\left(\sigma_2\right) = 12$ for some permutation $\sigma_2$. This means that the sum the first nine numbers in the permutation is at most $3 \cdot 12 = 36$ (three groups of thre numbers). Since the numbers sum to 45, the last number must be 9. Similarly, the first number must be 9. Contradiction.


MATHCOUNTS
function
number theory
modular arithmetic
relatively prime

John and Laurie sit on one teeter totter.  At noon, John's feet are touching the ground.  John kicks off immediately.  3 seconds later Laurie's feet are touching the ground.  She kicks off immediately.  4 seconds later John's feet are touching the ground again.  The two children kick off at the same rate for 21 minutes.

Susan and Sally are playing on the adjacent teeter totter.  At noon, Susan's feet are touching the ground.  Susan kicks off immediately.  4 seconds later Sally's feet are touching the ground.  Sally kicks off immediately.  5 seconds after that Susan's feet are touching the ground again.  Susan and Sally teeter back and forth at the same rate for 21 minutes.

During the 21 minutes, how many times are John and Sally touching the ground at the same time?

I tried to notice a pattern but it seemed like John ( I think it was him, I have lost my "work") was always behind Sally.  Can you give a hint as to how to solve this?

Do you know any modular arithmetic?  Can you describe the times that each of the children are touching the ground in terms of a modular congruence?  Can you describe them as parametric equations?

John is touching the ground at time [i]n[/i] where [i]n[/i] :=: 0 (mod 7).

Or, you could say John  is touching the ground at time 7j for any nonnegative integer j.

I don't know any modular arithmetic but I do know how parametrics work.  I can try to describe them as parametrics and see how it helps.

I am going to suggest learning about modular arithmetic.  It's one of the most useful problem solving tools that I have encountered.

That was fun.  I think it may also be the first time I've solved a problem like this.

[hide]

John is on the ground at noon, and returns to the ground every seven seconds after that.

Sally is on the ground four seconds after noon, and returns to the ground every nine seconds after that.

(Note:  I'm not sure if this is the "correct" notation for modulus, but it's what is used in C++ programming, so I'm familiar with it.  So, read X%Y as X mod Y, and if it isn't correct, please inform me)

Let t denote the number of seconds elapsed past noon.  0 :le: t :le: (21*60)

John touches the ground when t%7 = 0

Sally touches the ground when (t-4)%9=0  (this is easily verified)

So, when t%7=(t-4)%9=0, they are both touching the ground.

One way of solving this is to make a list of values where t%7 = 0 (multiples of 7), and another list of values where (t-4)%9=0 (multiples of 9, plus 4)

You'll find that at t=49, t%7=(t-4)%9=0.  This is the first time they are on the ground together.  Extend the lists farther, and you find that the next time they are on the ground simultaneously is t=112.

From this, it can be found that they are on the ground together once every 63 seconds, starting at second 49.  (112-49=63 seconds)  So we can set up the following equation to find floor(X), the number of times they are on the ground simultaneously in 21 minutes (the floor(X) function returns the greatest integer less than or equal to X.  Fractions of times on the ground do not make sense in the context of the problem)

49+63X = 1260

63X = 1211

X = 1211/63

X = 19.2222

floor(19.2222) = 19

They are on the ground simultaneously 19 times in 21 minutes.

[/hide]

Hope I did that correctly.

Notation for modular arithmetic typically goes like this:

t :equiv: 0 (mod 7) means that (t-0)/7 is an integer.

t :equiv: 4 (mod 9) means that (t-4)/9 is an integer.

The :equiv: sign means "is congruent to".

FYI -- Your solution is fine until the very end.  What values can X take? (the answer is not just the largest value for X.)

Otherwise, nicely done.

[quote="MCrawford"]
FYI -- Your solution is fine until the very end.  What values can X take? (the answer is not just the largest value for X.)

[/quote]

This is a very old post.  I was trying to find some new types of problems.  There were some very interesting problems around this date.

I looked at the discussion and can't seem to figure out what other value X can take.  Can someone take on the challenge?

Naga

I think he went back and edited his post to complete his solution after the post I made.

It looks to me like confuted had an "off-by-one" error.  Using confuted's terminology, X can take any integer value from 0 to 19.  So they are touching the ground together 20 times during the 21-minute period.

Oh, right, that must have been what I meant.

[hide="Something"]
Every seven seconds, and every nine seconds, plus four seconds.. 7 and 9 are relatively prime, so 63.

21(60)/63 = 20. Slightly less with the four seconds factored in, so 19.
[/hide]


Just wondering. If we want to ask something trivial like e.g. Why doesn't MSN start on my computer (love feedback for this  :P ) or something along those lines where would we post it?

Round table. MS sucks? Is this a new problem (i.e. you used MSN successfully before), or did you just download MSN and it won't run? If the former, what did you change about your comp. between last time it worked and now? Better go undo that...

[quote="Lazarus"]Reformat your entire computer and reinstall MSN.

:D[/quote]

Truth is that I'm so hopeless with computers I will probably destroy the computer before anything useful happens. I'm not sure what I changed because as far as I know I haven't changed anything about it.

This is irrelevant but a problem popped up yesterday after I re-installed MSN (yet again). Now my screen resolution oscillates between normal and then screws up after 5 minutes so I have to constantly re-set it.


topology
Functional Analysis
advanced fields
advanced fields unsolved

[b]If we have a Banach space X,and  X* its dual space;
is it true that a set C from X* which is compact in the strong topology(associated with the default norm on X*) is compact in the *-weak topology sigma(X*,X)??[/b]

Yes, it is. Every open set in the weak-* topology is open in the strong topology. Therefore, every cover of $C$ with sets open in the weak-* topology is a cover with sets open in the strong topology. Since $C$ is compact in the strong topology, there exists a finite subcover.

In fact, by the Banach-Alaoglu theorem every closed and bounded subset of a Banach space is compact in the weak-* topology.

[quote="Yury"]
In fact, by the Banach-Alaoglu theorem every [b]closed[/b] and bounded subset of [b]a Banach space[/b] is compact in the weak-* topology.[/quote]
You meant "a dual space of a Banach space" and  "weak*-closed", of course ;)

[quote="fedja"]You meant "a dual space of a Banach space" and  "weak*-closed", of course ;)[/quote]

Yes, I did  :D Thanks!


algebra
polynomial
trigonometry
modular arithmetic
complex numbers
Galois Theory

Solve the equation:

$ \text{tg}^2(2x)\plus{}2\cdot\text{tg}(2x)\cdot\text{tg}(3x)\minus{}1\equal{}0$

I expanded all the $ \tan(2x)$ and $ \tan(3x)$ terms into $ \tan(x)$ terms, and found that $ \tan(x)$ was a root of
\[ 7y^6\minus{}35y^4\plus{}21y^2\minus{}1,\]
if my algebra was correct.

[hide="Hint"] That polynomial is the imaginary part of $ (y \plus{} i)^7$. [/hide]

It's something you come to expect from working problems like this and using Euler's formula (equivalently, de Moivre) to simplify calculations.  The coefficients should ring a bell, as should the alternating pattern in the signs.

hello, by applying the addition formulas i have got the following equation
$ \frac{(\minus{}1\plus{}2\cos(2x)\minus{}2\cos(4x)\plus{}2\cos(6x))\csc(\frac{\pi}{4}\minus{}x)^2\csc(\frac{\pi}{4}\plus{}x)^2}{4(\minus{}1\plus{}2\cos(2x))}\equal{}0$
Sonnhard.

From t0rajir0u's hint, we have
$ Im (y\plus{}i)^7 \equal{} Im [(y^2\plus{}1)^{1/2}(\cos \theta \plus{} i\sin \theta)]^7$
$ \equal{} (y^2\plus{}1)^{7/2} \sin 7\theta$
where $ \theta \equal{} \arctan (1/y)$

So we have $ \sin 7\theta \equal{} 0 \implies \theta \equal{} \pi k/7; k\in \mathbb{Z}\minus{}\{z|z\in\mathbb{Z},z \equiv 0 \pmod7\}$ (excluding them as $ 1/y \ne 0$)
$ \tan \theta \equal{} 1/y \implies \cot \theta \equal{} y \equal{} \tan(\pi/2 \minus{} \theta) \equal{} \tan x$

So the solutions are $ x\equal{}\pi/2 \minus{} \pi k/7$.

t0rajir0u, so I had to expand and then factor into $ (y\plus{}i)^7$? Was there any easier way to notice it was the imaginary part of $ (y\plus{}i)^7$ without expanding?

Yes.  You could solve directly:
[hide="Solution"] $ \tan 3x \equal{} \frac {1 \minus{} \tan^2 2x}{2 \tan 2x} \equal{} \cot 4x \implies$
$ \tan 3x \tan 4x \equal{} 1 \implies$
$ \cot 7x \equal{} \frac {1 \minus{} \tan 3x \tan 4x}{\tan 3x \plus{} \tan 4x} \equal{} 0 \implies$
$ \cos 7x \equal{} 0$. [/hide]

Now that's a very good solution, thanks a lot t0rajir0u

Is there something wrong with my solution?  :maybe:

No, but expanding everything is quite tedious and usually the wrong approach on a competition problem (although it's certainly the most general approach, [i]although[/i] if you want general approaches to determining whether a polynomial has nice roots you'll need to be prepared to deal with all the machinery of Galois theory and that's just a mess).

Yeah, agreed; but sometimes there's so much time pressure that I can't even think... (I'm not used to quick problem solving)  :D


I am dressing up as a DE. I compiled my whole costume. How about you guys?

I'm going as myself. That's plenty costume enough!!!

Why do we have to dress up????

i don't know i always thought that! this is my 3rd year of dressing up and I am a 3rd year...

Eh I prolly won't go trick-or-treating but I will hand out candy to the little kiddies and try to scare the crap outta them

me and my friends are goin as the fanta girls!

I'm too tall to go trick or treating now. At 5-11, people are like, "Candy is for the kiddos you big brute!" and they act like I was the Grinch who stole Christmas or something :(

So, I am handing out candy or going to a party. I'm going as a member of the Blue Man Group, so I'm pretty much putting on a bald wig and painting all exposed skin blue :rotfl:

Same. I am 5'10 even though I just turned 14. I am going to go trick or treating lolz as Johnny Depp of Pirates of the Carribbean! wow how original! (not really) at a maze place in our city where lots of people not just kids go.

Have you gone trick or treating at such a large height? You'll get frowned upon a lot, unless they are nicer in Alabama than Wisconsin. I was wearing a hood that year, though, so they might have thought that I was a rowdy college student. (I live close a University famous for the amount of partying).

I'm wearing my AOPS shirt. :rotfl:

Practicing for cross-country by running around the neighborhood 10 times getting candy.

I probably won't dress up. But if I do, it'll be something mathy. The Math Club president here dressed up as a functor last year.  :o

since im too lazy to go trick or treating im just gonna stay home, and steal some of the candies that we're supposed to give out

i'm going to stay at home, 'cause everybody thinks that 17 year-olds are too old to do that.

but my class is going to school dresssed up in... i can't remember. i suggested two years ago that we should be "ordinary people who took the wrong turn and ended in school" but they chose something else.

Halloween is considered "fake" and an example of "forced Americanism" here.  There are a few kids here who trick or treat and they find that most people (including us) are either not prepared (no candy) or simply unaware that it's Halloween.

I would like to point out that for catholics, All Saints Day is on the first of November, and All Souls Day is the next day.  On the first day one should commemorate the saints and martyrs, on the second day one should remember (or even visit the grave of) close ones who have passed away.
While I myself am an atheist, I find the contrast between this serious, sad ritual and the activities across the ocean ...remarkable.

Here, it's just an excuse for free candy. It isn't religious at all anymore

I'm gonna be a random Chinese person with glasses.

I'm going around and getting candy and then selling it on EBAY. lol !!!!!

I'm going to put a massive bucket of candy on my front yard for kids to take candy from, and then I'm going to dress up as a mummy and hide in the bucket

and when the kids reach in, I go

ARGHHHH!!!!

I am going to put candy on the doorstep before I go to church and write, if you take one, you will have bad luck until you eat 100 slugs

Since I'm giving out candy, I'm going to try and trade extra of my candy for my favorite kinds that the little kids have :D .

haha everyone i but a nerd hat and wore it at our halloween swim meet

I'm dressing up as a school bus!!! :lol:  :rotfl:

hahaha
we should all dress up as unknowns and equations and join algebra
so that the friendly police who always give us glo sticks instead arrest us for being part of algebra

thats a brilliant idea!  :lol:  :rotfl:

I wonder which i shall be

They had this scare that went like this, it seriously scared the crap out of lots and lots of ppl:

Have one person sit motionless on the doorstep with a basket of candy in front of him.  Position another one in a hiding spot.  The trickortreater will nervously take a thing of candy.  Have the motionless person scream.  Trickortreating goes away shaken, thinking all is done.  Have the person in hiding run at the trickortreater screaming loudly.  THAT will scare them until they wet their pants.

I don't know about you, but I don't like people wetting my lawn with stuff that's not water

they will probably just run their heads off, not wet their pants , and perhaps birds poo too? lolz

Last year, I was an old man....... with a cape! :rotfl:

hahaha funcia, i'll go running around in a speedo in 50 degree weather

(swimmers =P)

50 F isn't bad - in Wyoming it always snows heaps, so you have to make a costume that fits over a snow suit.  We're in Australia over this Halloween though, and they don't do it here either, so nothing for me.  I would have been a skier at home though (not a big change from normal, just grab my poles and hat out of the garage).


[quote="B4k4ka"]So i had this really weird dream once... and i've decided to convert it to a math problem XP

An extremely fat dog weighs X pounds.
He excersizes for 30 minutes, and loses 5 pounds.
After the 30 minutes, his limbs fall off, making him lose 2/3 of his remaining weight.

a giant cricket nearby weighs 4 ounces.

Right before the dog lost his limbs (after the 30 minutes of excersize), his weight was about 160 times as much as the cricket.

How long would the dog need to excersize to be the same weight as the cricket. (Excersizing starting from X pounds)
Round your answer to the nearest second.[/quote]

[hide]According to your dream... the dog weighed $160*4=640$ ounces before he lost his limbs. Or 40 pounds. *shudder* That's a fat dog.

Then his limbs fall off... reducing him to $\frac{40}{3}=13\frac{1}{3}$ pounds.

Every minute he exercises, he loses $\frac{1}{12}$ of a pound. Every second he exercises, he loses $\frac{1}{720}$ of a pound. Every minute, he loses $\frac{4}{3}=1\frac{1}{3}$ of an ounce.

So, solving for X.. I found that X is equal to 45 pounds.

Saying he exercises for 1 hour, the dog will lose 10 pounds reducing him to $3\frac{1}{3}$ pounds, or $\frac{10}{3}$ pounds or $\frac{160}{3}=53\frac{1}{3}$ ounces.

$\frac{\frac{160}{3}}{\frac{4}{3}}=\frac{160}{3}*\frac{3}{4}=40$ So he will need to exercise for 40 more minutes.

Adding it all together, the dog exercised a total of $\boxed{130}$ minutes.[/hide]

Since I did that much arithmetic in my head... i think I probably did something wrong. If there is something wrong, can someone tell me please?

[hide]Okay, the dog weighs 40 pounds, and loses 5, to get 35.  Then he loses 2/3 of the remaining weight to get 35/3.  every minute he exercises, he loses 1/6 of a pound.  $\frac{\frac{35}{3}}{\frac{1}{6}}$ :D =70 :idea:  more minutes he will have to excersise.[/hide]

[hide]oops, i subtracted 5 pounds when i wasn't supposed to. :blush:
ok, 40 divided by 3=$\frac{40}{3}$ and he loses 1/6 of a pound every minute.  Divide$\frac{\frac{40}{3}}{\frac{1}{6}}$=80 :idea: our answer.[/hide]

[color=darkred][size=75][Edited by 236factorial to fix latex error][/size][/color]

[quote="buzzer11"][hide]80[/hide] :rotfl:[/quote]

Read the final post of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23214]this[/url] and follow directions.

[quote="4everwise"]Please refrain from using acronyms such as "omfg" in the middle school section.

If you feel that you need to post them in the College Playground, feel free to.  ;)[/quote]

Fixed  :D


induction
quadratics
number theory
algebra proposed
algebra

The sequence $ (a_n)_{n\ge 1}$ is defined by $ a_1 \equal{} 3$, $ a_2 \equal{} 11$ and $ a_n \equal{} 4a_{n\minus{}1}\minus{}a_{n\minus{}2}$, for $ n \ge 3$. Prove that each term of this sequence is of the form $ a^2 \plus{} 2b^2$ for some natural numbers $ a$ and $ b$.

I think it's more number theory than algebra . Clearly it suffices to prove that each of the terms has prime divisors of the form $ 8k\plus{}1$ or $ 8k\plus{}3$ . Then we can easily find by induction that $ x_{n\minus{}1}x_{n\plus{}1}\equal{}x_n^2\plus{}2$ so each prime divisor of
$ x_{n\minus{}1},\ x_{n\plus{}1}$ divides $ x_n^2\plus{}2$ so $ \minus{}2$ is a quadratic residue so this numbers are of the form $ 8k\plus{}1$ or $ 8k\plus{}3$
and we are done . :)

Why is from here ob the problem trivial? Because i dont see why every number of the form 8k+1 or 8k+3 is ekspresible as a^2+2b^2

Off topic: there is no country in the world that's called Yugoslavia for many years, so I doubt there is a National Olympiad of the same as well. I should know! Please change your subtitle...

Colud someone explain why each number with prime divisors of the form$ 8k\plus{}1$ and $ 8k\plus{}3$(only) can be expressed in the form $ a^2\plus{}2b^2$?

Is is possible to see all the steps in crazyfehmy's proof by induction? It doesn't seem straightforward to me.

Try first for 3 and 4, then assume it's true for 2n-2 and 2n-1, prove that it's true for 2n and 2n+1.

That's what I tried and it didn't seem straightforward which is why I asked to see the proof in full.

I'm still working on a proof by mathematical induction but haven't cracked it yet. Any ideas?

Ok, let show this proof;

First we will prove that  $ 2(1 \plus{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n) \equal{} a_{n \plus{} 1} \minus{} a_n$  by induction.

For $ n \equal{} 1$ , $ 2(1 \plus{} 3) \equal{} 11 \minus{} 3$ true. Let it's true for $ n \equal{} k$. Because of $ a_{k \plus{} 2} \equal{} 4a_{k \plus{} 1} \minus{} a_k \Longrightarrow$

$ 2(a_1 \plus{} a_2 \plus{} ... \plus{} a_k \plus{} a_{k \plus{} 1}) \equal{} a_{k \plus{} 1} \minus{} a_k \plus{} 2a_{k \plus{} 1} \equal{} a_{k \plus{} 2} \minus{} a_{k \plus{} 1}$.

Now we will prove that $ a_{2n} \equal{} a_n^2 \plus{} 2.(1 \plus{} a_1 \plus{} a_2 \plus{} ... \plus{} a_{n \minus{} 1})^2$ and $ a_{2n \plus{} 1} \equal{} a_n^2 \plus{} 2.(1 \plus{} a_1 \plus{} a_2 \plus{} ... \plus{} a_n)^2$ by induction.

For $ n \equal{} 1 , 11 \equal{} 3^2 \plus{} 2(1)^2$ and $ a_3 \equal{} 41 \equal{} 3^2 \plus{} 2(1 \plus{} 3)^2$ true.Let it's true for $ n \equal{} k \Longrightarrow$

$ a_{2k \plus{} 2} \equal{} 4a_{2k \plus{} 1} \minus{} a_{2k} \equal{} 3{a_k}^2 \plus{} 8(1 \plus{} a_1 \plus{} a_2 \plus{} ... \plus{} a_k)^2 \minus{} 2(1 \plus{} a_1 \plus{} a_2 \plus{} ... \plus{} a_{k \minus{} 1})^2 \equal{}$

$ [3a_k \plus{} 2(1 \plus{} a_1 \plus{} a_2 \plus{} ... \plus{} a_{k \minus{} 1})]^2 \plus{} 2(a_1 \plus{} a_2 \plus{} ... \plus{} a_k)^2 \equal{} {a_{k \plus{} 1}}^2 \plus{} 2(a_1 \plus{} a_2 \plus{} ... \plus{} a_k)^2$ and by same way;

$ a_{2k \plus{} 3} \equal{} {a_{k \plus{} 1}}^2 \plus{} 2.(1 \plus{} a_1 \plus{} a_2 \plus{} ... \plus{} a_{k \plus{} 1})^2$

[quote="AndrewTom"]I have tried to locate the source of this question but cannot find any reference to the Yugoslavia National Olympiad 2008 on the internet. Is there any such thing?[/quote]

As milin alrady pointed out, there's no such country in the world anymore. The problem is from [url=http://srb.imomath.com/zadaci/smo_2008_e.pdf]Serbian Mathematical Olympiad 2008[/url].

Wonderful crazyfehmy. Would you mind posting the steps for the case a(2k+3)?

Thanks. I've just done it but yours is shorter.

Having proved that a(n-2)a(n) = (a(n-1))^2 + 2, I am now looking at silouan's post and would like to see much more detail as I am presuming that this means -2 is a quadratic residue of a(n) but don't know why this means that the prime factors of a(n) are 1 or 3 (mod 8), although I've noticed that the first few terms, 3, 11, 41, 153, 571, 2131, 7953 and 29681 are, nor do I know why this means that  a(n) is of the form a^2 + 2b^2.

A generalization one .  :) . Let $ a_n$ be a sequence  in a [b]algebraically closed field[/b] defined by $ \ a_0 \equal{} 1, \ a_1 \equal{} B \minus{} 1,\ a_{n \plus{} 2} \equal{} Ba_{n \plus{} 1} \minus{} a_n$ .
Prove that : $ a_{2n} \equal{} a_n^2 \plus{} (B \minus{} 2)(a_0 \plus{} ... \plus{} a_{n \minus{} 1})^2,\ a_{2n \plus{} 1} \equal{} a_n^2 \plus{} (B \minus{} 2)(a_0 \plus{} ... \plus{} a_n)^2$
[hide]That's trivial if $ B \equal{} 2$ . If $ B \equal{} \minus{} 2$ it's easy to see that $ a_m \equal{} (2m \plus{} 1)( \minus{} 1)^m, \ a_0 \plus{} ... \plus{} a_m \equal{} (m \plus{} 1)( \minus{} 1)^m$ ,
and the relations ships are verified because of $ 4n \plus{} 1 \equal{} (2n \plus{} 1)^2 \minus{} 4n^2 , \ \minus{} (4n \plus{} 3) \equal{} (2n \plus{} 1)^2 \minus{} 4(n \plus{} 1)^2$
[b]Let's suppose now that $ B\neq \minus{} 2,\ B\neq 2$ [/b]
The solution of $ a_{n \plus{} 2} \equal{} Ba_{n \plus{} 1} \minus{} a_n$ is $ a_n \equal{} C_1x_1^n \plus{} C_2x_2^n$ , where  $ x_1\neq x_2$ are the roots of $ x^2 \minus{} Bx \plus{} 1 \equal{} 0$ ,
and the constants $ C_1,C_2$ are determined from $ C_1 \plus{} C_2 \equal{} a_0 \equal{} 1,\ C_1x_1 \plus{} C_2x_2 \equal{} a_1 \equal{} B \minus{} 1$.
Therefore  we have to verify the relations ships in following :
$ C_1x_1^{2n} \plus{} C_2x_2^{2n} \equal{} (C_1x_1^n \plus{} C_2x_2^n)^2 \plus{} (B \minus{} 2)(a_0 \minus{} C_1 \minus{} C_2 \plus{} C_1\frac {x_1^n \minus{} 1}{x_1 \minus{} 1} \plus{} C_2\frac {x_2^n \minus{} 1}{x_2 \minus{} 1})^2$,
$ C_1x_1^{2n \plus{} 1} \plus{} C_2x_2^{2n \plus{} 1} \equal{} (C_1x_1^n \plus{} C_2x_2^n)^2 \plus{} (B \minus{} 2)(a_0 \minus{} C_1 \minus{} C_2 \plus{} C_1\frac {x_1^{n \plus{} 1} \minus{} 1}{x_1 \minus{} 1} \plus{}$$ C_2\frac {x_2^{n \plus{} 1} \minus{} 1}{x_2 \minus{} 1})^2$
But $ \ C_1 \plus{} C_2 \equal{} a_0 , \ \frac {C_1}{x_1 \minus{} 1} \plus{} \frac {C_2}{x_2 \minus{} 1} \equal{} 0,\ (x_1 \minus{} 1)(x_2 \minus{} 1) \equal{} \minus{} B \plus{} 2,\ x_1x_2 \equal{} 1$, so  we have to verify  easier relations ships :
$ C_k \equal{} (C_k)^2 \plus{} (B \minus{} 2)(\frac {C_k}{x_k \minus{} 1})^2,\ C_kx_k \equal{} (C_k)^2 \plus{} (B \minus{} 2)(\frac {C_kx_k}{x_k \minus{} 1})^2$ .. and so on ... [/hide]
:cool:

I calculated some values and got the results
$a_1=(1,1)$
$a_2=(3,1)$
$a_3=(3,4)$
$a_4=(11,4)$
$a_5=(11,15)$
$a_6=(41,15)$
here $(x,y)$ denotes $x^2+2y^2$.
From these I guessed that the sequence works in the following way:
if $a_n=a^2+2b^2$ then $a_{n+1}$ is either $a^2+2(a+b)^2$ or $(a+2b)^2+2b^2$ and these two happen alternately.
This fact is easy to prove by induction. :lol:

Lemma :A number $K$ can be expressed $a^2+2b^2$ if and only if for primes $p \equiv 1,3 \pmod8$$v_p(K) \equiv 0 \pmod 2$
Proof: if part is trivial .For the only if part we should prove for a prime $p \equiv 1,3 \pmod8$ there exist positive integers $a ,b$ such that $p=a^2+2b^2$ (indeed these $a$ and $b$ are unique) .Becasue of $p \equiv 1,3 \pmod8$  there is a integer  $x^2  \equiv -2 \pmod p$ By thue's lemma there exist integers $a,b$ such that $0<a^2,b^2<p$ and $a \equiv xb \pmod p $ so $p|a^2+2b^2$ and $0<a^2+2b^2<3p$ then $p=a^2+2b^2$ or $2p=a^2+2b^2$ for first case we are done for second case $a$ is even $a=2a'$ then we are done again .$2=0^2+2.1^2$ for prime $p \equiv 1,3 \pmod8$ $p^2=p^2+2.0^2$
By Brahmagupta identity ,
$(x^2+2y^2)(a^2+2b^2)=(ax+by)^2+2(xb-ay)^2$ lemma is proven . $\square$

Beyond this theory problem is trivial .By induction
$a_n^2-4a_na_{n+1}+a_{n+1}^2=-2$ if $p|a_n$ then $p|a_{n+1}^2+2$ then $p \not\equiv 5 ,7 \pmod8$. $\blacksquare$


Prove that there is no integer between zero and one.

:rotfl:  :rotfl: is all I have to say.  I guess this might be considered spam, but...[hide="hint"]Prove by contradiction.  Let $x$ be the least integer such that $0<x<1$.[/hide]

[hide]
[b]Lemma 1[/b]. The number of integers between two integers, $a$ and $b$, where $b>a$ (exclusive), is $b-a-1$.
[i]Proof[/i]: The set is $S= \{ a, a+1, \ldots, b-1, b \}$. If we can make this set something like $\{1,2,3,\ldots,n\}$, then $n$ will be the answer inclusive. Subtract $a-1$ from all numbers. Then $S= \{ 1, 2, \ldots, b-a, b-a+1 \}$. Thus there $b-a+1$ inclusive, so $b-a-1$ exclusive.

The number of integers between $0$ and $1$ is $1-0-1=\boxed {0} \ \ \ \ \ \ \blacksquare$ Q.E.D.
[/hide]

 :rotfl:  :rotfl:

How do you know there's no integers between $a$ and $a+1$? How do you know that the successor of an integer $x$ is $x+1$? That's what we are trying to prove.

It's been a while since SWT...I don't remember what we can and can't use...
Let $x$ be the smallest integer greater than zero but less than one.
Then $1-x>x>0$.
We also have $1*x=x$.
Subtracting $x^2$ from both sides, we have $1*x-x^2=x-x^2$.
Thus $(1-x)x=x-x^2$.
Since $(1-x), x>0$, then $x-x^2>0$.
Finally $x>x^2$ which contradicts $x$ being the smallest positive integer since $x^2$ is an integer.

i dunno but it seems to me like its some sort of class or competion. :?

Maybe this will clear it up : http://www.nec.ac.uk/courses/category-browse?usca_p=t&category_id=802
[quote]South West Trains is a company that supports employee learning, and has recently set up a learning centre in Waterloo Station with courses being offered free of charge to employees.[/quote]
 :whistling:

Er actually SWT stands for Southwest Texas... SWT Honors Math Camp...

:rotfl: ok, thanks tarquin
I was way far from the actual meaning  :P

[quote="towersfreak2006"]
Finally $x^2>x$ which contradicts $x$ being the smallest positive integer since $x^2$ is an integer by contradiction.[/quote]

This is the key to the proof that I want to ask about.

Is this an axiom for integers? That $x^2 > x$?

Why can this be assumed?

And um, the claim doesn't make sense to me either. If $x^2 > x$, how does that (I'm assuming this is like a proof by WOP) $x$ is the smallest? You just found a larger one, not a smaller one by your claim.

Ah, great. That's a typo. It should be $x^2<x$, so we have found a smaller integer than $x$. It's fixed now. Sorry about the confusion.

K, thanks!  :D

Just wanted to make sure - I've never learned rigorous number theory before.


modular arithmetic
algebra unsolved
algebra

let $ a$ , $ b$ , $ c$, $ d$, $ m$ and $ n$ are naturals integers # 0.such that
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1989$ and $ a \plus{} b \plus{} c \plus{} d \equal{} m^2$ and $ n^2 \equal{} max (a,b,c,d)$
find $ m$ and $ n$.


AMC USA(J)MO USAMO AIME ARML geometry inequalities

Qualifiers are out! :-D :-D :-D

http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2004-ua/04usamoqualstate.html

[quote="Caroline"]Only around 20 ninth graders.. does that mean they'll choose an additional 10 for MOP, or they'll take the eighth graders?[/quote]

25 9th graders :-D *cheers for self*

Congrats for everyone else who made it!

I think they'll just take the 25...at least that's what the handbook implies...but who knows...

Personally, I hope they'll fill in the other 5 slots with other freshmen who did well on AMC/AIME :-)

[quote="MithsApprentice"]
25 9th graders :-D *cheers for self*

[/quote]

Ah yes, if I were able to count correctly perhaps I would have done better...

Congratulations =)

Congo Rats!

Wow. I wish I were a freshman right now...but if I'm as good as I profess to be :-p, I should do well on the USAMO right?

Again, congrats to everyone, and good luck!

This is weird. Why am I on the USAMO qualifier list?  :shock:  I mean, I do live in Canada. Did the USAMO people start to invite Canadians now or something? Does that mean that we can write it officially now, and be eligible for the prizes?

Well, not that it matters, since I'll be writing it in any case, under "Canadian juristication".

I thought the Canadians always could participate.  In fact, my senior year Jordan Ellenberg won 1st place at the USAMO and started his speech by thanking Ravi Vakil (a Canadian) for graduating the previous year.  Ravi, I think, had won the USAMO in 1988.

Yeah. Ravi is well known for that. But I believe that was the last year which Canadian could officially write the USAMO.
It would be weird though, having a Canadian winning the [b]USA[/b]MO.

Hey, congratulations to Wyoming for having a USAMO qualifier!  They often don't get anyone at AIME so they can't get anyone at USAMO.

[quote="probability1.01"]Does anyone happen to know what the cutoffs were?[/quote]

I believe it's a 7 for freshmen/sophomores...

Furthermore, the cutoff <i>index</i> must have been somewhere around 210. I got a 212.5 and qualified, but people that were slightly below my score did not.

for connecticut, hopkins had more qualifiers than choate! maybe it's time to reconsider my decision. :D  :D

[quote="yif man12"]Also, it says one person was in 13th grade. Is that just because he's from a canadian school?

Another thing. At the top of the page, where they put all the states, they don't have wyoming. you can tell i'm in a nitpicky mood[/quote]

Could be a PG (post-grad)

It seems like there were 15 8th-graders qualifying (I'm one! :D), and 25 9th-graders. Only one 7th-grader, apparently, and I don't even want to count the 10th graders.

You counted those?

[quote]Only one 7th-grader, apparently[/quote]
All the other years I've checked have 0 or 3. (mostly 0.)

[quote="Dnas"]You counted those?

[/quote]

Well, I used "Find" to find all the instances of "8" in the list, and I counted 15. Did the same with "9."

btw, congrats on making it and good luck.

I got a 7, so I just barely made it. But it's on the Statistics page at

http://www.unl.edu/amc

if you want to see.

[quote="probability1.01"]How did you 9th graders/8th graders/7th graders do on the AIME?[/quote]

Score: 8
Grade: 7

I got 11, but I think AIME is a little overrated.  Just a little... BTW, it seems that most of the 9th/8th/7th graders have skipped a grade or are young for their grade.  Wow... 8 is really good for a possibly young 7th grader.  Congrats!

Thanks!

[quote]seems that most of the 9th/8th/7th graders have skipped a grade or are young for their grade.[/quote]
Not my case. I'm too old if anything.

NC did very well, as it is apparently going to be fielding 4 freshmen (hrmph...upstarts) and 7 contestants overall. RDU alone has 4 entrants; this is not bad for a region not especially known for mathematics schools. I'm proud of my state :-D

Split to [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=11984]Age and Grade-Skipping in Math Contests[/url]

[quote="JP"]This is going to be the first time for me to take the USAMO.
What is the best way to prepare for it?
thanks[/quote]

There will be a MathJam regarding how to [i]take[/i] it on April 23.. I'm not sure if they're going into how to prepare.  It might be a good idea to attend, though.

[quote="JP"]This is going to be the first time for me to take the USAMO.
What is the best way to prepare for it?
thanks[/quote]

If you are new to proof-type problems in general, I doubt you are going to be able to get a lot done before the test...

Otherwise, work on past olympiads. A week or two is not enough time to make a big difference in your score, but at least you will have an idea of what to expect.

[quote="JP"]This is going to be the first time for me to take the USAMO.
What is the best way to prepare for it?
thanks[/quote]

Other than by going to the Math Jam, I would study geometry theorems, because they very well may help you in some angle chasing problem. Also, I would review inequalities, because they also come up a lot. Finally, I would try to find brute force ways to solve problems e.g. coordinates for geometry problems or LaGrange Multipliers for inequalities to get some cheap points. Since you have 4.5 hours you can probably actually finish a geometry problem with coordinates.

Yan Zhang

[quote="probability1.01"]How did you 9th graders/8th graders/7th graders do on the AIME?[/quote]

Score: 7
Grade: 9

By the way, Dnas, are you doing MathCounts as well? Did you qualify for nationals?

And while I'm at it, since I don't know where else to post this, does anyone know if Adam Hesterberg goes on this forum? Sorry for getting a little off topic...

[quote="picounts"]And while I'm at it, since I don't know where else to post this, does anyone know if Adam Hesterberg goes on this forum?[/quote]

The forum on which to ask about a Mathcounts participant is the [url=http://www.artofproblemsolving.com/Forum/viewforum.php?f=4]Mathcounts Forum[/url]. To answer your question, yes, he has posted here with his view of last year's Mathcounts championship, but I don't remember his screen name. (We usually try to keep it to screen names here, for online safety reasons, especially for the participants who are minors.)


what kind of spam do you get most irritated with?

[color=darkblue]{Please do not mention names, or indicate anyong who is spamming, as this can lead to flame wars and is completely irrelivent to G&FF. -7h3.D3m0n.117}[/color]

Am I the only person that finds flaming annoying?

irrelevent stuff like this thread(how ironic)

jk of course

DON'T read a calc book and do not at all costs visit my blog!!! :rotfl:

Definitely the calc book is the worst.

I hate it when someone says you're spamming, when they in actuality have spammed by saying that person is spamming.

[quote="ProtestanT"][quote="7h3.D3m0n.117"][quote="Tiaan"]What is "Flaming" spam?[/quote]

UR A NOOB FFS!!!! STUPID NOOB!!!

thats what flaming is

no offense =)[/quote]
he asked wut a flaming spam is, not what flaming is...but i think thats a good example.
i think a flaming spam is when someone is talking about one thing, then u flame at them for something completely irrelevant.[/quote]

Yeah people would yell "YOU'RE STUPID" for messing up some equations or something. It doesn't have to be irrelevant but it could insult any person(s) as a reply.

posting random spam just to increase post count when the post has no utter meaning is pretty annoying too

I know a person who still posts on AoPS who had about 200 posts in a week and is now on page 1 of the memberlist due to his extreme posting everyday.

[quote="now a ranger"]posting random spam just to increase post count when the post has no utter meaning is pretty annoying too

cough cough ------[/quote]

Well in G&FF posts don't count as your overall post count in case you haven't noticed  :D

[color=darkblue]{EDITTED BY MOD...Please do not mention names}[/color]

I also think that insults are pretty mind-boggling and have no point.

[quote="jli"][quote="Ignite168"][quote="CheeseIsGood"]Saying "read a calc book" is the most annoying spam.[/quote]

[color=darkred]QFT[/color][/quote]

what does that mean?

quit farting tom?[/quote]

Quoted for Truthery/Truthiness etc.

[quote="now a ranger"]posting random spam just to increase post count when the post has no utter meaning is pretty annoying too[/quote]

I was increasing my post count like I was mad, but I wasn't spamming. :wink: (I was in Navier stokes equations,  but then the Great Spam Massacre happened, and I was put in the rank of Poincare conjecture, and then I increased my posts like I was angry, by answering a bunch of Mathcounts problems. Since I've moved to imtermediate problems, my posts per day has really dropped.( :P ))

What if you say "I'M STUPID! I'M A COMPLETE NOOB!" as a minor note on some post that is constructive? Is that flaming spam?

[quote="1=2"]What if you say "I'M STUPID! I'M A COMPLETE NOOB!" as a minor note on some post that is constructive? Is that flaming spam?[/quote]

well here is the difference

"I'M STUPID! I'M A COMPLETE NOOB!" is considered spam as you are yelling and just randomly saying it

"sorry, i suck at math  :blush: " is not spam as it admits a mistake

[quote="7h3.D3m0n.117"]"sorry, i suck at math  :blush: " is not spam as it admits a mistake[/quote]

I'm going to have to disagree here. If every time you "take matters into your own hands", God kills a kitten, every time God hears that line, he kills a litter.


[quote="posabogdan"]In cate moduri putem alege doua submultimi disjuncte dintr-o n-multime?
Ps: daca se poate o alta solutie decat cea prezenta in Arthur Engel.[/quote]

e cumva 3^n moduri raspunsul corect?

[quote="turcas_c"][quote="Svejk"]La ce se refera denumirea de "n-multime" :huh:  ?[/quote]

Denumirea de $n$-multime se foloseste pentru o multime cu $n$ elemente.Raspunsul este $\frac{3^{n}+1}{2}-2^{n}$ moduri , dar eu am mai discutat problema la grupa' de excelenta...deci nu se pune' :)[/quote]

aa...atunci inseamna ca 3^n este numarul total in care se pot alege 2 submultimi nu neaparat disjuncte ale unei multimi :D

O n-multime reprezinta o multime cu n elemente...
Solutie prezenta in carte:pentru o pereche ordonata de submultimi disjuncte (A,B) definim functia caracteristica
f(x)=
1,daca x apartine lui A
       2,daca x apartine lui B
       o altfel
Functia f este un n-cuvant cu alfabetul {0,1,2}.Nr tuturor functiilor este $3^{n}$ .Exista $2^{n}$ cuvinte formate din multimea {0,2} si $2^{n}$cuvinte formate cu multime {0,1} si un sg cuvant numai din zerouri.De aici nr de perechi ordonate este $3^{n}$-$2^{n}$-$2^{n}$+$1$ deci nr de prechi neordonate este $(3^{n}+1)/2$ -$2^{n}$
O alta solutie daca mai exista...

[quote="mateivld"]Fixam k elemente ale multimii ale n-multimii.Avem $\binom{n}{k}$ de  astfel de submultimi.Cealalta multime poate fi vazuta ca o submultime nevida a unei multimi cu $n-k$ elemente.Asadar numarul cautat $\sum_{k=1}^{n-1}\binom{n}{k}(2^{n-k}-1)$ care se calculeaza usor.[/quote]
frumoasa solutie


[b] PenguinIntegral's Effort at being a unique and friendly poster Post #1[/b]

1a) Prove $2^n-1$ is composite for nonprime n.
1b) Prove $2^n+1$ is composite for n with an odd divisor
1c) Prove $2^{37}-1$ is composite.
1d) Prove if prime $p=4k-1$ and $2p+1$ is prime, then $2p+1$ divides $2^p-1$.

[quote="xxreddevilzxx"][quote="PenguinIntegral"][b] PenguinIntegral's Effort at being a unique and friendly poster Post #1[/b]

1b) Prove $2^n+1$ is composite for n with an odd divisor[/quote]

I don't quite get this one. Does odd divisor mean a number that has an odd divisor or a odd number of divisors?[/quote]

An odd divisor.  You answered your own question.  :D

[quote]I don't quite get this one. Does odd divisor mean a number that has an odd divisor or a odd number of divisors?[/quote]
This means that n has atleast one divisor (not counting 1) which is odd.
36 has odd divisor  (Eg 3) ,  32 does not.
[hide](A number which has no odd divisor can be written as $2^k$ ) (And for such values of n, $2^n+1$ is known as Fermat's numbers.[/hide]

[quote="Gyan"][quote="PenguinIntegral"]
1d) Prove if prime $p=4k-1$ and $2p+1$ is prime, then $2p+1$ divides $2^p-1$.[/quote]
p need not be prime, as long as $2p+1$ is prime, and $p$ is of the form $4k-1$ , $2p+1$ will divide $2^p-1$ :)[/quote]
Proof?

[hide="1b)"]If n is odd, $2^{odd}+1\equiv (-1)^{odd}+1 \equiv 0 \pmod{3}$.
If n is even, yet has an odd divisor, $2^n+1=2^{even}2^{odd}+1\equiv (-1)^{even}(-1)^{odd}+1\equiv 0\pmod{3}$. It is always divisible by 3, and is thus composite.[/hide]

[quote="pianoforte"]If n is even, yet has an odd divisor, $2^n+1=2^{even}2^{odd}+1\equiv (-1)^{even}(-1)^{odd}+1\equiv 0\pmod{3}$. It is always divisible by 3 [/quote]
Not quite true, eg n=6, $2^6+1$ is not divisible by 3.
$(2^{(even)(odd)}$  is NOT the same thing is $2^{even}2^{odd}$    whichis BTW would be $2^{even}2^{odd}=2^{even + odd}$

[quote="PenguinIntegral"]
Proof?[/quote]
[hide="Hint"]Here the prime number is     -1 mod 8   [hide="because"](2(4k-1)+1 = 8k-1) [/hide]
for any odd prime $p$, $2^{\frac{p-1} 2} is +1$ (when $p \equiv \pm 1 \bmod 8$)
[hide]     and  -1 (when $p \equiv \pm 3 \bmod 8$) [/hide]
[hide] (Any NT book would have proof   (when 2 is quadratic residue)   [/hide][/hide]


modular arithmetic
number theory
Congruences

Somebody incorrectly remembered Fermat's little theorem as saying that the congruence $a^{n+1} \equiv a \; \pmod{n}$ holds for all $a$ if $n$ is prime. Describe the set of integers $n$ for which this  property is in fact true.

Lemma: It's the set of those squarefree $n$ with the following property: if the prime $p$ divides $n$, then so does $p-1$.

Proof: If $p$ is some prime, $p^2 \nmid p^{n+1}-p$, thus $p^2 \nmid n$.
If $p|n$, then we take a primitive root $a \mod p$ and get that $a^{n} \equiv 1 \mod p$, thus $(p-1)|n$ by the order-lemma.
Conversely, if $n$ fulfills that properties and $p|n=(p-1)q$, then $a^{n+1} = a^{(p-1)q+1} \equiv a \mod p$ (even if $p|a$).

Claim: all such numbers $n$ are $1,2,6,42,1806$.

Proof: Testing these numbers to work using the lemma is easy, so we just show there are no others.
Assume there are more than we claim to be and let $n=p_1 p_2 ... p_k$ with $p_1 < p_2 < ... < p_k$ be the smallest exception. Then the number $n^\prime=p_1 p_2 ... p_{k-1}$ still has the property by the lemma (all requirements are fulfilled again) and is in $M$.
Additionally, $p_k-1|n^\prime$, leaving us to check that all numbers $d+1 $ with $d|1806=2\cdot 3 \cdot 7 \cdot 43$ and $d \not\in M$ are not prime.
This is left to the reader  :cool:


Let P be a point inside an acute triangle ABC. $ PA,PB,PC$ meet the circumcircles of the triangles $ PBC,PCA,PAB$ at $ A_1,B_1,C_1$ respectively. Prove that $ (1 \plus{} 2\frac {PA}{PA_1})(1 \plus{} 2\frac {PB}{PB_1})(1 \plus{} 2\frac {PC}{PC_1})\geq 8$

see [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=198471]here[/url]  :wink:


inequalities
geometry
trigonometry
circumcircle
inequalities proposed

In a triangle $ABC$, points $A_{1}$ and $A_{2}$ are chosen in the prolongations beyond $A$ of segments $AB$ and $AC$, such that $AA_{1}=AA_{2}=BC$. Define analogously points $B_{1}$, $B_{2}$, $C_{1}$, $C_{2}$. If $[ABC]$ denotes the area of triangle $ABC$, show that $[A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}] \geq 13 [ABC]$.

I think the const must be $10$ not $13$

\[[A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}]=[AB_{1}C_{2}]+[BC_{1}A_{2}]+[CA_{1}B_{2}]+[AA_{1}A_{2}]-2[ABC]\ge 10 [ABC] \]

\[\Leftrightarrow \frac{1}{2}(\sum (a+b)(a+c)\sin A+\sum a^{2}\sin A) \ge 12[ABC] \]
(Apply law of sin and $S=2R^{2}\prod\sin A$)
\[\Leftrightarrow \sum (\sin A+\sin B)(\sin A+\sin C)\sin A+\sum\sin^{3}A \ge 12\prod\sin A \]

\[\Leftrightarrow \sum\sin A(\sin A(\sum\sin A)+\sin B\sin C)+\sum\sin^{3}A\ge 12\prod\sin A \]

\[\Leftrightarrow \sum\sin^{3}A+(\sum\sin A)(\sum\sin^{2}A)\ge 12\prod\sin A \]
which is trivial by AM-GM.

If I am not mistaken, $ S \equal{}\left[A_{1}A_{2}B_{1}B_{2}C_{1}C_{2}\right] \equal{}\left[A_{1}BC_{2}\right]\plus{}\left[B_{1}CA_{2}\right]\plus{}\left[C_{1}AB_{2}\right]\plus{}\left[AA_{1}A_{2}\right]\plus{}\left[BB_{1}B_{2}\right]\plus{}\left[CC_{1}C_{2}\right]\minus{}2\left[ABC\right]$.

We need to prove that $ 30\left[ABC\right]\leq2(S\plus{}2\left[ABC\right]) \equal{}\sum_{cyclic}(a\plus{}b)^{2}\sin C\plus{}\sum_{cyclic}a^{2}\sin A \equal{} 12\left[ABC\right]\plus{}(a^{2}\plus{}b^{2}\plus{}c^{2})(\sin A\plus{}\sin B\plus{}\sin C)$.

Calling $ R$ the circumradius of $ \Delta ABC$, all we need to prove is that $ \frac{9abc}{2R}\equal{} 18\left[ABC\right]\leq\frac{(a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2})}{2R}$, which is obviously true because of the AMGM inequality applied to $ a,b,c$ and to $ a^{2},b^{2},c^{2}$, with equalitty iff the triangle is equilateral.


probability
function
pigeonhole principle
AoPS Books
graph theory
AMC

Which book goes into more depth, [url=http://www.amazon.com/exec/obidos/ASIN/0883856158/artofproblems-20]Mathematics of Choice[/url] or Intermediate counting and probability? or which book is written better?

oh, hehe i have mathematics of choice(i won it at a contest i pha1led :rotfl: )

erm, i haven't read it much but the little bit i skimmed over seemed like standard basic combinatorics though.

i suggest intermediate counting and probability. although i have not read it, the aops books are always extremely well written.

Well, not much information about the former, but I can tell you that Intro. to C&P will help you if you are getting started, and the Intermediate will definately help you if you have some background (if you've done Intro.)

i have never heard of "mathematics of choice" so yeah idk

That's not very helpful, cognos. :P

I'm not very qualified to answer this question, as I haven't done much of Intermediate Counting and Probability, and don't have Mathematics of Choice. However, by looking at the table of contents and previews on Amazon, it appears that over half the content in Mathematics of Choice is covered in Intro to Counting and Probability, while some of the topics, such as Generating Functions, are left to Intermediate Counting and Probability.

Intermediate Counting and Probability covers more material, but doesn't start with the basics. It covers many things that aren't directly related to combinatorics, in my opinion, such as graph theory and pidgeonhole, which I don't believe  Mathematics of Choice covers.

So, I'd say that Mathematics of Choice is in between the level of Intro to C+P and Inter. C+P. For a beginner, Mathematics of Choice would be better, but if you've already had some exposure to combinatorics, Intermediate would be best. I can't answer the question about which book is better written.

P.S. You can see the table of contents of Intermediate C+P in the bookstore, as well as read an excerpt. Perhaps that may aid you in making your decision.


I have a question about the [url=http://awesomemath.org/wp-content/uploads/2010/01/Personal_Information_Form_electronic2010.pdf]personal info form.[/url]

Where it says "Grade level which student will complete in spring 2009: ", does that mean "Grade level which student will complete in spring 2010:", or just the original?

Yeah, they just forgot to change the form. I guess you could just cross it out, put 2010, and put the grade you are currently in.  :)


algebra
polynomial
induction
real analysis
real analysis theorems

deduce Faulhaber's formula from the folowing serie 1^k+2^k+3^k+4^k+...+n^k

Series? Do you mean a sum? Can we assume that $k\in\mathbb{Z}_{+}$? In that case, the sum is always a polynomial in $n$ of degree $k+1$, so you can find the coefficients from the first $k+2$ values, either directly by Gaussian elimination, by Lagrange's interpolation formula, or by induction argument.


I always don't know how to do problems like these. A coin is flipped three times. Whats the probability of getting heads three times? Yeah, i have problems with these probability problems.

[hide="Hint"]
Each coin is an independent event. What is the probability of getting heads on one coin?
[/hide]

[hide="solution"]
[hide="Are you sure?"]
The probability of getting heads on 1 flip is 1/2.

Therefore, the probability of heads on all three is 1/2 * 1/2 * 1/2 = 1/8.

You just multiply the probabilities of the independent events.
[/hide]
[/hide]


Let $f(n)$ be a sum of all positive divisors of $n$.Prove that if $f(n)=2n+1$ then $n=a^2$ where $a$ is number odd.


Find all $n\in N$ s.t. $n^2+(n+1)^2$ is perfect square.

partial solution

$n^2 + (n + 1)^2 = k^2$
$2n^2 + 2n + 1 - k^2 = 0$
Discriminant is $2k^2 - 1$
A square is congruo $1$ or $0$ mod 4.
So it is $2k^2 \equiv 2$ and the solutions are for $k \equiv 1$

Well that partial solution wouldn't help so much :D :D

There is one trivial solution for $n=3$ and I think there's no more....

Try something via $Pithagors.triplets$

Can you please post hole solution because I don't know Pell equations

[i]rgep[/i] has already posted the whole solution :P This is the case of $X^2 - DY^2 = -1$.
See [url=http://mathworld.wolfram.com/PellEquation.html]this[/url] to find out more about Pell's equations.


I would like to know your opinions about the first set of problems ...

hey the problems were really challenging .i got two of them but cud not post the solutions as the computer at home wasnt working.and  just a question ,cud u please post the solution of Q3 as submitted by any contestant who got it right.i wud liek to see the solution.and it might not be able to take part in rd 2 and rd 3 as well as xams approaching.
keep up the good work
regards (hey,justa curiosity,how many did u score at imo 2002)

unfortunately just a small 19 ( 0 6 1 6 6 0 ) bad day(s) i guess - or a sever lack of cleverness :D

try to send the solutions from the place you are posting on the forum

at least 2 solutions at each of the problem are on-line at the archive section, go have a look


Hi,

I plan on self studying AP Statistics this year (soph) and I was wondering if anyone who's done this before, or someone who's taken Stats in school can recommend a textbook for me. A more challenging and overall better textbook would be nice. I just want to get an easy 5 on the test. A link would help, thanks!

A very good review of statistics books can be found at

http://www.statland.org/MAAFIXED.PDF

Use the books (in newer editions, as appropriate) that this reviewer recommends.

for the prep book, use Barron's. For stats, no other book comes close.

You might want to get your hands on a copy of the 4th printing (in 2000) of "The Practice of Statistics" by Daniel Yates, David Moore, and George McCabe. It is written specifically for the AP exam. The 15th (final) chapter of the book is even called "Post Exam topics" with "exam" meaning the AP exam. Because it is written for the AP exam, it was written based on the assumption the students have a TI-83 graphing calculator so it shows you how to do the problems by hand first and then it shows (very clearly i might add) the proper syntax to do the problem on your TI-83 graphing calculator. I am currrently in AP stats right now and my school uses this book and i know ALOT of other schools use this book for their AP stats classes as well.

As for prep books, go for Barron's or Princeton review. I haven't seen the Barron's book but other people say it's really good. This summer i started reading up on AP stats just so i'd have a head start in the school year and i used the Princeton review book. I like the Princenton Review book because, like my AP stats textbook, it shows you how to do problems by hand and with the TI-83 graphing calculator. It's also a quick and easy read.

Hope this info helps you and good luck in earning a 5 on the test!  :lol:

thanks

one more question....is stats hard?

my math teacher is telling me how i should take a course or something but our school is pretty small and self studying is uncommon.

I mean, it can't be much harder than understanding a concept and applying that concept right? There aren't any real problem solving skills like in the AOPS books?

It's not a subject that requires intense problem-solving skills. It does, however, require you to know a decent amount of material and some of the topics are difficult to grasp without solid teaching. I definitely underestimated stats; if you're going to self-study, make sure to start early (as opposed to 2 weeks before the test like me :fool:). I was VERY lucky to pull off a 5, I don't know how I did it at all.


I have found lately that when I click on the number 2 (or any other, for that matter) in the page list for long threads, it still takes me to the first page of the thread. Do I have a setting wrong? Thanks.

When you mouse over the "2", what URL do you see on the status bar?  When you actually click on the link, what URL is on the address bar?  There should be something like "&start=15" in the URL.

Which browser are you using?  Does the problem go away with other browsers?

Hey Ravi B those were the questions that I was supposed to ask :P:P ... now I'm out of questions  :blush:  :D

Hehe, that was my impersonation of an administrator.  I'll let you handle ComplexZeta's reply when it comes in. :)

The URL it brings up is the one it should bring up, but then it takes me to the wrong page. I'm using Firefox.

I have the latest one (the one that came out just a few days ago).

Let me get something straight here. When you say in the page's list of the threads, where is that actually? Is it in viewforum.php?f=14 for example? Or is it in viewtopic.php?t=13445 ?

If it's in viewforum.php that's probably because of the java script that makes a little hand when you cross the mouse over a topic's line (not only the topic link itself), so what happens is that you actually click very very near 2, but no on 2, on the topic line. Basically Firefox isn't that great at rendering java scripts yet, as far as I seen.

[quote="ComplexZeta"]I do mean viewforum. It doesn't seem to be a problem with Firefox 0.8, yet it doesn't work with 1.0[/quote]Well I am sorry then, but I feel that particular browser is the problem.

Try using IE 6.0, which at this point in time, if fully upgraded, is one of the best browsers around, and certainly better than Firefox.

[quote="ComplexZeta"]I got sick of getting infected with tons of spyware and viruses a while ago.[/quote]I use IE 6.0 for two years now, and I never got any spyware, trojans or viruses on my comp. I got the occasionaly e-mail worm, or some add-ware I myself downloaded from the net, but that's it.

Look at it this way: Firefox has just gotten out of beta, there are tons of stuff still wrong with it, but IE has been developed for some time now. In plus Firefox is freeware, so people do not barge in too much to help solve problems, as they do with IE (there are a lot of problems indeed, but Microsoft releases patches which solve the problem very quickly).

In plus most webpages are optimized for IE, and more, since Sp2, IE has also included pop-up blocker, ActiveX blocker, and all that stuff that helps you protect your computer.

Thinking that Firefox is invincible is just a clishee launched on the net to make the browser popular. If you really want something secure you should go for paid browsers like Opera or Mozzila.

I am using the new release of Firefox and am NOT having the problem described by ComplexZeta. I just tried a Fun and Games thread with eight pages and got the page I specified each time.

Uhh... just curious, but does it work if you change the viewing format?  If you usually view the site in AoPS style and change it to Mathlinks (or vice versa, as the case may be) does it work then?


logarithms
calculus
calculus computations

We know the Maclaurin series for $ \sqrt {1 \plus{} x}$.

If we let $ x \equal{} \frac { \minus{} 2s}{3} \plus{} \frac {2s^2}{4} \minus{} \frac {2s^3}{5} \plus{} \ldots$, or $ \displaystyle2\cdot\sum_{n \equal{} 3}^{\infty} \frac {( \minus{} 1)^n}{n}\cdot s^{n \minus{} 2}$, we can find the Maclaurin series in terms of this new $ x$.

My question is, what is the coefficient of the $ s^{k}$ term for any $ k$ in this series?

Not sure if it helps, but $ x \equal{} \displaystyle2 \sum_{n \equal{} 3}^{\infty} \frac {( \minus{} 1)^n}{n} s^{n \minus{} 2}$ is the MacLauren series for $ \frac{\minus{}2\ln(s\plus{}1)}{s^2} \plus{} \frac{2}{s} \minus{} 1$

So you want the coefficients of the MacLauren series for $ \sqrt{\frac{\minus{}2\ln(s\plus{}1)}{s^2} \plus{} \frac{2}{s}} \equal{} \frac{\sqrt{2}}{s}\sqrt{s\minus{}\ln(s\plus{}1)}$.

I had that, but the problem is the calculations... It's ridiculously annoying to calculate even the $ s^6$ term... and for a general term... well. It makes me cringe xD

There's a way you can generate the coefficients of the MacLauren series for $ \frac {\sqrt {2}}{s}\sqrt {s \minus{} \ln(s \plus{} 1)}$ recursively.

Let $ g(s) \equal{} \sqrt {2}\sqrt {s \minus{} \ln(s \plus{} 1)}$ (we can divide it by $ s$ later).

$ (g(s))^2 \equal{} 2s \minus{} 2\ln(s \plus{} 1) \equal{} \frac {2s^2}{2} \minus{} \frac {2s^3}{3} \plus{} \frac {2s^4}{4} \minus{} \frac {2s^5}{5} \plus{} \ldots$

If $ g(s)$ has a MacLauren series given by $ a_0 \plus{} a_1s \plus{} a_2s^2 \plus{} a_3s^3 \plus{} \ldots$, then $ (a_0 \plus{} a_1s \plus{} a_2s^2 \plus{} a_3s^3 \plus{} \ldots)^2 \equal{} s^2 \minus{} \frac {2}{3}s^3 \plus{} \frac {1}{2}s^4 \minus{} \frac {2}{5}s^5 \plus{} \ldots$

Therefore $ a_0 \equal{} 0$ and $ a_1 \equal{} 1$
$ 2a_1a_2 \equal{} \minus{} \frac {2}{3} \implies a_2 \equal{} \minus{} \frac {1}{3}$
$ 2a_1a_3 \plus{} a_2^2 \equal{} \frac {1}{2} \implies a_3 \equal{} \frac {7}{36}$
$ 2a_1a_4 \plus{} 2a_2a_3 \equal{} \minus{} \frac {2}{5} \implies a_4 \equal{} \minus{} \frac {73}{540}$
etc...

Well that is an observation I also made and it is quite nice actually; however, as you can see, as the powers increase, the terms become more and more complicated. Thanks for the help  :) but it turns out this problem is still unsolved in the general case.

You know, there [i]is[/i] a generic formula to compose two series.  It's called [url=http://en.wikipedia.org/wiki/Fa%C3%A0_di_Bruno%27s_formula]Faa di Bruno's formula[/url].  And it's likely you won't be able to simplify it much further.


limit
inequalities
LaTeX
induction
triangle inequality
real analysis
real analysis unsolved

I want to show that if $ a_{1}, a_{2}, a_{3}\cdots$ is a sequence of real numbers that converges to $ a$, then

$ \lim_{n\to\infty}(\frac{{\sum_{i = 1}^{n}}{a_{i}}}{n}) = a$.

This is what I have so far:

Since the original sequence converges to $ a$, for all $ \epsilon > 0$ there exists a natural number $ N$ such that $ |a_{n}-a| <\epsilon$ for all $ n > N$.

Let $ \frac{{\sum_{i = 1}^{n}}{a_{i}}}{n}= S_{n}$. Then, using algebra and the triangle inequality, we get:
${{{ {|S_{n}-a| =|\frac{1}{n}(a_{1}+\cdots+a_{n})-a| =|\frac{1}{n}\sum_{i = 1}^{n}}{a_{i}-a}|\leq\frac{1}{n}\sum_{i = 1}^{n}}{|a_{i}-a|}\leq\frac{1}{n}(|{a_{1}-a}|+\dots+|a_{n}-a|) <\frac{\epsilon}{n}+\frac{1}{n}\sum_{i = 1}^{n-1}}|{a_{i}-a}|}$

So: ${ {|S_{n}-a| <\frac{\epsilon}{n}+\frac{1}{n}\sum_{i = 1}^{n-1}}|{a_{i}-a}|}$
I don't know why it cut off to the second line.

Anyway, this is what I've done so far, and I feel like I went about it the wrong way. I feel like I started out correctly, but then missed an important step, since the last quantity doesn't really help me.

Any ideas, tips, hints, anything? Thanks

Expressions that are too long get split up by the site's $ \text{\LaTeX}$ parser- that's better than running off the screen as they do offline. You can spot this in the preview, and break it up yourself instead.

The key here: $ n$ and $ N$ are different, and you're not going to get anything by stopping at $ N+1$.
$ |S_{n}-a|\le\frac{1}n\sum_{i=1}^{n}|a_{i}-a|=\frac{1}n\sum_{i=1}^{N}|a_{i}-a|+\frac{1}n\sum_{i=N+1}^{n}|a_{i}-a|$
$ \le\frac{1}n\sum_{i=1}^{N}|a_{i}-a|+\frac{n-N}{n}\epsilon$
(I split that into two lines because it was a bit too long)

What can you do with that estimate?

The estimate tends to $ \epsilon$, but I'm not sure how to use that.

We want $ |S_{n}\minus{}a|<\epsilon$ for all $ n$ greater than a natural number $ N$.

I haven't had a chance to really sit down and work with it yet, but the only thing I can come up with is to use induction on $ n$, starting with $ n\equal{}N\plus{}1$ and show that it's less than $ c\epsilon$ for all $ n$. Seems overkill though.

If it goes to $ \epsilon$, that means there's some $ N'>N$ such that $ |S_{n}\minus{}a|<2\epsilon$ for all $ n>N'$. There was never any reason to expect you'd be able to use the same $ (\epsilon,N)$ pair for the sequence of averages as for the original sequence.

Use Stolz lemma for
$ x_{n}\equal{}\sum_{i\equal{}1}^{n}a_{i}$
$ y_{n}\equal{}n$
$ \lim_{n\to\infty}\frac{\sum_{i\equal{}1}^{n}}{n}\equal{}\lim_{n\to\infty}\frac{x_{n\plus{}1}\minus{}x_{n}}{n\plus{}1\minus{}n}\equal{}a$

jmerry already provided the "solution", so far we have $ \forall\epsilon >0\,\exists N$ such that $ \forall n\geq N:\,|S_{n}\minus{}a|\leq\frac{1}{n}\sum_{i\equal{}1}^{N}|a_{i}\minus{}a|\plus{}\frac{n\minus{}N}{N}\epsilon$

e.g. now choose $ N^{*}\geq N$ with $ \sum_{i\equal{}1}^{N}|a_{i}\minus{}a|<N^{*}\epsilon$, then $ \forall n\geq N^{*}: |S_{n}\minus{}a|\leq\left(\frac{N^{*}}{n}\plus{}\frac{n\minus{}N}{N}\right)\epsilon < 2\epsilon$

[quote="TTsphn"]Use Stolz lemma[/quote]
That's circular- this [i]is[/i] the theorem (Cesaro-Stolz is how I usually see it here).

[quote="hjbrasch"]e.g. now choose $ N^{*}\geq N$ with $ \sum_{i \equal{} 1}^{N}|a_{i} \minus{} a| < N^{*}\epsilon$, then $ \forall n\geq N^{*}: |S_{n} \minus{} a|\leq\left(\frac {N^{*}}{n} \plus{} \frac {n \minus{} N}{N}\right)\epsilon < 2\epsilon$[/quote]

Since $ N \le N^{*}$, why is it not $ |S_{n} \minus{} a|\leq\left(\frac {N^{*}}{n} \plus{} \frac {n \minus{} N}{N}\right)\epsilon \le \left(\frac {N}{n} \plus{} \frac {n}{N} \minus{} 1\right)\epsilon \le (2 \minus{} 1)\epsilon\equal{}\epsilon$ by the AM-GM inequality?


vector
linear algebra
matrix
induction
algebra
polynomial
binomial coefficients

We have $ n > 1$ lamps $ L_1, L_2... , L_n$ in a row, each of them being either on or off.
Every second we simultaneously modify the state of each lamp as follows:
if neighbours of the lamp $ L_i (1 < i < n)$ are in the same state, then $ L_i$ is switched on; otherwise, $ L_i$ is switched off. Initially the leftmost one lamp and the rightmost one  lamp are switched on.
Prove that if $ n \equal{} 2 ^ k \plus{} 1$ then all the lamps will eventually be on for an arbitrary initial position.

[b]Help![/b]

Using the matrix-trick of jmerry described above, we have to prove that the matrix of size $ (2^n+1)\times(2^n+1)$ with first and last row $ 0$, with $ 1$'s on the remaining entries of the upper and lower diagonal and with $ 0$'s elsewhere is nilpotent over the field of two elements.
Denote by $ \chi_n$ the characteristic polynomial modulo $ 2$ of this matrix. It is easily seen to be given by $ t^n$ for $ n\leq 2$ and by $ \chi_n=t\sum_{k=0}^{\lfloor(n-2)/2\rfloor}\chi_{n-1-2k}$ (working with coefficients in the field of two elements) otherwise. Setting $ \psi_n=\sum_{k=0}^{\lfloor(n-2)/2\lfloor}\chi_{n-2k}$ we have $ \chi_{n+1}=t\psi_n$ and we have to prove that $ \psi_{2^n}\equiv t^{2^n}$ for all $ n$. The recursion relation for $ \chi_n$ implies the recursion relation $ \psi_n=t\psi_{n-1}+\psi_{n-2}$ which defines the sequence of polynomials $ \psi_n$, together with the initial conditions $ \psi_n=t^n$ for $ n=1,2$.
Considered over the integers $ Z$, this yields the formula $ \psi_{2n+1}=\sum_{k=0}^n{n+k\choose n-k}t^{1+2k}$, $ \psi_{2n+1}=\sum_{k=0}^{n-1}{n+k\choose n-1-k}t^{2+2k}$. By an old result of Kummer, $ {n\choose k}$ is even except if the addition $ k+(n-k)=n$ has no carries in base $ 2$. This
shows that the binomial coefficients $ {2^{n-1}+k\choose 2^{n-1}-1-k}$ involved in $ \psi_{2^n}$ are all even except for $ k=2^{n-1}-1$ which yields the leading coefficient $ 1$. This ends the proof since it implies nilpotency of the relevant matrix over the field of two elements.

That's too much calculation for my taste. A simpler inductive solution:

Define $ M_n$ to be the $ n\times n$ matrix as in my post. The $ ij$-entry of $ M_{2n\plus{}1}^2$ is 1 if $ i\equal{}j\equal{}2,i\equal{}j\equal{}2n$, or $ |i\minus{}j|\equal{}2$ and $ 2\le i\le n$, and zero otherwise. Since all nonzero entries are either in an even row and an even column, we can change basis to $ e_1,e_3,e_5,\cdots,e_{2n\plus{}1},e_2,e_4,\dots,e_{2n}$; in this basis $ S$, $ S^{\minus{}1}M_{2n\plus{}1}^2S\equal{}\begin{bmatrix}A_n&0\\0&B_n\end{bmatrix}$ in block diagonal form. We immediately note that $ A_n\equal{}M_{n\plus{}1}$.
Now, $ B\equal{}\begin{bmatrix}1&1&0&\cdots&0&0\\1&0&1&\cdots&0&0\\ 0&1&0&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&1\\0&0&0&\cdots&1&1\end{bmatrix}$; the $ ij$-entry is $ 1$ if $ |i\minus{}j|\equal{}1$ or $ i\equal{}j\equal{}1$ or $ i\equal{}j\equal{}n$. Let $ C\equal{}\begin{bmatrix}B_n&0\\0&0\end{bmatrix}$, appending a single zero row and column. Consider the new basis $ v_1\equal{}e_1,v_2\equal{}e_1\plus{}e_2,v_3\equal{}e_2\plus{}e_3,\cdots,v_n\equal{}e_{n\minus{}1}\plus{}e_n, v_{n\plus{}1}\equal{}e_n\plus{}e_{n\plus{}1}$. We have $ Av_1\equal{}Ae_1\equal{}e_1\plus{}e_2\equal{}v_2$, $ Av_2\equal{}Ae_1\plus{}Ae_2\equal{}2e_1\plus{}e_2\plus{}e_3\equal{}v_3$, $ Av_j\equal{}Ae_{j\minus{}1}\plus{}Ae_j\equal{}e_{j\minus{}2}\plus{}e_j\plus{}e_{j\minus{}1}\plus{}e_{j\plus{}1}\equal{}v_{j\minus{}1}\plus{}v_{j\plus{}1}$ for $ 3\le j\le n\minus{}1$, $ Av_n\equal{}Ae_{n\minus{}1}\plus{}Ae_n\equal{}e_{n\minus{}2}\plus{}e_{n\minus{}1}\plus{}2e_n\equal{}v_{n\minus{}1}$, and $ Av_{n\plus{}1}\equal{}Ae_n\plus{}Ae_{n\plus{}1}\equal{}e_{n\minus{}1}\plus{}e_n\equal{}v_n$. In matrix form, that's $ V^{\minus{}1}AV\equal{}\begin{bmatrix}0&0&0&\cdots&0&0&0\\1&0&1&\cdots&0&0&0\\ 0&1&0&\cdots&0&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ 0&0&0&\cdots&0&1&0\\0&0&0&\cdots&1&0&1\\ 0&0&0&\cdots&0&0&0\end{bmatrix}\equal{}M_{n\plus{}1}$
Therefore, $ \begin{bmatrix}M_{2n\plus{}1}&0\\0&0\end{bmatrix}$ (one row and column added) is similar to $ \begin{bmatrix}M_{n\plus{}1}&0\\0&M_{n\plus{}1}\end{bmatrix}$. If $ M_{n\plus{}1}^n\equal{}0$, this gives that also $ M_{2n\plus{}1}^{2n}\equal{}0$; by induction, $ M_{2^k\plus{}1}^{2^k}\equal{}0$ for all $ k$.


[quote="gemath"]Given $x,y,z,a,b,c>0$ prove that
$ax+by+cz+\sqrt{(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})}\ge\frac{2}{3}(a+b+c)(x+y+z)$
P/s:Maybe there is an extension for 2n variables .[/quote]

just a little remark before I will try this problem. the equality holds when a=b=c or x=y=z. and more, if a>=b>=c and x>=y>=z is simple, use Cauchy then Cebasev

This (and also the generalised version) is in Vasc's book. I'm pretty sure it is somwhere in the forum.

[quote="gemath"]Given $x,y,z,a,b,c>0$ prove that
$ax+by+cz+\sqrt{(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2})}\ge\frac{2}{3}(a+b+c)(x+y+z)$
P/s:Maybe there is an extension for 2n variables .[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70625


AMC
USA(J)MO
USAMO
algebra unsolved
algebra

Let a1, a2, a3, ... be a sequence of positive integers with the following property:

for each n, aan equals the number of terms of the sequence that do not exceed n.

Prove that there are infinitely many values of n for which aan = n.

This problem was proposed (but not chosen) for USAMO 2002.

Wow, this problem got me scratching up my head :? ...and I see dandruffs falling off...

Damn, I don't manage to get anywhere with this problem :?


Two masses of 3kg and 1kg hang vertically at 2 ends of a light inelastic string passing over a fix pulley . Both the masses are originally at rest and 1m above the ground . The 3kg mass then falls from rest and hit the ground . Assuming the 3kg mass does not rebound after hitting the ground .Find

1)How much further will the 1kg mass rise after the impact of the 3kg mass on the ground ?

2)After the 1kg mass reaches maximum height , it will fall for some time and pull up the 3kg mass from the ground .Find this intially velocity of the 3kg mass rises up from the ground .

question 1 is doable but question 2 seems to be a bit weird . What I think is after the 1kg mass experiences free fall , it will exert a force of 10N and pull the 30N mass . But isn't that this force is not sufficient to pull the 3kg mass up ? Or there are some other force acting on it ?

Before collision,
     for 3kg mass 3g-T=3a
     for 1kg mass  T-g=a
solving we get,
                        a=g/2
hence velocity of the 1kg mass just before collision
                        v=g^1/2
hence max. height reached
                        h=0.5mts.
velocity of 1kg mass just before string becomes taut
                        =v=g^1/2
let the impulse imparted by the string on the blocks just when it becomes taut be I,then writing the impulse eqn. for just before and after the string becomes taut,
                         I=v-v'    (for 1kg mass)
                         I=3v'      (for 3kg mass)
[impulse imparted by gravity mg*dt is neglected due to its small value]
solving we get,
                          v'=v/4=[g^1/2]/4
It can be seen that the tension force at that small time interval is not 10nt. but much larger than that because the string is inextensible and instantly tries to move both the blocks with same velocity(as can be seen that the velocity of 3kg mass becomes v' in a very small time) or else may have rebound the 1kg mass if the other would have been too high similarly in a very short time by applying a large force.


Determine all continuous functions $f: \mathbb R \to \mathbb R$ such that
i) $ f(0) \ne 2$ and $f(2) \ne 0$, and
ii) For all reals $x$ and $y$, \[ f(x\plus{}y) \ne f(x) \cdot f(y) \equal{}f(xy)\plus{}f(x)\plus{}f(y).\]

[quote="Victory.US"]Determine all continuos function $ R\to R$ such that :
$ i)$ $ f(0) \ne 2;f(2) \ne 0$
$ ii)$ $ f(x \plus{} y) \ne f(x).f(y) \equal{} f(xy) \plus{} f(x) \plus{} f(y)$ ,$ \forall x,y \in R$[/quote]

Setting $ x\equal{}y\equal{}0$ in $ f(x).f(y) \equal{} f(xy) \plus{} f(x) \plus{} f(y)$, we get $ f(0)(f(0)\minus{}3)\equal{}0$ but $ f(0)\equal{}0$ would imply $ f(0\plus{}0)\equal{}f(0)f(0)$ which is supposed false. So $ f(0)\equal{}3$

Setting then $ y\equal{}0$ in $ f(x).f(y) \equal{} f(xy) \plus{} f(x) \plus{} f(y)$, we get $ 3f(x)\equal{}f(x)\plus{}6$ and so $ f(x)\equal{}3$, which indeed is a solution.

Hence the answer : $ \boxed{f(x)\equal{}3}$ $ \forall x\in\mathbb R$


Does anyone do this?  If so, can you give me any tips?  BTW: I am in $ 7^{th}$ grade so you know my difficulty level.

JINKIES (in place of harsher words)

there's activity in this forum!

ah yes, they're doing middle school later than the rest this year.

long term - kcatm really just sucks as a competition. it is very good at what they do, and that is to get interest in math. but compared to mathcounts/moems/amc8/10/12, it will seem trivially easy.

thus, imo, practice for the good competitions, because kcatm will be easy after that.

short term - do past tests on the website. if you do any past tests, do the most recent - why? because they are probably the most similar.

in fact, this year's high school test was the EXACT same test as 2006... and practically everyone noticed.

tips: do the easy problems first and watch the clock. that's all i can really say. bring sharpened pencils.

i'll say that my 8th grade scores were a huge improvement after doing mathcounts.

i'm in 10th grade btw.

kcatm is REALLY ez, but they suck.  i mean their answers to the probelms are all wrong in the tests on teh site!!!!

i wonder if they use the same answers for the real compitition.  that would suck, u get all teh questions right, but t=some weirdo guesses and gets them right cause the real answers are wrong!

Hey, who can tell me more about this KCATM contest?  Is it just for Kansas students?  Who runs it?  i'm in charge of the GPML contest series which runs the state championship, and i'd be interested in talking to these people about possibly helping them improve their tests (or even switch over to GPML format) since everyone seems not to like the current format..

http://07contest.kcatm.org/

Kansas City Area Teachers of Mathematics, hosts an annual math competition around march, comprising of grades 4-12 (I've competed 4 years). Of local competitions in the KC area, it is the biggest, especially at the lower levels. So I think both KS and MO schools compete, though it is KS heavy. There are 4 individual tests per grade, 4 concurrent team tests, and a Mathletics event.

My problems with the competition:

* Caps on number of individual tests that can be taken by a school per grade - this year, we had a deficit of juniors, but a lot of sophomores, and our coach (who has had a long history of working with the contest, proctoring, grading, coaching that kind of stuff) had to give some of the sophomores junior tests. Which, is quite a minor problem, and I can kind of see why they want to cap it. (For overall individual scores, for example, a 9th grader doing 10th grade tests, can win on the tests themselves, but does not win 10th grade overall, they can win 9th grade).

* The problems (the site has some practice tests) are essentially just classroom math with speed, no problem solving stuff comparable to mathcounts/amc/gpml. I have mixed feelings... in the lower levels, it's great to award those little math stars, but it never introduces people to problem solving math. Granted, mathcounts/moems/amc8 does a great job of that, but they're not nearly as widely competed in.

Mathletics is slightly better, but not much. Like easy state mathcounts sprint level.

It could also be that, imo, the level of competition in this geographic area is pretty low. Only the better people in the upper levels complain.

* Typos - probably 2-3 problems per test don't have right answer choices. The practice tests online have about 4-5 wrong answers.

* They reuse tests (???). Last years was an old test with different numbers. This years... was the EXACT same test as 06, which is the last year that they have online... So people happened to have printed them off to practice noticed they were the exact same. I was like "WTF?"

I hope I have been helpful. It would be great if they switched to GPML format (for me at least :P). But to be honest, the competition has been around for a while, and the test format is pretty rooted in tradition. And second, since it is 4-12, I don't think gpml format would benefit the lower levels. It DOES do a great job at recognizing potential in the lower levels, and they are very generous on awards, and the awards banquet at JCCC is pretty good.

I think getting more people to do GPML in KS would be better. Our school doesn't do it, though I competed once as a one man team in one of the qualifiers (perfect on sprint, and decent target and team scores), though I didn't go to KS State championships this year. GPML is great, but less people know about it. I also competed for MO ARML team last year.

I'm interested in ARML, is there any way to sign up?

Also can I compete if i'm in Middle School?

KS does not have a team, so you have to join MO (there were 6 Kansans on MO ARML last year). The site seems to down, please contact the head coach "RArmstrong@stlcc.edu" to join the mailing list. I think you are allowed to compete as a middle schooler, but just mention that you are in middle school when you contact him. An application is sent out in the spring, and there's a fee, like 200-300 dollars.