calculus
integration
quadratics
number theory unsolved
number theory

where would i find the proof for this? can anyone help?
thanks :)

For  what?

If you mean for integer solution of $p^2+q^2+r^2=7$ then is trivial, just go through all the cases $p^2=\{0, 1, 4\}$ etc.

rational numbers :) thanks and the problem is showing proof why 3 squares can never equal 7 in case i was unclear.

To get the rational case, multiply by a common denominator. Then you want to look for integral solutions of $x^2+y^2+z^2=7t^2$.
Looking $\mod 8$ shows you that this can only be possible when $2|x,y,z,t$, thus divide the equation by $4$ to get finally a contradiction by infinite descent: all four number are divisible by $2$ infinitely often.

[quote="ZetaX"]Looking $\mod 8$ shows you that this can only be possible when $2|x,y,z,t$[/quote]

How do you show this? :blush:

The quadratic residues $\mod 8$ are ${0,1,4}$ and the equation
is written $x^2+y^2+z^2+t^2 \equiv 0 \mod 8$ .

The only way to have this is when the quadratic residues are of the form :

$\\ \{ (0,0,0,0), \, (4,4,0,0), \, (4,4,4,4), \, (4,0,4,0), \, (4,0,0,4), \, (0,4,4,0), \, (0,4,0,4), \, (0,0,4,4) \}$

But $x^2 \equiv 4 \mod 8$ or $x^2 \equiv 0 \mod 8$ only when $x$ is even .


geometry
combinatorics unsolved
combinatorics

Prove that in every polygon there is a diagonal that cuts off a triangle and lies within the polygon.

[quote="ikap"]And this is what guarantees a triangulation for every simple polygon. For further challenge see problem #6 in the second session of iranian tst-s[/quote]

I think your solution is true when we know that we can triangulation a polygon but how we can prove this when we do not konw we can triangulation a polygon????(sorry for my bad english)>

Here's a sketch of how to prove both the result in the original problem and that any polygon has a triangulation.  There are three places in the argument I'm marking with (**) that you should try and convince yourself of in more detail.

Lemma: For any (simple) polygon there is a diagonal lying within the polygon (perhaps not cutting off a triangle)

Proof: Place the polygon arbitrarily in the plane, and let $ z_1$ be a vertex having maximal $ y$ coordinate.  Note that the angle at $ z_1$ must be less than $ 180$ degrees.  Let $ z_0$ and $ z_2$ be the adjacent vertices to $ z_1$.

Either the triangle $ z_0 z_1 z_2$ lies inside the polygon (in which case we are done already), or there must be other vertices of the polygon inside this triangle (**).  Let $ z$ be the vertex inside the triangle which is furthest from the line segment $ z_0 z_2$.   Then $ z z_1$ is our required diagonal, for if an edge crossed it one of its endpoints would be further from the segment (**).

Using this, we can inductively triangulate the polygon by first drawing a diagonal inside, then triangulating each of the two smaller polygons formed by this diagonal.  Since in any triangulation at least one triangle uses two adjacent sides from the polygon (**), this also proves the original problem.


This is from Aops Volume I: Show that the conjugate of $ z \plus{} w$ $ \equal{}$ the conjugate of $ z$ $ \plus{}$ the conjugate of $ w$ (how do you get the line above it in LaTex?)

Conjugate of $ z$ $ \plus{}$ conjugate of $ w$ $ \equal{}$ $ (z_1 \minus{} z_2i) \plus{} (w_1 \minus{} w_2i)$ $ \Rightarrow$  $ (z_1 \plus{} w_1) \minus{} (z_2 \plus{} w_2)i$ $ \Rightarrow$ conjugate of $ z \plus{}w$.

I can't figure out how they got from $ (z_1 \plus{} w_1) \minus{} (z_2 \plus{} w_2)i$ to the conjugate of $ z \plus{} w$.

Well, what is $ \overline{z \plus{} \omega}$? It's $ \text{Re}(z \plus{} \omega) \minus{} \text{Im}(z \plus{} \omega)\cdot i \equal{} \text{Re}(z) \plus{} \text{Re}(\omega) \minus{} \text{Im}(z)\cdot i \minus{} \text{Im}(\omega)\cdot i$ or in your notation $ z_1 \plus{} \omega_1 \minus{} z_2\cdot i \minus{} \omega_2\cdot i$, which you already have.

It is quite obvious, actually, so I'm not sure how to explain it.

And overline z plus omega is \overline{z+\omega} .

Okay thanks I get it now :)  Its just rewriting it in the general form.


algebra
polynomial
quadratics
Vieta
quadratic formula
Rational Root Theorem

$(x-1)(x-2)=x^{2}-3x+2$

we know that the roots of this polynomial is 1 and 2, say we knew one of the roots (for example $x=1$) and we didnt know the other, how could we use the $x=1$ to find the $x=2$? is there any relation between one root of a polynomial to the other roots of it? for all polynomial

In a quadratic, getting one root will give away the other, because the negative of the term with "x" is the sum of the roots. So in the example, the sum of the roots is 3, which would give the other root as x=2.

You can always divide to get a simpler polynomial to find the other roots. For example, in a cubic, given one solution, you could use synthetic or long division to obtain a quadratic, which can then be solved to get the other solutions.

[quote="binomial_4eva"] how could we use the $x=1$ to find the $x=2$?[/quote]
[hide="Method 1"]Use synthetic division to get the remaining polynomial.[/hide]
[hide="Method 2"]Use the fact that $3 = r_{1}+r_{2}$ and $2 = r_{1}r_{2}$ to solve if $r_{1}= 1$.[/hide]
[hide="Method 3"]Multiply out $(x-1)(x-r_{2})$ to get $x^{2}-x-r_{2}x+r_{2}$ and set the two polynomials equal.[/hide]

what about a polynomial with highest degree 11?
if we know one of its roots, can we find the other 10 roots using the one known root? i feel this has a compact formula, dunno if i am right or wrong.

Well, if we have a polynomial of degree 11, say $P(x)$, and we have one root, say $y$, we could divide by $x-y$ to get a tenth degree polynomial, say $Q(x)$, which has the same roots as $P(x)$, except $y$, unless there is another root y(double root) in $P(x)$.

In other words, if an 11th degree polynomial has roots a,b,c,....k, the 10th degree polynomial would have all of those roots except 1.

so can we use the one root of $P(x)$ to find the roots of $Q(x)$?

i dunno how to put it, i mean for example this polynomial $x^{3}-6x^{2}+11x-6$ we know that it has 3 roots, and they are x=1, x=2, and x=3.

lets say we only knew one of them $x=1$ , how could we get $x=2$ and $x=3$ stright away from $x=1$
.
what is the relation between $x=1$ , $x=2$ , and $x=3$ ?

my opinion $x_{1}=1$    $x_{2}=x_{1}+1$   $x_{3}=3(x_{1})$ or $x_{3}=x_{2}+1$

do you think there is a relation or connection,like this for roots of  any polynomial?,

Without even knowing any roots of a polynomial, you can easily find all the rational roots by using RRT and synthetic division.

[quote="binomial_4eva"]do you think there is a relation or connection,like this for roots of  any polynomial?[/quote] Of course not.  You've chosen the roots to be consecutive positive integers, and then conjectured that the roots will always be consecutive positive integers (or something).

As others have said: knowing one root of a polynomial allows you to reduce the degree by 1, by the Factor Theorem and division.  If you happen to have a small-degree polynomial, reducing the degree by 1 will make it easier to solve (i.e., reducing a cubic to a quadratic allows you to apply the quadratic formula, which is much easier than solving a cubic directly).  If you have a large-degree polynomial, knowing one of the roots will let you reduce to another large-degree polynomial which you won't (in general) be able to solve.  (Of course, if I put [i]you personally[/i] in charge of making polynomials, it appears that I will always be able to solve them by guessing that the roots are consecutive positive integers, or at worst rational numbers, in which case the Rational Root Theorem can be used.  Most polynomials don't have rational roots.)

The most general relations that hold between the roots and coefficients of a polynomial are known as Vieta's relations.  You can find lots of information about them and their uses by searching the forums, the Wiki or the internet.  236factorial and miyomo used essentially these relations in their solutions (in miyomo's case, in method 2; method 3 is essentially a derivation of the relations for quadratic equations).


function
logarithms
algebra unsolved
algebra

find all functions $f: \mathbb R\to \mathbb R$ such that \[f(x+y)=f(x)f(y)\quad \forall x,y\in \mathbb R.\]

We have $f=0$ or $g(x)=\ln f(x)$ is additive, but you need have add condition!

[quote="nayel"]like...? can you give some examples?[/quote]

Example : $f(x)=a^{x}$, with $a>0$

In fact : $f(\frac{x}{2}+\frac{x}{2})=f(\frac{x}{2})^{2}$ $\Rightarrow $ $f(x)\geq 0$ $\forall x\in \mathbb{R}$
If it exists $x_{0}$ such that $f(x_{0})=0$, then $f(x_{0}+(x-x_{0}))=f(x_{0})f(x-x_{0})$ $\Rightarrow $ $f(x)=0$ $\forall x\in \mathbb{R}$
If it does not exist $x_{0}$ such that $f(x_{0})=0$,then $f(x)>0$ $\forall x\in \mathbb{R}$ and it exist $g(x)=\ln(f(x))$
Then $g(x+y)=g(x)+g(y)$, well knowncase leading to $g(x)=\alpha x$ or $g(x)=$ one of the non-continuous solutions of Cauchy equation.

Hence three families of solutions :
1) $f(x)\equiv 0$
2) non null continuous ones : $f(x)=e^{\alpha x}$, $\alpha\in\mathbb{R}$
3) non null non continuous : $f(x)=e^{any\ non\ continuous\ solution\ of\ Cauchy\ additive\ equation}$

--
Patrick

[quote="nayel"]okey, from this can we conclude that $\exists x\in \mathbb R$ such that $f^{-1}(x)=1$ if $f$ is non-constant? if not, what condition should we restrict it to happen?[/quote]

I don't understand your question.
a) first, non continuous non constant $f(x)$ may be non-injective functions, and so $f^{-1}(x)$, if this is the inverse function of $f(x)$, is not always defined.
b) second, $f^{-1}(x)=1$ means $x=f(1)$ and it exist obviously one such $x$.

So what exactly are you asking for ?
May be are you looking for an $x$ such that $f(x)=1$ ?
In this case, we may say that it exist (with choice axiom) some non continuous non zero solutions such that it does not exist any $x$ such that $f(x)=1$.
In order to avoid these cases, you can add a lot of constraints :
 - f non zero and continuous
 - f strictly monotonous
 - it exist at least one x such that $f(x)=p$ for a given $p\in\mathbb{Q}^{*}$
 ...

--
Patrick


inequalities
linear algebra
matrix
vector
combinatorics proposed
combinatorics
Linear algebra in combinatorics

There are $n$ people living in the city.
Everyone belongs to at least one chess-club.
Each club has an odd number of members.
Every two clubs have an even number of members who belong to both of them.

Prove that there are at least $n$ clubs in the city.

i think the inequality must be the other way around.
 consider the $0$-$1$ matrix $A$ with the rows indexed by the $n$ people and the columns indexed by the $m$ clubs with $A(i,j)=1$ iff $i\in$ club $j$. Then over $\mathbb{Z}_2$ the given condition means that $A^{T}A=I_m\Rightarrow r(A)=m$.but then the number of rows must be atleast $m$.

It should be "there are no more than $n$ clubs in the city" and it was already discussed at least once.

I think Grobber provided link to superb paper with solution to this (and many more). It is short so:

We can see that $n$ is attainable - every person forms his own club.

Let's establish upper bound:  Instead of clubs will be talking about sets. Consider characteristic vectors of each set. It is not hard to prove (just suppose they are dependent and consider scalar product of $\sum \lambda _A X _A =0$ with $X _A$ where $X _A$ is characteristic vector of set $A$)  that they are independent over $\mathbb{F} _2 ^n$ hence number of clubs is less or equal than n.

[edit] I didn't see seshadri message when posting

seshandri is right, sorry for the typo...


Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}. Prove or disprove that if I select 10 distinct elements from the set A, that there are at least two pairs of elements whose sum will be exactly 17.

Suppose that the lab has n (where n > 8 and n is even) computers in it. Suppose that half the computers send messages to exactly 3 other computers in the lab and half the computers send messages to exactly 5 other computers in the lab. Prove or disprove that there is at least 1 computer in the lab that must have received messages from at least 4 other computers.

In poker, a flush is a 5-card hand where all the cards have the same suit. Recall that a deck of cards has 52 cards. There are 4 suits (hearts, clubs, diamonds and spades) and each suit has 13 cards. How many different flush hands can you have?

Suppose someone is standing at the point (0,0) and wants to go to the point (8,14). How many ways can this person travel from (0,0) to (8,14) if every time they take a step, they must either take one unit step in the positive X direction or take one unit step in the positive Y direction. For example, starting at (0,0), in one step a person can reach either (1,0) or (0,1).

for the first one, partition the set into the 8 sets $ \{1,16\}$, $ \{2,15\}$, ..., $ \{8,9\}$, where in each set the sum is 17. by the pigeonhole principle, 2 of the 10 elements you pick will be in one of the same parts of this partition, and we're done.

the second problem will also use pigeonhole. the other ones aren't too hard either.

[hide="1 Hint"]Pigeonhole with $ k$ holes of size 2 and $ k \plus{} 2$ pigeons means $ 2$ holes have more than one pigeon. (DER! Thanks JBL)[/hide]

[hide="2 Hint"]This problem is really weak. $ 3n/2 \plus{} 5n/2 \equal{} 4n$ total messages are sent. Is it supposed to say at least $ 1$ computer received $ 5$ messages? (I believe there is a counterexample then)[/hide]

[hide="3 Hint"]How many flush hands are there for one specific suit? Now multiply by four.[/hide]

[hide="4 Hint"]If $ U$ is a step up and $ R$ is a step right, we want to know how many ways there are to make a word with $ 14$ Us and $ 8$ Rs. The answer to this is a common counting function.[/hide]

[quote="MellowMelon"]$ k$ holes and $ k \plus{} 2$ pigeons means $ 2$ holes have more than one pigeon.[/quote]

This is not true in general, as I'm sure you know -- it just happens that in this case, the "sensible" holes all have size 2.


$a,\;b,\;c$ must all be positive, since if one of them, say $a$, was $a<0$, $b^{3}+c^{3}>\frac{1}{9}$, so it would exist some  $d\in \{ b,c\}$  such that  $d^{3}\geq \frac{1}{18}\rightarrow d>\frac{1}{2}$.

So  $a+b+c\geq 3\sqrt [3]{ \frac{ a^{3}+b^{3}+c^{3}}{3}}=\frac{3}{3}=1$.

By Cauchy:
$21(a^{2}+b^{2}+c^{2})+1\geq 7(a+b+c)^{2}+1$,  so, letting  $S=a+b+c$,  it is only left to prove   $7S^{2}-8S+1\geq 0$  which is true for all $S\geq 1$, as is the case.
Seems easy, but there could be a mistake in my reasoning.

[quote="Mimoide"]$d^{3}\geq \frac{1}{18}\Rightarrow d>\frac{1}{2}$.[/quote]

???

Yes, sorry; :oops_sign: . Ii is the most stupid thing I have ever written: there's a mistake almost in every line.

Its just $\sum\left(\left(\frac{1}{2}-a\right)\left(\frac{1}{3}-a\right)^{2}\right)\geq 0$


I thinks $\log_{3}\left(9^{x}-72\right)\;\text{has define}\Longleftrightarrow 9^{x}-72>0\Longleftrightarrow x>\log_{9}72>1$
And the inquation equivalent to:
\[\begin{cases}x>\log_{9}72\\ \log_{3}\left(9^{x}-72\right)\ge 0\\ \log_{3}\left(9^{x}-72\right)\le x\end{cases}\Longleftrightarrow\begin{cases}x>\log_{9}72\\ 9^{x}-72\ge 1\\ 9^{x}-72\le3^{x}\end{cases}\\ \Longleftrightarrow\begin{cases}x>\log_{9}72\\ x\ge\log_{9}73\\-8\le 3^{x}\le 9\end{cases}\Longleftrightarrow\log_{9}73\le x\le 2 \]
Is it true?

[quote="hikaru123"][quote="April"]Solve the inquation: $\log_{x}\left[\log_{3}\left(9^{x}-72\right)\right]\le 1$[/quote]
*if $x>1$ we have $log_{3}(9^{x}-72)\le 1$
So $9^{x}-72\le 3^{x}$
$\leftrightarrow-8\le 3^{x}\le 9$
$\rightarrow1<x\le 2$
*x=1 is a solution of the inquation
*if $0\le x<1$ we have $log_{3}(9^{x}-72)\ge 1$
Do the same as the first case we have
$3^{x}\le-8$ or $9\le 3^{x}$
$\rightarrow x\ge 2$ So there is no solution in this case.
So the answer is $1\le x\le 2$[/quote]
Your answer is not completely correct, but you can easily fix it.  Take for example $x=1$.


calculus
function
limit
induction
algebra
polynomial
real analysis

If $A_n = \sum_{k=1}^n k^3/3^k$ find the limit of $A_n$

it should be usefull to study $\sum \frac{x^k}{3^k}$ and to derivate ..

Denote by $f(x)=\sum \frac{x^k}{3^k}= \frac 1 {1- \frac x 3}$ for $x \le 3$
Following the next steps you get to the closed formula for $\sum \frac{k^3}{3^k}$ in terms of $f(1),f'(1),f''(1)$: differentiate,multiply by x, differentiate,multiply by x, differentiate.

There is also another way, which not envolves calculus...

Let me guess...  you use series $\sum \frac{k^2}{3^k}$ and $\sum \frac{k}{3^k}$ and $\sum \frac 1{3^k}$ for algebiraic manipulation?....

yes xirti, algebrical movements with those sequences :) [no calculus means here no functions, no derivates, just sequences]

Using the recurence from the my last reply, if denote $l_p(x)=\lim_{n\to\infty} \sum \limits_{k=1}^n k^p x^k$, where $p\in N$ and $x\in (0,1)$, then there is the following recurrence:
\[ \boxed {\ l_0=\frac{x}{1-x}\ \ ;\ \ \left(\forall\right)p\in N^*\ ,\ l_p(x)=\frac{x}{1-x}\cdot \left[1+\sum\limits_{k=0}^{p-1} C^k_p l_k(x)\right]\ }\ . \][b]Example:[/b] $x=\frac 13$ and $p=3\Longrightarrow l_0=\frac{\frac 13}{1-\frac 13}=\frac 12\ ;$
$l_1=\frac 12 (1+l_0)=\frac 12\left(1+\frac 12\right)=\frac 34,$ i.e.$\lim_{n\to\infty} \sum\limits_{k=1}^n \frac{k}{3^k}=\frac 34\ ;$
$l_2=\frac 12 (1+l_0+2l_1)=\frac 12\left(1+\frac 12+\frac 32\right)=\frac 32,$ i.e. $\lim_{n\to\infty} \sum_{k=1}^n\frac{k^2}{3^k}=\frac 32\ ;$
${l_3=\frac 12 (1+l_0+3l_1+3l_2)=\frac 12\left(1+\frac 12+\frac 94+\frac 92\right)=\frac{33}8},$ i.e. $\boxed {\ \lim_{n\to\infty} \sum_{k=1}^n \frac{k^3}{3^k}=\frac{33}{8}\ }\ .$

[quote="Virgil Nicula"]Let $x$ be a real number, $x\not \in \{0,1\}$. Then for any $p\in N^*$,
$A_p(x)\equiv \sum\limits_{k=1}^n k^px^k=\frac{x}{x-1}\left[(n+1)^px^n-1-\sum \limits_{m=0}^{p-1} C^m_p A_m(x)\right]$,
where $A_0(x)=\frac{x^{n+1}-x}{x-1}$ a.s.o.[/quote]

this can prove to be very usefull .... could you give some hints in proving it? induction?

[quote="v0id"]the initial limit can be calculated using derivates...not very hard[/quote]

Can you post your solution using derivates ? (I'm new to this technique)

Define for $ p\in \mathcal N^*$ the sum $ S_p(x) \equal{} \sum_{k \equal{} 1}^nk^px^k$ .

Observe that $ S_{p \plus{} 1}(x) \equal{} x\cdot \left(\sum_{k \equal{} 1}^nk^px^k\right)' \equal{} x\cdot S^{\ '}_p(x)$ .

Thus, $ \sum_{k \equal{} 1}^n\frac {k^2}{3^k}\equal{}S_2\left(\frac 13\right)$ .

Start with $ f_0(x) \equal{} \sum_n x^n \equal{} \frac1{1 \minus{} x}$. If $ f_k(x) \equal{} \sum_n n^kx^n$, $ f_k(x) \equal{} xf_{k \minus{} 1}^{\prime}(x)$; we can differentiate the power series termwise, and then multiply by $ x$ to get back to the original exponents.

Calculating the first few:
$ f_1(x) \equal{} \frac {x}{(1 \minus{} x)^2}$
$ f_2(x) \equal{} \frac {2x^2 \plus{} x(1 \minus{} x)}{(1 \minus{} x)^3} \equal{} \frac {x^2 \plus{} x}{(1 \minus{} x)^3}$
$ f_3(x) \equal{} \frac {3x^3 \plus{} 3x^2 \plus{} x(2x \plus{} 1)(1 \minus{} x)}{(1 \minus{} x)^4} \equal{} \frac {x^3 \plus{} 4x^2 \plus{} x}{(1 \minus{} x)^4}$

That last one should be $ x^3 \plus{} 4x^2 \plus{} x$ -- all these polynomials in the numerator have positive-integer coefficients (the Eulerian numbers).

Virgil's recurrence is not workable though. You couldn't derive it on the spot to solve this problem if you didn't know it beforehand. At least I can't.  :blush:

[quote="spix"][quote="Virgil Nicula"]Let $ x$ be a real number, $ x\not \in \{0,1\}$. Then for any $ p\in N^*$,
$ A_p(x)\equiv \sum\limits_{k \equal{} 1}^n k^px^k \equal{} \frac {x}{x \minus{} 1}\left[(n \plus{} 1)^px^n \minus{} 1 \minus{} \sum \limits_{m \equal{} 0}^{p \minus{} 1} C^m_p A_m(x)\right]$,
where $ A_0(x) \equal{} \frac {x^{n \plus{} 1} \minus{} x}{x \minus{} 1}$ a.s.o.[/quote]

this can prove to be very usefull .... could you give some hints in proving it? induction?[/quote]

I managed to prove it straightforward using Newton's formula when expanding $ (n\plus{}1)^p$.It's not that ugly as one would expect.It only took me a half of page.

[size=117][color=darkblue][b]Yes, I think that [u]Svejk[/u] proved this recurrence as mine (a half of page) ![/b][/color][/size]

[quote="Virgil Nicula"][color=darkred]Let $ x$ be a real number, $ x\not \in \{0,1\}$. Then

$ A_p(x)\equiv \sum\limits_{k \equal{} 1}^n k^px^k \equal{} \frac {x}{x \minus{} 1}\left[(n \plus{} 1)^px^n \minus{} 1 \minus{} \sum \limits_{m \equal{} 0}^{p \minus{} 1} C^m_p A_m(x)\right]$

for any $ p\in\mathcal N^*$ , where $ A_0(x) \equal{} \frac {x^{n \plus{} 1} \minus{} x}{x \minus{} 1}$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] $ A_p(x) \equal{} x \plus{} \sum_{k \equal{} 2}^nk^px^k \equal{}$ $ x \plus{} \sum_{k \equal{} 1}^{n \minus{} 1}(k \plus{} 1)^px^{k \plus{} 1}$ $ \implies$

$ A_p(x) \equal{} x \plus{} x\cdot\sum_{k \equal{} 1}^{n \minus{} 1}\sum_{m \equal{} 0}^pC_p^mk^mx^k \equal{}$ $ x \plus{} x\cdot \sum_{m \equal{} 0}^{p}C_p^m\sum_{k \equal{} 1}^{n \minus{} 1} k^mx^k$ $ \implies$

$ A_p(x) \equal{} x \plus{} x\cdot \sum_{m \equal{} 0}^{p}C_p^m\left(\sum_{k \equal{} 1}^{n} k^mx^k \minus{} n^mx^n\right)$ $ \implies$

$ A_p(x) \equal{} x \plus{} x\cdot\sum_{m \equal{} 0}^pC_p^mA_m(x) \minus{} x^{n \plus{} 1}\cdot\sum_{m \equal{} 0}^pC_p^mn^m$ $ \implies$

$ A_p(x) \equal{} x \plus{} x\cdot\left[\sum_{m \equal{} 0}^{p \minus{} 1}C_p^mA_m(x) \plus{} A_p(x)\right] \minus{} x^{n \plus{} 1}(1 \plus{} n)^p$ $ \implies$

$ (x \minus{} 1)\cdot A_p(x) \equal{} (n \plus{} 1)^px^{n \plus{} 1} \minus{} x \minus{} x\cdot\sum_{m \equal{} 0}^{p \minus{} 1}C_p^mA_m(x)$ $ \implies$

$ \blacktriangleright\ A_p(x) \equal{} \frac {x}{x \minus{} 1}\left[(n \plus{} 1)^px^n \minus{} 1 \minus{} \sum \limits_{m \equal{} 0}^{p \minus{} 1} C^m_p A_m(x)\right]$ .[/color]

In particular $ \sum_{k\equal{}0}^{\infty}{\frac{k^{4n}}{3^k}}$ is an integer for all n natural


modular arithmetic
number theory unsolved
number theory

[hide]
Taking it $\pmod{5}$, all the $2^m+3^m$ terms for $m$ odd disappear because they are divisible by $5$. It remains to find when

$(2^2+3^2)+(2^4+3^4)+\cdots+(2^{2n}+3^{2n})$

is divisible by $5$. Noting that $2^2+3^2 \equiv 3 \pmod{5}$ and $2^4+3^4 \equiv 2 \pmod{5}$ and this just repeats, we know that $n$ must be even for it to add up to $0 \pmod{5}$. Since $p$ has to be prime, it must be $p = 2n+1$ but $n$ is even, so $p$ is a prime of the form $4k+1$.[/hide]


I was unable to do this one:

  The  sum of the digits of a number of two digits is 10. If 18 be subtracted from it the digits in the remainder will be equal.Find the number.
Thank you.

pretty simple problem...
0 <= X, Y, Z <= 9
let the number after subtraction be ZZ
now ZZ + 18 = XY, X+Y = 10;

if there is no carry over,
we have X = Z+1, and Y = Z+8
X + Y = 10 => 2Z + 9 = 10 => 2Z = 1 (NOT POSSIBLE)

carry over
we have, X = Z+2, and Y = Z-2
X+Y = 10 => (Z+2)+(Z-2) = 10 => 2Z = 10 => Z = 5

original number = 55 + 18 = 73

Thank you.I was thinking of a similar method in the PT class in school.And i do not know how i could go wrong in this fashion.anyway,
thank you , again.


I am taking the AMC 8 this year. What problems did people like you miss? if any?

[quote="13375P34K43V312"]I took it unofficially, but I got a 25. I find that the most advanced students who are completely competent with the problems get questions wrong because of reading wrong. The test is multiple choice, and usually, the choices include the answer in a different form (feet instead of inches, etc.) that you will choose if you read the question wrong.[/quote]
You mean like nationals this year?
We all know nebula could've easily fullscored that :P
Do you know where I can get the AMC8?

Yeah, I got a 24 last year because I misread one problem. I think that when you get to a certain level, your sucess in math contests is based on your ability to not make sloppy mistakes as well as your skill. In National Mathcounts last year, there were 50-75 people who could have made the top 12 if they had been more careful. Looking back at Nationals, I can point out 3 points that I lost from ridiculously stupid mistakes, and 5 points that I lost on problems that I knew how to do, but for whatever reason didn't.

im taking geometry, yet i am also taking the AMC 8, do you think I will have any problems with rare formulas? or is it just to see how many stupid mistakes u can make?

AMC 8 is not about formulas, it's about thinking. Almost all of the concepts in it can be understood by someone just starting Algebra I. None of the problems are even supposed to require algebra.

im also preparing for this year's amc
Ive found that taking practice tests (easily obtainable off amc website) improves scores greatly
and also thinking about the questions
25s are no problem for me anymore after taking about 3 or 4of the practice tests and understanding the different types of problems thoroughly

[color=pink]I missed a lot, but i do not know the exact score, b/c no one ever told me... i know... that's sad.. but i'm taking it this year again and the logic's the same, sooo yeah... :oops_sign: [/color]

[quote="kimmystar94"]I missed a lot, but i do not know the exact score, b/c no one ever told me... i know... that's sad.. but i'm taking it this year again and the logic's the same, sooo yeah... :oops_sign: [/quote]
Could you please not post in color?  It's a bit distracting, and besides, this color is hard to read on this background.  Thanks.

[quote="now a ranger"]where is the amc website?[/quote]

It's here. [url]http://www.unl.edu/amc/index.shtml[/url]


abstract algebra
superior algebra
superior algebra unsolved

If $ K$ is a field, $ \Omega$ is an algebraic closure of $ K$, and $ E$, $ F$ are two extensions of $ K$ contained in $ \Omega$, then we can define the compositie extension of $ E$ and $ F$, beinq the subfield of $ \Omega$  generated by $ E$ and $ F$.
Now, if $ K\subset E$ and $ K\subset F$ are two field extensions (but we don't have $ \Omega$) can we define a composite extension field? If yes, how?

There is no canonical way to do this. But if at least one of the extensions are Galois, you could use the same defintion as above (choosing a embedding into the closure).

To see why there is no canonical way, consider $ \mathbb{R} \subset \mathbb{C}$ and $ \mathbb{R} \subset \mathbb{C}'$, where $ \mathbb{C} \equal{} \mathbb{R}[i]$ and $ \mathbb{C}'\equal{}\mathbb{R}[i']$, where $ \{i,\minus{}i\}$ and $ \{i',\minus{}i'\}$ are set-theoretically distinct elements whose square is $ \minus{}1$.  The compositum would need to contain four distinct roots of $ x^2 \plus{} 1$.

Probably the best we can do is the tensor product of field extensions, but in general this will not be a field.

ZetaX, LydianRain,
Thank you, I understand that there is no canonical way to do this. I'm not (yet) very familiar with Galois.
If I add the condition for $ K$ to be finite, does it make things better?

No; simply replace $ \mathbb{R}$ in my example with any field that does not have a square root of $ \minus{} 1$, for example.  As ZetaX says, we will need to consider embeddings in the algebraic closure, which will not necessarily maintain the set-theoretic properties of $ E$ and $ F$. (in other words, as in my example, $ i$ and $ i'$ might be mapped to the same element in such a construction)

I'm not sure if we need Galois--we may just need both extensions to be algebraic.  We may not even need that, if we start by breaking out the transcendent part. (this will be far from canonical, though, which might be what he means)

After choosing an embedding into a closure, the normal construction still applies. But if one of the extensions is Galois (in fact, normal is everything we need), it's image in the closure is unique (not the embedding itself), making the composition well defined up to isomorphism. This construction gives the standard one if the extensions are already embedded, but has the disadvantage of requireing on of the extensions to be normal (as otherwise the result is not even well defined up to isomorphism). Also, this construction is the same as considering $ E \otimes_K F$ which is a product of algebraic extensions of $ K$ and then taking one of these factors (they are each isomorphic as field extensions of $ K$).


algebra
polynomial
number theory unsolved
number theory

Is it  all the polynomial $\Phi_{n}(x)$ is irreducible?Could somebody give a proof or conterexample?Thanks

Cyclotomic ones¿ Yes, they are irreducible, see http://www.mathlinks.ro/Forum/viewtopic.php?t=15750

http://perso.orange.fr/yves.gallot/papers/cyclotomic.pdf

you have loads of useful information concerning cyclotomic polynomials here.

Wow, thinking of the cyclotomic polynomials as the minimal polynomial of the primitive roots of unity makes a lot of things clearer... thanks for the link, ZetaX!


calculus integration logarithms trigonometry function geometry rectangle

I had posted this problem as an indefinite integral sometime before.

But later when I checked with the integrator in mathworld, it gave a big expression with polylogs.

But I feel that the definite integral can be solved.Can someone help me?

The problem is Integral of Log(a+bcosx) dx from {0 to pi}

i think i've got this

let $u=log(a+bcosx)$ and $dv=dx$. Then $du=\frac{-bsinx}{(ln10)(a+bcosx)}$ and $v=x$. Integration by parts gives
$\displaystyle\int^\pi_0 log(a+bcosx)\,dx = [xlog(a+bcosx)]^\pi_0 + \frac{b}{ln10}\displaystyle\int^\pi_0 \frac{sinx}{a+bcosx}\,dx$.
For the remaining integral, let $w=a+bcosx$. So $dw=-sinx dx$ and our expression becomes
$[xlog(a+bcosx)]^\pi_0 - \frac{b}{ln10}[ln(a+bcosx)]^\pi_0= \frac{1}{ln10}[xlog(a+bcosx)-bln(a+bcosx)]^\pi_0= \frac{1}{ln10}[ln((a+bcosx)^{x-b})]^\pi_0= \frac{1}{ln10}((ln((a-b)^{\pi-b}))-(ln((a+b)^{-b})))= \frac{1}{ln10}(ln((a-b)^{\pi-b}(a+b)^{b})= log((a-b)^{\pi-b}(a+b)^{b})$

and we're done.

That doesn't work- the second integral should have a factor of $x$ in it.

You're not going to find this one by taking antiderivatives.

Here's a power series expression, if we reduce to the case $a=1$:

$\int_0^\pi\ln(1+b\cos x)\,dx=\pi\sum_{k=1}^\infty \binom{2k}{k}\frac1{2k}\left(\frac b2\right)^{2k}$

my bad. i had a feeling there was an error.

The answer, if I didn't make a mistake: $\int_0^\pi \ln(1+b\cos x)\,dx=\pi\ln\left(\frac{1+\sqrt{1-b^2}}2\right)$.

We assume $|b|\le 1$, so that the integral makes sense

[hide="The derivation"]
First, note that $\ln(1+b\cos x)=\int_0^b\frac{\cos x}{1+t\cos x}\,dt$. Rewrite the integral as
$\int_0^\pi\ln\left(\frac{1+\sqrt{1-b^2}}2\right)=\int_0^\pi\int_0^b\frac{\cos x}{1+t\cos x}\,dt\,dx$
$=\int_0^b\int_0^\pi\frac{\cos x}{1+t\cos x}\,dx\,dt$; the function we're integrating is continuous on the closed rectangle, so we can change the order.
Substituting $y=\tan\frac t2$,
$\int_0^\pi \ln(1+b\cos x)\,dx=\int_0^b\int_0^\infty\frac{\frac{1-y^2}{1+y^2}}{1+t\frac{1-y^2}{1+y^2}}\,\frac{2}{1+y^2}dy\,dt$
$=\int_0^b\int_0^\infty\frac{2(1-y^2)}{(1+y^2)(1+t+(1-t)y^2)}\,dy\,dt$
$=\int_0^b\int_0^\infty\frac{\frac2t}{1+y^2}-\frac{\frac2t}{(1-t)+(1-t)y^2}\,dy\,dt$
$=\int_0^b\frac{\pi}t-\frac{\pi}{t\sqrt{1-t^2}}\,dt= \pi\int_0^b\frac{\sqrt{1-t^2}-1}{t\sqrt{1-t^2}}\,dt= \int_0^b\frac{-t^2}{t\sqrt{1-t^2}(1+\sqrt{1-t^2})}\,dt$
$=\pi\left[\ln(1+\sqrt{1-t^2})\right]_0^b=\pi\ln\left(\frac{1+\sqrt{1-b^2}}2\right)$
[/hide]


geometry calculus integration trigonometry

ABCD is a square. AB=10. Find the area of  PQRS.(the dot area)
[img]http://i5.photobucket.com/albums/y194/jelly119/square.jpg[/img]

Sorry... I thought that I can delete my post ... :blush:

Can it be solved just by using geometry?

I doubt it.  Calculus seems to be the only obvious answer.  We have three simple regions which are easily seen to be bounded by functions.

All we have to do is find the area of RC(where the circle hits the base) and the rest will just fall out.

I think construct AC can do sth, but not so sure if we can solve the problem that way...

It seems to be very difficult.How have you got it?

It can be solved by using analytical geometry.

For one of the big circles, that goes from a to c i found the inregral for it. Not to nice though, (im trying to find the area of the tiny wee triangle at the bottom)

$ \int \sqrt{(x)(10-x)}=25 \sin ^{-1}({ \sqrt{ \frac{x}{10}})}- \frac{(5-x) \sqrt{(x)(10-x)}}{2}$

A solution is posted without using calculus:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=245128#245128


Identify all organic compounds with the following molecular formulas:

1) $ C_5H_{10}$

2) $ C_5H_8$

the DBE is 1 for the first one...
so....

1.Cyclo pentane
2.pentene
3.pent-2-ene
4.methyl cyclo butane.

You missed 2 methyl but-2-ene and dimethyl cyclopropane and 1,2 dimethyl cyclopropane :)

Both answers are incomplete. Don't forget this is just an 12th grade question.

u should not consider the unstable ones such as ethyl cyclo propane.....

[quote="hash_include"]i think there are 9 diff compounds for $ C_5H_{10}$ :

1. Cyclo pentane
2. methyl cyclo butane
3. ethyl cyclo propane
4. 1,2 dimethyl cyclopropane
5. pent-1-ene
6. pent-2-ene
7. 2 methyl but-1-ene
8. 3 methyl but-1-ene
9. 2 methyl but-2-ene

:)[/quote]

U missed 1,1 dimethyl cyclopropane  :huh:  :huh:

8 does have the molecular formula $ \ce{C5H10}$

Sorry was thinking abt cyclobutene  :oops:

The cyclo propane and its derivatives should be taken of from that list on account of their instability

PS-Carcul--did ya ask for the possible structures or stable structures????

[quote="Hash_include"]i think there are 9 diff compounds for $ \ce{C5H10}$:[/quote]

Incomplete answer!

[quote="Valeriummaximum"]u should not consider the unstable ones such as ethyl cyclo propane[/quote]

Why is that compound unstable?

[quote="Valeriummaximum"]The cyclo propane and its derivatives should be taken of from that list on account of their instability[/quote]

Again, why are cyclopropanes unstable? Aren't you making some confusion?

i think the qsn was quite clear. its ALL possible compounds, so syclopropane derivatives are included too.

and @rohit: yes, i 4got that. thx :)

Is it 10 compds incl 1,1 dimethyl cyclopropane?  :maybe:

All possible compounds with molecular formula $ \ce{C5H10}$ are stable, so let's not worry about that.

[quote="Madness"]Is it 10 compds incl 1,1 dimethyl cyclopropane?[/quote]

No, that's not correct.

I don't think I am giving the answer. People like you who are solving much harder problems like organic synthesis problems and other math/physics problems should be able to answer this with no problem.

just a general doubt to carcul:
isn't cycloproane unstable because of large angle strain (of about 50 degrees) of the C-C-C bond?

I think that there needs one more addition....

The cpompopund pent-2-ene shoes geometrical isomerism....
So i thnk i should consider the cis and the trans as separate compounds.......

[quote="Hash_include"]isn't cycloproane unstable because of large angle strain (of about 50 degrees) of the C-C-C bond?[/quote]

1) What is your definition of "unstable"?

2) Unstable compared to what?

3) To overcome the angle strain problem, the carbons are not sp3 but adopt an hybridization close to sp2.

3) Other compounds with three membered rings (like cyclopropenes, aziridines, and azirines) and which should have more strain are extensively used in synthesis.

carcul wrote:
[quote]3) To overcome the angle strain problem, the carbons are not sp3 but adopt an hybridization close to sp2.[/quote]

can u elucidate on this????

[quote="Carcul"]

1) What is your definition of "unstable"?
[/quote]

well, unstable should mean highly reactive even under standard conditions of pressure and temperature  :maybe:

[quote]
2) Unstable compared to what?
[/quote]

compared to higher cyclo compuounds i.e. cyclo butane, cyclopentane, etc.

[quote]
3) To overcome the angle strain problem, the carbons are not sp3 but adopt an hybridization close to sp2.

3) Other compounds with three membered rings (like cyclopropenes, aziridines, and azirines) and which should have more strain are extensively used in synthesis.[/quote]

hybridization close to sp2 as in??
delocalised pi electrons?!
 :|

For $ \ce{C5H10}$:
1. cyclopentane
2. methyl cyclobutane
3. 1,1-dimethyl cyclopropane
4. cis-1,2-dimethyl cyclopropane
5. trans-1,2-dimethyl cyclopropane
6. 1-pentene
7. (2Z)2-pentene
8. (2E)2-pentene
9. 2-methyl but-1-ene
10. 2-methyl but-2-ene
11. 3-methyl but-1-ene
12. ethyl cyclopropane

[quote="Chemrock"]12. 2,2-dimethyl propene[/quote]

That compound does not exist...

Yes sorry I remembered now only that's why I came online to edit that :oops:

However, you are right on this: there are 12 compounds with molecular formula $ \ce{C5H10}$.

Is the 11th and 12th trans-1,2-dimethyl cyclopropane d-form and l-form by any chance.

i have already mentioned that in my earlier posts.chemrock.......

is ur list complete

For $ \ce{C5H8}$:

1. 1-pentyne
2. 2-pentyne
3. 3-methyl but-1-yne
4. 3,3-dimethyl cyclopropene
5. d-2,3-dimethyl cyclopropene
6. l-2,3-dimethyl cyclopropene
7. 1,2-dimethyl cyclopropene
8. 3-ethyl cyclopropene
9. 1-ethyl cyclopropene
10. 3-methyl cyclobutene
11. cyclopentene
12. Spiro[2,2]pentane
13. bicyclo[2,1,0]pentane
14. 2-methyl Bicyclo[1,1,0]butane
15. 1-methyl cyclobutene
16. 1-methene 2-methyl cyclopropane
17. methene cyclobutene
18. 1,2-pentadiene
19. d-2,3-pentadiene
20. l-2,3-pentadiene
21. 1,3-pentadiene
22. 1,4-pentadiene
23. 2-methyl 1,3-butadiene
24. 3-methyl 1,2-butadiene

I wrote only for $ \ce{C5H8}$ madness. :roll:  :roll:

[quote="Chemrock"]16. 1-methene 2-methyl cyclopropane
17. methene cyclobutene[/quote]

It should be

16. 2-Methylmethylenecyclopropane
17. Methylenecyclobutane

Also, your list is incomplete.

25. Bicyclo[1,1,1]pentane

Also probably all the dienes can exist as cis and trans :maybe:  :maybe:

Still incomplete. As an hint, I could find 30 different compounds.


CEMC
geometry
geometric transformation
articles

In the page 8 of 2008 Euclid's Results, it says 14561 Canadian students registered for Euclid. Among them, 9336 from ON, 3327 from BC, so, the two provinces make up 87% of total registered students. I wonder why the participation numbers from other provinces are so low, especially QC which has a huge population......

[quote="ddziabenko"]I am just wondering, is the Euclid translated into French just like the APMO and CMO? If not, there goes your low participation. I have always been surprised by the low representation of Quebec in Canadian Math Competitions (in general). I don't really know why that is though.   :?:[/quote]
It should be also available in French.

I could think of two possibilities for the low representation of Quebecois:
1) Probably QC has its own math contests, and waterloo contests are not very popular there.
2) Students in QC don't like math as much as ON's or BC's. Probably they are more into liberal arts.

But how about other provinces? SK, Manitoba, and atlantic's?
Any ideas?

[quote="cnola"]2) Students in QC don't like math as much as ON's or BC's. Probably they are more into liberal arts.[/quote]

I think that's absurd, even if you replace "liberal arts" with "humanities"--for mathematics is indeed part of the liberal arts. I'm certain that your first reason is the true cause. While the CEMC does offer all of their contests in both English and French, for some reason their contests have never been popular in Quebec. Both sides are at fault for this. I have never lived in Quebec, so I don't know for sure why they haven't embraced the Waterloo contests, though I'm sure we can make good guesses. But on the other hand, the CEMC doesn't make much of an effort on its part. At the marking of the Canadian Open I don't think there were any native Francophones. The French papers were marked by Anglophones who felt confident enough in French to get enough of an idea of what the student was saying to assign a fair mark. While this might get the CEMC through their marking sessions, it makes it difficult to facilitate any two-way communication. I reckon that the CEMC has to jump through some translation hoops in order to respond to Francophone communications. To be fair, the CEMC will claim that they are fully bilingual and are able to respond competently in either language. I'm sure that this is true, but to engage Francophone Quebec the CEMC has to be more proactive. One thing worth noting is that the CEMC is run out of the University of Waterloo, and part of its (unofficial, of course) purpose is to encourage students who excel at mathematics to come to Waterloo. The CEMC would certainly find it more difficult to entice a Francophone Quebecer to come to Waterloo, to attend school in a foreign language and without the Quebec tuition rates.

All this said, I can't fault the CEMC. They have no obligation to run contests in French at all. It's not at all an obvious conclusion that Francophone students should participate in these contests or that the CEMC should spend lots of money to encourage them to. However, the CMS does have a duty to be fully accessible in its contests in both languages, since it is responsible for organizing the IMO team. For its part, the CMS made a laudatory stride in deciding to hold its National Camp, for Grades 9-10 students, at John Abbot College in Quebec. The first year the camp was held at John Abbot, in 2005, when I attended, 6 out of the 26 invited students were from Quebec. Out of these 6, only one attended, and he left after the first day. The camp was ostensibly to be held in both English and French, but since there no Francophone students were there past the first day, it was held only in English. I wondered how the organizers had planned to teach in French as well, considering that I had trouble communicating with some of my peers even in English. I hope the organizers had better luck in future years. However, I suspect not. I think that a lot of interest in math camps comes from word of mouth, from students who attended in previous years. Since so few Francophone students have ever attended these camps (including the Winter Camp and Waterloo Seminar), this effect is completely lost in Quebec.

Crux is also worth briefly mentioning. While most of the articles are in English, the problems are at least always in both languages. However, I think the CMS fails in its running of the correspondence program Olymon. On Dr. Barbeau's site, the last Olymon to be translated into French was the October 2006 set. Even when the sets are posted in French, I doubt Dr. Barbeau gets many responses, if any, in that language.

Who is losing out here? Perhaps the average Quebec student is not; that depends on Quebec's education system and their provincial system of math contests, if they have any. However, I suspect that many students who excel at mathematics are losing out on the chance to go to these camps and write olympiads, because they never get involved in the Waterloo system in the first place. Conversely the CMS is losing out on a potential rich source of olympians. I think the high numbers of Ontarians and British Columbians writing and succeeding at math contests is explained, not only by their high populations, but also by their immigrant centres of Vancouver and Toronto. (And as Calgary continues to grow in its economic importance, and hence attract more immigrants, watch it produce more competitive math students, like Jarno Sun and Danny Shi.) If this is any predictor, then we could be seeing some olympiad medallists coming out of Montreal, if Montrealers will participate more.

I think that the math contests in Quebec are called the AMQ contests (Association Mathematique du Quebec?) :|


quadratics
algebra
quadratic formula
special factorizations

how do you solve by completing the square?

r2 + 4r - 7 = 0

thanks for all help.

The quadratic equation is not what he's asking for...

[hide]You have:

$r^{2}+4r-7 = 0$.
Take the 7 to the other side:
$r^{2}+4r = 7$
Now, take the coefficient of the $r$ term, divide it by 2, square it, and add it to both sides:
$r^{2}+4r+4 = 7+4$
Which becomes
$(r+2)^{2}= 11$
Or $(r+2)^{2}-11 = 0$

So $(r+2) =$ [u]+[/u]$\sqrt{11}$
So $r =-2$ [u]+[/u]$\sqrt{11}$[/hide]

I do not understand the quadratic equation, so how do I use it to solve this problem without completing the square?

[hide="Quadratic Formula"]Given a quadratic equation in the form $ax^{2}+bx+c=0$, the solutions are $\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

In your equation, $a=1, b=4, c=-7$. Plugging into to the quadratic formula gives $\frac{-4\pm\sqrt{4^{2}-(4*1*-7)}}{2(1)}$ which, when simplified, gives that $r=-2\pm\sqrt{11}$[/hide]

[quote="now a ranger"]you can still solve it by quadratic equation, and then complete the square. The quadratic equation works every time...[/quote]

solve by quadratic and then complete the square? that makes no sense whatsoever because the quadratic and completing the square methods should yield the same results

Sure maybe one may be easier to you, but Im sure he was told to solve by completeing the square for a reason.

Whats more, the quadratic equation itself is derived from completing the square, so knowing how to complete the square is a valuable skill.


geometry
3D geometry
pyramid
ratio
trigonometry
rectangle

A square pyramid with base $ABCD$ and vertex $E$ has eight edges of length $4$. A plane passes through the midpoints of $\overline{AE}$, $\overline{BC}$, and $\overline{CD}$. The plane's intersection with the pyramid has an area that can be expressed as $\sqrt{p}$. Find $p$.

let those midpoints be P, Q, R, the midpoint of PQ connected to the midpoint of PR is parallel to the base, and through the plane PQR, and through the diagonal, so it intersects the lateral edges, this is a height of 1/4 [a midpoint of a midpoint wrt height]...anyway, consider the projection of the plane on ABCD, this is easy to find using side ratios to be something...anyway, get the cosine of the angle between the plane ABCD is and PQR using the dot product, it is sqrt{4/5} , and you get $\sqrt{80}$

the plain cut this pyramid to the pentagon. it has four side which is 2, and the base is 2sqrt2.
so area of this pentagon is the answer. Divide the pentagon into one rectangle and equillateral triangle.
finally, we can get sqrt 80

sorry for the revive, but is the way korlch did it actually right? how do you get sqrt80? and is it even possible for the 4 other sidelengths to have length 2, because if so, then the plane wouldn't be straight right?

[quote=vanstraelen]Midpoint $P$ of $AE$, midpoint $Q$ of $BC$, midpoint $R$ of $CD$.
Point $U \in EB$, point $V \in DE$.

$PV=PU=\sqrt{7}$ and $VR=UQ =\sqrt{3}$, already known $QR=2\sqrt{2}$.
Area $PUQRV = \sqrt{80}$.[/quote]

How did you compete the length of PU and UQ?


In solving a problem such as this:

Two numbers are chosen between 0 and 2.  What is the probability that the sum of the numbers is between 2 and 3?

you can use geometric probability in 2 dimensions, one for each variable, and find the area of the desired region / the area of the entire region.

Moving on to:

Three numbers are chosen, each between 0 and 2.  What is the probability that the sum of the numbers is between 2 and 3?

you simply increase to 3 dimensions and find the volume of the desired region / volume of entire region.

But how do you go about solving these problems for more variables?  Specifically-

Four numbers are chosen, each between 0 and 2.  What is the probability that the sum of the numbers is between 2 and 3?

Since I have no idea what the solution will look like, I had no idea which board this question was appropriate for.  Please move it where it belongs, if necessary.

Hey! This is actually a very interesting statistics question.

Well first I'm assuming that this is a uniform distribution.

So what happens is that you want to find the probability that the average is between certain values.

So you calculate the variance and standard deviation of one case, and use the fact that these are independent choices to find the probability.


Complete the square for the answer (It's what The Demon did, I believe)


parameterization
geometry
angle bisector
perpendicular bisector
geometry open

In triangle ABC  altitude AD,median AG ,angle bisector AL are given . How will you construct triangle from these three parameters?

1. Construct a line l passing through D which is perpendicular to AD.
2. Locate points L and G on l by the given lengths.
3. Construct a line n passing through G which is perpendicular to l. Let K be the intersection of n and AL.
4. Let O be the intersection of line n and the perpendicular bisector of AK.
5. Draw a circle with center O and radius AO.
6. B and C are the intersections of the circle and line l.


$\text{Find the range of the following three functions:}$

$(1): y=\frac{1+x}{\sqrt{1+x^2}}$

$(2): y=\sqrt{x^2+4}+\sqrt{x^2+2x+10}$

$(3): y=\sqrt{x^2-2x+5}-\sqrt{x^2+1}$

[hide="here is a hint"]
$|PQ|=\sqrt{(x_p-x_q)^2+(y_p-y_q)^2}$[/hide]

[hide="hint for 1"]
Think about this trig identities:
$\displaystyle \sin^2{x}+\cos^2{x}=1$
and its extention.
[/hide]


DO YOU THINK THAT MATHS HAS LOST ITS BEAUTY ( I AM NO ONE TO COMMENT) ?

Mathematics is timeless; it can't have something and then lose it.

That is why I also believe that disco will never die.

Math cannot be measured; Pi goes on forever, does it not? So, something that has infinite amount cannot lose its beauty, I reckon.

YOU'RE RIGHT. MATHS IS AS BEAUTIFUL AS EVER.

It's sad that many people perceive math as boring... One time, one of my classmates said "You're doing math on the bus!? OMG You're weird!!!!" when I was just working on generating functions...

My friends believe school stays at school, school shall not be brought home or on public places besides completing homework. Sadly, school involves math. You could have a very successful career if you did not believe that.

Also, please don't post in caps.

Broken caps lock, my friend?

[img]http://www.goodexperience.com/broken/i/04/02/capslock-s.jpg[/img]


In triangle $ABC$, $K$ and $L$ are points on the side $BC$ ($K$ being closer to $B$ than $L$) such that $BC \cdot KL = BK \cdot CL$ and $AL$ bisects $\angle KAC$. Show that $AL \perp AB.$

[hide]
$BC\cdot KL=BK\cdot CL\Longleftrightarrow\frac{BC}{BK}=\frac{CL}{KL}$

Since $AL$ bisects $\angle KAC$, $\frac{CL}{KL}=\frac{AC}{AK}$

$\therefore\frac{BC}{BK}=\frac{AC}{AK}$

This equation holds iff $AB$ bisects the exterior angle of $\angle KAC$.

Therefore, $\angle BAK+\angle KAL=\angle BAL=\frac{1}{2}(180^\circ)=90^\circ$, so $AL\perp AB$.[/hide]

Let $\angle BAK = \gamma, \angle KAL = \angle LAC = \alpha, \angle ACB = \beta$.
We know $BC \cdot KL = BK \cdot CL$

By [color=#960000]Sine Rule[/color],
$\frac{KL}{sin \alpha}=\frac{AK}{sin(\alpha+\beta)}$ and $\frac{CL}{sin \alpha}=\frac{AC}{sin(\alpha+\beta)}$
Again, $\frac{BK}{sin \gamma}=\frac{AK}{sin(2\alpha+\beta+\gamma)}$ and $\frac{BC}{sin 2\alpha+\gamma}=\frac{AC}{sin(2\alpha+\beta+\gamma)}$

Equating these, we get $\frac{KL}{CL}=\frac{AK}{AC}=\frac{\frac{BK}{sin \gamma}}{\frac{BC}{sin (2\alpha+\gamma)}}$

Note that we know $BC \cdot KL = BK \cdot CL \implies \gamma = 180-2\alpha-\gamma$ as $\gamma \neq 2\alpha+\gamma$.
Thus we get $\gamma=90^o-\alpha$
$\angle BAL = \gamma + \alpha = (90^o - \alpha)+\alpha = 90^o \implies AL \perp AB$.

Done :D

$\mathcal{H}(B,L;K,C)$ and $AL$ bisects $\angle KAC\implies AL\perp AB$

$BC.KL=BK.CL\implies\frac{BC}{BK}=\frac{CL}{KL}=\frac{AC}{AK}\implies\frac{AC}{BC}=\frac{AK}{BK}$. Let $Z$ be a point on $AB$ such thtat $KZ$ is the angular bisector of $\angle AKB$. So, $\frac{AK}{BK}=\frac{AZ}{BZ}=\frac{AC}{BC}$. Therefore $CZ$ is the angular bisector of $\angle ACB$. Now let $CZ\cap AL=M$. Let $\angle KAL=\angle LAC=\theta$. Let $\angle ACZ=\angle ZCB=\alpha$. Let $\angle AKZ=\angle ZKB=\beta$. Now $\angle KMC=90^\circ+\theta\implies\angle ZMK=90^\circ-\theta$. Now, $2\theta+2\alpha=2\beta\implies\theta+\alpha=\beta\implies \angle LAC+\angle ACZ=\angle AMZ=\angle AKZ=\beta$. Therefore $AMKZ$ is a cyclic quadrilateral. So, $\angle BAK=90^\circ-\theta\implies \angle BAL=90^\circ$. So, $AL\perp AB$.

Another Solution using Projective Geometry.

[b]Solution:[/b] Let $l$ be a line $\perp$ to $AL$ passing through $l$ and let $l\cap AK=P$ and $l\cap AC=Q$ and let $AB\cap QP=O$

Now $$-1=(C,K;B,L)\overset{A}{=} (Q,P;O,L)$$.

Now notice that $LQ=LP$ as $\triangle PAL\cong\triangle QAL$. So, $\frac{OQ}{OP}=1\implies OP=OQ$. So, $O$ must be the point at infinity. So, $AB\|PQ\implies AL\perp AB$.

@below no. :no:

Is [i]Projective Geometry[/i] allowed in RMO


[quote="tronghieu"]Hint: $ gcd(2^n\minus{}1,3^n\minus{}1)\equal{}1$ with n>1 .[/quote]
It is not true, for example $ (2^4\minus{}1,3^4\minus{}1)\equal{}5$.
$ n\equal{}1$ is not solution.
If $ n>1$ - odd, then $ 3^n\minus{}1\equal{}2\mod 8$, but $ 2^n\minus{}1$ -odd, therefore $ (2^n\minus{}1)(3^n\minus{}1)\equal{}2\mod 4$ is not square.
If $ n$ even, then $ 3|2^n\minus{}1$. It give $ n\equal{}2^k3^la, (a,6)\equal{}1$ and must be $ k\ge 2$ - even, $ l\ge 1$ - odd. Therefore $ 12|n$.
Consider prime factors $ (2^{12}\minus{}1)(3^{12}\minus{}1)\equal{}2^4*3^2*5^2*7^2*13^2*73$
Therefore or $ 36|n$ or $ 36\not |n, 12*73|n$,...

the key of my solution is applying the theorem of Pell's equation.

if $ m\equal{}0$, then $ n\equal{}0$.we claim that there is no other solutions
Assume the opposite, let $ (n,m)$ be a solution and $ m$ is a positive integer.
if $ n$ is odd, then $ 2\not |(2^n\minus{}1)$,$ 2|(3^n\minus{}1)$ and $ 4\not |(3^n\minus{}1)$, so $ 2|(2^n\minus{}1)(3^n\minus{}1)$ and $ 4\not |(2^n\minus{}1)(3^n\minus{}1)$.that means $ 2|m^2$ and $ 4\not |m^2$, a contradiction.
So $ n$ is even
Since $ (2^n\minus{}1)(3^n\minus{}1)\equal{}m^2$, we can assume that $ 2^n\minus{}1\equal{}da^2$ and $ 3^n\minus{}1\equal{}db^2$ ($ d,a,b$ are positive integers, $ d$ is square-free)

Then consider the Pell's Equation $ x^2\minus{}dy^2\equal{}1$, $ (2^{\frac {n}{2}}, a)$ and $ (3^{\frac {n}{2}}, b)$ are both solutions of it.
According to the theorem of Pell's equation, assume the solution of $ x^2\minus{}dy^2\equal{}1$ is $ (x_k,y_k)$, then $ x_0\equal{}1$, $ x_1\equal{}t$($ t$ is a positive integer),and $ x_{k\plus{}2}\equal{}2tx_{k\plus{}1}\minus{}x_k$ for any non-negetive integer $ k$.
So we have that $ 2^{\frac {n}{2}}$ and $ 3^{\frac {n}{2}}$ are both terms in the sequence $ {x_k}$

if $ t$ is odd, then every term of $ {x_k}$ is odd, while $ 2^{\frac {n}{2}}$ is even, a contradiction.
so $ t$ is even. then consider the recurrent relation we find that $ x_k$ is even iff $ k$ is odd.

if $ 3|t\minus{}1$, then every term of $ {x_k}$ is congruent to $ 1$ ($ mod 3$), while $ 3^{\frac {n}{2}}$ is a multiple of $ 3$, a contradiction.
if $ 3|t\minus{}2$, then $ x_k$ is congruent to $ 1$ ($ mod 3$) when $ k$ is even and $ x_k$ is congruent to $ 2$ ($ mod 3$) when $ k$ is odd.But $ 3^{\frac {n}{2}}$ is a multiple of $ 3$, a contradiction.
so $ t$ is a multiple of $ 3$. then consider the recurrent relation we find that $ x_k$ is a multiple of $ 3$ iff $ k$ is odd.

But for any $ x_k$, $ x_k$ is a multiple of $ 3$ iff $ k$ is odd iff $ x_k$ is even, so $ x_k$ can not be a power of 3, a contradiction,too.
Q.E.D.

[i]A very good solution . Congrats[/i] :first:

Please help me.I don't understand that why we have the recurence:$x_{k+2}=2tx_{k+1}-x_k$.Thank you

[quote=linboll]the key of my solution is applying the theorem of Pell's equation.

if $ m\equal{}0$, then $ n\equal{}0$.we claim that there is no other solutions
Assume the opposite, let $ (n,m)$ be a solution and $ m$ is a positive integer.
if $ n$ is odd, then $ 2\not |(2^n\minus{}1)$,$ 2|(3^n\minus{}1)$ and $ 4\not |(3^n\minus{}1)$, so $ 2|(2^n\minus{}1)(3^n\minus{}1)$ and $ 4\not |(2^n\minus{}1)(3^n\minus{}1)$.that means $ 2|m^2$ and $ 4\not |m^2$, a contradiction.
So $ n$ is even
Since $ (2^n\minus{}1)(3^n\minus{}1)\equal{}m^2$, we can assume that $ 2^n\minus{}1\equal{}da^2$ and $ 3^n\minus{}1\equal{}db^2$ ($ d,a,b$ are positive integers, $ d$ is square-free)

But can you assume that these guys have the same factor D?

Then consider the Pell's Equation $ x^2\minus{}dy^2\equal{}1$, $ (2^{\frac {n}{2}}, a)$ and $ (3^{\frac {n}{2}}, b)$ are both solutions of it.
According to the theorem of Pell's equation, assume the solution of $ x^2\minus{}dy^2\equal{}1$ is $ (x_k,y_k)$, then $ x_0\equal{}1$, $ x_1\equal{}t$($ t$ is a positive integer),and $ x_{k\plus{}2}\equal{}2tx_{k\plus{}1}\minus{}x_k$ for any non-negetive integer $ k$.
So we have that $ 2^{\frac {n}{2}}$ and $ 3^{\frac {n}{2}}$ are both terms in the sequence $ {x_k}$

if $ t$ is odd, then every term of $ {x_k}$ is odd, while $ 2^{\frac {n}{2}}$ is even, a contradiction.
so $ t$ is even. then consider the recurrent relation we find that $ x_k$ is even iff $ k$ is odd.

if $ 3|t\minus{}1$, then every term of $ {x_k}$ is congruent to $ 1$ ($ mod 3$), while $ 3^{\frac {n}{2}}$ is a multiple of $ 3$, a contradiction.
if $ 3|t\minus{}2$, then $ x_k$ is congruent to $ 1$ ($ mod 3$) when $ k$ is even and $ x_k$ is congruent to $ 2$ ($ mod 3$) when $ k$ is odd.But $ 3^{\frac {n}{2}}$ is a multiple of $ 3$, a contradiction.
so $ t$ is a multiple of $ 3$. then consider the recurrent relation we find that $ x_k$ is a multiple of $ 3$ iff $ k$ is odd.

But for any $ x_k$, $ x_k$ is a multiple of $ 3$ iff $ k$ is odd iff $ x_k$ is even, so $ x_k$ can not be a power of 3, a contradiction,too.
Q.E.D.[/quote]

But can you assume that these guys have the same factor D?

[quote=v_Enhance]I have a short solution using only modular arithmetic and lifting the exponent. :)

Assume that $n > 0$ is a solution. First, note that $2 \mid n$, otherwise the exponent of $2$ in LHS in odd.  From this (by lifting) we deduce that $n$ must be divisible by $3$ as well.

Let $n=6k$. Then \[ (2^{6k}-1)(3^{6k}-1) \equiv (2^k-1)(16^k-1) \equiv (2^k-1)^2 (2^k+1)(4^k+1) \pmod{31} \] must be a quadratic residue. We claim this implies $5 \mid k$.  Indeed, for $k \equiv 1,2,3,4 \pmod5$ we get $3 \cdot 5$, $5 \cdot 17$, $5 \cdot 13$, $17 \cdot 9$ are non-quadratic residues mod $31$, since of the primes here, only $5$ is a quadratic residue.

So let $n = 10c$. Then we have that $\left( (2^{10})^c - 1 \right)\left( (3^{10})^c - 1 \right)$ is a perfect square. But now by lifting the exponent on the prime $11$ we get a contradiction, because $\nu_{11}(2^{10}-1) = 1$ and $\nu_{11}(3^{10}-1) = 2$, thus the exponent of $11$ in the original is $3+2\nu_{11}(c)$.[/quote]

How did you get the idea to take mod11?

Because 11 is 3-Wieferich, i.e. $11^2 \mid 3^{10}-1$. This fact is the key ingredient of the proof.

[quote=shobber]Find all non-negative integers $m$ and $n$, such that $(2^n-1) \cdot (3^n-1)=m^2$.[/quote]

clearly $n=1$ is not an answer. now taking $mode(3)$ gives $2|n$ so $16|3^n-1$ which gives $4|n$.
rewrite as
$$(2^{4k}-1)(3^{4k}-1)=(6^{2k}-1)^2-(3^{2k}-2^{2k})^2=x^2$$
so $m^2+n^2=6^{2k}-1$ now taking $mode(4)$ gives that there are no solutions.

[quote=Mr.C][quote=shobber]
so $m^2+n^2=6^{2k}-1$ now taking $mode(4)$ gives that there are no solutions.[/quote]
$m^2+n^2=(6^{2k}-1)^2$

This problem and similar ones were solved in this paper of L. Szalay (2000):
[url]https://www.researchgate.net/profile/Laszlo-Szalay-2/publication/264956499_On_the_diophantine_equation_2_n_-13_n_-1x_2/links/588f3f7845851567c9405de3/On-the-diophantine-equation-2-n-13-n-1x-2.pdf[/url]


[hide]This is equivalent to asking for the complement that there are no common birthdays ( much less computation).  The first birthday can be anything, the second can be any of 364, the third of 363, and so on.  We have $\frac{365\cdot364\cdot363\cdots336}{365\cdot365\cdot365\cdots365}$.  This can be written as $\frac{364!}{365^{29}\cdot335!}\approx .2937$.  The desired probability is the complement, $1-.2937 \approx \boxed{.7063}$.[/hide]

[quote="hotnoodle"]What is the probability that there is at least one common birthday (month/day) in a class of 30 students?[/quote]

[hide]
\[{p= 1-\frac{ \dbinom{365}{30}}{ 365^{30}}}\approx .706 \] [/hide]
[b]Follow-up Question:[/b]

How many people are there in a class that has $p$ as close to $\frac{1}{2}$ as possible?

[quote="p4fn2w"]Follow up question -- what if you include leap years?
[hide="heres a start..."]
From wikipedia, 97 years out of 400 are leap years, so February 29 should have a weighted value of $\frac{97*1}{400*365+97}= \frac{97}{146097}$.

All other days have a weighted value of $\frac{400*365}{400*365+97}= \frac{146000}{146097}$

Using the same approach as illcrowflu, the first birthday can be anything. The second birthday can be any regular day [b]or[/b] February 29. Likewise, the third, fourth, fifth, and so on students can have the birthday on any regular day or February 29.

Computing this goes something like:
$\frac{146000 * 146000 * . . .}{146097 * 146097 * . . .}+\frac{146000 * 97 * . . .}{146097 * 146097 * . . .}+\frac{97 * 146000 * . . .}{146097 * 146097 * . . .}+\frac{97 * 97 * . . .}{146097 * 146097 * . . .}+. . .$

which can be expressed as:
$\sum_{i=0}^{30}\frac{(146000^{i})(97^{30-i})}{146097^{30}}$

But this must be wrong because it seems like this is greater than 1... Just $\frac{146000^{30}}{146097^{30}}$ alone is about 0.98... Which means that the complement of this is negative, so that cant be right.

What did I do wrong?[/hide][/quote]

[hide="Your summation comes to"]0.980924009858.

Apparently, the fact that you replace a 146000 with a 97 really brings down the number quickly.  So it's still less than 1.[/hide]

It also can't be right: including leap years shouldn't make the probability jump 20 percentage points.

Let's avoid ugly numbers until the end.  Let $p$ be the probability that an individual has a birthday on February 29 and $q$ be the probability that an individual has a birthday on each other day, so $365q+p = 1$.  The probability that no two people have the same birthday and no one has a birthday on February 29 is $(1-p)(1-p-q)(1-p-2q) \cdots (1-p-29q)$.  The probability that exactly one person has a birthday on February 29 and no two people have the same birthday is $30p(1-p)(1-p-q) \cdots (1-p-28q)$.  Summing these gives the complement of our desired probability, which is a really ugly rational number approximately equal to $0.29412338596341586$, so the number we want is approximately $0.7058766140365842$.  In other words, February 29 doesn't matter very much.


I am a contractor and oftem deal with final contract numbers that have a predetermined % markup on them.
This may be a really silly question but I am puzzled on how to figure this out with a calculator in hand.

I am trying to determine the cost of the labor and material before they are marked up with 15%. If I know that the final contract dollar amount is $\$$ 4,800.00, and there is a 15% markup on this number, what was the cost of the labor and materials?
Is there an easy way to figure this out with a caculator?
Every time I take (4800-15%) I get 4080. Than when I take (4080+15%) I get 4692. I just don't get it.

Thanks for taking a look at this for me. I apologize if this is a really silly question.

John

A simpler example, and a full explanation:

Suppose I have something that costs 100 dollars after a 10% markup.  You want to say that 100-10% = 90, which is true -- 100 dollars, marked down 10%, is 90 dollars.  However, it's clear that 90 dollars, marked up 10%, is not 100 dollars (it's only 99).  Why is this?  It's because the percentages are relative: when marking down 10%, you're marking down a percentage of a larger number, so it has a different effect.

For your problem:  Denote by $x$ the quantity you are trying to find -- the pre-markup price.  Then $4800 = x + x\cdot 0.15$.  Whatever the original price is, we must add 15% of it in order to get the final price.  So $4800 = x\cdot 1.15$ and you must divide 4800 by 1.15 to get $x = 4173.91$ (approximately), the pre-markup price.

JWach,you don't have to post the same quwestion in many forums ;)

JWach, I believe it is sufficient to post your question in getting started or intermediate topics; not getting started, intermediate topics and preolympiad.  ;)  ;)


function
quadratics
calculus
derivative
calculus computations

Find the values of $ a$ and $ b$ so that the function $ f(x)\equal{}x^2\plus{}ax\plus{}b$ has a local minimum at the point $ (6,\minus{}5)$.

a=-12, b=31
Differentiate, or use the fact that for a quadratic, the local extrema occurs mid way between the roots. Use the fact that (6,-5) is a point on the curve.

ya the ans is a= -12 and b = 31. find the first derivative equate to zero.we get x= -a /2 .equate -a/2 to 6.thus a= -12. substitute this in the equation and find b.


What is the largest factor of 54,321 that is less than 54,321?

[hide] The smallest factor of $ 54321$ other than $ 1$ is $ 3$. Thus, $ \frac{54321}{3}\equal{}\boxed{18107}$.[/hide]

[hide]We can't divide by one, and it is odd, so we can't divide by two.  We then divide  by three.  This works and we get an answer of [b]18107[/b][/hide]


Determine value of expression $ A\equal{}\sqrt{x(4\minus{}y)(4\minus{}z)}\plus{}\sqrt{y(4\minus{}z)(4\minus{}x)}\plus{}\sqrt{z(4\minus{}x)(4\minus{}y)}\minus{}\sqrt{xyz}$ where $ x,y,z$ are positive numbers satifying $ x\plus{}y\plus{}z\plus{}\sqrt{xyz}\equal{}4$

$ (4 \minus{} y)(4 \minus{} z) \equal{} 16 \plus{} 4( \minus{} y \minus{} z) \plus{} yz \equal{} 16 \plus{} 4(x \plus{} \sqrt {xyz} \minus{} 4) \plus{} yz$
$ \equal{} 4x \plus{} 4\sqrt {xyz} \plus{} yz \equal{} (2\sqrt {x} \plus{} \sqrt {yz})^2$
$ \sqrt {x(4 \minus{} y)(4 \minus{} z)} \equal{} 2x \plus{} \sqrt {xyz}$
$ A \equal{} (2x \plus{} \sqrt {xyz}) \plus{} (2y \plus{} \sqrt {xyz}) \plus{} (2z \plus{} \sqrt {xyz}) \minus{} \sqrt {xyz} \equal{} 2(x \plus{} y \plus{} z \plus{} \sqrt {xyz}) \equal{} \boxed{8}$


probability
trigonometry
ARML
Duke
college
geometry
3D geometry

This is my worst topic.. I don't know about you guys but I'm absolutely horrible at counting and/or probability. So, I'm going to do many problems on this topic by breaking it down into pieces.  The first piece is problems involving dice (or die).  Because probability questions vary so much, it's hard to have one big thread like in trigonometry so not all threads will have a lot of problems but anyway, enjoy!  :)

1. (AMC 12B 2002 #16) Juan rolls a fair regular octahedral die marked with the numbers 1 through 8. The Amal rolls a fair six-sided die.  What is the probability that the product of two rolls is a multiple of 3?
A) 1/12
B) 1/3
C) 1/2
D) 7/12
E) 2/3

2. (AHSME 1994 #30) When $ n$ standard 6-sided dice are rolled, the probability of obtaining of a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of $ S$. The smallest possible value of $ S$ is
A) 333
B) 335
C) 337
D) 339
E) 341

3. (ARML 1993 T7) The probability that 11 consecutive throws of a pair of ordinary fair die will each have a different total can be written in the form $ \displaystyle \frac{(a!)(b!)(c!)}{36^d}$ where $ a,b,c,$ and $ d$ are integers and $ a \leq b \leq c \leq d.$ Compute the ordered quadruple (a,b,c,d).

4. (Duke Math Meet 2002 T3) Paul rolls a standard 6-sided die, and records the results.  What is the probability that he rolls a 1 ten times before he rolls a 6 twice?

5. (ARML 1986 T6) [Note: An "icosahedral die" is shaped like a regular icosahedron (20 faces), with each face being a different integer from 1 through 20 inclusive. When it is thrown, one face will be uppermost, this is called the "top face."]
If a fair pair of icosahedral die are thrown, compute the probability that the sum of the numbers on the top faces will be less than 20.

first one is 1/2 using conditional probability

the duke one is 3/512 i think try 10 1s and 0 6 and 10 1s and 1 6.

the third one is more algebra then counting since you cant havea 36^11 in the denominator...I got (6!5!12!)/(36^11) because each has to be one of (2,3,...12) but that wont work since 12>11.

[quote="firecricket91"]first one is 1/2 using conditional probability
[/quote]

Got the same answer using PIE.

[quote="Silverfalcon"]This is my worst topic.. I don't know about you guys but I'm absolutely horrible at counting and/or probability. So, I'm going to do many problems on this topic by breaking it down into pieces.  The first piece is problems involving dice (or die).  Because probability questions vary so much, it's hard to have one big thread like in trigonometry so not all threads will have a lot of problems but anyway, enjoy!  :)
[hide="The Problems"]
1. (AMC 12B 2002 #16) Juan rolls a fair regular octahedral die marked with the numbers 1 through 8. The Amal rolls a fair six-sided die.  What is the probability that the product of two rolls is a multiple of 3?
A) 1/12
B) 1/3
C) 1/2
D) 7/12
E) 2/3

2. (AHSME 1994 #30) When $ n$ standard 6-sided dice are rolled, the probability of obtaining of a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of $ S$. The smallest possible value of $ S$ is
A) 333
B) 335
C) 337
D) 339
E) 341

3. (ARML 1993 T7) The probability that 11 consecutive throws of a pair of ordinary fair die will each have a different total can be written in the form $ \displaystyle \frac {(a!)(b!)(c!)}{36^d}$ where $ a,b,c,$ and $ d$ are integers and $ a \leq b \leq c \leq d.$ Compute the ordered quadruple (a,b,c,d).

4. (Duke Math Meet 2002 T3) Paul rolls a standard 6-sided die, and records the results.  What is the probability that he rolls a 1 ten times before he rolls a 6 twice?

5. (ARML 1986 T6) [Note: An "icosahedral die" is shaped like a regular icosahedron (20 faces), with each face being a different integer from 1 through 20 inclusive. When it is thrown, one face will be uppermost, this is called the "top face."]
If a fair pair of icosahedral die are thrown, compute the probability that the sum of the numbers on the top faces will be less than 20.[/hide][/quote]

[hide="1"]
The probability that Amal and Juan do [u]not[/u] throw multiples of $ 3$ are $ 1\minus{}\frac{2}{6} \equal{} \frac 23$ and $ 1\minus{}\frac{2}{8}\equal{}\frac 34$, respectively. Thus the probability that both do not throw multiples of $ 3$ is $ \frac{2}{3} \cdot \frac {3}{4} \equal{} \frac 12$, and the probability that at least one does is $ 1 \minus{} \frac 12 \equal{} \frac 12 \Rightarrow \mathrm{(C)}$.
[/hide]

[hide="2"]
I don’t have time to write up a proof now, but it seems apparent that we have to use symmetry. The smallest divisor of $ 6$ greater than $ 1994$ is $ 1998$, which can be obtained by rolling a minimum of $ 333$ dice. Thus by symmetry, the probability that sums of $ 333 \minus{} 1998$ are equal, and so on and we have that $ 337 \minus{} 1994$ are equal. Seems a bit simple though for a #30..
[/hide]

[hide="3"]
There are a total of $ 11$ possible sums for the dice. There are $ 11!$ different ways to arrange these possible sums, out of a total of $ (6 \times 6)^{11}$ possible rolls. Also, it is possible to get each sum in one or more different ways. Indeed, there is $ 1$ way to roll a 2, $ 2$ ways to roll a 3, and so forth until $ 6$ ways to roll a 7, $ 5$ ways to roll a 8, and so on.

Thus the total probability is $ \frac{11! \cdot 5! \cdot 6!}{36^{11}}$, and the answer is $ (5,6,11,11)$.
[/hide]

[hide="4"]
We ignore all other numbers besides 1 and 6 because they do not affect our data.

Depending upon how we interpret this, there are different answers (if Paul must throw 1 exactly $ 10$ times, or if Paul must throw 1 at least $ 10$ times). Either way, consider a string of the first 12 throws Paul makes which are either 1s or 6s. If Paul has to throw 1 exactly $ 10$ times, then the last throw must be a 6, and the other 6 may go in $ 11$ different slots. Hence the probability is $ \frac{11}{2^{12}} \equal{} \frac{11}{4096}$.

In the second scenario, we need the first $ 11$ characters of the string to have at least $ 10$ 1s, and the $ 6$ can go in $ 11$ different places or not at all appear. Then the probability is $ \frac{12}{2^{11}} \equal{} \frac{3}{512}$.

[/hide]

[hide="5"]
If the first dice rolls a $ 20$ or a $ 19$, then there are no possibilities. If it rolls a $ 18$, then there is one possibility for the second die; if it rolls a $ 17$, there are two possibilities for the second die, and so forth until for when it rolls a $ 1$, there are $ 18$ possibilities for the second die. Hence the probability is $ \frac{1\plus{}2\plus{}3\plus{}\cdots\plus{}18}{20^2} \equal{} \frac{171}{400}$.
[/hide]

oh wait it is 11!*6!*5!..wow im stupid i thought there were 12 for some reason..

I'm not good at probability and counting, and I am absolutely horrible with combinatorics and sequences, but here's some advice. I used to suck way more than I do now which was really scary. I would use the most retarded time consuming approaches.
Then one day we were doing mandelbrot and my friend had this insanely simple way to solve a counting problem. I actually got more pro at counting just by seeing it. Try looking around at some challenging problems, and if you can't get it read the solution, and try to use a synthesis of that method when solving future problems ;)

#2 should be 337.

You want to minimize n and then after that substitute as many 1s as possible to minimize S

so 1994 is 332 6s and 1 2s to minimize n as 333.

However, with 333 tosses there are a lot more ways to get 1994 also.

we can have 331 6s, 1 5s, 1 3s
or 331 6s, 2 4s
or 330 6s, 2 5s, 1 4s
or 329 6s, 4 5s

subtract another 6 we wont have enough to make a sum of 1994 so 5 cases

Then we want to substitute the 1s as 6s.  However, by doing that you cannot say we want from the first case 332 1s and 1 2s to minimize S.  Because if you do that the sum is 334 and with a sum of 334 theres no way 331x+1y+1z (second case) where they are integers form (1,2,3..6) can equall 334--x,y,z are distinct or if we have like x=y, it will be 332x+z and that changes the cases.  So we hvae to play around with it a bit to make it 332 1s and 15s to make it 337.  By doing that there is a unique solution for the second case (331 1s, 1 2s, 1 4s).  There is a unique solution for the third case (331 1s, 4 2s)..and so forth you will find there is a unique solution for all 5 cases so its 337.

[hide="1"]i drew a bipartite graph.

so the sample space is 48. and so you can have 3 and 6 for 1,2,4,5,7,8 and you can have 1-6 for 3 and 6. So you get $ \frac{12\plus{}12}{48}$, so that is $ \frac{24}{48}$ = $ \frac{1}{2}$.[/hide]

[hide="Sketch for Number 1"]
You could've just used $ P(A) \equal{} 1 \minus{} P(A')$. [/hide]

Number 2 is indeed the answers you guys had.

Same for Number 3.

[hide="Sketch for Number 4"]
Answer is indeed 3/512. Just consider the case of 11 numbers since this allows us to restrict cases for either at least 10 1's or at least 1 6's but NOT BOTH (note that this IS possible for 12 and bigger).

So, from here, it is just straightforward. [/hide]

Well done for Number 5 as well.

I guess this set of 5 numbers was too easy for you guys!  :roll: (or just one of those days I can't count...)

[quote="Zellex"][quote="miyukina"]
q2
1994 = 332*6 + 1*2
332*(7-6) + 1*(7-2)
= 332 + 5
= 337, C
[/quote]

can you elaborate?[/quote]

I think basically, he first minimized the way to form 1994. Then, since opposite faces of cubes add up to 7 (in standard sense), he found that 1's opposite of 6 and 5's opposite of 2 to get 1(332) + 5(1) = 337.

I could only do 1 and 5, and got 1/2 and 171/400, respectively


limit
integration
calculus
derivative
calculus computations

Calculate

$\lim_{n \to \infty} \sum_{m=1}^n (m+1)(m+2)r^m$　　$(0<r<1)$

Write this as

$S_n(x) = \sum_{m=0}^n (m+1)(m+2)x^m$.

This is even a finite sum, so there should be no concern about integrating inside the sum.

$\int_0^x\int_0^yS_n(t)dtdy = \sum_{m=0}^n x^{m+2} = \frac{x^2 - x^{n+3}}{1 - x}$

So, taking two derivatives with respect to $x$:

$\frac{(2x - (n+3)x^{n+2})(1-x) - (-1)(x^2 - x^{n+3})}{(1-x)^2}$

$\frac{(2-2x - (n+3)(n+2)x^{n+1} + (n+2)(n+3)x^{n+2})(1-x)^2 - (-2(1-x))(2x - x^2 - (n+3)x^{n+2} + (n+2)x^{n+3})}{(1-x)^4}$

This is $S_n(x)$, so we now take $x = r$ with $|r| < 1$, and let $n\to \infty$ so that all the terms with $n$ in the exponent go to 0.

$S(r) = \frac{(2-2r)(1-r)^2 + 2(1-r)(2r-r^2)}{(1-r)^4} = \frac{2(1-r)^2 + 2(2r-r^2)}{(1-r)^3} = \frac{2 - 4r + 2r^2 + 4r - 2r^2}{(1-r)^3}$

$S(r) = \frac{2}{(1-r)^3}$

I hope I made no mistake in the algebra... ;)

Does the series start at $m = 1$ or $m = 0$?

It does start at $m=1$, sorry. But that's not too much of a problem; we just have to subtract the erroneous $m=0$ term, which is 2.

The answer is $\frac{2r(r^2-3r+3)}{(1-r)^3}.$

You can solve with telescoping method as well.


prove

(1+1/n)^k <= 1+k/n + k^2/n^2

for all k<=n

n is natural and so is k

Let do an induction in $ k$:
For $ k\equal{}1$ is true.
Suppose that is true for $ k\equal{}j$

Then $ 1\plus{} \frac{k}{n} \plus{} \frac{k^2}{n^2}\geq(1\plus{}\frac{1}{n})^k$

So $ (1\plus{}\frac{k}{n}\plus{}\frac{k^2}{n^2})((1\plus{}\frac{1}{n})$

$ \geq(1\plus{}\frac{1}{n})^{k\plus{}1}$

And is true that $ 1\plus{} \frac{k\plus{}1}{n} \plus{} \frac{(k\plus{}1)^2}{n^2}$

$ \geq(1\plus{}\frac{k}{n}\plus{}\frac{k^2}{n^2})((1\plus{}\frac{1}{n})$

Then $ 1\plus{} \frac{k\plus{}1}{n} \plus{} \frac{(k\plus{}1)^2}{n^2}$

$ \geq(1\plus{}\frac{1}{n})^{k\plus{}1}$  as claimed


If $ n \cdot 1 \cdot \frac{1}{2} \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \frac{1}{5} \equal{} \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{1}{6} \cdot \frac{1}{8} \cdot \frac{1}{10}$, what is the value of $ n$? Express your answer as a common fraction.

Note how $ 1\equal{}2 \times \frac{1}{2}$, $ \frac{1}{2}\equal{}2 \times \frac{1}{4}$, etc.  Thus, the left side of the equation is $ 2^5\equal{}32$ times larger than the right without $ n$, so $ n\equal{}\frac{1}{32}$.


[quote="shyong"][quote="jelly119"][img]http://i5.photobucket.com/albums/y194/jelly119/05072401.jpg[/img]
Given the area of $\triangle{ABC}$=192.
BD:BC=1:4
BE:BA=1:3
Find the area of BDFE.[/quote]

First we see that the area of $\triangle ABD$=area of $\frac 14 \triangle ABC$=48
Then area of $BDFE=\frac 13 *\triangle ABD = 16$[/quote]

Actually, the second step doesn't work.  Think about it...

[quote="Bob"]Is there some kind of theorem that talks about the ratios AF:FD and EF:FC?[/quote]

You could use mass points to find that AF:FD = 8:3 and EF:FC = 2:9

[quote="riddler"]Is the answer $\frac{272}{11}$? :?[/quote]

I don't have the awnser... but I've got the same with you. :lol:

[quote="Bob"]Is there some kind of theorem that talks about the ratios AF:FD and EF:FC?[/quote]

Actually...yes!! Menelaus's Theorem is yelling at us!

Apply Menelaus's theorem to $\triangle{ABD}$ with respect to E,F, and C. We got
\[\frac{AF}{DF}\times\frac{DC}{BC}\times\frac{BE}{AE}=1\]
\[\Rightarrow\frac{AF}{DF}\times\frac34\times\frac12=1\]
\[\Rightarrow\frac{AF}{DF}=\frac83\]

[quote="mathfanatic"][hide][img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=176[/img]

Consider  $\triangle BCE$.  Its area is known as both $4X + Y$ and 192/3.

Consider  $\triangle ABD$.  Its area is known as both $X+3Y$ and 192/4.

Solve these two equations, and get $X = \frac{144}{11}, Y = \frac{128}{11}$.[/hide][/quote]

Cool...man! I've been thinking about a method like this ...neat and clear! Thanks! :coolspeak:

[quote="riddler"]so, Im wrong! ohhhhhh!
hey wait a minute the answer is same as mine because $X+Y=\frac{272}{11}$
Isnt it?
Or does $X$ and $Y$ stand for something else?[/quote]

You are correct . It is $\frac{272}{11}$ ,$X$ and $Y$ just divide for the area in $BDFE$ into $2$ parts  :P

okay then I am right, I also did the same thing. :P


linear algebra
matrix
function
linear algebra unsolved

Let $ A \equal{} [a_{i,j}]_{2009x2009}$ where $ a_{i,i} \equal{} a$ ($ a$ is equal to $ 1$ or $ \minus{} 1$) and $ a_{i,j} \equal{} b$ ($ b$ is equal to $ 2009$ or $ 0$).$ i$#$ j$
     Prove that:$ det(A)\neq 0$.

Reduce modulo 2 and you get $ \det A \neq 0$.

JC_math, since I am not very familiar with modular arithmetic, was does exactly mean in this context "reduce modulo 2"?

Actually you need to reduce modulo $ 7$ (or $ 41$,...).
Consider $ A\in \mathcal{M}_n(\mathbb{Z})$, and the natural morphism $ \varphi :  \mathcal{M}_n(\mathbb{Z}) \to \mathcal{M}_n(\mathbb{Z}/7\mathbb{Z})$, we have $ \det \varphi(A) \equal{} \pm 1$ because $ \varphi (A)$ is a diagonal matrix with diagonal entries equal to $ 1$ or $ \minus{}1$, hence $ \det A\neq 0$.

A couple of questions:

1) What does exactly mean on this context the word "morphism"?

2) What does exactly do your function $ \varphi$ to each matrix in $ M_n(\mathbb{Z})$?

[quote="Carcul"]1) What does exactly mean on this context the word "morphism"?
[/quote]
http://en.wikipedia.org/wiki/Morphism
[quote="Carcul"]
2) What does exactly do your function $ \varphi$ to each matrix in $ M_n(\mathbb{Z})$?[/quote]
It takes each integer entry modulo $ 7$.

One further question: why $ \det \varphi (A) \neq 0 \equal{}> \det(A) \neq 0$?


Several positive numbers each not exceeding 1 are written on a circle. Prove that one can divide the circle into three arcs so that the sums of the numbers on any two arcs differ by no more than 1.

[quote="grobber"]Positive [i]integers[/i] not exceeding $1$?[/quote]

Funny typo, huh?  :D

Edited.

I figured that was supposed to be the real problem, but still.. :) Anyway, thanks for fixing it so fast.

Consider the division of our circle in three arcs which achieves the smallest absolute difference between the arc $A$ with the largest sum and the arc $a$ with the smallest sum, and assume that this difference is $>1$. Now start moving the border point between the $A$ and $a$ into $A$, until one of the points in $A$ crosses over to $a$. Since we still have $A>a$ (short for "the sum of the points in $A$ is larger than that of the points in $a$"), there are three possibilities:

(1) $a,A$ are still the smallest, and, respectively, largest arcs, but this time $A-a$ is strictly lower than what it was before, contradicting the minimality property of the previous situation;

(2) $a,A$ are the smallest and middle arcs now. The difference between the other arc $b$ and $a$ is less than what it was before, and then it was at most $A-a$, so again, we have a contradiciton;

(3) $a,A$ are the middle and largest arcs respectively. This is analogous to (2).


Here you have the problems of the "Centro". The Olympiad was held in San Salvador, El Salvador, from June 19th to June 25th.

There are 12 countries participating in Centro: México, Guatemala, Honduras, El Salvador, Nicaragua, Costa Rica, Panamá, Venezuela, Colombia, Cuba, Puerto Rico and  Dominican Republic.

I must accept Centro has always been a very easy olympiad. In particular, this year problem 1 was disappointing...   :? I think the order or difficulty of the problems is (from the easiest one to the hardest one): 1, 2, 4, 5, 6, 3. The median score in problem 3 was zero :)

Problem 1: http://www.mathlinks.ro/Forum/viewtopic.php?t=46026

Problem 2: http://www.mathlinks.ro/Forum/viewtopic.php?t=46028

Problem 3: http://www.mathlinks.ro/Forum/viewtopic.php?p=291333#p291333

Problem 4: http://www.mathlinks.ro/Forum/viewtopic.php?p=291331#p291331

Problem 5: http://www.mathlinks.ro/Forum/viewtopic.php?t=46029

Problem 6: http://www.mathlinks.ro/Forum/viewtopic.php?p=291327#p291327

I hope you find the problems interesting  :P

mmmm let's see... If I don't miss, the bronze, silver and gold cut-offs were 14, 25 and 34, respectively. The gold medals were awarded to two mexicans and one colombian (I don't remember their exact names) :)


geometry
circumcircle
geometric sequence
geometry unsolved

[quote="Ahiles"][color=darkred]Let $ G$ be the centroid of $ ABC$ and $ R_1$, $ R_2$ and $ R_3$ be the circumradii of $ AGB$, $ BGC$ and $ CGA$, respectively. Prove that if $ R_1$, $ R_2$, $ R_3$ form a geometric progression then the circumradius of $ ABC$ equals one of the numbers $ R_1$, $ R_2$, $ R_3$.[/color][/quote]

It's not very nice, but still a solution.
Suppose $ R_1^2 \equal{} R_2\cdot R_3$.
It's obviuos that $ S(AGB) \equal{} S(BGC) \equal{} S(CGA) \equal{} \dfrac{1}{3}S(ABC)$. Then since $ R_1 \equal{} \dfrac{AG\cdot GB\cdot AB}{4S(AGB)}$, we know from the identity that:

$ m_a\cdot m_b\cdot c^2 \equal{} m_c^2\cdot ab \space (*)$

We will show that $ R_1 \equal{} R\iff 4m_am_b \equal{} 3ab$, so because of $ (*)$ it's enough to show that: $ 4m_c^2 \equal{} 3c^2\iff a^2 \plus{} b^2 \equal{} 2c^2$.

Let $ a^2 \equal{} x,b^2 \equal{} y$ and $ c^2 \equal{} z$.

$ m_a\cdot m_b\cdot c^2 \equal{} m_c^2\cdot ab\iff (2y \plus{} 2z \minus{} x)(2x \plus{} 2z \minus{} y)z^2 \equal{} (2x \plus{} 2y \minus{} z)^2xy$

$ \iff 2z^4 \plus{} z^3(x \plus{} y) \minus{} z^2(x \minus{} y)^2 \plus{} 2xyz(x \plus{} y) \minus{} 2xy(x \plus{} y)^2 \equal{} 0$

$ \iff (x \plus{} y \minus{} 2z)[ \minus{} z^3 \minus{} z^2(x \plus{} y) \minus{} 2xyz \minus{} 2xy(x \plus{} y)] \equal{} 0$

so $ 2z \equal{} x \plus{} y\iff 2c^2 \equal{} a^2 \plus{} b^2$. The conclusion follows.


logarithms
quadratics
algebra
quadratic formula

The values of $ a$ in the equation:  $ \log_{10}(a^2 \minus{} 15a) \equal{} 2$ are:

$ \textbf{(A)}\ \frac {15\pm\sqrt {233}}{2} \qquad\textbf{(B)}\ 20, \minus{} 5 \qquad\textbf{(C)}\ \frac {15 \pm \sqrt {305}}{2}$
$ \textbf{(D)}\ \pm20 \qquad\textbf{(E)}\ \text{none of these}$

[hide]

Expressing this exponentially, $ a^2\minus{}15a\equal{}100$. Rearranging and applying the quadratic formula gives:

$ a\equal{}\frac{\minus{}15\pm25}{2}\equal{}20, \minus{}5\Longrightarrow\textbf{ (B) }$

[/hide]


Nice.

LOL.

Dieting seems to be so much about carbohydrates and protein. It is, but I think it is just the calorie count that worries people.

I mean, people don't even know why proteins are important - it is so sad. Protein is essential for growth and development. It provides the body with energy, and is needed for the manufacture of hormones, antibodies, enzymes, and tissues. It also helps maintain the proper acid-alkali balance in the body. When protein is consumed, the body breaks it down into amino acids, the building blocks of all proteins. Some of the amino acids are designated nonessential. This does not mean that they are unnecessary, but rather that they do not have to come from the diet because they can be synthesized by the body from other amino acids. Other amino acids are considered essential, meaning that the body cannot synthesize them, and therefore must obtain them from the diet.

I've heard people say - I'm not eating protein because it makes me fat. That is so much bogus - protein does not make you fat, and without it you could not survive. When they eat vegetables, they think it makes them thin. Not so. Vegetables reduces your tissue growth in your muscles because some vital amino acids are missing from the diet. That is why you are thinner. You are not getting "fit" when you diet without precaution.

Atkins Diet  =  solution for people who are too lazy to eat healthily and exercise.

[quote="white_horse_king88"]When they eat vegetables, they think it makes them thin. Not so. Vegetables reduces your tissue growth in your muscles because some vital amino acids are missing from the diet. That is why you are thinner. You are not getting "fit" when you diet without precaution.[/quote]

I agree with everything you said, with the exception of this point.  The way you put it, vegetables sound like a filler food that doesn't do anything except make you less hungry.  But I disagree; vegetables have important vitamins and minerals.  They are a good source of fiber, which helps ``clean you out'', if you know what I mean.  It is true that vegetables do not make you thin, but they don't necessarily prevent you from getting amino acids.

[quote]Atkins Diet  =  solution for people who are too lazy to eat healthily and exercise.[/quote]

Indeed.

Um, so anyone actually [b]try [/b]this??? It actually do work, even better than Atkins. It kicks in a lot faster too :P

Hello,

I really think that the Atkins diet is unhealthy. I have lost 25 punds in two months (After I gain 15 in the hollidays!!), and there is no real secret about that, everything I have done is eat less and continue excercising.

In January 2004 I said that it was crazy that with 23 years I had 100 pounds of overweigh. I started doing excercise regularly (started with 30 minutes three times a week, and by July it was 1:30 to 2 hours, 3 to 5 times a week, and it stayed there for the rest of the year, because I didn't have much time left) Now, it was great, I was doing excercise, but I was kraving al the time and I couldn't help it, I ate all the time, and often more than what I should. Then, OK, I wasn't gaining more weight for what I ate, but I wasn't loosing it either.

This January (Yes, the new year own proposals) I said that I was already doing excercise, and that I was not going to stop, but I needed to change my eating habits, and that is what I did. I started measuring how many calories I ate during two weeks, and not surprisingly, I was eating more than 3000 calories a day!!!, then, I started by trying one week 2500 a day, the next 2000, and now I am between 1500 to 1800 most of the day (some days I can't help it and eat a little more, but not that much, and not more than once a week). It is true, the first three weeks, even the first five that you do this, you are starving most of the time, because it is not that your body needs it, but it is use to it, then it miss it. Now, I still have 80 pounds to loose, but I believe that for the end of this year I will be at least 60 pounds thinner than right now.

My advise for people dieting:

1 - Close your mouth a little. Eat less, but eat responsably, don't starve yourself to dead. You can research by yourself how to eat properly, there are a bunch of good sites out there, but if you don't have the time to do it (or are too lazy to do it by yourself) go to a nutricionist, in most of the countries I have been, I know they are affordable most of the time.

2 - Excercise. If you are overweight, chances are that a big percentage of that overweight is fat, and fat is energy accumulated, therefore, you have to get rid of that energy. Excercise is the best way. Also, it helps to all your body to be in good shape. If you don't know how to do excercise, look for someone who knows, you can hurt yourself if you don't know how to do something!

3 - Spect (and be happy) with small rewards. Half pound a week is good if it is a constant rate. One pound is great, two pounds even better, more than that is [b]VERY[/b] hard to get, and specially, to mantain. Also, I have always hear that when you are in the last 20 pounds of overweight, it start being a little slower, may be a doctor or nutricionist could say if this is just psicologic or there is a medical reason for that.

4 - Go Slow. Realize that you are going to spend a good time loosing weight, specially if you want to keep it off. Then, if you do it by your own, start measuring your dialy income of calories for a good time, and start substracting 100 calories a week (if you do 500, as I did, it turns ugly). If you haven't done excercise in a good amount of time, do it slowly. I started walking (not running, I wasn't even capable of that) for thirty minutes, and a couple of weeks after that, I added 5 minutes of cardio, then 10, then I was walking 20 minutes, running 5 and doing 20 minutes of cardio, and after a time I wasn't walking anymore, I was doint 45 minutes of cardio, 10 minutes of slow running and one hour of weights. Now, I know that I have lost just around a 20 to 25% of the total of the weight I want to loose, but I have also gained a lot of muscle, and I am more healthy now.

Just some advises.

Best regards,

[quote="cats..."]djimenez: i dont know how much you weigh or how active you are, but given that you exercise 1.5-2 hrs on most days, it sounds like 1500-1800 is so far below your maintenance calories that on the scale you'll be losing a lot of weight, but a lot of that will also be muscle.  Ideally to lose fat and preserve the most muscle possible, it's usually recommended to be about 300-500 less than your maintenance calories.  Also, a lot of peole advocate having a cheat meal or cheat day once a week to help them feel like they're not being starved, so that's okay if it'll help you stay with your diet.  cause a lot of people if they stay with a strict cutting diet for a long time will eventually crash and decide to binge eat which will cause much more problems than having one cheat meal / week.  just some things to consider, good job keep up the work.[/quote]

The excercise that I do is extensive, but not 100% all the 2 hours. I practice martial arts (specifically Hapkido), and usually it is 40 minutes of a middle to hard warm up (a few laps running, 200 push ups and 200 of sit ups, usually in sets of 50-50, and a few similar things) and after that, it depends on the activity of the day. If the class that day is a lot of kicking and rolling, or wrestling, well, it is two hours of pure activity, but otherwise the other part of the class is phisically demanding, but not THAT demanding. Any way, given that I don't want to loose any muscle, I try to keep my diet rich in protein and calcium.

Of course, I understand that there are draw backs, specially because I am not an expert in nutricion (I am far from it) I may be making some mistakes, but I try to be as efficient as I want with the information I have.

On the other hand, thinking about the loose of muscle, I tried for a week to have 300 more in average, and I didn't loose any weight that week, then probably I should stay where I am.

Thank you for your advise.

Best,

Sum ergo edo... er... edo ergo sum... mmm food!..  Effective Dieting means  eat lot of food, at least, that's what  my  Labrodor Retirever says ..

If you want a good diet, just put a bunch of salt on your food  :D

[quote="Tare"]Instead of that unhealthy Atkins b******* try my dieting method! Simply put a mouthful of water, look up, gargle, and while gargling try to stick your tongue as far skywards (up) as possible. Just do seven sets a day, before and after you eat breakfast lunch and dinner and right before you go to sleep, three gargle a set and you'll be trim in no time! Try it today!

*[size=59]Tare is not responsible for any health damages caused or induced because of this dieting method. FDA [b]not [/b]approved.[/size][/quote]

xD.

That makes you throw up :P . Bullsh-*COUGH* bulimia.

[quote="Valentin Vornicu"][quote="bleumoose"]why? because then you wont eat it because it's gross?[/quote]Maybe he ment sal[b]ad[/b] :? :P[/quote]

No, bleumoose is right.

To be honest, the most important thing about diet is just exercise. I do not like all these fad diets: (now its protein, before it was carbs, before that: who knows?)

Still, I do not need to diet: it just does not seem fair. I can eat anything, and not gain a pound. Still, I exercise almost every day, but not that much. I know people that have to work hard to maintain weight and not gain weight. Oh well.

Yes, it is usually unfair for some people.

I am eating, not too few, but not that much (less than 2000 calories a day) and excercise 3 to 5 times a week, around 2 hour, but I can't loose more than 2 pounds a week, and most of the weeks it is around 1 pound (and I DO have a lot of weight to loose) But I have a friend, he wants to gain weight, because accordingly to him, when he wrestle, his weight is a big disadvantage (any way he wrestle me, he wins, but he has been in class two years more than me) He works out basically the same than I, he eats, really eats a lot, but eitherway, he doesn't gain a single punnd!!!

Some times nature is not totally fair. I remember that I watch a program about the Venezuelan Beauty Queens, and they have a big school. 99% of the girls that begin in the school have to basically kill themselves doing a lot of excercise and eating halve portions of salad and nothing else, to stay in an "acceptable weight", but once in a while, there are those girls that go there and have to start eating a lot more than what they use to, chicken, pork, chocolate, etc, because they are TOO slim! (really TOO SLIM)

Best,

[quote="Valentin Vornicu"]Haha, that's funny. I never heard a girl complaining about her being too slim :D  :lol:[/quote]

Well Valentin, may be you haven't asked enough (or haven't paid that much attention to some female conversation), I am sure it is not the most common situation, but it happens more often than what you should spected. I say so because for around two years I worked as a High School teacher, and you see girls in that situation. Specially before the prompt, they start worring about that because they look as a stick in the dresses they want to use!!

Best regards,

Trust me I know a lot of girl, and have never ever heard such a complaint! :) Even if the girl was slim already, they all still say they need to loose some weight because they think they look fat (although they obviously don't - but that's what they always say - well at least here in Romania :) ).

[quote="bleumoose"]v- the reason skinny girls complain about their skinniness is because it comes with a lack of breasts (being that those are mostly fat). so maybe now it will seem a little more believable? in america, having a large chest size is pretty important to a lot of women.[/quote]

Not when they are fake  ;) But that option is only available to women, not girls.

Hope this post is not inappropriate.

Let's get back on dieting and live sillicons and such aside, shall we? :D


Well ive taken a few classes now, and i find that its difficult to concentrate for 1.5 hours straight.  It gets very tiring, especially in a tough class.

Do you guys think that maybe a 5 minute break in the middle would be of help?

Take a five minute break yourself :D

Then come back and read what was kept during your break.  :P

Are you in AIME B class?

If so, I would say you'd better improve your ability of concentration, since the real AIME is even longer and tougher, and nobody would lead you to the solution...

But sometimes it's a little harder to focus in the evening, so taking a break yourself is a good idea. It's not that bad to read the transcript, I promise, since I often do that due to my damned *BRAND NEW* vista...

[quote="1=2"][quote="unimpossible"]Do you guys think that maybe a 5 minute break in the middle would be of help?[/quote]

The only class that I've seen done that is WOOT.[/quote]

Only cuz our classes are two hours long.  :wink:


How many perfect squares less than 1000 have a ones digit of 2, 3 or 4?

Well no perfect square can have units digit 2 or 3, so it's just the perfect squares ending in a 4.  For a square to end in a 4, the square root must end in 2 or 8.  Since $ 32^{2}>1000$, we have $ 2^{2}$, $ 12^{2}$, and $ 22^{2}$, for those ending in 2.  Ending in 8, we have $ 8^{2}$, $ 18^{2}$, and $ 28^{2}$.  Hence there are $ \boxed{6}$ in total.


[b]I need to find all twin primes p, p+2 so that the mid term p+1 to be triangular.[/b]

Thank you

let $ p \plus{} 1 \equal{} a(a \plus{} 1)/2$

then $ p \equal{} a(a \plus{} 1)/2 \minus{} 1 \equal{} (a^{2} \plus{} a \minus{} 2)/2 \equal{} (a \plus{} 2)(a \minus{} 1)/2$

Hence, a=2or3 implying that p=2or5
Since 4 is composite, p=5 is the only solution