Define f(n) by:

f(n)= n/2 (if n is even),

         (n+1023)/2 (if n is odd)

Find the least positive integer n such that f(f(f(f(f(n)))))= n

Can someone explain how to do this problem?

I would try different cases for n being odd or even. If n is even then:
f(f(f(f(n/2))))=n then try n/2 is even:
f(f(f(n/4)))=n and so on.
Then, what you get should be linear so just solve for n. There are only 2^5 or 32 combinations (if you're desparate...) There's probably a better way that I don't know.

Polymorphic, can you please explain your reasoning more clearly? If you plug 1 into the equation, it does not equal n.

well, if you plug 1 into that equation, you get n= 528, but i bet you could get a smaller n since my reasoning isn't correct.

yea, that was stupid, i'm pretty sure n = 33.  Basically as i said before f(f(f(f(f(n)))) = n/32 + something.  THat something clearly has a minimum value of 1023/32. (n is only odd at the beginning.  After that, each application of the function keeps it even.)  Sorry that wasn't very clear, someone else can probably explain it better.     [/hide]

Okay a trick which makes things easier  :)

One of the ways  to look at it is  to look at the inverse function... let is calll it g

It is  'two valued'

that is              g(n) = 2*n
                   or          = 2*n -1023

we can choose any one of the above to [i]work it to our advantage.. [/i]

to make sense: fo n = (g......(n) ) we musht have

so    n = g(g(g(g(g(n))))) = (2^5)* n -1023

or n = 32n -1023
or n = 1023/31 =[b] 33. [/b] :)

BTW: Can any one tell me what will be the next  such n ..
Hint (It is not hard)  you can actually easily find the general value of such n .

[quote="keta"]Let's see, is the next n [b]341[/b]?[/quote]

No.(341  does not work - check it out )  Hint: There are only  (2^ 5) or 32   such  numbers (Including 0 and 1023)  and answer is fairly simple once you think about it.

Added later: BTW 341 [b] is  [/b] the smallest positive number for which f(f(n))=n. .. (but not  f(f(f(f(f(n))))))[/b]

Interesting point, If you have not noticed it uptill now:

Let us only consider numbers between 0 and 1023.

Assume the number is written as a binary 10 digit number (padding with zeros on the left if needed)

The  'f' now just means   right shift  -circular - (eg each bit shifts to right and the right most  bit becomess the left most).

Now "five times  f" just means shift the bits  5 times ..

Think about it and the answer becomes really trivial..


inequalities
calculus
calculus computations

For postive integer $ n$, prove that:

$ \left(\frac {2n \minus{} 1}{e}\right)^{\frac {2n \minus{} 1}{2}} < 1\cdot 3\cdot 5\cdots (2n \minus{} 1) < \left(\frac {2n \plus{} 1}{e}\right)^{\frac {2n \plus{} 1}{2}}$


LaTeX
integration
calculus
derivative
calculus computations

Use intergration to find a general solution to the differential equation y'=x(x+1)^(1/2)

thank you

Since you didn't use $\LaTeX$, I am assuming that you have

\[ y'=x\sqrt{x+1}.  \]

In this case, when you integrate you get

\[ y=\int x\sqrt{x+1}\,dx=2\int\left(t^{4}-t^{2}\right)\,dt=\frac{2\left(x+1\right)^{\frac{3}{2}}\left(3x-2\right)}{15}.  \]

Here I just made the substitution $x+1=t^{2}$.

Oh yes, I forgot about the constant.  It wouldn't make a difference if you wanted to check your answer (take the derivative), but I suppose it's a good idea if you don't have an initial condition.


There might be something on the Mathcounts website.

In 2004, the format of the Countdown Round at the national competition changed to a single elimination bracket. The top 12 scorers on the written round advance to the Countdown Round. In the first round, 5th place challenges 12th place, 6th place challenges 11th place, etc.   The top 4 get a bye in the first round.  For more details, go to page 12 of the document on mathcounts.org found at

http://www.mathcounts.org/webarticles/articlefiles/510-C&SGenInstruct.pdf

From what I know:
>Its a regular bracket formula, sort of like the March Madness thing.
>One on One
>First rounds are best of 5, but in semis and finals its first to 4.
>Top 12 participate
>45 seconds to answer the Q, but can buzz in before then (obviously)
>After they announce your name, you have 3 sec to answer question
>If you get wrng, other kid has remainder of 45 sec to answer Q
>If after 5 (in beginnign part) they are tied, sudden victory ensues (first right wins)


If $ a$ and $ b$ are positive numbers, determine the minimum value of \[ \sqrt{49 \plus{} a^2 \minus{} 7\sqrt{2}a} \plus{} \sqrt{a^2 \plus{} b^2 \minus{} \sqrt{2}ab} \plus{} \sqrt{50 \plus{} b^2 \minus{} 10b}\] Any hints for this one before solutions are offered? I see nice patterns in the numbers given but no nice way to proceed..


I'll start with one which is pretty funny

hmm...
i'm not sure how this is gonna turn out

i'll keep this going for now though
if anything starts getting bad, it will be locked (just a heads up)

I cleaned up some unnecessary posts. If this continues, I will recycle this topic.

[quote="myyellowducky82"][quote="Ttocs45"][quote="AIME15"]Why does something make me think that all the guests are THECHAMP?[/quote]
Their not.
I was one. (Second place :trampoline: )
Ac-king was one.
One was a real guest.
27r was five.  :rotfl:[/quote]
Just wondering but how can someone be 5 GUESTS???[/quote]

Multiple tabs.

Nice post, mewto...I think you're trying to reach your goal too soon...

Anyway, I noticed that now, you can play a game:
3 person
5 problems
any time, rank system

and if you win, the leaderboard counts the game as a win!
Despite the fact that the leaderboard states
* Number of games won is only included for games with at least [b]10 problems[/b] and 3 players.

i can view the nonviewable----------- that's pretty unusual.

He takes advantage of some glitch that lets him view unviewable games...and now hes braggin abotu it...and he also won't tell levans :P

I'm pretty sure that viewing the unviewable isn't that hard...I think I might have a way to do it.  Not that viewing an unviewable game serves any purpose.

Wow, three lines of games!  And made by all different people!

EDIT: I told somebody to make a cd so it would be 3 lines, but it was already two.

That's not very unusual.

Back in the day, 3 lines of games would be normal.  The most # of different games started by different people that had started was 13. (that I saw, that is)

"back in the day"?  Summer?

Well now we have school...i haven't even been seeing two lines..

[quote="lotsofmath"][/quote][quote="myyellowducky82"][quote="Ttocs45"][quote="AIME15"]Why does something make me think that all the guests are THECHAMP?[/quote]
Their not.
I was one. (Second place :trampoline: )
Ac-king was one.
One was a real guest.
27r was five.  :rotfl:[/quote]
Just wondering but how can someone be 5 GUESTS???[/quote][quote="lotsofmath"]

Multiple tabs.[/quote]
haha I acctually had about ten guests open but couldn't get them all in the game before it started.
it crashed my computer soon though.

Actually, it became 15-player countdown.

edit: 16-player countdown.
edit: 18-player countdown.
edit: 19-player countdown.
edit: 21-player countdown.

Yeah champis good at the stuff. AS of now its 21. How do you post screenshot in text?

lol i can make more-than-2 people cd too...

and its not a hack...

i cant even make a program for my calc...

[quote="isabella2296"]Wow, 6-player countdown![/quote] isn't countdown supposed to be only 2 players?

Yes, THECHAMP found a way to get more people in.


Well as many of you know state chess tournaments for missouri are being held. This year it is on Oct. 25 and is in Columbia, MO. So who's going?

I've never spent too much time on chess.

Yeah but its still fun to be able to pwn random noob kids who think that because they played chess once they can beat you and then say "Ha I'm smarter than ****** *******!!!" Not that that's happened to me :rotfl:


inequalities
floor function
ceiling function
ratio
AMC
AIME
probability

[hide="Solution"]For $k>0$ we can write $A_k$ as

$A_k={1000\choose k}{1\over {5^k}}={{1000\cdot 999\cdot 998\dots(1001-k)}\over k!}{1\over {5^k}}={1000\over{5\cdot 1}}\cdot{999\over{5\cdot 2}}\cdot{998\over{5\cdot 3}}\dots{{1001-k}\over{5k}}$.

It follows that $A_k=A_{k-1}{{1001-k}\over{5k}}$. Hence, $A_k$ will increase as long as ${{1001-k}\over{5k}}\geq 1$. Solving the inequality gives $k\leq 166{5\over 6}$. Therefore, the largest $A_k$ is $A_{166}$.[/hide]

that is a very nice solution, all the others are quite long. was that your first solution?

Interesting remark: we know definitely it's not $A_{1000}$ because you can't bubble that on AIME :rotfl: (the first thought I have for this problem)\

my solution is almost exactly the same as Farenhajt's, so I will not bother to post it.

Sorry for a very long revive, but I have a intuitive, highly nonrigorous approach.

Dividing each term by $ 1.2^{1000}$ will preserve the $ A_k$ that is maximum, so we may equivalently consider the expansion of $ (\frac{5}{6} \plus{} \frac{1}{6})^{1000}\equal{}B_0\plus{}B_1\plus{}B_2\plus{}\cdots\plus{}B_{1000}$. Then $ B_k$ is the probability that if we flipped a coin weighted to land on heads $ \frac{5}{6}$ of the time and tails $ \frac{1}{6}$ of the time of getting exactly $ k$ tails. Intuitively, since tails land $ \frac{1}{6}$ of the time, we should expect the most likely result to be $ \frac{1000}{6}$, or $ 166\frac{2}{3}$. This should limit us to the choices of $ A_{166}$ or $ A_{167}$, which we can easily compare.

Sorry for the revive, but how did (for the first solution posted) did he simplify (1000 choose k) times 1/5 ^ k to the expression that he simplified to?

[quote="Farenhajt"]$\begin{aligned}
A_k={1000\choose k}{1\over {5^k}}&={{1000\cdot 999\cdot 998\cdots(1001-k)}\over k!}{1\over {5^k}}\\
&={1000\over{5\cdot 1}}\cdot{999\over{5\cdot 2}}\cdot{998\over{5\cdot 3}}\cdots{{1001-k}\over{5k}}\\
\end{aligned}$
[/quote]
There are $k$ factors on the numerator, there are $k$ factors in $k!$ and there are also $k$ factors in $5^k$... so he packed one of each factors into an individual fraction...

Could you please elaborate on how he placed the factors together? Also, where did (1001-k) come from...?

Let's elaborate one thing at a time: the 1001-k is basically $1001-k=1000-k+1$.

Now, for every $A_k$, we got

$\begin{aligned}
\frac{1000\cdot999\cdots(1000-(k-1))}{k!\cdot5^k}&=\frac{\underbrace{(1000-(1-1))(1000-(2-1))\dots(1000-(k-1))}_{k\text{ factors}}}{\left(\underbrace{1\cdot2\cdots k}_{k\text{ factors}}\right)\cdot\left(\underbrace{5\cdot5\cdots5}_{k\text{ factors}}\right)}\\
&\phantom{=}\vphantom{}\\
&=\underbrace{\frac{(1000-(1-1))}{1\cdot5}\cdot\frac{(1000-(2-1))}{2\cdot5}\cdots\frac{(1000-(k-1))}{k\cdot5}}_{k\text{ factors}}
\end{aligned}$

to find the greatest numerical term in $(1+x)^n$  where $n$ is a natural number and $x$ is real.

calculate $m = \frac{|x|(n+1)}{|x|+1}$

if $m$ is an integer, then the $m$th terms and the $(m+1)$th term are both the greatest terms.

if $m$ is not an integer, then the $([m] + 1)$th term is the greatest term.

here $[m]$ means the greatest integers less than or equal to $m$.

applying this to the problem,

$m = \frac{(0.2)(1001)}{1.2} = 166.83$

since, $m$ is not an integer, the greatest term would be the $167$th term.

that is $\binom{1000}{166}(0.2)^{166}$

[quote="jimhu"][hide]
Notice that we want to find the point where $A_i$ stop increasing.
The ratio of $A_i$ to $A_i+1$ is $R=\frac{1000-k}{k}$. We want $\left\lfloor R \right\rfloor=5$.
So: $R=\frac{1000-k}{k}=5\Rightarrow k=166+2/3\Rightarrow k'=\boxed{166}$.[/hide][/quote]
oops

I had the ratio as $\frac{1000-k}{k+1}$, because, for example, $\binom{1000}{1}\cdot \frac{999}{2}=\binom{1000}{2}$ and $1000-999$ is one less than $2$.  What is wrong with this reasoning?  It gave me $k=167.5$ :(

Can somebody explain the first line of solution 1 on the wiki page?

http://artofproblemsolving.com/wiki/index.php/1991_AIME_Problems/Problem_3

I'm not quite sure how it got $ \frac{N!}{k!(N-k)!}x^k = \frac{(N - k + 1)!}{k!}x^k$

@above:$$\frac{N!}{(N-k)!}=\frac{1\cdot2\ldots(N-k-1)(N-k)(N-k+1)\ldots(N-1)N}{1\cdot2\ldots(N-k-1)(N-k)\phantom{\cdot(N-k+1)\ldots(N-1)N}}\bold{\ne}(N-k+1)!$$
That doesn't seem right . . .

p5: We consider the ratio of consecutive terms, which is $\frac{\binom{1000}{k+1}(0.2)^k \cdot 0.2}{\binom{1000}{k} \cdot 0.2^k} = \frac{\frac{1000-k}{(k+1)!} \cdot 0.2}{\frac{1}{k!}} = \frac{1000-k}{k+1} \cdot 0.2 \geq 1$. Solving and rounding down gives $k \geq 165$, so the greatest term is $k=165+1=\boxed{166}$.


I saw this problem somewhere and didn't know how to do it...maybe I saw wrong and its not possible. Anyways, i'd like to know how to do it if it is possible. What value times $ x\plus{}y$ equals $ x^5\plus{}y^5?$

Factor $ x^5\plus{}y^5$. What do you notice?

$ a^{2n\plus{}1} \plus{} b^{2n\plus{}1} \equal{} (a \plus{} b)(a^{2n} \minus{} a^{2n\minus{}1}b \plus{} a^{2n\minus{}2}b^2 \plus{} \cdots \plus{} ab^{2n\minus{}1} \plus{} b^{2n})$

[quote="mathswimmer"]What value times $ x \plus{} y$ equals $ x^5 \plus{} y^5?$[/quote]

Yup, you can factor this.

[hide]
$ x^3 \plus{} y^3 \equal{} (x \plus{} y)(x^2 \minus{} xy \plus{} y^2)$
$ x^5 \plus{} y^5 \equal{} (x \plus{} y)(x^4 \minus{} x^3y \plus{} x^2y^2 \minus{} xy^3 \plus{} y^4)$
$ x^7 \plus{} y^7 \equal{} (x \plus{} y)(x^6 \minus{} x^5y \plus{} x^4y^2 \minus{} x^3y^3 \plus{} x^2y^2 \minus{} xy^5 \plus{} x^6)$

etc... There's a pattern.[/hide]

Ok...thanks that makes a lot of sense. Then what about for $ x^5 \minus{} y^5$? Would you do use x-y and then do the same but does the order of +/- reverse? Just a guess...

Nope, it's actually nicer:

$ x^5\minus{}y^5\equal{}(x\minus{}y)(x^4\plus{}xy^3\plus{}x^2y^2\plus{}x^3y\plus{}y^4)$


geometry
3D geometry
calculus
integration
inequalities
geometric series
algebra

Let $F(x)$ represent the reciprocal of the cube root of $x$. Without using calculators or computers find the integral part of
$F(4)+F(5)+F(6)+F(7)+\cdot \cdot \cdot+F(999999)+F(1000000)$

Misread.

How'd you get your first inequality, and how does the second follow from your first?  :maybe:


calculus
integration
limit
logarithms
function
calculus computations

Salutations,

I have had some troubles with improper integrals. After the lesson with my teacher and reading the math book, I still don't understand precisely what to do. I'm using James Stewart's [i]Calculus: Early Transcendentals[/i] I have a question on how to evaluate both continuous and discontinuous improper integrals. Here are two examples from the book I'm using.

Continuous: Evaluate the integral x*e^x with respect to x from negative infinity to zero.

Discontinuous: Evaluate the integral 1/(x-1) with respect to x from zero to three if possible.

Any advice, guidance, or help on this would be much appreciated. I think I might be missing one key thing, but I am not certain.

Thank you in advance,
Brandon[/i]

I haven't done improper integrals yet, but I'll give it a shot based on what I know from reading ahead. You should probably wait until someone more adept comes along. I also used integration by parts which we haven't covered in class either.
[hide]1.
$\int^0_{-\infty}xe^x\;dx=\lim_{n\to -\infty}\int^0_n xe^x\;dx$
$=\lim_{n\to -\infty}\displaystyle\left(\displaystyle\left.xe^x\right]^0_n-\int^0_n e^x\;dx\right)$
$=\lim_{n\to -\infty}\displaystyle\left(0-ne^n\right)-\lim_{n\to -\infty}\displaystyle\left(e^0-e^n\right)$
$=\lim_{n\to -\infty}\left(e^n-ne^n-1\right)$
$=-1$
2.
$\frac{1}{x-1}$ is symmetric about (1,0), so $\int^0_2 \frac{1}{x-1}\;dx=0$.
$\int^0_3 \frac{1}{x-1}\;dx=\int^2_3 \frac{1}{x-1}\;dx=\ln 2$. [/hide]

Thank you very much,

The first one you were very correct on. The second one was not correct, but I don't understand where you really went wrong.

Brandon

[hide]
[quote]Discontinuous: Evaluate the integral 1/(x-1) with respect to x from zero to three [b]if possible[/b]. [/quote]
Remember that its not true in an integral like this that $-\infty + \infty=0$. Try calculating the bit that you are cancelling out, and you will see that the integral does not exist.
[/hide]

A lot depends on the definition you want to use here. For the case of finite singularity s inside the interval [a,b] of integration you may want to undestand the improper integral as the limit of the sum of integrals from a to s-d1 and from
s+d2 to b when d1,d2>0 and tend to 0 independently. If so, then the integral, indeed, does not exist. Another possibility is to consider the so-called principal value when you take d1=d2=d>0 and let d tend to 0. In this case the limit exists and is ln2.

First: you can post questions like this over in the "Caclulus calculations and tutorials" forum under "College playground."

Fedja mentioned the principal value. I'd mention two caveats:

1. It's not a default of the notation. If you intend to be talking about the principal value, you'd better say that somewhere. If you just say $\int_0^3\frac{dx}{x-1}$ without further comment, then you do not mean the principal value.

2. I just looked up "principal value" in the index of Stewart. (Not the current edition, but that probably doesn't matter.) It's not there. There's no doubt that Stewart did not intend this problem to involve the concept.

Here's a framework for looking at this in the case of any individual integral: for how many distinct reasons is this integral improper? $\int_{-\infty}^0xe^{-x}dx$ is improper for one reason: the $-\infty.$ But $\int_0^3\frac{dx}{x-1}$ is improper for two reasons: the left side of 1 and the right side of 1.

The rule is that you must separate the integral into pieces each of which is improper for only one reason. In order to claim convergence, you must separately have convergence on each piece. In the case of $\int_0^3\frac{dx}{x-1}$ one piece is $-\infty$ and the other is $+\infty$, so the integral is divergent.

Another example: $\int_{-\infty}^{\infty}\frac x{x^2+1}\,dx$ is improper for two reasons. Since $\int_0^{\infty}\frac x{x^2+1}\,dx=\infty$ (Note comparison to the integral of $\frac1x$ or the harmonic series), the whole integral diverges despite its symmetry.

And any student who tries to tell me that $\int_0^3\frac{dx}{(x-1)^2}=-\frac32$ will be asked to explain how they think a positive function could give a negative integral.

[quote="Kent Merryfield"]First: you can post questions like this over in the "Caclulus calculations and tutorials" forum under "College playground."
[/quote]

I actually tried moving this thread there but it doesn't let me.   It says:

[quote]
The target forum you choose is a category or a link, so can not contain topics.[/quote]

So maybe this is an error that the Admins can try to fix...(yes I know, they have a plethora of things to do, so it's not that urgent).

[quote="Kent Merryfield"]
Fedja mentioned the principal value. I'd mention two caveats:

1. It's not a default of the notation. If you intend to be talking about the principal value, you'd better say that somewhere. If you just say $\int_0^3\frac{dx}{x-1}$ without further comment, then you do not mean the principal value.
[/quote]
That's right: you should write $p.v.$ in front of the integral to emphasize that you use this notion. I brought the concept in just to show that Bictor717 is not as wrong as one may think. After all, the whole concept of the improper integral appeared out of the desire to make sense of integrals of functions that are not Riemann integrable. So why not to try to go a little bit further and introduce one more concept that allows to integrate even more functions?

[quote]
2. I just looked up "principal value" in the index of Stewart. (Not the current edition, but that probably doesn't matter.) It's not there. There's no doubt that Stewart did not intend this problem to involve the concept.
[/quote]
Sure. The correct answer to write on a test in the class based on Stewart's book is "doesn't exist". But the principal value is so useful in higher level analysis that  I couldn't resist the temptation to throw it in after Bictor717 essentially reinvented it. I apologize if it confused anybody.
[quote]
And any student who tries to tell me that $\int_0^3\frac{dx}{(x-1)^2}=-\frac32$ will be asked to explain how they think a positive function could give a negative integral.[/quote]
Come on Kent! Nobody in this thread did anything like that! But if you change $x$ to $z$...  ;)

Anyway, I agree to Kent in one important respect: when learning new stuff, one  should go one step a time and first master the concepts that are in the book exactly as they are there. I just didn't want a nice and really useful idea that already surfaced to sink without any further comment on the ground that it is not in the textbook. I again apologize for any confusion my posts might cause here.     :blush:


Let a,b,c>0 such that a-b-c \geq abc
Prove that : a 2 -b 2 -c 2  \geq 2 \sqrt 2abc

We have c \leq (a-b)/(ab+1), so it suffices to prove that a+b \geq 2 \sqrt 2ab/(1+ab)+(a-b)/(1+ab) 2 . Of course, it suffices to have (1+ab)(a+b) \geq 2 \sqrt 2 ab+a-bw hich comes down to ab+( \sqrt 2-a) 2  \geq 0.

Since c \leq (a-b)/(1+ab) we have b^2+c^2+2 \sqrt2 abc \leq b^2+(a-b)^2/(1+ab)^2+2 \sqrt 2*ab(a-b)/(1+ab). So, we should prove that this is smaller than a^2. Since a^2-b^2=(a-b)(a+b), we can divide with a-b to obtain a+b \geq 2 \sqrt 2 *ab/(1+ab)+(a-b)/(1+ab)^2. But (a-b)/(1+ab)^2 \leq (a-b)/(1+ab). So, it suffices to show that a-b+2  \sqrt 2 ab \leq (a+b)(ab+1) and this is easy.


function
algebra unsolved
algebra
functional equation

Find all real functions $f$ definited on positive integers and satisying:

(a) $f(x+22)=f(x)$,

(b) $f\left(x^{2}y\right)=\left(f(x)\right)^{2}f(y)$

for all positive integers $x$ and $y$.

There are more solutions.
From (a), we know that $f$ is periodic. We know that $x \equiv y \pmod {22} \implies f(x)=f(y)$.
We will extend this function to $\mathbb{Z} \rightarrow \mathbb{R}$ with (a).
It is easy to see that this extension also satisfies (b) as well. (If uncomfortable, you can just replace $f(0)$ with $f(22)$)
From (b), $f(x^{2n}y)=f(x)^2f(x^{2n-2}y)= \cdots = f(x)^{2n}f(y)$. Setting $x=y$, we have that $f(x)^{2n+1}=f(x^{2n+1})$.
From Euler's Theorem, we have $x^{11} \equiv x \pmod {22} \implies f(x)^{11}=f(x^{11})=f(x) \implies f(x) \in [-1,0,1]$

Case 1. $f(0)=\pm 1$
$f(0)=f(0^2x)=f(0)^2f(x)=f(x)$, so $f(x)$ is a constant function.
We have Solution 1: $f(x) \equiv -1$, Solution 2: $f(x) \equiv 1$.

Case 2. $f(\pm 1)=0$
$f(x)=f({\pm 1}^2x)=f(\pm1)^2f(x)=0$. We have Solution 3: $f(x) \equiv 0$.

Now consider the cases where $f(0)=0, f(\pm1) \not= 0$.
From Euler's Theorem we have that if $(x,22)=1$, $x^{10} \equiv 1 \pmod{22}$
Therefore, $f(1)=f(x^{10}\cdot 1) =f(x)^{10}f(1) \implies f(x)=\pm1$ for $(x,22)=1$

Consider the following four sets: $A_p = [1,3,5,9,15], A_n = [7,13,17,19,21], B_p = [2,6,8,10,18], B_n = [4,12,14,16,20]$
$A_p$ is a set of quadratic residues $\pmod {22}$ that is relatively prime with $22$.
$A_n$ is a set of relatively prime numbers that are not in $A_p$.
$B_p$ is a set of numbers that can be written as $2m^2$ $\pmod {22}$.
$B_n$ is a set of even numbers that are not in $B_p$.
Note that we can consider $A$ as odd, $B$ as even, $p$ as positive, $n$ as negative.
Also, note that we can each element in the same set has the same output as well.
Now, we just need to consider $f(0),f(11),f(A_p),f(A_n),f(B_p),f(B_n)$.
We already know that $f(0)=0, f(A_p) \not= 0, f(A_n) \not= 0$.
From $f(0)=f(44)=f(2)^2f(11)$, we have either $f(2)$ or $f(11)$ is $0$.

Case 3. $f(2)=0$.
We have $f(4)=f(2)^2f(1)=0$ so $f(B_p)=f(B_n)=0$.
If $f(11) = 0$ as well, we have $(f(0),f(11),f(A_p),f(A_n),f(B_p),f(B_n))=(0,0,1,1,0,0),(0,0,-1,1,0,0),(0,0,-1,1,0,0),(0,0,-1,-1,0,0)$.
If $f(11) \not= 0$, we have if $y$ is odd, $f(11)=f(11^2y)=f(11)^2f(y)=f(y) \implies f(A_p)=f(A_n)=f(11)$.
Also, we have $f(0) = f(\pm2 \cdot 11^2)=f(11)^2f(\pm2)=f(\pm2) \implies f(B_p)=f(B_n)=0$.
Therefore, we have $(f(0),f(11),f(A_p),f(A_n),f(B_p),f(B_n))=(0,1,1,1,0,0),(0,-1,-1,-1,0,0)$

Case 4. $f(11)=0, f(2) \not= 0$.
We have that if $y$ is odd, $f(y+11)=f(12y)=f(2)^{10}f(y)=f(y)$
Therefore, we have $f(A_p)=f(B_p), f(A_n)=f(B_n)$
$(f(0),f(11),f(A_p),f(A_n),f(B_p),f(B_n))=(0,0,1,1,1,1),(0,0,1,-1,1,-1),(0,0,-1,1,-1,1),(0,0,-1,-1,-1,-1)$

Therefore, there are $13$ possible solutions.


quadratics
geometry
similar triangles
algebra
quadratic formula

Sorry made a mistake. Deleted my solution.

[hide]By similar triangles,
AB x BC = AD x DE
So, DE = 150/7

Draw a tangent line from A to the circle above the line AC.
Call the point of tangency X. Draw another line connecting A and the center O. Call the point of intersection with the circle Y.
By similar triangles,
AX^2 = (24)(25) = AY(r) --> AY = 600/r
AX^2 = ((AY+r)^2 + r^2) by right angle of triangle AXO.
600 = (3600/r^2 + r^2)
0 = r^4 - 600r^2 + 3600
Then just use quadratic formula to get

r^2= 60(2(6)^(1/2) + 5)

Then take square root, and

r = about 24.4 [/hide]

[quote="trex"]which is the development that allows me to arrive to some of the following alternatives:
notices: AD=28 cm, the radius is requested
http://img290.imageshack.us/img290/4397/p5kt5.jpg
a)33cm
b)16.5 cm
c)25 cm
d)12.5 cm
e)17 cm[/quote]

I believe not to agree with some things

first to calculate DE would be due:

ABxAC=ADxAE
this generates DE=14cm

If I understand your explanation well the drawing it would be this way

http://img501.imageshack.us/img501/9052/newcy8.jpg

your you say

AX^2 = (24)(25) = AY(r)--> AY = 600/r

it should not be

AX^2 = 24x(24+25)=(AY-r)x(AY+r)

here you say

AX^2 = ((AY+r)^2 + r^2) by right angle of triangle AXO.

according to me
the right angle this in X, I don't find that it is a Pythagorean relationship

and also according to your calculations if AY=600/r

in the previous relationship it is

AX^2 = ((600/r+r)^2 + r^2)

very different from 3600/r^2

of there one doesn't eat from now on to continue

you could revise

(it excuses for my English use a translator)

Ok. I'm editing my solution to account for the mistake. Thanks.


Έστω $ a,b$ θετικοί ακέραοι τέτοιοι ώστε $ gcd(a,b) \equal{} 1$. Δείξτε ότι η εξίσωση $ ax \plus{} by \equal{} n$ έχι μη αρνητικές ακέραιες λύσεις $ (x,y)$ για όλους τους ακέραιους $ n > ab \minus{} a \minus{} b$. Τι ισχύει όταν έχουμε $ n \equal{} ba \minus{} a \minus{} b$?? :)

if it's what i think it is, it's pretty basic. see my post at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180036

Θα δώσω (επιτέλους) μια σωστή απόδειξη στη θεωρία αριθμών. (Από χθες άρχισα να τη μελετάω μεθοδικά και είμαι ακριβώς σε αυτό το σημείο!)

Ορίστε η λύση μου:

Υπάρχουν $ k , m$ τέτοια ώστε $ ak \plus{} bm \equal{} gcd(a , b) \equal{} 1$

Η διοφαντική εξίσωση $ ax \plus{} by \equal{} n$ έχει $ gcd(a , b) \equal{} 1$ άρα έχει τις λύσεις:
$ x_0 \equal{} kn , y_0 \equal{} mn$ άρα όλες οι λύσεις της θα είναι της μορφής
$ x \equal{} x_0 \plus{} bt , y \equal{} y_0 \minus{} at < \equal{} > x \equal{} kn \plus{} bt , y \equal{} mn \minus{} at$ , $ t \in \mathbb{Z}$

έστω ότι κάποιες λύσεις είναι μη αρνητικές τότε για κάποιες τιμές του t (βασικά τις αντίστοιχες τιμές του) πρέπει:
$ kn >\equal{} \minus{} bt, mn >\equal{} at < \equal{} > \frac {kn}{ \minus{} b} <\equal{} t <\equal{} \frac {mn}{a}$
άρα (επειδή αρκεί να δείξουμε ότι υπάρχει t, το οποίο είναι ακέραιος) αρκεί να δείξουμε ότι το παρπάνω διάστημα περιέχει ακεραίους, δηλαδή ότι
$ \frac {kn}{ \minus{} b} <\equal{} \frac {mn}{a} \plus{} 1 < \equal{} >$
$ \minus{} kna <\equal{} mnb \plus{} ab < \equal{} > 0 <\equal{} n(ka \plus{} mb) \plus{} ab < \equal{} > 0 <\equal{} n \plus{} ab$,
που ισχύει γιατί ισχύει $ n \plus{} ab >\equal{} 2ab \minus{} a \minus{} b \equal{} a(b \minus{} 1) \plus{} b(b \minus{} 1) >\equal{} 0$ επειδή $ a, b >\equal{} 1$

Στην περίπτωση του ίσον έχουμε μοναδική λύση (εκτός αν $ \frac {mn}{a}$ είναι ακέραιος οπότε θα είχαμε δύο ακριβώ λύσεις).

Επειδή μπορεί να έχω ασάφειες (η λύση είναι πάντως σίγουρα σωστή) λόγω έλειψης εμπειρίας διορθώστε με αν βρείτε καμία.

Ωραία άσκηση. Πού τη βρήκες;

Πιο βιβλιο δαβασες τελευταία και εκανες ςξάσκηση στη θεωρία αριθμών?? Εγώ πάντως ου προτείνω το βιβλιο του Andreescu γνστό ως ''104 number theoy problems from the trainig of yhe USA IMO team" αυτός είναι ο τίτλος. Αν το εχεις έντάξει, εαν όχι πάλι στείλε μου με personal message το mail σου για να σου στείλω το pdf αρχείο του. οκ? τα λεμε... :)

Διαβάζω αυτό που είναι στη σελίδα της μαθηματικής εταιρίας γιατί αγνοώ μεγάλος μέρος της θεωρίας. Για εξάσκηση, κυρίως ασκήσεις από διαγωνισμούς στο internet και από το βιβλίο Advanced Number Theory του Harvey Cohn (αν και τις βρίσκω κάπως υπερβολικά εύκολες). Σου στέλνω το mail μου με Personal Message ώστε να μου στείλεις το βιβλίο του Andreescu όποτε βρεις χρόνο.


limit
real analysis
real analysis unsolved

Let $ (x_{n})_{n\geq 0}$ be a convergent sequance and  $ y_{n}\equal{}\frac{1}{2^n}\sum_{k\equal{}0}^n C_{n}^k \cdot x_{k}$
Prove that $ (y_{n})_{n\geq 0}$ is convergent and $ \lim_{n\minus{}> \infty}y_{n}\equal{}\lim_{n\minus{}> \infty}x_{n}$

[hide="obs:"] $ C_{n}^k \equal{} \binom{n}{k}$ [/hide]

Let $ \alpha_k^{(n)}\equal{}\frac{1}{2^n}\binom{n}{k}$. The only properties of $ \alpha_k^{(n)}$ that you need here are:
(1) $ \alpha_k^{(n)}\ge 0$
(2) $ \sum_{k\equal{}0}^n \alpha_k^{(n)}\equal{}1$
(3) $ \lim_{n\to\infty}\sum_{k\equal{}0}^m \alpha_k^{(n)}\equal{}0$ for any $ m$

I suppose $ m$ is fixed. I don't see how this helps.Could you be more specific ?

You could do this:

Without loosing the generality let us consider that $ x_{n}\rightarrow 0$. Let $ \epsilon>0$ be fixed.

There is $ n(\epsilon)$ such that $ |x_{n}|<\epsilon$ for all $ n>n(\epsilon)$.

We have that

$ x_{n}\equal{}\frac{1}{2^{n}}\sum\limits_{k\equal{}0}^{n_{\epsilon}}\binom{n}{k}x_{k}\plus{}\frac{1}{2^{n}}\sum\limits_{k\equal{}n_{\epsilon}\plus{}1}^{n}\binom{n}{k}x_{k}$.
The first sum converges to $ 0$ and the second one is bounded by $ \epsilon$ since, denote the second sequence by $ y_{n}$,

$ |y_{n}|<\frac{1}{2^{n}}\sum\limits_{k\equal{}n_{\epsilon}\plus{}1}^{n}\binom{n}{k}|x_{k}|<\epsilon$

and hence the desired result holds.


Hey guys,

I'm planning to hold a math talk in my website (in a chatroom) soon in the topic of algebra. The level will be from hard mathcounts questions up to AMC 10 (thus, this is why it's placed in High School Basics). The talk, with discussion along in it, will last probably for an hour and half.

Would anyone be interested? If so, sign up here.

Day: This Sunday
Time: TBA

I would, but it'd depend on the timing of it.
Cool idea though  :lol:

[quote="Silverfalcon"][quote="hasan4444"]Can you inform us about the time?[/quote]

Yup, edited to include it. I usually don't edit time until I see couple "might be"s.  :)[/quote]

Thanks! But sadly, I will not be able to make it.  :(

Sorry I can't participate; this would be 4:00 AM to 5:30 AM  :rotfl: Next Time  :)

I'm certainly interested and will attend if I can.  Oh, and are you doing anything that would required high-speed internet?  I've got dial-up :( so that might be a problem.

I am also preparing for AMC's and the like, and I would like to join, so can you PM me the details? Time, chatroom, etc.

There is a slim possiblity that this time range may not work for me if I don't have internet access to my lap top. It's slim, but possible. So, I'll make an official time by sometime later tomorrow.

Meanwhile, provide time and day that would work best for most of you guys. (Eastern Time Zone please so we don't have any confusion)

[quote="bengal"]It might be a little inconvenient, but maybe you could have 2 dif. sessions so more people could attend. If enough people attend on the 1st session, I guess you could cancel the 2nd session, but this is just a suggestion...[/quote]

There will be transcripts afterward available for download for private use.

If this session goes well, there will be more to come.

[quote="Urc"]Any good times for me would be:
Wednesday/Thursday anytime
Sunday evening
Next Monday, Wednesday - Friday evening.

Thanks  :)[/quote]

What time on Sunday evening would work?

[quote="TZF"]I'm just curious -- is your chatroom LaTeX-enabled?[/quote]

I don't think so.

I'd be interested, and I think that I can make the time. :)

[quote="nikeballa96"]i'm gonna be at ARML!
i'd like to come though.[/quote]

That's a great point that I did not realize.

It's going to be sometime on Sunday at the earliest. I don't want to have anyone miss the opportunity due to ARML (if you are going to ARML, have fun and good luck!).

Please suggest specific time on Sunday (5/31/09).

How about 4:00 - 5:30 PM EST (or maybe an hour or two ahead)? I will be able to attend now because it's on Sunday because I will also be attending ARML.

Any time range from 5-9 is good for me. But (in term of times) I'm probably less important than arml-people.

Hey SilverFalcon,

I am also going to be at ARML on Friday. I think that would be a great topic. So, if you could change it to Sunday or Monday , that would be great. Thanks :)

[quote="nikeballa96"]wait, is this going to be strictly amc 10/mc level? will it go higher?[/quote]

This talk will only be AMC 10/hard MathCounts questions. For future ones, there may be higher or various topics, however.

The chat will be on Sunday, 7:30 to 9:00 PM EST.

[b]Access to Chat Room[/b]
To access the chat room, you must go to my website ([url=http://www.hiddendiscoveries.com]Hidden Discoveries[/url]) and click Lectures and follow the guidelines there.
Because this type of lecture will continue throughout this summer, I suggest that you register user name.
I will be [b]Hidden Discoveries[/b] in terms of user name.

[quote="hasan4444"]Great Job Silverfalcon but can you record a script for it, I mean like a recored of the chat?[/quote]

There will be a transcript.

err, I know this is lame, but where do you go and what do you click to get the transcript?

How can there be a transcript if there was no chat yet?  :wink:

No because I'll be busy when you're having the chat, so can I know where to access the transcript?

I will put the transcript up on the website.

REMINDER: THE CHAT TODAY AT 7:30! Like in three-five minutes!

The transcripts are now available on my website under the Lectures. Please visit the Home page though so you can increase my "Hit" count.  :D

Thanks to all coming!

For anyone wanting to know further sessions like this, contact me via PM.

O.K., Thanks alot! The chat was quite wonderful apparently... :oops:

[quote="ThinkFlow"]Here's the [url=ftp://www.hiddendiscoveries.com/Transcript%20Lecture%20Algebra.pdf]transcript[/url], which you can access at Hidden Discoveries. :)[/quote]

This URL does not work. It requires password.

My FTP account is restricted, so to do this, you should go to the website directly and click under Lectures.

http://www.hiddendiscoveries.com

[quote="batteredbutnotdefeated"]I missed the chat :( , but is the transcript similar to an AoPS transcript?[/quote]

Yes, the lecture was run very similar to AoPS classes. The only difference (a major one) is that the chatroom is not LaTeX enabled. But other than that (and that instructors are much better in AoPS), it's very similar. I put up the problems in the link and so was the image.

Also, the chatroom wasn't moderated.  :)


conics
ellipse
geometry unsolved
geometry

Prove that the midpoints of the diagonals of a complete quadrilateral are collinear.

I wrote in 2003 a little book "Méthodes et techniques en Géométrie, A propos de la droite de Newton" (Jean-Lois Ayme, Ellipses, Paris) where 30 different proofs of your question are proposed.
Sincerely


rotation
analytic geometry
combinatorics proposed
combinatorics

We must tile the plane with square tiles, all of the same size. There is a finite number of tile families. The sides of the tiles are labeled by letters of an alphabet, and all the tiles of one family have the same labels, thus are indistinguishable. Tiles may not be rotated or reflected, and the labels are specified for their north side, south side, and so on. There is an infinite supply of tiles of each family. The tiles may be put one next to another, the sides converging only if these two sides have the same labels.
Prove that if it is possible to tile the entire upper-right quadrant, then it is possible to tile the entire plane with this family of tiles!
If the number of tile families is not restricted to be finite, one can find sets of families for which the upper-right quadrant is tileable, while the whole plane is not.

For the second question (the easy one :)), do the following:

Label the families $0,1,2,\ldots$. Cover the lattice square with lower left corner of coordinates $(i,j)$ with the family $i+j$, and color the edges of a square in family $n$ as follows: the south and west sides with color $a_n$, and the north and east sides with color $a_{n+1}$, where $(a_n)_{n\ge 0}$ is a sequence of distinct colors.

It's easy to see that as we go to the left, for example, the indices of the families which cover the squares decrease strictly, so we cannot go arbitrarily far to the left (i.e. we can't cover an entire row, and, as a consequence, we can't cover the entire plane).

This is a really nice problem, and I have no idea why I didn't see this sooner :).

The only important thing here is the name of the family, so let's simply denote the families by $1,2,\ldots,n$.

Among $1,2,\ldots,n$ there is a family (call it $c_1$) appearing infinitely often on the diagonal $x=y$ (I think you understand what I mean, even though it's not that clear :)). Place a square $c_1$ in the plane. Now, among the infinitely many $3\times 3$ squares with center on $x=y$ and belonging to the family $c_1$, there are infinitely many identical ones. Call the family configuration which covers all of these $3\times 3$ squares $c_2$. Place the other $8$ squares from $c_2$ around the square $c_1$ already placed in the plane. Continue like this, each time finding a larger configuration $c_k$ which appears infinitely often among the configurations of $(2k-1)\times(2k-1)$ squares with center on the diagonal $x=y$, and place the families on the border of $c_k$ around the $c_{k-1}$ configuration which we already have  in the plane. This way, at each step, the region of the plane we have covered gets larger and larger, in all four directions.

I'm sorry, again, for any ambiguities, and I hope it's correct.

Ok, this is the correct idea :)
The way you proved it is just proving Konig's Infinity Lemma for this problem. Konig's Infinity lemma is the one we had to prove at Romanian National Mathematical Olympiad for $10^{th}$ classes last year :)


function
calculus
derivative
analytic geometry
graphing lines
slope
algebra unsolved

Prove that if we have function f satisfying the condition
f(x+h)-f(x)=h*f'(x)
where h is a constant
then we have f=ax+b
the same question with
f(x+h)-f(x)=h*f'(x+h/2)
My class has discussed many times but all solutions seem so long
i look forward to hearing from you!
 :oops:

Look at the definition of differentiation.

f(x+h) - f(x) = f'(x) (x+h-x) = hf'(x)

In the definition of differentiation, h must be approaching zero in order to find the tangent slope. However, your problem makes no point of that, so the slope must be constant in order to present the same slope for any number h.

Hence, it is a line, y=mx+b.

Same proof for the second case.


SMO 2007, Open #23

Find the number of 9-digit numbers formed by some or all of the digits 0, 1, 2 and 3 that contains no sequences of 21 or 12 in the number.

[hide]
Consider the first digit (far left digit) of this 9-digit number. Since it cannot be 0, there are 3 possible digits it may be.

Now consider the digit to the right of the previous one considered. It may have the values 0, 1, 2, or 3 so the total possibilities so far must be 3*4 = 12. BUT we need to subtract two from this number because there two illegal positions to discount. Namely, if the first digit is 1 the second digit cannot be 2, and if the first digit is 2 the second digit cannot be 1. So there are 12-2 = 10 possible positions for the first couple digits.

Considering the third digit, there are 10*4 - 2 = 38 (since there are again two possibilities which are illegal we must subtract 2) possibilities.

Continuing the process:

$number.of.legal.positions = ...(((3*4)-2)*4)-2...$

for nine steps. My calculations show that the # of legal positions are 152918.
[/hide]

Is there some undercounting with the 3-digit numbers? The possible 2-digit numbers are 10, 11, 13, 20, 22, 23, 30, 31, 32, 33.

So the number of possible 3-digit numbers with the first two digits chosen from the above set is indeed $10 \times 4$ but many numbers have the sequence 12 or 21 in the second or third digits, namely 221, 321, 312, etc. So we have to subtract more than 2 from 40 possibilities, right?

Yeah, you do need to subtract more than 2. Unfortunately, the number you need to subtract is difficult to predict and merits its own recursive sequence.

[hide="Solution with that method"]Let $a_{n}$ be the number of possibilities for an $n$ digit number. We define $a_{1}= 3$ so that we don't allow $n$ digit numbers beginning with 0. Let $b_{n}$ be the number of these possibilities ending with a $1$ or $2$. So $b_{1}= 2$.

We propose the following:
$b_{n+1}= 2a_{n}-b_{n}$
For every $n$ digit number not ending in a $1$ or $2$ ($a_{n}-b_{n}$ numbers exist), we have two ways of adding a new digit such that the new number has a $1$ or $2$ on the end. But for every $n$ digit number ending in a $1$ or $2$ ($b_{n}$ numbers), we have only one way. So $b_{n+1}= 2(a_{n}-b_{n})+b_{n}= 2a_{n}-b_{n}$

Also:
$a_{n+1}= 2a_{n}+b_{n+1}$
We only need to add the numbers not ending in $1$ or $2$ to $b_{n+1}$. Basically any number ending in $0$ or $3$. But this is always 2 times $a_{n}$, the number of $n$ digit numbers.

$b_{1}= 2, a_{1}= 3$
$b_{2}= 4, a_{2}= 10$
$b_{3}= 16, a_{3}= 36$
$b_{4}= 56, a_{4}= 128$
$b_{5}= 200, a_{5}= 456$
$b_{6}= 712, a_{6}= 1624$
$b_{7}= 2536, a_{7}= 5784$
$b_{8}= 9032, a_{8}= 20600$
$b_{9}= 32168, a_{9}= \fbox{73368}$[/hide]
I'm pretty confident in the construction of the sequences, but the computations at the end may contain errors.

The sequence looks correct, i actually got $a_{n}= 3a_{n-1}+2a_{n-2}$ which can be obtained by combining the recursive relations and eliminating terms in $b$. However, sad to say, i miscounted at $a_{9}$ and only multiplied $20600$ by $2$ before adding to $2 \times 5784$  :blush: . We can't use calculators in this contest and with ten combinatorics problems, it's easy to miss out cases, overcount or undercount.


Prove that:
  $ \sum cos^2\frac{A}{2}\minus{}2\leq\frac{1}{4}\prod cos\frac{A\minus{}B}{2}$

[quote="karis"]Prove that:
  $ \sum cos^2\frac {A}{2} \minus{} 2\leq\frac {1}{4}\prod cos\frac {A \minus{} B}{2}$[/quote]
remember that:
$ 2\prod sin\frac{A}{2}\equal{}\sum {cos\frac{A}{2}}^2\minus{}2$
so we need to prove that:
$ 8\prod sin\frac{A}{2}\leq \prod cos\frac{A\minus{}B}{2}$
the inequality becomes:
$ \sum cosAcosB\plus{}\sum sinAsinB\plus{}9 \geq 8\sum cosA$
but
$ sinAsinB\equal{}cosAcosB\plus{}cosC$
so we need to prove that:
$ 2\sum xy\plus{}9\geq 8\sum x$    (A)(x=cosA..)
I use Lagrange mutipliers to solve this A ineq


algebra
polynomial
function
induction
algebra unsolved

Let be given two real polynomials $ P(x)$ and $ Q(x)$ satisfy that:
         $ P(x_0)$ is a perfect square $ \Longleftrightarrow Q(x_0)$ is a perfect square.
Prove the existence of a real polynomial $ f(x)$ so that:
$ P(x).Q(x) = f^2(x)$ for all $ x \in R$

Yes, a very nice problem. I remember to have solved it when seeing it in an nold shortlist. Difficult one!

it means "old shortlist" i guess.
however, the problem has to have an extra condition. just take two arbitrary polynomials which are always negative. it is unlikely that their product is always a square(e.g. $P(x)=-x^2-1$, $Q(x)=-x^2-2$).
i'm just searching through kalva for the problem.

Peter

Hi darkmaster!
You can see the solution at page 7 ,the 190 Mathematical and Youth Magazine/April/1993/Problem T6\187 ;)

Thanks, pigfly.
I think that we must add the condition: the the leading coefficient of P(x) and Q(x) must be positive.

[quote="pigfly"]
You can see the solution at page 7 ,the 190 Mathematical and Youth Magazine/April/1993/Problem T6\187 ;)[/quote]

Can you post the solution because I don't have this book.

thx

First,if deg P>0, deg Q>0 then Q(x) = a 2 for all x.Thus, P(x) is square for all x, which is impossible because P(x) is continous function.
Now, assume deg P>0,degQ>0,$\exists[N;\infty] $such that: P(x),Q(x) are strictly increasing function for $x\in[N;\infty]$
Let $x_0\geqq $such that:$ P(x_0)=a^2; Q(x_0)=b^2$
Because P(x) is continous and strictly increasing,$\exists\{x_n\}:N\leqq_0<x_1<...<x_n$ such that:
                          $ P(x_n)=(a+n)^2(n\geq0)$
We will prove that $Q(x_n)=(b+n)^2 (1)$ by induction
- For n=0 we are done
-Assume that it holds for some n.because $P(x_{n+})$ is square,$Q(x_{n+1})$ is square i.e. $Q(x_{n+1})=(b+m)^2$
If m=n+1 we are done
If m>n+1 then there exists x' such that:$Q(x')=(b+n+1)^2$. We have :
                      $ Q(x_n)< Q(x')<Q(x_{n+1})\rightarrow x_n<x'<x_{n+1}\rightarrow P(x_n)< P(x')<P(x_{n+1})\rightarrow (a+n)^2<P(x')<(a+n+1)^2$.
This is impossible because P(x') is square(Q(x') is square)
In every case (1) is right.We have:
$\sqrt {P(x_n)} - \sqrt {Q(x_n)} = a - b \rightarrow P(x_n)+Q(x_n)-\sqrt{ P(x_n)Q(x_n)}=(a-b)^2\rightarrow
 P(x_n)Q(x_n)=[\frac {P(x_n)+Q(x_n)-(a-b)^2}{2}]^2$
If we put $\phi(x)=[\frac {P(x)+Q(x)-(a-b)^2}{2}]$ then $P(x_n)Q(x_n)=[\phi(x_n)]^2\forall n=1,2,3,...$
Thus $ P(x)Q(x)=[\phi(x)]^2$,as desire
What do you think,Moubinool? :P


What is the shortest distance to the line y = 3x+4 from the point (6,2)?

Please explain.

Also, I was wondering...
Is there a general technique for finding the distance from a point to a line?

I will give you a hint:

[hide="hint"]
The shortest distance to a line is a perpendicular line.
[/hide]

AIME15's hint should be sufficient.

[hide="Do not look at this until you have solved the problem!"]
In general, the shortest distance from a point $ (m,n)$ to a line $ Ax\plus{}By\plus{}C\equal{}0$ is $ \frac{|Am\plus{}Bn\plus{}C|}{\sqrt{A^2\plus{}B^2}}$.
[/hide]

Where the standard form of an equation is $ Ax \plus{} By \plus{} C \equal{} 0$, and the point $ (x_0,y_0)$, the formula is:

[hide]
$ \frac {|Ax_0 \plus{} By_0 \plus{} C|}{\sqrt {A^2 \plus{} B^2}}$
[/hide]

You can prove this by drawing perpendiculars.

Edit: Beaten, fixed

Um, your formula is also incorrect: it should be $ A^2\plus{}B^2$ under the radical--probably a typo though.

I solved the problem (without the formula  :) ) and got 2 root 10 as my answer. I don't have the answer, so if anyone could verify it that would be great.

Another way to solve it might be to come up with the equation of a perpendicular line passing through the given point, find the point of intersection between the two lines, and then find the distance between the two points.

The formula gives me $ \frac{21}{\sqrt{10}}$, but I'm often wrong.

The numerator should be $ 20$, so the answer is $ 2\sqrt{10}$. And that method works, but is usually tedious.


please help me solve this problem
 In a quadrilateral ABCD prove that AB+BC+CD+CD<2(AC+BD) the sum of four sides is less than twice the sum of its diagonals

use the idea that sum of two sides of a triangle is greater than the third side.
diagonals form four triangles in the quadrilateral

Proofs should be moved to the High school Basics Forum

This forum is not for proofs

Let the intersection of the diagonals be $ E$.

$ AE \plus{} BE > AB$
$ BE \plus{} CE > BC$
$ CE \plus{} DE > CD$
$ AE \plus{} DE > AD$

$ AE \plus{} CE \equal{} AC$
$ BE \plus{} DE \equal{} BD$

$ 2(AE \plus{} BE \plus{} CE \plus{} DE) > AB \plus{} BC \plus{} CD \plus{} AD$
$ 2(AC \plus{} BC) > AB \plus{} BC \plus{} CD \plus{} AD$


This problem is posed and inspired from a nice one of VASC (people say that) but Nguyen Viet Anh (my brother) shown a very nice solution

\[\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}}\ge \frac{10}{(a+b+c)^{2}}.\]

Notice that

\[10=2\cdot3+4\]
\[32=2\cdot 14+4.\]

Arqady? What's your idea?


Let $ K$, $ A$, and $ L$ all be functions of time. Let $ L'/L\equal{}n$, $ A'/A\equal{}g$. Also, let $ k\equal{}K/AL$. Making all substitutions, find $ \frac{dk}{dt}$

$ k(t)\equal{}K(t)[A(t)L(t)]^{\minus{}1}$

$ k'\equal{}\frac{K'}{AL}\minus{}\frac{KA'}{A^2L}\minus{}\frac{KL'}{AL^2}\equal{}[\frac{K'}{K}\minus{}g\minus{}n]k(t)$

if you got a letter for $ \frac{K'}{K}$ this would look a lot nicer..


inequalities
function
inequalities proposed

Let$ a,b,c>0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$ Prove that
$ ab \plus{} bc \plus{} ca\le abc \plus{} 2$ :lol:

[quote="Allnames"]Let$ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$ Prove that
$ ab \plus{} bc \plus{} ca\le abc \plus{} 2$ :lol:[/quote]
You yourself  usuelly give  hints. :D

Let $ a \equal{} \sqrt {\frac {4yz}{(x \plus{} y)(x \plus{} z)}},$ $ b \equal{} \sqrt {\frac {4xz}{(x \plus{} y)(y \plus{} z)}},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Hence, $ c \equal{} \sqrt {\frac {4xy}{(x \plus{} z)(y \plus{} z)}}$ and $ ab \plus{} ac \plus{} bc\leq abc \plus{} 2\Leftrightarrow$
$ \Leftrightarrow2\sum_{cyc}\frac {x}{y \plus{} z}\sqrt {\frac {yz}{(x \plus{} y)(x \plus{} z)}}\leq\frac {4xyz}{(x \plus{} y)(x \plus{} z)(y \plus{} z)} \plus{} 1\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}2x\sqrt {y(x \plus{} z)\cdot z(x \plus{} y)}\leq\sum_{cyc}(x^2y \plus{} x^2z \plus{} 2xyz),$ which is true by AM-GM.

[quote="arqady"][quote="Allnames"]Let$ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$ Prove that
$ ab \plus{} bc \plus{} ca\le abc \plus{} 2$ :lol:[/quote]

.[/quote]
mistakes,sorry

The following inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\plus{}2abc\equal{}5.$ Prove that
\[ 2(ab\plus{}ac\plus{}bc)\leq5\plus{}abc\]

Dear Arqady I have a question concerning Your solution. You have written something like this:
[quote]Let $ a \equal{} \sqrt {\frac {4yz}{(x \plus{} y)(x \plus{} z)}}$[/quote] and similarly for b.

And I have a question how did You find such substitution?

Is there any general useful method for solving inequalities similar to this one?

thanks for reply

[quote="matex (L)(L)(L)"]

Is there any general useful method for solving inequalities similar to this one?

[/quote]
No, there is no a method! At least I don't see a method here. There is an identity only.
I think for the condition $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 5$ we need  find something another.

Arquady's solution is strange for me  :wink:  (I'm thinking how to you find out the indentity  :maybe: )
I have another solution by Mixing Varible method  :wink:
Without loss of generally we can assume that $ a \equal{} min\{a,b,c\}$
Easy to see that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4 \ge 3a^2 \plus{} a^3 \iff 0\le a\le 1$
We call:
\[ f(a,b,c) \equal{} abc \plus{} 2 \minus{} ab \minus{} bc \minus{} ca
\]
We will prove that:
\[ f(a,b,c)\ge f(a,\sqrt {2 \minus{} a},\sqrt {2 \minus{} a})
\]
In fact, we have:
\[ f(a,b,c)\ge f(a,\sqrt {2 \minus{} a},\sqrt {2 \minus{} a})
\]

\[ \iff (a \minus{} 1)(bc \minus{} 2 \plus{} a) \minus{} a(b \plus{} c \minus{} 2\sqrt {2 \minus{} a})\ge 0
\]

\[ \iff \frac {(1 \minus{} a)(b \minus{} c)^2}{2 \plus{} a} \plus{} \frac {a(2 \minus{} a)(b \minus{} c)^2}{(2 \plus{} a)(b \plus{} c \plus{} 2\sqrt {2 \minus{} a})}\ge 0
\]

\[ \iff \frac {(b \minus{} c)^2}{2 \plus{} a}\left(1 \minus{} a \plus{} \frac {a(2 \minus{} a)}{b \plus{} c \plus{} 2\sqrt {2 \minus{} a}}\right)\ge 0,\forall a\le 1
\]
So we only need prove that:
\[ f(a,\sqrt {2 \minus{} a},\sqrt {2 \minus{} a}\ge 0
\]
But it is quite obvious because we have
\[ f(a,\sqrt {2 \minus{} a},\sqrt {2 \minus{} a} \equal{} a(\sqrt {2 \minus{} a} \minus{} 1)^2\ge 0
\]
Which is always correct for all $ a\in [0,1]$
The inequality holds on $ a \equal{} b \equal{} c \equal{} 1$
We are done. I think it isn's a hard problem  :wink:

About Argady's problem, I think we can use the following hint
\[ f(a,b,c)\equal{}RHS\minus{}LHS\ge f\left(a,\sqrt{\frac{5\minus{}a^2}{2(a\plus{}1)}},\sqrt{\frac{5\minus{}a^2}{2(a\plus{}1)}}\right)\]

Yes, zaizai-hoang! It is the idea!  :lol:

here is another solution..........with triangle fuction .... :)

[quote="ye109"][quote="can_hang2007"][quote="Allnames"]
But $ (1 \minus{} cosA)(1 \minus{} cosB)(1 \minus{} cosC)\le cosAcosBcosC$  (1)(very well-know)
Expand (1) and use $ \sum cosA\ge \frac {3}{2}$we have done
[/quote]
Sure?[/quote]

this is very well know ....
$ cosAcosBcosC\le (1 \minus{} cosA)(1 \minus{} cosB)(1 \minus{} cosC)$[/quote]
I know, my friend. But he wrote it in the reverse form and applied this reversed for his solution!

[quote="Allnames"]Let$ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \equal{} 4$ Prove that
$ ab \plus{} bc \plus{} ca\le abc \plus{} 2$ :lol:[/quote]

Without loss of generality $ (a \minus{} 1)(b \minus{} 1)\ge 0$(Because Pigeonhole's Principle),
$ abc \plus{} 2\ge ab \plus{} ac \minus{} a \plus{} 2\ge ab \plus{} ac \plus{} bc$
$ (\because if a \plus{} bc > 2\rightarrow 4 \equal{} abc \plus{} a^2 \plus{} b^2 \plus{} c^2 \equal{} a(a \plus{} bc) \plus{} b^2 \plus{} c^2 > 2(a \plus{} bc) > 4)$


I think that this should be moved to a  different forum

There is no correct answer. One item might end with 200 points, or 1 might win with 1 point.

[quote="CA Math"]Sorry for not knowing much about this stuff, but when does the game "end?"[/quote]
When everyone runs out of points except for one person.

@chenhsi:
[quote]what is the [u]maximum[/u] time needed[/quote]

This is not really much of a math problem, but more of a way expressing frustration at all of the hurt and heal games that are going on right now.

[hide]well, if we had ffc this would be all solved. :P[/hide]

[quote="andersonw"]8 hurt or heal games are made in the fun factory, each starting out with 10 "things", each with 20 points. Whenever someone makes a post in a hurt or heal thread, the person gives one "thing" a point and takes away 2 points from another "thing". Assume that 1 post is made every ten minutes in the fun factory. If only 70% of these posts are in hurt or heal threads, what is the maximum time needed (in hours) for all of the games to finish? Round your answer to the nearest tenth.

Extra credit: What is the minimum amount of time needed (in seconds) before everyone becomes annoyed with all of the hurt or heal games?[/quote]

Simply each thread has a "point value" of 200, every post decreases the "point value" by 1. And minimum "pt val" = 1. So there is 1592 posts necessary to forcewin every game.
If 70% of the posts are in H&H, then 2274.28 posts are necessary.
Then it takes about 22743 minutes, or 379.1 hours, or 15.8 days.

For EC: 1 second

This should be a mathcounts problem or MOEMS section.

Heh...just noticed someone has answered this problem.

james4l: Your solution isn't entirely correct. Hint: The minimum number of points someone can have at the end isn't 1.
But you do get extra credit, so that makes up for it.  :lol:

[quote="ernie"]Ok, FFC is the [b]STUPIDEST[/b] thing ever made.[/quote]

What is FFC?

Is that like the mods that lock all of our posts?

No, it is jjx1's futile attempt to make a committee.

Yay!  My 2000th post overall!

It was designed to control what games go on the FF. That's basically what Mods are for. And FFC might turn into a Senate, and that would just slow the process down. So Mods are best.

Back on topic, the original post touches on a very important issue: the fact that Hurt&Heals are becoming ubiquitous, which would just take up site space. My original suggestion is to contain all of the hurt&heals into one topic and have one at a time.


Clearly there are many economic issues, and have been for a long time, in the United States today with poverty, education, etc. But how much can the government help with these issues? Anything the government does is dulled in its effectiveness by the various levels of authority money has to pass through, as well as the governments inability to cater to specific needs. But then again, economy without government is impossible, because the competition generated in free markets is not enough to cover over all ills; just look at the industrial revolution. So where should we balance? How much should the government do?

There is a certain point past which the government's attempt at economic rectification is both futile and encumbering, since market forces will often solve these problems on their own.  However, the same market forces can perpetuate economic problems, notably in the "robber baron" period of U. S. history where trusts and other major businesses were able to expand enough to influence market forces enough to destroy almost every instance of competition.  This hardly follows Smith's economic theory (since the "invisible hand" was basically rendered powerless), and clearly needed rectification both because of public outcry and corruption within these businesses.  Ultimately, I think history proves that an effective balance of government intervention and natural economic forces is essential to economic prosperity.

[quote="hello"]Well, pure capitalism involves no intervention by the government at all right? But the Keynesian theory makes government intervention possible in crucial circumstances like the Great DEpression which resulted in the formation of macroeconomic policy. I think that the market should probably work according to laws of supply and demand with government intervation in crises situation as it happend after the great depression with President Roosevelt's Social Welfare projects, also monetary policy of the federal reserve has a lot of control over the economy.[/quote]

I think history does not weigh on the side that emergence from the Great Depression was due to Keynesian policy.  As technology advances, economies go through growing pains, sometimes to extreme proportions.  Capital and labor takes time to find their new, efficient homes.  That plus a war brought the U.S. out of depression.

The problem with Keynesian government controls (Countercyclical Demand Theory) is that solution via government fiat stands in the way of economic arbitrage of market forces -- when the entrepreneurs have a chance to solve the problem.

Additionally, government control over interest rates and money supplies have been historically used to tax citizens without approval (or representation) in a manner few people understand.  When the government prints money, no economic value is created beyond marginal liquidity (which is really a production concern, not a paper money concern).  However, every pre-existing dollar is devalued, hence the government takes value from capital holders.

The unfortunate side of the Fed's setting of interest rates such as bond rates and the overnight call rate are that our markets are perfectly capable of functioning without them and discrete adjustments fly in the face of economic continuity.  Why should the value of all capital in the U.S. (and really the world since U.S. bonds are the gold standard) undergo immense volatility every few months according to the point at which Greenspan & Co. decide that it's time?  Volatility alone destroys the value of capital (a la Sharpe Ratios).

I'm also convinced that great fiscal power of the government creates greater incentive for thieves and tools to enter the political arena, thus creating a host of negative externalities.

Unfortunately discussions such as this are moot as government economic control can only be dismantled in times of great efficiency and prosperity and the stage to apply that standard is growing globally.  The only real way to put an end to bad government economic policies is to reduce the need for economic stimulus and that means starting companies (employing people) to do what needs to be done -- something that is very hard to do these days.

Sorry that this post is somewhat rambling and meandering, but I didn't want to take the time out to write a more cogent essay.

I don't think I am good enough in economics to comprehend everything that you have wrote, Mr. Crawford, but in my economics textbook, it shows that macroeconomic policies were developed as a result of the Great Depression, so they did not exist before? Complexzeta told me you studied economics in college, where did you go? Just curious, how come you did not want to work in economic policy?

I went to Washington University in St. Louis but never took an Economics class there, but I read as much Economics as any other subject while in school.

The history of macroeconomic policy is nearly always presented in an overly simplistic way -- even through 100 level college courses.  I think this is partially due to the fact that modern economics (with understandings of incentives, technology, and mathematical tools such as calculus and game theory) doesn't support macroeconomic policy except insofar as "the establishment has power, so it's practical to go along with it."

The best macroeconomic analysis comes almost universally from a microeconomic perspective.  Every person, family, company, industry, etcetera is highly unique.  In such a way economics can be a powerful social science -- perhaps the one which affects our lives the most.  In that regard, macroeconomics of the 30's through 70's is little more enlightening than Astrology and serves only to bolster the major political religions.

[quote="MCrawford"]
The best macroeconomic analysis comes almost universally from a microeconomic perspective.  Every person, family, company, industry, etcetera is highly unique. [/quote]

Undoubtedly true, but are such analyses practical or even possible?  It seems like analyzing any modern large nation's economy today would be too complex to study from a microeconomic perspective.

I do, however, agree that
[quote="MCrawford"]In that regard, macroeconomics of the 30's through 70's is little more enlightening than Astrology and serves only to bolster the major political religions.[/quote]

as macroeconomic theories of this time period (both keynesian and supply side) are clearly not totally reliable.

[quote="miraculouspostmaster"][quote="MCrawford"]
The best macroeconomic analysis comes almost universally from a microeconomic perspective.  Every person, family, company, industry, etcetera is highly unique. [/quote]

Undoubtedly true, but are such analyses practical or even possible?  It seems like analyzing any modern large nation's economy today would be too complex to study from a microeconomic perspective.
[/quote]

To say that they are not practical would be to say that national economic policy is necessarily blind to the same degree.

Unfortunately it seems that a democratic state must fly in the dark in regards to economic policy unless the population is educated enough not be afraid of the light -- no such nation yet exists.

A certain amount of microeconomic analysis is certainly impractical, but its not at all clear that there is need of such government analysis as it hasn't yet been pointed out how the government can do much of anything to improve the macroeconomic environment beyond protecting citizens.

Fortunately, people and business go on with or without somebody else's macroeconomic analysis.

[quote="hello"]Once again I am not sure I understood everything. But are you trying to say that most of the policies that economists suggest, like nobel laureats not good policymaking?[/quote]

That's not really what I'm suggesting, though I do think many past Nobel winners are out of date.  Ideas that won the Nobel Prize 3 decades ago are now either ordinary (mostly because there are more mathematically savvy Economists and financiers) or outdated.  Current winners are often politically chosen and the merit of their work up to debate.

I do think there is some truth to the fact that some academics fail to test their theories in legitimate real world scenarios, but often they can't be blamed as the experiments would have to take on very large proportions.  It's highly unusual that somebody like Milton Friedman gets a chance to rewrite the economic policies of an entire nation (the Chicago Boys got a rare opportunity to advise in nearly every facet of economic policy in Chili).

Many of the best Economists in the world such as John Nye are ignored in the political arena and this is largely due to the fact that the mathematics of artificial intelligence and game theory grew up allowing academics to see that government barriers to economic activity are almost never in the best interest of a self-organizing economic system.

None of this is easy to explain -- a solid foundation of economics is built from years of studying mathematics and economic applications, but I recommend reading introductions to game theory such as The Prisoners Dilemma and Moral Calculations.  I think the next generation of economics and finance students will be largely well-schooled in game theory and artificial intelligence as these are the appropriate tools for chiseling out a finer understanding of economics -- particularly in a technological world.

Thanks for your response. I took introductory course in microeconomics where I learned the basic ideas of game theory which include the Priosoner's deliemma that you mentioned and also Nash Equilibrium. I did not take any advanced classes yet. I just listened to president Bush speak on the conference and a lot of his talk was about social security, I cant really comprehend what is that is gonna be changed if it will? Will people have to buy stocks and bonds?

The Bush administration discussed allowing younger workers to divert some of their FICA tax into personal savings accounts, but they backed off.  These would be investment accounts, yes, but nobody would be forced to use them.

Personally, I think it's the single best proposal of the current administration in regards to any economic issue and I'm sad to see them back off.

The problem with the current system is that it's a Ponzi scheme that requires a heavy tax bailout.  In essence, the government cannot provide all the promised benefits without dipping into the general income tax pool.  In a sense they've been doing this all along as social security is invested (the portion that is invested at all) in treasury bills -- the interest from which comes from the general tax pool.

While social security is not the single largest economic failure of the government, it's one of the larger ones that will have to be dealt with at some point.  Somebody must necessarily get hurt and it's about time we bit the bullet and took on at least some of that pain, but politicians are very afraid that suggesting taking on some pain now will mobilize opposition against them.  Don't expect the Republicans to back anything controversial now that they've finally won the hard fought battle of filling congressional rolls with their own along with a sitting president.

The irony is that the only members of Congress willing to support the status quo system are the tools who tote the party banners and are either too stupid or too unwilling to tell the truth about much of anything.  Everybody knows it's a problem, but nobody wants to hold the potato except the already reelected president.

[quote="hello"]So is Bush's social security plan then a good idea? So what he is suggesting is that people invest in the stock market=personal savings accounts? I dont understand, if average people invest in the stock market, isnt there a very good chance that they will lose a lot of money? Sorry for showing my incomprehensiveness.[/quote]

Over a long period of time, the stock market is not much riskier than the treasury bond market.  Besides, people could still simply invest in T-bills like the current social security account is invested.  Give that for most people the money invested has a long investment horizon, the returns are likely to be higher.

There are some problems with this idea.  People might choose to invest in junk stocks.  The government might then limit possible account investments, but that opens up a certain amount of moral hazard on the end of the government and business ties.  Such risk is mitigated by the fact that one investment tends to float all investments to some degree.  In other words, if large investors added a lot of capital to a certain stock without any change in the investment climate, other investors might simply take their gains as the stock goes up and invest in something else looking for higher returns.  This is because relative values should theoretically stay nearly the same (except for the minute value of increase in stock liquidity from the increase in the investor pool).

If you read about what happened in Chili, nearly all young people chose private savings accounts and that system has been mostly successful and probably bolsters their private economy with investment capital.  The problem replicating such success here is that our social security system is old and very large, so the economic hiccup of changing policy won't be as small.

Overall I think Bush's proposed changes are good, though that's an extremely rare instance in which you'll see me suggest that a Bush economic policy is a good idea.  Still, like most good ideas in government, it's probably not going to happen until the entire system is on its death bed and many people have lost their retirement funds.

Not sure I understand everything as I did not have a any advanced classes yet but thanks for your replies nevertheless. How come you did not want to work in economic policy?

Helping to build a company that focuses on helping students build economically viable skills [i]is[/i] my idea of working in economic policy.

[quote="MCrawford"]
While social security is not the single largest economic failure of the government, it's one of the larger ones that will have to be dealt with at some point. Somebody must necessarily get hurt and it's about time we bit the bullet and took on at least some of that pain, but politicians are very afraid that suggesting taking on some pain now will mobilize opposition against them. Don't expect the Republicans to back anything controversial now that they've finally won the hard fought battle of filling congressional rolls with their own along with a sitting president.

[...]

If you read about what happened in Chili, nearly all young people chose private savings accounts and that system has been mostly successful and probably bolsters their private economy with investment capital.  The problem replicating such success here is that our social security system is old and very large, so the economic hiccup of changing policy won't be as small.[/quote]

Yes, so true, the social security system must be ammended. It was a brilliant implementation on FDR's part, but at this point, I think the economy has outgrown it. With the baby boomer population entering retirement, the next working population faces big issues of how they must pay for it.

Raising taxes to pay for the social security benefits is not really an option because by doing so, the U.S. would be facing the problem that plagues many developed countries: excessively high taxes creating a disincentive to work/hire work. This not only decreases the workforce participation figure (unemployment, essentially) but it also jeopardizes future growth and can cause economic stagnation. Also, it can result in a loss of talent; if workers cannot find an incentive to work in the U.S., the mobile ones will leave the country and seek economic opportunities elsewhere. Furthermore, the burden of increased taxes often falls upon the shoulders of the lower and middle classes, thereby further increasing the socio-economic gap between the rich and poor. This, as we know, can create monsterous potential problems as well.

The U.S. economy is driven largely by consumption and is notorious for their low savings rate. While this may be nice now, it poses great problems for future investment and growth. Furthermore, it makes the country extremely vulnerable during the upturns and downturns of the business cycle. It places the U.S. at risk of undesired inflation during an upturn and can cause things such as housing bubbles, which can exacerbate the savings dilema because real estate is a form of savings in the U.S. During a downturn, the damage is obvious and can lauch the nation into recession.

As we have now learned, small tax cuts are not enough to increase savings. Rather, they merely increase disposable household income and drive consumption futher, creating a country that lives on credit. But large tax cuts are unfeasible because it's hard to adjust government spending accordingly.

To solve this problem, the U.S. needs to eliminate the social security system as much as possible because there will always be a smaller working population that will struggle to pay for the larger aging population. Without a social security system, people will find more incentive to save, and hopefully, this will be able to improve future investment and growth. But, as MCrawford pointed out, this isn't going to happen until everything comes crashing down. So in the meantime, we need to ensure that the forms of saving that the country [i]does[/i] engage in remains stable.

Sorry if this post is rambling. Maybe I'll edit later.

[quote="MCrawford"]Additionally, government control over interest rates and money supplies have been historically used to tax citizens without approval (or representation) in a manner few people understand.  When the government prints money, no economic value is created beyond marginal liquidity (which is really a production concern, not a paper money concern).  However, every pre-existing dollar is devalued, hence the government takes value from capital holders.

The unfortunate side of the Fed's setting of interest rates such as bond rates and the overnight call rate are that our markets are perfectly capable of functioning without them and discrete adjustments fly in the face of economic continuity.  Why should the value of all capital in the U.S. (and really the world since U.S. bonds are the gold standard) undergo immense volatility every few months according to the point at which Greenspan & Co. decide that it's time?  Volatility alone destroys the value of capital (a la Sharpe Ratios).

I'm also convinced that great fiscal power of the government creates greater incentive for thieves and tools to enter the political arena, thus creating a host of negative externalities.

Unfortunately discussions such as this are moot as government economic control can only be dismantled in times of great efficiency and prosperity and the stage to apply that standard is growing globally.  The only real way to put an end to bad government economic policies is to reduce the need for economic stimulus and that means starting companies (employing people) to do what needs to be done -- something that is very hard to do these days.[/quote]

Yeah, good points. But these are all the things we have to put up with in the name of government.

From a consumer's perspective, I guess the only defense is knowledge.

I've decided that my previous posts are so convoluted, I have absolutely no desire to edit them. But the more I think about this, the more it bothers me...(and at this point, I'm probably just talking to myself here.)

[quote="MCrawford"]Additionally, government control over interest rates and money supplies have been historically used to tax citizens without approval (or representation) in a manner few people understand.  When the government prints money, no economic value is created beyond marginal liquidity (which is really a production concern, not a paper money concern).  However, every pre-existing dollar is devalued, hence the government takes value from capital holders. [/quote]
Also, when a government creates inflation by altering the money supply, constituents end up paying higher taxes (in real terms) if the tax bands are not adjusted accordingly. And all along, the government gets the advantage of paying back debt much more easily.

[quote="MCrawford"]The unfortunate side of the Fed's setting of interest rates such as bond rates and the overnight call rate are that our markets are perfectly capable of functioning without them and discrete adjustments fly in the face of economic continuity.  Why should the value of all capital in the U.S. (and really the world since U.S. bonds are the gold standard) undergo immense volatility every few months according to the point at which Greenspan & Co. decide that it's time?  Volatility alone destroys the value of capital (a la Sharpe Ratios).[/quote]
Yes, this may be true. A developed country is perfectly capable of functioning without constant monetary and fiscal policy if the interest rates are set at an appropriate level. But if successful, prudent monetary and fiscal policy can help reduce unemployment. The question is, how much good does this reduction in unemployment do from a purely economic (not political) perspective? Are the opportunity costs justified? And as the little guy in a big world, how do we ensure that all these policies do not create dire consequences? After all, while knowledge can soften the blow, inflation and economic instability doesn't discriminate. [/quote]

To respond directly to the OP, of course some government is needed to help the economy, espcially in the times of recession.  However too much government regulation, restrictions and red-tape will hinder the economy and slow down economic growth.  Different varieties of capitalism is what leads to the growth of an economy- always.  Pure capitalism itself is a bad idea, as the corporations would rule the people- but variations are the only way that an economy can see good economic growth.

Take the Nordic Model- while it does have more social benefits than the US per se, it is still capitalist.  Even China- an authoritarian state was able to see good growth when it started to free the market which led to the huge economic growth and the bustling economy of China we see today.

Overall in my opinion the government should be the least involved in the economy as possible, but a variation of capitalism is necessary for sucess.  While there are good countries with strong social benefits, overall- freer markets is the best way to go.

(And sorry for the 15 year bump.)

Since China was a capitalistic communist, it prospered and is the largest economy as per GDP PPP(very reliable measure) and has no chance of collapsing like the 1990 Soviet Union Collapse..... the US should learn from them and increase spending on Artificial Intelligence and invest in entrepreneurship.....US should implement laws against monopoly and encourage startups. If only 5 companies namely Google, Microsoft, Apple,
Facebook and Amazon play the game in the AI field and buy all startups, they should be sued.....another thing the U.S. can do is to call companies overseas into the U.S.
So, the Government will only help the economy if they implement good economic policies and reduce its fiscal deficit or it can also hinder the economy if it passes flawed economic policies. So far I see some policies of the U.S. government flawed like the ban on visas which will improve the economy of the U.S. and I see Trump's and Biden's economic policy of made in America which is absolutely great and will reduce the US fiscal deficit :)

I found this video great:

[url]https://www.youtube.com/watch?v=IKrTqah6XWc[/url]


Given $ \theta_i \in [0,2\pi) \minus{} \{\pi\}$ for $ 1\leq i \leq n$,no two of them being the same (i.e. $ \theta_i \neq \theta_j$  $ \forall i,j$ )
and $ \frac{\theta_i}{\pi}$ is irrational for all $ 1\leq i \leq n$
Define
$ f(k) \equal{} \sum_{i\equal{}1}^n A_i \cos{k\theta_i}$ for all $ k \in \mathbb{N}$
where
$ \sum_{i\equal{}1}^n A_i \equal{} 0$

Show that there exist $ k_1,k_2$ such that $ f(k_1) < 0$ and $ f(k_2) > 0$.

Does the same result hold when one of the $ \theta_i$ is a rational multiple of $ \pi$?

Take,
 Aicosk1θi=j1
 Aicosk2θi=j2

f(k1)=summationAicosk1θi
f(k2)=summationAicosk2θi
Therefore,
f(k2)=summationj2
f(k1)=summationj1

Therefore we can see that by taking k2>k1,
it is possible f(k1)<0 and f(k2)>0


with $H_{2}SO_{4}$
a.$2-Butene$ or $iso-Butylene$
b.$1-pentene$ or $iso-Butylene$

[hide="i am"]confuse whether steric factors will take over carbocation stability or it's the other way round[/hide]

also
what will be the product of addition of $HI$ with $2,3-dimethyl,1-Pentene$

Iso-butylene should react faster in both cases, since it forms a tertiary carbocation and so the activation energy will be lower.

The product of addition of HI with 2,3-dimethyl-1-pentene should be 2-iodo-2,3-dimethylpentane.

isn't there an equal probability of getting $2,3-dimethyl,3-iodo pent-1-ene$


(8h^3)(2)^2/-4h^4v^6

i can't get it to -6/(hv^6) (teacher answer) Help needed, please show some working as possible. Thanks.

hello, do you mean $ \frac{8h^3\cdot2^2}{\minus{}4h^4v^6}$?
Sonnhard.

If so, then you can easily cancel out $ h^3$.

$ \frac{32}{\minus{}4hv^6}$

Divide by $ \frac{\minus{}4}{\minus{}4}$.

$ \frac{\minus{}8}{hv^6}$

Hmmm... did I make a mistake?  I don't think so...

Yes, your solution is correct. The question is whether there was an error in the problem, or the solution?


function
topology
real analysis
real analysis theorems

Let $ A \subset \mathbb{R}^n$, $ B \subset \mathbb{R}^m$, $ C \subset \mathbb{R}^l$ be open.

Let $ f: A \times B \to C$ be a function. Is it true that f is continuous if and only if $ f(a,b_0)$ is continuous on $ A$ for every fixed $ b_0$ in B?

No, and it's not enough even if you require continuity along slices in both directions. One version of a counterexample: $ f(x,y)\equal{}\frac{xy}{x^2\plus{}y^2}$, $ 0$ at the origin.

It is false even if you require continuity on each line though the counterexample is a bit more complicated: put a series of humps accumulating to $ 0$ between $ y\equal{}x^2$ and $ y\equal{}2x^2$. On the other hand, it raises an interesting question worth investigating: what is the smallest (or close to the smallest) class of curves such that continuity on every curve in the class implies continuity in $ \mathbb R^n$? It is essentially equivalent to asking what is the smallest class of curves such that for every sequence $ x_k\to 0$, one can draw a curve from this class that contains infinitely many $ x_k$. I suspect that the answer is $ C^1$ (i.e., continuity on every $ C^1$ curve implies continuity in $ \mathbb R^n$ and for every $ \alpha>1$, one can find a function $ f$ continuous on every $ C^\alpha$ curve but discontinuous at the origin).

I see. The "only if" part is true though. If O is an open set in $ C$, then $ f^{ \minus{} 1} (O)$ is an open set in $ A \times B$ which means it is open in every cross section $ A \times b_0$ by the definition of the subspace topology which is the same as the product topology on $ A \times b_0$. And a set is open in a space cross a point if and only if it is open in the space.


What is sum of the postive integer factors of 225?

Can you find a differnet way than listing all the factors and adding them up?[/hide]

[hide="Solution"] Prime factorization of 225: $3^2\times5^2$
The formula: $(3^0+3^1+3^2)(5^0+5^1+5^2)=(1+3+9)(1+5+25)=13\times31=\boxed{403}$. :D This formula works because when distributed, it has all the factors. [/hide]


Define the sum $a+_Bb$ as the map $a+_Bb=f(a,b)=a+b+1$, where $+_B$ is the 'Bush' addition binary relation and $+$ is the 'correct' addition binary relation from the ordinary arithmetics.

[quote="cardwacko9999"]NoTe: Let's call this Bush arithmetics (no offense to Bush but ppl say that he thinks 1+1=10)[/quote]

1+1 does equal 10 in binary.

this is just lyk a problem that ( i think used ) to be on miller mathcounts!

(course we turned it into a spam thing.... :D )

[quote="gameworld7"]question: how could 1+1=10? shouldn't it equal 2?[/quote]
Binary, h_s_potter2002 said. Base 2. When the rightmost digit contributes itself to the number, the second-rightmost digit contributes itself times 2, the third-rightmost digit contributes itself times 2^2=4, etc. If you want a better explanation, ask about it in the Getting Started forum.

[quote="cardwacko9999"]What happens if we assume the following:

1+1=3
2+2=5
3+3=7, etc.

Does this contradict itself?

NoTe: Let's call this Bush arithmetics (no offense to Bush but ppl say that he thinks 1+1=10)[/quote]
this means
x+x=y
y+y=z
z+z=a
a+a=a[1]
...
and they are all just variables
so no

One Solution...

[hide]1+1=3

2+2=5

3+3=7

What if-

1->1, 2->1.5, 3->2, etc.?

So y=(x+1)/2, where x is bush arith # while y is regular base10 #.

I wonder how long it would take me if I were doing it on the AMC8 or Mathcounts.... (I might skip this problem)[/hide]

If anyone thought the above solution presented by me was right, then you need to reread the problem.

New wacky solution!

1+1=3
2+2=5
3+3=7

Since 1+1=3, then 1=2 so 1+2=5 and 1+2=4 and 2+2=5...

[quote="bob123"]0+0=1, 1+1=3, 0+0+1+1=4, 1+1=4

So we have 1+1=3 and 1+1=4.....[/quote]

u cant cancel out the 0's in bush arith.- they stand for 1/2 in real arith.  :)

I know....

Actually, they could also stand for 1, or 2, or in fact, any number greater than 0.

I say there is no contradiction because if you assigned 0 to be 0.5, 1 to be 1, 2 to be 1.5... then you can convert it to the regular arithmetic we use, and there is obviously no contradiction there.

?????????????!!!!! seriously! this doesn't make any sense at all!

Did anyone look at my replying post above? You will find discrepancies if you mix up the 'Bush' binary relation with our ordinary binary relation.

BRUH IM THE ONLY ONE THAT DID THE POLL

BUMPITY BUMP BUMP  :rotfl:


vector
rotation
trigonometry
linear algebra
matrix
analytic geometry
geometry

Is it possible to calculate the resulting vector after rotating a given vector about some axis by a given angle? For example:
I have a vector v(0,0,-1). Then i rotate v about y-axis by 20deg. How can i calculate the resulting vector?

Thank you

I'm pretty sure you can do it, but you need to use trigonometry with the angle you're given. There might also be a pretty matrix way of doing it...

Sure, it's just an extension of the 2D version.  The only difference is that the $y$ coordinate doesn't change here; the $x$ and $z$ coordinates change.  The matrix is

$\textrm{Rotation}_{\textrm{y}}\left[\theta\right]=\begin{bmatrix}\cos\theta&0&-\sin\theta \\ 0&1&0 \\ \sin\theta&0&\cos\theta\end{bmatrix}.$

...which you can see comes from

$\textrm{Rotation}\left[\theta\right]=\begin{bmatrix}\cos\theta&-\sin\theta \\ \sin\theta&\cos\theta\end{bmatrix}.$

The derivation for the 2D version is simple if you draw out a figure, and the others follow.


calculus integration search function LaTeX articles

Occasionaly people who come across a new math concept by its symbol rather than it's name. For example, when I was in 6th grade I was wondering what the integral symbol meant (of course i didn't realize they were called integrals) so I had to wander around for several months before randomly finding it online. Why not have a capability in the AoPS Wiki search function to include LaTeX functions?

For example-

Search "\int" will provide the same article as if you search "integral"

[quote="JBL"]If you know the Latex symbol for something, you almost certainly already know what it means.[/quote]

Well some LaTeX symbols are abreviated, like $\pm$ (\pm).

One approach that would accomplish what you mean is to have a 'Math Symbols' page in the wiki.  List symbols there, grouped roughly by subject, with brief descriptions and links to suitable entries in the wiki.  Sounds like a useful idea for new students.  I remember being terrified the first time I saw $\equiv$ in a solution as a student.  I had never heard of mods and was totally lost.  I wouldn't have known where to start to look it up, either.

[quote="JBL"]If you know the Latex symbol for something, you almost certainly already know what it means.[/quote]

On the forums, in particular, it's very possible for someone to see a symbol they don't know in LaTeX.  And then they can figure out the LaTeX command for it by hovering their mouse over it or clicking.

Btw, \nabla's another example of something which isn't too intuitive.

You could also make a bunch of pages in the wiki such as '\frac' or '\equiv' so that when people search for the term (s)he gets the explanation.

[quote="rrusczyk"]One approach that would accomplish what you mean is to have a 'Math Symbols' page in the wiki.  List symbols there, grouped roughly by subject, with brief descriptions and links to suitable entries in the wiki.[/quote]

That's a better idea than mine  :D . And yes, many times when I used mods in math class the other students assume that is an equal sign and their eyes are playing tricks on them.

hmmm...methinks that rusczyk's idea is the frontrunner.  mayhaps we should put a link to the symbols guide on the main page?

their eyes are playing tricks on them - a ring just became a field (modulo a prime, of course...)


calculus
integration
inequalities
trigonometry
calculus computations

Prove that the following inequality holds for each natural number $ n$.
\[ \int_0^{\frac {\pi}{2}} \sum_{k \equal{} 1}^n \left(\frac {\sin kx}{k}\right)^2dx < \frac {61}{144}\pi\]

I think that students being exposed to this test are supposed to be not aware of the value that $ \sum_1^\infty \frac{1}{k^2}$ converges to,
instead we can use the fact that $ \sum_{k\equal{}0}^{n\minus{}1}\frac{1}{(k\plus{}1)^2}\leq \sum_{k\equal{}0}^{n\minus{}1}\frac{1}{k^2\plus{}1}\leq  \sum_{k\equal{}0}^{n\minus{}1}\int_{k\minus{}1}^{k}\frac{1}{x^2\plus{}1}\,dx$.
Taking the limit on right hand side we can deduce something related to $ \pi$.


Consider a $\triangle{ABC}$ with $AC \perp BC$, and $\frac{AC}{BC}=3$. Consider a point $O$ in $AB$ such that $AO=OB$. $P$ is a point in $AC$ such that $OP \perp AB$. $Q$ is a point in $BC$ such that $PQ \parallel AB$. Finally, $R$ is a point in $AB$ such that $QR \perp AB$. Find $\frac{PQ}{RQ}$.

Let $BC=a$. Then $AC=3a$ and $AB=a\sqrt{10}$.

$OA=\frac{a\sqrt{10}}{2}$

$OP: OA=BC: CA\implies OP=\frac{a\sqrt{10}}{6}$

$AP=\sqrt{OP^{2}+OA^{2}}={5a\over 3}$

$PC=AC-AP={4a\over 3}$

$PQ: PC=AB: AC\implies PQ=\frac{4a\sqrt{10}}{9}$

$RQ=OP\implies{PQ\over RQ}={8\over 3}$

[hide="Here is the answer."]$\frac{8}{3}$[/hide][hide="Here is a generalization."][color=darkblue] Consider a $\triangle{ABC}$ with $AC \perp BC$ and $\frac{AC}{BC}=k$ and a point $O$ in $AB$ such that $AO=m\cdot OB$ and $m<k^{2}$. $P$ is a point in $AC$ such that $OP \perp AB$. $Q$ is a point in $BC$ such that $PQ \parallel AB$. Finally, $R$ is a point in $AB$ such that $QR \perp AB$. Prove that $\frac{QP}{QR}=\frac{k^{2}-m}{km}$, i.e. $\boxed{\ \frac{QP}{QR}+\frac{CB}{CA}=\frac{CA}{CB}\cdot \frac{OB}{OA}\ }\ .$[/color][/hide]


[quote="CircleSquared"]In the 1st one, the second term should be 2*3, right?[/quote]
yea sorry i changed it

[quote="CircleSquared"]Umm.... What's hockey stick?[/quote]

Best way is to see this: [url]http://www.artofproblemsolving.com/Wiki/index.php/Combinatorial_identity[/url]

[quote="asdff"]1. what is 1*2+2*3+3*4+..+n(n+1)+...+100*101

2. what is 1*2*3+2*3*4+..+n(n+1)(n+2)+...100*101*102[/quote]

[hide="1."]  Each number is 2 times a triangular number.  Therefore the sum is $2\times\sum_{n=1}^{100}\frac{n(n+1)}{2}$.  The formula for the sum of the first n triangular numbers is $\binom{n+2}{3}$.  Applying this, we find that the part with the summation symbol is 171700.  The whole sum is 343400.

Extension: Derive the formula for the sum of the first n triangular numbers.
[/hide]

[hide="2."]  We can write this as $\sum_{n=1}^{100}n(n+1)(n+2)$.  Multiplying out the binomials and stuff, we have $\sum_{n=1}^{100}n^{3}+3n^{2}+2n$.  This is also $\sum_{n=1}^{100}n^{3}$+$3\sum_{n=1}^{100}n^{2}$+$2\sum_{n=1}^{100}n$.
$\frac{(100(101))^{2}}{4}+3\times\frac{(100)(101)(201)}{6}+2\times\frac{100(101)}{2}$
$25502500+1015050+10100$
$26527650$[/hide]

[quote="AstroPhys"][quote="alanchou"] Therefore the sum is $2(\sum_{n=1}^{1}00 \frac{n(n+1)}{2}$.  [/quote]
 :huh: That would make it zero, probably a typo.[/quote]

It is most probable he meant $2 \left( \sum_{n=1}^{100}\frac{n(n+1)}{2}\right)$

Yeah, I changed it.

I'm not completely used to $\LaTeX$ yet.  Oh well.

[quote="t0rajir0u"]Can anyone generalize?  Specifically, to $\prod_{n=1}^{N}n(n+1)...(n+(k-1))$.

[hide="Hint 1"] Divide by $k!$. [/hide]
[hide="Hint 2"] Hockeystick. [/hide][/quote]
[hide]
$\frac{n(n+1)...(n+(k-1))}{k!}=\binom{n}{k}$
$\binom{1}{k}+\binom{2}{k}+\binom{3}{k}+...+\binom{n}{k}=\binom{n+1}{k+1}$?[/hide]


probability
inequalities
triangle inequality

In each of three boxes are four straws with integer length 1 cm through 4 cm, inclusive. A straw is randomly selected from each box. What is the probability that a triangle can be formed with the three straws chosen? Express your answer as a common fraction.

The Triangle Inequality Theorem says that if you have two side lengths, $ \alpha$ and $ \beta$, then the third side length, $ \gamma$, must fit the following requirements:

$ \alpha \plus{} \beta > \gamma$

$ \gamma \plus{} \beta > \alpha$

$ \gamma \plus{} \alpha > \beta$

So let's make a list of all the possible side lengths of a triangle.

1-1-1
1-2-2
2-2-2
2-2-3
2-3-4
2-4-5
3-3-3
3-3-5
3-4-6
4-4-4

There are 10 ways in total to make a nondegenerate triangle. The number of ways to choose any 3 lengths, one from each box, is $ 4^3 \equal{} 64$. Thus our answer is $ \frac{10}{64} \implies \boxed{\frac{5}{32}}$.

1-1-1
1-2-2
1-3-3
1-4-4
2-2-2
2-2-3
2-3-3
2-3-4
2-4-4
3-3-3
3-3-4
3-4-4
4-4-4

Here are the thirteen ways.

For four ways:
1-1-1
2-2-2
3-3-3
4-4-4
The combination can only be picked through one permutation.

for 8 ways:
1-2-2
1-3-3
1-4-4
2-2-3
2-3-3
2-4-4
3-3-4
3-4-4
The combination can be picked through three permutations.

for 1 ways:
2-3-4
the combination can be picked through six permutations.

Thus the number of ways to choose a nondegenerate triangle is $ 4\cdot1\plus{}8\cdot3\plus{}1\cdot6\equal{}4\plus{}24\plus{}6\equal{}34$.

Since there are $ 64$ ways to choose your three lengths, the answer is $ \frac{34}{64}\equal{}\boxed{\frac{17}{32}}$

[b]CA Math[/b]'s method and answer is correct.  @Isabella, you forgot to account for the different ways each triangle could be made, for example, a $2-2-3$ triangle can be made by $2-2-3$, $2-3-2$, or $3-2-2$.


group theory
abstract algebra
superior algebra
superior algebra solved

Uh.. $S_{3}$ is a subgroup of order 6, embedded as the permutations which fix one of the points.

[quote="rustam"]Hi!
Please help me with the problem:
Prove that $S_{4}$ has no subgroup of order 6.[/quote]
Hi Rustam!
Jmerry is right, something is wrong with that question.
You probably meant $A_{4}$ didn't you?

Suppose there is one of order six, called $H$
Well, $A_{4}$ has 12 elements, so a subgroup of order 6 would be a normal subgroup as it would have index 2.(that is a theorem, subgroups with index 2, are normal subgroups)

So that would mean that for every $a\in A_{4}$ : $(a H)^{2}= H$,

or thus that $a^{2}\in H$

But that would mean that $\forall a,b,c \ ( a b c)^{2}= (a c b) \ in H$ and there are at least eight elements like that--> contradiction


Let $ a,b,c$ be three positive numbers. Prove that
$ \frac {ab}{c(c\plus{}a)}\plus{} \frac {bc}{a(a\plus{}b)}\plus{}\frac {ca}{b(b\plus{}c)} \ge \frac {a}{a\plus{}b}\plus{}\frac {b}{b\plus{}c}\plus{}\frac {c}{c\plus{}a}$

[quote="nguyenvanhuyen"]Let $ a,b,c$ be three positive numbers. Prove that
$ \frac {ab}{c(c \plus{} a)} \plus{} \frac {bc}{a(a \plus{} b)} \plus{} \frac {ca}{b(b \plus{} c)} \ge \frac {a}{a \plus{} b} \plus{} \frac {b}{b \plus{} c} \plus{} \frac {c}{c \plus{} a}$[/quote]

clearing of fractions, equivalent to

$ a^2b^3\plus{}b^2c^3\plus{}c^2a^3 \ge abc(ab\plus{}bc\plus{}ca)$

or

$ \sum_{cyc}{\left(\frac{4}{7}a^2b^3\plus{}\frac{1}{7}b^2c^3\plus{}\frac{2}{7}c^2a^3\right)} \ge \sum_{cyc}{a^2b^2c},$

which is true by Am-Gm.

[quote="kuing"]
clearing of fractions, equivalent to

$ a^2b^3 \plus{} b^2c^3 \plus{} c^2a^3 \ge abc(ab \plus{} bc \plus{} ca)$

[/quote]

$ a^2b^3 \plus{} b^2c^3 \plus{} c^2a^3 \ge abc(ab \plus{} bc \plus{} ca)$

$ \Leftrightarrow$

$ \frac{ab^2}{c}\plus{}\frac{bc^2}{a} \plus{}\frac{ca^2}{b} \geq ab \plus{}bc\plus{}ac$

for this inequality :
$ \frac{ab^2}{c} \plus{}ac \plus{} ab \geq 3ab$ (by [b]AM-GM[/b])

$ \frac{ca^2}{b} \plus{}ac \plus{} bc \geq 3ac$

$ \frac{bc^2}{a} \plus{}bc \plus{} ab \geq 3bc$

Add up these inequalities and enjoy !  :P


Let $P$ be a point on the circumcircle of a triangle $A_{1}A_{2}A_{3}$,and let $H$ be the orthocentre of the triangle.The feet $B_{1}$,$B_{2}$,$B_{3}$ of the perpendiculars from $P$ to $A_{2}A_{3}$,$A_{3}A_{1}$,$A_{1}A_{2}$ lie on aline.Prove that this line bisects the segments $PH$.

I think it's a known fact that the Simson line bisects $PH$. Note that the intersection is on the nine-point circle.  :)

Sorry for not providing a diagram, but there were too many lines to be drawn. :D

Notice that if two lines are perpendicular to the same line, they are parallel. Using this, we see that $PH$ and the Simson Line are two diagonals of a parallelogram, and therefore bisect each other.


I think you mean http://www.mathmatix.com/ by joml88.

Does Cicatriz have a site  :huh: ?  (obviously he does...but are we allowed to see it?)

Nice site, thoobik ;) :)

Right now it's hosted on my computer, so I don't want a large public exposure for security and slowdown. I'm planning to get a real host soon.

Congrats to thoobik and joml. How many people ended up entering?

Now I feel as though I actually did something worthwhile this summer.  (My guess is 3, Cicatriz ;))

[quote="thoobik"]So.. uh.. prizes?[/quote]

Did you do as it says [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35707]here?[/url]


Let $ABC$ be an acute triangle with circumcenter $O$. The circumcircle of $\triangle ABO$ meets $AC,BC$ at $M,N$ respectively. Prove that the circumcircle of triangles $ABO,MNC$ have equal radii.

$\angle{MNO}=\angle{MAO}=\angle{A}-\angle{BAO}=\angle{A}+\angle{C}-90^{o}$.

Similarly, we have $\angle{NMO}=\angle{B}+\angle{C}-90^{o}$, thus we have $\angle{MON}=180^{o}-\angle{C}$. Therefore $2R_{ABO}\cdot \sin{\angle{MON}}=MN=2R_{MCN}\cdot \sin{C}\Longrightarrow R_{ABO}=R_{MCN}$.

$\widehat {MNO}=\widehat{MAO}=90^\circ-\widehat{CMN}\iff NO\bot AC$, similarly $MO\bot BC$, hence $O$ is the orthocenter of $\Delta CMN$, or $\odot (MON)$ is a [b]Carnot circle[/b] of $\Delta CMN$, done.

Best regards,
sunken rock

[quote="sunken rock"]$\widehat {MNO}=\widehat{MAO}=90^\circ-\widehat{CMN}\iff NO\bot AC$, similarly $ {\color{blue} NO\bot BC }$, hence $O$ is the orthocenter of $\Delta CMN$, or $\odot (MON)$ is a [b]Carnot circle[/b] of $\Delta CMN$, done.

Best regards,
sunken rock[/quote]

Shouldn't i be "similarly $ MO \bot BC$"? (cant seem to highlight the erroneous part lol)