The volume of a cube is 729 cm$ ^3$. How many square centimeters are in the area of one of its faces?
[asy]draw(unitsquare); size(90); defaultpen(linewidth(0.8));
draw((0,1)--(0.5,1.5)--(1.5,1.5)--(1.5,0.5)--(1,0)^^(1.5,1.5)--(1,1));[/asy]

You find the cube root of 729 which is 9. Since the faces are squares, you just square 9 so the answer is

                                                      81.


if $E$ and $F$ two sets of real numbers in a way that for every x$\in E$ there exists $y\in F$ : $x\le y$, prove that $supE\le supF$

An important intermediate fact: if $t<\sup E$, there exists $y\in F$ with $y>t$.
Can you see how to prove this? Can you see how to use this to prove that $\sup F>t$ and $\sup F\ge\sup E$?

Sorry it's a little difficult for me to prove these steps  :blush:

If you're trying to prove something about a least upper bound (supremum), you [b]must[/b] know and use the precise definition of that concept - else, how can you prove anything at all?

First, write down the precise definition of supremum.

Now look at jmerry's "intermediate fact." That's really very close to being the definition, isn't it?

OK I think I found it. For every $x \in E$ there is a $y_0 \in F :$ $x\le y_0 \le supF$, so $x\le supF$. $supF$ is an upper bound for all $x$, but $supE$ is the least upper bound for all $x$ so $supE \le supF$

Is that true?

in your first post did you mean $x \in E$? if so, you can reason by contradiction.


can you find the sum ? :D

$ \frac{1}{1\cdot 2\cdot 3}\plus{}\frac{1}{4\cdot 6\cdot 7}\plus{}\frac{1}{8\cdot 9\cdot 11}\plus{}\frac{1}{12\cdot 13\cdot 14}\plus{}\frac{1}{16\cdot 17\cdot 18}\plus{}\frac{1}{19\cdot 21\cdot 22}\plus{}\;\ldots\;\;\;\ldots$

(i.e. all multiples of $ 5$ are excluded within the products in the denominator)

$ P_{n}\equal{}\frac{1}{\left(K_{n}\plus{}1\right)\left(K_{n}\plus{}2\right)\left(K_{n}\plus{}3\right)}\plus{}\frac{1}{\left(K_{n}\plus{}4\right)\left(K_{n}\plus{}6\right)\left(K_{n}\plus{}7\right)}\plus{}\frac{1}{\left(K_{n}\plus{}8\right)\left(K_{n}\plus{}9\right)\left(K_{n}\plus{}11\right)}\plus{}\frac{1}{\left(K_{n}\plus{}12\right)\left(K_{n}\plus{}13\right)\left(K_{n}\plus{}14\right)}$

$ K_{n}\equal{}15 n$

$ S \equal{}\sum_{n\equal{}0}^{\infty }P_{n}$

After Some lot of simplification (???) we get a big sum:

I dont see what is your trick.


function
limit
trigonometry
real analysis
real analysis unsolved

$\ f: \mathbb{R}\to\mathbb{R}$ is a differential function such that $\ f(0)=0$ and $\ f(x)\geq x$ for every $\ x\in(0,\infty)$.
[b]a)[/b] Prove that $\ f'(0)\geq 1$.
[b]b)[/b] Prove that there is a sequence $\ (x_n)_{n\geq 1}\to 0$ such that $\ x_n>0$ and $\ f'(x_n)\geq 1$.
[b]c)[/b] Find a function f, such that there exists $\ (a_n)_{n\geq 1}\to 0$ a decreasing sequence such that $\ f'(a_n)<1$ for every $\ n\geq 1$.

[hide]

a) $f'(0) = \lim_{x \to 0} \frac{f(x)-f(0)}{x-0} \ge \lim_{x \to 0} \frac{x}{x} = 1$.

b) Hmm... I don't know if this works. Take $g(x) = f(x)-x$. Then $g'(x) = f'(x)-1$. Since we have $g(0) = 0$ and $g(x) \ge 0$ for all $x \in (0, \infty)$, there must be some interval in which $g$ is increasing or constant. In particular, since $g'(0) = 0$, there must be an interval starting at $0$. If we just take all $x_n$ in that interval (in a decreasing manner), they should satisfy the desired properties.[/hide]

Here's my solution for b): we take a sequence $\ y_n\to 0$ and use Lagrage's Theorem on $\ [0,y_n]$. We get another sequence $\ x_n$ such that $\ f'(x_n)=\frac{f(y_n)}{y_n}\geq 1$ and $\ x_n\in(0,y_n)$.
And for c) i gave [hide]$\ f(x)=x^3\sin\left(\frac1x\right)+x^3+x$.[/hide]

[hide]For a) , More carefully $f(x) \ge x$ on $(0,\infty )$ and we know nothing about $( - \infty ,0)$ we obtain that $f'^ + (0) = \mathop {\lim }\limits_{x \to 0 + } \frac{{f(x) - f(0)}}{{x - 0}} = \mathop {\lim }\limits_{x \to 0 + } \frac{{f(x)}}{x} \ge \mathop {\lim }\limits_{x \to 0 + } \frac{x}{x} \ge 1$
Since $f$ is differentiable on R 	then  $f'(0) = f'^ + (0) \ge 1$[/hide]


Does the equation $ x^5\plus{}y^5\plus{}z^5\equal{}4$ have a solution?

No

By Fermat's little theorem $ x^{10}\equiv 0\text{ or }1\mod 11\implies x^5\equiv 0\text{ or }\pm 1 \mod 11\ \ \forall \ x\in \mathbb{Z}$

So $ x^5 \plus{} y^5 \plus{} z^5\equiv 0\text{ or }\pm 1\text{ or }\pm 2\text{ or }\pm 3\mod 11\ \ \forall \ (x,y,z)\in \mathbb{Z}^3$
Hence $ x^5 \plus{} y^5 \plus{} z^5\neq 4\mod 11\ \ \forall \ (x,y,z)\in \mathbb{Z}^3$

QED


How many integers, n, can be written as a+b+c where a, b, and c are distinct odd integers between 2 and 20?

[color=white]the lowest sum is 3+5+7, which is 15. The highest is 15+17+19, which is 51. 51-15=36. 36/2=18
18+1 is 19[/color]

How do you know you can get every single one?

[hide]First we start with 3+5+7 then make 7 bigger until its biggest possible by adding 2;s gradually, then we add the 2's to the 5, then to the 3. Thus we didn't miss any numbers.[/hide]

Rockin'.  Always an important thing to check for such problems -- what the maximum is, and then whether or not you can actually get to it.


function
algebra
domain
calculus
derivative
functional equation

Solve $ f(x\plus{}t)\minus{}f(x\minus{}t)\equal{}4xt$.

I remember reading in AOPS about arbitrary functions, so I reorganized it as $ f(x\plus{}t)\minus{}2xt\equal{}f(x\minus{}t)\plus{}2xt$, but I'm stuck.  I also know that substitution for $ x\equal{}0$ also suggests the function is even.  Where can I go from here?

Generally speaking, you should also define the domain of the function.
Anyway, try substituting \[ \displaystyle x\equal{}t\equal{}\frac{y}{2}\].

I don't think they gave a domain restriction, so I will assume that it is for all real numbers.

Thanks for the substitution!  I still have one question though.

[hide]Substitution for both values yields $ f(y)\minus{}f(0)\equal{}y^2$.  A new substitution of $ y\equal{}\minus{}y$ yields $ f(\minus{}y)\minus{}f(0)\equal{}y^2$.  Subtraction suggests that $ f(y)\equal{}f(\minus{}y)$.  Does this imply that any even function would work, or would there be some restrictions?[/hide]

From your first equation, $ f(y) \equal{} y^2 \plus{} f(0)$. You can stop now because that's all of the solutions right there. Whether all even functions work can be answered by plugging them into the equation and seeing if both sides cancel.

No; substituting that expression into the original equation shows that $ f(0)$ can be chosen arbitrarily.

When you plug in "specific" substitutions into a functional equation, you get an expression that is true because of the functional equation, but you cannot use it to conclude that any function that satisfies that expression satisfies the functional equation unless the substitution you used is reversible.  That is, all your last line tells you is that if $ f$ satisfies the functional equation, it's even, and not the other way around.  An example of a reversible substitution is to let $ a \equal{} x \plus{} t, b \equal{} x \minus{} t$ to get

$ f(a) \minus{} f(b) \equal{} a^2 \minus{} b^2$

and now there's not much left to do.

Thanks, t0rajir0u! I solved some func. equations and got solutions like that and then smashed my head with the question: (in this example) What's the value of f(0)? So I can't figure it out. Thanks again!

Let $ g(x)\equal{}f(x)\minus{}x^2$. Then

$ 0\equal{}f(x\plus{}t)\minus{}f(x\minus{}t)\minus{}4xt\equal{}g(x\plus{}t)\plus{}(x\plus{}t)^2\minus{}g(x\minus{}t)\minus{}(x\minus{}t)^2\minus{}4xt\equal{}g(x\plus{}t)\minus{}g(x\minus{}t)$

It follows that $ g(x)$ is a constant.

So it is necessary that $ f(x)\equal{}x^2\plus{}c$. Plugging this into the original functional equation, we see that any value of $ c$ works.

(Comment: it is sloppy to say that $ f(x)\equal{}x^2\plus{}f(0)$ or something because $ f(0)$ isn't defined directly.)

f(x) = x^2 seems to work.

in general f(x) = x^2 + c works.

also,

[f(x+t) - f(x-t)/2t] = 2x

the LHS is the derivative of f,

f'(x) = 2x

=> f(x) = x^2 + c.

so that nails it!  :)

How do you know that $ f$ is differentiable?

Oops!, sorry I forgot that it may not be differentiable. :blush:


[quote="soruz"]Substitution for  condition is

$ x_1 \equal{} \frac {y_1}{y_2 \plus{} y_3 \plus{} ... \plus{} y_{2007}}$
$ x_2 \equal{} \frac {y_2}{y_1 \plus{} y_3 \plus{} ... \plus{} y_{2007}}$
.
.
.
$ x_{2007} \equal{} \frac {y_{2007}}{y_1 \plus{} y_2 \plus{} ... \plus{} y_{2006}}$.[/quote]

I think it is wrong,their sum is 2007,not 2006. :wink:


Given  that $ n\in \mathbb {N}$ , and $ n \ge 2 ,a_1,a_2,a_3,\cdots,a_n \in (0,1)$.

Find the max. value of $ \boxed{\sum_{i \equal{} 1}^n \sqrt [6] {a_i(1 \minus{} a_{i \plus{} 1})}}$, where $ a_{n \plus{} 1} \equal{} a_1$

[hide="ans."]

[color=blue]max. vaue[/color]=$ {n\over \sqrt [3] {2} }$ when $ a_1 \equal{} a_2 \equal{} \cdots \equal{} a_n \equal{} {1\over 2}$[/hide]

Sorry, the problem was edited.

[quote="stephencheng"][quote="PKMathew"]Given  that $ n\in \mathbb {N}$ , and $ n \ge 2 ,a_1,a_2,a_3,\cdots,a_n \in (0,1)$.

Find the max. value of $ \sum_{i = 1}^n \sqrt [n] {a_i(1 - a_{i + 1})}$, where $ a_{n + 1} = a_1$

[hide="ans."]

[color=blue]max. vaue[/color]=${ \{n\over \sqrt [3] {2} }$ when $ a_1 = a_2 = \cdots = a_n = {1\over 2}$[/hide][/quote]

$ \dfrac{\sum_{i = 1}^n [a_i(1 - a_{i + 1})]^{\frac {1}{n}}}{n}$
$ \le (\dfrac{\sum_{i = 1}^n [a_i(1 - a_{i + 1})]^{\frac {1}{2}}}{n})^{\frac {2}{n}}$ by Power Mean
$ \le (\dfrac{\sum_{i = 1}^n [a_i + (1 - a_{i + 1})]}{2n})^{\frac {2}{n}}$
$ = (\dfrac{1}{2})^{\frac {2}{n}}$,

So Max. value $ = n(\dfrac{1}{2})^{\frac {2}{n}}$

$ a_1 = a_2 = ... = a_n = \frac {1}{2} \Rightarrow$ equality holds for all $ n \ge 2$.

For $ n = 2$, $ (a_1,a_2) = (k,1 - k)$ also gives the equality case.[/quote]

Dear stephencheng,

It's a typo in the  question , and was edited.

Apol. for your inconvenience caused.


linear algebra
matrix
combinatorics proposed
combinatorics

In the Duma there are 1600 delegates, who have formed 16000 committees of 80 people
each. Prove that one can nd two committees having no fewer than four common members.

This topic is very old but seeing as how nobody replied, I'll do it here rather than make a new topic.

Let the delegates be $ d_1, d_2, \ldots d_{1600}$. Let the committees be $ c_1, c_2, \ldots , c_{16000}$. Create a grid numbering the delegates along the horizontal axis and the committees along the vertical axis, with a $ 1$ in $ (i,j)$ if delegate $ d_j$ is in committee $ c_i$ and a $ 0$ otherwise.
\[ \begin{bmatrix}
& d_1 & d_2 & \ldots & d_{1600} \\
c_1 & 1 & 1 & \ldots & 0 \\
c_2 & 0 & 1 & \ldots & 0 \\
. & . & . & . & . \\
. & . & . & . & . \\
. & . & . & . & . \\
c_{16000} & 0 & 0 & \ldots & 1 \\
\end{bmatrix}
\]

Each of the $ 16,000$ committees has $ 80$ ones in that row, so there are $ 1280000$ ones total in the array. Let's count the total number of pairs of ones in the same column. Supposing that no two committees have fewer than four common members, any two rows share at most $ 3$ ones in the same column. This means that the total number is at most
\[ 3\binom{16000}{2} \equal{} 383976000
\]
On the other hand, suppose that there are $ a_1$ ones in column $ 1$, $ a_2$ ones column in column $ 2$, etc.  Then we have
\[ \sum_{k\equal{}1}^{1600} \binom{a_i}{2}
\]
pairs of ones in the same column. Since the $ a_i$ sum to $ 1280000$ (an average of $ 800$), this is greater than or equal to
\[ \sum_{k\equal{}1}^{1600} \binom{800}{2} \equal{} 511360000
\]
We thus have that the total number of pairs of ones is greater than $ 383976000$ but less than $ 5113600000$, which is a clear contradiction. Phew. I don't know how I would have done that without a calculator.


algebra
polynomial
abstract algebra
number theory
prime numbers
algebra unsolved

For which $ p$ prime numbers, there is an integer root of the polynominal $ 1 \plus{} p \plus{} Q(x^1)\cdot\ Q(x^2)\ldots\ Q(x^{2p \minus{} 2})$ such that $ Q(x)$ is a polynominal with integer coefficients?

Come back after long inactive time!

My answer is p has the form 4*k+1. You could prove that for any polynomial Q and an arbitrary integer number x, we have the product Q(x).Q(x^2)... is congruent module p
with a perfect square ( thus groups the 2p-2 terms in pairs). From here we got p divides C^2+1 for certain C. Hence p=4k+1

Now for p has the form 4k+1. We could use the Wilson theorem to give an example

[quote="stephencheng"]I don't know whether my solution is correct, as a Problem 4 should not be so easy.

It's easy to see that none of $ Q(x),Q(x^2),...,Q(x^{2p \minus{} 2})$ are of the same parity and they are not equal to zero.
So for $ p \ge 3$ being odd, $ (p \plus{} 1)$ is even $ \Rightarrow 2^{2p \minus{} 2}|(p \plus{} 1)$, impossible for $ p \ge 3$.So the only possible $ p$ is $ 2$, and it's easy to find a $ Q(x)$ and $ x$, for example, $ Q(x) \equal{} x^2 \minus{} 4x \plus{} 3$,$ x \equal{} 2$ that satisfies the conditions.[/quote]

Sorry I have made a typo, it should be [b]$ Q(x),Q(x^2),...,Q(x^{2p \minus{} 2})$ are of the same parity [/b]. The words [b]''none of''[/b] should be deleted.

Also, [b]mto[/b], it seems that you and I got different results . Can you give an example for $ p\equal{}5$? :D

stephencheng, your result is true. Besides, the fact that this is very easy for a TST problem is true but as I learned from our teachers this solution hadn't recognised before the contest. If we replace p^p by p in the initial question, it may be more interesting or at least closer to the expected solution :D

I cant understand the problem :s is for everything polinomial ?
because if Q(x ) =x, then $ 1\plus{}p\plus{}x^{(p\minus{}1)(2p\minus{}1)} \equal{} P (x)$, then , there exist $ a$ integer such that P(a)=0, then if p : odd p-1 = even, -> P(x) >0, if p=2, 3+x^3=0, there isnt solution in integer.

[quote="aev5peru"]I cant understand the problem :s is for everything polinomial ?
because if Q(x ) =x, then $ 1 \plus{} p \plus{} x^{(p \minus{} 1)(2p \minus{} 1)} \equal{} P (x)$, then , there exist $ a$ integer such that P(a)=0, then if p : odd p-1 = even, -> P(x) >0, if p=2, 3+x^3=0, there isnt solution in integer.[/quote]

The problem is : for which primes $ p$ does there exist $ Q(x)$ so that $ 1\plus{}p\plus{}Q(x^1) \cdot Q(x^2) \cdot ... \cdot Q(x^{2p\minus{}2})\equal{}0$ has integer root(s)?

it is very easy problem
I think it can be good for Junior Balkam M.O. :)  :lol:

Why do we have that $ 2^{p\minus{}2}|{p \plus{} 1}$ ?

Hello

Do you mean 'Why do we have that $ 2^{2p \minus{} 2}|{p \plus{} 1}$ ?

It is because $ Q(x),Q(x^2),...$ all have to be even.

Hope that helps

[quote="stephencheng"]I don't know whether my solution is correct, as a Problem 4 should not be so easy.

It's easy to see that none of $ Q(x),Q(x^2),...,Q(x^{2p\minus{}2})$ are of the same parity and they are not equal to zero.
So for $ p \ge 3$ being odd, $ (p\plus{}1)$ is even $ \Rightarrow 2^{2p\minus{}2}|(p\plus{}1)$, impossible for $ p \ge 3$.So the only possible $ p$ is $ 2$, and it's easy to find a $ Q(x)$ and $ x$, for example, $ Q(x)\equal{}x^2\minus{}4x\plus{}3$,$ x\equal{}2$ that satisfies the conditions.[/quote]
for $p=2$ there exist a polynomial of degree $1$.$Q(x)=2x-5$ and the equation have $x=2$ as a solution.

The only answer is $p=2$. The key lemma is if $x \equiv y \pmod{m}$ then $Q(x) \equiv Q(y) \pmod{m}$.

Firstly assume that $p \equiv 3 \pmod{4}$. FTSOC let there be an integer root $x$. Then we have $\prod_{i=1}^{2p-2} Q(x^i) \equiv \prod_{i=1}^{p-1} Q(x^i)^2 \Rightarrow \prod_{i=1}^{p-1} Q(x^i)^2+p+1 \equiv y^2+1 \equiv 0 \pmod{p}$ for some integer $y$. Contradiction since $p \equiv 3 \pmod{4}$.

Now assume that $p \equiv 1 \pmod{4}$. FTSOC let there be an integer root $x$. We will reach contradiction by $\pmod{4}$.

[b]1) $x$ is odd. [/b] Then we would have $x^k \equiv x^{k+2} \pmod{8} \Rightarrow \prod_{i=1}^{2p-2} Q(x^i)+p+1 \equiv Q(x)^{p-1}Q(x^2)^{p-1}+2 \equiv y^2+2 \equiv 0 \pmod{4}$ for some integer $y$. Contradiction since $2$ is not a QR for $\pmod{4}$.

[b]2) $x$ is divisible by $4$ [/b] Using $Q(x^i) \equiv Q(0) \pmod{4}$, we have $\prod_{i=1}^{2p-2} Q(x^i)+p+1 \equiv Q(0)^{2p-2}+2 \equiv y^2+2 \equiv 0 \pmod{4}$ which is again a contradiction because of the same argument.

[b]3) $x \equiv 2 \pmod{4}$ [/b] We have $\prod_{i=1}^{2p-2} Q(x^i)+p+1 \equiv Q(0)^{p-1}Q(2)^{p-1}+2 \equiv y^2+2 \equiv 0 \pmod{4}$ which is again a contradiction because of the same argument.

For $p=2$ we need $Q(x)Q(x^2)=-3$ for some polynomial $Q$. Trying small cases gives us $x=-1$ is a solution where $Q(x)=2x-1$.

[quote=stephencheng]I don't know whether my solution is correct, as a Problem 4 should not be so easy.

It's easy to see that none of $ Q(x),Q(x^2),...,Q(x^{2p\minus{}2})$ are of the same parity and they are not equal to zero.
So for $ p \ge 3$ being odd, $ (p\plus{}1)$ is even $ \Rightarrow 2^{2p\minus{}2}|(p\plus{}1)$, impossible for $ p \ge 3$.So the only possible $ p$ is $ 2$, and it's easy to find a $ Q(x)$ and $ x$, for example, $ Q(x)\equal{}x^2\minus{}4x\plus{}3$,$ x\equal{}2$ that satisfies the conditions.[/quote]

I don't know how you get from $ (p\plus{}1)$ is even to $ 2^{2p\minus{}2}|(p\plus{}1)$

Because 1+p+Q(x)….=0, so 1+p=-Q(x)…..
Here the RHS is divisble by 2^2p-2, so the LHS has to be too


calculus
integration
function
derivative
parameterization
factorial
calculus computations

[b]find all functions f defined from IR to IR;bounded and continious such that for evry x belonging to IR we have: [/b]

it was already posted (you can try to find ), but there no one post solution since it's not interesting problem and maybe looks like homework .Maybe you   just need to calculate
$ \frac {d \int _{x^{2} \minus{} 1}^{x^{2} \plus{} 1}f(2t)dt}{dx}$ ?
Do you know more general formula like this :
\[ \frac {d \int\limits_{\alpha(x)}^{\beta(x)} f(t)dt}{dx} \equal{} f(\beta(x))\beta'(x) \minus{} f(\alpha(x))\alpha'(x) \plus{} \int\limits_{\alpha(x)}^{\beta(x)} f'(t)dt
\]
I assume that there is $ f'(x)$ or $ f \in C^{1}$maybe it can be explaind easy why.

do you know if $ f$ is differentiable? just because it is continuous does not mean it is differentiable.
see here for an example
http://planetmath.org/encyclopedia/WeierstrassFunction.html

I know, therefore I wrote :

[quote="Extremal"]
I assume that there is $ f'(x)$ or $ f \in C^{1}$maybe it can be explaind easy why.[/quote]

yeah i saw that but what does $ f\in{C^1}$ mean?

it's mean that exist $ f'$ and $ f'$ is continious.

This post is going to be long, technical, and disappointing. Needless to say, Extremal's remarks are completely off track again. I doubt this problem could be given as a regular homework to anybody except somebody like grobber during his graduate school years. Anyway, what I am going to do is to exhibit a huge family of such functions though I am far from claiming that my construction exhausts all possible solutions.

[b]General idea[/b] The equation can be rewritten as $ f(x) \equal{} \frac 12\int_{2x^2 \minus{} 2}^{2x^2 \plus{} 2}f(t)\,dt \equal{} Tf(x)$. The operator $ T$ is, clearly, linear and its image is contained in the space of even functions. Also, if $ f$ is $ 4$-periodic and the integral of $ f$ over its period equals $ 0$, then $ Tf \equal{} 0$. Note that $ Tf$ expands rather than contracts ($ Tf$ is roughly speaking twice larger than $ f$ in reasonable norms  and high derivatives get larger as well), so it makes sense to try the following. Fix some function $ f_0$ such that $ Tf_0 \equal{} 0$. Now define $ f_n$ recursively by $ Tf_{n} \equal{} f_{n \minus{} 1}$. At last, put $ f \equal{} \sum_{n\ge 0}f_n$. To make sense out of it, we need to show that the recurrency is solvable and the series converges in $ C(\mathbb R)$ to a bounded function.

[b]Choice of $ f_0$[/b]. Take any $ C^\infty$-function that is even, $ 4$-periodic, has zero integral over the period, satisfies $ |f^{(k)}|\le M_{k,0}$ for all $ k\ge 0$ where $ M_{k,0}$ grows not too fast (we'll see the exact condition later), and vanishes on $ [ \minus{} 1,1]$.
Put $ u_0 \equal{} 1$.

[b]Recursion[/b]. Assume that $ h\in C^{\infty}([0, \plus{} \infty))$ is even and vanishes on $ [ \minus{} 1,1]$ Define $ g(t) \equal{} 2h(\sqrt {\tfrac{t \minus{} 2}2})$ for $ t\ge 2$, extend it by $ 0$ to $ [0,2]$, and put
\[ Sh(x) \equal{} \sum_{0\le \ell < x/4}g'(x \minus{} 4\ell)
\]
for $ x > 0$ and $ Sh(x) \equal{} Sh( \minus{} x)$ for $ x < 0$. Note that $ S$ is a linear operator and that, if $ h$ vanished on $ [ \minus{} u,u]$, then $ g$ and $ Sh$ vanish on $ [ \minus{} 2u^2 \minus{} 2,2u^2 \plus{} 2]$. Now it is easy to see that $ TSh \equal{} h$. This allows us to put $ f_n \equal{} Sf_{n \minus{} 1}$ and observe that $ f_n$ vanishes on $ [ \minus{} u_n,u_n]$ where $ u_0 \equal{} 1$, $ u_n \equal{} 2u_{n \minus{} 1}^2 \plus{} 2$. Also, $ Sh\in C^\infty$.

[b]Derivative estimates[/b].
Our first task will be to estimate the derivatives of $ g$ in terms of the derivatives of $ h$ and the parameter $ u$. Note that the linear change of variable $ t\mapsto \frac {t \minus{} 2}2$ does not increase the derivatives, so it is enough to bound the derivatives of $ G(x) \equal{} h(\sqrt x)$. Now notice that, when you differentiate the composition, you have 2 choices at each step: either to differentiate the power outside $ h$ or to differentiate $ h$ and add a factor of $ \frac 1{2\sqrt x}$. Thus, we can show that if $ |h^{(k)}|\le M_k$, then $ |G^{(k)}(x)|\le 2^k k! x^{ \minus{} k/2}(M_1 \plus{} M_2 \plus{} \dots \plus{} M_k)$ for $ x > 1$ (recall that $ G \equal{} 0$ on $ [0,1]$, so we do not need to make any estimates there) and, obviously, $ |G|\le 2 M_0$. This immediately allows us to make an estimate for the derivatives of $ Sh$ of the second and higher order:
\[ |Sh^{(k)}|\le \left[\sum_{0\le\ell < \frac {x}4} (\tfrac x2 \minus{} 2\ell \minus{} 1)^{ \minus{} (k \plus{} 1)/2}\right]2^{k \plus{} 1}(k \plus{} 1)!(M_1 \plus{} \dots \plus{} M_{k \plus{} 1})\le C_k u^{ \minus{} (k \minus{} 1)}(M_1 \plus{} \dots \plus{} M_{k \plus{} 1})
\]
where $ C_k$ grow roughly speaking like $ k!$.

The estimate for $ |Sh|$ and $ |(Sh)'|$ is a bit subtler. The series will no longer converge, so we need some trick. The idea is that $ g'(t) \equal{} \frac 14[g(t \plus{} 2) \minus{} g(t \minus{} 2)]$ up to an error that is dominated by the maximum of the third derivative of $ g$ on $ [t \minus{} 2,t \plus{} 2]$. So,
\[ |Sh(x)|\le \frac 14|g(x \plus{} 2)| \plus{} C_3 u^{ \minus{} 1}(M_1 \plus{} M_2 \plus{} M_3)\le \frac 12M_0 \plus{} C_3 u^{ \minus{} 1}(M_1 \plus{} M_2 \plus{} M_3)
\]
and, similarly,
\[ |Sh(x)|\le \frac 12M_1 \plus{} C_4 u^{ \minus{} 1}(M_1 \plus{} M_2 \plus{} M_3 \plus{} M_4)\,.
\]
Thus, we get the recursions for $ M_{k,n} \equal{} \sup|f_{n}^{(k)}|$:
\[ \begin{aligned} M_{0,n} & \le\frac 12{M_{0,n \minus{} 1}} \plus{} C_0 u_{n \minus{} 1}^{ \minus{} 1} (M_{1,0} \plus{} \dots \plus{} M_{3,0}) \\
M_{1,n} & \le\frac 12{M_{1,n \minus{} 1}} \plus{} C_1 u_{n \minus{} 1}^{ \minus{} 2} (M_{1,0} \plus{} \dots \plus{} M_{4,0}) \\
M_{k,n} & \le C_k u_{n \minus{} 1}^{ \minus{} (k \minus{} 1)} (M_{1,0} \plus{} \dots \plus{} M_{k \plus{} 1,0})\quad\text{ for }k\ge 2 \end{aligned}
\]
[b]Convergence of the series[/b].
Let $ \sigma_k(n) \equal{} M_{0,n} \plus{} M_{1,n} \plus{} \dots \plus{} M_{k,n}$. The recurrence relations above (together with the trivial observation that $ u_n\ge 1$ for all $ n$) imply that
\[ \sigma_k(n)\le \frac 12\sigma_2(n \minus{} 1) \plus{} D_k u_{n \minus{} 1}^{ \minus{} 1}\sigma_{k \plus{} 1}(n \minus{} 1)\,.
\]
where $ D_k$ still grow like $ k!$.
Now fix large $ A$ and let $ n$ be the least index such that $ \sigma_2(n) > A\cdot 2^{ \minus{} n/2}$ (if it exists; if there is no such index, we are clearly done). Note that then we must have
\[ \sigma_3(n \minus{} 1) > 0.1 A D_3^{ \minus{} 1}u_{n \minus{} 1}\,.
\]
If the number on the right is much bigger than $ A$, we can derive from here that $ \sigma_4(n \minus{} 2) > 0.1 A D_3^{ \minus{} 1}D_4^{ \minus{} 1}u_{n \minus{} 1}u_{n \minus{} 2}$ and so on. So, if $ u_{n \minus{} 1} > 10 D_1D_2\dots D_{n \plus{} 2}$ (which is true for all large $ n$ because $ u_n$ grows like double exponent and $ D_k$ are just pitiful factorials), we will go all the way down to
\[ \sigma_{n \plus{} 2}(0) > 0.1 A (D_1D_2\dots D_{n \plus{} 2})^{ \minus{} 1}u_{n \minus{} 1}.
\]
Now the number on the right is essentially double exponential in $ n$ (the factor $ A$ only makes it bigger) and the number on the left is just $ M_{0,0} \plus{} \dots \plus{} M_{n \plus{} 2,0}$. Recalling standard theorems about quasianalytic classes, we see that this sum can grow just like $ n^{2n}$ if $ f_0$ was chosen appropriately. Thus, it cannot happen that $ n$ is large. But if $ A$ is large, it cannot be small either.

[b]Why is $ f$ not identically $ 0$[/b]. This is quite clear because $ f_0$ is not identically $ 0$ on $ [ \minus{} 2,2]$ and $ f_1,f_2,\dots$ vanish there.

The end.

[b]Conclusion[/b] I doubt the problem is solvable in any reasonable terms. The OP told me that "this exercise was suggested by his(her?) professor in the preparatory (sic!) classes as a challenge". Well, this makes one wonder "preparatory" for what? (On the other hand, I should confess that I like to play tricks like that on my students too, giving them a problem I cannot solve myself for extra credit :P).

I hope I didn't make too many misprints when typing this post. If anybody has any questions about it, feel free to ask :).

oh my god,I spent several days trying solve this problem and using some poor tricks and some classical technique and I discover in the end that I was loosing my time,by the way thank's Mr fedja for this very nice and very difficult solution. :)


Let $ S_1$ denote the sequence of positive integerts 1,2,3,4,5,6,..., and define the sequence $ S_{n\plus{}1}$ in terms of $ S_n$ by adding 1 to those integers in $ S_n$ which are divisible by $ n$. Thus, for example, $ S_2$ is 2,3,4,5,6,7,..., $ S_3$ is 3,3,5,5,7,7,.... Determine those integers $ n$ with the property that the first $ n\minus{}1$ integers in $ S_n$ are $ n$.

[quote="iniquitus"]n=1,n=2,n=3 are the only solutions.

After n=3, no n will make the first two integers go up by one; thus the first two integers will always be three.[/quote]

I think this is wrong, since $ S_3 \equal{} \{3,3,5,5,7,7,9,9,11,11,13,13,...\}$. And clearly the first 2 terms go up by one ...

The right conjecture should be (I think) if and only if $ n$ is a prime. I can not make that rigourous.


What number is one third of the way from $ \frac14$ to $ \frac34$?

$ \textbf{(A)}\ \frac{1}{3} \qquad
\textbf{(B)}\ \frac{5}{12} \qquad
\textbf{(C)}\ \frac{1}{2} \qquad
\textbf{(D)}\ \frac{7}{12} \qquad
\textbf{(E)}\ \frac{2}{3}$

[hide="Answer"]
Divide the distance from $ \frac{1}{4}$ to $ \frac{3}{4}$ into thirds: $ \frac{\frac{1}{2}}{3}\equal{}\frac{1}{6}$

Add $ \frac{1}{6}\plus{}\frac{1}{4}\equal{}\frac{5}{12} \implies \boxed{B}$
[/hide]


Would anyone advise me not to take an AP Physics course through EPGY during my Junior year of high school when I took a non-AP phys course Sophomore year (AP Physics isn't offered @ my high school)? The course I'm taking now, as a sophomore, moves extremely slow and because of that there will be a significant portion of intro physics that won't be covered (we'll barely be touching on waves by the end). If it is the case that this would not be a good idea, then I have no problem self studying the material, although I would still prefer to take the class.

Thanks.

If the price of the EPGY course is not a problem (and be sure to check for the generous financial aid EPGY offers), then I would advise taking that course along with whatever you are taking. We may be doing something like that ourselves next year.

It would most likely be better to take the course somewhere (EPGY) than to self-study.  However, you could always try to take a college physics course at a local university (if you live near one).

If I were to take a course through a university, would I then be considered a transfer student when applying to collegeS?

No.

This text is to make my message long enough to post.

You are a transfer student if you enroll in a degree program (matriculate) at a previous college, but lots of college applicants have college credits on their "high school" transcripts and can still apply as first-year applicants.

It makes you dual-enrollment, so you're still a freshman when you apply to colleges ("first-time student" that is). DO NOT even THINK about trying to study AP Physics on your own time. This is really not a good self-study course. It's hard and confusing and a lot of the AP books are really lax on their explanations. I recommend a teacher of some type (E-whatever, or dual enrollment, if you can do that). But for the love of god, don't try and study AP Phys by yourself. Unless, maybe AP Phys B. That might be easier. AP Phys C was a terrible mistress.

Haha bleu; I'm sorry if I conveyed that message, but the thought hadn't crossed my mind. Perhaps the wording of my first post was confusing. I intended to ask whether I should take an AP Phys course online or study physics on my own (which wouldn't involve taking an AP test).

To everyone who posted: thanks for the advice.

Me and a friend have essentially taught ourselves AP Physics over the past 3 months, and although we have a teach he doesnt teach the stuff. However i have to say learning the stuff on my own took a tremendous amount of diligence, I basically read a textbook, an AP book, and did most of the past multiple choice and free response. So if your up to it go for it, if you cant put in the work i wouldnt suggest it.


I'm trying to start writing up my solutions, but every time that I go to Build-Current File-Build and View, I get this error:

[quote][Doc:Open("%bm.pdf")][FileOpen("%bm.pdf")]

Cannot execute the command.[/quote]

Does anyone know how this can be fixed?

Is your pdf file open when you are trying to build?

Additionally, the outut 'bar' shows 0 Errors, 0 Pages.

This might be important too, but I'm not sure  :maybe:

Did you write any content yet, even "hello world"?

Yes, I wrote my name in the Real Name brackets.

[quote="Ravi B"]Try writing something after \begin{document}.[/quote]

It worked!  :lol:

Thanks a lot.  I really appreciate it.


Here are the Numbers: 8, 10, 23, 45

The first person who posts an answer that is correct must also give a new 24 problem -- list four more numbers that you can make 24 out of. If you post 4 numbers, you must know how to make 24 out of them!

Go!!!!!!!!!!!!!!!

There's already a 24 Game marathon [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=209924]here[/url].


trigonometry
function
HMMT
algebra
polynomial
quadratics
binomial theorem

If $ \tan 15^{\circ} \tan 25^{\circ} \tan 35^{\circ} \equal{}\tan \theta$ and $ 0^{\circ}\leq \theta \leq 180^{\circ}$, find $ \theta$.

I was thinking about having 15=30/2, 25=30-5, and 35=30+5, but i couldnt' find a way to do it.

Any thoughts?

[quote="The Zuton Force"]You're on the right track in that you've expressed 15, 25, and 35 in terms of the nicer set of values, 30 and 5. However, half-angle is not the prettiest approach. Consider this variation:

[hide="Solution"]
We can express $ 25 \equal{} 30 \minus{} 5, \; 35 \equal{} 30 \plus{} 5, \; 15 \equal{} 3 \cdot 5$. We know $ \tan 30 \equal{} \frac {1}{\sqrt {3}}$, and let's set $ \tan 5 \equal{} x$.

First, let's simplify the product of the two similar-form tangents:
$ \tan 25 \tan 35 \equal{} \left( \frac {\frac {1}{\sqrt {3}} \plus{} x}{1 \minus{} \frac {1}{\sqrt {3}} x} \right) \left( \frac {\frac {1}{\sqrt {3}} \minus{} x}{1 \plus{} \frac {1}{\sqrt {3}} x} \right) \equal{} \frac {\frac {1}{3} \minus{} x^2}{1 \minus{} \frac {1}{3} x^2} \equal{} \frac {1 \minus{} 3x^2}{3 \minus{} x^2}$

That's good enough to deal with. We can use the tan-addition formula twice to find the other tangent:
$ \tan 10 \equal{} \frac {2x}{1 \minus{} x^2} \implies \tan 15 \equal{} \frac {x \plus{} \frac {2x}{1 \minus{} x^2}}{1 \minus{} \frac {2x^2}{1 \minus{} x^2}} \equal{} \frac {3x \minus{} x^3}{1 \minus{} 3x^2}$

Multiplying these two results conveniently gives:
$ \tan 15 \tan 25 \tan 35 \equal{} \left( \frac {1 \minus{} 3x^2}{3 \minus{} x^2} \right) \left( \frac {3x \minus{} x^3}{1 \minus{} 3x^2} \right) \equal{} x \equal{} \tan 5$

Thus, $ \boxed{\theta \equal{} 5^{\circ}}$
[/hide][/quote]

nice!  How did you think of expressing 15 as 3 times 5? 5 isnt your everyday angle used in competition problems  :P .

This is a contrived problem. I believe the most motivated solution is to say: let $ z\equal{}\text{cis}\, 5^\circ$. Then you can express the product in terms of $ z$. Then you have to simplify and match it to the other side.

[quote="unimpossible"]nice!  How did you think of expressing 15 as 3 times 5? 5 isnt your everyday angle used in competition problems  :P .[/quote]

Well, we can express 25 and 35 nicely in terms of 30 and 5 (note how we got a difference of squares when we multiplied these in my solution), so it is reasonable to want to express 15 in terms of 30 and/or 5 as well. We can do this in one of two basic ways: 15 = 30/2 or 15 = 3*5. You chose the first, I chose the second.

In general if you can express a lot of angles in terms of a standard angle and a nonstandard angle, it often helps to set the trig function of the nonstandard angle as a variable and see how far it simplifies, hoping that you do not have to compute it out.  :wink: I first used this technique at HMMT on number 8 for algebra this year  :)

[quote="serialk11r"]On a side note I think you can find the tan of 5 degrees in a messy way, expand it out using cis like so:
cis30=cis5*6. So cis30=(use binomial theorem, blah to the 6th power), seperate real and imaginary solve for tan5... But its extremely ugly so...[/quote]Not really, you'll end up solving a 5th degree polynomial which unfortunately does not have 'nice' radical solutions. All degrees that are multiples of 3 are either rationals or quadratic irrationals due to 75-72=3. So you can't really find $ \tan 5^\circ$ in radical form (it's not a constructible number).


inequalities
LaTeX
calculus
derivative
inequalities proposed

Prove that if $ a^2\plus{}b^2\plus{}c^2\equal{}1$ then:

$ a^6\plus{}b^6\plus{}c^6\plus{}a^3c\plus{}c^3b\plus{}b^3a \ge 3a^2b^2c^2\plus{}a^3b\plus{}b^3c\plus{}c^3a$

we only need to prove:$ suma^6 \plus{} suma^5c \plus{} suma^3b^2c > \equal{} 3a^2b^2c^2 \plus{} suma^3bc^2 \plus{} suma^5b$
$ suma^6 > \equal{} suma^5b.$
$ 1/2(a^5c \plus{} c^3ab^2) > \equal{} a^3bc^2$
$ 1/2(suma^5c \plus{} suma^3b^2c) > \equal{} ?a^2b^2c^2$
proof.
I'm not wrong?

Dear Cai Ni, I don't understand your solution very well because it is not written in $ Latex$, but thanks for it anyway. So I ask someone else to check if it is correct. If you want to compare it with mine, check this:

[hide]The inequality is equivalent with:

$ a^6\plus{}b^6\plus{}c^6\minus{}3a^2b^2c^2\plus{}a^3c\plus{}c^3b\plus{}b^3a\minus{}a^3b\minus{}b^3c\minus{}c^3a\ge 0$

$ \Leftrightarrow (a^2\plus{}b^2\plus{}c^2)(a^4\plus{}b^4\plus{}c^4\minus{}a^2b^2\minus{}b^2c^2\minus{}c^2a^2)\plus{}a^3c\plus{}c^3b\plus{}b^3a\minus{}a^3b\minus{}b^3c\minus{}c^3a\ge 0$

$ \Leftrightarrow a^4\plus{}b^4\plus{}c^4\minus{}a^2b^2\minus{}b^2c^2\minus{}c^2a^2\plus{}a^3c\plus{}c^3b\plus{}b^3a\minus{}a^3b\minus{}b^3c\minus{}c^3a\ge 0$

$ \Leftrightarrow a(a\minus{}b)^2(a\plus{}b)\plus{}b(b\minus{}c)^2(b\plus{}c)\plus{}c(c\minus{}a)^2(c\plus{}a) \ge 0$[/hide]

It's nice, I had the same solution using the factorization $ x^3\plus{}y^3\plus{}z^3\minus{}3xyx\equal{}(x\plus{}y\plus{}z)(x^2\plus{}y^2\plus{}z^2\minus{}2xy\minus{}2yz\minus{}2xz)$

Sorry,I first go to this forum the day before yesterday.
 :oops:
We need to prove$ \sum\[a^6 + \sum\[a^5c + \sum\[a^3b^2c \ge 3a^2b^2c^2 + \sum\[a^3bc^2 + \sum\[a^5b$.
And
$ \sum\[a^6\ge\sum\[a^5b$.
$ (1/2)*(a^5c + ab^2c^3)\ge a^3bc^2$
.${{ (1/2)*(\sum\{a^5c} + \sum\{a^3b^2c})\ge?{a^2b^2c^2}$.
proof.

[quote="moldovan"]The inequality is equivalent with:

$ a^4 \plus{} b^4 \plus{} c^4 \minus{} a^2b^2 \minus{} b^2c^2 \minus{} c^2a^2 \plus{} a^3c \plus{} c^3b \plus{} b^3a \minus{} a^3b \minus{} b^3c \minus{} c^3a\ge 0$[/quote]For non-negative $ a,b,c$, we have a stronger inequality:

$ a^4 \plus{} b^4 \plus{} c^4  \minus{} b^2c^2 \minus{} c^2a^2 \minus{} a^2b^2 \ge 2\left(b^3c \plus{}c^3a\plus{}a^3b \minus{}bc^3\minus{}ca^3\minus{}ab^3\right),$

with equality if $ a\equal{}0,b\equal{}2,c\equal{}\sqrt{5}\minus{}1.$

[quote="dgreenb801"]It's nice, I had the same solution using the factorization $ x^3 \plus{} y^3 \plus{} z^3 \minus{} 3xyx \equal{} (x \plus{} y \plus{} z)(x^2 \plus{} y^2 \plus{} z^2 \minus{} 2xy \minus{} 2yz \minus{} 2xz)$[/quote]

Isn't it $ \sum x^3 \minus{} 3xyz \equal{} (\sum x)(\sum x^2 \minus{} \sum xy)$. Sorry for pointing this out if it appears silly but you mihgt have got a wrong solution had by by mistake used the equality you typed.

O yeah, I made a mistake (I think a typo)  :blush: , but it still works out  :lol:

[quote="Ji Chen"][quote="moldovan"]The inequality is equivalent with:

$ a^4 \plus{} b^4 \plus{} c^4 \minus{} a^2b^2 \minus{} b^2c^2 \minus{} c^2a^2 \plus{} a^3c \plus{} c^3b \plus{} b^3a \minus{} a^3b \minus{} b^3c \minus{} c^3a\ge 0$[/quote]For non-negative $ a,b,c$, we have a stronger inequality:

$ a^4 \plus{} b^4 \plus{} c^4 \minus{} b^2c^2 \minus{} c^2a^2 \minus{} a^2b^2 \ge 2\left(b^3c \plus{} c^3a \plus{} a^3b \minus{} bc^3 \minus{} ca^3 \minus{} ab^3\right),$

with equality if $ a \equal{} 0,b \equal{} 2,c \equal{} \sqrt {5} \minus{} 1.$[/quote]
$ f\equal{}a^4 \plus{} b^4 \plus{} c^4 \minus{} b^2c^2 \minus{} c^2a^2 \minus{} a^2b^2\minus{}2\left(b^3c \plus{} c^3a \plus{} a^3b \minus{} bc^3 \minus{} ca^3 \minus{} ab^3\right)$
the global derivative
$ [f]\equal{}\sum{4a^3}\minus{}2\sum{a(b^2\plus{}c^2)}\minus{}6\sum{a^2b}\plus{}6\sum{a^2c}$
$ \equal{}4\sum{a(a\minus{}b)^2} \geq 0$
when $ a\equal{}0$
$ f\equal{}b^4\plus{}c^4\minus{}b^2c^2\minus{}2b^3c\plus{}2bc^3$
$ \equal{}(b^2\minus{}bc\minus{}c^2)^2 \geq 0$
with equality if $ a \equal{} 0,b \equal{} 2,c \equal{} \sqrt {5} \minus{} 1$


The key is to find the Weierstrass Factorizations for $ \cosh z$ and $ \sinh z$.

I get:

$ \cosh z = \prod_{n=1}^{\infty}\left( 1+\frac{4z^{2}}{(2n-1)^{2}\pi^{2}}\right)$

and

$ \sinh z = \prod_{n=1}^{\infty}\left( 1+\frac{z^{2}}{n^{2}\pi^{2}}\right)$.

Therefore, your infinite product is equal to:

$ \frac{\frac{\pi^{2}}{4}\cosh 1}{\frac{\pi^{2}}{4}\sinh 1}=\frac{\cosh 1}{\sinh 1}=\frac{e^{2}+1}{e^{2}-1}$.


LaTeX
inequalities
inequalities unsolved

[quote="navid"]If a,b,c>=0 Prove that:
                 ( ∑a)(∑(a/(b.^2+c.^2)))>=4.5[/quote]
It's wrong: $a=b=1$ and $c=0.$ :wink:

Yes, the better problem is
\[\sum \frac{a}{b^{2}+c^{2}}\ge \frac{4}{a+b+c}\]
and
\[\sum \frac{a}{b^{2}+c^{2}+d^{2}}\ge \frac{4}{a+b+c+d}. \]
Prove by Cauchy.

[quote="GacKiem"]Yes, the better problem is
\[\sum \frac{a}{b^{2}+c^{2}}\ge \frac{4}{a+b+c}\]
[/quote]
It's very easy:
$\sum \frac{a}{b^{2}+c^{2}}\geq\frac{(a+b+c)^{2}}{\sum( a^{2}b+a^{2}c)}\geq\frac{4}{a+b+c}.$ :wink:

[quote="navid"]Dear arqady:
Yes it is very easy but the problem i purpose is ,made when
you substitue :4 by 4.5.
                                                  Regards.[/quote]

Sorry, it doesn't matter even if a,b,c are positive.

Counter example: $a=0.01$, $b=c=1$

[quote="GacKiem"]A similar one
\[\frac{a}{a^{2}+bc}+\frac{b}{b^{2}+ca}+\frac{c}{c^{2}+ab}\ge \frac{3}{a+b+c}. \]
To navid. Latex here
\[\frac{a(a+b+c)}{b^{2}+c^{2}}+\frac{b(a+b+c)}{c^{2}+a^{2}}+\frac{c(a+b+c)}{a^{2}+b^{2}}\ge \frac{9}{2}. \]
[/quote]

Who can solve the first inequality I post by classical ways? I am curious.


geometry
rectangle
3D geometry
pyramid
frustum
sphere
number theory

ABCD, a rectangle with AB = 12 and BC = 16, is the base of pyramid P,
which has a height of 24. A plane parallel to ABCD is passed through P,
dividing P into a frustum F and a smaller pyramid P’. Let X denote the
center of the circum-sphere of F, and let T denote the apex of P. If the
volume of P is eight times that of P’, then the value of XT can be
expressed as m/n, where m and n are relatively prime positive integers.
Compute the value of m + n.

i need help on how to do the prob

[hide="answer"]177[/hide]

err...Mildorf wrote up solutions documents for all of his mock tests...

[hide="a solution"]
$ P$ is $ 8$ times the volume of $ P'$, which means it was cut at a height of $ 12$. The top of frustum is a $ 6 x 8$ rectangle. Imagine the two bases of the frustum inscribed in on two circles on a sphere, with radii of $ 5$ and $ 10$ (By pythagoras). Using a graph of a circle as a representation, $ y$ being the radius of the cross-section of circles on the sphere, we have the equation:
$ x^2 \plus{} y^2 \equal{} z^2$. Plugging values in, we get:
$ a^2 \plus{} 10^2 \equal{} z^2$ and $ (a \plus{} 12)^2 \plus{} 5^2 \equal{} z^2$
Solving for $ a$ we obtain $ a \equal{} \minus{} 23/8$. The center of sphere being at $ (0,0)$, we have center of frust to top of frust to be $ \minus{} 23/8 \plus{} 12 \equal{} 73/8$. That plus the height of pyramind, $ 73/8 \plus{} 12 \equal{} \mathbf{169/8} \Rightarrow 169 \plus{} 8 \equal{} \mathbf{177}$
[/hide]


Prove that for every integer $ n$ we have:

$ \sum_{k = 1}^{n^2}{\sqrt {k} - \sum_{k = 1}^{n^2}[{\sqrt {k}]\geq\frac {n^2 - 1}{2}}}$

The claim is false. Try $ n \equal{} 2$.

But I think the converse can be proved if you use the following inequality:

$ \sqrt {n^2 \plus{} t} \plus{} \sqrt {(n \plus{} 1)^2 \minus{} t} \le 2n \plus{} 1 \plus{} \frac{1}{2n}$.


trigonometry
geometry
trig identities
Law of Sines
geometry unsolved

Suppose that there is a point $ C$ on the side $ AD$ of a triangle $ ABD$ such that $ \angle ABC \equal{} 90^{\circ}, \angle CBD\equal{}30^{\circ}$ and $ AB\equal{}CD\equal{}1$. Find $ AC$.

I have a brute force way, I will try to find a nicer way.

[hide="Solution"]By the law of sines, $ \frac {1}{2} \equal{} \frac {sin C}{BD}$. We also have $ sin C \equal{} \frac {1}{AC}$. Thus, $ AC \cdot BD \equal{} 2$.
By Pythag, $ BC^2 \equal{} AC^2 \minus{} 1$
By Stewart's theorem, $ 1 \plus{} BD^2 \cdot AC \equal{} (AC^2 \minus{} 1)(AC \plus{} 1) \plus{} AC(AC \plus{} 1)$
Let $ AC \equal{} x$. Then $ BD \equal{} \frac {2}{x}$. Substituting and simplifying yields
$ x^3 \plus{} 2x^2 \minus{} 2x \minus{} 4 \equal{} 0$
$ (x^3 \minus{} 2)(x \plus{} 2) \equal{} 0$
So the only real positive root is $ AC \equal{} \sqrt [3] {2}$[/hide]


Never ahead, ever behind,
Yet flying swiftly past;
For a child I last forever,
For adults I'm gone too fast.

Who or what corresponds to "I" referred to in the problem?


I submitted my Round 4 solutions about a week and a half ago, but there is still no green checkmark by "Round 4" in my score report on My USAMTS (the one that indicates that my solutions were received).  Is this because the checkmarks won't appear until after the submission deadline?

Wow, you finished quickly.

You should get a checkmark soon, or maybe at the deadline.

[quote="BOGTRO"]Wow, you finished quickly.[/quote]
Don't be too impressed.  I did a LOT less than you think  :wink: .
I have other school/aops stuff going on, so I decided not to spend any more time on USAMTS than what I've submitted.

Maybe they check it after the deadline.

You could email them or something if you are really concerned.

BOGTRO, I dont think its that when people turn it in early, they finish quickly.  I think its because they dont procrastinate.  They probably took the same amount of time as everyone else, but it seems way shorter (since its earlier) because most people dont start until the last week.

I don't know about that, I've been laboring over round 4 and I've only got 3 problems.

Maybe that's because I'm 6th grade, idk.

I'm the only 6th grader on the leaderboard BTW  :lol:

meh, i don't really procrastinate. i look at the problems, take a few minutes to make solution outlines, and actually write the solutions on the last day when i've forgotten all my solution outlines

Oh my bad guys.

I was talking basically about myself.  I work on all of the problems the last 2-3 days because I procrastinate (too much school stuff, other competitions)...so it seems like when someone turns it in early, they worked on it a lot...but its just because I dont do any work for the first couple of weeks.

yeah, but why turn it in early?

when you think about a problem, dont get it, and leave it for a while. the solution may come to you.

I submitted my solutions about three and half weeks ago by now, and I still haven't gotten a checkmark.  Should I contact USAMTS?  I still have a copy of my solutions.

If you submitted by mail, I don't believe that we have processed any of them yet.  We will likely not start to do this until Tuesday (March 10).

Does anyone usually know how long priority mail takes?

Im think Im just overreating, but I sent my solutions last week (Wednesday or Thursday), and I did not get a checkmark today (the first day for marking solutions).

I dont think it is a big deal, but the last time I sent the solutions, they were received in 2 days...its been 6 days now.  Again, its probably not a big deal, but how long is too long for priority mail?

Thanks.

Our office manager had jury duty so we have not processed any solutions yet.

If you sent mid-last-week via Priority Mail, we almost certainly have received it.

At this point, there's nothing you can do anyway.  Besides, it doesn't matter when we receive it, so long as the postmark was on time.  Try not to obsess over the checkmark.


At Beaumont High School, there are 12 players on the baseball team.  All 12 players are taking at least one of biology or chemistry.  If 7 players are taking biology and 2 players are taking both sciences, how many players are taking chemistry?

By PIE, $ 7\plus{}c\minus{}2\equal{}12\implies c\equal{}\boxed7$.

You can also draw a venn diagram to solve this problem. There are seven people taking biology, so if you shade the region that have people taking biology, that means the unshaded region is the other five people. This is the region containing the people that just take chemistry. All you have to do is add $ 2$ to get the total number of people who take chemistry. The answer is $ 7$.

[hide]use this

n(a U b) = n(a) + n(b) - n(a intersection b)

hence
12=7+x-2
x=7[/hide]

shouldn't the problem say "if 7 players are taking biology only"?. To me, "7 players are taking biology and 2 are taking both" read as 7 are taking biology in total, which can include the 2 are are also taking chemistry. Am not sure the wording is 100% clear?

[hide=Using a venn diagram][img]https://i.ibb.co/NS7hVS0/aaaaaaaaaaaaaaaaaaaaaaaaaaaaa.png[/img][/hide]

[hide=Explanatory Solution]
There are $7$ players taking biology, and that includes the $2$ players who are taking both sciences. So $7-2=5$ players are taking only biology. This must mean that there are $12-5-2=5$ players taking only chemistry. Tacking on the $2$ players who are taking both sciences gives a final answer of $\boxed{7}$.
[/hide]


Let $ (I_a)$ be a A-excircle of $ ABC, (I_a)$ touch $ BC,CA,AB$ at $ D,E,F$ and $ P$ is apoint on line $ EF$ let $ PB\cap AC\equal{}M,PC\cap AB\equal{}N,FD\cap AC\equal{}I,De\cap AB\equal{}K,PK\cap AC\equal{}L,PI\cap AB\equal{}G,I_aG\cap ME\equal{}X,I_aN\cap FL\equal{}Y,I_aN\cap ME\equal{}S,FL\cap I_aG\equal{}T$ prove that $ XSYT$ is cyclic quadrilateral.

dear encyclopedia:
Is it true?I can't see the result from my figure , maybe I am wrong
could you please draw a figure  :(

Sorry dear friend, my problem has typos , I correct it here

Let $ (I_a)$ be a A-excircle of $ ABC, (I_a)$ touch $ BC,CA,AB$ at $ D,E,F$ and $ P$ is apoint on line $ EF$ let $ PB\cap AC \equal{} M,PC\cap AB \equal{} N,FD\cap AC \equal{} I,DE\cap AB \equal{} K,PK\cap AC \equal{} L,PI\cap AB \equal{} G, I_aN\cap EG \equal{} X,I_aN\cap FL \equal{} Y,I_aM\cap EG \equal{} S,I_aM\cap FL \equal{} T$ prove that $ XSYT$ is cyclic quadrilateral.

[hide="Hint"]$ I_aN\perp FL,I_aM\perp EG$ good luck[/hide]


:o  What a long solution!

[hide="Try this instead"] Let's reduce $\bmod (x^{2}+x+1)$.  Since we have $(x^{2}+x+1)(x-1) = x^{3}-1 \equiv 0$ we have $x^{3}\equiv 1$.  Therefore

$f(x) \equiv x^{0}+2x^{1}+3x^{1}-x^{1}+x^{2}-x^{0}+x^{2}+2x-1 \equiv 2x^{2}+6x-1$

Reducing further, we get $\boxed{4x-3}$. [/hide]


Find the last three digits of the largest multiple of 8 which has no two digits the same.

(AIME B 2003 #2)

[hide]That largest number with no repeating digits is 9876543210. Let's arrange it so that the last three digits are divisible by 8. 9876543120 works fine. The last three digits is 120. [/hide]

The divisibility rule for $8$ is if the last three digits are divisible by $8$, so knowing that the largest number that has no repeating digits is $9876543210$, so you know that the last digit must be either $8, 4, 2, 6, 0$ so the last digit should be $0$ and from there just try the smallest digits left which are $1$ and $2$, $210$ is not divisible by $8$ but $120$ is, therefore, the largest number is $\boxed{9876543120}$.

Here's a question, what's  the smallest $10$ digit number in which no digit repeats which is divisible by $8$?

Just use the same technique.... :lol:

[quote="chess64"]Um, if that's the question, then $1000000000$ ;)[/quote]

Sorry, I meant, where no digit repeats :blush:  :rotfl:

[hide]1023457968? After some guessing and checking... [/hide]

:mad:  Sometimes my mind really wanders. I started thinking about the greatest number at the end :wallbash: :furious:

[quote="joml88"]Shobhit, you don't even know the answer to your own problem :!: :D (just kidding with you)

[hide]
The smallest possible such 10 digit numbers would look like 1023456_ _ _.  We wish to put 7, 8 and 9 in the blanks (the divisibility rule for 8 requires looking at the last three numbers so we focus on them).  Is there a multiple of 8 with only the digits 7, 8 and 9 used and used once?  No.

So we go on to 1023457_ _ _.  Are there any multiples of 8 with the digits 6, 8, and 9?  896 and 968 work but 896 is smaller and thus the answer is $1023457896.$[/hide][/quote]

Sorry, I forgot to work on it :P


Define:
\[ q(n)\equal{} \left\lfloor \frac{n}{\lfloor \sqrt{n} \rfloor} \right\rfloor \qquad (n\equal{}1,2,\ldots)\]
Determine (with proof) all positive integers $ n$ for which $ q(n) > q(n \plus{} 1)$.

[hide="Solution"]Suppose there exist consecutive perfect squares $ a,b$, such that $ a \le n < n \plus{} 1 < b$. Then $ \lfloor \sqrt {n} \rfloor \equal{} \lfloor \sqrt {n \plus{} 1} \rfloor \equal{} \sqrt {a}$.

Then $ \frac {n}{\sqrt {a}} < \frac {n \plus{} 1}{\sqrt {a}}$, so $ q(n) \le q(n \plus{} 1)$.

This means that the only possiblility is when $ n \plus{} 1$ is a perfect square itself. Let $ n\plus{}1\equal{}x^2$ In this case we have:

$ q(n) \equal{} \frac {x^2\minus{}1}{x\minus{}1} \equal{} x\plus{}1$

$ q(n \plus{} 1) \equal{} \frac {x^2}{x} \equal{} x$

So $ q(n)\equal{}q(n\plus{}1)\plus{}1$, and $ q(n) > q(n\plus{}1)$

So the answer is whenever $ n$ is exactly one less than a perfect square.[/hide]

This is sequence [url=http://www.research.att.com/~njas/sequences/A079643]A079643[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].


Let be $ p\in \mathbb{N}\ ,\ p\ge 2$ , fixed . Find $ a\in \mathbb{R}^*$ such that :

$ \lfloor x\rfloor \plus{} \left\lfloor x \plus{} \frac {1}{a} \right\rfloor \plus{} \left\lfloor x \plus{} \frac {2}{a}\right\rfloor \plus{} ... \plus{} \left\lfloor x \plus{} \frac {p \minus{} 1}{a} \right\rfloor \equal{} \lfloor ax\rfloor\ ,\ (\forall)x\in \mathbb{R}$

For $ p\equal{}2$ such $ a$ doesn't exist, since we get $ (\forall x) \left\lfloor x\plus{}{1\over a}\right\rfloor\equal{}0$ which is clearly impossible. It's not difficult to discuss the general case too. Are you sure the statement is correct?

My bad  :oops: Try to solve it now , Farenhajt  :)

[hide="Solution"]Put $ x = 0$:

$ \left[{1\over a}\right] + \dots + \left[{p - 1\over a}\right] = 0\qquad (1)$

Obviously, $ a$ can't be negative (since otherwise all the terms on the LHS would be negative), and also neither of the terms on the LHS can reach or exceed $ 1$, hence all the fractions under the integer part sign are strictly between $ 0$ and $ 1$. Therefore for integer $ n$ and $ k = \overline{1,p - 1}$ we have $ \left[n + {k\over a}\right] = n\quad (*)$

Now put $ x = 1$ and use $ (*)$:

$ p = [a]$

That yields $ p\leqslant a < p + 1\quad (2)$

Now put $ x = - 1$ and use $ (*)$:

$ - p = [ - a]$

That yields $ - p\leqslant - a < - p + 1\iff p - 1 < a\leqslant p\quad (3)$

From $ (2)$ and $ (3)$ we get $ p\leqslant a\land a\leqslant p\implies a = p$

Now we have to show that the identity indeed holds for $ a = p$:

To that end, write $ x$ as $ x = n + {k\over p} + \phi$, where $ n, k$ are integer, $ \phi$ is real, $ 0\leqslant k\leqslant p - 1$ and $ 0\leqslant\phi < {1\over p}$ (in effect, we're "squeezing" $ x$ within a subinterval of the length $ {1\over p}$)

Now the first $ p - k$ terms on the LHS will have the value $ n$, and the remaining $ k$ terms will have the value $ n + 1$, which gives

\begin{eqnarray*}(p - k)n + k(n + 1) & = & pn - kn + kn + k \\
& = & pn + k \\
& = & [pn + k + p\phi] \\
& = & \left[p\left(n + {k\over p} + \phi\right)\right] \\
& = & [px]\end{eqnarray*}

End of proof.[/hide]

[quote="Farenhajt"][hide]Put $ x = 0$:

$ \left[{1\over a}\right] + \dots + \left[{p - 1\over a}\right] = 0\qquad (1)$

Obviously, $ a$ can't be negative (since otherwise all the terms on the LHS would be negative), and also neither of the terms on the LHS can reach or exceed $ 1$, hence all the fractions under the integer part sign are strictly between $ 0$ and $ 1$. Therefore for integer $ n$ and $ k = \overline{1,p - 1}$ we have $ \left[n + {k\over a}\right] = n\quad (*)$

Now put $ x = 1$ and use $ (*)$:

$ p = [a]$

That yields $ p\leqslant a < p + 1\quad (2)$

Now put $ x = - 1$ and use $ (*)$:

$ - p = [ - a]$

That yields $ - p\leqslant - a < - p + 1\iff p - 1 < a\leqslant p\quad (3)$

From $ (2)$ and $ (3)$ we get $ p\leqslant a\land a\leqslant p\implies a = p$

Now we have to show that the identity indeed holds for $ a = p$:

To that end, write $ x$ as $ x = n + {k\over p} + \phi$, where $ n, k$ are integer, $ \phi$ is real, $ 0\leqslant k\leqslant p - 1$ and $ 0\leqslant\phi < {1\over p}$ (in effect, we're "squeezing" $ x$ within a subinterval of the length $ {1\over p}$)

Now the first $ p - k$ terms on the LHS will have the value $ n$, and the remaining $ k$ terms will have the value $ n + 1$, which gives

\begin{eqnarray*}(p - k)n + k(n + 1) & = & pn - kn + kn + k \\
& = & pn + k \\
& = & [pn + k + p\phi] \\
& = & \left[p\left(n + {k\over p} + \phi\right)\right] \\
& = & [px]\end{eqnarray*}

End of proof.[/hide][/quote]

Nice solution , Farenhajt  :thumbup:

[quote="bpgbcg"]For a=p it's just the Hermite Identity.[/quote]

Yes ,  i knew . That's why i said the answer it's obvious , but you can't only say that in a contest


calculus
derivative
function
trigonometry
real analysis
real analysis unsolved

is it possible to find a function $ f: \mathbb{R}\minus{}>\mathbb{R}$ differentiable, with $ f'$ discontinuos?

\[ f(x)\equal{}\left\{\begin{array}{ll}x^2\sin x & x>0 \\ 0 & x\equal{}0 \end{array}\right.\]

That should be $ x^2\sin(1/x)$, of course.

Given a bounded sequence $ \{x_n\}_{n\ge 1}$, if you let $ g_n(x)\equal{}(x\minus{}x_n)^2\sin\left (\frac{1}{x\minus{}x_n}\right )$ for all $ x\neq x_n$ and $ g_n(x)\equal{}0$ if $ x\equal{}x_n$, then the function $ g: \mathbb{R}\to \mathbb{R}$ denoted by

$ g\equal{}\displaystyle\sum_{k\equal{}1}^\infty 2^{\minus{}k}g_k(x)$

is continous and differentiable on $ \mathbb{R}$, but $ g'$ is not continuous at all points in the sequence $ \{x_n\}$.


geometry
rhombus
trapezoid
parallelogram
trigonometry
rectangle
area of a triangle

[hide="Isosceles Trapezoid (2 bases, leg)"]$a$ and $b$ are the bases and $x$ is a leg.

The area of a trapezoid is its height times the average of its bases. In this case, the height can be found using the Pythagorean Theorem. It is $\sqrt{x^{2}-\left(\frac{b-a}{2}\right)^{2}}$. The average of the bases is $\frac{a+b}2$.

$A=\frac{a+b}{2}\sqrt{x^{2}-\left(\frac{b-a}{2}\right)^{2}}$[/hide]

[hide="Isosceles Trapezoid (2 bases, angle)"]$a$ and $b$ are the bases and $\theta$ is the acute angle.

Again, we need to express the height in terms of what we know.

$\tan\theta=\frac{h}{\frac{b-a}{2}}$

$h=\frac{b-a}{2}\tan\theta$

$A=\frac{b-a}{2}\cdot\frac{a+b}{2}\tan\theta=\frac{b^{2}-a^{2}}{4}\tan\theta$[/hide]

[hide="rhombus"]
some geometry theorems state that the diagnals of a rhombus are perpendicular to each other and bisect each other suppose we call one diagnal $a$ and the other one $b$ by SAS congruency theorem, we see that the diagnals cut the rhombus into four congruent right triangles, each with $\frac{a}{2}$ as one leg and $\frac{b}{2}$ as the other. thus we can use the area of a triangle formula to get each triangle to be $\frac{a \cdot b}{8}$ multiply this by $4$ to get the formula for the area of a rhombus:
$\frac{a \cdot b}{2}$ where $a$ and $b$ are the diagnals
[/hide]

to prove rnwang's formula, note that the area of a triangle formed by the diagonals can also be expressed as $\frac{ab \sin C}{8}$ supposing $a$ and $b$ are the diagnals. drawing in these diagonals, you get two pairs of similar traingles. now we must prove that the different pairs still have the same area. since their sides(the diagnals divided by 2) are the same, and since the angles(formed by the diagonals) add up to 180, then the $\sin C$ part would also be the same for the triangles. thus the four triangles have equal areas so multiply the formula by 4 to get what you want. i'm sorry that this is such a bad explanation, but im a bad explainer.

[quote="rnwang2"]The general formula for the area of a parallelogram is

$A=\frac{1}{2}\, d_{1}d_{2}\sin\theta$,

where $d_{1}$, $d_{2}$ are the lengths of the diagonals, and $\theta$ is the measure of the angle formed by the diagonals.

Note that $\theta=90^{\circ}$ for a rhombus.[/quote]
That is one way to look at it, but you could also look at it as (the height of the parrelelogram)*(the corresponding side).  Here is why:
[hide] A parrelelogram is a rectangle, with one triangle taken out of one end and a congruent triangle added to the other end.  If you "shift" the attached congruent triangle to the end which is missing the triangle, the parrelelogram will turn it into a rectangle.  We know the area of this is base*height, so we have that the area of the rectangle is base*(corresponding height).   Sorry, this probably made little sense.  [/hide]

[quote="Interval"]How does one derive the formulas for geometric areas?

Derive the following:

(1) area of rhombus

(2) area of isosceles trapezoid

(3) area of parallelogram[/quote]

1) draw lines between each of the opposite vertices.

2) in terms of what?

may i ask why you split this into 2 topics?

probably to gain an extra post..............
if you haven't realized, he/she posts in the high school basics with many questions, some not even high school level.
I suggest to you, Inteval, to think before posting a question that is not the right level.

[quote="Interval"]To gain an extra post here?

Are you serious?[/quote]

Yeah, he is serious.

Have you actually answered any other questions other people have produced?

[quote="davidlizeng"]Generally, people post problems they are having trouble with, or post answers to problems of other people. the forum is not just for posting random questions from a math book. judging by the fact that you have only joined for <2 months and already you have >600 posts, every single one from posting random problems, yes, I am serious......[/quote]

well I checked his posts from like when he had like 100 or so posts. About 25 were answering other threads. I have asked this before, about why he is spamming these forums, but I couldn't ask it in a polite way. Apparently, he wants to study hard.

he probably isnt doing it for the post count.

i mean cmon, hes 40 years old.  im pretty sure there are better things to do in life than try to get the top poster on a math forum.

anyhow, it appears no one has anymore to say about the actual question so LOCKED.


algebra
binomial theorem
number theory unsolved
number theory

Prove or disprove: For any positive integer $k$ there exists an integer $n > 1$ such that the binomial coeffcient $\binom{n}{i}$  is divisible by $k$ for any $1 \leq i \leq n-1.$

The statement is false.  The case $k = 4$ is a counterexample.  If $n$ is a power of 2, then set $i = n / 2$.  Then we can show that $\binom{n}{i}$ is not a multiple of $4$.  One method is to use [url=http://planetmath.org/encyclopedia/KummersTheorem.html]Kummer's theorem[/url].  Otherwise, suppose that when $n$ is written in binary (base 2), it has a 1 in the $j$th position from the right.  Then set $i = 2^{j}$.  This time, we can show that $\binom{n}{i}$ is odd, and so not a multiple of 4.  Again, we can use Kummer's theorem.

(There is a typo in the linked page: Where it says add $n$ and $n-m$, it means add $m$ and $n-m$.)

for $k=4$
 if $n$ is a power of $2$, $i=\frac{n}{2}$
 if for some $k$, $2^{k}<n<2^{k+1}$, $i=2^{k}$ doesn't work.

It is false if k is not square free ($k\not =rad(k)$).
If $k=rad(k)$ work $n=k^{2}$.

Here's another approach to show that $k = 4$ is a counterexample.  By the Binomial Theorem, we have
\[\sum_{i = 1}^{n-1}\binom{n}{i}= 2^{n}-2. \]
For $n > 1$, the right-hand side is not divisible by 4.  Hence one of the individual terms $\binom{n}{i}$ is not divisible by 4.

to Ravi. Read my post. If $k\not = rad(k)$ it is false.

Yes, I read your post, and agree with what it says.  But since I didn't see a proof there, I thought I would write a simple proof for the case $k = 4$.  You are welcome to write a proof for the general case.

Obviosly $k|n=C_{n}^{1}.$
If $4|k$ work Ravi's proof.
Let $p^{2}|k$ p-odd prime. If $n=p^{k}$ we get \[C_{p^{k}}^{p^{k-1}}=p(mod \ p^{2}).\]
If $n=p^{k}m, p\not |p$, then $p\not | C_{n}^{p^{k}}$.
It proof, that $k=rad(k)=p_{1}...p_{m}$, were $p_{i}$ distinct prime divisors.
If $m>1$ we get $p_{1}\not |C_{n}^{p_{1}}$.
Therefore $k=p$ prime number and $n=p^{k},k\in N$.

This also follows very easy from here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=133601


quadratics trigonometry algebra polynomial

Is there a cubic equational formula for cubic equation? Like Quadratic Equation for second degree equation?

It would be nice if there's any. I asked my trig teacher and she said there is one but it's really long..

And also, sum and product formula too.

yes, there is one thats really long, long enough that its pretty much useless

[img]http://www.math.vanderbilt.edu/~schectex/courses/cubic/cubic.gif[/img]

Don't waste your time memorizing it, but the derivation consists of some beautiful mathematics (google for cubic formula).  By sum product and product formula I presume you mean the sum and product of the roots.  You could play around with the formula (and be ready for tedious algebra calculations) to determine this.

I also recently posted a nice method to solve cubics using trigonometry in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24356]this thread.[/url]

[quote="themonster"][img]http://www.math.vanderbilt.edu/~schectex/courses/cubic/cubic.gif[/img][/quote]Is it just me or should there be three solutions? That looks like one.

i think the squareroot actually is +/- square root?

[quote="AntonioMainenti"]Is it just me or should there be three solutions? That looks like one.[/quote]
I think the given equation gives you a real root.  Once you know a real root, the cubic equation can be reduced to a quadratic and solved in the usual ways.

Formulas for all three roots can be found here:  [url]http://planetmath.org/encyclopedia/CubicFormula.html[/url]

As far as the sum and product of the roots.  As rep123max says, there is a formula that works for all polynomials of any degree.  You would find it instructive to multiply out the roots of a quartic to see for yourself.  Suppose the quartic has roots a, b, c, and d.  Multiply out $(x-a)(x-b)(x-c)(x-d)$ and see the pretty coefficients you get for $x^4,\ x^3,\ x^2,\ x^1,\ \rm{and}\ x^0$.


If $ x,y,z$ are positive real numbers with $ x,y,z < 2$ and $ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 3$ prove that $ \frac{3}{2}<\frac{1\plus{}y^{2}}{x\plus{}2}\plus{}\frac{1\plus{}z^{2}}{y\plus{}2}\plus{}\frac{1\plus{}x^{2}}{z\plus{}2}<3$

[quote="vo thanh van"]If $ x,y,z$ are positive real numbers with $ x,y,z < 2$ and $ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 3$ prove that $ \frac {3}{2} < \frac {1 \plus{} y^{2}}{x \plus{} 2} \plus{} \frac {1 \plus{} z^{2}}{y \plus{} 2} \plus{} \frac {1 \plus{} x^{2}}{z \plus{} 2} < 3$[/quote]
I think there is something joking here. Since with $ x^2\plus{}y^2\plus{}z^2\equal{}3$, then we can easily get that $ x,y,z \le \sqrt{3}<2$. Also, we can even prove that
\[ \sum \frac{x^2\plus{}1}{z\plus{}2} \ge 2\]
Indeed, by AM-GM Inequality and Cauchy Schwarz Inequality, we have
\[ \sum \frac{x^2\plus{}1}{z\plus{}2} \ge 2\sum \frac{x^2\plus{}1}{z^2\plus{}5} \ge \frac{2(x^2\plus{}y^2\plus{}z^2\plus{}3)^2}{\sum (x^2\plus{}1)(z^2\plus{}5)} \equal{}\frac{72}{\sum x^2y^2\plus{} 33} \ge \frac{72}{\frac{1}{3}(x^2\plus{}y^2\plus{}z^2)^2\plus{}33} \equal{}\frac{72}{3\plus{}33} \equal{}2\]

[quote="micf"]I know that is so easy to you, but I just started learning math and I don't know how to use AM-GM here:
$ \sum \frac {x^2 \plus{} 1}{z \plus{} 2} \ge 2\sum \frac {x^2 \plus{} 1}{z^2 \plus{} 5}$[/quote]
Dear friend,
it is
\[ \frac {x^2 \plus{} 1}{z \plus{} 2} \ge \frac {x^2 \plus{} 1} {\frac {z^2 \plus{} 1}{2} \plus{} 2} \equal{} \frac {2(x^2 \plus{} 1)}{z^2 \plus{} 5}
\]
:)


function
integration
real analysis
real analysis unsolved

Let $\alpha$ be a non-decreasing function on $[a, b]$. Define $R(\alpha)$ the set of Riemann-integrable functions on $[a,b]$. In this set, we define a metric $\parallel f-g\parallel = \sqrt{\int_{a}^{b}{|f-g|}^{2}d \alpha}$. Show that the set $C[a,b]$ of continuous functions defined on $[a,b]$ is dense in $R(\alpha)$.

Where's $\alpha$?

As written, this is the $L^{2}$ norm, and we have the classic results of the Lebesgue theory that "nice" functions such as the continuous functions are dense in $L^{p}(I)$ for $1\le p<\infty$ and any compact $I$.
Since Riemann integrable functions are bounded and measurable, they are a subset of each $L^{p}$.

Which means you're really talking about the Riemann-Stieltjes integral.


see : http://pagesperso-orange.fr/gery.huvent/articlespdf/Autour_primitive.pdf
page 14  :lol:


01-1． 李庄的每一位男孩所认识的所有女孩都彼此认识，而每一个女孩所认识的男孩都比她所认识的女孩多。证明：李庄的男孩不少于女孩。

01-2．数列1，9，8，2，…自第5项起，每一项都等于它前面四项之和的个位数。试问，是否在该数列中有连续的四项为3，0，4，4。

01-3． 试画出一个图形，使得用它不能覆盖一个半径为1的半圆；但用两个这样的图形却可以覆盖一个半径为1的圆。

01-4． 在国际象棋棋盘上画圆，使圆周只经过黑色方格。试问，这种圆的半径最大可能为多少？

01-5． 有一个插座，其边缘上均匀地分布着6个插孔；还有一个与之相应的插头。插座上的插洞被按某种方式编为1至6号，插头上的插栓也作了相应的编号。证明：可以把插头插入插座，使得所有插栓都不插在对号的插孔内。如果插孔和插栓的数目是7啦？

01-6．  一个图具有n个顶点，每个顶点的度数*不超过5。证明：可以把每个顶点都染为3个颜色之一，使得两个端点同色的边的数目不多于n/2。
*注： 度数，是又给定顶点所连出的边的数目，称为该顶点的度数

01-7． 宇宙空间中飞行着2005架航天飞机，每一架航天飞机上有一位天文学家。航天飞机之间的距离各不相等，每位天文学家都在观察离自己最近的另一架航天飞机。证明：至少有一架航天飞机无人观察。

01-8． （1）桌面上放着一些直径相同的硬币。证明：其中有的硬币至多与3枚其他硬币相切。（2）对于直径不尽相同的硬币，证明：有的硬币至多与5枚其他硬币相切。

01-9． 若干条线段覆盖了区间[0，1]。可以从这些线段中选出若干条互不相交的线段来，它们的长度之和不小于1/2。

01-10．眼花缭乱：正方形的各边9等分，过各分点及顶点在正方形内作另一边及两对角线的平行线，在这个图中有三角形和梯形的个数分别是多少？如果n等分它们的该数分别又是多少？

02-1。 能否将正方形分为若干个钝角三角形？

02-2。如何将一个任意三角形分成三部分，使得可用他们拼成一个矩形？

02-3。 黑板上写着1个1。每一秒钟都将黑板上的数增加上它的各位数字之和。试问，能否经过有限一端时间后，在黑板上出现数123456？

02-4。（1）试找出一个七位数，它的各位数字各不相同，它还可被自己的每一位数整除。 （2） 是否存在这样的八位数？

02-5。求出19^100的各位数字之和，再求出所得之和得各位数字之和，并一直如此进行下去，直至得到一个一位数。该一位数是多少？

02-6。 30把椅子排成一行，随时又人走过来坐在空椅上。但是每当又一个人坐下时，便有一个坐在他邻座上的人起身离开（只要他的邻座上原来有人）。试问，最多可以有多少把椅子上同时坐着人，如果：1） 原来30把椅子全是空的； 2）原来有10把椅子坐着人。

02-7。 给定一张20x30的方格纸，能否作一条直线穿过纸上50个方格的内部？

02-8。 已知自然数a,b,c,d都可以被ab-cd整除。证明：ab-cd 等于1或-1。

02-9。 一堆火柴共3001根。 每人每次可从中取走p^n根(n=0,1,2,... ),p为小于3001任意一个给定质数。谁取得最后一根火柴，就算谁赢，谁有必胜的策略？

02-10。在（1）19x91；（2）19x92的方格中，每人每次涂黑1个或多个小方格，使之成为正方形。已被涂黑的方格不许再涂。谁涂黑了最后一个方格，就算谁赢，谁有必胜的策略？

[*]03-1.求出Fibonacci 小数的和是多少?第n个Fibonacci 小数是把第n个Fibonacci数右移n位所得的数,如下所示:

0.1
0.01
0.002
0.0003
0.00005
0.000008
………………………
0.112359… = ?

[*]03-2.找出一个最小整数,使得其首位为7,当将首位调到末位时,其值为原数的1/3.

[*]03-3.试在平面上标出7个点，使得其中任何3个点都有某两个点的距离等于1。

[*]03-4.6枚5分硬币方在桌面上形成一个‘闭链’。另有1枚5分硬币贴着‘闭链’的外侧边缘作无滑动的转动，依次以外切状态转过链上的所有硬币。它需要转过多少转后回到初始状态？

[*]03-5.试求所有的自然数n，使得n^n+1为不超过19位的质数。

[**]03-6. 数列a(n)具有如下性质：a(1)=1,而对任何n，有a(n+1)-a(n)等于0或1。现知对某个n，有a(n)=n/1000. 证明：存在某个m，使a(m)=m/500.

[**]03-7.一个圆由两个正方形所完全覆盖,正方形的边长为1米,问圆的最大直径为多少?

04-1.[★★]从29x29的方格纸上剪下了99个正方形，它们都由4个方格组成。证明：从这张方格纸上还可以再剪出一个这样的正方形。

04-2[★★] 在平面上给定一个半径为1的圆和5条与圆相交的直线。平面上有一点x与圆心的距离为11.1。证明：如果作x关于第一条直线的对称点x1,再作x1关于第二条直线的对称点x2，如此下去，最终得到x5，则x5不可能位于给定的圆内。

04-3. [★★] 自然数y是由重排自然数x的各位数字后得到的。今知x+y=10^200,证明：x是50的倍数。

04-4[★★] 在直线上给定若干条线段，其中任何两条线段都相交。证明：必有一个点属于所有这些线段。

04-5[★★★] ΔABC的3个顶点都位于坐标平面的整点之上（两个坐标都为整数的点称为整点，又称格点）。现知|AB|>|AC|。证明：|AB|-|AC|>1/p， p是ΔABC的周长。

04-6[★★] 平面被染为（1）两种；（2）3种；（3）100种不同颜色。证明：从中可以找出各个顶点都为同一颜色的矩形。

04-7[★★]. 黑板上写有1,2,...,n。每分钟小李擦掉两个数a,b,然后写上a+b+ab。显然n-1分钟后，黑板上只剩下一个数。问最后剩下的数和操作顺序有关吗？对怎样的n最后的得数是奇数？

04-8.[★★★] 设a(1)=3，且对n≥1有a(n+1)=(3a(n)^2+1)/2 – a(n)，如果 n是3的方幂, 证明n∣a(n) 。

04-9.[★★★] 一个domino是一个1x2或2x1的矩形, 平面某一区域的domino拼铺是全用domino无重叠地覆盖这一区域。问一个3xN的矩形有多少种domino拼铺？

04-10.[ ★★★★] 设 S={1,2,…,n}, T是S的所有非空子集构成的集合。若函数f: T→S满足：对任意A,B∈T，如果A是B的真子集，则有f(A)≠f(B)，则称f为garish的。问有多少garish的函数，并证明你的结论。

05-1.[★]对于三角形ABC，其内切圆与外接圆半径分别为r和R、圆心分别为P与O，求P与O的距离。

05-2[★★]注意到 959^2=919681,919+681=40^2; 960^=921600, 921+600=39^2; 961^2=923521,923+521=38^2. 试对于这种情况建立一个一般的规律。

05-3[★★] 解函数方程： f(x+2f(y))=f(x)+y+f(y)

05-4[★★]如果圆上有16个点，问以这些点为顶点最多可以有多少个锐角三角形？

05-5[★★★]设S1，S2，S3，…，是一列两两不交的整数集，每个集合中恰有2个元素，并且，Sn中的元素的和恰好是n。证明：有无穷多个n，使得Sn中存在一个大于13n/7的元素。

05-6[★★★]一个2n 位的数字的牌照号码，如果前n位数字之和等于后n位数字之和，我们就称这2n位数字牌照号为幸运数。牌照号可以以0开头。

1） 计算6位幸运数的个数；
2） 计算8位幸运数的个数；
3） 计算10位幸运数的个数；
4） 计算2进制的2n位幸运数的个数；
5） 计算m进制的2n位幸运数的个数。

05-7[★★★]考虑平面中 3n个不同的点组成的集合M，任两点之间的距离不超过1。证明：

1） 对M的任何四点中，至少有两个点，其距离不超过√2 / 2。
2） 如果n=2，对于任何ε>0，存在六个点的构型，使其12个距离属于区间（1-ε，1），但不存在这样的构形，使至少13个距离属于（√2/2，1）。
3） 存在一个半径不超过√3/3的圆，含了所有的点。
4） M中存在两点，其距离不超过4/（3√n - √3）。


modular arithmetic
number theory unsolved
number theory

Find all $7$-digit numbers which use only the digits $5$ and $7$ and are divisible by $35$.

On first glance, this problem seems like a brute force whacking-type of problem and isn't very interesting, especially when there are a finite cases to check to get all the possible solutions. Well, by such nature, brute force is unavoidable, but surprisingly, there's a solution I got which needs to consider only $4$ cases, so the 'brute force' is quite minimised.

Before we present the solution, we set up some notation for our convenience. Restricting $X_0, X_1, \ldots, X_k$ to integers from the set $\{ 0, 1, \ldots , 9 \}$, we define $[X_kX_{k-1} \ldots X_1X_0] = \sum^k_{i=0} {10}^iX_i$. For example, letting $X = 2, Y = 3, Z = 5$, we have $[ABC] = 235, [BAC] = 325, [CBA] = 532$.

Back to the question. We want to find all $n = [ABCDEFG] \equiv 0 \pmod{35}$. Obviously $G=5$, so we need to find all $n = [ABCDEF5] \equiv 0 \pmod{7}$. Using the divisibility rule for $7$, which states that $[X_kX_{k-1} \ldots X_1X_0] \equiv 0 \pmod{7}$ iff $[X_kX_{k-1} \ldots X_1] - 2X_0 \equiv 0 \pmod{7}$, it is equivalent to finding all $x = [ABCD(E-1)F] \equiv 0 \pmod{7}$. Using the fact that $1000 \equiv -1 \pmod{7}$, this reduces to $[ABC] \equiv [D(E-1)F] \pmod{7}$, or equivalently, $[ABC] \equiv [DEF] - 3 \pmod{7}$.

Now, we note that for $[X_kX_{k-1} \ldots X_1X_0]$, changing a digit $X_t$ from $5$ to $7$ or from $7$ to $5$ is equivalent to adding or subtracting $2 \cdot {10}^t$ respectively to the original number, thus for the case of $3$-digit integers, we consider all possible combinations of $200a_1 + 20a_2 + 2a_3$, where $a_1, a_2, a_3 \in \{ -1, 0, 1 \}$. We see that out of all these combinations, only $+200, +20-2, -200+20+2, -200-20$ are $\equiv -3 \pmod{7}$ Thus back to $[ABC] \equiv [DEF] - 3 \pmod{7}$, we have four cases:

Case 1: $[ABC] \equiv [DEF] + 200 \pmod{7}$
Since $\{ A, B, C, D, E, F \} \in \{5, 7 \}$, it follows that $D=5, A=7, B = E, C=F$. Thus from this case, the possible solutions for $n$ are $n = 7555555, 7575575, 7755755, 7775775$.

Case 2: $[ABC] \equiv [DEF] +20 - 2 \pmod{7}$
Since $\{ A, B, C, D, E, F \} \in \{5, 7 \}$, it follows that $C=E=5, B=F=7, A=D$. Thus from this case, the possible solutions for $n$ are $n = 5755575, 7757575$.

Case 3: $[ABC] \equiv [DEF] -200 +20 + 2 \pmod{7}$
Since $\{ A, B, C, D, E, F \} \in \{5, 7 \}$, it follows that $A=E=F=5, B=C=D=7$. Thus from this case, the only possible solution for $n$ is $n = 5777555$.

Case 4: $[ABC] \equiv [DEF] -200 - 20 \pmod{7}$
Since $\{ A, B, C, D, E, F \} \in \{5, 7 \}$, it follows that $A=B=5, D=E=7, C=F$. Thus from this case, the possible solutions for $n$ are $n = 5557755, 5577775$.

Thus, from the above 4 cases, there are a total of nine possible numbers satisfying the conditions, namely, $7555555, 7575575, 7755755, 7775775, 5755575, 7757575, 5777555, 5557755, 5577775$.

Nevertheless, the solution is still quite ugly, with the consideration of quite a few cases, but I can't think of any way else to minimise the 'brute force'. Anyone else got better ideas?

[hide=It's not that bashy guys...]
Consider the number \(\overline{a_6a_5a_4a_3a_2a_15}\). This is congruent to
\[10(\overline{a_6a_5a_4a_3a_2a_1}) - 5 \pmod{7}\]
so we must have
\begin{align*}
\overline{a_6a_5a_4a_3a_2a_1} &\equiv 3 \pmod{7} \\
\implies (a_1 - a_4) + 3(a_2 - a_5) + (2a_3 + a_6) &\equiv 3 \pmod{7}
\end{align*}

Observe the following table:
$$\begin{array}{|c | c|}
\hline
a_1 - a_4 & 0, 0, 2, 5 \\
\hline
3(a_2 - a_5) & 0, 0, 1, 6 \\
\hline
2a_3 + a_6 & 0,1,3,5 \\
\hline
\end{array}$$
The right column shows the possible values of \(x = a_1 - a_4\), \(y = 3(a_2 - a_5)\) and \(z = 2a_3 + a_6\). Thus, we want
\[x + y + z \equiv 3 \pmod{7}\]
which is possible iff \((x,y,z)\in \{(0,0,3),(2,0,1),(2,1,0),(5,0,5)\}\). This gives a total of
\[2\cdot 2 \cdot 1 + 1\cdot 2 \cdot 1 + 1\cdot 1 \cdot 1 + 1 \cdot 2 \cdot 1 = \boxed{9}\]
such numbers.
[/hide]


given two reals, r1, r2, r1>r2, prove that there exists a rational q, such that r1>q>r2

use only some basic axioms, for example, the assumtion that all reals have a unique decimal expansion is way too complicated....

The basic property you need is the Archimedean property; there exists a positive integer $ N$ such that $ N > \frac{1}{r_1 \minus{} r_2}$.  Work from here.

but how would someone prove that? would it be helpful for me to post the axioms i am allowed to use?

That depends on where you got this problem from.  In Rudin the Archimedean property can be cited without proof; it is in some sense a defining property of the reals, along with the least upper bound property.

(1)x+y=y+x
(2)x*y=y*x
(3)(x+y)+z=x+(y+z)
(4)(xy)z=x(yz)
(5)there exists a unique additive identity denoted by 0.
(6)there exists a unique multiplicative identity denoted by 1.
(7)for all x, there exists a unique additive inverse denoted -x such that x+(-x)=0
(8)blah blah blah unique nonzero multiplicative inverse if x doesnt equal 0.
(9)a(b+c)=ab+bc
(10)exactly one of the following is true: x<y, x>y, x=y.
(11)x<y implies x+z<y+z
(12)x<y, 0<z, xz<yz
(13) if a set has an upper bound, it has a least upper bound.

these hold for reals, with the appropriate quantifiers.

i think i forgot one... but hopefully these are enough(sorry!)

Ah, my apologies.  According to Rudin (Theorem 1.20), it is possible to prove the Archimedean property from the least upper bound property.

[b]Archimedean property:[/b]  For every $ x, y \in \mathbb{R}$ with $ x > 0$, there exists a positive integer $ n$ such that $ nx > y$.

[i]Proof.[/i]  Suppose there did not exist such $ n$; then $ y$ is an upper bound on $ X \equal{} \{ nx : n \in \mathbb{Z} \}$.  But then this set has a least upper bound $ \alpha$ such that $ \alpha \minus{} x$ is not an upper bound on $ X$, so there exists $ n$ with $ nx > \alpha \minus{} x \Leftrightarrow (n\plus{}1)x > \alpha$; contradiction.

i'm sorry, i don't quite see how the archimedean property helps...

i've never really done proofs just with the axioms so...yeah.

No worries.  It can be difficult to get into this mindset if you're not used to it.

First, by the Archimedean property (setting $ x \equal{} r_1 \minus{} r_2$ and $ y \equal{} 1$) there exists a positive integer $ N$ such that $ N > \frac{1}{r_1 \minus{} r_2}$.  Rewrite this as $ Nr_1 > Nr_2 \plus{} 1$.  In other words, the interval $ (Nr_1, Nr_2)$ has length strictly greater than $ 1$.

Can you see how to prove that this interval must have an integer in it?  (One way is by contradiction.)

Try and prove that if $ x\plus{}1<y$, then there is a natural number such that $ x<n<y$.

ok, thanks i think that is sufficient for what this problem is for.

Thanks for all the help!!


Prove if $n = p_1 ^{\alpha _1 } ...p_s ^{\alpha _s }$ and $n>1$
$\tau (n)\varphi (n) \ge n$

In other words, prove \[ \prod_{j=1}^s\left((1+\alpha_j)\left(p_j^{\alpha_j-1}(p_j-1)\right)\right)\ge n \]

[quote="tµtµ"]I'll use (which are in fact harder to prove than is result ;) ) :

$\sum_{d|n}\varphi (d) = n$
$d|n \Rightarrow \varphi (d) | \varphi (n)$

Then $\sum_{d|n}\varphi (n) = \tau (n)\varphi (n) \geq \sum_{d|n}\varphi (d) = n$[/quote]
I like your solution, I forgot about multiplicative property of Euler totient  :blush:


graph theory
combinatorics unsolved
combinatorics

[quote="Book"]Problem: Suppose $G$ is a simple graph on $10$ vertices that is not connected. Prove that $G$ has at most $36$ edges. Can equality occur?[/quote]I don't get it... what with a full graph on 9 vertices plus 1 single vertex... that's not connected, has $10$ vertices, but has $9!>36$ edges to my count...

Does anyone have an idea what they mean by this? In the solutions they give this "hint": [quote]Let the vertex set $V$ be the disjoint union of $V_1$ and $V_2$, with no edges from $V_1$ to $V_2$. What is the maximum number of edges possible?[/quote]I'd answer the question again by $9!$... :? weird problem.

[ this is problem 2 in the course, so is most likely intended to be trivial ]

How do you come to the faculty¿
The complete graph on $9$ vertives has $\frac{9 \cdot(9-8)}{2}=36$ edges only (one edge between any pair of vertices).

d'oh :fool: sorry, I have no clue where I got the $9!$... :?


An object with a constant mass rests on a smooth and perfectly horizontal table. If a horizontal force F is applied, acceleration a results. If F is doubled without changing the direction, what will be the effect(s) on the acceleration?
1. The acceleration will remain the same
2. The acceleration will be doubled
3.The acceleration will decrease
4.The acceleration will increase but not double
Here's the answer's choice:
(A)1,2,3 only
(B)1 and 2 only
(C) 2 and 4 only
(D)4 only
(E)None of the above
After reading the book's solution, it says that F is the only force that acts on this object(no friction here because of smooth table) and acceleration is proportional to the unbalanced force that acts on an object. So I think the answer must be: The acceleration is doubled, so I choose (E). But the answer from the book is C, which contains 2 contradictory statements. How do you guys think of this question as well as its answer?

You are correct, the acceleration will double so 2 would be the only correct statement, and hence none of the choices are correct.  The book is definitely wrong since it does indeed contain two contradicting statements.  You are correct, so don't worry about it.


quadratics
algebra
linear equation
calculus
calculus computations

Why is it not possible for a line in three-space to be represented by a scalar equation.

That's, uh, not true, unless "scalar equation" means "linear equation."  For example, the graph of the solutions to the equation $ (x \minus{} y)^2 \plus{} (y \minus{} z)^2 \equal{} 0$ is of the line parametrized by $ x \equal{} y \equal{} z \equal{} t$, $ t \in \mathbb{R}$, and probably you can figure out how to modify this trick to get a (quadratic) equation whose solutions are a given line of your choice.

If "scalar equation" does mean "linear equation," then one way of phrasing the answer is to note that, generically speaking, the solution space of a system of $ k$ linear equations in $ n$ variables has dimension $ n \minus{} k$ (each equation kills off one degree of freedom), and that a line is a 1-dimensional object in 3-dimensional space and so needs two linear equations in order to describe it.


let a,b&c a reels such that:
  a+1/b=b+1/c=c+1/a=p
  prov that:
   abc=-p

why do you repost a problem??  :?

http://www.mathlinks.ro/Forum/viewtopic.php?p=341948#p341948


I have three coins A, B, C. I play A, it has 40% chance of getting heads (I lose) and 60% tails (I win), if comes up head I play A again, if I get tails flip B. This second coin has 60% chance of getting heads and 40% chance to tails. If head comes up I picked up A and start all over again. If I get tails, I flip C - 80% chance of getting heads and 20% chance to tails. Finally, if tails comes up I win. When I was flipping the coins depending on the results of heads, the chance to get tails increases and how?
I am in doubt because, for example, to get CC or CK with a fair coin, CK is 75% and CC 25%, then I would like to know in this case.

Question: [quote="Suou"]When I was flipping the coins depending on the results of heads, the chance to get tails increases and how?[/quote]

The game is not well described.  I think this is the question:

There are 3 coins, A, B, and C, with probability of coming up tails 60%, 40%, and 20%.  Start by flipping A.

-When flipping A: If it is tails, flip B.  If A is heads, you lose.
-When flipping B: If B comes up tails, flip C, else go back and flip A.
-When flipping C: If C is tails, you WIN.  else go back and flip B.

The question: at each point in the game, what is the probability of winning the game?

Answer:
Let's start from the end and work backwards.  At the point where the player flips coin C, he has a $ .2$ chance of winning immediately, and a $ .8$ chance of having to flip B.  Let $ P(A)$, $ P(B)$, and $ P(C)$ be the probabilities of winning once the player has to flip A, B, or C, respectively.  So we have $ P(C) \equal{} .2\plus{}.8P(B)$.

Similarly, $ P(B) \equal{} .4P(C)\plus{}.6 P(A)$ and $ P(A) \equal{} .6P(B)$.

So it reduces to algebra.  The last equation allows us to substitute for $ P(A)$, yielding $ P(B) \equal{} .4P(C) \plus{} .36P(B)$, i.e. $ .64P(B) \equal{} .4P(C)$ so $ 1.6P(B) \equal{} P(C)$.  Plugging this in to the first equation I got, $ 1.6P(B) \equal{} .2 \plus{} .8P(B)$, so $ P(B) \equal{} .2/.8 \equal{} .25$.  This yields $ P(C) \equal{} 1.6*.25 \equal{} .4$ and $ P(A) \equal{} .6*.25 \equal{} .15$.

So the player has a 15% chance of eventually winning when they first start the game (or when flipping A later
), 25% if they are flipping coin B, and 40% if they are flipping C.

Did I interpret your question right?

edit:
I see one error...
[quote="facis"]The game is not well described.  I think this is the question:

-When flipping C: If C is tails, you WIN.  else go back and flip B.

Did I interpret your question right?[/quote]
It should be :
- When fliiping C: If C is tails, you WIN, else go back and flip [color=red]A[/color]

I remake, see if is correct:
$ P(C) \equal{} 0,2 \plus{} 0,8P(A)$, $ P(B) \equal{} 0,4P(C) \plus{} 0,6P(A)$ and $ P(A) \equal{} 0,6P(B)$. I got $ P(A) \equal{} 0,062$, $ P(B) \equal{} 0,1$ and $ P(C) \equal{} 0,25$
I think that understood it, but I can not understand why is not $ (0,6)(0,4)(0,2)$ for $ P(A)$...is like to have 3 fair coins, the probability for 3 heads is $ (\frac {1}{2})^3$.

That probability is the probability you win [i]by flipping tails on your first three flips[/i].  However, there are lots of ways to win this game that don't require you to win on your first three flips: for example, you could flip THTTT, or TTHTTT, or lots of other things.

Earlier, you had a question about expected value that you eliminated in the edit.  We can do expected value exactly the same way: let $ E(A)$ be the expected number of steps remaining if you're starting with $ A$, etc.  Then we have for example that $ E(A) \equal{} 1 \plus{} \frac{3}{5} \cdot E(B) \plus{} \frac{2}{5} \cdot 0$, etc.  Solving the resulting system is straightforward.

(These solutions are an example of the use of "state diagrams.")


the number [(17+12(2^1/2))]^1/2 is an example of a number expressed in "embedded radical form" because of the (2^1/2) under the radical.  What is the equivalent value of this number expressed in simplest radical form a+b(c^1/2) where a and b are integers and c is a positive integer containing no perfect sqare factors greater than 1?

Sorry about the weird notation of the expression; i'm still trying to figure out LaTex.  The whole ^1/2 is meant to equal the square root of the number.

Thanks for the help.

I'm just going to put it in LaTeX so I can figure it out:
$ [(17 + 12{({2}^\dfrac{1}{2}))}]^{\dfrac{1}{2}}$

For my answer, I just got $ 17 \plus{} 12\sqrt{2}$

[quote="jonathanchou711"]For my answer, I just got $ 17 \plus{} 12\sqrt {2}$[/quote]
The original problem asked for an [b]embedded radical[/b] simplified to a [b]nonembedded radical[/b]. :wink: Therefore $ \sqrt {17 \plus{} 12\sqrt {2}}$ (the [b]original[/b] embedded radical expression) is not the same as $ 17 \plus{} 12\sqrt {2}$, so what you put down can't be the answer.

[quote="gauss1181"]No, it's supposed to be

$ \sqrt {17 \plus{} 12\sqrt {2}}$. (EDIT: meh, whatever. I just put it in radical form for the general convenience of the problem.)

[hide="Solution"]We want to put the embedded radical in $ a \plus{} b\sqrt {c}$ form. (if we let $ x$ be the embedded radical, we end up having a square root of a squared radical expression in the a-b-c standard form) To do this, we first note that we need $ c \equal{} 2$ and $ ab \equal{} 6$. That means it must be either $ 3 \plus{} 2\sqrt {2}$ or $ 2 \plus{} 3\sqrt {2}$. Checking both reveals that the first expression squared gives the radicand of the original embedded radical.

$ (3 \plus{} 2\sqrt {2})^2 \equal{} 9 \plus{} 12\sqrt {2} \plus{} 8 \equal{} 17 \plus{} 12\sqrt {2}$

So the equivalent value of the number expressed in the form $ a \plus{} b\sqrt {c}$ is $ \boxed{3 \plus{} 2\sqrt {2}}$. :D  :D  :D [/hide][/quote]

I appreciate the answer but i do not really understand how you knew that ab=6 and that c=2.

Thanks gaussy! But how is ab=6???????  :huh:

When you expand a squared expression, in this case, you get $ (a\plus{}b\sqrt{c})^2\equal{}a^2\plus{}2ab\sqrt{c}\plus{}b^2c$. Then when using corresponding coefficients, you get the fact that $ c\equal{}2$ and $ 2ab\equal{}12$, which, in simplified form, is $ ab\equal{}6$. Also $ a^2\plus{}2b^2\equal{}17$, so the only case is with $ a\equal{}3$ and $ b\equal{}2$. :)


continued fraction
algebra unsolved
algebra

Given $ a,b,c \in Z$, not all are zero.  and $ |a| , |b| , |c| < 1000$. Prove that:
$ | a\plus{} b \sqrt{2} \plus{} c\sqrt{3} | > 10^{\minus{}21}$

$ (a\plus{}b\sqrt 2 \plus{}c\sqrt 3)(a\plus{}b\sqrt 2\minus{}c\sqrt 3)(a\minus{}b\sqrt 2\plus{}c\sqrt 3)(a\minus{}b\sqrt 2\minus{}c\sqrt 3)\in Z$.
Therefore $ |a\plus{}bsqrt 2\plus{}c\sqrt 3|\ge \frac{1}{|8abc\sqrt 6|(1\plus{}\epsilon )}$.

[quote="long14893"]Given $ a,b,c \in Z$, not all are zero.  and $ |a| , |b| , |c| < 1000$. Prove that:
$ | a \plus{} b \sqrt {2} \plus{} c\sqrt {3} | > 10^{ \minus{} 21}$[/quote]

Suppose we have $ a,b,c\in\mathbb Z$ such that $ |a\plus{}b\sqrt 2 \plus{}c\sqrt 3|<10^{\minus{}21}$
1) suppose $ bc\neq 0$
$ \implies$ $ |a|\minus{}10^{\minus{}21}<|b\sqrt 2 \plus{} c\sqrt 3|<|a|\plus{}10^{\minus{}21}$
$ \implies$ $ a^2\minus{}2|a|10^{\minus{}21}\plus{}10^{\minus{}42}<2b^2\plus{}3c^2\plus{}2bc\sqrt 6<a^2\plus{}2|a|10^{\minus{}21}\plus{}10^{\minus{}42}$
$ \implies$ $ N\minus{}2|a|10^{\minus{}21}\plus{}10^{\minus{}42}<2bc\sqrt 6<N\plus{}2|a|10^{\minus{}21}\plus{}10^{\minus{}42}$

$ \implies$ $ |\sqrt 6 \minus{}\frac{|N|}{|2bc|}|<\frac{|a|}{|bc|}10^{\minus{}21}\plus{}\frac{1}{2|bc|}10^{\minus{}42}$

$ \implies$ $ |\sqrt 6 \minus{}\frac{|N|}{|2bc|}|<10^{\minus{}17}$
And so we have a $ 10^{\minus{}17}$ rational approximation of $ \sqrt 6$ whose denominator is less than $ 2,000,000$

But the continued fraction of $ \sqrt 6$ is $ [2;\overline{2,4}]$ and $ u\equal{}[2;2,4,2,4,2,4,2,4,2,4,2,4,2,4]\equal{}\frac{20721118}{8459361}$
And this $ u$ is the best approximation of $ \sqrt 6$ with denominator less than $ 8459361$ and this approximation is such that $ |\sqrt 6 \minus{} u|>10^{\minus{}15}$
Hence the result.

2) if $ b\equal{}c\equal{}0$, $ a\neq 0$ and the result is obvious

3) if $ b\equal{}0$ and $ c\neq 0$, the same method than in 1) but with $ \sqrt 3$ continued fraction gives the result.

4) if $ b\neq 0$ and $ c\equal{}0$, the same method than in 1) but with $ \sqrt 2$ continued fraction gives the result.

[hide=" Using weak bounds, but quick"]
Let $ P \equal{} (a \plus{} b\sqrt 2 \plus{} c\sqrt 3)$, $ Q \equal{} (a \plus{} b\sqrt 2 \minus{} c\sqrt 3)$, $ R \equal{} (a \minus{} b\sqrt 2 \plus{} c\sqrt 3)$, $ S\equal{}(a \minus{} b\sqrt 2 \minus{} c\sqrt 3)$, and $ N\equal{}PQRS$.

It is easy to see that $ N \equal{} a^4\minus{}4 a^2 b^2\minus{}6 a^2 c^2\plus{}4 b^4\minus{}12 b^2 c^2\plus{}9 c^4 \in \mathbb{Z}$, and that $ N \neq 0$.

[hide="Formal Proof that N is not equal to 0"]If $ N\equal{}0$, then one of $ P,Q,R,S$ is $ 0$.

If $ a \pm b\sqrt{2} \pm c\sqrt{3} \equal{} 0$, then we have:

$ a \pm b\sqrt{2} \equal{} \pm c\sqrt{3} \Rightarrow a^2 \pm 2ab\sqrt{2} \plus{} 2b^2 \equal{} 3c^2 \Rightarrow$ $ \sqrt{2} \equal{} \pm \dfrac{3c^2\minus{}a^2\minus{}2b^2}{2ab} \text{ \ OR \ } ab \equal{} 0$

$ a \pm c\sqrt{3} \equal{} \pm b\sqrt{2} \Rightarrow a^2 \pm 2ac\sqrt{3} \plus{} 3c^2 \equal{} 2b^2 \Rightarrow$ $ \sqrt{3} \equal{} \pm \dfrac{2b^2\minus{}a^2\minus{}3c^2}{2ac} \text{ \ OR \ } ac \equal{} 0$

$ a \equal{} \pm b\sqrt{2} \pm c\sqrt{3} \Rightarrow a^2  \equal{} 2b^2 \pm 2bc\sqrt{6} \plus{} 3c^2 \Rightarrow$ $ \sqrt{6} \equal{} \pm \dfrac{a^2\minus{}2b^2\minus{}3c^2}{2bc} \text{ \ OR \ } bc \equal{} 0$

Since $ \sqrt{2},\sqrt{3},\sqrt{6} \not\in \mathbb{Q}$, we have $ ab \equal{} ac \equal{} bc \equal{} 0$.

Clearly, we must have two of $ a,b,c$ equal to $ 0$.

If $ a\equal{}b\equal{}0$, then $ a \pm b\sqrt{2} \pm c\sqrt{3} \equal{} 0 \Rightarrow \pm c\sqrt{3} \equal{} 0 \Rightarrow c\equal{}0$

If $ a\equal{}c\equal{}0$, then $ a \pm b\sqrt{2} \pm c\sqrt{3} \equal{} 0 \Rightarrow \pm b\sqrt{2} \equal{} 0 \Rightarrow b\equal{}0$

If $ b\equal{}c\equal{}0$, then $ a \pm b\sqrt{2} \pm c\sqrt{3} \equal{} 0 \Rightarrow a\equal{}0$

Thus, if $ a \pm b\sqrt{2} \pm c\sqrt{3} \equal{} 0$ then $ a\equal{}b\equal{}c\equal{}0$. This is a contradiction. Therefore, $ N \neq 0$. [/hide]

Thus, $ |N| \ge 1$.

Also, $ |a \pm b\sqrt{2} \pm c\sqrt{3}| \le 1000(1\plus{}\sqrt{2}\plus{}\sqrt{3}) \le 5000$.

Therefore, $ |a \plus{} b\sqrt 2 \plus{} c\sqrt 3| \equal{} \dfrac{|N|}{|Q|\cdot|R|\cdot|S|} \ge \dfrac{1}{5000^3} \equal{} 8 \times 10^{\minus{}12}$.
[/hide]

[quote="pco"][quote="Rust"]$ (a \plus{} b\sqrt 2 \plus{} c\sqrt 3)(a \plus{} b\sqrt 2 \minus{} c\sqrt 3)(a \minus{} b\sqrt 2 \plus{} c\sqrt 3)(a \minus{} b\sqrt 2 \minus{} c\sqrt 3)\in Z$.
Therefore $ |a \plus{} bsqrt 2 \plus{} c\sqrt 3|\ge \frac {1}{|8abc\sqrt 6|(1 \plus{} \epsilon )}$.[/quote]

and ?[/quote]
Isn't it very straightfoward?
First you have to prove that $ a \plus{} b\sqrt {2} \plus{} c\sqrt {3} \equal{} 0 \iff a \equal{} b \equal{} c \equal{} 0$. [hide="Proof"]
Of course $ a \plus{} b\sqrt {2} \plus{} c\sqrt {3} \equal{} 0 \in \mathbb{Z} \Rightarrow b\sqrt {2} \plus{} c\sqrt {3} \in \mathbb{Z}$ $ \Rightarrow 2b \plus{} 3c \plus{} 2bc\sqrt {6} \in \mathbb{Z} \iff bc \equal{} 0$. And it's obvious.
[/hide]
Then $ (a \plus{} b\sqrt 2 \plus{} c\sqrt 3)(a \plus{} b\sqrt 2 \minus{} c\sqrt 3)(a \minus{} b\sqrt 2 \plus{} c\sqrt 3)(a \minus{} b\sqrt 2 \minus{} c\sqrt 3) \equal{} n \in \mathbb{Z}, |n| \ge 1$.
So $ |a \plus{} b\sqrt {2} \plus{} c\sqrt {3}| \ge \frac {1}{|a \plus{} b\sqrt 2 \minus{} c\sqrt 3| \cdot |a \minus{} b\sqrt 2 \plus{} c\sqrt 3| \cdot |a \minus{} b\sqrt 2 \minus{} c\sqrt 3|}$
But $ |a \pm b\sqrt {2} \pm c\sqrt {3}| \le |a| \plus{} |b\sqrt {2}| \plus{} |c\sqrt {3}| < 10^3(1 \plus{} \sqrt {2} \plus{} \sqrt {3}) < 10^4$.
Hence $ |a \plus{} b\sqrt {2} \plus{} c\sqrt {3}| > \frac {1}{10^{12}} > \frac {1}{10^{21}}$

[quote="pco"][quote="Mathias_DK"] ...Isn't it very straightfoward?[/quote]

I'm sorry, but no, it was not very straightfoward for me.
I'm surely rather narrowminded and of a very little level according to all the other high-level members of this forum.

I beg your great pardon for this poor situation.[/quote]
I didn't want to be rude. Are you joking? :) It's very hard to see on text - also because I don't know you personally. But when I read "of a very little level according to all the other high-level members" I can't help to think that you are joking :)

If $ a\plus{}b\sqrt 2 \plus{}c\sqrt 3\equal{}x, |x|<\epsilon$, then $ a\plus{}b\sqrt 2\minus{}c\sqrt 3\equal{}\minus{}2c\sqrt 3\plus{}x$,
$ a\minus{}b\sqrt 2\plus{}c\sqrt 3\equal{}\minus{}2b\sqrt 2\plus{}x, a\minus{}b\sqrt 2\minus{}c\sqrt 3\equal{}2a\minus{}x$.
Therefore $ n\equal{}x(\minus{}2c\sqrt 3 \plus{}x)(\minus{}2b\sqrt 2 \plus{}x)(2a\minus{}x)$
$ |x|>\frac{1}{8|abc|\sqrt 6 \plus{}4|x|(|bc|\sqrt 6\plus{}|ac|\sqrt 3 \plus{}|ab|\sqrt 2)\plus{}2x^2(|a|\plus{}|b|\sqrt 2\plus{}|c|\sqrt 3)\plus{}|x|^3}$.
If $ \minus{}N\le a,b,c\le 1000$, then $ |x|>\frac{1}{8N^3\sqrt 6\plus{}1}$

[quote="pco"][quote="Mathias_DK"] I didn't want to be rude. Are you joking? :) It's very hard to see on text - also because I don't know you personally. But when I read "of a very little level according to all the other high-level members" I can't help to think that you are joking :)[/quote]

Np. I was.

And the solution is nice. But I really did not see it at first glance.
Thanks. :)[/quote]
thx all :D I wonder if there is any solution uses Dirichlet lemma:D ?


superior algebra
superior algebra unsolved

Let $G$ be a finite group such that $H\leq G$ and
          $\forall x\notin H,$     $H\cap x^{-1}Hx=\left\{e_{G}\right\}$
Prove that $\left|H\right|$ and $\left|G: H\right|$ are relatively prime.

$G$ operates by conjugation on $G \setminus \{e\}$. By assumption, $H \setminus \{e\}$ is an orbit with stabilizer $H$, so $|H|-1 = |G: H|$.

edit: this is wrong

By assumption, $H \setminus \{e\}$ is a union of orbits with stabilizers $H' \leq H$, so $|H|-1 = |G: H|\cdot k$.


Eighty percent of adults drink coffee and $ 70\%$ drink tea. What is the smallest possible percent of adults who drink both coffee and tea?

80% of adults drink coffee, and 70% of adults drink tea.

If you add them, the result should be 100%, but it isn't.

So there must be adults who drink both coffee and tea.

70+80-100=150-100=50.

Answer : 50


[i]POST REMOVED: DO NOT PLACE COMMERCIAL SPAM ON BOARDS[/i]

sorry, not really. this site has it all.

btw, this post is more in the "round table" forumesque category. since this forum is dedicated to math problems.

Actually it's a referral program where he gets some kind of incentive for drawing people to sign up for their services. So it's not really appropriate for the Round Table either.

Locked.


hi . i solved them in a few lines using MA-MG ,but i didn't write the solution because i don't know how to write symbols  :blush:.If somebody tells me how to do that i will post them :lol:

[quote="Svejk"]hi . i solved them in a few lines using MA-MG ,but i didn't write the solution because i don't know how to write symbols .If somebody tells me how to do that i will post them [/quote]
See http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php  :wink:

how can you applicate MA-MG here? I can just see the second is $64 (a^{2}+bc) (b^{2}+ca) (c^{2}+ab) \le (a+b+c)^{6}$. Has that been posted before?

This is a problem of mine, in the contest I gave before the day my book is published in 2006. The general inequality is (very hard, in this contest, too)

\[n^{2n-2}(1+a_{1}^{2})(1+a_{2}^{2})...(1+a_{n}^{2}) \le 2^{n}(a_{1}+a_{2}+...+a_{n})^{2n-2}.\]


inequalities
trigonometry
trig identities
Law of Cosines
geometry solved
geometry

Prove that in every triangle ABC we have:
 \sum a^2>=4S*sqrt3+ \sum (a-b)^2

This is known as the Hadwiger-Finsler inequality.
From the law of cosines, we have :
a 2  = b 2  + c 2  - 2bccos(A) = (b-c) 2  + 2bc(1-cos(A)) = (b-c) 2  + 4Stan(A/2).
Summing similar relations for b 2  and c 2 , we obtain :
a 2  + b 2  + c 2
 = (b-c) 2  + (c-a) 2  + (a-b) 2  + 4S(tan(A/2) + tan(B/2) + tan(C/2))
 \geq  (b-c) 2  + (c-a) 2  + (a-b) 2  + 4Sx3tan( \pi /6)  (from Jensen's inequality, since tan is convex on [0, \pi /2[)
= (b-c) 2  + (c-a) 2  + (a-b) 2  + 4S \sqrt 3.

Pierre.


[quote="E^(pi*i)=-1"]www.artofproblemsolving.com/Forum/viewtopic.php?t=110509[/quote]

Is everyone able to get to that thread?

[quote="Carlos Jacob Rubio Barrio"]Let a,b and n be positive integers. Show that if n divides
a^n-b^n, then n also divides (a^n-b^n)/(a-b).

Here, x^n means, x*x*x ... *x, n times.[/quote]
Let $n=md$, were $d=(n,a-b),m=n/d$. From $n|(a^{n}-b^{n})$ we have $m|\frac{a^{n}-b^{n}}{a-b}$.
We have a=b+cd and $\frac{a^{n}-b^{n}}{d}=\sum_{k=1}^{n}C_{n}^{k}b^{n-k}d^{k-1}.$
If prime p|d and $ord_{p}(n)=r$ then $ord_{p}(C_{n}^{k}p^{k-1})\ge r$. Therefore it is true.


[quote="perfect628"]An isosceles triangle is inscribed in a circle of radius 1. The ratio the length of its base to the length of one of its legs is 1/3. What is the triangle's area?[/quote][hide]Let the length of the base be $2a$. So the length of one of the legs is $6a$. Hence the height of the triangle is $\sqrt{35}a$. So the area of the triangle is $\frac{1}{2}\times 2a \times \sqrt{35}a = \sqrt{35}a^{2}$. Since area of the triangle inscribed in a circle of radius $R$ is $A = \frac{abc}{4R}$, we find $A = \frac{2a \times 6a \times 6a}{4}= 18a^{3}$. Hence $a = \frac{\sqrt{35}}{18}$ and the area of the triangle is $A = \sqrt{35}a^{2}= \boxed{\frac{35\sqrt{35}}{324}}$.[/hide]


Prove that there exist infinitely four positive numbers $ x,y,u,v$ such that all the numbers :
$ xy+1,xu+1,xv+1,yv+1,yu+1,uv+1$ are perfect squares.