In the following set of inequalities, each is drasticaly harder to prove than the previous ones, most of which are there just to point out what comprises the difficulty of the former problem. They all pertain to three positive real numbers $a,b,c\geq 0$ such that $a+b+c=1.$

(a) Prove the inequality
\[\frac{1}{27}\geq\left(a^{2}+b^{2}+c^{2}\right)\left(ab+bc+ca\right)^{2}\geq abc. \]
(b) Prove the inequality
\[\frac{1}{27}-\frac{1}{3}\left(\frac{1}{9}-ab-bc-ca\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right)\geq abc. \]
(c) Prove the inequality
\[\left(a^{2}+b^{2}+c^{2}\right)\left(ab+bc+ca\right)^{2}\geq \frac{1}{27}-\frac{1}{3}\left(\frac{1}{2}-ab-bc-ca\right)\left(a^{2}+b^{2}+c^{2}-ab-bc-ca\right). \]
Note that the right side of $(a)$ is almost trivial, whereas the left side has a very nice factorization, which is the base of the solution to $(c).$ $(b)$ has no connection to the other two at all, it rather serves as a reason of why is $(c)$ stronger than $(a).$ It may in fact be the hardest of the three, as the only solution I have uses bunching.

[color=blue]EDIT: Bound in (c) improved.[/color]

They are nice, and you say that (b) is most difficult. So I will prove (b) first.  I use SOS.

Notice that
\[(a+b+c)^{3}-27abc=\sum a^{3}+3\sum a^{2}(b+c)-21abc \]

\[=\sum (a+b-c)(a-b)^{2}+4\sum c(a-b)^{2}=\sum (a+b+3c)(a-b)^{2}. \]
We need to prove that
\[2(1-27abc) \ge (1-9\sum ab)(\sum (a-b)^{2}) \]
or
\[(a+b+5c+9\sum ab)(a-b)^{2}\ge 0. \]
However, this is clearly true. I think the following one is better
\[(d)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{27}-\frac{1}{3}\left(\frac{1}{9}+\frac{1}{3}(ab+bc+ca)\right)(a^{2}+b^{2}+c^{2}-ab-bc-ca) \ge abc. \]
For (a), the left-hand inequality is just AM-GM directly. The right-hand inequality comes from
\[(ab+bc+ca)^{2}\ge 3abc(a+b+c) \]
and
\[(a^{2}+b^{2}+c^{2}) \ge \frac{1}{3}. \]
For $(c)$, I feel suspicious.

[quote="GacKiem"]
For $(c)$, I feel suspicious.
[/quote]

I checked once again and you are right, the bound needs to be improved. The following inequality, however, is valid:
\[(a^{2}+b^{2}+c^{2})(ab+bc+ca)^{2}\geq \frac{1}{27}-\frac{1}{3}\left(\frac{1}{2}-ab-bc-ca\right)(a^{2}+b^{2}+c^{2}-ab-bc-ca) \]
I will hide the solution in case you want to try to solve the problem.

[hide="Solution"]
The main idea behind the solution is the following factorization:
\[\frac{(a+b+c)^{6}}{27}-(a^{2}+b^{2}+c^{2})(ab+bc+ca)^{2}= \] \[= \frac{1}{27}(a^{2}+b^{2}+c^{2}-ab-bc-ca)^{2}(a^{2}+b^{2}+c^{2}+8ab+8bc+8ca). \]
Using this identity we obtain that the original inequality is equivalent to
\[\frac{3}{2}-3(ab+bc+ca)\geq (a^{2}+b^{2}+c^{2}-ab-bc-ca)(a^{2}+b^{2}+c^{2}+8ab+8bc+8ca). \]
Now by putting $p=ab+bc+ca$ and having in mind that $a^{2}+b^{2}+c^{2}=1-2p$ we want to prove that
\[\frac{3}{2}-3p\geq (1-3p)(1+6p)=1+3p-18p^{2}\]
or
\[1+36p^{2}\geq 12p \]
which is self-evident from $AM-GM$.

[b]Remark. [/b] We can prove that
\[(a^{2}+b^{2}+c^{2})(ab+bc+ca)^{2}\geq \frac{1}{27}-p\left(q-ab-bc-ca\right)(a^{2}+b^{2}+c^{2}-ab-bc-ca) \]
where $q=\frac{1}{24}\left(2+\frac{3}{p}+3p\right)$ for all positive $p$ (the solution is analoguous to the one above). Although we can't reach the bound for which $(b)$ holds this way, we can get infinitely close $(p\rightarrow 0)$. Moreover, the inequality is not much weaker as $(b)$ holds in some cases and in some doesn't.
[/hide]


distru prop
using division
remmeber the power rules

dont get what you mean i tyed but dont get it ;)  :|


The problem reads:
A man has a fortune that increases at a rate proportional to the square of his present wealth.  If he had 1 million dollars one year ago and has 2 million today, how much will he have at the end of six months?  at the end of two years?  Comment on the plausibility of this model.

I interpert this as:
W(t) = wealth at the end of year t
dW/dt = kt^2
then W(t) = k(t^3)/3 + c  with W(0) = 2 and W(-1) = 1
solving for k and c, k = 3, and c = 2.  Then to answer the question, at the end of six months W(.5) = 2.125 million and after two years W(2)= 10 million.  My conlcusion is that this model is absurd by real world standards.  Is this correct?

You wrote $\frac{dW}{dt}=kt^{2}.$

That is not what the problem said. It didn't say anything about the square of the the time; it spoke of the square of his present wealth. In other words, the equation is
\[\frac{dW}{dt}=kW^{2}\]
Solutions to that "blow up": they tend to infinity in a finite amount of time. I'd say that's a pretty serious offense against plausibility.

So then W(t) = -1/(kt) + c right?
But then how do I solve for c and k since W(0) is infinity?

You missed the parentheses in the denominator; that $c$ is part of the denominator.

It's a little more illuminating to write the answer as

$W(t)=\frac{1}{c-kt}.$ Put in an intial condition and we have $W(t)=\frac{W_{0}}{1-W_{0}kt}.$

And that solution is only valid for $t<\frac{1}{W_{0}k}.$

Thanks for your help Kent, I really appreciate.  As a note, this problem is not out of a book, it is a problem (presumably concocted by my professor) on a handout.  I really don't think he would give us a problem with this type of solution.  Do you think that "proportional to the square of his present wealth" might refer to the given constant of 2 million?  So that dW/dt = 4kt?

[quote]Comment on the plausibility of this model. [/quote]
I think your professor did mean $W'=kW^{2}$ and that he knows full well that this is absurd. He wants you to find the absurdity for yourselves.

I understand that, but I don't understand why he would ask how much at the end of six months and how much at the end of two years if there is no real solution.
Edit:
Oops.  :blush:  Nevermind I was still trying to solve it without c in the denominator.


[tex]

\bfseries{If you are not in Math mode, you can use \backslash bfseries}
[/tex]

for bold stuff in text mode, \textbf{bold stuff}. If you're in math mode (as on this site), you'll need \mathbf{bold stuff}. For example:

[tex]\det\mathbf{A}=\det\mathbf{A}^\text{T}[/tex]

[tex]\text{here is some \textbf{bold} text}[/tex]

is it different for fractions?  I can't seem to make them bold.

[quote="cats..."]hmm here is my code:

$S$=$\frac{a(r^n-1)}{r-1}$

When i want to make it bold,
$S$=\mathbf{$\frac{a(r^n-1)}{r-1}$}    ?

Is that right?[/quote]

No: \mathbf works only in math environment, so it should be inside the pair of $$.

$S = \mathbf{\frac{a(r^n - 1)}{r - 1}}$


Hi, im new in this forum and I'm sorry if this isnt the right place to put this,
I wanted to know, how can I train geometry and have good results.
Maybe my method of doing excercises isnt very well, because I can´t do a lot of the excercises and I get frustrated.
Plz help me to get better results :D

Greetings

please help me! how can I train for geometry and get better results? :blush:

Dear Mr Ham and Mathlinkers,
you can visit some sites like
http://www.cip.ifi.lmu.de/~grinberg/   where you can find a great number of articles and a quite complete link
Sincerely
Jean-Louis


Let n be a integer > 5 and m be the greatest positive integer such that $\frac{m(m-3)}{2}<n$
Prove that there exists a m-element subset S of {1,2,...,n} such that no sums (a+b)/$a,b\in{S}$;${a}\neq{b}$ are equal
:)

you've already posted it  :mad:

http://www.mathlinks.ro/Forum/viewtopic.php?p=580491&search_author=tmbtw&show_results=posts&sort_dir=desc&return_chars=200#580491

continue discussion there


vector
linear algebra
matrix
inequalities
real analysis
real analysis unsolved

Let $ h$ be a positive, symmetric bilinear form on a vector space $ V$.
Show that: $ |h(x,y)|^2\leq h(x,x) h(y,y)$

I suppose it is over $ R$. Then one can change basis so that the matrix representation of $ h$ is diag(1,..1,-1,...,-1,0,...,0). Since h is positive definite there is no -1. The inequality is then the consequence of Cauchy Schwarz inequality.


If  \[ f(x\plus{}y\plus{}1)\equal{}\left(\sqrt{f(x)}\plus{}\sqrt{f(y)}\right)^{2}\] for all reals $x$ and $y$ and $f(0)=1$, determine $f(x)$.

Maybe $ f\left(x\right)\equal{}\left(x\plus{}1\right)^{2}$, but I haven't proven it!

plz someone start it......just give me a hint atlest...
i am getting f(x)=1   :P

[quote="eureka-123"]easy question..but answer is not matching :(

If  $ f(x \plus{} y \plus{} 1) \equal{} (\sqrt {f(x)} \plus{} \sqrt {f(y)})^{2}$ and f(0)=1 for all x,y $ \epsilon R$.Then determine f(x)[/quote]

Obviously $ f(x)\geq 0$ $ \forall x\in \mathbb R$
Let then $ f(x)\equal{}g(x)^2$. We get $ |g(x\plus{}y\plus{}1)|\equal{}|g(x)|\plus{}|g(y)|$ $ \forall x,y$

Let then $ h(x)\equal{}|g(x\minus{}1)|$. We get $ h(x\plus{}y\plus{}2)\equal{}h(x\plus{}1)\plus{}h(y\plus{}1)$ And so $ h(x\plus{}y)\equal{}h(x)\plus{}h(y)$ and $ h(x)\geq 0$. This last constraint implies that $ h(x)$ is non decreasing and so that the solution is $ h(x)\equal{}cx$. But then, we need to have $ c\equal{}0$ in order to have $ h(x)\geq 0$ $ \forall x$

So $ f(x)\equal{}0$ $ \forall x$
And so no solution exists since we have the constraint $ f(0)\equal{}1$

[quote="pco"]We get $ |g(x \plus{} y \plus{} 1)| \equal{} |g(x)| \plus{} |g(y)|$ $ \forall x,y$ [/quote]it should be $ |g(x \plus{} y \plus{} 1)|^2 \equal{} |g(x)| \plus{} |g(y)|$
my solution:
1. it's easy to prove that $ f(\minus{}1)\equal{}0$
2. substitution $ y\equal{}\minus{}x\minus{}2$
so $ f(x)\equal{}0$ =contradiction

[quote="Dumel"][quote="pco"]We get $ |g(x \plus{} y \plus{} 1)| \equal{} |g(x)| \plus{} |g(y)|$ $ \forall x,y$ [/quote]it should be $ |g(x \plus{} y \plus{} 1)|^2 \equal{} |g(x)| \plus{} |g(y)|$
[/quote]

No, we have $ g(x \plus{} y \plus{} 1)^2 \equal{} (|g(x)| \plus{} |g(y)|)^2$ and so $ |g(x \plus{} y \plus{} 1)| \equal{} |g(x)| \plus{} |g(y)|$

Btw, your solution is good
... and better than mine :)

[quote="eureka-123"]so whats the final answer???????????

ia m getting confused  :([/quote]

The answer is ... no such function exists.

kkk...............thanx pico and everyone else.. :)


1. three landscape views of a building made from cubes are shown below. How many cubes were used to make the building?  l shaped front view using 4 cubes 6 cubes in the form of a square and two in front top-sown view view from the right side is 4 cubes on the bottom 2 in the middle and 1 on top  how do i set this problem up

[quote="MCNational!"]wait is there supposed to be a figure?

 :huh:[/quote]

can you post the figure? :roll:

The person described it, but without any punctuation and poor grammar, so it is not very comprehendable.  Yay!!  I used a 14-letter word!

spelled wat wrong?? comprehendable?? he spelled it rite!!  :o

That was trivial and should be considered  :spam: NO SPAM ON AOPS!!!!  :mad:


ratio AMC USA(J)MO USAMO USAMTS Support function

I don't completely understand "proving" yet... Here is a problem and my solution. Am I missing something in my solution or unclear at any point? Thanks if you can help me out.  :)

Bob plays a game. Every round, he keeps on flipping a fair coin until he gets tails. If he plays 2008 rounds, how many times in total is he expected to flip the coin?

My Solution:
[hide]4016[/hide]

My Solution: If possible, could you rate my solution from 1 to 10? Thanks
[hide]
First, let's find out how many coins Bob is expected to flip in one tound. There is a $ \frac {1}{2}$ chance of flipping a tail on the first flip, thus flipping $ 1$ time. There is a $ \frac {1}{4}$ chance of flipping a head then a tail, flipping the coin $ 2$ times. For each one more flip, there is a half chance of getting that. To find the expected number of flips per round, we get: $ \frac {1}{2} \cdot 1 \plus{} \frac {1}{4} \cdot 2 \plus{} \frac {1}{8} \cdot 3 \plus{} \dots$ or $ \frac {1}{2} \plus{} \frac {2}{4} \plus{} \frac {3}{8} \plus{} \dots$ Let this expression equal $ x$.

To solve this expression, we can have a seperate expression: $ \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \dots$.
Solving this one is much easier, let $ y$ equal the expression. $ y \equal{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \dots$ ==> $ 2y \equal{} 1 \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \dots$ ==> $ 2y \equal{} 1 \plus{} y$ ==> $ y \equal{} 1$.

We can subtract the two equations that equal x and y...:
$ x \equal{} \frac {1}{2} \plus{} \frac {2}{4} \plus{} \frac {3}{8} \plus{} \frac {4}{16} \plus{} \dots$
$ y \equal{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} \frac {1}{8} \plus{} \frac {1}{16} \plus{} \dots$
...to get this. Also, I replaced y with 1:
$ x \minus{} 1 \equal{} \frac {1}{4} \plus{} \frac {2}{8} \plus{} \frac {3}{16} \plus{} \dots$
Now we multiply both sides by 2...
$ 2(x \minus{} 1) \equal{} \frac {1}{2} \plus{} \frac {2}{4} \plus{} \frac {3}{8} \plus{} \dots$
The expression on the right side of the equation above is just $ x$! :) Therefore,
$ 2(x \minus{} 1) \equal{} x$
$ 2x \minus{} 2 \equal{} x$
$ x \equal{} 2$
So Bob flips an average of $ 2$ flips per round. In $ 2008$ rounds, he flips an average of $ 2008 \cdot 2 \equal{} \boxed{4016}$ flips.
[/hide]

This isn't the kind of problem that usually requires "proof," so this isn't the best example.  Your [i]calculations[/i] are correct.  [hide="Would you like to know the properties youre implicitly citing?"]

- Linearity of expected value
- Convergence of geometric series with $ |r| < 1$
- Convergence of series that pass the ratio test

You may not be sure what those mean.  The first one just means that the expected value of $ 2008$ rounds is $ 2008$ times the expected value of one round, which you use in the last step.  The second is required to know that $ \frac {1}{2} \plus{} \frac {1}{4} \plus{} ...$ is finite.  The third is required to know that your expected value is finite.

Here is an example of why you need the third:  Bob is playing another game.  In this game, he flips a fair coin until he gets tails, and if he flips $ h$ coins then he gets $ 4^h$ dollars.  How much money is he expected to make?

By the same reasoning, this is $ \frac {4}{2} \plus{} \frac {4^2}{2^2} \plus{} ... \equal{} 2 \plus{} 2^2 \plus{} 2^3 \plus{} ...$.  Let this equal $ x$.  Then

$ \frac {x}{2} \equal{} 1 \plus{} 2 \plus{} 2^2 \plus{} ... \equal{} x \plus{} 1 \implies$
$ x \equal{} \minus{} 2$

but this is absurd.  Bob seems to be gaining so much money by playing this game, yet by this argument he is [b]losing[/b] money with every play.

The problem is that the expected value we calculated above is [b]infinite.[/b] [/hide]
In most proof problems, you can assume simple things like that, but certain other things that aren't as obvious you'll have to prove.

Generally, an actual proof problem will ask you to prove a statement that's true no matter what variables (integers or reals or whatever) are substituted into the statement.  For example,

[b]Prove:[/b]  $ 1 \plus{} 3 \plus{} 5 \plus{} ... \plus{} (2n \minus{} 1) \equal{} n^2$.

Other types of "proof problems" (such as you might find on competitions) might ask you to find all numbers satisfying some property and then the proof is that no other numbers satisfy that property.

Oh ok thanks.

So is the USAMO about proof problems or just problems that require solutions?

The USAMO is a proof competition; you won't get any points if you give the answer without giving the proof (since the proof is much harder than the answer).  If you want some practice, try the USAMTS; you need to support your answers to get points there, but the standards aren't as rigorous and the problems aren't as hard.

Here's an example that should help you understand why just the answer is unsatisfactory:  suppose I told you that the [b]only[/b] solutions $ (x, n)$ in positive integers to the equation

$ x^2 \plus{} 7 \equal{} 2^n$

are $ (1, 3), (3, 4), (5, 5), (11, 7), (181, 15)$, and that there [b]aren't any others.[/b]  Wouldn't you want to know why I knew that?

Here are some problems:

[hide="Problem 1"]
Show that there is no real-valued function $ f$ on the reals such that $ \frac{f(x)\plus{}f(y)}{2}\ge f(\frac{x\plus{}y}{2})\plus{}|x\minus{}y|$ for all $ x,y$.
[/hide]

[hide="Problem 2"]
$ p(x)$ is a polynomial of degree $ n$ with real coefficients and leading coefficient $ 1$. Show that we can find two polynomials $ q(x)$ and $ r(x)$ which both have degree $ n$, all roots real and leading coefficient $ 1$, such that $ p(x)\equal{}\frac{q(x)}{2}\plus{}\frac{r(x)}{2}$.
[/hide]

[hide="Problem 3"]
Non-negative reals $ x,y,z$ satisfy $ x^2\plus{}y^2\plus{}z^2\plus{}xyz\equal{}4$. Show that $ xyz\le xy\plus{}yz\plus{}zx\le xyz\plus{}2$.
[/hide]

These are all USAMO so they might be a little hard.

Problem #2 seems trivial. Just let $ q(x)$ and $ r(x)$ equal $ p(x)$. Done. QED. Blah blah. :?:

No, it says that $ q(x)$ and $ r(x)$ must have only real roots, and $ p(x)$ does not necessarily meet this criteria.

[quote="l Bob l"]... Bob plays a game. Every round, he keeps on flipping a fair coin until he gets tails. If he plays 2008 rounds, how many times in total is he expected to flip the coin?[/quote]

You've got a geometric distribution with a probability of heads, $ p \equal{}\frac12$ for a fair coin.

For a geometric distribution, $ E(X) \equal{}\frac{1}{p}$. In this case, $ E(X) \equal{} \frac{1}{\frac12} \equal{} 2$.

That is you expect to flip the coin twice to obtain heads. For 2008 rounds, $ \#\text{ of coin flips} \equal{} 2008 E(X) \equal{} 2008\cdot 2 \equal{} 4016$.


inequalities calculus derivative function

i got a very ugly solution, i was looking for some simple solution. give it a try.

Hm, I'm sure if I take the derivative of the function, the derivative will be less than 1 as n approaches infinity...

how about an elementary proof, without calculus.

Meh, taking the derivative of (2^n)^0.5^n shouldn't be too tough, then all I have to do is re-differentiate it to find out a local maximum and plugging in values where the first derivative equals 1...

I don't know how to interpret this for negative numbers, so I'll prove it for the whole numbers, $n \in \mathbb W$.

[hide="Click me."]LHS is $2^{\Sigma^n_{i=1} \frac{i}{2^i}}$. As $n$ approaches infinity, the sum approaches 2.
So the inequality is true for $n \ge 3$, since the sum is never more than 4.
For $n=2$, $\sqrt 2 \cdot \sqrt 2=2\le 2+1$.
For $n=1$, $\sqrt 2 \le 1+1$.
For $n=0$, $1^1=0+1$.[/hide]

[quote="Bictor717"]LHS is $2^{\Sigma^n_{i=1} \frac{i}{2^i}}$. As $n$ approaches infinity, the sum approaches 2.
So the inequality is true for $n \ge 3$, since the sum is never more than 4.
For $n=2$, $\sqrt 2 \cdot \sqrt 2=2\le 2+1$.
For $n=1$, $\sqrt 2 \le 1+1$.
For $n=0$, $1^1=0+1$.[/quote]
That is what I meant :)


For positive integer $ n$, find all positive integers $ a$ such that $ 0<a<10^n$ and $ 10^n\plus{}a|10^{n\plus{}1}\plus{}a$ in terms of $ n$.

Consider $ n > 2$
$ 10^{n \plus{} 1} \plus{} a \equal{} k(10^{n} \plus{} a)$
$ 10^{n}\frac {(10 \minus{} k)}{(k \minus{} 1)} \equal{} a$
$ k$ can take values from 2 to 9.
Since $ a < 10^{n}$ it implies $ 10 \minus{} k < k \minus{} 1$
This implies $ k$ can take values from $ 6$ to $ 9$. Since $ n > 2$ the only values which satisfy $ k$ are $ 6,7,9$
So $ a$ can take values $ (2^{n \plus{} 2} \cdot 5^{n \minus{} 1}) (2^{n \minus{} 1} \cdot 5^{n}) (2^{n \minus{} 3} \cdot 5^{n})$


induction
function
algebra unsolved
algebra

Given a sequence $\{a_n\}$ of natural numbers s.t. $a_1=1$, $a_n$ is the minimal number s.t. $a_n$ doesn't coincide with $a_1,...,a_{n-1}$ and $n\mid a_1+...+a_n$. Prove that $a_{a_n}=n$.

By observation we see that
$a_n=k(n-1)$ or $n+k(n-1)$, where $k(n)=\frac{a_1+...+a_n}{n}$
Use induction on this fact and the fact $a_{a_n}=n$ and we are done.(There are some steps after this, but not difficult then)
Note that $k(n)=k(n-1)$ if $a_n=k(n-1)$ and $k(n)=k(n-1)+1$ otherwise.

By the way, this problem can be asked so:
Find a function $f:N \to N$ such that
1. $f(f(n)=n$
2. $n\mid f(1)+...+f(n)$

could you write it out in full? i observed those few things, but the induction is not obvious (to me at least).

In fact, the facts used for induction should be:
1. For all i=1,2,...,n-1 either $a_i\geq n$ or $a_{a_i}=i$.
2. For all i=2,...,n-1 either $a_i=k(i-1)$ or $a_i=i+k(i-1)$.
Others either come from these two or can be proved directly.

Let's observe one more things. Firstly, if $k(n-1)$ doesn't appear before, then $a_n=k(n-1)$. This fact can be proved immediately. So all the integers less than k(n) must appear in $a_1,...,a_{n-1}$

Secondly, call $n$ is in case 1 when $a_n=k(n-1)$ and in case 2 when $a_n=n+k(n-1)$. Since we cannot have $a_n=a_{n+1}=k(n)=k(n-1)$, we see that either $a_n$ or $a_n+1$ is in case 2 for every integer n(If they obey the laws I gave).

Now use induction. If $a_i=n$ for some $i<n$, we want to show that $a_n=i$. Now we have two possiblilities:

First possiblility : $i-1$ is in case 2. Then $a_{i-1}=i-1+k(i-2)=i+k(i-1)-2=a_i-2=n-2$. So $a_{n-2}=i-1$. Hence $k(n-2)=k(n-3)=i-1$. Now $a_{n-1}$ does not equal to $a_{n-2}$ so $a_{n-1}=n-1+k(n-2)$ and so $k(n-1)=k(n-2)+1=i$ and then $a_n=i$
Second one : $i-1$ is not in case 2 but $i-2$ is in case 2. Then similarly we have $a_{i-2}=n-3$, $a_{n-3}=i-2$ and that leads to $a_n=i$ in a similar way.

If n does not appear before, we want to show that $a_n=n+k(n-1)$. Note that since n and k(n) are both increasing, $a_n=n+k(n-1)$ never appears before so we only have to prove that $a_n=k(n-1)$ is not the choice. If $a_{k(n-1)}<n$, then it's ok. If not, then we also have two possiblility:

First one : $k(n-1)-1$ is in case 2. Then $a_{k(n-1)-1}=a_{k(n)}-2\geq n-1$. Since k(n-1) appears before. We have $a_{n-1}=k(n-1)-1=k(n-1)$ and thus contradiction.
Second one : $k(n-1)-1$ is not in case 2 but $k(n-1)-2$ is in case 2. Then $a_{k(n-1)-2}=a_{k(n)}-3\geq n-2$. Similarly that leads to contradiction.

So we are done.
Sorry for giving a not-so-complete proof. But I've gotta go now. Hope you could understand.

Let $w=\frac 2 {1+\sqrt 5}$ (golden number).
I'm 90 % sure that $a_1+\cdots+a_n=n(1+[wn])$
After noticing this it becomes easy

It is a problem of a brilliant peterburgian mathematician Oleg Izhboldin (1963--2000). It was proposed on 2004 SPb olympiad. Or do you, Myth, know it from other source?


Let $ ABC$ be an isosceles triangle with $ AB\equal{}AC$.The segment $ BC$ and $ AC$ are used as hypotenuses to construct three right triangle $ BCM,BCN$ and $ ACP$.
Prove that,if $ \angle{ACP}\plus{}\angle{BCM}\plus{}\angle{BCN} \equal{} 90$,then the triangle MPN is isosceles.

the problem seems wrong to me  :maybe:  see the picture.
can you send your solution ,please?

[quote="apollo"]$ \angle{ACP} \plus{} \angle{BCM} \plus{} \angle{BCN} \equal{} 90$[/quote]

Should be $ \minus{} \measuredangle{ACP} \plus{} \measuredangle{BCM} \plus{} \measuredangle{BCN} \equal{} 90 \mod \pi$.
all angle are directedangle mod 180.
Let $ X$ be midpoint of $ BC$. Since $ XN \equal{} XM$, $ PN \equal{} PM$ is equivalent to $ PX\perp MN$. From $ \angle APC \equal{} \angle AXC \equal{} 90$,we obtain $ \measuredangle PXC \equal{} \measuredangle PAC \equal{} 90 \minus{} \measuredangle ACP$

$ \measuredangle (PX;MN)$
$ \equal{} \measuredangle (PX;BC) \plus{} \measuredangle (BC,NM)$
$ \equal{} \measuredangle PAC \plus{} (\measuredangle (BC,BN) \plus{} \measuredangle (BN,NM))$
$ \equal{} (90 \minus{} \measuredangle ACP) \plus{} (90 \plus{} \measuredangle BCN) \plus{} \measuredangle BCM \equal{} 90$
Hence $ PX\perp MN$


Use determinants to determine the number of solutions for each of the following systems.

(a) 3x-y=-6
2x=3y+2

(b) 3x-y=7
6x-2y=10

there are no solutions iff |A|=0 where A is the matrix of the coefficients, (like the one in the denominator in cramer's rule)

[quote="Farenhajt"]Wrong. There is no solution IFF the system determinant is zero and at least one of the variable determinants is non-zero.

If all the determinants are zero, the system has infinitely many solutions.[/quote]

woops...forgot about that


geometry
incenter
circumcircle
geometry unsolved

Sorry, I don't understand, for I get that the problem is incorrect.

Let $ E_p$ be the $ P$-excenter of $ \triangle PAD$, $ E_q$ be the $ Q$-excenter of $ \triangle QAB$, $ I_q$ be the incenter of $ \triangle QDC$, and $ I_p$ be the incenter of $ \triangle PBC$.  Now, this means that since the triplets $ (P, I_p, E_p)$, $ (Q, E_q, I_q)$, $ (A, E_q, E_p)$, $ (C, I_q, I_p)$, $ (D, I_q, E_p)$, $ (B, E_q, I_p)$ are collinear, we have that $ M$ is the intersection between $ E_qE_p$ and $ I_qI_p$.  Furthermore, $ L$ is the intersection between $ I_pE_p$ and $ E_qI_q$.  Finally, $ N$ is the intersection between $ E_qI_p$ and $ I_qE_p$.  Clearly, if we take any quadrilateral $ ABCD$, let $ X$ be the intersection of $ AD$ and $ BC$, let $ Y$ be that of $ AB$ and $ CD$, and let $ Z$ be that between $ AC$ and $ BD$, $ X$, $ Y$, and $ Z$ cannot be collinear.  However, applying this to $ E_qE_pI_qI_p$, we see that $ M$, $ N$, and $ L$ cannot collinear unless they coincide (meaning that $ ABCD$ is circumscriptible).

Where do I make a mistake?

[geogebra]1607b6d3a01b0472412d325c5a42ab50beb43171[/geogebra]
The three points definitely do not look collinear.

[quote="jbmorgan"]In quadrilateral $ ABCD$, let $ AB$ and $ CD$ cut at $ P$, and $ BC$ and $ DA$ at $ Q$. Let $ M$ be the intersection of the (internal) angle bisectors at $ A$ and $ C$, and $ N$ be the intersection of the (internal) angle bisectors at $ B$ and $ D$, and $ L$ be the intersection of the (internal) angle bisectors at $ P$ and $ Q$. Prove that $ M,N,L$ are collinear.[/quote]

Assuming $ ABCD$ is convex, let $ L$ be intersection of the [color=red]external[/color] angle bisectors at $ P, Q.$ With this correction, $ M, N, L$ are collinear.
_______________________

Let $ k_A, k_B, k_C, k_D$ be the internal and $ l_A, l_B, l_C, l_D$ the external bisectors of the angles at $ A, B, C, D,$ forming 2 (cyclic) quadrilaterals $ XYZT$ and $ X_eY_eZ_eT_e,$ where $ X \equiv k_A \cap k_B,\ Y \equiv k_B \cap k_C,\ Z \equiv k_C \cap k_D,\ T \equiv k_D \cap k_A$ and similarly for the pairwise intersection $ X_e, Y_e, Z_e, T_e$ of $ l_A, l_B, l_C, l_D.$

Let $ k_P, k_Q$ be the internal bisectors of the angles at $ P, Q$ intersecting at $ K.$ Points $ P, Y, T, Y_e, T_e \in k_P$ are collinear as incenters/P-excenters of $ \triangle PBC, \triangle PDA$ and points $ Q, X, Z, X_e, Z_e \in k_Q$ are also collinear as incenters/Q-excenters of $ \triangle QAB, \triangle QCD.$ Consequently, the quadrilaterals $ XYZT$ and $ X_eY_eZ_eT_e$ have common diagonals intersecting at $ K.$

Let $ l_P, l_Q$ be the external bisectors of the angles at $ P, Q$ intersecting at $ L.$ The intersections $ I_a \equiv k_A \cap l_B \equiv TX \cap X_eY_e,$ $ I_b \equiv k_B \cap l_A \equiv XY \cap T_eX_e,$ $ I_c \equiv k_C \cap l_D \equiv YZ \cap Z_eT_e,$ $ I_d \equiv k_D \cap l_C \equiv ZT \cap Y_eZ_e$ are all on $ l_Q$ as A-, B-, C-, D-excenters of $ \triangle QAB, \triangle QCD.$

$ M \equiv k_A \cap k_C \equiv TX \cap YZ,$ $ N \equiv k_B \cap k_D \equiv XY \cap ZT$ are defined as intersections of the opposite sides of $ XYZT.$ Similarly, let $ M_e \equiv l_A \cap l_C \equiv T_eX_e \cap Y_eZ_e,$ $ N_e \equiv l_B \cap l_D \equiv X_eY_e \cap Z_eT_e$ be intersections of the opposite sides of $ X_eY_eZ_eT_e.$

Projecting the line $ l_Q \equiv I_aI_bI_cI_d$ to infinity makes parallel the corresponding sides of the projected quadrilaterals $ X^*Y^*Z^*T^*, X_e^*T_e^*Z_e^*Y_e^*$ with common diagonals, they become centrally similar with center $ K^*$ and the projected lines $ M^*N^* \parallel N_e^*M_e^*$ become parallel, which means that the original lines $ MN, N_eM_e$ intersect on the line $ l_Q.$ In exactly the same way, by projecting the line $ l_P$ to infinity, we can show that the lines $ MN, N_eM_e$ intersect on the line $ l_P.$ As a result, both $ MN$ and $ M_eN_e$ go through the intersection $ L \equiv l_P \cap l_Q.$ QED.
_______________________

Interesting consequence: Circumcircle $ \odot(KPLQ)$ is perpendicular to the circumcircles $ \odot(XYZT),\ \odot(X_eY_eZ_eT_e).$


ye forume test bara in kara hast : http://www.mathlinks.ro/Forum/index.php?f=224


algebra
polynomial
superior algebra
superior algebra unsolved

For f(X) in F[X] of degree n, if K/F is the splitting field of f(X) over F, prove that [F:K]<= n!

Brute force: adjoin all $ n$ roots of the polynomial, one by one. The $ k$th root $ x_k$ satisfies a polynomial $ \frac{f(x)}{(x\minus{}x_1)(x\minus{}x_2)\cdots (x\minus{}x_{k\minus{}1})}$ of degree $ n\minus{}k\plus{}1$ with coefficients in the previous field.

because splitting fields are essentially unique, this can be done by looking at the explicit construction of a splitting field, namely: take an irreducible factor, and divide it out of the polynomial ring. if it splits there, you're done - otherwise continue with an irreducible factor of the new factorization. now the degree of K[x]/f over K is the degree of f. this actually shows, that the degree of F/K divides n!.


quadratics
conics
parabola
trigonometry
analytic geometry
algebra
system of equations

$(x-2)(x+3)=0$

what do you mean by a real-life situation for this?

the answers are 2 and -3.
Is it to find a question that will give an answer both $2$ or $-3$?

Imagine you had uh..

I dont think this has too much of a real life connetion to real life but algebra does. Because this results in a quadratic it results in 2 answers (because a negitive squared is positive.) You have to make a judgement call which is the correct answer

[quote="tex"]Hello everyone,

I'm new to the forum and may be this kind of a problem has been posted before, but I couldn't find anything in the archives, so here we go:

Find a real life situation that fits the formula (x-2)(x+3)=0

Thanks![/quote]

$(x-2)(x+3)=0$

The solutions are $\{2,-3\}$

You do that:

$x-2=0$
$x=0$

OR

$x+3=0$
$x=-3$

So those are the two solutions. :)

Thanks everyone, I appreciate your efforts to help.

When I say "real life situation" I mean that (x-2)(x+3)=0 must be the mathematical solution of an event that is hapening in our everyday lives, or in the industry etc.

Let  me give an example:

Question: Find a real life situation for the following system of  equations:
2x-y=40
x-y=10

Solution: I sell and buy stocks. Yesterday I sold 2 shares of stock x and bought 1 share of stock y and my balance is 40 dollars. Today, at the same prices, I sold 1 share of stock x and bought 1 share of stock y and my balance is 10 dollars. What is the price of each stock?

Hope this helps in clarifying my question.

:idea: o so like were making this equation into a "word problem" ?

[quote="Ihatepie"]?!?!?!?! how did u think of that?[/quote]

well, that equation is quadratic, meaning that if graphed, it will appear as a parabola, or arch. now the original poster came with the question:

$(x-2)(x+3)$ so you would use FOIL and get the quadratic equation. the rest is purely imaginal

Another real life problem. A formula for the height of a certain projectile (ft.) is $x^{2}+x$ after x seconds (for positive values of x). After how many seconds is its height 72?
With the quadratic you can find that the answer is 8.

[quote="vishalarul"]I'm not sure if this is the kind of word problem you had in mind...

A group of people have decided to build an arch in their city to attract more tourists. They will build it in the form of a parabola (it actually would be hyperbolic cosine, but that's close enough to a parabola) with equation $y=-(x^{2}+x-6)$ (the original parabola would be shaped like an U, so I flipped it upside down to make the arch). They need to know how tall it will be and how wide it will be.

The solution is:
[hide="Click me!"]$y=-(x-2)(x+3)$.

Thus, it intersects the x-axis (the ground) at $x= 2,-3$. Therefore, the width is the positive difference, or $|2-(-3)|=5$. The vertex of the parabola is $( \frac{1}{2}, \frac{21}{2})$. Thus, the height is the y-coordinate of the vertex, or $\frac{21}{2}$.

Even though this might not be the exact equation of the parabola, the townspeople will know how to alter this parabola
to make a pretty good arch. [/hide]
In general, builders often use a parabola to construct arches.[/quote]One little reminder... this is classroom math and not high school intermediate topics. How about giving a little simpler equation or story problem :wink:


1. In base $n$, the number 25 is doubled when its digits are reversed. Find $n$.

2. If the logarithrm of 27 to the base $b$ is 1.5, evaluate $b$.

[quote="4everwise"]Those were high school probs?[/quote]

I'm not sure.. They're from Michian Math Prize Competition.. And I don't live in Michigan so I don't know. :)


Call $ a$ the set of numbers $ \{1,2,3,4,5,5\}$. What is the sum of all 5-digit numbers that can be attained by using the numbers in set $ a$? (Each number can only be used the number of times it appears in $ a$)

My way:
Consider 2 cases, the case where there are two 5s in the numbers, and the case where there is only one five in the number.

My question is, is there is an easier way to do the problem than my way?
However (if there isn't an easier way), how would you add the numbers that have two 5s without using brute force? (I know how to add the ones that only have one five)

[hide="I am not sure if this is right"]Look at every digit separately.

In the ones place: There are $ \frac {5!}{2!} \equal{} 60$ ways to have a one in the ones place. This is also true for all numbers $ 1$ through $ 4$. For $ 5$, there are $ 5!$ ways to have it in the ones digit. The sum of these is $ 60 \cdot 2 \plus{} 60 \cdot 3 \plus{} 60 \cdot 1 \plus{} 60 \cdot 4 \plus{} 120 \cdot 5 \equal{} 1200$.

The same is true for every other digit place, except that you have to multiply by that digit place's value. Thus, the sum becomes:

$ 1200 \plus{} 10 \cdot 1200 \plus{} 100 \cdot 1200 \plus{} 1000\cdot 1200 \plus{} 10000\cdot 1200 \equal{} 11111 \cdot 1200 \equal{} 13 333 200$.
[/hide]

2000th post!

I'm not too sure (I understand your solution, but I'm a bit hesitant).... Can anybody else confirm this?

for the ones that have 2 5's...

each digit different from 5 occurs exactly $ \binom{4}{2}\cdot 4\cdot 3\equal{}72$ times in each  position, and thus these digits contribute $ 11111\cdot (1\plus{}2\plus{}3\plus{}4)\cdot 72 \equal{}7999920$ to the sum. 5 occurs $ 4\cdot 4\cdot 3\cdot 2\equal{}96$ times in each position and thus contributes $ 11111\cdot 5\cdot 96\equal{}5333280$. so the total is $ 13333200$


hello,i suppose both $ H\plus{}$ dissociate in $ H_2SO_4$

[url=http://wiki.answers.com/Q/What_are_all_the_strong_acids_and_strong_bases]Read the note at the bottom[/url]

It says that only one dissociates.

:\ I can't find any error in this solution. What's the correct answer?

I don't know the correct answer. :(
The web program I am using is not accepting 12.43.

I have no idea how to do the problem, but shouldn't the answer be to 3 sig figs?

We have to find the chemical quantity of H+ and HO- from each acid and base, then these ions will react according to H+ + HO- ---> H2O, and the ion that remains in excess (if that happen) will determinate the pH of the solution.

The concentration of H+ in a solution of sulphuric acid with concentration C is given by

$ [H^\plus{}] \equal{} \frac{\sqrt{C^2 \plus{} 6KC \plus{} K^2} \plus{} C \minus{} K}{2}$,

where K is the Ka for HSO4-. On this case, we get $ [H^\plus{}]_{H_2SO_4} \equal{} 0.11\,\,M$, and $ n(H^\plus{}/H_2SO_4) \equal{} 0.11 \cdot 72.0 \times 10^{\minus{}3} \equal{} 7.92 \times 10^{\minus{}3}$.

Can you give me a link where I can find that formula? Thanks!

I tried my solution using 7.92E-3 instead of 0.00720 moles for H2SO4, and I got 12.35, which still doesn't work...

As for the sig figs, maybe you're right, but its not the sig figs that are messing up my solution; the website would tell me that my sig figs are off if my answer was right, but my sig figs were incorrect.

ok i just looked up the Ka value of HSO4- and it was 0.012

From the first dissociation of H2SO4 we get 0.00720 mol HSO4- and H+ (still 0.100 M concentration of both)
From the second dissocation, we get an equilibrium expression of (0.100+x)(x)/(0.100-x)=0.012

Solving for x, x=0.00985 so [H+]=0.100+0.00985=0.10985

0.10985*0.0072L=7.91*10^-3 mol H+ from H2SO4-> confirms Carcul's answer. :\

Other than the H2SO4 contribution to the [H+], is there anything else that could be wrong? Nobody I have asked yet could find anything problematic.  :maybe:

[quote="buzzer11"]I'm not sure where I'm going wrong with my solution.

Consider a solution formed by mixing 72.0 mL of 0.100 M H2SO4, 27.0 mL of 0.100 M HOCl, 25.0 mL of 0.200 M NaOH, 25.0 mL of 0.100 M Ca(OH)2, and 10.0 mL of 0.150 M KOH. Calculate the pH of this solution.

[hide="My Solution"]
H2SO4: Strong acid.
Moles H+ = 0.0720L * 0.100M = 0.00720 moles
Only one of the H+'s dissociate.

HOCl: Weak acid. ICE.
Ka = 3.5E-8 = (x^2)/(0.0027-x)
Moles H+ = x = 9.7E-6

TOTAL MOLES H+ = 0.00720 + 9.7E-6 = 0.00720 moles

NaOH: Strong base.
Moles OH- = 0.0250L * 0.200M = 0.005 moles

Ca(OH)2: Strong base.
Moles OH- = 0.0250L * 0.100M = 0.0025 moles
Both OH-'s dissociate.
Moles OH- = 0.0025 * 2 = 0.005 moles

KOH: Strong base.
Moles OH- = 0.0100L * 0.150M = 0.0015 moles

TOTAL MOLES OH- = 0.005 + 0.005 + 0.0015 = 0.0115 moles

0.00720 moles H+, 0.0115 moles OH-
Then, net moles OH- = 0.0115-0.00720 = 0.0043 moles
[OH-] = 0.0043 moles/(0.072+0.027+0.025+0.025+0.010)L = 0.027M

pOH = -log[OH-] = 1.57
pH = 14-pOH = 12.43

[/hide]

help please![/hide][/quote]

i think you cant write like this because there's "weak acid", if there were all strong acids and bases then it can be true(i think)...but there's weak acid...

[quote="buzzer11"]Then how would one do it differently? I ICEed the weak acid separately..[/quote]

sorry i made a mistake... Ka of HOCl is 2.88x10^-8?not 3.5x10^-8 :)?

Maybe... my book says 3.5E-8, but when I searched for it online, it said 2.9E-8, like you said... hm.

Either way, it contributes an insignificant amount of [H+], so I end up with the same answer.

The correct answer should be something like pH = 11.74.

Hm, that's a different number from what I got, but my teacher said that the correct answer was in the 7-8 range.  :|

Interesting. In that case could your teacher enlighten us with his/her reasoning?

I'm going to ask her about it soon, I will put up the solution when she tells me. :)

The problem is that you are assuming that each weak acid is dissociating as if it were in water by itself. I calculate that you have 10 millimoles of hydroxide ion from strong bases, along with 7.2 millimoles of hydrogen (or hydronium) ion from the first ionization of sulfuric acid. That would give you 2.8 moles oif excess hydroxide ion, which in a total of about 160 mL water would give you a pH of above 12, if no other acid or base is present.
However, that excess hydroxide ion still has to neutralize the 10 millimoles of weak acids present (you have 7.2 millimoles of bisulfate ion and 2.7 millimoles of hypochlorous acid), and there is not enough base around for a complete neutralization. I believe the mix will be slightly acidic.
The dissociation equilibrium of a weak acid is not static. In a solution of weak acid, there is a lot of undissociated acid, but when you start neutralizing it with a strong base, the equilibrium shifts as the acid dissociates more and more liberating those hydrogens that will combine with the hydroxide ions. The final pH is on the basic side, but during the middle of that titration you have a buffer solution with an acidic pH. At half titration pH would be about equal to -log(Ka), with Ka being the dissociation constant of the acid.

1) The book dissociated both H+ ions from the H2SO4.
2) When all the dissociating and reacting was finished, a weak acid and its salt remained. This meant we had to do a buffer problem!


calculus
integration
calculus computations

the integral is $2\xi(3)$.
Use the power series expansion of the denominator in terms of $e^{-x}$ and you should get it.
Didi

Yep, that was mine solution. For ksi you of course mean zeta :)?


modular arithmetic
search
pigeonhole principle

prove that the product of five consecutive integers cannot be the square of a positive integer...

hint: you will need to consider the divisors of consecutive numbers. sumthing tells me that the product of $r$ consecutive numbers in a row is divisible by $r!$

Bleh, this isn't rigorous, but it's what I thought immediately after reading the problem.

[hide]Let's say that the sequence is $n$, $n+1$, $n+2$, $n+3$, and $n+4$.

Notice $n$ cannot be divisible by any integer greater than $4$.

If this is so, if $n | p$, where $p > 4$, $ n \pmod{p} \not\equiv n + 1 \pmod{p} \ldots \not\equiv n + 4 \pmod{p}$. So, no perfect square.

We can see that $n$ must be divisible by 1, 2, 3, and/or 4.

If $n | 2^m$, then $(n + 2) | 2^{m \pm 1}$, and $(n + 4)  | 2^m$. Thus, the number will have an odd number of 2 factors.

So, $n$ cannot be even.

If $n | 3$, then $(n+3) | 2^q$. So, $(n + 1) | 2^{q \pm 1}$. Odd number of 2 factors. And no perfect square.

Proven?  :blush: [/hide]

[quote="white_horse_king88"][hide]Notice $n$ cannot be divisible by any integer greater than $4$.

If this is so, if $n | p$, where $p > 4$, $ n \pmod{p} \not\equiv n + 1 \pmod{p} \ldots \not\equiv n + 4 \pmod{p}$. So, no perfect square.[/hide][/quote]That's what I thought at first, but what if $p^{2m}\mid{n}$ (I think you might be using $a\mid{b}$ where you mean $b\mid{a}$ but I might have it backwards) and the $2m\text{th}$ is the greatest such power of $p$. Then $p^{2m}\mid{n\cdot(n+1)\cdots(n+4)}$ but not for any power greater than $2m$ so it could still be a square.

symettry? (spelling?)
(x-2)(x-1)x(x+1)(x+2?
and then work from there?

BTW -  this  problem and has been discussed  here before (Do a search)

Assuming that  "integers" here mean positive integers :) ...(else 0.1.2.3.4  would give you 0 :)

Here is an old post by me here: (Cut and paste) :

[hide]Solution inspired by a famous hindi song which goes "kabootar kabootar ").. (kabootar  = pigeon BTW).

Get five kabootars and tie one number to each of them. $(n, n+1, n+2, n+3, n+4)$
Get 4 kabootar-pijaras and number them "1", "2","3", "6"

Now each kabootar would know that his number can be written as $(a \cdot x^2)$  where $a$ is sqaure free. furthermore, as pointed out before ( none of the numbers have a common prime factor greater than 3) "$a$" must be one of the kabootar-pijara number.  (That is $a$  must be one of  the numbers of the set{1,2,3,6} )

Let every kabootar fly and get to the corrrosponding pijara.

4 kabootar-pijara's, 5 kabootars .. at least one pijara would contain two (or more) kabootars...

For $n>1$  , this is obviously a contradiction ( difference $ a (x_1^2 -x _2^2)$  would be greater than 4). (one can easily check n=1 case (120 is not a  perfect square) .

QED!
P.S. (The above solution will work even if you do not know that kabootar is pigeon in hindi and even if kabootars can't do math)

PPS - Using similar logic you can prove the more general case of any number (say k) of consecutive integers. (With somve work one can prove that all a's would be distinct..etc) ( Narumi proved that this is true for k's less than 200 and Erdos proved for all the numbers .. in fact in 1975 he proved, as pointed out by pbornsztein above, that the product is not equal to any power too.[/hide]

Gyan, great proof. Did you make that solution (with the kabootars and whatnot)?

Thanks. This method , of course, is known as Pigeon-hole  principle (or Dirichlet's box principle) (Also known as kabootar kabootar in our family,  and yes, there is really a old hindi film song - with rather very silly lyrics - by this name. :)

[quote="ddziabenko"]My solution is as follows: Exactly one of $5$ consecutive integers is $0(mod5)$, but then their product is divisible by $5$ [b]but not by $25$, [/b]and cannot possibly be a perfect square.  :D[/quote]

hmm... why? a number can both be divisible by 5 and 25 ... (eg  (21,22,23,24,25) only one number is 0 mod 5 yet the product is divisible by 25  (We know  the product is not a perfect square but   NOT because of the factors of 5)

(All you can say is , [b] "that" number (which is 0 mod 5) must have  an even power of 5 as a  divisor[/b])
Hope this helps.

(BTW - The problem is  an old favorite and  not  considered  that easy ... it has appeared in  NMO type contests)

[quote="ddziabenko"]My solution is as follows: Exactly one of $5$ consecutive integers is $0(mod5)$, but then their product is divisible by $5$ but not by $25$, and cannot possibly be a perfect square.  :D[/quote]That's the first thought that occurred to me as well, but it is wrong (see what I said above in my first post).

[hide="not a smooth solution"]well we can say the smallest number is $x-2$

then it is $(x-2)(x-1)x(x+1)(x+2)$
$ = (x^2+x-2)(x^2-x-2)x$

denote $x^2-2 = y$

then it becomes $(x+y)(x-y)x$

$(x-y)(x+y) = x^2 - y^2$ can be a square (Pythagorean Triple), but it cannot be a square if multiplied by x.

i dont think i proved it very well. I had no other way to think, except maybe modulus but you guys alreayd had it. This was another thinking for a better and more elegant solution.... ;)
[/hide]

[quote="xxreddevilzxx"][hide="not a smooth solution"]well we can say the smallest number is $x-2$

then it is $(x-2)(x-1)x(x+1)(x+2)$
$ = (x^2+x-2)(x^2-x-2)x$

denote $x^2-2 = y$

then it becomes $(x+y)(x-y)x$

[quote]$(x-y)(x+y) = x^2 - y^2$ can be a square (Pythagorean Triple), but it cannot be a square if multiplied by x.[/quote]

i dont think i proved it very well. I had no other way to think, except maybe modulus but you guys alreayd had it. This was another thinking for a better and more elegant solution.... ;)
[/hide][/quote]

Nice work but it still requires some work:
simple:
[quote]$(x-y)(x+y) = x^2 - y^2$ can be a square (Pythagorean Triple), but it cannot be a square if multiplied by x.[/quote]
is not true for arbitary value of x and y ..  for example if x=25 and  y=24    $x^2-y^2 = 7^2 $ and if multipled by $x$ it is still a  perfect square of $35$

But a little logic and some work  you can prove what you want to prove  when $y=(x^2-2)$

([hide="Hint"](For example you can   start with showing  first that $x^2-y^2$  has to be of the form  $k^2$ or $2k^2$ and have to do some more work to show  that each case is impossible. Can be done but would be some work. try it)
[/hide]


integration
limit
algebra unsolved
algebra

The sequence $ a_{n}$ is denoted by $ a_{k}\equal{}\frac{1}{k}(1\plus{}/frac{1}{2}\plus{}..\plus{}\frac{1}{k})$. Find the limit of $ a_{n}$ as $ n$ tends to infinity.
    P.S. If anyone can prove that the limit is not $ 0$, please post. I need this to complete the solution of one problem about probability.
Thanks in advance. :)

[quote="lasha"]The sequence $ a_{n}$ is denoted by $ a_{k} \equal{} \frac {1}{k}(1 \plus{} \frac {1}{2} \plus{} .. \plus{} \frac {1}{k})$. Find the limit of $ a_{n}$ as $ n$ tends to infinity.
    P.S. If anyone can prove that the limit is not $ 0$, please post. I need this to complete the solution of one problem about probability.
Thanks in advance. :)[/quote]

Classical computation : $ \sum_{k \equal{} 1}^n\frac 1k < 1 \plus{} \sum_{k \equal{} 2}^n\int_{k \minus{} 1}^k\frac {dx}x$ $ < 1 \plus{} \int_1^n\frac {dx}x$ $ \equal{} 1 \plus{} \ln(n)$. So $ 0<a_n < \frac {1 \plus{} \ln(n)}n$ and so $ \lim_{n\to \plus{} \infty}a_n \equal{} 0$

Another classical result much more general than your question: if $ \lim_{n\to \infty} a_n \equal{} 0$ then
\[ \lim_{n\to \infty} \frac{\sum_{k\equal{}1}^n a_k}{n}\equal{} 0\]


[quote="zhaobin"]for $a,b,c>0,p \ge 1$.
prove:
$\frac{(a^p+b^p+c^p)^2}{a^pb^p+b^pc^p+c^pa^p} \ge \frac{(a+b+c)^2}{ab+bc+ca}$[/quote]
$\frac{(a^p+b^p+c^p)^2}{a^pb^p+b^pc^p+c^pa^p} \geq \frac{(a+b+c)^2}{ab+bc+ca}$
$\Leftrightarrow \frac{a^{2p}+b^{2p}+c^{2p}}{a^pb^p+b^pc^p+c^pa^p}\geq\frac{a^2+b^2+c^2}{ab+bc+ca}$
$\Leftrightarrow (a^{2p}+b^{2p}+c^{2p})(ab+bc+ca)\geq (a^2+b^2+c^2)(a^pb^p+b^pc^p+c^pa^p)$
$\Leftrightarrow \sum_{sym}a^{2p+1}b+\sum_{cyc}a^{2p}bc\geq\sum_{sym}a^{p+2}b^p+\sum_{cyc}a^pb^pc^2$
which is true by Muirhead Inequality. :)

thanks,leepakhin,
I used it to prove another inequality,but it also a nice one isn't it?


I formulated this question on my own and lets see if you can get it, its more mathematical than anything else.

Consider a particle of mass m and charge q at the origin of co-ordinates. An electric field $ \vec E \equal{} E_0\vec i$ is present alongwith a magnetic field $ \vec B \equal{} B_0 \vec j$. the particle is projected with a velocity $ \vec v \equal{} v_0(\vec j \plus{} \vec k)$ at time t=0. Also, the particle is subjected to a force, which varies with the y co-ordinate as $ \vec F \equal{} \minus{} F_0y \vec j$. Trace out the trajectory of the particle. Assume $ E_0$ is very small.

The most fun part of this question is that you need not even write down equations, but just try and draw out the path the particle follows as time passes. It is really very interesting.
 :)

What is $ \vec i \;, \vec j \; , \vec k$??

$ \vec \i \;, \vec \j \; , \vec k$ are "orts", or unit vectors corresponding to Cartesian system axii $ x,y,z$ respectively.

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

This is quite known problem.Symmetry is just nice!First we get oscillation of z and x components of velocity,then integrate in order to get trajectory.

[quote="hell_ever"]I formulated this question on my own and lets see if you can get it, its more mathematical than anything else.
[/quote]
This is strikingly similar to the question of FIITJEE AITS Part Test-II. :wink:

But I don't neglect the possibility that it might be pure coincidence.!! :wow:

[quote="hell_ever"]Believe me, i am not a big fan of fiitjee, so i don't even bother to look at their questions... :wink:[/quote]

Then you and me have a lot in common except that thing about looking at their questions. Their questions are interesting although not easily solvable, atleast not by me! :D

[hide="Some thoughts"]From Newton's 2nd law:

\[ m\vec{a} \equal{} \vec{F} \plus{} q(\vec{E} \plus{}\vec{v}\times \vec{B}) \\ \\
\implies \vec{a} \equal{} \minus{}\frac{F_0}{m} y\, \mathbf{\hat{j}} \plus{}\frac{q}{m}(E_0 \minus{} v_0B_0)\,\mathbf{\hat{i}}\]

Therefore, to calculate the particle's path as a function of time:

\[ \vec{v}(t) \equal{} \int_{0}^{t} a(\lambda)\,\mathrm{d}\lambda \plus{}\vec{v}(0)\]

and likewise,

\[ \vec{s}(t) \equal{} \int_{0}^{t} v(\mu)\, \mathrm{d}\mu \plus{} \vec{s}(0)\]

With the substitution that $ y \equal{} v_0 t$ then

\[ \vec{v}(t) \equal{} \minus{}\frac{F_0 v_0}{2m} t^2\, \mathbf{\hat{j}} \plus{} \frac{q}{m}(E_0 \minus{} v_0B_0) t\,\mathbf{\hat{i}} \plus{} v_0\, \mathbf{\hat{j}} \plus{} v_0\, \mathbf{\hat{k}}\]

Also,

\[ \vec{s}(t) \equal{} \minus{}\frac{F_0 v_0}{6m} t^3\, \mathbf{\hat{j}} \plus{} \frac{q}{2m}(E_0 \minus{} v_0B_0) t^2\,\mathbf{\hat{i}} \plus{} v_0 t\, \mathbf{\hat{j}} \plus{} v_0 t\, \mathbf{\hat{k}} \plus{} \vec{s}(0)\]

Since $ \vec{s}(0) \equal{} (0, 0, 0)$ then

\[ \boxed{\vec{s}(t) \equal{} \frac{q}{2m}(E_0 \minus{} v_0B_0) t^2\,\mathbf{\hat{i}} \plus{}v_0 (t \minus{}\frac{F_0}{6m} t^3)\, \mathbf{\hat{j}} \plus{} v_0 t\, \mathbf{\hat{k}}}\]

[/hide]


MATHCOUNTS
geometry
3D geometry
rotation
function
similar triangles

Post any protests, solutions, questions, or discussion here.
I'll get a link of the test again here

[quote="jhredsox"]is this anirudh's test?

-jorian[/quote]
No, that's his mock...

[quote="jhredsox"]out of curiousity, how do you put #27 in terms of pi?

-jorian[/quote]
It forms two cones when rotated about the y-axis.

haha i don't get those revolving around junk.

Doe sanyone have a clear explanation
?

[quote="Fanatic"]Not that it matters, but I only count 8 wrong for myself when checking my sentbox answers with the given answers.

ETA:

I'm probably wrong, but can I challenge #25? For example, if the function is (x-4)(x-8), squaring the function would make it (x-4)(x-4)(x-8)(x-8). The only two zeroes would still be 8 and 4. So 2/21. So someone tell me where I went wrong logistically, because I'm pretty sure I did somewhere Razz .

Also:

Could I get a solution for #24 and #30? Or should I just make new threads for this? TIA[/quote]
the problem is that you are simply squaring x in the function, not the function itself.

the question stated that the zeroes were 4 and 9. we can then write the function as $f(x)=(x-4)(x-9)$. when we square the x this turns into $f(x^{2})=(x^{2}-4)(x^{2}-9)=(x-3)(x+3)(x-2)(x+2)$ the zeroes are 2,-2,3,-3 so the answer is 4/21.

these are giveaway hints
24-similar triangles
30-the medians divide the triangle into 6 triangles of equal area

Thanks for the clearup, ragnarok.

Well the miscue with my score won't matter now.

ok this is pratik here.
I just talked with eli and he told me to regrade the papers. as I regrade them, I will post the new (possibly) scores up here and the countdown competitors will be decided.

anirudh, eli told me to take off 5 points from your total score cuz you were late.
SCORES:
1. Ubemaya-[b]44[/b](28,16)
2. SplashD-[b]43[/b](29,14)
3. davidli-[b]35[/b](23,12)
4. Anirudh-[b]33[/b](26,12)ANIRUDH WINS TB FOR SPRINT    SCORE
5. Fanatic-[b]33[/b](21,12)
6. bpms-[b]31[/b](23,8) 7 OUT OF LAST 10 BPMS WINS TB
-----------------------------------------------
7. sonny-[b]31[/b](23,8) 6 OUT OF LAST 10
8. davidlizeng-[b]29[/b](25,4)
9. redcomet-[b]28[/b](20/8) REDCOMET WINS TB FOR SPRINT
10.humblemastermind-[b]28[/b](18/10)
11. funcia-[b]8[/b](6/2)

[quote="sonny"][quote="Walk Around The River"]ok this is pratik here.
I just talked with eli and he told me to regrade the papers. as I regrade them, I will post the new (possibly) scores up here and the countdown competitors will be decided.

anirudh, eli told me to take off 5 points from your total score cuz you were late.
SCORES:
Ubemaya-[b]44[/b](28,16)
SplashD-[b]43[/b](29,14)
davidli-[b]35[/b](23,12)
Fanatic-[b]33[/b](21,12)
bpms-[b]31[/b](23,8)
davidlizeng-[b]29[/b](25,4)
humblemastermind-[b]28[/b](18/10)[/quote]

What about me?? :huh:[/quote]
i'm adding them as I grade them
I just put yours up

[quote="davidli"]What was the point of regrading the papers? They're all the same scores.[/quote]
just to be safe, doesn't matter I'm done

COUNTDOWN HAS BEEN DETERMINED

eh top 8 is fine I guess eli will probably be ok with it

eli's aops is being stupid so he can't come now, I'll talk with him later on aim

[quote="anirudh"]Wait...I'm getting confused

I thought Eli was Walk Around the River

Isn't he :|[/quote]
yeah he is but he let me get on his account because his aops wasn't working

You still havent announced what format the cd is going to be

well, i do remember the actual davidli being in the cd...
-jorian

[quote="bpms"][quote="jhredsox"][quote="davidlizeng"]could you possibly stretch the
countdown # to 8 .
since i am 7th, then i'll be able to
get in.. heh heh.........
you probably won't  though :([/quote]

wait, you don't happen to be the same person as davidli, do you?

-jorian[/quote]
I'm getting a vibe, it's telling me no.
Then again, maybe I'm wrong.[/quote]

You're vibe is correct. We're different people. Although it did kinda scare me when I first saw his name...

[quote="jhredsox"]lol, that reminds me...how DO you change your username? :D

ok, got that davidli thing out of my head :lol:
-jorian[/quote]
Changing your username should only be done in very, very extreme cases. You have to ask an admin to do it, but the admins will stop if they get too many requests for username changes.

[quote="anirudh"]Okay. Is the countdown bracket or ladder?[/quote]

its ladder.  no complaints. and this is actually eli. my grandma has some dial-upm **** that takes about 30 seconds to load each page.

since we made countdown because people wanted it, the questions arent amazing, so we didnt want to put too much weight on the standings on countdown, so were using ladder.

Yay :lol:

What time? Anytime today is fine for me...

it has been posted about 8 times that the

COUNTDOWN ROUND IS JANUARY 7th at 7 PM (EST)

Please send out remainder PM's on January 6th. Thanks.

[quote="Ubemaya"]May I ask what your reasoning is to accept 2 answers for #2? In my opinion the question is completely clear, what makes it unclear?[/quote]

It is because the question doesnt state that if Owais wins, then that still counts as a win for Eli.
The question was poorly worded (our fault) and it could easily be intepreted in two ways.

[quote=jhredsox]out of curiousity, how do you put #27 in terms of pi?

-jorian[/quote]

yes the problem literally says to express your answer in terms of pi


Let $(I)$ and $(O)$ be the incircle and circumcircle of triangle $ABC$,respectively.$(I)$ is tangent to $BC$ at $D$.A circle $(O_{1})$ ia tangent $(O)$ at $T$ on arc $BC$(arc $BC$ have $A$) and $(O_{1})$ is tangent to $BC$ at $D$.Prove that $\hat{ATI}=90^{o}$

By using the inversion having $D$ for pole and $\bar{DB}.\bar{DC}$ for constant, easy to prove that $T,D$ and the midpoint $S$ of the arc $BC$ doesn't contain $A$ are collinear.
Let $f$ is the inversion having $S$ for pole and $SI^{2}$ for constant. $AO$ intersects $(O)$ again at $K$. We have $f( (O) )$ is the line $BC$,and $f(K)$ is the intersection $P$ of $SK$ and $BC$. So $f(IK)$ is the circle passes through $I,S$ and $P$. Note that $\hat{ISP}= \hat{IDP}= 90^{o}$ so this circle pass through $D$. It means that $SD, BC$ and $(IDP)$ is concurent at $D$ , so $SD, (O)$ and $IK$ is concurent at $T$, which ends the proof  :wink:

[quote="toan hoc"]I can't prove that $T,D$ and the midpoint $S$ of the arc $BC$ doesn't contain $A$ are collinear.[/quote]
Let $f$ is the inversion having $D$ for pole and $\bar{DB}.\bar{DC}$. The circle $(O_{1})$ passes through $D$ so $f(O)$ is a line. $(O_{1})$ touches $BC$ at $D$ and touches $(O)$ at $T$ so $f(O)$ is the line which touches $(O)$ at $f(T)$ and parallel with $BC$. So, $f(T)$ is the midpoint of the arc $BC$ doesn't contain $A$

$TD$ meets $(O)$ at $E$. Because $(O)$ and $(O1)$ are tangent at $T$ so $O1D//OE$ so $OE \bot BC$
Thus $E$ is the midpoint of arc $BC$. So $AE$ is the bisector of $\angle BAC$
We have $\angle EBC=\angle EAC=\angle EAB=\angle BTE$ so triangle $EBD$ and $ETB$ are similar. This means $EB^{2}=ED*ET$
Otherwise $EB=EI$ so $EI^{2}= ED*ET$ so triangle $EID$ and $ETI$ are similar. So $\angle ETI=\angle EID=\angle IEO=\angle EAO=\angle EAF$ ($AO$ meets $TI$ at $F$)
Thus $A,T,E,F$ are cyclic
So $AF$ is the diameter of $(O)$
This means $\angle ATI=90$ $q.e.d$


linear algebra
matrix
linear algebra unsolved

let $ A$ be a p x n matrix and $ C$ a p x q matrix. show that there exists a unique n x q matrix $ B$ such that 1) $ AB\equal{}C$ and

                                                                                2)$ B^Tv\equal{}0$ whenever $ Av\equal{}0$

kouboy,
I think you need more asumptions. For instance take $ n\equal{}p\equal{}q$, $ C\equal{}I_n$, and $ A$ non invertible, then you can't find such $ B$.


give a,b,c >0 and $ a^{3} \plus{}b^{3}\plus{}c^{3}\leq \frac{1}{\sqrt{3}}$
Tìm minimum :
$ \sum (a\minus{}\frac{1}{a})(b\minus{}\frac{1}{b})$
It's one problem very pleasant ! :wink:

[quote="iqtimo"]give a,b,c >0 and $ a^{3} \plus{} b^{3} \plus{} c^{3}\leq \frac {1}{\sqrt {3}}$
Tìm minimum :
$ \sum (a \minus{} \frac {1}{a})(b \minus{} \frac {1}{b})$
It's one problem very pleasant ! :wink:[/quote]
It's easy because $ \frac{(1\minus{}a^2)(1\minus{}b^2)(1\minus{}c^2)}{abc}\geq\frac{8}{3\sqrt{3}}$(easy by Am-Gm)


[quote="Ignite168"]Find the smallest positive integer n such that (19360)(n) will be the cube of an integer.[/quote]
[hide]In order to find the smallest integer to make it into a cube if a number we need to break down 19360 to see what numbers it needs. When you prime factorize it you get $2^{5}\cdot5\cdot11^{2}$. So we need three of each number. We need one more 11, two more 5's and one more 2 to make it into a oerfect cube. Our final answer is $2\cdot5^{2}\cdot11=550$.
You can check this on a graphing calculator or a scientific calculator.[/hide]


With a conductor, the charge is spread on the surface so there is no charge inside, hence no electric field.  You can't "remove" gravity.

charge is zero inside doesn't necessarily imply that the electric field is zero inside :noo: . but we are given that the field [i]is[/i] zero inside.
the question is, calculating both E and g at a point inside require the same type of summation. so, why is E zero whereas g is not?

The reference points are different. For the electric field, the reference point $ \mathbf r_0$ is inside the conductor, while the reference point for gravtitational field is outside the conductor, at the center of the Earth.

[quote="Attila the Hun"]the electric field strength inside a conductor of any shape with charge on its surface is 0, but the gravitational field strength inside a shell of matter is not zero. how is this possible given that both forces are inverse square laws? where is the analogy incomplete?[/quote]

The gravitational case is analogous to that of a charged insulating shell, not that of a charged conducting shell.  In the insulating case, the charges are held rigid due to the material, which exerts forces to counterbalance those from the electric field (unlike in the conducting case, in which the charges are free to move around and attain equilibrium).  Same with the gravitational case - the shell is rigid, so mass is not free-flowing.

I was thinking along the same lines. in a conductor, charge does not distribute itself evenly. it concentrates at sharp points. in a shell of matter, mass is distributed evenly. so the difference arises there.


AMC
AIME
geometry
perimeter
Pythagorean Theorem
perpendicular bisector
number theory

The points A, B, C lie on a line in that order with AB = 9, BC = 21. Let D be a point not on AC such that AD = CD and the distances AD and BD are integral. Find the sum of all possible n, where n is the perimeter of triangle ACD.

Hey, I got this one! (celebration)

[hide]

If we draw out the triangle, we notice that it is isoceles.  We draw the altitude D to AC call this point E.  Label side AD as Z, BD as Y, and ED as X.

By the pythagorean theorem we get 2 equations.

6 :^2: +x^2=y^2

15 :^2: +x^2=z^2

Subtracting these we get 15 :^2: - 6 :^2: =z^2-y^2=(z-y)(z+y)

-->  189=3 :^3: * 7=(z-y)(z+y)

We do the combinations of the factors of 189:  (1,189); (3,63); (9,21); (27,7).

With each of these, we solve z and y to get : (17,10);(15,6);(33,30);(95,94).  Of course we should also test these to make sure they work.   Of course (15,6) does not work....

To get the perimeters, we do 30*3+17*2+33*2+95*2=380

TA DA!

[/hide]

[hide]AD=CD iff D is on the perpendicular bisector of AC, so DX  :perp: AC, where X is on AC and AX=XC.  Since AC = 30, AX = XC = 15.  Also, since AB = 9, BX = 6.

Thus we form a right triangle AXD with base of length 15 and a smaller right triangle BXD inside of AXD, with BX = 6.

By the Pythagorean Theorem, since AXD is right, AX^2 + XD^2 = AD^2, so 15^2 + XD^2 = AD^2, so XD = sqrt(AD^2-225).

Applying the Pythagorean Theorem to BXD, BX^2 + XD^2 = BD^2, so 36 + AD^2 - 225 = BD^2.  Simplifying, we have AD^2 - 189 = BD^2, or AD^2 - BD^2 = 189.

Factoring this, we have (AD+BD)(AD-BD) = 189.  The prime factorization of 189 is 3^3*7, so pairs of integers that multiply to 189 are (1,189), (3,63), (7,27), and (9,21).

In the first case,

AD+BD=1

AD-BD =189

Adding the two together yields 2AD = 190

In the second case,

AD+BD=3

AD-BD=63

Adding the two together yields 2AD = 66

In the third case,

AD+BD=7

AD-BD=27

Adding the two together yields 2AD = 34

In the fourth case,

AD+BD=9

AD-BD=21

2AD = 30

So our possible values of AD are 95, 33, 17, 15.  But if AD=15, D is on AC, which is undesired.  Thus the only possible values of AD are 95, 33, 17.

(Check: if AD=95, BD=66.  If AD=33, BD=30.  If AD=17, BD=10.  Yay.)

The perimeter of ACD is AC + AD + CD = 30 + 2AD.  So the perimeters of triangles with 2AD=190,66,34 are 220, 96, 64.  The sum is 380.[/hide]


posted before:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=931557#931557[/url]

Also at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150776 for those who only speak English.


we put a chain of length $ L$ straightened out on a horizontal table, so that the end of the chain is just at the edge of the table and it lies perpendicular to it. We push the chain with velocity $ v_0$ and it goes over the edge. The part of the chain which has left the table is always vertical, and the part on the table is horizontal. The chain is totally flexible and non elastic. all friction is negligible. Find the length of the vertical part of the chain(the part which has left the table) as a function of time, and find the velocity as a function of time.

You should be able to do this using energy conservation.

There are some problems of this category where energy conservation cannot be used (the string is piled up, not laid out etc.), but here energy is conserved.

Energy conservation just makes the whole thing difficult (you get a first order difficult DE). Try using the force equation (Here you get a somewhat simpler second order DE) :D  (I already know the solution, I just posted so I see how you guys would solve it :D )

:huh:
For energy conservation the equation
$ x^2g\plus{}Lv_0^2\equal{}L\dot{x}^2$ is variable-seperable.

For Force the equation is
$ L\ddot{x}\minus{}\dot{x}^2\equal{}xg$

Your force equation is wrong! :wink:

it should be:
$ \ddot x \minus{}\frac{g}{L} x \equal{}0$

Which is much easier.

:oops: Ok, that's right. If you notice, mine doesn't satisfy the initial conditions.

is this problem from the textbook Introduction to Mechanics by Kleppner and Kolenkow??? I did this problem before in my physics course~ you can simply use newton's law to derive a relationship and get a differential equation and solve.


Ten days before I get away from UW! Good riddance exams!!

i'm very excited. After years of wanting to get out of highschool, and into university, I can't believe its actualy here!

So you will post every day and write down the number of days left? :lol:
LOL, that's funny.
Good luck and best wishes in UW to the many Aopsers who are going there this year! :)

[quote]So you will post every day and write down the number of days left?
LOL, that's funny. [/quote]

Hehe, no, I won't be posting every day, but you're welcome to count for me too!  :P I'll try though!

1102$_{(3)}$ DAYS!!!!

Heh, gotta vary the numbers, or I'll get bored!

SSSSOOOOOOO EXCITED!!!!!

HEheheheheheh  38 days...awesome :D

Unrelated question for Elyot:

Will the DDR machine be functioning by August 5th?  XD

I really know nothing about the Campus Cove DDR machine... my guess is that it'll be back sometime around frosh week, but who knows.  Now that zappers is closed for renovations, you'll just have to call me up and come over to my place for DDR!

And yeah... EXAMS NEXT WEEK FOR ME!!  :S

I can't wait until they're over though :D.

[quote="lowest_reynolds"]Yeah...I'm going to UWaterloo next week for...something.

I haven't played DDR in a while.  I'm hungry for some Dance Dance Revolution.

HOOK ME UP.[/quote]

note to all: all the cool kids simply call it "dance dance"  :D

Oh wow, its been a while since i've played dance dance, at least a few months. Even if I got back to the arcade, you guys would stomp me. I'm not built for the dance dance, I'm six foot five...!
I always get REALLY wierd looks when I play at the arcade.  :lol:

[quote="rzsolt"]all the cool kids simply call it "dance dance" [/quote]

I've never heard of "Dance dance", but it sounds good.  (I say stuff like "shiny shiny" all the time.)

that was actually posted at 6.15 my time. I didn't end up falling asleep 'til 7  :(

Okay okay, maybe I am getting a bit anxious over this whole thing  :ninja:

Hehehehe I'm in waterloo now...wow it's hard to believe.

I have the smallest darkest room imaginable.  It's awesome.   :D

[quote="Sunny"]Hehehehe I'm in waterloo now...wow it's hard to believe.[/quote]

Welcome!! FINALLY!!!!!!! I'm back, so call me sometime! I can't find your phone number!!

[quote="Sunny"]I have the smallest darkest room imaginable.  It's awesome.   :D[/quote]

lol! It suits you, eh?

[quote="jpark"][quote="Sunny"]I have the smallest darkest room imaginable.  It's awesome.   :D[/quote]

lol! It suits you, eh?[/quote]

OOH!!!  Burn!  :furious:

Are you gonna take that kinda crap, Sarah?!?!

[quote="jpark"]what crap? I never said anything offensive :P[/quote]

Sure you did.  That was REALLY, REALLY offensive!!!

[size=17](the only way we'll ever be able to beat them is to turn them against one another!)[/size]

AHHHHHHHH
I leave tomorrow morning
AHHHHHHHHH

Hahahaha none offense taken Jenn.

Room actually looks like a room now.  :D :D :D

[quote="WindingFunction"]You guys are SOOOOOOO first year it's not even funny.  Don't worry, it's a good thing.  :D[/quote]

LOL... yeah, I know what you mean.  Just wait until they realize how much work it is and they start whining...

[quote="Sunny"]Hahahaha none offense taken Jenn.

Room actually looks like a room now. :D:D:D[/quote]

Yea... actually Sarah, your room looks quite nice.  I have seen smaller, darker dorm rooms than V1 singles (although, not by a lot).  At least your dungeon is well-decorated ;)

I liked your room! I've seen rooms much, much worse (not that I've seen many :P)

oh well, orientation starts tomorrow!!!


function
inequalities
algebra proposed
algebra

Find all functions $ f: \mathbb N\rightarrow \mathbb N$, for which we have:

$ f(n)+f(n+1)=f(n+2)\cdot f(n+3)-1996$

$ f(n)\plus{}f(n\plus{}1) \equal{} (f(n\plus{}2))(n\plus{}3)\minus{}1996 ,$$ \ f(n\plus{}1)\plus{}f(n\plus{}2) \equal{}( f(n\plus{}3))(n\plus{}4)\minus{}1996$, therefore $ f(n)\equal{}(f(n\plus{}2)\minus{}f(n\plus{}3))(n\plus{}4)$
$ f(n)$ is positive so $ f(n\plus{}2)$ is decreasing . Finally there is no fonction satisfying the statement .
 :cool:

[quote="Inequalities Master"][quote="Diogene"]$ f(n) \plus{} f(n \plus{} 1) \equal{} (f(n \plus{} 2))(n \plus{} 3) \minus{} 1996 ,$$ \ f(n \plus{} 1) \plus{} f(n \plus{} 2) \equal{} ( f(n \plus{} 3))(n \plus{} 4) \minus{} 1996$, therefore $ f(n) \equal{} (f(n \plus{} 2) \minus{} f(n \plus{} 3))(n \plus{} 4)$
$ f(n)$ is positive so $ f(n \plus{} 2)$ is decreasing . Finally there is no fonction satisfying the statement .
 :cool:[/quote]

You're wrong.[/quote]
Are you so sure  :P  :!:  Could you show where is my fault  :?:

[quote="Inequalities Master"][quote="Diogene"] Finally there is no fonction satisfying the statement .
 [/quote]

I'm no functions expert, but it's just enough to read this line to understand that your solution is wrong or incomplete. That's because there are $ 4$ functions already satisfying the given condition. An example would be $ f(n) \equal{} 2$, if $ n \equal{} 2k$ and $ f(n) \equal{} 1998$, if $ n \equal{} 2k \plus{} 1$.[/quote]

That's strange from an "Inequalities Master", because your exemple is manifesty false  : $ f(2k)\plus{}f(2k\plus{}1)\equal{}2000 ; (f(2k\plus{}2))(2k\plus{}3)\minus{}1996 \equal{}4k\minus{}1990$  :D

[quote="BG Yoda"]It must be

[quote="Inequalities Master"]Find all functions $ f: \mathbb N\rightarrow \mathbb N$, for which we have:

$ f(n) \plus{} f(n \plus{} 1) \equal{} f(n \plus{} 2)\cdot f(n \plus{} 3) \minus{} 1996$[/quote]

[b]Inequalities Master[/b] just wrote [i]"\cdotf"[/i] instead of [i]"\cdot f"[/i] = $ \cdot f$ and because of this $ f(n \plus{} 3)$ was typed only $ (n \plus{} 3)$.[/quote]

Yeah, you're right ! I really didn't notice that. Thank you very much, [b]BG Yoda[/b]. Indeed the original problem is:

[color=darkblue]Find all functions $ f: \mathbb N\rightarrow \mathbb N$, for which we have:

$ f(n) \plus{} f(n \plus{} 1) \equal{} f(n \plus{} 2)\cdot f(n \plus{} 3) \minus{} 1996$[/color]

Unfortunately, I can't edit the initial post anymore. My excuses to [b]Diogene[/b].

[quote="Diogene"]

That's strange from an "Inequalities Master" [/quote]

Unfortunately, I am not an "Inequalities Master", am too far away from it, this is just my Mathlinks ID. :(  But I will surely try to become one, so I could be worthy of this name.

Let $ a_n \equal{} f(n)$ for every nature $ n$.

$ (a_{n \plus{} 2} \plus{} a_{n \plus{} 1}) \minus{} (a_{n \plus{} 1} \plus{} a_{n}) \equal{} (a_{n \plus{} 3}a_{n \plus{} 4} \minus{} 1996) \minus{} (a_{n \plus{} 2}a_{n \plus{} 3} \minus{} 1996)$

$ \Rightarrow a_{n \plus{} 2} \minus{} a_n \equal{} a_{n \plus{} 3}(a_{n \plus{} 4} \minus{} a_{n \plus{} 2}) \equal{}$ $ \Pi_{i \equal{} 1}^{k}a_{n \plus{} 1 \plus{} 2i}(a_{n \plus{} 2k \plus{} 2} \minus{} a_{n \plus{} 2k})$ for some $ k\in \mathbb{N}$.

Therefor we get that $ \Pi_{i \equal{} 1}^{k}a_{n \plus{} 1 \plus{} 2i}|a_{n \plus{} 2} \minus{} a_n$ for infinitely large $ k$ and for every $ n$, which means either there exist infinitely many consecutive members of the sequence $ a_{t_0 \plus{} 2t} \equal{} 1$ for some $ t_0$ or means that $ a_{2n} \equal{} a_{2}$ and $ a_{2n \plus{} 1} \equal{} a_{1}$.

I think from the first statement we get $ a_{2n \minus{} 1} \equal{} 1$; $ a_{2n \plus{} 2} \equal{} a_{2n} \plus{} 1997$ and $ a_{2n} \equal{} 1$; $ a_{2n \plus{} 1} \equal{} a_{2n \minus{} 1} \plus{} 1997$.

From the second we get $ a_{2k} \plus{} a_{2k \plus{} 1} \equal{} a_{2k}.a_{2k \plus{} 1} \minus{} 1996$.

Let $ a_{2k} \equal{} a$ and $ a_{2k \plus{} 1} \equal{} b$.
$ 1996 \equal{} ab \minus{} a \minus{} b \Leftrightarrow (a \minus{} 1)(b \minus{} 1) \equal{} 1997$ from where $ (a,b) \equal{} (2,1998)$ or $ (a,b) \equal{} (1998,2)$.

...  :huh:

[b]Inequalities Master[/b], [b]Diogene[/b], please, check if I am right?


inequalities
geometry
3D geometry
area of a triangle
Heron\u0027s formula
inequalities solved

I have some inequalities problems unsolved:
P1:Prove that
(x^2 -x+1)*(y^2-y+1)*(z^2-z+1)>=(xyz)^2 +xyz +1.
for all real numbers x, y, z.
p2: Prove that
(a+b-c)^2*(b+c-a)^2*(c+a-b)^2 >= (a^2+b^2-c^2)*(b^2+c^2-a^2)*(a^2+c^2-b^2)
for all real numbers a,b,c.

hello!

P1 is false! take x=y=z=1 and you get if I am right: 1>=3 false!!

perhaps it is:(x^2 -x+1)*(y^2-y+1)*(z^2-z+1)>=(xyz)^2 -xyz +1.
I don't know... cheers! :D

P1 is correct
Prove that
3*(x^2 -x+1)*(y^2-y+1)*(z^2-z+1)>=(xyz)^2 +xyz +1.
for all real numbers x, y, z.
thanks Lagrangia

this is my soln: :D
Put S=3(x2-x+1)(y2-y+1)(z2-z+1).
We'll prove inequality for x,y,z>0 .Applying Cauchy's, we get: x2+1>=2x then 3(x2-x+1)>=(x2+x+1). Similar, we have
3(y2-y+1)>=(y2+y+1), 3(z2-z+1)>=(z2+z+1)
Then S>=1/9(x2+x+1)(y2+y+1)(z2+z+1).
Case 1:  xyz>=1, put a=(xyz)1/3, then a>=1 and (xyz)2 +xyz+1 =a6 + a3 +1
Applying Minkowski's, we get: ((x2+x+1)(y2+y+1)(z2+z+1))1/3 >= (xyz)2/3 + (xyz)1/3 +1 =a2 +a +1.
Then S>=1/9(a2 +a+1)3.
We'll prove 1/9(a2 +a+1)3>=a6 + a3 +1
Iff 3a4(a+1)2(a2+1) >=8(a6 + a3 +1) Which is true because:
Applying Cauchy's, we get a6 +1>=2a3, then 2(a6 + a3 +1)<=3a6 +3<=3a4(a2+1)
Then 8(a6 + a3 +1) <=12a4(a2+1)<=3a4(a+1)2(a2+1).
The proof of this case complete.
Case 2: xyz<=1, put x=1/x', y=1/y', z=1/z', then x'y'z'>=1,this case is true by applying case 1 and dividing (x'y'z')2
The proof for otherwise is very easy.

P2:
Hint: prove inequality in the case: a,b,c>0
Try to solv :D

[quote="hxtung"] Applying Minkowski's, we get:
((x2+x+1)(y2+y+1)(z2+z+1))1/3 >= (xyz)2/3 + (xyz)1/3 +1 [/quote]

Is this really Minkowski ? I don't recognise it here ...
Can someone explain a bit ?

p2 was given in Poland in 1992 and a solution can be found in [b]Valentin Vornicu's book[/b]!

OK but I don't have Valentin Vornicu's book ... :?

So can someone tell me why this is an application of Minkowski's inequality ?
I really don't recognise it here.

actually I have two solutions for this problem in my book - the one I like more ( and the first I found ) is using the simillarlyness between the given inequality and the extended Heron's formula, that is

16S2 = - \sum a4  + 2\sum a2b2

.... give it a try :)

In HXTUNG solution of inequality P1 I have not understood
the inequality

1/9(a^2 +a+1)^3  \geq a^6 +a^3 +1

I think it is not correct !!!!

It is clear we must prove inequality for x,y,z > 0.

Suppose WLOG x \geq y \geq z.

Suppose z<1 and x \geq 1.

And now main trick:
3(1-y+y^2) > 1+y+y^2
(1-z+z^2)(1+y+y^2)-(yz)^2-yz-1 = (z-y)(z-1)(y+1) > 0
(1-x+x^2)(1+yz+(yz)^2)-(xyz)^2-xyz-1=(x-yz)(x-1)(yz+1) > 0  !!!!

So there are two alternatives
Case 1. x,y,z \geq 1
Case 2. x,y,z \leq 1
But Case 1 is consequence of Case 2 due to (x,y,z)->(1/x,1/y,1/z) (see hxtung proof)

Let us prove Case 2.

Suppose x,y,z is optimal set (i.e. LHS-RHS=min) from cube K={0 \leq x,y,z \leq 1}.

Suppose x>y. Consider new set (x-eps,y+eps,z), where eps is a small positive parameter.
It is clear xyz<(x-eps)(y+eps)z.
Let us show
(1) (x^2-x+1)(y^2-y+1) > ((x-eps)^2-(x-eps)+1)((y+eps)^2-(y+eps)+1).
Actually since eps is SMALL (1) is consequence of
(2x-1)(y^2-y+1)-(2y-1)(x^2-x+1) > 0
or
(x-y)(2xy-x-y-1) < 0 - it is true because 1 \geq x>y.

Thus if initial set (x,y,z) (x \geq y \geq z) contains unequal values then (x,y,z) is not optimal set.

So x=y=z.

3(x^2-x+1)^3 \geq x^6+x^3+1
You can prove it without my help.

[quote=Ji Chen]For all real numbers $x, y, z$ prove that

$3(x^{2}-x+1)(y^{2}-y+1)(z^{2}-z+1)$
$\geq x^{2}y^{2}z^{2}+x^{2}(y+z)+y^{2}(z+x)+z^{2}(x+y)-5xyz+1$,

with equality if $x=0,y=z=1$.[/quote]

See also here:
http://www.artofproblemsolving.com/community/c6h1270987

Let us provide a (sort of a) brute-force solution to P1: Let $P(t) = t^2 - t + 1.$ Then the inequality may be rewritten in the form

$$ (3P(y)P(z)-y^2z^2)x^2 - (3P(y)P(z) + yz)x +(3P(y)P(z) - 1) \ge 0.$$

This is a quadratic equation in $x,$ so, in order for this (as a function of $x$ alone) to be positive, its leading coefficient has to be $\ge 0$ and its discriminant $\le 0.$

(i) For the leading coefficient, it is pretty trivial, as $P(t) \ge \frac{3}{4} t^2,$ so $3P(y)P(z) \ge \frac{27}{16}y^2 z^2 > y^2z^2.$
(ii) For the discriminant, one sees, after some calculations, that
$$\Delta_{y,z} = 4(1+y^2z^2)P(y)P(z) - (P(y)P(z) - yz)^2 - 8P(y)^2P(z)^2.$$
We will prove a slightly stronger fact that $2P(y)^2P(z)^2 \ge (1+y^2 z^2)P(y)P(z).$ This is equivalent to
$$ 2P(y)P(z) \ge (1+y^2 z^2).$$
We proceed to perform the same analysis as before: we write $LHS - RHS$ as
$$ (2P(z) - z^2)y^2 - 2P(z) y +(2P(z) -1) \ge 0.$$
The leading coefficient is $z^2 -2z + 2 \ge 1,$ so we are OK if we prove that the discriminant is non-positive. This is achieved by realizing that
$$ \Delta_z = 4P(z)^2 - 4(2P(z) - 1)(2P(z)-z^2)= -4(z-1)^4.$$
This proves all inequalities we wished to prove. Notice that, for equality, we [i]must[/i] have $\Delta_{y,z} = \Delta_z = 0.$ The latter implies $z=1,$ which then implies $y=1$ as well. By plugging back in, we see that $x=1$ as well in order to have equality, an thus the equality case is only $x=y=z=1.$

After we show the inequality for $x=y=z$ we have : $3(x^2-x+1)^3\ge x^6+x^3+1$\\
we only need to prove the inequaliy when $x,y,z\ge 0$\\
$27(x^2-x+1)^3(y^2-y+1)^3(z^2-z+1)^3\ge (x^6+x^3+1)(y^6+y^3+1)(z^6+z^3+1)\ge ((xyz)^2+xyz+1)^3$\\
where the last inequality holds by holder


Well, as long as $n \not= \infty$, $S < \frac{\pi^2}{6}$.
The LHS of the inequality simplifies to $\frac{2n(2n-1)}{(2n+1)^2}\left(\frac{\pi^2}{6}\right)$ amd the RHS simplifies to $\frac{4n(n+1)}{(2n+1)^2}\left(\frac{\pi^2}{6}\right)$.
Since $n > 0$, the RHS is always greater than the LHS $4n^2+4n > 4n^2-2n$.
I could not get any further.

[quote="Myth"]I have rewritten it. Please, check![/quote]
It's OK!


inequalities
function
induction
algebra unsolved
algebra

For any two rational numbers $ p$ and $ q$ in the interval $ (0,1)$ and function $ f$, there is always $ \displaystyle f \left( \frac{p\plus{}q}{2} \right) \leq \frac{f(p) \plus{} f(q)}{2}$. Then prove that for any rational numbers $ \lambda, x_1, x_2 \in (0,1)$, there is always:

\[ f( \lambda x_1 \plus{} (1\minus{}\lambda) x_2 ) \leq \lambda f(x_i) \plus{} (1\minus{}\lambda) f(x_2)\]

[quote="shobber"]For any two rational numbers $ p$ and $ q$ in the interval $ (0,1)$ and function $ f$, there is always $ \displaystyle f \left( \frac {p \plus{} q}{2} \right) \leq \frac {f(p) \plus{} f(q)}{2}$. Then prove that for any rational numbers $ \lambda, x_1, x_2 \in (0,1)$, there is always:
\[ f( \lambda x_1 \plus{} (1 \minus{} \lambda) x_2 ) \leq \lambda f(x_i) \plus{} (1 \minus{} \lambda) f(x_2)
\]
[/quote]
 First, prove that $ f(\frac{{\sum\limits_{i \equal{} 1}^n {a_i } }}{n}) \le \frac{{\sum\limits_{i \equal{} 1}^n {f(a_i )} }}{n}$
then $ f(\lambda x_1  \plus{} (1 \minus{} \lambda )x_2 ) \le \frac{{\lambda f(x_1 ) \plus{} (1 \minus{} \lambda )f(x_2 )}}{{\lambda  \plus{} 1 \minus{} \lambda }} \equal{} \lambda f(x_1 ) \plus{} (1 \minus{} \lambda )f(x_2 )$ :) Is that right?

[hide="Solution"]

[b]Lemma 1.[/b] For all positive integers $n$, and rational numbers $x_1$,..., $x_{2^n}$ in the interval $(0,1)$, the inequality \[f(\frac{x_1+...+x_{2^n}}{2^n})\le \frac{f(x_1)+...+f(x_{2^n})}{2^n}\]holds.

[b]Proof.[/b] The proof is by induction on $n$.

[i]Base Case.[/i] For $n=1$, note that \[f(\frac{x_1+x_2}{2})\le \frac{f(x_1)+f(x_2)}{2}\]by the given condition.

[i]Inductive Step.[/i] Suppose the claim holds for $n=k$. Then, for $n=k+1$,\[f(\frac{x_1+...+x_{2^{k+1}}}{2^{k+1}})\le f(\frac{\frac{x_1+...+x_{2^k}}{2^k}+\frac{x_{2^k+1}+...+x_{2^{k+1}}}{2^k}}{2})\le \frac{f(\frac{x_1+...+x_{2^k}}{2^k})+f(\frac{x_{2^k+1}+...+x_{2^{k+1}}}{2^k})}{2} \le \frac{f(x_1)+...+f(x_{2^{k+1}})}{2^{k+1}}.\]Thus, by induction, the desired inequality holds for all positive integers $n$. $\blacksquare$

[b]Lemma 2.[/b] If \[f(\frac{x_1+...+x_{n+1}}{n+1})\le \frac{f(x_1)+...+f(x_{n+1})}{n+1},\]holds for all rational numbers $x_1$,..., $x_{n+1}$ in the interval $(0,1)$, then \[f(\frac{x_1+...+x_n}{n})\le \frac{f(x_1)+...+f(x_n)}{n}.\]

[b]Proof.[/b] Let \[x_{n+1}=\frac{x_1+...+x_n}{n}\]and note that $x_{n+1}\in (0,1)$ is a rational number. Also, note that \[\frac{x_1+...+x_{n+1}}{n+1}=\frac{x_1+...+x_n}{n}.\]Thus, \[f(\frac{x_1+...+x_n}{n})=f(\frac{x_1+...+x_{n+1}}{n+1})\le \frac{f(x_1)+...+f(x_{n+1})}{n+1}=\frac{f(x_1)+...+f(x_n)+f(\frac{x_1+...+x_n}{n})}{n+1}.\]Rearranging gives the desired inequality. $\blacksquare$

Lemma 1 and Lemma 2 imply that for every positive integer $k$ and rational numbers $a_1$,..., $a_k$ that are in the interval $(0,1)$, \[f(\frac{a_1+...+a_k}{k})\le \frac{f(a_1)+...+f(a_k)}{k}.\]Now, suppose $\lambda=\frac{p}{q}\in (0,1)$ for positive integers $p$ and $q$. Since $p<q$, $q-p>0$. Let $x_1$ and $x_2$ be rational numbers in the interval $(0,1)$. Let $k=q$, $a_1=...=a_p=x_1$, and $a_{p+1}=...=a_{q}=x_2$. Then, \[f(\lambda x_1+(1-\lambda) x_2 )=f(\frac{px_1+(q-p)x_2}{q}) \le \frac{pf(x_1)+(q-p)f(x_2)}{q}=\lambda f(x_i)+(1-\lambda) f(x_2).\]

[/hide]

Both the condition given are for convex function therefore we can prove it by usual induction


What do you use (or what would you use) to write your final solutions in the USAMO: pen, or pencil?

I'm not sure what I should use...

I have a tendency to write darkly with a pencil, so I will use a pencil.  This way, I can erase if I need to.

I agree with dts.  No wait, other--chisel :D.

[quote="joml88"]Well, I'm using the advice given [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWriteWrite.php]here (#6)[/url] :P[/quote]

#1 choice: Black Pilot Varsity.  Cheapest fountain pen I can find :D
#2 choice: Black Bic Cristal.
#3 choice: Blue BIc Cristal.
#4 choice: Black Pilot G2
...
#infinity choice: pencil

i would definitely [b]not [/b]use a erasable pen, those are notoriously unreliable and can make smears

i think if you write dark enough, pencil is perfectly fine (and i don't tend to do very much planning before i start writing the proof, so a good eraser is helpful).

[quote="joml88"]I will most likely use a black pen.  You have to fax your solutions I think, so it will show up better.[/quote]

You have to fax your solutions? Don't we take USAMO at schools?
BTW, I dont even know how to use a fax machine.

Look [url=http://www.unl.edu/amc/d-publication/d1-pubarchive/2004-5pub/05TM-USAMO/html-05U/08-05usamo-sii.html]here[/url] and once it again it is number 6 on that page (what a coincidence!).

[quote="USAMO Teacher's Mannual"]You will be allowed 4.5 hours on Tuesday April 19, 2005 (12:30 p.m. –5:00 p.m. EDT or equivalent in your time zone) for Problems 1, 2 and 3, and 4.5 hours on Wednesday April 20, 2005 (12:30 p.m. –5:00 p.m. EDT or equivalent in your time zone) for Problems 4, 5 and 6.[/quote]

What time zone is EDT? and what time is it in eastern time?

umm...EDT [i]is[/i] eastern time (the D is because it is daylight savings time instead of the S for standard time)

[quote="jmadsen"]---One horse is the same color as itself.
--If any set of k horses are the same color, then k+1 horses are the same color because any k of them are the same color.
--By induction, all horses are the same color.[/quote]

You know, there is an example just like this in AoPS volume I. The author wants you to find whats wrong with the proof.

I'm going to use a black pen.

BTW, I LOVE your signature!  BUT[hide]The flaw is that you would have to prove that given any set of k horses that are the same color, if you added another horse to make k+1 horses, they would all be the same color.  You assumed that all sets of k horses are the same color, which is exactly what your trying to prove.  Basically, it be like saying...

Prove: All numbers are less than 100.  1 is less than one hundred, and given that any set of k numbers has every number less than 100, that means that any set of k+1 numbers has every number less than 100, because any set of k number has every number less than 100.  QED.  [/hide]

It's not worded very well in my signature, because of the text limit, but the real proof is better. It's something like this:

One horse is the same color as itself. Assume any set of k horses are the same color. Then add another horse so there are k+1 horses. This new horse must be the same color as the first k-1 horses, since any set of k horses are the same color. It also must be the same color as the last k-1 horses, and in fact any set of k-1 of the horses. Thus, the set of k+1 horses are all the same color. By induction, all horses are the same color.

Of course, the error you pointed out is still there, it's just concealed a little better.

Even though I put pencil, I would bring both... for scratchwork, pencil works nicely for me... for actual work, pen is a must.

I still think pencil is better, because you can erase.  Maybe I will go over what I write in pen.  Or maybe that will just make it worse :(.

As for the horses, induction fails at k = 1, since 1 - 1 = 0.  So if you could start with "every two horses are the same color" instead of "every horse is the same color as itself," then you would have a proof.  But it is not true.

also, by induction, all you know is that k works.  that does not gaurantee k+1 works.  the goal is to show given k, k+1 works, but k+1 itself is not given.

Pencil for geometry problems. Pen for rest. It's annoying to mess up diagrams even in the final proof, and it happens.

You could always put a pen in your compass :D .  jk...for a geometry problem I would probably use pencil also.

[quote="rcv"]Whether you use pen or pencil, you might ask your proctor if the school's copy machine is in excellent working order, and if he or she is willing to fax a photocopy of your solutions.

This gives your proctor the opportunity to adjust lightness/darkness to accomodate your handwriting density, and gives a high-density black and white copy a Fax machine should handle well, provided you didn't write too small.[/quote]This seems like reasonable advice; however:
[quote="AMC website"]Neither the problems [i]nor the student’s solutions[/i] should be machine copied. The USAMO is copyrighted.[/quote](emphasis added)

I strongly recommend using a black pen.  We receive faxed pencil-written stuff here all the time, and it's very difficult to read.

My (very strong) advice would be to use pen. (If you have to use pencil because  you are so poor that you can't afford a pen - don't use eraser ) . If you have to make a corection, just draw a simple line (or cross it out) rather than smudge the whole thing ( you never know, you may have to see  fall back of  what you wrote earlier - if you suddenly realize that error which you thought wasn't an error ..)

Write clearly (not only for the examiner but you may misread your own  hand writing and make silly mistake)

etc..etc..
(and above all answer all  questions  with really fault-free proof)

Agreed with Gyan.  Use pen.  Single line strikethrough your errors.  As I understand it, you're faxing your solutions.  We get stacks and stacks of homeworks in pencil that are very hard to read.

Pen.

And even if you use a pencil, I agree with Gyan - no eraser.  Pencil + erase over + rewrite + erase again + rewrite again + fax = no one will be able to read it = sad grader = lower grade.


[quote="236factorial"]If x and y are integers from 1-9, what is the the smallest value of $x+y+\frac{4}{x}+\frac{1}{y}$?[/quote]

[hide]Seperating the equation into the x and y terms...

$(x+\frac{4}{x})+(y+\frac{1}{y})$

To minimize both terms, we must make $x$ and $y$ the smallest they can be.

Let's try to see what it becomes if $x=1$ and $y=2$:

$1+\frac{4}{1}+2+\frac{1}{2}=7\frac{1}{2}$

Let's reverse it and try it ($x=2$ and $y=1$)

$2+\frac{4}{2}+1+\frac{1}{1}=6$

So the smallest value is when $x=2$ and $y=1$[/hide]

How did you come up with the end of the first line?  :?

the AM-GM inequality $\frac{a+b}{2}\geq\sqrt{ab}$
(because $a-2\sqrt{ab}+b=(\sqrt{a}-\sqrt{b})^2\geq0$)

I thought you used that  :P . It's a little too hard for the basics forum  ;)


I would like to thank everything mod for ftw thank you for keeping us safe without you chaos would occur and ftw would not be as fun.  :thumbup:
Please be respectful in your replies.

Monkeygirl12, your appreciation is greatly appreciated. It is for people like you that we continue on with our difficult, thankless task.


I'm trying to write a script to solve a problem, and I need the Mod (% in C++ and many other languages) to do it. Does JS support this feature?

I don't know any Javascript, but googling for 'javascript mod' tells me that % works just fine.

All right. For anyone that knows JS, why doesn't this work then?

[code]
<HTML>
<HEAD>
<SCRIPT LANGUAGE="JavaScript">
<!--
for (i=1; i<10000; i++) {
if (13 % i == 0 &&
12 % i == 1 &&
11 % i == 1 &&
10 % i == 1 &&
9 % i == 1 &&
8 % i == 1 &&
7 % i == 1 &&
6 % i == 1 &&
5 % i == 1 &&
4 % i == 1 &&
3 % i == 1 &&
2 % i == 1) {
alert("One is: " + i ) ;
     }
}
// -->
</SCRIPT>
<TITLE>
Question 30
</TITLE>
</HEAD>
<BODY>
This is my program! Good luck! -Archive
</BODY>
</HTML>
[/code]

[quote="TripleM"]It works fine. But a%b means the remainder when you divide a by b, and I'm guessing you meant i%13==0, i%12==1, etc.

(edit) oh yeah, and you may also want to know that if you did mean what I said above, the smallest solution to that set of equations is 83161, so you'll need to increase that 10000 if you want it to do anything.[/quote]

Yeah, I actually need the three lowest values. And thanks for the tips!

EDIT: Got the other two. 443521 and 803381. Thanks alot!


limit
trigonometry
calculus
calculus computations

We have:$S_n  = \sum\limits_{k = 2}^n {\frac{1}{k}} ctg\frac{\pi }{{2k}},\forall n \in N,n \ge 2,$and
$
P _n  = (\sum\limits_{k = 0}^{n - 2} {2^k } tg\frac{\pi }{{2^{n - k} }})^{ - 2} ,\forall n \in N,n \ge 2.
$

Compute:$\mathop {\lim }\limits_{n \to \infty } S_n ,\mathop {\lim }\limits_{n \to \infty } P _n $

$S_{n}$ obviously diverges to +$\infty$,because $\frac{1}{n}\cot{\frac{\pi}{2n}}{\sim}\frac{2}{\pi}$,when $n$ is sufficently large.
$P_{n}{\leq}(2^{n-2}\tan{\frac{\pi}{4}})^{-2}{\rightarrow}0$.


1.) You and your coworker together make $16 per hour. You know your coworker earns 10 percent more than you do. Your hourly wage is $. 7.62
After taking Math 1010 your hourly wage is raised to $12. This is a raise of %. _________
After returning to work you can't help mentioning casually to your coworker that now you make_______ % more than he does.
2.)You can paint a certain room in 7 hours. Your brother can do it in 6 hours. How long does it take the two of you working together? (hours)
3.)How much 90 percent vinegar do you have to add to a gallon of 5 percent vinegar to get 20 percent vinegar?  (gallons)
4.)Enter an algebraic expression that gives the area of a circle with a circumference.
5.)You buy a house including the land it sits on for $155000. The real estate agent tells you that the land costs $24000 more than the house.
The price of the house is $____  and the price of the land is $_____ .

6.)You figure that you will buy property many times and you endeavor to work out a formula that will tell you the price of the land and the price of the buildings any time.

So suppose the price for the land and buildings combined is  (for "sum") and the land costs  (for "difference") more than the buildings.

Then the price of the buildings is _____   dollars
and the price of the land is ______  dollars.

7.)Suppose the next time you buy a house including the land it sits on for 120000 Dollars. The real estate agent tells you that the land costs 25000 more than the house.

You apply your formula derived in the previous problem.
and you find out that the price of the house is ____  dollars and the price of the land is _____ dollars.

8.)The next time you buy a house including the land for 156000 Dollars. This time the real estate agent tells you that the land costs 13000 Dollars less than the house.

You realize that you can still apply your formula. You set S=____  dollars and D=____  dollars.
and you find out that the price of the house is _____dollars and the price of the land is____dollars.

Sorry for all the problems i cant seem to figure them out


function
LaTeX
number theory unsolved
number theory

i cant solve the following problem (general version of 1978/3)
Let $f(n),g(n)$ be functions from $\mathbb{N}\to \mathbb{N}$ that are strictly increasing and together covers the set of integers exactly (each integer $k$ occurs exactly once in the sequence $f(1),g(1),f(2),g(2)$. etc). Also $g(n) = f(f(n)+1$ (its easy to see $f(f(n))+1 = n+f(n))$.
explicitly find $f(n)$ and $g(n)$ as succinctly as possible.

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