binomial coefficients
combinatorics
counting
function
symmetry
IMO Shortlist

:) I was looking through some old threads.

$c(n,k)$ is, in fact, the number of vector subspaces of dimension $k$ in a vector space of dimension $n$ over $\mathbb F_2$ (the field with $2$ elements).

Anyway, we can show, by induction, that the explicit formula for $c(n,k)$ is $\frac{u_n(2)}{u_k(2)\cdot u_{n-k}(2)}$, where $u_n(a)=(a^n-1)(a^{n-1}-1)\ldots(a-1)$. It's obvious from the formula that the conclusion holds.

[quote="kunterbunt"]Show that

\[ c(n,k)=\frac{\prod_{l=1}^n (2^l-1)}{\prod_{l=1}^k (2^l-1)\prod_{l=1}^{n-k} (2^l-1)} \]

by plugging it into the recurrence.

This is a special case of the $q$-binomial coefficient.

But you can get this closed form from scratch by trying to set up an induction (over $n$) proof, where you realize that you need the relation $c(n,k)= \text{(some factor) } c(n,k-1)$. When you iterate this relation and use $c(n,0)=1$, you get the closed form above which is readily proved by the recurrence.[/quote]

@kunterbunt,I don't think its easy enough to realise to use as induction hypothesis.

For the induction step you want to link:

\[c(n+1,k)= 2^k c(n,k) + c(n,k-1)\]

and  \[c(n+1,n+1-k) = 2^{n+1-k}c(n,n+1-k) + c(n,n-k) = 2^{n+1-k}c(n,k-1) + c(n,k).\]

Therefore, the missing identity is
\begin{align*}
2^k c(n,k) + c(n,k-1) &=2^{n+1-k}c(n,k-1) + c(n,k) \\
 \Leftrightarrow (2^k-1) c(n,k) &=(2^{n+1-k}-1)c(n,k-1) \\\Leftrightarrow c(n,k) &=\frac{2^{n+1-k}-1}{2^k-1}c(n,k-1)\end{align*}

which leads you directly to the conjecture of the product form of $c(n,k)$ in my above post.

[u][b]Another nice proof[/b][/u]:

Since it is required to show that $c(n,k)=c(n,n-k)$ we search for a function of the form $c(n,k)=\frac{F(n)}{F(k)F(n-k)}$ where $F(j)=f(1)f(2) \dots f(j)$ where $f:\mathbb{N}\rightarrow \mathbb{R}$ is a function.If such $f$ exists we are done!!
Plugging it into the recurrence we get $f(n+1)=2^kf(n-k+1)+f(k)$ and taking $k=1$ we get the simple recursion $f(n+1)=2f(n)+f(1)$.This is like the famous Tower of Hanoi recursion the solution of which is $f(n)=2^n-1$.Thus we got such an $f$ and so the problem is solved.

Storage
[quote=1998 A4]For any two nonnegative integers $n$ and $k$ satisfying $n\geq k$, we define the number $c(n,k)$ as follows:

- $c\left(n,0\right)=c\left(n,n\right)=1$ for all $n\geq 0$;

- $c\left(n+1,k\right)=2^{k}c\left(n,k\right)+c\left(n,k-1\right)$ for $n\geq k\geq 1$.

Prove that $c\left(n,k\right)=c\left(n,n-k\right)$ for all $n\geq k\geq 0$.[/quote]

We claim that $c(k+r, k)=\frac{(2^{k+1}-1) \dots (2^{k+r}-1)}{(2^1-1)\dots (2^r-1)}$ for all $k, r \geqslant 1$. Indeed, this holds for base cases, and for the induction step, one simply verifies the identity (for $n=k+r$) $$2^k \cdot \left( \frac{(2^{k+1}-1) \dots (2^{k+r}-1)}{(2^1-1)\dots (2^r-1)}\right)+\left(\frac{(2^{k}-1) \dots (2^{k+r}-1)}{(2^1-1)\dots (2^r-1)(2^{r+1}-1)} \right)=\left(2^k+\frac{2^k-1}{2^{r+1}-1} \right) \cdot \left(\frac{(2^{k+1}-1)\dots (2^{k+r}-1)}{(2^1-1)\dots (2^r-1)} \right)
$$ which indeed satisfies the recursion.

[b]Remark.[/b] Though this feels unmotivated, I think the easiest way to get this solution is to calculate $c(k+r, k)$ for small values of $r$. For $r=2$, I got a suspicious $3$ in the denominator and a factorised numerator, but when for $r=3$ I got $3 \cdot 7$ in the denominator and same factors continued in the numerator with the additional $(2^{k+3}-1)$ factor, it made sense that all factors on both sides should be of the form $2^{m}-1$ and the range for $m$ is just $1$ through $r$. By plugging this in, we indeed get a solution, as shown above. This is a classic example of small cases being really useful :)

This seems to be the first purely combinatorial proof in this thread:

It's inductively easy to see that $c(n,k)$ is equal to the number of $\{L,R\}$-sequences containing $k$ $R$s and $n-k$ $L$s, where each $L$ can be connected to an arbitrary subset of previous $R$s. Given any such sequence, we can flip all $L$s to $R$s and vice versa. Reading the result backwards gives a valid sequence for $c(n,n-k)$, hence the two quantities are equal.

Dude why i didnt knew the existence of this problem before, its awsome!
[b][color=#f00]The construction:[/color][/b] Call $P(n,k)$ the assertion of that sequence. With $P(n,1)$
$$c(n+1,1)=2c(n,1)+1 \implies c(n,1)=2^n-1 \; \forall n \in \mathbb Z^+$$
Becuase of the induction and becuase $c(1,1)=1$
Now if we try $P(n,2)$ and do some induction xd
$$c(n+1,2)=4c(n,2)+2^n-1$$
Well now we need to take that $2^n$ out hence we let $c(n,2)=b_n-2^{n-1}$ to get that $b_{n+1}=4b_n-1$ and since $c(2,2)=1$ we have that $b_2=3$ and hence by easy induction we get that $b_n=\frac{2^{2n-1}+1}{3}$ which means that $c(n,2)=\frac{(2^n-1)(2^{n-1}-1)}{3}$ and now we can sus that $c(n,k)=\frac{(2^n-1)(2^{n-1}-1)...(2^1-1)}{((2^k-1)(2^{k-1}-1)...(2^1-1))((2^{n-k}-1)(2^{n-k-1}-1)...(2^1-1))}$ for all $n \ge k \ge 1$ and since this definition follows what is the problem asking us, we will prove that its true by induction.
So assume that it holds for $c(n,k-1)$ hence we get
$$c(n+1,k)=2^kc(n,k)+\frac{(2^n-1)(2^{n-1}-1)...(2^1-1)}{((2^{k-1}-1)(2^{k-2}-1)...(2^1-1))((2^{n-k+1}-1)(2^{n-k}-1)...(2^1-1))}$$
Now i make the replace $c(n,k)=\frac{(2^n-1)(2^{n-1}-1)...(2^1-1)}{((2^k-1)(2^{k-1}-1)...(2^1-1))((2^{n-k}-1)(2^{n-k-1}-1)...(2^1-1))}+L(n,k)$ and now by replacing on the equation and a lot of bashing i'm not gonna write for the sake of my sanity.
$$L(n+1,k)=2^kL(n,k) \implies L(n,k)=0 \; \forall n \ge k$$
Becuase we have that by the first condition of the problem $L(k,k)=0$
Hence our hypotesis was true meaning that we are done  :blush:

Induct on $n$ using the fact that \[\frac{c(n,k)}{c(n,k-1)} = \frac{2^{n+1-k}-1}{2^k-1}\]

lol this is just Gaussian binomial coefficient for $q=2$
Since the starting term and recurrence is same therefore $c(n,k)$ is  Gaussian binomial coefficient for $q=2$. since $ c(n,k)=\frac{\prod_{i=1}^n (2^i-1)}{\prod_{i=1}^k (2^i-1)\prod_{i=1}^{n-k} (2^i-1)} $ we're done


So I just found out that the State took out the Chemistry and Physics EOC's.  What do you guys think?  I know people here wouldn't have trouble with it anyways, but looking at the big picture good or bad?

What do you think?  Would you rather endure 4 hours of WAY too easy testing or skip it and do more fun stuff?  (I'm not in high school, but...really, if there's one person who hates EOCs, it's me).

good

then the actual CLASS is the grade, and the EOC doesn't matter much. It's too trivial anyways...

also i second professordad's point about the wasting 4 hours

In theory, the EOCs are useful, since they tell if students have a basic knowledge of the subject.  However, in the presence of SAT IIs (which don't test at an incredibly high level, but still do a better job than the EOCs), the EOCs are not particularly useful and promote poor testing.

However, the problem with removing EOCs is that a lot of the poorer teachers probably will not have much motivation to teach anything in particular, so there is the potential to have students learning nothing...  I guess the argument for the EOCs would be that at least students are learning [i]something[/i] in the class.

I totally agree with b-flat.  Especially since, due to the nature of physics, some teachers get confused themselves.  As a result, they do a really perfunctory job teaching that part.  About half of the people in my physics class understand the subject like this: "Find a formula, use it, don't ask y".  That sorta thinking might be harmful on the EOC.

Chemistry.... I don't know.  It's more about memorizing facts... Can't really say much on that one.

@professordad: Lucky you.  At my school, if one person isn't finished with any EOC test, the whole school waits in the classroom.  lol


linear algebra
matrix
induction
linear algebra unsolved

Prove that for any complex matrix $A$ of determinant 1 there exist unitary matrices $U,V$ such that $UAV$ is upper triangular, with 1 on the main diagonal.

We can use induction on dimension.

Clearly, the initial assertion is equivalent to the apparently stronger statement that any $n\times n$ matrix with determinant $d$ can be turned into an upper triangular matrix with $|d|^{\frac 1n}$ on the diagonal, so in the induction step I'll use this fact (call it $\mathcal P$) for smaller dimensions.

First of all, make the matrix upper triangular by unitary similarity, and let $d_1,\ldots,d_n$ be its eigenvalues. Multiply it with suitable block diagonal unitary matrices to make it upper triangular again, with $a_1=|d_2\ldots d_n|^{\frac 1{n-1}}$ on its last $n-1$ entries of the diagonal (possible by the induction hypothesis for dimension $n-1$), then do it again to get $a_2=|d_1a_1^{n-2}|^{\frac 1{n-1}}$ on positions $1,3,\ldots,n$ of the diagonal, then again to get $a_3=|a_1a_2^{n-2}|^{\frac 1{n-1}}$ on positions $1,2,4,\ldots,n$, and so on.

Clearly, the diagonals of the sequence of matrices that we get tend to $(1,1,\ldots,1)$, and since these diagonal entries are achieved by multiplication with sequences $(U_k)_k,(V_k)_k$ of unitary matrices to the left and right, we can extract convergent subsequences of $(U_k)_k,(V_k)_k$ which tend to $U,V$, also unitary, s.t. $UAV$ is upper triangular and has all its diagonal entries equal to $1$.


geometry
3D geometry
algorithm
vector
USAMTS

How to find all positive integers $ \leq 10^{50}$ such that the sum of its digits raised to some integer power is the same as the number?

No one's going to write the whole program for you. Show us some examples of what you have accomplished already and someone may help.

That number looks so big so I was wondering what would be easier method than check $ 10^{50}$ special cases. I guess that it is easier to compute first squares, then cubes and so on until I got a number larger that $ 50$ digits. I have still to think the case where power is one.

But long long is not a data type long enough to save the numbers so probably I need to program my own division and multiplication routines and save the numbers as linked lists.

Is this a correct method?

Your method will never finish because Sqrt[10^50]==10^25, which is going to finish when the universe blows up
It is much faster to try different sums of digits ($ 2 < \equal{} x < \equal{} 450$), and keep exponenting it until it surpasses 10^50
This needs a multiplication algorithm, and you should probably store $ x^n$ to calculate $ x^{n \plus{} 1}$ so that you don't have to write an exponentiation algorithm
A vector is enough to store the digits, which should probably be stored as unsigned long longs in base $ 10^{16}$
(approx $ \frac {2^{64}}{450}$)
For those of you who selected mathematica as a usamts prize and never used it:
[code]{#[[1]] + 1, #[[2]] - 1} & /@
 Position[
  Table[# == i & /@
    Total /@
     IntegerDigits /@
       NestList[#*i &, 1, Floor[50 Log[i, 10]]],
     {i, 2, 450}
  ],
True][/code]
Running time: 0.047s (compiled) : mathematica is impressive, my guess is that they used assembly
My own code: 0.083s using vectors in c++
0.060s using arrays in c
0.050s by replacing sprintf
just to verify, there are 104 of them (including 1 and not including 0)


Is there a way to automatically label a point without having to type in the coordinates? For example, can I  label the intersection point of (5,5)--(-5,-5) and (-4,4)--(4,-4) directly as (0,0) without finding the coordinates?

I'm not sure if that's what you ask, but the [b]intersectionpoint(P,Q)[/b] command gives you the intersection point of the paths $ P$ and $ Q$. So, if you write
[code]
path P=(5,5)--(-5,-5), Q=(-4,4)--(4,-4);
pair R=intersectionpoint(P,Q);
draw(P);
draw(Q);
dot(R);
label("$A$",R,S);
[/code]
you'll get
[asy]size(100);
path P=(5,5)--(-5,-5), Q=(-4,4)--(4,-4);
pair R=intersectionpoint(P,Q);
draw(P);
draw(Q);
dot(R);
label("$A$",R,S);[/asy]

PS. After reading your question more attentively, I think you may want something like this:
[asy]size(100);
path P=(5,5)--(-5,-4), Q=(-4,4)--(4,-4);
pair R=intersectionpoint(P,Q);
draw(P);
draw(Q);
dot(R);
string u="%#.1f";
label( "$("+format(u,R.x)+","+format(u,R.y)+")$", R,S,fontsize(8) );[/asy]
(I specially changed the lines to make the coordinates ugly and introduced the format string to cut decimal precision to 1 decimal place; the available formats are the same as for printf command in C++).


ratio
geometric series
geometric sequence

A flea jumps 1/2 ft right, then 1/4 ft, 1/8 ft right, and so on. To what point is the flea zooming in?

I know this is an infinite geometric series problem but am having trouble incoporating the change of direction.

Wait, do you mean 1/2 right, 1/4 left, 1/8 right, 1/16 left,..., since you said "change of direction"

If so...

[hide]
The first term is $ \frac {1}{2}$, the ratio is $ \minus{} \frac {1}{2}$, so

$ \frac {\frac {1}{2}}{\frac {3}{2}}$

$ \equal{} \frac {1}{3}$[/hide]

Yes that is what i meant. Sorry for the error.

[hide]Consider the jumps two at a time.  First he jumps 1/2 foot right and 1/4 foot left.  So add these to get a total of 1/4 foot right.  The next two are 1/8 foot right and 1/16 foot left.  Add these to get a total of 1/16 foot right.  Continuing in this manner, we see we want the sum of the infinite geometric sequence:

$ \frac {1}{4} \plus{} \frac {1}{16} \plus{} \frac {1}{64} \plus{} ...$.  This is equal to $ \frac {a}{1 \minus{} r}$, where a is the first term, and r is the ratio.  So $ \frac {a}{1 \minus{} r} \equal{} \frac {\frac {1}{4}}{1 \minus{} \frac {3}{4}} \equal{} \boxed{\frac {1}{3}}$

[hide="Alternatively"]
We could break it into the difference of two sequences:

$ (\frac {1}{2} \plus{} \frac {1}{8} \plus{} \frac {1}{32} \plus{} ...) \minus{} (\frac {1}{4} \plus{} \frac {1}{16} \plus{} \frac {1}{64} \plus{} ...)$ This is equal to $ \frac {2}{3} \minus{} \frac {1}{3} \equal{} \boxed{\frac {1}{3}}$[/hide][/hide]


inequalities
trigonometry
inequalities proposed

[quote="rogue"]Let $ A,B,C$ be angles of acute triangle. Prove the inequalities:
a) $ \dfrac{\cos A}{\sin B\sin C} \plus{} \dfrac{\cos B}{\sin C\sin A} \plus{} \dfrac{\cos C}{\sin A\sin B}\ge 2;$ b) $ \dfrac{\cos A}{\sqrt {\sin B\sin C}} \plus{} \dfrac{\cos B}{\sqrt {\sin C\sin A}} \plus{} \dfrac{\cos C}{\sqrt {\sin A\sin B}}\le\sqrt {3}.$ (A. Kukush, M. Rozhkova)[/quote]
The first is en equality, the second is obvious after substitution $ a^2 \plus{} b^2 \minus{} c^2 \equal{} z,...$ and Cauchy-Schwartz.

[quote="rogue"]Let $ A,B,C$ be angles of acute triangle. Prove the inequalities:
a) $ \dfrac{\cos A}{\sin B\sin C} \plus{} \dfrac{\cos B}{\sin C\sin A} \plus{} \dfrac{\cos C}{\sin A\sin B}\ge 2;$ b) $ \dfrac{\cos A}{\sqrt {\sin B\sin C}} \plus{} \dfrac{\cos B}{\sqrt {\sin C\sin A}} \plus{} \dfrac{\cos C}{\sqrt {\sin A\sin B}}\le\sqrt {3}.$ (A. Kukush, M. Rozhkova)[/quote]
we can let:
$ cosA\equal{}\sqrt{\frac{ab}{(a\plus{}c)(b\plus{}c)}}$ => $ sinA\equal{}\sqrt{\frac{c*(a\plus{}b\plus{}c)}{(a\plus{}c)(b\plus{}c)}}$
a)LHS=$ \sum \frac{b\plus{}a}{a\plus{}b\plus{}c}\equal{}2$
b)$ LHS\equal{}\sum \sqrt[4]{\frac{(ab}{(a\plus{}c)(b\plus{}c)}}*\sqrt{\frac{a\plus{}b}{a\plus{}b\plus{}c}}$
$ LHS^2 \leq (\sum \sqrt{\frac{ab}{(a\plus{}c)(b\plus{}c)}})(\sum \frac{a\plus{}b}{a\plus{}b\plus{}c}) \leq 2(\sum \frac{1}{2}*(\frac{a}{a\plus{}c}\plus{}\frac{b}{b\plus{}c}))\equal{}3$
so $ LHS \leq 3$
QED

[quote="rogue"]Let $ A,B,C$ be angles of acute triangle. Prove the inequalities:
a) $ \dfrac{\cos A}{\sin B\sin C} \plus{} \dfrac{\cos B}{\sin C\sin A} \plus{} \dfrac{\cos C}{\sin A\sin B}\ge 2;$ b) $ \dfrac{\cos A}{\sqrt {\sin B\sin C}} \plus{} \dfrac{\cos B}{\sqrt {\sin C\sin A}} \plus{} \dfrac{\cos C}{\sqrt {\sin A\sin B}}\le\sqrt {3}.$ (A. Kukush, M. Rozhkova)[/quote]
a) $ \dfrac{\cos A}{\sin B\sin C} \plus{} \dfrac{\cos B}{\sin C\sin A} \plus{} \dfrac{\cos C}{\sin A\sin B}=2\iff$
$ -\dfrac{\cos (B+C)}{\sin B\sin C} -\dfrac{\cos (C+A)}{\sin C\sin A} - \dfrac{\cos (A+B)}{\sin A\sin B}=2\iff$
$cotBcotC+cotCcotA+cotAcotB=1 \iff$
$ tanA+tanB+tanC=tanAtanBtanC.$
b) $\left(\dfrac{\cos A}{\sqrt {\sin B\sin C}} \plus{} \dfrac{\cos B}{\sqrt {\sin C\sin A}} \plus{} \dfrac{\cos C}{\sqrt {\sin A\sin B}}\right)^2$
$\le \left( \dfrac{\cos A}{\sin B\sin C} \plus{} \dfrac{\cos B}{\sin C\sin A} \plus{} \dfrac{\cos C}{\sin A\sin B}\right)\left(cosA+cosB+cosC \right)$
$=2\left(cosA+cosB+cosC \right)\le 3 \Rightarrow$
$ \dfrac{\cos A}{\sqrt {\sin B\sin C}} \plus{} \dfrac{\cos B}{\sqrt {\sin C\sin A}} \plus{} \dfrac{\cos C}{\sqrt {\sin A\sin B}}\le\sqrt {3}.$


Have I misinterpreted something?
[hide] The only number in the set that fulfills the condition is $ 175$. Hence, the answer is A.[/hide]

[quote="RobRoobiks"]it could mean
$ (1,2,3,4...,177,178)$
In that case it would be $ E 18$[/quote]

$ \left \lfloor \frac {178}{5} \right \rfloor \minus{} \left \lfloor \frac {178}{10} \right \rfloor \equal{} 18$

I corrected the problem, assuming "..." is wanted in the list.

[quote="sri340"]{1,2,3,..., 175,176,177,178 }
 How many numbers in the set above have 5 as a factor but do not have 10 as a factor?

(A) 1   (B) 3   (C) 4   (D) 17  (E) 18[/quote]
[img]http://www.mathmontada.net/vb/uploaded/24_1262687376.gif[/img]


Find all positive integer $n>1$ such that every integer $1<k<n$ satisfying $\gcd{(k,n)}=1$, is a prime.

Must be $p|n$ for all prime p, suth that $p^{2}<n$.
Therefore solutions are $n=2,3,4,6,8,12,18,24,30$.

[quote="Rust"]Must be $p|n$ for all prime p, suth that $p^{2}<n$.[/quote]
I am agree!
[quote]
Therefore solutions are $n=2,3,4,6,8,12,18,24,30$.[/quote]
But this, Why?  :oops:

If $p^{2}<n$, then $gcd(p^{2},n)\not =1$, because $p^{2}$ is not prime.
It mean $p|n$ for all prime, suth that $p^{2}<n$.

I know it. But why we have only $n=2,3,4,6,8,12,18,24,30$ ?  :oops:

Following candidate is $n=60$. But if $n\ge 60$, then $\prod_{p<\sqrt n }p >n$.
We can prove, that $\prod_{p<x}p >2^{x}$ for x>7. And $2^{x}>x^{2}$.
Let $p_{1}=2,p_{2}=3,p_{3}-5,p_{4}=7,...$ sequence of primes and $P_{k}=\# p_{k}=\prod_{i=1}^{k}p_{i}$ . Yoy need $P_{k}>p_{k+1}^{2}$ for k>3.


If there are 240 socks in a box. 80 socks in blue, 80 socks in red and 80 socks in white. Then, 120 socks are left foot sock and 120 socks are right foot sock. If a left foot sock and a right foot sock are in the same colour is considered one pair of sock. What is the minumum pairs of sock in the box?

extra question: what sub-forum should i post this kind level of question?

[hide="Solution"]Let the number of left-foot socks of the respective colors be $ a,b,c$.  Then $ 0\le a,b,c\le 80$, and $ a \plus{} b \plus{} c \equal{} 120$.  The number of pairs of socks, which we want to minimize, is

$ \text{min}(a, 80 \minus{} a) \plus{} \text{min}(b, 80 \minus{} b) \plus{} \text{min}(c, 80 \minus{} c)$

WLOG, assume that at least two of these minimums are in $ \{a,b,c\}$, otherwise we could have defined $ a,b,c$ as the number of right-foot socks of the respective colors and then at least two of our minimums would be in this set.  If the minimums are $ a,b,c$, then we get $ a \plus{} b \plus{} c \equal{} 120$ pairs of socks.  If the minimums are $ a,b,80 \minus{} c$, then the number of pairs of socks is $ a \plus{} b \plus{} 80 \minus{} c \equal{} 120 \minus{} c \plus{} 80 \minus{} c \equal{} 200 \minus{} 2c$.  But since $ c\le 80$, we have at least $ 200 \minus{} 2\cdot 80 \equal{} 40$ pairs of socks.  This is achieved when $ a \equal{} 0, b \equal{} 40, c \equal{} 80$ (or any $ a \plus{} b \equal{} 40$, $ c \equal{} 80$), so 40 is indeed our minimum.[/hide]

This is probably more of an Intermediate problem, but it's not a big deal if you post it here.


What do you think is the best out of these?

Madriva
Fedora
Xandros
Mepis
SuSE

Hm, never heard of any of those, but then I've only heard of Red Hat and Mandrake.

Mandrake changed its name to Mandriva a while ago.
Fedora is made by Red Hat Inc.

Red hat is the best out of those, but SuSe is the more powerful one. I also wa tinking about that. if you want a lower download, jsut download SuSE, which costs about $99 for pro and maybe free for personal, whhile you can donwload the feudora projecto fof red hat for free, but i believe it is 5 cds. YOu can downlaod SuSe personal for free [url=http://linuxiso.org/viewmirrors.php/593]here[/url]. If it's good, just tlel me at http://4aunlimited.co.nr. click on index and it will take you to my forum. tell me tehre, so themods/admins don't hink we are spamming. also, if u have a hi speed connection, you can donload the 4 disc feudora on cds or a 1 dsk dvd [url=http://linuxiso.org/distro.php?distro=64]here[/url]

Personally, I prefer Slackware. But take in mind that I only have tried Red Hat, Knoppix, and SuSE. And that it was a few years since I had something else than Slackware installed on my machine. Take a look at [url]http://distrowatch.com/[/url]!

Ubuntu is about the same level as SuSE. SuSE 9.3 Pro is free over at opensuse.org.

Thank You so much solidefarms. Now i can have a good version of linux on my machine. also, are you sure that its free, or is it just a trial? I am currently on XP, and my computer got attacked by 1300 hackers yesterday.  :)

Do you even know what linux is?
Do you know how to program?
http://en.wikipedia.org/wiki/Linux
I feel like this to you now--->[img]http://img.photobucket.com/albums/v251/Anna_Z/domokun.gif[/img]

Ummmmm....please back up your stuff before doing ANYTHING else. I'm scared we might lose you as a valuable and respected member of AoPS if your hardware doesn't work with Linux.

[quote="anirudh"]is that a faulty site. My computer now got attacked by 6000 hackers in 2 days. My computer is dying. It keeps on getting attacked by about 1 hacker in 3 seconds.   ;) :(  :(  :([/quote]
Wikipedia isn't faulty!
Do you have a firewall?
Virii/spyware scanners?

[quote="anirudh"]is that a faulty site. My computer now got attacked by 6000 hackers in 2 days. My computer is dying. It keeps on getting attacked by about 1 hacker in 3 seconds.   ;) :(  :(  :([/quote] By your first figure, you are getting attacked by 25/12 hackers per minute. By your second, you are getting attacked by 57600 hackers in 2 days. Now maybe you just forgot a 0 in the first number :rotfl:

there's some problem. Well, i started downloading 2 days ago. After downloaddig disk 1(after 1 day), my comp. was atttacked by 2300 hackers. Now it says that it has been attacked by 6600  :D

I am running ZoneAlarm and it tells that it protected my computer from (right now 7112) hackers. And i wasnt' talking about wikipedia, since I have been there myself before. I am talking about whetehr http://www.opensuse.org is faulty.  ;)

No. I know why you are asking. You pressed on the ip button on one of my posts in the MATHCOUNTS Forum(since you are a moderator there), and it was different than my other ones.  That's because, SBC Yahoo! DSL gives dynamic IP addresses, which annoys me when I set up a webserver on my computer(i just did).

It shouldn't take a day to download a SuSE disk; I used my DSL to get all 5 in 3 hours. And no, I do not know your IP address; I do not have that much power!

cUZ I'm on 384KBPS DSL and it took me a day to download all 5 discs.  :)

Out of those, Fedora is probably the best.

My favorite Linux distro is probably Gentoo, though it is very hard to set up.


Here is an old problem of mine posted previously on the Romanian forum:

Let $ x,y,x>0$ and $ \frac {1} {x^2}\plus{}\frac {1} {y^2}\plus{}\frac {1} {z^2}\equal{}1$

Prove that:

$ \frac {1} {x\plus{}y\plus{}z} \ge \frac {1} {x^3\plus{}y^3\plus{}xyz} \plus{} \frac {1} {y^3\plus{}z^3\plus{}xyz} \plus{} \frac {1} {z^3\plus{}x^3\plus{}xyz}$

[quote="moldovan"]Here is an old problem of mine posted previously on the Romanian forum:

Let $ x,y,x > 0$ and $ \frac {1} {x^2} \plus{} \frac {1} {y^2} \plus{} \frac {1} {z^2} \equal{} 1$

Prove that:

$ \frac {1} {x \plus{} y \plus{} z} \ge \frac {1} {x^3 \plus{} y^3 \plus{} xyz} \plus{} \frac {1} {y^3 \plus{} z^3 \plus{} xyz} \plus{} \frac {1} {z^3 \plus{} x^3 \plus{} xyz}$[/quote]
It 's quite easy

$ RHS \equal{} \frac {1} {x^3 \plus{} y^3 \plus{} xyz} \plus{} \frac {1} {y^3 \plus{} z^3 \plus{} xyz} \plus{} \frac {1} {z^3 \plus{} x^3 \plus{} xyz} \le\ \frac {1} {xy(x \plus{} y) \plus{} xyz} \plus{} \frac {1} {yz(y \plus{} z) \plus{} xyz} \plus{} \frac {1} {zx(z \plus{} x) \plus{} xyz}$

$ RHS \le\ (\frac {1}{xy} \plus{} \frac {1}{yz} \plus{} \frac {1}{zx})\frac {1}{x \plus{} y \plus{} z}$

And follow AM-GM we have :

$ \frac {1} {x^2} \plus{} \frac {1} {y^2} \plus{} \frac {1} {z^2} \ge\ \frac {1}{xy} \plus{} \frac {1}{yz} \plus{} \frac {1}{zx}$

We have done!


Source: TJ ARML practice, 3/30/06

Find the area of an equiangular octagon, the lengths of whose sides are alternately $1$ and $\sqrt{2}$.

[hide="Solution"]We notice that a square with side length 3 can be drawn around the octagon, and thus the area is ${3^2} - 4(\frac{1}{2}) = \boxed{7}$[/hide]


sinx + sin3x + sin5x = 0

Domain is:
[0 , 360]

Solve in degrees.

Please list all of your steps.

[quote="AdorableAddict"]One obvious solution is $x = 0$, and I think that's the only solution.  :maybe:[/quote]  not a good idea to assume one solution when dealing with high degree polynomials/nx trig functions..

$(\sin{x})+(-4\sin^{3}{x}+3\sin{x})+(16\sin^{5}{x}-20\sin^{3}{x}+5\sin{x})=0$

$16\sin^{5}{x}-24\sin^{3}{x}+9\sin{x}=0$

$(\sin{x})(4\sin^{2}{x}-3)^{2}=0$

$\sin{x}=0, \pm\frac{\sqrt{3}}{2}$

$x=0^\circ, 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ$

$f(x)$ graph's where $f(x)=sinx+sin3x+sin5x$ in $[0..2\pi]:$

[quote="pkothari13"][quote="Aznwithabrain"]

Solving using normal methods will give you the same answer as jhcreinhardt, except I think you need to include 360, given the specified domain, even though usually it is $[0,360)$.[/quote][/quote]

yea, i assumed he didnt mean to include 360

btw, your solution is considerably better than mine...

[quote="pkothari13"][quote="Aznwithabrain"]sinx + sin3x + sin5x = 0

Domain is:
[0 , 360]
[/quote]

$(\sin x+\sin5x)+\sin3x =(2 \sin3x \cos x)+\sin3x = \sin3x(2 \cos x+1) = 0$

This means $\sin3x = 0$ or $2\cos x+1=0$

Solving using normal methods will give you the same answer as jhcreinhardt, except I think you need to include 360, given the specified domain, even though usually it is $[0,360)$.[/quote]

One thing.  How did you get from the first step to the second step?

Sum-to-product formulas (Prosthaphaeresis Identities):

[quote="AoPSWiki"]
$\sin \theta \pm \sin \gamma = 2 \sin \frac{\theta\pm \gamma}2 \cos \frac{\theta\mp \gamma}2$

$\cos \theta+\cos \gamma = 2 \cos \frac{\theta+\gamma}2 \cos \frac{\theta-\gamma}2$

$\cos \theta-\cos \gamma =-2 \sin \frac{\theta+\gamma}2 \sin \frac{\theta-\gamma}2$

[/quote]


Using each digit only once, place the digits $ 7, 8$ and $ 9$, one in
each box, so that the product is as close to $ 6$ as possible. What
is the product $ (\Box.\Box)(0.\Box)$? Express your answer as a
decimal to the nearest hundredth.

There are only 6 possible combinations so it's not hard to check each of them.  We get:
$ 7.8\times .9\equal{}7.02$
$ 7.9\times .8\equal{}6.32$
$ 8.7\times .9\equal{}7.83$
$ 8.9\times .7\equal{}6.23$
$ 9.7\times .8\equal{}7.76$
$ 9.8\times .7\equal{}6.86$
So we see that $ 8.9\times .7\equal{}6.23$, which makes sense since $ 8.9$ is pretty close to 9 and $ .7$ is pretty close to $ .666$.


You flip two fair coins A and B. You flip A 2 times and B 4 times. What is the probability that in the final outcome, the number of heads for A is equal to the number of heads for B?

Hm.  Is there any way to do this besides looking case-by-case?

[hide]There are $2^6$ total ways to flip the coins.

There is 1 way for A and B to each come up with 0 heads.
There are 8 ways for A and B to each come up with 1 head.
There are 6 ways for A and B to each come up with 2 heads.

There's a total of 15 ways, making the probability $\frac{15}{64}$.[/hide]


[quote="tpulak"]Heres the problem:
"What is the ones digit when 9 to the power of 55 is expressed as an integer?"[/quote]

[hide]
We are basically looking for $ 9^{55}$ in mod 10, so
$ 9^{55}\equiv (\minus{}1)^{55}\equal{}\minus{}1\equiv 9 (\mbox{mod 10})$

So the answer is 9
[/hide]

At every even power, it's 1. Otherwise, it's 9.

So the ones digit is 9 when 9 to the power of 55 is expressed as an integer.

Here's a nice way to solve the problem without mods.

[hide]
$ 9^{55}$

Factor accordingly:

$ (9^{2})^{27} \times 9^{1}$

Because $ 9^2$ yields a units digit of $ 1$, you have

$ 1^{27} \times 9^1$

$ 1 \times 9$

So the units digit is $ 9$
[/hide]

That is basically the same thing, but without writing out the mod part

[quote="tpulak"]Heres the problem:
"What is the ones digit when 9 to the power of 55 is expressed as an integer?"[/quote]
Just in case if you don't know mod, here's an easy explanation.

[hide]There a patteren.

They tend to repeat every 4 sequence.

$ 9^1\equal{}9$
$ 9^2\equal{}1$
$ 9^3\equal{}9$
$ 9^4\equal{}1$

But in this case, they tend to repeat in every 2 sequence, which is fine.

So $ 55\div2\equal{}27 ~r\equal{}1$

So answer is $ \boxed9$[/hide]

[quote="tpulak"]I have a question: What's a mod?[/quote]

Which is exactly why I posted a solution [i]without[/i] mods.

[url=http://en.wikipedia.org/wiki/Modular_arithmetic]Wikipedia[/url] has a basic guide.

Also, [url=http://www.artofproblemsolving.com/Books/Vol1/exc1.pdf]AoPS[/url] has an excerpt from their Volume 1 book relating to mods (page 4 on the pdf).

[quote="tpulak"]I have a question: What's a mod?[/quote]
Modular arithmetic.
Which is basically the remainder of a number when divided by another.
So...
$ 55 \equiv 1 \mbox{(mod 3)}$

[quote="7h3.D3m0n.117"][quote="ZhangPeijin"]$ 56 \equiv 1 \mbox{(mod 3)}$[/quote]

That doesn't even make sense.[/quote]
 :oops:
Wait, I thought I wrote 55.
 :mad:

[quote="space cadet"]There's a pattern to powers of numbers...

[hide]for 9...
9^1=9
9^2=81
9^3=729
9^4=6561

all odd powers end in 9, all evens in one.
therefore 9^55 ends in 9[/hide][/quote]

eh...That's exactly what I put...

I think this has been solved enough, don't you think? :wink:

Yes.

I think this should be locked now, as everyone is seeming to get it now.


How can I express long addition with base numbers in latex?

Not exactly sure what you need but you could set things out in an array. Something like:
In base 7, 1234 + 3456 = 5032 so
$ \begin{tabular}{rrrrr} & 1 & 2 & 3 & 4 \\
\plus{} & 3 & 4 & 5 & 6 \\
\cline{2 \minus{} 5} & 5 & 0 & 2 & 3 \end{tabular}$
Using \renewcommand{\tabcolsep}{2pt} will reduce the width of each column.

Or you could use the power of the xlop package to lay things out.
Normally to add 1234 + 3456 using xlop you would say:
[code]\documentclass{article}
\usepackage{xlop}
\begin{document}
\opadd[carryadd=false]{1234}{3456}
\end{document}[/code]
but if the numbers are in base 7 then you need to suppress the result and use 5032 instead:
[code]\documentclass{article}
\usepackage{xlop}
\begin{document}
\newcommand{\numd}[1]{5}
\newcommand{\numc}[1]{0}
\newcommand{\numb}[1]{2}
\newcommand{\numa}[1]{3}
\opadd[carryadd=false,resultstyle.1=\numa,resultstyle.2=\numb,resultstyle.3=\numc,resultstyle.4=\numd]{1234}{3456}
\end{document}[/code]which gives a ncely laid out addition.

You can learn more about xlop [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=xlop]here[/url] (the documentation is in French)

$ %Error. "opadd" is a bad command.
[carryadd=false]{1234}{3456}$

Evidently it's not installed here.

Basically what you did with the tabular is what I needed.  But how would I show carrying, with smaller numbers at the top?

Can I show long division like that?  I know about longdiv, but that doesn't work with base numbers.

Just add rows and columns to the table and fill in your numbers. For example, in base 7:
\begin{tabular}{rrrr}
& & 1 & 2 \\
\times & & 3 & 4 \\
\cline{2-4}
& 3 & 6 & 0 \\
+ & & 5 & 1 \\
\cline{2-4}
& 4 & 4 & 1
\end{tabular}


Just a reminder that the next Columbia Math Circle will take place this Saturday from 1pm to 5pm at Hammond School. The topic will be on techniques for solving difficult probability problems. We will also work on preparing students for the upcoming AMC contests. Hope to see you all there!

Feel free to use this thread to discuss any of the problems from the last Math Circle, or to ask for answers/hints/solutions to any of the problems.

Can anybody post these problems? I really want to do those. I need to improve my counting, so share some "food" with me pleased. :)


Let $ a$ and $ b$ be positive integers and let the sequence $ (x_n)_{n \ge 0}$ be defined by $ x_0 \equal{} 1$ and $ x_{n\plus{}1} \equal{} ax_n\plus{}b$ for all non-negative integers $ n$. Prove that for any choice of $ a$ and $ b$, the sequence $ (x_n)_{x \ge 0}$ contains infinitely many composite numbers.

How do I do this question? I just started self-learning number theory as recreation so yeah if anyone can show me how to solve it, I would really appreciate it.

$ x_n\equal{}\frac{a^n(b\plus{}a\minus{}1)\minus{}b}{a\minus{}1}$. Chose $ p>a\minus{}1$, suth that $ p|a^k(b\plus{}a\minus{}1)\minus{}b$, then for any $ n\equal{}k\plus{}m(p\minus{}1)$ we get $ p|x_n$.


What's are good examples of the following?

a) A subset of $ \mathbb R$ which has exactly 5 boundary points but no cluster points.

b) A compact subset of $ \mathbb R$ which has exactly 5 cluster points.

c) A non-compact subset of $ \mathbb R$ which has exactly 5 cluster points.

d) A subset of $ \mathbb R^{2}$ which has exactly 5 cluster points.

The definitions of cluster, boundary points and compact sets are pretty clear, but I just need examples to "see" it.

b) the set $ A\equal{}\{0\}\cup\{1/n\colon n\equal{}2,3,\dots\}$ is compact and has one cluster point. Then $ A\cup (A\plus{}1)\cup (A\plus{}2)\cup (A\plus{}3)\cup (A\plus{}4)$ has five cluster points.

[quote="mstrip"]
a) A subset of $ \mathbb R$ which has exactly 5 boundary points but no cluster points.
[/quote]
How about $ S \equal{}\{ 1 , 2 , 3, 4, 5\}$ any point in $ S$ is a boundary point because if we draw a small point around any of those points we get non-trivial intersection in $ S$ and $ S^{c}$.
No point can be an accumulation point because the set is finite.

Maybe you can do the rest now.  :maybe:


function
limit
real analysis
real analysis unsolved

if $ f(x) \geq 0$, $ \lim_{x\to\infty}f(x)\neq 0$, then there exist $ \epsilon >0$ and a sequence $ (x_n)_{n\geq 1}$ such that $ x_n\geq n$ and $ f(x_n)>\epsilon$.
why ?
thanks

This is a reasonably straightforward negation of the definition of $ \lim_{x \to \infty} f(x) \equal{} 0$.  Write down that definition first, then write down how it can fail.

Incidentally, this has nothing to do with continuity.  Please try to pick more informative titles for your posts.


i am not sure what year this is from but i have a question:
A billiard ball is hit at a 45-degree angle from a corner of a  6  foot by  8  foot billiards table. How many times will the ball rebound off an edge of the table before landing in a corner?

EDIT:
another one:
Beginning with the letter M at the top of the triangle and reading diagonally downward to an adjoining letter, in how many ways is it possible to read the word  MATHCOUNTS ?

         M
        A A
       T T T
      H H H H
     C C C C C
    O O O O O O
   U U U U U U U
  N N N N N N N N
 T T T T T T T T T
S S S S S S S S S S

#1

Picture: http://www.talpink.com/public/08/08/13/billiardtablee4de2.jpg

I put question-mark near number 3, couse it's a middle of table (COULD BE ANOTHER HOLE)

for number two, i believe the answer given is incorrect

i think the answer is 512

the answer is related to pascal's triangle

we have....
M
A A
T T T
H H H H
C C C C C
O O O O O O
U U U U U U U
N N N N N N N N
T T T T T T T T T
S S S S S S S S S S

there is only one way to get to each of the 2 A's from the M, so from that point, there are 2 paths
to go to the T's, there is one way to get to the T's on the end, but two ways to get to the T in the center, giving us 4 paths
to get to the H, there is one way to get to the ones on the end, but 3 ways to get to them in the center, giving us 8 paths
there is one way to get to the C's on the end, 4 ways to get the ones that's in one more, and 6 ways to get  to the center, givings us 16 paths

extending this we finally get our answer at 512, either continuing this list, using pascal's triangle, or multiplying to 2 until we get the line with the S's

#2 again but differently

Its pascal's triangle. Basically from M to the S on the left its, 9 choose 0. And continuing going right is 9c1,9c2,9c3,9c4...... so when you add them up it equals 2^9=512. The left most is 9c0 because from the M to the S, it takes 9 moves and you have to choose  0 of them to go right. The one next to the the S I just mentioned, there are 9 moves and you have to choose 1 of them to go right....

oh, misread #2.

didnt see 'diagonally downward to adjoining letter'.

it is $ 2^9\equal{}512$


Each integer $ 1$ through $ 9$ is written on a separate slip of paper and all nine slips are put into a hat.  Jack picks one of these slips at random and puts it back.  Then, Jill picks a slip at random.  Which digit is most likely to be the units digit of the [b]sum[/b] of Jack's integer and Jill's integer?

$ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ \text{each digit is equally likely}$

[hide="My Solution"]
The highest possible sum is $ 9 \plus{} 9 \equal{} 18$ and the lowest sum is $ 1 \plus{} 1 \equal{} 2$. Obviously, each digit is not equally likely since there are $ 0$ ways to get $ 1$. We also notice that as we list out all the possible ways you can find two positive integers to sum to a number $ n$, there are $ n \minus{} 1$ ways to get it. So it encourages one to answer with $ (D)$. The problem is, you can only get a units digit of $ 9$ if the numbers sum to a $ 9$. You can't sum to $ 19$. So $ 8$ will have a greater number of pairs that sum to it. The ways to get a units digit of $ 8$ are sums up to $ 8$ and $ 18$. There are $ 8 \minus{} 1 \plus{} 18 \minus{} 1 \equal{} 24$ which is greater than any other combo (try yourself). Thus, the answer is $ \boxed{(C)}$.
[/hide]

EDIT: Ignore my post. I forgot that you only have numbers from $ 1 \minus{} 9$.

The answer is 0. It occurs $ \frac {9}{81}\equal{}\frac {1}{9}$ of the time. Everything else occurs $ \frac {8}{81}$ of the time. To see this, draw a table (similar to a multiplication table) and list the possibilities.

What is wrong with this solution:

Since the expected value of each draw is 4.5, the expected sum is 9, our our answer is C? :maybe:

Read the question. Each integer [b]1 through 9[/b]. Not 0 through 9.

:blush: I see. So expected value would work: 5 + 5 = 10 $ \Rightarrow \boxed{0}$. Thanks!

Nope, that's not why. The only reason is that for every number Jack chooses, there is a number that Jill can choose to make the units digit $ 0$, but this doesn't hold for any other units digit.


geometry
trapezoid
circumcircle
geometry unsolved

3-There is given a trapezoid $ ABCD$  in the plane with $ BC\parallel{}AD$.We know that the angle bisectors of the angles of the trapezoid are concurrent at $ O$.Let $ T$ be the intersection of the diagonals $ AC,BD$.Let $ Q$ be on $ CD$ such that $ \angle OQD \equal{} 90^\circ$.Prove that if the circumcircle of the triangle $ OTQ$ intersects $ CD$ again at $ P$ then $ TP\parallel{}AD$.

The trapezoid has incircle $ (O)$ and $ X,Y,Q$ are the tangency point of $ (O)$ with $ BC,AD,CD,$ respectively. By Newton's theorem, the diagonal intersection $ T$ lies on $ XY,$ which is perpendicular to its bases. Since the quadrilateral $ PQTO$ is cyclic, it follows that $ \angle OQP \equal{} \angle PTO \equal{} 90^{\circ}$ $\Longrightarrow$ $ TP \perp XY$ $ \Longrightarrow$ $ TP \parallel AD, $ as desired.


[quote="caubetoanhoc94"][quote="quykhtn-qa1"][quote="caubetoanhoc94"]Let $ a;b;c \ge 0$.Prove that:
$ \frac {1}{a \plus{} b} \plus{} \frac {1}{b \plus{} c} \plus{} \frac {1}{c \plus{} a} \plus{} \frac {8}{a \plus{} b \plus{} c} \ge \frac {6}{\sqrt {ab \plus{} bc \plus{} ca}}$
It's only use Am-Gm  :P[/quote]
This inequality by author hungkhtn  :wink:[/quote]

I don't know:This inequality by author hungkhtn  :(
Can you slove it ???[/quote]
Assume: $ ab\plus{}bc\plus{}ca\equal{}1$,we have:
$ \frac{1}{a\plus{}b}\equal{}\frac{c(a\plus{}b)\plus{}ab}{a\plus{}b}\equal{}c\plus{}\frac{ab}{a\plus{}b} \ge c\plus{}\frac{ab}{a\plus{}b\plus{}c}$
then,we hve: $ LHS \ge a\plus{}b\plus{}c\plus{}\frac{9}{a\plus{}b\plus{}c} \ge 6 \equal{}RHS$
We have Q.E.D :)

[quote="quykhtn-qa1"][/quote][quote="caubetoanhoc94"][quote="quykhtn-qa1"][quote="caubetoanhoc94"]Let $ a;b;c \ge 0$.Prove that:
$ \frac {1}{a + b} + \frac {1}{b + c} + \frac {1}{c + a} + \frac {8}{a + b + c} \ge \frac {6}{\sqrt {ab + bc + ca}}$
It's only use Am-Gm  :P[/quote]
This inequality by author hungkhtn  :wink:[/quote]

I don't know:This inequality by author hungkhtn  :(
Can you slove it ???[/quote][quote="quykhtn-qa1"]
Assume: $ ab + bc + ca = 1$,we have:
$ \frac {1}{a + b} = \frac {c(a + b) + ab}{a + b} = c + \frac {ab}{a + b} \ge c + \frac {ab}{a + b + c}$
then,we hve: $ LHS \ge a + b + c + \frac {9}{a + b + c} \ge 6 = RHS$
We have Q.E.D :)[/quote]
Very nice  :P
It's case $ k = 8$
We can prove that with $ k \ge 3$
$ \frac {1}{a + b} + \frac {1}{b + c} + \frac {1}{c + a} + \frac {k}{a + b + c} \ge 2\frac {\sqrt {k + 1}}{\sqrt {ab + bc + ac}}$

[quote="caubetoanhoc94"][/quote][quote="quykhtn-qa1"][/quote][quote="caubetoanhoc94"][quote="quykhtn-qa1"][quote="caubetoanhoc94"]Let $ a;b;c \ge 0$.Prove that:
$ \frac {1}{a + b} + \frac {1}{b + c} + \frac {1}{c + a} + \frac {8}{a + b + c} \ge \frac {6}{\sqrt {ab + bc + ca}}$
It's only use Am-Gm  :P[/quote]
This inequality by author hungkhtn  :wink:[/quote]

I don't know:This inequality by author hungkhtn  :(
Can you slove it ???[/quote][quote="quykhtn-qa1"]
Assume: $ ab + bc + ca = 1$,we have:
$ \frac {1}{a + b} = \frac {c(a + b) + ab}{a + b} = c + \frac {ab}{a + b} \ge c + \frac {ab}{a + b + c}$
then,we hve: $ LHS \ge a + b + c + \frac {9}{a + b + c} \ge 6 = RHS$
We have Q.E.D :)[/quote][quote="caubetoanhoc94"]
Very nice  :P
It's case $ k = 8$
We can prove that with $ k \ge 3$
$ \frac {1}{a + b} + \frac {1}{b + c} + \frac {1}{c + a} + \frac {k}{a + b + c} \ge 2\frac {\sqrt {k + 1}}{\sqrt {ab + bc + ac}}$[/quote]
It 's true with $ k \ge -1$ :wink:


calculus
integration
function
algebra
polynomial
binomial coefficients
binomial theorem

Say you have an equation like this: $ a \plus{} b \plus{} c \plus{} d \equal{} 18$.  You can use Balls + Urns to find the answer (assuming that you have only non-integral or positive solutions).  However, what if each number had an upper-bound of $ 6$?  (I'm just making this up.)  Then, how would you solve it?

One way is to use PIE to subtract the solutions where any of $ a, b, c, d$ are greater than the upper bound.  If the upper bound is, say, $ 16$ then you won't need too many terms to do this, but for $ 6$ this is going to be more tedious.

Anyway, there's a pretty slick way to do this computation using negative binomial coefficients, but I'm having trouble translating it back into a combinatorial argument.

[hide="Sketch"] You want to find the coefficient of $ x^{18}$ in $ (1 \plus{} x \plus{} ... \plus{} x^6)^4 \equal{} \frac {(1 \minus{} x^7)^4}{(1 \minus{} x)^4}$, which by the binomial theorem is

$ \left( \sum_{k \equal{} 0}^{4} {4 \choose k} ( \minus{} x^7)^k \right) \left( \sum_{k \ge 0} { \minus{} 4 \choose k} x^k \right)$

so the coefficient of $ x^{18}$ is $ {4 \choose 0} { \minus{} 4 \choose 18} \minus{} {4 \choose 1} { \minus{} 4 \choose 11} \plus{} {4 \choose 2} { \minus{} 4 \choose 4}$.  [/hide]

I believe this technique is in AoPS Vol. 2.  Now that I think about it, it might just be a particularly slick way to do the PIE argument.

Wait what, [b]negative[/b] combinations?  Sorry, I don't think I've ever seen one of those...

Use generating functions.

If it's non-negative, then the coefficient of $ x^{18}$ in $ (x^0\plus{}x^1\plus{}x^2\plus{}\ldots\plus{}x^6)^4$.  If it's positive, get rid of $ x^0$ term.

I think it would take awhile though :)[/i]

[quote="AIME15"]I think it would take awhile though :)[/quote]

Yes, I kinda noticed that.  Any simpler way?

$ {\minus{}n \choose k}$ just means $ \frac{(\minus{}n)(\minus{}n\minus{}1)...(\minus{}n\minus{}(k\minus{}1))}{k!}$, which is $ (\minus{}1)^k$ times the number of non-negative integer solutions to $ x_1 \plus{} ... \plus{} x_k \equal{} n$.  You might recognize this from regular balls-and-urns; this can also be proven by a generating functions argument and what's called the generalized binomial theorem.

AIME15:  It doesn't.  If you factor like I did, the final answer has three terms and can be interpreted as follows:

First we ignore the upper bound.  That gets us $ {4 \choose 0} {21 \choose 3}$ by the usual balls-and-urns formula (and this is the first term in the generating functions solution).

Second, we subtract the case where one of $ a, b, c, d$ is greater than or equal to $ 7$.  Then we're counting solutions to (WLOG) $ (a\minus{}7) \plus{} b \plus{} c \plus{} d \equal{} 11$, which gets us $ {4 \choose 1} {14 \choose 3}$ (and this is the second term in the generating functions solution).

Finally, we add back the case where two of $ a, b, c, d$ are greater than or equal to $ 7$.  Then we're counting solutions to (WLOG) $ (a\minus{}7) \plus{} (b\minus{}7) \plus{} c \plus{} d \equal{} 4$, which gets us $ {4 \choose 2} {7 \choose 3}$ (and this is the third term in the generating functions solution).

[hide="would this work?"] Let $ a \equal{} 6 \minus{} a_2$, $ b \equal{} 6 \minus{} b_2, c \equal{} 6 \minus{} c_2, d \equal{} 6 \minus{} d_2$.

Substituting in, we have $ 24 \minus{} (a_2 \plus{} b_2 \plus{} c_2 \plus{} d_2) \equal{} 18$ so $ a_2 \plus{} b_2 \plus{} c_2 \plus{} d_2 \equal{} 6$, where each term on the left hand side is nonnegative and less than or equal to six (otherwise one of a, b, c, or d would be less than zero). From balls and urns, we have $ \binom{9}{3} \equal{} 84$ solutions. (Luckily, none can be more than six).[/hide]

That's if you can have zeroes. Otherwise it's basically like rolling four dice.

Fixed the subscripts.

That works, but for the values from $ 7$ to $ 17$ it doesn't generalize as well as t0rajir0u's method does.

It's still nice.  What this amounts to is observing that powers of symmetric polynomials are symmetric, so the coefficients of $ (1 \plus{} x \plus{} ... \plus{} x^6)^4$ are symmetric and hence the coefficient of $ x^{18}$ is also the coefficient of $ x^{6}$.

t0rajir0u:  I'm not seeing your method very clearly.  Can you perhaps make it a little more basic?  I'm not familiar with generating functions...

[quote="t0rajir0u"]The argument I gave in my second post doesn't use generating functions at all.  Do you understand how I'm using inclusion-exclusion?[/quote]

Oops, I see where I went wrong.  I accidentally added all three instead of subtracting.  But now I see it.  Thanks t0rajir0u! :)

Also, is this method including or excluding zeroes?  And how do you get the other way?

Including.  If you want to exclude zeroes, that's the same thing as solving $ (a \minus{} 1) \plus{} (b \minus{} 1) \plus{} (c \minus{} 1) \plus{} (d \minus{} 1) \equal{} 14$ where $ a\minus{}1, b\minus{}1, c\minus{}1, d\minus{}1$ are nonzero.  In generating function terms, we can just divide out by $ x^4$.


Are there any olympiads that are available for the younger ones out there in middle school that are on an international basis? I'd be interested to see who comes out top when the kids are of different age.

I don't really know, but one of my friends from junior high school went to an international abacus competition when he was in 7th grade I think. He took private lessons for abacus. I think he got 2nd or 3rd place, which (to say the least) is not bad for an international competition.

I've heard that Mathcounts might bifurcate into high school competition.  Is that true, or is that a rumor?

-interesting_move

Never heard of that one, but it's possible - especially since they have a strong set of sponsors (CNA, NCTM, NSPE, TI, GM, Lockheed Martin, NASA< etc.) and will be able to reach more kids this way...

Oh, I hope so.  I haven't heard that, but that would be awesome! :D I'm going into 9th grade and still getting over the end of mathcounts.  I never really made it that far.

Idiot -> I think (am not sure, don't know a whole lot about it) that Tournament of the Towns might be relevant to what you're looking for.

Isn't MC already kinda international? I heard that there were people from England and DoD/DoS schools from around the world, plus peeps from Puerto Rico. I dont know......but it would be kewl. :shock:

What is the tournament of the towns?

Thanks.

Info is [url=http://www.amt.canberra.edu.au/wwwroot/imtot.html]here[/url].

I don't know much about this contest except that whenever I've seen a Tournament of Towns problem in a book or something, it's been a very interesting problem.  Has anyone ever participated in it?

I did a few years ago, they're pretty good problems. If you're really lucky, you may be invited to Russia for a week or so, its not a competition but they just have a few different topics and sets of questions on each of them, which you can pretty much do whenever you want. No tests, no anything, just do some problems, go out for a walk, .. real interesting. They sort of get harder and harder from there and end in some open questions, probably as a result of someones research.


geometry parallelogram trapezoid MATHCOUNTS rhombus

no a trapezoid is a quadrilater with exactly one pair of paralell sides.

[quote="13375P34K43V312"]Actually, different people define it different ways as far as I know, but in any decent contest problem, if they mention trapezoid, and it matters at all, they will distinguish whether or not a parallelogram is a trapezoid or not.[/quote]

im pretty sure what indianamath said was correct. usually theres something else that goes with it though. for example, on one test they said the trapezoid has 2 right angles. so you can somewhat draw it out. still, a trapezoid HAS to have only 1 pair of parralel sides

Where's your proof? Here's mine. First 2 paragraphs will do.

[url]http://en.wikipedia.org/wiki/Trapezoid[/url]

Also, a similar statement is somewhere in Introduction to Geometry, I'm pretty sure. I'm too lazy to dig out the book and look for it, however.

Yeah, it's one of those things...  See Introduction to Geometry - I think Mr. Rusczyk talked about it.

Some people define a trapezoid as having exactly one pair of parellel sides and/or having adjacent base angles both acute or both obtuse, but the problem will usually clarify if it matters at all.  MATHCOUNTS has this wonderful thing called protest  :D .

[quote="Some people"]A trapezoid is a quadrilateral with only ONE pair of similar sides.[/quote]

[quote="Other people"]A trapezoid is a quadrilateral with AT LEAST ONE pair of similar sides.[/quote]

I think if it is not explicitly stated in the problem, you should assume that Some people are correct. But then again, it may not matter to the problem.

i remember doing a mathcounts problem involving counting trapezoids. i inlcude parallelograms and i got it wrong.

[quote="archimedes1"][quote="buzzer11"][quote="firecricket91"]On a related note, would a rhombus be considered a type of kite?[/quote]

i think it is considered a kite. a kite is any quadrilateral with perpendicular diagonals.[/quote]

Um isn't that (your definition of a kite) an orthodiagonal quadrilateral?  I though kites had adjacent sides congruent, too.[/quote]

I believe the definition of a kite is that adjacent sides are congruent, which implies that it is also an orthodiagonal quadrilateral - though not necessarily the other way around... I have seen an orthodiagonal trapezoid.

[quote="Anaxerzia"][/quote][quote="archimedes1"][quote="buzzer11"][quote="firecricket91"]On a related note, would a rhombus be considered a type of kite?[/quote]

i think it is considered a kite. a kite is any quadrilateral with perpendicular diagonals.[/quote]

Um isn't that (your definition of a kite) an orthodiagonal quadrilateral?  I though kites had adjacent sides congruent, too.[/quote][quote="Anaxerzia"]

I believe the definition of a kite is that adjacent sides are congruent, which implies that it is also an orthodiagonal quadrilateral - though not necessarily the other way around... I have seen an orthodiagonal trapezoid.[/quote]

i have too. in my geometry class, the definition of a kite was just that, an orthodiagonal quadrilateral with congruent adjacent sides.

[quote="mihail911"][/quote][quote="Anaxerzia"][/quote][quote="archimedes1"][quote="buzzer11"][quote="firecricket91"]On a related note, would a rhombus be considered a type of kite?[/quote]

i think it is considered a kite. a kite is any quadrilateral with perpendicular diagonals.[/quote]

Um isn't that (your definition of a kite) an orthodiagonal quadrilateral?  I though kites had adjacent sides congruent, too.[/quote][quote="Anaxerzia"]

I believe the definition of a kite is that adjacent sides are congruent, which implies that it is also an orthodiagonal quadrilateral - though not necessarily the other way around... I have seen an orthodiagonal trapezoid.[/quote][quote="mihail911"]

i have too. in my geometry class, the definition of a kite was just that, an orthodiagonal quadrilateral with congruent adjacent sides.[/quote]

Oops, you guys are right. I forgot the adjacent sides are congruent part. Also a kite must be convex. a concave kite turns into a dart.


geometry
3D geometry
calculus
MATHCOUNTS

How do I find the volume of a cone if I know the slant height and diameter/radius?

I know the surface area by using this formula: $ \pi  rs\plus{}\pi r^2$  Where r is the raduis and s is the slant height.  Is there a formula like this for the volume?  Or can it be done by without a formula?  If so, here is the probelm:

A cone has a diameter 16 and slant height 17.

What is its height?

What is its volume to the nearest hundredth?

What is its surface area to the nearest hundredth?

[hide="Answers"]Height=15, Volume= 1005.31, Surface area= 628.32[/hide]

Thanks. :coolspeak:

1) $ V\equal{}\frac{1}{3}\pi r^{2}h$ where $ r$ is the radius and $ h$ is the height.  If you know the slant and the radius (which you can find from the diameter obviosuly), just use Pythagoras to solve for the missing $ h$.

2) By Pythagoras, $ h^{2}\plus{}8^{2}\equal{}17^{2}\Rightarrow h^{2}\equal{}225\Rightarrow h\equal{}15$

3) $ V\equal{}\frac{1}{3}\pi r^{2}h\equal{}\frac{1}{3}\pi \cdot 64\cdot 15\approx \frac{3015.93}{3}\approx 1005.31$

4) $ S\equal{}\pi r(s\plus{}r)\equal{}8\pi(25)\equal{}200\pi\approx 628.32$

You can also derive the formula yourself:

Just let $ h \equal{} \sqrt{s^2 \minus{} r^2}$ where $ h$ is the height, $ s$ is the slant height, and $ r$ is the radius.

Then you have
$ V \equal{} \frac{1}{3}\pi r^2\sqrt{s^2 \minus{} r^2}$

[quote="shentang"]But, it is kind of hard to figure out that the cone is 1/3 of a cylinder.[/quote]

Alas, it's another one of those formulas you have to memorize.

[b]Disclaimer:[/b] Ok, ok, you don't have to memorize it... you can derive it using calculus, but since this is a MATHCOUNTS forum, I'm not supposed to post calculus, right? :wink:

Thanks guys.  Right after I posted it, in the bathroom (don't ask) i figured out you could use pythogerean thearom.  So it would be 15 and everything is downhill form there!

[size=75][color=red][Calculus is prohibited in the MATHCOUNTS Forum.  The original post, along with the (numerous) quotations of the post are deleted.  -- moderator rcv][/color][/size]


In the 19th Century, Kolbe proposed the following total synthesis of acetic acid:

1. $ C\plus{}S_8$ and $ \Delta$ to get $ CS_2$

2. $ CS_2 \plus{}Cl_2$ and $ hf$ to get $ CCl_4$

3. $ \delta$ to get $ Cl_2 C \equal{} C Cl_2$

4. $ OH^{\minus{}}$ to get $ CHCl_2COOH$

5. $ Zn$ to get Acetic Acid.

Can someone explain what these steps are (e.g. 2. is probably, not sure, a free radical substitution)?

Yes, number 2 is very likely a radical reaction. Number 4 is a very nice reaction:

1) $ Cl_2C\equal{}CCl_2 \plus{} HO^\minus{} \longrightarrow Cl_2C\equal{}CClOH \plus{} Cl^\minus{}$

2) $ Cl_2C\equal{}CClOH \plus{} HO^\minus{} \longrightarrow Cl_2C\equal{}C\equal{}O \plus{} H_2O \plus{} Cl^\minus{}$

Now we have the hydrolysis of a ketene:

3) $ Cl_2C\equal{}C\equal{}O \plus{} HO^\minus{} \longrightarrow  Cl_2\minus{}C\equal{}C(OH)\minus{}O^\minus{}$

4) $ Cl_2\minus{}C\equal{}C(OH)\minus{}O^\minus{} \plus{} H_2O \longrightarrow  Cl_2CHCO_2H \plus{} HO^\minus{}$.

In number 5, the reaction of zinc envolves probably an organo zinc species -C-Zn which is then protonated to -C-H (similarly to Grignard reagents).

What kind of reaction is number 4? Is it derived from a general reaction? Is it a well known reaction? Does it have a name?

Yes, number 4 is a general reaction: treatment of organic halides with zinc and then hydrolysis to get C-H bonds. However, I have never heard of a special name for this reaction.

What do you mean by that? We just add metalic zinc to perform the reaction.

Yes, it is the fifth reaction. In my second post above, I should have written "number 5" instead of "number 4". But its obvious that I was referring to reaction 5. Was that your doubt?

So what's the fourth reaction?

Thanks for clearing up the confusion.

The fourth reaction is in [i]your[/i] first post above, and my comments about it are in [i]my[/i] first post above.


function
real analysis
real analysis unsolved

Fix $ k\geq 1$ and define $ f: \ell_{\infty}\rightarrow\mathbb{R}$ by $ f(x)\equal{}x_{k}$.  Is $ f$ continuous? [Hint: $ f$ is Lipschitz].

Of course, any Lipschitz function is continuous. The hint helps because the Lipschitz property (when it holds) is usually easier to prove than continuity. Especially when you can guess the Lipschitz constant beforehand.

My guess is that the Lipschitz constant of $ f$ is $ 1$.

Let's assume it is 1.  Then we would have
$ (Kd(x,y))^{2}\equal{}(d(x,y))^{2}\equal{}(x\minus{}y)_{1}^{2}\plus{}(x\minus{}y)_{2}^{2}\plus{}\cdots\plus{}(x\minus{}y)_{k}^{2}$ and
$ (\rho(f(x),f(y)))^{2}\equal{}(\rho(x_{k},y_{k}))^{2}\equal{}\sup_{k}\left|x_{k}\minus{}y_{k}\right|^{2}$.
Then we have
$ \sup_{k}\left|x_{k}\minus{}y_{k}\right|^{2}\leq(x\minus{}y)_{1}^{2}\plus{}(x\minus{}y)_{2}^{2}\plus{}\cdots\plus{}(x\minus{}y)_{k}^{2}\Rightarrow \rho(f(x),f(y))\leq d(x,y)$,
so $ f$ is Lipschitz and hence continuous.

QED?

$ |x_k|\leq \sup\{|x_1|,\dots,|x_n|,\dots\}\equal{}\|x\|_{\infty},\forall x\equal{}(x_1,\dots,x_n,\dots)\in \ell_{\infty}$, this solves all your problems.

Could you please clarify?  Is what I did correct?

I don't understand what you did, this part in particular
[quote="JRav"]
$ \sup_{k}\left|x_{k} \minus{} y_{k}\right|^{2}\leq(x \minus{} y)_{1}^{2} \plus{} (x \minus{} y)_{2}^{2} \plus{} \cdots \plus{} (x \minus{} y)_{k}^{2}$
[/quote]
this doesn't seem to be valid ( unless I missinterpreted the notation ).
$ \forall x,y\in \ell_{\infty}     |f(x) \minus{} f(y)| \equal{} |x_k \minus{} y_k|\leq \|x \minus{} y\|_{\infty}$, from this follows that $ f$ is Lipschitz with Lipschitz konstant $ 1$.

@JRav: You misinterpreted the metric. On $ \ell^\infty$ the natural metric is the sup metric. The metric you used is the natural one for $ \ell^2$.

I think I used the wrong metric for $ \mathbb{R}$ and went in reverse.  I did $ \d\leq K\rho$, instead of the other way around.

$ \rho(f(x),f(y))=\left\|f(x)-f(y)\right\|_{1}=\left|f(x)-f(y)\right|=\left|x_{k}-y_{k}\right|\leq\sup_{k}\left|x_{k}-y_{k}\right|=\left\|x-y\right\|_{\infty}=d(x,y)$

Thanks.  Could you have a look at my other problem please?  I've done most of the work, I'm just looking for an error.


trigonometry
inequalities proposed
inequalities

Let a,b,c>0 such that   $ a^2\plus{}b^2\plus{}c^2\plus{}2abc\equal{}1$
Prove that
$ (1\minus{}a)(1\minus{}b)(1\minus{}c) \geq abc$

Just make a trigonometrical substitution to $ \cos$ and the rest is well-known...


Hey all... i have a math exam tomorrow...can any of u out there help me with these problems really quickly please?Please show ur working...

a)A man has 3 sons of ages 5,10 and 15. He puts 80,250 in the bank at the time of his death,paying 5% simple interest.If each gets the same amount at the age of 20,find their shares at the time of their father's death.
b)A man lent two equal sums of money. One at 10% per annum simple interest for 5 years and the other at 15% for 3years 4months.If the difference between the  two sums of interest earned is 200,find both the sums of money lent out.(I think this question is wrong but the text book has an answer...)
Thanx a lot...

[hide]
If you're asking what simple interest is, think about this:

A man puts $\$$ 100 in the bank at 10% simple interest.  This means that he earns 10 dollars every year, regardless of the amount of money he has now in his account.  So, when he has $\$$ 110, he will not start earning $\$$11. That's compound interest.  He will continue to earn $\$$10.[/hide]

Simple interest is when the interest remains the same each year...eg if u borrow 100 and the rate is 10% then u repay 110 after 1yr,120 after 2yrs,130 after 3yrs... etc..
Well when I tried the 2nd one by taking the sums as x and y, I got the two interests as equal although the question says that the difference is 200. That's what I can't figure out....

Let's save work by not duplicating results.  See [url]http://www.mymathforum.com/viewtopic.php?t=126[/url].

[quote="moogra"]hm... the first problem looks very familiar. I think I did that for homework once. Where did you get this?[/quote]
My math textbook which is Concise Mathematics for Class 9-Selina Publishers. Author-R.K.Bansal.
U may not know of him unless u take ICSE!! :lol:
I needed help real quick and that's why u'll see the post on the other site too...but finally they didn't give us any such sum!!I was so shocked!!  :lol: They gave us  all terribly easy ones and I finished a 3hr paper in 1 hr! Anyway...thanx everyone ad by the way the answers are 22,500, 26,250, and 31,500.

Note: those answers are all for part a.  For what I consider to be the most likely answer to part b, see the link that I gave in my last post.


trigonometry
geometry
perpendicular bisector
trig identities
Law of Sines

Let $ M$ denote a set of pairwise distinct points on the plane having integer coordinates. Find the maximal possible number of points in $ M$ under the condition that the distance between any point in $ M$ and the origin is less than the distance between this point and any other point in $ M$.

[quote="Nightmare124"]rofl thats too hard to answer[/quote]
If the forum is above your level, read and don't talk. AOPS/ML is not intended to be a standard free-for-all message board; posts will require a substantial amount of thought.

[hide="I think this method works"]If $ M$ consists of the vertices of a regular pentagon with center at the origin, then the condition is satisfied.

To show $ 6$ or more doesn't work, let the origin be $ O$ and suppose we have a point $ A \in M$. Draw the circle $ \omega$ centered at $ A$ intersecting the origin. We can't put any points in this circle. Construct distinct points $ B,C$ on $ \omega$ so that $ \angle AOB \equal{} \angle AOC \equal{} 60$. Draw rays $ OB$ and $ OC$.

We show that any point in the region $ R$ of the plane on the side of those two rays containing point $ A$ doesn't satisfy the conditions because either the point is inside $ \omega$ or closer to $ A$ than $ O$, both not allowed by the conditions. Note that a point is closer to $ A$ than $ O$ iff it lies above the perpendicular bisector of $ OA$. But both $ B$ and $ C$ lie on this perpendicular bisector. Since the segments $ OB$ and $ OC$ lie entirely within $ \omega$, all points in $ R$ and not in $ \omega$ will be above the perpendicular bisector, and this is sufficient to show no point in $ R$ will work.

What this means is that if $ A$ and $ B$ are two points in $ M$, then $ \angle AOB > 60$. So it is impossible to have $ 6$ or more points.[/hide]

[hide="Possibly Quicker"]Consider two points $ A$ and $ B$ in $ M$.  Then $ OA < AB$ and $ OB < AB$.  Therefore, $ AB$ is the longest side of $ \triangle OAB$, so $ \angle AOB$ is the largest angle of $ \triangle OAB$.

I think this is well-known enough to cite (right?) but I'll prove it anyway.  Suppose we have $ \triangle ABC$ with $ a>b,c$.  By Law of Sines, $ \sin A>\sin B,\sin C$.  If the triangle is acute, then $ A$ is clearly larger than $ B$ or $ C$.  Now suppose for the sake of contradiction that the triangle is right/obtuse, and right/obtuse angle is not $ \angle A$ (WLOG let it be $ \angle B$).  We have $ \sin A>\sin (180\minus{}B)$, and since these are both acute/right angles, $ \angle A>180\minus{}\angle B$.  Then $ \angle A\plus{}\angle B\plus{}\angle C>180\minus{}\angle B\plus{}\angle B\plus{}\angle C\equal{}180\plus{}\angle C>180$, a contradiction.

Therefore, $ \angle AOB > 60$ since the conditions in the problem are strict, so we can't have six or more points.

Here's a construction for five points (basically the strategy here is to make the pentagon with these points as vertices as regular as possible):

$ (3,0),(1,3),(1, \minus{} 3),( \minus{} 3,2),( \minus{} 3, \minus{} 2)$.

Actually I suppose you could argue that you can get arbitrarily close to a regular pentagon as you use larger and larger numbers, and a regular pentagon easily satisfies the conditions in the problem, but this isn't too rigorous.[/hide]


geometry
circumcircle
cyclic quadrilateral
angle bisector
geometry proposed

Let $ ABCD$ be a cyclic quadrilateral and $ O$ be the intersection of diagonal $ AC$ and $ BD$. The circumcircles of triangle $ ABO$ and the triangle $ CDO$ intersect at $ K$. Let $ L$ be a point such that the triangle $ BLC$ is similar to $ AKD$ (in that order). Prove that if $ BLCK$ is a convex quadrilateral, then it has an incircle.

Let $ P \equiv OK \cap BC.$ From the cyclic quadrilaterals $ ABKO,DCKO$ it's easy to see that $ KP$ bisects $ \angle BKC$ since $ \angle BKP \equal{} \angle BAO \equal{} \angle ODC \equal{} \angle CKP.$ Let the angle bisectors of $ \angle KBL$ and $ \angle KCL$ cut $ LC,LB$ at $ B',C'.$

$ \angle KCL \equal{} \angle KCB \plus{} \angle ADK \equal{} \angle ACB \minus{} \angle KCO \plus{} \angle ADB \plus{} \angle ODK$

$ \Longrightarrow  \angle KCL \equal{} 2\angle BCA \equal{} \angle BCC' \equal{} \angle KCA$

Similarly, $ \angle CBB' \equal{} \angle KBD,$ which implies that $ CA,CC'$ and $ BD,BB'$ are isogonals WRT $ \angle KCB$ and $ \angle KBC,$ respectively. Hence if $ I \equiv BB' \cap CC',$ then $ O,I$ are isogonal conjugates WRT $\triangle BKC,$ but angle bisector $ KO$ of $ \angle BKC$ is self-isogonal $ \Longrightarrow$ $ I \in KP$ $\Longrightarrow$ internal angle bisectors of the convex quadrilateral $ BLCK$ concur at $I.$


[hide]
Am I wrong in thinking that by picking 51 numbers, the first 50 you choose may be not divisible, but the 51st will either be the half or double of one already chosen?

[/hide]

Yes, you are wrong.  For example, say you took all the odd numbers and the number 4--there would be no halves or doubles; however, there would be other multiples.

I've heard the name pigeonhole before, but what is it actually?

[quote="chess64"]Suppose 51 numbers are chosen from $1,2,3,\ldots,99,100$. Show that there are two such that one divides the other.[/quote]

As a side note, you must choose at least 25 so that one divides another since there are 24 primes.

[quote="quantum leap"]ahh, the pidgeonhole principle (aka, Dirichlet principle, box principle)

"If there are $n$ boxes and we have $n+1$ pidgeons we want to put in the boxes, then there must be at least one box stuffed with 2 or more pidgeons." or something like that[/quote]

A slight strengthening:  If we have $n$ boxes and $kn+1$ pidgeons then at least one box must have at least $k+1$ pidgeons.  Very useful, for example in this problem:

Prove that among $ab + 1$ people we must have either $a+1$ of them not descended from one another or $b+1$ of them, each of which are descended from one of the others in sequence.

[quote="t0rajir0u"][hide="Well-known application of Pidgeonhole"]
Write each number in the form $n = 2^k m$, where $m$ is the largest odd divisor of each number.  There are $50$ odd numbers in the range, but we have picked $51$ numbers, so two of them must share a largest odd factor.  Then clearly the smallest one divides the largest one.
[/hide][/quote]

Correct! :)

I was wondering, though.  I can't seem to pick $50$ numbers such that none of them divide the other, but I can't prove that this will generally be the case.  So:

What is the smallest integer $f(2n)$ such that one may choose an $f(2n)$-sized subset of $S = \{ 1, 2, 3, ... 2n \}$ such that none of the members of this subset divide each other?  The problem seems to imply $f(2n) = n$ but I can't find such a set for $n = 50$.   :huh:

would it help to know that there are only 40 numbers les than or equal to 100 that are relativly prime?  :?:

[quote="scorpius119"][quote="t0rajir0u"] The problem seems to imply $f(2n)=n$ [/quote]

[hide]
Well, it's obvious that such a set can't contain 1 (unless $n=1$) and similarly, other small values appearing in the subset would create problems. So at least the elements in the subset should be very big. This suggests using as many as the largest elements of possible.

Guess what? $\{ x\in \mathbb{Z}|n<x\leq 2n \}$ has $n$ elements and no two divide each other! Because if $a$ and $b$ were in the subset and $b$ was divisible by $a$, then $b>a$. But $b\leq 2n<2a$ so $1<(b/a)<2$, a contradiction. So there exists a set of $n$ elements such that no two divide each other.
[/hide][/quote]

Oh.  I'd constructed that set, but thought it had $n-1$ elements.   :blush:

This does not bode well for my USAMO performance :P


A bakery owner turns on his doughnut machine at 8:30 AM. At 11:10 AM the machine has completed one third of the day's job. At what time will the doughnut machine complete the job?

$ \textbf{(A)}$ 1:50 PM $ \qquad
\textbf{(B)}$ 3:00 PM $ \qquad
\textbf{(C)}$ 3:30 PM $ \qquad
\textbf{(D)}$ 4:30 PM $ \qquad
\textbf{(E)}$ 5:50 PM

[hide]It takes $ \frac{8}{3}$ hours for $ \frac{1}{3}$ the job, so it will take $ 8$ hours for the whole job.

$ 4: 30$ P.M. (D)[/hide]

4:30 pm, i set it up like 2hrs and 40 mins= 1/3x and just multiply and you get 8 hours= x so u add 8 hours to 8:30= 4:30

sorry if this is a spam..

hehe. I missed this one.. I failed :-(


inequalities
function
inequalities proposed

Are you sure the inequality is correct? $a=b=c=2$ yields the summation being equal to $\frac{3}{13}$, which is a contradiction.

[quote="silouan"]Here is my ungly solution .
We put $a=e^{x}$ ,$b=e^{y}$, $c=e^{z}$ .
Now the inequality becomes
$\sum\frac{1}{4e^{2x}-2e^{x}+1}\geq 1$ and we consider the function
$f(x)=\frac{1}{4e^{2x}-2e^{x}+1}$

Now we can find that $f''(x)>0$ so by Jensen
$\sum\frac{1}{4e^{2x}-2e^{x}+1}\geq 1$  and we are done[/quote]
Sorry, siluan. Please forgive me if i have any mistakes, you should note that equality occurs in the cases: $a=b=c=1$ or $a \rightarrow \infty, b\rightarrow \infty,c\rightarrow 0$. But in you proof, i think equality only occurs in the case $a=b=c=1$. :(

[quote="toanhocmuonmau"]Let $a,b,c$ be positive real numbers satisgy $abc=1$. Prove that
\[\sum\frac{1}{4a^{2}-2a+1}\ge 1 \]
:)[/quote]
\[\frac{1}{4a^{2}-2a+1}\ge \frac 1{2a^{3}+1}\]
Nice! :)

Mr Vasc very nice . :)  . Mr Vasc my solution above is wrong ?I think it is true  :maybe:

To mathman
Supose:$a \le b \le c$
**If $0<a \le \frac{1}{2}$ then $0<4a^{2}-2a+1 \le 1$ then $\sum \frac{1}{4a^{2}-2a+1}\geq \frac{1}{4a^{2}-2a+1}>1$
**If $a \geq \frac{1}{2}$ then : $\frac{1}{4x^{2}-2x+1}\geq \frac{2}{3x}-\frac{1}{3}$ ; $\forall x \geq \frac{1}{2}$ [hide]( hệ số bất định)[/hide]then $\sum\frac{1}{4a^{2}-2a+1}\geq \sum \frac{2}{3a}-1 \geq 2\sqrt[3]{\frac{1}{abc}}-1=1$ [hide](use AM_MG) [/hide]
 $\to$ Brother Cẩn (Toanhocmuonmau)
[hide]Em them cach : nữa dài hơn
Chì cần đk : $abc\le 1$[/hide]


Given $ x;y;z$ are positive real numbers satisfying $ x\plus{}y\plus{}z \leq \frac{3}{2}$
Prove that
$ \frac{x^6\plus{}2x^2y\plus{}1}{x^4}\plus{}\frac{y^6\plus{}2y^2z\plus{}1}{y^4}\plus{}\frac{z^6\plus{}2z^2x\plus{}1}{z^4} \geq \frac{243}{4}$

[quote="chien than"]Given $ x;y;z$ are positive real numbers satisfying $ x \plus{} y \plus{} z \leq \frac {3}{2}$
Prove that
$ \frac {x^6 \plus{} 2x^2y \plus{} 1}{x^4} \plus{} \frac {y^6 \plus{} 2y^2z \plus{} 1}{y^4} \plus{} \frac {z^6 \plus{} 2z^2x \plus{} 1}{z^4} \geq \frac {243}{4}$[/quote]
$ \frac {x^6 \plus{} 2x^2y \plus{} 1}{x^4} \plus{} \frac {y^6 \plus{} 2y^2z \plus{} 1}{y^4} \plus{} \frac {z^6 \plus{} 2z^2x \plus{} 1}{z^4} \equal{} (x^2 \plus{} y^2 \plus{} z^2) \plus{} (\frac {y}{x^2} \plus{} \frac {z}{y^2} \plus{} \frac {x}{z^2}) \plus{} (\frac {1}{x^4} \plus{} \frac {1}{y^4} \plus{} \frac {1}{z^4})\ge 3(\sqrt [3]{x^2y^2z^2} \plus{} \frac {2}{\sqrt [3]{xyz}} \plus{} \frac {1}{\sqrt [3]{x^4y^4z^4}})$
We have
$ \sqrt [3]{x^2y^2z^2} \plus{} \frac {1}{8\sqrt [3]{xyz}} \plus{} \frac {1}{8\sqrt [3]{xyz}}\ge \frac {3}{4}$
and $ \frac {3}{2} \ge x \plus{} y \plus{} z \ge 3\sqrt [3]{xyz}$
So ineq was proved  :blush:

[quote="FOURRIER"][quote="Mather"]
We have
$ \sqrt [3]{x^2y^2z^2} \plus{} \frac {1}{8\sqrt [3]{xyz}} \plus{} \frac {1}{8\sqrt [3]{xyz}}\ge \frac {3}{4}$
and $ \frac {3}{2} \ge x \plus{} y \plus{} z \ge 3\sqrt [3]{xyz}$
So ineq was proved  :blush:[/quote]

No, it wasnt !
we have $ (xyz)^{\frac {2}{3}}$ $ \plus{}$ $ \frac {1}{4(xyz)^{\frac {1}{3}}}$ $ \geq \frac {1}{4}$

So it reminds to prove:
$ \frac {7}{4(xyz)^{\frac {2}{3}}} \plus{} \frac {1}{(xyz)^{\frac {3}{4}}}\geq \frac {1}{4}$
Wich Is Not True
Please check your answer before posting it  :mad:[/quote]
I don't thing it isn't true
$ \sqrt [3]{x^2y^2z^2} \plus{} \frac {1}{8\sqrt [3]{xyz}} \plus{} \frac {1}{8\sqrt [3]{xyz}}\ge \frac {3}{4}$
And $ \frac{1}{2}\ge \sqrt[3]{xyz}$
$ \frac{1}{16}\ge \sqrt[3]{x^4y^4z^4}$
So $ 3(\sqrt[3]{x^2y^2z^2}\plus{}\frac{2}{\sqrt[3]{xyz}}\plus{}\frac{1}{\sqrt[3]{x^4y^4z^4}})\ge 3(\frac{3}{4}\plus{}\frac{7}{2}\plus{}16)\equal{}\frac{243}{4}$
 :roll:


Find a ten-digit number whose first digit is the number of ones in the number, whose second digit is the number of twos in the number, whose third digit is the number of threes in the number, and so on, up to the tenth digit, which is the number of zeros in the number.

[hide]I try to keep as many zeros as possible to keep it simpler, so I start with 1000000000.
Then it changes into 1000000008 (there are 8 0s),1000000108 (there is 1 8), 2100000108 (2 1s,1 2), then lastly 2100010006 (now there are only 6 0s). The number is 2100010006.[/hide]

[hide]Start with 1000000000, but there are 8 0's, so it is 1000000108, but now there are 2 ones so you make it 2000000108, but now there is only one 1 and there is a 2 so it becomes 2100000108, but now there are only 6 0's, so you change it to 2100010006[/hide]

try to keep as many 0's as possible
       2100010006


Hello!?!?!

I am just trying to find some advice on creating a math team that in the future after lots and lots of practice will be competitive at least at the county and state level.

Does anyone out there know how this can be done? :?:

You know that the USAMO allows for 1.5 hours per proof. So being able to do a proof in 20 minutes is very very good.

Get a team of motivated and smart people.
Stay afterschool twice a week minimum.
Solve difficult problems and make sure all of them pick up techniques and strategies and understand the problems.

Speaking from experience  :D

[quote="3ric"]hmmm...  get some people together who are good in math i guess...  and make sure you can work under pressure and in limited time.  you can be the most brilliant mathematician in the world, but if you have to spend 20 minutes doing a proof or something, its hard to be competitive.[/quote]

In addition to what Phelpedo said, if you were the most brilliant mathematician in the world, you would be very very very fast at doing proofs anyway...

No necessarily. The most genius mathematician likes the long-term thinking. He will look at a problem and that problem will trigger other problems and so on. He will be completely sidetracked. :rotfl:

Talent, time and teamwork

Get some people together who are naturally good at math, who like math, who math comes easily too. I know this sounds biased but your math team is going to be better if they pick things up better. Find the talent.

Spend time. Lots of time. Once a week may be enough if enough talent is there, but more is always better. don't spend your time doing past competitions or anything, assign that as 'homework' and spend yuor time learning from each other.

When you get a team, make sure you can work together. Take this example:
in my math club, where I am a freshman, we were doing a partners seed round where there was a bonus for being the first one with the correct answer. A team of two people who were both much better at math than either my partner or I would come up with an answer as much as 10 seconds before us, yet they would invariably argue about it for long enough for us to get the first-answer bonus.
Naturally, that's an example of what not to do.
Another reason you spend time in practices is to build this teamwork.

Addidionally, find a coach who knows a lot more about math than you do and wants you to do well, enough that he/she spends as much or more time than you do and teaches you things.

So starting a math team is not an easy thing to do.


Is there a way to make all my seperate align environments line up their = in a document?

You'll only want to do this for each page (it would not be detectable between pages) and it only makes sense for similar equations.

What you can do is interrrupt a single align environment with text by using the \intertext command, which will allow you to then carry on as before. Thus[code]\begin{align}
1+1&=2 \\
2+2 &=4
\intertext{and we can add text here without disturbing the alignment}
3+3 &=6
\end{align}[/code]gives
\begin{align}1+1&=2 \\ 2+2 &=4 \intertext{and we can add text here without disturbing the alignment}2 &\neq 4 \end{align}


The number 121 is a palindrome, because it reads the same
backwards as forward. How many integer palindromes are
between 100 and 500?

$ \ 100$ has $ \ 101, 111, 121, 131, 141, ... 191$ as palindromes. The same applies for $ \ 200, 300, 400$.
 However, we are asked how many palindromes are BETWEEN $ \ 100$ and $ \ 500$ so 500 does not apply. Therefore, we have $ \ 4 \cdot 10 \equal{} \boxed {40}$ palindromes.


search
function
number theory proposed
number theory

Prove or disprove: No four perfect squares form a non-constant arithmetic sequence.

Posted so many times before..please use the search function


How do you post LaTeX messages in the forum? I have the image in the pdf file, but I can't put it in the body of my message. Please help.  :maybe:

Just click 'add an attachment'; Next to 'message body'.


Sa se arate ca pentru $\forall{p}\in\mathbb{N}$, $p\ge{2}$ si orice interval nevid, deschis,${I}$, $\exists{\; a_{1},a_{2},\cdots,a_{p}}\in{I}\setminus\mathbb{Q}$,
astfel incat $a_{1}^{n}+a_{2}^{n}+\cdots+a_{p}^{n}\in\mathbb{Q}$, $\forall{n}\in\mathbb{N}$.

Da. Vezi [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=titularization&t=45860]topicul ăsta[/url]. Acolo se construie un polinom ireductibil, cu toate radacinile reale, deci irationale.

Apoi folosim [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=106013]formulele astea[/url] pentru a demonstra că toate sumele sunt raţionale.


complex analysis
complex analysis unsolved

Sorry for writing the problem in that way:

find the value of

summation from infinity to infinity of 1/(z+n)^2 + a^2
thx

I suspect you're missing a pair of parentheses.

Do you actually mean $\sum_{n=-\infty}^{\infty}\frac1{(z+n)^2+a^2}$ ?

yes u are right, it was not my fault but it is written in that way but i doubled check it and u are right as u wrote it


function
number theory open
number theory

If $\mathbb{P}$ is the set of primes, define the function $f: \mathbb{P}\to\mathbb{Z}^{+}$ thus:

$f(\overline{a_1a_2\dots a_{n-1}a_n})=\overline{a_na_{n-1}\dots a_2a_1}$

Is the set $\{p|f(p)\in\mathbb{P}\}$ infinite?

(Examples of such numbers: $13,17,37,101,107\dots$)

No solution available.

[quote="Farenhajt"]No solution available.[/quote]
Then I claim that Proposed & Own Problems is the wrong forum ;)
Open Questions seems more appropriate...

Did you create this problem, or is this an unsolved problem?  Just curious. :)

I created it (which doesn't mean that someone else couldn't create it too before me, but I have no knowledge of it).

[quote="ZetaX"]This problem is indeed not new, and no approach or good partial results are known to me...[/quote]
IMHO,the problem of this kind is practically unsolvable by the current status of math of prime numbers.
There's no analytical formula for strict characterization of primes just on the basis of digits they are consisted of..
Maybe one can give a try with some deep results regarding Mersene numbers or such, but I doubt it would be enough to proove the hypotesis.
However,my gut feeling tells me yes, the  set of prime palindromes is infinite .
What do you think ZetaX?

Yes I think there are infinetely many.
A sharper question would be if there are $\infty$ primes of type $10^n+10^{n-1}+...+100+10+1=1111...111$. Some checking gave me that it is prime for  $n=1,18,22$, but I found no others till now ($n$ up to $500$). But a proof of the (in)finity of such primes would with possibly also solve the problem for Mersenne numbers (since they are the same in base $2$).


[quote="sharkman"]A basketball player made 9 shots, which is 75 percent of the number he attempted.  The coach tells this player to keep making foul shots until the total number made is 90 percent of the total he attempted.  What is the minimum number of additional foul shots the basketball player needed to make?

MY WORK:

9 foul shots = 75 percent

x additional foul shots = 90 percent

I then crossed-multiplied and got:

8.1 = .75x

After solving for x, I got about 11 additional foul shots.

The book's answer is 13 additional foul shots.

How can I get 13?[/quote]
[hide="My answer"]So 9 is 75%. So the player took 12 shots in total. Then he also took x amount of shots. So we have the equation is $\frac{9+x}{12+x}=\frac{9}{10}$. Cross multiplying we get $90+10x=108+9x$. Solving for x we get that x=18. [/hide]
hmm, I seem to get a different answer. Can somebody tell me what I did wrong?

The book's answer 13 additional foul shots.

I got 11 and you said 18.

How do I get 13?

I can see that the textbook has a typo.

I also understood the fractional equation set up.

However, is there a way to reason out this sort of question without creating an equation and solving for x?


Case 1 x = 0.
Then LHS and RHS are not the same. Therefore x = 0 is not a solution.
Case 2 $x\neq0$
Rationalization:
$\frac{14x}{\sqrt{7x^2+9x+13}-\sqrt{7x^2-5x+13}}=7x$
Therefore, $\sqrt{7x^2+9x+13}-\sqrt{7x^2-5x+13}=2$
Adding to the original equation: $2\sqrt{7x^2+9x+13}=7x+2$
Squaring both sides and solving: $x=\frac{12}{7}$ or $x=-\frac{4}{3}$.
After checking: $x=\frac{12}7$ is the only solution.

[quote="José"][quote="shobber"]Solve for $x \in R$:
\[ \sqrt{7x^2+9x+13}+\sqrt{7x^2-5x+13}=7x  \][/quote]

Can anyone help me with this???:

 Consider $y=7x^2+9x+13$.

Then, $\sqrt{y} + \sqrt{y-14x}=7x.$ I saw a lot of problems solved by considering y instead of some equation, generally when there are similar equations, and then calculate y and solve the final equation....[/quote]

That is often used for solving higher degree polynomials. In this case it would not help you much except for making it easier to write and harder to make a silly mistake.

With what you did, you could procede like this:
$\sqrt{y-14x}+\sqrt{y}=7x$ $\sqrt{y-14x}-\sqrt{y}=7x-2\sqrt{y}$
Multiply to get $-14x=7x(7x-2\sqrt{y})$
Since x is not zero, $-2=7x-2\sqrt{y}$; $7x+2=2\sqrt{y}$ etc etc etc

Question: Is this part of the middle school math [i]curriculum [/i]in China?

thanks for showing an example you both........

[quote="Virgil Nicula"]$u=\sqrt{7x^2+9x+13}>0\ ,\ v=\sqrt{7x^2-5x+13}>0\Longrightarrow$
$u+v=7x\ ,\ u^2-v^2=14x\Longrightarrow u+v=7x\ ,\ u-v=2\ (u>v\Longrightarrow x>0)$
$2u=7x+2\Longrightarrow 4(7x^2+9x+13)=49x^2+28x+4\Longrightarrow$
$21x^2-8x-48=0\Longrightarrow x_1=\frac{12}{7}\ ,\ x_2=-\frac 43 <0\ .$ Therefore, $\boxed {x=\frac{12}{7}}\ .$[/quote]

OMG that was beautiful!!

[quote="The Zuton Force"]
Question: Is this part of the middle school math [i]curriculum [/i]in China?[/quote]

Guessing Middle School Olympiad.  Curriculum=insanity :)


how do you factor  (step by step plz)

\[ 20x^3\plus{}37x^2\plus{}15x\equal{}0\]

[quote="grn_trtle"]You can always just find the roots with the quadratic formula if you're stuck, and then work backwards to factor it.[/quote]

in this case you would have to [hide]take out the x, and you'll notice that x can be 0 and a root[/hide].
Then do the quadratic formula


trigonometry
induction
search
algebra proposed
algebra

Given $a_{0}> 0$ and $c > 0,$ the sequence ${a_{n}}$ is defined by

$a_{n+1}= \frac{a_{n}+c}{1-ca_{n}}$

for $n = 0, 1, \cdots.$ Is it possible that $a_{0}, a_{1}, \cdots, a_{1989}$ are all positive but $a_{1990}$ is negative?

[hide="Solution"]

[b]Claim: It is possible to construct such a sequence[/b]
[b]Proof:[/b]

Let us substitute $\tan x_{n}= a_{n}$ and $\tan y = c.$ In this case, we can rewrite the recurrence relation as follows:

$\tan x_{n+1}= \frac{\tan x_{n}+\tan y}{1-\tan x_{n}\tan y}= \tan (x_{n}+y)$

We can then find $a_{n+2}= \tan x_{n+2}$ as follows:

$a_{n+2}= \tan x_{n+2}= \tan{x_{n+1}+y}= \tan{(x_{n}+y)+y}= \tan{x_{n}+2y}$

By induction, it follows that $a_{n+k}= \tan x_{n+k}= \tan (x_{n}+ky).$ If we let $n = 0,$ we then have $\tan x_{k}= \tan(x_{0}+ky).$

So, we need $\tan (x_{0}+ky) > 0 \Leftrightarrow 0 < x_{0}+ky < \frac{\pi}{2}$ for $k= 0, 1, \cdots, 1989$ and $\frac{\pi}{2}< x_{0}+1990y < \pi.$

Now, if we let $x_{0}= y = \frac{\pi}{2\cdot 1990}= \frac{\pi}{2\cdot 1990},$ we get $x_{0}+1989y = \frac{\pi}{2}.$ Also, then $x_{0}+1990y = \frac{1991\pi}{3980}.$ Since $\frac{1991\pi}{3980}-\frac{\pi}{2}= \frac{\pi}{3980},$ we can subtract a positive value, $\epsilon$, smaller than $\frac{\pi}{3980}$ from all the terms and still have a sequence that satisfies the conditions. Since we have constructed an appropriate sequence involving $x_{0}$ and $y,$ we can appropriately construct a sequence with $a_{n}$ and $c.$ $\blacksquare$

[/hide]

Could someone grade my solution out of seven points and/or give feedback? I would really appreciate it, thanks  :)


Has anyone played the game Settlers of Catan?

http://en.wikipedia.org/wiki/The_Settlers_of_Catan

i've played it before, and it's really fun.
The extension, seafarers is cooler though.

Cool, how do you play Seafarers?  I've played Settlers of Catan before, though. I enjoy the strategy and all, but I don't play it very often because the setup takes forever.

I've played it. I like Starfares better, but I barely ever get to play any version :(

Very fun.
It deals with stratagy but although it takes a lot of time to set up and a lot of time to play, it is so "mind breaking".


modular arithmetic
number theory
relatively prime

it's wrong: for $ (a,m,n) \equal{} (3,3,7)$, both expressions are divisible by 5. what you meant to say is $ a^{2^m} \plus{} 2^{2^m}$ and $ a^{2^n} \plus{} 2^{2^n}$.

we may assume WLOG that $ m < n$. let $ p|a^{2^m} \plus{} 2^{2^m}$, where $ p$ is prime. since $ a$ is odd, $ p$ is odd. then $ a^{2^m}\equiv \minus{} 2^{2^m} \pmod{p}$. repeatedly squaring both sides, we eventually get $ a^{2^n}\equiv 2^{2^n}\not\equiv \minus{} 2^{2^n}\pmod{p}$, since $ p$ is odd. this shows that the sets of primes dividing the two expressions are disjoint, implying the assertion.

[quote=hasan4444]Let $a$ be an odd integer. Prove that $a^{2^m}+2^{2^m}$ and $a^{2^n}+2^{2^n}$ are relatively prime for all positive integers $n$ and $m$ with $n\not= m$.[/quote]

[hide="My solution"]
Without loss of generality, assume that $n> m.$ it is easy to see that, $a^{2^m}\equiv -2^{2^m} \pmod{p}$, for every prime $p$ dividing $a^{2^m}+2^{2^m}.$ So we deduce that,
$$a^{2^{n}} \equiv \left(a^{2^m} \right)^{2^{n-m}} \equiv \left(-2^{2^m} \right)^{2^{n-m}} \equiv 2^{2^n} \pmod{p}$$
This mean that, $p \mid a^{2^n}-2^{2^n}$ for every positve integer $n >m.$ Now, Assume that $\gcd \left(a^{2^m}+2^{2^m},a^{2^n}+2^{2^n} \right) \neq 1$, For some positive integer $n >m.$ This, implied that there exist a prime divisor of $a^{2^m}+2^{2^m}$ devide $a^{2^n}+2^{2^n}$ so this divisor must devide $\left(a^{2^n}+2^{2^n}\right)-\left(a^{2^n}-2^{2^n} \right)=2^{2^{n}+1}.$ Since, $a$ is an odd integer we deduce that, $a^{2^m}+2^{2^m}$ is an odd positive integer. Thus, $p \neq 2$ which mean that $p \nmid 2^{2^n+1}$ for any positive integer $n.$ So we deduce the contradiction. Therefore, by symmetric we deduce that, $\gcd \left(a^{2^m}+2^{2^m},a^{2^n}+2^{2^n} \right)=1$  for all positive integers $n$ and $m$ with $n\not= m.$
[/hide]


I just made this up. It's pretty dumb.

Given any number $x$, what is the easiest way, in your opinion, to generate an infinite sequence that converges to it? There's more than one answer of course.

This is mine: [hide]\[x=\sum_{k=0}^\infty{x}\cdot\max(0,1-2k)\] :D [/hide]

:huh:  :what?:  Is this problem at the appropriate level for this forum? :ninja: :alien:

[hide=""My dumb contribution""]\[x=\sum_{k=0}^\infty x^{e^{2k\pi i}}\sin (2^k\pi)\] :cool:  :P  :D  :rotfl: :first:[/hide]

[quote="3X.lich"][hide=""My dumb contribution""]\[x=\sum_{k=0}^\infty x^{e^{2k\pi i}}\sin (2^k\pi)\] :cool:  :P  :D  :rotfl: :first:[/hide][/quote]Mine's easier ($x+0+0+0+0+\ldots$) :D... 'cause I have no clue what you're talking about  :(.

[quote="joml88"]His is just $(x+0+0+0+\cdots)$ also!  :D

The $\sin(2^k\pi)$ term will be $0$ for every $k$ in the domain except $k=0$ because when $k=0$ you get $\sin \pi$ and everytime else you get a multiple of $2\pi$ which makes its sine just $0$.

The $x^{e^{2\pi ki}}$ scared me at first  :P[/quote]Ah I see. I didn't notice that. Now I must be missing something very obvious, but wouldn't $\sin(2^k\pi)=0$ for all $k\ge0$? $\sin\pi=\sin2\pi=\sin4\pi=\ldots=0$.

Whoops...if you change it to $2^{k-1}$ i think it should work then.

[quote="Tokyo"]i dun even know wat that E means, can some1 explain it to me?[/quote]$\Sigma$ is the capital Greek letter sigma, and it is used to shorten (on paper) summations that are either long and tedious to write out, contain a variable number of terms, or are infinitely long (i.e. impossible to write). Here are some examples to explain the notation..

$\sum_{i=m}^n{a_i}=a_m+a_{m+1}+a_{m+2}+\ldots+a_{n-1}+a_n$.

The $i$ is called the $\text{index}$. Its starting value is defined by $i=m$ ($m$ is an integer) and it increases by $1$ as each term is added until it reaches $n$.

$f(x)=\sum_{n=0}^\infty\frac{(x-2)^n}{n!}=1+(x-2)+\frac{(x-2)^2}{2!}+\frac{(x-2)^3}{3!}+\frac{(x-2)^4}{4!}+\ldots$

Here $n$ ranges from $0$ to $\infty$; there is an infinite number of terms being added.

$\sum_{\text{prime }p}\frac{1}{p}=\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}+\frac{1}{13}+\ldots$

(This summation has no value; it is infinite.) Sometimes there will only be stuff written on the bottom. It will give specific conditions that must be met for a term to be included in the summation. In this case, $\frac{1}{p}$ is added if and only if $p$ is a prime number. Another example of this:

$\sum_{2\le{j}\le4;0\le{k}\le1}x^jy^k=x^2y^0+x^3y^0+x^4y^0+x^2y^1+x^3y^1+x^4y^1=(1+y)(x^2+x^3+x^4)$

This summation adds all possible terms $x^jy^k$ such that both $2\le{j}\le4$ and $0\le{k}\le1$. And finally:

$\sum_{\text{people}}1=\text{the total number of people in the world,}$

because the summation has the term $1$ for every person in the world, so this represents the world population.


I have a solution for part 1:
i want to show $ MX^2 \equal{} MT^2$.whit law of cos we have:(angle $ XBM \equal{} \alpha$)
\[ \begin{array}{l} XM^2 \equal{} BX^2 \plus{} BM^2 \minus{} 2BM.BX.COS\alpha \\
MT^2 \equal{} TC^2 \plus{} MC^2 \minus{} 2.MC.TC.COS\alpha \\
\end{array}
\]
then enough to show that:
\[ BX^2 \minus{} 2BM.BX.COS\alpha \equal{} TC^2 \minus{} 2.MC.TC.COS\alpha
\]
then:
\[ (BX \minus{} TC)(BX \plus{} TC) \equal{} BC.COS\alpha (BX \minus{} TC)
\]
then:
\[ BX \plus{} TC \equal{} BC.COS\alpha
\]
and last part is true because:
\[ BX \plus{} TC \equal{} BD.COS\alpha \plus{} CD.COS\alpha \equal{} BC.COS\alpha
\]
and my prove is complete.we can repeat this way to prove $ MY \equal{} MZ$.
i think this XM=MY maybe fales :!:

[b]m bahador[/b] notice that the fact $ MX \equal{} MT$ and $ MY \equal{} MZ$ is obvious cause M is midpoint of $ BC$ .

for the rest notice that $ \angle ABY \equal{} \angle AXY$ and $ \angle AZT \equal{} \angle ACT$ but $ \angle ABY \equal{} \angle ACT$ hence $ \angle AXY \equal{} \angle AZT$ hence $ M$ is center of circcumcircle of $ XYZT$

[quote="Virgil Nicula"]Consider in the triangle $ ABC$ two points $ D$ , $ E$ on the side $ [BC]$ so that $ \widehat{BAD}\equiv\widehat {CAE} = \phi$ .

Denote $ \|\begin{array}{ccc} M\in (BC) & , & MB = MC \\
 \\
\{X,T\}\subset AD\ & , & \begin{array}{c} BX\perp AD \\
\ CT\perp AD\end{array} \\
 \\
\{Y,Z\}\subset AE\ & , & \begin{array}{c} BZ\perp AE \\
\ CY\perp AE\end{array}\end{array}\|$ . Prove that $ \|\begin{array}{c} MX = MY = MZ = MT \\
 \\
m(\widehat {XMY}) = 180^{\circ} - 2\phi \\
 \\
m(\widehat {ZMT}) = 180^{\circ} - 2(A - \phi )\end{array}$ .

[b]Remark (only for [u]Ahiles[/u] !).[/b] I hope that this problem is mine and only mine.

Althrough in geometry it  is very hard for a problem to be original.[/quote]
[hide="Solution"]
Let $ XZ$ meet $ AC$ at $ K$ and let the midpoint of $ AB$ be $ P$.  Since $ \angle AZB = 90 = \angle AXB$, we have that $ AZXB$ is cyclic.  Since $ \angle AZB = 90$ and $ AP = BP$, we see that $ P$ is the center of the circumcircle of $ AZXB$.  Therefore, $ PZ = PX$.  Now,\[ \angle AXK = \angle AXZ = \angle ABZ = 90 - \angle BAZ = 90 - \angle XAK\]It follows that $ \angle XKA = 180 - \angle AXK - \angle XAK = 90$, so $ XZ\perp AC$.  Yet, $ MP$ is a midline of $ \triangle ABC$, so $ MP\parallel AC$, which means that $ MP\perp XZ$.  Yet, $ P$ is on the perpindicular bisector of $ XZ$ since $ PZ=PX$, so $ M$ is on the perpindicular bisector of $ XZ$, so $ MX = MZ$.  Similarly, $ MT = MY$.

Notice that $ \angle XAB = \angle YAC$ and $ \angle AXB = 90 = \angle AYC$, so $ \triangle AXB\sim \triangle AYC$.  Therefore, $ \frac {AX}{AY} = \frac {AB}{AC}$.  Similarly, $ \triangle AZB\sim \triangle ATC$, so $ \frac {AZ}{AT} = \frac {AB}{AC} = \frac {AX}{AY}$, which implies that $ AZ\cdot AY = AT\cdot AX$, so $ XZYT$ is cyclic.  Since $ MX = MZ$, we have that $ M$ on the perpindicular bisector of $ XZ$.  Similarly, it lies on the perpindicular bisector of $ TY$, so $ M$ is the circumcenter of $ XZYT$.  It follows that $ MX = MZ = MY = MT$, as desired. [/hide]


limit
integration
function
real analysis
calculus
real analysis unsolved

[b]Problem.[/b] Let $ f: [0,\infty) \rightarrow \mathbb{R}$, ${ g: [0,1] \rightarrow \mathbb{R}}$ so that ${ \lim_{x \rightarrow \infty}f(x)=L} \in \mathbb{R}$, $ f$ is integrable and $ g$ is continuous. Prove that $ \lim_{n \rightarrow \infty}\frac{1}{n}\int_{0}^{n}f(x)g(\frac{x}{n})dx=L\int_{0}^{1}g(x)dx$

[b] My proof.[/b]
If we use the substitution $ t=\frac{x}{n}$ then we have $ \frac{1}{n}\int_{0}^{n}f(x)g(\frac{x}{n})dx=\int_{0}^{1}f(nt)g(t)dt.$
Because $ \lim_{x \rightarrow \infty}f(x)=L$ it means that for any sequence $ (x_n)_{n \in \mathbb{N^*}}$, with $ \lim_{n \rightarrow \infty}x_n=\infty$, we have $ \lim_{n \rightarrow \infty}f(x_n)=L$. So $ \lim_{n \rightarrow \infty}f(nt)=L$, $ t \in [0,1]$
$ \Rightarrow \forall \epsilon >0, \exists n_{\epsilon}$ so that $ \left|f(nt)-L \right|<\epsilon$, $ \forall n \geq n_{\epsilon}$. So there is a $ n_{\epsilon_1}\in\mathbb{N}$, with the condition that $ \forall n \geq n_{\epsilon_1}$, $ \left|f(nt)-L\right|<\frac{\epsilon}{\left|\int_{0}^{1}g(t)dt\right|}$

We have $ \left|\int_{0}^{1}f(nt)g(t)dt-\int_{0}^{1}Lg(t)dt\right|=\left|\int_{0}^{1}g(t)(f(nt)-L)dt\right|<\left|\int_{0}^{1}g(t)\cdot \epsilon\right|<$, $ \forall n \geq n_{\epsilon_1}$.  And from here I think it's obvious.
I asked about my solution because the solution presented in the book where I found this problem is complicated and long, in comparison with this one.

[quote="Aalexandru"]...
$ \Rightarrow \forall \epsilon > 0, \exists n_{\epsilon}$ so that $ \left|f(nt) \minus{} L \right| < \epsilon$, $ \forall n \geq n_{\epsilon}$. So there is a $ n_{\epsilon_1}\in\mathbb{N}$, with the condition that $ \forall n \geq n_{\epsilon_1}$, $ \left|f(nt) \minus{} L\right| < \frac {\epsilon}{\left|\int_{0}^{1}g(t)dt\right|}$

...[/quote]
it's true if $ t$ is fixed, but when you  integrate function $ f(nt)$ on interval $ (0,1)$ hence when $ t \approx 0$ then  you can't find common such $ n_{\epsilon}$ for all $ t \in [0,1]$.

it's simple exercises about Lebesgue dominated convergence theorem .

Thank you [b]Extremal[/b] for your observation, it is clear now what my mistake was.

Can this problem be solved using Lebesgue dominated convergence theorem, but without actually refering to it? :lol:  I don't know if I made myself clear but I ask this because I don't think we are allowed to use this theorem in the national competion (Romania) without proving it.

you should to fix some $ \epsilon$  such that $ \int_{0}^{\epsilon} f(nt)g(t) dx < \delta$  since $ g$ is continuous on $ [0,1]$ and hence bounded by some constant $ C$, and $ f$ is integrable on $ [0,\infty)$ and bounded at infinity  (note that integral is absolutely continuous). Now you can find $ n_{\epsilon}$ for $ t \in [\epsilon,1]$ and  hence write above integral by this:
\[ \int_{0}^{1} ... \equal{} \int_{0}^{\epsilon} ... \plus{} \int_{\epsilon}^{1}...
\]
and estimate each of them easly.