[hide="1"]
$n^{2}-n=n(n-1)$. Since $n$ and $n-1$ are consecutive integers, one of them is even and so $2|n(n-1)$[/hide]
[hide="2"]
$n^{3}-1=n(n^{2}-1)$. If $3$ does not divide $n$ then $n=3k+1$ or $3k-1$. In either case $3|n^{2}-1$. Also, $2|n$ or $2|n^{2}-1$. Hence $6|n(n^{2}-1)$[/hide]

[hide="3"]\begin{eqnarray*}n^{5}-n &=& n(n^{2}-1)(n^{2}+1)\\ &=& (n-1)n(n+1)(n^{2}-4+5)\\ &=& (n-2)(n-1)n(n+1)(n+2)+5(n-1)n(n+1)\end{eqnarray*}

Among five consecutive integers, at least one is divisible by $2$, at least one by $3$, exactly one by $5$, hence the first term is divisible by $30$

Among three consecutive integers, at least one is divisible by $2$ and exactly one by $3$, hence the second term is divisible by $30$.[/hide]

[quote="Astronum"]5. If n is odd and not a multiple of 3, then $6|n^{2}-1$[/quote]

[hide="5"]
test values mod 6... $n^{2}-1$ is 0 mod 6 when n is 1 or 5 mod 6

that or just realize that $n^{2}-1$ = $(n+1)(n-1)$ and one of them has to be 0 mod 6 in order for it to be divisible by 6 (one divisible by 2 and one divisible by 3 won't work because they both have to be of the same parity)... I know I'm not explaining it well but it isn't too difficult if you play around with it.
[/hide]


Albert Pujols had 250 at-bats during the first half of a baseball season.  In those at-bats his hitting average was .404.  His batting
average during the second half of the season was .320.  If Albert's
total season batting average was .360, how many at-bats did Albert have during the second half of the season?

[b]Note: A batting average is calculated as the ratio of the number of hits a player had to the total number of at-bats the player had.[/b]

How come the total season batting average isn't the hitting average of the first season plus the hitting average of the second season divided by 2? (It says .360 but I came out with .362)
I don't watch baseball, so I'm missing out on all the basises.

[quote]How come the total season batting average isn't the hitting average of the first season plus the hitting average of the second season divided by 2? (It says .360 but I came out with .362)
[/quote]

Because he may have had a different number of at-bats in ech season. For example, if a batter had 200 at-bats during the first season with a batting average of .100, and a batting average in the second season of .900, but only 1 at-bat, would his average be .500?

[quote="Fierytycoon"]For example, if a batter had 200 at-bats during the first season with a batting average of .100, and a batting average in the second season of .900, but only 1 at-bat, would his average be .500?[/quote]

Well, if he had 1 at-bat, his batting average couldn't be .900  :wink:

[quote]Well, if he had 1 at-bat, his batting average couldn't be .900 [/quote]

Hehe, whoops.  :oops: Well, let's say that he had 10 at-bat's then.

TheSouthpaw wrote:HINT: USE ALGEBRA

[hide]
I do believe the answer is 275[/hide]

If you could, please show the algebra you used behind spoiler text.  This is a nice little introductory statistics problem and I would like people who have not seen a problem like this one to see a well-engineered solution.

One last question (and it's a stupid one, as usual): Is the total batting average calculated by dividing the sum of each of the batting average by its number of hit by the total hit number?[/hide]

1) Your questions aren't stupid.
2) The total batting average is the number of hits divided by the number of bats. If we know that the batting average from the first a bats is x, and the batting average from the next b bats is y, can you figure out what the total batting average is?


modular arithmetic
number theory
relatively prime
number theory unsolved

Determine all positive integers $n$ such that if $a, b \in \mathbb N$ and if $n$ divides $a^2b + 1$ then $n$ divides and $a^2 + b$

Answer: $n = 2^i 3^j 5^k$, where $i \in \{0, 1, 2, 3, 4\}, j \in \{0, 1\}, k\in \{0, 1\}$.

Proof:

First note that to achieve the congruence $a^2b \equiv -1 \pmod n$, we're free to choose $a$ to be any value relatively prime to $n$; there is then a $b$ (unique) that makes the congruence true.

Let $p$ be a prime.

$a^2b \equiv -1 \pmod {p^k} \Longrightarrow a^2 \equiv -b^{-1} \pmod {p^k} \Longrightarrow a^2+b \equiv -b^{-1} + b \pmod {p^k},$

and

$-b^{-1}+b \equiv 0 \pmod {p^k} \Longleftrightarrow b \equiv b^{-1} \pmod {p^k} \Longleftrightarrow b^2 \equiv 1 \pmod {p^k} \Longleftrightarrow (-a^{-2})^2 = a^{-4} \equiv 1 \pmod {p^k} \Longleftrightarrow a^4 \equiv 1 \pmod {p^k}.$

now, in the case of $p \neq 2$, if $k > 1$ or if $p > 5$ then $\phi(p^k) = p^{k-1}(p-1) > 4$, so we can choose $a$ to be a primitive root $\pmod {p^k}$ and in that case $a^4 \neq 1 \pmod {p^k}$. so if $n$ has the required property, $n$ cannot be divisible by any prime $> 5$ and can have at most one factor of each of $3, 5$. on the other hand, since $a^4 \equiv 1 \pmod 3$ and $a^4 \equiv 1 \pmod 5$ if $\gcd(a, 3) = \gcd(a, 5) = 1$, $n$ can indeed be divisible by $3$ or $5$ (but not by $3^2$ or $5^2$).

next observe that if $\gcd(a, 2) = 1$ then $a^4 \equiv 1$ modulo any of the numbers $2, 2^2, 2^3, 2^4$. So $n$ can have any of these factors as well. On the other hand, if $n$ were divisible by $2^5$, we could choose $a = 29 \equiv - 3 \pmod {32}$, for which $a^4 \equiv (-3)^4 = 81 \neq 1 \pmod {32}$, so $n$ can't be divisible by $32$.


Given (x,y) is a point in the equation (x^2)+(y^2)=1, which would be the graph of the set consisting of all points (2x,y)?

(a)  an ellipse containing points (1/2, 0), (-1/2,0), (0,1), (0,-1)
or
(b)  an ellipse containing points (2,0), (-2,0), (0,1), (0,-1)

If you plug in the point $ (2x,y)$ into the equation you get $ 4x^{2}\plus{}y^{2}\equal{}1$, from which it is clear that the first option is correct.

JRav is completely wrong.  $ (1, 0)$ is on the circle, so $ (2, 0)$ is on the "stretched circle" that you've got there.

A brief look at the equation and the options shows that option $ (b)$ is the only one which satisfies the conditions. If $ 2x \equal{} \frac {1}{2}$, then $ x \equal{} \frac {1}{4}$ and $ x^{2} \equal{} \frac {1}{16}$, so $ y \neq 0$, and option $ (a)$ is incorrect.

[quote="Differ"]In other words, you divide by four and not multiply.[/quote]

Yes, and here's how you could have figured it out on your own (if you weren't given points on the curves, say):  We have two curves, one satisfying $ x^2 \plus{} y^2 \equal{} 1$ and one satisfying $ (t, w) \equal{} (2x, y)$.  How do we find out the equation that $ (t, w)$ should satisfy?  Well, we have $ t \equal{} 2x$, so $ x \equal{} \frac {t}{2}$.  And we have $ w \equal{} y$.  And we know an equation that relates $ x$ and $ y$.  So substitute in to see that $ (t, w)$ must satisfy the equation $ \left(\frac {t}{2}\right)^2 \plus{} w^2 \equal{} 1$.

If you want, at the end you can replace $ t$ by $ x$ and $ w$ by $ y$ again, but it's a [i]very[/i] good idea to use a second set of variables for the computation -- if you start off with $ (x, y) \mapsto (2x, y)$ or something, you're liable to get very confused.  Using other variables makes it clear who needs to be substituted in for what, etc.


Prove that

$ (\frac{1}{m_a}\plus{}\frac{1}{m_b}\plus{}\frac{1}{m_c})(\frac{1}{w_a}\plus{}\frac{1}{w_b}\plus{}\frac{1}{w_c})\geq\frac{18}{s}$

Dimensionless or Scale Unbalanced.
There are such cases,Often.
BQ


inequalities
logarithms
function
limit
real analysis
real analysis unsolved

Find the value of $a,c$ and determine the range of $b$ such that the following inequality holds for $x\geq 0$.

\[\frac{3x}{2x+3}\leq \frac{x+a}{bx+c}\leq \ln (1+x)\]

$a=0 , c=1 , \frac 12 \leq b \leq   \frac 23$
 :cool:  not too bad for a Medical University ! :cool:

You are right, Diogene.

I want you to show your solution. :)

kunny

[quote="kunny"]I want you to show your solution. :) [/quote]
$x=0 \Longrightarrow 0 \leq \frac{a}{c}\leq \0 \Longrightarrow a=0$
$\frac{3x}{2x+3}\leq \frac{x}{bx+c}\Longleftrightarrow \frac{3}{2x+3}\leq \frac{1}{bx+c},\forall x > 0 \Longleftrightarrow  b \leq \frac 23 ,c\leq 1$
$\frac{x}{bx+c} \leq ln(1+x) \leq x \Longrightarrow \frac{1}{bx+c}  \leq 1,\forall x > 0 \Longrightarrow  c \geq 1 $, So $c=1$
$\frac{x}{bx+1} \leq ln(1+x) \Longleftrightarrow b  \geq \frac{1}{ln(1+x)}-\frac{1}{x},\forall x > 0$
But the function $\ \frac{1}{ln(1+x)}-\frac{1}{x}\ $ is decreasing and $\ \displaystyle \lim_{x\rightarrow 0}(\frac{1}{ln(1+x)}-\frac{1}{x})=\frac 12 $, so $b \geq \frac 12$
Conclusion : $a=0 , c=1 , \frac 12 \leq b \leq \frac 23 $
:cool:

That's true.Thank you, Diogene. :surf:

kunny


vector
inequalities proposed
inequalities

$ a_1\plus{}a_2\plus{}a_3\plus{}a_4\plus{}a_5\equal{}0$ and $ |a_i\minus{}a_n|\leq1$ for $ i,n\equal{}1,2,3,4,5$
Find an upper-bound for $ S\equal{}a_1^2\plus{}a_2^2\plus{}a_3^2\plus{}a_4^2\plus{}a_5^2$
I am able to prove that $ S\leq2$ but there should be a better bound since the largest example I can find is $ 0.6^2\plus{}0.6^2\plus{}(\minus{}0.4)^2\plus{}(\minus{}0.4)^2\plus{}(\minus{}0.4)^2\equal{}1.2$
Larger examples and proves for bounds are welcome.

[quote="panag"]$ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5 \equal{} 0$ and $ |a_i \minus{} a_n|\leq1$ for $ i,n \equal{} 1,2,3,4,5$
Find an upper-bound for $ S \equal{} a_1^2 \plus{} a_2^2 \plus{} a_3^2 \plus{} a_4^2 \plus{} a_5^2$
I am able to prove that $ S\leq2$ but there should be a better bound since the largest example I can find is $ 0.6^2 \plus{} 0.6^2 \plus{} ( \minus{} 0.4)^2 \plus{} ( \minus{} 0.4)^2 \plus{} ( \minus{} 0.4)^2 \equal{} 1.2$
Larger examples and proves for bounds are welcome.[/quote]

Wlog we can suppose $ a_1\ge\dots\ge a_5$. For a vector $ \vec a$ where $ S$ is maximal, there is obviously a $ 1\le k<5$ such that $ a_1\equal{}\dots\equal{} a_k\equal{}: a$ with $ 0<a<1$ and $ a_{k\plus{}1}\equal{}\dots\equal{} a_5\equal{}a\minus{}1$.  Carry on from there and you will see that your bound is best possible. :)


calculus
integration
logarithms
function
calculus computations

is it possible to calculate this integral:
$ \displaystyle\int{\frac{e^x}{x}dx}$

if it is, what is the result and why?
if it isn't, why not?

[b]Solution:[/b]
[hide]There is no simple solution to it, but you can do something like:

$ \int \frac{e^x}{x} ~dx \equal{} \int \frac{1\plus{}x\plus{}\frac{x^2}{2!}\plus{}\frac{x^3}{3!} \plus{} \cdots}{x} ~dx$
$ \equal{}\int \left( \frac{1}{x}\plus{}1\plus{}\frac{x}{2!}\plus{}\frac{x^2}{3!} \plus{} \cdots \right) ~dx$
$ \equal{}\ln |x| \plus{} x \plus{} \frac{x^2}{2 \cdot 2!} \plus{} \frac{x^3}{3 \cdot 3!} \plus{} \cdots \plus{} C$
$ \equal{}\boxed{\ln |x| \plus{} \sum_{n\equal{}1}^{\infty} \left( \frac{x^n}{n \cdot n!} \right) \plus{} C}$[/hide]

The antiderivative is not an elementary function- this is one of many well-known examples of that phenomenon.


number theory
relatively prime
algebra
partial fractions

I was looking on one math problem and all I need to do was evaluate this:

$\sum_{n=4}^{2004}  \frac {2}{2n+n(n-3)}$ and the result equals to $\frac {m}{n}$ where m and n are relatively prime integers.

Can anyone tell me how to evaluate this in both calculator and without calculator?

Thanks so much!

An important method in manipulating fractions is partial fractions:
[hide]
Notice that [tex]\displaystyle\frac {2}{2n+n(n-3)}[/tex] can be written as [tex]\displaystyle\frac{A}{n}+\frac{B}{n-1}[/tex], so we have solve for the numerators:

[tex]\displaystyle An-A+Bn=2[/tex]
[tex]\displaystyle -A=2\rightarrow A=-2[/tex]
[tex]\displaystyle -2n+Bn=0n\rightarrow B=2[/tex]

[tex]\displaystyle \sum_{n=4}^{2004} \frac {2}{2n+n(n-3)}=\displaystyle \sum_{n=4}^{2004}\frac{-2}{n}+\frac{2}{n-1}[/tex]
[tex]\displaystyle \sum_{n=4}^{2004} \frac {2}{2n+n(n-3)}=(\frac{-2}{4}+\frac{2}{3}+\frac{-2}{5}+\frac{2}{4}+...+\frac{-2}{2004}+\frac{2}{2003}[/tex]

The summation is a telescopes, with the only remaining terms becoming [tex]\frac{2}{3}+\frac{-2}{2004}[/tex]

The desired answer is [tex]\frac{667}{1002}[/tex].
[/hide]

[hide="Code for Ti-89"]
F3--->4::Sigma: ( sum
2/(2*x+x*(x-3)),x,4,2004)

So that the line reads:
:Sigma: (2/(2*x+x*(x-3)),x,4,2004)
[/hide]

You need to know partial fraction decomposition.  Despite the long name it's a fairly easy process to learn.  I suggest googling it.  If you still don't get it ask again on the forum.  But it would also have been helpful to write the sum as:

\[\frac{2}{n(n-1)}.\]

Basically partial fraction decomposition is the opposite of adding fractions i.e. you are decomposing them.

[quote="kool_dudy"]When I was looking at the solution, I didn't understand and how towers came up with

[hide]$\frac{A}{n}$ + $\frac{B}{n-1}$[/hide]

to solve for the solution.[/quote]

[hide="Here Is How"]$\frac {2}{2n+n(n-3)}=\frac{2}{n^2-n}=\frac{2}{n(n-1)}=\frac{2n-2n+2}{n(n-1)}=\frac{2n-2(n-1)}{n(n-1)}=\frac{-2}{n}+\frac{2}{n-1}$[/hide]


linear algebra
matrix
linear algebra unsolved

As we know ,there are three different elementary transformations of matrix, they are "[color=red]change two distinct rows[/color]", "[color=red]multiply a non-zero number to a row of the matrix[/color]","[color=red]add a k times of a row to another row,where k is a number in the field and non-zero[/color]".

Obviously,we can use the last two to get the first elementary transformation("change two distinct rows"), [color=red]so whether the first elementary transformation is unnecessary?[/color]
Another question is : [color=red]what is the geometric meaning of these elementary transformations of matrix? [/color]

Could you carefully show us the "obvious" bit?

Let the rows be $ a$ and $ b$. We proceed from $ a$ and $ b$ to $ a$ and $ a\plus{}b$ (add first row to second row), which leads to $ \minus{}b$ and $ a\plus{}b$ (add negative second row to first row), which leads to $ \minus{}b$ and $ a$ (add first row to second row), which leads to $ b$ and $ a$ (multiply first row by $ \minus{}1$).

Row interchanges aren't "necessary" in that we can simulate row interchanges with the above more complicated procedure. However, they make such a procedure more easy to follow; the above procedure doesn't really have much use if our final goal is just to interchange the rows.


Show that $ \phi(n) \equal{} n \minus{} 1$, then $ n$ is prime.

[hide]There are a total of $ n\minus{}1$ positive integers less than $ n$. And from the given condition we conclude that none of them share any factor with $ n$. Thus by the definition of a prime number we conclude that $ n$ is prime.[/hide]


Prove that any odd-degree polynomials has at least one real root.

[hide="hint"]let  $P(x)$ be the polynomial of odd degree then $lim_{x\to\-\infty}P(x)=-\infty$and$lim_{x\to\+\infty}P(x)=+\infty$ hence use intermediate value theroem  where the first $\infty$ is $-\infty$[/hide]

Yes, I know. But there are other solutions? An how to prove $lim_{x\to\-\infty}P(x)=-\infty$?

$p(x)=\sum_{i=0}^{n}a_{i}x^{i}=x^{n}\sum_{i=0}^{n}a_{i}x^{i-n}$
where n is the degree of the polynomial


superior algebra
superior algebra unsolved

Find all finite groups which have exactly two conjugacy classes.

[hide=" ex:"] $ \mathbb{Z}_2$[/hide]

And that's it. The number of elements in the conjugacy class of $ g\in G$ is the index of the centralizer of $ g$, which is a divisor of $ |G|$. And $ |G| \minus{} 1$ does not divide $ |G|$ unless $ |G| \equal{} 2$.

thnx. Thats what I thought but my book seems to have written " find all groups"  and the plural bothered me.

Using the singular form would be giving the answer away.  :)


I think $ n$ must be positive interger, $ k$ is positive  and $ RHS \equal{} n( \sqrt [n]{\frac {k \plus{} 1}{k}} \minus{} 1)$
$ \Leftrightarrow \frac { kn \plus{} 1}{kn} \plus{} \frac {kn \plus{} 2}{kn \plus{} 1} \plus{} ....\frac {(k \plus{} 1)n}{ (k \plus{} 1)n \minus{} 1} \ge n \sqrt [n]{\frac {k \plus{} 1}{k}}$

Which is trivial.


This is my situation. My friend let me borrow Art and Craft by Zeitz, albeit no solutions. But then Ive heard a lot of good things about Engel's book (gogo StellarSupremacy and Kalva). My birthday is coming up, so its not a problem getting it. Im just interested in knowing which one people here (who have tried both of em) consider to be better. And even if you havent used both of them, list the pros and cons of the one you have used please. Also, I only plan on devoting myself to one of them, because in the other time, I also have a full schedule of AP courses + other math to do.
Thanx in advance.

I really like Zeitz, although I haven't seen Engel.  I like the general problem solving philosophy stuff at the beginning of Zeitz, and then it has plenty of good problems to back it up.  I think the standard criticism of Zeitz is that it's thin on geometry, but you can always get that somewhere else.  I'm not sure how well I can judge it, because I already had a decent background in problem solving when I got Zeitz.  And it's a really pretty book :-)  I think it is pretty expensive, although I don't know about the competition.

if you want to do lots of questions and checking solutions, pick Engel.  If you dont want that, pick Zeitz.  If you dont care... hmm... pick Engel cuz eventually you will want to do some questions and check there solutions.

Jelyman,

Keep in mind that Engel is rather challenging. When I first picked it up, I had a heck of a time just digesting the first chapter on invariance.

In the long run though, Engel is certainly worth your time, because it builds up olympiad level skills. Its my training book for the USAMO :D


For $a,b,c$ positive real numbers prove that

$\frac{(a^5+b^5+c^5)-(a+b+c)^5}{(a^3+b^3+c^3)-(a+b+c)^3} \leq \frac{10}{9}(a+b+c)^2 $

Rearrange the given to get:

$9a^5 +9b^5 + 9c^5 - 9(a+b+c)^5 \leq 10(a+b+c)^2(a^3+b^3+c^3)-10(a+b+c)^5$

$9a^5 +9b^5 + 9c^5 + (a+b+c)^5 \leq 10(a+b+c)^2(a^3+b^3+c^3)$

After expanding and rearranging we have:

$\displaystyle \sum_{sym}{15a^4b} \geq \sum_{sym}{30a^2b^2c}$

This clearly follows from Muirhead since $(4,1,0)$ majorizes $(2,2,1)$. Thus the inequality is proven. $\blacksquare$

We can prove that
        $(a+b+c)^3 - a^3 - b^3 - c^3 = 3(a+b)(b+c)(c+a)
          (a+b+c)^5 - a^5 - b^5 - c^5 = (a+b)(b+c)(c+a)(k(a^2+b^2+c^2)+l(ab+bc+ca) $
   Plug 2 triples (a,b,c) to find k and l  and we will obtain a simpler inequality. :D


[quote="N.T.TUAN"][quote="Inspired By Nature"]Happy Year to everyone! :lol:[/quote]

Thank you very much !  :D[/quote]

I LOVE NEW YEAR!!

HAPPY NEW YEAR EVERYONE  :lol:


Παιδιά το παρακάτω πρόβλημα δώθηκε σην Α Λυκείου και μου το μετέφερε ένας μαθητής του τμήματος. Αν δεν υπάρχει κανένα πρόβλημα στα δεδομένα ( δεν την έχω κοιτάξει ακόμη ) πρόκειται για μια υπέροχη άσκηση ... :)

Δείτε το attach

ps.  Δεν ξέρω αν η θέση της είναι εδώ ή στις σχολικές ασκήσεις αλλά πιστεύω ότι θα μπορούσε να είναι και θέμα σε κάποιο διαγωνισμό ...
Ο κύριος Μπάμπης ή ο Σιλουανός μπορούν να τη τοποθετήσουν όπου πιστεύουν ότι ταιριάζει περισσότερο .

πολυ ωραια ασκηση.βρηκα μια λυση με τριγωνομετρια.
[hide]λοιπον εστω οτι το τριγωνο ΑΒΓ εχει γωνιες $ A_1, B_1,C_1$,$ x\equal{}BD$,$ y\equal{}DC$ και  $ A_2$ και $ A_3$ οι γωνιες που αντιστοιχουν στο ορθογωνιο ΑΓΘΙ.
εχουμε ομως οτι η γωνια C1 ειναι ιση με την Α2 ως εντος ενναλαξ αρα
$ \cos A_3\equal{}\frac{x}{AL} < \equal{} > \sin C_1 \equal{}\frac{x}{AL}$
ομοια $ \sin B_1 \equal{}\frac{y}{AK}$ διαρω τις δυο σχεσεις κατα μελη και παιρνω οτι
: $ \frac{\sin C_1}{\sin B_1} \equal{}  \frac{xAK}{yAL}$(1)
φερνωντας το υψος στο τριγωνο ΑΒΓ ΑΖ εχουμε $ \sin C_1\equal{} \frac{AZ}{AC}$
και $ \sin B_1 \equal{} \frac{AZ}{AB}$ διαρωντας περνουμε $ \frac{\sin C_1}{\sin B_1} \equal{} \frac{AB}{AC}$(2)
στηριζομενος λοιπον στις (1) και (2) και στο θεωρημα των διχοτομων $ \frac{x}{y} \equal{} \frac{AB}{AC}$ περνουμε
$ \frac{AB}{AC} \equal{}  \frac{xAK}{yAL}$
$ <\equal{}>$
$ (AL)\equal{}(AK)$ που ειναι και το ζητουμενο.[/hide]

[hide]Έστω τα σημεία $ K'\equiv AC\cap ZH,$ $ L'\equiv AB\cap EF$ και $ P\equiv EF\cap ZH.$

Στο παραλληλόγραμμο $ PK'AL',$ οι διαγώνιες $ PA,\ K'L'$ αλληλοδιχοτομούνται και επομένως αρκεί να αποδειχθεί ότι $ K'L'\parallel KL.$

Αυτό όμως αληθεύει γιατί $ \frac {AK'}{AL'} \equal{} \frac {AH}{AF} \equal{} \frac {DC}{DB} \equal{} \frac {AC}{AB}$ $ ,(1)$

$ ($ από όμοια ορθογώνια τρίγωνα $ \bigtriangleup HAK'$ και $ \bigtriangleup FAL'$ και από τη διχοτόμο $ AD$ της γωνίας $ \angle A$ $ ).$

Από $ (1)$ $ \Longrightarrow$ $ \frac {AK'}{AC} \equal{} \frac {AL'}{AB}$ $ \Longrightarrow$ $ K'L'\parallel KL$ και η πρόταση έχει αποδειχθεί.

$ \bullet$ Αποδεικνύεται επίσης εύκολα, ότι η ευθεία $ PA$ περνάει από το μέσον της πλευράς $ BC$ και άρα οποιοδήποτε ευθύγραμμο τμήμα $ KL,$ που έχει τα άκρα του $ K,\ L,$ επί των ευθειων $ ZH,\ EF$ αντιστοίχως και είναι παράλληλο προς την πλευρά $ BC,$, διχοτομείται από τη διάμεσο  $ AM$ του δοσμένου τριγώνου $ \bigtriangleup ABC.$[/hide]
Κώστας Βήττας.

Πολύ ωραίες λύσεις και οι δυο ...

Κώστα ευχαριτώ και για το σχήμα και για την επιπλέον παρατήρηση που έκανες   :)

Όπως είπα και στον Ηλία αυτό τον καιρό πνίγομαι ... Ελπίζω όταν τελειώσω με το τρέξιμο να γίνω πιο τακτικός γιατί είχα να μπώ κανα δυο μήνες  :(


Aliens Exist

Put your evidence here!

You should watch UFO Hunters if you can (I should watch sometime too).

My friends stay things like random abductions and evidence on hieroglyphics about the visiting of "godly" beings, where their drawings of these beings resemble aliens.

gasp

Put your evidence here and start a controversy!

They dont...or atleast they dont live anywhere close to Earth. Even if someone claims to see them, it was probably their imagination. Also, if aliens existed in any planet nearby we would know. Aliens may exist in a planet far away from here, but we may never know, cuz it would be impossible to travel a distance that large. Scientists have a large radio that can detect signals from outer space, and no evidence has been found at all. Humans exist because the Earth is in a perfect distance away from the Sun. Who knows, there may be no other planet in a convenient place like Earth. Many people lie that they've seen UFOs and such, but thats just their imagination. There is no such thing as Area 51, by the way. Ah...I love science...

Sorry if you think that was spam i am trying to keep some discussion in this thread, because there is no way 3 posts are enough.

Yes, that was complete spam and had no contribution to the discussion whatsoever. I deleted that post to save some people from 5 year's work.

King!: Google up "Roswell, New Mexico". Read about crop fields. Then read about Area 51.

well, this thread seems short. we need some discussion in here. as an alternative, a mod can lock it.

[color=green]dragon96: If the thread won't extend, please don't spam just to get discussion if it won't be getting any discussion.[/color]

... I'm not thinking aliens exist. My family drove past Area 51 on a cross-country trip, and we did see a plane taking off... Um... that didn't help my case, did it?

you only showed that you did not prove the existence of aliens. you did not disprove the existence, however.

im going to be a believer here:

aliens can exist because, since our technology has not advanced to the point that we cannot see past a couple lightyears, and since we cannot even see what is on surfaces (or atmospheres) beyond the asteroid belt, there is still a chance that aliens exist on other planets, or even asteroids. anyway, since the visible universe is approx. 23 billion lightyears across, with trillions of stars with their own planets.
when we define "aliens," we think of intelligent beings that reside and thrive on other planets. however, i will take it as any living organism that resides in any planet in the universe. life, being a delicate and complex idea, only happens on chance. but, with that many possibilities in store, the chance becomes considerably bigger. trillions of chances, there has got to be one other planet that provides the many needs of life.
the reason that other people think that there are no aliens is because they are usually narrow-minded. remember, impossibility is only the limits of human imagination. since we have only closely studied about 3 different celestial bodies: mars, the moon, and our own blue marble, the earth. also, despite all the scifi movies, the aliens could be less technological oriented as we are. so star trek is a very distant possibility.
but, all in all, since we still do not have the right tech, we might as well sit and hope for the best, whatever you believe.

thank you.

good point. what if the aliens are wondering if we exist? could that be a possibility? maybe they started a forum and made this thread in their own fun factory :rotfl:

Right now they might be thinking "Should we blow them up or not?" which is why we should be keeping up appearances. Which we are failing at, miserably.

According to Know1ng, we're all gonna die tomorrow because of this massive solar flare that we didn't detect because of our wasting our time killing each other instead of trying to improve the human race so that we could build a Detect-a-Flare 9000.

"Sometimes I think the surest sign that intelligent life exists elsewhere in the universe is that none of it has tried to contact us."

Bill Watterson

Lol, wasn't that from a Calvin and Hobbes comic strip? BTW, take a look at this link. Very statistical.

[url]http://www.thekeyboard.org.uk/Extraterrestrial%20life.htm[/url]

Why wouldn't they exist? Life exists on earth, why not anywhere else?

Personally, I don't believe in aliens. I've never heard of anything that couldn't be proved to be either something normal or someone just playing a hoax.

well, what if we will never make contact? wont that be just as bad as them not existing?

pretty much. if we never make contact, then all the conspiracies will be for nothing, even though if aliens do exist (which i do not believe, despite my lecture earlier before  :| ). because either way, it doesnt matter.


In triangle $ ABC$, let $ X,Y,Z$ be points on sides $ BC,CA,AB$ respectively. Assume that the perpendicular to $ BC$ at $ X$, the perpendicular to $ CA$ at $ Y$, and the perpendicular to $ AB$ at $ Z$ are concurrent. Prove that the perpendicular from $ A$ to $ YZ$, the perpendicular from $ B$ to $ ZX$, and the perpendicular from $ C$ to $ XY$ are concurrent.

Let $P$ be the concurrency point of the perpendiculars from $X,Y,Z$ to $BC,CA,AB.$ Let $M$ be the orthogonal projection of $A$ onto $YZ.$ Note that $AYPZ$ is cyclic on account of  $\angle PYA=\angle PZA=90^{\circ}.$ Therefore, $\angle ZAM=\angle PZY=\angle PAY$ $\Longrightarrow$ Rays $AP,AM$ are isogonal WRT $\angle BAC.$ By similar reasoning, we conclude that perpendiculars from $A,B,C$ to $YZ,ZX,XY$ concur at the isogonal conjugate of $P.$

Wasn't that the Carnot Theorem?

Best regards,
sunken rock

I think Carnot theorem states something different: Given three points on the plane of $\triangle ABC,$ then the perpendiculars from $X,Y,Z$ to $BC,CA,AB$ concur $\Longleftrightarrow$
\[(XB^2-XC^2)+(YC^2-YA^2)+(ZA^2-ZB^2)=0\]


conics parabola algebra polynomial Vieta geometry parameterization

$l$ is a line that is parallel to $y=x$. $l$ and the parabola $y^2=4x$ intersect at points $P$, $Q$. Find the locus of the midpoints of $\overline{PQ}$.

cool problem...
[hide="hint"]you don't need the solutions directly, use vieta on a polynomial solving for $x$...[/hide]

since this is a graphing locus, you should give a relation that is all the points, ex it could be a circle (it is not though...)

[quote="kunny"][hide]The half-line $x=2\ (y<3).$[/hide][/quote]

i agree with part of it...
[hide]i think it is $y>2\sqrt{2}$, since you do the intersection with the locus, and the parabola, that that will be as far as you can go, before there will be no intersections[/hide]

I must say my answer differs from both of yours.

[hide="It is"]$Y=2, X\geqslant 1$

From $y^2=4x$ and $y=x+a$ where $a$ is a parameter, we get $(x+a)^2=4x\iff x^2+(2a-4)x+a^2=0$. From Vieta's we get $X=\frac{x_1+x_2}{2}=-a+2$ and $Y=\frac{y_1+y_2}{2}=\frac{x_1+a+x_2+a}{2}=X+a=2$. Also, the discriminant must be non-negative, hence $(2a-4)^2-4a^2\geqslant 0\iff a\leqslant 1$. That gives $-a\geqslant -1\iff -a+2\geqslant 1$. Therefore the locus is the ray $X\geqslant 1, Y=2$[/hide]

Oops, I was wrong. Farenhajt is almost correct.

Let $M$ be the midpoint of the given segment, I should have set $M(Y,X).$ ;)

[quote="chess64"]Sorry, I'm new to loci... by find the locus, do you mean the area or what?[/quote]

A locus is the set of all points satisfying the given condition (for example, a circle, a line, a plane.... etc.)

[quote="Altheman"]wow, that was pretty dumb of me, your post confused me that it was x=2, where it is actually, y=2, and we can just do the intersection with y^2=4x, to get x=1, as fahrenheit said, but perhaps the problem implies that P and Q are distinct, so we cannot have the tangent point, giving $y=2$ for $x>1$[/quote]

Possible, but in my practice, the degenerate segments (i.e. single points) are usually taken into account in such problems.


[quote="mdk"]Alice and Bob are playing a game. Alice thinks of a secret polynomial p(x) with non-negative integer coefficients. Bob chooses an integer a and Alice tells Bob the value p(a). Then Bob chooses an integer b and Alice tells Bob the value p(b).
Now Bob can tell Alice the polynomial p(x) correctly. What is Bob’s strategy for determining the polynomial p(x)?[/quote]
Let's say $ p(x)=(x+1)^{2}+3$. Bob says $ 0$, and Alice replies $ 4$. Bob says $ 1$, and Alice replies $ 7$.

How can Bob determine that it's $ p(x)=(x+1)^{2}+3$ when it could also be $ p(x)=3x^{2}+4$ and so many other things?

[quote="i_like_pie"]
How can Bob determine that it's $ p(x)=(x+1)^{2}+3$ when it could also be $ p(x)=3x^{2}+4$ and so many other things?[/quote]
Maybe Bob is choosing the wrong numbers and using the wrong strategy, but I don't see anything other than Bob saying $ 0$ and $ 1$ which gets $ a_{0}$ the sum of $ a_{i}$, respectively.  If he chose $ -1$ and $ 1$, he could get the sum of $ a_{i}$ where $ i$ is odd, and the sum when $ i$ is even, but...  it's not enough.

Seen this elsewhere, but I don't remember the actual solution, just bits of the method.

The important part of this problem is Bob picking his second number based on the answer he gets the first time. Just as you need to carefully pick what the first number is. Don't forget it's integer coefficients; I think this problem may be as much number theory as algebra.

[quote="0714446459923"]I think my solution is correct. There is no way to determine p(x) from the information given[/quote]

Dude, mind reading isn't cheating.  You're wrong.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1160819217&t=152900[/url]

This problem has already been discussed before.
Just search
"polynomial game".
Please don't post questions already answered before, especially those posted only two months ago.

[hide]Suppose the polynomial is $ P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}\cdots a_{1}x+a_{0}$.
So $ P(1)=a_{n}+a_{n-1}\cdots a_{1}+a_{0}$. Let $ P(1)=A$.

Plug in $ P(10^{A})$. In the resulting number, the trailing digits of $ a_{i}$ will appear in the $ (10^{Ai)}+1)^{(th)}$ digit from the left. Since $ 10^{A}>a_{i}$ for any $ i$, the coefficients will never "intersect" and carry over. We can now figure out the polynomial.

[u][b]Example[/b][/u]:

Say Alice picks the polynomial $ 3x^{2}+4$. Bob plugs in $ 1$, getting $ A=7$. So then Bob plugs in 10^14, getting $ 300000000000004$. The first digit from the left is $ 4$, the eight digit from the left is $ 0$, the fifteenth digit from the left is $ 3$, so Bob knows the polynomial will be $ 3x^{2}+4$.[/hide]


Let $\displaystyle ABC$ and $\displaystyle DBC$ be isosceles triangle with the base $\displaystyle BC$. We know that $\displaystyle \measuredangle ABD = \frac{\pi}{2}$. Let $\displaystyle M$ be the midpoint of $\displaystyle BC$. The points $\displaystyle E,F,P$ are chosen such that $\displaystyle E \in (AB)$, $\displaystyle P \in (MC)$, $\displaystyle C \in (AF)$, and $\displaystyle \measuredangle BDE = \measuredangle ADP = \measuredangle CDF$. Prove that $\displaystyle P$ is the midpoint of $\displaystyle EF$ and $\displaystyle DP \perp EF$.

I hope the following is right.

Let $S$ the point of intersection of the lines ED, BC .
We have that the points A,D,M are collinear as AB=AC and BD=DC
We have that $\widehat{EDB}=\widehat{CDF}$  so the triangles BDE and CDF are congruent.
So the triangle DEF is isosceles . Now we have that
$\widehat{ADE}=90-\widehat{DSC}=90-\widehat{SBD}-\widehat{SDB}=\widehat{ ABC}-\widehat{ SDB}$

So $\widehat{ EDP}=\widehat{ABC}$ .
Also the quadrilateral ABDC is cyclic so  $\widehat{ADC}=\widehat{ABC}$
and so $\widehat{EDP}=\widehat{PDF}$  so the triangles EDP and PDF are congruent . So $\widehat{ EPD}=\widehat{ DPF}$
But the quadrilateral EBDP is cyclic as   $\widehat{EDP}=\widehat{ABC}$  so
$\widehat{EDP}=90=\widehat{PDF}$

and the conclusion is obvious
Please tell me if it is correct.

$1.\blacktriangleright x=m(\widehat {BDE})=$ $m(\widehat {ADP})=$ $m(\widehat {CDF})$ and $y=m(\widehat {ADE})=m(\widehat {CDP})$
$\Longrightarrow$ $\boxed {m(\widehat {PDE})=m(\widehat {PDF})=x+y}\ .$

$2.\blacktriangleright DB=DC$, $\widehat {BDE}\equiv \widehat {CDF}$, $m(\widehat {DBE})=m(\widehat {DCF})=90^{\circ}$[b] (a.s.a.) [/b]$\Longrightarrow \triangle BDE\equiv\triangle CDF\Longrightarrow \boxed {DE=DF}\ .$

$3.\blacktriangleright$ Apply the Menelaus' theorem to the transversal $\overline {EPF}$ for the triangle $ABC$:
$\frac{FC}{FA}\cdot \frac{EA}{EB}\cdot \frac{PB}{PC}=$ $\frac{DC\sin x}{DA\sin (2x+y)}\cdot \frac{DA\sin y}{DB\sin x}\cdot \frac{DB\sin (2x+y)}{DC\sin y}=1\Longrightarrow$ $\boxed {P\in EF}\ .$

$(1)\ \wedge\ (2)\ \wedge\ (3)\Longleftrightarrow$ the triangle $EDF$ is isosceles in the vertex $A$, $P\in (EF)$
and the ray $[DP$ is the bisector of the angle $\widehat {EDF}$ $\Longrightarrow$ $PE=PF\ .$

By using the cyclic cuadrylaterals $FAED$ and $FCPD$ we get: $\measuredangle DFE=\measuredangle DFP$, so E,F,P are collinear

Then we have $DE=DF$ and $\measuredangle EDP=\measuredangle PDF$, so: $EP=EF$ and $DP \perp EF$

By spiral similarity centered at $D$ we get:
$B ===> E$
$M ===> P$
$C ===> E$
Hence, $P$ is the midpoint of $EF$.


1)In the interior of each side of a square we choose n different points.
Find the number of all triangles whose vertices are these points
_____________________________________________________
2)In how many ways can one choose three of the vertices of a regular
n-gon A_1A_2...A_n , n \geq 4 such that they form the vertices of an obtuse triangle ?
______________________________________________________
3)Let the convex n-gon G (n \geq 6) be given.Find the number of all
triangles with vertices among those of G and with sides lying on diagonals of G ?
______________________________________________________
4) We are given a finite set S of n different points in the plane (n \geq 3)
of wich no three lie on a line.Show that there are at least n(n-2)/3 triangles whose vertices are among the points of S and such that they
do not contain any other points of S.

[i]unknown sources[/i]

For 1) :
I assume that you only consider nondegenerate triangles. So, you form a triangle if and only if you do not choose three points which belong to the same side of the square. There are C(4n,3) ways to choose 3 points, and 4*C(n,3) wyas to choose them on the same side of the square. Thus, the desired number of triangles is C(4n,3) - 4*C(n,3).

For 2) :
- First, suppose that n = 2k+1 is odd.
The triangle has an obtuse angle at A_1 if and only if its vertices are A_1,A_iA_j, where i is in {2,...,k+1} (by symetry) and j is in {k+1+i,...,2k+1} (that is A_1 belongs to the shorter arc A_iA_j). Thus, for a fixed i, it gives exactly k+1-i values for j, and then there are exactly k(k-1)/2 obtuse triangles in A_1. By symetryn, there are the same number which are obtuse in A_i for each i. Note that there cannot be two different obtuse angles, thus these triangles are pairwise distincts. It follows that the total number of obtuse triangles is O(2k+1) = k(k-1)(2k+1)/2.

- Now suppose that n = 2k+2 is even.
Exactly the same reasoning (but with care to avoid right-angled triangles), gives the total of obtuse triangles in that case to be O(2k+2) = k(k-1)(2k+2)/2 = k(k-1)(k+1).

For 3) :
Well, this one is trivial since for each choice of three vertices among those of G, you will have a triangle satisfying your condition. Thus, the number of triangles here is C(n,3), and the condition n  \geq 6 is useless.

For 4) (I like this one) :
Let A be any given point in S. From an arbitrary origin, we label the other points of S as M_1,...,M_(n-1) according to the rays [AM_i) occurs in the trigonometric sense (that is where we use that there are no three collinear points in S).
Let a_i be the value in degree of angle <M_iAM_(i+1) where M_n = M_1.
Note that at most one of the a_i's is greater than or equal to 180. Moreover, if a_i < 180 then the triangle M_iAM_(i+1) does not contains any other point from S than its vertices. Thus, there are at least n-2 such triangles with vertice A.
Since A is arbitrary, and since a triangle has three vertices, the total number of such triangles is at least n(n-2)/3.

Pierre.


well,, it's hard for me to understand these two theorems :
multiplication theorem & remainder theorem

[i]please[/i] give me easier difinitons of them and
also some simple examples and how to solve them

thanks in advance!

This is definitely [b]not[/b] MOEMS level.  I think it should be moved somewhere else.
In the meantime there is always Wikipedia!

ok, well if im not mistaken the remainder theorem states that the remainder when polynomial $ P(x)$ is divided by $ (x \minus{} c)$ the remainder is just $ P(x)$ evaluated at $ c$

Im not sure about the multiplication theorem though, ive never heard of it!

yeah, but I don't think that the remainder theorem would ever be on any MOEMS test

Agreed, i just learned in Algebra II, and from by vauge knowlege of MOEMS, i have never seen the remainder theorem on a test.  I only use it to find zeroes of a polynomial equation, havent seen this on any test for that matter (other than my math tests at school!).

as for the multiplication theorem, there's one for the gamma function, another for the polygamma function, and many more, so... yeah

either way, it's definitely not middle school level

As zolojetto said, the remainder theorem states that given a polynomial, $ P(x)$, the remainder when $ P(x)$ is divided by $ (x\minus{}c)$ is $ P(c)$.  However, the only multiplication theorems that I know of are more advanced than HSB, so they are probably not what you are looking for.

[url=http://en.wikipedia.org/wiki/Multiplication_theorem]if you want them anyway[/url]


analytic geometry
number theory
greatest common divisor
modular arithmetic

Let $A,B,C$ be $3$ points on the plane with integral coordinates. Prove that there exists a point $P$ with integral coordinates distinct from $A,B$ and $C$ such that the interiors of the segments $PA,PB$ and $PC$ do not contain points with integral coordinates.

What is the interior of a segment?  Do you just mean that on each of the lines $ PA, PB, PC$ there are no points with integral coordinates between $ P$ and $ A, B, C$ respectively?

Yes, the interior of segment PA is the set of points between P,A (except P,A). I did use original words in Singapore TST 2007.

Try this.  It's not pretty, but I think it works.

Let the points be $(a_i,b_i)$, for $i=1,2,3.$  The problem is equivalent to showing that there exists a lattice point $(r,s)$ such that gcd$(r-a_i,s-b_i) = 1$, for $i=1,2,3$.

Consider the unit square with bottom left vertex the origin.  Each of the four vertices represent a different combination modulo 2, so at least one of them must differ modulo 2 from each of the $(a_i,b_i)$.  Call that point $(m,n)$.  So $2 \nmid gcd(m-a_i,n-b_i)$, for $i=1,2,3$.  Consider $(m+2k,n+2l)$, for $k,l \in \{0,1,2\}$.  Those 9 points range over all combinations modulo 3, so at least one is different modulo 3 from the $(a_i,b_i)$.  Call it $(m',n')$.  So $2,3 \nmid gcd(m'-a_i,n'-b_i)$, for $i=1,2,3$.

Now take all the points of the form $(m'+6k,n'+6k)$.  Those points represent 5 different pairs modulo 5, so again we can find one $(m",n")$ such that $2,3,5 \nmid gcd(m"-a_i,n"-b_i)$, for $i=1,2,3$.  Continuing on this way, for any T, we can find a point $(m'+A,n'+A)$ such that $p \nmid gcd(m'+A-a_i,n'+A-b_i)$, for $i=1,2,3$, where p is any prime less than T.

No suppose q is a prime such that $q \mid (m'+A-a_i)$ and $q \mid (n'+A-b_i)$ for some i.  Then $q \mid (m'-n'+b_i-a_i)$.  But this is impossible if we choose T large enough, so we have a contradiction, and $(m'+A,n'+A)$ is the point we are looking for.

[quote="polya78"]Now take all the points of the form $(m'+6k,n'+6k)$.  Those points represent 5 different pairs modulo 5, so again we can find one $(m",n")$ such that $2,3,5 \nmid gcd(m"-a_i,n"-b_i)$, for $i=1,2,3$.  Continuing on this way, for any T, we can find a point $(m'+A,n'+A)$ such that $p \nmid gcd(m'+A-a_i,n'+A-b_i)$, for $i=1,2,3$, where p is any prime less than T.

No suppose q is a prime such that $q \mid (m'+A-a_i)$ and $q \mid (n'+A-b_i)$ for some i.  Then $q \mid (m'-n'+b_i-a_i)$.  But this is impossible if we choose T large enough, so we have a contradiction, and $(m'+A,n'+A)$ is the point we are looking for.[/quote]
That needs a little clarification. Once we got past the primes $2$ and $3$, assume we found $M,N$ such that $2,3,\ldots, p_{t-1} \nmid \gcd(M-a_i,N-b_i)$, for all $i=1,2,3$. We now build $M_k = M + (p_t-2)!k$, $N_k = N + (p_t-2)!k$, $k=0,1,2,3$, for the next prime $p_t$. By Wilson's theorem we have $(p_t - 2)! \equiv 1 \pmod{p_t}$, so $M_k \equiv M + k \pmod{p_t}$ and $N_k \equiv N + k \pmod{p_t}$, $k=0,1,2,3$. At least for one $k$ we will have the property that $2,3,\ldots, p_{t} \nmid \gcd(M_k-a_i,N_k-b_i)$, for all $i=1,2,3$. Now we can conclude, with having reached far enough, that no prime $q$ will divide the final $\gcd$'s.

[hide = Solution]

We will first show the following lemma:

$\mathbf{Lemma.}$ Let $x_0, y_0$ be positive integers satisfying $nx + my = c$, for some $c \neq 0$ and relatively prime $m, n \in \mathbb{Z}$, and let $(x_k, y_k) = (x_0 +mk, y_0 - nk)$ for $k \in \mathbb{N}.$ Then, there exists some $k \in \mathbb{N}$ with $\gcd(x_k, y_k) = 1.$

$\mathbf{Proof.}$ Notice that $nx_k + my_k = c$, for all $k \ge 0$. Therefore, if $p|x_k, y_k$, then we must have $p|c$. Let $p_1, p_2, \cdots, p_t$ be the distinct prime divisors of $c$. Then, for each $p_i$, there clearly exists some $a_i \in \mathbb{N}$ with $p_i \nmid \gcd(x_{a_i+p_ik}, y_{a_i+p_ik})$. Hence, it only suffices to solve the system of equations with CRT to produce some $k \in \mathbb{N}$ with $p_i \nmid \gcd(x_k, y_k)$ for all $p_i$'s.

$\blacksquare$

With this lemma, we may now return to the problem. Let the line through $A, B$ have slope $\frac{-m}{n},$ for some relatively prime positive integers $m, n$. (If the line is vertical, let $m = -1, n=0$ and vice versa if horizontal). Call that line $nx +my = c$. Now, consider the lines $nx+my = c \pm 1$. Clearly, one of them must not contain $C$, and neither of them can contain $A$ or $B$. Furthermore, for any lattice point on either of these lines, say $P$, neither $PA$ nor $PB$ contains another lattice point. Therefore, assume WLOG by symmetry that $nx+my = c+1$ doesn't contain $A, B, C.$ Letting $(x, y)$ be on this line and $(x_c, y_c)$ be the coordinates of $C$, it only suffices to apply the lemma for $(x_0, y_0) = (x - x_c, y - y_c)$, noting that $nx_0 + my_0 \neq 0$ because $C$ isn't on the line.

$\square$

[/hide]


Factor: $\cos 4 \alpha + \cos 5\alpha + \cos 6 \alpha$

[hide]$\cos 6a + \cos 4a + \cos 5a = 2\cos 5a*\cos a + \cos 5a = \cos 5a(2\cos a + 1)$[/hide]

[quote="Jan"]Yes, that's what i got too, but I was wondering if we could factor $2 cos \alpha +1$[/quote]

[hide="Sure thing"]

$\cos 5a (2 \cos a + 1) = 2 \cos 5a \left( \cos a + \frac{1}{2} \right) = 2\cos 5a \left( \cos a + \cos \frac{\pi}{3}\right)$[/hide]


AMC
AMC 12
AIME
AMC 10
USA(J)MO
USAMO
AMC 12 A

Could someone give me a definitive answer on this question?  I want to know if someone in 10th or less grade can take the AMC 12 twice.  It would seem that this would be acceptable and I have been told by some people that it is ok.  But the AMC website says otherwise in their FAQ's:

[quote]Q: Can a student take both contests?

A: Yes, as long as a student is eligible to take the appropriate contests, that is

    * A 10th grader or below can take the AMC 10 A on February 10, and the AMC 12 B on February 25, or
    * A 10th grader or below can take the AMC 10 A on February 10, and the AMC 10 B on February 25 or
    * A 10th grader or below can take the AMC 12 A on February 10, and the AMC 10 B on February 25 or
    * An 11th or 12th grader can take the AMC 12 A on February 10, and the AMC 12 B on February 25

However, in order to do this the school would have to register for both dates and order contest bundles to have the contests on hand.[/quote]

They don't explicitly say that a 10th or below can in that but neither do they confirm that they cannot.  I know the AMC website is unreliable (not to mention that the FAQ still refers to 2004!) so I could use a good response from someone.

Thanks

Btw, I'm an 11th grader so that does not apply directly to me but rather to a few of my teammates.

According to my math teacher a 10th grader can take the AMC12 both times.

I am taking it both times, and I did last year as well.

In my opinion if you are taking the AMC 12 in the first place then you probably don't need to take two AMCs to qualify in the first place :P.

[quote="MithsApprentice"]Then neither of you are in compliance with the rules.

Sophomores and below can take the AMC 12 a maximum of one time per year. They can take the AMC 10 both times though.

You might want to get a direct answer from the AMCDirector, but I believe the rules are quite clear.[/quote]

Hmm can we really trust the website?

Sometimes the website have false information on it(i.e AIME floor value, they say only ppl who take AMC10 can qualify from AIME floor value system, while actually all 10th grader/below can).

I can honestly tell that I know somebody taking the AMC-12 both times as an 8th grader, so I really doubt that rule.

Also, don't they encourage people to take AMC12? So this rule seems strange.

The AMC website is completely inaccurate, so I wouldn't ever count on it.  It ought to and seems to be legal for someone 10th grade or under to take the AMC 12 twice.  However, I do not see any particular value in this, as you shouldn't be worrying about your USAMO index anyway as a 7th/8th/9th/10th grader.

[quote="MithsApprentice"]You might want to get a direct answer from the AMCDirector, but I believe the rules are quite clear.[/quote]

Unfortunately, the rules aren't clear.  They didn't say "the following are the ONLY ways of taking two tests" - for all we know (and as probably is the case) they just forgot to mention it.

It makes no sense to have a rule like "you can only take the 12 once if you're < 10th grade" because it would discriminate against younger kids for no apparent reason.

And, on a side note, if such a rule exists but isn't enforced, then compliance doesn't matter, eh?   ;)

Yes, a student in grade 10 and below can take the AMC 12 A and then take the AMC 12 B.  There is no rule that prevents it.

The examples quoted from the Web page were meant as representative examples.  I had no idea people would take examples so literally, or I would have written in every possible combination.

Please refer to the current year's Teachers Manual (which is mailed to each participating school with the contest packages) and is also reproduced directly from its source material on the AMC Website for the current rules, policies and procedures.  The Teacher's Manual is the definitive source for the rules and policies.

AMC Director
Steve Dunbar


calculus
integration
trigonometry
parameterization
calculus computations

Evaluate $\int_{0}^{2\pi}\frac{\sin^{2}\theta}{a-b\cos\theta}d\theta$. $a > b >0$.

We use the substitution $\tan \frac{\theta}{2}=t$

And get: $8\int^{2\pi}_{0}\frac{t^{2}dt}{(1+t^{2})^{2}((a-b)+(a+b)t^{2})}$
We break integral into more simple integrals and get result.

[quote="prithwijit"]Evaluate $\int_{0}^{2\pi}\frac{\sin^{2}x}{a-b\cos x}\ dx$, where $a > b >0$.[/quote]
[color=darkblue]The limits of this integral are correctly because $a-b\cos x=b\left(\frac{a}{b}-\cos x\right)>b(1-\cos x)\ge 0$ $\Longrightarrow$ $a-b\cos x>0$ $\Longrightarrow$ $a-b\cos x\ne 0\ .$
But the Duginov's substitution is, [u]in the my opinion[/u], incorrectly ! And now ...[/color] \[\boxed{\ I\equiv\int_{0}^{2\pi}\frac{\sin^{2}x}{a-b\cos x}\ dx\ }\] $I=\int_{-\pi}^{\pi}\frac{\sin^{2}x}{a+b\cos x}\ dx=$ $\int_{0}^{\pi}\left(\frac{\sin^{2}x}{a+b\cos x}+\frac{\sin^{2}x}{a-b\cos x}\right)\ dx=$ $2a\int_{0}^{\pi}\frac{\sin^{2}x}{a^{2}-b^{2}\cos^{2}x}\ dx=$ $2a\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\cos^{2}x}{a^{2}-b^{2}\sin^{2}x}\ dx=$ $4a\int_{0}^{\frac{\pi}{2}}\frac{\cos^{2}x}{(a^{2}-b^{2})\sin^{2}x+a^{2}\cos^{2}x}\ dx=$ $\frac{4a}{a^{2}-b^{2}}\int_{0}^{\frac{\pi}{2}}\frac{\cos^{2}x}{\sin^{2}x+k^{2}\cos^{2}x}\ dx$, where $\boxed{\ \frac{a}{\sqrt{a^{2}-b^{2}}}=k>1\ }\ .$ Therefore, $I=\frac{4k^{2}}{a}\int_{0}^{\frac{\pi}{2}}\frac{1}{k^{2}+\tan^{2}x}\cdot \frac{1}{1+\tan^{2}x}\cdot (\tan x)'\ dx=$ $\frac{4k^{2}}{a}\int_{0}^{\infty }\frac{1}{(t^{2}+1)(t^{2}+k^{2})}\ dx=$ $\frac{4k^{2}}{a(k^{2}-1)}\int_{0}^{\infty}\left(\frac{1}{t^{2}+1}-\frac{1}{t^{2}+k^{2}}\right)\ dx=$ $\frac{4a}{b^{2}}\left[\left(\arctan t-\frac{1}{k}\cdot\arctan\frac{t}{k}\right)\right]_{0}^{\infty}=$ $\frac{4a}{b^{2}}\cdot\frac{\pi}{2}\cdot \frac{k-1}{k}$ $\Longrightarrow$ $\boxed{\ I=\frac{2\pi}{b^{2}}\cdot \left(a-\sqrt{a^{2}-b^{2}}\right)\ }\ .$

Not easier, but entertaining:

Note that $\frac1{a-b\cos x}=\sum_{n=0}^{\infty}\frac{b^{n}}{a^{n+1}}\cos^{n}x$

Also, $\int_{0}^{2\pi}\sin^{2}x\cos^{n}x\,dx=0$ if $n$ is odd, while

$\int_{0}^{2\pi}\sin^{2}x\cos^{2k}x\,dx= 2\pi\cdot\frac12\cdot\frac34\cdots\frac{2k-1}{2k}\cdot \frac1{2k+2}$

So we can put that together and interchange the sum and the integral to get

$\int_{0}^{2\pi}\frac{\sin^{2}x}{a-b\cos x}\,dx= 2\pi\sum_{k=0}^{\infty}\frac{b^{2k}}{a^{2k+1}}\cdot \frac12\cdot\frac34\cdots\frac{2k-1}{2k}\cdot\frac1{2k+2}$

Now we have a messy exercise in trying to recognize the binomial series.

There is the residue theorem. I'll get it started; someone else can finish if they have the patience.

Consider $z=e^{i\theta},$ the standard parameterization of the unit circle.

Then $dz=ie^{i\theta}\,d\theta$ or $d\theta=\frac{dz}{iz}.$

$\sin\theta=\frac1{2i}\left(z-\frac1z\right)$ so $\sin^{2}\theta=-\frac14\left(z-\frac1z\right)^{2},$ and $\cos\theta=\frac12\left(z+\frac1z\right).$

If I did the algebra correctly, we get

$\int_{o}^{2\pi}\frac{\sin^{2}\theta}{a-b\cos\theta}\,d\theta= \frac12\int_{C}\frac{z^{4}-2z^{2}+1}{z^{2}\left(bz^{2}-2aiz+b\right)}\,dz$

where $C$ is the the unit circle, once counterclockwise.

There are one singularities inside the unit circle: a simple pole at $z=\frac{\left(a-\sqrt{a^{2}+b^{2}}\right)i}b$ and a pole of order 2 at zero.

Double-check the algebra to get there (I wouldn't put the odds of it being right at any better than 50%), then calculate those residues, and you should get the answer.

[quote="prithwijit"]Are the limits correct?[/quote]
I was wrong.


vector
real analysis
real analysis unsolved

Are you using $\parallel \cdot\parallel$ in the $\parallel (x,y)\parallel =\sqrt{x^{2}+y^{2}}$ sense? If so,

1. Yes, this is an open disk connected to a line, so any two points in this set can be connected by a polygonal curve completely within the set itself.

2. No, this is an open disk and a line, but there is nothing at the point $(1,0)$, so although the two separate, these are the two components that are connected.

3. Yes, this is an open annulus. Same reasoning as 1.

4. No, the circle $\{(x,y)\in\mathbb{R}^{2}: \parallel (x,y)\parallel =1\}$ separates the two regions, but the two regions are themselves connected so they are the components.

It's not very important- all of these have the same answers in any of the standard norms.
I'd still assume the Euclidean norm.

Ah, that's right. All norms over a finite dimensional vector space is equivalent.

I said [i]standard[/i] norms- if $\|(1,0)\|\neq 1$, it breaks in the first two cases.

Yes, it is the Euclidean norm: $\parallel (x,y) \parallel = \sqrt{x^{2}+y^{2}}$


Let $x,y,z$ be positive reals s.t $x+y+z=1$. Prove the following inequality :

$(x+1)\cdot  \sqrt{\frac{1-x}{x} }+(y+1)\cdot  \sqrt{\frac{1-y}{y} }+(z+1)\cdot  \sqrt{\frac{1-z}{z} }\geq
\frac{3\sqrt{3}}{4}\cdot \frac{(1+x)(1+y)(1+z)}{\sqrt{(1-x)(1-y)(1-z)}}$

Hint : try to transform it in a geometric inequality!
oxi pos egine poli pio eukoli tora...

Galeos' words mean " I don't think that it became easier after my hint. "

To Galeos :
Bre c to mono pou mporeis na kaneis gia na paikseis mazi tis einai metasximatismous sto trigono. basika ekeino to $\frac{\sqrt{3}}{4}$ se odigei ekei. alla pali i anisotita den douleuete einai poli diskoli. Ego gemisa merikes selides prakseis kai tin paratisa.

i will send you a private message with a more clarifying hint...

by the way, the source is not me but Crux!!!


Example where n = 3:

<---A---B---C--->  (points on a line)

A curve in one dimension consists of either 1) one
or more points, 2) one or more line segments
(they may be adjacent to each other or not) which
have points as endpoints, or 3) any combination of the above.

The points are all distinct, however the positive distances
between/among them is immaterial.

Example of category 1) A
Example of category 2) AC
Example of category 3) A & BC

Total number of curves where n = 3 is 12.

As n gets larger, the relation between the numbers
of cases of line segments and points together
to be counted is relatively more difficult to work out.

Is there a formula for the total number of curves on a
line for general n?

Your description is a little ambiguous, but I gather from the fact that you want the answer to be 12 in the case $ n \equal{} 3$ that you are considering two of these objects as the same if the sets of points included in each are the same, regardless of how you arrived at them.  (Meaning, for three points, one can arrive at the entire segment $ \overline{AC}$ in several ways, e.g. as $ \overline{AC}$, as $ \overline{AC}\cup \{B\}$ or as $ \overline{AB}\cup\overline{BC}$, and I'm assuming that you treat these three as the same.)

Let $ f(n)$ be one more than the number you're interested in.  (Why one more?  Because I'd like to include the empty curve.)  Let our points be $ A_1, \ldots, A_n$, occuring left to right.  Consider whether your curve contains $ A_1$ or not.  There are $ f(n \minus{} 1)$ possible curves that don't contain $ A_1$, and there are another $ f(n \minus{} 1)$ curves that contain $ A_1$ as an isolated point (i.e., not as the endpoint of a segment).  Otherwise, it contains $ A_1$ as part of a segment $ \overline{A_1A_{n \minus{} i}}$ for some $ 0 \leq i \leq n \minus{} 2$.  In that case, we have $ i$ points remaining on which we may construct an arbitrary curve structure, so we get $ f(i)$ from each of these cases.  It follows that $ f(n) \equal{} 2f(n \minus{} 1) \plus{} \sum_{i \equal{} 0}^{n \minus{} 2} f(i)$.

We can calculate that the first few values of our sequence (starting with $ n \equal{} 0$) are 1, 2, 5, 13, 34, 89, ....  (The 13 is the same as your 12 -- recall that we're including the empty curve.)  An intrepid solver of combinatorial problems might now conjecture that $ f(n) \equal{} F_{2n}$ where $ F_i$ is the $ i$th Fibonacci number, defined with initial values $ F_0 \equal{} F_1 \equal{} 1$.  Our result is then reduced to a pretty standard identity on the Fibonacci numbers, one that can easily be proven in a number of ways (e.g. inductively or using the closed form for the Fibonacci sequence).

Yes, the manner in which each curve is arrived at is unimportant.  The final appearance of each curve is what "counts" or matters.


analytic geometry
pigeonhole principle
combinatorics unsolved
combinatorics

A convex polygon on the coordinate plane contains at least $ m^2\plus{}1$ points with integer coordinates in its interior. Show that some $ m\plus{}1$ of these points lie on a line.

From the pigeonhole principle, there are two of the integer points in the polygon, say $ A(a,b)$ and $ B(c,d)$, such that $ a \minus{} c \equal{} b \minus{} d \equal{} 0 \mod[m]$.
Let $ M_k (a \plus{} \frac k m (a \minus{} c),b \plus{} \frac k m (b \minus{} d))$ for $ k \equal{} 0,1,\cdots, m$.
Then, each $ M_k$ is an integer point which belongs to the line segment $ [AB]$. Since the polygon is convex, each of these points belongs to the polygon, and we are done.

Pierre.


calculus
integration
trigonometry
logarithms
calculus computations

$ \int \frac {\cos^4 x}{\sin^3 x} dx$

Ans- $ \minus{} \frac {\cos^3 x}{2\sin^2 x} \minus{} \frac {3\cos x}{2} \plus{} \frac {3}{4}.\log \frac {1 \plus{} \cos x}{1 \minus{} \cos x} \plus{} c$

$ \cos^4x \equal{} \left(1 \minus{} \sin^2x\right)^2 \equal{} 1 \minus{} 2 \sin^2 x \plus{} \sin^4x$, and so

$ \int \frac{\cos^4x}{\sin^3x}\,dx \equal{} \int \frac{1 \minus{} 2 \sin^2 x \plus{} \sin^4x}{\sin^3x}\,dx \equal{} \int \csc^3x\,dx \minus{} 2\int\csc x\,dx \plus{}$

$ \plus{} \int\sin x\,dx \equal{} P \minus{} 2 \ln \left|\tan \frac{x}{2}\right| \minus{} \cos x \plus{} C$

and

$ P \equal{} \int \csc^3x\,dx \equal{} \minus{}\cot x \csc x \minus{} \int \cot^2x \csc x\,dx \equal{} \minus{}\cot x \csc x \minus{}$

$ \minus{} \int(\csc^2x \minus{} 1) \csc x\,dx \equal{}> P \equal{} \minus{}\cot x \csc x \minus{} P \plus{} \ln \left|\tan \frac{x}{2}\right|$

$ \equal{}> P \equal{} \minus{}\frac{1}{2} \cot x \csc x \plus{} \frac{1}{2} \ln \left|\tan \frac{x}{2}\right|$

and so

$ \int \frac{\cos^4x}{\sin^3x}\,dx \equal{} \minus{}\frac{1}{2} \cot x \csc x \minus{} \frac{3}{2} \ln \left|\tan \frac{x}{2}\right| \minus{} \cos x \plus{} C$.


I'm thinking of taking USAMTS next year when I did several of AIMEs and built better skill of math than what I'm now.

I'm hoping to qualify through AIME using USAMTS. But just two questions (maybe be posted before but any kind answers would be grateful)

A) If I qualify for AIME, where do I take test? In my school?

B) Perhaps this is really stupid question but how are the difficulty of USAMTS for each round? Is it getting harder?

B) look at the current rounds and previous ones yourself to determine yourself cuz some problems may be harder to some people than others...

A) I know if you qualify for AIME through the AMC, it is generally  taken at your AMC testing site.  If it is not being offered at your school, I would think that you could take it at a school nearby that is already giving it to people.  To be sure, I would contact your math team coach or math department head.

What are the requirements for qualifying for AIME through USAMTS?

Recently, its been a score of 68 or more thru the first three rounds.

[quote]The minimum score is determined as a percentile of the highest scoring students. I also know that qualifying for the aime throught USAMTS greatly hurts your chances of advancing to the USAMO. If you receive a 90 on the USAMTS, that is equivalant to having received a 90 on the AMC 12. You will need nearly a perfect score on the AIME to advance to the USAMO.[/quote]

This is only true for juniors and seniors. Also keep in mind that not all four rounds are considered in AIME qualification (that is, only the first three count.)


inequalities
geometry
incenter
Pythagorean Theorem
geometry proposed

Let $ I$ be the incenter of triangle $ ABC$,prove that:
$ IA^2\plus{}IB^2\plus{}IC^2 \ge \frac{AB^2\plus{}AC^2\plus{}BC^2}{3}$

Draw incircle and denote sidelengths as $ x\plus{}y$, $ y\plus{}z$ and $ z\plus{}x$ in an usual way.

Then $ r\equal{}\sqrt{\frac{xyz}{x\plus{}y\plus{}z}}$.

From Pythagorean theorem we easily calculate: $ AI^{2}\equal{}r^{2}\plus{}x^{2}$, $ BI^{2}\equal{}y^{2}\plus{}r^{2}$ and $ CI^{2}\equal{}r^{2}\plus{}z^{2}$.

Our inequality becomes equivalent to :

$ \frac {9xyz}{x\plus{}y\plus{}z}\plus{}3x^{2}\plus{}3y^{2}\plus{}3z^{2} \geq (x\plus{}y)^{2}\plus{}(y\plus{}z)^{2}\plus{}(z\plus{}x)^{2}$ which can be easily reduced to its equivalent:

$ x^{3}\plus{}y^{3}\plus{}z^{3}\plus{}3xyz \geq x^{2}y\plus{}xy^{2}\plus{}x^{2}z\plus{}xz^{2}\plus{}y^{2}z\plus{}yz^{2}$   which is true by Schur's inequality.

Equality is achieved when $ x\equal{}y\equal{}y$ / when triangle $ \triangle ABC$ is equilateral.

[color=darkblue]By [u][b]Leibnitz[/b][/u] $ IA^2 \plus{} IB^2 \plus{} IC^2 \equal{} 3IG^2 \plus{} GA^2 \plus{} GB^2 \plus{} GC^2 \equal{} 3IG^2 \plus{} \frac {1}{3}(AB^2 \plus{} BC^2 \plus{} CA^2)$, so $ IA^2 \plus{} IB^2 \plus{} IC^2 \ge \frac {1}{3}(AB^2 \plus{} BC^2 \plus{} CA^2)$ with equality if $ I \equal{} G$ or triangle $ ABC$ is equilateral.[/color]

[color=darkred][b][u]Note.[/u] This inequality is true not only for $ I$, but every point in the plane of triangle.[/b][/color]


geometry function algebra domain perimeter symmetry

I went to this "junior" (A.K.A <=grade 10) math battle, and was stunned at how hard the questions were. Here are some that I still don't get:

2. Among 24 coins, identical in appearance, two are counterfeit: one being heavier and one being lighter than a real one. Find the minimal number of weightings needed to decide whether the total weight of the two counterfeit coins equals to the total weight of two real ones? ( find with proof)

4. Integer A has B (including A) divisors, while B has A/2 divisors. Find the number of divisors of the number (A+2B) (find with proof)

7. Each member of the Club hates exactly three other members. The Club President wants to split the members into n groups so that in each group nobody hates each other. Find the minimal number of n, when this is possible. (I think that the answer was 7?)

8. An angle and a point outside of it are given. Construct a line through this point which intersects the angle, forming a triangle with a given perimeter.

(Add on)
9. In a summer camp, the sport field has the shape of convex quadrilateral. The director of the camp split the field into four parts by two straight lines, connecting the midpoints of opposite sides and asked four students (A, B, C, D) to get rid of the poison ivy. When the job is over, it occurs that A, B, C cleaned the parts with areas 100,200,250, respectively. What is the area of the part cleaned by Daniel?

Peng Shi

Anyone check my solution for #4 ?

I haven't found a nicer solution, but here it is:

[hide]

First define D(N) as the number of divisors of N.

let X = a1 * a2 * a3 * ... * an, where a1, a2 ... an are distinct primes.

Now let Y = D(X) = 2^n.

1. If n>= 3:

an >= 5 with equality occurs only when X = 2 * 3 * 5.

So X/2 >= a1 * a2 * a3 * ... * an/2 > 2 * 2 * 2 * ... * 2 (n 2's)

Leading to X/2 > 2^n = Y -->  Y < X/2.

If A = N * X, where N is any positive integer with only the prime factors of X.  It

means A has at least 3 factors.

Also, from a simple proof, we know that D(A) <= N * D(X).

meaning B = D(A) <= N * D(X).

We know D(X) = Y < X/2, so B <N * X/2 = A/2.

However, from our question, D(B) = A/2 --> B>= A/2.  Contradiction with B < A/2.

From this, we know A has less than or equal to 2 factors.

2. A has 1 factor, n = 1.

Because D(B) = A/2 is an integer, A is even.

So A's only factor is 2.  A = 2^c.

We can test that when c<=2, only c=2 works.

Where A = 2^2 = 4, B = D(A) = 3. D(B) = 2 = A/2.

When c>2, meaning A is multiple of 8.

However, consider even integers x and 2x that are multiples of 4.

Let's say x has divisors 1, 2, 4, d1, d2, d3 ... x

So 2x has divisors 1, 2, 4, 8, 2d1, 2d2, ... 2x

By creating two sets in which set 1 contains all the divisors of x and set 2 contains

two times the divisors of x (which are divisors of 2x), we find that set 2 and set 1

has at least two common number - 2 and 4.  It means that

Under worst condition, 2D(x) > D(2x) + 1.

So when A = 4, 2D(A) >= D(2A) + 1 --> 2B >= D(2B) + 1.

B = D(2A) = D(2*4) < 2D(4) - 1 = 5

It leads to B <= 4.  But B >= A/2 = 4.  So B = 4, which doesn't satisfy D(B) = A/2.

Similarly, for any c>3, we can find that D(B) < A/2. by D(2x) < 2D(x) - 1.  Contradiction.

Thus A = 4 in this case, where B = D(A) = 3.  D(A + 2B) = 4.

3. A has 2 factors.

Now X>= 2 * 3.

It's the only case when Y = D(X) = 4 > X/2 = 2 * 1.5.

Now A = 2^f * 3^g.

By similar arguments as case 2, we know that A can only be 6.

Now B = D(A) = 4.  D(A + 2B) = 4.

We conclude that D(A + 2B) = 4.

[/hide]

Any essential errors?

[quote="billzhao"][quote="pengshi"]7. Each member of the Club hates exactly three other members. The Club President wants to split the members into n groups so that in each group nobody hates each other. Find the minimal number of n, when this is possible. (I think that the answer was 7?)[/quote]

During the contest, I thought that hatred was mutual. In which case 4 groups is enough. But the solution is practically the same in either interpretation.

I'm just going to prove for sufficiency. You can find a example to show necessity yourself.

Let us first arbitrarily put the members in the 7 groups. Let d denote the number of pairs of members that are (1) in the same group and (2) one hate the other. If d>0, then we can choose such a pair. Since the hated person hate/is hated by no more than 5 other people, there is a group where no person hates/is hated by the aforementioned person. By moving that person into the aforementioned group, we can reduce d.

After a finite number of such steps, we can get d to be zero, and we are done.[/quote]

Why can't there by > 5 other people hating the person?  How many people are there?  We're assuming that hatred is not mutual, right?

[quote="schulmannerism"]Solution for #2:
Answer: 6[/quote]

No, we've shown that 3 weighing is good enough.
Note that the problem asks whether the two counterfeits add up to two real ones. Not which ones are counterfeits.

I think that anipoh's solution is right, but I find all this talk with N unneccessary and confusing, here's my version of your solution (basically the same idea):

Define D(x) as the number of divisors of integer x. If x=p1^a1 * p2^a2 * p3 ^ a3 ..... Then D(x)= (a1+1)(a2+1)(a3+1)
Also, we know from the question that A is divisble by 2

     Lemma 1: D(x)<=x for all integer x.
Proof: pi^ai - ai - 1>=2^ai - ai -1
When ai = 0, 2^ai-ai -1=0.
Since the function 2^y-y-1 is increasing when y>=0, we have p^ai >= ai+1 . Taking the product of this over all ai's, the result follows.

     Lemma 2: All prime divisors of A are less than 5.
Proof: By lemma 1, A >=B >=A/2
==> (a1+1)(a2+1)... >= (1/2) * 2^(a1) * p2^a2...
if on the RHS, there's some pk >=5, then pk^ak - 2ak - 2 >= 5^ak - 2ak - 2>0 (By a similar argument as in Lemma 1)
==> (1/2)(pk^ai) > ak + 1
Since for all i's ai + 1 <= pi^ai, then multiplying all terms, we have LHS < RHS, contradiction.

This means that A can only be in the form 2^a * 3^b, considering that
B=(a+1)(b+1) >= 2^(a-1) * 3^b
Since 2^(a-1) * 3^b - (a+1)(b+1) is an increasing function in the desired domain for both a and b, we can obtain the maximum values of a or b for which this function is negative by substituting in the smallest possible values for b or a. Also since, a>=1, we get that (a, b) can only be one of  (1, 1) , (2, 0), (3, 0). This implies that A can only be 4, 6, 8.

Testing, we find that A can not be A so (A, B) = (4 , 3) or (6, 4). In both cases A + 2B has 4 divisors.

Peng Shi

I meant 3 weighings, that's what I proved it for, I changed it in the original post.

Nice revision of my solution.  I dunno why I forgot to apply "the divisor formula" for calculating B, which would make the solution considerably easier to understand.

[quote="pengshi"](Add on)
9. In a summer camp, the sport field has the shape of convex quadrilateral. The director of the camp split the field into four parts by two straight lines, connecting the midpoints of opposite sides and asked four students (A, B, C, D) to get rid of the poison ivy. When the job is over, it occurs that A, B, C cleaned the parts with areas 100,200,250, respectively. What is the area of the part cleaned by Daniel?
[/quote]
I solved this question for fun because it really easy to me (maybe I'm wrong).  Since you are dealing with midpoints, you just connect the intersection of those two lines,  say O, with all four vertices.  Next, use the areas of the triangles to get three equations, (we only have 4 variables here) thus leading to an answer.  (the total area of two of the triangles)

But how do you know they are the same height?

[quote="anipoh"]Um, but those triangles share a vertex right?[/quote]

Sorry, I thought you were thinking of bases, and I wasn't paying attention.  Yeah, that's how you do it (at least I think...).

[quote="pengshi"]4. Integer A has B (including A) divisors, while B has A/2 divisors. Find the number of divisors of the number (A+2B) (find with proof)
[/quote]

This problem becomes simpler and simpler after thinking about it.

Except for n = 1 or 2, [tex]\tau(n)<n[/tex], which is obvious since all divisors of n are less or equal to n.

So after checking the cases where [tex]A=\tau(A)[/tex] or [tex]A/2=\tau(A)[/tex], we must have [tex]A > \tau(A) > A/2[/tex]. But no divisors are A are in [tex](A/2, A)[/tex]. It follows that the divisors are A are 1, 2, ..., A/2, A. And it's not hard to deduce that A must be 4 or 6.

[quote="probability1.01"]Could you clarify how knowing that the divisors are 1, 2, ... A/2, A helps?  I'm a bit slow :-)[/quote]

It's easy from there, even if you want to do some case-cracking.

Note that [tex]\frac{A}{2}[/tex] and [tex]\frac{A}{2}-1[/tex] are relatively prime. Hence [tex]\frac{A(A-2)}{4}[/tex] is a factor of A. So [tex]\frac{4}{A-2}[/tex] is an integer. So [tex]A-2[/tex] can be either 4, 2, or 1. But A is even. So A=4 or 6.

[quote="pengshi"][quote="billzhao"]Let us first arbitrarily put the members in the 7 groups. Let d denote the number of pairs of members that are (1) in the same group and (2) one hate the other. If d>0, then we can choose such a pair. Since the hated person hate/is hated by no more than 5 other people, there is a group where no person hates/is hated by the aforementioned person. By moving that person into the aforementioned group, we can reduce d.[/quote]

Why is the hated person hated by no more than 5 other people?[/quote]

Good question.

Maybe hate is, in fact, mutual, but then the answer wouldn't be 7... maybe there are only a limited number of people?

[quote="pengshi"]

How do you know that the answer can't be
200+100-250=50
or 200+250-100=350

Since you don't know the relative positions of the 4 areas?

Peng[/quote]

I guess it's three cases then.  Didn't think too carefully about that one.

[quote="pengshi"][quote="anipoh"]
I solved this question for fun because it really easy to me (maybe I'm wrong).  Since you are dealing with midpoints, you just connect the intersection of those two lines,  say O, with all four midpoints.  Next, use the areas of the triangles to get three equations, (we only have 4 variables here) thus leading to the answer 150.  (the total area of two of the triangles)[/quote]

How do you know that the answer can't be
200+100-250=50
or 200+250-100=350

Since you don't know the relative positions of the 4 areas?

Peng[/quote]

I think that there can be only 1 answer, and that we are supposed to prove the other two to be impossible. Or else this problem would be too trivial, which dosn't fit the whole test, which was HARD.

[quote="schulmannerism"]What exactly is a math battle?

Solution for #2:
Answer: 3
We're going have to have every pair of coins on different sides of the scale at some point. First find the minimum number of sets of coins such that each coin is not a member of the exact same group of sets as any other coin. Say each coin has has a binary number written on it representing which sets it's a member of. If each coin has an n-digit number, there are 2^n possible unique numbers. Thus 24 coins requires log[size=75]2[/size]24 = 5 sets. However, we can only produce an even number of sets by doing weighings. Therefore 6 sets becomes a lower bound, meaning 3 weighings since each weighing produces two sets. Here's how we doing the weighing: we give each coin a binary number from 0-31, and on the kth weighing we put all coins with a 0 in the 2k-1 place on the right, and all coins with a 0 in the 2k place on the left.[/quote]

It's a Cool idea, but what if the coin has a number like 8, which dosn't have 0's in its binary representation, therefore is neglected? Also, how do you interpret the results of this kind of weighings to tell if the two counterfeit coins have the same total weight as two real ones? Furthermore, this dosn't guarantee that there is the same number of coins on each side during each during. Without this, weighing is pointless because the side with more coins will definitely be heavier.

[quote="pengshi"][quote="probability1.01"]Ah, ok.  I missed the 1, 2, 3, ... A/2 - 2, A/2 - 1, A/2 part.[/quote]
Let d(n) denote the number of divisors of n.
Since d(A)>A/2, and because the only possible divisors of A must be in the range {1,2,3,4,....A/2-1,A/2,A}, it follows that A must have all these A/2+1 divisors.[/quote]

I missed it but I got it after I re-read ;-).  Thanks anyway, though.

Can anyone post #5 and #6 as well if they are not too easy for advanced section?  I may have left the paper with the questions at somewhere in U of T and I haven't looked at #5 and #6 yet since they were not assigned to my team.
Thanks!

Alright, here's #5 and #6.

5. In how many ways one can choose a subset of the set {1,2,...n} so that there are no three consecutive numbers among the chosen? for:
a) n is arbitrary
b) n=13

6. A magician has one cat and many mice. He can turn some mice into cats or some cats into mice according to the rule: if the number of cats and mice are not the same, then the smaller number of animals must be doubled. After applying this rule 17 times, for the first time the number of cats becomes twice the number of mice. Find the total number of the animals.

Peng Shi

I'm pretty sure my solution works. It makes sure that for each pair of coins there is a weighing where they aren't in the same place.

However, what you said about there sometimes being more on one side than the other is true, and there was another problem with my method--it sometimes required that the same coin be on two sides at the same time! However, both of these problems can be solved by numbering the coins cleverly. I just wrote down all of the numbers that didn't have 0's in both the 1st and 2nd place, 3rd and 4th place, or 5th and 6th place. This solved the problem of the coin having to be in two places at once. Then we were left with  27 numbers, but luckily these numbers satisfied the requirement that the same amount would be on each side in each weighing. I had to remove three, but luckily this was possible to do without messing up everything else. I removed 111111 since it was never used anywhere, and the two numbers 111101 and 111110 since they are each used once on opposite sides of the scale.

010101,010110,010111,011001,011010,011011,011101,011110,
011111,100101,100110,100111,101001,101010,101011,101101,
101110,101111,110101,110110,110111,111001,111010,111011

for #5, would the graders accept a recursive solution, because using recursion trivializes the problem considerably.

In fact, recursing is what I used for #5. It turned out that f(n)=f(n-1)+f(n-2)+f(n-3), if f(n) is the number of ways of choosing subsets of {1,2,...n} without picking three consecutive numbers.

Peng

for problem 4, my solution is as follows:
if $d(n)$ is the number of divisors (of n) then clearly $d(n) \leq \frac{n}{2}+1$ (all divisors are between 1 and $\frac{n}{2}$ and we also count n itself)
Now, $B \leq \frac{A}{2}+1$ and $\frac{A}{2} \leq \frac{B}{2}+1$ or $B \leq 4$ and we are left with the pairs (6,4) and (4,3), in both cases the number of divisors is 4.

for problem 8 (with the given perimeter), the construction is as follows (proof is easy)
(1) Construct circle with radius p/2 and centre in the vertex (of the angle) and let it intersect the two straight lines (forming the angle) at A and B respectively.
(2) Construct a circle tangent to the two lines at A and B.
(3) Construct tangent to the last circle constructed (of course through the given point). The circle becomes  the excircle of the triangle which is formed by the last line)
The perimeter of the resulting triangle is clearly p/2 + p/2 (which to my knowledge equals to p)
 :P

Is hating a symmetry relation or not ? if it is i think that it is trivial that $4$ group is enough !! but if it isn't again it s trivial that $7$ group is enough !!  :D


Για $ a,b,c$ θετικούς πραγματικούς αριθμούς με $ a\plus{}b\plus{}c\equal{}1$ να αποδείξετε ότι:

$ \frac{a\plus{}b}{c}\plus{}\frac{b\plus{}c}{a}\plus{}\frac{a\plus{}c}{b}\plus{}2\geq 24(a^2\plus{}b^2\plus{}c^2)$

mim ευχαριστώ για την αφιέρωση . Η άσκηση είναι πολύ καλη .

Να η λύση μου .
Η ανισότητα είναι ισοδύναμη με $ \sum (a\minus{}b)^2(\frac{1}{ab}\minus{}8)\geq 0$

Αν ονομάσουμε $ S_a\equal{}\frac{1}{bc}\minus{}8$ κτλ

Τότε προφανώς αν χωρίς βλάβη της γενικότητας υποθέσουμε ότι $ a\geq b\geq c$ τότε
$ S_a\geq 0$ και $ S_b\equal{}\frac{(a\plus{}b\minus{}3c)^2\plus{}9(b\minus{}c)}{ac}\geq 0$

Επομένως αρκεί νδο $ S_c(a\minus{}b)^2\plus{}S_b(a\minus{}c)^2\geq 0$

Όμως $ a\minus{}c\geq a\minus{}b$ άρα αρκεί νδο $ (a\minus{}b)^2(S_b\plus{}S_c)\geq 0$

Όμως $ S_b\plus{}S_c\equal{}\frac{1}{a}(\frac{1}{b}\plus{}\frac{1}{c})\minus{}16\geq\frac{1}{a}\frac{4}{1\minus{}a}\minus{}16\equal{}\frac{4(2a\minus{}1)^2}{(1\minus{}a)a}\geq 0$

άρα qed  :D

2 μπράβο Σιλουανέ  :thumbup:  Ενα γιατί έλυσες την άσκηση κι ένα γιατί δεν έμεινες στη λύση με mixing variables αλλά βρήκες άλλη πολύ πιο elementary :wink:

Βάζω κι εγώ τη λύση που έιχα υπόψη μου η οποία χρησιμοποιεί το θεώρημα-ανισότητα ABC.

Η ανισότητα ισοδυναμεί με την:

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\minus{}24(a^2\plus{}b^2\plus{}c^2)\geq 1$  ή ισοδύναμα

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\minus{}24(1\minus{}2(ab\plus{}bc\plus{}ca))\geq 1$  ή ισοδύναμα

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\plus{}48(ab\plus{}bc\plus{}ca)\geq 25$

Θετουμε $ ab\plus{}bc\plus{}ca\equal{}\frac{1\minus{}q^2}{3}$  με $ 0\leq {q}\leq 1$.

και σύμφωνα με το θεώρημα που προανέφερα ισχύει: $ abc\leq\frac{(1\minus{}q)^2(1\plus{}2q)}{27}$

Αρα έχουμε:

$ LHS\equal{}\frac{1\minus{}q^2}{3r}\plus{}16(1\minus{}q^2)\geq\frac{9(1\plus{}q)}{(1\minus{}q)(1\plus{}2q)}\plus{}16(1\minus{}q^2)\equal{}$

$ \equal{}\frac{2q^2(4q\minus{}1)^2}{(1\minus{}q)(1\plus{}2q)}\plus{}25\geq 25$

[b]Παρατήρηση[/b]: Εχει ενδαφέρον το γεγονός ότι η ισότητα παραπάνω απιτυγχάνεται εκτός απ' την περπίπτωση που $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ και στην περίπτωση που $ a\equal{}\frac{1}{2}, b\equal{}c\equal{}\frac{1}{4}$

Καλή άσκηση mim με πολύ ωραίες λύσεις  :) . Βάζω και γω επιγραμματικά τη λύση μου, η οποία επίσης χρησιμοποιεί την ιδέα των mixing variables, αλλά σε πιο κλασσική μορφή  :wink: .

Όπως και στη λύση του mim η ανισότητα είναι ισοδύναμη με
\[ f(a,b,c) \equal{} \sum \frac {1}{a} \plus{} 48\sum ab \minus{} 25 \geq 0\ .
\]
Επειδή η $ f(a,b,c)$ είναι συμμετρική ως προς τα $ a,b,c$ υποθέτουμε χωρίς βλάβη της γενικότητας $ a \geq b \geq c$ και έστω $ m \equal{} \frac {b \plus{} c}{2}$. Τότε μπορούμε εύκολα να δείξουμε ότι $ f(a,b,c) \geq f(a,m,m)$. Όμως
\[ f(a,m,m) \equal{} f(1 \minus{} 2m,m,m) \equal{} \frac {2(3m \minus{} 1)^2(4m \minus{} 1)^2}{m(1 \minus{} 2m)} \geq 0,
\]
που είναι το ζητούμενο.

Καλά δεν το συζητάω ότι έχετε απίστευτες γνώσεις και απίστευτες λύσεις και οι 3.  :D
Απλά επειδή βλέπω συνεχώς mixing variable και δεν έχω ιδέα τι είναι  :huh:  αν και περίπου καταλαβαίνω τη μεθοδολογία :lol:  μπορεί κανείς να με διαφωτίσει λίγο σύντομα ή να με παραπέμψει κάπου? :P

Κώστα, σίγουρα δεν είμαι ο πιο κατάλληλος για να απαντήσω σε αυτή την ερώτηση.
Εξάλλου ελάχιστα ξέρω αυτή τη θεωρία και δεν έχω λύσει ποτέ πρόβλημα χρησιμοποιώντας τη :(

Πάντως νομίζω ότι η ανισότητα mixing variables είναι αυτή που περιγράφεται εδώ:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=113248[/url] αν και η αλήθεια είναι ότι έχω μπερδευτεί ως προς το αν αυτή είναι η ανισότητα στην οποία αναφερόμαστε ή είναι διαφορετική επειδή αυτή είναι "strong(er)".

Εξάλλου όλες αυτές οι ανισότητες (SMV, EV, MV, n-1 EV) έχουν πολύ παρόμοιες ονομασίες αλλά είναι αρκετά διαφορετικές (νομίζω...).

Πάντως εν αναμονή και της απάντησης του Σιλουανού (που είναι Master στις ανισότητες) ή του Μιχάλη ή όποιου άλλου γνωρίζει σε παραπέμπω και σε αυτό το αρχείο που αναφέρεται στην μέθοδο S.M.V.
[url]http://reflections.awesomemath.org/2006_6/2006_6_mixing.pdf[/url]


analytic geometry
graphing lines
slope
Pascal\u0027s Triangle

The attachment is what the slope  looks like.
The question is that how many ways are there from the top to the bottom. We've been doing some problems with Pascal's Triangle lately.

Lets just say a math teacher... :cursing:

it is really similar to pascal's triangle.

just label the diagram like pascal's triangle and then find what number is on the bottom point.

it's

[hide]200 i think[/hide]

Hm... Its kinda messed up though, Not many of the dots align. I mean the intersections. You do it the same regular way right? The way where you list the numbers of pascal out?

I got 200 too:
[img]http://www.part.lt/img/aa1a538b96271da662a8e5adfc363bcc920.JPG[/img]


[quote="catcurio"]Source: 2005 AMC-10B #18

All of David's telephone numbers have the form $555-abc-defg$, where $a$, $b$, $c$, $d$, $e$, $f$, and $g$ are distinct digits and in increasing order, and none is either $0$ or $1$. How many different telephone numbers can David have?

$\text{(A)} \ 1 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ 7 \qquad \text{(D)} \ 8 \qquad \text{(E)} \ 9$[/quote]

[hide]I got (C) $7$
$a$ either starts with $2$ or $3$. If $a$ starts with $2$ then there are $6$ possibilities because each digit can be changed resulting in the other digits determined, and if $a$ starts with 3, there is one combination. $6+1=7$[/hide]

[hide]
I got something a little different.
we start with either 2 or 3. Starting with 3, there's only $1$ possibility.
Starting with 2..
2,(3,4,5,6,7,8,9)
from the 7 numbers in the brackets choose 1 that isnt included. 7C1 = 7 + $1$ = 8
[/hide]

[hide]

we only have 2 to 9 which is 8 numbers . Since there are seven space to fill , we have to give up one number . There are $8C1$ ways to give up . So answer is $\boxed{D}$
[/hide]

[hide]Possible numbers are 2 through 9. 8 possible numbers and only 7 spots means one digit has to be left out. 8 possible number to leave out, so there are 8 different seven digit numbers. $\boxed{D}$[/hide]


Let $ a,b,c$ be the three length sides of a triangle. Does the following inequality hold or not:

$ \large 12\sum_{cyc}^{a,b,c}ab(a\plus{}b)\geq a^{3}\plus{}b^{3}\plus{}c^{3}\plus{}57abc$

It is second half of the inequality from Vietnam, mathematics and young friends magazine,  please don't post a reply. :mad:.
@ Mashimaru, please does not do a same things again

No, the inequality ${ \large\sum_{cyc}^{a,b,c}\frac{a}{\sqrt{a^{2}+3bc}}\geq\frac{3}{2}}$ (where $ a,b,c$ are 3 lengths side of a triangle) on Mathematics and Youths can be easily proved by declaring and applying CBS and Schur's inequality. This one is not, it is my homework. It is even stronger than that Ineq >.<

(Anyway, I see, if you apply Holder's ineq to the left hand side of the Ineq on Mathematics and Youths, you will receive the ineq I've posted. So please do not give the solution too soon, I'll try proving it myself >.<)


Duke
college
trigonometry
ARML
number theory
relatively prime

Obviously, East Chapel Hill won the thing.  SC did decent, but we didn't place.  I got a 4, (problems 1, 2, 5, 8).  How did other people do individually (I assume Arnav got a 7?).

Well I only got a [size=59]5[/size]...

And my team sucks, lol

TJ had first place individual and then got owned

Vitek has gotten a lot better. OMFG AN 8

[quote="MysticTerminator"]dude note that numbers 1 through 4 on the individual were like second-grade computation I was all like wtf I have forgotten how to multiply
[/quote]

hm, so basically you have to be an idiot to get less than a 4?

[quote="MysticTerminator"]dude note that numbers 1 through 4 on the individual were like second-grade computation I was all like wtf I have forgotten how to multiply

power round was pretty nice though I'd prefer it if they'd kept it in the original form, team round was pretty ok, I will pass over the individual besides the comment above, relays were just like eh relays w/e although the second one was pretty funny since the last person only had 3 possible answers, that last thing was like wtf[/quote]

let me guess the problem you got wrong was

[quote]Paul rolls a standard 6-sided die and records the results. What is the probability that he rolls a 1 ten times before he rolls a 6 twice?[/quote]

[quote="jb05"][size=150]Concerning the number 7[/size]

Arnav got a 7 and was the only person at East Chapel Hill who got at least 7. I got 7, John Pardon got 7, Xantos C. Guin got 7... that's all I know who got 7s. They really should have at least acknowledged us (I bet there weren't many of us).

Ok, so... I know how to do number 7 (I figured it out in the last minute but didn't have time and didn't think of the really trivial way), but I really want to know if anybody got it without guessing.[/quote]

dude what you've got arctangents I would suggest taking the tangent it seems like a pretty natural thing to do

hm i can see why a lot of ppl got it wrong. i just did the problem right now, and i carelessly put $0$, but that would only work if it was $\tan{(\arctan{1}+\arctan{2}+\arctan{3})}$, but the answer should be $\tan{S}=0$, (where S is the sum)...so yea, it was easy to make trivial mistakes...is that the one you missed arnav?

Arnav, I assume you missed #3 or #7?  Shobhit: I got a 4, and so did Fangda and Jim.  That was the SC Team High Score.  They didn't give out grade-level high scorers, so idk if there were any freshman 5s.

wtf? all you south carolinians need to go back from whence you came

it is not my fault if you have a contest without prizes that wants to kill you all

aha it all makes sense now

so do you want those 2 USD or what

[quote="mysmartmouth"]Arnav, I assume you missed #3 or #7?  [/quote]

Are you kidding? there was a reason the whole room laughed when we started the fourth set of problems, and that's cause #7 is TOO easy.

I got a six. Added wrong on #2 and totally misread #6 (thought they were relatively prime only to adjacent terms). Oh well - we had a 7 on our team, and I figured they were doing Lightning with 7's & 8's. Too bad.

Team #10 was cool.

#7 usually is easy on ARML style tests (#7 on ARML last year was even easier).  Ideally #6 and #8 should be the two hardest (due to the style of the test and how it's timed).

Uh, #7 on last year's ARML was the right difficulty; about half a minute or so, where you actually have to work to find the right answer. This year's Duke #7 was way too easy; I didn't even finish reading the Q before I put down pi.

[quote="jb05"][quote]dude what you've got arctangents I would suggest taking the tangent it seems like a pretty natural thing to do[/quote]

Um... I did take the arctangent. Like in the first ten seconds. And I got 0 and came back right after he said a minute and started stupidly plugging in 2-pi and 3-pi for those tangents. I didn't have enough time. I should have realized that it had to be a multiple of pi anyway-then I would have been done.[/quote]

Yes, that's what I did, but I did not realize it was wrong until I only had about 30 seconds left...

@Chess64:  That problem was from an old exam.

yeah I know but arnav was pretending he didn't know how to do those

[quote="not_trig"]ARML 7 was much easier.[/quote]

There's something hard about taking the tangent of two arctangents?

if you have not seen a similar type of problem before, duke #7 is not easy

arml #7 on the other hand, if you know basic log rules, you should get

I don't see how someone who has a chance of doing well hasn't seen enough math not to take tans when you have two arctans

edit> oh yea what's up with the drivers in NC? They are TERRIBLE

I just saw results...

8: Daniel Vitek, William Drobny, Vivek Bhattacharya, Haitao Mao
7: Ryan Muller, John Pardon, Arnav Tripathy, Yevgeniy Grechka, John Berman, Santosh Karnik, Noah Blach

1. East Chapel Hill A
2. Georgia A
3. Vestavia Hills
4. Enloe A
5. TJ A
6. NCSSM A
7. Georgia B
8. Winstom-Salem A
9. Hoggard A (yay!)
10. TJ B
11. Enloe B
12. SC

12th Place.  That's really not all of SC, that's one school in SC, plus one last minute replacement from a different school.  I was hoping for higher, but we got beat by some pretty darn good schools ( a majority from NC).


Two $ n$-digit integers (leading zero allowed) are said to be equivalent, if one is a permutation of the other. For instance, 10075, 01075 and 00751 are equivalent 5-digit integers.
i. Find the number of 5-digit integers such that no two are equivalent.
ii. If the digits 5,7,9 can appear at most once, how many non-equivalent 5-digit integers are there?

i. in other words, the number of 5-element multisets with elements in $ \{0,1,2,...,9\}$... and that's just the number of non-negative integer solutions to $ x_1\plus{}x_2\plus{}\cdots\plus{}x_{10}\equal{}5$, which is $ \binom{14}{5}$.

ii. the number of non-negative integer solutions to $ x_1\plus{}x_2\plus{}\cdots\plus{}x_{10}\equal{}5$ with $ 0\leq x_5,x_7,x_9\leq 1$...

if all three of these variables is 0 (which happens in 1 way), the number of solutions is $ \binom{11}{5}$. if exactly one of them is 1 (which can happen in 3 ways), the number of solutions is $ \binom{10}{4}$. if exactly two of them are 1 (which can happen in 3 ways), the number of solutions is $ \binom{9}{3}$. and if all three of them are 1 (which happens in 1 way), the number of solutions is $ \binom{8}{2}$. so the answer is $ \binom{11}{5}\plus{}3\binom{10}{4}\plus{}3\binom{9}{3}\plus{}\binom{8}{2}$.


AMC
AIME
USA(J)MO
USAMO
Pascal\u0027s Triangle
SFFT
special factorizations

in the AIME solution packets, i've been seeing a lot of solutions involving factoring equations. sometimes it's hard for me to see what to factor them into.

for example:

xy - x(n/2) - y(n/2) = 0
(x-n/2)(y-n/2) = n :^2: /4

and another:

mn + 29m - 29n = 0
(m-n)(n+29) = -29 :^2:

i always understand it after seeing it done, but i can never 'see' it on my own.

i'd appreciate tips and/or practice questions

thanks

mathkid...

the trick to knowing how to factor is doing a lot of problems, failing a lot, and learning from your mistakes. You recognize popular forms very quickly after doing hundreds of problems.

However, there are certain tricks to factoring...

-Always look for roots. Even variable roots, such as x=y or x=y^2 to know that (x-y) and (x-y^2) may be factors. FTOA is always good.

-Look for completion of squares. Such a simple idea is sometimes very cleverly disguised in a specific problem. Squares are nice because the solution is elegant but the problem looks tough.

-Factoring by grouping. Not my favorite method, but a good option.

-Memorize Pascal's triangle up to n = 6. This could help you recognize some forms...

-Learn some generating functions. These are definately good practice excercises.[/hide]

Heres an example of why you may work out how to (partially) factorise something like mn + 29m - 29n = 0.

If we want to factor something like that, lets put it in two brackets. We want a mn term, and nothing else has mn's in it, so we may as well put m and n in opposite brackets. (m....)(n....)
Now how do we get a 29m term? (m.....)(n + 29)
And how do we get a -29n term? (m-29)(n+29).
Then the other term that we get is -29^2.

Just think about what the two (or more brackets) will have to look like. That may help a bit.

[quote="TripleM"]Heres an example of why you may work out how to (partially) factorise something like mn + 29m - 29n = 0.

If we want to factor something like that, lets put it in two brackets. We want a mn term, and nothing else has mn's in it, so we may as well put m and n in opposite brackets. (m....)(n....)
Now how do we get a 29m term? (m.....)(n + 29)
And how do we get a -29n term? (m-29)(n+29).
Then the other term that we get is -29^2.
[/quote]

Yeah, this factoring method is known as "Simon's favorite factoring trick"  :mrgreen:

lol, I'm just kidding. But some people here will know what I mean. ;)

If you say ``Simon's favorite factoring trick'' on the USAMO, you will automatically get 8 points on that question.

Explain the best way to factor these:

y :^2: + 2xy + 40x - 400 = 0
y :^2: - 2xy + 40x - 400 = 0

Try using the same method as I just did before..

Also, it only works if the numbers you are looking for are integers.

Well not in this case.. in this case, it can be for all reals.

[quote="Idiot_without_a_village"]Yo, I believe x^4+4 factors funkily too. . . what's the deal with that? Can we like generalize something from that. . .[/quote]

Sophie Germain's identity. YOu can factor a^4 + 4b^4. Just write it as (a^2 + 2b^2)^2 - 4a^2b^2 and you see that it is a difference of two squares.


consider numbers written in base b, where 2<b<9.
                                                           -  -
clearly indicate how you can tell quickly whether the number b is odd or even.

examples
13269b(base b) = 996 is even number
2039b(base b) = 165 is odd number
21245b(base b)=289 is odd number
1226b(base b)= 50 is even number

any help would be greatly appreciated.

Start the other way around:
Assume you know the base $ b$, find a way to describe if a given number in base $ b$ is even or not.

That helps but I have tried that and this problem has had me stumped for 2 weeks. I am just mad at myself I know there is probably a simple answer and that upsets me.

Is the question to decide whether $ b$ is odd or even or to decide whether $ n_b$ is odd or even?

whether the number b is odd or even and how you could quickly tell that it was so.

Two hints:  first, rewrite everything in terms of $ b$.  So the examples become

$ 9 \plus{} 6b \plus{} 2b^2 \plus{} 3b^3 \plus{} b^4$ is even,
$ 9 \plus{} 3b \plus{} 2b^3$ is odd,
$ 5 \plus{} 4b \plus{} 2b^2 \plus{} b^3 \plus{} b^4$ is odd,

and so forth.  Second, note that if $ b$ is odd then so are $ b^2, b^3, b^4, ...$ and if $ b$ is even then so are $ b^2, b^3, b^4, ...$.

wow thank you so much! I feel dumb but at least I learned a new approach to this problem that I will not soon forget thank you so much.


A friend asked me a question, and I thought it was nice :blush:
Find all natural number satisfy:When I delete one figure, this natural number will lessen in 71 instalments ? And we don't know how many number what this natural number have.
 [hide]Answer: $ \overline{71\underbrace{00...00}_{n \minus{} 1}}$, and the figure deleted is 7. So don't type only result, please, thanks  :blush: [/hide]
If it used to have, I'm sorry because I don't know.

I don't understand that question at all.

err....... a pointless question given that it makes no sense whatsoever... unless you mean that when one digit of the number is deleted, the number becomes $ \frac{1}{71}$ of the original number, but that's not right, because then... there would be infinite...

yes i would like to know the same thing, otherwise the post makes no sense at all.  Or maybe he means that the number decreases by a multiple of 71, or atleast that is how i interperite the question.

Oh I'm sorry, My English so bad  :(.  Wait everybody sympathetically. This is my question.
[b]Take all of natural numbers so that when erase those a natural number number of strokes letterpress there will cool off 71 times. [/b]
It is tenor in my dictionary  :blush:

[quote]To mean to be
We Call the natural number need to find is A. After erasing one digit of A we have a natural number is called B and $ B \equal{} \frac {1}{71}A$. Find A [/quote]

Do you understand what I say? :maybe:

Actually It makes perfect sense now; But there's infinitely many of them... $ A\equal{}71,B\equal{}1 A\equal{}71*10^k,B\equal{}10^k$ all work.

Uh, the problem is easy to understand, but there are infinetly many solutions:

The problem is asking that given some natural number A, there is a number created B such that B is A, except with one deleted digit, and also that B=$ \frac{1}{71}$A.

The question:

We wish to find all natural numbers A such that if we erase exactly one digit of A, the resulting number is $ \frac{1}{71}$ of the original number.

Example: Let A=71.  Then if we erase the $ 7$, we have 1, which is $ \frac{71}{71}$

Oh.......... I was talking about, at first the general solution B = 71 x 10^x, but that was cuz I saw 71 x 10x. I'm good now, don't worry.

Yea, this task is very easy discovering is resulting, but I need perfect key from you  :)