Country A, B, and C are fighting each other. Country A is stronger than B or C, but the combined forces of B and C can beat A.
If the countries can choose to form alliances with each other (they could just fight alone), what are the possibility of Country A winning this war?
Can we generalize this for x number of countries fighting the war? (One country is stronger than any other country alone, but if [b]all[/b] the countries unite against the stronger country, then the alliance will win.)

Is there an equal probability of each country siding or not siding with A?

Yes. Also note that
A.) The countries may all choose not to form a coalition (fight alone)
B.) The members of the coalition will fight against each other if there are no more countries outside of the coalition

This is a more complicated question than it might seem, and I think you need to be more specific about the probabilties.

Specifically, does A have to be the sole victor in order to satisfy the conditions?  Do all the countries HAVE to fight?  Does order matter?  (Say, if A beats B, then C vs if A beats C, then B.)

It seems to me that country A wins the war only if B and C don't side together.  These are the possibilities: (A wins? Y/N in parens)

B and C side together to defeat A (N)
B sides with A to defeat C (Y)
C sides with A to defeat C (Y)
B and C don't side together, but fight A individually (Y)
A and B don't side together, but fight C individually (Y)
A and C don't side together, but fight B individually (Y)
B and C fight against each other.  Then A defeats the victor.  (Y)

I'm sure there are more, so I won't venture to say that I have listed all of them.  Again, you need to define the rules more clearly for us to analyze it properly.

OK, here goes.
1.) There are x number of countries
2.) An alliance may be formed between 2 to x-1 number of countries
3.) There can only be one alliance at a time (For example, if A and B form an alliance then C and D can't form an alliance at the same time)
4.) Country A is stronger than any other country. The only way to win against Country A is if every other country (x-1 number of countries) form an alliance
5.) The alliance, once every country outside of the alliance is destroyed, fights within the alliance
6.) When the members of a former alliance is fighting each other, they may choose to form another alliance (rule number 3 still stands)
7.) If none of the countries choose to form an alliance, it's "every man for himself" and is considered [b]1[/b] possibility. (There are no "order" in which the countries would fight. All of them will fight simulteneously)
8.) Every single country [b]must[/b] fight. However, there are cases where some survives.
9.) It doesn't matter who wins as long as you know if Country A won or not.
10.) Every country excluding Country A has equal strength, thus having the equal probability of winning/losing
Any questions?

I think I've got this.

[hide]There is only one possiblity in which A loses, when every other countries allies against it.

So the chance that country A wins is (a-1)/a, where a=total # of possiblities.

I found that:

a= :Sigma: (x-y)^y+(x-2y)^y+....+z, or

a=2^x-x-1

where y>z :ge: 0 and y=(1 to x-1)[/hide]


Assume that the function $ f : \mathbb{N}\rightarrow \mathbb{N}$ satisfies $ f(n+1) > f(f(n))$, for every $ n \in \mathbb{N}$. Prove
that $ f(n) = n$ for every $ n$.

see here http://www.mathlinks.ro/Forum/viewtopic.php?t=75980  :wink:


The incircle of triangle  $ABC$  has center $I$ and touches sides  $BC , CA , AB$  at points  $D ,E,F$  respectively .The bisector  of angle $BIC$ meets side  $BC$  at  $M$. If $AM$  intersects $FE$  at  $P$ , prove that  $DP$ is the bisector of angle $FDE$.

   [u]Babis[/u]

 [i]I think I have seen this problem again , solved , but I can not find where.[/i]

This is nice since now we know an older source of part [b](a)[/b] from http://www.mathlinks.ro/Forum/viewtopic.php?t=24699 .

  darij


Prove that for positive reals $ a,b,c$ ,
$ \frac{a^2\plus{}ac}{2b\plus{}a\plus{}c} \plus{} \frac{b^2\plus{}ba}{2c\plus{}a\plus{}b} \plus{}\frac{c^2\plus{}cb}{2a\plus{}b\plus{}c} \ge 2(a\plus{}b\plus{}c)$
when does the equality hold?

[quote="geniusbliss"]Prove that for positive reals $ a,b,c$ ,
$ \frac {a^2 \plus{} ac}{2b \plus{} a \plus{} c} \plus{} \frac {b^2 \plus{} ba}{2c \plus{} a \plus{} b} \plus{} \frac {c^2 \plus{} cb}{2a \plus{} b \plus{} c} \ge 2(a \plus{} b \plus{} c)$
when does the equality hold?[/quote]

The inequality inversely holds

$ \frac {a^2 \plus{} ac}{2b \plus{} a \plus{} c} \plus{} \frac {b^2 \plus{} ba}{2c \plus{} a \plus{} b} \plus{} \frac {c^2 \plus{} cb}{2a \plus{} b \plus{} c} \le 2(a \plus{} b \plus{} c)$

$ \Longleftrightarrow  20\,{b}^{2}{a}^{2}\plus{}13\,a{c}^{3}\plus{}11\,{a}^{3}c\plus{}20\,{a}^{2}{c}^{2}\plus{}13\,b{a}^{3}\plus{}11\,{b}^{3}a\plus{}13\,c{b}^{3}$$ \plus{}20\,{c}^{2}{b}^{2}\plus{}50\,a{c}^{2}b\plus{}50\,ac{b}^{2}\plus{}50\,{a}^{2}cb\plus{}11\,{c}^{3}b\plus{}2\,{a}^{4}\plus{}2\,{b}^{4}\plus{}2\,{c}^{4}\geq 0$

are you sure ?? i am positive that i posted the question correctly and if i had done so you are wrong..
i dont know about those calculations you have made .. just check yours again...
still if i any doubts i will post my solution..

For $ a\equal{}b\equal{}c\equal{}1$, $ RHS\equal{}\frac32$, $ LHS\equal{}6$.

i am very sorry you two are right there was a grievous mistake in my calculations..very sorry

[quote="hedeng123"][quote="geniusbliss"]Prove that for positive reals $ a,b,c$ ,
$ \frac {a^2 \plus{} ac}{2b \plus{} a \plus{} c} \plus{} \frac {b^2 \plus{} ba}{2c \plus{} a \plus{} b} \plus{} \frac {c^2 \plus{} cb}{2a \plus{} b \plus{} c} \ge 2(a \plus{} b \plus{} c)$
when does the equality hold?[/quote]

$ \frac {a^2 \plus{} ac}{2b \plus{} a \plus{} c} \plus{} \frac {b^2 \plus{} ba}{2c \plus{} a \plus{} b} \plus{} \frac {c^2 \plus{} cb}{2a \plus{} b \plus{} c} \ge k(a \plus{} b \plus{} c)$

$ k_{max} \equal{} \frac {1}{2}$

assume  $ a \equal{} max{(a,b,c)}$

$ \frac {a^2 \plus{} ac}{2b \plus{} a \plus{} c} \plus{} \frac {b^2 \plus{} ba}{2c \plus{} a \plus{} b} \plus{} \frac {c^2 \plus{} cb}{2a \plus{} b \plus{} c} \ge \frac {1}{2}(a \plus{} b \plus{} c)$

$ \Longleftrightarrow( a \plus{} b \plus{} c )(((b \plus{} 2a \plus{} 5c)(a \minus{} b)(a \minus{} c) \plus{} (4a \plus{} 2b \plus{} 2c)(b \minus{} c)^2)\geq 0$

obvious true[/quote]
you are right and here is my proof for $ k\equal{}\frac{1}{2}$ -
since the inequality is homogenous assume - $ a\plus{}b\plus{}c\equal{}1$,
then,to prove,
$ \sum_{cyclic}\frac{a^2\plus{}ac}{b\plus{}1} \ge \frac{1}{2}$,
or,
add 1 to both sides,
$ \sum_{cyclic}\frac{a^2\plus{}ac\plus{}ab\plus{}a}{b\plus{}1} \ge \frac{3}{2}$
or,
$ \sum_{cyclic}\frac{2a}{b\plus{}1} \ge \frac{3}{2}$,
or equivalently,
$ \sum_{cyclic}\frac{4a^2}{2ab\plus{}2a} \ge {3}{2}$,
$ \sum_{cyclic}\frac{4(a\plus{}b\plus{}c)^2}{2(ab\plus{}bc\plus{}ca\plus{}1)} \ge \frac{3}{2}$ as it is equivalent to -
$ (a\plus{}b\plus{}c)^2\ge 3ab\plus{}3bc\plus{}3ca$ which is trivial,
with equality for $ a\equal{}b\equal{}c$
$ Q.E.D$


The Klein bottle $ K^2$  is a square where the opposite vertical edges are identified in the opposite direction and the horizontal edges are identified in the same direction.

Consider the space $ \mathbb{R}P^2 \# \mathbb{R}P^2$  resulting from an annulus by identifying antipodal points on the outer circle,  and also identifying antipodal points in the inner circle. Show that $ K^2 \cong \mathbb{R}P^2 \# \mathbb{R}P^2$, i.e. $ K^2$  is homeomorphic to $ \mathbb{R}P^2 \# \mathbb{R}P^2.$

well by the classification of surfaces the klein bottle results as a sphere where two disks are replaced by mobius strips. the projective plane is when we replace one disk with a mobius strip.
It's clear, look at the annulus as a sphere with two missing disks.

Consider the fundamental polygon of the figures.


The silver number when you do a problem doesn't match with the number in the home page of Alcumus, and it doesn't match with the sum of the number of problems right, wrong, and give up in the report.  It's like 30 off for me, is it supposed to be this way?

Different questions are different point values

Ah....thanks...I was confused by that, but if it were question number, that would be kind off more helpful...I guess.

We don't have question numbers. If you want to see how many problems you have done, view your Report page.


geometry
analytic geometry
trigonometry
geometry proposed

Consider a right-angled triangle $ABC$ with the hypothenuse $AB=1$. The bisector of $\angle{ACB}$ cuts the medians $BE$ and $AF$ at $P$ and $M$, respectively. If ${AF}\cap{BE}=\{P\}$, determine the maximum value of the area of $\triangle{MNP}$.

Let $C$ be the origin of the coordinate system, $A(2a,0)$ and $B(0,2b)$. Then: $E(a,0)$, $F(0,b)$ and $\displaystyle{a^2+b^2=\frac{1}{4}}$. The equation of the line $BE$ is:
\[ BE: y=\frac{-2b}{a}x+2b.  \]
The equation of the line $AF$ is:
\[ AF: y=\frac{-b}{2a}x+b.  \]
Obviously the bisector of angle $C$ has the equation $y=x$. Thus:
\[ \displaystyle{P(\frac{2ab}{a+2b},\frac{2ab}{a+2b})}  \]
\[ \displaystyle{M(\frac{2ab}{2a+b},\frac{2ab}{2a+b})}  \] and
\[ \displaystyle{N(\frac{2a}{3},\frac{2b}{3})}.  \]

Therefore the area of $\triangle{MNP}$ is:
\[ f(t)=\frac{t(1-8t)}{3(1+10t)}  \] where $0<t=ab\leq{\frac{1}{8}}$.
Observe that:
\[ f'(t)=\frac{(1-20t)(4t+1)}{3(1+10t)^2}  \]
Since $f(\frac{1}{8})=0$ we conclude that
\[ max(S_{\triangle{MNP}})=f(\frac{1}{20})=\frac{1}{150}.  \]

Answer: The maximum are is: $\displaystyle{\frac{1}{150}}$.

This problem was given in our competition in 2002 :)
It's not very tough anyway(since I coped with it). There`s another Solution using only Trigonometry but I'm too lazy to post it :)

:!: I think that I understood what is $N$.

It is O.K. Soon I will offer the my solution.

[color=darkred][b]Lemma.[/b] Let $ABC$ be a triangle. For the points $M\in (BC)$, $N\in (CA)$, $P\in (AB)$ define the intersections $X\in BN\cap CP$, $Y\in CP\cap AM$, $Z\in AM\cap BN$. Denote $\frac{MB}{MC}=x$, $\frac{NC}{NA}=y$, $\frac{PA}{PB}=z$. Then \[ \boxed {\frac{[XYZ]}{[ABC]}=\frac{(1-xyz)^2}{(1+x+xy)(1+y+yz)(1+z+zx)}}\ .  \][/color]
[color=darkblue][u]Remark.[/u] For the our particular case $S=\frac{bc}{2}$, $b^2+c^2=1$, $x=\frac cb$,
$y=z=1$ we obtain $[XYZ]=$ $\frac{bc(b-c)^2}{6(b+2c)(2b+c)}=$ $\frac{bc(1-2bc)}{6(2+5bc)}$.
Therefore, the proposed problem becomes $\max_{b^2+c^2=1}\ \frac{bc(1-2bc)}{6(2+5bc)}$.
Denote $t=bc$. Because $2bc\le b^2+c^2=1$ results $t\in \left(0,\frac 12\right]$.
Thus $[XYZ]=\frac 16\cdot f(t)$, where $f: \left(0,\frac 12\right]\rightarrow R$, $f(t)=\frac{t-2t^2}{2+5t}$.
But $f'(t)\ .s.s.\ 1-5t\Longrightarrow f(t)\le f\left(\frac 15\right)$ $\Longrightarrow$ $\max [XYZ]=\frac{1}{150}$.
We have equality if and only if $b^2+c^2=1$ and $5bc=1$ a.s.o.[/color]

[b]Generalization.[/b] $[XYZ]\le \frac{\left(3-2\sqrt 2 \sin \frac A2\right)\sin A}{6\left(3+2\sqrt 2\sin \frac A2\right)^2}\ .$ We have equality
[u]if and only if[/u] $bc=\frac{1}{\sin \frac A2\left(3\sqrt 2+4\sin^2 \frac A2\right)},\ b^2+c^2=1+2bc\cos A\ .$


Find $ \sup\{m~|~\forall n \in \mathbb{N} \prod_{k\equal{}1}^m (n\plus{}k) \mbox{ is not a perfect square}\}$

Obviosly $ \sup\{m~|~\forall n \in \mathbb{N} \prod_{k \equal{} 1}^m (n \plus{} k) \mbox{ is not a perfect square}\}\equal{}\infty$.

Rust,
I probably didn't write my inital question properly, trying to be short :(
Denote by $ A$ the set of postitive integer $ m$, s.t. for all $ n \in \mathbb{N}$, $ \prod_{k\equal{}1}^m (n\plus{}k)$ is never a perfect square.
What is $ \sup A$?
If this is obvious (which I doubt), what else could we say about $ A$?

$ sup A \equal{} \infty$, because $ A \equal{}\{n|n>1\}\equal{} N\minus{}  \{1\}$.


I am asking this question because I want to confirm if my concept of power is correct.

The circuit is two batteries and a resistor, all in parallel (so both their positive poles are pointing in the same direction). Battery 1 has emf E1 and internal resistance r1, and Battery 2 has emf E2 with internal resistance r2. The resistor in the circuit has resistance R. Call the current through the first battery I1, current through the second I2, and current through the resistor I2 (going in opposite direction as I1 and I2)

I found this problem in my textbook, and though finding the power wasn't the exact question, the answer was reached based on the assumption that the power delivered in the circuit is equal to [(I3)^2]*(R). But isn't that not taking into account the power lost through the internal resistors? In fact, instead of calculating the power lost through resistors, wouldn't an easier expression simply be the sum of the powers delivered by each battery, i.e. Power = (I1)*(E1) + (I2)*(E2)?

Thank you! Please let me know if my concept of power is correct or not!

If I am in no mistake, I think you are confusing [i]total power[/i] of a source with the [i]power supplied[/i] by the source to the [i]circuit[/i]. For each of the batteries, the power balance is given by

$ \epsilon_{i}I \equal{} VI\plus{}r_{i}I^{2}$,

where $ \epsilon_{i}$ is the emf of battery i, V the voltage across its terminals (in this circuit it is the same, because they are in parallel), I the current through each one, and $ r_{i}$ the internal resistance of each battery. The term on the left represents the total power that the source generates inside it, but only the term VI is effectively supplied to the circuit, because the power $ r_{i}I^{2}$ is dissipated inside it by Joule effect.
So, for your circuit we have

$ E_{1}I_{1}\equal{} VI_{1}\plus{}r_{1}I_{1}^{2}$

$ E_{2}I_{2}\equal{} VI_{2}\plus{}r_{2}I_{2}^{2}$

and so the total power supplied to the circuit is $ P \equal{} VI_{1}\plus{}VI_{2}\equal{} V(I_{1}\plus{}I_{2})$. But applying KCL we have $ I_{1}\plus{}I_{2}\equal{} I_{3}$, while applying OL to the resistor we have $ V \equal{} RI_{3}$. Hence: $ P \equal{} RI_{3}(I_{3}) \equal{} RI_{3}^{2}$.

(Note: KCL = Kirchhoff's Current Law, OL = Ohm's Law)

[quote="Bhh198"]wouldn't an easier expression simply be the sum of the powers delivered by each battery, i.e. Power = (I1)*(E1) + (I2)*(E2)?[/quote]

That is total power that both batteries can supply, but because they have internal resistances part of it is dissipated.

Thank you very much. I see what you're saying now - the energy lost through internal resistance is energy lost in the source, not in the circuit. Thanks again.


trigonometry
calculus
calculus computations

Let $ f_1(x)\equal{}\pi \sin x$, for $ n\equal{}2,\ 3,\ 4,\ \cdots$, define functional sequeneces $ f_2(x),\ f_3(x),\ f_4(x),\ \cdots$ such that $ f_n(x)\equal{}f_1(f_{n\minus{}1}(x))$.
Express the number of $ x$ in terms of $ n$ such that $ f_n(x)$ has extrema in the interval of $ 0<x<\pi$.

[hide="maybe?"]$ 0\equal{}f_n'(x)\equal{}f_1'(f_{n\minus{}1}(x))f_{n\minus{}1}'(x)\equal{}\pi \cos (f_{n\minus{}1}(x)) f_{n\minus{}1}'(x)$ iff

$ \cos (f_{n\minus{}1}(x)) \equal{}0,$ or $ 0\equal{} f_{n\minus{}1}'(x)$

$ \to f_{n\minus{}1}(x)\equal{}\frac{\pi}{2},$ or x is a critical point of $ f_{n\minus{}1}$

so $ C_n \equal{} \{ x\in (0, \pi) : f_n ' (x)\equal{}0\}$ must have $ C_n \subset C_{n\plus{}1}$

in fact $ C_{n\plus{}1} \equal{}C_n \cup f_n^{\minus{}1}\left( \frac{\pi}{2}\right)$
also,

$ |f_n^{\minus{}1}\left( \frac{\pi}{2}\right)| \equal{} 2 |f_{n\minus{}1}^{\minus{}1}\left( \frac{\pi}{2}\right)| \to
 |f_n^{\minus{}1}\left( \frac{\pi}{2}\right)| \equal{}2^n$
and $ f_n^{\minus{}1}\left( \frac{\pi}{2}\right) \cap f_m^{\minus{}1}\left( \frac{\pi}{2}\right) \equal{} \emptyset$ when $ n \ne m$

so, $ |C_{n\plus{}1}| \equal{}|C_n|\plus{}2^n \to |C_n| \equal{} \sum_{k\equal{}0}^n 2^k \equal{} 2^{n\plus{}1}\minus{}1$[/hide]


function
number theory
prime factorization

is there a quick and easy way to find divisors?

the sum of the divisors of an integer is 0...

the sum of the [b]positive[/b] divisors of an integer has a formula, which I'll let you prove on your own, but if the integer's prime factorization is $ p_{1}^{e_{1}}p_{2}^{e_{2}}p_{3}^{e_{3}}\cdots p_{n}^{e_{n}}$, then the sum of its factors is, and hopefully you'll understand the notation, is
\[ \prod_{i \equal{} 1}^{n}\left(\sum_{j \equal{} 0}^{e_{i}} p_{i}^{j}\right)
\]

why the product function? thats eg $ 2^{x_1} 5^{x_2} 7^{x_3}$ thats (sum of powers of 2)(sum of powers of 5)(sum of powers of 7) which doesnt seem right....or did i intepret it wrong?

i think it's right...you need add to the powers of each prime factor...(e.g. for 2^6, we have 1 + 2 + 4 + 8...+ 64)
Then you multiply them...

this formula has been a lifesaver!  :)

for 45, we have (3^2)(5^1)

so we have (1+3+9)(1+5) = 78.

this formula is explained in aops vol.2. basically, it works since after disctributing everything, you get all possible combinations of prime factors(check for yourself if you want).
which is synonomous to summing all of the factors


Given a table (see the figure). It is allowed to permute its rows and to permute its columns (in any order). How many distinct tables can be obtained from the given table by these operations?

\[ \left[
\begin{array}{ccccccc}
1 & 2 & 3 & 4 & 5 & 6 & 7 \\
7 & 1 & 2 & 3 & 4 & 5 & 6 \\
6 & 7 & 1 & 2 & 3 & 4 & 5 \\
5 & 6 & 7 & 1 & 2 & 3 & 4 \\
4 & 5 & 6 & 7 & 1 & 2 & 3 \\
3 & 4 & 5 & 6 & 7 & 1 & 2 \\
2 & 3 & 4 & 5 & 6 & 7 & 1
\end{array}
\right]\]

This is the last problem of Senior O-Level Paper, Spring 2008.

[quote="NapoleonXIV"]We can get all permutations by just changing the columns (or rows), and each column has $ 7!$ permutations. Now its just the case how many permutations there are in $ 7$ columns. This is like arranging $ 7$ sings, and each sign has $ 7!$ possibilities.
Therefore there are $ (7!)^7$ permutations in total[/quote]

I think that this is wrong, because we can get no more than $ (7!)^2$ different tables, columns in $ 7!$ permutations, and for each permutation of columns exactly $ 7!$ permutations of rows, but I don't how many times did we get the same table. I think that solution is $ 6! \cdot 7!$ (each table we have counted 7 times), but I am not sure, so I need your help, thanks  :)

I think $ 6!*7!$ is right, as can be seen if you prove the following two observations.

1. For any given first row/column each containing the numbers from $ 1$ to $ 7$ exactly once, there is at least one permutation of the table with the correct entries in those places.

2.  For any permuted table $ T'$ we have $ T'(i,j) \plus{} T'(1,1) \equal{} T'(1,j) \plus{} T'(i,1)$ (where $ T'(i,j)$ is the entry in the $ i^{th}$ row and $ j^{th}$ column of the permuted table), so the table is uniquely determined by its first row and column.


Euler
quadratics
number theory unsolved
number theory

Prove that :
a) Odd prime number $p$ can be representable in the form $x^2+2y^2 \iff p=8k +1$ or $p=8k+3$
b) Odd prime number $p$ can be representable in the form $x^2-2y^2 \iff p=8k +1$ or $p=8k+7$
c) Prime number $p$ can be representable in the form $x^2+3y^2 \iff p=6k+1$

I think you meant " Euler " . But I don't know Euler's solution ! :(

Your c is wrong - supposed to be p=3k+1. Here's an outline of a proof.
Use quadratic forms, quadratic law of reciprocity, and Guass' lemma.
Consider a quadratic form ax^2+bxy+cz^2. Let h be its dicriminant = b^2-4ac. Then, we have a number n is expressible as that form if

n=? (mod 4h), which you solve this by law of reciprocity.

You can also use reduced forms, fundamental discriminants and genus theory to help out.

As a help out, here's a harder problem
Find all k from 1 to 20 such that if p=1+kn is prime, then p=a^2+kb^2 for some integers a,b.

Bomb

Another way would be just to use quadratic reciprocity to show that a corresponding equation $x^2 \equiv a \mod p$ can be solved and then to use the lemma of Thue.
But all these problems were posted before, so take a search.


Let $a,b,c$ be positive numbers. Prove that
\[\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geqslant \frac{a+b+c}{2}\cdot\left(\frac{2}{ab+bc+ca}+\frac{1}{a^{2}+b^{2}+c^{2}}\right).\]

Assuming that $a+b+c=1$ and denoting $S=ab+bc+ca$ ($3S\leq 1$), we can rewrite the inequality as
$S^{2}(1-4S)+(2-3S)abc \geq 0$.
Since $1-4S \geq-9abc$ (Schur's Inequality), we need to show that
$abc(1-3S)(2+3S) \geq 0$,
which clearly is true. :)
Nice inequality, Rzeszut ! :lol:

If $a,b,c$ are positive number such that $\frac 1{a+b}+\frac 1{b+c}+\frac 1{c+a}=\frac 3{2}$, then
$\frac 3{a+b+c}\geq \frac 2{ab+bc+ca}+\frac 1{a^{2}+b^{2}+c^{2}}$.

[quote="Rzeszut"]Let $a,b,c$ be positive numbers. Prove that
\[\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\geqslant \frac{a+b+c}{2}\cdot\left(\frac{2}{ab+bc+ca}+\frac{1}{a^{2}+b^{2}+c^{2}}\right). \]
[/quote]

[quote="Vasc"]
If a,b,c are positive number such that $\frac 1{a+b}+\frac 1{b+c}+\frac 1{c+a}=\frac 3{2}$, then
$\frac 3{a+b+c}\geq \frac 2{ab+bc+ca}+\frac 1{a^{2}+b^{2}+c^{2}}$.
[/quote]
I don't understand what is the different between 2 problems?

I think there is no difference,but may be Vasc thought that it is a little diffecult than the original one.

In my opinion, the last is only a more beautiful form of the same inequality. :P

We may naturally assume $a+b+c=2$, then mix it. I think it's nicer with the condition,
\[{1 \over a+b}+{1 \over b+c}+{1 \over c+a}\geq{1 \over a^{2}+b^{2}+c^{2}}+{2 \over ab+bc+ca}. \]


Find  the greast value of the expression
$\frac{a}{b}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}-\frac{abcd}{(ab+cd)^{2}}$
where $a;b;c;d$ are distinct real numbers satisfying conditions $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$ and $ac=bd$

This problem is currently in Problem Solving Contest of MYM (Vietnam). The topic should be closed. Thanks.


The equation

$x^{10}+(13x-1)^{10}=0$
has 10 complex roots $r_1$, $\overline{r_1}$,$r_2$, $\overline{r_2}$,$r_3$, $\overline{r_3}$, $r_4$,$\overline{r_4}$,$r_5$,$\overline{r_5}$, where the bar denotes complex conjugation. Find the value of

$\frac 1{r_1\overline{r_1}}+\frac 1{r_2\overline{r_2}}+\frac 1{r_3\overline{r_3}}+\frac 1{r_4\overline{r_4}}+\frac 1{r_5\overline{r_5}}.$


Show that there exists an equiangular hexagon in the plane, whose sides measure 5,8,11,14,23, and 29 units in some order.

[hide]The hexagon with side lengths $8$, $29$, $5$, $14$, $23$, $11$, in this order, is an equiangular hexagon.

Proof:

Consider equilateral triangle $XYZ$ with side length $42$.
Take points $A$ and $B$ on $XZ$ and $YZ$ respectively such that $AB\parallel XY$ and $ZA=ZB=8$.
Take points $C$ and $D$ on $YZ$ and $XY$ respectively such that $CD\parallel XZ$ and $YC=YD=5$.
Take points $E$ and $F$ on $XY$ and $XZ$ respectively such that $EF\parallel YZ$ and $XE=XF=23$.

Note that $\triangle ZAB$, $\triangle YCD$, and $\triangle XEF$ are equilateral triangles.
By "cutting out" these three small equilateral triangles, we obtain a hexagon with side lengths $AB=8$, $BC=29$, $CD=5$, $DE=14$, $EF=23$, and $FA=11$.
Since we have cut out small equilateral triangles from a large equilateral trianlge, each interior angle of the resultant hexagon is $120^\circ$, and thus hexagon $ABCDEF$ is an equiangular hexagon.[/hide]


number theory
relatively prime
prime numbers
number theory unsolved

Let $n$ be an integer of the form $a^2 + b^2$, where $a$ and $b$ are relatively prime integers and such that if $p$ is a prime, $p \leq \sqrt{n}$, then $p$ divides $ab$.  Determine all such $n$.

Since $(a,b)=1$ then $(ab, a^{2}+b^{2})=1$, so for any $p\leq\sqrt{n}$, $p$ does not divide $n$. So $n$ is a prime. Let $p_{1}=2, \ldots p_{k}\leq sqrt{a^{2}+b^{2}}$ be the prime numbers less than $\sqrt{n}$. So $ab$ is divisible by$p_{1}\cdot p_{2} \ldots p_{k}$, and $ab\geq p_{1}p_{2}\ldots p_{k}$. Suppose $\sqrt{n}\geq 16$. Then using Bertrand, we have that there exista at least one prime number between $\frac{\sqrt{n}}{16}$ and $\frac{\sqrt{n}}{8}$, one prime number between $\frac{\sqrt{n}}{8}$ and $\frac{\sqrt{n}}{4}$, one prime number between $\frac{\sqrt{n}}{4}$ and $\frac{\sqrt{n}}{2}$, and one prime number between $\frac{\sqrt{n}}{2}$ and $\sqrt{n}$.
Since $ab$ divides the product of these 4 numbers, then $ab\geq \frac{\sqrt{n}}{8}\cdot\frac{\sqrt{n}}{4}\cdot\frac{\sqrt{n}}{2}\cdot\sqrt{n}=\frac{n^{2}}{1024}$. Well, $ab\leq \frac{a^{2}+b^{2}}{2}=\frac{n}{2}$. So, $\frac{n}{2}\geq\frac{n^{2}}{1024}$, or $n\leq 512$. If $\sqrt{n}\leq 16$ we obtain $n\leq 256$. So treating the case $n\leq 512$ includes $n\leq 256$. It implies $a,b\leq22$. If $n \geq 13^{2}$, then $ab \vdots 13$. Since $a,b\leq 22$, then WLOG (without loss of generality) $a = 13$. Then $22\geq b\vdots 2\cdot3\cdot5\cdot7\cdot11$, impossible. So $n\leq168$. So $a,b\leq12$. If $n\geq 121$, then $ab\vdots 11$. WLOG $a=11$. Then $12\geq b\vdots 2\cdot3\cdot5\cdot7$, impossible. So $n\leq120$. We proceed analogously with the case $n\geq 7^{2}$. We get $n\leq48$. And analogously with casse $n\geq25$. We get $n\leq24$. Since $n$ is a prime, an easy checking gives $n=13$, with $a=2, b=3$. So the only solution is $n=13$.

Well, this is [url=http://mathworld.wolfram.com/BertrandsPostulate.html]Bertrand's postulate[/url], and it is fairly hard to prove..
But! look at this wonderful proof due to Erdös : http://en.wikipedia.org/wiki/Proof_of_Bertrand's_postulate .  :cool:

[quote=jin]I use a theorem that there is at least a prime between n and 2n(n>1).
But how to prove this one?[/quote]

Revived this old topic because I encountered this problem again and saw that the solution uses Bertrand's postulate unnecessarily. Here is a clean solution without Bertrand's Postulate using only some known facts, and this way we can also remove the constraint $\gcd(a,b)=1$ if one doesn't get bored of tedious work.

Let $k$ be the unique positive integer such that $p_k^2\leq n<p_{k+1}^2$ where $p_i$ is the $i$th prime. Since any prime $p\leq\sqrt n$ divides $ab$, we must have $\prod_{i=1}^kp_i|ab$ and thus $ab\geq\prod_{i=1}^kp_i$. From [url=https://en.wikipedia.org/wiki/Bonse%27s_inequality]Bonse's Inequality[/url], $\prod_{i=1}^kp_i>p_{k+1}^2$ for $k\geq4$. Thus, using A.M-G.M
\[n=a^2+b^2\geq2ab\geq2\prod_{i=1}^kp_i>2p_{k+1}^2>2n\]
if $k\geq4$. Thus $k\leq4$ and $\sqrt n<7$ or $n<49$.

Also, if $\gcd(a,b)=1$, no prime of the form $4k+3$ divides $a^2+b^2$. Otherwise, $\dfrac{p-1}2$ is odd, so $p|(a^2)^{\frac{p-1}2}+(b^2)^{\frac{p-1}2}=a^{p-1}+b^{p-1}$. But from Fermat's Little Theorem, $p|1+1=2$, contradiction. Thus $n$ has only prime factor $2,5,13,17,29,37,41$.

Let's consider two cases. If $n$ is prime, then $n=2,5,13$ are valid according to the statement. $17=4^2+1,29=2^2+5^2,37=6^2+1,41=4^2+5^2$ and in no representation $ab$ is divisible by $3$.

Now, $n$ is composite. Notice that $n$ can not have prime factor greater than $17$ since $29\cdot2=58>49$. If $13$ divides $n$, then $n$ must be $2\cdot13$ since $3$ doesn't divide $n$ and $4\cdot13=52>49$. But $26=5^2+1$, doesn't follow the rules. Similarly, if $17$ divides $n$ then $n=34=5^2+3^2$ but same problem. If $2$ divides $n$, then $4$ can not divide $n$. Because then $a$ and $b$ both have to be even. Therefore $n$ must be $10=1^2+3^2$ but this is not possible either. If $5$ divides $n$ then $n$ must be $25=3^2+4^2$ which is not valid either. Thus, $n=\boxed{2,5,13}$ are all solutions.

WLOG $1 \leq b \leq a$.
[hide=a-b=0]
 $a$ and $b$ are relatively prime$\implies a = b = 1\implies \boxed{n = 2}$
[/hide]
[hide=a-b=1]
$b=1\implies b = 1$ and $a=2\implies \boxed{n = 5}$.
$b=2\implies b = 2$ and $a=3\implies \boxed{n = 13}$.
$2b^2+2b+1 - (b^2+4b+4)= (b-1)^2 - 4 \geq 0(for$ $ b \geq 3)$
$b\geq 3\implies 1< b-1 < a+1 = \sqrt{b^2+4b+4}\leq\sqrt{2b^2+2b+1}=\sqrt{a^2+b^2}.$
$\implies gcd(b-1,b)>1$ or $gcd(b-1,a)>1\implies gcd(b-1,a) = 2\implies 2|a$.
$gcd(a+1,b)>1$ or $gcd(a+1,a)>1\implies gcd(a+1,b) = 2\implies 2|b$ - contradiction.
[/hide]
[hide=a-b>1]
Denote by $p$ any prime divisor of $a-b$ .
$p\leq a-b= \sqrt{a^2-2ab+b^2}<\sqrt{a^2+b^2}=\sqrt{n} \implies p|ab$ and $p|(a-b) \implies p|a$ and $p|b$ - contradiction.
[/hide]
[hide=answer]
2, 5, 13
[/hide]

Since $p \le \sqrt{n}$ and $\gcd(a,b)=1$ either $p|a$ or $p|b$
Since $(a,b)=1$ then $(ab, a^{2}+b^{2})=1$, so for any $p\leq\sqrt{n}$, $p$ does not divide $n$. So $n$ is a prime and $\equiv 1 \pmod 4$(other than $n=2$ of course)
Denote $p_1,p_2,p_3....$ be the primes less than $\sqrt{n}$
Using Bonse's inequality yeilds $ab \ge \prod{p_k}>p_{k+1}^2>$,a contradiction so $k \le 3$ and $\sqrt{n}<7$
$n \ge 24$ is ruled out by size arguments and checking all remaining possibilities gives $n=[2,5,13]$


MIT college Harvard geometry algebra polynomial

dont u think too much hype or in other words too much standard is given to iit
as compred to its international rankings.u c considering MIT,CALTECH etc.........
see its ok as its the best in india but dont u think its not the best and also know the admission procedure becoming strict and new iit's being opened!

[hide]
Wait, IIT is in INDIA???? OMG!!!! my frinds dad went to that school!!!!!! sorry that's off topic, are u mad at me? :maybe: [/hide]

yes it is true that IITs are not so good if you compare with other institutions

but the fact remains that the students in IITs are certainly world's best intelligentsia if you look as per this aspect
admission criteria is tough(due to heavy population pressure),questions are tougher than any exam(i say 'any exam' as in olympiads,only one subject is judged)

I'ld like to quote Narayan Murthy(ex CEO infosys):"My son went to Cornell to study CSE because he couldn't study CSE in any IIT.I know many persons who got good scholarships in Harvard and MIT who couldn't clear IIT"

this is standard of IIT(and certainly AIIMS also)

though they lack in infrastructure if compared to their foreign counterparts,they have something special

IIT is not a school but a graduation ,PG and PhD college
ranked 2 in Asia and its entrance exam is recognized as toughest in the world

usually tough is in terms of competition(Like for AIIMS only comp is tough exam is tooooo easy) :rotfl:

But for IIT- well--when u r expected to do 66 MCQ's in 3hrs u can urself judge the level of diff

Getting into an institute named "IIT" may now b easy --and in future IIt may bcome just a hype
But the CSE studs of IIT B,D.M,k wud still remain a much better class than even MIT

getting adm in MIT,IVY is diff  for Middle class in india bcause of dist and cost.... :o  :(

so one more question arises
then y is the ranking of iit not so good as in terms of others
u c standing at 553 in the world according to webometrics is it only infrastructure that makes IIT a little less from MIT and CALTECH.

it is onlyyyy infrastructure and facilities- nothing else :oops:
ofc i m talking abt main stream iits

there even the "else" factors r not better :rotfl:  :D

wat else factors i didnt understand
and tell me where would u like u to go MIT or CALTECH or IIT!

no he is not interested in going there

he'ld be probably interested in any medical colleges there

i don't know the names

perhaps HARVARD

but i won't prefer nething before India :mad:

thats a good choice !
wat do u think is the basic difference between studying abroad and in india?
plz be frank!

i may be able to answer ur question if i knew wat abroard exactly meant, how how u want it to mean.

hey actually see if u were asked to study at a place away from ur country or IN UR HOMELAND  then wat would u prefer

Hmmmmm. toughie..........i want a great colledge.oxford in london... i dunno about india......
i never really though of ny college options in indi.... gosh my head hurts from thinking too hard.

i wanna favour
plz help
me
with my rating
(O but still i dont care)

i'll help u if u help me!!!! in the same area!!! i'm only a one star...but it' okay if u dont

rating and stars are different
and rating is not even visible to others
y 2 care 4 tat?? :rotfl:

:rotfl:  :rotfl:  :rotfl:
wait till Da vince sees this

plz be serious ppl.............
post some math probelms or do some imp...discussions
and this applies mainly to me!

of course, serious that's what i love about the world, isn't being serious sooooo fun :huh: \
so math question simple algebraic critical thinking. [b]Find two binomials whose product is a binomial;a trinomial; a four-term Polynomial?[/b]

hey any one coukd tell me hows challenges and thrills of pre college mathematcs book ?
is it good?

it is a superb book
simply awesome!!!!!!!! :D

haven't heard abt that but challenges and thrill is a very good one
i used to do it in class X :)


[quote]
Old russian one?
[/quote]
Yes, Kurlyandchik & Merkuriev, All-Russian Math Olympiad, 1985.

The best constant is 2.

Hint for constant 4: reorder $a_{k}$ by increasing.


Hi, I'm wondering if anyone can help me here... I'm using TeXnicCenter 1 beta 7.01, MikTex 2.7 and Adobe Reader 7.0. When trying to use the "Build and View" option, it seems that TeXnicCenter is sending the .tex file instead of the .pdf file to Adobe. The .pdf file is still created and is fine otherwise but it's rather inconvenient.

If it helps, this started after I "fixed" another error I was having. I'd found someone on another message board with the same error (sorry that I don't remember the details) and fixing it involved making some changes under Build --> Define Output Profiles...

The exact error message that Adobe gives me is this:
[quote]Adobe Reader could not open 'filename.tex' because it is either not a supported file type or because the file has been damaged (for example, it was sent as an email attachment and wasn't correctly decoded).[/quote]

Thanks in advance!

In Build --> Define Output Profiles --> LaTex=>PDF click on the viewer tab. In two places DDE Command should be chosen and underneath it should say:
[DocOpen("%bm.pdf")][FileOpen("%bm.pdf")]
In the final command box near the bottom of the page it should say:
[DocClose("%bm.pdf")]


A fair six-sided die is rolled n times. If the standard deviation of the number of times a 5 comes up is 10, determine the value of n.

is it:
A. 400
B. 625
C. 720
D. 800

There is a summation formula for standard deviation that I don't remeber.  If you plug your data in, you should get an answer.

The formula for the standard deviation of a binomial probability distribution is $ \sqrt {np(1\minus{}p)}$ where n is the number of trials and p is the probability of success in each trial.

If you're in high school, you probably aren't expected to be able to prove this. If you are studying statistics at a more advanced level, however, it is possible that your professor expects you to be able to prove this result.

The formula I was referring to is
$ \sigma\equal{}\sqrt{\frac{1}{N}\sum_{i\equal{}1}^{N}(x_{i}\minus{}\bar{x})^{2}}$
where $ N$ is the number of samples taken.


Let $x,y,z>0$ such that $x+y+z=3$.Prove that:

       $\frac{9}{x^2+y^2+z^2}\leq\sqrt{x}+\sqrt{y}+\sqrt{z}\leq\frac{9}{xy+yz+zx}$ ;)

Just observe that

[tex](xy+xz+yz)\leq \frac{(x+y+z)^2}{3}=3[/tex]

and

[tex](\sqrt{x}+\sqrt{y}+\sqrt{z})\leq 3\sqrt{\frac{x+y+z}{3}}=3[/tex]

For the first part:

[tex]\left(\frac{x^2+y^2+z^2}{3}\right)\left(\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}\right)\geq  \left(\frac{x^{\frac{5}{4}}+y^{\frac{5}{4}}+z^{\frac{5}{4}}}{3}\right)^2\geq \left(\frac{x+y+z}{3}\right)^{\frac{5}{2}}=1[/tex]
By Cauchy and Power-Mean


[quote="Interval"]First of all, what is a frustum?

If you take a cone of a given size and cut it from the small end to
the large end, then lay it out on a flat surface, what will the inside
and outside sizes of this flattened-out cone be?

Example:
   x = diameter of small end of cone   x is 3"
   y = diameter of large end of cone   y is 6"
   z = length of cone                  z is 2"[/quote]
[url]http://en.wikipedia.org/wiki/Frustum[/url]
Basically, you are asking what is the surface area of the cone (not including the flat sides).

So, do I use the long equation in the provided link to find the volume?

Is this what is being sought--the volume?


Why or why aren't you watching the big game?

I am not allowed because I already watched a pistons clevelend game today.

I'm with New England and since Colts won over them (stupid last minute interception) i couldn't care less about who will win...

What kind of name is colts? Seriously...

You should add an option that says I could care less :)

gosh i cannot STAND football  :P

however, i am watching the commercials

[size=200][color=darkblue][b]COLTS[/b][/color][/size]

I'm watching for the game and the commercials :D

im going for the colts because im a chargers fan, and they lost to the patriots, and the colts beat the patriots which made me happy.

Lions and Tigers and Bears, OH MY!

I mean come on! What's the point of watching innocent souls getting their limbs broken just to throw a ball through a gate? Who CARES?? There are more important things than football!!!

[quote="K81o7"]I liked the slapping commercial.  "Dude, the fist pound is out."  "Then what is in?"  Slap.  "Cool!"

"Good job!  You saved the account!"  Slap.[/quote]

ROFL  :rotfl:

i liked the commercial with the giant map monster

as well as the one where that guy comes to mess with your stuff while your asleep

I liked the beer ones particularly.

Some commercials were too long though. but my my i had a blast

oh and there was the one with the hitchhiker with the axe which he claimed was a "can-opener", and then there was the guy with the chainsaw :rotfl:


What are some good introductory economics books?

Thanx!

I just picked up a copy of "A Textbook of Economic Theory" by Alfred W. Stonier & Douglas C. Hague $4^{th}$ edition. I have just started it but it seems really good,

There is Baumol and Blinder. Also, Greg Mankiw

I just read "Economics for Dummies" by Sean Masaki Flynn , and it was a good overview of Micro and Macro and other general concepts. I hadn't done economics in a long time and it really was a good introduction.

As I said in another thread, read "Essential Economics", which is like a dictionary of economic terms. Read the economist every week to get a picture of what's going on.


limit
irrational number
real analysis
real analysis unsolved

Let $ f : \mathbb R \to \mathbb R^{+}$ be a continuous, periodical with irrational period function. If $M = \max f$, then find \[\lim_{n \to \infty}\frac{f(1) f(2) \ldots f(n)}{M^{n}}.\] [i]Marian Ionescu[/i]

PS: I know it's kind of easy (if my ideas are correct, that is), but...

en... let us assume that f is not a constant.
then all we have to do is to show that there is an $\epsilon$, s.t. infinite many $f(n)/M\leq\epsilon$.  Assuming $min{f}=f(x_{0})=m<M$, take a neighbourhood of $x_{0}$, e.g. $U$, then there are infinitely many $f(n)$ lies in $\cup U+k, k=1,2,\ldots$, which implies the conclusion.

It is hard to understand your solution... but from what I understand of your solution, this is my problem with it:

Suppose $x$ is the irrational period. If $U = (a,b)$ we must be sure that there are infinitely many integers $k$ such that $(a+kx,b+kx)$ contains an integer. To me this is not obvious.

edit: I found an argument for the above: Let $E = b-a$. We assume $E < 1$ or it is trivial. According to Dirichlet's approximation theorem we can find integers $n > 0$ and a $p > 0$ such that $\left| nx-p \right| < E$. Therefore, the set $(a+nx-p, b+nx-p)$ will be shifted less than $E$ from $U$, i.e. the two sets overlap. I don't want to prove it, but if you consider the sets $(a+mnx, b+mnx)$ you realize that for some integer $m>0$,  $(a+mnx, b+mnx)$ will cover an integer. To realize it draw the sets $(a+mnx-mp, b+mnx-mp)$. Depending on the sign of $\left| nx-p \right| < E$ they will either creep up to or down to an integer as $m$ increases.  So now we've found one such $k$ you can construct another one in the same way, and so on.

It is a simple consequence of Weyl equidistribution theorem. Consider $(a+kx-j, b+kx-j)$ where we choose integer $j$ so that $a+kx-j \le 0$ but if $j$ were smaller it would be positive. Then $a+kx-j$ is uniformly distributed in $(-1, 0]$ and $b-a$ is not 0 so Weyl tells us out of $N$ intervals we should have about $(b-a)N$ integers contained in our intervals.

Thank you Kalle and Xevarion.  Actually the solution in my mind is like this:  consider $(a+b)/2$.  As is know to all that in $\mathbb{R}/\mathbb{Z}, (a+b)/2+nx$ is dense. Then the conclusion follows when we take a subsequence above letting it tends to 0 or 1.

Ok. Unfortunately I don't know enough math to understand either of the last two posts.

The following is (I believe) equivalent to what everyone else has done, but I think it's a bit more transparent.

Let $p$ be the period of $f$ and let $g(x) = f(xp)$, so $g$ is periodic with period 1, and $M$ is still the maximum of $g$, so we want to find $\lim_{n\to\infty}\frac{g(\frac1p)g(\frac2p)\cdots g(\frac{n}{p})}{M^{n}}$.  Now, pick some closed subinterval $I$ of $[0, 1]$ on which $g$ does not attain the value $M$ and let the maximum of $g$ on this interval be $m$.

Here's the important theorem: for any irrational number $q$ (in our case we want $q = \frac{1}{p}$), the fractional parts of $nq$ for integers $n$ form a dense subset of $[0, 1]$.  This means that we can find infinitely many integers $n$ such that $\{nq\}\in I$ and so that $g(nq) \leq m$.  But $\frac{g(nq)}{M}$ is bounded by 1 everywhere, so we're dealing with a product which never increases but steps down by (at least) a factor of $\frac{m}{M}$ infinitely often.  Certainly, this product must go to 0.


trigonometry
inequalities
LaTeX
inequalities unsolved

hello, so it is $\LaTeX$
Prove that if $ \frac{\pi}{n} <x < \pi-\frac{\pi}{n}$ then
$\frac{sinx}{3}+\frac{sin{nx}}{2n} \geq 0$
Sonnhard.


geometry
circumcircle
geometric transformation
reflection
trigonometry
parallelogram
perpendicular bisector

Let [i]C[/i] be a semicircle with diameter $AB$. On the same half-plane containing the semicircle, construct two rays $Ax$ and $By$. Let $C, D$ be two points lying on [i]C[/i] and $Q$ be the midpoint of $CD$. Denote by $P$ and $R$ the midpoints of $AC$ and $BD$. Through $P$, draw the line perpendicular to $QA$ that intersects the ray $Ax$ at $S$. Through $R$, draw the line perpendicular to $QB$ that intersects the ray $By$ at $T$.

Show that $ST // CD$.

Read it again, please. This cannot be true. Think about it: if you keep everything fixed and start moving $Ax$, then $C,D,T$ stay fixed, while $S$ moves on a line through $P$ which, in general, will be different from $T$. So the parallelism cannot be conserved.

I 'm really sorry. I missed one important condition. The rays $Ax$ and $By$ are the tangents of the semicircle at $A$ and $B$.  :oops:

Let the perpendicular bisectors of AQ, BQ meet the tangents Ax, Bx at E, F and let the perpendicular bisector of AB meet EF at N. The perpendicular bisectors of AQ, BQ, AB meet at the circumcenter O of the triangle $\triangle ABQ.$ Since N is the midpoint of EF, ON is the O-mediann of the triangle $\triangle EFO.$ Since $OE \perp AQ,\ ON \perp AB,$ $\angle EON = \angle QAB \equiv \angle A$ and similarly, since $OF \perp BQ,\ ON \perp AB,$ $\angle FON = \angle QBA \equiv \angle B.$

Let Q' be the reflection of Q in the midpoint M of AB (the center of the circle $\mathcal C$). The triangles $\triangle Q'AM \cong \triangle QBM$ are congruent by SAS, hence $\angle Q'AM = \angle QBM \equiv \angle B$ and M is the midpoint of QQ'. It follows that the triangles $\triangle EFO \sim \triangle QQ'A$ are similar. Since their corresponding medians $AM \perp ON$ are perpendicular, their corresponding sides $QQ' \perp EF$ are also perpendicular. Obviously, $QQ' \equiv QM \perp CD,$ hence $EF \parallel CD.$

Let K, L be the midpoints of AQ, BQ and U, V the feet of normals from P, R to AQ, BQ, respectively. $PK \parallel CQ$ is a midline of the triangle $\triangle AQC,$ while $RL \parallel DQ$ is a midline of the triangle $\triangle BQD,$ so that $PK = RL = \frac{CD}{4}.$ Since $PU \perp AQ,\ PK \perp QM,$ $\angle UPK = \angle AQM$ and similarly, since $RV \perp BQ,\ RL \perp QM,$ $\angle VRL = \angle BQM,$ so that

$UK = PK \sin \widehat{UPK} = \frac{CD}{4} \sin \widehat{AQM},\ \ \ VL= RL \sin \widehat{VRL} = \frac{CD}{4} \sin \widehat{BQM}$

Since $EK \perp AQ,\ EA \perp AB,$ $\angle AEK = \angle QAB \equiv \angle A$ and similarly, since $FL \perp BQ,\ FB \perp AB,$ $\angle BFL = \angle QBA \equiv \angle B,$ so that

$SE = \frac{UK}{\sin \widehat{AEK}} = \frac{CD}{4} \frac{\sin \widehat{AQM}}{\sin A},\ \ \ TF = \frac{VL}{\sin \widehat{BFL}} = \frac{CD}{4} \frac{\sin \widehat{BQM}}{\sin B}$

$\frac{SE}{TF} = \frac{\sin \widehat{AQM}}{\sin A} \cdot \frac{\sin B}{\sin \widehat{BQM}}$

From the sine theorem for the triangles $\triangle QAM,\ \triangle QBM,$

$\frac{\sin \widehat{AQM}}{\sin A} = \frac{AM}{QM} = \frac{BM}{QM} = \frac{\sin \widehat{BQM}}{\sin B}$

and consequently, SE = TF. Since also $SE \parallel TF,$ it follows that SEFT is a parallelogram and $ST \parallel EF.$ Combining with $EF \parallel CD$ shown previously, we get $ST \parallel CD.$

Let the feet of perpendicular from $P$ to $AQ$ be $U$ and the feet of perpendicular from $R$ to $AQ$ be $V$. we should prove $OQ\bot ST$ so we should prove :$OT^2-OS^2 = QT^2-QS^2$ .we have :
$OT^2-OS^2=OB^2+BT^2-OA^2-AS^2=BV^2+VT^2-AU^2-US^2$ and
$QT^2-QS^2=QV^2+VT^2-QU^2-US^2$ so we should prove :
$BV^2-AU^2=QV^2-QU^2$ $\Longleftrightarrow$ $BR^2-RV^2-AP^2+PU^2=QR^2-RV^2-QP^2+PU^2$ $\Longleftrightarrow$ $AD^2+BD^2=AC^2+BC^2$ and we are done .

[color=darkblue][b]Generalization.[/b] Let $ABCD$ be a convex quadrilateral inscribed in the circle $w$. Denote the middlepoints $P,Q,R$ of the sides $[AD],[DC],[CB]$ respectively and the points: $S\in AA$, $SP\perp AQ$: $T\in BB$, $TR\perp BQ$. Then $ST\parallel CD$ if and only if $DC\parallel AB$ or the side $[AB]$ is a diameter of the circle $w$.

[b]Remark.[/b] $XX$ is the tangent-line in the point $X\in w$ to the circle $w$.[/color]

[quote="Virgil Nicula"][color=darkblue]Very nice your prove, Behzad ! I remark that the intersection of the lines $SP$ and $TR$ belongs to the bisector of the diameter $[AB]$.[/color][/quote]

Very thanks dear "Virgil Nicula". :)


calculus
integration
logarithms
function
algebra
polynomial
calculus computations

Can you solve this question which I found in another forum?

[url=http://www.imagehosting.com][img]http://www.imagehosting.com/out.php/i1821707_TTintegral7708.GIF[/img][/url]

Well, IMHO this is nowhere near a "nice integral". It's tedious... So I used the machine power:

$ 1 \minus{} \frac {1} {40} (\sqrt {10 \minus{} 2\sqrt 5}\ln (11 \minus{} 4\sqrt 5 \minus{} 2\sqrt {50 \minus{} 22\sqrt 5}) \plus{} \sqrt {10 \plus{} 2\sqrt 5}\ln (11 \plus{} 4\sqrt 5 \minus{} 2\sqrt {50 \plus{} 22\sqrt 5}) \minus{} 2 (1 \plus{} \sqrt 5)\pi)$

Are you sure you didn't mean

$ \int_{0}^{1} \frac {x^9}{x^{10} \plus{} 1}dx$?

If the answer is yes... then note: $ x^9 \equal{} \frac {1}{10}\frac{d(x^{10} \plus{} 1)}{dx}$...

And the final answer is barely simple: $ \frac {\ln 2}{10}$.

The standard technique for integrating rational functions is partial fraction decomposition.  On a polynomial like $ x^{10} \plus{} 1$ this quickly gets very tedious, and from the looks of hsiljak's answer it doesn't look like there are any particularly nice shortcuts.

[quote="hsiljak"]Well, IMHO this is nowhere near a "nice integral". [/quote]
 :rotfl:

What does IMHO mean?  :huh:
And which forum did you find that integral?  Just curious.


Simplify \[ \frac{1}{\log_a abc}\plus{}\frac{1}{\log_b abc}\plus{}\frac{1}{\log_cabc},\] here $ a,b,c>0$ and $ a,b,c,abc\not \equal{}1$.

[hide]\[ \frac {1}{\log_a abc} \plus{} \frac {1}{\log_b abc} \plus{} \frac {1}{\log_cabc}\equal{}\log_{abc} a \plus{}\log_{abc} b \plus{}\log_{abc} c\equal{}\log_{abc}abc\equal{}1\][/hide]


metru, I don't understand how you got to each of your steps so easily  :?

[quote="4everwise"]The sum $\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ equals
$\text{(A)} \ \frac 32 \qquad \text{(B)} \ \frac{\sqrt[3]{65}}{4} \qquad \text{(C)} \ \frac{1+\sqrt[6]{13}}{2} \qquad \text{(D)} \ \sqrt[3]{2} \qquad \text{(E)} \ \text{none of these}$[/quote]

[hide]
We'll let $a + b = 5 + 2\sqrt{13}$ and $a - b = 5 - 2\sqrt{13}$

So we have that
\begin{eqnarray*} (a + b)^{1/3} + (a - b)^{1/3) &=& x\\ \Big((a+b)^{1/3} + (a- b)^{1/3}\Big)^3 &=& x^3\\ 2a + 3(a + b)^{1/3}(a -b)^{(1/3)}(x) &=& x^3\\ 10 - 9x &=& x^3\\ }\end{eqnarray*}
So uh, $E$ ? None of the answers work when you substitute into this.

[/hide]

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=cube+root&t=57189#img_5951

I guess Silverfalcon forgot to link it ... :D

[hide=Solution]Let $a=\sqrt[3]{5+2\sqrt{13}}$ and $b=\sqrt[3]{5-2\sqrt{13}}$. We know that \[a^3+b^3=10\] and that \[a^3b^3=-27\] implying that $ab=-3$. As \[10=a^3+b^3=(a+b)^3-3(a+b)ab=(a+b)^3+9(a+b).\] As $a+b=1$ is a valid solution, $\boxed{1}$ is the answer.[/hide]


Having taken the various placement tests for the courses, and coming out, as I expected, weakest in geometry (though still over the threshold for it), I've decided that I'd like to take the course.  However, there is a problem...

I currently attend a boarding school (Lawrenceville, I'm not one of those Exonian USIMO team members =p), and am behind firewalls, proxies, and god knows what other security measures.  That having been said, during math jams, I have attempted to connect to the classroom in the past through both the normal and firewall link, and have tried Safari and Camino on my Mac and Internet Explorer and Firefox on a friend's PC; none worked (The classroom loaded the frames, but no actual chat/whatnot).  My question is, if anyone has the info, what ports/whatnot do I need to pester the IT folks to unblock or find workarounds for in order to connect to the classroom?  If more information is needed about the setup on this end before you can tell me what I need unblocked, please ask for the said info, and I'll get back with it as soon as I can.

Thanks in advance!

Note:  I wasn't sure weather to post this in the "Support" or "Classes Information" forum.  However, since this pertains to the classroom, and the Support forum seems somewhat inactive (though that's possibly just due to a lack of problems), and the start of Olympiad Geometry is coming up, I figured that I should probably post here.

I assume that the Java applet on the classroom entry page works fine, so that the problem is not that you don't have Java installed.

If using Windows XP, are you logged in as an Administrator?

I'm not exactly sure what's going on...I suspect you are right to conclude that if it doesn't work from multiple browsers and multiple computers, that something is set on your network firewall that's blocking our classroom.

The first time that you connect, you have to accept our applet's security certificate -- perhaps your network is rejecting this?  But (at least on my computer), if you click on the firewall link on the classroom page, it loads without needing the certificate.

What you may want to do is try to attend the Math Jam tonight at 8 PM Eastern (if your schedule permits, if not there's another one tomorrow night at 7:30 PM Eastern), and if you can't get in, give us a call here at (619) 659-1612 and we'll try to walk you through whatever isn't working.

Amazingly enough, I tried entering tonight's math jam and... it worked.  I couldn't stay, since I'm somewhat busy, but I guess something changed over the past two months such that I'm now able to connect, awesome.  Thanks a bunch for your time, and sorry to have taken it up for what turned out to be a non-issue.


irrational number
algebra
number theory
IMO Shortlist
equation

$ \forall n > 0, n \in \mathbb{Z},$ there exists uniquely determined integers $ a_n, b_n, c_n \in \mathbb{Z}$ such

\[ \left(1 \plus{} 4 \cdot \sqrt[3]{2} \minus{} 4 \cdot \sqrt[3]{4} \right)^n \equal{} a_n \plus{} b_n \cdot \sqrt[3]{2} \plus{} c_n \cdot \sqrt[3]{4}.\]

Prove that $ c_n \equal{} 0$ implies $ n \equal{} 0.$

Hint

General Term : $ \frac{n!}{p! q! r!} 1^p \left( 4 \cdot \sqrt[3]{2} \right)^q \left( 4 \cdot \sqrt[3]{4} \right)^r \equal{} \frac{n!}{p! q! r!} 4^{q\plus{}r} \left( \sqrt[3]{2} \right)^q \left( \sqrt[3]{4} \right)^r$
$ p\plus{}q\plus{}r\equal{}n \ (p \ge 0, \ q \ge 0, \ r \ge 0)$


can someone help me with problem 6? it's located in the following PDF.

thanks.

http://www.ams.jhu.edu/~eyink/Applied-Analysis/hw01.pdf

In the special case $A=I$ you can write the inverse of $I+B$ as a series $(I+B)^{-1}=I-B+B^{2}-B^{3}+\dots$ which converges because $\parallel B\parallel <1$. The general case can be reduced to this one.


In an experiment, the equation $ P \equal{} \sqrt{x \plus{} 1} \minus{} \sqrt{2x}$ is used. Find $ x$ when $ p$ is $ \minus{}1$

I feel that the difficulty level can't be that high. In theory. Meh.
When I first saw this, my first gut instinct was to use a conjugate. Of the right side.
Then it was to square everything.
My friends all told me to guess and check....
Is the answer 0? What is the $ \sqrt{0}$, anyways?
How do I do this problem?

The square root of $ 0$ is $ 0$ because $ 0^2 \equal{} 0$.

[hide="Discussion"] We want to solve the equation $ \sqrt {x \plus{} 1} \minus{} \sqrt {2x} \equal{} \minus{} 1$.  The squaring method is not only straightforward but works.  Squaring gives

$ x \plus{} 1 \plus{} 2x \minus{} 2 \sqrt {2x(x \plus{} 1)} \equal{} 1 \implies$
$ 2 \sqrt {2x(x \plus{} 1)} \equal{} \minus{} 3x \implies$
$ 8x(x \plus{} 1) \equal{} 9x^2 \implies$
$ x^2 \minus{} 8x \equal{} 0 \implies$
$ x \equal{} 0, 8$.

Plugging back in, $ x \equal{} 0$ gives $ P \equal{} 1$, but $ \boxed{x \equal{} 8}$ gives the correct answer.  (When we square an equation to solve it, we introduce [i]extraneous roots[/i] that have to be checked.) [/hide]

Hmm. Why do I always miss the most straightforward of solutions? Huh...

Thanks a lot! This really helped. Psh. Extraneous. I knew 0 didn't work. :lol:

Another approach:
[hide]
$ \minus{}1\plus{}\sqrt{2x}\equal{}\sqrt{x\plus{}1}$
$ 1\minus{}2\sqrt{2x}\plus{}2x\equal{}x\plus{}1$
$ x\equal{}2\sqrt{2x}$
$ x^2\equal{}8x$
$ x(x\minus{}8)\equal{}0$
$ x\equal{}8, 0$
$ 0 doesn't work$
$ 8$[/hide]

Hmm. Today, we went over this problem in class, and my teacher said that 0 [i]worked[/i]. This was her reasoning: (Could you tell me if it is true?)

[hide="Reasoning"]

After we had solved for 0 and 8, we checked for extraneous solutions. This was when we first plugged in $ 0$ for $ x$.

$ \minus{} 1 \equal{} \sqrt {x \plus{} 1} \minus{} \sqrt {2x}$
$ \minus{} 1 \equal{} \sqrt {(0) \plus{} 1} \minus{} \sqrt {2(0)}$
$ \minus{} 1 \equal{} \sqrt {1} \minus{} \sqrt {0}$
$ \minus{} 1 \equal{} \pm 1 \minus{} 0$
$ \minus{} 1 \equal{} \pm 1$

If plus one:

-1 is not equal to 1

If minus one:

-1 is equal to 1

My teacher then concluded that 0 was a valid solution since it could be plus or minus one, and that you could [i]choose[/i] which one it was to make it true. Her reasoning doesn't sound sound to me. Could someone explain?

[/hide]

P.S. Could someone should me how to make plus or minus signs in LaTeX, and how to make a "not equal" sign, a slash through an equal sign? Thanks! :)

[color=darkblue]Moderator says: Fixed it. Use the LaTeX code \pm to denote plus or minus.[/color]

[quote="The_Scintillator"]Hmm. Today, we went over this problem in class, and my teacher said that 0 [i]worked[/i]. This was her reasoning: (Could you tell me if it is true?)

[hide="Reasoning"]

After we had solved for 0 and 8, we checked for extraneous solutions. This was when we first plugged in $ 0$ for $ x$.

$ \minus{} 1 \equal{} \sqrt {x \plus{} 1} \minus{} \sqrt {2x}$
$ \minus{} 1 \equal{} \sqrt {(0) \plus{} 1} \minus{} \sqrt {2(0)}$
$ \minus{} 1 \equal{} \sqrt {1} \minus{} \sqrt {0}$
$ \minus{} 1 \equal{} \plus{} or \minus{} 1 \minus{} 0$
$ \minus{} 1 \equal{} \plus{} or \minus{} 1$

If plus one:

-1 is not equal to 1

If minus one:

-1 is equal to 1

My teacher then concluded that 0 was a valid solution since it could be plus or minus one, and that you could [i]choose[/i] which one it was to make it true. Her reasoning doesn't sound sound to me. Could someone explain?

[/hide][/quote]
No, your teacher's reasoning is off. The square root is only used for the principal square root, which must be positive.
[quote="The_Scintillator"]P.S. Could someone should me how to make plus or minus signs in LaTeX, and how to make a "not equal" sign, a slash through an equal sign? Thanks! :)[/quote]
\pm makes $ \pm$, \mp makes $ \mp$.
\neq or \not= makes $ \neq$

I should've know that...thanks. :D

But, hypothetically, say that you [i]did[/i] have a $ \pm$ after plugging in to check for roots. Would that root then be considered extraneous or a solution? Or both?

Thanks for the code! I would've never guessed that...
...and what is the $ \mp$ used for?

[quote="The_Scintillator"]But, hypothetically, say that you [i]did[/i] have a $ \pm$ after plugging in to check for roots. Would that root then be considered extraneous or a solution? Or both?[/quote]
If you don't square anything in your solution and get an expression with $ \pm$ in it, that means both (or as many as there are) roots work. Squaring is what adds the extraneous root.

[quote="The_Scintillator"]Thanks for the code! I would've never guessed that...[/quote]
You don't have to: [[LaTeX]]. :)

[quote="The_Scintillator"]...and what is the $ \mp$ used for?[/quote]
In some equations, both a $ \pm$ and a $ \mp$ is present. This means the $ \plus{}$ in the $ \pm$ should be used with the $ \minus{}$ in the $ \mp$ and the $ \minus{}$ in the $ \pm$ should be used with the $ \plus{}$ in the $ \mp$.

An example of this is $ \cos\left(A\pm B\right) \equal{} \cos{A}\cos{B}\mp\sin{A}\sin{B}$:
\[ \cos\left(A \plus{} B\right) \equal{} \cos{A}\cos{B} \minus{} \sin{A}\sin{B} \\
\phantom{\qquad\qquad\qquad\qquad}\text{and} \\
\cos\left(A \minus{} B\right) \equal{} \cos{A}\cos{B} \plus{} \sin{A}\sin{B}
\]


Find all nonnegative integers $m,n$ so that \[\sum_{k=1}^{2^{m}}\lfloor \frac{kn}{2^{m}}\rfloor\in \{28,29,30\}\]

[quote=Takeya.O]I'v solved by casework.Are there any another solution?

It is easy to show that $n=0,1$ doesn't satisfy the condition.And $n=2$ case, obviously $m\ge 1$.Then $\sum_{k=1}^{2^m}[\frac{k}{2^{m-1}}]=2^{m-1}+2\in \{ 28,29,30\}$.
Since $m$ is nonnegative integer, this is absurd.From now, we consider $n\ge 3$ case.[b]Let $P(n)$ denotes that sum is bigger than $30$, and $Q(n)$ denotes that sum is smaller than $28$[/b].Take care that if $P(k)$, $P(l)(\forall l\ge k)$, and if $Q(k)$, $Q(l)(\forall l\le k)$.

[b]Case1)$m=0$[/b]
$n=28,29,30$

[b]Case2)$m=1$[/b]
It is easy to check that $n=19,20$ satisfies the condition, and $Q(18)$, $P(21)$.Therefore $n=19,20$.

[b]Case3)$m=2$[/b]
It is easy to check that $n=12$ satisfies the condition, and $Q(11)$, $P(13)$.Therefore $n=12$.

[b]Case4)$m=3$[/b]
It is easy to check that $n=7$ satisfies the condition, and $Q(6)$, $P(8)$.Therefore $n=7$.

[b]Case5)$m=4$[/b]
It is easy to check that $n=4$ satisfies the condition, and $Q(3)$, $P(5)$.Therefore $n=4$.

[b]Case6)$m\ge 5$[/b]
It is easy to check that $P(3)$.Therefore there exists no n which satisfies the condition.

From above case,

$\boxed{(m,n)=(0,28),(0,29),(0,30),(1,19),(1,20),(2,12),(3,7),(4,4)}$.[/quote]

(4;4) is not a solution


complex numbers
real analysis
real analysis unsolved

Define a complex number $z$ to be algebraic if there exist integers $a_{0},\ldots,a_{n}$, not all zero, such that
\[a_{0}z^{n}+\cdots+a_{n-1}z+a_{n}= 0 \].
Prove that the set of all algebraic numbers is countable.

(Hint given: For each positive integer $N$ there are only finitely many equations with $n+|a_{0}|+\cdots+|a_{n}| = N$.)

For all $N$ denote set of all equations $a_{0}x^{n}+...+a_{n-1}x+a_{n}=0(a_{i}\in\mathbb{Z}, n+|a_{0}|+...+|a_{n}|=N)$ is $E(N)$  and set of all complex numbers is a root of at least one element of $E(N)$ is $R(N)$, then $R(N)$ is finite and $S\subseteq \cup_{N=1}^{\infty}R(N)$. Done! :)

Note: $S$ is set of all algebraic numbers.


On each line in a plane we chose a segment. Prove that there exists a triangle such that each point on its boundary lies on one of these segments.

Here it is:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=set+of+point&t=4740[/url]

oh i see. as myth have said , [b]pbornsztein[/b] you are really smart, wonderfull solution.
but as pbornsztein have said its useless for the plane , any idea :?:


find the major product when u treat a mixture of 1 mol $ 1,3 Buta diene$ and 1 mol
a.$ 1-butene$
b.$ 1,4 penta diene$
c.$ 2-methyl 1,3 butadiene$
d.$ 1,3 penta diene$

with
1.$ HCl$
2.$ BrCCl_{3}$ and $ \Delta$

[hide="Answer"]Part 1 (HCl) refers to electrophilic additions, and part 2 (BrCCl3, heat) to radical additions.The key to solve this problem is to realize that the more stable the intermediate produced in the first step (a carbocation in 1 and a radical in 2), more favorable will the reaction be, and then to relate stability with structure: conjugated dienes (1,3-butadiene, 2-methyl-1,3-butadiene, 1,3-pentadiene) react more readily than simple alkenes (1-butene) or non-conjugated dienes (1,4-pentadiene), because in the first case the carbocation/radical formed is stabilised by ressonance. Also, any substituent that increases electron density will favour reaction (case of 2-methyl-1,3-butadiene).[/hide]

sorry for the elimantion
but how do u decide between 1,3buta diene and 1 butene because one forms allylic and the secondary which are of nearly same stability


Which of the following divisions has the smallest remainder?

A) 4002 ÷ 4
B) 503 ÷ 5
C) 604 ÷ 6
D) 75 ÷ 7

[hide]A is 2
B is 3
C is 4
D is 5
A[/hide]

lol  I was helping my 4th grade sister study for some test by CTY called SCAT and a few days back she had this problem in some Math league contest. I think it was 4th grade...?  :lol:

CTY is Center for Talented Youth.  It's a summer camp and they have courses like: science and engineering, biology, flight science, the edible world, crime scene investigation,etc..
I've been to CTY twice.  I did Logic and Mathematical reasoning and Science and Engineering.
SCAT is the test you have to take in order to qualify for CTY, i think.  (I'm getting better at typing :) )


linear algebra
matrix
vector
algebra
polynomial
linear algebra unsolved

PDF attached, written by me, so sorry for the bad english.
HUGE thanks in advance, just need the direction, but feel free to solve it.
THANKS.

perhaps someone will come by to make your questions readable :)

What part you don't understand?
anyway, do you have time to help me?
Thanks.

I suggest you that you should explain some words like as: Diagonalizable matrix, regular matrix, have the same degree(number of non 0 rows) (in question B), question3. I think all questions are easy but maybe I don't understand your problems.

Regular- nonsingular
Diagonalizable matrix- http://en.wikipedia.org/wiki/Diagonalizable_matrix
degree= number of nonzero and non proportional rows
question 3:
IMT  is the image(vectors from B when transformation is from A to B) of the transformation T.
KertelT is the group of all vectors that the transformation directs them to the 0 vector.
When transformation is from A to B, the kertel is a group of special vectors from A.
dim= dimension= number of vectors in a basis of a vector space= http://en.wikipedia.org/wiki/Hamel_dimension
whereever i wrote base change it to basis.

For question A: If $ a\equal{} c$ then by a direct calculating $ (A\minus{}aI)^{2}\equal{}0$ and $ A\minus{}aI\neq0$ so it's not diagonalizabl. If $ a\neq c$ then also again we get $ (A\minus{}aI)(A\minus{}cI)\equal{}0$ so $ A$ is diagonalizable.

For $ a\equal{}2,b\equal{}0$ and $ c\equal{}1$ try to eingein vectors with eigein values $ a$ and $ c$. We find a basic and this implies the matrix $ P$

Question B_2: Yes! If $ Ax \equal{} tx$ then $ A^{m}x \equal{} t^{m}x$ so $ t^{m}x \equal{} 0$. This implies $ t \equal{} 0$ for $ x$ is non-zero vector.

Question 3: $ dim ker T \equal{} n\minus{}1 > 0$ so there is $ x\neq0$ such that $ Tx \equal{} 0$ so $ t \equal{} 0$ is an aigenvalue. Also cause $ dim ker T < n$ so its characteristic polynomial $ det(P\minus{}tI) \equal{} t^{n\minus{}1}(at\plus{}b)$ where $ a,b\neq0$ it means $ T$ has an eigenvalue $ k\neq0$.
For $ n \equal{} 3$ take a base with eigenvectors of $ T$, two non proportional vectors for $ t \equal{} 0$ and one for $ k\neq0$.

The last question is trivial: take any $ x\in ker (T\minus{}tI)$ then $ Tx\equal{}tx$. We have $ T(Sx)\equal{}S(Tx)\equal{}S(tx)\equal{}tS(x)$ so $ Sx\in ker(T\minus{}tI)\equal{}W$.

[quote="pluricomplex"]For question A: If $ a \equal{} c$ then by a direct calculating $ (A\minus{}aI)^{2}\equal{} 0$ and $ A\minus{}aI\neq0$ so it's not diagonalizabl. If $ a\neq c$ then also again we get $ (A\minus{}aI)(A\minus{}cI) \equal{} 0$ so $ A$ is diagonalizable.

For $ a \equal{} 2,b \equal{} 0$ and $ c \equal{} 1$ try to eingein vectors with eigein values $ a$ and $ c$. We find a basic and this implies the matrix $ P$[/quote]

Can you please explain this to me?
Thanks pal, you are a good man.

[quote]For $ a \equal{} 2,b \equal{} 0$ and $ c \equal{} 1$ try to eingein vectors with eigein values $ a$ and $ c$. We find a basic and this implies the matrix $ P$[/quote]
But this matrix has only one eigenvalue which has one eigenvector, and P must be a square matrix, i am confused and the time for giving those homework is coming soon, please help.


A rectangular solid has each of its three dimensions doubled. How many times bigger is the surface area of the new solid?

A) 2 B) 4 C) 8 D) 16 E) 32

If each side multiplies by 2, the surface area simply multiplies by $ 2^2\equal{}4$. So the answer is $ \boxed{\text{B}}$.

By the way, please refrain from posting one problem per thread; that is spam, and it only worthlessly increases your post count. If they're all algebra/geometry problems (or problems of the same kind), why not just post them all in one thread?

Who knows, he is in his own world right now. Countless amount have people told him already. I think he might be a bot.


$E=$εμβαδόν τριγώνου με πλευρές $a,b,c$.

Να δειχθεί ότι:

$\frac{9E^n}{(a^n+b^n+c^n)^2}\leq \displaystyle\left(\frac{\sqrt{3}}{4}\right)^n$

για κάθε $n\geq 1$ .

_________________

$\frac{9E^n}{(a^n+b^n+c^n)^2}\leq \displaystyle\left(\frac{\sqrt{3}}{4}\right)^n$

ή ισοδύναμα $9\cdot 4^nE^n\leq (\sqrt 3)^n(a^n+b^n+c^n)^2$  ή ισοδύναμα
$9\cdot 4^n\frac{(abc)^n}{(4R)^n}\leq (\sqrt 3)^n(a^n+b^n+c^n)^2$ ή ισοδύναμα
$9(abc)^n\leq (R\sqrt 3)^n(a^n+b^n+c^n)^2$
ή ισοδύναμα
$9\cdot 3^n(abc)^n\leq (3R\sqrt 3)^n(a^n+b^n+c^n)^2$

όμως $3R\sqrt 3\geq a+b+c$  . Άρα αρκεί να δείξουμε ότι
$9\cdot 3^n(abc)^n\leq (a+b+c)^n(a^n+b^n+c^n)^2$  (1)
Όμως από την AM-GM έχουμε
$a^n+b^n+c^n\geq 3\sqrt[3]{(abc)^n}$  άρα υψώνοντας στο τετραγωνο
$(a^n+b^n+c^n)^2\geq 9\sqrt[3]{(abc)^{2n}}$  (2)
Από την AM-GM πάλι
$(a+b+c)^n\geq (3\sqrt[3]{abc})^n$  (3)
Πολλαπλασιάζουμε την (2) και τη (3)  και παίρνουμε την ζητούμε νη (1)

Ευκολα αποδεικνυουμε την ανισοτητα για n=1 και για n=2 χρησιμοποιωντας την ανισοτητα του Weitzenbock ( εχει πεσει ΔΜΟ 1961 2ο θεμα).Υποθετουμε οτι ισχυει για n,n>2, ara arkei na δειξουμε οτι ισχυει για n+2. Πραγματι πολ/με κατα μελη την ανισοτητα για n=2 kai n (αφου ισχυουν και οι δυο και ειναι θετικοι ολοι). τοτε αρκει να αποδειξουμε μια ανισοτητα η οποια βγαινει, αλλα without latex.... :blush:

Στο ποστ μου παραπάνω πάτα quote και θα σου βγει πώς έγραψα το Latex και σιγά σιγά θα το μάθεις απέξω ;)
Ας πούμε θες να γράψεις το κλάσμα τρία τέταρτα .
Πατάς αρχικά το σήμα του δολλαρίου , μετά γράφεις \frac{3}{4}
και μετά πάλι δολλάριο ;)
Κάνε μια προσπάθεια

Πολύ ωραία!!! Να και η λύση μου:

Για $n\geq 1$ και $x,y,z\geq 0$ ισχύει :

$\displaystyle\left[\frac{16xyz(x+y+z)}{3}\right]^n\leq \left[\frac{(x+y)^n+(y+z)^n+(z+x)^n}{3}\right]^4$

(υπάρχει και γενίκευση  ;)  ...).

Θέτοντας $x=\frac{a+b-c}{2}$ , $y=\frac{b+c-a}{2}$ και $z=\frac{c+a-b}{2}$ προκύπτει το ζητόυμενο .

Stelios...


Κάλπη περιέχει n μπάλες αριθμημένες από το 1 έως το n και r από αυτές είναι άσπρες.Επιλέγουμε στην τύχη k χωρίς επανατοποθέτηση με k<r.Ποια η πιθανότητα και οι k να είναι άσπρες;
Στο πρόβλημα αυτό βρήκα δύο τρόπους που βγάζουν διαφορετικά αποτελέσματα αλλά δεν μπορώ να βρω ποιος είναι ο λάθος
1ος τρόπος: έστω A το ενδεχόμενο και οι k μπάλες να είναι άσπρες και S ο δειγματικός χώρος.Προφανώς $ |S| \equal{} n(n \minus{} 1)....(n \minus{} k \plus{} 1)$.Για το Α επιλέγουμε την πρώτη έχουμε r επιλογές για τη δεύτερη r-1 κ.ο.κ άρα $ |A| \equal{} r(r \minus{} 1)...(r \minus{} k \plus{} 1)$ και τελικά $ P(A) \equal{} \frac {r(r \minus{} 1)...(r \minus{} k \plus{} 1)}{n(n \minus{} 1)....(n \minus{} k \plus{} 1)}$
2ος τρόπος.Θεωρούμε το $ A^c$ δηλαδή μία τουλάχιστον μπάλα δεν είναι άσπρη.Για αυτήν έχουμε n-r επιλογές ενώ για όλες τις άλλες έχουμε (n-1)(n-2)....(n-k+1) επιλογές δηλαδή $ |A^c| \equal{} (n \minus{} r)(n \minus{} 1)....(n \minus{} k \plus{} 1)$ άρα $ P(A) \equal{} 1 \minus{} P(A^c) \equal{} 1 \minus{} (\frac {n \minus{} r}{n}) \equal{} \frac {r}{n}$  :o
Μπορεί κάποιος να με βοηθήσει να βρω το λάθος please :?: Ευχαριστώ για την κάθε βοήθεια

Δεν πιανω την 2ερη περιπτωση αλλα και εγω το που το εβαλα κατω βγαζω αυτο που βγαζεις στην Α περιπτωση και δεν βλεπω λαθος. Αρα μαλλον η Β ειναι λαθος. :wink:

Ο δεύτερος τρόπος είναι λάθος. Αντί να υπολογίσεις την πιθανότητα "τουλάχιστον μια μπάλα δεν είναι άσπρη", υπολόγισες την πιθανότητα "η πρώτη μπάλα δεν είναι άσπρη", που προφανώς είναι $ (n\minus{}r)/r$.

Σωστός...ευχαριστώ πολύ είχα κολλήσει και τελικά ήταν απλούστατο. :D


inequalities
geometry
angle bisector
triangle inequality

Triangle $ ABC$ has integer-length sides, and $ AC\equal{}2007$. The internal bisector of $ \angle BAC$ meets $ BC$ at $ D$. Given that $ AB\equal{}CD$, determine $ AB$ and $ BC$.

[hide]By Angle Bisector Theorem, $ \frac {AB}{BD} \equal{} \frac {2007}{CD} \Longrightarrow AB^2 \equal{} 3^2 223 BD$. Since sides have integer lengths, we let $ BD \equal{} 223k^2, AB \equal{} 669k$, where $ k$ is a positive integer. By Triangle Inequality, $ 2007 < 2 \times 223k^2 \plus{} 669k$, and $ 223k^2 \plus{} 669k < 223k^2 \plus{} 2007$, and $ 223k^2 < 223k^2 \plus{} 669k \plus{} 2007$, which imply respectively $ k > 3/2$, $ k < 3$ (the third is always satisfied). Thus $ k \equal{} 2 \Longrightarrow \boxed{AB \equal{} 1338, BC \equal{} 2230}$.[/hide]


The Hessian is not negative, so it is convex and maximum is obtained at boundary. So we only need to test the boundary cases, and equality holds when two of them are 1 or three of them are 1.


logarithms
limit
calculus
calculus computations

Same as [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=758718#758718]here[/url], you should prove that $f'(x)\leq 1$ when $x\geq 1$ where $f(x)=1+\ln\frac{x(x^{2}+3)}{1+3x^{2}}$. That is not hard, just some calculations.


hallo,
find all positive intergers $x, y$ such that
$x^2+(x+1)^2=y^4+(y+1)^4$.


Unfortunately,I am stuck with the following inequality:

 If $ x_1,x_2,...,x_n$ are positive real numbers such that $ x_1x_2...x_n\equal{}1$ then prove the following inequality:

    $ \frac{1}{n\minus{}1\plus{}x_1}\plus{}\frac{1}{n\minus{}1\plus{}x_2}\plus{}...\plus{}\frac{1}{n\minus{}1\plus{}x_n} \leq 1$

Since I am still working on this problem,I woudn't like any full solution at this point but some hints(if that's possible) that could
help me find the solution.Thanks for your help :)

[hide="hint1"]let $ x_i\equal{}a_i^n$[/hide]
[hide="hint2"]$ x_i\equal{}\frac{a_i^{n\minus{}1}}{a_1a_2\ldots a_{i\minus{}1}a_{i\plus{}1}\ldots a_n}$[/hide]

It's from Romania 1999 anyway.


modular arithmetic
number theory unsolved
number theory

Given is a Pell's equation $ (24n\plus{}7)^{2}\minus{}48y^{2}\equal{}1$ , where $ n$ and $ y$ are positive integers.

Can we conclude that $ n$ is divisible by 7. Give a proof or find counter example.

$ 48y^2 \equal{} (24n \plus{} 7)^2 \minus{} 1^2$
$ y^2 \equal{} (4n \plus{} 1)(3n \plus{} 1)$
$ gcd((4n \plus{} 1),(3n \plus{} 1)) \equal{} 1$
So $ 4n \plus{} 1 \equal{} a^2$
and $ \frac {a^2 \minus{} 1}{4} \equal{} b^2$
$ a^2 \minus{} 1\not \equal{} (2b)^2\ \forall a > 1$
Thus $ a \equal{} 1$ and $ n \equal{} 0$ and $ y \equal{} 1$ is the only solution.

Edit:There is no solution. $ n$ must be positive.

There are $ \infty$ many solutions. I don't see how you got $ \frac {a^2 \minus{} 1}{4} \equal{} b^2$.

To solve the problem, just consider the minimal solution $ (x,y)\equal{}(7,1)$ of $ x^2\minus{}48y^2\equal{}1$, then use the standard fact's about pell equations to find all solutions. Comparing, you get the result.

Can anyone give a complete proof for this question.

[quote="Akashnil"][quote="ZetaX"]I don't see how you got $ \frac {a^2 \minus{} 1}{4} \equal{} b^2$.[/quote]

I'm sorry.It should be $ \frac {3a^2 \plus{} 1}{4} \equal{} b^2$[/quote]

Which can the be written $ 3a^2\equal{}(2b\plus{}1)(2b\minus{}1)$ where $ 2b\plus{}1,2b\minus{}1$ relatively prime. So there exist integers $ p,q$ such that $ 2b\plus{}1\equal{}3p^2,2b\minus{}1\equal{}q^2\Rightarrow 3p^2\minus{}q^2\equal{}2$. (The other case fails modulo 3.)

From this we have
\[ n\equal{}\frac{b^2\minus{}1}{3}\equal{}\frac{(\frac{q^2\plus{}1}{2})^2\minus{}1}{3}\equal{}\frac{(q^2\plus{}3)(q^2\minus{}1)}{12}\]
\[ \equiv \frac{(q\minus{}1)(q\plus{}1)(q\minus{}2)(q\plus{}2)}{12}\pmod{7}\]
So the question is reduced to showing $ q\not\equiv 0,\pm 3\pmod{7}$. For if $ q\equiv 0\pmod{7}$, then $ 3p^2\equiv 2\pmod{7}$ and if $ q\equiv \pm 3\pmod{7}$, then $ 3p^2\equiv 4\pmod{7}$. These have no solutions. So $ n$ must necessarily be divisible by 7.


What is the sum of the mean and the median of the set $ \{15,17,19,21,23\}$?

Quicker way to find the mean - note that 15 is 4 away from 19, and 23 is also 4 away. Thus the mean must be 19.

However, this will only be true if $ 17$ and $ 21$ are the same "distance" away from $ 19$, which happens to be true.

Well yes, but before calculating the mean, just quickly look and see if this is the sort of case - it should only take a second or so.


trigonometry quadratics function algebra

[quote="Phelpedo"]Use the quadratic equation on $x+\frac{1}{x}= 2\cos(a)$ to find that $x= \cos a \pm i \sin a = e^{\pm i a}$. Of course, we know that the two roots correspond to $x$ and $\frac{1}{x}$, so $x^{n}+\frac{1}{x^{n}}= e^{ina}+e^{-ina}$, which is $2 \cos na$.[/quote]
That's pretty cool, noticing 1/x+x has roots x and 1/y and y...

[quote="PenguinIntegral"]Show that if $x+1/x=2\cos a$, then for any integer n,

$x^{n}+1/x^{n}=2 \cos na$.[/quote]
The function $f(x)=x+1/x$ is clearly 1 to 1 on $-1,1$, the range of the cos function. Thus there is exactly one $x$ which satisfies the first equation.
$e^{ia}=\cos a+i\sin a\implies 2\cos a=e^{ia}+e^{-ia}$ after substituting negatives and adding equations so they cancel. Notice $e^{ia}+e^{-ia}=f(e^{ia})\implies e^{ia}=x$ is the single value for $x$ as proven above. Now compute $x^{n}+1/x^{n}$ and it is clearly $2 \cos na$ after returning to the $\cos$ equation above and substituting.


Find the $n$ th term of the sequence $\{a_{n}\}$ such that $a_{1}=1,\ a_{n+1}=na_{n}+n-1\ (n=1,\ 2,\ 3,\ \cdots).$


calculus
derivative
inequalities proposed
inequalities

$a, b, c \in \mathbb{R}^+ \, | \, a \le b + c;\, b \le c + a;\, c \le a + b$. Show  that

$\frac{b+c-a}{\sqrt{9a^2 + 7\left(b-c\right)^2}} + \frac{c+a-b}{\sqrt{9b^2 + 7\left(c-a\right)^2}} + \frac{a+b-c}{\sqrt{9c^2 + 7\left(a-b\right)^2}} \ge 1.$

and determine when equality holds.

[quote="Megus"]It is equivalent to:

$\sum _{cyc} \frac{x}{\sqrt{4y^2+yz+4z^2}} \geq 1$

where $x=b+c-a$ and so on...

I have no strenght to finish it though  :P[/quote]

Does this work?

Ok, set $A=4(y^2+z^2)+yz$, etc, and WLOG we have $x+y+z=1$.  Note that $f(t) = t^{-\frac{1}{2}}$ has second derivative $f''(t)=\frac{3}{4}t^{-\frac{5}{2}} > 0$, so we may apply Jensen's:

\[\sum_{\text{cyc}} \frac{x}{\sqrt{A}} \geq \frac{1}{\sqrt{xA+yB+zC}},\]

so it suffices to prove that

$xA+yB+zC \leq 1 \iff 4\sum_{\text{sym}} x^2y+3xyz \leq (x+y+z)^3$

$\iff \sum_{\text{sym}} x^2y \leq x^3+y^3+z^3+3xyz,$

which of course, is just Schur's!


inequalities
LaTeX
inequalities unsolved

[quote="altusi"]the fourth:
i hope i haven't  made any mistake because it has been  done in few minutes:

by cauchy we have:

LHS>= (a+b+c)² /(a+b+c+ab+bc+ac)
the new inequality is equivalent to:

2(a²+b²+c²)+(ab+bc+ac)>=3(a+b+c)
or: 6+ (ab+bc+ac)>=3rac(3+2(ab+bc+ac))

wich is obviously true by putting x=ab+bc+ac

 :D[/quote]
What is 3rac(3+2(ab+bc+ac)?

[quote="altusi"]the fourth:
6+ (ab+bc+ac)>=3rac(3+2(ab+bc+ac))

 :D[/quote]
The fourth is very easy. But in your solution I don't know what is "rac(3+2(ab+bc+ac))"

Is it $\sqrt{3+2(ab+bc+ac)}$?
If it is $\sqrt{3+2(ab+bc+ac)}$ your solution is quite nice.
But no one solve another ineq ??? :(

sorry ,i didn't use Latex, but it was the square as you have gessed :D

The sign of the first and the second are reserved

 For the third one :
  we have : $3 \geq a+b+c >\sqrt{3}$
  Note that : $a-\frac{a}{1+b^2}=\frac{ab^2}{1+b^2}$
  Apply AM-GM : $1+b^2 \geq 2b$
 It's enough to prove : $ab+ac+bc \geq 2(a+b+c)-3$
                                $\iff (a+b+c)^2-4(a+b+c)+3 \leq 0$
                                $\iff (a+b+c-1)(a+b+c-3) \leq 0$ , which is obviously true

nice as well! although there is a small typo in the 6th line. ;)


I'm amazed, how do you do it? On most state tests I can't score higher than mid-20's while some other person here (no names) was mad because he made a 35. Is there some baisc thing I'm missing here!?

Well, not really, here's a checklist.

1. Have you studied AoPS book #1 at least once, and reviewed it at least twice?

2. Have you done at least 6 mathcounts competitions of that level?

3. Do you study 5 hrs a day?

Have a check on all of these, and you should have a 40. (like me :) yummy 45's)

_________________It's known as practice.
______________Practice with problems, ducks, ...
__________Practice at home, at school, car, on a duck....
________Practice by your self, with mom, dad, friends, a duck....
____Pracitice when it's sunny, when it rains, when ducks come, when...
__Practice on Monday, Tuesday, Wednesday, Duckday, Friday, ....everyday!

They're right.  The two things that help you on this level are practice and new learning.  I'd recommned continuing to work on new stuff while reviewing the old; the AoPS books are really good for that because most of the problems require you to use stuff you already know as well as the new ideas.

I have just recently bought Volume 1 of the AoPS books, and it's pretty nice, but I'm still going over it. ABout the number of state tests I've taken, I've only taken a few. I wish I had access to some more states, but I can't find any that go back farther than 2001.

I do have a couple years of handbooks though :).

P.S. Does anyone have any copies of older State tests I can have? PM me if so.

1. Have you studied AoPS book #1 at least once, and reviewed it at least twice?

2. Have you done at least 6 mathcounts competitions of that level?

3. Do you study 5 hrs a day?

-----------------------------------------

#1 isn't necessary for mathcounts STATE. I finished about half of it and collapsed on the geometry portion. If I were to go back to it I could finish it NOW, but at the time when I was working through it (very badly), I could still get low 40's at states. In fact I think I squeaked a perfect on 95-96, a feat that I have yet to duplicated even though I'm LOADS better.

I can get 40's on state rounds and the only one I follow is #2.

#3 I sometimes do, but there's NO WAY I could possibly do that on weekdays... get home at 4:30, eat, homework -- how is that even possible?

Who said anything about not doing it during school?

Throwing away other subjects just for math isn't smart, and how are you going to get away with that stuff?

"Lucky" you, your teachers must not be too bright. Either that or you have like a two hour study hall.

It WOULD be the end of the world for me if I didn't WIN nationals -- but I'm just too much of a perfectionist in things that don't really matter...

Also, I've spent 2 minutes, 10 minutes, 30 minutes,  and 1.5 hours in preparation for timed history essays (history = my best SCHOOL subject) -- and gotten the exact same grade on them all. We've got this huge timed essay with huge expectations coming up and I'm only going to study for a bit to see if I get the same grade AGAIN (it's a pretty satisfactory grade even if it isn't perfect so I'd be content with it).

On sprint round, most of the problems have some kind of shortcut, so after doing a lot of problems, your score's probably going up. It's important to understand everything though. Like I did a ton of past mathcounts problems and I raised my score 12 points from last year.


Quadrilateral $ ABCD$ is a square with area 16 square inches. The figure represents the pieces of a Chinese tangram in which all the triangles are isosceles and piece ``e'' is a square. What is the area of the gray piece, in square inches?

[asy]fill((2,0)--(4,2)--(4,0)--cycle,gray(.7));
draw((0,0)--(0,4)--(4,4)--(4,0)--cycle,linewidth(1));
draw((0,0)--(4,4),linewidth(1));
draw((1,1)--(2,0)--(4,2),linewidth(1));
draw((0,4)--(3,1)--(3,3),linewidth(1));
label("$A$",(0,4),W);
label("$B$",(4,4),E);
label("$C$",(4,0),E);
label("$D$",(0,0),W);
label("e",(2,1));[/asy]

It's countdown.  Just eyeball it, and we get $ \boxed{\frac{1}{8}}$.

Actually the answer is 2, cause the whole area is 16.


The quarters, dimes, nickels, and pennies in Jeremy's pocket total $ \$2.35$. He has as many quarters as pennies, as many dimes as nickels, and fewer quarters than dimes. How many dimes does he have?

Let $ q$ be the number of quarters, $ p$ be the number of pennies, $ n$ be the number of nickels, and $ d$ be the number of dimes.

(1) $ 25q \plus{} p \plus{} 5n \plus{} 10d \equal{} 235$

(2) $ q \equal{} p$

(3) $ d \equal{} n$

(4) $ q < d$

Use (2) and (3) to substitute as shown:

$ 25q \plus{} q \plus{} 5d \plus{} 10d \equal{} 235$

$ 26q \plus{} 15d \equal{} 235$

I'm not sure where to continue from here..

He must have a multiple of 5 pennies. Lets see what happens with

0: 0p 0q Also, d=15.66666 not happening.

5: 5p 5q=1.30 Thus he is has 1.05 in d and n, or 1.05/.15 or 7 of each.

Answer: 7.


geometry
geometric transformation
rotation
analytic geometry
vector
algebra
linear equation

The point A (2, 1, -1) is rotated 90 degrees about the line x −1 = (y+2)/2= -z/2 to
become point B. Which of the following could be the coordinates of point B?

a) (5/3, 1/3, -2/3)
b) (10/3,-1/3, -5/3)
c) (-2,2,-2)
d) (5/3, 4/3, -7/3)

[hide="Sketch"]We have $ x - 1 = y/2 + 1 = - z/2 = t \iff \left \{\begin{array}{c} x = t + 1 \\
y = 2t - 2 \\
z = - 2t \end$.

Translate this so that it goes through the origin by subtracting $ \mathbf{u} = (1, -2, 0)$, giving $ \mathbf{x} (t) = (1, 2, - 2)t$ we will undo this later.

Now, let $ \mathbf{a} - \mathbf{x} (t) = \mathbf{r}(t)$. We want the $ t$ value for which $ \mathbf{r} (t)$ is perpendicular to $ \mathbf{x} (t), \; t \neq 0$. We solve the linear equation $ \mathbf{r} (t) \cdot \mathbf{x} (t) = 0$, and find $ t = t_o$ for some number $ t_o$.

Let $ \mathbf{x} = \mathbf{x} (t_o)$, and $ \mathbf{r} = \mathbf{r} (t_o)$.

We seek a vector $ \mathbf{v}$ such that $ \| \mathbf{v} \| = \| \mathbf{r} \|$, and such that $ \mathbf{v}$ is perpendicular to $ \mathbf{r}$ and $ \mathbf{x}$. Hence, $ \mathbf{v} = k (\mathbf{r} \times \mathbf{x})$.

Use $ \| \mathbf{v} \| = \| \mathbf{r} \|$ to find the positive value of $ k$, and thus one value of $ \mathbf{v}$.

Finally, the rotation of the original point is equal to $ \mathbf{b} = \mathbf{u} + \mathbf{x} \pm \mathbf{v}$.
[/hide]


Determine does there exist a bounded function f:R-->R  such as f(1)>0 and for each x,y from real

 [f(x+y)]^2>=[f(x)]^2 + 2f(xy) + [f(y)]^2

The answer is No.
Let $ M\equal{}\sup_{x\ne0}f(x)^2<\infty$ and $ c\equal{}f(1)>0$.
Now we have, for $ x\ne0$.
$ f(x)^2\equal{}f(x\plus{}\frac1x)^2\minus{}f(\frac1x)^2\minus{}2c\le M\minus{}2c<M$,
or
$ M\equal{}\sup_{x\ne0}f(x)^2< M$. Contradiction!


From four distinct points A,B,C,D each triplet of points is non-collinear. The lines AB and CD intersect in point E and BC and AD intersect in point F. We construct the circles with diameters AC, BD and EF. Prove that or this 3 circles have a common point or each pair do not have common pints.

That's because these 3 circles are coaxal (with 2 common points or one common point or no common points). See [url]http://www.mathlinks.ro/viewtopic.php?t=207108[/url] for details.


Duke
college
MATHCOUNTS
geometry
AoPS Books

Im going to be taking a course called Mathematical Problem Solving for the Duke TIP thingy...has anyone taken it?

I hope its not all simple algebra...HELP!!!

Is it good for Mathcounts?

I believe my brother and sister take it (but not for the TIP).... but it doesnt seem mathcounts level.... somewhat simple stuff. I took the TIP this year and got a 30 on math (weird, what I usually get on practice nats...).

I think it's okay. When I was at Davidson last year, I took Geometry believing it would help me more (Geometry was kinda my weak point for MC). My roommate took Math Problem Solving, though.

Math Problem Solving:
Focuses on strategies and proofs
Not necessarily material-oriented
More AMC, etc. than Mathcounts-style

Alg I/Geom/Alg II:
Focuses on material
Condensed course
First two helps for MC, but only if you can remember them (imo material is more easily forgettable than a year-long course)

But really, I don't know that much about MPS. You should try it if it sounds better to you.

Term 1, sorry. I'm not going this year. Oh yeah...they used AoPS 1 and another book...

thnks...
ya im going term 1

ive done a lot of the AoPS books...1 and 2, intro to Geometry and Algebra

Maybe I shoulda signed up for geometry....

I believe AoPS user diophantient was took this class, he went to Nationals twice. You may want to ask him about it.

teng!

my advice.. since I never did the duke tip thing (didnt want to pay 72 dollars to take a test then 3000 dollars for a few weeks of camp..)

ditchduketipandletmetutoryouforanoverpriced20bucksanhour. just kidding. ;)

[quote="Anaxerzia"][quote="academicsrule"][quote="Anaxerzia"]DUDE WHAT TERM WERE YOU AT DAVIDSON? I was there term 2...

I'll be going again for Bio at Duke West. I took problem solving last year and it was a pretty good introduction into the subject. MC/AMC level... They use AoPS books :)[/quote]
OMG I was at Davidson in Term II last year too! I took creative writing though...
And this year I got waitlisted.
Fortunately, I'm spending 5 weeks at Choate anyway, so w/e. I can't go.[/quote]

DUDE!! What was your name? Mine was Ravi...[/quote]
I KNOW WHO YOU ARE!
I think. Do you know Kimberly? She was in my RC group.
Anyway, My name is Cathy.


[quote="ayush_2008"]how many 7 digit numbers divisible by 3 is possible using 0,1,2,3,4,5,6.repetition is allowed.

there are too many cases when i solve this question,pl suggest some another way.[/quote]

Let $ P(n)$ be the number of n-digit numbers $ \overline{a_1a_2...a_n}$ divisible by 3 using only $ 0,1,2,3,4,5,6$

So for $ \overline{b_1b_2...b_nb_{n \plus{} 1}}$, where $ b_1, b_2,..., b_n$ is fixed,

[b]Case 1:[/b] $ 3 \nmid (b_1 \plus{} b_2 \plus{} ... \plus{} b_n)$, then there are two choices for $ b_{n \plus{} 1}$ to make $ \overline{b_1b_2...b_nb_{n \plus{} 1}}$ divisible by $ 3$
[b]Case 2:[/b] $ 3 | (b_1 \plus{} b_2 \plus{} ... \plus{} b_n)$, then there are three choices for $ b_{n \plus{} 1}$ to make $ \overline{b_1b_2...b_nb_{n \plus{} 1}}$ divisible by $ 3$

So $ P(n) \equal{} (6 \times 7^{n \minus{} 2}) \times 2 \plus{} P(n \minus{} 1)$ with $ P(1) \equal{} 2$ gives

$ P(7) \equal{} 12 \times 7^5 \plus{} 12 \times 7^4 \plus{} ... \plus{} 12 \times 7 \plus{} 12 \times 1 \plus{} 2 \equal{} 2 \times 7^6$

[quote="stephencheng"][quote="ayush_2008"]how many 7 digit numbers divisible by 3 is possible using 0,1,2,3,4,5,6.repetition is allowed.

there are too many cases when i solve this question,pl suggest some another way.[/quote]

Let $ P(n)$ be the number of n-digit numbers $ \overline{a_1a_2...a_n}$ divisible by 3 using only $ 0,1,2,3,4,5,6$

So for $ \overline{b_1b_2...b_nb_{n \plus{} 1}}$, where $ b_1, b_2,..., b_n$ is fixed,

[b]Case 1:[/b] $ 3 \nmid (b_1 \plus{} b_2 \plus{} ... \plus{} b_n)$, then there are two choices for $ b_{n \plus{} 1}$ to make $ \overline{b_1b_2...b_nb_{n \plus{} 1}}$ divisible by $ 3$
[b]Case 2:[/b] $ 3 | (b_1 \plus{} b_2 \plus{} ... \plus{} b_n)$, then there are three choices for $ b_{n \plus{} 1}$ to make $ \overline{b_1b_2...b_nb_{n \plus{} 1}}$ divisible by $ 3$

So $ P(n) \equal{} (6 \times 7^{n \minus{} 2}) \times 2 \plus{} f(n \minus{} 1)$ with $ P(1) \equal{} 2$ gives

$ P(7) \equal{} 12 \times 7^5 \plus{} 12 \times 7^4 \plus{} ... \plus{} 12 \times 7 \plus{} 12 \times 1 \plus{} 2 \equal{} 2 \times 7^6$[/quote]
good night  from India SIR.
SIR  pl explain in a simple manner.what is p(n).everything is at par for  me.

1. This is a double post.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=287105

2. Only clarification I can see is that $ P(n \minus{} 1)$ was meant instead of $ f(n \minus{} 1)$. Did you actually read his post? $ P(n)$ is exactly what he said: the number of n-digit numbers $ \overline{a_{1}a_{2}...a_{n}}$ divisible by 3 using only $ 0,1,2,3,4,5,6$. Is some part of this confusing?


if X points are equally spaced on a circle, find the probability that 3 points chosen at random will form a right triangle.[/code]

[hide]
To have a right triangle in a circle, you have to have one of the sides a diameter.  If $X$ is odd, then this can't happen...  If $X$ is even ($X\ge4$), then there are $\frac{X(X-2)}{2}$ ways to choose a right triangle (there are $\frac{X}{2}$ ways to choose a diameter, and $X-2$ ways to choose another point of the circle).  Then the probability would be $\frac{X(X-2)}{2\binom{X}{3}}$ (I think).
[/hide]

[quote="fireforce"]100% since no three points are colinear on a circle. Unless of course there are less than 3 pts. then its DNE (question does not exist).[/quote]

it says right triangle.

$0$ (Or to be more precise,  $\frac{1}{\infty}$)

Edit: nevermind,  I was assuming infinite points.

I think that:

there are X/2 points directly across from another. then select 2 of them
after that, select an additional one from what's left.

so it's X/2 choose 2,  multiply that buy (X-2)

That should be the Numerator.

The denominator is simply X choose 3.

 :bruce:

I have no idea if I'm doing this right ...
[hide]
For it to be a right triangle, two of the points must for a diameter. If X is odd, then the probability is 0, so X must be even and greater than 3. There are $x/2$ sets of diameters, and for each set, any of the remaining $x-2$ points will make the triangle right. In all, there are $\binom{x}{3}$ ways of choosing any 3 points. The probability that the triangle is right is:
\begin{eqnarray*}\frac{\frac{x}{2}(x-2)}{\binom{x}{3}} &=& \frac{x(x-2)}{2}\cdot \frac{3!(x-3)!}{x!} \\ &=& \frac{x(x-2)}{2}\cdot \frac{3!(x-3)!}{x(x-1)(x-2)(x-3)!}=\boxed{\frac{3}{x-1}} \end{eqnarray*}
This assuming that X is greater than 3 and is an even number.
[/hide]

Yeah I see my mistake now...

Using Thales's theorem we can see that we only need to find the probability that two of the points are opposite sides of the circle from each other.


[quote="ritchjp"]I don't know if this question has ever been posted before, but i'll share it anyway.

How many ways are there to make change for a dollar?[/quote]

What types of coins are allowed? Quarters, dimes, nickels, pennies, half dollars, golden dollars, a dollar bill, etc.

[quote="Treething"][quote="Ignite168"][quote="ritchjp"]I don't know if this question has ever been posted before, but i'll share it anyway.

How many ways are there to make change for a dollar?[/quote]

What types of coins are allowed? Quarters, dimes, nickels, pennies, half dollars, [b]golden dollars, a dollar bill[/b], etc.[/quote]

omg...[/quote]

Seriously, can you really make CHANGE for a dollar with a dollar bill?

Any denominations less than a dollar.
From pennies all the way up to half dollar pieces.
And I am talking about only American currency.

Hmmm if that appeared on Mathcounts, I don't believe you would have the chance to brute force a list of 292 possibilities for an hour.
Kevin, since you're so much smarter than us, would you mind actually posting a solution?

The solution is to go to http://www.teachnet.com/lesson/math/293changedollar.html .

 :P

[quote="meenamathgirl"]I did it the long way, listing everything.  It took me almost an hour.  I used an ordered list.  I got...

[hide]
292
[/hide][/quote]

You must be really bored.


Sorry for this spam, but does anyone have the IMO 2007 logo in the Corel format (or in any other vector-graphic format)? I've sent an email to imo2007@hn.vnn.vn but there's no response.

EDIT: Whops, this should be in the IMO 2007 section, if there's any difference...

Only the organizers can have access to the vector file (unless they've released it publicly on the web, which I think they didn't).

PS I've moved the thread.

Thanks Valentin. I will wait for the response from the organizers, but I doubt that there will be any...


How do you make a header on top of every page?
For instance, if I wanted to put the name of my article at the top of each page...


The statement "if $ k\ge m$ then $ k<n$" is wrong. What is the correct one?

Well first, I have no idea what $ n$ is.  If $ k\geq m$, then either $ k>m$ or $ k\equal{}m$.  I'm not really sure what else you're asking.

I would go with: "if $ k\ge{m} \longrightarrow k\le{n} \lor  k > n$."  :P


Ring Theory
superior algebra
superior algebra solved

$A$ a commutative banach $\mathbb{C}$-algebra. Prove that any maximal ideal is a closed hyperplane.

The closure of an ideal is an ideal, so every maximal ideal $M$ of our algebra $A$ must be closed. $A/M$ is a field and it can only be $\mathbb C$, i.e. $M$ has codimension $1$, and the conclusion follows.

yes  :)  (according to Gelfand-Mazur theorem.)


:help:

7. How many distinct sums can be obtained by adding three different numbers from the set {-2, -1, 1, 2, 3, 4, 5}?

How do you solve this other than enumeration?

Thanks

[hide]

Anirudh's Great Theorem of Distinct Sums:

[i]The number of distinct sums that can be obtained from a given set of numbers, are the numbers you cannot get, subtracted from the range of the numbers that can be acquired.[/i]   :rotfl:

So, by following that, you know that you cannot get $-2$. Therefore, you subtract that from the range of $12$. Therefore, your answer is $11$.[/hide]

[quote="frost13"][hide]doesn't (-2)+(-1)+1 = -2??????[/hide][/quote]

Yep it does.

oooh. whoops. I read the problem wrong. I thought it said 2 numbers.  :blush:

Anirudh, misreading aside I did the problem the same way you did :lol:

would the answer be 15?

because the range is from 12 to -2.

Yes, there are 15 distinct sums, I've posted this problem before. Nice theorem anirudh.

Anirudh, your theorem only works if the numbers are consecutive, as far as I know.

No. Not necessarily. Re-read it.  :wink:

Thank You, All!
Thanks for teaching me how to do this!

The answer is 15.


1) Find all functions:R->R such that f(1)=1 and
     $ f(f(x)y \plus{} \frac {x}{y}) \equal{} xyf(x^{2} \plus{} y^{2})$
2) R* denote the set of nonzero real numbers. Find all functions f:R*->R* such that
 $ f(x^{2} \plus{} y) \equal{} (f(x))^{2} \plus{} \frac {f(xy)}{f(x)}$
for every pair of nonzero real numbers x,y with $ x^{2} \plus{} y < > 0$

Posted before,
1, http://www.mathlinks.ro/Forum/viewtopic.php?t=148711

2, http://www.mathlinks.ro/Forum/viewtopic.php?t=151360


The sum of the reciprocals of three distinct natural numbers is $ \frac{31}{30}$. What is the least possible product of three such natural numbers?

$ \frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}\equal{}\frac{ab\plus{}bc\plus{}ca}{abc}\equal{}\frac{31}{30}$

oh hey 2, 3, 5 works

answer : 30

Is there a way to solve it without using brute force?

idk...

but isnt brute force the easiest way to solve this??

just one try : 2, 3, 5

I suppose, but there has got to be some other way.

I thought about it a while... even looked at Vieta's formulae for inspiration.

I'm tentatively coming to the conclusion that there is no straightforward mathematical explanation.

I think these "if they look at it naively they'll find the answer quickly" questions are annoying in the sense they don't really need mathematical thought, however they do teach you to keep your mind open for shortcuts.

This wasn't really brute force so much as "it's easy enough to look at and get... 30 2,3,5, that probably works, here we go..."  This is pretty much shooting an arrow at a target which is one giant bullseye.

Haha, it makes me so sad and angry to hear people say "the easiest way is brute force".

I can just imagine them as kids, ramming triangle peg into the square hole, then the circle, then the hexagon...


A square is inscribed in a circle. The area of the square is 64 square inches. How many square inches are in the area of the
circle? Express your answer in terms of $ \pi$.

side of square=8

radius of circumscribed circle=1/2 diagonal=1/2 8sqrt2=4sqrt2

4sqrt2^2(pi)=32pi

I know for a fact the answer isn't $ 32\pi$, but I'm not sure what it is.

Actually, the answer is $ 32\pi$.  It might not be on FTW, however.

[asy]draw(unitsquare);
draw((0,0)..(1,0)..(1,1)..(0,1)..cycle);
draw((0.5,0.5)--(1,1));
label("$8$", (0.5,1), N);
label("$4\sqrt{2}$", (0.75, 0.75), SE);[/asy]

The answer is $ 32\pi$.

(!--Made mz's solution more clear--)

The side length of the square is 8, so the diagonal is $ 8\sqrt{2}$.
Half the diagonal would also be the radius of the circle, which is $ 4\sqrt{2}$.
Thus, the circle's area is $ (4\sqrt{2})^2\cdot\pi \equal{} 32\pi$.

[quote="isabella2296"]But it's not $ 32\pi$. When I type $ 32\pi$, it was counted as incorrect.[/quote]

typo?

No; I've entered $ 32\pi$ multiple times...I wouldn't typo 10 times. I think it may be a bug.

Did you check the answer? What did the answer say?

I forgot to, but I'll be sure to check next time I see it.

[quote=bever209]the answer is 64 pi but y[/quote]

If the area of the square is 64, then the side length is 8. Then the diagonal from one edge to another is $8\sqrt{2}$ which is also the diameter of the circle. Then the radius is $4\sqrt{2}$ so the area is $32\pi$... How is it 64 pi?

Probably a bug?


find all of the functions $f: \mathbb{R^+}\rightarrow\mathbb{R^+}$ if x,y element of $R^+$ we have $\frac{1}{2}\ xf(x)+yf^-1(y) \geq\ xy$