linear algebra
matrix
limit
linear algebra unsolved

Let $ M_1,\ldots ,M_k$ be $ n\times n$ positive Hermitian matrices whose largest eigenvalue is 1.
Show that $ \lim_{m\to \infty} (M_1\cdots M_k)^m$ exists and equals 0 $ \iff \ker(M_1 \cdots M_k \minus{} I_n) \equal{} \{0\}$.

Let$ \parallel{}.\parallel{}$ be the $ 2$-norm.
i) for all $ i,x$ $ \parallel{}M_ix\parallel{}\leq\parallel{}x\parallel{}$ and $ \parallel{}M_ix\parallel{}=\parallel{}x\parallel{}$ implies $ M_ix=x$.
Proof: $ \parallel{}M_ix\parallel{}^2=x^T{M_i}^2x\leq{x^Tx}$ with equality only if $ x$ is in the eigenspace associated to the eigenvalue $ 1$ of $ M_i^2$.
ii) ${ \parallel{}M_1\cdots{M_k}x\parallel{}=\parallel{}x\parallel{}}\Rightarrow{x\in\bigcap_iker(M_i-I)}$.
Proof: $ \parallel{}M_1\cdots{M_k}x\parallel{}\leq{\parallel{}M_2\cdots{M_k}x\parallel{}}\leq{\parallel{}M_{k-1}M_kx\parallel{}}\leq{\parallel{}M_kx\parallel{}}\leq{\parallel{}x\parallel{}}$ with equality if $ \parallel{}M_kx\parallel{}=\parallel{}x\parallel{}$ that is $ M_kx=x$ and if $ \parallel{}M_{k-1}x\parallel{}=\parallel{}x\parallel{}$ that is $ M_{k-1}x=x$ and if $ \cdots$.
iii) $ ker(M_1\cdots{M_k}-I)=\bigcap_iker(M_i-I)$.
Proof :$ x\in{ker}(M_1\cdots{M_k}-I)\Leftrightarrow{M_1\cdots{M_k}x=x}\Rightarrow{\parallel{}M_1\cdots{M_k}x\parallel{}=\parallel{}x\parallel{}}$
$ \Rightarrow{x\in\bigcap_iker(M_i-I)}\Rightarrow{M_1\cdots{M_k}x=x}$.
iv) $ (M_1\cdots{M_k})^m$ tends to $ 0$ iff $ ker(M_1\cdots{M_k}-I)=\{0\}$.
Proof: $ (M_1\cdots{M_k})^m$ tends to $ 0$ iff $ \{$for all eigenvalues $ \lambda$ of $ M_1\cdots{M_k}$ :  $ |\lambda|<1\}\Rightarrow{ker(M_1\cdots{M_k}-I)=\{0\}}$.
Conversely let $ \lambda$ be an eigenvalue of $ M_1\cdots{M_k}$ s.t. $ |\lambda|\geq{1}$.
By the proofs of ii),iii) $ |\lambda|=1$, $ \lambda=1$ and  $ ker(M_1\cdots{M_k}-I)\not=\{0\}$.


[hide="a"]the string of 1's is $\frac{10^{2002}-1}{9}$. Division by 9 won't affect divisibility, and since 2003 is prime, $10^{2002}\equiv 1\pmod{2003}$ by Fermat's Little Theorem. Then the string is divisible by 2003.[/hide]

[quote="agolsme"][quote="Scrambled"]Given that we have a number with a decimal representation with only 1s (ex: 1 is a string of one 1, 111 is a string of three ones, 111111 is a string of 6 ones, etc.) then

a) prove that a string of 2001 1s will be divisible by 2003
b) prove that there is a smaller string of 1s that will be divisible by 2003[/quote]

[hide="a"]

$2003$ is prime. so we can rewrite the string of 2001 1s as:
$\frac{10^{2002} - 1}{10-1}$ we don't care about the 9 on the bottom because it doesn't divide 2003 so if 2003 divides $10^{2002} -1$ then we're set, and that's trivially true by fermat's little theorem.
[/hide]

[hide="b"]
$10^d \equiv 1 \pmod{2003}$ for $d | 2002$ because  you have $(10^d)^{\text{oddpower}} \equiv 1 \pmod{2003}$
[/hide][/hide][/quote]

For b, d would have to divide 2002. But that does not necessarily happen (what about d=1?)


linear algebra
matrix
calculus
calculus computations

This is a worked example

The objective is to prove

$ - 1 \leq \rho(X,Y) \leq 1$

Then the book uses this formula...

(2) $ 0 \leq Var( \frac {X}{\sigma_x} + \frac {Y}{\sigma_y} )$

(3) $ = \frac {Var(X)}{{\sigma_x}^2} + \frac {Var(Y)}{{\sigma_y}^2} + \frac {2Cov(X,Y)}{\sigma_x \sigma_y}$

The question is, how does 2 lead to 3? Namely, how does $ Var(\frac {X}{\sigma_x} ) = > \frac {Var(X)}{{\sigma_x}^2}$?

Also, how does one get the idea to use formula (2) to prove (1)? It doesn't seem like a natural step

Variance scales quadratically- just look at the definition, and think of units.

The key fact we know if we want to prove (1) is that the covariance matrix is positive definite; to use this, we either study the matrix directly and calculate its determinant, or look at the variance of various linear combinations.

oh true

actually I got it, due to the simple Var(aX) = a^2 Var(X) formula. I encountered it before. my attention lapses are bad... =/


For any set $S$, let $|S|$ denote the sum of elements in $S$, and let $n(S)$ be the number of subsets of $S$, including the empty set and the set $S$ itself. If $A,B$ and $C$ are sets for which

$n(A)+n(B)+n(C) = n(A \cup B \cup C) \ \text{and} \ |A| = |B| = 100,$

then what is the minimum possible value of $|A \cap B \cap C|$?

$\text{(A)} \ 96 \qquad \text{(B)} \ 97 \qquad \text{(C)} \ 98 \qquad \text{(D)} \ 99 \qquad \text{(E)} \ 100$

Obviously $n(S) = 2^{|S|}$.  The equation boils down to $2^{101} + 2^{|C|} = 2^{|A\cup B \cup C|}$.

For the LS to be a power of two, it must be the case that $|C| = 101$; then $|A\cup B \cup C| = 102$.

We can have $|A\cap B\cap C| = 97$, $|A\cap B|, |B\cap C|, |C\cap A| = 1,2,2$.  If we had 96, we would need to use 6 more elements to achieve a gain of 6 in C and 4 in A and B.  It's not possible.

Answer B.

Silverfalcon, I think you have a typo...$|S|$ should be the NUMBER of elements in $S$, NOT  the SUM of the elements in $S$.  Anyways, this is a fun problem in AoPS Volume II :lol:  :lol:


What is the maximal number of points in $R^n$ such that the distance between any two is an odd integer?

Not long ago i was solving this problem for $n=2$.In this case $n< 4$.

Yes, and even that one is hard in my humble opinion. But for general n, I have no idea. I thing I can play a little bit arround with the idea for $n=2$ in case $n$ has a special form modulo 8, but even there I cannot solve it completely.


Se dau numerele n, n+1, n+2 si n+3, n natural.
a)   Sa se arate ca exista un singur trapez si numai unul avand ca lungimi ale laturilor sale aceste numere.
b)   Aflati n astfel incat trapezul sa fie dreptunghic.
c)   Aflati n astfel incat inaltimea trapeuzlui sa fie egala cu baza mica a trapezului.

Haide, haide, putina inviorare!Chiar nu vrea nimeni sa o faca?Sau e prea simpla?Sau prea grea?

Da, s-a dat la a VII-a.Problema e cam grea totusi pentru clasa a VII-a. :ninja: Ai cumva o solutie?Btw, care e numele tau? :)

[hide="Hint"]Se arata, prin reducere la absurd, ca bazele au lungimile $n$ si $n+3$.[/hide]

Da, domnule profesor.Asa e.Dar eu am pierdut cam o jumatate de ora sa gasesc solutia asta.M-am chinuit sa incerc toate variantele posibile,ca, poate gresesc altcumva.Si cred ca dupa asta problema devine cam simpla...asa cred.Problema asta mi-a dat-o doamna profesoara a mea pentru selectia la Concursul Pitagora.

Problema asta am rezovat-o pe vremuri :lol:  cand eram a VIIa.

ar trebui sa o rezolve cei de a VIIa...dar nu se prea inghesuie nimeni...whatever...

o idee asemanatoare

[hide]Daca ABCD este trapezul cerut, cu AB baza mare, se duce paralela CE la AD. Se aplica inegalitatea triunghiului in triunghiul BCE si se vede luand toate cazurile ca AB=n+3, CD=n. Punctele b) si c) sunt usoare.[/hide]


geometry
3D geometry
calculus
derivative
calculus computations

A right circular cone has base radius $ r$, height $ h$ and slant length $ l$. Its volume $ V$, and the area $ A$ of ts curved surface, are given by

$ V\equal{} \frac{1}{3}\pi r^{2}h, \,A \equal{} \pi rl$

(i) Given that $ A$ is fixed and $ r$ is chosen so that $ V$ is at its stationary value, show that $ A^{2} \equal{} 3\pi^{2}r^{4}$ and that $ l\equal{}\sqrt{3}r$.

(ii) Given instead, that $ V$ is fixed and $ r$ is chosen so that $ A$ is at its stationary value, find $ h$ in terms of $ r$.

[hide="(i)"]
$ l \equal{} \sqrt{r^2 \plus{} h^2}$, so $ A \equal{} \pi r \sqrt{r^2 \plus{} h^2}$ gives $ h \equal{} \frac1r\sqrt{\frac{A^2}{\pi^2} \minus{} r^4}$.

$ \frac{dV}{dr}\equal{} \frac{d}{dr}\left[ \frac13 \pi r \sqrt{\frac{A^2}{\pi^2} \minus{} r^4}\right] \equal{} \frac{\pi}3 \left[\sqrt{\frac{A^2}{\pi^2} \minus{} r^4} \plus{} \frac12 (\minus{} 4r^3) r \frac1{\sqrt{\frac{A^2}{\pi^2}\minus{} r^4}}\right]$
$ \equal{} \frac{\pi}3 \sqrt{\frac{A^2}{\pi^2} \minus{} r^4}\left[1 \minus{} \frac{2 r^4}{\frac{A^2}{\pi^2} \minus{} r^4}\right] \equal{} 0$

Gives $ 0 \equal{} 1 \minus{} \frac{2 r^4}{\frac{A^2}{\pi^2} \minus{} r^4}$ so $ A^2 \equal{} 3 \pi^2 r^4$.

And then also $ l \equal{} \sqrt{3} r$ as then $ A^2 \equal{} \pi^2 r^2 l^2 \equal{} \pi^2 r^2 3 r^2 \equal{} 3 \pi^2 r^4$.
[/hide]

[hide="(ii)"]
$ h \equal{} \frac{3 V}{\pi r^2}$.

So $ A \equal{} \pi r \sqrt{r^2 \plus{} h^2} \equal{} \pi r \sqrt{r^2 \plus{} \frac{9 V^2}{\pi^2 r^4}} \equal{} r^2 \sqrt{\pi^2 \plus{} \frac{9 V^2}{r^6}}$.

so $ \frac{dA}{dr} \equal{} 2 r \sqrt{\pi^2 \plus{} \frac{9 V^2}{r^6}} \minus{} \frac1{2 r^5} \frac{45 V^2}{ \sqrt{\pi^2 \plus{} \frac{9 V^2}{r^6}}}$
$ \equal{} \frac{2}{r^2} \sqrt{\pi^2 r^6 \plus{} 9 V^2} \minus{} \frac1{2r^2} \frac{45 V^2}{ \sqrt{\pi^2 r^6 \plus{} 9 V^2}} \equal{} 0$
gives
$ \pi^2 r^6 \plus{} 9 V^2 \equal{} \frac{45}4 V^2$ (or $ r \equal{} 0$).

This then implies $ r \equal{} \left(\frac{3 V}{2 \pi}\right)^{\frac13}$.

$ h \equal{} \frac{3 V}{\pi r^2} \equal{} \frac{3 V}{\pi}  \left(\frac{2 \pi}{3 V}\right)^{\frac23} \equal{} 3 \left(\frac{4 V}{\pi}\right)^{\frac13}$.
[/hide]

Actually this is wrong. I apparently thought the derivative of $ \frac1{r^6}$ was $ \minus{} 5 \frac1{r^7}$ at some point.

We instead get $ \pi^2 r^6 \plus{} 9 V^2 \equal{} \frac {9 \cdot 6}4 V^2$ which gives $ r \equal{} \left(\frac {9 V^2}{2 \pi^2}\right)^{\frac13}$ and then $ h \equal{} \left(\frac {3 V}{2 \pi}\right)^{\frac13}$.


For $\{1,2,3,...,n\}$ and each of its nonempty subsets, a unique [b]alternating sum[/b] is defnined as follows: Arrange the numbers in the subset in decreasing order and then, beginning with the largest, alternately subtract and add successive numbers. (For example, the alternating sum for {1,2,4,6,9} is 9-6+4-2+1=6 and for {5} it is simply 5.) For each $n$, find a formula for the sum of all the alternating sums of all the subsets.

[hide="hint"]consider pairing a set $S$ that does not have n, with the set that has all the elements of $S$ and 7[/hide]

[hide]Recursion would be another way.[/hide]


Για καθε n φυσικο να βρεθει ο μεγιστος φυσικος k ετσι ωστε $ 2^k|[(3 + \sqrt{11}){}^2{}^n{}^-{}^1]$

Πολύ ωραία και δύσκολη άσκηση. :!: Τη λύση αυτής βρήκα καθώς ξεφύλλιζα το βιβλίο "Παγκόσμια Θεματογραφία Ολυμπιάδων Μαθηματικής Ανάλυσης" του Γιάννη Μπαϊλάκη. Μία υπόδειξη:[hide]Θεωρούμε την ακολουθία με ν-οστό όρο $ (3+\sqrt{11})^n$+$ (3-\sqrt{11})^n$, για την οποία είναι εύκολο να βρούμε αναδρομικό τύπο. Επίσης αποδείξτε ότι το $ [(3+\sqrt{11}){}^{2}{}^{n}{}^{-}{}^{1}]$ είναι ο $ 2n-1$ όρος της ακολουθίας αυτής. Επαγωγικά δείχνουμε ότι το $ 2^n$ διαιρεί τους όρους $ 2n-1$ και $ 2n-2$ της παραπάνω ακολουθίας και ότι το $ 2$^(n+1) δεν διαιρεί τον $ 2n-1$ όρο αυτής.[/hide]

Ένας τρόπος να λύσουμε την άσκηση αυτή είναι ο εξής:

- Καταρχάς θέτουμε $ a\equal{}3\plus{}\sqrt{11}$  και παρατηρούμε ότι ο αριθμός $ a^{2n\minus{}1}$ δεν είναι φυσικός. (αν ήταν φυσικός τότε το διώνυμο του Νεύτωνα μας δίνει εύκολα ότι ο $ \sqrt{11}$ είναι ρητός)

- Βρίσκουμε μία έκφραση για το ακέραιο μέρος του $ a^{2n\minus{}1}$. Αυτό γίνεται δουλεύοντας με τον αριθμό $ b\equal{}3\minus{}\sqrt{11}$. Ο αριθμός
αυτός βρίσκεται μεταξύ -1 και 0. Το ίδιο ισχύει και για τον $ b^{2n\minus{}1}$. Στη συνέχεια το διώνυμο του Νεύτωνα μας δίνει ότι ο
αριθμός $ a^{2n\minus{}1}\plus{}b^{2n\minus{}1}$ είναι φυσικός και συνεπώς έχουμε την εξής κατάσταση. Προσθέσαμε ένα αριθμό μεταξύ -1 και 0 σε
ένα θετικό αριθμό και βρήκαμε ένα φυσικό αριθμό. Ο φυσικός αυτός αριθμός πρέπει να είναι το ακέραιο μέρος του θετικού αριθμού που
προαναφέραμε.

- Στη συνέχεια θα μελετήσουμε την ακολουθία $ x_n\equal{}a^n\plus{}b^n$. Προκειται για μια ακολουθία φυσικών αριθμών η οποία περιγράφεται από την αναδρομική σχέση:

\[ x_0\equal{}2\quad x_1\equal{}6\quad x_{n\plus{}2}\equal{}6x_{n\plus{}1}\plus{}2x_n\]

Η θεωρία των ακολουθιών αυτών είναι στοιχειώδης και συναντάται και στη μελέτη της ακολουθίας Fibonacci. Μια ακολουθια με την
παραπάνω αναδρομική σχέση έχει πάντα τη μορφή $ \lambda_1\omega_1^n\plus{}\lambda_2\omega_2^n$ όπου $ \omega_1$ kai $ \omega_2$ είναι
οι λύσεις της χαρακτηριστικής εξίσωσης $ \omega^2\minus{}6\omega\minus{}2\equal{}0$. Εδώ φυσικά έχουμε $ \omega_1\equal{}a$ και $ \omega_2\equal{}b$.

- Στο τελευταίο στάδιο της λύσης θα χρησιμοποιήσουμε τη συνάρτηση δυαδικής τίμησης. Στην προκειμένη περίπτωση μπορούμε να
τη θεωρήσουμε σα μια συνάρτηση με πεδίο ορισμού τους φυσικούς και σύνολο τιμών τους μη αρνητικούς ακεραίους. Αν $ n$ είναι
ένας φυσικός αριθμός τότε ορίζουμε $ v(n)$ ως το μέγιστο μη αρνητικό ακέραιο που είναι τέτοιος ώστε $ 2^{v(n)}|n$. Για παράδειγμα
$ v(1)\equal{}0$, $ v(2)\equal{}1$, $ v(3)\equal{}0$, $ v(4)\equal{}2$ κλπ. Θα χρειαστούμε δύο βασικές και στοιχειώδεις ιδιότητες της συνάρτησης αυτής. Η πρώτη είναι ότι $ v(nm)\equal{}v(n)\plus{}v(m)$ για οποιουσδήποτε δύο φυσικούς αριθμούς $ m$ kai $ n$ και η δεύτερη είναι ότι $ v(m\plus{}n)\geq \min(v(m),v(n))$, και αν $ v(m)\neq v(n)$ τότε $ v(m\plus{}n)\equal{} \min(v(m),v(n))$. Αξίζει να σχολιάσουμε λίγο τη δεύτερη ιδιότητα. Ας υποθέσουμε ότι έχουμε ενα πολλαπλάσιο του 8 και ένα ακριβές πολλαπλάσιο του 4. Με αυτό εννοούμε ότι ο δεύτερος αριθμός διαιρείται με το 4 αλλά όχι με το 8. Τότε το άθροισμά τους είναι ένα ακριβές πολλαπλάσιο του 4.

- Ο σκοπός της άσκησης είναι να υπολογίσουμε το $ v(x_{2n\minus{}1})$. Χρησιμοποιώντας την αναδρομική σχέση μπορούμε να δούμε ότι:

\[ x_0\equal{}2\quad x_1\equal{}6 \quad x_2\equal{}40 \quad x_3\equal{}252\]

απ' όπου λαμβάνουμε:

\[ v(x_0)\equal{}1\quad v(x_1)\equal{}1\quad v(x_2)\equal{}3 \quad  v(x_3)\equal{}2\]

Θα χρησιμοποιήσουμε επαγωγή για να δείξουμε ότι:

\[ v(x_{4k})\geq 2k\plus{}1 \quad v(x_{4k\plus{}1})\equal{}2k\plus{}1 \quad v(x_{4k\plus{}2})\geq 2k\plus{}2 \quad v(x_{4k\plus{}3})\equal{}2k\plus{}2\]

για κάθε $ k$ μη αρνητικό ακέραιο.

Πραγματικά αν ισχύουν οι παραπάνω σχέσεις θα δείξουμε ότι ισχύουν αν το $ k$ αντικατασταθεί με $ k\plus{}1$. (H $ k\equal{}0$ είναι προφανής
από τον αρχικό λογαριασμό.)

Ξέρουμε ότι $ x_{4k\plus{}4}\equal{}6x_{4k\plus{}3}\plus{}2x_{4k\plus{}2}$. Η επαγωγική υπόθεση συνεπάγεται ότι
$ v(6x_{4k\plus{}3})\equal{}2k\plus{}3$ και $ v(2x_{4k\plus{}2})\geq 2k\plus{}3$. Η δεύτερη ιδιότητα της συνάρτησης τίμησης συνεπάγεται ότι $ v(4k\plus{}4)\geq 2k\plus{}3$.

Ο συλλογισμός για την επόμενη σχέση είναι παρόμοιος:

Ισχύει $ x_{4k\plus{}5}\equal{}6x_{4k\plus{}4}\plus{}2x_{4k\plus{}3}$. Από τις  $ v(6x_{4k\plus{}4})\geq 2k\plus{}4$ και $ v(2x_{4k\plus{}3})\equal{} 2k\plus{}3$, έπεται $ v(x_{4k\plus{}5})\equal{}2k\plus{}3$.

Παραλείπουμε τις δυο τελευταίες σχέσεις των οποίων η απόδειξη είναι παρόμοια.

Αυτό που μας ενδιαφέρει είναι ότι:

\[ v(x_{4k\plus{}1})\equal{}2k\plus{}1 \quad v(x_{4k\plus{}3})\equal{}2k\plus{}2\]

Οι δύο αυτές σχέσεις συνοψίζονται στην $ v(x_{2n\minus{}1})\equal{}n$ η οποία είναι και η απάντηση στο ερώτημα της άσκησης.


trigonometry
rotation
geometry
trig identities
Law of Sines

Two lighthouses are located on a north-south line.  From lighthouse A, the baring of a ship 3742 m away is 129 degrees and 43 minutes.  From lightouse B, the bearing is 39 degrees and 43 minutes.  Find the distance between the lighthouses.

I'm not sure how to set up the diagram.  I drew a y-axis, and put lighthouses A and B on it.  Then I rotated 129 degrees and 43 minutes clockwise.  I'm not sure if I'm drawing it right.  Will someone please help me?

i believe that this is a law of sines problem, try drawing the lighthouses, and then drawing another point for the ship so they form those angles

A
|
|
|
............Ship
|
|
|B


let $ A\subset \mathbb{R}^2$ be a countable subset , prove that $ \mathbb{R}^2\setminus A$ is path-connected ?

the hint in the book is : how many lines pass through a point in $ \mathbb{R}^2$ ?

But I don't really get it, could some one point it out for me ? thanks

Fix $ x\in\mathbb{R}^{2}\setminus A$, and consider the set of all (straight) lines through $ x$. There are uncountably many of these and they meet in the single point $ x$ , so not all of them contain a point of $ A$ . Choose one that doesn't and move along it: your distance from $ p$ takes on uncountably many values, and hence at some point this distance $ r$ from $ p$ is not shared by any point of $ A$ . The whole of the circle with radius $ r$ and center $ p$, lies in $ \mathbb{R}^{2}\setminus A$ so we may move around it freely.

Consider all lines through $ p$: these all intersect this circle, and there are uncountably many of them so we may choose one, say $ L$ , that contains no point of $ A$. Moving around the circle until we meet $ L$ and then following it inwards completes our path from $ x$ to $ p$.

The one-step version: to get from $ A$ to $ B$, consider all circular arcs from $ A$ to $ B$; there are two for each possible center, and they're all disjoint. There are uncountably many of these, so some miss all of the points in $ A$.

Let $ x,y\in\mathbb{R}^{2}\setminus A$; then, as was alraedy said, there is an uncountable number of lines passing through x that doesn't contain any points from $ A$ (with the possible exception of x, but this doesn't matter); if one of these lines also contains $ y$, then we're done. If not, consider the similar lines passing through $ y$; if one of them intersects another passing through $ x$, we're done again; if not, they must be paralell, so pick a point $ z$ in one of them, a line through z that doesn't intersect  $ A$ and is different from both lines previously chosen; this line must intersect the other lines and, from the three, you may construct a path from $ x$ to $ y$.


Putnam
superior algebra
superior algebra unsolved

Let $G$ be a infinite group with the following propertis:
(I) :if $a\in{G}$ and $n\in{\mathbb{N}}$ and $a^{n}=e$ then $a=e$
(II):if $x,y\in{G}$ then $(xy)^{2}=(yx)^{2}$
prove that $G$ is commutative.

[quote="ali666"]can you tell me why  we have:$(y^{-1}x^{-1})^{2}(yx)^{2}=(y^{-1}x^{-1}yx)^{2}$[/quote]

Associativity.  At least, I hope it's just associativity.

t0rajir0u...that's not associativity... in order to have that should know that the group is comutative :maybe:

[quote="me@home"]Is there a problem source because this problem looks familiar, possibly from an old Putnam?[/quote]
 sorry,I don't know the original source.

[quote="sina rezazadeh baghal"]Hi my friend  :)
Let$z = xyx^{-1}$ then $z^{2}= (xyx^{-1})(xyx^{-1})=xy^{2}x^{-1}$,
but$z^{2}= (xyx^{-1})^{2}= (x^{-1}xy)^{2}= y^{2}$
This implies$xy^{2}= y^{2}x$ for all x and y lie in $G$
Now assume $xy = w(yx)$ then we conclude that:

$(xy)^{2}= (wyx)(xy) =wyx^{2}y=wy^{2}x^{2}=(yx)^{2}=(yx)(yx)$
$wy^{2}x=yxy$
$wxy^{2}=yxy$
$wxy = yx$
$w^{2}(yx)=yx$
$w^{2}= e$
then $w=e$ and$xy = yx$[/quote]
very nice!
Thx a lot my friend :roll:


We have an equilateral triangle with the edge sized 2.
in each generation,we will erect a new equilateral triangle inside every existing triangle ,by drawing lines between the midpoints of it's edges...,so,let n be the number of generation,then:
if n = 1 , we have 1 triangle
if n = 2 , we will have 5 triangle (4 triangles with the edge sized 1,and 1 triangle with the edge sized 2)
if n = 3 , we will have 27 triangles (16 (1/2)-sized-edge triangles,7  1-sized-edge triangles,3 (3/2)-sized-edge triangles,and 1 4-sized-edge triangle)
...
determine the number of all triangles in the generation number n,in terms of n.

Well..if I do understand, you just want to determine the number of equilateral triangles (of any size) formed when each side of an equilateral triangle is divided by by points into $ n$ equal parts, and when we draw lines parallel to the sides of the triangle and joining these points by pairs?

Pierre.

That appears to be the question, although the O.P. has asked only for the answer in the case $ n$ is a power of $ 2$.  I suggest splitting into the "right side up" and "upside down" cases -- the former are easy while the latter are a little more complicated.

This is sequence number [url=http://www.research.att.com/~njas/sequences/A002717]A002717[/url] in the [url=http://www.research.att.com/~njas/sequences/]OEIS[/url].

yes pbornsztein,that's what i was trying to state... :)
can you please post your solutions too? (i mean not just the answer)


geometry
parallelogram
geometric transformation
reflection
similar triangles
geometry unsolved

Given a parallelogram $ ABCD$ ($ BD>AC$). A circle with diameter $ AC$ cut $ BD$ at $ P,Q$ ($ BQ>BP$) A tangent to that circle at $ C$ meet $ AB,AD$ at $ X,Y$ respectively. Prove that $ P,Q,X,Y$ are concyclic.

Let $ O$ be the center of the parallelogram and $ M \equiv{} DB \cap XY.$ $ C'$ is the reflection of $ C$ about $ Y.$ Then $ AC' \parallel DB$ and let $ P_{\infty}$ be their infinite point. Since $(D,B,O,P_{\infty})=-1,$ it follows that $( X,C,Y,C')=-1$ $\Longrightarrow$ $ MC^2 \equal{} MY \cdot MX,$ but  $ MC^2$ is the power of $ M$ to the circle with diameter $\overline{AC}$ $\Longrightarrow$ $ MY \cdot MX \equal{} MP \cdot MQ.$

P.S. We also can go for a couple of similar triangles to get the same relation.

[quote] We also can go for a couple of similar triangles to get the same relation.[/quote]
I don't think it's that easy. Can you show it?

$\triangle MBY \sim \triangle MDC \Longrightarrow \ \frac {MB}{MD} \equal{} \frac {MY}{MC} \ \ (1)$

$\triangle MBC \sim  \triangle MDX \Longrightarrow \ \frac {MD}{MB} \equal{} \frac {MX}{MC} \ \ (2)$

Combining $(1)$ and $(2)$ yields $ MC^2 \equal{} MX \cdot MY,$ and the conclusion follows.


geometry
quadratics
algebra
polynomial
superior algebra
superior algebra unsolved

Suppose that F is a subfield of the field of real numbers and that c is a real number whose square root is not in F. Show that any number d in the square root extension field F[c] is a root of a linear or quadratic polynomial P(x) with coefficients in F.

[quote="t0rajir0u"]This is not (really) geometry, Olympiad, or an Inequality.  It belongs [url=http://www.artofproblemsolving.com/Forum/index.php?f=61]here[/url].[/quote]

I am new and I could not find this place for my posting, but I am not sure if I got an answer from you or anyone else yet. Please advise. Thanks

Well, I assume that that should be $ \sqrt{c}$. It is indeed a general fact that if $ \alpha$ is a root of an irreducible polynomial $ p$ over $ K$, every $ x\in K(\alpha)$ is a root of some polynomial over $ K$ with degree dividing the degree of $ p$.
In this quadratic case, you can explicitly construct a polynomial that $ a\plus{}b\sqrt{c}$ satisfies.

[quote="ZetaX"]This is standard algebra, and true for any field $ F$ and any solution $ c$ of a quadratic equation in $ F$.[/quote]

Do you have a more explicit proof, because that is what I need. Thanks

[quote="jmerry"]$ (x \minus{} (a \plus{} b\sqrt {c}))(x \minus{} (a \minus{} b\sqrt {c})) \equal{} x^2 \minus{} 2ax \plus{} (a^2 \minus{} b^2c)$.[/quote]

Did I get it right here?
I assume that the number d should be sqrt(c). It is a general fact that if d is a root of an irreducible polynomial p over F, every element of F(d) is a root of some polynomial over K with degree dividing the degree of p.
In this case, we can explicitly construct a quadratic polynomial satisfied by a+b*sqrt(c) and its conjugate, where a, b, and c are in F, based on the fact that P(x)= (x-1st root)*(x-2nd root).
P(x) = [x-(a+b*sqrt(c))]*[x-(a-b*sqrt(c))]=x^2-2ax+(a^2-(b^2)*c), which is the quadratic polynomial P(x) with coefficients 1, -2a, and a^2-(b^2)*c in F.
Thanks jmerry.

$ d$ is one of the two roots; it doesn't matter which one. You do have to verify that all numbers in the field have that form.


The remainder when $x^5-3x^2+ax+b$ is divided by $(x-1)(x-2)$ is $11x-10$. Find $a, b$.

[hide]dividing $x^5-3x^2+ax+b$ by $x^2-3x+2$ we obtain the result by $a-14=36-11$ $b-24=-10 \rightarrow a=-11$ $b=14$ looking at the remainders in the division[/hide]

[hide="hint"]
Let $g(x)$ be the quotient of the division. Then by the division algorithm we have \[ x^5-3x^2+ax+b=g(x)\cdot (x-1)(x-2)+(11x-10). \] and plug in $x=1,2$.
[/hide]


A sequence $ a_1,a_2,a_3$ is formed by selecting $ a_1$ at random from $ [1,5]$, $ a_2$ at random from $ [11,15]$, $ a_3$ at random from $ [21,25]$.

What is the probability that $ a_i$ is either an arithmetic or geometrical sequence?

[hide]For an arithmetic:

Put $ a_1\equal{}1\plus{}x, a_2\equal{}11\plus{}y, a_3\equal{}21\plus{}z$ where $ x,y,z\in[0,4]$. Then we must have

$ a_1\plus{}a_3\equal{}2a_2\iff x\plus{}z\equal{}2y$

So, within a cube defined by $ x,y,z\in[0,4]$ the space of favorable outcomes is a subset of a [i]plane[/i] $ x\minus{}2y\plus{}z\equal{}0$. However, the volume of a planar figure is zero, so is our probability.

For a geometric:

Put $ a_1\equal{}x, a_2\equal{}11y, a_3\equal{}21z$ where $ x\in[1,5], y\in[1,{15\over 11}], z\in[1,{25\over 21}]$ to get

$ a_1a_3\equal{}a_2^2\iff 21xz\equal{}121y^2$

and apply similar argument.

Therefore, since the probability of both events is zero, the probability of their disjunction is also zero.[/hide]

Do you mean that $ a_1, a_2, a_3$ are integers?  As Farenhajt describes, the problem is trivial otherwise.

Yes, trivialization. Sorry, I should have been more clear.  :oops:


Well for positive x,

$(x^{2})^{2}< x^{4}+x^{3}+x^{2}+x+1 < (x^{2}+1)^{2}$

Do something similar for negative to get that the only answer is 0.

after a few steps i have $x^{4}+x^{3}+x^{2}+x=8k,k\in\mathbb{N}$. i'm stuck there.

here's a guess: $x=3$.

[quote="Jan"]Sorry, that's not true...

$x^{4}+x^{3}+x^{2}+x+1 < (x^{2}+1)^{2}$ implies that $x^{3}+x<x^{2}$ which is obviously false.[/quote]

Computational error on my part. Nonetheless, I think it might be solvable by the same idea (squeeze the expression in between two perfect squares).

x4 + x3 + x2 + x + 1
x3(x + 1) + x(x+1) + 1
(x3 + x)(x+1) + 1
x(x2 + 1)(x+1) + 1

This expression has to equal a perfect square so you could start by setting it equal to one to cancel the other one on the other side of the equation.

x(x2 + 1)(x+1) + 1 = 1
x(x2 + 1)(x+1) = 0

Now x can equal zero or negative one.
At this point I’m stuck.

[quote="ch1n353ch3s54a1l"][quote="Jan"]This question was proposed by Belgium  :)

Find all numbers $x \in \mathbb{Z}$ for which the number \[x^{4}+x^{3}+x^{2}+x+1\] is a perfect square.

Enjoy!  :lol:[/quote]
[hide="Hmm..."]$x^{4}+x^{3}+x^{2}+x+1=n^{2}$, so $x=0$
$x(x+1)(x^{2}+1)=(n-1)(n+1)$
No term on the LHS can equal $(n-1)(n+1)$ because if it did then the other two terms would have to multiply to 1 which is impossible.
You can divide the terms on the LHS into 3 groups..(1) $x(x+1), x^{2}+1$, (2) $x(x^{2}+1), x+1$, and (3) $x, (x+1)(x^{2}+1)$. Each group has two cases where one term $=(n+1)$ and the other $=(n-1)$

So $n+1=x^{2}+1, n-1=x(x+1)$ which has solution $x=-1$

$n+1=x(x+1), n-1=x^{2}+1$ which has solution $x=3$

$n+1=x+1, n-1=x(x^{2}+1)$ which has solution $x=-1$

$n+1=x(x^{2}+1), n-1=x+1$ which has no integer solution

$n+1=x, n-1=x^{3}+x^{2}+x+1$ which simplifies to $x^{2}(x+1)=-3$,  since 3 is prime then $x^{2}=1,3$, $x^{2}=1$, $x=1,-1$ which has no solution

$n+1=x^{3}+x^{2}+x+1, n-1=x$ which simplifies to $x^{3}+x^{2}-1$ which has no solution
So the solutions are $\boxed{x=-1}, \boxed{x=3}, \boxed{x=0}$.

I dont know if this method works and it's kind of ugly anyways :dry:[/hide][/quote]

A flaw in your reasoning? Just because $x(x+1)(x^{2}+1) = (n-1)(n+1)$ doesn't mean that the factorization has to fit into one of your three cases... For example, consider $n-1=12$. Well, then we could have $x(x+1) = 2, (x^{2}+1) = 6*13$, or $x(x+1) = 3, (x^{2}+1) = 4*13$, etc...

We can also do the same thing for x. Who says that x has to be one of the factors? Say x is even. Then why can't we do
$\frac{x}{2}= n-1$, $2(x+1)(x^{2}+1) = n+1$. Obviously, some of these examples are impossible, but the point is that the logic doesn't work.

In general, it's a common mistake for people to see $abc = xy$ and say, then $ab=x$ and $c=y$, or $ac=x$ and $b=y$ etc. but this isn't necessarily true.

This question (for positive $x$) was a PROMYS application question, and it is not so bad.

[hide="Case 1"] $x$ is even.  $(x^{2}+\frac{x}{2})^{2}< x^{4}+x^{3}+x^{2}+x+1 < (x^{2}+\frac{x}{2}+1)^{2}$ [/hide] [hide="Case 2"] $x$ is odd.  $(x^{2}+\frac{x}{2}+\frac{1}{2})^{2}< x^{4}+x^{3}+x^{2}+x+1 < (x^{2}+\frac{x}{2}+\frac{3}{2})^{2}$ for $x > 3$.  Equality on the left for $x = 3$.  [/hide]

The cases for $x$ negative (edit:) is similar.


geometry
analytic geometry
circumcircle
Pythagorean Theorem

I didnt understand can you give a figure please

[quote="Andreas"]Three different points A, B, C lie on a circle [i]o[/i]. The tangent lines to [i]o[/i] at the points A and B intersect in the point P. The tangent line to o at C intersects the line AB in the point Q.
Prove that  PQ^2 = PB^2 + QC^2.[/quote]

I have solution using coordinate geometry

Let the equation to the circle is $x^2 + y^2=r^2$
P is $(x_1, y_1)$  and Q is $(x_2, y_2)$
From the given information, AB is the chord of contact of P w.r.t. the Circle.
so, its equation is given by
$xx_1+yy_1=r^2$.  Since it passes through Q, we have

$x_2x_1+y_2y_1=r^2$....(I)

$PB^2$ and $QC^2$ are the lengths of tangents from P and Q respectively, hence
are given by $x_1^2 + y_1^2-r^2$ and $x_2^2 + y_2^2-r^2$ respectively.

So, $PQ^2=(x_2-x_1)^2+(y_2-y_1)^2$ gives
$x_1^2 + y_1^2+x_2^2 + y_2^2-2(x_2x_1+y_2y_1)$
Using (I), we can write it as

$x_1^2 + y_1^2-r^2+x_2^2 + y_2^2-r^2$ which is equal to $PB^2+QC^2$

hence the proof.

Let O be the circumcenter of the triangle $\triangle ABC$, the center of the circle $o \equiv (O)$. The line OP cuts the chord AB at its midpoint F and $OP \perp AB$. Using Pythagorean theorem for the right angle triangles $\triangle PQF, \triangle PBF$,

$PQ^2 = PF^2 + QF^2,\ \ PB^2 = PF^2 + BF^2$

Subtracting and using the fact that F is the midpoint of the chord AB, i.e., AF = BF,

$PQ^2 - PB^2 = QF^2 - BF^2 = (QF + BF)(QF - BF) =$

$= (QF + BF)(QF - AF) = QB \cdot QA$

This is the power of the point Q to the circle $o \equiv (O)$, hence,

$PQ^2 - BF^2 = QB \cdot QA = QC^2$

$PQ^2 = PB^2 + QC^2$

[quote="mathclass"][quote="Palytoxin"]here is a pic that i think it describes the geometry above

[img]http://www.mathlinks.ro/Forum/album_page.php?pic_id=20[/img][/quote]

which program did you use to make that drawing? also, do you know why those pics don't display...? mine don't display either x__x[/quote]
haha i use cheap program MSPaint  :blush:
i will try to get Geometry Sketchpad
and yea, Valentin said he will come back to fix it after IMO2005


USAMTS LaTeX AMC AIME email geometry Asymptote

I have a few questions:

1) Will the answer to a problem ever be "no solutions"?

2) If you wish to use LaTeX, do you have to own it?

3) [quote="The person who made the USAMTS website"]The USAMTS is one of the ways to enter the process of selecting the USA Mathematical Olympiad team, which participates in the International Mathematical Olympiad. Participants who score well after 3 rounds will be invited to participate in the American Invitational Mathematics Examination (AIME), the second step in the process of selecting the USA Mathematical Olympiad team.[/quote]

How well do you have to score to "score well" and make it to the AIME?

Thanks in advance! :)

EDIT:

4) One last one: If we plan to use LaTeX, can we send the solutions via mail? If so, we must use the USAMTS solutions form, correct?

1. Sometimes - For an example, see the last problem set from year 18, the answer to part of the first question was that it had no solution, and the same for others. Just make sure you prove that there are no solutions.

2. No, it is a free download.

3. Last year the cutoff was 68 points after the 3rd round, it might be different this year, but that should give a general suggestion to how well.

4. I don't know.

3. It has indeed been 68 for the past two years.

4. I believe you'd be allowed to if you print the LaTeX file that you would have e-mailed. Of course, if you're going to type up your solutions on the computer, you might as well send it by e-mail.

[quote="buzzer11"]1) Will the answer to a problem ever be "no solutions"? [/quote]
[quote="monkeymatt"]1. Sometimes - For an example, see the last problem set from year 18, the answer to part of the first question was that it had no solution, and the same for others. Just make sure you prove that there are no solutions.[/quote]
monkeymatt is correct. However, note that Problem 1/4/18 part (a) that he used as an example never said that there had to be a solution. It just asked the reader to find all solutions, and in that case, the set of all solutions was the empty set. And part (b) did have solutions. In contrast, if a problem asks a reader to find a particular solution, it will have a solution.

The goal of the USAMTS is to challenge the participants mathematically. Posting misleading phrasing would interfere in that goal, so we try to be as clear and honest as possible.

Erin Schram
USAMTS grader

[quote="buzzer11"]Yes, I am a newbie. :) To USAMTS, at least.

How exactly does the email system work? I have downloaded all the files and put them in a single folder... then what do we do?[/quote]

Are those two separate questions?

How exactly does the email system work?
Email your .tex file, .pdf file, and any other pictures that you used in your solution to solutions@usamts.org.

I have downloaded all the files and put them in a single folder... then what do we do?
Start writing your solutions! In the solutions template, there are directions on where you should put your solutions. If you experience the "I can't write on the file solutions.tex" while you compile, you may need to close any existing pdf files.

Yes, those were two seperate questions, and thank you.

Does anyone have any idea how to reference points in the 18-gon when writing our solutions?

Include a diagram.  Geometry solutions without diagrams make graders sad.

do they have the diagram as they grade? for instance, if we say, "the point under X", will they know which point it is?

or i'll just include a diagram... :)

They might, but then they might not and take off.

I have another question to add.  I were to be working on the second part of a question, are all the results from the first part assumed?

[quote="buzzer11"]i would think so. im sure they wont make you prove the first part all over again.

to attach a diagram, is all you have to do make one and then attach it to the email? or do we have to reference it in some way in our solutions.tex file?[/quote]

What do you mean by "reference it in some way"?

I recommend that you make your picture a .png file, and then use the command:

\includegraphics{file.png}

if you saved it as file.png [b]in the same folder as you solutions[/b]. For more detail on how to put a picture in LaTeX, visit:
[url]http://www.artofproblemsolving.com/Wiki/index.php/LaTeX:Pictures[/url]

I don't like making separate pictures, though. I prefer asymptote.

how do we convert a file into a .png file? will the \includegraphics{} command work for other types of files (.jpg, .gif, etc)? and do we have to attach this image to the email when we send in the solutions?

It will work for a .jpg:

\includegraphics[scale=.5]{Problem2Diagram3.jpg}

And I believe you do have to attach all images you include.  It says: http://www.usamts.org/Problems/U_ProblemsSoln.php to include all image files you use.

it does it for you... if you use the solutions.tex template.

[quote="Tenoreoz"]If you're emailing solutions, do you need to post your username or name on every page?[/quote]Regardless of how you are submitting your solutions -- whether by mail, email, or fax -- you should have your name and ID# on every page.

hm i am using the solutions.tex template but it's still placing my information only on the first page. How do I fix this?

[quote="le00327146"]how do i convert a picture i made with asymptote into a .png?[/quote]

Well, if you're going to use asymptote, you might as well do it in TeXnicCenter, instead of converting them into .png format (the advantage of this is that you don't need to make your pictures separately). If you would like to learn how this works, click below:

[url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#Using_Asymptote_in_LaTeX[/url]

But, if you really want to make your pictures separately (the only thing that's nice with this is that LaTeX takes less time to compile):

[url]http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Advanced_Configuration#PDF_output_configuration[/url]

and download ImageMagick.


On the popular game show [i]The Price is Right[/i], in the showcase showdown, players spin a big wheel in turn.  This wheel has all multiples of 5 from 5 to 100.  There are 3 players, twice during the game, and each player gets to spin twice if they want.  The object is to get the closest to 100, without going over.  The two winners' scores are compared to see who will get to choose the showcase first.  What is the expected value of this person's score?

[quote="tenniskidperson3"]each player gets to spin twice if they want.[/quote]

What's the probability of this? Does each player follow their optimal strategy?

I suppose they would, so they would spin any time they get under 47.5.   :)

In order to talk about expected values we need to first fix their strategies.  There is really only one type of strategy for this game, "the $ n$ strategy": choose some $ n$.  If your first spin is less than $ n$, spin again.  (You didn't state this, but I assume that if you spin again, we add the two values that you've spun, and if you get a sum larger than 100, we set it equal to 0.  Also, that the second player has no knowledge of how the first player has done.  Otherwise, the strategies depend rather much on the order of spins -- AABB will be quite different than ABAB from the point of view of player A.)  So, we can then ask, suppose we have two players, one of whom follows the $ m$ strategy and one of whom follows the $ n$ strategy.  What is the expected value of the score of the winning player?  Is this along the lines of what you're asking?

Well, the order actually is AABBCC more games DDEEFF.  And yes, the other players know how previous players have played.  You are right in suspecting the 105-and-up rule, that it goes to 0.

[quote="tenniskidperson3"]There are 3 players, twice during the game, and each player gets to spin twice if they want.  [/quote]
What does "twice during the game" mean? :huh:


This is an open marathon very much like the 24 game marathon.

What you do is post an equation with one total on one side and 4-10 numbers that add, subtract, multiply, or divide to the total, but you take out all the signs, and the solution will be the equation with signs like this:
4  1  1  3  2 = 0

Solution: 4+1+1-3x2=0

Order of ops always applies. The answer must be with a new problem.

I'll start easy, with 3  5  6  4  2 = 10

$ 4 \cdot 2 \cdot 4 \cdot 2 \plus{} 4 \plus{} 2 \div 4 \cdot 2 \equal{} 69$, I think.

1 2 3 4 5 6 = 1

Are parentheses allowed? If so, $ (5\plus{}5) \cdot (3\minus{}2) \minus{}6\equal{}4$.

3 6 4 4 8 5 = 10

(6*5*4)^2=14400

My problem wasn't solved, you people got 47

8 2 5 10 3 5 1 2=48

[quote="ValentineA"]3 6 100 75 50 25 ~ 952[/quote]
Nobody?

[hide][youtube]6mCgiaAFCu8[/youtube]

So it’s $ [3\times(6 \plus{} 100)\times75 \minus{} 50]/25$ – and the numbers I myself gave are in their correct order. Hopefully I’m playng by the rules of the game.[/hide]


inequalities
LaTeX
calculus
integration
inequalities open

Can anyone show me a proof of the inequlaity:

0<s<r     =>

[code]
\displaystyle{  {(\sum_{i=1}^m (\sum_{j=1}^n ({A_{i,j}}^s)^{r/s})}^s
                \leq  {(\sum_{j=1}^n (\sum_{i=1}^m ({A_{i,j}}^r)^{s/r})}^r  }[/code]
What would be the condition for equality?

and i thought my latex skills were bad... :P

Theorem

0<s<r    =>   M_s^n ( M_r^m A_{i,j} ) \leq M_r^m (M_s^n A_{i,j} )

where  M_r^m  {x_k}  = (\sum_{i=1}^m (x_i)^r)^{1/r)

In particulr  when     s= -\infinity , r= +\infinity  we have the Minimax Inequality

The proof is already done by myself! :lol:

Plus:  Thanks for someone remind the inequality follows from Minkowski Inequality  :roll:

Note that the inequality can also be written into the form

${0<s<r  \Rightarrow ( {1\over m}\sum_{i=1}^m
({1\over n} \sum_{j=1}^n (A_{i,j})^s)^{r\over s} )^{1\over r}  \leq
( {1\over n}\sum_{j=1}^n
({1\over m} \sum_{i=1}^m (A_{i,j})^r)^{s\over r} )^{1\over s}  }$

The integral version of the above is equivalent to Holder's Inequality :roll:


Mr. A owns a home worth $ \$ 10000$. He sells it to Mr. B at a $ 10\%$ profit based on the worth of the house. Mr. B sells the house back to Mr. A at a $ 10\%$ loss. Then:

$ \textbf{(A)}\ \text{A comes out even} \qquad\textbf{(B)}\ \text{A makes }\$ 1100\text{ on the deal}\qquad \textbf{(C)}\ \text{A makes }\$ 1000\text{ on the deal}$ $ \textbf{(D)}\ \text{A loses }\$ 900\text{ on the deal} \qquad\textbf{(E)}\ \text{A loses }\$ 1000\text{ on the deal}$

[hide="Solution"]
Because Mr. A gets a $ 10\%$ profit, he sold his house for $ \$10000 \plus{} \$1000 \equal{} \$11000$. Now Mr. B sells it back to Mr. A at a $ 10\%$ loss which means he sells it back for $ 10\%$ less. So he sold it back at $ 11000 \minus{} 1100 \equal{} 9900$. This means that Mr. A makes $ 11000 \minus{} 9900 \equal{} \$1100$ out of the deal, so the answer is $ \boxed{(B)}$.
[/hide]


Find the one hundrdeth smallest positive integer that can be written using only the digits 1, 3, 5, 7, and 9 in base 11

in order, there are 5 that are 1-digit, then 25 that are 2-digit, then 50 3-digit numbers that start with 1 or 3, then 20 3-digit numbers that start with 51, 53, 55, or 57 -- the last of which is 579 (base 11) and is the 100th number.

[quote="pluristiq"]in order, there are 5 that are 1-digit, then 25 that are 2-digit, then 50 3-digit numbers that start with 1 or 3, then 20 3-digit numbers that start with 51, 52, 53, or 54 -- the last of which is 545 and is the 100th number.[/quote]

Don't we have $ 20$ $ 3$-digit numbers that start with $ 51,53,55,57$, with the last being $ 579_{11}\equal{}691_{10}$?


factorial
floor function
number theory
prime numbers
Divisibility Theory

Let $n$ be a positive integer with $n>1$. Prove that \[\frac{1}{2}+\cdots+\frac{1}{n}\] is not an integer.

[quote="nicetry007"]Let $ 2^{k}\;\leq \; n \;<\; 2^{k+1}$.

It is to be noted that every number $ m \;\neq\; 2^{k}$ and $ m \;\leq\;n$ has a power of $ 2$ strictly smaller than $ k$.

Suppose not. Let $ m \; = \; 2^{l}\cdot s$ where $ l \;\geq \;k$ and $ s$ is odd and $ s \;\geq\; 1$.

We have $ m \; = \; 2^{l}\cdot s \;\leq \;n\;< \;2^{k+1}\;\Rightarrow s \;< \;2^{k+1-l}\;\leq \; 2\;$ ( as $ l \;\geq\; k$) $ \Rightarrow s = 1$ (as $ s$ is odd ) $ \Rightarrow m \; = \; 2^{l}\;<\; 2^{k+1}\;\Rightarrow \; l \;\leq \; k$.

But $ l\;\geq \;k$. Hence, $ l \; = \; k \;\Rightarrow \; m \; = \; 2^{k}$, which is a contradiction as $ m \;\neq\; 2^{k}$.

Thus, we can conclude that every $ m \; \neq \; 2^{k}$ and $ m \;\leq \; n$ has a power of $ 2$ strictly smaller than $ k$.

The denominator of the sum is the $ lcm(2\;,\;3\;,\;\cdots\;,\;n)$ which has $ 2^{k}$ as a factor. Thus, when we take the sum of all the unit fractions, the numerator of every fraction other than $ \frac{1}{2^{k}}$
gets multiplied by a power of $ 2$ and the numerator of $ \frac{1}{2^{k}}$ gets multiplied by an odd number. Hence, the numerator of the sum of all the fractions is an odd number as $ odd+even \; = \; odd$.

Therefore, the sum can never be an integer as the numerator is odd and the denominator is even.[/quote]

[quote="aham"]Sorry to revive this but I remember there was a proof using Bertrand's postulate and
factorials , but I can't remember the whole thing .Can anyone post it?[/quote]

Consider

$\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{n}=\frac{\frac{n!}{1}+\frac{n!}{2}+\cdots+\frac{n!}{n}}{n!}$

By Bertrand's postulate, there exists a prime $p$ such that $\lfloor{\frac{n}{2}}\rfloor <p<n$

If $2p\le n$, we have $\lfloor{\frac{n}{2}}\rfloor\:<p \le \frac{n}{2}$ which is a clear contradiction.

So, $2p>n$

Or in other words, $p\nmid \frac{n!}{p}$

Now Note that $p|\frac{n!}{k} \forall k \in \mathbb{N}, 1\le k \le n, k\ne p$

So, ${p\nmid \frac{n!}{1}+\frac{n!}{2}+\cdots+\frac{n!}{n}}$ but $p|n!$

$\Longrightarrow \frac{\frac{n!}{1}+\frac{n!}{2}+\cdots+\frac{n!}{n}}{n!}$ is never an integer.

Let $p$ be the largest prime among 2,3,....,n. Then by Bertrand's postulate, $p<2n$. Then we factorize each $m\in [2,n]$ as a product of prime numbers: $m=p_1p_2\cdots p_r$ (here the $p_i$ may duplicate); then $p$ appears in the factorizations only once (when we factorize $p$ itself). Let $p_1,p_2,\cdots,p_k$ be all the prime numbers between 2 and n, excluding the aformentioned $p$. Then we multiply $\frac 12+\frac 13+\cdots+\frac 1n$ with $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$, and we raise the powers $e_i$ large enough to make $\frac 1m p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ an integer for all $m\neq p$. But $p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}\frac 1p$ is not an integer! So comes the conclusion.

[quote="MeKnowsNothing"][quote="gouthamphilomath"]By Bertrand's postulate, there exists a prime $p$ such that $\left\lfloor{\frac{n}{2}}\right\rfloor <p<n$
[/quote]
Be careful! There is no prime $p$ such that $\left\lfloor{\frac{1}{2}}\right\rfloor <p<1$.[/quote]

To apply Bertrand's postulate, $n \ge 7$ the others are trivial.

[quote="bobthesmartypants"]Suppose that the numerator when simplified is even. This means that the denominator is odd. This can only happen if no fraction in the original list of fractions that are added has an even denominator; this means that the list of fractions added must be a single fraction $\dfrac{1}{a}$ where $a$ is an odd number. However, the numerator would be $1$ which is odd, so we are done.[/quote]

I proved that the numerator must be odd for any sum $\dfrac{1}{a_1}+\dfrac{1}{a_2}+\cdots +\dfrac{1}{a_n}$ for consecutive integers $a_1,a_2,\cdots a_n$. However, I don't see how this can immediately be applied to this problem.

Well, we have to use nicetry007's proof for the denominator to be even?

By Bertrand's postulate,there exists a prime $p$ such that $\lfloor \frac{n}{2}\rfloor <p\le 2\lfloor \frac{n}{2}\rfloor\le n$.If $2p\le n$,then $\lfloor \frac{n}{2}\rfloor <p\le \frac{n}{2}$ which is absurd.Thus if $2\le k\le n$ and $k\neq p$,then $p\nmid k$.So ∃$a,b\in \mathbb N,p\nmid a$ s.t. $\sum_{k=2,k\neq p}^{n}\frac{1}{k}=\frac{b}{a}$. Then
$\sum_{k=2}^{n}\frac{1}{k}=\frac{b}{a}+\frac{1}{p}=\frac{pb+a}{pa}$.
Since $p\nmid pb+a$,$\frac{pb+a}{pa}$ isn't positive integer.Therefore the proof is completed.$\blacksquare$ :coolspeak:


The incircle of triangle $ABC$ touches sides $BC,CA,AB$ at $A',B',C'$. Are $AA',BB',CC'$ necessarily concurrent?

[quote="pohoatza"]yes, it`s called Gergonne`s point, a short demonstrations is with ceva[/quote]

[hide]
Using the tangent property, we have $AC' = AB'$, $B'C=CA'$, $A'B=BC'$. Thus, by Ceva, we have:

$\frac{AC'}{C'B}*\frac{A'B}{CA'}*\frac{B'C}{AB'}= 1$, so they must be concurrent.
[/hide]


Suppose $f$ is a real function such that for all $x,y$ we have $|f(x)+f(y)|=|f(x+y)|$. Then $f$ is additive.

[quote="harazi"]Suppose $ f$ is a real function such that for all $ x,y$ we have $ |f(x) \plus{} f(y)| \equal{} |f(x \plus{} y)|$. Then $ f$ is additive.[/quote]

For any pair $ (x,y)$, either $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$, either $ f(x\plus{}y)\equal{}\minus{}f(x)\minus{}f(y)$

It's rather immediate to see that $ f(0)\equal{}0$ and that $ f(\minus{}x)\equal{}\minus{}f(x)$ $ \forall x$.

Let then $ x,y$ such that $ f(x\plus{}y)\equal{}\minus{}f(x)\minus{}f(y)$
We have either $ f(x\plus{}(\minus{}2x))\equal{}f(x)\plus{}f(\minus{}2x)$, either $ f(x\plus{}(\minus{}2x))\equal{}\minus{}f(x)\minus{}f(\minus{}2x)$
1) If $ f(x\plus{}(\minus{}2x))\equal{}\minus{}f(x)\minus{}f(\minus{}2x)$, it means $ f(\minus{}x)\equal{}\minus{}f(x)\minus{}f(\minus{}2x)$ and, since $ f(\minus{}x)\equal{}\minus{}f(x)$ : $ f(2x)\equal{}0$ and so $ f(x)\equal{}0$
2) If $ f(x\plus{}(\minus{}2x))\equal{}f(x)\plus{}f(\minus{}2x)$, then :
We have then $ f(x\plus{}y\plus{}(\minus{}2x))\equal{}\epsilon_1f(x\plus{}y)\plus{}\epsilon_1f(\minus{}2x)\equal{}\minus{}\epsilon_1f(x)\minus{}\epsilon_1f(y)\plus{}\epsilon_1f(\minus{}2x)$ where $ \epsilon_1\in\{\minus{}1,\plus{}1\}$
We also have $ f(x\plus{}y\plus{}(\minus{}2x))\equal{}\epsilon_2f(x\plus{}(\minus{}2x))\plus{}\epsilon_2f(y)\equal{}\epsilon_2f(x)\plus{}\epsilon_2f(y)\plus{}\epsilon_2f(\minus{}2x)$ where $ \epsilon_2\in\{\minus{}1,\plus{}1\}$

And so, subtracting the two equalities, we have :
$ (\epsilon_1\plus{}\epsilon_2)(f(x)\plus{}f(y))\plus{}(\epsilon_2\minus{}\epsilon_1)f(\minus{}2x)\equal{}0$
Then :
2.1) either $ \epsilon_1\equal{}\epsilon_2$ and we have $ f(x)\plus{}f(y)\equal{}0$
2.2) either $ \epsilon_1\equal{}\minus{}\epsilon_2$ and we have $ f(\minus{}2x)\equal{}0$ and so $ f(x)\equal{}0$

So, $ f(x\plus{}y)\equal{}\minus{}f(x)\minus{}f(y)$ implies either $ f(x)\equal{}0$, either $ f(x)\plus{}f(y)\equal{}0$
With the same demonstration, using $ f(y\plus{}(\minus{}2y))$ instead of $ f(x\plus{}(\minus{}2x))$, we have :
So, $ f(x\plus{}y)\equal{}\minus{}f(x)\minus{}f(y)$ implies either $ f(y)\equal{}0$, either $ f(x)\plus{}f(y)\equal{}0$

Then $ f(x\plus{}y)\equal{}\minus{}f(x)\minus{}f(y)$ implies always $ f(x)\plus{}f(y)\equal{}0$ (since $ f(x)\equal{}0$ and $ f(y)\equal{}0$ imply too  $ f(x)\plus{}f(y)\equal{}0$).

So $ f(x\plus{}y)\equal{}\minus{}f(x)\minus{}f(y)$ implies always $ f(x\plus{}y)\equal{}0$ and so $ f(x\plus{}y)\equal{}\minus{}f(x\plus{}y)\equal{}f(x)\plus{}f(y)$

So we always have $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and $ f(x)$ is additive.


The minute hand of the tower clock is 12 inches long, and the hour
hand is 9 inches long. On this clock, what is the positive difference
between the number of inches traveled by the tip of the minute hand and the tip of the hour hand in one hour? Express your answer as a decimal in terms of $ \pi$.

The circumference of a circle is the diameter times pi. The length of the clock hand is the radius of the circle, which is half the diameter. The circle traveled by the minute hand is (12+12)pi=24pi. The circle traveled by the hour hand is (9+9)pi=18pi. However, in a one hour period, the hour hand only moves 1/12 of the way around the circle.

18pi/12=1.5pi
24pi-1.5pi=22.5pi

also, how do you get the pi symbol?

[quote="isabella2296"]Do you mean in latex or on the FTW?

$ \text{\LaTeX}$ : $ \pi =$ \pi

FTW: either pi or \pi[/quote]
[code]$\pi$[/code] like that?
and in FTW do u have to use the dollar signs?

On the FTW, you don't have to, but on the forum, you do.

Like on the FTW, instead of: [dollar sign]22.5\pi[dollar sign]  you could just put "22.5pi".


Does anybody know anything about 1st ROMANIAN MASTER IN MATHEMATICS 2008 ?
It  will be held in Bucharest, in February 2008.
I know that many teams are invited - I think United Kingdom, Bulgaria, Moldova, Poland,... and my Serbia.
So, who is coming ? :)

Moldova is coming (well, waiting for visas) :) Our team is:
1. Andrei Frimu (me)
2. Stefan Sanduleanu
3. Iulian Gramatki
Me and Iulian Gramatki attended IMO 2007 in Vietnam; and me and Stefan Sanduleanu attended Balkan MO 2007, in Greece; so we might meet old friends there :)
It's weird the teams have only 3 contestants each. A lot of teams are needed for organizing a high-level contest.
I heard from our leader that Russia and Hungary might come too...

See you there!

[quote="freemind"]I heard from our leader that Russia and Hungary might come too...[/quote]

Hungary? I can only hope you're right..  :maybe:

Hi - the UK is sending a team:
Tim Hennock, Jonathan Lee and Dominic Yeo (that's me). Look forward to seeing you there in two days!
Dom

Who has got the Shortlist for the first Romanian Master in Mathematics? I know one of them was used in the Romanian IMO Team Selection Test in 2008.


geometry
circumcircle
reflection
cyclic quadrilateral
IMO Shortlist
geometry solved
mixtilinear incircle

A not very nice solution uses coordinates: taking origin at M(0,0), C(2a,0),B(-2a,0), A(0,2h) then let X(p,q) with p>0 and q>0 because the minor arc MA is in the first quadrant. If T(u,v) then up+vq=0 because MX is perpendicular to MT.
After a lot of computations one get
tan(MTB-CTM) = a/h, which in fact does not depends on X(p,q).

Denote E be the midpoint of BT, then the quadrilateral XMET inscribed the circle whose diameter is XT. Since ME || CT,
$ \angle MTB \minus{} \angle MTC \equal{} \angle MTE \minus{} \angle EMT \equal{} \angle MXE \minus{} \angle EXT \equal{} \angle MXE \minus{} \angle EXB \equal{} \angle MXB \equal{} \angle MAB$

Let the perpendiculers from $ B$ and $ C$ to $ XM$ and $ TM$ respectively intersect at $ P$.
$ \angle BPC \equal{} 90^0\implies BM \equal{} CM \equal{} PM$. So $ P$ is the reflection of $ C$ at $ TM$ and of $ B$ at $ XM$.
$ XT \equal{} XB \equal{} XP\implies$ $ X$ is the circumcenter of $ TBP$.
$ \angle MTB \minus{} \angle CTM \equal{} \angle MTB \minus{} \angle PTM \equal{} \angle BTP \equal{} \frac {1}{2}\angle BXP \equal{}$
$ \equal{} \angle BXM \equal{} \angle BAM\ \ \text{(Const.)}$

Νotice here that the condition AB=AC is absolutely redundant.... Hence the problem can be generalized to any triangle ABC,using the same method of proof as mr.danh's idea,which is also the same as mine.

Cheers,
Nick

Sorry for the revival; why is XMET cyclic in mr.danh's solution?

[quote="mr.danh"]then the quadrilateral XMET inscribed the circle whose diameter is XT[/quote]

[quote="AwesomeToad"]Sorry for the revival; why is XMET cyclic in mr.danh's solution?
[/quote]

$TX=BX$
so, $\triangle TXB$ is isosceles
since $E$ is the midpoint
$XE$ is the median as well as altitude of the triangle
So, $\angle TEX=90^\circ$
it is given $\angle TMX=90^\circ$
So, $TEMX$ is cyclic with $TX$ being the diameter

[hide="Solution"]
Let $N$ be the midpoint of $BT$. Since $M$ is the midpoint of $BC$, we have $MN||CT$. Therefore, $\angle{CTM}=\angle{TMN}$.

Since $TX=BX$, $\triangle{BXT}$ is isosceles, so $\angle{TXN}=\angle{BXN}$ and $XN\perp BT$. Then $\angle{TNX}=90=\angle{TMX}$, so $XMNT$ is cyclic. Thus, $\angle{MTB}=\angle{MXN}$ and $\angle{CTM}=\angle{TMN}=\angle{TXN}=\angle{BXN}$. Finally, observe that from cyclic quadrilateral $ABMX$,
\[ \angle{BAM}=\angle{BXM}=\angle{MXN}-\angle{BXN}=\angle{MTB}-\angle{CTM}.\]
Because $\angle{BAM}$ is independent of $X$, $\angle{MTB}-\angle{CTM}$ is also not dependent on the choice of $X$. // [/hide]

Let $C' $ be the symetrical of $C$ wrt $TM$  and $L=AM\cap TC'$ and $Y$ the reflexion of $X$ wrt the midpoint of $AB$. and $E$ the midpoint of $TB$
then $ \angle{MTB}-\angle{CTM} = \angle LTB$
it is enough to show that $ATLB$ is cyclic to conclude that $\angle{MTB}-\angle{CTM} = \angle MAB$
we have: $\angle ALB=\frac{\pi}{2}-\angle TMB - \angle MTC = \frac{\pi }{2}- \angle AYX -\angle TME = \frac{\pi}{2}-\angle AYX -\frac{\angle TXB}{2}=\angle XBT -\angle ABX =\angle ABT $
Hence $ATBL$ is cyclic which leads to
\[ \angle{MTB}-\angle{CTM}= \angle LTB = \angle MAB \]

Here is an analytic solution:

We use complex numbers and set $(ABM)$ to be the unit circle. Furthermore, let $a=-1,b=1$. Then it is clear that $c=2m-1$. Now since $TM \perp XM$, we have \[\frac{t-m}{x-m}=-\frac{\bar t - \frac{1}{m}}{\frac{1}{x}-\frac{1}{m}} \implies t-m=mx \bar t - x \implies  \bar t = \frac{t-m+x}{mx}\] Now since $XT=XB$, we have \[(t-x)(\bar t - \tfrac{1}{x})=(1-x)(1-\tfrac{1}{x}) \implies t \bar t - \frac{t}{x}- \bar t x + x +\frac{1}{x} -1=0\] Plugging in our first equation into our second, we get \[t\left(\frac{t-m+x}{mx}\right)-\frac{t}{x}-\left(\frac{t-m+x}{m}\right) + x + \frac{1}{x} -1=0 \implies t^2 - 2mt + (m-1)x^2+m=0\] Now let $r$ be the root that lies within angle $AMB$ and call the other root $s$. It is clear that these roots are reflections about $M$ (and thus $r,m,s$ are collinear), so now consider the following complex number which has argument $\angle MTB - \angle CTM$ \[\frac{\frac{s-r}{1-r}}{\frac{2m-1-r}{s-r}}=-\frac{(s-r)^2}{(1-r)(1-s)}=\frac{4m^2-4((m-1)x^2+m)}
{(m-1)(1-x^2)}=\frac{4(m-x^2)}{1-x^2}\] Now notice that since $|x|=1$, $x^2$ lies on the unit circle, and so we see that the argument of this complex number is $\angle MXB=\angle MAB$ which is independent of $X$, as desired. $\Box$

We do not use complex numbers

[hide=Solution]
Let $N$ be the midpoint of $BT$. Since $XB=XT$, $\angle XNT=90^{\circ}=\angle XMT$, so $XMNT$ is cyclic, yielding $\angle BTM=\angle NTM=\angle NXM$.
Now $MN$ is a midline in $\triangle BCT$, so
\[ \angle MTC=\angle BTC - \angle MTC=\angle BNM-\angle MXN=\angle MXT-\angle MXN=\angle NXT=\angle NXB \]
Hence, $\angle BTM-\angle CTM=\angle BXN - \angle MXN=\angle BXM=\angle BAM$ which is fixed.
[/hide]

When I started working on this problem, I misread it as $TB=TX$. If we characterize $T$ as such, can we find any interesting properties of this new configuration? All my progress so far for this configuration is rather trivial.

[hide=Solution]
Let $N$ be the midpoint of $BT$. Then $MN \parallel CT$. Also, $XN \perp NT$, so $XMNT$ is cyclic. Then \[\begin{aligned} \angle MTB - \angle CTM &= \angle MTB - (\angle CTB - \angle MTB) \\ &= 2\angle MTB - \angle CTB \\ &= 2\angle MTN - \angle MNB \\ &= 2\angle MXN - (\pi - \angle MNT) \\ &= 2\angle MXN - \angle MXT \\ &= 2\angle MXN - (\angle MXN + \angle NXT) \\ &= \angle MXN - \angle NXT \\ &= (\angle MXB + \angle BXN) - \angle NXT \\ &= (\angle MXB + \angle NXT) - \angle NXT \\ &= \angle MXB \\ &= \angle MAB \end{aligned}\] which does not depend on $X$.
[/hide]

My solution:(similar MStang and mr.danh)
Let $Q$ be the midpoint of $BT,$ then $\angle XQT =90,$ also $\angle XMT=90\to XMQT$ is cyclic.
Also $MQ$ is the midline of $\triangle CBT\to MQ\parallel CT.$
Then $\angle MTB-\angle CTM=\angle MXQ-\angle QMT=\angle MXQ-\angle QXT=\angle MXQ-\angle BXQ=\angle MXB=\angle MAB.$ As desired.

Let $N$ be the midpoint of $BT$. Now since $M$ is the midpoint of $BC$, we hace $MN||CT$ so $\angle CTM = \angle TMN$. Also $\angle TNX = 90 = \angle XMT$, hence $\omega_{TNMX}$ exists. So coupled with this, we get $\angle CTM = \angle TMN  =  \angle TXN$ (call this $\spadesuit$)

Now, $\angle BTM $ $$= \angle BTX - \angle XTM = \angle BTX + \angle TXM - 90 = \angle TXM + (\angle BTX - 90)$$ $$= \angle TXM - \angle TXN = \angle NXM = \angle NXB + \angle BXM = \angle TXN + \angle BXM \overset{\spadesuit}{=} \angle CTM + \angle BAM$$, hence $\angle BTM - \angle CTM = \angle BAM$, independent of $X$, as desired.

Let $N$ be [i]not[/i] the midpoint of $BT$. ;)

[hide=Solution]Let $N$ be the reflection of $T$ across point $M$ and note that $BNCT$ is a parallelogram. Let $\angle{TNX}=\beta$ and $\angle{BNT}=\angle{NTC}=\alpha$. Then since $XT=XN$ and $M$ is the midpoint of $TN$, $\angle{XTM}=\angle{XNM}=\beta$. Since $XT=XB=XN$ is the radius of the circumcircle of $\triangle{TBN}$, $\angle{BXT}=2\angle{BNT}=2\alpha$. Since $\triangle{BTX}$ is isosceles, $\angle{BTX}=90^{\circ}-\alpha$ so $\angle{BTN}=\angle{BTX}-\angle{NTX}=90^{\circ}-\alpha-\beta$. So we want to find the value of $\angle{BTN}-\angle{CTN}=90^{\circ}-2\alpha-beta$.

Since $\triangle{XNT}$ is a right triangle, $\angle{TXN}=90^{\circ}-\beta$ and $\angle{BTX}=2\alpha$ so $\angle{BXM}=\angle{TXN}-\angle{BTX}=90^{\circ}-\beta-2\alpha=\angle{BAM}$ as $BMXA$ is cyclic. Therefore, $\angle{BTM}-\angle{CTM}=\angle{BAM}$, which is independent of $X$.[/hide]

[quote=delegat]Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$.

Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$.

[i]Author: Farzan Barekat, Canada[/i][/quote]
Let $N$ be the midpoint of $AB.$ Define $\omega,\Omega$ to be the circles $(AMB), \mathcal{C}(X,XB).$ Let $Y$ be the antipode of $X$ in $\omega.$ Clearly $T \in \overline{MY}, T \in \Omega$ by the conditions. Also, $\measuredangle XBY=\pi/2$ implies that $YB$ is tangent to $\Omega.$

[color=#f00]Claim:[/color] $\measuredangle YBT=\measuredangle CTM.$
[color=#00f]Proof:[/color] Let $L$ be the reflection of $T$ over $M.$ Then $TBLC$ is a parallelogram. Further, since $TM=ML$ and $\measuredangle XMT=\pi/2,$ hence $TX=XL.$

So, $L \in \Omega.$ Since $YB$ is tangent to this circle, hence $\measuredangle YBT=\measuredangle BLT=\measuredangle CTL.$ $\square$
-------
Thus we have
$$\measuredangle MTB-\measuredangle CTM=\measuredangle MTB-\measuredangle YBT=\measuredangle TYB=\measuredangle MAB$$
which is independent of the position of $X.$ $\blacksquare$
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[hide=Solution]Spent Whole day solving $BX=BT$  :wallbash:
[quote=ISL 2007 G2]Denote by $ M$ midpoint of side $ BC$ in an isosceles triangle $ \triangle ABC$ with $ AC = AB$. Take a point $ X$ on a smaller arc $ \overarc{MA}$ of circumcircle of triangle $ \triangle ABM$. Denote by $ T$ point inside of angle $ BMA$ such that $ \angle TMX = 90$ and $ TX = BX$.

Prove that $ \angle MTB - \angle CTM$ does not depend on choice of $ X$.
[/quote]
[b][color=#000]Solution:[/color][/b] Let $T'$ be reflection of $T$ over $M$ $\implies$ $X$ is center of $\odot (BTT')$. Hence, $$ \angle BTM-\angle CTM=\angle BXM-\tfrac{1}{2} \angle BXT-\angle BT'T=\angle BXM=\angle BAM \qquad \blacksquare$$
[/hide]

The structure of the most un-elegant solution in this topic.

I will compute these step by step:

1) Let $a,b,c$ be the side lengths of $\Delta ABC$.We can compute $AM$ by Pythagoras.

2) Let $AX=x$, We can compute $BX$ by Pythagoras.

3) We can find $MX$ by Ptolemy over $ABMX$.

4)We can compute $MT$ by Pythagoras over $\Delta MTX$($TX=BX$ will be useful here.)

5) Note that $\angle XMA =\angle TMB$. The side lengths of $\Delta MXA$ is known, hence both the angles is known.

6) By cosine rule we can find BT and TC and hence $\angle BTM \& \angle CTM$.

7) If we subtract these angles, we will get a relation in which there is not $x$ ! That means it is independent of $X$.

I know it is near impossible to do but It is the first thing that came to my mind :-D (Geo is a lie without bash)

[asy]
size(9cm);
defaultpen(fontsize(10pt));
defaultpen(linewidth(0.4));

pair A = dir(55), B = dir(235), M = dir(305), C = 2M-B, X = dir(355), D = dir(175), T = intersectionpoint(D--M,arc(X,abs(B-X),90,225)), E = 2M-T;

dot("$A$", A, dir(55));
dot("$B$", B, dir(235));
dot("$M$", M, dir(270));
dot("$C$", C, dir(20));
dot("$X$", X, dir(5));
dot("$D$", D, dir(175));
dot("$T$", T, dir(210));
dot("$E$", E, dir(280));

draw(A--B--C--A--M--X^^unitcircle, purple);
draw(T--X^^E--X^^B--X, blue);
draw(T--B--E--C--T, orange);
draw(X--A--D--2B-D^^D--E, heavygreen);
draw(arc(circumcenter(T,B,E),abs(circumcenter(T,B,E)-T),165,315), magenta);
[/asy]
Let $\overline{MT}$ meet $(ABM)$ again at $D$, and $E$ be the reflection of $T$ over $M$.

[b]Claim:[/b] $X$ is the circumcenter of $(TBE)$.

[i]Proof.[/i] Note that $\triangle XMT \cong \triangle XME$ by SAS, so $TX = EX$. Also we are given that $TX = BX$ so we are done. $\square$

In addition, $\angle DBX = \angle DMX = 90^\circ$, so $\overline{DB}$ is tangent to $(TBE)$. It follows that \begin{align*}\angle MTB - \angle CTM &= \angle MTB - \angle BEM = \angle MTB - \angle DBT \\ &= \angle MDB = \angle BAM,\end{align*} which does not depend on $X$ as desired. $\blacksquare$

Solution from [i]Twitch Solves ISL[/i]:

Construct parallelogram $CTBS$ whose diagonals meet at $M$. Also, let $N$ be the midpoint of $\overline{BT}$.
[asy] pair A = dir(180); pair B = dir(0); pair M = dir(35); pair C = 2*M-B; pair X = dir(100);

filldraw(A--B--C--cycle, invisible, red); draw(A--M, red); filldraw(unitcircle, invisible, blue);

pair Y = -X; draw(CP(X, B), mediumgreen); pair T = IP(M--Y, CP(X, B)); pair S = 2*M-T; draw(S--T, mediumgreen); draw(B--X--M, mediumgreen);

pair N = midpoint(B--T); draw(B--T, deepgreen); draw(T--X--S, mediumgreen); draw(X--N, deepgreen);

filldraw(circumcircle(X, M, T), invisible, dotted+deepgreen);

dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$M$", M, dir(M)); dot("$C$", C, dir(C)); dot("$X$", X, dir(X)); dot("$T$", T, dir(T)); dot("$S$", S, dir(S)); dot("$N$", N, dir(N));

/* TSQ Source:

A = dir 180 B = dir 0 M = dir 35 C = 2*M-B X = dir 100

A--B--C--cycle 0.1 lightred / red A--M red unitcircle 0.1 cyan / blue

Y := -X CP X B mediumgreen T = IP M--Y CP X B S = 2*M-T S--T mediumgreen B--X--M mediumgreen

N = midpoint B--T B--T deepgreen T--X--S mediumgreen X--N deepgreen

circumcircle X M T 0.1 yellow / dotted deepgreen

*/ [/asy]
We first eliminate $C$ from the diagram by noting that \[ \angle CTM = \angle MSB = \angle TSB = \frac{1}{2} \angle TXB = \angle NXB.  \] Also, noting that $XMTN$ are cyclic (as $\angle XMT = \angle XNT = 90^{\circ}$), we have \[ \angle MTB = \angle MTN = \angle NXM. \] Thus $\angle MTB - \angle CTM = \angle NXM - \angle NXB = \angle MXB$ which is fixed.

Let $N$ be the midpoint of $BT$. Observe that since $\angle XMT = \angle XNT = 90$, the quadrilateral $XMNT$ is cyclic. Furthermore, $NM \parallel CT$.

On the one hand, note that $$\angle MTB = \angle MTN = MXN.$$

On the other hand, note that $$\angle CTM = \angle NMT = \angle NXT = \angle NXB.$$ Now, note that $$\angle MTB - \angle CTM = \angle MXN - \angle NXB = \angle BXM = \angle BAM$$ which does not depend on $X$, as desired.

[asy]
		unitsize(3cm);
		pair A, B, C, M, T, T1, X;

		A = dir(90);
		B = dir(210);
		C = dir(330);
		M = (B + C) / 2;
		X = rotate(20, (A + B) / 2) * M;
		T = IP(CP(X, B), Line(M, rotate(90, M) * X, 10), 0);
		T1 = 2 * M - T;

		draw(A--B--C--A^^unitcircle, heavyred);
		draw(circumcircle(A, B, M), heavygreen);
		draw(CP(X, T), blue);
		draw(X--T^^X--B^^X--M^^X--T1^^T--T1, orange);

		dot("$A$", A, N);
		dot("$B$", B, SW);
		dot("$C$", C, SE);
		dot("$M$", M, S);
		dot("$X$", X, E);
		dot("$T$", T, NW);
		dot("$T'$", T1, SE);
[/asy]

Let $T'$ be the reflection of $T$ over $M$. Note that $TBT'C$ is a parallelogram. Additionally, as $\angle TMX = 90^\circ$, $\triangle XTT'$ is isosceles, so we have $XT = XB = XT'$. Hence, $X$ is the circumcenter of $\triangle BTT'$. Now,
\begin{align*}
	\angle BTM - \angle CTM &= \angle BTT' - \angle CTT' \\
	&= \angle BTT' - \angle BT'T \\
	&= \frac{1}{2}(\angle BXT' - \angle BXT) \\
	&= \frac{1}{2}((\angle BXM + \angle MXT) - (\angle MXT - \angle BXM)) \\
	&= \frac{1}{2}(2\angle BXM) \\
	&= \angle BAM,
\end{align*}
which is fixed, as desired. $\Box$

[quote=AMN300]We do not use complex numbers

[hide=Solution]
Let $N$ be the midpoint of $BT$. Since $XB=XT$, $\angle XNT=90^{\circ}=\angle XMT$, so $XMNT$ is cyclic, yielding $\angle BTM=\angle NTM=\angle NXM$.
Now $MN$ is a midline in $\triangle BCT$, so
\[ \angle MTC=\angle BTC - \angle MTC=\angle BNM-\angle MXN=\angle MXT-\angle MXN=\angle NXT=\angle NXB \]
Hence, $\angle BTM-\angle CTM=\angle BXN - \angle MXN=\angle BXM=\angle BAM$ which is fixed.
[/hide] yes I think the locus of T is a circle passing through $A,M$ and midpoint of $AB$
 ;)

When I started working on this problem, I misread it as $TB=TX$. If we characterize $T$ as such, can we find any interesting properties of this new configuration? All my progress so far for this configuration is rather trivial.[/quote]

Dunno if this has been posted before.

Complex bash with $a=-1, b=1$ so that $|m|=|x|=1$ and $c=2m-1$. Since the foot from $T$ to $\overline{MX}$ is $M$, $$m=\frac{1}{2}(t+m+x-mx\overline{t}) \implies \overline{t}=\frac{t-m+x}{mx}.$$ Since $\overline{TX}=\overline{BX}$ we should have $\frac{t-x}{b-x}$ on the unit circle, and upon substituting our expression for $\overline{t}$ $$\frac{t-x}{b-x}=\frac{\frac{1}{b}-\frac{1}{x}}{\overline{t}-\frac{1}{x}} \implies t^2-2mt+x^2m+m-x^2.$$ Let $r$ be the unwanted root. By Vieta's, it follows that the argument of $\angle{MTB}-\angle{CTM}$ is (for $k \in \mathbb{R}$) $$\frac{\frac{m-t}{1-t}}{\frac{2m-1-t}{m-t}}=k_1 \cdot \frac{\frac{r-t}{1-t}}{\frac{r-1}{r-t}}=k_1 \cdot \frac{(r-t)^2}{-(rt-r-t+1)} = k_1 \cdot \frac{4m^2-4m-4x^2m+4x^2}{-(x^2m-x^2-m+1)} = k_2 \cdot \frac{m-x^2}{1-x^2}$$ or $\frac{m-x^2}{1-x^2}$. Since $x^2 \in (ABM)$, this is equivalent to $\angle{MXB}=\angle{MAB}$ which does not depend on $X$, done.

Construct $N$ the midpoint of $TB$. Since $\overline{MN} \parallel \overline{TC}$, we can calculate \begin{align*} \angle MTB - \angle MTC = \angle MTN - \angle TMN = \angle MXN - \angle NXB = \angle MXB = \angle MAB, \end{align*} which is constant done yay kill me now

misread this problem for a good 50 minutes :D

Solved with [b][url=https://artofproblemsolving.com/community/user/599649]MrOreoJuice[/url][/b] and [b][url=https://artofproblemsolving.com/community/user/785149]Fakesolver19[/url][/b].
[asy]
//meh asy, as always...
size(210);
pair A=(0,6.5),B=(-3,0),C=(3,0),M=(0,0),T=(-3.246,3.0711);
pair X=(1.828,1.9321), TT=(3.246,-3.0711);
draw(A--B--C--cycle, orange);
draw(circumcircle(A,B,M), orange);
draw(T--B--TT--C--cycle, magenta);
draw(X--T, red);
draw(X--B, red);
draw(X--TT, red);
draw(X--M, orange);
draw(circumcircle(B,T,TT), red);
draw(T--TT, magenta);
dot("$A$",A,N);
dot("$B$",B,W);
dot("$C$",C,E);
dot("$M$",M,S);
dot("$T'$",TT,S);
dot("$X$",X,NE);
dot("$T$",T,NW);
[/asy]
Let $T'$ be the reflection of $T$ over $M$. Note that $TBT'C$ is a parallelogram and $X$ is circumcenter of $(TBT')$.
\begin{align*}
    \angle MTB - \angle CTM &= \angle T'TB - \angle TT'B \\
    &= \dfrac{\angle T'XB - \angle TXB}{2} \\
    &= \dfrac{\angle TXT' - 2\angle TXB}{2} \\
    &= \dfrac{2\angle TXM - 2\angle TXB}{2}\\
    &= \dfrac{2\angle BXM}{2}\\
    &= \angle BAM
\end{align*}
which does not depend on $X$ as desired. $\blacksquare$

Let $N$ be the midpoint of $BT$. We have that $\angle XMT=\angle XNT=90$ so $XMNT$ is cyclic. Also $CT\parallel MN$ thus $\angle CTM=\angle TMN$. In turn $\angle TMN=\angle TXN=\angle BXN$. Also $\angle MTB=\angle MXN$, so we can get that $$\angle MTB-\angle CTM= \angle MXN-\angle BXN = \angle MXB=\angle MAB$$ which doesn't depend on $X$.

Denote by $N$ midpoint of $BT.$ Clearly $CT\parallel MN, XN\perp BT,$ hence $$\angle BTM-\angle CTM=\angle NTM-\angle NMT=\angle NXM-\angle NXT=\angle BXM=\angle BAM.$$

Let $N$ be midpoint of $TB$. $BXT$ is isosceles so $\angle XNT = 90 = XMT$ so $XMNT$ is cyclic.
$\angle MTB - \angle MTC = \angle MXE - \angle EXT = \angle MXE - \angle BXE = \angle MXB = \angle MAB$ and $\angle MAB$ is fixed and independent from $X$.
we're Done.

Let $N$ be the midpoint of $\overline{BT}$ and note $MNTX$ is cyclic and $\overline{MN}\parallel\overline{CT}.$ Hence, $\angle BTM=\angle BXM$ and $$\angle MTC=\angle NMB=\angle TXN=\angle NXB$$ so $$\angle MTB-\angle CTM=\angle BXM=\tfrac{1}{2}\angle A.$$ $\square$

All angles are in degrees.

Let $MT$ intersect the circle again at $U$. Note that $\angle XBU = \angle XMU = 90$, thus $AXBU$ is a rectangle. Since $\angle XBM + \angle TMB = \angle XBM + \angle UMB = \angle XBM + \angle ABX = \angle MBA$, we have $$\angle MTB = 180 - \angle TBX - \angle XBM - \angle TMB = 180 - \angle MBA - \angle TBX.$$

Let $N$ be the midpoint of $BT$. Notice that $\angle XMT = \angle XNT = 90$, so $XMNT$ is cyclic. Since $MN\parallel CT$, we have $$\angle MTC = \angle TMN = \angle TXN = 90 - \angle XBT.$$ This implies that $\angle BTM - \angle MTC = 90 - \angle MBA$, a constant value.

[img width=50]https://media.discordapp.net/attachments/986049049834713139/986428810968825886/ISL2007G2.png[/img]Let $N$ denote the midpoint of $BT.$ Then, $\angle TNX=\angle TMX=90^\circ$ so $TNMX$ is cyclic. Then, $\angle MTB-\angle CTM=\angle NTM-\angle TMN$ because $NM || TC.$ Now, $\angle NTM-\angle TMN=\angle NXM-\angle TXN=\angle BXM=\angle BAM$ which is fixed so we're done.

Let $T_1$ be the reflection of $T$ in $M$, ray $MT$ meet $(ABM)$ again at $Y$, and the circle centered at $X$ with radius $XB = XT$ meet $(ABM)$ again at $Q$.

Because $XM \perp TT_1$, we know $XT = XT_1$, so $BTQT_1$ is cyclic. Now, observe $$\angle YBX = \angle YMX = \angle TMX = 90^{\circ}$$ which implies $YB$ is tangent to $(BTQT_1)$. Similarly, we deduce that $YQ$ is tangent to $(BTQT_1)$, so $\overline{TT_1Y}$ is the $T$-Symmedian of $BQT$.

Now, since $M$ is the midpoint of $TT_1$, we know it's the $T$-Dumpty point of $BQT$, yielding $MBT \overset{+}{\sim} MTQ$. Thus, because $BTCT_1$ is clearly a parallelogram, $$\angle MTB - \angle CTM = \angle MQT - \angle BT_1T$$ $$= \angle MQT - \angle BQT = \angle MQB = \angle MAB$$ which finishes. $\blacksquare$

[b]Config Stuff:[/b] Let ray $XT$ meet $(ABM)$ again at $P$. The Incenter-Excenter Lemma implies $T$ is the incenter of $BPQ$. Moreover, since $\angle XMT = 90^{\circ}$, we know the $P$-Mixtilinear Incircle of $BPQ$ touches $(BPQ)$ at $M$. In fact, $MBT \overset{+}{\sim} MTQ$ is actually a well-known mixtilinear lemma.

Let $D$ be the midpoint of $BT$. So $TDMX$ is cyclic and $MD\parallel CT$. So the desired angle quantity is $\angle BAM$ which is fixed.

Let $E = \overline{TM} \cap (X)$. Notice that $\angle MTB - \angle CTM = \angle MTB - \angle BET = \frac 12(\widehat{BE} - \widehat{TB})$ because $TCEB$ is a parallelogram. However, setting $D = \overline{XM} \cap (X)$ as the arc midpoint of $\widehat{TE}$, we have $$\frac 12(\widehat{BE} - \widehat{TB})= \widehat{BD} = \angle BXM = \angle BAM$$ is fixed regardless of $X$, as needed.

time to go through my writeup list :P

Let $N$ be the midpoint of $BT$, with $XMNT$ cyclic (since $\angle TNX=\angle TMX=90^\circ$) and $MN \parallel TC$ (since $MN$ is the $B$-midsegment in $\triangle BTC$). Now we angle chase:
\[\angle MTC=\angle NMT=\angle TXN=\angle NXB\]
and $\angle BTM=\angle NXM$, so we are done since $\angle BTM - \angle MTC = \angle BXM = 90 - \angle ACM$. $\square$

[quote]When I started working on this problem, I misread it as $TB=TX$. If we characterize $T$ as such, can we find any interesting properties of this new configuration? All my progress so far for this configuration is rather trivial.[/quote]
LOL i made the exact same mistake on ggb, even though i READ TX=BX I didn't construct it properly and constructed it as on the circle with radius BX but with center B, so i was like "why is this not working"

My solution is the same as above, but without typos: @above I think it should be MTC=NMT=..., and it should also be BTM=NXM, hence BTM-MTC=BXM=BAM, which is independent.

Let $\overline{MT}$ intersect $(ABM)$ again at $X'$ (the $X$-antipode) and the circle $\omega$ centered at $X$ passing through $B$ and $T$ at $D$. Then since $\overline{XM} \perp \overline{TD}$, $M$ is the midpoint of chord $\overline{DT}$, hence $BDCT$ is a parallelogram. On the other hand, since $\angle X'BX=90^\circ$, $\overline{X'B}$ is tangent to $\omega$, hence $\angle X'BT=\angle BDT=\angle CTM$, hence $\angle MTB-\angle CTM=\angle BX'M=\angle BAM$ which is fixed. $\blacksquare$

Let $S$ be such that $SBTC$ is a parallelogram. Then
\[2(\angle MTB-\angle CTM)=\angle BXS-\angle BTX=(\angle BXM+\angle MXS)-(\angle MXT-\angle BXM)=2\angle BXM=2\angle BAM\]
done.

Let Omega be circle with center X and radius XB.
Let TM intersect Omega at Y.
Let P  be point on Omega(Arc TBY) such that Angle PTY=CTY
After that observe TY||BP (Angle chase for this)
We are done


Suppose that there are n people, and they each have their favourite numbers between 1 and n. Person #1 starts at 1 and moves towards n, and takes his favourite number. Person #2,... repeats the steps. If his favourite number is already taken, then he would move on and take the next available number. However, if all the numbers between the person's favourite number and n are taken, the person gets angry and leaves. Find the number of pairs of the (person, favourite number) so that no one gets angry and every number gets taken.

[Edit] Note that, two people may have the same favourite number. For instance, the arrangement (n, f(n)), where f(n) is the favourite number of nth person, may be (1,5), (2,5), ..., (6,5), (7,5) (assuming n=10). Then this arrangement is not possible, because the 2nd person would take 6, 3rd would take 7, etc., and 6th taking 10. Then the 7th person has nothing to take, so he leaves for good.

I hope I have made the problem more understandable.

Sorry, I haven't been checking on my problems lately.

Maxal, do you want to share your solution? The reason I posted this problem is because it has an ingenious way of solving it, which I can't remember anymore (maybe only vaguely).

As soon as I remember mine, I'll post it as well, in maybe a week or so, so that others can try it as well.


A wooden raft (density=775 kg/m^3) is supporting a 72.7 kg man entirely above the water when he stands on it at Lake Travis. Find the minimum volume of the raft.

I'm so happy to finally see an easy problem! :)


It might be a little early for me to be thinking about this, but should I go straight to grad school after college?  Is it all right to work for maybe 2 years, and then apply for grad?  Or will it be more difficult that way?

I have read some graduate stats for schools in the ivy league and such and it seems that quite a few people take time off before continuing on to graduate school.  I would assume this is the same throughout most schools.  What are you studying, by the way?

I might be doing Physics and CS...
College will (hopefully) determine which field I want to go into, though.  I'm very indecisive.

You have (at least) 4 years -- don't worry about it yet.


[quote="Vutang"]Given $a, b, c$ are three real numbers such that : $a <b<c$ ; $a+b+c=6$;$ab+bc+ac=9$. Prove that: $0<a<1<b<3<c<4$
[/quote]

[hide]As $a<b<c$ it is clear that $c>2$
$a+b=6-c$
$ab+c(a+b)=9$ we see that $a,b$ satisfy
$a+b=6-c$ and $ab=(c-3)^2$ so $a$ and $b$ are the roots of $f(x)=x^2+(c-6)x+(c-3)^2=0$ the discriminant is $D=-3c(c-4)$ $D>0$ when $c<4$
$c$ is outside the roots of $f(x)=0$ so $f(c)>0$ but $f(c)=3(c-1)(c-3)$ so $c>3$
As $f(1)=(c-1)(c-4)$ we see that $f(1)<0$ so $a<1<b$
$ab=(c-3)^2>0$ and $b>1$ we conclude that $a>0$ and as $a+c>3$ $b<3$
[/hide]


For each positive integer $k>1$, define the sequence ${a_{n}}$ by
$a_{0}=1$ and $a_{n}=kn+(-1)^{n}a_{n-1}$ for each $n \geq 1$.

Determine all values of $k$ for which 2000 is a term of the sequence.

I think that [hide="k can also equal"]87[/hide] :maybe:

You're right.  I'll outline my solution:
[hide]So you can prove by induction or whatever that if
$\ n \equiv 0 \mod 4 \rightarrow a_{n}=nk+1$
$\ n \equiv 1 \mod 4 \rightarrow a_{n}=k-1$
$\ n \equiv 2 \mod 4 \rightarrow a_{n}= (n+1)k-1$
$\ n \equiv 3 \mod 4 \rightarrow a_{n}= 1$

So, clearly when $\ n \equiv 3 \mod 4$ $\ a_{n}$ can never be 2001.
Also, if $\ n \equiv 0 \mod 4$ then $\ 2000=4mk+1$ however $\ 4$ does not divide $\ 1999$ so $\ 2000$ will not appear when $\ n$ is divisible by $\ 4$.
So, when $\ n \equiv 1 \mod 4$ then $\ a_{n}= k-1$ so if $\ k=2001$ $\ 2000$ will appear in the series.
When $\ n$ is in the form $\ 4m+3$ then we can right $\ 2000=(4m+3)k-1$ or $\ 2001=(4m+3)k$.  Now the factors of $\ 2001$ which appear in the form $\ 4m+3$ are $\ 3, 23, 87, 667$ and the corresponding $\ k$ values are $\ 667, 87, 23, 3$
So, $\ k= 3, 23, 87, 667, 2001$
 [/hide]


function
calculus
algebra
domain
linear algebra
calculus computations

Let g: A - B, f: B - C, f o g: A - C be functions.
If f and f o g are one to one, does it follow that g is one to one?

Suppose $g$ is not 1-1. Then there exists $x,y\in A$ such that $g(x)=c=g(y)$ for some $c\in B$. So $f(g(x))=f(c)$ and $f(g(y))=f(c)$, but this is a contradiction since $f\circ g$ is 1-1. Therefore $g$ is 1-1.

...I didn't use $f$ is 1-1. Do you only need $f\circ g$ to be 1-1 to get $g$ 1-1?

Yes, you only need $f \circ g$ to be injective.
Proof: choose $a_{1}, a_{2}\in A$. Suppose $g(a_{1}) = g(a_{2})$, we wish to show that $a_{1}= a_{2}$. Applying f to $g(a_{1}) = g(a_{2})$, and use the fact that this composition is injective, we get $a_{1}= a_{2}$, as it was required. So you did not need f to be injective.

This is clearly not calculus. This is elementary "functionspeak." Problems of this ilk also sometimes show up in the linear algebra forum. I'd move them, except I'm stumped as to where to move them to.

At my university at least, the first time you would see this material and this terminology would be in a second year analysis course (it is also covered in a third-year abstract algebra course).

I think it's fine where it is?

[quote="blahblahblah"]At my university at least, the first time you would see this material and this terminology would be in a second year analysis course (it is also covered in a third-year abstract algebra course).[/quote]
An abstract algebra course is a bit late in the game for this. I'm using some of this language right now teaching a (college sophomore level) linear algebra course; we also quite explicitly deal with it in a sophomore level "transition" course (discrete structures/introduction to proof and abstraction.) Some questions one might ask - say about counting certain kinds of functions between finite sets - would properly belong to the combinatorics forum here. Ideally, students should come to a basic understanding of this language, and be able to write short proofs and construct counterexamples, before they even start calculus - but in the real world, that's a lot to ask.

It's really a matter of vocabulary: relation, function, domain, codomain, range, injective (1-1), surjective (onto), composition, identity map, inverse function.


LaTeX
number theory proposed
number theory

Let $ k_1,k_2,...,$k_304 be the positive integers satisfying $ k_1^3\plus{}k_2^3\plus{}...\plus{}$(k_304)^3≤2003.
Finh the largest value of A=$ k_1\plus{}k_2\plus{}k_3\plus{}...\plus{}$k_ 304

[quote="New player"]:blush: The answer is 304*3√2003/304[/quote]
New player, use $ LaTeX$
Also $ k_1, k_2, \dots \dots ,k_{304}$ are positive integers.


Each of seven boys went to a cafe 3 times, such that every pair met there. Prove that at some moment 3 of the boys were at the cafe.

Take boy 1.
He meets the 6 other boys in his 3 visits to the cafe.
So in at least one of his visits, he meets more than one of the boys.
And at that time, there are 3 boys at the cafe.

Or have i misread the question? Do you mean [b]exactly[/b] three boys in the cafe?

You understood the problem well, but the thing is, I thought of the same solution and it's wrong. Namely, while one of the 3 boys is in the cafe, one of the other 2 might have come and left before the third one arrived.
______       ___________
     ____________  Here it is visually, reperesented on time line, where easch of these 3 lines represents the time each of the three boys were present.

Consider the intervals when boys are in cafe as line-segments on a line. Consider the 21 left ends of this segments. If two boys meet they also meet in some of this 21 points(not hard to prove). If some of this points coincide then we may think that there are 20 points with this condition. If not take the leftmost point. Obviously there is no meeting there so again we may think the points are 20. But there are at least $ \binom{7}{2}= 21$ meetings so some two meetings coincide, i.e. some 3 boys are in the cafe at the same time.


summer program
PROMYS
Ross Mathematics Program

Ive got into both PROMYS, SSP, and the ROSS (unfortunately no RSI).
which one should it be?

[quote="xxreddevilzxx"]Ive got into [b]both[/b] PROMYS, SSP, and the ROSS (unfortunately no RSI).
which one should it be?[/quote]

 :P

PROMYS and ROSS are more math related than SSP, ( i think ;)  ),

go to the one that gives you more money, so you have to pay less :lol:

Well Qiaochu will be at PROMYS, Edward at Ross, so maybe you'll see one of them there.

well im still undecided to attend PROMYS or ROSS.
i was wondering where Young will go?

Young is going to PROMYS I believe.  They've offered him a good chunk of money.

Yeah...I'm going somewhere else.  You Washingtonians have fun!!!  And say no to drugs!  :ninja:

Actually i might be going somewhere else. My schedule just totally changed 360 degrees due to some personal reasons.

[quote="xxreddevilzxx"]somewhere else. [/quote]

as in? :P

Ewww... I don't want anybody like Young or Amir to be at Ross with me. They're just icky. :P And we all know that QC's a loser for choosing PROMYS over SIMUW, since SIMUW rules over all else.  :D

[quote="Uh oh!"]Ewww... I don't want anybody like Young or Amir to be at Ross with me.  :P [/quote]

err...you're the one who made me apply in the first place!! :D


induction
inequalities
strong induction
number theory unsolved
number theory

the problem like following is far from my ability maybe masters would help :)
Put $a_k=p_k^2-p_k,k\geq 2$.($p_k$ is kth prime number)then there exists an absolute constant $A$ so that every even integer greater than $A$ is the sum of distinct $a_k$'s.

Wow, that's too beautiful! Thanks for the dedication, but this problem seems far too much for my poor head.

Here's a way to reduce the problem to a finite problem.  Suppose we can find an even integer $A$ and an integer $k_0 \ge 2$ with the following properties:

(1) $A+2$, $A+4$, $A+6$, ..., $A + a_{k_0}$ are representable (as the sum of distinct $a_k$'s).

(2) $2 a_k > a_{k+1} + A$ for every integer $k \ge k_0$.  (Presumably we can prove such a bound using the Prime Number Theorem or something similar.)

Then I claim that every even integer $n$ bigger than $A$ is representable.  The proof is by strong induction on $n$.  The cases $n \le  a_{k_0} + A$ follow from property 1.  So assume $n > a_{k_0} + A$.  Let $j$ be the maximum integer such that $a_j + A < n$.  We know that $j \ge k_0$.  By Property 2 and maximality, we have
\[
  2 a_j > a_{j+1} + A \ge n \, .
\]
By induction, $n - a_j$ is representable.  Because $2 a_j > n$, the representation of $n - a_j$ cannot use $a_j$.  Thus $n = (n - a_j) + a_j$ is representable.  That completes the inductive proof.

Alas, I haven't done the computations needed to find $A$ and $k_0$.

Okay, I did some computations on an Excel spreadsheet.  The choice $A = 3106$ and $k_0 = 19$ seems to work.  We have $p_{k_0} = 67$ and $a_{k_0} = 4422$.  The spreadsheet says that 3108, 3110, 3112, ..., 7528 are representable, so Property 1 is true.  The inequality $2 a_k > a_{k+1} + 3106$ is true for the first few $k \ge 19$, and can presumably be shown for all $k \ge 19$ because $p_{k+1} / p_k$ (and thus $a_{k+1}/a_k$) approaches 1.

Edit: I found a way to weaken Property 2 (at the expense of strengthening Property 1).  That should make it easier to prove Property 2.  Here are the new properties.

(1') The numbers $A+2$, $A+4$, $A+6$, $A + a_{k_0}$ are representable with the numbers $a_2$, ..., $a_{k_0 - 1}$.

(2') $a_{k+1} \le 2 a_k$ for all $k \ge k_0$.

According to my spreadsheet, the new Property 1' is true for $A = 3106$ and $k_0 = 17$.  So we just need to prove Property 2' that $a_{k+1} \le 2 a_k$ for all $k \ge 17$.  In fact, the inequality seems to be true for all $k \ge 5$.  We can probably prove it by using inequalities such as $p_{k+1} \le 1.3 p_k$ for all sufficiently large $k$.


function
algebra
floor function
number theory open
number theory

[quote="ertanrock"]If $ x\in R$ then find all solution of the given floor - function $ [\frac {2x \minus{} 1}{3}] \plus{} [\frac {4x \plus{} 1}{6}] \equal{} \frac {5x \minus{} 4}{3}$[/quote]

$ 5x\equal{}3 ([\frac {2x \minus{} 1}{3}] \plus{} [\frac {4x \plus{} 1}{6}] ) \plus{}4\equal{}a \in \mathbb{Z} \Rightarrow x\equal{}\frac{a}{5}$
$ a\equal{}3 ([\frac {4a\minus{}10}{30}] \plus{} [\frac {4a\plus{}5}{30}]) \plus{}4$

$ a\equal{}30m\plus{}n$ with $ 0 \le n < 30$
$ 30m\plus{}n\equal{}3 ([\frac {4(30m\plus{}n)\minus{}10}{30}] \plus{} [\frac {4(30m\plus{}n)\plus{}5}{30}]) \plus{}4$
$ 30m\plus{}n\equal{}24m\plus{}3 ([\frac {4n\minus{}10}{30}] \plus{} [\frac {4n\plus{}5}{30}]) \plus{}4$
$ 6m\plus{}n\equal{}3 ([\frac {4n\minus{}10}{30}] \plus{} [\frac {4n\plus{}5}{30}]) \plus{}4$
taking mod $ 3$ then $ n \equiv 1 (3)$ therefore only solutions for $ n \in \{1,4,7, 10,28 \}$
Thus $ x \in \{\frac{1}{5},\frac{4}{5},\frac{7}{5},2,\minus{}\frac{2}{5} \}$

[quote="mszew"][quote="ertanrock"]If $ x\in R$ then find all solution of the given floor - function $ [\frac {2x \minus{} 1}{3}] \plus{} [\frac {4x \plus{} 1}{6}] \equal{} \frac {5x \minus{} 4}{3}$[/quote]

$ 5x \equal{} 3 ([\frac {2x \minus{} 1}{3}] \plus{} [\frac {4x \plus{} 1}{6}] ) \plus{} 4 \equal{} a \in \mathbb{Z} \Rightarrow x \equal{} \frac {a}{5}$
$ a \equal{} 3 ([\frac {4a \minus{} 10}{30}] \plus{} [\frac {4a \plus{} 5}{30}]) \plus{} 4$

$ a \equal{} 30m \plus{} n$ with $ 0 \le n < 30$
$ 30m \plus{} n \equal{} 3 ([\frac {4(30m \plus{} n) \minus{} 10}{30}] \plus{} [\frac {4(30m \plus{} n) \plus{} 5}{30}]) \plus{} 4$
$ 30m \plus{} n \equal{} 24m \plus{} 3 ([\frac {4n \minus{} 10}{30}] \plus{} [\frac {4n \plus{} 5}{30}]) \plus{} 4$
$ 6m \plus{} n \equal{} 3 ([\frac {4n \minus{} 10}{30}] \plus{} [\frac {4n \plus{} 5}{30}]) \plus{} 4$
taking mod $ 3$ then $ n \equiv 1 (3)$ therefore only solutions for $ n \in \{1,4,7, 10,28 \}$
Thus $ x \in \{\frac {1}{5},\frac {4}{5},\frac {7}{5},2, \minus{} \frac {2}{5} \}$[/quote] great solution.


I'm taking Sequences and Series at nationals, so I was studying and I've found that certain number sequences appear like Pentagonal numbers, Catalan numbers, and Lucas numbers.  If someone knows offhand several types of numbers and the formulas to find them, it'd be greatly appreciated.  Or at least compiling a list of different types of numbers

Try searching on [url=http://mathworld.wolfram.com/]MathWorld[/url].


Topsoil costs $ \$6$ per cubic foot. What is the cost, in dollars, of
5 cubic yards of topsoil?

Well there are 3x3x3=27 cubic feet in a cubic yard. Thus there are 5x27=135 cubic feet, so its 6x135=810.


geometry
trigonometry
function
algebra
polynomial
quadratics
analytic geometry

Hey guys,
i'm 8th grade and about preparing for the euclid,
i have these books:
"AoPS 2"
"Challenging Problems in Geometry"
Alfred S. Posamentier
"Contest Problem Books (ahsme) 1-4"
How to Solve It"
G. Polya
"Five Hundred Mathematical Challenges"
Edward J. Barbeau
"Mathematics of Choice"
Ivan Morton Niven
"Introduction to Geometry"
Coxeter

Are they helpful?

And of course i did the old tests and my marks ranged from 90-100 and an 86. (I know on the real contest, i'll do crap :( )

Anyway, thanks for helping :)

i'm sure thats enough..lol..u'll probably do fine if u can tackle most of the questions in past euclid contests

I found out that I can write the euclid as well as the hypatia  :D

But what would you guys think the best topics to review would be, theorems that kinda thing?

[quote="max_tm"]For Euclid? This is the official list of topics (and it's pretty thorough IMO):

-Euclidean and Analytic Geometry
-Trigonometry, including functions, graphs, identities, sine and cosine laws
-Exponential and logarithmic functions
-Functional notation
-Systems of equations
-Polynomials, including relationships involving the roots of quadratic and cubic equations, the remainder theorem.
-Sequences and series
-Simple counting problems
-Properties of numbers

Basically anything you'd see in the high school basics forum extending into a bit of intermediate forum stuff.[/quote]
i dont get anything by looking at the topics :rotfl:

i already knew the topics and saw through old contests and most of the difficulty of the topics were kind of basic.

thanks anyway


[quote="Quevvy"][quote="myspacesucks"]A clock chimes two times 15 minutes after the hour, four times 30 minutes after the hour and 6 times 45 minutes after the hour.  The clock also chimes eight times on each hour in addition to chiming the number of times equal to the hour.  (So at 2:00 p.m. the clock chimes 8+2=10 times).  Starting at 12:05 a.m., how many times does the clock chime in a 24 hour period?[/quote]
[hide="There is probably a shorter way..."]For every hour (x:00-x:59), the number of chimes is $2+4+6+8+x=20+x$
Then you solve for each value (1-12)
The values are then 21,22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32
$21+22+23+24+25+26+27+28+29+30+31+32=6(21+32)=6(53)$
Multiply by 2 for am/pm to get a final answer of $318*2=\boxed{636}$[/hide]
That's an annoying clock![/quote]
[hide="I have a much easier way"]Each hour, excluding the clock chiming for the number of hours in the time, there are $2+4+6+8=20$ chimes. Since there are $24$ hours, there are $480$ chimes. Next we account for the clock chiming for the number of hours in the time. It is not hard to check that it is: $2(1+2+...+12)=156$. Add these two up to get a final answer of $636$. [/hide]

[quote="Ubemaya"]You do realize that your ways are pretty much the same, right?[/quote]

I'm just curious, but: Ubemaya, which method would you use?

I realize that they are practically the same ways, but I'm more of a "casework" kind of person.

Uhh... well if I didn't read your solutions I probably couldn't have solved this.

You guys are right, the answer is 636, Vishalarul, your method is pretty much the same with a minute difference in the middle when you add them together.  Nice job you guys.